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Premium quality villa built to the highest standard is situated along the coastal road less than 10 minutes drive from Ibiza town in the quiet and exclusive residential area of Can Fita. It offers open countryside and panoramic sea views, Salinas Natural Park, Dalt Vila and neighbour island of Formentera. You can also immerse yourself into the stunning sunset views every evening.
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{
"redpajama_set_name": "RedPajamaC4"
}
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Here are a few of the reptiles we've rescued since we started listing our adoptable pets on PetFinder.com. PetFinder is a great site to find all kinds of adoptable pets, both warm and cold-blooded. Check out some of their links and search boxes on this page as you enjoy pictures of a handful of the animals we've found homes for so far.
©2016 Forgotten Friend Reptile Sanctuary.
|
{
"redpajama_set_name": "RedPajamaC4"
}
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\section{Introduction}
Concernant les variétés ayant beaucoup de points rationnels, une question naturelle est combien il y en a de hauteur bornée et comment ils sont distribués.
Dans des années 1990, Batyrev et Manin ont conjecturé une formule qui donne une prédiction pour l'ordre de croissance du cardinal de l'ensemble des points de hauteur bornée. Concernant la distribution globale des points rationnels sur $X$, Peyre a ensuite reformulé et raffiné leur conjecture sous une forme faisant intervenir des mesures, que nous énonçons comme suit. On note $X(\mathcal{A}_\QQ)^{\operatorname{Br}}$ l'ensemble des points adélique de $X$ pour lesquels l'obstruction de Brauer-Manin à l'approximation faible est triviale.
\begin{conjecture}[Batyrev-Manin-Peyre]\label{conj:bmp}
Il existe un ouvert $U\subseteq X$ tel qu'en notant
\begin{equation}\label{eq:countmeasure}
\delta_{U,B}=\sum_{\substack{P\in U(\QQ)\\H(P)\leqslant B}} \delta_P,
\end{equation}
on ait
$$\frac{1}{B (\log B)^{\operatorname{rg}(\operatorname{Pic}(X))}}\delta_{U,B} \longrightarrow \mu_{X}^{\operatorname{Br}},\quad B\to \infty$$
au sens de convergence faible pour certaine mesure $\mu_{X}^{\operatorname{Br}}$ déduite d'un produit de mesures $\prod_{\nu\in \operatorname{Val}(\QQ)} \mu_{\nu}$ par restriction à $X(\mathcal{A}_\QQ)^{\operatorname{Br}}$.
\end{conjecture}
Remarquons qu'en fait, la partie $\infty$-adique $\mu_{\infty}$ est une mesure à densité continue relativement à la mesure de Lebesgue sur $X(\QQ_\nu)$. Tandis que la conjecture de Batyrev-Manin est vérifiée pour beaucoup de variétés presque de Fano voire singulières malgré quelques contre-exemples connus, cette conjecture concernant la distribution est rarement abordée dans la littérature, en dehors des cas classiques. Cependant le travail de Batyrev et Tschinkel \cite{B-T1} semble être en faveur d'elle pour les variétés toriques.
\begin{comment}
Parmi les prédictions des points rationnels sur les variétés de Fano $X$, le principe de Batyrev-Manin \cite{batyrev-manin} et sa version raffinée de Peyre \cite{Peyre1} conjecture qu'il existe un ouvert $U$ dense pour la topologie de Zariski tel que
\begin{equation}\label{eq:maninpeyre}
\sum_{P\in U(\QQ):H(P)\leqslant B} \delta_P\sim B (\log B)^{r-1} \mu_X,\quad B\to \infty,
\end{equation}
où $H$ est une hauteur de Weil associée au fibré anticanonique ample $\omega_X^{-1}$ et $\mu_X$ est une mesure définie sur $X(\mathcal{A}_{\QQ})$, appelée la mesure de Tamagawa. L'existence de formules asymptotiques (c'est-à-dire le terme principal espéré) Toutefois outre que des cas classiques, l'existence de la mesure de Tamagawa reste encore rarement aboutie (cf. \cite{Peyre3}), sachant qu'elle fournit des informations plus fines.
\end{comment}
Nous nous demandons dans quelle mesure une version plus forte de ce principe soit valide. À savoir pour un ouvert analytique de la variété $D=D(B)$ dont la taille dépend de $B$, pourrions-nous espérer qu'il existe une mesure $\mu_X^\prime$ telle que
\begin{equation}\label{eq:generalquestion}
\sum_{P\in D(B)\cap X(\QQ):H(P)\leqslant B}\delta_P\sim \operatorname{Vol}(D(B)) B (\log B)^{r-1} \mu_X^\prime,\quad B\to \infty?
\end{equation}
Une motivation de considérer ce problème est l'étude de la \emph{distribution locale} des points rationnels. Elle fut considérée en premier par S. Pagelot \cite{pagelot}, où il a pris pour $D(B)$ des boules de rayon $\asymp B^{-\frac{1}{r}}$ et où il a constaté que la distribution des points autour d'un point rationnel fixé est différente (i.e. la mesure $\mu_X^\prime$) et cette différence n'est pas reflétée par la distribution globale (i.e. la mesure $\mu_X^{\operatorname{Br}}$) sur certaines surfaces.
Pour établir l'existence de $\mu_X^\prime$, souvent il faut retirer certaines sous-variétés \emph{localement accumulatrices} et bien choisir la constante $r$. Les travaux de D. McKinnon et M. Roth \cite{McKinnon2007} et \cite{McKinnon-Roth1} sur l'approximation diophantienne sur les variétés fournissent une constante de nature arithmétique et géométrique, appelée \emph{constante d'approximation} et notée $\alpha$. S. Pagelot définit dans \cite{pagelot} \emph{la constante essentielle} $\aess$ qui caractérise l'approximation diophantienne générique. En prenant $r$ entre $\alpha$ et $\aess$, nous pouvons récupérer plus d'informations sur la distribution des points rationnels de hauteur bornée qui sont \og négligées\fg{} dans la considération \eqref{eq:countmeasure}. Voir par exemple des explications et des illustrations dans \cite{huang2}.
Dans cet article on considère la surface torique $Y_3$ d'éventail suivant.
\begin{figure}[h]\label{fg:y3}
\centering
\includegraphics[scale=0.8]{Y4Y3p1p1.mps}
\caption{Les éventails de $Y_4,Y_3$ et $\PP^1\times\PP^1$}
\end{figure}
Admettant $\PP^2$ et $\PP^1\times \PP^1$ comme modèles minimaux, elle est une surface de del Pezzo généralisée (cf. \cite{derenthal2014}) (c'est-à-dire la désingularisation minimale d'une surface de del Pezzo singulière de degré $5$).
C'est un fait empirique que la difficulté d'établir l'existence de la mesure limite augmente lorsque le degré de la surface baisse. Ceci est en parallèle avec la conjecture de Batyrev-Manin sur la répartition globale.
Nous fixons tout au long ce travail le point central $Q=[1:1]\times [1:1]$. Toutes les constantes d'approximation et tous les degrés seront calculés par rapport au fibré anti-canonique. C'est la première fois que les courbes cuspidales entrent dans l'histoire. Notre premier résultat principal, concernant l'approximation de $Q$ par d'autres points rationnels sur la variété définie sur $\QQ$, énonce qu'avec les courbes nodales qui sont déjà des objets centraux d'étude pour la surface $Y_4$ \cite{huang2}, elles donnent en même temps des meilleurs approximants en dehors d'un fermé de Zariski. (cf. Sections \ref{se:cusp}, \ref{se:nodal})
\begin{theorem}[cf. Proposition \ref{po:lowerbound1}, Corollaire \ref{co:aess}]\label{th:main1}
Nous avons
\begin{itemize}
\item $\alpha(Q,Y_3)=2$. Elle s'obtient sur les trois courbes rationnelles lisses $l_i,(1\leqslant i\leqslant 3)$ de degré minimal passant par $Q$. Ces courbes sont localement accumulatrices;
\item $\alpha_{\operatorname{ess}}(Q)=\frac{5}{2}$. Elle peut être calculée sur des courbes nodales et des courbes cuspidales passant par $Q$.
\end{itemize}
\end{theorem}
Les courbes $l_i,1\leqslant i\leqslant 3$ sont en fait les (transformations strictes des) sections de degré $(1,1)$ joignant $Q$ et l'un des $3$ points éclatés dans $\PP^1\times\PP^1$ (cf. \eqref{eq:curvesmalldegree}). Il s'en suit que la surface $Y_3$ vérifie la conjecture de D. McKinnon \cite[Conjecture 2.9]{McKinnon2007}. Bien que les courbes nodales et des courbes cuspidales aient la même constante d'approximation, le point $Q$ est approché de nature radicalement différente suivant ce type de courbes (cf. Section \ref{se:compcuspnodal}). Le premier type correspond à l'approximation d'un nombre quadratique et le nombre des points rationnels entrant dans le zoom critique est faible. Il n'existe pas de mesure décrivant la distribution locale. Alors que pour le deuxième on approche un point rationnel sur le corps de base $\QQ$. C'est le cas où l'on peut établir l'existence d'une mesure locale limite.
Le fait intéressant ici est que les points sur les courbes cuspidales forme une partie \emph{mince} de type II contenu dans un ensemble \emph{semi-algébrique} pour la topologie réelle, qui domine son complémentaire pour le zoom critique. Mais il n'est pas clair \emph{a priori} que les mesures de dimension $1$ sur chaque courbe cuspidale se combinent en une mesure de dimension $2$.
En dépit de la difficulté pour démontrer l'existence de mesure limite au vu du paramétrage donné par des courbes nodales révélée dans \cite{huang2}, nous pouvons néanmoins démontrer l'existence d'une mesure locale pour le zoom critique en retirant cette partie mince.
Pour une partie $W$ de $Y_3$, nous notons $\{\delta_{W,Q,B,r}\}$ la famille de mesures de zoom de facteur $r$ comptant les points rationnels sur $W$ de hauteur $\leqslant B$ (cf. \S\ref{se:zoomoper}).
\begin{theorem}[cf. Théorèmes \ref{th:principalthm}, \ref{th:S2}, \ref{th:S3}]\label{th:main2}
Soient $Z=\cup_{i=1}^3 l_i, U=Y_3\setminus Z$.
\begin{enumerate}
\item Pour $2\leqslant r<\frac{5}{2}$, nous avons $$\frac{1}{B^{1-\frac{1}{r}}}\delta_{Y_3,Q,B,r}\to \delta_r,$$ où $\delta_r$ est une mesure de dimension de Hausdorff $1$ à support dans $Z$;
\item Pour $r=\aess=\frac{5}{2}$, il existe une partie mince, qui est la réunion de $Z$ du type I et $M$ du type II, telle qu'en notant $V=Y_3\setminus (Z\cup M)$, nous ayons
$$\frac{1}{B^\frac{1}{5}}\delta_{V,Q,B,\frac{5}{2}}\longrightarrow \delta_{\frac{5}{2}},$$
où $\delta_{\frac{5}{2}}$ est une mesure de dimension de Hausdorff $2$.
\end{enumerate}
\end{theorem}
Il n'est pas ardu d'établir une majoration uniforme d'ordre de grandeur $B^{\frac{1}{5}+\delta}$ (cf. Proposition \ref{po:uniformupperbound}) pour $r=\frac{5}{2}$. Cependant, démontrer des formules asymptotiques ainsi que la convergences de mesures de zoom est beaucoup plus délicat. Nous renvoyons au théorème \ref{th:principalthm} pour l'énoncé précis sur la forme de la mesure $\delta_{\frac{5}{2}}$ et le terme d'erreur. Mentionnons que cette mesure fait apparaître les trois courbes qui sont localement accumulatrices. Ceci est en fait analogue du résultat pour la surface $X_3$ (cf. \cite{huang1}).
\begin{theorem}[cf. Proposition \ref{po:lowerboundmeasure}, Lemmes \ref{le:lowerboundM} et \ref{le:upperboundM}, Corollaire \ref{co:rab}]
Soit $M$ la partie mince comme ci-dessus. Soit $\chi(\varepsilon)$ la fonction caractéristique de la boule de rayon $\varepsilon>0$. Nous avons que pour toute $\mathcal{C}$ courbe cuspidale,
$$\delta_{M,Q,B,\frac{5}{2}}(\chi(\varepsilon))\gg B^\frac{1}{5}\delta_\mathcal{C}(\chi(\varepsilon)),$$
où $\delta_{\mathcal{C}}$ est une mesure de dimension de Hausdorff $1$ à support dans $\mathcal{C}$. De plus
$$\delta_{M,Q,B,\frac{5}{2}}(\chi(\varepsilon))\gg\ll_\varepsilon B^{\frac{1}{5}}\log B,$$
pour tout $\varepsilon>0$ suffisamment grand.
En particulier, la partie mince $M$ est localement faiblement accumulatrice.
\end{theorem}
Les parties minces, dont la contribution a été considérée comme négligeable dans plupart de cas (cf. \cite{Serre} et \cite{browningloughran}), se montre parfois problématique dans le programme de Batyrev-Manin. Elles ont été reconsidérées depuis le premier contre-exemple de V. V. Batyrev et Yu. Tschinkel \cite{batyrevtschinkel1996}. Mais c'est rare dans la littérature qu'on soit capable de contrôler le cardinal de la partie obtenue en retirant l'ensemble mince. La seule réussite de ce point de vue est le résultat du Rudulier \cite{LeRudulier} (cf. aussi \cite{Peyre3}) où elle a considéré des schémas de Hilbert des points sur des surfaces. Le théorème ci-dessus montre que les parties minces ont aussi des influences non-négligeables pour le problème de distribution locale. Il nous fournit ainsi un autre exemple sur la gestion de ces ensembles.
L'heuristique directe du principe de Batyrev-Manin-Peyre \eqref{conj:bmp} (cf. aussi \eqref{se:heuristicbm}) pour le problème \eqref{eq:generalquestion} suggère que le terme principal dans le zoom critique devrait être $B^\frac{1}{5}(\log B)^4$. Mais en réalité il est $B^\frac{1}{5}$ et donc cette heuristique mérite d'être améliorée. Nous proposons alors, compte tenu des exemples connus, une formule prédisant l'ordre de grandeur de la famille de mesures de zoom. Soit $X$ une \og bonne \fg{}\ variété. Nous désignons par
$N_1(X)$ le groupe de Chow des $1$-cycles modulo l'équivalence algébrique, et par
$l(X)$ le rang du sous-groupe de $N_1(X)$ engendré par les classes des courbes rationnelles $C$ vérifiant $\alpha(Q,C)=\aess$.
\begin{fconjecture}[version faible]
En dehors d'une partie mince $M$, nous avons que pour tout $\varepsilon>0$,
\begin{equation}\label{eq:weak}
\delta_{X\setminus M,Q,B,\aess}(\chi(\varepsilon))=O_\varepsilon\left(B^{1-\frac{\dim X}{\aess}} (\log B)^{l(X)-1}\right).
\end{equation}
\end{fconjecture}
\begin{fconjecture}[version forte]
Si $\aess>\dim X$, en dehors d'une partie mince $M$, nous avons
\begin{equation}\label{eq:conj}
\frac{1}{B^{1-\frac{\dim X}{\aess}} (\log B)^{l(X)-1}}\delta_{X\setminus M,Q,B,\aess}\to \delta_{\aess}.
\end{equation}
\end{fconjecture}
\begin{comment}
La troisième conjecture est en conséquence avec la formule empirique de Batyrev-Manin. On espère que quand le facteur de zoom est suffisamment grand, l'ordre de grandeur du dénombrement de zoom devrait tendre vers celui sans opération de zoom.
\begin{conjecture}[version folle]
Pour $r\geqslant\aess(>\dim X)$, en dehors d'une partie mince $M$,
\begin{equation}\label{eq:conjstrong}
\frac{1}{B^{1-\frac{\dim X}{r}} (\log B)^{l(r,X)-1}}\delta_{X\setminus M,Q,B,\aess}\to \delta_{r}.
\end{equation}
où $l(r,X)$ est le nombre de classes des courbes rationnelles $C$ vérifiant $\aess\leqslant\alpha(Q,C)\leqslant r$.
\end{conjecture}
\end{comment}
Esquissons d'où vient le facteur $l(X)\leqslant \operatorname{rg}(N_1(X))=\operatorname{rg}\operatorname{Pic}(X)$.
Nous verrons dans l'appendice que les surfaces toriques $\PP^2,X_1,X_2,X_3,\PP^1\times \PP^1, Y_3$ vérifient la formule forte \eqref{eq:conj} hormis $Y_4$, la dernière étant toutefois conforme à la formule faible \eqref{eq:weak} (nous avons qu'en fait $\aess(Q)=\dim Y_4=2$ pour $Y_4$).
En effet, pour la surface $Y_3$ étudiée dans ce papier, toutes telles courbes appartiennent aux classes $m[\omega_{Y_3}^{-1}],m\in\NN_{\geqslant 1}$. Cependant il n'y qu'une seule classe des courbes rationnelles très libres $C$ réalisant $\alpha(Q,C)=\aess(Q)$ pour les surfaces $\PP^2,X_1,X_2$.
Détaillons la situation pour la surface $X_3$ étudiée dans \cite{huang1}. Rappelons qu'elle est l'éclatement de $\PP^2$ en trois points en position générale. Fixons $Q$ un point dans l'orbite ouverte. Nous savons que $\aess(Q)=3$ et que la classe $C_1$ des droites générales et celle $C_2$ des coniques passant par les trois points éclatés atteignent la constante essentielle. Or les cubiques singulières en $Q$ de degré $6$ (comme par exemple les courbes nodales $\cab$ \eqref{eq:nodal} et les courbes cuspidales $\rab$ \eqref{eq:cusp}) passant par les points éclatés contribuent également à $\aess(Q)$. Mais elles toutes appartiennent à la classe $C_1+C_2$ ainsi qu'au sous-groupe engendré par $C_1$ et $C_2$. Toutes les courbes qui jouent un rôle dans le zoom critique doivent être des combinaisons positives de $C_1$ et $C_2$. Ceci est similaire à ce qui se passe sur $\PP^1\times\PP^1$.
\textbf{Heuristique géométrique.} Pour soutenir ces formules empiriques, nous allons raconter l'analogue géométrique du principe de Batyrev-Manin-Peyre (Conjecture \ref{conj:bmp}).
Il s'agit d'une sorte de \og distribution\fg{}\ des courbes rationnelles sur une variété algébrique. De ce point de vue, cela fournit aussi une interprétation des ordres de grandeur sur $B$ et surtout sur $\log B$.
À la lumière d'un théorème de McKinnon (Théorème \ref{th:singularities}), en prenant en compte seulement des courbes ayant des multiplicités suffisamment grandes (moralement elles croissent de façon proportionnelle par rapport au degré), il s'avère que le terme principal indique les ordres de grandeur actuels pour tous les résultats connus. Nous verrons les détails dans la section \ref{se:heuristicbatyrev}.
L'interprétation empirique déduites de ces exemples est que pour que la dimension de Hausdorff de la mesure limite du zoom critique soit grande, il faudrait \og suffisamment\fg{}\ de familles de courbes rationnelles. En effet, parmi celles qui admettent une mesures limite pour le zoom critique, les surfaces $\PP^1\times\PP^1, X_3$ et $Y_3$ donnent des mesures de dimension de Hausdorff $2$. Alors que pour les autres la dimension est $1$.
Pour $X=\PP^2,X_1,X_2$, comme l'a souligné avant, la seule famille de courbes très libre réalisant $\aess(Q)$ est en fait lisse et \emph{très libres} de degré minimal. Quant à $\PP^1\times\PP^1,X_3$ et $Y_3$, nous pouvons construire une infinité de familles, dans lesquelles chaque courbe rationnelle donne toujours $\aess$. Bien entendu le degré augmente autant qu'on veut, de même pour la multiplicité.
Nous manquons encore d'évidences pour dire plus sur le facteur arithmétique qui apparaît dans le terme principal.
Expliquons comment le zoom critique se traduit en des problèmes arithmétiques et comment la partie mince intervient.
En considérant une famille de courbes nodales $\{\cab\}$ de paramètres $(a,b)\in\NN_{\operatorname{prem}}^2$, nous sommes amené à considérer les équations de Pell-Fermat
$$ax^2-by^2=c,\quad (x,y)\in\ZZ_{\operatorname{prem}}^2,$$
de sorte que l'application rationnelle
\begin{align*}
\PP^1\times\PP^1&\dashrightarrow Y_3\\
[a:b]\times[x:y]&\longrightarrow \cab\hookrightarrow Y_3.
\end{align*}
donne un paramétrage des points rationnels sur $Y_3$ par les $\cab$.
L'idée clef ici est que plutôt que de dénombrer les points sur chaque courbe $\cab$, c'est-à-dire fixer le coupe $(a,b)$ en comptant $(x,y)$, ce qui était la méthode adoptée dans \cite{huang2}, on compte directement les couples $(a,b)\times (x,y)$.
Plus précisément, il s'avère qu'il faut se concentrer sur les solutions des équations du type comme par exemple,
\begin{equation}\label{eq:introduction}
ax^2-by^2=b-a, \quad b-a\mid x-y,\quad b>a,\quad \pgcd(a,b)=1,
\end{equation}
qui équivaut à
$$a(x^2+1)=b(y^2+1),\quad b-a\mid x-y.$$
Nous en déduisons que
$$a\mid y^2+1,\quad b\mid x^2+1,$$
puisque $\pgcd(a,b)=1$.
Nous arrivons donc au paramétrage
$$x^2=bk-1,\quad y^2=ak-1,\quad k\in\NN,$$
et nous voyons que $\pgcd(x^2+1,y^2+1)=k=k(x,y)$ et $$a=\frac{y^2+1}{k(x,y)},\quad b=\frac{x^2+1}{k(x,y)}.$$
La condition $b-a\mid x-y$ s'écrit $(x^2-y^2)/k(x,y)\mid x-y$, qui est équivalente à $x+y \mid k(x,y)$.
Alors la résolution de l'équation se transforme en un problème de congruence $$k(x,y)\equiv 0[x+y]\Leftrightarrow x^2\equiv y^2\equiv -1[x+y].$$
Ce dénombrement correspond à un problème de distribution des racines de congruence, pour lequel nous appliquons un résultat de Hooley (cf. \cite[Theorem 3]{Hooley1}, \cite[Theorem 2]{Hooley2}).
La partie mince entre dans cette histoire en changeant le signe de l'équation \eqref{eq:introduction} comme suit.
$$ax^2-by^2=a-b,\quad b-a\mid x-y,\quad \pgcd(a,b)=1,\quad b>a.$$
Dans ce cas nous sommes amené au problème de congruence
$$x^2\equiv y^2\equiv 1 [x+y].$$
Ceci n'est plus quadratique mais linéaire puisque nous avons de manière équivalente,
$$(x+1)(x-1)\equiv 0[x+y],$$
qui admet en général plus de solutions.
\textbf{Organisation du texte.} Nous rappelons la définition des constante approximation et de l'opération de zoom dans la section \ref{se:approx}. On y trouve aussi une comparaison entre l'heuristique géométrique globale et celle locale (\S\ref{se:heuristicbatyrev}). Une discussion de la géométrie de $Y_3$ occupe la section \ref{se:geometry}, où la famille de courbes nodales et celle de courbes cuspidales sont données en détails (\S\ref{se:curvesony3}). Nous démontrons le théorème \ref{th:main1} dans la section \ref{se:appconstmaj} et nous donnons une majoration uniforme naïve (Proposition \ref{po:uniformupperbound}) dont la preuve est courte, qui est un résultat plus faible que le théorème \ref{th:main2}. Nous donnons le paramétrage des points rationnels par les courbes nodales dans la section \ref{se:parametrization}. La définition de la partie mince est ensuite rappelée dans la section \ref{se:thinsets}. En particulier pour la surface $Y_3$ nous démontrons que l'ensemble des points se situant sur les courbes cuspidales forme une partie mince (\S\ref{se:thinsetiny3}). Dans la section \ref{se:congHooley}, nous réétudions, en suivant Erdös et Hooley, des problèmes de congruence polynomiales dans les cas irréductibles (\S\ref{se:irred}) et scindés (\S\ref{se:splitcase}), avec une référence spéciale sur la équirépartition modulo $1$ des racines de congruence (\S\ref{se:uniformdistmodulo1}). Le dénombrement se déroule dans la section \ref{se:counting}. Nous établions la convergence vague de la famille de mesures de zoom (\S\ref{se:criticalmeasure}) à condition que la partie mince soit retirée (\S\ref{se:countthethinpart}). À la fin, quelques perspectives et questions seront discutés. Les appendices contiennent des rappels de faits généraux sur la théorie de déformation (Annexe \ref{se:deformation}) et sur les courbes rationnelles des variétés toriques (Annexe \ref{se:ratcurveontoric}) et a pour but principal (\S\ref{se:compatibility}) de démontrer la compatibilité des résultats pour les surfaces étudiées auparavant avec les formules \ref{eq:weak} et \ref{eq:conj}.
\textbf{Notations:}
$\mu(\cdot)$ désigne la fonction le Möbius, $\tau$ désigne la fonction donnant le nombre de diviseurs. $p$ est réservé pour les nombres premiers, la condition La lettre $p^\nu \| n$ signifie que $p^\nu\mid n$ et $p^{\nu+1}\nmid n$. On désigne par $a=\square$ pour un nombre réel $a$ si $\sqrt{a}\in\QQ$.
Soit $E\subset\ZZ^n$, on note $E_{\operatorname{prem}}$ le sous-ensemble de $E$ définie par
$$E_{\operatorname{prem}}=\{(x_1,\cdots,x_n)\in E:\pgcd(x_i,1\leqslant i\leqslant n)=1\}.$$ Soient $b,c,m\in\ZZ$. L'expression $b\equiv c[m]$ signifie que $m\mid b-c$.
Définissons une fonction arithmétique $g:\NN_{\geqslant 1}\to\NN_{\geqslant 1}$ donnée par
\begin{equation}\label{eq:functiong}
g(n)=\prod_{p}p^{\lceil \frac{v_p(n)}{2}\rceil},\quad n\in\NN.
\end{equation}
Pour un intervalle $I\subset \RR$, $\textbf{1}_I(\cdot)$ désigne sa fonction caractéristique.
\section{Constante d'approximation et opération de grossissement}\label{se:approx}
\subsection{Constantes d'approximation et constantes essentielles}
La notion de constantes d'approximation fut introduite en premier par D. McKinnon dans \cite{McKinnon2007}. Il s'agit d'une généralisation des notions classiques comme mesure d'irrationalité dans l'approximation diophantienne classique. Elle apparaît déjà dans plusieurs problèmes de distribution locale des points rationnels considérés en premier par S. Pagelot dans \cite{pagelot}. Elle est reprise et étudiée systématiquement par D. McKinnon et M. Roth dans \cite{McKinnon-Roth1}. Nous résumons quelques propriétés dont nous aurons besoin ici brièvement et référons le lecteur à \cite[\S2]{McKinnon-Roth1}, \cite[\S1.1.3]{huang1} et \cite[\S2]{huang2} pour des détails et d'autres propriétés.
Soient $X$ une variété projective irréductible définie sur $\QQ$, $Q\in X(\bar{\QQ})$ un point rationnel fixé et $L$ un fibré en droite gros sur $X$ tel qu'il induise une application birationnelle qui est un morphisme sur un ouvert contenant $Q$. Nous y associons une hauteur de Weil exponentielle $H$ et nous fixons une distance archimédienne $d(\cdot,\cdot)$.
\begin{definition}[McKinnon-Roth, cf. \cite{huang2}, Définition 2.2]
Soit $V$ une partie constructible de $X$. \emph{La constante d'approximation} $\alpha(Q,V)$ est le supremum des $\delta>0$ tels que l'inégalité du type de Liouville suivante
$$\exists C(\delta)>0,\quad \forall y\in V(\QQ)\setminus \{Q\},\quad d(Q,y)^\delta H(y)>C(\delta),$$
soit valide.
\end{definition}
D'après cette définition, la constante $\alpha(Q,V)$ est indépendante du choix de la distance et de la hauteur. Une petite valeur de la constante correspond à de meilleures approximations, contrairement à la mesure d'irrationalité (puisque nous avons déplacé la puissance de la hauteur à la distance).
\begin{definition}[Pagelot]
Soit $V$ une partie constructible de $X$. \emph{La constante essentielle} (par rapport à $V$) est
$$\alpha_{\operatorname{ess}}(Q,V)=\sup_{\substack{Y\subseteq V \text{partie constructible}\\ \text{dense pour la topologie de Zariski induite}}} \alpha(Q,Y).$$
Par convention, $\aess(Q)=\aess(Q,X)$.
$V$ est dit \emph{localement accumulateur} s'il vérifie
$$\alpha_{\operatorname{ess}}(Q,V)<\alpha_{\operatorname{ess}}(Q).$$
\end{definition}
La constante essentielle mesure donc la constante d'approximation générique en le point $Q$ sur $V$, et les variétés localement accumulatrices contiennent des points plus proches de $Q$.
Elle peut être utilisée pour interpréter le principe célèbre dans l'approximation diophantienne: toute approximation suffisamment \og bonne\fg{}\ (i.e. dont la constante d'approximation est plus petite que la constante essentielle) devrait être faite sur certaines sous-variétés fermées strictes.
Les courbes rationnelles jouent un rôle très important dans l'approximation diophantienne. Leur constantes d'approximation sont calculées de la façon suivante. Elles sont sensibles aux singularités.
\begin{theorem}\label{th:singularities}
Soient $C$ une courbe rationnelle définie sur $\QQ$ et $L$ un faisceau inversible ample sur $C$ et $Q\in C(\bar{\QQ})$. Soit $\phi:\PP^1\to C$ le morphisme de normalisation. Alors
$$\alpha(Q,C)=\aess(Q,C)=\min_{P\in\phi^{-1}(Q)}\frac{d}{m_P r_P},$$
où
$d=\deg_C(L)$, $m_P$ est la multiplicité de la branche de $C$ passant par $Q$ correspondant à $P$ et
\begin{equation*}
r(P)=\begin{cases}
0 \text{ si } k(P)\not\subset \RR;\\
1 \text{ si } k(P)=\QQ;\\
2 \text{ sinon},
\end{cases}
\end{equation*}
où par convention, $r_P=0$ signifie que $\frac{d}{m_Pr_P}=\infty$.
\end{theorem}
Nous soulignons ici que ce théorème implique que la seule façon d'augmenter le degré en préservant la constante d'approximation est d'augmenter simultanément \emph{la multiplicité} de la courbe en ce point. Car un point \emph{multiple} donne un certain nombre de branches, ce qui finalement ne contribue qu'à un facteur de division $2$.
\subsection{Opération de grossissement}\label{se:zoomoper}
\subsubsection{Énoncé du problème}
On résume la définition de l'opération de zoom. Pour des détails, voir \cite[\S2.2]{huang2}
On note $\mathcal{C}_{Q}^{\operatorname{b}}(X)$ l'espace vectoriel des fonctions continues à support compact définie sur $(T_Q X)_\RR$ à valeurs réelles.
En ayant fixé un difféomorphisme local
$$\varrho:X(\RR)\dashrightarrow (T_QX)_\RR,$$
pour $U$ une partie constructible de $X$, la famille de mesures de zoom $\{\delta_{U,Q,B,r}\}$ de facteur $r$ par rapport à $U$ est définie par
\begin{equation}\label{eq:zoommeasure}
\int f \operatorname{d}\delta_{U,Q,B,r}=\sum_{x\in U(\QQ):H_L(x)\leqslant B} f(B^{\frac{1}{r}}\rho(x)),\quad \forall f\in\mathcal{C}_{Q}^{\operatorname{b}}(X).
\end{equation}
Plus le facteur de zoom $r$ est grand, plus le zoom est faible (c'est-à-dire on compte plus de points).
Si $r<\alpha(U,Q)$, alors on a $\delta_{U,Q,B,r}\to \delta_Q$ la mesure de Dirac concentré sur $Q$ (\cite[Proposition 2.8]{huang2}). Ce zoom est \og trop fort\fg.
Supposons que $U$ est un ouvert dense de $X$ tel que
$$\alpha(Q,U)=\aess(Q)<\infty.$$
Le zoom plus intéressant est appelé \emph{critique} avec le facteur de zoom $r=\aess(Q)$ car il semble exister souvent un \og saut\fg{}\ de dimension du support de mesures de zoom quand le facteur $r$ traverse $\aess$. On espère aussi que pour $r>\aess(Q)$,
les points se distribuent de façon plus uniforme sans \og trou\fg{} autour de $Q$.
\begin{problem}
Existe-t-il $\beta=\beta(r),\gamma=\gamma(r)\geqslant 0$ et $\delta_r$ une mesure sur $(T_QX)_\RR$ tels que
$$\frac{\delta_{U,Q,B,r}}{B^\beta(\log B)^\gamma}\longrightarrow \delta_r$$
au sens de \emph{convergence vague}?
\end{problem}
\begin{remark}
La formulation de la série \eqref{eq:zoommeasure} dépende \emph{a priori} de la hauteur et du difféomorphisme choisis. La dernière est évidemment fonctorielle: un changement d'une carte locale devrait produire une mesure limite qui est la composée de l'ancienne avec cette application. Compte-tenu de tous les exemples connus, le choix d'une hauteur équivalente ne contribue \emph{a posteriori} qu'à une légère modification sur la mesure limite, à savoir \emph{les conditions de seuils}. La partie de densité reste invariante et donc semble être une propriété intrinsèque et géométrique (cf. aussi la discussion dans la section \ref{se:perspectives})
\end{remark}
\subsubsection{Heuristique de Batyrev-Manin}\label{se:heuristicbm}
La conjecture de Batyrev-Manin-Peyre \ref{conj:bmp} donne une prédiction sur les ordres de grandeur $\beta,\gamma$.
\begin{align*}
\int \chi(\varepsilon) \operatorname{d} \delta_{U,Q,B,r} &= \int \chi(\varepsilon B^{-\frac{1}{r}}) d\delta_{U,B}
\asymp_\varepsilon \operatorname {Vol}(\mathbb{B}(0,\varepsilon B^{-\frac{1}{r}})) B (\log B)^{k-1}\\
&\asymp_\varepsilon B^{-\frac{n}{r}}\times B (\log B)^{k-1}=B^{1-\frac{n}{r}}(\log B)^{k-1}.
\end{align*}
Il se trouve qu'au moins pour $r=\aess$, la prédiction
$$\beta(\aess(Q))=1-\frac{n}{\aess(Q)}$$
est en accord avec tous les exemples connus. Ceci n'est plus le cas pour celle de $\gamma(r)$.
\subsubsection{Heuristique géométrique}\label{se:heuristicbatyrev}
Étant parallèle au principe de Batyrev-Manin pour les variétés définies sur un corps global, sa version géométrique est concernée par la distribution des courbes rationnelles de degré borné sur les variétés, que nous pouvons interpréter comme le dénombrement des points rationnels définis sur la corps de fonction de $\PP^1$. Ceci fut l'objet d'étude de \cite{Peyre4} et \cite{Bourqui} (cf. aussi \cite{LehmannTanimoto}).
Étant donnée une \og belle\fg{}\ variété $X$ définie sur un corps $k$,
i.e. $X$ lisse, projective, géométriquement intègre, le groupe de Néron-Severi $\operatorname{NS}(X)$ est libre de rang fini avec $\operatorname{rg}\operatorname{Pic}(X)=\operatorname{rg}\operatorname{NS}(X)$, et le cône $\mathcal{C}_{\operatorname{eff}}(X)$ est engendré par un nombre fini de classes de diviseurs effectifs.
\begin{comment}
\end{comment}
\noindent\textbf{Principe géométrique global:}
On désigne par $\Mor(\PP^1,X)$ l'espace des morphismes $\PP^1_k\to X$.
Nous avons
$\Mor(\PP^1,X)(k)=X(k(\PP^1))$.
\begin{comment}
Pour tout $\mathcal{C}\in \Mor_k(\PP^1,X)(k):\PP^1\to X$, définissons
\begin{align*}
&h_{\omega_X^{-1}}(\mathcal{C})=\deg (\mathcal{C}^*(\omega_X^{-1})) \quad\text{ hauteur logarithmique},\\
\end{align*}
\end{comment}
Pour $y\in \operatorname{NS}(X)^\vee$, $d\in\NN_{\geqslant 1}$ et $U$ une partie,
$$\Mor_U(\PP^1,X,y)=\{\mathcal{C}\in\Mor(\PP^1,X)(k):\PP^1\to X:\mathcal{C}_*(\PP^1)=y,\operatorname{Im}(\mathcal{C})\not\subset X\setminus U\},$$
$$\Mor_U(\PP^1,X,d)=\{\mathcal{C}\in\Mor(\PP^1,X)(k):\PP^1\to X:\h_{\omega_{X}^{-1}}(\mathcal{C})=d,\operatorname{Im}(\mathcal{C})\not\subset X\setminus U\}.$$
Fixons $q\in\RR_{>1}$. Considérons la fonction zêta globale (cf. \cite[\S6]{LehmannTanimoto})
\begin{equation}\label{eq:heightzetafunction}
\mathcal{Z}_U(s)=\sum_{d\geqslant 1}\sum_{\substack{W \text{ comp. irréd. de}\\ \Mor_U(\PP^1,X,d)}} q^{-sd+\dim W}.
\end{equation}
On devrait y penser comme un analogue de la fonction zêta des hauteurs en caractéristique positive (i.e. $k=\mathbf{F}_q$ un corps fini et $q$ une puissance d'un nombre premier):
$$\sum_{\mathcal{C}\in X(k(\PP^1)) }\frac{1}{H(\mathcal{C})^{s}},$$
où l'on associe à chaque $1$-cycle $\mathcal{C}$ la hauteur exponentielle:
$$H_{\omega_X^{-1}}(\mathcal{C})=q^{\deg_{\omega_X^{-1}}(\mathcal{C})}.$$
Supposons que les hypothèses suivantes sont vérifiées:\\
\emph{Hypothèse I}: Il existe un partie dense $U$ tel que $X\setminus U$ est une partie mince et pour tout $y\in \operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee$ de degré suffisamment grand, l'espace $\Mor_U(\PP^1,X,y)$ est de dimension espérée, i.e.
$$\dim \Mor_U(\PP^1,X,y)=\omega_X^{-1}\cdot y+\dim X.$$
\emph{Hypothèse II}: Le nombre de composantes irréductibles de l'espace $\Mor_U(\PP^1,X,d)$ stabilisent pour tout $d$ suffisamment grand. C'est-à-dire
$$\#\{\text{composantes irréductibles de } \Mor_U(\PP^1,X,d)\}\sim P(d)$$ pour certain polynôme $P(T)\in\RR[T]$, dont le coefficient du terme dominant est noté $c(X)\in\RR$.
\begin{comment}
\emph{Hypothèse III}: Stabilité homologique. C'est-à-dire que pour tout $y\in \operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee$ de degré anti-canonique suffisamment grand,
$$\# \Mor_U(\PP^1,X,y)(k)\sim q^{\omega_{X}^{-1}\cdot y +\dim X}.$$
\end{comment}
Remarquons que si $X$ est une variété \emph{homogène}, les hypothèses I et II sont vérifiées (cf. \cite{Perrin}). De plus tout $y\in \operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee$ est la classe d'une courbe rationnelle libre. Ceci implique aussi que $\deg P(T)=\operatorname{rg}\operatorname{Pic}(X)-1$.
Avec les hypothèses ci-dessus, on devrait avoir l'estimation suivante du terme principal de \eqref{eq:heightzetafunction}. Pour $\Re(s)$ suffisamment grand, en notant $l_0=\operatorname{rg}\operatorname{Pic}(X)$,
\begin{align*}
\mathcal{Z}_U(s)&\sim c(X)\sum_{d\geqslant 1} q^{-sd}\times q^{d+\dim X}\times d^{l_0-1}\\
&=c(X)\sum_{d\geqslant 1} q^{-(s-1)d+\dim X}d^{l_0-1}\\
&=\frac{c(X)q^{\dim X}}{(1-q^{s-1})^{l_0}}H(s),
\end{align*}
où $H$ est une fonction holomorphe en $\Re(s)>\sigma_0\in]0,1[$. Ce calcul nous dit que la fonction $\mathcal{Z}_U(s)$ a un pôle d'ordre $l_0$ en $s=1$, d'où un raisonnement taubérian implique que \emph{la fonction de comptage associée}
$$\#\{\mathcal{C}\in \operatorname{NS}(X)^\vee,q^{\deg_{\omega_X^{-1}} (\mathcal{C})}\leqslant B\}\sim c^\prime(X) B(\log B)^{l_0-1}.$$
\begin{comment}
\begin{align*}
\# \Mor_U(\PP^1,X,d)(k)&=\sum_{\substack{y\in\operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee\\ \omega_{X}^{-1}\cdot y=d}}\# \Mor_U(\PP^1,X,y)(k)\\
&\sim \sum_{\substack{y\in\operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee\\ \omega_{X}^{-1}\cdot y= d}} q^{\omega_{X}^{-1}\cdot y+\dim X}\\
&\sim q^d\# \{y\in \operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee:\omega_{X}^{-1}\cdot y=d\}\\
&\sim q^d d^{\operatorname{rg}\operatorname{Pic}(X)-1}.
\end{align*}
Si l'on note $B=q^d$, alors $d\sim \log B$. On retrouve alors le terme principal bien connu $B(\log B)^{\operatorname{rg}\operatorname{Pic}(X)-1}$ sous cette identification.
\end{comment}
\noindent\textbf{Principe géométrique local:}
Désormais nous basculons vers le sous-espace paramétrant les courbes rationnelles ayant des \emph{multiplicités restrictives}. Nous référons à l'annexe A (Définition \ref{def:restrictmul}) pour les détails.
Fixons $Q\in U(k)$. Moralement une courbe $f:\PP^1\to X$ a la multiplicité \emph{restrictive} $\mathcal{M}_Q^{\operatorname{res}}=N\in\NN$ en $Q\in \operatorname{Im}(f)(\bar{k})$ si soit elle est de multiplicité cuspidale $N$ en $Q$, soit $2\mid N$ et elle est multi-branchée est en certaines branches $Q$ la multiplicité cuspidale est $\frac{N}{2}$. La raison pour laquelle nous nous bornons à ces types de courbes vient du théorème \ref{th:singularities}. Ce sont les courbes dont le degré divisé par $\mathcal{M}_Q^{\operatorname{res}}$ pourraient atteindre la constante d'approximation plus petite que possible ainsi que entrent dans certain zoom de facteur donné.
Nous considérons
$$\Mor_Q(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant N)=\{\mathcal{C}\in \Mor_U(\PP^1,X,d):Q\in \operatorname{Im}(\mathcal{C})(\bar{k}),\mathcal{M}_Q^{\operatorname{res}}\geqslant N\}.$$
Soit $r>0$, où on devrait penser comme le facteur de zoom donné.
\noindent\emph{Hypothèse I'}: Supposons qu'il n'y a pas d'obstruction à la déformation de toute classe de courbes de degré $d\gg 1$ en fixant le point $Q$ et conservant sa multiplicité restrictive.
Nous avons d'après le théorème \ref{th:expdim} et la proposition \ref{prop:expecteddim}, si $\Mor_Q\left(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant\frac{d}{r}\right)\neq\varnothing$, il est de dimension espéré
$$\dim \Mor_Q\left(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant \frac{d}{r}\right)=d\left(1-\frac{\dim X}{r}\right)+O(1),$$
où la constante implicite est toujours bornée en terme de la dimension de $X$
Notons $l_r(X)$ le rang du sous-groupe de $\operatorname{NS}(X)^\vee$ engendré par les classes des courbes dans $\Mor_Q\left(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant\frac{d}{r}\right)$. Supposons de plus que cet espace et sa classe de courbes satisfont à
\noindent\emph{Hypothèse II'}:
$$\#\{\text{composantes irréductibles de } \Mor_U(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant\frac{d}{r})\}\sim R(d),\forall d\gg 1$$
où $R(T)\in\RR[T]$ est degré $l_r(X)-1$.
Nous remarquons que le cône engendré par les classes de courbes dans $\Mor_U(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant\frac{d}{r}),d\geqslant 1$ est une \emph{face} de dimension $l_r(X)$ du cône polyédral engendré par $y\in\operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee$.
Considérons maintenant la fonction zêta locale
\begin{equation}\label{eq:heightzetalocal}
\mathfrak{Z}_{U,r}(s)=\sum_{d\geqslant 1}\sum_{\substack{W \text{ comp. irréd. de}\\ \Mor_U(\PP^1,X,d,\mathcal{M}_Q^{\operatorname{res}}\geqslant\frac{d}{r})}} q^{-sd+\dim W}.
\end{equation}
Nous devrions avoir qu'il existe $e(r,X,q)>0$ tel que
\begin{align*}
\mathfrak{Z}_{U,r}(s)&\sim e(r,X,q)\sum_{d\geqslant 1} q^{-sd}\times q^{d\left(1-\frac{\dim X}{r}\right)} \times d^{l_r-1}\\
&=e(r,X,q)\sum_{d\geqslant 1} q^{-d(s-1+\frac{\dim X}{r})} d^{l_r-1}\\
&=\frac{e(r,X,q)}{(1-q^{s-1+\frac{\dim X}{r}})^{l_r}}G(s),
\end{align*}
où $G(s)$ est holomorphe. On en conclut que $\mathfrak{Z}_{U,r}(s)$ devrait admettre un pôle en $s=1-\frac{\dim X}{r}$ d'ordre $l_r$.
\begin{comment}
\begin{align*}
&\# \Mor_Q\left(\PP^1,X,d,\geqslant\frac{d}{r}\right)(k)\\
&\sim q^{d\left(1-\frac{\dim X}{r}\right)}\# \{y\in \operatorname{NS}(X)^\vee\cap \mathcal{C}_{\operatorname{eff}}(X)^\vee:\exists \mathcal{C}\in\Mor_Q\left(\PP^1,X,d,\geqslant\frac{d}{r}\right),\mathcal{C}_*(\PP^1)=y, \omega_{X}^{-1}\cdot y=d \}\\
&\sim q^{d\left(1-\frac{\dim X}{r}\right)} d^{l_r(X)-1}.
\end{align*}
\end{comment}
L'heuristique qu'on obtient de cette estimation est qu'en outre la puissance $1-\frac{\dim X}{r}$ sur $B$, celle sur $\log B$ devrait être $l_r(X)-1$, plus petite que l'exposante $\operatorname{rg}\operatorname{Pic}(X)-1$ provenant de l'heuristique \ref{se:heuristicbm}.
\subsubsection{Variétés localement faiblement accumulatrices}
L'opération de zoom permet de détecter des sous-variétés possédant localement beaucoup de points dont le nombre domine celui sur le complémentaire même si leur constantes d'approximation ne soient pas plus petites.
\begin{definition}[\cite{huang2}, Définition 2.7]
Supposons que $\aess(Q)<\infty$. Un sous-ensemble $W\subset Y_3(\QQ)$ est dit \emph{localement faiblement accumulateur} si $\aess(Q,W)=\aess(Q)$ et qu'il existe un ouvert $U$ dense de $X$ vérifiant $\alpha(Q,U)=\aess(Q)$ tel que pour toute partie dense $V$ de $W$ (pour la topologie induite) satisfaisant à $\alpha(Q,V)=\aess(Q)$ et pour tout $\varepsilon>0$ suffisamment grand, on ait
$$\int \chi(\varepsilon)\operatorname{d}\delta_{U\setminus W,Q,B,\aess(Q)}=o\left(\int \chi(\varepsilon)\operatorname{d}\delta_{V,Q,B,\aess(Q)} \right),\quad B\to\infty.$$
\end{definition}
\subsection{Comparaison entre les courbes nodales et les courbes cuspidales}\label{se:compcuspnodal}
Comparons maintenant le zoom critique pour des courbes nodales et celui pour des courbes cuspidales.
Le premier est essentiellement l'approximation des nombres \emph{quadratiques irrationnels} par des nombres rationnels lorsque les tangents en le point nodal sont irrationnelles. Le théorème classique sur la relation entre la résolution des équations de Pell-Fermat et l'approximation diophantienne des nombres quadratiques (cf. \cite[Chapitres 3, 4]{huang2} pour les détails) nous permet de déduire que les meilleurs approximants s'obtiennent de façon discrète. Mais l'approximation sur les courbes nodales est justement l'approximation des nombre \emph{rationnels} si la tangente en le point de rebroussement est rationnelle.
Par simplicité nous prendrons des courbes cubiques affines d'équation simple se plongeant dans $\PP^2$, toutes les courbes de singularités simples étant homéomorphes localement aux deux cas suivants.
On munit le fibré $\mathcal{O}(1)$ d'une hauteur de Weil absolue exponentielle $H$.
Choisissons le point $Q=[0:0:1]$ à approcher et
définissons la distance
$$d([u:v:w])=\max(\left|\frac{u}{w}\right|,\left|\frac{v}{w}\right|).$$
\noindent \textbf{Cas I.} Considérons la courbe nodale $\mathcal{C}:y^2=x^3+ax^2$ pour $a\in\QQ_{\geqslant 0},a\neq\square$.
Les tangentes en points $(0,0)$ sont de pentes $\pm\sqrt{a}$.
Un paramétrage est donné par
$$k\longmapsto (k^2-a,k(k^2-a)).$$
On voit alors que la distance induite est $|k^2-a|$, qui est équivalente à
$$\min(|k-\sqrt{a}|,|k+\sqrt{a}|).$$
Donc pour approcher l'origine $(0,0)$, il faut que $k$ approche $\pm \sqrt{a}$.
La constante d'approximation est égale au degré (par rapport à $\mathcal{O}(1)$) divisé par $2$ ($r_P=2,\forall P$ dans le théorème \ref{th:singularities}).
\begin{theorem}[\cite{huang2} Théorème 1.3]
Nous avons $\alpha(Q,\mathcal{C})=\alpha_{\operatorname{ess}}(Q,\mathcal{C})=\frac{3}{2}$, et
$$\delta_{\mathcal{C},Q,\frac{3}{2},B}(\chi(\varepsilon))=O_\varepsilon(1).$$
Il n'existe pas de mesure limite sur cette courbe.
\end{theorem}
\noindent \textbf{Cas II.} Considérons la courbe cuspidale $\mathcal{C}^\prime:y^2=x^3$ avec le point de rebroussement de première espèce $Q$ ayant la tangente $y=0$.
Le point de clef est que quand on approche l'origine en suivant cette courbe, la distance diminue à la puissance $2$. En effet, le paramétrage usuel pour cette courbe est
$$t \longmapsto (t^2,t^3).$$
D'où la distance induite se calcule comme $|t^2|$.
La constante d'approximation est alors divisée par $2$ ($m_P=2$ dans le théorème \ref{th:singularities}).
\begin{theorem}[Pagelot, \cite{pagelot}]\label{th:Pagelot}
Nous avons $\alpha(Q,\mathcal{C}^\prime)=\aess(Q,\mathcal{C}^\prime)=\frac{3}{2}$ et que
$$\frac{1}{B^\frac{3}{4}}\delta_{\mathcal{C}^\prime,Q,\frac{3}{2},B}(\chi(\varepsilon))\to \delta_{\frac{3}{2}},$$
une mesure de dimension de Hausdorff $1$ à support dans $\mathcal{C}^\prime$.
\end{theorem}
\section{Géométrie et courbes rationnelles sur $Y_3$}\label{se:geometry}
Nous considérons dans cet article la surface torique obtenue en éclatant $3$ des $4$ points invariants de $\PP^1\times\PP^1$.
Nous désignerons par $Y_3$ cette surface puisque $Y_4$ a été réservée pour la surface étudiée dans \cite{huang2}.
Sans perte de généralité, on peut supposer que $Y_3$ est l'éclatement en
$$P_1=[1:0]\times [1:0],\quad P_2=[0:1]\times[1:0],\quad P_3=[1:0]\times[0:1].$$
\subsection{Géométrie de $Y_3$}
Nous utilisons les coordonnées $[x:y]\times[s:t]$ de $\PP^1\times\PP^1$ pour les points différents de $P_i,1\leqslant i\leqslant 3$.
Rappelons l'éventail de $Y_3$ (Figure \ref{fg:y3}). Les relations primitives minimales centrées sont
$$\rho_1+\rho_4=0,\quad \rho_2+\rho_6=0,\quad \rho_3+\rho_7=0.$$
Elles représentent les classes des (transformations strictes des) courbes rationnelles $l_i,1\leqslant i\leqslant 3$ définies par les équations
\begin{equation}\label{eq:curvesmalldegree}
y=x,\quad t=s,\quad yt=xs.
\end{equation}
On a trois diviseurs exceptionnels $E_i~(1\leqslant i\leqslant 3)$. On note $Z=\cup_{i=1}^3 l_i$ et $U=Y_3\setminus Z$.
Le diviseur anti-canonique dans ce cas là est $$\omega_{Y_3}^{-1}=\mathcal{O}(2,0)+\mathcal{O}(0,2)-E_1-E_2-E_3,$$
avec une base $S$ de ses sections globales données par
\begin{equation}\label{eq:basisofglobalsec}
x^2 st,\quad y^2 st,\quad t^2 xy,\quad s^2 xy,\quad xyst,\quad y^2t^2.
\end{equation}
On identifie localement $T_Q Y_3$ sur la carte $(y\neq 0)\cap (t\neq 0)$ à l'espace affine $\mathbf{A}^2$ via
\begin{equation}\label{eq:diffeomorphism}
\varrho:[x:y]\times [s:t]\longmapsto (w,z)=\left(\frac{x}{y}-1,\frac{s}{t}-1\right),
\end{equation}
sur lequel on utilise la distance
\begin{equation}\label{eq:distance}
d((w,z),\varrho(Q))=\max(|w|,|z|).
\end{equation}
La courbe $yt=xs$ s'écrivant $wz+w+z=0$ (une hyperbole) sous ce difféomorphisme divise $\mathbf{A}^2$ en deux régions
\begin{equation}\label{eq:s1s2}
R_1=\{(w,z):wz+w+z>0\},\quad R_2=\{(w,z):wz+w+z<0\}.
\end{equation}
\subsection{Calcul de hauteur}\label{se:height}
Nous utilisons la hauteur de Weil absolue associée à $\omega_{Y_3}^{-1}$ définie par
$$H(P)=\frac{\max_{f\in S}(|f(P)|)}{\pgcd(f(P),f\in S)}, \quad P\in (Y_3\setminus \cup_{i=1}^3 E_i)(\QQ).$$
Pour un point $P=[x:y]\times[s:t]$ avec $$\pgcd(x,y)=\pgcd(s,t)=1,$$ on a
\begin{align}
&\pgcd(x^2 st,y^2 st,t^2 xy,s^2 xy,xyst,y^2t^2)\nonumber\\
&=\pgcd(\pgcd(x,s)\pgcd(x,t)\pgcd(y,s)\pgcd(y,t),y^2 t^2)\nonumber\\
&=\pgcd(x,t)\pgcd(y,s)\pgcd(y,t) \label{eq:pgcd}
\end{align}
\subsection{Symétries}
De la structure de l'éventail de $Y_3$, nous constatons qu'il est symétrique par rapport à la droite engendrée par le $3$-ième rayon. Cette surface admet donc l'automorphisme étendant celui de $\PP^1\times\PP^1$ s'écrivant en coordonnées homogènes comme suit.
\begin{equation}\label{eq:symmetry}
\Phi:[x:y]\times[s:t]\longmapsto [s:t]\times[x:y].
\end{equation}
Il fixe le diviseur exceptionnel $E_1$ et échange $E_2$ avec $E_3$
et il préserve le diviseur anti-canonique $\omega_{Y_3}^{-1}$ ainsi que la hauteur associée définie dans la section \ref{se:height}.
Dans les coordonnées $(w,z)$ \eqref{eq:diffeomorphism}, l'application $\Phi$ n'est rien d'autre que la permutation des coordonnées
$$(\varrho\circ \Phi\circ\varrho^{-1})(w,z)=(z,w).$$
Cette symétrie nous permet de se ramener l'étude aux régions
\begin{equation}\label{eq:S1}
S_1=\{(w,z)\in\RR^2:w>0,z>w\},
\end{equation}
\begin{equation}\label{eq:S2}
S_2=\{(w,z)\in\RR^2:w<0,wz+w+z>0\},
\end{equation}
\begin{equation}\label{eq:S3}
S_3=\{(w,z)\in\RR^2:w>0,wz+w+z<0\},
\end{equation}
\begin{equation}\label{eq:S4}
S_4=\{(w,z)\in\RR^2:w<0,z<w\}.
\end{equation}
On a
$$S_1\cup S_2\subset R_1,\quad S_3\cup S_4\subset R_2.$$
\subsection{Courbes nodales et courbes cuspidales sur $Y_3$}\label{se:curvesony3}
\subsubsection{L'espace $\Mor_{5,Q}(\PP^1,Y_3)$}
On note $\Mor_{5,Q}(\PP^1,Y_3)$ l'espace module des courbes rationnelles qui passent par $Q$ de degré anticanonique $5$. Il a plusieurs composantes irréductibles. Nous nous intéressons particulièrement à la composante $\mathfrak{Mor}$ qui correspond à la relation $$\mathcal{P}:\rho_2+\rho_3+\rho_4+\rho_6+\rho_7=0.$$
Notons $\mathfrak{Comp}$ l'image de $\mathfrak{Mor}$ dans $\operatorname{Chow}(X)$.
Nous avons
\begin{align*}
\dim \mathfrak{Comp}=\deg_{\omega_{Y_3}^{-1}}(\mathcal{P})-\dim(\operatorname{Aut}(\PP^1))=5-3=2.
\end{align*}
Ceci dit moralement que les courbes rationnelles appartenant à $\mathfrak{Comp}$ forment une famille de dimension $2$.
En effet, si l'on considère $H^0(Y_3,\omega_{Y_3}^{-1})$, il contient précisément les sections de $\omega_{\PP^1\times\PP^1}^{-1}$ qui s'annule en les points éclatés. Cela impose $3$ conditions linéaires et on vérifie donc que
\begin{align*}
\dim H^0(Y_3,\omega_{Y_3}^{-1})&=\dim H^0(\PP^1\times\PP^1,\omega_{\PP^1\times\PP^1}^{-1})-3 \\
&=3\times 3-3=6.
\end{align*}
En ajoutant la condition que la courbe passe par $Q$ et soit singulière en $Q$, on obtient $3$ conditions linéaires (dont $2$ correspondent à l'annulation des dérivées partielles en $Q$) de plus sur les coefficients. Notons $\mathfrak{V}$ ce sous-espace de $H^0(Y_3,\omega_{Y_3}^{-1})$. La dimension de la projectivisé $\PP(\mathfrak{V})$ est $2$. Puisque le lieu des zéros d'une section irréductible devient une équation cubique dans les coordonnées $(w,z)$, une section irréductible dans $\mathfrak{V}$ définit une courbe cubique singulière en $Q$, donc rationnelle.
Nous identifions donc les courbes dans $\mathfrak{Comp}$ aux membres de $\PP(\mathfrak{V})$ qui déterminent les courbes à $3$ paramètres $c,d,e\in\ZZ^3_{\operatorname{prem}}$ d'équations
$$c(y-x)^2st+d(t-s)^2xy+eyt(y-x)(t-s)=0.$$
Elles correspondent à des courbes cubiques dans les coordonnées $(w,z)$ singulières en $Q$.
L'annulation de la matrice Hessienne donne que les tangentes en $Q$ coïncident si et seulement si $e^2=4cd$, ce qui revient à la condition qu'il existe $(a,b)\in\ZZ^2_{\operatorname{prem}}$ tel que
\begin{equation}\label{eq:coeffcusp}
c=a^2,\quad d=b^2,\quad e=-2ab.
\end{equation}
\subsubsection{Famille de courbes cuspidales}\label{se:cusp}
Nous obtenons donc la famille de courbes cuspidales
\begin{equation}\label{eq:cusp}
\rab:b^2(y-x)^2st+a^2(t-s)^2xy- 2abyt(y-x)(t-s)=0.
\end{equation}
Dans les coordonnées $(w,z)$, elles s'écrivent
\begin{equation}\label{eq:cuspplane}
(az-bw)^2+zw(a^2z+b^2 w)=0.
\end{equation}
Nous avons (rappelons l'automorphisme $\Phi$ \eqref{eq:symmetry}) $\Phi(R_{a,b})=R_{b,a}$ et la pente de la tangente en $Q=(0,0)$ est $\frac{b}{a}$.
Le paramétrage donné par les paramètres $u,v,a,b$ n'est pas surjectif. L'image est un ensemble \emph{mince} de la région $S_2$. Nous en discutons plus loin dans la section \ref{se:thinsets}.
\subsubsection{Famille de courbes nodales}\label{se:nodal}
Si les coefficients $c,d,e$ ne vérifient pas la condition \eqref{eq:coeffcusp}, nous obtenons en général une famille de courbes nodales à $3$ paramètres. Celles qui nous intéresse le plus est la sous-famille suivante, également utilisée pour l'étude de la surface $Y_4$ dans \cite{huang2}.
\begin{equation}\label{eq:nodal}
\cab: axy(s-t)^2=bst(x-y)^2,\quad (a,b)\in\ZZ_{\operatorname{prem}}^2.
\end{equation}
Les tangentes en $Q=[1:1]\times[1:1]$ ont les pentes $\pm\sqrt{\frac{b}{a}}$. Elles sont donc irrationnelles si et seulement si $ab\neq \square$.
\section{Déduction des constantes d'approximation et majoration naïve}\label{se:appconstmaj}
\subsection{Bornes inférieures uniformes}
Nous allons démontrer que la meilleure constante d'approximation est $2$ par une estimation directe.
\begin{proposition}\label{po:lowerbound1}
Nous avons
$$\alpha(Q,Y_3)=2.$$
\end{proposition}
\begin{proof}
Pour $P=[x:y]\times[s:t] \neq Q$, supposons que $s\neq t$. On a alors
\begin{align*}
\h_{\omega_{Y_3}^{-1}}(P) d(P)^2 &\geqslant\frac{|t^2 xy|}{\pgcd(x,t)\pgcd(y,s)\pgcd(y,t)} \left(\frac{s}{t}-1\right)^2\\
&=\frac{|xy|}{\pgcd(x,t)\pgcd(y,s)\pgcd(y,t)}(s-t)^2\\
&\geqslant 1.
\end{align*}
Cela montre que $\alpha(Q,Y_3)\geqslant 2$. Mais les droites spéciales $l_i,1\leqslant i\leqslant 3$ de \eqref{eq:curvesmalldegree} vérifient $\alpha(Q,l_i)=2$. D'où
$\alpha(Q,Y_3)\leqslant \alpha(Q,l_i)=2$.
Ceci clôt la démonstration.
\end{proof}
\subsection{Constante essentielle}
Ensuite on déduit une borne inférieure pour les points généraux dans $Y_3$. On va l'utiliser pour obtenir la constante essentielle.
\begin{proposition}\label{po:lowerbound2}
Pour $P=[x:y]\times[s:t]$ n'appartenant pas aux trois droites spéciales, à savoir
\begin{equation}\label{eq:tmp}
x\neq y, \quad s\neq t,\quad xs\neq yt,
\end{equation}
avec
$$d(P)\leqslant C,\quad C>0,$$
il existe $D=D(C)>0$ tel que
$$ \h_{\omega_{Y_3}^{-1}}(P) d(P)^\frac{5}{2}\geqslant D.$$
\end{proposition}
\begin{proof}
Commençons par l'estimation
$$\left|\frac{xs}{yt}-1\right|=\left|\frac{x}{y}\left(\frac{s}{t}-1\right)+\frac{x}{y}-1\right|\leqslant (2+C)\max\left(\left|\frac{x}{y}-1\right|,\left|\frac{s}{t}-1\right|\right)=(2+C)d(P).$$
Rappelons l'hypothèse \eqref{eq:tmp}, nous avons alors
\begin{align*}
&\h_{\omega_{Y_3}^{-1}}(P) d(P)^\frac{5}{2}\\
&\geqslant \frac{|xyst|}{\pgcd(x,t)\pgcd(y,t)\pgcd(y,s)}\left|\frac{y}{x}-1\right|\left|\frac{t}{s}-1\right|\left|\frac{xs}{yt}-1\right|^\frac{1}{2}(2+C)^{-\frac{1}{2}}\\
&=\left(\frac{|y|}{\pgcd(y,t)\pgcd(y,s)}\frac{|t|}{\pgcd(y,t)\pgcd(x,t)}\frac{|yt-xs|}{\pgcd(x,t)\pgcd(y,s)}\right)^\frac{1}{2} |x-y||t-s|(2+C)^{-\frac{1}{2}}\\
&\geqslant (2+C)^{-\frac{1}{2}}
\end{align*}
\end{proof}
\begin{corollary}\label{co:aess}
Nous avons
\begin{equation}\label{eq:tmp2}
\aess(Q)=\frac{5}{2}.
\end{equation}
Les courbes \eqref{eq:curvesmalldegree} sont les variétés localement accumulatrices.
\end{corollary}
\begin{proof}
Les courbes nodales $C_{a,b},ab\neq \square$ vérifient $\alpha(C_{a,b},Q)=\frac{5}{2}$ et leur réunion est dense dans l'ouvert $U=Y_3\setminus (\cup_{i=1}^3 l_i)$. D'où $$\aess(Q)\leqslant \alpha(Q,C_{a,b})=\frac{5}{2}.$$
D'après la proposition \ref{po:lowerbound2}, nous savons
$$\aess(Q)\geqslant \alpha(Q,U)\geqslant\frac{5}{2}.$$
Cela démontre \eqref{eq:tmp2}.
Alors que les courbes $l_i,1\leqslant i\leqslant 3$ de \eqref{eq:curvesmalldegree} sont lisse et de degré $2$ et donc vérifient que $\alpha(Q,l_i)=2$ (théorème \ref{th:singularities}).
\end{proof}
\subsection{Majoration uniforme}
Cette section est consacrée à la démonstration d'une majoration pour le zoom critique, qui est en faveur de l'heuristique de Batyrev-Manin.
\begin{proposition}\label{po:uniformupperbound}
Pour $\varepsilon>0$ fixé, on a que, pour $B$ suffisamment grand et pour tout $\delta>0$,
$$\delta_{U,Q,B,\frac{5}{2}}(\chi(\varepsilon))\ll_{\delta,\varepsilon }B^{\frac{1}{5}+\delta}.$$
\end{proposition}
\begin{proof}
Écrivons la condition de zoom
$$\h_{\omega_{Y_3}^{-1}}(P)\leqslant B,\quad d(P)\ll_\varepsilon B^{-\frac{2}{5}}.$$
Introduisons les notations
\begin{equation}\label{eq:ei}
e_1=\pgcd(y,t),\quad e_2=\pgcd(x,t),\quad e_3=\pgcd(y,s);
\end{equation}
\begin{equation}\label{eq:fi}
f_1=\frac{|y|}{e_1e_3},\quad f_2=\frac{|t|}{e_1e_2}, \quad f_3=\frac{|yt-xs|}{e_2e_3};
\end{equation}
\begin{equation}\label{eq:gi}
g_1=\frac{|x|}{e_2},\quad g_2=\frac{|s|}{e_3}.
\end{equation}
Reprenons la démonstration de la proposition \ref{po:lowerbound2},
$$1\gg_\varepsilon\h_{\omega_{Y_3}^{-1}}(P)d(P)^\frac{5}{2}\gg_\varepsilon(f_1f_2f_3)^\frac{1}{2} |x-y||t-s|.$$
On voit que
\begin{equation}\label{eq:m0}
|x-y|,\quad |t-s|,\quad f_1,\quad f_2,\quad f_3\quad \ll_\varepsilon 1.
\end{equation}
Donc le dénombrement des points $[x:y]\times[s:t]$ est équivalent à celui des paramètres $e_1,e_2,e_3,g_1,g_2$.
On va en déduire des encadrements des paramètres $e_i,g_i$,
afin de démontrer qu'ayant fixé $g_1,g_2,e_1$, il n'y a qu'un nombre fini (dépendant de $\varepsilon$) de choix pour $e_2,e_3$.
La condition de zoom
$$\left|\frac{x-y}{y}\right|\ll_\varepsilon B^{-\frac{2}{5}},\quad \left|\frac{s-t}{t}\right|\ll_\varepsilon B^{-\frac{2}{5}}$$
implique que
\begin{equation}\label{eq:m1}
|t|\gg_\varepsilon B^{\frac{2}{5}},\quad |y|\gg_\varepsilon B^{\frac{2}{5}}.
\end{equation}
D'où l'on déduit
\begin{equation}\label{eq:m2}
|s|\gg_\varepsilon B^{\frac{2}{5}},\quad |x|\gg_\varepsilon B^{\frac{2}{5}}.
\end{equation}
La condition que la hauteur soit bornée implique que $$|xs|e_1^2e_2e_3 \leqslant|xs|f_1f_2e_1^2e_2e_3= |xyst|\leqslant Be_1e_2e_3.$$
D'où l'on déduit, grâce à \eqref{eq:m2}
\begin{equation}\label{eq:m3}
e_1\ll_{\varepsilon}B^{\frac{1}{5}}.
\end{equation}
D'après \eqref{eq:m0}, $$f_3=\frac{|yt-xs|}{e_2e_3}=|f_1f_2e_1^2-g_1g_2|\ll_{\varepsilon} 1,$$
on voit qu'une fois que $f_1$ et $e_1$ sont choisis, il n'y a qu'un nombre fini de valeurs possibles pour le produit $g_1g_2$.
Puisque
\begin{equation}\label{eq:m4}
\# \{(n_1,n_2)\in\NN_{\geqslant 1}^2:n_1n_2=n\}=\tau(n)=O(n^\delta)
\end{equation}
pour tout $\delta>0$,
on a donc démontré qu'ayant fixé $f_1,e_1$, le nombre de possibilités pour le couple $(g_1,g_2)$ est $O_{\delta,\varepsilon}(B^\delta)$.
Il nous reste à déterminer $e_2,e_3$. Soit $C_2(\varepsilon)>0$ tel que $$\max(|y-x|,|t-s|)\leqslant C_2(\varepsilon).$$ Fixons $e_1$ et $f_1$ ainsi que $g_1$ et $g_2$. Comme $yt\neq xs$, on a $$f_1f_2e_1^2\neq g_1g_2.$$
Démontrons que le cardinal des $(e_2,e_3)$ possibles est majoré par (compte tenu des signes possibles de $x,y,s,t$)
$$4\#\{(u,v)\in\ZZ_{\operatorname{prem}}^2:|ug_1-f_1e_1v|\leqslant C_2(\varepsilon)\text{ et }|vg_2-f_2e_1u|\leqslant C_2(\varepsilon)\},$$
où on compte les points primitifs sur un réseau du déterminant $|f_3|$ dans un carré d'aire $4C_2(\varepsilon)^2$ dont l'intérieur contient l'origine.
On en conclut qu'il n'y a qu'un nombre fini de choix pour $e_2,e_3$ une fois que $f_1,e_1,g_1,g_2$ sont choisis (\cite[Lemma 2]{Heath-Brown2}).
La majoration qui fallait démontrer résulte de \eqref{eq:m0}, \eqref{eq:m3} et \eqref{eq:m4}.
\end{proof}
\section{Paramétrage par des courbes nodales}\label{se:parametrization}
On rappelle brièvement le paramétrage donné par la famille de courbes nodales \eqref{eq:nodal} utilisé de manière cruciale dans \cite{huang2}.
L'équation de $\cab$ s'écrit sous les coordonnées $(w,z)$ comme
$$a(1+w)z^2=b(1+z)w^2.$$
Le paramétrage rationnel que l'on utilise est (cf. \cite[5.3.1]{huang2})
\begin{equation}\label{eq:paracab}
\Psi:[a:b]\times[u:v]\longmapsto[x:y]\times[s:t]= \left[\frac{bv(u-v)}{D_1d_2d_3}:\frac{u(bv-au)}{D_1d_2d_3}\right]\times \left[\frac{au(u-v)}{d_1D_2d_3}:\frac{v(bv-au)}{d_1D_2d_3}\right],
\end{equation}
où
\begin{equation}\label{eq:D1D2}
\begin{split}
d_1=\pgcd(u,b),\quad d_2=\pgcd(v,a),\quad d_3=\pgcd(u-v,b-a),\\
D_1=\pgcd(u^2,b),\quad D_2=\pgcd(v^2,a).
\end{split}
\end{equation}
Nous avons (cf. \cite[5.12]{huang2})
\begin{equation}\label{eq:pgcd2}
\pgcd(x,t)=\frac{v}{d_2},\quad \pgcd(y,s)=\frac{u}{d_1},\quad \pgcd(y,t)=\frac{bv-au}{d_1d_2d_3}
\end{equation}
En le composant avec le difféomorphisme $\varrho$, nous obtenons donc le paramétrage des points dans les régions $S_i,1\leqslant i\leqslant 4$ comme suit:
\begin{equation}\label{eq:wz}
(\varrho\circ\Psi) \left((a,b)\times(u,v)\right)=(w,z)=\left(\frac{au^2-bv^2}{u(bv-au)},\frac{au^2-bv^2}{v(bv-au)}\right).
\end{equation}
En particulier nous voyons que $\frac{z}{w}=\frac{u}{v}$,
ce qui signifie que $\frac{u}{v}$ est la pente du point $(w,z)$.
Soient
\begin{equation}\label{eq:T1}
T_1=\left\{(a,b)\times(u,v)\in\NN_{\operatorname{prem}}^2\times\ZZ_{\operatorname{prem}}^2:b>a>0,u>v>0,\sqrt{\frac{b}{a}}<\frac{u}{v}<\frac{b}{a},\right\},
\end{equation}
\begin{equation}\label{eq:T2}
T_2=\left\{(a,b)\times(u,v)\in\NN_{\operatorname{prem}}^2\times\ZZ_{\operatorname{prem}}^2:b>a>0,u>-v>0,-\frac{u}{v}>\sqrt{\frac{b}{a}}\right\},
\end{equation}
\begin{equation}\label{eq:T3}
T_3=\left\{(a,b)\times(u,v)\in\NN_{\operatorname{prem}}^2\times\ZZ_{\operatorname{prem}}^2:b>a>0,u>-v>0,-\frac{u}{v}<\sqrt{\frac{b}{a}}\right\},
\end{equation}
\begin{equation}\label{eq:T4}
T_4=\left\{(a,b)\times(u,v)\in\NN_{\operatorname{prem}}^2\times\ZZ_{\operatorname{prem}}^2:b>a>0,u>v>0,\frac{u}{v}<\sqrt{\frac{b}{a}}\right\},
\end{equation}
\begin{lemma}\label{le:au2-bv2}
Pour $P=(a,b)\times (u,v)\in \cup_{i=1}^4 T_i$, nous avons $au^2-bv^2<0$ si et seulement si $P\in T_3\cup T_4$.
\end{lemma}
\begin{proof}
$$au^2-bv^2<0\Leftrightarrow -\sqrt{\frac{b}{a}}<\frac{u}{v}<\sqrt{\frac{b}{a}}\Leftrightarrow (a,b)\times (u,v)\in T_3\cup T_4.$$
\end{proof}
\begin{proposition}
L'application $\varrho\circ\Psi$ donne une bijection entre l'ensemble $T_i$ et l'ensemble des $(w,z)\in\QQ^2$ satisfaisant à $\max(|w|,|z|)<1$ dans $S_i$ pour tout $1\leqslant i\leqslant 4$.
\end{proposition}
\begin{proof}
(cf. la démonstration du lemme 5.4 dans \cite{huang2}) Étant donné $(w,z)\in \cup_{i=1}^4 S_i$ tel que $\max(|w|,|z|)<1$., les égalités
\begin{equation}\label{eq:inter1}
\frac{b}{a}=\frac{z^2(1+w)}{w^2(1+z)}\quad(>0),\quad \frac{u}{v}=\frac{z}{w}
\end{equation}
détermine de façon unique les couples $(a,b)\in\NN_{\operatorname{prem}}^2$ et $(u,v)\in(\NN\times\ZZ)_{\operatorname{prem}}$.
Réciproquement, les égalités dans \eqref{eq:inter1} nous permettent de résoudre $(w,z)$ pour $(a,b)\times (u,v)\in \NN_{\operatorname{prem}}^2\times (\NN\times\ZZ)_{\operatorname{prem}}$ et la solution est donnée précisément par $\varrho\circ\Psi$ \eqref{eq:wz}.
De cette manière, nous avons que si $\max(|w|,|z|)<1$,
\begin{align*}
&\frac{b}{a}=\frac{z^2(1+w)}{w^2(1+z)}>1 \\
\Leftrightarrow &(z-w)(z+w+zw)>0\\
\Leftrightarrow &z>w \text{ et } z+w+zw>0 \quad \text{ou}\quad z<w \text{ et } z+w+zw<0,\\
\Leftrightarrow &(z,w)\in \cup_{i=1}^4 S_i.
\end{align*}
De plus,
$$(w,z)\in S_1\Rightarrow \sqrt{\frac{1+w}{1+z}}\frac{z}{w}<\frac{z}{w}<\frac{z^2(1+w)}{w^2(1+z)}\Leftrightarrow \sqrt{\frac{b}{a}}<\frac{u}{v}<\frac{b}{a},$$
$$(w,z)\in S_2\Rightarrow \sqrt{\frac{1+w}{1+z}}\left(-\frac{z}{w}\right)<-\frac{z}{w}\Leftrightarrow -\frac{u}{v}>\sqrt{\frac{b}{a}},$$
$$(w,z)\in S_3\Rightarrow \sqrt{\frac{1+w}{1+z}}\left(-\frac{z}{w}\right)>-\frac{z}{w}>1\Leftrightarrow 1<-\frac{u}{v}<\sqrt{\frac{b}{a}},$$
$$(w,z)\in S_4\Rightarrow \sqrt{\frac{1+w}{1+z}}\frac{z}{w}>\frac{z}{w}>1\Leftrightarrow \sqrt{\frac{b}{a}}>\frac{u}{v}>1.$$
Nous avons donc établi $$\varrho\circ\Psi(T_i)=\{(w,z)\in\QQ^2:(w,z)\in S_i,\max(|w|,|z|)<1\}.$$
\end{proof}
La distance \eqref{eq:distance} se calcule sur $\cup_{i=1}^4 S_i$ par
\begin{equation}\label{eq:distance2}
d((w,z),(0,0))=\max\left(\left|\frac{au^2-bv^2}{u(bv-au)}\right|,\left|\frac{au^2-bv^2}{v(bv-au)}\right|\right)=\left|\frac{\frac{u^2}{v^2}-\frac{b}{a}}{\frac{b}{a}-\frac{u}{v}}\right|
\end{equation}
puisque nous avons supposé $|\frac{b}{a}|>1$ et d'où $|\frac{u}{v}|=\left|\frac{z}{w}\right|>1$.
\begin{comment}
\begin{lemma}\label{le:au2-bv2}
Nous avons $au^2-bv^2<0$ si et seulement si soit $(a,b)\times(u,v)\in T_1$ et $(\varrho\circ\Psi) \left((a,b)\times(u,v)\right)\in S_4$, soit $(a,b)\times(u,v)\in T_2$ et $(\varrho\circ\Psi) \left((a,b)\times(u,v)\right)\in S_3$.
\end{lemma}
\begin{proof}
Démontrons l'équivalence sur $S_3$. Par la définition \eqref{eq:S3}, pour $(w,z)\in S_3$ nous avons $w>0>z$. Au vu de \eqref{eq:wz}, nous avons $$\frac{au^2-bv^2}{u(bv-au)}>0>\frac{au^2-bv^2}{v(bv-au)}.$$
Puisque pour paramétrer les points dans $S_3$ nous avons supposé que $u>0>v$ et $b,a>0$ (cf. \eqref{eq:T2}), nous en concluons que
$$\frac{au^2-bv^2}{bv-au}>0.$$
Or $bv-au<0$, ceci implique que nécessairement $au^2-bv^2<0$.
\end{proof}
\end{comment}
\section{Parties minces}\label{se:thinsets}
\subsection{Définition}
Des parties minces jouent le rôle dominant dans ce problème de comptage. Le phénomène d'accumulation globale est déjà constaté dans plusieurs travaux. Brièvement ils sont des sous-ensembles accumulateurs (souvent denses) contenant des points de hauteur bornée non-négligeable par rapport au complémentaire. L'énoncé original de Batyrev-Manin nécessite alors de renforcement par des conditions plus fortes (cf. par exemple \cite{Peyre3}).
\begin{definition}(cf. \cite{Serre} \S9.1)
Soit $V$ une variété intègre sur $\QQ$. Une \emph{partie mince} $M$ est un sous-ensemble de $V(\QQ)$ vérifiant qu'il existe une variété $X$ et un morphisme $f:X\to V$ tels que
\begin{enumerate}
\item $M\subseteq f(X(\QQ)) $;
\item Le morphisme $f$ est génériquement fini et il n'admet pas de section rationnelle.
\end{enumerate}
Avec toutes ces notations, la partie mince $M$ est dite du \emph{type I} si la fibre générique est vide.
Elle est du \emph{type II} si la variété $X$ est intègre et le morphisme $f$ est dominant.
\end{definition}
Une partie mince du type I est donc contenue dans l'ensemble des points rationnels d'un fermé de Zariski. Alors que celle du type II est dense dans $V$ pour la topologie de Zariski. En général une partie mince est une réunion finie de celles du type I et II.
\subsection{Ensemble mince dans $Y_3$}\label{se:thinsetiny3}
Considérons l'ensemble
\begin{equation}\label{eq:thinsetM}
M=\{(w,z)\in\QQ^2:-w-z-wz=\square\neq 0\}\subset S_2.
\end{equation}
C'est un ensemble se trouvant dans la partie au-dessous de l'hyperbole $wz+w+z=0$ et dense dans $S_2$ pour la topologie réelle.
D'après le paramétrage $\Psi$ \eqref{eq:paracab},
$$-w-z-wz=\square\Leftrightarrow -\frac{(au^2-bv^2)(b-a)}{(bv-au)^2}=\square\Leftrightarrow \left(\frac{u^2}{v^2}-\frac{b}{a}\right)\left(1-\frac{b}{a}\right)=\square,$$
Soit $X$ la sous-variété de $\mathbf{A}^3_{x_1,x_2,x_3}$ d'équation
$$x_3^2=\left(x_2^2-x_1\right)\left(1-x_1\right).$$
Considérons l'application rationnelle
$\pi:X \dashrightarrow Y_3$
donnée par
$$\pi\left(x_1=\frac{b}{a},x_2=\frac{u}{v},x_3\right)=\Psi((a,b)\times(u,v)).$$
Elle est donc génériquement de degré $2$.
Alors nous avons que $M\subset \operatorname{Im}(\pi)(\QQ)$.
D'où $M$ est une partie mince dans $Y_3$.
\begin{proposition}\label{po:thincusp}
Un point $P=(w,z)\in\QQ^2$ est dans $M$ si et seulement s'il existe une courbe cuspidale $R_{a,b}$ passant par $P$.
\end{proposition}
\begin{proof}
Une des courbes \eqref{eq:cuspplane}, vue comme une équation en $\lambda=\frac{b}{a}$, s'écrit comme
\begin{equation}\label{eq:zwlambda}
w^2(1+z)\lambda^2- 2zw\lambda+z^2(1+w)=0,
\end{equation}
avec le discriminant $$\triangle=4w^2z^2(1-(1+w)(1+z))=-4w^2z^2(wz+w+z).$$
Donc un point $(w,z)\in\QQ^2$ est sur $R_{a,b}$ seulement si
\begin{equation}\label{eq:square}
\triangle=\square \Leftrightarrow -(wz+w+z)=\square.
\end{equation}
Réciproquement, si un point $(w,z)\in M$, alors l'équation \eqref{eq:zwlambda} admet deux solutions rationnelles distinctes, qui correspondent à deux courbes cuspidales passant par $M$.
\end{proof}
\section{Congruences polynomiales en moyenne et équidistribution, d'après Erdös et Hooley}\label{se:congHooley}
Nous considérons ici une version de congruences polynomiales à résidu fixé à la Hooley. Elle diffère de plupart de méthodes utilisées dans le comptage des points entiers sur les torseurs universels principalement pour démontrer la conjecture de Batyrev-Manin-Peyre pour les surfaces de del Pezzo (cf. par exemple \cite{baierderenthal} et les références là-dedans).
Nous serons concerné par des dénombrements analogues au cardinal de l'ensemble des $(l,m)\in\NN^2$ tels que
\begin{equation}\label{eq:conditionF}
\lambda_2 B^\frac{1}{5}\leqslant m\leqslant \lambda_1 B^\frac{1}{5},\quad \tau_2 \leqslant \frac{l}{m}\leqslant \tau_1,\quad F(l)\equiv 0[m],
\end{equation}
où $0<\lambda_2<\lambda_1,0<\tau_2<\tau_1\leqslant 1$ et $F(X)\in\ZZ[X]$ est un polynôme de degré $\geqslant 2$.
Nous allons distinguer la discussion en deux cas séparément: $F(X)$ est irréductible ou non dans $\ZZ[X]$. En particulier au cas où $\deg F(X)=2$, ceci est distingué par regarder son discriminant $\Delta(f)=\square$ dans $\QQ$ ou non.
Étant donné $F(X)\in\ZZ[X]$, on définit la fonction
\begin{equation}\label{eq:rhoF}
\varrho_F(n)=\#\{0\leqslant k\leqslant n-1:F(k)\equiv 0[n]\}.
\end{equation}
C'est une fonction arithmétique multiplicative.
$$\varrho_F(k_1k_2)=\varrho_F(k_1)\varrho_F(k_2),\quad \text{si }\pgcd(k_1,k_2)=1.$$
Mais en général elle n'est pas complètement multiplicative.
Elle admet la majoration (comme par exemple cf. \cite[VIII, 8.4]{Hardy}) que pour certain $A_F>1$,
\begin{equation}\label{eq:upperboundofvarF}
\varrho_F(n)\ll A_F^{\omega(n)}\ll n^\varepsilon,\quad \forall \varepsilon>0.
\end{equation}
On y associe la série de Dirichlet (définie pour $\Re(s)$ suffisamment grand)
\begin{equation}\label{eq:DFs}
D_F(s)=\sum_{n=1}^{\infty} \frac{\varrho_F(n)}{n^s}.
\end{equation}
\subsection{Ordre Moyen}\label{se:irred}
On discute premièrement le cas où $F(X)$ est irréductible sur $\ZZ[X]$. Une observation qui remonte à Erdös dit que la série $D_F(s)$ se comporte de façon similaire à la fonction zêta de Dedekind associée au corps de nombres engendré par une racine de $F(X)$. En conséquence on en déduit l'ordre moyen de $\varrho_F$ (Proposition \ref{le:erdos}), à l'aide de méthode d'analyse complexe standard.
\begin{lemma}[Erdös, \cite{Erdos}]\label{le:erdos}
Supposons que $F(X)\in\ZZ[X]$ est irréductible. Soit $\theta$ une racine algébrique de $F(X)$ et notons $K_F=\QQ(\theta)$ le corps de nombres qu'il génère. Alors nous avons
$$D_F(s)=\zeta_{K_F}(s)\varPsi(s),$$
où $\zeta_{K_F}(s)$ est la fonction zêta de Dedekind du corps $K_F$ et $\varPsi(s)$ est méromorphe et borné en le demi-plan $\Re(s)>\frac{1}{2}+\varepsilon,\forall \varepsilon>0$. Par conséquent, la série $D_F(s)$ admet un pôle simple en $s=1$.
\end{lemma}
\begin{proof}
(cf. \cite{Erdos}, Lemma 10) Notons $d_{K_F}$ le discriminant de l'extension $K_F/\QQ$. Pour tout idéal premier $\mathcal{P}\triangleleft \mathcal{O}_{K_F}$, soit $p$ le nombre premier divisé par $\mathcal{P}$. On sait qu'il existe $f_\mathcal{P}\in\NN_{\geqslant 1}$ tel que $$N_{K_F/\QQ}(\mathcal{P})=p^{f_{\mathcal{P}}}.$$
Tout d'abord nous rassemblons quelques propriétés basiques de la fonction $\varrho_F$. Nous avons $\forall p\nmid d_{K_F}$,
\begin{enumerate}
\item $\varrho_F(p^\nu)=\varrho_F(p),\quad \forall\nu\geqslant 1$;
\item $\varrho_F(p)=\#\{\mathcal{P}\triangleleft\mathcal{O}_{K_F}:\mathcal{P}\mid p, f_\mathcal{P}=1\}\leqslant \deg F(X)$.
\end{enumerate}
L'énoncé (1) est classique. Voir par exemple \cite[VIII, 8.4]{Hardy}. Le deuxième découle de la factorisation explicite de $F(X)\mod p$, cf. \cite[I. \S8]{Lang}.
Rappelons que pour un corps de nombres $K$, notons $N_{K/\QQ}(\cdot)$ la norme de $K$ sur $\QQ$, et la fonction de Dedekind est définie par
$$\zeta_K(s)=\prod_{\mathcal{P}\triangleleft \mathcal{O}_K}\left(1-N_{K/\QQ}(\mathcal{P})^{-s}\right)^{-1}=\prod_{p}\prod_{\mathcal{P}\mid p} \left(1-N_{K/\QQ}(\mathcal{P})^{-s}\right)^{-1},$$
où $\mathcal{P}$ parcourt l'ensemble des idéaux premiers de l'anneau des entiers $\mathcal{O}_K$.
Écrivons formellement
\begin{equation}\label{eq:Dirichletseries}
D_F(s)=\prod_{p} A_p(s), \quad A_p(s)=\sum_{n=0}^{\infty}\frac{\varrho_F(p^n)}{p^{ns}}.
\end{equation}
Définissons maintenant
$$\varPsi_1(s)=\prod_{p\mid d_{K_F}}A_p(s)\prod_{\mathcal{P}\mid p}(1-N_{K/\QQ}(\mathcal{P})^{-s}),\quad \varPsi_2(s)=\prod_{p\nmid d_{K_F}}A_p(s)\prod_{\mathcal{P}\mid p}(1-N_{K/\QQ}(\mathcal{P})^{-s}).$$
Nous voyons que $\varPsi_1$ est bien holomorphe en $\Re(s)>0$. Quant à $\varPsi_2$, d'après (2), en notons
$$\varPsi_3(s)=\prod_{p\nmid d_{K_F}}\prod_{\mathcal{P}\mid p:f_\mathcal{P}\geqslant 2}\left(1-p^{-sf_\mathcal{P}}\right)^{-1},$$
qui est évidemment holomorphe, bornée et sans zéros en $\Re(s)>\frac{1}{2}+\varepsilon$,
nous avons
\begin{align*}
\varPsi_2(s)&=\exp\left(\sum_{p\nmid d_{K_F}}\left(\log A_p(s)+\sum_{\mathcal{P}\mid p}\log\left(1-p^{-sf_\mathcal{P}}\right)\right)\right)\\
&=\exp\left(\sum_{p\nmid d_{K_F}}\left(\log\left(1+\varrho_F(p)\sum_{n=1}^{\infty}p^{-ns}\right)+\varrho_F(p)\log\left(1-p^{-s}\right)+\sum_{\mathcal{P}\mid p:f_\mathcal{P}\geqslant 2}\log\left(1-p^{-sf_\mathcal{P}}\right)\right)\right)\\
&=\varPsi_3(s)^{-1}\exp\left(\sum_{p\nmid d_{K_F}}\left(\log\left(1+\varrho_F(p)\sum_{n=1}^{\infty}p^{-ns}\right)+\varrho_F(p)\log\left(1-p^{-s}\right)\right)\right)\\
&=\varPsi_3(s)^{-1}\exp \left(O\left(\sum_{p\nmid d_{K_F}}\frac{\deg F(X)}{p^{2\Re(s)}}\right)\right).
\end{align*}
D'où nous déduisons que $\varPsi_2$ est holomorphe et bornée en $\Re(s)>\frac{1}{2}+\varepsilon,\forall \varepsilon>0$.
Au final observons que
$$D_F(s)=\varPsi_1(s)\varPsi_2(s)\zeta_{K_F}(s),$$
la fonction $\varPsi=\varPsi_1\varPsi_2$ vérifie bien l'hypothèse souhaitée.
\end{proof}
\begin{proposition}\label{le:varrhoF}
Il existe $\lambda\in]0,1[$ dépendant du polynôme $F$ tel que
$$\sum_{n\leqslant X} \varrho_F(n)=Z_F X+O(X^{\lambda}),$$
où, avec les notations du lemme \ref{le:erdos},
$$Z_F=\varPsi(1)\lim_{s\to 1} (s-1)\zeta_{K_F}(s).$$
\end{proposition}
\begin{proof}
(cf. aussi \cite[Lemma 2.1]{Tenenbaum1})
Notons $S(X)=\sum_{n=1}^{X}\varrho_F(n)$. Nous utilisons la formule de Perron (cf. \cite[Lemma 5.4]{Browning}). Soient $T>0$ et $0<a<1<b$. Supposons que $X\notin\NN$. Soit $\lambda_0>0$ tel qu'en notant $\lambda_1=1+\lambda_0,\lambda_2=1-\lambda_0>\frac{1}{2}$, nous avons
\begin{align*}
S(X)=&\frac{1}{2\pi i}\int_{\lambda_1-iT}^{\lambda_1+iT} D_F(s)\frac{X^s}{s}\operatorname{d}s+O\left(\frac{X^c}{T}+\sum_{aX\leqslant k\leqslant bX}\varrho(k)\min\left(1,\frac{1}{T|\log \frac{X}{k}|}\right)\right)\\
&=\frac{1}{2\pi i}\int_{\lambda_1-iT}^{\lambda_1+iT}D_F(s)\frac{X^s}{s}\operatorname{d}s+O\left(\frac{X^{1+\varepsilon}}{T}\right),
\end{align*}
puisque $\varrho_F(k)\ll_\varepsilon k^\varepsilon$ \eqref{eq:upperboundofvarF}.
D'après le théorème des résidus, en choisissant le rectangle $\gamma$ joignant les points $\lambda_2-iT,\lambda_2+iT,\lambda_1+iT,\lambda_1-iT$, nous avons
$$\int_\gamma D_F(s)\frac{X^s}{s}\operatorname{d}s =\operatorname{Res}_{s=1}\left(D_F(s)\frac{X^s}{s}\right)=Z_F X.$$
D'après le théorème de Phragmén-Lindelöf (\cite[II. Théorème 1.16]{Tenenbaum}) on a l'estimation de sous-convexité suivante. (cf. \cite[II \S3.4]{Tenenbaum} pour la fonction zêta de Riemann. En général on sait que $|\zeta_K(\frac{1}{2}+it)|\ll |t|^{[K:\QQ]/6}$ \cite{Heath-Brown3} et la convergence absolue en $\Re(s)>1$) Il existe $c_0>0$ tel que pour tout $\lambda_0$ suffisamment petit,
\begin{equation*}
\zeta_{K_F}(\sigma+it)\ll_\varepsilon |t|^{c_0(1-\sigma)+\varepsilon}, \quad |t|\geqslant 1,\sigma\in\mathopen[\lambda_2,1\mathclose].
\end{equation*}
Compte tenu du fait que $\varPsi$ est bornée, dans la région contournée par $\gamma$, nous obtenons que pour $T\geqslant 1$,
$$\int_{\lambda_2\pm iT}^{\lambda_1\pm iT} \left|\zeta_{K_F}(s)\varPsi(s)(s)\frac{X^s}{s}\right|\operatorname{d}s\ll_\varepsilon \int_{\lambda_2}^{\lambda_1} \frac{X^{\sigma} }{T^{1-c_0\lambda_0-\varepsilon}}\operatorname{d}\sigma\ll \frac{X^{\lambda_1}}{T^{1-c_0\lambda_0-\varepsilon}},$$
et que
$$\int_{\lambda_2-iT}^{\lambda_2+iT} \left|\zeta_{K_F}(s)\varPsi(s)\frac{X^s}{s}\right|\operatorname{d}s \ll X^{\lambda_2}\int_{-T}^T \frac{|\zeta(\lambda_2+it)|}{t}\operatorname{d}t\ll_\varepsilon X^{\lambda_2}\int_{-T}^T |t|^{c_0\lambda_0-1+\varepsilon}\operatorname{d}t\ll_\varepsilon X^{\lambda_2}T^{c_0\lambda_0+\varepsilon}.$$
Nous choisissons $T=X^{\mu_0}$ avec $\mu_0>0$ vérifiant
$$\lambda^\prime=\max(\lambda_1-\mu_0(1-c_0\lambda_0-\varepsilon),\lambda_2+\mu_0(c_0\lambda_0+\varepsilon))<1.$$
Ceci est achevé en prenant $\lambda_0$ suffisamment petit, $\varepsilon<\lambda_0$ et $$\frac{\lambda_0}{1-\lambda_0(c_0+1)}<\mu_0<\frac{1}{1+c_0}.$$
En résumé, nous avons démontré que
$$S(X)=Z_F X +O\left(X^{\lambda}\right),$$
où $$\lambda=\max(\lambda^\prime,1+\varepsilon-\mu_0)<\max(\lambda^\prime,1+\lambda_0-\frac{\lambda_0}{1-\lambda_0(c_0+1)})<1.$$
\end{proof}
\begin{remark}
Une formule asymptotique pour l'ordre moyen de $\varrho_F$ est également obtenue par Hooley \cite{Hooley3}, \cite{Hooley1} dans le cas quadratique par une méthode concrète mais complètement différente.
\end{remark}
\subsection{Équirépartion modulo $1$}\label{se:uniformdistmodulo1}
Nous continuons à supposer dans cette section que le polynôme $F(X)$ est irréductible. Hooley \cite{Hooley2} a démontré le résultat suivant sur les résidus de congruences polynomiales.
Ce théorème a été énoncé pour les polynômes primitifs, mais la même preuve applique pour tous les polynômes non-nécessairement primitifs.
\begin{theorem}[Hooley \cite{Hooley2}, Theorem 1]\label{th:Hooley}
Soient $X>1,h\in\NN_{\geqslant 1},d=\deg F(X)\geqslant 2$ et
$$R(h,X)=\sum_{k\leqslant X}\sum_{\substack{F(v)\equiv 0 [k]\\ 0\leqslant v\leqslant k-1}}\operatorname{exp}\left(\frac{2\pi ih v}{k}\right).$$
Alors
$$R(h,X)=O_F\left(\frac{h^\frac{1}{2}X(\log\log X)^{\frac{1}{2}(d^2+1)}}{(\log X)^{\delta_d}}\right)\quad \text{où}\quad \delta_d=\frac{d-\sqrt{d}}{d!}.$$
(N.B. Le résultat original de Hooley omet l'ordre de grandeur de $h$. Mais on le récupère facilement de sa preuve.)
\end{theorem}
Le but de cette section est de démontrer:
\begin{proposition}\label{prop:discrepancy}
Pour tout intervalle $I\subseteq[0,1[$, nous avons
$$\sum_{k\leqslant X}\sum_{\substack{F(v)\equiv 0 [k]\\ 0\leqslant v\leqslant k-1}}\mathbf{1}_I\left(\frac{v}{k}\right)=Z_F |I| X+O\left(X\frac{(\log\log X)^{\alpha_d}}{(\log X)^{\beta_d}}\right),$$
où $$\alpha_d=\frac{1}{3}(d^2+1),\quad \beta_d=\frac{1}{3}\delta_d.$$
\end{proposition}
Nous notons $(s_n)$ la suite construite en numérotant les nombres rationnels $\frac{v}{k}\in\mathopen[0,1\mathclose[$ tels que $F(v)\equiv 0[k]$ par rapport à l'ordre croissant des dénominateurs. C'est-à-dire $\frac{v_1}{k_1}\leqslant \frac{v_2}{k_2}$ si et seulement si
$$k_1\leqslant k_2,\quad \text{ou} \quad k_1=k_2 \text{ et } v_1\leqslant v_2.$$
Le théorème de Hooley implique que la suite $(s_n)$ est équirépartie modulo $1$.
Nous avons besoin d'une estimation de la discrépance $D_N(s_n)$ de cette suite.
Pour la définition de \emph{la discrépance}, voir par exemple \cite{K-N}.
Les outils sont l'inégalité de Koksma-Denjoy (cf. \cite[p. 143]{K-N}) et celle de Erdös-Turán (cf. \cite[Theorem 2.5 p. 112]{K-N}).
\begin{theorem}[Koksma-Denjoy]\label{th:koksma-denjoy}
Soient $(x_n)$ une suite équirépartie modulo $1$ et $N\geqslant 1$. Soit $\phi$ une fonction mesurable à variation bornée définie sur $\mathopen[0,1\mathclose]$ (on note $V(\phi)$ la variation totale de $\phi$). Alors
$$\left|\frac{1}{N}\sum_{1}^{N}\phi(\{x_n\})-\int_{0}^{1}\phi\right|\leqslant V(\phi)D_N(x_n).$$
\end{theorem}
\begin{theorem}[Erdös-Turán]
Soient $x_i$ $(1\leqslant i\leqslant N)$ des nombres réel. Alors pour tout $m\in\NN_{\geqslant 1}$, on a
$$D_N(x_n)=O\left(\frac{1}{m}+\sum_{h=1}^{m}\frac{1}{h}\left|\frac{1}{N}\sum_{n=1}^{N}\operatorname{exp}(2\pi i h x_n)\right|\right),$$
où la constante implicite est absolue.
\end{theorem}
\begin{corollary}\label{co:estimationofdiscrepancy}
Avec les notations ci-dessus, nous avons
$$D_N(s_n)=O\left(\frac{(\log\log N)^{\frac{1}{3}(d^2+1)}}{(\log N)^{\frac{2}{3}\delta_d}}\right).$$
\end{corollary}
\begin{proof}
Fixons $N\in\NN_{\geqslant 1}$ et notons
$$S(h,N)=\sum_{n=1}^{N}\exp\left(2\pi i h x_n\right).$$
Soit $M$ le dénominateur de $x_N$. D'après la Proposition \ref{le:varrhoF} et la relation suivante
$$\sum_{k<M}\varrho_F(k)<N\leqslant \sum_{k\leqslant M}\varrho_F(k),$$
nous en concluons qu'il existe $C_1,C_2$ deux constantes absolues positives telles que
$$C_1 M\leqslant N\leqslant C_2M.$$
Nous avons aussi la comparaison suivante (cf. \eqref{eq:upperboundofvarF})
$$S(h,N)=R(h,M)+O(\varrho_F(M))=R(h,M)+O(M^\varepsilon),\quad \forall\varepsilon>0,$$
d'où nous calculons que pour tout $m\in\NN_{\geqslant 1}$,
\begin{align*}
D_N(s_n)&=O\left(\frac{1}{m}+\sum_{h=1}^{m}\frac{1}{h}\left|\frac{1}{N}S(h,N)\right|\right)\\
&=O\left(\frac{1}{m}+\sum_{h=1}^{m}\frac{1}{h}\left|\frac{R(h,M)+M^\varepsilon}{M}\right|\right)\\
&=O\left(\frac{1}{m}+\sum_{h=1}^{m}\frac{1}{h}\left(h^\frac{1}{2}\frac{(\log\log M)^{\frac{1}{2}(d^2+1)}}{(\log M)^{\delta_d}}+\frac{1}{M^{1-\varepsilon}}\right)\right)\\
&=O\left(\frac{1}{m}+\sqrt{m}\frac{(\log\log M)^{\frac{1}{2}(d^2+1)}}{(\log M)^{\delta_d}}+\frac{\log m}{M^{1-\varepsilon}}\right)
\end{align*}
En prenant $$m=\left\lfloor\left(\frac{(\log M)^{\delta_d}}{(\log\log M)^{\frac{1}{2}(d^2+1)}}\right)^\frac{2}{3}\right\rfloor,$$
nous obtenons que
$$D_N(s_n)=O\left(\frac{(\log\log M)^{\frac{1}{3}(d^2+1)}}{(\log M)^{\frac{2}{3}\delta_d}}\right)=O\left(\frac{(\log\log N)^{\frac{1}{3}(d^2+1)}}{(\log N)^{\frac{2}{3}\delta_d}}\right),$$
la majoration souhaitée.
\end{proof}
\begin{proof}[Démonstration de la Proposition \ref{prop:discrepancy}]
Posons $$N=N(X)=\sum_{k\leqslant X}\varrho_F(k).$$
D'après la Proposition \ref{le:varrhoF}, $N\sim Z_F X$. Alors compte tenu de l'inégalité de Koksma-Denjoy (théorème \ref{th:koksma-denjoy}) et du Corollaire \ref{co:estimationofdiscrepancy},
\begin{align*}
\sum_{k\leqslant X}\sum_{\substack{F(v)\equiv 0 [k]\\ 0\leqslant v\leqslant k-1}}\mathbf{1}_I\left(\frac{v}{k}\right)&=\sum_{n=1}^{N}\mathbf{1}_I(s_n)\\
&=N\int_{0}^{1}\mathbf{1}_I(x)\operatorname{d}x+O\left(ND_N(s_n)\right)\\
&=N|I| +O\left(\frac{N(\log\log N)^{\frac{1}{3}(d^2+1)}}{(\log N)^{\frac{2}{3}\delta_d}}\right)\\
&=Z_F |I| X +O\left(X\frac{(\log\log X)^{\frac{1}{3}(d^2+1)}}{(\log X)^{\frac{2}{3}\delta_d}}\right).
\end{align*}
\end{proof}
\subsection{Cas scindés}\label{se:splitcase}
Nous nous intéresserons exclusivement dans cette section aux polynômes quadratiques \emph{scindés}, i.e. réductible sur $\ZZ[X]$.
Fixons $a\in\NN_{\geqslant 1},c\in\ZZ_{\neq 0}$. Définissons $$F(X)=aX^2+c,\quad G(X)=X^2+ac.$$
Rappelons \eqref{eq:rhoF} la fonction $\varrho_F$ (resp. $\varrho_G$).
La fonction $\varrho_G$ vérifie que pour $p$ un nombre premier tel que $p\nmid 2ac$, $$\varrho_G(p^r)=1+\left(\dfrac{-ac}{p}\right),\quad \forall r\geqslant 1,$$
(cf. \cite[VIII, 8.4 Example (3)]{Hardy}, rappelons qu'ici $\left(\dfrac{\cdot}{\cdot}\right)$ désigne le symbole de Jacobi).
Une observation basique due à McKee \cite{McKee2} nous permet de relier $\varrho_F$ à $\varrho_G$.
\begin{lemma}\label{le:mckee}
Nous avons
$$\varrho_F(p^r)=\varrho_G(p^r),\quad \forall p\nmid 2a,r\in\NN_{\geqslant 1},$$
et donc $$\varrho_F(p^r)=1+\left(\dfrac{-ac}{p}\right),\quad \forall p\nmid 2ac,r\in\NN_{\geqslant 1}.$$
\end{lemma}
\begin{proof}
Si $x_0\in\ZZ$ est une solution de $F(x)\equiv 0[p^r]$, alors $y_0=ax_0$ vérifie
$$G(y_0)=aF(x_0)\equiv 0[p^r].$$
et donc il est une solution de $G(y)\equiv 0[p^r]$. Réciproquement, soit $q\in \NN$ tel que $aq\equiv 1[p^r]$. Si $y_0\in\ZZ$ est une solution de la congruence $G(y)\equiv 0[p^r]$, alors $x_0=y_0 q$ satisfait à $F(x)\equiv 0 [p^r]$. Il en suit que $\varrho_F(p^r)=\varrho_G(p^r)$.
\end{proof}
Concernant les cas où
$$a/\pgcd(a,c),-c/\pgcd(a,c)\in\square,$$ ou de manière équivalente, $$F(X)=\pgcd(a,c)(bX+d)(bX-d)$$ dans $\ZZ[X]$ avec $$b^2=\frac{a}{\pgcd(a,c)},\quad d^2=\frac{-c}{\pgcd(a,c)}.$$
La fonction donnant le nombre de congruents $\varrho_F$ ainsi que la série Dirichlet $D_F(s)$ sont de nature différente que celles dans les cas non-déployés. Le résultat principal est la proposition \ref{po:splitcong}.
En effet il s'agit ici de problèmes de diviseurs additifs au lieu d'être quadratiques. Ceci est déjà considérés en premier par Ingham \cite{Ingham}, citant aussi des améliorations antérieures \cite{Estermann}, \cite{Heath-Brown}, \cite{D-I}, où des formules asymptotiques de la forme suivante sont établies.
\begin{equation}\label{eq:Ingham}
\sum_{n\leqslant X} \tau(n)\tau(n+h)\sim c(h)X(\log X)^2.
\end{equation}
Cela fournit presque ce dont nous avons besoin, à savoir l'ordre moyen du type $$\sum_{n\leqslant X} \tau((\lambda n+\sigma)(\lambda^\prime n+\sigma^\prime)).$$
\begin{comment}
\begin{lemma}\label{le:splitcasevarrho}
Pour tout nombre premier $p\nmid 2ac$, nous avons $\varrho_F(p^r)=2$ pour tout $r\in\NN_{\geqslant 1}$.
Alors que si $p\mid 2ac$, nous avons $\varrho_F(p^r)=O_{a,c}(1)$.
\end{lemma}
\begin{proof}
Les cas où $p\nmid 2ac$ découle directement du lemme \ref{le:mckee}.
Par contre si $p\mid 2ac$, une analyse plus fine sur les entiers $a,c,b,d$ similaire à la preuve du lemme \ref{le:dirichletseries} nous permet de se ramener à la discussion précédente.
\end{proof}
\end{comment}
\begin{lemma}
Nous avons
$$D_F(s)=\zeta(s)^2\varPhi(s),$$
où $\varPhi(s)$ est holomorphe, bornée et sans zéros en $\Re(s)>\frac{1}{2}+\varepsilon,\forall \varepsilon>0$.
En particulier la série $D_F(s)$ admet en $s=1$ un pôle d'ordre $2$ et converge absolument en $\Re s>1$.
\end{lemma}
\begin{proof}
Conservons les notations \eqref{eq:Dirichletseries} dans la preuve du lemme \ref{le:erdos}.
Si $p\nmid 2ac$, nous avons $\varrho_F(p^r)=2$ pour tout $r\in\NN_{\geqslant 1}$. Alors pour un tel $p$,
$$A_p(s)=1+\sum_{r=1}^{\infty}\frac{2}{p^{rs}}=\left(1-\frac{1}{p^s}\right)^{-1}\left(1+\frac{1}{p^s}\right)=\left(1-\frac{1}{p^s}\right)^{-2}\left(1-\frac{1}{p^{2s}}\right).$$
On définit
$$\varPhi(s)=\left(\prod_{p\mid 2ac}A_p(s)\left(1-\frac{1}{p^s}\right)^2\right)\times \left(\prod_{p\nmid 2ac}\left(1-\frac{1}{p^{2s}}\right)\right).$$
Ceci est évidemment une fonction holomorphe en le demi-plan $\Re(s)>\frac{1}{2}$.
\end{proof}
\begin{proposition}\label{po:splitcong}
Il existe une constante $C_F>0$ telle que
$$\sum_{n\leqslant X}\varrho_F(n)\sim C_F X\log X.$$
\end{proposition}
\begin{proof}
La série $D_F(s)$ est bien à coefficients positifs.
Puisque nous n'avons pas besoin de préciser le terme d'erreur, sachant le demi-plan de convergence et l'ordre du pôle, le terme principal découle d'un théorème taubérien bien connu (voir par exemple, \cite{Narkiewicz}).
\end{proof}
\section{Dénombrement d'approximants}\label{se:counting}
\subsection{Finitude des paramètres}
Le lemme suivant donne tous les paramètres possibles qui sont de nombre finis quand on regarde le zoom dans un voisinage borné fixé.
\begin{lemma}\label{le:finitenessofpara}
Soient $\varepsilon>0$. Nous avons que pour tout $P=[a:b]\times [u:v]\in U(\QQ)$ tel que $d(P)^\frac{5}{2}\h_{\omega_{Y_3}^{-1}}(P)\leqslant \varepsilon$ (rappelons les notations \eqref{eq:D1D2}),
$$D_1,\quad D_2,\quad \frac{|b-a|}{d_3},\quad \frac{|au^2-bv^2|}{D_1D_2d_3}\ll_\varepsilon 1.$$
\end{lemma}
\begin{proof}
Reprenons les notations \eqref{eq:D1D2} et rappelons encore la démonstration de la proposition \ref{po:lowerbound1}. Compte tenu du paramétrage dans \S.\ref{se:parametrization}, nous voyons que les quantités
$$|x-y|=\frac{|au^2-bv^2|}{D_1d_2d_3},\quad |t-s|=\frac{|au^2-bv^2|}{D_2d_1d_3},$$
$$f_1=\frac{|y|}{e_1e_3}=\frac{d_1^2}{D_1},\quad f_2=\frac{|t|}{e_1e_2}=\frac{d_2^2}{D_2},$$
$$f_3=\frac{|yt-xs|}{e_2e_3}=\frac{|(b-a)(au^2-bv^2)|}{D_1D_2d_3^2}.$$
sont finies dans tout voisinage borné. Il en résulte alors que les quantités
$$\frac{|b-a|}{d_3},\quad \frac{|au^2-bv^2|}{D_1D_2d_3}$$
sont finies.
De plus, l'égalité
$$|x-y|\times f_2=\frac{|au^2-bv^2|}{D_1d_2d_3}\times \frac{d_2^2}{D_2}=\frac{|au^2-bv^2|}{D_1D_2d_3}\times d_2,$$
et la finitude ci-dessus impliquent que $d_2$ ainsi que $D_2$ sont finies grâce à la relation $d_2\leqslant D_2\leqslant d_2^2$.
De même pour $d_1$ et $D_2$ en considérant $|t-s|\times f_1$.
\end{proof}
\subsection{Réduction aux problèmes de congruences}
Le lemme suivant est une généralisation de la transformation faite dans l'introduction du texte pour les équations \eqref{eq:introduction}.
\begin{lemma}\label{le:parametrizationofpell}
Pour $C_3\in\NN_{\geqslant 1},D\in\ZZ_{\neq 0}$ avec $\pgcd(C_3,D)=1$,
fixons $(u,v)\in\ZZ_{\operatorname{prem}}^2$ tel que \begin{equation}\label{eq:uv}
\min(u^2,v^2)>-\frac{D}{C_3},
\end{equation}
alors il existe $(a,b)\in\NN_{\operatorname{prem}}^2$ tel que $C_3\mid b-a$ et
\begin{comment}
l'application
\begin{align*}
\tilde{\Psi}:\ZZ_{\operatorname{prem}}^2 \longrightarrow& \NN_{\operatorname{prem}}^2\times \ZZ_{\operatorname{prem}}^2\\
(x,y)\longmapsto &\left(\frac{C_3 y^2+D}{k(x,y)},\frac{C_3 x^2+D}{k(x,y)}\right)\times (x,y),
\end{align*}
où
est une bijection sur
l'ensemble $A_{C_3,D}$ des $(a,b)\times (u,v)\in\NN_{\operatorname{prem}}^2\times\ZZ_{\operatorname{prem}}^2$ vérifiant
\end{comment}
\begin{equation}\label{eq:eqpellfermatgen}
\mathcal{E}_{C_3,D}:au^2-bv^2=\frac{b-a}{C_3}\times D
\end{equation}
si et seulement si
\begin{equation}\label{eq:ab}
(a,b)=\left(\frac{C_3 v^2+D}{k(u,v)},\frac{C_3 u^2+D}{k(u,v)}\right),
\end{equation}
où
$$ k(x,y)=\pgcd(C_3x^2+D,C_3y^2+D).$$
\end{lemma}
\begin{proof}
Étant donné un $(a,b)\in\NN_{\operatorname{prem}}^2$ vérifiant $\mathcal{E}_{C_3,D}$ \eqref{eq:eqpellfermatgen}, nous obtenons que
$$(C_3u^2+D)a=(C_3v^2+D)b,$$
et cela donne bien \eqref{eq:ab}.
Réciproquement, nous vérifions qu'avec l'expression \eqref{eq:ab},
le couple $(a,b)\times(u,v)$ vérifie l'équation \eqref{eq:eqpellfermatgen}.
D'après la définition de $k(u,v)$ la co-primalité de $C_3$ avec $D$, on a $\pgcd(C_3,k(u,v))=1$, nous concluons
de
$$k(u,v)\mid (C_3u^2+D)-(C_3v^2+D)=C_3(u^2-v^2),$$
que
$$k(u,v)\mid u^2-v^2.$$
D'où $$C_3\mid \frac{C_3(u^2-v^2)}{k(u,v)} =\frac{C_3u^2+D}{k(u,v)}-\frac{C_3v^2+D}{k(u,v)}=b-a.$$
\end{proof}
\begin{lemma}\label{le:pgcd}
Pour tout $(a,b)\times (u,v)\in\NN_{\operatorname{prem}}^2\times\ZZ_{\operatorname{prem}}^2$ vérifiant \eqref{eq:eqpellfermatgen}, nous avons $$\pgcd(u^2,b)\pgcd(v^2,a)\mid D,\quad \pgcd(u-v,b-a)\mid \frac{b-a}{C_3}.$$
\end{lemma}
\begin{proof}
(cf. la démonstration du Théorème 5.6 dans \cite{huang2}.)
Puisque $$au^2-bv^2=a(u^2-v^2)-(b-a)v^2,$$ nous avons
$$\pgcd(u^2,b)\pgcd(v^2,a)\pgcd(u-v,b-a)\mid au^2-bv^2.$$
\end{proof}
L'observation suivante est l'étape cruciale de la réduction du comptage aux problèmes de congruences.
Introduisons la notation
\begin{equation}\label{eq:c3}
c_3=\frac{b-a}{d_3}=\frac{b-a}{\pgcd(u-v,b-a)}.
\end{equation}
Remplaçons désormais $d_3$ par $c_3$.
Attention à la différence de $c_3$ et $C_3$. Nous avons en fait
\begin{equation}\label{eq:c3C3}
C_3\mid c_3,
\end{equation}
grâce à l'équation \eqref{eq:eqpellfermatgen} et le lemme \ref{le:pgcd},
$$\pgcd(u-v,b-a)=\frac{b-a}{c_3}\mid \frac{b-a}{C_3}.$$
\begin{proposition}\label{le:keytranslation}
Fixons $q\in\NN_{\geqslant 1}$. Pour tout $(a,b)\times (u,v)\in\NN_{\operatorname{prem}}^2\times \ZZ_{\operatorname{prem}}^2$ vérifiant \eqref{eq:uv} et \eqref{eq:eqpellfermatgen}, la condition
\begin{equation}\label{eq:quadraticcong}
u+v\mid qk(u,v).
\end{equation}
équivaut à
\begin{equation}\label{eq:W}
\frac{c_3}{C_3}\mid q\Leftrightarrow \frac{b-a}{C_3q}\mid \pgcd(u-v,b-a).
\end{equation}
\end{proposition}
\begin{proof}
Adaptons les notations introduite dans le lemme \ref{le:parametrizationofpell}.\\
``$\Longrightarrow$'' Écrivons $$qk(u,v)=\lambda(u+v)$$ pour $\lambda\in\NN_{\geqslant 1}$. Définissons
$$q^\prime=\frac{q}{\pgcd(q,\lambda)},\quad \lambda^\prime=\frac{\lambda}{\pgcd(q,\lambda)}.$$
Nous avons $$q^\prime k(u,v)=\lambda^\prime (u+v),\quad \pgcd(\lambda^\prime,q^\prime)=1.$$
D'où nous déduisons que $$q^\prime\mid u+v,\quad \lambda^\prime\mid k(u,v).$$
Cela nous permet d'écrire
$$\frac{b-a}{C_3}=\frac{u^2-v^2}{k(u,v)}=\frac{q^\prime(u-v)}{\lambda^\prime},$$
et nous avons donc $\lambda^\prime\mid u-v$.
Alors
\begin{align*}
\pgcd(u-v,b-a)&=\pgcd\left(u-v,\frac{C_3(u^2-v^2)}{k(u,v)}\right)\\
&=\pgcd\left(u-v,\frac{C_3q^\prime(u-v)}{\lambda^\prime}\right)\\
&=\frac{u-v}{\lambda^\prime}\pgcd(\lambda^\prime,C_3q^\prime)\\
&=\frac{b-a}{C_3q^\prime},
\end{align*}
puisque $\pgcd(k(u,v),C_3)=1$ et $\lambda^\prime\mid k(u,v)$.
D'où $$\frac{b-a}{C_3\pgcd(u-v,b-a)}=q^\prime\mid q.$$\\
``$\Longleftarrow$'' Soit $q^\prime\mid q$ tel que
\begin{align*}
\frac{b-a}{C_3q^\prime}=\frac{u^2-v^2}{k(u,v)q^\prime}&=\pgcd(u-v,b-a)\\
&=\pgcd\left(u-v,\frac{C_3(u^2-v^2)}{k(u,v)}\right)\\
&=\pgcd\left(\frac{ k(u,v)}{u+v}\times \frac{u^2-v^2}{k(u,v)},C_3\times \frac{u^2-v^2}{k(u,v)q^\prime}\right),
\end{align*}
alors nous en déduisons que $u+v\mid q^\prime k(u,v)$ et \emph{a fortiori} $u+v\mid qk(u,v)$.
\end{proof}
La prochaine proposition révèle que les courbes cuspidales change radicalement le dénombrement.
\begin{proposition}\label{le:splitpoly}
Rappelons l'ensemble mince $M$ \eqref{eq:thinsetM} et notons $A_{C_3,D}$ l'ensemble des $(a,b)\times (u,v)\in\NN_{\operatorname{prem}}^2\times \ZZ_{\operatorname{prem}}^2$ vérifiant \eqref{eq:uv} et \eqref{eq:eqpellfermatgen}.
Nous avons
$(\varrho\circ \Psi)(A_{C_3,D})\cap M\neq\varnothing$
si et seulement si $C_3,-D\in\square$.
En particulier dans ce cas nous avons la factorisation suivante de polynômes dans $\ZZ[X]$: $$C_3X^2+D=(\sqrt{C_3}X+\sqrt{|D|})(\sqrt{C_3}X-\sqrt{|D|}).$$
Par conséquent, $\varrho\circ \Psi$ restreignant à $$\bigsqcup_{\substack{(C_3,D)\in\NN_{\geqslant 1}\times \ZZ_{< 0}\\\pgcd(C_3,D)=1}}A_{C_3,D}\cap (T_3\cup T_4)\to (T_Q Y_3)_\RR$$ est un paramétrage de l'ensemble mince $$\{(w,z)\in M\cap(S_3\cup S_4):\max(|w|,|z|)<1\}.$$
Les composantes cuspidales des courbes $\{R_{a,b}\}$ vivent seulement dans la région $R_2$.
\end{proposition}
\begin{proof}
Prenons $(w,z)\in\QQ^2\cap (\cup_{i=1}^4 S_i)$.
Rappelons le paramétrage pour les coordonnées $(w,z)$ \eqref{eq:wz}.
Nous calculons
\begin{align*}
wz+w+z&=\frac{(au^2-bv^2)^2}{uv(bv-au)^2}+\frac{au^2-bv^2}{u(bv-au)}+\frac{au^2-bv^2}{v(bv-au)}\\
&=\frac{(au^2-bv^2)(b-a)}{(bv-au)^2}.
\end{align*}
Par la définition de l'ensemble $M$, $\varrho\circ \Psi(\cup_{i=1}^4 T_i)\cap M\neq \varnothing$ si et seulement si
$$(au^2-bv^2)(b-a)=-\square.$$
En particulier nous voyons que $au^2-bv^2<0$. Nous déduisons du lemme \ref{le:au2-bv2} que l'image d'un tel point est $\in (S_3\cup S_4)$. De l'équation $\mathcal{E}_{C_3,D}$, la condition ci-dessus est équivalente à
$$-(b-a)^2\frac{D}{C_3}=\square\Leftrightarrow -\frac{D}{C_3}=\square.$$
Le résultat en suit directement puisque nous avons imposé que $D,C_3$ soient premiers entre eux et que $C_3>0$.
Le dernier énoncé suit de la proposition \ref{po:thincusp}.
\end{proof}
\subsection{Région $R_1$}\label{se:criticalmeasure}
Comme révélé précédemment, il n'y a pas de composantes cuspidales des courbes \eqref{eq:cusp} qui rentrent dans la région $R_1$. Donc cette région n'intersecte pas l'ensemble mince $M$.
\subsubsection{Région $S_1$}
Pour $\varepsilon_1>\varepsilon_2>0,\theta_1>\theta_2>1$ fixés, on note
\begin{equation}\label{eq:chi}
R(\varepsilon_i,\theta_j)=\{(w,z)\in S_1: \varepsilon_2<z\leqslant\varepsilon_1,\theta_2w<z<\theta_1w\},
\end{equation}
une région trapézoïdale et $\chi_{\varepsilon_i,\theta_j}(\cdot)=\textbf{1}_R(\cdot)$ la fonction caractéristique. Elle sert à une fonction de \og test\fg{}.
\begin{theorem}\label{th:principalthm}
On a
$$\delta_{\varrho^{-1}(S_1)\cap U,Q,B,\frac{5}{2}}(\chi_{\varepsilon_i,\theta_j})= B^\frac{1}{5}\left(\int \chi_{\varepsilon_i,\theta_j}(w,z) \frac{\mathds{E}(wz\sqrt{w+z}) }{wz\sqrt{w+z}}\operatorname{d}w\operatorname{d}z +O_{\varepsilon_i,\theta_j}\left(\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)\right),$$
où
$\mathds{E}:\mathopen]0,\infty\mathclose[\to\mathopen[0,\infty\mathclose[$ est une fonction croissante en escalier.
\end{theorem}
\begin{remark}
La mesure apparaissant dans la minoration fait souligner les trois lignes qui sont localement accumulatrices et la somme de leur puissances est égale exactement à la constante essentielle. Ceci est en analogue avec le résultat \cite{huang1} pour la surface torique de del Pezzo de degré $6$.
\end{remark}
\subsubsection{Déroulement de comptage}
Comme pour $P=[x:y]\times [s:t]$ tel que $\varrho\circ \Psi(P)\in S_1$,
\begin{equation}\label{eq:maxheight}
\max(|x^2 st|,| y^2 st|,| t^2 xy|,|s^2 xy|,|xyst|,|y^2t^2|)=|s^2 xy|,
\end{equation}
La formule (cf. Section \ref{se:height}) pour calculer la hauteur par rapport aux paramètres $a,b,u,v$ est donnée par, grâce à \eqref{eq:paracab}, \eqref{eq:pgcd} et \eqref{eq:pgcd2}, pour $(a,b)\times (u,v)\in T_1$,
\begin{align*}
\h((\varrho\circ\Psi) (a,b)\times (u,v))&=\frac{|s^2xy|}{\pgcd(x,t)\pgcd(y,s)\pgcd(y,t)}\\
&=\frac{a^2bu^2(u-v)^3}{D_1^2D_2^2d_3^3},
\end{align*}
où les paramètres $d_3,D_i$ sont définis par \eqref{eq:D1D2}.
La distance restreinte \eqref{eq:distance} est
\begin{equation}\label{eq:maxdistance}
d((\varrho\circ\Psi) (a,b)\times (u,v))=\frac{\frac{u^2}{v^2} -\frac{b}{a}}{\frac{b}{a}-\frac{u}{v}}.
\end{equation}
Pour le reste de cette section, $D_1,D_2$ désigneront des paramètres au lieu d'avoir le sens \eqref{eq:D1D2}.
L'équivalence établie dans la proposition \ref{le:keytranslation} nous permet de faire lien avec le problème de congruences.
Nous allons faire la partition des paramètres $(a,b)\times (u,v)\in T_1$ selon la famille des équations $(\mathcal{E}_{C_3,D})_{C_3\in\NN_{\geqslant 1},D\in\ZZ_{\neq 0}}$ \eqref{eq:eqpellfermatgen}. D'après le lemme \ref{le:au2-bv2}, $D>0$.
\begin{comment}
Pour $q,C_3,D\in\NN_{\geqslant 1},\pgcd(C_3,D)=1$, soit $A_{C_3,D}(q)$ l'ensemble des points $P=(a,b)\times(u,v)\in T_1$ vérifiant l'équation $\mathcal{E}_{C_3,D}$ \eqref{eq:eqpellfermatgen}, la condition de divisibilité \eqref{eq:quadraticcong}.
Les points dans $$\bigsqcup_{\substack{C_3,D\in\NN_{\geqslant 1}\\\pgcd(C_3,D)=1}}\left(\bigcup_{q\in\NN_{\geqslant 1}}A_{C_3,D}(q)\right)$$ paramètrent les points dans la région $R_1$.
Nous remarquons l'identité évidente
\begin{equation}\label{eq:interA}
A_{C_3,D}(q)\cap A_{C_3,D}(q^\prime)=A_{C_3,D}(\pgcd(q,q^\prime)),\quad q\qed^\prime\in \NN_{\geqslant 1}.
\end{equation}
\end{comment}
Pour $D_1,D_2\in\NN_{\geqslant 1}, D_1D_2\mid D,\pgcd(D_1,D_2)=1$ et $W\in\NN_{\geqslant 1},\pgcd(W,D)=1$, nous définissons $E^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)$ l'ensemble des $(a,b)\times(u,v)\in T_1$ satisfaisant à l'équation $\mathcal{E}_{C_3,D}$ \eqref{eq:eqpellfermatgen} et vérifiant \eqref{eq:D1D2div}, \eqref{eq:Wdiv} et \eqref{eq:conditionE} suivantes.
\begin{equation}\label{eq:D1D2div}
D_1=\pgcd(u^2,b),\quad D_2=\pgcd(v^2,a),
\end{equation}
\begin{equation}\label{eq:Wdiv}
\pgcd(u-v,b-a)=\frac{b-a}{C_3W},
\end{equation}
et
\begin{equation}\label{eq:conditionE}
\theta_2<\frac{u}{v}<\theta_1,\quad \varepsilon_2B^{-\frac{2}{5}}<\frac{\frac{u^2}{v^2} -\frac{b}{a}}{\frac{b}{a}-\frac{u}{v} }\leqslant \varepsilon_1 B^{-\frac{2}{5}},
\end{equation}
\begin{equation}\label{eq:conditionEH}
\frac{a^2bu^2(u-v)^3}{D_1^2D_2^2d_3^3}=\frac{a^2bu^2(u-v)^3C_3^3W^3}{(b-a)^3D_1^2D_2^2}\leqslant B.
\end{equation}
On a donc
\begin{equation}\label{eq:decomp1}
\begin{split}
\delta_{\varrho^{-1}(S_1)\cap U,Q,B,\frac{5}{2}}(\chi_{\varepsilon_i,\theta_j})&=\# \left(\bigsqcup_{\substack{C_3,D,W\in\NN_{\geqslant 1},\pgcd(C_3W,D)=1\\D_1D_2\mid D,\pgcd(D_1,D_2)=1}}E^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)\right)
\end{split}
\end{equation}
La réunion dans \eqref{eq:decomp1} est en fait finie puisque la démonstration du lemme \ref{le:finitenessofpara} et l'identité
$$c_3=C_3W$$
nous révèle que $$C_3,D,W=O_{\varepsilon_1}(1).$$
\subsubsection{Conditions de seuils}\label{se:threshold}
Nous allons trouver dans cette section pour chaque point $P=(a,b)\times (u,v)\in E^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)$ une condition pour qu'il soit dénombré. Soit $(w,z)=\varrho\circ\Psi (P)$. Cette condition s'avérai être une \emph{équation de seuil}
\begin{equation}\label{eq:threshold1}
zw\sqrt{z+w}>\frac{D^\frac{5}{2}C_3^\frac{1}{2}W^3}{D_1^2D_2^2}.
\end{equation}
Elle nous montre \emph{a posteriori} que la condition de hauteur sert à certaine condition de seuils et celle de zoom joue le vrai rôle dans le dénombrement.
\begin{comment}
Prenons la borne supérieure dans \eqref{eq:condzoom}:
\begin{equation}\label{eq:condzoomu}
\varepsilon_1y(C_3xy-D)\geqslant B^{\frac{2}{5}}D(x+y).
\end{equation}
Nous commençons par un encadrement des paramètres $x,y$.
\begin{lemma}\label{le:boundforxy}
Pour $\varepsilon_1>0,\theta_1>1$ fixés, il existe des constantes $\lambda_1,\lambda_2>0$ dépendant de $\varepsilon_1,\theta_1,C_3,D$ telles que les $(x,y)\in\NN_{\operatorname{prem}}^2$ satisfaisant aux conditions \eqref{eq:condheight}, \eqref{eq:condzoomu} et
$$z\leqslant\varepsilon_1,\quad 1<\frac{z}{w}<\theta_1,$$
où $(w,z)=\varrho\circ\Psi\circ\tilde{\Psi}^{-1}(x,y)$ vérifient
$$\min(x,y)\geqslant \lambda_1B^\frac{1}{5},\quad \max(x,y)\leqslant \lambda_2 B^\frac{1}{5}.$$
\end{lemma}
\begin{proof}
Nous déduisons de \eqref{eq:condheight} et de \eqref{eq:condzoomu} que
$$\varepsilon_1 C_3 xy^2\geqslant B^\frac{2}{5}D(x+y),\quad x^4y^4\leqslant BD^2(x+y)^3.$$
Ces deux bornes (inférieure et supérieure) impliquent
$$\frac{D(1+\theta_1)}{\varepsilon_1 C_3\theta_1}B^\frac{1}{5}<y<D^\frac{2}{5}\theta_1^\frac{1}{5}(1+\theta_1)^\frac{3}{5}B^\frac{1}{5}.$$
D'où un encadrement sur $x$ et $y$.
\end{proof}
\end{comment}
\begin{lemma}\label{le:threshold0}
Pour tout $P=(a,b)\times (u,v)\in T_1$ qui vérifie \eqref{eq:conditionEH}, \eqref{eq:eqpellfermatgen}, \eqref{eq:D1D2div}, \eqref{eq:Wdiv}, notons
\begin{equation}\label{eq:w0z0}
(w_0,z_0)=B^{\frac{2}{5}}\varrho\circ\Psi(P)=(B^{\frac{2}{5}}w,B^\frac{2}{5}z).
\end{equation}
Alors nous avons
\begin{equation}\label{eq:threshold}
z_0w_0\sqrt{z_0+w_0}> \frac{D^\frac{5}{2} C_3^\frac{1}{2}W^3}{D_1^2D_2^2}.
\end{equation}
\end{lemma}
\begin{proof}
Tout d'abord d'après la définition de zoom, nous avons
\begin{equation}\label{eq:condzoomu1}
z_0=B^\frac{2}{5}z=B^\frac{2}{5}\frac{D(u+v)}{v(C_3uv-D)}.
\end{equation}
Nous obtenons que
\begin{equation}\label{eq:b1}
\frac{z_0C_3 uv^2 }{D(u+v)}>B^\frac{2}{5}.
\end{equation}
D'après le lemme \ref{le:parametrizationofpell},
$$a=\frac{C_3 v^2+D}{k(u,v)},\quad b=\frac{C_3 u^2+D}{k(u,v)}.$$
La condition que la hauteur soit bornée s'écrit
\begin{equation}\label{eq:condheight}
W^3(C_3v^2+D)^2(C_3u^2+D)u^2< B(u+v)^3D_1^2D_2^2.
\end{equation}
Cette borne supérieure nous donne la majoration
\begin{equation*}\label{eq:b2}
\frac{C_3^3W^3 u^4v^4}{D_1^2D_2^2(u+v)^3}< B.
\end{equation*}
Ces deux inégalités donnent
$$\left(\frac{C_3^3W^3}{D_1^2D_2^2}\right)^\frac{2}{5}\frac{u^\frac{8}{5}v^\frac{8}{5}}{(u+v)^\frac{6}{5}}<\frac{z_0C_3 uv^2 }{D(u+v)},$$
qui elle-même implique
$$\frac{C_3^\frac{1}{5}W^\frac{6}{5}D}{D_1^\frac{4}{5}D_2^\frac{4}{5}}< z_0\left(1+\frac{v}{u}\right)^\frac{1}{5}\left(\frac{v}{u}\right)^\frac{2}{5}.$$
D'après la définition du paramétrage $\Psi$ et du difféomorphisme $\varrho$ \eqref{eq:diffeomorphism}, nous avons la relation entre les paramètres $(u,v)$ et les coordonnées $(w,z)$,
$$\frac{u}{v}=\frac{z}{w}=\frac{z_0}{w_0},$$
d'ou nous obtenons
$$\frac{DC_3^\frac{1}{5}W^\frac{6}{5}}{D_1^\frac{4}{5}D_2^\frac{4}{5}}< z_0\left(1+\frac{w_0}{z_0}\right)^\frac{1}{5}\left(\frac{w_0}{z_0}\right)^\frac{2}{5}=z_0^\frac{2}{5}w_0^\frac{2}{5}(z_0+w_0)^\frac{1}{5}.$$
En prenons la puissance $\frac{5}{2}$, nous obtenons la borne inférieure cherchée puisque $w_0,z_0>0$ dans la région $S_1$.
\end{proof}
\begin{lemma}\label{le:threshold1}
Conservant la notation \eqref{eq:w0z0}. Il existe $\mu_0>0$ ne dépendant que de $C_3,D,W,\varepsilon$ tel que pour tout $P=(a,b)\times (u,v)\in E^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)$ vérifiant \eqref{eq:D1D2div}, \eqref{eq:Wdiv} et
\begin{equation}\label{eq:w0z01}
z_0w_0\sqrt{z_0+w_0}\geqslant\frac{D^\frac{5}{2}C_3^\frac{1}{2}W^3}{D_1^2D_2^2}+\mu_0 B^{-\frac{2}{5}}
\end{equation}
avec les notations \eqref{eq:w0z0},
alors \eqref{eq:conditionEH} est vérifiée pour tout $B\gg_{C_3,D,W,\varepsilon} 1$.
\end{lemma}
\begin{proof}
Notons $$\theta_0=\frac{u}{v}=\frac{z_0}{w_0}>1.$$
Nous avons
$$\frac{z_0^\frac{5}{2}}{\sqrt{z_0+w_0}}=\frac{\theta_0^\frac{5}{2}}{\sqrt{1+\theta_0}}\leqslant \frac{z_0^\frac{5}{2}D_1^2D_2^2}{D^\frac{5}{2}C_3^\frac{1}{2}W^3}\ll_{C_3,D,W,\varepsilon} 1,$$
et d'où \begin{equation}\label{eq:b3}
\theta_0\ll_{C_3,D,W,\varepsilon} 1.
\end{equation}
Rappelons l'identité \eqref{eq:condzoomu1}, nous avons $$B^{\frac{2}{5}}=\frac{z_0v(C_3uv-D)}{D(u+v)}.$$
La condition de hauteur \eqref{eq:condheight} est déduite d'une inégalité du type
$$C_3^3 W^3 u^4v^4+O_{C_3,D}(u^4v^2)\leqslant \left(\frac{z_0C_3uv^2}{u+v}-O_{C_3,D,W}(1)\right)^\frac{5}{2}(u+v)^3D_1^2D_2^2,$$
elle-même étant déduite de, compte-tenu de \eqref{eq:b3},
$$\frac{W^3C_3^3}{D_1^2D_2^2}\frac{u^4v^4}{(u+v)^3}(1+\mu_1 B^{-\frac{2}{5}})\leqslant \left(\frac{z_0C_3uv^2}{u+v}\right)^\frac{5}{2}(1-\mu_2B^{-\frac{2}{5}}),$$
où $\mu_1,\mu_2>0$ dépendant de $C_3,D,W,\varepsilon$.
La preuve déroule essentiellement en suivant celle précédente.
\end{proof}
Nous concluons des lemmes \ref{le:threshold0} et \ref{le:threshold1} dans cette section qu'en prenant une fonction de test \eqref{eq:chi}, nous pouvons \og oublier\fg{}\ la condition de hauteur (i.e. la condition \eqref{eq:conditionEH}) en rajoutant la condition de seuil \eqref{eq:threshold1}.
À savoir,
\begin{equation}\label{eq:seuilreplaceheight}
\begin{split}
\# E^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)&=\# \tilde{E}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)+O(\# \operatorname{Er}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i))
\end{split}
\end{equation}
où $\tilde{E}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)$ consiste en les $(a,b)\times(u,v)\in T_1$ vérifiant \eqref{eq:eqpellfermatgen}, \eqref{eq:D1D2div}, \eqref{eq:Wdiv}, \eqref{eq:conditionE}, \eqref{eq:threshold} et $\operatorname{Er}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)$ est l'ensemble des $(a,b)\times(u,v)\in T_1$ satisfaisant les mêmes conditions précédentes sauf \eqref{eq:threshold} est remplacée par (cf. \eqref{eq:w0z01})
\begin{equation}\label{eq:errorterm}
\frac{D^\frac{5}{2} C_3^\frac{1}{2}W^3}{D_1^2D_2^2}\leqslant z_0w_0\sqrt{z_0+w_0}\leqslant \frac{D^\frac{5}{2} C_3^\frac{1}{2}W^3}{D_1^2D_2^2}+\mu_0 B^{-\frac{2}{5}}.
\end{equation}
\subsubsection{Réduction du comptage}
Premièrement nous éliminons la condition de pgcd \eqref{eq:Wdiv}.
Compte-tenu de la proposition \ref{le:keytranslation}, une inversion de Möbius nous donne
\begin{equation}\label{eq:decomp2}
\# \tilde{E}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)=\sum_{q\mid W} \mu\left(\frac{W}{q}\right)\# B_{C_3,D}^{\varepsilon_i,\theta_j}\left(D_i,q\right)
\end{equation}
où pour $q, D_1,D_2\in\NN_{\geqslant 1}$ fixés, $B_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_i,q)$ est constitué des $(a,b)\times (u,v)\in T_1$ vérifiant \eqref{eq:eqpellfermatgen}, \eqref{eq:quadraticcong}, \eqref{eq:D1D2div}, \eqref{eq:conditionE}, \eqref{eq:threshold}.
Nous allons désormais nous concentrer sur un seul ensemble $B_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_i,q)$ pour $C_3,D,W,D_1,D_2,q$ fixés.
\begin{comment}
et nous le décomposons comme
\begin{equation}\label{eq:decomp2}
B_{C_3,D}^{\varepsilon_i,\theta_j}(q)=\bigsqcup_{\substack{D_1D_2\mid D\\\pgcd(D_1,D_2)=1}}B_{C_3,D}^{\varepsilon_i,\theta_j}(D_i,q),
\end{equation}
où pour $D_1,D_2$ fixés,
Cela nous amène à considérer un seul ensemble $B_{C_3,D}^{\varepsilon_i,\theta_j}(D_i,q)$.
\end{comment}
Nous avons, grâce à la co-primalité pré-supposée,
$$D_1=\pgcd(u^2,b)=\pgcd(u^2,au^2-bv^2)=\pgcd\left(u^2,\frac{b-a}{C_3}D\right)=\pgcd(u^2,D).$$
De même manière,
$$D_2=\pgcd(v^2,a)=\pgcd(v^2,D).$$
Donc nous pouvons appliquer l'inversion de Möbius pour éliminer ces pgcd. À l'aide du lemme \ref{le:parametrizationofpell}, nous pouvons éliminer les paramètres $a,b$. Nous copions la condition de zoom et de pente dans \eqref{eq:conditionE} de ce point de vue ici.
\begin{equation}\label{eq:condzoom1}
\varepsilon_2 v(C_3uv-D)\leqslant B^\frac{2}{5}D(u+v)\leqslant \varepsilon_1 v(C_3uv-D),\quad \theta_2<\frac{u}{v}<\theta_1.
\end{equation}
\begin{lemma}
\begin{equation}\label{eq:decomp3}
\# B_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_i,q)=\sum_{\substack{h_1,h_2\in\NN_{\geqslant 1}\\D_1h_1,D_2h_2\mid D\\ \pgcd(h_1,h_2)=1}} \mu(h_1)\mu(h_2)\# \mathcal{B}_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_ih_i,q)
\end{equation}
où $\mathcal{B}_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_ih_i,q)$ est l'ensemble des $(u,v)\in \NN_{\operatorname{prem}}^2$ vérifiant \eqref{eq:quadraticcong}, \eqref{eq:condzoom1}, \eqref{eq:threshold} et
\begin{equation}\label{eq:condxyr1}
h_1D_1\mid u^2,\quad h_2D_2\mid v^2\quad \Longleftrightarrow \quad g(h_1D_1)\mid u,\quad g(h_2D_2)\mid v,
\end{equation}
\end{lemma}
La condition de divisibilité \eqref{eq:quadraticcong} maintenant s'écrit,
\begin{equation}\label{eq:condxyr2}
q(C_3u^2+D)\equiv 0 [u+v],\quad q(C_3v^2+D)\equiv 0[u+v].
\end{equation}
L'une entre ces deux conditions implique l'autre.
La restriction à la région $S_1$ demande $\frac{u}{v}>1$ et impose donc l'unicité de couples de solutions $(u,v)$.
Nous allons désormais nous concerner sur $u$ et $u+v$.
Introduisons les notations $m,n$ de sorte que
\begin{equation}\label{eq:changeofvar1}
u=g(h_1D_1)n,\quad u+v=m.
\end{equation}
Alors la condition de co-primalité de $(u,v)$ implique celle de $(m,n)$:
\begin{equation}\label{eq:coprime1}
\pgcd(u,v)=1\Longleftrightarrow\pgcd(m,n)=1 \text{ et} \pgcd(m,g(h_1D_1))=1.
\end{equation}
Maintenant \eqref{eq:condxyr1} et \eqref{eq:condxyr2} s'écrivent
\begin{equation}\label{eq:condxys1}
m-g(h_1D_1)n\equiv 0[g(h_2D_2)],\quad q(C_3g(h_1D_1)^2 n^2+D)\equiv 0[m].
\end{equation}
Puisque $\pgcd(h_1D_1,h_2D_2)=1$, soient $1\leqslant \gamma_1<g(h_2D_2)$ tel que $$\gamma_1g(h_1D_1)\equiv 1[g(h_2D_2)].$$
Nous obtenons
$$\gamma_1 m-\gamma_1 g(h_1D_1)n\equiv \gamma_1 m-n\equiv 0[g(h_2D_2)].$$
Puisque $v=m-g(h_1D_1)n\geqslant 1$,
$$\gamma_1\leqslant \gamma_1 m-\gamma_1 g(h_1D_1)n\leqslant \gamma_1 m-n.$$
Il existe donc un entier $l\in\NN_{\geqslant 1}$ tel que
\begin{equation}\label{eq:congrln}
\gamma_1m-g(h_2D_2)l=n,
\end{equation}
et la condition de congruence dans \eqref{eq:condxys1} maintenant devient
\begin{align}\label{eq:condxys2}
q(C_3g(h_1D_1h_2D_2)^2l^2+D)\equiv 0[m]
\end{align}
avec la condition de co-primalité pour $(m,l)$ déduite de \eqref{eq:coprime1}:
\begin{equation}\label{eq:coprime2}
\pgcd(m,g(h_1D_1h_2D_2))=\pgcd(m,l)=1.
\end{equation}
Nous faisons une dernière inversion de Möbius qui élimine ces dernières conditions de pgcd.
Soient $$ e_1\mid\pgcd(m,g(h_1D_1)),\quad e_2\mid \pgcd(m,g(h_2D_2)),\quad e_3\mid \pgcd(m,l),$$ tels que $\pgcd(e_3,g(h_1D_1h_2D_2))=1$. Alors $e_1e_2e_3\mid qD$ sinon la congruence \eqref{eq:condxys2} n'a pas de solution. Écrivons
\begin{equation}\label{eq:lprimemprime}
m=e_1e_2e_3m^\prime,\quad l=e_3l^\prime.
\end{equation}
Nous pouvons enfin réécrire \eqref{eq:condxys2} comme
\begin{equation}\label{eq:condxyss}
\frac{qC_3g(h_1D_1h_2D_2)^2e_3}{e_1e_2}(l^\prime )^2+\frac{qD}{e_1e_2e_3}\equiv 0[m^\prime].
\end{equation}
Les relations entre $(l^\prime,m^\prime)$ et $(u,v)$ sont, d'après \eqref{eq:changeofvar1}, \eqref{eq:congrln} et \eqref{eq:lprimemprime}
\begin{equation}\label{eq:xylm}
u+v=e_1e_2e_3m^\prime,\quad u= g(h_1D_1)e_3(\gamma_1e_1e_2m^\prime-g(h_2D_2) l^\prime).
\end{equation}
En résumé, nous avons établi la formule suivant.
\begin{lemma}
\begin{equation}\label{eq:decomp4}
\# \mathcal{B}_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_ih_i,q)=\sum_{\substack{e_1e_2\mid g(h_1D_1h_2D_2),\pgcd(e_1,e_2)=1\\e_1e_2e_3\mid qD,\pgcd(e_3,g(h_1D_1h_2D_2))=1}}\left(\prod_{j=1}^{3}\mu(e_j)\right)\# \mathcal{C}_{C_3,D}^{\varepsilon_i,\theta_j}(D_ih_i,e_j,q),
\end{equation}
où $\mathcal{C}_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_ih_i,e_j,q)$ est l'ensemble des couples $(l^\prime,m^\prime)\in\NN_{\geqslant 1}\times\NN_{\geqslant 1}$ vérifiant les conditions \eqref{eq:xylm} et \eqref{eq:conditionE} et
\begin{equation}\label{eq:condcongtocount}
\mathcal{F}(l^\prime)\equiv 0[m^\prime],
\end{equation}
où
\begin{equation}\label{eq:formF}
\mathcal{F}(X)=\mathcal{F}_{C_3,D,h_iD_i,e_j,q}(X)=\frac{qC_3g(h_1D_1h_2D_2)^2e_3}{e_1e_2}X^2+\frac{qD}{e_1e_2e_3}\in\ZZ[X].
\end{equation}
\end{lemma}
\begin{comment}
La condition de coprimalité $\pgcd(x,y)=1$ équivaut à, avec les notations $(l,m)$, $\pgcd(l,m)=1$.
Écrivons
\begin{equation}\label{eq:mprime}
m=\hat{m} q.
\end{equation}
Nous obtenons que \eqref{eq:condxys2} s'écrit comme
\begin{equation}\label{eq:condxys3}
C_3\lambda_1\lambda_2 l^2+\mu\equiv 0[\hat{m}],\quad \pgcd(l,\hat{m})=\pgcd(l,q)=1.
\end{equation}
Appliquons une inversion de Möbius encore et écrivons pour $d\mid \mu,\pgcd(d,q)=1$,
$$l=\hat{l} d,\quad\hat{m}= \tilde{m} d,$$
nous pouvons simplifier \eqref{eq:condxys3} comme
\begin{equation}\label{eq:condxys4}
C_3\lambda_1\lambda_2 d\hat{l}^2+\frac{\mu}{d}\equiv 0[\tilde{m}],\quad \pgcd(\hat{l},q)=1.
\end{equation}
Soient $d^\prime |q$ et $\hat{l}=l^\prime d^\prime$,
alors \eqref{eq:condxys4} devient maintenant
\begin{equation}\label{eq:condxys5}
C_3\lambda_1\lambda_2 d(d^\prime)^2 (l^\prime)^2+\frac{\mu}{d}\equiv 0[\tilde{m}].
\end{equation}
Notons
\begin{equation}\label{eq:D}
D=\pgcd\left(C_3\lambda_1\lambda_2 d(d^\prime)^2 \frac{\mu}{d}\right),\quad A_1=\frac{C_3\lambda_1\lambda_2 d(d^\prime)^2}{D},\quad A_2=\frac{\mu}{dD},
\end{equation}
et définissons
\begin{equation}\label{eq:Fmprime}
F(X)=F_{C_3,e,F_i,q_i,d,d^\prime}(X)=A_1 X^2+A_2,\quad m^\prime=\frac{\tilde{m}}{D}.
\end{equation}
Par définition, $F(X)$ est un polynôme irréductible, primitif et de discriminant négatif.
La condition de hauteur \eqref{eq:condheight} et la condition de zoom \eqref{eq:condzoom} donnent la condition de \og seuil\fg:
\begin{equation}\label{eq:seuil}
(z,w)\in S:zw\sqrt{z+w}\geqslant \frac{C_3^3q_3^3(D)^\frac{5}{2}}{D_1^2D_2^2}.
\end{equation}
\end{comment}
\subsubsection{Autre régions}\label{se:otherregions}
Des les régions $S_2,S_3,S_4$, la situation est similaire, sauf un changement de hauteur qui finalement résulte en un changement mineur des seuils. Nous n'allons pas donner tous les détails.
Néanmoins nous remarquons que pour les régions $S_2,S_3$, les polynômes \eqref{eq:formF} changent de signe. Avec les notations précédentes, nous avons $D<0$ (Lemme \ref{le:au2-bv2}). Puisqu'en écrivant
$$\frac{qC_3g(h_1D_1h_2D_2)^2e_3}{e_1e_2}X^2+\frac{qD}{e_1e_2e_3}=\frac{q}{e_1e_2e_3}\left(C_3(g(h_1D_1h_2D_2)e_3X)^2+D\right)=\frac{q}{e_1e_2e_3}\mathcal{G}(X),$$
le (non-)scindage du polynôme $C_3X^2+D$ équivaut à celui de $\mathcal{G}(X)$.
Le lemme \ref{le:splitpoly} nous dit qu'en dehors de la partie mince $M$ \eqref{eq:thinsetM}, au moins un des $C_3,-D\not\in \square$. Donc ces polynôme restent non-scindés et nous conduisent toujours au problème du congruence quadratique.
\subsubsection{Une étape de clef}
Nous commençons par démontrer une formule asymptotique suivante. Les notations dans cette proposition et sa preuve sont indépendantes de celles utilisées avant.
\begin{proposition}\label{po:centralcouting}
Soient $A,X>0$ et $0<\vartheta_2<\vartheta_1<1$ vérifiant $A>\vartheta_1$.
Soit $G:\mathopen]0,\vartheta_1\mathclose]\to \RR_{>0}$ une fonction continue.
Soit $F(X)\in\ZZ[X]$ un polynôme irréductible de degré $\geqslant 2$. Rappelons $\alpha_d,\beta_d$ dans la Proposition \ref{prop:discrepancy}. Alors nous avons
$$\sum_{\substack{l,m\in\NN,F(l)\equiv 0[m]\\m^2G(A-\frac{l}{m})\leqslant X}}\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l}{m}\right)=Z_F X^\frac{1}{2}\int_{\vartheta_2}^{\vartheta_1}\frac{\operatorname{d}x}{\sqrt{G(x)}}+O\left(X^\frac{1}{2}\frac{(\log\log X)^{\frac{\alpha_d}{2}}}{(\log X)^{\frac{\beta_d}{2}}}\right).$$
\end{proposition}
\begin{proof}
Fixons $\alpha=\frac{\alpha_d}{2},\beta=\frac{\beta_d}{2}$. Divisons l'intervalle $]\vartheta_2,\vartheta_1[$ en $O\left( \frac{(\log X)^\alpha}{(\log \log X)^\beta}\right)$ sous-intervalles $\{J_k\}$ où
$$J_k=\left[r_k,r_{k+1}\right[,\quad r_{k+1}=r_k+\frac{(\log \log X)^\beta}{(\log X)^\alpha}$$
et définissons la fonction en escalier $H$ par
$$H(x)=\min_{y\in J_k}G(y),\quad x\in J_k.$$
Puisque $G$ est uniformément continue sur $[\vartheta_2,\vartheta_1]$, il existe $c_0>0$ une constante absolue telle que
\begin{equation}\label{eq:interestimate1}
0\leqslant \sup_{x\in ]\vartheta_2,\vartheta_1[}\left(\sqrt{G(x)^{-1}}-\sqrt{H(x)^{-1}}\right)<c_0\frac{(\log \log X)^\beta}{(\log X)^\alpha}.
\end{equation}
\begin{comment}
Pour tout $k$, notons $$N_k(X)=\sum_{m\leqslant \frac{X}{\sqrt{H(r_k)}}}\varrho_F(m),$$
et $\{s_i\}_{i=1}^{N_k(X)}$ la suite des solutions de congruences
$$F(l)\equiv 0[m],\quad 0\leqslant l<m,\quad m\leqslant \frac{X^\frac{1}{2}}{\sqrt{{H(r_k)}}}.$$
Nous avons, d'après le corollaire \ref{co:estimationofdiscrepancy},
\begin{equation}\label{eq:interestimate0}
D_{N_k(X)}(s_i)=O\left(\frac{(\log\log X)^{\frac{5}{3}}}{(\log X)^{\frac{2-\sqrt{2}}{3}}}\right),
\end{equation}
où la constante implicite ne dépende que de $F$ et elle est indépendante de $k$.
\end{comment}
Si l'intervalle $]A-r_{k+1},A-r_k[$ ne contient aucun entier, on définit $M_{k}=\lfloor A-r_{k+1}\rfloor$ et l'intervalle $$I_k=]A-r_{k+1}-M_k,A-r_k-M_k[\subset]0,1[.$$ Cependant si $]A-r_{k+1},A-r_k[$ contient un entier $\tilde{M}_k$, définissons les intervalles
$$I_{1,k}=[0,A-r_k-\tilde{M}_k[,\quad I_{2,k}=]A-r_{k+1}-(\tilde{M}_k-1),1[,\quad I_k=I_{1,k}\cup I_{2,k}.$$
Alors $|J_k|=|I_k|$ et les solutions $(l,m)\in\ZZ^2$ du problème de congruence
$$F(l)\equiv 0[m],\quad A-\frac{l}{m}\in J_k$$
sont en bijection avec celles $(l^\prime,m)\in\ZZ^2$ de
$$F(l^\prime)\equiv 0[m], \quad \frac{l^\prime}{m}\in I_k.$$
Nous avons, grâce à la Proposition \ref{prop:discrepancy} et \eqref{eq:interestimate1},
\begin{align*}
\sum_{\substack{l,m\in\NN_{\geqslant 1},F(l)\equiv 0[m]\\m^2H(A-\frac{l}{m})\leqslant X}}\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l}{m}\right)&= \sum_{\substack{l,m\in\NN_{\geqslant 1},F(l)\equiv 0[m]\\m^2H(A-\frac{l}{m})\leqslant X}}\sum_{k}\textbf{1}_{J_k}\left(A-\frac{l}{m}\right)\\
&=\sum_{k} \sum_{\substack{l^\prime,m\in\NN_{\geqslant 1},F(l^\prime)\equiv 0[m]\\m\leqslant \sqrt{\frac{X}{H(r_k)}} }} \textbf{1}_{I_k}\left(\frac{l^\prime}{m}\right)\\
&=Z_F X^\frac{1}{2} \left(\sum_{k}\frac{|I_k|}{\sqrt{H(r_k)}}\right)+O\left( \sum_{k} X^\frac{1}{2}\frac{(\log\log X)^{\alpha_d}}{(\log X)^{\beta_d}}\right)\\
&=Z_F X^\frac{1}{2}\int_{\vartheta_2}^{\vartheta_1}\frac{\operatorname{d}x}{\sqrt{G(x)}}+O\left(X^\frac{1}{2}\frac{(\log\log X)^{\alpha}}{(\log X)^{\beta}}\right).
\end{align*}
Or nous avons aussi,
\begin{align*}
&\sum_{\substack{l,m\in\NN_{\geqslant 1},F(l)\equiv 0[m]\\m^2G(A-\frac{l}{m})\leqslant X}}\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l}{m}\right)\\
=&\sum_{\substack{l,m\in\NN_{\geqslant 1},F(l)\equiv 0[m]\\m^2H(A-\frac{l}{m})\leqslant X}}\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l}{m}\right)+O\left(\sum_{\substack{X^\frac{1}{2}/(\sqrt{H\left(A-\frac{l}{m}\right)}+c_0\frac{(\log \log X)^\beta}{(\log X)^\alpha})<m\leqslant X^\frac{1}{2}/\sqrt{H\left(A-\frac{l}{m}\right)}\\F(l)\equiv 0[m]}}\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l}{m}\right)\right)\\
=&Z_F X^\frac{1}{2}\int_{\vartheta_2}^{\vartheta_1}\frac{\operatorname{d}x}{\sqrt{G(x)}}+O\left(X^\frac{1}{2}\frac{(\log\log X)^{\alpha}}{(\log X)^{\beta}}\right)\\
&+O\left(\left(\sum_{\substack{m\leqslant X^\frac{1}{2}/\sqrt{H\left(A-\frac{l}{m}\right)}\\F(l)\equiv 0[m]}}-\sum_{\substack{m\leqslant X^\frac{1}{2}/(\sqrt{H\left(A-\frac{l}{m}\right)}+c_0\frac{(\log \log X)^\beta}{(\log X)^\alpha})\\F(l)\equiv 0[m]}}\right)\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l}{m}\right)\right)\\
=&Z_F X^\frac{1}{2}\int_{\vartheta_2}^{\vartheta_1}\frac{\operatorname{d}x}{\sqrt{G(x)}}+O\left(X^\frac{1}{2}\frac{(\log\log X)^{\alpha}}{(\log X)^{\beta}}\right).
\end{align*}
Ceci achève la formule énoncée.
\end{proof}
\subsubsection{Dénouement}
On désigne par $\mathfrak{C}$ la courbe
$$\{(w,z)\in S_1:zw\sqrt{z+w}=\frac{D^\frac{5}{2} C_3^\frac{1}{2}W^3}{D_1^2D_2^2}\}.$$
Supposons pour la suite que la région $R(\varepsilon_i,\theta_j)$ \eqref{eq:chi} vérifie que $\operatorname{dist}(\mathfrak{C},R(\varepsilon_i,\theta_j))\geqslant\eta>0$ et donc l'ensemble $B_{C_3,D}^{\varepsilon_i,\theta_j}(D_i,q)$ vérifie la condition de seuil donné par les équations de seuils dans les lemmes \ref{le:threshold0} et \ref{le:threshold1} pour $B\gg_{C_3,D,\varepsilon_i,\theta_j,} 1$. Il vérifie aussi \eqref{eq:threshold1}.
\begin{corollary}
Il existe une constante $\mathcal{E}=\mathcal{E}_{C_3,D}(D_ih_i,e_j,q)>0$ telle que
\begin{equation}\label{eq:decomp5}
\# \mathcal{C}_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_ih_i,e_j,q)=B^{\frac{1}{5}}\mathcal{E}Z_{\mathcal{F}}\int\int\chi_{\varepsilon_i,\theta_j}\frac{\mathfrak{E}_{C_3,D,D_i,W}(zw\sqrt{z+w})\operatorname{d}z\operatorname{d}w}{zw\sqrt{z+w}}+O\left(B^\frac{1}{5}\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right),
\end{equation}
où $$\mathfrak{E}_{C_3,D,D_i,W}(x)=\textbf{1}\left\{x:x>\frac{D^\frac{5}{2}C_3^\frac{1}{2}W^3}{D_1^2D_2^2}\right\},$$
$\chi_{\varepsilon_i,\theta_j}$ est définie par \eqref{eq:chi}, $\mathcal{F}(X)=\mathcal{F}_{C_3,D,h_iD_i,e_j,q}(X)$ est le polynôme \eqref{eq:formF} et la constant $f_{\mathcal{F}}(1)$ est définie dans la proposition \ref{le:varrhoF}.
\end{corollary}
\begin{proof}
Avec le changement de paramètres \eqref{eq:xylm}, nous avons
$$v=e_1e_2e_3m^\prime-u=e_3g(D_1h_1D_2h_2)l^\prime-e_1e_2e_3(\gamma_1g(D_1h_1)-1)m^\prime,$$ et d'où
\begin{equation}\label{eq:lprimemprime2}
\frac{C_3}{D}\frac{uv^2}{u+v}=\frac{C_3e_3^2g(h_1D_1h_2D_2)^3}{De_1e_2}(m^\prime)^2\left(\frac{e_1e_2\gamma_1}{g(h_2D_2)}-\frac{l^\prime}{m^\prime}\right)\left(\frac{e_1e_2}{g_1g_2}-\left(\frac{e_1e_2\gamma_1}{g(h_2D_2)}-\frac{l^\prime}{m^\prime}\right)\right)^2.
\end{equation}
Quitte à rajouter un terme d'erreur d'ordre grandeur $O(1)$, nous pouvons réécrire \eqref{eq:condzoom1} comme
\begin{equation}\label{eq:lprimemprime1}
\frac{e_1e_2\gamma_1}{g(h_2D_2)}-\frac{e_1e_2}{g_1g_2}\frac{\theta_1}{1+\theta_1}<\frac{l^\prime}{m^\prime}<\frac{e_1e_2\gamma_1}{g(h_2D_2)}-\frac{e_1e_2}{g_1g_2}\frac{\theta_2}{1+\theta_2};
\end{equation}
$$\varepsilon_2 (m^\prime)^2 G\left(A-\frac{l^\prime}{m^\prime}\right)\leqslant B^\frac{2}{5}\leqslant \varepsilon_1 (m^\prime)^2 G\left(A-\frac{l^\prime}{m^\prime}\right).$$
Remarquons $e_1e_2\mid g_1g_2$ et $\gamma_1g(h_1D_1)\geqslant 1$. Nous avons donc
$$\frac{e_1e_2\gamma_1}{g(h_2D_2)}-\frac{l^\prime}{m^\prime}\subset \left]\frac{e_1e_2}{g_1g_2}\frac{\theta_2}{1+\theta_2},\frac{e_1e_2}{g_1g_2}\frac{\theta_1}{1+\theta_1}\right[\subset\mathopen]0,1\mathclose[,$$
$$\frac{e_1e_2\gamma_1}{g_2}\geqslant \frac{e_1e_2}{g_1g_2}>\frac{e_1e_2}{g_1g_2}\frac{\theta}{1+\theta},\quad \forall \theta\geqslant 0.$$
Nous appliquons la proposition \ref{po:centralcouting} avec
$$G(X)=\frac{C_3e_3^2g(h_1D_1h_2D_2)^3}{De_1e_2}X\left(\frac{e_1e_2}{g_1g_2}-X\right)^2,\quad F(X)=\mathcal{F}_{C_3,D,h_iD_i,e_j,q}(X)$$
$$A=\frac{e_1e_2\gamma_1}{g_2},\quad \vartheta_i=\frac{e_1e_2}{g_1g_2}\frac{\theta_i}{1+\theta_i},\quad (i=1,2).$$
Il en suit que
\begin{align*}
\# \mathcal{C}_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_ih_i,e_j,q)&=\left(\sum_{\substack{l^\prime,m^\prime\in\NN,F(l^\prime)\equiv 0[m^\prime]\\(m^\prime)^2\varepsilon_2G(A-\frac{l^\prime}{m^\prime})\leqslant B^\frac{2}{5}}}-\sum_{\substack{l^\prime,m^\prime\in\NN,F(l^\prime)\equiv 0[m^\prime]\\(m^\prime)^2\varepsilon_1G(A-\frac{l^\prime}{m^\prime})\leqslant B^\frac{2}{5}}}\right)\textbf{1}_{\vartheta_2,\vartheta_1}\left(A-\frac{l^\prime}{m^\prime}\right)\\
&=B^{\frac{1}{5}}\frac{(De_1e_2)^\frac{1}{2}Z_\mathcal{F}}{(C_3e_3^2g(h_1D_1h_2D_2)^3)^\frac{1}{2}}\left(\frac{1}{\sqrt{\varepsilon_2}}-\frac{1}{\sqrt{\varepsilon_1}}\right)\int_{\vartheta_1}^{\vartheta_2}\frac{\operatorname{d}x}{\sqrt{x\left(\frac{e_1e_2}{g_1g_2}-x\right)^2}}+O\left(B^\frac{1}{5}\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)\\
&=B^{\frac{1}{5}}\frac{D^\frac{1}{2}Z_\mathcal{F}}{2C_3^\frac{1}{2}e_3g(D_1h_1D_2h_2)}\int_{\varepsilon_2}^{\varepsilon_1}\int_{\theta_2}^{\theta_1}\frac{\operatorname{d}z\operatorname{d}\theta}{z^\frac{3}{2}\sqrt{\theta(1+\theta)} }+O\left(B^\frac{1}{5}\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)\\
&=B^{\frac{1}{5}}\frac{D^\frac{1}{2}Z_\mathcal{F}}{2C_3^\frac{1}{2}e_3g(D_1h_1D_2h_2)}\int\int_{\substack{\varepsilon_2\leqslant z\leqslant\varepsilon_1\\\theta_2<\frac{z}{w}<\theta_1}}\frac{\operatorname{d}z\operatorname{d}w}{zw\sqrt{z+w}}+O\left(B^\frac{1}{5}\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right).
\end{align*}
Nous obtenons donc la formule énoncée avec $$\mathcal{E}_{C_3,D}(D_ih_i,e_j,q)=\frac{D^\frac{1}{2}}{2C_3^\frac{1}{2}e_3g(D_1h_1D_2h_2)}.$$
\end{proof}
Il nous reste à traiter le terme d'erreur introduit dans \eqref{eq:seuilreplaceheight}. Le même raisonnement comme précédemment nous donne l'estimation
\begin{equation}\label{eq:errortermcontrol}
\begin{split}
&\# \operatorname{Er}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)\ll_{C_3,D,W,\varepsilon_i}\\
&B^{\frac{1}{5}}\int\int \chi\left(R(\varepsilon_i,\theta_j)\cap \{(w,z):\frac{D^\frac{5}{2}C_3^\frac{1}{2}W^3}{D_1^2D_2^2}\leqslant zw\sqrt{z+w}\leqslant \frac{D^\frac{5}{2}C_3^\frac{1}{2}W^3}{D_1^2D_2^2}+\mu_0 (B^{-\frac{2}{5}})\}\right)\frac{\operatorname{d}z\operatorname{d}w}{zw\sqrt{z+w}}\\
&+O\left(B^\frac{1}{5}\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)=O\left(B^\frac{1}{5}\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right).
\end{split}
\end{equation}
Rassemblons ensemble des égalités \eqref{eq:decomp1}, \eqref{eq:decomp2}, \eqref{eq:decomp3}, \eqref{eq:decomp4}, \eqref{eq:decomp5} et \eqref{eq:errortermcontrol}, nous sommes prêt à démontrer le théorème \ref{th:principalthm}.
\begin{proof}[Démonstration du théorème \ref{th:principalthm}]
Dans ce qui suit, pour éviter des formules superflues, on écrira le terme d'erreur sous la forme non-explicite $o(B^\frac{1}{5})$ car il serai clair d'où viennent ces contributions.
Soient $C_3,D,W,D_1,D_2,q\in\NN$ tels que
$$\pgcd(C_3W,D)=\pgcd(D_1,D_2)=1,\quad D_1D_2\mid D,\quad q\mid W.$$
En les fixant et en sommant sur les formules \eqref{eq:decomp3}, \eqref{eq:decomp4}, \eqref{eq:decomp5}, nous obtenons la formule pour le cardinal de l'ensemble $B_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_i,q)$ comme suit.
Il existe une constante
$$Z_{C_3,D,W}(D_i,q)=\sum_{\substack{h_1,h_2\in\NN_{\geqslant 1}\\D_1h_1,D_2h_2\mid D\\ \pgcd(h_1,h_2)=1}} \mu(h_1)\mu(h_2)\sum_{\substack{e_1e_2\mid g(h_1D_1h_2D_2),\pgcd(e_1,e_2)=1\\e_1e_2e_3\mid qD,\pgcd(e_3,g(h_1D_1h_2D_2))=1}}\left(\prod_{j=1}^{3}\mu(e_j)\right)\frac{Z_\mathcal{F}D^\frac{1}{2}}{2C_3^\frac{1}{2}e_3g(D_1h_1D_2h_2)}$$ telle que
\begin{align*}
\# B_{C_3,D,W}^{\varepsilon_i,\theta_j}(D_i,q)&=B^\frac{1}{5}Z_{C_3,D,W}(D_i,q)\int\int\chi_{\varepsilon_i,\theta_j}\frac{\mathfrak{E}_{C_3,D,D_i,W}\left(zw\sqrt{z+w}\right)}{zw\sqrt{z+w}}\operatorname{d}z\operatorname{d}w+o(B^\frac{1}{5}).
\end{align*}
Nous concluons de \eqref{eq:decomp2} qu'en fixant $C_3,D,W,D_1,D_2$,
\begin{align*}
\# \tilde{E}^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)&=\sum_{q\mid W} \mu\left(\frac{W}{q}\right)B_{C_3,D,W}^{\varepsilon_i,\theta_j}\left(D_i,q\right)\\
&=B^\frac{1}{5}\int\int\chi_{\varepsilon_i,\theta_j}\frac{\mathbf{E}_{C_3,D,D_i,W}\left(zw\sqrt{z+w}\right)}{zw\sqrt{z+w}}\operatorname{d}z\operatorname{d}w+o(B^\frac{1}{5}),
\end{align*}
où $$\mathbf{E}_{C_3,D,D_i,W}(x)=\sum_{q\mid W}\mu\left(\frac{W}{q}\right)Z_{C_3,D,W}\left(D_i,q\right)\mathfrak{E}_{C_3,D,D_i,q}(x).$$
En reportant dans \eqref{eq:decomp1} et \eqref{eq:seuilreplaceheight}, rappelons l'estimation \eqref{eq:errortermcontrol}, nous obtenons finalement
\begin{align*}
\delta_{\varrho^{-1}(S_1)\cap U,Q,B,\frac{5}{2}}(\chi_{\varepsilon_i,\theta_j})&=\sum_{\substack{C_3,D,W\in\NN_{\geqslant 1}\\\pgcd(C_3W,D)=1}}\sum_{\substack{D_1D_2\mid D\\\pgcd(D_1,D_2)=1}}\# E^{\varepsilon_i,\theta_j}_{C_3,D,W}(D_i)\\
&=B^\frac{1}{5}\int\int\chi_{\varepsilon_i,\theta_j}(w,z)\frac{\mathds{E}\left(zw\sqrt{z+w}\right)}{zw\sqrt{z+w}}\operatorname{d}z\operatorname{d}w+o(B^\frac{1}{5}).
\end{align*}
où \begin{equation}\label{eq:E}
\mathds{E}(x)=\sum_{\substack{C_3,D,W\in\NN_{\geqslant 1}\\\pgcd(C_3W,D)=1}}\sum_{\substack{D_1D_2\mid D\\\pgcd(D_1,D_2)=1}}\mathbf{E}_{C_3,D,D_i,W}(x).
\end{equation}
\end{proof}
\subsubsection{Résultat pour d'autres régions.}
Esquissons la La méthode et les calculs sont presque pareils. Nous ne rentrerons pas dans des détails.
\noindent\textbf{Région $S_2$.}
Rappelons \eqref{eq:S2}. Les points dans cette région sont (dans le voisinage $\max(|w|,|z|)<1$) paramétrés par les couples $(a,b)\times(u,v)\in T_2$. Nous avons toujours \eqref{eq:maxheight} et \eqref{eq:maxdistance}, comme $0<-w<z$. Le paramètre $D$ reste positif en vertu du lemme \ref{le:au2-bv2}.
\begin{theorem}\label{th:S2}
$$\delta_{\varrho^{-1}(S_2)\cap U,Q,B,\frac{5}{2}}(\chi_{\varepsilon_i,\theta_j})= B^\frac{1}{5}\left(\int\chi_{\varepsilon_i,\theta_j}(w,z) \frac{\mathds{E^\prime}(-wz\sqrt{z+w})}{-wz\sqrt{z+w}}\operatorname{d}w\operatorname{d}z +O_{\varepsilon_i,\theta_j}\left(\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)\right),$$
où $\mathds{E}^\prime(\cdot)$ est certaine fonction en escalier définie de façon analogue à $\mathds{E}(\cdot)$ \eqref{eq:E}.
\end{theorem}
\noindent\textbf{Régions $S_3$ et $S_4$.}
Pour ces deux régions il y une modification légère sur le calcul de la hauteur et celui de la distance en comparant avec \eqref{eq:maxheight} et \eqref{eq:maxdistance}. Par exemple dans $S_4$,
$$\max_{f\in S}|f|=|y^2t^2|,$$
$$d((\varrho\circ\Psi) (a,b)\times (u,v))=-z=-\frac{\frac{u^2}{v^2} -\frac{b}{a}}{\frac{b}{a}-\frac{u}{v}}.$$
Des calculs similaires montrent qu'elle ne donne pas de différence essentielle. C'est en fait la partie mince $M$ qui obscurcit l'équidistribution des approximants locaux.
Rappelons la proposition \ref{le:splitpoly} et l'observation dans \ref{se:otherregions}, l'ensemble mince $M$ correspond aux paramètres $-C_3D\in\square$. En dehors de $M$, autrement dit, en ne considérons que la famille d'équations $\mathcal{E}_{C_3,D}$ \eqref{eq:eqpellfermatgen} avec $-C_3D\not\in\square$, nous pouvons démontrer:
\begin{theorem}\label{th:S3}
$$\delta_{\varrho^{-1}(S_i)\cap U\setminus M,Q,B,\frac{5}{2}}(\chi_{\varepsilon_i,\theta_j})= \begin{cases}
B^\frac{1}{5}\left(\int \chi_{\varepsilon_i,\theta_j}(w,z)\frac{\mathds{E^{\prime\prime}\mkern-1.2mu}(-wz\sqrt{-(w+z)}) }{-wz\sqrt{-(w+z)}} \operatorname{d}w\operatorname{d}z +O_{\varepsilon_i,\theta_j}\left(\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)\right),& (i=3);\\
B^\frac{1}{5}\left(\int \chi_{\varepsilon_i,\theta_j}(w,z)\frac{\mathds{E^{\prime\prime}\mkern-1.2mu}(wz\sqrt{-(w+z)}) }{wz\sqrt{-(w+z)}} \operatorname{d}w\operatorname{d}z +O_{\varepsilon_i,\theta_j}\left(\frac{(\log\log B)^\frac{5}{6}}{(\log B)^\frac{2-\sqrt{2}}{6}}\right)\right),& (i=4).
\end{cases}$$
où $$\mathds{E}_{S_i}^{\prime\prime}\mkern-1.2mu(\cdot)=\sum_{\substack{C_3,W\in\NN_{\geqslant 1},D\in\ZZ_{< 0}\\\pgcd(C_3W,D)=1,-DC_3\not\in\square}}\sum_{\substack{D_1D_2\mid D\\\pgcd(D_1,D_2)=1}}\mathbf{E}^{\prime\prime}\mkern-1.2mu_{S_i,C_3,D,D_i,W}(\cdot)$$
est définie de façon analogue à des quantité dans la preuve du théorème \ref{th:principalthm}.
\end{theorem}
\subsection{Décompte de la partie mince}\label{se:countthethinpart}
Dans cette partie, nous dénombrons les points dans la partie mince $M$ qui rentrent dans le zoom critique.
Rappelons-nous que $M$ est entièrement dans la région $R_2$. Dans l'esprit de l'équidistribution globale, il ne serait pas raisonnable à croire que la distribution locale autour de $Q$ soit concentrée sur une partie semi-algébrique qui n'est pas dense pour la topologie réelle.
\subsubsection{Une minoration sur la région $R_2$}
Dans la régions $R_2$, grâce à la famille des courbes cuspidales \eqref{eq:cusp}, on est capable de localiser des points dans le grossissement ainsi qu'une minoration pour la famille de mesure $\{\delta_{\varrho^{-1}(R_2)\cap U,Q,B,\frac{5}{2}}\}$. Cependant, à cause de la complexité des conditions de divisibilité fournies par son paramétrage, nous n'allons pas l'utiliser ni donner des détails.
\begin{proposition}\label{po:lowerboundmeasure}
Pour tout fonction positive $f$, on a
$$\delta_{\varrho^{-1}(R_2)\cap U,Q,B,\frac{5}{2}}(f)\geqslant B^{\frac{1}{5}} \delta_{\rab}(f),$$
où $\delta_{\rab}$ est la mesure limite de la famille $\{\delta_{\rab,Q,B,\frac{5}{2}}\}$.
\end{proposition}
\begin{proof}
Observons
$$\delta_{\varrho^{-1}(R_2)\cap U,Q,B,\frac{5}{2}}(f)\geqslant \delta_{ \rab,Q,B,\frac{5}{2}}(f).$$
La minoration résulte directement du théorème \ref{th:Pagelot}.
\end{proof}
\subsubsection{Minoration et majoration globale}
\begin{lemma}\label{le:lowerboundM}
Pour $\varepsilon\gg 1$, nous avons
$$\delta_{M,Q,B,\frac{5}{2}}(\chi(\varepsilon))\gg_\varepsilon B^\frac{1}{5}\log B.$$
\end{lemma}
\begin{proof}
Nous nous bornons à la région $S_4$.
Nous avons
$$\max_{f\in S}|f|=|y^2t^2|.$$
Compte-tenu du paramétrage \eqref{eq:T1}, rappelons que $au^2-bv^2<0$ (lemme \ref{le:au2-bv2}) et $bv-au>0$. Nous considérons $P=(a,b)\times (u,v)\in T_2$ vérifiant l'équation avec la divisibilité
$$au^2-bv^2=a-b,\quad b-a\mid u-v.$$
Nous avons alors
$$D_1=D_2=1,\quad d_3=b-a.$$
La condition de zoom $B^\frac{2}{5}d(P)\leqslant \varepsilon$ et la condition de hauteur se traduisent maintenant en
\begin{equation}\label{eq:conditionE1}
-\frac{\frac{u^2}{v^2} -\frac{b}{a}}{\frac{b}{a}-\frac{u}{v} }\leqslant \varepsilon B^{-\frac{2}{5}},\quad \frac{uv(bv-au)^3}{D_1^2D_2^2d_3^3}=\frac{uv(bv-au)^3}{(b-a)^3}\leqslant B
\end{equation}
Comme indiqué au début du texte, nous sommes ramené au problème de congruence
\begin{equation}\label{eq:conditionE3}
(u+1)(u-1)\equiv 0[u+v],\quad a=\frac{v^2-1}{k(u,v)},\quad b=\frac{u^2-1}{k(u,v)}.
\end{equation}
Avec cela nous réécrivons \eqref{eq:conditionE1} comme
\begin{equation}\label{eq:conditionE2}
\varepsilon v(uv+1)\geqslant B^\frac{2}{5}(u+v),\quad uv(uv+1)^3\leqslant B(u+v)^3.
\end{equation}
Nous précédons comme l'argument dans la démonstration du lemme \ref{le:threshold1}. Ces deux inégalités donnent
$$\frac{u^4v^4}{(u+v)^3}\leqslant B\leqslant \varepsilon^\frac{5}{2}\frac{u^\frac{5}{2}v^5}{(u+v)^\frac{5}{2}}\left(1+O_\varepsilon(B^{-\frac{2}{5}})\right),$$
ce qui implique que pour $B$ suffisamment grand, $$\varepsilon\frac{v}{u}\sqrt{1+\frac{v}{u}}\geqslant 1-O_\varepsilon(B^{-\frac{2}{5}})\geqslant \frac{1}{2}.$$
Soit $\theta>1$ tel que $$\frac{\theta^3}{1+\theta}=4\varepsilon^2,\quad \frac{2(1+\theta)^\frac{7}{2}}{\theta^\frac{5}{2}}> 2^\frac{16}{5}.$$
Un tel $\theta$ existe lorsque $\varepsilon$ est suffisamment grand.
Alors tout $(u,v)$ vérifiant \eqref{eq:conditionE2} satisfait nécessairement à
$$1+\frac{u}{v}\leqslant 4\varepsilon^2\left(\frac{u}{v}\right)^3,$$ d'où
$\frac{u}{v}\leqslant\theta$.
Nous en concluons que la condition
$$\frac{(1+\theta)^3}{\varepsilon\theta}B^\frac{2}{5}\leqslant (u+v)^2\leqslant 2^{\frac{16}{5}} B^\frac{2}{5}$$
implique \eqref{eq:conditionE2}. De plus, la condition $\pgcd(u,v)=1$, qui est équivalente à $\pgcd(u,u+v)=1$, est une condition nécessaire pour que la congruence \eqref{eq:conditionE3} ait des solutions. Le dénombrement du zoom est donc minoré par le cardinal des $(l,m)\in\NN_{\geqslant 1}^2$ tels que
$$C_1(\varepsilon)B^\frac{1}{5}\leqslant m\leqslant C_2(\varepsilon)B^\frac{1}{5},\quad l^2-1\equiv 0[m],\quad 1\leqslant l\leqslant m<2l.$$
Remarquons que si $(l,m)$ est une solution, le couple $(m-l,m)$ l'est aussi mais elle vérifie $1\leqslant m-l<l$. Ceci signifie que les solutions apparaissent en couples et exactement une nous convient.
La majoration cherchée est donc la conséquence de la proposition \ref{po:splitcong}:
$$\delta_{M,Q,B,\frac{5}{2}}(\chi(\varepsilon))\geqslant\frac{1}{2} \sum_{C_1(\varepsilon)B^\frac{1}{5}\leqslant m\leqslant C_2(\varepsilon)B^\frac{1}{5}}\varrho_{X^2-1}(m)\sim_\varepsilon B^\frac{1}{5}\log B.$$
\end{proof}
\begin{comment}
sur l'ensemble $E(B)$, définie par
\begin{equation}\label{eq:bijEF}
(x,y)\longmapsto [u:v]\times[a:b]=[x:y]\times\left[\frac{y^2+1}{k}:\frac{x^2+1}{k}\right],\quad k=\pgcd(x^2+1,y^2+1).
\end{equation}
Grâce au raisonnement précédent, on voit que la troisième condition dans \eqref{eq:conditionF} implique la condition \eqref{eq:abba}.
Elle est en fait équivalente à dire que $x,y$ sont les deux solutions de l'équation de congruence quadratique
$$v^2\equiv -1 [q],\quad 0\leqslant v\leqslant q-1.$$
Réciproquement toute telle paire $(x,y)\in \NN^2$ vérifie la condition de co-primalité puisque $\pgcd(x,y)\mid -1$.
Compte-tenu de \eqref{eq:bijEF}, les dernières deux conditions dans \eqref{eq:conditionE} sont équivalentes à
\begin{equation}\label{eq:equivE}
\varepsilon y(xy-1)\geqslant (x+y)B^{\frac{2}{5}},\quad (y^2+1)^2 (x^2+1)x^2\leqslant B(x+y)^3.
\end{equation}
Soient
$$\tau_1=\frac{\theta_2}{1+\theta_2},\quad \tau_2=\frac{\theta_1}{1+\theta_1},$$
et $\varepsilon>0$ tel que $$\frac{(\theta_2+1)^{16}}{\theta_2^8}>\frac{2^{11}(\theta_1+1)^{15}}{\varepsilon \theta_1^5}.$$
Puis choisissons $\lambda_1,\lambda_2>0$ tels que
$$\frac{2^{5}(\theta_1+1)^{15}}{\varepsilon \theta_1^5}\leqslant\lambda_2^{10}<\lambda_1^{10}\leqslant\frac{(\theta_2+1)^{16}}{2^6 \theta_2^8}.$$
Démontrons-nous maintenant que les encadrements sur $l,m$ dans \eqref{eq:conditionF} implique la condition \eqref{eq:conditionE} pour $B\gg 1$. Supposons que $y+x\leqslant \lambda_1 B^{\frac{1}{5}}$. Nous avons
\begin{align*}
(y^2+1)^2 (x^2+1)x^2&\leqslant \left(\frac{(x+y)^2}{(\theta_2+1)^2}+1\right)^2 (\tau_1^2(x+y)^2+1)\tau_1^2 (x+y)^2\\
&\leqslant \frac{8 \tau_1^4}{(\theta_2+1)^4}(x+y)^8\\
&=\frac{8 (\theta_2+1)^8 }{\theta_2^4}(x+y)^8\\
&\leqslant \frac{8 \lambda_1^5 (\theta_2+1)^8 }{\theta_2^4} B (x+y)^3\\
&\leqslant B(x+y)^3.
\end{align*}
De façon similaire nous pouvons montrer que la deuxième implication.
Choisissons $\lambda_2,\tau_2>0$ suffisamment petit tels que
$$\theta_2<\frac{\lambda_2}{\tau_2}<\frac{\theta_2(1+\theta_1)}{1+\theta_2},\quad 8\lambda_2^4 \tau_2^4\leqslant (\lambda_2+\tau_2)^3.$$
Le choix du quotient $\frac{\lambda_2}{\tau_2}$ implique que
$$\theta_1\tau_2-\lambda_2>\frac{\lambda_2}{\theta_2}-\tau_2.$$
Nous pouvons donc choisir $\lambda_1>\lambda_2$ et $\tau_1>\tau_2$ tels que
$$\tau_1<\frac{\lambda_2}{\theta_2},\quad \lambda_1<\theta_1\tau_2,\quad \tau_1-\tau_2>\lambda_1-\lambda_2.$$
Montrons que pour $\varepsilon>0$ tel que
$$\varepsilon \lambda_2\tau_2^2\geqslant (\lambda_1+\tau_1),$$
la condition \eqref{eq:conditionF} implique la condition \eqref{eq:conditionE}.
\end{comment}
\begin{lemma}\label{le:upperboundM}
Pour tout région $R$ à support compact dans $S_3\cup S_4$, nous avons
$$\delta_{M,Q,B,\frac{5}{2}}(\chi(R))\ll_R B^\frac{1}{5}\log B.$$
\end{lemma}
\begin{proof}
Chaque équation \eqref{eq:eqpellfermatgen} contribue des points d'ordre de grandeur $O_{C_3,D}(B^{\frac{1}{5}}\log B)$ (À cette fin nous pouvons oublier la condition de zoom et celle de coprimalité). En effet, compte-tenu de la proposition \ref{le:splitpoly}, une telle contribution est majorée par
$$\sum_{k\leqslant CB^\frac{1}{5}}\varrho_{C_3X^2+D}(k)\sim_{C_3,D}B^\frac{1}{5}\log B,$$
où $C=C(C_3,D)>0$.
D'après un argument similaire pour obtenir les équations de seuil comme \eqref{eq:threshold1} dans les régions $S_3$ et $S_4$, soit $\varepsilon>0$ tel que $R\subset \mathbb{B}(0,\varepsilon)$,
nous n'avons qu'un nombre de paramètres $C_3,D$ possibles paramétrant localement $M$ (cf. Proposition \ref{le:splitpoly}). D'où la majoration énoncée.
\end{proof}
\begin{corollary}\label{co:rab}
La partie mince $M$ est localement faiblement accumulatrice.
\end{corollary}
\begin{proof}
Ceci suit du fait que
$$\alpha(M,Q)=\alpha(\rab,Q)=\frac{5}{2}>\alpha(Q,Y_3)=2$$
pour toute courbe cuspidale $\rab$ et des deux lemmes précédents.
\end{proof}
\section{Perspectives et questions}\label{se:perspectives}
\begin{comment}
\subsection{Courbes rationnelles contribuant des meilleurs approximants génériques}
Dans l'heuristique du calcul de la dimension de l'espace de courbes rationnelle plan de degré donné, on espère que ce seraient les courbes rationnelles singulières qui contribuent à l'approximation générique à partir des surfaces de degré $5$. De plus, les courbes cubiques cuspidales ne jouent un rôle que pour des surfaces de degré $5$. On espère de plus qu'à partir de celles de degré $4$ jusqu'à celles de degré $2$, la constante essentielle serait toujours $2$ et les approximants se trouveraient dans des courbes nodales.
\subsubsection{Courbes rationnelles très libres}
Si une composante irréductible $\mathfrak{Comp}$ de $\Mor(\PP^1,X)$ paramètre des courbes rationnelles \emph{très libres} sur $X$, alors nous avons une application rationnelles dominantes
$$\pi :\mathfrak{Comp}\times \PP^1\to X.$$
Les cas plus intéressant est $\mathfrak{Comp}$ contient des courbes $\mathcal{C}$ de sorte que $=\alpha(C,Q)=\aess(Q)$ et l'application $\pi$ est un paramétrage des points rationnels sur $X$. Fixons une hauteur $\h$ sur $X(\QQ)$. Alors nous obtenons à l'aide de tiré en arrière par $\pi$, une hauteur sur $\mathfrak{Comp}\times \PP^1$. Ce serait intéressant de traduire des informations diophantiennes venant de $X$ à $\mathfrak{Comp}$ et d'obtenir des correspondances utiles pour meilleur comprendre le rôle qui jouent par les courbes rationnelles dans l'approximation diophantienne.
\end{comment}
\subsection{Lien avec la notion de liberté}
Avant la découverte des résultats pour $Y_3$ et $Y_4$, la constante d'approximation des sous-variétés donne un indicateur fidèle pour détecter les points qui doivent être exclus pour l'existence d'une mesure du zoom critique. Suite à l'apparence de variétés localement faiblement accumulatrices et même de parties mince, nous ne pouvons plus espérer l'existence d'une distribution localement en simplement supprimant des sous-variétés strictes.
Nous manquons un critère vis-à-vis aux points eux-même.
Dans \cite{Peyre3}, s'inspirant des études des courbes rationnelles sur les variétés, E. Peyre a introduit la notion de \emph{liberté} pour les points rationnels originalement due à Bost. Ceci a pour but d'améliorer le principe de Batyrev-Manin. La liberté est définie par chaque point rationnel.
L'espérance générale est que les points qui sont susceptible d'être accumulateurs ainsi qu'être exclus, ont la liberté réduite.
Comme déjà constaté par E. Peyre dans un travail non-publié, pour la surface $\PP^1\times\PP^1$ où il n'y a pas d'accumulation globale, la liberté peut caractériser les deux droites (horizontale et verticale) contenant asymptotiquement les points plus proches du point $Q$.
Il a d'espoir que ce phénomène puisse s'étendre aux cas plus généraux. Seulement des études plus approfondies nous permet donner des évidences à trancher.
\subsection{Interprétation de la mesure critique}
Comme premier exemple de variété torique non complètement symétrique, la fonction de densité, qui a des pôles définies par les équations (asymptotiques) de $3$ droites localement accumulatrices, met en lumière de certaine \og non-régularité\fg{}\ dans l'uniformité de la distribution des points rationnels. Plus précisément, la concentration proche de chaque droite n'est pas pareille. Ceci est reflétée par la puissance sur chaque équation. Dans un travail à venir, compte-tenu d'une autre variété torique obtenue en contractant le $6$-ième rayon de l'éventail de $Y_3$, qu'on va appeler $Y_2$, un phénomène similaire est également constaté. La puissance sur les $2$ droites localement accumulatrices dans ce cas est respectivement $1$ et $2$. Une explication \emph{à priori} pourrait être \textbf{la configuration} des points éclatés. Elle façonne des courbes rationnelles qui réalisent l'approximation essentielle et finalement résulterait le mis en forme de la mesure limite.
Un phénomène surprenant est que pour la fonction de densité apparaissant dans \emph{tous les} exemples, la somme des ordres de pôles vaut exactement la constante essentielle (le facteur de zoom que l'on a pris). En fait il apparaît déjà dans le raisonnement pour trouver les équations de seuils \eqref{eq:threshold1} (cf. \S\ref{se:threshold}). Ce phénomène devrait être partiellement de nature géométrique et il serait intéressant de trouver des interprétations ultérieures.
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| 3,921
|
<?php
namespace App\Builder;
use App\Entity\Document;
class DocumentBuilder
{
/** @var string */
private $cacheDir;
/** @var \Twig_Environment */
private $environment;
public function __construct(\Twig_Environment $environment, $cacheDir)
{
$this->environment = $environment;
$this->cacheDir = $cacheDir;
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public function build(Document $document, $section)
{
$isAutoReload = $this->environment->isAutoReload();
$this->environment->enableAutoReload();
$template = $document->getDocumentTemplate();
$company = $document->getCompany();
$method = 'get'.ucfirst($section);
//
// $file = $section . '.twig.html';
// $dir = sprintf('%s/companies/%d/templates/%d', $this->cacheDir, $company->getId(), $template->getId());
// $path = $dir . '/' . $file;
//
// @mkdir($dir, 0777, true);
// file_put_contents($path, $template->$method());
$content = $this->environment->createTemplate($template->$method());
if (!$isAutoReload) {
$this->environment->disableAutoReload();
}
return $content->render(
[
'document' => $document,
]
);
}
}
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{
"redpajama_set_name": "RedPajamaGithub"
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| 1,190
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Charles Antoine Gidel, född 5 mars 1827 i Gannat, död 31 oktober 1899 i Paris, var en fransk litteraturhistoriker.
Gidel ägnade sig åt lärarkallet och förestod flera lyceer. Bland hans arbeten är: Imitations faites en grec depuis le douzième siècle de nos andens poèmes de chevalerie (1864), Étude sur Saint-Evremond (1866), Études sur la littérature grecque moderne (1866–1878), Discours sur J. J. Rousseau (1868), Les français du XVII:e siècle (1873) och Histoire de la littérature française depuis son origine jusqu'à nos jours (1874–1891) samt upplagor av Boileau m.fl.
Källor
Franska litteraturvetare
Litteraturvetare under 1800-talet
Franska forskare under 1800-talet
Födda 1827
Avlidna 1899
Män
Ugglan
|
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| 8,384
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{"url":"http:\/\/clay6.com\/qa\/34039\/the-temperature-gradient-in-a-rod-0-5-m-long-is-80-c-m-the-temperature-of-t","text":"# The temperature gradient in a rod 0.5 m long is 80 c\/m . The temperature of the hot end of the rod is $\\;30^{0}C\\;$. Then the temperature of the cooler and is\n$(a)\\;40^{0}C\\qquad(b)\\;10^{0}C\\qquad(c)\\;-10^{0}C\\qquad(d)\\;15^{0}C$","date":"2021-04-13 18:53:35","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5684041380882263, \"perplexity\": 229.5539104352924}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-17\/segments\/1618038074941.13\/warc\/CC-MAIN-20210413183055-20210413213055-00024.warc.gz\"}"}
| null | null |
Showing all the cultural awareness of an Outer Mongolian hammering on Morrisons on Christmas Day demanding a tin of spam, I pitched up to Mumtaz (or 'The World Famous Mumtaz' as it hails itself) in the middle of Ramadan at sunset.
So it should have been no surprise that the place was absolutely heaving with most of the local Muslim community getting their Iftihar on. Luckily, the Bradford behemoth hasn't got where it is today by not being prepared, and we were hastily sent upstairs for a free starter while all hands were on deck.
And what a shiny deck it is. With granite on every available surface not made of glass, the place positively dazzles. Far from being an authentic Asian dining experience, Mumtaz is authentically late twentieth century, more redolent of aspirational Far East architecture than anything else. Young enough to dictate its own heritage, you'll find no flock wallpaper here.
On the night we went, the restaurant was a real mix of people, from young couples and families to animated groups of friends. Despite the throng, our waiter shimmied over and took our order, which we chose from a heady list of delicious-sounding dishes. Delicious sounding and deliciously described. Before the food had arrived, it was clear that some serious marketing has gone on here. The Mumtaz logo is on everything, so you are never in any doubt as to where you are. And the menu describes itself in glowing terms. Velvety flavours and marriages made in heaven; how could we not enjoy it?
Our free starters were a nice mix of lamb kebab, chicken wings and chick-pea stew. For my main course, though, I chose the large chicken keraki dopiaza (£13), with cardamom, cinnamon, cloves, pepper, cumin seed, red dry bullon chillies and coriander leaves. I ordered mild to be on the safe side and it was very mild indeed, though seriously delicious. My chicken was tender and the dopiaza was tasty enough. I paired it with a small portion of pilau rice (£2.25). My companion had a regular-sized kerahi lamb sookha bouhna (£7.85) with lamb on the (segmented) bone, featuring tomatoes, onion, garlic, ginger and spices. He declared the sauce nice and rich, not overpowering but a bit "gravyish".
We shared a garlic naan (£1.95) and it was thin and flavoursome, well designed for scooping up food, unlike the rolled-up pillows you often get, and we both had lassis (mine a strawberry and his a mango at £3.95).
The dessert menu was a refreshing collection of real puddings and not frozen ready dishes. I tried and tried not to have my favourite ras malai (£2.75) but I couldn't resist. It was worth it though, a really moist dish of milky dumplings with pistachio. My partner had rassogula (milky balls) (£2.50), nice and textured and creamy.
The waiting staff were unfailingly attentive and polite and we only had to so much as look up from the table to get instant service. There also appeared to be a young boy whose sole job it was to offer people extra water. Possibly the most boring job in the world but also the easiest and with least chance of complaint. There was some excitement as a deeply embarrassed boyfriend on the table behind us received a birthday cake complete with sparkler. He could barely cope with the humiliation but his girlfriend was delighted as the waiter half-sung the first verse of Happy Birthday then realised no-one else was joining in and skulked off.
My – as advertised – creamy and velvety latte (£2.25) was a nice way to round everything off and as we looked around we realised that the rush had passed and there were just a few handfuls of people having coffee and chilling out. We stayed awhile, talking and taking in the peace and quiet but didn't feel like we were in the way. We left, via the shiny foodhall, very happy customers.
|
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| 8,384
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Косуха () — коротка шкіряна куртка зі звуженою талією і блискавкою навскоси. Саме від цієї косої «блискавки», що іменується в молодіжному жаргоні «трактором», куртка і отримала свою назву. Таке розташування «блискавки» (косуха — в її чоловічому варіанті — застібається від лівого стегна до правого плеча) надавало куртці схожість з військовими мундирами часів Громадянської війни в США.
Джерела
Speace, Gery: Leather Jacket, St. James Encyclopedia of Popular Culture, Gale 2000. Vaatii HighBeam-tilauksen.
Куртки
|
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| 7,532
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\section{Introduction}
\begin{figure}
\includegraphics[width=\textwidth]{size5.jpg}
\hbox{\hspace{0.035em} \includegraphics[width=0.9925\textwidth]{speed5.jpg}}
\caption{We obtained the following graphs from a computer simulation of the critical $1$-d MDLA. The first picture shows the size as a function of time of 5 independent aggregates running for time $300,000$. The second picture shows the speed of growth of the blue aggregate in the first picture. Theorem~\ref{thm:main theorem 2} states that the speed converges to the solution of $dZ_t=2Z_t^{5/2}dB_t$ with initial condition $Z_0=\infty $.}\label{fig:simulation}
\end{figure}
We study a variant of the DLA model called the multi-particle Diffusion Limited Aggregation (MDLA), where a cloud of particles diffuse simultaneously before adhering to a growing aggregate. This model was first studied in the physics \cite{voss84mdfa, voss1984multiparticle, meakin98book,meakin1988multiparticle,meakin83} and later in the mathematics literature \cite{sidoravicius2019multi,sidoravicius2017one,kesten2008problem,sly2016one,dembo2019criticality,kesten2008positive}.
The model in dimension $d \ge 1$ and density $\lambda >0$ is defined as follows. For each time $t>0$ the aggregate is a set of vertices $\mathcal A _t \subseteq \mathbb Z ^d$. Initially, $\mathcal A _0=\{0\}$ and on each vertex $v \neq 0$ there is a random number of particles distributed as $ \mathrm{Poisson} (\lambda )$, independently of the other vertices. At time $t=0$ each particle starts to move according to a simple, continuous time random walk, independently of other particles. The aggregate $\mathcal A _t$ grows according to the following rule: If at time $t$ one of the particles at $v \notin \mathcal A_{t-} $ attempts to jump into the aggregate, it freezes in place, together with all the other particles at $v$ and the aggregate grows by $\mathcal A _t =\mathcal A_{t-}\cup \{v\}$. Frozen particles do not move for the rest of the process and cannot make the aggregate grow.
Perhaps surprisingly at first sight, only the one dimensional model is expected to exhibit a phase transition, while in higher dimensions the aggregate is expected to grow linearly for all initial densities $\lambda>0$ which has been confirmed only for large enough $\lambda$ by Sidoravicius and Stauffer~\cite{sidoravicius2019multi} and also in~\cite{sly2016one}. From now on we focus only on the one dimensional case. In this case the aggregate is simply a line segment and the processes on the positive and negative axes are independent so we simply restrict our attention to the rightmost position of the aggregate at time $t$ which we denote $X_t$. If at time $t$ a particle at $X_{t^-} + 1$ attempts to take a step to the left, the aggregate grows by one: $X_t=X_{t^-}+1$.
Kesten and Sidoravicious \cite{kesten2008problem} proved that in the subcritical regime when $\lambda <1$, $X_t=\Theta (\sqrt{t})$ with high probability. Moreover, using the results of Dembo and Tsai \cite{dembo2019criticality}, one can show that in fact $X_t =(c_-(\lambda) +o(1))\sqrt{t}$ for an explicit $c_-(\lambda) >0$. In the supercritical regime, the third author proved the existence of a phase transition~\cite{sly2016one} by showing that when $\lambda >1$, $X_t$ grows linearly and moreover that $t^{-1}X_t \to c_+(\lambda) >0$ almost surely.
It was widely conjectured that in the critical case, when $\lambda =1$, the aggregate grows like $t^{2/3}$. Sidoravicious and Rath~\cite{sidoravicius2017one} gave a heuristic explanation for this prediction via PDEs. In \cite{dembo2019criticality}, Dembo and Tsai proved an upper bound of $ O(t^{2/3})$ by studying a modified ``frictionless'' variant of the model which stochastically dominates the aggregate (see Section~\ref{subsec:intro:relatedworks} for details). We verify the conjectured $t^{2/3}$ growth in the following theorem and determine the scaling limit.
\begin{theorem}\label{thm:main theorem 2}
Let $V_t$ be the Bessel process with dimension $\frac{8}{3}$ given by
\begin{equation}
d V_t=\frac{5}{6} \frac{dt}{V_t} + d B _t, \quad V_0 = 0,
\end{equation}
and let $Z_t = (3V_t)^{-2/3}$. Then,
\begin{equation}
\big\{ t^{-\frac{2}{3}}X_{st} \big\} _{s>0} \overset{d}{\longrightarrow } \Big\{ \int _0^s Z_x dx \Big\} _{s>0},\quad t \to \infty.
\end{equation}
\end{theorem}
In particular $ t^{-2/3}X_{t}$ has a limiting distribution which is positive almost surely. The limiting speed process $Z_t$ is a local martingale which is $(-1/3)$-self-similar which means $\{Z_{st}\}_{s\geq 0} \stackrel{d}{=} \{t^{1/3} Z_s\}$. While this representation is perhaps surprising, any $(-1/3)$-self-similar diffusion must be a $(-2/3)$ power of a Bessel process. An alternative representation, which can easily be verified by It\^o's Formula, is that
\begin{equation}\label{eq:SDE equation in main theorem}
d Z_t = 2Z_t^{5/2} dB_t,
\end{equation}
with $Z_0=\infty.$
The initial condition $Z_0= \infty$ can be understood by the limit of the diffusion satisfying~\eqref{eq:SDE equation in main theorem} with finite initial condition $Z_0$ sent to infinity. See Sections \ref{subsubsec:outline:scaling limit} and \ref{sec:scalinglimit} for details.
\subsection{Related models}\label{subsec:intro:relatedworks}
The MDLA model originated from the study of Diffusion Limited Aggregation (DLA). In the DLA model, particles come from infinity one by one by a diffusion and adhere to the aggregate. The fundamental problem is determining the asymptotic growth and shape of the aggregate. The model was introduced by Witten and Sanders~\cite{witten1981diffusion}, and describes physical phenomena such as mineral growths and electrodeposition. Although it has been studied extensively (e.g., \cite{witten83dla,meakin83,vicsek84,kesten87hitprob, kesten1987long, kesten1990upper, benjamini17}), there has been limited progress in obtaining rigorous results.
The frictionless model, which is a variant of MDLA, was introduced and successfully studied by Dembo and Tsai~\cite{dembo2019criticality}. The model is defined in one dimension as follows. Like the MDLA, there is a growing aggregate on the set of positive integers and a cloud of particles that perform a continuous time random walk but the aggregate grows by the number of particles it absorbs rather than 1. In the supercritical case it explodes in finite time.
The size of the aggregate is exactly equal to the number of particles absorbed, which is called in \cite{dembo2019criticality} the ``flux condition" allowed Dembo and Tsai to derive the asymptotic behavior of the model based on PDE techniques, deriving a discontinuous scaling limit. We note that the flux condition does not hold in the MDLA model but the front stochastically dominates it and thus gives an upper bound.
Another way to interpret the MDLA is to view it as a special case of a two-type particle system (so-called the $A$-$B$ model), which is sometimes used to model the spread of a rumor or a disease in a network. In the $A$-$B$ model, there are two types of particles, the $A$-type (``healthy individuals") and the $B$-type (``infected"). All the particles perform a continuous time random walk on $\mathbb Z ^d $ in which the $A$ particles jump with rate $D_A\ge 0$ and the $B$ particles jump with rate $D_B \ge 0$. When $A$ and $B$ particles coincide, the $A$ particle turns into a $B$ particle.
Theses models were studied extensively and we refer to \cite{kesten2012asymptotic,berard2010large,comets2009fluctuations,kesten2008shape,richardson1973random,kesten2005spread}.
In the case $D_B=0$ and $D_A>0$ (that is, only the $A$ particles move) the transition rule is modified such that when an $A$ particle tries to jump into a vertex with a $B$ particle, the jump is suppressed and the $A$ particle, together with all other $A$ particles on the same vertex turn into $B$ particles. It is clear that when $D_A=1$ and when initially there is one $B$ particle in the origin this corresponds to the MDLA.
The opposite case $D_A=0$ and $D_B>0$ is sometimes referred to in the physics literature as the Stochastic combustion process and it is used to model the burning of propellant material \cite{ramirez2004asymptotic,comets2007fluctuations}. In the mathematics literature it is sometimes called the frog model \cite{alves2002phase,alves2002shape}.
\subsection{Further directions and open problems}
\subsubsection{Near critical behavior}
For the one-dimensional MDLA with initial density $\lambda $ very close to $1$, Sidoravicius and Rath \cite{sidoravicius2017one} asked to determine the critical exponent of the aggregate size as $\lambda \to 1$. To be precise, when $\lambda >1$, they predicted that the constant $c_+(\lambda) := \lim_{t\to\infty} t^{-1} X_t$ is linear in $(\lambda -1)$ as $\lambda\searrow 1$, and conjectured that $c_+(\lambda) \sim \frac{1}{2}(\lambda-1)$ from a heuristic argument counting the sites in the aggregate with more than one particle. On the other hand, for the subcritical case, they expected that the constant $c_-(\lambda):= \lim_{t \to \infty} t^{-1/2}X_t$ satisfies $ c_-(\lambda) \sim (1-\lambda)^{-1/2}.$
In a forthcoming companion paper~\cite{ens20nearcrit} building on the techniques we develop here, we give an affirmative answer to the latter conjecture, but disprove the first one by establishing
\begin{equation}\label{eq:nearcrit:super}
\lim_{\lambda\searrow 1} \frac{c_+(\lambda)}{\lambda-1} = \frac{1}{3}, \quad \quad \lim_{\lambda\nearrow 1} (1-\lambda)^{1/2} c_-(\lambda) = 1.
\end{equation}
The heuristic calculation from \cite{sidoravicius2017one} for the slightly supercritical regime leads to a wrong answer as it assumed that the speed of the growth is concentrated as $t\to \infty$. Instead, it converges to a nondegenerate diffusion upon appropriate rescaling as $\lambda \searrow 1$.
We remark that the second identity of \eqref{eq:nearcrit:super} can be obtained by the results of \cite{dembo2019criticality} from a completely different approach.
When $\lambda$ is very close to $1$, it is intuitively clear that when $t$ is not very large, the aggregate will look the same as the critical case $\lambda =1$. This leads to the question of how large must $t$ be (in terms of $\lambda $) in order for the model to ``feel" that it is slightly supercritical or subcritical, and is related to the rescaling of the process mentioned in the previous paragraph. In \cite{ens20nearcrit}, we show that the answer to this question is $t\asymp |\lambda -1|^{-3} $, and give a more precise description on the scaling limit of $X_t$ as follows: Let $U_t$ and $R_t$ be the solutions to the SDEs
\begin{equation}
dU_t = 2U_t^3 dt + 2U_t^{5/2} dB_t,\quad dR_t = -2R_t^3 dt + 2R_t^{5/2} dB_t,
\end{equation}
with initial conditions $U_0 = R_0 = \infty.$ The methods from the current paper can be extended to prove that
the aggregate size $X_t = X_t(\lambda)$ satisfies
\begin{equation}
\big\{ (\lambda -1)^2 X_{s(\lambda -1)^{-3}} \big\}_{s>0} \overset{d}{\longrightarrow } \Big\{ \int _0^s U_x dx \Big\}_{s>0}, \quad \lambda \searrow 1.
\end{equation}
Similarly, in the slightly subcritical case we have
\begin{equation}
\big\{ (1-\lambda )^2 X_{s(1-\lambda )^{-3}} \big\}_{s> 0} \overset{d}{\longrightarrow } \Big\{ \int _0^s R_x dx \Big\}_{s> 0}, \quad \lambda \nearrow 1.
\end{equation}
One can see that $U_t$ converges to a nondegenerate stationary process as $t\to \infty$, and a standard calculation tells us that its stationary measure is an inverse gamma distribution whose mean is $\frac{1}{3}$, implying \eqref{eq:nearcrit:super}.
On the other hand, in the subcritical case, we have almost surely that $2 \sqrt{t} R(t)\to 1$ as $t\to \infty $, since the diffusive term of the SDE becomes negligible compared to the drift as $t$ increases. This leads to the second equation of \eqref{eq:nearcrit:super}.
\subsubsection{Different initial distribution}
One can consider the one-dimensional MDLA with an initial distribution that is not Poisson. Suppose that at time $0$, the number of particles on vertex $i$ is $Z_i$ where \begin{equation}\label{eq:initial condition}
Z_1,Z_2,\dots \ \text{are i.i.d.}, \quad \mathbb E Z_1=\lambda , \quad \mathrm{Var} (Z_1) =\sigma ^2.
\end{equation}
In case where all the particles just perform independent random walks on $\mathbb Z$ without a growing aggregate, it is clear that for all $\lambda >0$, i.i.d.~Poisson$(\lambda)$ on each point of $\mathbb Z$ is a stationary distribution for the dynamics. Thus, one might expect that the aggregate with the initial condition \eqref{eq:initial condition} grows exactly like the original MDLA with the Poisson initial profile. However, we expect that in the critical case, the aggregate grows faster than the time it takes for large intervals to approach stationarity. Therefore, the scaling limit in the critical case is expected to depend on the initial distribution of particles. More precisely, we conjecture formulate the following conjecture.
\begin{conj}
Let $X_t$ be the size of the aggregate with initial condition as in \eqref{eq:initial condition}. Then, with high probability
\begin{enumerate}
\item
If $\lambda <1$, then $X_t=\Theta (\sqrt{t} )$.
\item
If $\lambda >1$, then $X_t =\Theta (t)$.
\item
If $\lambda =1$, then $X_t=\Theta ( t^{2/3} )$. Morover, we have
\begin{equation}
\big\{ t^{-\frac{2}{3}} X_{st} \big\} _{s>0} \overset{d}{\longrightarrow } \Big\{ \int _0^s Z_xdx \Big\}_{s>0} ,\quad t \to \infty,
\end{equation}
where $Z_t$ is the solution to the equation
\begin{equation}\label{eq:sde general initial condition}
dZ_t=(4 \sigma ^2 -4)Z_t^4dt +2 \sigma Z_t^{5/2}dB_t,
\end{equation}
with $Z_0=\infty $.
\end{enumerate}
\end{conj}
\begin{remark}
We elaborate several aspects of the conjecture as follows.
\begin{enumerate}
\item
Theorem~\ref{thm:main theorem 2} is compatible with this conjecture. Indeed, in the Poisson case with $\lambda =1$, we have that $\sigma ^2 =1$ and therefore the drift term vanishes.
\item
For the deterministic initial condition where each vertex has exactly one particle in the beginning, we see that the diffusive term vanishes and we get a deterministic scaling limit of the form $X_t=(c_0+o(1)) t^{2/3}$ with high probability.
\item
Note that $Z_t$ from \eqref{eq:sde general initial condition} is $(-\frac{1}{3})$-self-similar and hence is a power of a Bessel process. Also, recall that the Bessel process with dimension $n \le 2$ is recurrent. Applying It\^o's formula,
it turns out that the corresponding Bessel process for $Z_t$ has dimension $(\frac{4}{3} + \frac{4}{3\sigma^2})$. Therefore, $Z_t$ explodes in finite time if $\sigma ^2 \ge 2$ while it tends to $0$ if $\sigma ^2 <2$. Thus, we expect that the scaling limit of $X_t$ is differentiable when $\sigma ^2 <2$ and is not differentiable when $\sigma ^2 \ge 2$.
\end{enumerate}
\end{remark}
\section{Approximations of the speed and proof outline}\label{sec:outline}
In this section, we define the speed of the process, which is our main object of study. As the speed is a complicated process by itself, we give several approximations of the speed by more tractable processes. Based on those expressions, we also give an overview of the proof of Theorem \ref{thm:main theorem 2}.
\begin{definition}
The speed $S(t)$ of the aggregate at time $t$ is $\frac{1}{2}$ times the conditional expectation of the number of particles at position $X_t +1$, given $\{X_s\}_{0\le s \le t}$.
\end{definition}
This corresponds to the infinitesimal jump rate of $X_t$; note that the particles at position $X_t+1$ are equally likely to jump to $X_t$ and $X_t+2$, hence the multiplication by $\frac{1}{2}$. To give a more explicit expression of $S(t)$, we follow the approach of \cite{sly2016one} and define
\begin{equation}\label{eq:def:Yt}
Y_t(s):= \begin{cases}
X_t - X_{t-s} & \textnormal{if } 0\le s \le t;\\
\infty & \textnormal{if } s>t.
\end{cases}
\end{equation}
The following observation was first made in \cite{sly2016one}, and this is the starting point of our work as well. Here, we consider a general MDLA with initial density $\lambda>0$, i.e., having i.i.d. Poisson$(\lambda)$ number of particles at each position in the beginning.
\begin{prop}\label{prop:speed:basicdef}
Let $\{X_t\}_{t\ge 0}$ be the 1D MDLA with density $\lambda>0$ and let $W(s)$ denote an independent continuous-time simple random walk starting at $0$. Then, we have
\begin{equation}\label{eq:speed:basic expression}
S(t) = \frac{\lambda}{2}\ \mathbb{P} \left( \left.\sup_{s\ge 0 } \{ W(s)-Y_t(s)\} \le 0 \
\right| \ Y_t\right) .
\end{equation}
Moreover, $\{X_t\}_{t\ge 0}$ is a Poisson process with rate $\{S(t) \}_{t\ge 0}.$
\end{prop}
In \cite{sly2016one}, \eqref{eq:speed:basic expression} was used to prove that $X_t$ grows linearly in $t$ if $\lambda>1$. The approach was to show that the speed has the property of mean reversion. If the growth rate of $X_t$ becomes too small, say, $\alpha$, for a period of time, then $S(t) \approx\lambda \alpha$, corresponding to a faster growth rate. However, at criticality, the same analysis would give $S(t) = \alpha(1+o(1))$, and thus understanding the lower order terms becomes necessary. Indeed, our ultimate goal is to deduce the scaling limit of the \textit{smoothed} speed by conducting a more refined and quantitative analysis on such error terms.
In Section \ref{subsec:speed:speedproc}, we introduce the notion of the \textit{speed process} which generalizes \eqref{eq:speed:basic expression} and explain how they are approximated by the objects we can quantitatively deal with. Then, we see how the speed of the critical aggregate is described in Section \ref{subsec:speed:speedagg}. Furthermore, in Section \ref{subsec:speed:Lt}, we give an overview on how to obtain its scaling limit by using a \textit{smoothed} version. In Section \ref{subsec:speed:outline}, we describe an outline of the proof. In the final subsection, Section \ref{subsec:Kestim:intro}, we illustrate the necessary estimates on the important quantities that appear frequently in the article.
\subsection{The speed process and its approximation}\label{subsec:speed:speedproc}
In the remainder of this section, $Y$ denotes a \textit{unit-step function} which is defined as follows.
\begin{definition}[Unit-step function and its speed process]\label{def:unitstepfunc}
$Y:\mathbb{R}_{\ge 0} \to \mathbb{N}\cup \{\infty\}$ is called a \textit{unit-step function} if there exist either $x_1 < x_2 < \ldots $ such that
\begin{equation}
Y(s) = \sum_{i\ge 1} I\{x_i < s\},
\end{equation}
or $n \in \mathbb{N}$ and $x_1<x_2<\ldots < x_n < x$ such that
\begin{equation}
Y(s) = \begin{cases}
\sum_{i=1}^n I\{ x_i < s\} & \textnormal{if } s\le x;\\
\infty & \textnormal{if } s>x.
\end{cases}
\end{equation}
Note that in both cases $Y(0)=0$, $Y$ is increasing and left-continuous. Furthermore, the \textit{speed process} of $Y$ is defined as
\begin{equation}\label{eq:def:speedproc}
S(Y) : = \frac{1}{2} \ \mathbb{P}\left(\left. \sup_{s\ge 0} \{W(s)-Y(s)\} \le 0 \ \right| \ Y \right),
\end{equation}
where $W(s)$ denotes the continuous-time simple random walk starting at $0$.
\end{definition}
We let $\mathbb{P}_\alpha$ denote the probability with respect to $Y$ being a rate-$\alpha$ Poisson process. We will proceed by considering the formula $S(Y)$ under $\mathbb{P}_\alpha$. In doing so we obtain a formula for $S(Y)$ described in Proposition~\ref{prop:speed:Dt0thorder} below which enables us to conduct a refined quantitative study on the speed process. While our starting point is a Poisson $Y$, the formula will hold for all unit-step functions $Y$.
We define the stopping time $T$ as
\begin{equation}\label{eq:def:T:basic form}
T = T(Y) := \inf \{s>0: W(s) > Y(s) \}.
\end{equation}
and set
\begin{equation}
U(s) := Y(s) - W(s) +1.
\end{equation}
which is a continuous time random walk with drift under $\mathbb{P}_\alpha$. Setting $\xi = (1+2\alpha)^{-1}$ we have that $\xi^{U_s}$ is a martingale, and hence
\begin{equation}\label{eq:speed:Ut mg conv}
\xi^{U_{s\wedge T}} \overset{\textnormal{in }L^2 }{\underset{s\to \infty}{\longrightarrow}} \ \ I\{T<\infty \}.
\end{equation}
Thus, an application of the optimal stopping theorem gives
\begin{equation}
\mathbb{P}_\alpha ( T<\infty) = \xi.
\end{equation}
Thus, we obtain that
\begin{equation}
\mathbb{E}_\alpha [ S(Y)] = \frac{\alpha}{1+2\alpha}.
\end{equation}
Coming back to the general case, let $Y$ be a given unit-step function, let $\alpha>0$ and set $\xi = \frac{1}{1+2\alpha}$. We define $D_t$ to be
\begin{equation}
D_t := \mathbb{E}_\alpha \left[\left.\xi^{U_{t\wedge T}} \right| Y_{\le t}\right],
\end{equation}
where $\mathbb{E}_\alpha [ \ \cdot \ | Y_{\le t}]$ denotes the expectation over $W$ and the unit-step function obtained by the following procedure: take the profile of $Y$ in $[0,t]$ and generate the rest beyond $t$ by a rate-$\alpha$ Poisson process. In such a case, $\{\xi^{U_{t' \wedge T}} \}_{t'\ge t}$ forms a martingale in $t'$ which converges to $I\{T<\infty \}$ as $t'\to \infty$. Thus, we also know that
\begin{equation}\label{eq:speed:Dt is prob T finite}
D_t = \mathbb{P}_\alpha ( T<\infty \ | \ Y_{\le t}).
\end{equation}
The proposition below tells us a way to understand $D_t$ from its infinitesimal differences. In what follows, $dY$ denote the additive combination of the Dirac measures at the points of discontinuity, that is, under the notation of Definition \ref{def:unitstepfunc}, we define
\begin{equation}
dY(s) := \sum_{x_i} \delta_{x_i}(s).
\end{equation}
\begin{prop}\label{prop:speed:Dt0thorder}
Let $\alpha, t>0$, and $ Y$ be a unit-step function that is finite on $[0,t]$. Denoting $d\widehat{Y}(s):= dY(s) - \alpha ds$, we have
\begin{equation}\label{eq:speed:Dt0thorder}
D_t = \frac{1}{1+2\alpha} - \frac{2\alpha}{1+2\alpha} \intop_0^t
H_s \ d\widehat{Y}(s),
\end{equation}
where $H_s=H_s(Y_{\le s}, \alpha):= \mathbb{P}_\alpha ( s<T<\infty \ | \ Y_{\le s}).$
\end{prop}
\begin{proof}
Set $\xi = \frac{1}{1+2\alpha}$ as before. For $Y$ which is a unit-step function, we write
$$\frac{dY(s)}{ds} = Y(s+)-Y(s). $$
Then, observe that
\begin{equation}
\begin{split}
\lim_{\Delta \searrow 0} \frac{1}{\Delta} \Big\{\mathbb{E}\left[D_{s+\Delta} | \mathcal{F}_s \right]
-\mathbb{E}\left[D_s | \mathcal{F}_s \right]
\Big\}
&=
\mathbb{E}_\alpha \left[\left.\xi^{U_{s\wedge T}} I\{s<T \}\left\{ (\xi-1)\frac{dY(s)}{ds} + \left(\frac{\xi+\xi^{-1}}{2}-1 \right) \right\} \ \right| \ Y_{\le s} \right]\\
&=
-\frac{2\alpha}{1+2\alpha} \left(\frac{dY(s)}{ds} - \alpha \right) \mathbb{E}_\alpha \left[\left.\xi^{U_{s\wedge T}} I\{s<T \}\ \right| \ Y_{\le s} \right]\\
&=
-\frac{2\alpha}{1+2\alpha} \left(\frac{dY(s)}{ds} - \alpha \right)
\mathbb{P}_\alpha (s<T<\infty \ | \ Y_{\le s}),
\end{split}
\end{equation}
where the last identity can be deduced similarly as \eqref{eq:speed:Dt is prob T finite}: for a fixed $s>0$,
\begin{equation}
\left\{ \xi^{U_{s'\wedge T}} I\{s<T \} \right\}_{s'\ge s}
\end{equation}
is a martingale in $s'$, considering $Y$ as a rate-$\alpha$ Poisson process on $[s,\infty)$. Since this converges to $I\{s<T <\infty \}$ as $s'\to \infty$, the optimal stopping theorem tells us that
\begin{equation}
\mathbb{E}_\alpha \left[\left.\xi^{U_{s\wedge T}} I\{s<T \}\ \right| \ Y_{\le s} \right]
=
\mathbb{P}_\alpha (s<T<\infty \ | \ Y_{\le s}).
\end{equation}
Then, we can conclude the proof by observing that $D_0 = \xi.$
\end{proof}
From Proposition \ref{prop:speed:Dt0thorder}, we can also deduce the value of $\intop_0^\infty \mathbb{E}_\alpha H_s ds$ as follows.
\begin{cor}\label{cor:speed:integralofH}
Let $\alpha>0$. For $H_s=H_s(Y_{\le s}, \alpha)$ defined as above, we have
\begin{equation}
\intop_0^\infty \mathbb{E}_\alpha [H_s] ds = \frac{1}{\alpha(1+2\alpha)}.
\end{equation}
\end{cor}
\begin{proof}
Let $\alpha'>0$ (different from $\alpha$), and let $Y$ be the rate-$\alpha'$ Poisson process. Recall the expression for $D_\infty(Y)$ from Proposition \ref{prop:speed:Dt0thorder} and take the expectation over $Y$:
\begin{equation}
\frac{1}{1+2\alpha'}=\mathbb{E}_{\alpha'} D_\infty(Y) = \frac{1}{1+2\alpha} - \frac{2\alpha}{1+2\alpha} \mathbb{E}_{\alpha'} \left[\intop_0^\infty H_s(Y_{\le s},\alpha) d\widehat{Y}(s)\right].
\end{equation}
Here, note that $d\widehat{Y}(s)=dY(s)-\alpha ds$ does not depend on $\alpha'$.
Thus, differentiating each side with $\alpha'$ and then plugging in $\alpha'=\alpha$ gives
\begin{equation}
-\frac{2}{(1+2\alpha)^2} = -\frac{2\alpha}{1+2\alpha} \intop_0^\infty \mathbb{E}_\alpha[ H_s(Y_{\le s},\alpha) ]ds,
\end{equation}
which gives the conclusion after rearranging.
\end{proof}
The above corollary motivates us to define the \textit{branching density} $K_\alpha$, which is one of the main objects that we need to understand very precisely.
\begin{definition}[The branching density]\label{def:Kalpha:main}
For each $\alpha>0$, the function $K_\alpha : \mathbb{R}_{\ge 0} \to \mathbb{R}_{\ge 0}$, which we call the \textit{branching denstiy}, is defined as
\begin{equation}
K_\alpha (s ):= \alpha(1+2\alpha)\mathbb{E}_\alpha[H_s] = \alpha(1+2\alpha) \mathbb{P}_\alpha (s<T<\infty).
\end{equation}
\end{definition}
Corollary \ref{cor:speed:integralofH} tells us that $\intop_0^\infty K_\alpha(s)ds = 1$, and hence $K_\alpha$ can be understood as a probability density function. It turns out that $K_\alpha$ describes the density of the branch lengths of a certain branching process which locally approximates $X_t$ (see Remark \ref{rmk:speed:branchingdensity}), and hence we call it the branching density.
Next, we introduce a similar decomposition as \eqref{eq:speed:Dt0thorder} that works for $H_s$, in order to approximate $H_s$ by $\mathbb{E}_\alpha[H_s]$, or $K_\alpha(s)$. Here, we stress that the expectation $\mathbb{E}_\alpha[H_s] = \mathbb{E}_\alpha[H_s(Y_{\le s},{ \alpha})]$ is taken over the randomness of $Y_{\le s}$ as the rate-$\alpha$ Poisson process. For each $u>0$, define the stopping time $T_u$ to be
\begin{equation}\label{eq:def:Tu}
T_u= T_u(Y) := \inf \{s>0: Y(s)-W(s)+ I\{s>u\} <0 \}.
\end{equation}
In other words, it is the first time when $W$ exceeds $Y$, with an additional unit jump added to $Y$ at $u$. Then, the decomposition for $H_s$ is given by the following proposition.
\begin{prop}\label{prop:speed:Dt1storder}
Suppose that $Y$ is finite in $[0,t]$ and let $d\widehat{Y}$, $H_s$ and $T_u$ be as above. Define
\begin{equation}\label{eq:def:J:basic form}
{\textnormal{\textbf{J}}}_{u,s}= {\textnormal{\textbf{J}}}_{u,s}(Y; \alpha) := \mathbb{P}_\alpha (s<T_u<\infty \ | \ Y_{\le u}) - \mathbb{P}_\alpha(s<T<\infty \ | \ Y_{\le u}).
\end{equation}
Then, we have
\begin{equation}
H_s = \mathbb{E}_\alpha[H_s] + \intop_0^s {\textnormal{\textbf{J}}}_{u,s} \ d\widehat{Y}(u).
\end{equation}
In particular, we can rewrite the formula \eqref{eq:speed:Dt0thorder} as
\begin{equation}\label{eq:Dt:1storderexpansion}
D_t = \frac{1}{1+2\alpha} -\frac{1}{(1+2\alpha)^2} \intop_0^t K_\alpha(s) \ d\widehat{Y}(s) - \frac{2\alpha}{1+2\alpha} \intop_0^t \intop_0^s {\textnormal{\textbf{J}}}_{u,s} \ d\widehat{Y}(u)d\widehat{Y}(s).
\end{equation}
\end{prop}
\begin{remark}
We use bold alphabet ${\textnormal{\textbf{J}}}_{u,s}$ to define \eqref{eq:def:J:basic form} to emphasize its stochastic nature, since it depends on $Y_{\le u}$ which is going to be random. This will prevent confusion from its ``averaged'' form introduced in \eqref{eq:def:J:expected val}.
\end{remark}
\begin{proof}[Proof of Proposition \ref{prop:speed:Dt1storder}]
For $u<s$, observe that we have the following expression for an infinitesimal difference of $H_s$:
\begin{equation}
\lim_{\Delta \searrow 0}\frac{1}{\Delta} \Big\{ \mathbb{E}_\alpha \left[ H_s \ | \ Y_{\le u+\Delta} \right]-\mathbb{E}_\alpha \left[ H_s \ | \ Y_{\le u} \right] \Big\}
= {\textnormal{\textbf{J}}}_{u,s} \left(\frac{d{Y}(u)}{du} - \alpha \right).
\end{equation}
Thus, we directly obtain the conclusion by noticing that $H_s = \mathbb{E}_\alpha[H_s \ | \ Y_{\le s}]$.
\end{proof}
Combining \eqref{eq:speed:Dt is prob T finite}, Corollary \ref{cor:speed:integralofH} and Proposition \ref{prop:speed:Dt1storder}, we obtain the following expression for the speed process:
\begin{equation}\label{eq:speedproc:1storder main}
\begin{split}
\mathbb{E}_\alpha&[S(Y) \ | \ Y_{\le t}]
= \mathbb{P}_\alpha ( T =\infty \ | \ Y_{\le t})
\\
&=\frac{\alpha}{1+2\alpha} + \frac{1}{(1+2\alpha)^2} \intop_0^t K_\alpha(s) \left\{dY(s) - \alpha ds \right\} + \frac{\alpha}{1+2\alpha} \intop_0^t \intop_0^s {\textnormal{\textbf{J}}}_{u,s} \ d\widehat{Y}(u)d\widehat{Y}(s) \\
&= \frac{2\alpha^2}{(1+2\alpha)^2} + \frac{1}{(1+2\alpha)^2}\intop_0^t K_\alpha(s) dY(s)
+
\frac{\alpha}{1+2\alpha} \intop_0^t \intop_0^s {\textnormal{\textbf{J}}}_{u,s} \ d\widehat{Y}(u)d\widehat{Y}(s).
\end{split}
\end{equation}
This formula is one of the two primary formulas which lies at the heart of our work, and sometimes we refer to this as \textit{the first-order expansion of the speed process}.
In the next subsection, we discuss in more detail that how we express the actual speed of the aggregate in the form of \eqref{eq:speedproc:1storder main}.
\begin{remark}\label{rmk:speed:branchingdensity}
We will later see that $\intop_0^t K_\alpha(s) dY(s)$ is the leading order in the expression \eqref{eq:speedproc:1storder main}, which is of order $\alpha$ while other terms are of smaller order. Moreover, this term can be understood as a branching process, where each jump in Y at distance $s$ results in a new jump to branch out at rate $K_\alpha(s)$. More detailed illustrations are given in the following subsections (see, e.g., \eqref{eq:def:Rt:basic form} and the discussion below).
\end{remark}
The second main formula gives an additional order of approximation of the speed process. In \eqref{eq:speedproc:1storder main}, it turns out that the double integral of ${\textnormal{\textbf{J}}}_{u,s}$ has a non-negligible order in the derivation of the scaling limit of (the smoothed version of) the speed. Thus, we need another layer of approximation to control its size more accurately. To this end, we first define the stopping time $T_{v,u}$ for each $u>v>0$ similarly as \eqref{eq:def:Tu}.
\begin{equation}
T_{v,u} = T_{v,u}(Y) := \inf \{s>0: Y(s)-W(s)+I\{s>v\} + I\{s>u\} < 0 \}.
\end{equation}
In other words, it is the first time when $W$ exceeds $Y$, with two additional unit jumps added to $Y$ at $v$ and $u$.
\begin{prop}\label{prop:speed:Dt2ndorder}
Suppose that $Y$ is finite on $[0,t]$ and let $d\widehat{Y}$, ${\textnormal{\textbf{J}}}_{u,s}$, $T_{v,u}$ and $T_u$ be as above. Define
\begin{equation}\label{eq:def:Q:basic form}
\begin{split}
{\textnormal{\textbf{Q}}}_{v,u,s}= {\textnormal{\textbf{Q}}}_{v,u,s}(Y;\alpha):=& \mathbb{P}_\alpha(s<T_{v,u}<\infty \ | \ Y_{\le v}) -\mathbb{P}_\alpha(s<T_{v}<\infty \ | \ Y_{\le v})\\
&-\mathbb{P}_\alpha(s<T_{u}<\infty \ | \ Y_{\le v})+\mathbb{P}_\alpha(s<T<\infty \ | \ Y_{\le v}).
\end{split}
\end{equation}
Then, we have
\begin{equation}
{\textnormal{\textbf{J}}}_{u,s} = \intop_0^u {\textnormal{\textbf{Q}}}_{v,u,s} \ d\widehat{Y}(v).
\end{equation}
In particular, the speed process admits the following expression:
\begin{equation}\label{eq:speedproc:2ndorder main}
\begin{split}
\mathbb{E}_\alpha[S(Y) | Y_{\le t}]
=&
\frac{2\alpha^2}{(1+2\alpha)^2} + \frac{1}{(1+2\alpha)^2}\intop_0^t K_\alpha(s)dY(s) \\
&+
\frac{\alpha}{1+2\alpha} \intop_0^t \intop_0^s J_{u,s}^{(\alpha)} d\widehat{Y}(u)d\widehat{Y}(s)+ \frac{\alpha}{1+2\alpha}\mathcal{Q}_t ,
\end{split}
\end{equation}
where $J_{u,s}^{(\alpha)} $ is defined as
\begin{equation}\label{eq:def:J:expected val}
J_{u,s}^{(\alpha)} := \mathbb{E}_\alpha [{\textnormal{\textbf{J}}}_{u,s}],
\end{equation}
and $\mathcal{Q}_t$ denotes
\begin{equation}
\begin{split}
\mathcal{Q}_t = \mathcal{Q}_t(Y;\alpha):= \intop_0^t \intop_0^s \intop_0^u {\textnormal{\textbf{Q}}}_{v,u,s} \ d\widehat{Y}(v)d\widehat{Y}(u)d\widehat{Y}(s).
\end{split}
\end{equation}
\end{prop}
\begin{proof}
Proof follows analogously as Proposition \ref{prop:speed:Dt1storder}, by observing that
\begin{equation}
\lim_{\Delta \searrow 0} \frac{d}{d\Delta} \mathbb{E}_\alpha [{\textnormal{\textbf{J}}}_{u,s} \ | \ Y_{\le v+\Delta}] = {\textnormal{\textbf{Q}}}_{v,u,s} \left(\frac{d{Y}(v)}{dv} - \alpha \right).
\end{equation}
\end{proof}
We sometimes refer to the formula \eqref{eq:speedproc:2ndorder main} as \textit{the second-order expansion of the speed process}. Later, it turns out that the triple integral of ${\textnormal{\textbf{Q}}}_{v,u,s}$ is negligible for the purpose of obtaining the scaling limit. Moreover, the terms $K_\alpha(s)$ and $J_{u,s}^{(\alpha)} $ no longer possess dependence on $Y$ and in fact they can be studied very explicitly. These facts eventually allow us to analyze \eqref{eq:speedproc:2ndorder main} very precisely, which leads to the derivation of its scaling limit.
\subsection{The speed of the aggregate}\label{subsec:speed:speedagg}
The goal of this section is to see how the actual speed of the aggregate is described in terms of the first- and second-order expansion. Recall the definition of $Y_t$ \eqref{eq:def:Yt} and the formula \eqref{eq:speed:basic expression} for $S(t)$. In the aggregate, a new particle is added at rate $S(t)$. In other words, if we let $\Pi$ be the Poisson point process on the plane with the standard Lebesgue intensity, we can write $X_t = |\Pi_S [0,t]|$, where
\begin{equation}
\Pi_S[0,t] = \{x \in[0,t] : \ (x,y)\in \Pi \textnormal{ for some } y \in[0, S(x)] \}.
\end{equation}
Based on this notation, we can write $dY_t(s)$ as
\begin{equation}
dY_t(s) = d\Pi_S(t-s),
\end{equation}
which is the notation we mainly use throughout the rest of the paper. From this identity, we can recover $\{Y_t(s)\}_{s\ge 0 }$ from $\Pi_S[0,t]$ as
\begin{equation}\label{eq:def:Y:from S}
Y_t(s)= |\Pi_S[t-s,t]|.
\end{equation}
In principle, $S(t)$ can be written as \eqref{eq:speedproc:1storder main} and \eqref{eq:speedproc:2ndorder main} by plugging in $Y_t$ in the place of $Y$. However, there are several aspects that we need to keep in mind when making such a substitution.
\begin{itemize}
\item[(a)] $Y_t$ reads off the aggregate backwards in time, and hence as $t$ increases, the jump is added at the origin as a particle sticks to the aggregate.
\item[(b)] $Y_t(s)$ is infinite for $s>t$, and on this regime Propositions \ref{prop:speed:Dt0thorder}, \ref{prop:speed:Dt1storder} and \ref{prop:speed:Dt2ndorder} are not applicable.
\end{itemize}
Furthermore, in order to give valid approximations on the speed, we introduce several technical assumptions on the aggregate that will be justified in the proof section.
\begin{itemize}
\item[(A1)] Let $t_0>0$ be a large enough number. In some time interval $[t_0, t_0+\Delta]$, $S(t)$ stays ``close'' to some small enough constant $\alpha>0$. Here, one may understand $\alpha$ as the average of $S(t)$ over $t\in[t_0,t_0+\Delta]$.
\item[(A2)] There exists $0<t_0^-<t_0$ such that $S(t)$ is ``sufficiently close'' to
\begin{equation}\label{eq:def:Sprime:basic form}
S'(t)= S'(t;t_0^-,\alpha) := \mathbb{E}_\alpha [ S(Y_t) \ | \ (Y_t)_{\le t-t_0^-} ],
\end{equation}
uniformly on $t\in[t_0,t_0+\Delta],$ where we denote $(Y_t)_{\le t-t_0^-} = \{Y_t(s)\}_{s\le t-t_0^-}$ as before.
\end{itemize}
Note that we can avoid the issue raised in (b) if we work with $S'(t)$ defined in (A2). Moreover, (A1) will tell us that the first- and second-order approximations on $S'(t)$ are valid. Both assumptions (A1) and (A2) will be a part of the notion that we call \textit{regularity} of the speed. This will be discussed in more detail in Section \ref{subsec:speed:outline}.
Having the issue (a) in mind, we can express $S'(t)$ as \eqref{eq:speedproc:1storder main} and \eqref{eq:speedproc:2ndorder main} by plugging in $Y_t$, or $\Pi_S$, in the place of $Y$. We first define
\begin{equation}\label{eq:def:S1:basic form}
S_1(t) = S_1(t;t_0^-,\alpha) := \intop_{t_0^-}^t K_\alpha(t-s) d\Pi_S (s),
\end{equation}
Then, the first-order expansion of the speed for $t\in[t_0, t_0+\Delta]$ is given as
\begin{equation}\label{eq:speed:1storder main}
\begin{split}
S'(t) = \frac{2\alpha^2}{(1+2\alpha)^2} +
\frac{S_1(t)}{(1+2\alpha)^2}+\frac{\alpha}{1+2\alpha} \intop_{t_0^-}^t \intop_{t_0^-}^s
{\textnormal{\textbf{J}}}_{t-s,t-u; t}^{\Pi_S} \ d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(s),
\end{split}
\end{equation}
where $d\widehat{\Pi}_S(s) := d\Pi_S(s) - \alpha ds$. We also call $S_1$ \eqref{eq:def:S1:basic form} the \textbf{first-order approximation of the speed}. One important change we want to emphasize is the definition of ${\textnormal{\textbf{J}}}_{t-s,t-u;t}^{\Pi_S}$ which is slightly different from \eqref{eq:def:J:basic form}: due to the time-reversed nature of $Y_t$, it is defined as
\begin{equation}\label{eq:def:J:agg}
\begin{split}
{\textnormal{\textbf{J}}}_{t-s,t-u;t}^{\Pi_S} &= {\textnormal{\textbf{J}}}_{t-s,t-u;t}^{\Pi_S}(\alpha)\\
&:= \mathbb{P}_\alpha(t-u<T_{t-s}<\infty \ | \ \Pi_S[s,t])- \mathbb{P}_\alpha(t-u<T<\infty \ | \ \Pi_S[s,t]).
\end{split}
\end{equation}
From \eqref{eq:def:Y:from S}, note that conditioning on $\Pi_S[s,t]$ is equivalent to conditioning on $\{ Y_t(s')\}_{0 \le s'\le t-s}$.
Further, the second order expansion is written as
\begin{equation}\label{eq:speed:2ndorder main}
\begin{split}
S'(t) = \frac{2\alpha^2}{(1+2\alpha)^2} +
\frac{S_1(t)}{(1+2\alpha)^2}+\frac{\alpha}{1+2\alpha} \intop_{t_0^-}^t \intop_{t_0^-}^s
J_{t-s,t-u}^{(\alpha)} \ d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(s) + \frac{\alpha}{1+2\alpha} \mathcal{Q}_t,
\end{split}
\end{equation}
where we define $J_{t-s, t-u}^{(\alpha)} $ as \eqref{eq:def:J:expected val} and $\mathcal{Q}_t$ as
\begin{equation}\label{eq:def:Qt:agg}
\begin{split}
\mathcal{Q}_t = \mathcal{Q}_t(t_0^-, \alpha):= \intop_{t_0^-}^t \intop_{t_0^-}^s \intop_{t_0^-}^u {\textnormal{\textbf{Q}}}_{t-s,t-u,t-v;t}^{\Pi_S} \ d\widehat{\Pi}_S(v)d\widehat{\Pi}_S(u)d\widehat{\Pi}_S(s),
\end{split}
\end{equation}
with ${\textnormal{\textbf{Q}}}_{t-s,t-u,t-v;t}^{\Pi_S}$ given by
\begin{equation}\label{eq:def:Q:agg}
\begin{split}
{\textnormal{\textbf{Q}}}_{t-s,t-u,t-v;t}^{\Pi_S}=&{\textnormal{\textbf{Q}}}_{t-s,t-u,t-v;t}^{\Pi_S}(\alpha)
\\
:=& \mathbb{P}_\alpha(t-v<T_{t-s,t-u}<\infty \ | \ \Pi_S[s,t])- \mathbb{P}_\alpha(t-v<T_{t-s}<\infty \ | \ \Pi_S[s,t])\\
&- \mathbb{P}_\alpha(t-v<T_{t-u}<\infty \ | \ \Pi_S[s,t])+\mathbb{P}_\alpha(t-v<T<\infty \ | \ \Pi_S[s,t]).
\end{split}
\end{equation}
Further, we define the \textbf{second-order approximation of the speed} as
\begin{equation}\label{eq:def:S2:basic form}
S_2(t) = S_2(t;t_0^-,\alpha) := \frac{2\alpha^2}{(1+2\alpha)^2} + \frac{S_1(t)}{(1+2\alpha)^2} + \frac{\alpha}{1+2\alpha}\intop_{t_0^-}^t \intop_{t_0^-}^s J_{t-s,t-u}^{(\alpha)} d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(s).
\end{equation}
\begin{remark}
We will need both the first- and second-order approximations \eqref{eq:def:S1:basic form}, \eqref{eq:def:S2:basic form} in order to understand the speed precisely enough for our purpose. Although the first-order approximation often turns out to be enough in many cases, the second-order approximation is essential in the derivation of the scaling limit of the speed. See Sections \ref{subsec:speed:Lt}, \ref{subsubsec:outline:reg} and \ref{subsubsec:outline:EJ} for more discussion.
\end{remark}
Suppose that we define a process $R(t)$ by
\begin{equation}\label{eq:def:Rt:basic form}
R(t) =R(t;t_0^-,t_0,\alpha):= \begin{cases}
\alpha & \textnormal{if } t < t_0,\\
\intop_{t_0^-}^t K_\alpha(t-s) d\Pi_{R}(s ) & \textnormal{if } t \ge t_0.
\end{cases}
\end{equation}
We can interpret $R(t)$ as an age-dependent critical branching process: the average number of offspring that a particle at $t>t_0$ produces is $\intop_t^{\infty} K_\alpha(t'-t) dt = 1$. In this branching process, a particle at $t>t_0$ generates Pois$(1)$ number of offspring, and each of them will be located at $t+s$ with probability $K_\alpha(s)ds$ independently.
The critical branching process $R(t)$ is easier to understand than $S(t)$, and our study on $S(t)$ largely depends on comprehending $R(t)$. A more detailed outline of the plan is discussed in Section \ref{subsec:speed:outline}.
\subsection{Smoothing the speed and the scaling limit}\label{subsec:speed:Lt}
In this subsection, we give the heuristics on how the speed converges to its scaling limit. Recall the critical branching process $R(t)$ defined in the previous subsection. Since its branching is age-dependent, $R(t)$ itself is not a martingale in $t\ge t_0$ (unlike the discrete time critical Galton-Watson branching process where the number of particles at each generation forms a martingale). To build a martingale based on $R(t)$, we first define the following \textit{renewal process} of $K_\alpha$:
\begin{equation}\label{eq:def:Kstar}
K_\alpha^*(s) = \sum_{j\ge 1 } K_\alpha^{*j}(s),
\end{equation}
where $K_\alpha^{*j}$ is the convolution of $K_\alpha$ with itself taken $j$ times. At the moment, assume that $K_\alpha^*$ is well-defined, and moreover that the limit
\begin{equation}
K_\alpha^* = \lim_{s\to \infty} K_\alpha^*(s)
\end{equation}
exists. These facts are explained in Section \ref{subsec:Kestim:intro}.
Then, we consider a ``smoothed'' version of the speed, which is defined for $t \ge t_0$ as
\begin{equation}\label{eq:def:Lt:basic form}
L(t) = L(t;t_0^-,\alpha) := \intop_{t_0^-}^t \intop_t^\infty K_\alpha^* \cdot K_\alpha(x-s) dx d\Pi_S(s).
\end{equation}
In words, $L(t)$ is the expected branching rate at infinity when we start the critical branching process from time $t$ with initial points given by $\Pi_S[t_0^-,t]$. For details on this process, see Section \ref{sec:criticalbranching}.
We choose to study the scaling limit of $L(t)$ rather than $S(t)$, since it has a more tractable structure described as a martingale added with some perturbation. On the other hand, the behavior of $S(t)$ has a very spiky nature: $S(t)$ jumps significantly whenever a new point arrives at $\Pi_S$. Furthermore, our ultimate interest is to obtain $X_t$, which is closely related with the integral of $S(t)$. From this perspective, we would not lose anything from studying $L(t)$ since it is essentially a smoothed version of $S(t)$.
In order to obtain the scaling limit of $L(t)$, we need to understand the mean and variance of the increment $L(t_0+\Delta)-L(t_0)$ (conditioned on $\mathcal{F}_{t_0})$. Observe that
\begin{equation}\label{eq:Lt diff:basic}
dL(t) = K_\alpha^* d\Pi_S(t) - \intop_{t_0^-}^t K_\alpha^* \cdot K_\alpha(t-s) d\Pi_S(s) dt = K_\alpha^* \left\{d\Pi_S(t) - S_1(t)dt \right\}.
\end{equation}
which gives
\begin{equation}\label{eq:Lt:increment}
\begin{split}
L({t_0+\Delta}) - L({t_0}) =
K_\alpha^* \intop_{t_0}^{t_0+\Delta} \big(d\Pi_S(t) - S(t)dt\big) + K_\alpha^* \intop_{t_0}^{t_0+\Delta} \big(S(t)-S_1(t)\big)dt.
\end{split}
\end{equation}
As seen in the formula, we can decompose this increment into a ``martingale part'' and a ``drift part''.
Computations of its mean and variance is the most important ingredient in establishing the main theorem. To this end, we need to assume that the processes $S(t)$ and $S_1(t)$ behaves nicely enough, otherwise the terms such as $S(t)-S_1(t)$ might become too large and the increment would not scale as expected. At the moment, we state this assumption on \textit{regularity} informally as follows.
\begin{itemize}
\item[(A3)] In addition to (A1) from Section \ref{subsec:speed:speedagg}, the processes $S(t)$ and $S_1(t)$ both stay ``close'' to $\alpha$ and to each other in the interval $[t_0, t_0+\Delta]$.
\end{itemize}
Formulating (A1), (A2) and (A3) into a rigorous, quantitative condition of \textit{regularity} and proving it are two of the core endeavours of this article, and the methods are outlined in the next subsection.
Under these assumptions, we can state the estimates on the increment informally as follows.
\begin{thm}[Informal]\label{thm:speed:increment:informal}
Suppose that (A1), (A2) and (A3) are satisfied in the interval $[t_0, t_0+\Delta]$ for a sufficiently small $\alpha>0$. Then, we have
\begin{equation}\label{eq:increment:informal}
\begin{split}
\mathbb{E} [L(t_0+\Delta)- L(t_0) \ | \ \mathcal{F}_{t_0}] &= o(\alpha^4)\Delta;\\
\mathrm{Var}[L(t_0+\Delta)- L(t_0) \ | \ \mathcal{F}_{t_0}]
&= (1+ o(1))4\alpha^5 \Delta.
\end{split}
\end{equation}
\end{thm}
\begin{proof}[Sketch of proof]
Recall the expression \eqref{eq:Lt:increment}. We can first see that
\begin{equation}
\mathbb{E} [L(t_0+\Delta)-L(t_0) \ | \ \mathcal{F}_{t_0}] = \mathbb{E} \left[\left.K_\alpha^* \intop_{t_0}^{t_0+\Delta} (S(t)-S_1(t)) dt\ \right| \ \mathcal{F}_{t_0} \right].
\end{equation}
Later, we will see that $\int_{t_0}^{t_0+\Delta}|S(t)-S_1(t)|dt = o(\alpha^2\Delta)$ when $S(t)$ is \textit{regular}. Moreover, it turns out that (see Section \ref{subsec:Kestim:intro})
\begin{equation}
K_\alpha^* = \left(\intop_0^\infty xK_\alpha(x)dx \right)^{-1} = \frac{2\alpha^2}{1+2\alpha}.
\end{equation}
Thus we obtain the first equation of \eqref{eq:increment:informal}. For the second one, we will show that
\begin{equation}
\begin{split}
\mathrm{Var}[L(t_0+\Delta)-L(t_0)\ | \ \mathcal{F}_{t_0}] &\approx \mathrm{Var}\left[\left.
K_\alpha^* \intop_{t_0}^{t_0+\Delta} (d\Pi_S(t)- S(t)dt)
\ \right| \ \mathcal{F}_{t_0} \right]\\
&=
(K_\alpha^*)^2\mathbb{E}\left[\left. \intop_{t_0}^{t_0+\Delta} S(t)dt \ \right| \ \mathcal{F}_{t_0} \right] = (1+o(1))4\alpha^5 \Delta.
\end{split}
\end{equation}
Note that the last equality will follow from the assumption (A1).
\end{proof}
To obtain the scaling limit of $L(t)$ from the above, we need to rescale $L(t)$ by $\alpha^{-1}$, and then we can see that the time $\Delta$ should be rescaled by $\alpha^{-3}$. In fact, roughly speaking, we will later see that
\begin{equation}\label{eq:conv to sde:informal}
\left\{M^{1/3} L(t_0 + Ms)\right\}_{s\ge 0} \underset{M\to\infty}{\longrightarrow} \{Z_s\}_{s\ge 0},
\end{equation}
where $\{Z_s\}_{s\ge 0}$ satisfies the following stochastic differential equation:
\begin{equation}\label{eq:sde:main}
dZ_s = 2Z_s^{5/2} dB_s.
\end{equation}
Here, $t_0$ can be understood as a large time parameter where we start to see $S(t)$ being \textit{regular} with high enough probability. A more detailed description is given in Section \ref{subsubsec:outline:scaling limit}.
Note that $Z_t$ given by \eqref{eq:sde:main} is $(-\frac{1}{3})$-self-similar: $Z_t' := s^{1/3} Z_{st}$ has the same law as $Z_t$ for any $s>0$. This suggests that the decay of $Z_t$ (and hence, $S(t)$) in a larger time scale is asymptotically $t^{-1/3}$, and consequently the asymptotic growth of $X_t$ should be of order $t^{2/3}$.
\begin{remark}\label{remark:on bessel}
Since the only $\frac{1}{2}$-self-similar processes are Bessel processes, $(Z_t)^{-3/2}$ should be a constant multiple of a Bessel process. One can see by applying It\^o's formula that $Z_t = (3V_t)^{-2/3}$, where $V_t$ is the $\frac{8}{3}$-Bessel process.
\end{remark}
\subsection{Proof outline}\label{subsec:speed:outline}
Before delving into the actual proof, we devote this subsection to outlining the proof step by step. The following describes the main steps that we go through in order.
\begin{itemize}
\item $O(t^{3/4+\epsilon})$ upper bound and reduction to the fixed rate processes.
\item Analysis of the fixed rate process and its perturbation.
\item The age-dependent critical branching process.
\item Regularity of the speed.
\item Refined computation of the lower order terms.
\item Scaling limit of the speed.
\end{itemize}
\noindent In the following subsections, we explain the purpose of each step and the main ideas to achieve it.
\subsubsection{$O(t^{3/4+\epsilon})$-upper bound and reduction to the fixed rate processes}\label{subsubsec:outline:simpler upper bound}
All the discussions from the previous subsections assumes $\alpha>0$ to be small enough. In order to apply such arguments, we at least need to know that the speed tends arbitrarily close to $0$ in finite time.
We obtain this by a simpler but weaker argument that gives an $O(t^{3/4+\epsilon})$-upper bound on $X_t$. This is done by counting the number of \textit{excess particles}, which is the difference between the total number of particles attached to the aggregate up to time $t$ and the size $X_t$ of the aggregate.
Then, we relate $S(t)$ with the fixed rate processes as follows: since $S(t)$ drops to an arbitrary low scale in finite time, we show that $S(t)$ can be bounded above and below by arbitrary small constants for a while. Thus, our later analysis focuses on the study of the speed process generate by those fixed constants, which will converge to the same scaling limit.
\subsubsection{Analysis of the fixed rate process and its perturbation}\label{subsubsec:outline:fixedrate}
In studying \eqref{eq:speed:1storder main} and \eqref{eq:speed:2ndorder main}, bounding the sizes of double and triple integrals in the expressions
turns out to be essential, because they are the error terms of the first- and second-order approximations (equations \eqref{eq:def:S1:basic form} and~\eqref{eq:def:S2:basic form}), respectively. However, these integrals are in terms of $d\widehat{\Pi}_S$, which is difficult to understand since we do not have a good understanding on $S(t)$. Furthermore, for instance the term ${\textnormal{\textbf{J}}}_{t-u, t-s; t}^{\Pi_S}$ in \eqref{eq:speed:1storder main} is only measurable with respect to $\mathcal{F}_t$, meaning that the double integral in \eqref{eq:speed:1storder main} would not have a martingale structure. Instead, we choose to work with the integrals over the fixed rate Poisson process, namely,
\begin{equation}\label{eq:speed:fixedrate:med}
\intop_{0}^t \intop_{0}^s
{\textnormal{\textbf{J}}}_{u,s} \ d\widehat{\Pi}_\alpha(u) d\widehat{\Pi}_\alpha(s),
\end{equation}
upon reversing the time axis, and the corresponding analogue for $\mathcal{Q}_t$. Here, $\Pi_\alpha$ denotes the rate-$\alpha$ Poisson process. A precise form of \eqref{eq:speed:fixedrate:med} will be given in Section \ref{sec:fixedrate}. These integrals can be understood using Azuma-type concentration inequalities for martingales.
In addition, we also need an appropriate estimate on ${\textnormal{\textbf{J}}}_{u,s}$ (and ${\textnormal{\textbf{Q}}}_{v,u,s}$). Recalling its definition in \eqref{eq:def:J:basic form}, we couple the events $\{s<T_u<\infty \}$ and $\{s<T<\infty \}$ and rely on fundamental hitting time estimates of random walk. The same but more complicated coupling argument can be carried out for ${\textnormal{\textbf{Q}}}_{v,u,s}$ as well.
To translate the estimates for the fixed rate process to those for $S(t)$, we analyze the Radon-Nykodym derivative of $d\Pi_S$ with respect to $d\Pi_\alpha$. We show that it does not become too big with high probability under an appropriately mild assumption on $S(t)$ which is included in the definition of regularity. On the other hand, the double integral over $d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(s)$ would stay small with high probability if the Radon-Nykodym derivative is smaller compared to the reciprocal of the probability that \eqref{eq:speed:fixedrate:med} becomes too large. Details will be discussed in Section \ref{sec:fixedrate}.
\subsubsection{The age-dependent critical branching process}\label{subsubsec:critical branching}
As discussed in Section \ref{subsec:speed:speedagg}, studying the speed $S(t)$ directly is rather complicated. Thus, we work with the age-dependent critical branching process $R(t)$ \eqref{eq:def:Rt:basic form} in Section \ref{sec:criticalbranching} as an intermediate step to understand the speed, which is particularly useful in understanding the evolution of the fixed rate process from Section \ref{subsubsec:outline:simpler upper bound}.
The main property of $R(t)$ we deduce here is that it stays close to $\alpha$ by the right distance, roughly of order $\alpha^{\frac{3}{2}}$. We obtain this by applications of martingale concentration inequalities and relating it with the standard Galton-Watson critical branching processes.
\subsubsection{Regularity of the speed}\label{subsubsec:outline:reg}
As discussed in the previous subsection, a notion of regularity is introduced to deduce Theorem \ref{thm:speed:increment:informal} and thus to obtain the scaling limit of the speed. Its definition provides a quantitative description of the assumptions (A1), (A2) and (A3) from Sections \ref{subsec:speed:speedagg} and~\ref{subsec:speed:Lt}, and is given by various stopping times: Each stopping time will denote the first time when a certain desired property is violated, and $S(t)$ will be called regular at time $t$ if the minimum of all those stopping times is larger than $t$ by a certain amount. For instance, the first times when $S(t)$, $|S(t)-S_1(t)|$ become larger than they are supposed to be are included in the definition. In Section \ref{sec:reg:intro}, we give a thorough overview on the notion of regularity and explain why such conditions need to be understood.
The main purpose of defining the regularity using stopping times is to effectively understand how all the conditions we impose are intertwined with each other. If $\tau = \min_i \tau_i < t'$, then there should exist a $j$ such that $\tau_j = \tau <t'$, meaning that the property described by $\tau_j$ is violated for the first time among all others. Thus, we can show that $\tau\ge t'$, that is, the speed is regular with high probability, by proving $\tau_i=\tau<t'$ happens with very small probability for every $i$. In Section \ref{sec:reg:conti of reg}, we explain the details of this argument.
Another major issue is the variation of the frame of reference $\alpha$, to complete building the inductive framework of the argument. After we prove that the speed is regular from time $t_0$ to $t_1=t_0+\Delta$ with respect to $\alpha$, we want to argue that the speed continues to be regular in $[t_1,t_1+\Delta']$ with high probability, with respect to a different value $\alpha'$ to take account of the change of $L(t)$ during the previous time interval. This means that we generate the processes $S_1(t)$ and $S_2(t)$ with $\alpha'$, and show that they still satisfy the conditions of regularity as long as the change from $\alpha$ to $\alpha'$ is small. To this end, we require a refined understanding on analytic properties of $K_\alpha(s)$, $K_\alpha^*(s)$ and $J_{s,t}^{(\alpha)}$, as well as a combination of various tools mentioned in the previous subsections.
The computations to establish the regularity turn out to be long and complicated, and hence we divide their details into three sections, Sections \ref{sec:reg:intro}, \ref{sec:reg:conti of reg} and \ref{sec:reg:next step}.
\subsubsection{Refined computation of the lower order terms}\label{subsubsec:outline:EJ}
In the computation of the mean of the increment in Theorem \ref{thm:speed:increment:informal}, we need to understand the mean of $S(t)-S_1(t)$, which can be expressed, by using \eqref{eq:speed:2ndorder main}, as
\begin{equation}\label{eq:speed:SminusS1:2nd order:basic}
\begin{split}
S(t)-S_1(t) =&
\frac{2\alpha^2}{(1+2\alpha)^2} +
\frac{4(\alpha+\alpha^2)}{(1+2\alpha)^2}S_1(t)\\
&+\frac{\alpha}{1+2\alpha} \intop_{t_0^-}^t \intop_{t_0^-}^s
J_{t-s,t-u}^{(\alpha)} \ d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(s) + \frac{\alpha\mathcal{Q}_t}{1+2\alpha} .
\end{split}
\end{equation}
Although we have derived an appropriate bound on the double integral part which is essentially sharp, it is an estimate that holds with high probability that does not give the precise information on the mean
\begin{equation}\label{eq:speed:mean of doubleint}
\mathbb{E} \left[ \intop_{t_0^-}^t \intop_{t_0^-}^s
J_{t-s,t-u}^{(\alpha)} \ d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(s) \right].
\end{equation}
To precisely caculate \eqref{eq:speed:mean of doubleint}, we switch the integral over $d\widehat{\Pi}_S(u)d\widehat{\Pi}_S(s)$ with simpler ones that only involves $K_\alpha^*(x)$ and $\Pi_\alpha$,
and show that the error arising from the change is of negligible compared to the whole integral. After such modification, we carry out an analytic study to compute the precise expected value.
Combining with the previous analysis on regularity, we also discuss the formal version of Theorem \ref{thm:speed:increment:informal}.
\subsubsection{Obtaining the scaling limit}\label{subsubsec:outline:scaling limit}
Given all the works mentioned in Sections \ref{subsubsec:outline:simpler upper bound}--\ref{subsubsec:outline:EJ}, we are almost ready to deduce the scaling limit of $L(t)$ based on Theorem \ref{thm:speed:increment:informal}. However, when we consider the process $M^{1/3} L(t_0 + sM)$, the initial condition at $s=0$ blows up as $M$ tends to infinity, if $t_0$ is a fixed value. The classical results on the convergence to the limiting diffusion (which we mention in the next subsection) would not be applicable in such a case.
Thus, our choice is to make $t_0$ be a random time $\tau(M)$ depending on $M$, which is roughly
\begin{equation}
\tau(M):= \inf\{t>0: L(t) \le CM^{-1/3} \textnormal{ and }L(t) \textnormal{ is regular} \},
\end{equation}
where $C$ is a fixed, large enough constant. Then, we need to show that $\tau(M) \le \epsilon M$ with high probability, with a constant $\epsilon>0 $ that can be arbitrarily small by setting $C$ to be large, and this is equivalent to show that $X_t = O(t^{2/3})$ with high probability.
This is done by a multi-scale analysis on $L(t)$ based on Theorem \ref{thm:speed:increment:informal}. To ensure it is regular, we first drop $L(t)$ to a small enough scale using the results from Section \ref{subsubsec:outline:simpler upper bound}. From then, we show that $L(t)$ is more likely to halve than to double its size. Further, we argue that either event is going to happen after an appropriate amount of time
that makes $L(t)$ scale like $t^{-1/3}$. The main technical tools we use to achieve this goal are the martingale concentration inequalities such as the Azuma-Bernstein inequality.
When the above issue on the initial regularity is resolved, we can appeal to the classical result of Helland \cite{helland1981minimal} and Theorem \ref{thm:speed:increment:informal} to show that
\begin{equation}\label{eq:speed:conv to sde:outline}
\left\{ M^{1/3} L(\tau(M) + sM) \right\}_{s\ge 0} \underset{M\to \infty}{\longrightarrow} \{A_s\}_{s\ge 0},
\end{equation}
where $\{A_s\}_{s\ge 0}$ is the solution of
\begin{equation}
dA_s = 2A_s^{5/2} dB_s, \quad A_0 = C.
\end{equation}
The convergence \eqref{eq:speed:conv to sde:outline} is in Stone topology, which is good enough to ensure the convergence of the integrated process
\begin{equation}
\left\{ M^{-2/3} \intop_{0}^{Mt} L(\tau(M)+s)ds\right\}_{t\ge 0}
\underset{M\to \infty}{\longrightarrow} \left\{ \intop_0^t A_s ds \right\}.
\end{equation}
Finally, the main result is obtained by increasing the level of $C$ to infinity, and then observing that $\intop_0^{Mt} L(s)ds$ is close enough to $X_{Mt}$ based on the conditions of regularity.
\subsection{Estimates on fundamental quantities}\label{subsec:Kestim:intro}
As mentioned before, understanding the analytic properties of the branching density $K_\alpha(s)$ and the renewal process $K_\alpha^*(s)$ will be crucial throughout the paper. In this subsection, we review the main estimates we need, while their proofs are deferred to Appendix \ref{sec:fourier and renewal}. They can be obtained by studying the moment generating function of $T$ via Fourier analysis.
To begin with, the branching density $K_\alpha(s)$ (Definition \ref{def:Kalpha:main}) satisfies the following property.
\begin{lem}\label{lem:estimate on K:intro}
There exist absolute constants $c, C>0$ such that for any $t>0$ and $0<\alpha<\frac{1}{2}$,
\begin{equation}
K(t) \le \frac{C\alpha }{\sqrt{t+1}}e^{-c \alpha ^2 t}.
\end{equation}
Furthermore, its derivative in $t$ satisfies
\begin{equation}
|K_\alpha'(t)| \le C\alpha (t+1)^{-\frac{3}{2}} e^{-c\alpha^2 t}.
\end{equation}
\end{lem}
Moreover, the properties of $K_\alpha^*(s)$ \eqref{eq:def:Kstar} are summarized by the following lemma.
\begin{lem}\label{lem:estimat for K tilde:intro}
Let $K^*_\alpha:= \frac{2\alpha^2}{1+2\alpha}$. There exist absolute constants $c, C>0$ such that for any $t>0$ and $0<\alpha<\frac{1}{2}$,
\begin{equation}
\left|K^*_\alpha (t) - K^*_\alpha \right| \le \frac{C\alpha}{ \sqrt{t+1}}e^{-c\alpha^2 t}.
\end{equation}
Furthermore, its derivative in $t$ satisfies
\begin{equation}
|(K_\alpha^*)'(t)| \le C\alpha (t+1)^{-\frac{3}{2}} e^{-c\alpha^2 t}.
\end{equation}
\end{lem}
Finally, we state a similar bound on $J_{s,t}^{(\alpha)}$ \eqref{eq:def:J:expected val}.
\begin{lem}\label{lem:bound on deterministic J}
There exist absolute constants $c,C>0$ such that for any $t>s>0$ and $0<\alpha<\frac{1}{2}$,
\begin{equation}
\left|J_{s,t}^{(\alpha)} \right| \le \frac{Ce^{-c\alpha^2 t}}{\sqrt{(s+1)(t+1)}}.
\end{equation}
\end{lem}
\section{An a priori upper bound and the induction base}\label{sec:inductionbase}
The first step of our proof is establishing that the speed drops down to arbitrary small level in finite time. In Section \ref{subsec:simpleupbd}, we achieve this by showing an $t^{\frac{3}{4}}$polylog$(t)$-bound on $X_t$ which follows from counting the excessive particles in the aggregate. Based on this result, we prove in Sections \ref{subsec:sandwich1}, \ref{sec:Sandwich} that when $t$ gets sufficiently large (but finite), the speed stays between arbitrary small constants for a long enough time. Such a sandwiching argument enables us to analyze the speed process that evolves from particles given by a fixed rate process in the later sections.
\subsection{Simple upper bound of $t^{\frac{3}{4}}$}\label{subsec:simpleupbd}
\begin{lem}\label{lem:three quarter upper bound}
For any $t>0$ we have that
\begin{equation}
\mathbb P (X_t\ge t^{\frac{3}{4}}\log ^{2}t) \le C t^{-10}
\end{equation}
\end{lem}
\begin{proof}
Denote by $N_t$ the total number of particles that were absorbed into the aggregate by time $t$. We start by bounding $N_t$. To this end, note that the event $\mathcal A:=\{ X_t \le t\}$ satisfies $\mathbb P (\mathcal A )\ge 1-e^{-ct}$. Indeed, we have with probability one that $ S(x)\le 1/2$ for all $x>0$ and therefore $ X_t\precsim \text{Poisson}(t/2)$.
Next, for any $l \ge 0$, define the random variable $Y_l$ to be the number of particles that at some point, before time $t$, were inside the interval $[0,l]$. Define the event $\mathcal B _l :=\{ N_l \le l+\sqrt{t}\log ^2 t \}$. It is clear that on the event
\begin{equation}
\mathcal B := \mathcal A \cap \bigcap _{l=1}^{\lceil t \rceil } \mathcal B _l
\end{equation}
we have that
\begin{equation}\label{eq:bound on N}
N_t \le X_t + \sqrt{t} \log ^ 2 t.
\end{equation}
Indeed, a particle that was absorbed into the aggregate before time $t$ was, at some point, inside the interval $[0,X_t]$.
We turn to show that $\mathcal B $ happens with high probability. For any $
l \le t $, let $\mathcal D _l $ be the event that any particle which was initially outside the interval $[0, l +\frac{1}{2}\sqrt{t} \log ^2 t ]$ did not reach the interval $[0,l ]$ and let $\mathcal E _l $ be the event that the number of particles initially in the interval $[0, l + \frac{1}{2} \sqrt{t} \log ^2 t ]$ is at most $l + \sqrt{t}\log ^2 t $. It is clear that $\mathcal D _l \cap \mathcal E _l \subseteq \mathcal B _ l $ and that $\mathbb P (\mathcal D _l )\ge 1-Ce^{-c \log ^2 t }$. Since the number of particles that were initially in $[0, l + \frac{1}{2} \sqrt{t} \log ^2 t ]$ is distributed by $\text{Poisson} (l + \lfloor \frac{1}{2} \sqrt{t} \log ^2 t \rfloor )$ we have for any $l \le t$ that $\mathbb P ( \mathcal E _l) \ge 1- C e^{-c \log ^2 t }$. Thus we have that $\mathbb P ( \mathcal B _l) \ge 1- C e^{-c \log ^2 t }$ and therefore, from a union bound we get $\mathbb P ( \mathcal B ) \ge 1- C e^{-c \log ^2 t }$.
Next, let $Z_t $ be the number of particles at $X_t+1$ at time $t$, and recall that the conditional distribution of $Z_t$ given $\{X_s, \ s\le t \}$ is $ \text{Poisson}( 2S(t) )$.
The aggregate size, $X_t$, increases by $1$ with rate $S(t)$ and therefore $X_t -\int _0^t S(x)ds $ is a martingale. Similarly, for each $j\ge 1 $, $N_t $ increases by $j$ with rate $ \frac{1}{2} j \cdot \mathbb P (Z_t =j \ | \ X_s, \ s\le t )$ and therefore
\begin{equation}
N_t -\frac{1}{2} \intop _0^t \sum _{j=1}^{\infty } j^2 \cdot \mathbb P (Z_x =j \ | \ X_s, \ s\le x )dx= N_t-\frac{1}{2}\intop _0^t \mathbb E \left[ Z_x^2 \ | \ X_s , \ s\le x \right] dx= N_t-\intop _0^t S(x)+2S(x)^2 dx
\end{equation}
is a martingale. By Corollary~\ref{cor:concentration of integral} with $M=t$ and $f=1$ (using the bounds $S,S+2S^2 \le 1$) we have probability at least $1-C e^{-c \log ^2 t }$ that
\begin{equation}\label{eq:azuma for X and N}
\left| X_t-\intop _0^t S(x)dx \right| \le \sqrt{t} \log ^2 t,\quad \left| N_t-\intop _0^t S(x)+2 S(x)^2dx \right| \le \sqrt{t} \log ^2t.
\end{equation}
Denote by $\mathcal C$ the event where these inequalities hold. On $\mathcal B \cap \mathcal C $ we have that
\begin{equation}
2\intop _0^t S(x)^2 dx \le -X_t +\sqrt{t} \log ^2 t +\intop _0^t S(x)+2 S(x)^2dx \le N_t-X_t +2 \sqrt{t}\log ^2 t \le 3 \sqrt{t} \log ^2 t,
\end{equation}
where in the first two inequalities we used \eqref{eq:azuma for X and N} and in the last inequality we used \eqref{eq:bound on N}. Thus, on $\mathcal B \cap \mathcal C $, by Cauchy Schwartz inequality we have
\begin{equation}
X_t- \sqrt{t}\log ^2t \le \intop _0^t S(x)dx\le \sqrt{t} \cdot \Big( \intop _0^t S(x)^2dx\Big) ^{\frac{1}{2}} \le 2 t^{\frac{3}{4}}\log t.
\end{equation}
This finishes the proof of the lemma.
\end{proof}
The next corollary follows immediately from Lemma~\ref{lem:three quarter upper bound}.
\begin{cor}
We have that
\begin{enumerate}
\item
$\mathbb P (X_t =O(t^{\frac{3}{4}}\log ^2t) )=1$.
\item
For all $t\ge 2$, $\mathbb E X_t \le C t^{\frac{3}{4}} \log ^2 t$.
\end{enumerate}
\end{cor}
\begin{proof}
The first part follows immediately from Lemma~\ref{lem:three quarter upper bound} and the Borel-Cantelli lemma. We turn to prove the second statement. By Lemma~\ref{lem:three quarter upper bound} and the fact that $X_t\precsim \text{Poisson}(t/2)$ we have for sufficiently large $t$,
\begin{equation}
\mathbb E X_t \le t^{\frac{3}{4}} \log ^2t +t\cdot \mathbb P ( t^{\frac{3}{4}} \log ^2t \le X_t \le t) +\sum _{k=1}^{\infty } (k+1)t \cdot \mathbb P (X_t\ge kt) \le Ct^{\frac{3}{4}} \log ^2t,
\end{equation}
as needed.
\end{proof}
\subsection{Stochastic domination and bound on rate}\label{subsec:sandwich1}
The process $Y_t$ is itself a function valued Markov chain and it will be useful at time to initialize it from different starting states. We define the generalized aggregate with initial condition $Y_{0}$ for some $t_0\ge 0$.
\begin{definition}\label{def:aggregate with initial condition}
Let $Y_0:\mathbb R_+\to \mathbb N \cup \{\infty \}$ be a monotone random step function and let $\Pi $ be an independent $2$-D Poisson process. The rate of the aggregate with initial condition $(Y_0,t_0)$ is defined by
\begin{equation}
S(t):=\frac{1}{2}\mathbb P \left( W(s) \le Y_t(s) \ | \ Y_t\right), \quad t \ge t_0,
\end{equation}
where $Y_t$ is defined by
\begin{equation}\label{eq:generalised Y}
Y_t(s)=\begin{cases}
\quad \quad X_t-X_{t-s}\quad \quad \quad \quad s \le t-t_0 \\
Y_t(t-t_0)+Y_0(s-(t-t_0)) \ \quad s >t-t_0
\end{cases}
\end{equation}
As before, the size of the generalised aggregate will be $X_t:= \Pi _{S}[t_0,t]$. We say that $(Y_t ,S(t), X_t)$ is the aggregate with initial condition $(Y_0,t_0)$ driven by the Poisson process $\Pi $.
\end{definition}
The following claim follows immediately from the above definition.
\begin{claim}\label{claim:aggregate with initial condition}
Let $Y_0^{(1)}, Y_0^{(2)}:\mathbb R_+\to \mathbb N \cup \{\infty \}$ be monotone random step functions and let $\Pi$ be an independent Poisson process. Let $(Y_t^{(1)},S^{(1)} (t), X_t^{(1)} )$ and $(Y_t^{(2)},S^{(2)} (t), X_t^{(2)} )$ be the aggregates with initial conditions $(Y_0^{(1)},0)$ and $(Y_0^{(2)},0)$ respectively, both driven by $\Pi $. We also let $(Y_t,S(t),X_t )$ be the usual aggregate.
\begin{enumerate}
\item
Suppose that $Y_0^{(1)}\equiv \infty $ with probability one. Then
\begin{equation}
\{S^{(1)}(t)\} _{t \ge 0}\overset{d}{=} \{S(t)\} _{t \ge 0} \quad \{ X_t^{(1)}\} _{t \ge 0}\overset{d}{=} \{X_t\} _{t \ge 0}.
\end{equation}
\item
Let $t_0>0$ and suppose that $Y_0^{(1)}\overset{d}{=} Y_{t_0}$. Then
\begin{equation}
\{S^{(1)}(t)\} _{t \ge 0}\overset{d}{=} \{S(t+t_0)\} _{t \ge 0} \quad \{ X_t^{(1)}\} _{t \ge 0}\overset{d}{=} \{X_{t+t_0}\} _{t \ge 0}.
\end{equation}
\item
On the event $\{ Y_0^{(1)}(s)\le Y_0^{(2)}(s) \text{ for all }s \}$ we have for all $t>0$
\begin{equation}
S^{(1)}(t) \le S^{(2)}(t),\quad X_t^{(1)} \le X_t^{(2)}, \quad \forall s>0,\ Y_t^{(1)}(s)\le Y_t^{(2)}(s).
\end{equation}
\end{enumerate}
\end{claim}
\begin{cor}\label{cor:stochastic domination}
We have that $S(t)$ is stochastically decreasing in the sense that, if $t_1\le t_2$ then there is a coupling of the processes $\{S(t+t_1)\}_{ t \ge 0}$ and $\{ S(t+t_2)\} _{t\ge 0}$ such that the first process is larger than the second.
\end{cor}
\begin{proof}
Let $Y_0^{(1)}\overset{d}{=}Y_{t_2-t_1}$ and $Y_0^{(2)}\equiv \infty$ and let $\Pi $ be a Poisson process independent of everything. By parts (1) and (2) of Claim~\ref{claim:aggregate with initial condition} we have that
\begin{equation}
\{S^{(1)}(t)\} _{t \ge 0}\overset{d}{=} \{S(t)\} _{t \ge 0} \quad \text{and} \quad \{S^{(2)}(t)\}_{t \ge _0 }\overset{d}{=}\{S(t+t_2-t_1)\}_{t \ge 0}.
\end{equation}
Since $Y_0^{(1)}(s)\le Y_0^{(2)}(s)$ for all $s$ almost surely, we have by part (3) of Claim~\ref{claim:aggregate with initial condition} that $S^{(1)}(t)\le S^{(2)}(t)$ for all $t>0$. Thus, the processes $\{S^{(1)}(t+t_1) \}_{t \ge 0}$ and $\{S^{(2)}(t+t_1) \}_{t \ge 0}$ give the desired coupeling between $\{S(t+t_1)\}_{t \ge 0}$ and $\{S(t+t_2)\}_{t \ge 0}$.
\end{proof}
In the following lemma we show that the rate of the aggregate gets arbitrarily close to $0$ for long periods of time.
\begin{lem}\label{lem:bound on rate}
For any $\alpha >0$ sufficiently small the following holds. For all $t\ge \alpha ^{-20}$ we have that
\begin{equation}
\mathbb P \left( \sup _{t-\alpha ^{-3} \le s \le t } S(s) \le \alpha \right) \ge 1- \alpha .
\end{equation}
\end{lem}
\begin{proof}
Let $t\ge \alpha ^{-20}$ and let $I_k:=[t- k \alpha ^{-3}, t -(k-1) \alpha ^{-3}]$ for all $1\le k \le \lfloor \alpha ^3 t \rfloor $.
Define the random sets
\begin{equation}
\begin{split}
A_1:=\left\{ 1\le k\le \lfloor\alpha ^3 t \rfloor-1 \text{ odd }: \substack{ \text{the aggregate grew in }\\ \text{the time interval } I_{k+1}\cup I_{k} } \right\}, \\
A_2:=\left\{ 1 \le k\le \lfloor\alpha ^3 t \rfloor-1 \text{ even }: \substack{ \text{the aggregate grew in}\\ \text{the time interval } I_{k+1}\cup I_{k} } \right\}
\end{split}
\end{equation}
and
\begin{equation}
B:= \left\{ 1\le k\le \lfloor\alpha ^3 t \rfloor-1 \ : \ \sup _{s\in I_k}S(s)> \alpha \right\}.
\end{equation}
It is clear that $|A_1|\le X_{t}$ and $|A_2|\le X_{t}$. Next, we claim that $B \subseteq A_1\cup A_2$. Indeed suppose that $k \notin A_1\cup A_2$ and let $s \in I_k$. By the definition of $A_1,A_2$, the aggregate did not grow in the time interval $[s-\alpha ^{-3} ,s]$ and therefore
\begin{equation}
S(s)=\frac{1}{2} \mathbb P \left(\forall x>0 , \ W(x) \le Y_s(x) \ | \ Y_s \right) \le \mathbb P \left( \forall x \le \alpha ^{-3}, \ W(x)\le 0 \right) \le C \alpha ^{\frac{3}{2}} \le \alpha ,
\end{equation}
where the last two inequalities hold for sufficiently small $\alpha $. Since the last bound holds for all $s \in I_k$ we get that $k \notin B$.
Finally, by Corollary~\ref{cor:stochastic domination} we have that $\mathbb P (k \in B) \ge \mathbb P (1 \in B)$ and therefore using Lemma~\ref{lem:three quarter upper bound} we obtain
\begin{equation}
(\lfloor\alpha ^3 t \rfloor-1 ) \mathbb P (1 \in B) \le \sum _{k=1}^{ \lfloor\alpha ^3 t \rfloor-1} \mathbb P (k\in B) =\mathbb E |B| \le \mathbb E |A_1|+\mathbb E |A_2| \le 2 \mathbb E X_{t} \le C t ^{\frac{3}{4}} \log ^5 t .
\end{equation}
Thus, using that $t \ge \alpha ^{-20}$ and that $\alpha $ is sufficiently small we get
\begin{equation}
\mathbb P \left( \sup _{t-\alpha ^{-3} \le s \le t } S(s) \ge \alpha \right) =\mathbb P (1 \in B) \le \alpha
\end{equation}
as needed.
\end{proof}
\subsection{Bounding with fixed rate processes}\label{sec:Sandwich}
In this section we show that the aggregate can be bounded from above and below by regular processes. Recall that $\Pi $ is the $2-D$ Poisson process that is driving the aggregate.
For $\alpha , t >0 $ and define the random functions
\begin{equation}\label{eq:def:sandwichfixed}
\underline{Y} _{t,\alpha } (s):= \Pi _{\alpha }[t -s , t ] ,\quad \overline{Y} _{t,\alpha }(s):=
\begin{cases}
\Pi _{ \alpha }[t-s,t], \quad s \le \alpha ^{-3} \\
\quad \quad \infty,\quad \ \quad \quad s > \alpha ^{-3}
\end{cases}\!\!\!\!
\end{equation}
for all $s>0$. In terms of the speed process, we can write
\begin{equation}\label{eq:def:sandwich speed}
\underline{S}_{t,\alpha}(s) \equiv \alpha, \quad \overline{S}_{t,\alpha}(s) = \begin{cases}
\alpha, & t-\alpha^{-3}\le s < t;\\
\infty, & s<t-\alpha^{-3}.
\end{cases}
\end{equation}
The following lemma follows immediately from Lemma~\ref{lem:bound on rate} and the fact that $S(t) \ge c t^{-\frac{1}{2}}$ for all $t>0$ almost surely.
\begin{lem}\label{lem:aggregate is bounded}
Let $\alpha >0$ sufficiently small.
\begin{enumerate}
\item
For all $t \ge \alpha ^{-20}$
\begin{equation}
\mathbb P \left( \forall s>0, \ Y_{t}(s) \le \overline{Y} _{t,\alpha }(s) \right) \ge 1-\alpha
\end{equation}
\item
For all $t \le 100 \alpha ^{-2}$
\begin{equation}
\mathbb P \left( \forall s>0, \ Y_{t}(s) \ge \underline{Y} _{t,\alpha }(s) \right)=1
\end{equation}
\end{enumerate}
\end{lem}
In the next sections we'll show that the processes with initial conditions $\overline{Y} _{t,\alpha }$ and $\underline{Y} _{t,\alpha }$ satisfy some regularity properties which will allow us to apply the inductive argument on them. We then conclude that the same results hold for the usual aggregate as these processes sandwich the aggregate by Lemma~\ref{lem:aggregate is bounded}
\section{Fixed rate process and its perturbation}\label{sec:fixedrate}
Recall the first- and second-order expansions of speed \eqref{eq:speed:1storder main}, \eqref{eq:speed:2ndorder main} and the definitions \eqref{eq:def:J:agg}, \eqref{eq:def:Qt:agg}. In this section, our objective is to understand the multiple integral
\begin{equation}
\intop_{0}^t \intop_{0}^s
{\textnormal{\textbf{J}}}_{t-s,t-u;t}^{\Pi_g} \ d\widehat{\Pi}_g(u) d\widehat{\Pi}_g(s),
\end{equation}
both when $g$ is a fixed rate $g\equiv \alpha$ and when it is slightly perturbed, and also the corresponding analogue for $\mathcal{Q}_t$. Consequently, we further derive estimates on the error of the first- and second-order approximations of the speed \eqref{eq:def:S1:basic form}, \eqref{eq:def:S2:basic form}. Throughout this section, $g(s)>0$ denotes a random function which represents the rate of the Poisson process at time $s$.
To begin with, we introduce some notations as follows. As before, $\Pi$ denotes the Poisson point process on the plane with the standard Lebesgue intensity, and let $\Pi_g$ be the collection of the $x$-coordinate locations of points lying below $g$, that is,
\begin{equation}
\Pi_g[0,t] = \{x \in[0,t] : \ (x,y)\in \Pi \textnormal{ for some } y \in[0, g(x)] \},
\end{equation}
we define $\pi_i(t;g)$ to be the distance in $x$-axis from $t$ to the $i$-th closest point to $t$ in $\Pi_g[0,t]$. For instance,
\begin{equation}\label{eq:def:pi closest points}
\pi_1(t;g) =\pi_1(t;\Pi_g) = t- \max\{x: x\in \Pi_g[0,t] \}.
\end{equation}
Moreover, for convenience, we denote
\begin{equation}\label{eq:def:sig closest points}
\sigma_i(t;g):= (\pi_i(t;g)+1)^{-\frac{1}{2}},
\end{equation}
and abbreviate the products among them by
\begin{equation}
\sigma_1\sigma_2(t;g):= \sigma_1(t;g)\sigma_2(t;g), \quad \textnormal{and} \quad \sigma_1\sigma_2\sigma_3(t;g):= \sigma_1(t;g)\sigma_2(t;g)\sigma_3(t;g).
\end{equation}
Further, $\alpha,C>0$ denote small enough and large enough constants, respectively, and we set
\begin{equation}\label{eq:def:horizon}
\hat{h} = \hat{h}(\alpha,C):= \alpha^{-2} \log^C(1/\alpha).
\end{equation}
Then, the goal of this section is to establish the following statements.
\begin{prop}\label{prop:fixed perturbed:double int}
Let $\epsilon>0$ be arbitrary, $\alpha>0$ be a sufficiently small constant depending on $\epsilon,C
$, and recall the definition of ${\textnormal{\textbf{J}}}$ \eqref{eq:def:J:agg}. Let $\tau$ be a stopping time, and $\{g(s)\}_{s \ge 0}$ be a positive stochastic process progressively measurable with respect to $\Pi_g$, and suppose that it satisfies the following three conditions almost surely:
\begin{equation}\label{eq:fixed perturbed:assumption}
\begin{split}
\intop_0^{\hat{h}\wedge \tau} (g(s)-\alpha)^2 ds \le \alpha^{1-\frac{\epsilon}{400}},\quad
\intop_0^{\hat{h}\wedge \tau} (g(s)-\alpha)^2 g(s)ds \le \alpha^{2-\frac{\epsilon}{400}},\quad
\sup_{s\le \hat{h}\wedge \tau}g(s) \le \alpha^{1-\frac{\epsilon}{400}}.
\end{split}
\end{equation}
Furthermore, denoting $d\widehat{\Pi}_g(s) = d\Pi_g(s) - \alpha ds$ as before, we define for $t'\le t$ that
\begin{equation}\label{eq:def:Jint:general}
\mathcal{J}[t;\Pi_g]:= \intop_{0}^{t} \intop_{0}^s
{\textnormal{\textbf{J}}}_{t-s,t-u;t}^{\Pi_g} \ d\widehat{\Pi}_g(u) d\widehat{\Pi}_g(s).
\end{equation}
Then, there exist $c_\epsilon,\alpha_0(\epsilon,C)>0$ such that for any $\alpha\in (0,\alpha_0)$, we have
\begin{equation}
\mathbb{P} \left(-\alpha^{\frac{1}{2}-\epsilon}\sigma_1(t;g)\le \mathcal{J}[t;\Pi_g] \le \alpha^{-\epsilon}\sigma_1\sigma_2(t;g),\ \forall t\le \hat{h}(\alpha,C)\wedge \tau \right)
\ge 1-
\exp\left(-\alpha^{-c_\epsilon} \right).
\end{equation}
\end{prop}
\begin{prop}\label{prop:fixed perturbed:triple int}
Recall the definition of $Q$ \eqref{eq:def:Q:agg}, and define $\mathcal{Q}[t;\Pi_g]$ as
\begin{equation}\label{eq:def:Qint:general}
\mathcal{Q}[t;\Pi_g] := \intop_{0}^{t} \intop_{0}^s \intop_0^u
{\textnormal{\textbf{Q}}}_{t-s,t-u,t-v;t}^{\Pi_g} \ d\widehat{\Pi}_g(v) d\widehat{\Pi}_g(u) d\widehat{\Pi}_g(s).
\end{equation}
Then, under the same setting as Proposition \ref{prop:fixed perturbed:double int}, we have
\begin{equation}
\mathbb{P} \left(\left| \mathcal{Q}[t;\Pi_g] \right| \le \alpha^{-\epsilon}\sigma_1\sigma_2\sigma_3(t;g), \ \forall t \le \hat{h}(\alpha,C)\wedge \tau \right) \ge 1- \exp\left(-\alpha^{-c_\epsilon}\right).
\end{equation}
\end{prop}
Based on these results, we discuss the error estimates for the first- and second-order approximations. Formal statements and details will be given in Section \ref{subsec:fixed:error}.
\begin{remark}
The assumption \eqref{eq:fixed perturbed:assumption} will be revisited in Section \ref{sec:reg:intro} when we discuss regularity. In fact, it will be shown that the speed $S(t)$ of the aggregate actually satisfies \eqref{eq:fixed perturbed:assumption} with high probability.
\end{remark}
\begin{remark}
In Proposition \ref{prop:fixed perturbed:double int}, observe that the lower bound of $\mathcal{J}[t;\Pi_g]$ is stronger than the upper bound. This comes from the nature of $\textbf{J}$ who can be (positively) larger compared to the (negative) lower bound, which will be clear in Section \ref{subsec:fixed:JandQ}. Having a stronger lower bound on $\mathcal{J}$ will play an important role in the discussion of regularity, in Sections \ref{subsec:reg:conseq} and \ref{sec:double int}.
\end{remark}
As mentioned in Section \ref{subsubsec:outline:fixedrate}, the proofs consist of four major steps which are discussed one by one in detail in the following subsections:
\begin{itemize}
\item The Azuma-type martingale concentration lemmas: Section \ref{subsec:fixed:mgconcen}.
\item Controlling the sizes of ${\textnormal{\textbf{J}}}_{u,s}$ and ${\textnormal{\textbf{Q}}}_{v,u,s}$: Section \ref{subsec:fixed:JandQ}.
\item Proving the propositions with respect to the fixed rate process $\Pi_\alpha$: Section \ref{subsec:fixed:fixed}.
\item Converting the results into a general form: Section \ref{subsec:fixed:perturbed}.
\item The error estimate on the first- and second-order approximations of the speed: Section \ref{subsec:fixed:error}.
\end{itemize}
\subsection{The martingale concentration lemmas}\label{subsec:fixed:mgconcen}
We begin with developing lemmas on martingale concentration used in the proof of Propositions \ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int}. Moreover, the lemmas we announce here will appear again frequently in Sections \ref{sec:criticalbranching}--\ref{sec:reg:next step}.
For the Poisson point process $\Pi$ with standard Lebesgue intensity, let $\mathcal F _t$ denote the $\sigma$-algebra generated by $\Pi$ up to time $t$ (including $t$). For any positive stochastic process $g(t)$ which is predictable with respect to $\mathcal F _t$, we write $\Pi _g$ to denote the points in $\Pi[0,t]$ below the function $g$ as before. In particular, for a constant $\alpha>0$, $\Pi _\alpha $ refers to the rate-$\alpha$ Poisson process. We note that for any such function $g$ the process $\widetilde{ \Pi}_ g(t):= |\Pi _g[0,t]| -\int_0^t g(s)ds $ is a martingale.
We begin with the simplest form of concentration lemma for martingales. The statement includes a stopping time which is introduced for a later purpose.
\begin{lem}\label{lem:concentration of integral}
Let $a,M,\lambda > 0$, let $g(t)>0$ and $f(t)$ be stochastic processes predictable with respect to $\mathcal F _t $, and let $\tau$ be a stopping time. Suppose that
\begin{equation}\label{eq:concentration:conditions:basic}
\lambda |f(x\wedge \tau )|\le 1, \quad \text{and} \quad \intop _{0}^{t \wedge \tau }f(x)^2g(x)dx \le M, \quad a.s.
\end{equation}
Then
\begin{equation}
\mathbb P \left(\sup _{s \le t\wedge \tau } \left| \intop _0^s f(x) d \widetilde{\Pi }_g(x) \right|\ge a\sqrt{M} \right) \le Ce^{\lambda ^2 M-a\lambda \sqrt{M}}.
\end{equation}
\end{lem}
\begin{proof}
For $0 \le s \le t $ let
\begin{equation}
\begin{split}
M_s&:=\exp \left( \lambda \intop _0^s f(x) d \Pi _g (x)-\intop _0^s \left(e^{\lambda f(x)}-1\right) g(x)dx \right) \\
L_s &:=\exp \left( \intop _0^s \left(e^{\lambda f(x)}-1\right) g(x)- \lambda f(x)g(x) dx \right).
\end{split}
\end{equation}
Note that $M_s$ is a martingale and that, using the inequality $e^x\le 1+x+x^2$ for $|x|\le 1$ we get
\begin{equation}
L_{s\wedge \tau } \le \exp \left( \lambda ^2 \intop _0^{s\wedge \tau } f(x)^2 g(x)dx\right) \le \exp \left( \lambda ^2 \intop _0^{t\wedge \tau } f(x)^2 g(x)dx\right) \le e^{\lambda ^2 M}.
\end{equation}
Define the stopping time
\begin{equation}
\tau _1:= \inf \left\{ s>0\ : \ \intop _0^s f(x) d \widetilde{\Pi }_g(x) \ge a\sqrt{M} \right\},\quad \tau _2:= \inf \left\{ s>0\ : \ \intop _0^s f(x) d \widetilde{\Pi }_g(x) \le - a\sqrt{M} \right\}
\end{equation}
and let $\tau ' =\tau \wedge \tau _1$. We have
\begin{equation}
e^{\lambda a\sqrt{M}}\cdot \mathbb P \left( \tau _1\le \tau \right) \le \mathbb E \left[ \lambda \exp \left( \intop _0^{t \wedge \tau '} f(x) d \widetilde{\Pi }_g(x) \right) \right] =\mathbb E \left[ M_{t\wedge \tau '}\cdot L_{t\wedge \tau '} \right] \le e^{\lambda ^2 M} \mathbb E [M_{t\wedge \tau '}]= e^{\lambda ^2 M},
\end{equation}
which gives the desired bound for $\tau _1$. In order to get the same bound for $\tau _2$ we swich $f$ by $-f$.
\end{proof}
In particular, taking $\lambda =1/ \sqrt{M}$ in Lemma \ref{lem:concentration of integral} gives the following corollary.
\begin{cor}\label{cor:concentration of integral}
Let $a,M > 0$, let $g(t)>0$ and $f(t)$ be predictable processes with respect to $\mathcal F _t $ and let $\tau$ be a stopping time. Suppose that
\begin{equation}
|f(x\wedge \tau )|\le \sqrt{M} \quad \text{and} \quad \intop _{0}^{t \wedge \tau }f(x)^2g(x)dx \le M, \quad a.s.
\end{equation}
Then
\begin{equation}
\mathbb P \left(\sup _{s \le t\wedge \tau } \left| \intop _0^s f(x) d \widetilde{\Pi }_g(x) \right|\ge a\sqrt{M} \right) \le Ce^{ -a}.
\end{equation}
\end{cor}
The following corollary deals with the case $f\equiv 1$, which is useful when estimating the size of the Poisson process itself.
\begin{cor}\label{cor:concentration:numberofpts each interval}
Let $a,h,\Delta > 0$, $M\geq 1$, let $g(t)>0$ a process with respect to $\mathcal F _t $ and let $\tau$ be a stopping time. Suppose that
\begin{equation}
\intop _{(t-\Delta)\wedge \tau}^{t \wedge \tau }g(x)dx \le M, \quad a.s.,
\end{equation}
and define the stopping time
\begin{equation}
\tau'=\inf\{t>0: |\Pi_g[(t-\Delta)\vee 0, t]|\geq 2M+2a\sqrt{M}+2 \}\wedge h.
\end{equation}
Then, we have $\mathbb{P}(\tau'\le \tau)\le \left(\frac{h}{\Delta}\right)e^{-a} $.
\end{cor}
\begin{proof}
For each $t\in [0,h],$ we apply Corollary \ref{cor:concentration of integral} to the quantity
\begin{equation}
P(t):= |\Pi_g[t\wedge\tau-\Delta, t\wedge\tau]|,
\end{equation}
and obtain that
\begin{equation}
\mathbb{P} \left(P(t)\geq M+a\sqrt{M} \right) \le e^{-a}.
\end{equation}
Then, we take a union bound over all $t\in[0,h]$ of the form $t=k \Delta $, $k\in \mathbb{Z}$:
\begin{equation}
\mathbb{P} \left(P(\Delta)\leq M+a\sqrt{M}, \ \ \forall k\Delta \in[0,h], \ k\in\mathbb{Z} \right) \ge 1-\left(\frac{h}{\Delta} \right)e^{-a}.
\end{equation}
Under the event described inside the above probability, the intervals $[t\wedge\tau-\Delta, t\wedge\tau]$ should contain at most $2M+2a\sqrt{M}$ points for all $t\in[0,h]$, implying that $\tau'> \tau$.
\end{proof}
Unfortunately, Lemma \ref{lem:concentration of integral} often turns out to be insufficient due to several reasons, and we require more involved versions of it.
To motivate the formulation of the more complicated lemmas below, we briefly explain the setting. Suppose that $f_t(x)$, $g(x)$ are predictable processes with respect to $\mathcal{F}_x$, where $f_t$ is defined for each $t\in[0,h]$. We want to control the size of
$$\intop_{ \tau_-}^{t\wedge \tau} f_t(x)d\widetilde{\Pi}_g(x), $$
where $\tau_-,\tau\ge 0$ are two given stopping times. Our desired estimate should take account of the following traits, which causes additional complication compared to the previous lemma.
\begin{itemize}
\item We want our bound to hold for all $t\in[0,h]$, which would require a union bound followed by a continuity argument.
\item We need to deal with the cases where $|f_t(x)|$ is not bounded uniformly in $x$ as in the first condition of \eqref{eq:concentration:conditions:basic}.
\item The integral starts from $\tau_-$, which potentially causes the second bound of \eqref{eq:concentration:conditions:basic} to be $\tau_-$-measurable (which is random).
\end{itemize}
Having these aspects in mind, the device we need is stated as follows. Due to its complicatedness, we suggest the reader to skip this lemma for the moment and come back later when it is actually applied (e.g., in Section \ref{subsec:fixed:fixed}).
\begin{lem}\label{lem:concen of int:conti:forwardtime}
Let $h, \Delta, D>0$ be given constants, and let $g(x)>0$ and $f_{t}(x)$ be stochastic processes predictable with respect to $\mathcal{F}_x$, where $f_{t}:[0,t] \to \mathbb{R}$ is defined for each $t\in [0, h]$. Suppose that there exist stopping times $\tau_-, \tau$, and random variables ${\bf M},{\bf A}>0$ that satisfy the following conditions:
\begin{itemize}
\item ${\bf M},{\bf A}$ are $\tau_-$-measurable.
\item $f_t$ and $g$ satisfy
\begin{equation}\label{eq:concen:condition:forwardtime}
|{f}_t(x)| \le \sqrt{{\bf M}} \ \forall x \ge \Delta, \ \ \intop_0^\Delta |{f}_t(x)|g(x) dx \le {\bf A}, \ \ \intop_0^{t\wedge \tau} {f}_t(x)^2 g(x)dx \le {\bf M}, \ \ a.s. \ \forall t\in[0,h].
\end{equation}
\item With probability one, we have $|\partial_t f_t(x)| \vee |\partial_x f_t(x)| \le D$ for all $t\in[0,h]$, $x\le t$.
\item We define $\bar{f}_t(x)\ge 0$ and $\underline{f}_t(x) \le 0$ to be random functions that are decreasing and increasing, respectively, and satisfy the following conditions:
\begin{itemize}
\item For all $x\le t$, $\bar{f}_t(x)$ and $\underline{f}_t(x)$ are $\tau_-$-measurable.
\item For all $x\le t \le \tau$, $\bar{f}_t(x) \ge \sup_{y\ge x} f_t(y)$ and $\underline{f}_t(x)\le \inf_{y\ge x} f_t(y)$ a.s..
\end{itemize}
\end{itemize}
Furthermore, for given constants $N,\eta>0$, define stopping times
\begin{equation}\label{eq:def:tau:concenofint:conti}
\begin{split}
\tau' &:= \inf\{t>\Delta : |\Pi_g[t-\Delta,t]| \ge N \},\\
\tau'' &:= \inf \{t>0: |g(t)| \ge \eta \},\\
\tau_0&:= \tau \wedge \tau' \wedge \tau'',
\end{split}
\end{equation}
and let $\delta>0$ be a constant satisfying a.s.~that
\begin{equation}\label{eq:concen:condition:delta:forward}
\delta \le \frac{{\bf A}}{D(2N\Delta+TN+T\eta \Delta)}.
\end{equation}
Then, denoting the closest point to $0$ in $\Pi_g[0,t]$ by $p_1$, we have for all $a>0$ that
\begin{equation}
\begin{split}
\mathbb{P} \left( \sup_{0\le s \le t\wedge \tau} \intop_{\tau_-}^s {f}_t(x) d\widetilde{\Pi}_g(x)
\le 2N \bar{f}_t(p_1) + 3{\bf A} + a\sqrt{{\bf M}}, \ \forall t\in[0,h]
\right)
\ge
1- \left(\frac{Ch}{\delta\Delta} \right)e^{-a};\\
\mathbb{P} \left( \inf_{0\le s \le t\wedge \tau} \intop_{\tau_-}^s {f}_t(x) d\widetilde{\Pi}_g(x)
\ge 2N \underline{f}_t(p_1) - 3{\bf A} - a\sqrt{{\bf M}}, \ \forall t\in[0,h]
\right)
\ge
1- \left(\frac{Ch}{\delta\Delta} \right)e^{-a}.
\end{split}
\end{equation}
with an absolute constant $C>0$.
In the integral, we regard $\intop_a^b f = 0$ if $a\ge b$.
\end{lem}
Its proof is based on a union bound applied to Lemma \ref{lem:concentration of integral} to cover a discretized points in the interval $[0,h]$ and a continuity argument to cover the whole interval. Distinction of upper and lower bounds on the quantity is necessary to deduce Proposition \ref{prop:fixed perturbed:double int}. Due to its technicality, the proof is deferred to Appendix \ref{subsec:app:mgconcen}.
The following corollary will not be used in Section \ref{sec:fixedrate}, but they will be applied frequently later in Sections \ref{sec:criticalbranching}--\ref{sec:reg:next step}. It is an analogue of Lemma \ref{lem:concen of int:conti:forwardtime}, but with \textit{deterministic} $f_t(x)$ which is \textit{increasing} in $x$, and also simpler due to the absence of $\tau_-$, allowing ${\bf M}, {\bf A}$ above to be deterministic.
\begin{cor}\label{lem:concentrationofint:continuity}
Let $h,M,\Delta,D,A>0$ be given, let $g(x)>0$ be a predictable process with respect to $\mathcal F _x $ and let $\tau$ be a stopping time. For each $t\in[0,h]$, Let $f_t:(-\infty,t]\to \mathbb{R}_{\ge 0}$ be a deterministic, increasing function that satisfies
\begin{equation}
|f_t(x)|\leq \sqrt{M} \ \ \forall x\leq t-\Delta,\quad \intop_{t- \Delta}^t |f_t(x)|g(x)dx \leq A,\quad |\partial_t f_t(x)| \vee |\partial_x f_t(x)| \leq D \ \ \forall 0\le t\leq h, \ x<t.
\end{equation}
Suppose that for each $t\in[0,h]$,
\begin{equation}
\intop _{0}^{ t\wedge \tau }f_t(x)^2g(x)dx \le M, \quad a.s.
\end{equation}
Furthermore, for given constants $N,\eta>0$, let the stopping times $\tau',\tau'',\tau_0$ and the constant $\delta>0$ be as \eqref{eq:def:tau:concenofint:conti}, \eqref{eq:concen:condition:delta:forward}.
Then, denoting the closest point to $t$ in $\Pi_g[0,t]$ by $p_1(t)$, we have
\begin{equation}
\begin{split}
\mathbb P \left( \intop _0^{t\wedge \tau_0} f_t(x) d \widetilde{\Pi }_g(x) \le 2Nf_t(p_1(t)) +3A+ a\sqrt{M}, \ \forall t\in[0,h] \right) \ge 1- \left(\frac{Ch}{\delta \Delta} \right)e^{-a};\\
\mathbb P \left( \intop _0^{t\wedge \tau_0} f_t(x) d \widetilde{\Pi }_g(x) \ge-3A- a\sqrt{M}, \ \forall t\in[0,h] \right) \ge 1- \left(\frac{Ch}{\delta \Delta} \right)e^{-a}.
\end{split}
\end{equation}
\end{cor}
Note that the absence of the term $Nf_t(p_1(t))$ in the lower bound comes from the assumption that $f_t$ is nonnegative. It turns out that this can be proven analogously as Lemma \ref{lem:concen of int:conti:forwardtime}, and we omit the details (see Appendix \ref{subsec:app:mgconcen}).
\subsection{Estimating ${\textnormal{\textbf{J}}}_{u,s}$ and ${\textnormal{\textbf{Q}}}_{v,u,s}$}\label{subsec:fixed:JandQ}
Recall the definitions of ${\textnormal{\textbf{J}}}_{u,s}$ \eqref{eq:def:J:basic form} and ${\textnormal{\textbf{Q}}}_{v,u,s}$ \eqref{eq:def:Q:basic form}. W study its quenched version ${\textnormal{\textbf{J}}}_{u,s}$ and show that it satisfies a similar but weaker bound than $J_{u,s}^{(\alpha)} $ from Section \ref{subsec:Kestim:intro}. Moreover, we derive an analogous result for ${\textnormal{\textbf{Q}}}_{v,u,s}$.
The goal is to establish the following estimates on ${\textnormal{\textbf{J}}}_{u,s}$ and ${\textnormal{\textbf{Q}}}_{v,u,s}$.
Throughout this section we fix $C_0>0$ and let $\alpha >0$ sufficiently small depending on $\epsilon $ and $C_0$. We also let $\hat{h}:= \alpha ^{-2} \log ^{C_0}(1/ \alpha )$.
\begin{prop}\label{prop:Jbound:quenched}
For all $\epsilon >0$ there exists $c_\epsilon >0$ such that
\begin{equation}
\mathbb{P}_\alpha \left( -\frac{\alpha ^{1-\epsilon }}{\sqrt{s+1}} \le {\textnormal{\textbf{J}}}_{u,s} \le \frac{\alpha^{-\epsilon}}{\sqrt{(u+1)(s+1)}}, \ \forall 0\le u<s\le \hat{h} \right) \ge 1- \exp\left(-\alpha^{-c_\epsilon } \right).
\end{equation}
\end{prop}
\begin{prop}\label{prop:Qbound:quenched}
For all $\epsilon >0$ there exists $c_\epsilon >0$ such that
\begin{equation}
\mathbb{P}_\alpha \left( |{\textnormal{\textbf{Q}}}_{v,u,s}| \le \frac{\alpha^{-\epsilon}}{\sqrt{(v+1)(u+1)(s+1)}}, \ \forall 0\le s\le \hat{h} \right) \ge 1- \exp\left(-\alpha^{-c_\epsilon } \right).
\end{equation}
\end{prop}
Proposition \ref{prop:Jbound:quenched} can be established by a coupling argument of the events $\{ s<T_u<\infty \}$ and $\{s< T <\infty \}$, and using hitting time estimates of random walks. Proof of Proposition \ref{prop:Qbound:quenched} is based on the same idea, but the coupling for the events defining ${\textnormal{\textbf{Q}}}_{v,u,s}$ is more technical and we defer the proof to Appendix \ref{subsec:app:Qbound}.
As the first step in verifying Proposition \ref{prop:Jbound:quenched}, we address the following elementary lemma on simple random walk.
\begin{lem}\label{lem:basic random walk estimates}
Let $W_t$ be a continuous time random walk with $W_0=0$. Define the maximum process $M_t:= \max _{s \le t} W_s$. For any $t >0$ we have:
\begin{enumerate}
\item
for any $A\ge 1$
\begin{equation}
\mathbb P (M_t \le A) \le C \frac{A}{\sqrt{t+1}}
\end{equation}
\item
For any integer $k \ge 0 $
\begin{equation}
\mathbb P \left( M_t \le 0, \ W_t=-k \right)\le \frac{C}{t+1}
\end{equation}
\end{enumerate}
\end{lem}
\begin{proof}
By the reflection principle we have
\begin{equation}
\mathbb P (M_t \le A) \le \mathbb P (|W_t |\le A) \le C\frac{A} {\sqrt{t+1}}
\end{equation}
where in the last inequality we used that $\mathbb P (W_t =k) \le C/ \sqrt{t+1}$ for any $k \in \mathbb Z$ which follows from the local central limit theorem.
The second part of the claim also follows from the reflection principle.
\end{proof}
Based on the above observation, we deduce the following estimate on $T$ defined in \eqref{eq:def:T:basic form}.
\begin{lem}\label{lem:bound on H}
For all $\epsilon >0$ there exists $\delta > 0$ such that on the event
\begin{equation}\label{eq:def of C delta}
\mathcal C =\mathcal C _\delta :=\{ \forall x>0, \ Y_x \le \alpha ^{-\delta } +2 \alpha x \}
\end{equation}
we have for all $s\le \hat{h}$
\begin{equation}\label{eq:bound on C}
\mathbb{P}(T \ge s \: | \: Y) \le \frac{\alpha^{-\epsilon}}{\sqrt{s+1}}.
\end{equation}
In particular there exist $c_\epsilon >0$ such that
\begin{equation}
\mathbb{P}_\alpha \left[ \mathbb{P}(T \ge s \: | \: Y) \le \frac{\alpha^{-\epsilon}}{\sqrt{s+1}}, \ \forall 0\le s \le \hat{h} \right] \ge 1- \exp\left(-\alpha^{-c_\epsilon } \right).
\end{equation}
\end{lem}
Note that the outer probability $\mathbb{P}_\alpha$ considers the probability over the rate-$\alpha$ Poisson process $Y$, and the inner probability $\mathbb{P}$ is taken over the simple random walk conditioned on $Y$.
\begin{proof}
Let $\delta >0$ sufficiently small depending on $\epsilon $, and $\alpha >0$ sufficiently small depending on all other parameters. It is easy to check that $\mathbb P (\mathcal C ) \ge 1- \exp (-\alpha ^{-c_\delta })$ and therefore the second statement in the lemma follows from the first one.
On $\mathcal C$ we have
\begin{equation}
\mathbb P \left( T \ge s \ \big| \ Y \right) = \mathbb P \left(\forall x \le s, \ W_x \le Y_x \ \big| \ Y \right)\le \mathbb P \left(\forall x \le s, \ W_x \le \alpha ^{-\delta } +2 \alpha x\right)
\end{equation}
Denote the last probability by $p_s$. If $s \le \alpha ^{-1-\delta }$ by the first part of Lemma~\ref{lem:basic random walk estimates} we have
\begin{equation}
p_s \le \mathbb P \left(M_s \le 3 \alpha ^{-\delta } \right) \le C \frac{\alpha ^{-\delta }}{\sqrt{s+1}}.
\end{equation}
If $\alpha ^{-1-\delta } \le s \le \alpha ^{-\frac{3}{2}- \delta }$ we have
\begin{equation}
\begin{split}
p_s &\le \mathbb P \left( M_{\alpha ^{-1}} \le 2 \alpha ^{-\delta }, \ M_s \le 2 \alpha ^{-\frac{1}{2}- \delta } \right) \\
&\le \mathbb P \left( M_{\alpha ^{-1}}\le 2 \alpha ^{-\delta } , \ W_{\alpha ^{-1}} \ge -\alpha ^{-\frac{1}{2}-\delta } , \ \max _{\alpha ^{-1} \le x \le s } W_x - W_{\alpha ^{-1}} \le 3 \alpha ^{-\frac{1}{2}-\delta }, \right)+\mathbb P \left( W_{\alpha ^{-1}}\le -\alpha ^{-\frac{1}{2}-\delta } \right) \\
&\le \mathbb P \left(M_{\alpha ^{-1}} \le 2 \alpha ^{-\delta } \right) \mathbb P \left( M _{s - \alpha ^{-1 }} \le 3 \alpha ^{-\frac{1}{2}- \delta }\right) +C e^{-\alpha ^{-\delta }}\le C\frac{\alpha ^{-2 \delta }}{\sqrt{s+1}}
\end{split}
\end{equation}
if $\alpha ^{-\frac{3}{2}- \delta } \le s \le \alpha ^{-\frac{7}{4}- \delta }$ by the same arguments we have $p_s \le C _{\delta }\alpha ^{-3 \delta } / \sqrt{s+1}$. We can repeat this process $C\log (1/\delta )$ times to obtain the bound $p_s \le C _{\delta }\alpha ^{-C\delta \log (1/ \delta ) } / \sqrt{s+1}$ for any $s \le \alpha ^{-2-\frac{\delta }{2} } $. Taking $\delta >0$ sufficiently small depending on $\epsilon $ we get $p_s \le \alpha ^{-\epsilon }/ \sqrt{s+1} $ for any $s\le \alpha ^{-2 -\frac{\delta }{2}}$.
\end{proof}
Let $f_Y$ be the probability density of $T $ given $Y$. That is
\begin{equation}
f_Y(s):= \frac{d}{ds} \mathbb P (T \le s \ | \ Y).
\end{equation}
In the following lemma we give a bound on $f$ that folds with high probability.
\begin{lem}\label{lem:bound on density}
. For all $\epsilon >0$ there exists $c_\epsilon >0$ such that
\begin{equation}\label{eq:the equation in the lemma on the density}
\mathbb{P}_\alpha\left[ f_Y(s)\le \frac{\alpha^{-\epsilon}}{(s+1)^{\frac{3}{2}}}, \ \forall 0\le s\le \hat{h} \right] \ge 1- \exp\left(-\alpha^{-c_\epsilon } \right).
\end{equation}
\end{lem}
\begin{proof}
Let $\delta >0$ sufficiently small such that Lemma~\ref{lem:bound on H} holds and recall the definition of $\mathcal C= \mathcal C _\delta $ in \eqref{eq:def of C delta}. It is clear that $f_Y$ is given by
\begin{equation}
f_Y(s)=\frac{1}{2} \cdot \mathbb P \left( \forall x \le s , \ W_x \le Y_x, \ W_s=Y_s \ | \ Y \right).
\end{equation}
Thus, on $\mathcal C$ we have
\begin{equation}
\begin{split}
f_Y(s)&=\frac{1}{2} \sum _{k \le Y(s/2)}\mathbb P \left( \forall x \le s , \ W_x \le Y_x, \ W ( s/2 ) =k, \ W_s=Y_s \ | \ Y \right)\\
&\le \sum _{k \le Y(s/2)}\mathbb P \left( \forall x \le s/2, \ W_x \le Y_x \ | \ Y \right) \mathbb P \left( W(s/2) =k \ \big| \ \forall x \le s/2, \ W_x \le Y_x , \ Y \right) \\
&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cdot \mathbb P \left( \forall s/2 \le x \le s, \ W_x \le Y_s, \ W_s=Y_s \ \big| \ W(s/2 )=k , \ Y \right)\\
&\le \frac{C \alpha ^{-\epsilon }}{\sqrt{s+1}}\sum _{k \le Y(s/2)}\mathbb P \left( W(s/2) =k \ \big| \ \forall x \le s/2, \ W_x \le Y_x , \ Y \right) \\
&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cdot \mathbb P \left( \forall x\le s/ 2, \ W_x \le Y_s, \ W(s/2 )=k \big| \ W_0=Y_s , \ Y \right)\\
&= \frac{C \alpha ^{-\epsilon }}{\sqrt{s+1}}\sum _{k \le Y(s/2)}\mathbb P \left( W(s/2) =k \ \big| \ \forall x \le s/2, \ W_x \le Y_x , \ Y \right) \\
&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \cdot \mathbb P \left( M_{\frac{s}{2}}=0, \ W(s/2 )=k-Y_s \ \big| \ W_0=0 , Y \right) \\
&\le \frac{C \alpha ^{-\epsilon }}{(s+1)^{\frac{3}{2}}}\sum _{k \le Y(s/2)}\mathbb P \left( W(s/2) =k \ \big| \ \forall x \le s/2, \ W_x \le Y_x \right) \le \frac{C \alpha ^{-\epsilon }}{(s+1)^{\frac{3}{2}}}\le \frac{ \alpha ^{-2\epsilon }}{(s+1)^{\frac{3}{2}}},
\end{split}
\end{equation}
where in the second inequality we used Lemma~\ref{lem:bound on H} in order to bound the first factor and we shifted and inverted time in the third factor. In the third inequality we used Lemma~\ref{lem:basic random walk estimates}. The last bound finishes the proof as $\epsilon $ is arbitrary.
\end{proof}
Let $f_u$ be the density of $T$ conditioned on $\mathcal F _u :=\sigma (Y_{\le u })=\sigma (Y_x , \ x\le u )$. That is
\begin{equation}
f_u(s):=\mathbb E [f_Y(s) \ | \ \mathcal F _u ]= \frac{d}{ds} \mathbb P (T \le s \ | \ \mathcal F _u ).
\end{equation}
In the following corollary we show, using Lemma~\ref{lem:bound on density} and Doob's martingale inequality that a bound similar to \eqref{eq:the equation in the lemma on the density} holds for $f_u$ as well.
\begin{cor}\label{cor:conditioning on less information}
For all $\epsilon >0$ there exists $c_\epsilon >0$ such that
\begin{equation}
\mathbb{P}_\alpha\left[ f_u(s)\le \frac{\alpha^{-\epsilon}}{(s+1)^{\frac{3}{2}}}, \ \forall 0\le u, s\le \hat{h} \right] \ge 1- \exp\left(-\alpha^{-c_\epsilon } \right).
\end{equation}
\end{cor}
\begin{proof}
Let $\mathcal D $ be the event inside the probability in \eqref{eq:the equation in the lemma on the density}. $\mathbb P (\mathcal D ^c \ | \ \mathcal F _u )$ is a martingale and therefore by Doob's martingale inequality we have
\begin{equation}
\mathbb P \Big( \sup _{ u\le \hat{h} } \mathbb P (\mathcal D ^c \ | \ \mathcal F _u ) \ge \alpha ^3 \Big) \le \alpha ^{-3 } \mathbb P (\mathcal D ^c) \le \exp (-\alpha ^{- c_\epsilon }).
\end{equation}
Next we claim that $f_Y(s)\le 1$. Indeed, on the event $\{ s < T \le s+\delta \}$ we have that $\{ W_{s+\delta }\neq W_s \}$ which happens with probability $1-e^{-\delta }\le \delta $ and therefore $f_Y(s) =\frac{d}{ds} \mathbb P (T\le s \ | \ Y) \le 1$. Thus, on the event $\{\sup _{ u\le s } \mathbb P (\mathcal D ^c \ | \ \mathcal F _u ) \le \alpha ^3 \}$ we have for all $u,s\le \hat{h}$
\begin{equation}
\begin{split}
f_u(s) = \mathbb E [f_Y (s) \ | \ \mathcal F _u ] &= \mathbb E [ f_Y (s) \cdot \mathds 1 _{\mathcal D } \ | \ \mathcal F _u ] +\mathbb E [f_Y (s) \cdot \mathds 1 _{\mathcal D^c } \ | \ \mathcal F _u ] \\
&\le \frac{\alpha ^{-\epsilon}}{\sqrt{s+1}} +\mathbb P( \mathcal D ^c \ | \ \mathcal F _u)\le \frac{\alpha ^{-\epsilon}}{\sqrt{s+1}} +\alpha ^ 3 \le \frac{\alpha ^{-2\epsilon}}{\sqrt{s+1}},
\end{split}
\end{equation}
where in the first inequality we used the definition of $\mathcal D $ and the fact that $f_Y (s) \le 1 $. This finishes the proof as $\epsilon $ is arbitrary.
\end{proof}
Now we are ready to conclude the proof of Proposition \ref{prop:Jbound:quenched}.
\begin{proof}[Proof of Proposition~\ref{prop:Jbound:quenched}]
Let $u<s<\hat{h}$. We have
\begin{equation}\label{eq:coupeling form of J}
{\textnormal{\textbf{J}}}_{u,s}= \mathbb{P}_\alpha (s<T_u<\infty \ | \ Y_{\le u}) - \mathbb{P}_\alpha(s<T<\infty \ | \ Y_{\le u}) = \mathbb P \left( \mathcal A _{u,s} \ | \ Y_{\le u} \right)-\mathbb P \left( \mathcal B _s \ | \ Y_{\le u} \right)
\end{equation}
where
\begin{equation}
\mathcal A _{u,s}:=\left\{ u \le T <s,\ s\le T _u <\infty \right\},\quad \mathcal B _s :=\left\{s \le T <\infty ,\ T _u =\infty \right\}.
\end{equation}
Let $\delta >0$ sufficiently small such that Lemma~\ref{lem:bound on H} holds and let $\mathcal D $ be the event inside the probability in Lemma~\ref{lem:bound on density}. Define
\begin{equation}
\mathcal E := \mathcal D \cap \left\{ \forall u \le \alpha ^{-3} ,\ \forall x>u, \ Y_x- Y_u \le \alpha ^{-\delta } +2 \alpha (x-u) \right\}
\end{equation}
From Lemma~\ref{lem:bound on density} it is clear that $\mathbb P (\mathcal E ) \ge 1 - \exp (-\alpha ^{-c_\epsilon })$.
We start by bounding the first term on the right hand side of \eqref{eq:coupeling form of J}. To this end we first condition of all of $Y$ and then use the same arguments as in Corollary~\ref{cor:conditioning on less information} to get the same bound when conditioning only on $Y _{\le u }$. By integrating over the different values $x$ that $T$ can take we get that on $\mathcal E$,
\begin{equation}\label{eq:A_us probability}
\begin{split}
\mathbb P \left(\mathcal A_{u,s} \ \big| \ Y \right) &\le \intop _u^s f_Y (x) \cdot \mathbb P \left( \forall x \le y \le s, \ W_y \le Y_y +1 \ \big| \ W_x=Y_x+1 , \ Y \right) dx\\
& \le \intop _u^s \frac{\alpha ^{-\epsilon }}{(x+1)^{\frac{3}{2}}} \mathbb P \left( \forall y \le s-x, \ W_y \le Y_{y+x}-Y_x \ \big| \ W_0=0 , \ Y \right) dx\\
&\le \intop _u^s \frac{\alpha ^{-\epsilon }}{(x+1)^{\frac{3}{2}}} \frac{\alpha ^{-\epsilon }}{\sqrt{s-x+1}} dx\le \frac{C \alpha ^{-2 \epsilon }}{\sqrt{s+1}\sqrt{u+1}},
\end{split}
\end{equation}
where the second inequality is by the definition of $\mathcal D $ and the third inequality is by Lemma~\ref{lem:bound on H}. Indeed, on $\mathcal E $ the process $Y'_y :=Y_{y+x}-Y_x$ satisfies $Y'_y\le \alpha ^{-\delta }+2\alpha y$ and therefore by Lemma~\ref{lem:bound on H} the bound \eqref{eq:bound on C} holds when we replace $Y$ by $Y'$. The last inequality is by Claim~\ref{claim:some claim}.
Thus, as $\epsilon $ is arbitrary we get
\begin{equation}
\mathbb P \left( \mathbb P \left( \mathcal A _{u,s} \ \big| \ Y \right) \ge \frac{\alpha ^{-\epsilon }}{ \sqrt{s+1} \sqrt{u+1}} , \ \forall 0 \le u<s \le \hat{h} \right) \ge 1-\exp ( -\alpha ^ {-c_\epsilon } ).
\end{equation}
By the same arguments as in Corollary~\ref{cor:conditioning on less information} we get that
\begin{equation}\label{eq:bound on first term}
\mathbb P \left( \mathbb P \left( \mathcal A _{u,s} \ \big| \ Y_{\le u} \right) \ge \frac{\alpha ^{-\epsilon }}{ \sqrt{s+1} \sqrt{u+1}} , \ \forall 0 \le u<s \le \hat{h} \right) \ge 1-\exp ( -\alpha ^ {-c_\epsilon } ).
\end{equation}
We turn to bound the second term in \eqref{eq:coupeling form of J}. By the same arguments as in \eqref{eq:A_us probability} we have on $\mathcal E $,
\begin{equation}
\begin{split}
\mathbb P \left( \mathcal B _s \ \big| \ Y \right) &\le \intop _s^\infty f_Y(x) \mathbb P \left( \forall y \ge x,\ W_y \le Y_y+1 \ \big | \ W_x=Y_x+1, \ Y \right)dx \\
& \le \intop _s^\infty \frac{\alpha ^{-\epsilon }}{(x+1)^{\frac{3}{2}}}
\mathbb P \left( \forall y \le \hat{h} ,\ W_y \le Y_{y+x}-Y_x \ \big | \ W_0=0, \ Y \right)dx \\
&\le \intop _s^\infty \frac{\alpha ^{1-2 \epsilon }}{(x+1)^ {\frac{3}{2}} } \le \frac{\alpha ^{1-3 \epsilon }}{\sqrt{s+1}} ,
\end{split}
\end{equation}
where in the third inequality we used Lemma~\ref{lem:bound on H}. Since $\epsilon $ is arbitrary and by the same arguments as in Corollary~\ref{cor:conditioning on less information} we get that
\begin{equation}\label{eq:bound on the second term}
\mathbb P \left( \mathbb P \left( \mathcal B _s \ \big| \ Y_{\le u} \right) \ge \frac{\alpha ^{1-\epsilon }}{ \sqrt{s+1}} , \ \forall 0 \le s \le \hat{h} \right) \ge 1-\exp ( -\alpha ^ {-c_\epsilon } ).
\end{equation}
Substituting \eqref{eq:bound on first term} and \eqref{eq:bound on the second term} into \eqref{eq:coupeling form of J} finishes the proof of the proposition.
\end{proof}
\subsection{Multiple integrals over the fixed rate process}\label{subsec:fixed:fixed}
Based on the results we obtained in the previous subsections, we prove Propositions \ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int} in the case of $\Pi_\alpha$, the fixed rate Poisson process. Switching $\Pi_g$ into $\Pi_\alpha$ not only makes the underlying process be simpler, but more importantly, enables us to interpret the integral as martingales. Previously, this was impossible since $g$ is progressively measureable with respect to $\Pi$ and ${\textnormal{\textbf{J}}}_{t-s,t-u;t}^{\Pi_g}$ is measureble only in terms of $\Pi_g[s,t]$, which essentially requires to revealing the entire information of $\Pi_g$.
Let $\Pi_\alpha^{(t)}$ denote the backwards-time point process of $\Pi_\alpha$ with respect to $t$, defined as
\begin{equation}
\Pi_\alpha^{(t)}[0,s] := \Pi_\alpha[t-s,t], \ \textnormal{ with } d\widehat{\Pi}_\alpha^{(t)}(s ) = d\Pi_\alpha^{(t)}(s)-\alpha ds.
\end{equation}
Then, we can see that $\mathcal{J}$ defined in \eqref{eq:def:Jint:general} becomes
\begin{equation}\label{eq:def:Jint:rev time}
\mathcal{J}[t;\Pi_\alpha] = \intop_{0}^t \intop_{0}^s {\textnormal{\textbf{J}}}_{u,s}^{\Pi_\alpha^{(t)}} \ d\widehat{\Pi}_\alpha^{(t)}(u) d\widehat{\Pi}_\alpha^{(t)}(s),
\end{equation}
where $ {\textnormal{\textbf{J}}}_{u,s}^{\Pi^{(t)}_\alpha}:= \mathbb{P}_\alpha(s<T_u<\infty\, | \, \Pi_\alpha^{(t)}[0,u] ) - \mathbb{P}_\alpha(s<T<\infty\, |\, \Pi_\alpha^{(t)}[0,u])$ follows the previous definition \eqref{eq:def:J:basic form}, with respect to $Y$ generated by $\Pi_\alpha^{(t)}$. Similarly, we can write $\mathcal{Q}$ in \eqref{eq:def:Qint:general} by
\begin{equation}\label{eq:def:Qint:rev time}
\mathcal{Q}[t;\Pi_\alpha] := \intop_{0}^{t} \intop_{0}^s \intop_{0}^u
{\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} \ d\widehat{\Pi}_\alpha^{(t)}(v) d\widehat{\Pi}_\alpha^{(t)}(u) d\widehat{\Pi}_\alpha^{(t)}(s),
\end{equation}
where ${\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}}$ is defined as \eqref{eq:def:Q:basic form}, namely,
\begin{equation}\label{eq:def:Q:fixed rate}
\begin{split}
{\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} := & \mathbb{P}_\alpha(s<T_{v,u}<\infty\ | \ \Pi_\alpha^{(t)}[0,v] ) - \mathbb{P}_\alpha(s<T_v<\infty\ |\ \Pi_\alpha^{(t)}[0,v])\\
&- \mathbb{P}_\alpha(s<T_u<\infty\ | \ \Pi_\alpha^{(t)}[0,v] ) + \mathbb{P}_\alpha(s<T<\infty\ |\ \Pi_\alpha^{(t)}[0,v]).
\end{split}
\end{equation}
Furthermore, the points $\pi_i(t;\alpha)$ from \eqref{eq:def:pi closest points} can now be interpreted as the $i$-th closest point from the origin in $\Pi_\alpha^{(t)}$.
Based on these identities, our goal is to establish the following estimates.
\begin{prop}\label{prop:fixed perturbed:fixed rate}
Let $\epsilon, C>0$ be given, and set $\hat{h}$ as \eqref{eq:def:horizon}. Then, there exists $\alpha_0=\alpha_0(\epsilon,C)>0$ such that for all $0<\alpha<\alpha_0$, we have
\begin{equation}\label{eq:fixed perturbed:fixed result}
\begin{split}
&\mathbb{P} \left(-\alpha^{\frac{1}{2}-\epsilon}\sigma_1(t)\le \mathcal{J}[t;\Pi_\alpha] \le \alpha^{-\epsilon} \sigma_1\sigma_2(t), \ \forall t \le \hat{h} \right) \ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{200}} \right);\\
&\mathbb{P} \left(|\mathcal{Q}[t;\Pi_\alpha]| \le
\alpha^{-\epsilon} \sigma_1\sigma_2\sigma_3(t),\ \forall t\le \hat{h} \right) \ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{200}} \right).
\end{split}
\end{equation}
\end{prop}
Since everything dealt in this subsection is within the fixed rate Poisson process $\Pi_\alpha$, we use the simplified notation $\pi_i(t) = \pi_i(t;\Pi_\alpha)$ and $\sigma_i(t) = \sigma_i(t;\Pi_\alpha).$
We discuss the proof only for the second one, since the first one can be derived by analogous but simpler arguments; Although there is a difference that the first one has distinct upper and lower bound while the other is not, they can be obtained from the same proof as we will see in Remark \ref{rmk:fixed perturbed:Janalog}. The main idea is applying results from Section \ref{subsec:fixed:mgconcen} to control the triple integral \eqref{eq:def:Qint:rev time} based on the bound we got from Proposition \ref{prop:Qbound:quenched}. However, there are a couple of major differences we need to keep in mind:
\begin{enumerate}
\item We want to estimate each integral in \eqref{eq:def:Qint:rev time} one by one, starting from the inner one. This entails understanding an appropriate continuity property (in $u,s,t',t$) of the inner integrals, which we will obtain using Lemma \ref{lem:derivative of Q and J}.
\item ${\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}}$ is not small enough for small $v,u,s$, and we need to control such cases relying on Lemma \ref{lem:concentrationofint:continuity}, not on Lemma~\ref{lem:concentration of integral}. This is why an upper bound of $\alpha^{-\epsilon} \sigma_1\sigma_2\sigma_3$ is necessary.
\end{enumerate}
The following estimate on the derivative of ${\textnormal{\textbf{Q}}}_{v,u,s}$ gives the desired continuity property later.
\begin{lem} \label{lem:derivative of Q and J}
Let ${\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}}$ be defined as \eqref{eq:def:Q:fixed rate}. With probability $1$, we have
\begin{equation}
\begin{split}
\left|\frac{\partial}{\partial u} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} \right| \vee
\left|\frac{\partial}{\partial s} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} \right|
\leq 4
\end{split}
\end{equation}
for all $0\leq v <u<s\le t.$
\end{lem}
\begin{proof}
Let $\delta>0$ be a small number such that $s+\delta<t$ and recall the definition \eqref{eq:def:Q:fixed rate}. We investigate each of the four terms in \eqref{eq:def:Q:fixed rate} separately. To study the partial derivative of the first term, we observe that
\begin{equation} \label{eq:tau prob diff}
\begin{split}
& \left| \,\mathbb{P}_\alpha (s\le T_{v,u+\delta}<\infty \,|\, \Pi_\alpha^{(t)}[0,v] )
-\mathbb{P}_\alpha (s\le T_{v,u}<\infty \,|\, \Pi_\alpha^{(t)}[0,v] )\,\right| \\
&\quad =
\mathbb{P}( u\leq T_{v,u+\delta}<u+\delta,\; t \leq T_{v,u}<\infty \,|\,\Pi_\alpha^{(t)}[0,v] ),
\end{split}
\end{equation}
since the only possible way to have $T_{v,u+\delta}\neq T_{v,u}$ is when $u\leq T_{v,u+\delta}<u+\delta \leq T_{v,u}$. Then, we can write down a crude bound such that
\begin{equation}
\mathbb{P}( s\leq T_{v,u+\delta}<u+\delta,\; t\leq T_{v,u}<\infty \,|\,\mathcal{F}_v)
\leq \delta,
\end{equation}
since $s\leq T_{u,s+\delta} <s+\delta$ implies that the random walk $W$ has performed a jump between times $s$ and $s+\delta$. The same argument holds for $\delta<0$ as well, and hence we obtain that
\begin{equation}
\left| \frac{\partial}{\partial u} \mathbb{P}_\alpha (s\le T_{v,u}<\infty \,|\, \mathcal{F}_v ) \right| \leq 1.
\end{equation}
Applying this argument to the three other terms of \eqref{eq:def:Q:fixed rate}, we get the desired bound for the $s$-partial derivative of $\textbf{Q}_{v,u,s}$.
The derivative with respect to $s$ can be estimated analogously, by noticing that (for $\delta>0$)
\begin{equation}
|\,\mathbb{P}_\alpha (s+\delta\le T_{v,u}<\infty \,|\, \mathcal{F}_v )-
\mathbb{P}_\alpha (s\le T_{v,u}<\infty \,|\, \mathcal{F}_v )\,|
=
\mathbb{P}(s\leq T_{v,u} <s+\delta \,|\,\mathcal{F}_v).
\end{equation}
\end{proof}
A straight-forward generalization gives the analogous estimate on $\partial_t J(s,t\,|\,\mathcal{F}_s)$, which we record in the following corollary.
\begin{cor} \label{cor:derivative of J}
With probability $1$, we have
\begin{equation}
\left|\frac{\partial}{\partial s} {\textnormal{\textbf{J}}}_{u,s}^{\Pi_\alpha^{(t)}} \right| \leq 2,
\end{equation}
for all $0\leq u< s\le t.$
\end{cor}
Next, we introduce stopping times that provides a fundamental advantage on dealing with the issue (2) mentioned above.
\begin{equation} \label{eq:def:tau:Qint bound}
\begin{split}
&\tau_{\textnormal{f}1}^{\textnormal{rev}} (t)= \tau_{\textnormal{f}1}^{\textnormal{rev}}(t,\alpha) := \inf \left\{s>\alpha^{-1}: \left|\Pi_{\alpha}^{(t)}[s-\alpha^{-1},\, s]\right| \geq \alpha^{-\frac{\epsilon}{20}} \right\};\\
&\tau_{\textnormal{f}2}^{\textnormal{rev}} (t) =\tau_{\textnormal{f}2}^{\textnormal{rev}}(t,\alpha) := \inf\left\{{s>\alpha^{-1-\frac{\epsilon}{10}}: \left|\Pi_\alpha^{(t)}[s-\alpha^{-1-\frac{\epsilon}{20}},\,s] \right|=0 }\right\},\\
&\tau_{\textnormal{f}3}^{\textnormal{rev}}(t) =\tau_{\textnormal{f}3}^{\textnormal{rev}}(t,\alpha) :=
\inf \left\{s>0: \sup_{v,u: \, v<u<s} \left\{\left|{\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}}\right| - \frac{\alpha^{-\frac{\epsilon}{20}}}{\sqrt{(v+1)(u+1)(s+1)}}\right\} >0
\right\},\\
&\tau_{\textnormal{f}}^{\textnormal{rev}}(t) =\tau_{\textnormal{f}}^{\textnormal{rev}}(t,\alpha) := \tau_{\textnormal{f}1}^{\textnormal{rev}}(t)\wedge\tau_{\textnormal{f}2}^{\textnormal{rev}}(t)\wedge\tau_{\textnormal{f}3}^{\textnormal{rev}}(t).
\end{split}
\end{equation}
In the proof of Proposition \ref{prop:fixed perturbed:fixed rate}, $\tau_{\textnormal{f}1}^{\textnormal{rev}}$ provides fundamental control on the size of the Poisson process, and $\tau_{\textnormal{f}2}^{\textnormal{rev}}$ is needed to ensure that $\pi_1, \pi_2, \pi_3 \leq 4\alpha^{-1-\frac{\epsilon}{10}}$. The purpose of $\tau_{\textnormal{f}3}^{\textnormal{rev}}$ is obvious in controlling the integral \eqref{eq:def:Qint:rev time}.
\begin{lem} \label{lem:Qint:taubase}
Let $\hat{h}$ be as \eqref{eq:def:horizon}. Under the above definition, there exists $\alpha_0 = \alpha_0(\epsilon,C)$ such that for all $0<\alpha<\alpha_0$, $$\mathbb{P}\left(\inf_{t\le \hat{h}} \left\{ \tau_{\textnormal{f}}^{\textnormal{rev}}(t,\alpha)\right\} \ge \hat{h} \right) \geq 1- \exp(-\alpha^{-\epsilon/30}).$$
\end{lem}
\begin{proof}
To estimate $\tau_{\textnormal{f}1}^{\textnormal{rev}}$, we divide the interval $[0,\hat{h}]$ into subintervals of length $\alpha^{-1}$. Each interval $[k\alpha^{-1}, (k+1)\alpha^{-1}]$ contains more than $\alpha^{-\frac{\epsilon}{20}}/2$ points with probability less than $\exp(-\alpha^{-\frac{\epsilon}{25}})$, and we take union bound over $k=0,1,\ldots, \alpha^{-2}$. Note that if each interval $[k\alpha^{-1}, (k+1)\alpha^{-1}]$ contains at most $\alpha^{-\frac{\epsilon}{20}}/2$ points, then every interval $[t-\alpha^{-1},\,t]$ has at most $\alpha^{-\frac{\epsilon}{20}}$ points.
Controlling $\tau_{\textnormal{f}2}^{\textnormal{rev}}$ is similar, where we divide $[0,\hat{h}]$ into subintervals of length $\alpha^{-1-\frac{\epsilon}{20}}/4$. Each of these subintervals contains at least one point with probability $1-\exp(-\alpha^{-\frac{\epsilon}{25}} )$.
Finally, we will see that $\tau_{\textnormal{f}3}^{\textnormal{rev}} \geq \hat{h}$ with very high probability from Proposition \ref{prop:Qbound:quenched}.
\end{proof}
\begin{lem} \label{lem:Qint 1st}
Let $\hat{h}$ be as \eqref{eq:def:horizon}, and for each $t\in[0,\hat{h}]$ let $\tau_{\textnormal{f}}^{\textnormal{rev}}(t)$ be as \eqref{eq:def:tau:Qint bound}. Then, we have
\begin{equation}\label{eq:Qint:result:1stint}
\begin{split}
\mathbb{P} \left(\sup_{u'\le u\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)} \left|\intop_0^{u' } {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{ \Pi}_\alpha^{(t)}(v) \right| \leq \frac{5\alpha^{-\frac{\epsilon}{8}}\sigma_1(t) }{\sqrt{(u+1)(s+1)}},\;\forall 0<u<s \le t\leq \hat{h} \right) \\\geq 1-\exp\left(-\alpha^{-\frac{\epsilon}{160}} \right).
\end{split}
\end{equation}
\end{lem}
\begin{remark} \label{rmk:Qint 1st}
Another way to phrase Lemma \ref{lem:Qint 1st} is the following: for each $t\in[0,\hat{h}]$, define the stopping time $\tau_1^{\textnormal{int}}(t)$ as
\begin{equation} \label{eq:def:sig1:Qint}
\tau_1^{\textnormal{int}}(t) := \inf\left\{s>0: \exists u\leq s \textnormal{ s.t. }
\sup_{u'\le u}\left|\intop_0^{u'} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{ \Pi}_\alpha^{(t)}(v) \right| \geq \frac{5\alpha^{-\frac{\epsilon}{8}}\sigma_1(t)}{\sqrt{(u+1)(s+1)}} \right\}.
\end{equation}
Then, Lemmas \ref{lem:Qint:taubase} and \ref{lem:Qint 1st} imply that \begin{equation}
\mathbb{P}_\alpha \left( \bigcap_{t\le \hat{h}}\left\{\tau_1^{\textnormal{int}}(t) \wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t) \ge t \right\} \right) \geq 1-2\exp\left(-\alpha^{-\frac{\epsilon}{160}} \right).
\end{equation}
\end{remark}
\begin{proof}[Proof of Lemma \ref{lem:Qint 1st}]
We first establish \eqref{eq:Qint:result:1stint} for fixed $s$ and $t$, then extend the result to the form of \eqref{eq:Qint:result:1stint}.
Let $s<t\le \hat{h}$ be fixed, and set
\begin{equation}
f_u(x) := {\textnormal{\textbf{Q}}}_{x,u,s}^{\Pi_\alpha^{(t)}},\quad \bar{f}_u(x):= \frac{\alpha^{-\frac{\epsilon}{20}}}{\sqrt{(x+1)(u+1)(s+1)}}, \quad g(x) \equiv \alpha.
\end{equation}
We apply Lemma \ref{lem:concen of int:conti:forwardtime} to these functions, setting the parameters in the lemma as follows:
\begin{itemize}
\item We set $\tau_- \equiv 0 $ be deterministic, and let $\tau= \tau(t) = \tau_{\textnormal{f}}^{\textnormal{rev}}(t)$.
\item Set $h=\hat{h}$, $\eta = 2\alpha$, $\Delta=\alpha^{-1}$, and $N=\alpha^{-\frac{\epsilon}{20}}$. Under these values, we let $\tau' = \tau_{\textnormal{f}1}^{\textnormal{rev}}(t)$, and $\tau'' = \infty$.
\item Letting $$D = 4, \quad M = \frac{\alpha^{1-\frac{\epsilon}{8}}}{(u+1)(s+1)}, \quad A = \frac{3\alpha^{-\frac{1}{2}-\frac{\epsilon}{20}}}{\sqrt{(u+1)(s+1)}},$$ (where we set $M$ to be deterministic as $\tau_- \equiv 0$) we see that the conditions given in \eqref{eq:concen:condition:forwardtime} are satisfied.
\item Set $\delta=\alpha^{10}$ which satisfies \eqref{eq:concen:condition:delta:forward}, and let $a = \alpha^{-\frac{\epsilon}{100}}$.
\end{itemize}
Applying Lemma \ref{lem:concen of int:conti:forwardtime} under this setting gives
\begin{equation}
\begin{split}
\mathbb{P} \left( \sup_{u'\le u\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)}
\left|
\intop_0^{u'} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{\Pi}_\alpha^{(t)}(v)
\right| \le \frac{\alpha^{-\frac{\epsilon}{8}}\sigma_1(t) }{\sqrt{(u+1)(s+1)}}
, \ \forall u \in [0,s]
\right) \\
\ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{150} }\right),
\end{split}
\end{equation}
and this holds for all small enough $\alpha>0$. Note that we used $\pi_1(t) \le \alpha^{-1-\frac{\epsilon}{20}}$, which comes from $\tau_{\textnormal{f}2}^{\textnormal{rev}}(t)$, to simplify the bound inside the probability.
What remains is to extend this bound to hold for all $s$ and $t$. To this end, we first take a union bound over $s<t$ in a discretized interval such that
\begin{equation}
s,t \in \mathcal{T}:= \{x\in[0,\hat{h}]: x= k\delta, \ k \in \mathbb{Z} \},
\end{equation}
where $\delta = \alpha^{10}$.
This gives
\begin{equation}
\begin{split}
\mathbb{P} \left( \sup_{u'\le u\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)}
\left|
\intop_0^{u'} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{\Pi}_\alpha^{(t)}(v)
\right| \le \frac{\alpha^{-\frac{\epsilon}{8}}}{\sqrt{(\pi_1(t)+1)(u+1)(s+1)}}
, \ \forall u \in [0,s], \ s,t\in \mathcal{T}, \ s\le t
\right) \\
\ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{160} }\right).
\end{split}
\end{equation}
We first extend this to all $s<t$ and $t\in \mathcal{T}$. For any $s<t$ with $t\in \mathcal{T}$, let $s_\delta \ge s$ such that $s_\delta \in \mathcal{T}$ and $s_\delta \le s+\delta$. Using the estimate
\begin{equation}
\left| {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} -{\textnormal{\textbf{Q}}}_{v,u,s_\delta}^{\Pi_\alpha^{(t)}} \right| \le 4\delta= 4\alpha^{10},
\end{equation}
and the fact that $\int_0^{u\wedge{\tau_{\textnormal{f}}^{\textnormal{rev}}(t)}} \left| d\widehat{\Pi}_\alpha^{(t)} (v)\right| \le \alpha^{-1-\frac{\epsilon}{10}}$ which comes from the definition of $\tau_{\textnormal{f}1}^{\textnormal{rev}}(t)$, we obtain that
\begin{equation}
\begin{split}
\mathbb{P} \left( \sup_{u'\le u\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)}
\left|
\intop_0^{u'} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{\Pi}_\alpha^{(t)}(v)
\right| \le \frac{2\alpha^{-\frac{\epsilon}{8}}}{\sqrt{(\pi_1(t)+1)(u+1)(s+1)}}
, \ \forall u <s \le t, \ t\in \mathcal{T}
\right) \\
\ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{160} }\right).
\end{split}
\end{equation}
Finally, we extend this result to hold for all $t\le \hat{h}$. Let $t_\delta \ge t$ be $t_\delta \in \mathcal{T}$ and $\delta' := t_\delta -t\le \delta$. Here, we need to consider the difference between measure $\Pi_\alpha^{(t)}$ and $\Pi_\alpha^{(t_\delta)}$. Note that for any $u'\le u<s\le t$ such that $u'\le \tau_{\textnormal{f}}^{\textnormal{rev}}(t)$,
\begin{equation}
\intop_0^{u'} {\textnormal{\textbf{Q}}}_{v, u,s}^{\Pi_\alpha^{(t)}} d\widehat{\Pi}_\alpha^{(t)}(v)
= \intop_{\delta'}^{u'+\delta'} {\textnormal{\textbf{Q}}}_{v-\delta',u,s}^{\Pi_\alpha^{(t_\delta)}} d\widehat{\Pi }_\alpha^{(t_\delta)}(v).
\end{equation}
We can then proceed by switching ${\textnormal{\textbf{Q}}}_{v-\delta',u,s}^{\Pi_\alpha^{(t_\delta)}}$ to ${\textnormal{\textbf{Q}}}_{v,u,s}^{(t_\delta)}$, and noting that on the event $\inf_{t\le \hat{h}} \{\tau_{\textnormal{f}}^{\textnormal{rev}}(t) \} \ge \hat{h}$ which holds with high probability from Lemma \ref{lem:Qint:taubase}, we have
\begin{equation}\label{eq:Qint:1st:conti:estim}
\begin{split}
&\left| \intop_{\delta'}^{(u'+\delta') \wedge u} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t_\delta)}} d\widehat{\Pi }_\alpha^{(t_\delta)}(v)
+ \intop_{u}^{u'+\delta'} {\textnormal{\textbf{Q}}}_{v-\delta',u,s}^{\Pi_\alpha^{(t_\delta)}} d\widehat{\Pi }_\alpha^{(t_\delta)}(v)\right|\\
&\le
\left|\intop_{0}^{(u'+\delta')\wedge u} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t_\delta)}} d\widehat{\Pi }_\alpha^{(t_\delta)}(v)\right|
+
\left|\intop_{0}^{\delta'} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t_\delta)}} d\widehat{\Pi }_\alpha^{(t_\delta)}(v)\right| + \alpha^{-\frac{\epsilon}{20}}\bar{f}_u(\pi_1(t)).
\end{split}
\end{equation}
Thus, we obtain
\begin{equation}
\begin{split}
\mathbb{P} \left( \sup_{u'\le u\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)}
\left|
\intop_0^{u'} {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{\Pi}_\alpha^{(t)}(v)
\right| \le \frac{5\alpha^{-\frac{\epsilon}{8}}}{\sqrt{(\pi_1(t)+1)(u+1)(s+1)}}
, \ \forall u <s \le t \le \hat{h}
\right) \\
\ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{160} }\right),
\end{split}
\end{equation}
concluding the proof.
\end{proof}
\begin{remark}\label{rmk:fixed perturbed:Janalog}
We stress that the same method can be applied to deduce the corresponding bound for the integral of $\textnormal{\textbf{J}}.$ Namely, changing the definition of $\tau_{\textnormal{f}3}^{\textnormal{rev}}$ into the one that has the bound from Lemma \ref{prop:Jbound:quenched} instead of \ref{prop:Qbound:quenched}, we obtain
\begin{equation}
\begin{split}
\mathbb{P} \left(\sup_{s'\le s\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)} \intop_0^{s' } {\textnormal{\textbf{J}}}_{u,s}^{\Pi_\alpha^{(t)}} d\widehat{ \Pi}_\alpha^{(t)}(u) \leq \frac{\alpha^{-\frac{\epsilon}{8}}\sigma_1(t) }{\sqrt{(s+1)}},\;\forall 0<s \le t\leq \hat{h} \right) \geq 1-\exp\left(-\alpha^{-\frac{\epsilon}{160}} \right);\\
\mathbb{P} \left(\inf_{s'\le s\wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)} \intop_0^{s' } {\textnormal{\textbf{J}}}_{u,s}^{\Pi_\alpha^{(t)}} d\widehat{ \Pi}_\alpha^{(t)}(u) \geq -{\alpha^{1-\frac{\epsilon}{8}}\sigma_1(t) }-\frac{\alpha^{\frac{1}{2} -\frac{\epsilon}{8} }}{\sqrt{s+1}} ,\;\forall 0<s \le t\leq \hat{h} \right) \geq 1-\exp\left(-\alpha^{-\frac{\epsilon}{160}} \right).
\end{split}
\end{equation}
(Note that unlike the upper bound, the two terms in the lower bound cannot be unified together since one is not always larger than the other.)
Based on this observation, we similarly obtain the analogue of Corollary \ref{cor:Qint 2nd} for $\textnormal{\textbf{J}}$ to deduce the first conclusion of Proposition \ref{prop:fixed perturbed:fixed rate}.
\end{remark}
Next step, we derive an analogous result after integrating once more. For convenience we define
\begin{equation}
F_1(u,s,t):= \intop_0^u {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} d\widehat{ \Pi}_\alpha(v).
\end{equation}
\begin{cor}\label{cor:Qint 2nd}
For each $t$, let $\tilde{ \tau}_1^{\textnormal{int}}(t):= \tau_1^{\textnormal{int}}(t) \wedge \tau_{\textnormal{f}}^{\textnormal{rev}}(t)$ with $\tau_1^{\textnormal{int}}(t) $ given in Remark \ref{rmk:Qint 1st}. Then, we have
\begin{equation}\label{eq:Qint 2nd:main ineq:inlem}
\begin{split}
\mathbb{P} \left( \sup_{s'\le s\wedge \tilde{\tau}_1^{\textnormal{int}}(t) } \left|\intop_0^{s'} F_1(u,s,t) d\widehat{ \Pi}_\alpha^{(t)}(u) \right| \leq \frac{\alpha^{-\frac{\epsilon}{4}}\sigma_1\sigma_2(t)}{\sqrt{s+1}},\;\forall 0\le s\le t\leq\hat{h} \right) \\
\geq 1-\exp\left(-\alpha^{-\frac{\epsilon}{160}} \right).
\end{split}
\end{equation}
\end{cor}
\begin{remark} \label{rmk:Qint 2nd}
Similarly as Remark \ref{rmk:Qint 1st}, define
\begin{equation} \label{eq:def:sig2:Qint}
\tau_2^{\textnormal{int}}(t):= \inf \left\{s>0:
\sup_{s'\le s}\left| \intop_0^{s'} F_1(u,s,t) d\widehat{ \Pi}_\alpha^{(t)}(u) \right|\geq
\frac{\alpha^{-8\epsilon}\sigma_1\sigma_2(t)}{\sqrt{s+1}}
\right\}.
\end{equation}
Then, Lemma \ref{lem:Qint 1st} and Corollary \ref{cor:Qint 2nd} imply
\begin{equation}
\mathbb{P}\left( \bigcap_{t\le \hat{h}} \left\{ \tau_2^{\textnormal{int}}(t) \wedge \tilde{\tau}_1^{\textnormal{int}}(t) \ge t \right\} \right) \geq 1-3\exp\left(-\alpha^{-\frac{\epsilon}{160}} \right).
\end{equation}
\end{remark}
\begin{proof}[Proof of Corolloary \ref{cor:Qint 2nd}]
The proof goes similar to that of Lemma \ref{lem:Qint 1st}, and hence we describe the argument more concisely. We first work with fixed $t$, and then extend the result to the desired conclusion. For each $s\le t \le \hat{h}$, define
\begin{equation}
f_s(x):= F_1(x,s,t), \quad \bar{f}_s(x):= \frac{5\alpha^{-\frac{\epsilon}{8}}\sigma_1(t) }{\sqrt{(x+1)(s+1)}}, \quad g(x)\equiv \alpha.
\end{equation}
We again apply Lemma \ref{lem:concen of int:conti:forwardtime} to these functions, with the other parameters set to be as follows.
\begin{itemize}
\item Let $\tau_- = \tau_-(t):= \pi_1(t)$, and let $\tau = \tau(t) = \tilde{\tau}_1^{\textnormal{int}}(t)$.
\item Set $h=\bar{h}, \ \eta = 2\alpha, \ \Delta = \alpha^{-1},$ $N=\alpha^{-\frac{\epsilon}{20}}$, $\tau' = \tau_{\textnormal{f}1}^{\textnormal{rev}}(t)$, and $\tau''=\infty$ as in the proof of Lemma \ref{lem:Qint 1st}.
\item Similarly before, the parameters
\begin{equation}
D=\alpha^{-2},\quad M = \frac{\alpha^{1-\frac{\epsilon}{3}} \sigma_1(t)^2}{s+1}, \quad A = \frac{\alpha^{-\frac{1}{2}-\frac{\epsilon}{7}} \sigma_1(t)}{\sqrt{s+1}},
\end{equation}
satisfy the conditions in \eqref{eq:concen:condition:forwardtime}. Note that $D$, the parameter for the derivative can be controlled deterministically by
\begin{equation}
|\partial_u F_1(u,s,t)| \le {\textnormal{\textbf{Q}}}_{u,u,s}^{\Pi_\alpha^{(t)}} \cdot (1+\alpha)
+
\intop_{0}^u \left|\partial_u {\textnormal{\textbf{Q}}}_{v,u,s}^{\Pi_\alpha^{(t)}} \right|\cdot \left| d\widehat{\Pi}_\alpha^{(t)} (v)\right| \le \alpha^{-2},
\end{equation}
for all $s\le \tau \le \tau_{\textnormal{f}1}^{\textnormal{rev}} $. $|\partial_s F_1(u,s,t)| $ can be estimated in the same but simpler way, since the regime of integral is independent of $s$.
\item Set $\delta = \alpha^{10}$ as before which satisfies \eqref{eq:concen:condition:delta:forward}, and let $a=\alpha^{\frac{\epsilon}{100}}.$
\end{itemize}
Then, Lemma \ref{lem:concen of int:conti:forwardtime} implies that for each $t\in[0,\hat{h}]$,
\begin{equation}
\begin{split}
\mathbb{P} \left(
\sup_{s'\le s\wedge \tilde{\tau}_{1}^{\textnormal{int}}} \left|
\intop_{\tau_-}^{s'} F_1(u,s,t)d\widehat{\Pi}_\alpha^{(t)}(u)
\right|
\le
\frac{\alpha^{-\frac{\epsilon}{4}}\sigma_1\sigma_2(t)}{\sqrt{s+1}}, \ \forall s\in [0,t]
\right)\\
\ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{150}} \right),
\end{split}
\end{equation}
which holds for all small enough $\alpha>0.$ Note that we used $\pi_2(t) \le 3 \alpha^{-1-\frac{\epsilon}{20}}$, which comes from $\tau_{\textnormal{f}2}^{\textnormal{rev}}(t)$, to simplify the bound inside the probability.
Moreover, note that the definition $\tau_- = \pi_1(t)$ gives that with probability one,
\begin{equation}
\begin{split}
\left|\intop_{0}^{\tau_-} F_1(u,s,t) d\widehat{\Pi}_\alpha^{(t)}(u) \right|
&\le
\intop_0^{\alpha^{-1-\frac{\epsilon}{20}}}
\intop_0^u \frac{\alpha^{-\frac{\epsilon}{20}} \alpha^2 dvdu}{\sqrt{(v+1)(u+1)(s+1)}} \\
&= \frac{2\alpha^{1-\frac{\epsilon}{10}}}{\sqrt{s+1}}
\le
\frac{\alpha^{-\frac{\epsilon}{4}} \sigma_1\sigma_2(t)}{\sqrt{s+1}},
\end{split}
\end{equation}
where the first and the last inequality hold if $s\le \tau_{\textnormal{f}}^{\textnormal{rev}}(t)$. Thus, we combine the two to obtain
\begin{equation}
\begin{split}
\mathbb{P} \left(
\sup_{s'\le s\wedge \tilde{\tau}_{1}^{\textnormal{int}}} \left|
\intop_{0}^{s'} F_1(u,s,t)d\widehat{\Pi}_\alpha^{(t)}(u)
\right|
\le
\frac{2\alpha^{-\frac{\epsilon}{4}}\sigma_1\sigma_2(t)}{\sqrt{s+1}}, \ \forall s\in [0,t]
\right)\\
\ge 1- \exp\left(-\alpha^{-\frac{\epsilon}{150}} \right),
\end{split}
\end{equation}
Extending this to the desired result is analogous as the argument in the proof of Lemma \ref{lem:Qint 1st}. Namely we take a union bound over $t$ in the discretized interval $\mathcal{T}$, and deduce an estimate corresponding to \eqref{eq:Qint:1st:conti:estim} appealing to the fact that the derivatives of $F_1$ are bounded. The details are omitted and left for interested readers.
\end{proof}
Now we are only left with the outer integral. The corollary below follows analogously as the proof of Lemma \ref{lem:Qint 1st} and Corollary \ref{cor:Qint 2nd}. Thus, we state the result and omit the details of its proof.
\begin{cor} \label{cor:Qint 3rd}
Let $\tilde{\tau}_2^{\textnormal{int}}(t) := \tau_2^{\textnormal{int}}(t) \wedge \tilde{\tau}_1^{\textnormal{int}}(t)$ with $\tau_2^{\textnormal{int}}(t), \tilde{\tau}_1^{\textnormal{int}}(t) $ given in Remark \ref{rmk:Qint 2nd}, Lemma \ref{cor:Qint 2nd}, respectively, and define
\begin{equation}
F_2(s,t):= \intop_0^s F_1(u,s,t) d\widehat{ \Pi}_\alpha^{(t)} (u).
\end{equation}
Then, we have
\begin{equation}
\begin{split}
\mathbb{P} \left( \sup_{t'\leq t \wedge \tilde{\tau}_2^{\textnormal{int}}(t) } \left| \intop_0^{t'} F_2(s,t) d\widehat{ \Pi}_\alpha^{(t)} (s) \right| \leq \alpha^{-\epsilon} \sigma_1\sigma_2\sigma_3(t), \ \forall t\le\hat{h} \right) \geq 1-\exp \left( -\alpha^{-\frac{\epsilon}{160}}\right).
\end{split}
\end{equation}
\end{cor}
We conclude Section \ref{subsec:fixed:fixed} by establishing Proposition \ref{prop:fixed perturbed:fixed rate}.
\begin{proof}[Proof of Proposition \ref{prop:fixed perturbed:fixed rate}]
Combining Remark \ref{rmk:Qint 2nd} and Corollary \ref{cor:Qint 3rd} implies that
\begin{equation}
\mathbb{P} \left(|\mathcal{Q}[t;\Pi_\alpha]| \le
\alpha^{-\epsilon} \sigma_1\sigma_2\sigma_3(t),\ \forall t\le \hat{h} \right) \ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{200}} \right),
\end{equation}
proving the second statement of \eqref{eq:fixed perturbed:fixed result} in Proposition \ref{prop:fixed perturbed:fixed rate}.
The first inequality in \eqref{eq:fixed perturbed:fixed result} can be obtained analogously by developing Lemma \ref{lem:Qint 1st}, Corollary \ref{cor:Qint 2nd}, Remarks \ref{rmk:Qint 1st} and \ref{rmk:Qint 2nd}, based on the bound on ${\textnormal{\textbf{J}}}_{u,s}^{\Pi_\alpha^{(t)}}$ given in Proposition \ref{prop:Jbound:quenched}. We omit the details which are left for interested readers.
\end{proof}
To conclude this subsection, we introduce the following result which controls the integral of the error bound given in Proposition \ref{prop:fixed perturbed:fixed rate}
\begin{lem}\label{lem:fixed perturbed:error int:fixed}
let $\epsilon, C >0$ be given, and set $\hat{h}$ as \eqref{eq:def:horizon}. Then, there exists $\alpha_0=\alpha_0(\epsilon,C)>0$ such that for all $0<\alpha<\alpha_0$, we have
\begin{equation}
\begin{split}
\mathbb{P}_\alpha \left( \left|\intop _{0}^{\hat{h}} \sigma_1\sigma_2(t )dt \right| \le \alpha ^{-1-\epsilon } \right) &\ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{20}} \right);\\
\mathbb{P}_\alpha \left( \left|\intop _{0}^{\hat{h}} \sigma_1\sigma_2\sigma_3(t )dt \right| \le \alpha ^{-\frac{1}{2}-\epsilon } \right) &\ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{20}} \right).
\end{split}
\end{equation}
\end{lem}
\begin{proof}
Let $N:=\Pi _\alpha [0,\hat{h}]$ and let $\pi _1 \le \pi _2\le \cdots $ be the points of the process. For convinence we let $\pi _0:=0$ and $\pi _{N+1}=T$. Note that with very high probability $\alpha ^{-1+\epsilon }\le N\le \alpha ^{-1-\epsilon }$ and for any $0 \le i \le N$, $\pi _{i+1}-\pi _i\le \alpha ^{-1-\epsilon }$. We start with the first inequality. We have with very high probability
\begin{equation}
\intop _{0}^{T} \frac{dt}{\sqrt{(\pi _1(t)+1)(\pi _2(t) +1)}} \le \intop _{0}^{T} \frac{dt}{\pi _1(t)+1} =\sum _{i=0}^{N} \intop _{\pi _i}^{\pi _{i+1}} \frac{dt }{t-\pi _i+1}\le C N \log(1/ \alpha ) \le \alpha ^{-1-\epsilon }
\end{equation}
We turn to prove the second inequality. For $k \le \log _2 (1/\alpha )$ and $\alpha ^{-1+\epsilon } \le n\le \alpha ^{-1-\epsilon }$ define the sets
\begin{equation}
I_{k}^n:=\left\{ i\le n \ \big| \ \pi _{i}-\pi _{i-1}\in [2^k,2^{k+1}]\right\}
\end{equation}
and $I_{k}:=I_{k}^N$. We have that $\mathbb P (i \in I _{k}^n)\le \alpha 2^k$ and therefore with very high probability for all $\alpha ^{-1+\epsilon } \le n \le \alpha ^{-1-\epsilon }$ we have $|I_{k}^n|\le \alpha ^{-\epsilon }n\alpha 2^k \le \alpha ^{-2\epsilon }2^k$. Thus with very high probability $|I_{k}|\le \alpha ^{-2\epsilon }2^k$.
Thus, if we let $I':=[N]\setminus \bigcup _{k}I_{k}$ we have
\begin{equation}
\begin{split}
\intop _{0}^{T}& \frac{dt}{\sqrt{(\pi _1(t)+1)(\pi _2(t) +1)(\pi _3(t) +1)}} \le \intop _{0}^{T} \frac{dt}{\sqrt{(\pi _1(t)+1)}(\pi _2(t)+1)} \\
&= \sum _{i = 0}^N \intop _{\pi _i}^{\pi _{i+1}} \frac{dt}{\sqrt{(t- \pi _i +1)}(\pi _2(t) +1)} \le C\alpha \sum _{i \in I'} \intop _{\pi _i}^{\pi _{i+1}} \frac{1}{\sqrt{t- \pi _i +1 }}+ C \sum _{k=1 }^{\log _2(1/\alpha )} 2^{-k} \sum _{i \in I _k} \intop _{\pi _i}^{\pi _{i+1}} \frac{1}{\sqrt{t-\pi _i +1}} \\
&\le C \alpha ^{\frac{1}{2}-\epsilon }|I'| + C\alpha ^{-\frac{1}{2}-\epsilon } \sum _{k=1}^{\log _2(1/\alpha )} 2^{-k}|I_k| \le CN\alpha ^{\frac{1}{2}-\epsilon } +C\alpha ^{-\frac{1}{2}-3 \epsilon }\log (1/\alpha ) \le C \alpha ^{-\frac{1}{2}-4 \epsilon }.\qedhere
\end{split}
\end{equation}
\end{proof}
\subsection{Perturbation of the underlying Poisson processes}\label{subsec:fixed:perturbed}
In this subsection, we establish Propositions \ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int}. As mentioned before, our method is to derive an estimate on the Radon-Nykodym derivative of $\Pi_g$ with respect to $\Pi_\alpha$ and use Proposition \ref{prop:fixed perturbed:fixed rate}.
Let $\mathcal{P}_g^t$ denote the law of $\Pi_g[0,t]$. Using this notation, we will write, for instance,
\begin{equation}
\mathbb{P} \left( \mathcal{J}[t;\Pi_g] \in \mathcal{E} \right)
=\int_{\Pi} I\{\mathcal{J}[t;\Pi] \in \mathcal{E} \} \mathcal{P}_g^t(d\Pi).
\end{equation}
Similarly, we let $\mathcal{P}_\alpha^t$ to be the law of $\Pi_\alpha[0,t]$. The main observation to establish Propositions \ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int} is the following: for a stopping time $\tau$ in terms of $g$ and $\Pi$, the Radon-Nykodym derivative of $\mathcal{P}_g^\tau$ with respect to $\mathcal{P}_\alpha^\tau$ can be written as
\begin{equation}\label{eq:def:radon deriv:fixed perturbed}
r_g^{\tau}( \Pi) := \frac{d\mathcal{P}_g^{\tau} }{d\mathcal{P}_\alpha^{\tau} }(\Pi) = \exp\left( -\intop_0^{\tau} (g(y)-\alpha) dy \right) \prod_{x\in \Pi[0,{\tau}]} \frac{g(x)}{\alpha}.
\end{equation}
This comes from the fact that $g$ is predictable with respect to $\Pi$.
From now on, let $\tau$ be the stopping time given in Proposition \ref{prop:fixed perturbed:double int}, and for convenience we set $${\hat{\tau}}:=\hat{h}\wedge \tau.$$ The next lemma shows how to control the size of $r_g^\tau(\Pi)$, when $\Pi$ is given by $\Pi_g \sim \mathcal{P}_g^\tau$.
\begin{lem}\label{lem:fixed perturbed:radon deriv bd}
Under the setting of Proposition \ref{prop:fixed perturbed:double int}, we have
\begin{equation}
\mathbb{P}_{\Pi \sim \mathcal{P}_g^{\hat{\tau}}}\left( r_g^{\hat{\tau}}(\Pi) \ge \exp\left(\alpha^{-\frac{\epsilon}{250}} \right) \right) \le \exp\left(-\alpha^{\frac{\epsilon}{2000}} \right).
\end{equation}
\end{lem}
\begin{proof}
From \eqref{eq:def:radon deriv:fixed perturbed}, we use $1+x\le e^x$ to obtain that
\begin{equation}
\begin{split}
r_g^{\hat{\tau}}(\Pi )\le \exp \left( \alpha ^{-1} \intop_{0}^{\hat{\tau}} (g(x)-\alpha ) \, d\Pi (x)-\intop_{0}^{\hat{\tau}} (g(x)-\alpha ) \, dx \right).
\end{split}
\end{equation}
Thus, for $\lambda := \alpha^{\frac{\epsilon}{300}}$ we have $r_g^\tau(\Pi)^\lambda \le M_\tau\cdot L_\tau$ where
\begin{equation}
\begin{split}
M_t:&=\exp \left( \lambda \alpha ^{-1} \intop_{0}^t (g(x)-\alpha ) \, d\Pi _g (x)-\intop_{0}^t g(x)(e^{\lambda \alpha ^{-1} (g(x)-\alpha )}-1) \, dx \right), \\
L_t:&=\exp \left( \intop_{0}^t g(x)(e^{\lambda \alpha ^{-1} (g(x)-\alpha )}-1) - \lambda (g(x)-\alpha ) \, dx \right).
\end{split}
\end{equation}
We start by bounding $L_{ {\hat{\tau}} }$. We have that
\begin{equation}
\begin{split}
L_{{\hat{\tau}} }&\le \exp \left( \intop_{0}^{{\hat{\tau}} } \left\{g(x)\left( \lambda \alpha ^{-1} (g(x)-\alpha )+\lambda ^2 \alpha ^{-2} (g(x)-\alpha )^2 \right) - \lambda (g(x)-\alpha ) \right\}\, dx \right)\\
& \le \exp \left( \intop_0^{{\hat{\tau}}} \left\{ \lambda \alpha^{-1} (g(x)-\alpha)^2 + \lambda^2\alpha^{-2} (g(x)-\alpha)^2g(x) \right\} dx \right)\le 2
\end{split}
\end{equation}
where in the first inequality we used $\lambda \alpha ^{-1} |g(x)-\alpha |\le \lambda \alpha ^{-1}(g(x)+\alpha) \le 1$ from the third assumption of \eqref{eq:fixed perturbed:assumption}, along with the fact that $e^y\le 1+y+y^2$ for $y\le 1$. In the last inequlity we used the first two assumptions of \eqref{eq:fixed perturbed:assumption}.
Thus, we obtain that
\begin{equation}
\begin{split}
\exp\left(\lambda \alpha^{-\frac{\epsilon}{250}} \right)\cdot \mathbb{P}_{\Pi \sim \mathcal{P}_g^{\hat{\tau}}}\left( r_g^{\hat{\tau}}(\Pi) \ge \exp\left(\alpha^{-\frac{\epsilon}{250}} \right) \right) &\le
\mathbb E_{\Pi\sim \mathcal{P}_g^{\hat{\tau}}} \left[ r_{g }^{{\hat{\tau}}}(\Pi)^\lambda \right]\\
& \le \mathbb E \left[ M_{ {\hat{\tau}} }\cdot L_{ {\hat{\tau}} } \right]\le 2 \mathbb E \left[ M_{ {\hat{\tau}} } \right]=2,
\end{split}
\end{equation}
where in the last equality we used that $M_t$ and therefore $M_{t\wedge \tau }$ are martingales and that $M_{0}=1$. This concludes the proof of the lemma.
\end{proof}
Now, We are ready to establish Propositions \ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int}.
\begin{proof}[Proof of Propositions~\ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int}]
Define the events
\begin{eqnarray}
\mathcal A_1(t) :=\left\{ \Pi : -\alpha^{\frac{1}{2}-\epsilon}\sigma_1(t)\le \mathcal{J}[t;\Pi ] \le \alpha ^{-\epsilon }\sigma_1\sigma_2(t) \right\}, &
\mathcal A_2(t) :=\left\{ \Pi : |\mathcal{Q}[t;\Pi ] | \le \alpha ^{-\epsilon }\sigma_1\sigma_2\sigma_3(t) \right\}, \\
\mathcal{A}(t):= \mathcal{A}_1(t)\bigcap \mathcal{A}_2(t) ,\qquad \qquad \qquad \qquad &
\mathcal B := \left\{ \Pi : r_g^{{\hat{\tau}}} (\Pi )\ge \exp\left(\alpha^{-\frac{\epsilon}{250}} \right) \right\}. \quad
\end{eqnarray}
Moreover, we set $\mathcal{A}_{\le t} := \cap_{s\le t} \mathcal{A}(s)$. Then, we have
\begin{equation}\label{eq:fixed perturbed:radon dec}
\mathbb{P}_{\Pi \sim \mathcal{P}_g^{\hat{\tau}}} \left( \left(\mathcal{A}_{\le {\hat{\tau}}}\right)^c \right)
\le \mathbb{P}_{\Pi \sim \mathcal{P}_g^{\hat{\tau}}} \left( \left(\mathcal{A}_{\le {\hat{\tau}}}\right)^c \bigcap \mathcal{B}^c \right) + \mathbb{P}_{\Pi \sim \mathcal{P}_g^{\hat{\tau}}} \left( \mathcal{B} \right).
\end{equation}
Lemma~\ref{lem:fixed perturbed:radon deriv bd} tells us that
the second term in the RHS is bounded by $\exp(-\alpha^{-\frac{\epsilon}{2000}})$, and the first term can be estimated by
\begin{equation}
\begin{split}
\mathbb{P}_{\Pi \sim \mathcal{P}_g^{\hat{\tau}}} \left( \left(\mathcal{A}_{\le {\hat{\tau}}}\right)^c \bigcap \mathcal{B}^c \right)
&\le
\int_{\Pi} I \left\{\left(\mathcal{A}_{\le \hat{h}}\right)^c \bigcap \mathcal{B}^c \right\} r_g^{\hat{\tau}}(\Pi) \mathcal{P}_\alpha^{\hat{\tau}}(d\Pi)\\
&\le
\exp\left(\alpha^{-\frac{\epsilon}{250}} \right) \int_{\Pi}I\{\left(\mathcal{A}_{\le \hat{h}}\right)^c \} \mathcal{P}_\alpha^{{\hat{h}}}(d\Pi)\\
&\le
\exp\left(-\alpha^{-\frac{\epsilon}{250}} \right),
\end{split}
\end{equation}
where the last inequality comes from Proposition \ref{prop:fixed perturbed:fixed rate}. Plugging the two bounds into \eqref{eq:fixed perturbed:radon dec} deduces the desired results.
\end{proof}
The same proof as above can be applied to generalize Lemma \ref{lem:fixed perturbed:error int:fixed}. Due to its similarity, we omit the proof of the following result.
\begin{lem}\label{lem:fixed perturbed:error int:perturbed}
Let $\epsilon>0$ be arbitrary, $\alpha>0$ be a sufficiently small constant depending on $\epsilon$, and $\hat{h}$ be as \eqref{eq:def:horizon}. Let $\tau$ be a stopping time, and $\{g(s)\}_{s\ge 0} $ be a positive stochastic process progressively measurable with respect to $\Pi_g$, and suppose that they satisfy \eqref{eq:fixed perturbed:assumption}. Then, we have
\begin{equation}
\begin{split}
\mathbb{P} \left(\left|\intop_0^{\hat{h}} \sigma_1\sigma_2(t;g) dt\right| \le \alpha^{-1-\epsilon} \right)& \ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{3000}} \right);\\
\mathbb{P} \left(\left|\intop_0^{\hat{h}} \sigma_1\sigma_2\sigma_3(t;g) dt\right| \le \alpha^{-\frac{1}{2}-\epsilon} \right) &\ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{3000}} \right).
\end{split}
\end{equation}
\end{lem}
\subsection{The error estimates of the speed}\label{subsec:fixed:error}
In this subsection, we translate the main estimates into the error estimates of the first- and second-order approximations (\eqref{eq:def:S1:basic form} and \eqref{eq:def:S2:basic form}). The results are direct consequences of Propositions \ref{prop:fixed perturbed:double int} and \ref{prop:fixed perturbed:triple int}, and we can state them as follows.
\begin{prop}\label{prop:fixed perturbed:error}
Let $\epsilon,C>0$ be arbitrary, $\alpha>0$ be a sufficiently small constant depending on $\epsilon,C
$, and let $t^, \hat{t}$ be $t^-<\hat{t}$ satisfying $\hat{t}-t^- \le \alpha^{-2} \log^C(1/\alpha)$. Define $S_1(s)=S_1(s;t^-,\alpha)$ and $S_2(s)=S_2(s;t^-,\alpha)$ as \eqref{eq:def:S1:basic form} and \eqref{eq:def:S2:basic form}, respectively, and recall the notation $\sigma_i(s;S)$ from \eqref{eq:def:sig closest points}. Suppose that there is a stopping time $\tau$ satisfying the following conditions almost surely:
\begin{equation}\label{eq:error:assumptions}
\begin{split}
&\intop_{t^-}^{\hat{t}\wedge \tau} (S(s)-\alpha)^2 ds \le \alpha^{1-\frac{\epsilon}{400}}; \qquad \intop_{t^-}^{\hat{t}\wedge \tau} (S(s)-\alpha)^2 S(s)ds \le \alpha^{2-\frac{\epsilon}{400}};\\ & \sup_{t^-\le s\le \hat{t}\wedge\tau }\{ S(s) \vee S_1(s )\} \le \alpha^{1-\frac{\epsilon}{400}}.
\end{split}
\end{equation}
Further, let $S'(s):= S'(s;t^-,\alpha)$ be defined as \eqref{eq:def:Sprime:basic form}.
Then, we have that
\begin{equation}
\begin{split}
\mathbb{P}\left(\left. -\alpha^{\frac{3}{2}-\epsilon}\sigma_1(s;S)\le S'(s) - S_1(s) \le \alpha^{1-\epsilon} \sigma_1\sigma_2(s;S), \ \forall s\in[t^-, \hat{t}\wedge \tau]\, \right| \, \mathcal{F}_{t^-} \right) &\ge 1-2\exp\left(-\alpha^{-\frac{\epsilon}{3000}} \right);\\
\mathbb{P} \left( \left.|S'(s)-S_2(s)| \le \alpha^{1-\epsilon}\sigma_1\sigma_2\sigma_3(s;S), \ \forall s\in [t^-, \hat{t}\wedge \tau]\,\right| \, \mathcal{F}_{t^-} \right)
&\ge
1-2\exp\left(-\alpha^{-\frac{\epsilon}{3000}} \right),
\end{split}
\end{equation}
where this holds for any given $\mathcal{F}_{t^-}$, the sigma-algebra generated by $\Pi[0,t^-]$ and $\{S(s)\}_{s\le t^-}$.
\end{prop}
\begin{proof}
Note that the second statement follows directly from the assumptions, Proposition \ref{prop:fixed perturbed:triple int}, and the formula \ref{eq:speed:2ndorder main}. The first inequality follows similarly: we first claim that
\begin{equation}
\mathbb{P}\left( \pi_2(s;S) \le \alpha^{-1-\frac{\epsilon}{10}}, \ \forall s\in [t^-, \hat{t}\wedge\tau] \right) \ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{3000}} \right).
\end{equation}
This follows from the analogous argument as Lemma \ref{lem:Qint:taubase} used to establish bounds on $\tau_{\textnormal{f}2}^{\textnormal{rev}}$, in our case relying on the assumption $\sup_{t^-\le s\le \hat{t}\wedge\tau} S_1(s) \le \alpha^{1-\frac{\epsilon}{400}}$. This implies
\begin{equation}\label{eq:fixed perturbed:prop:med}
\mathbb{P} \left( \frac{4\alpha+4\alpha^2}{(1+2\alpha)^2} S_1(s) \le \frac{1}{2}\alpha^{\frac{3}{2}-\epsilon}\sigma_1(s;S) \le \frac{1}{2}\alpha^{1-\epsilon}\sigma_1\sigma_2(s;S), \ \forall s\in[t^-,\hat{t}\wedge\tau] \right) \ge 1-\exp\left(-\alpha^{-\frac{\epsilon}{3000}} \right),
\end{equation}
and combining this with Proposition \ref{prop:fixed perturbed:double int} and the formula \eqref{eq:speed:1storder main} gives the conclusion.
\end{proof}
We remark that the assumptions \eqref{eq:error:assumptions} are actually satisfied by the speed with high probability with appropriate parameters $t^-, \hat{t}$ and the stopping time $\tau.$ The formal definitions and their proofs will be discussed in Sections \ref{sec:reg:intro}--\ref{sec:reg:next step}.
\section{The age-dependent critical branching process}\label{sec:criticalbranching}
In Section \ref{sec:inductionbase}, we saw that the speed drops down to an arbitrarily low size and can be sandwiched by two fixed rate processes.
To continue our investigation, we need to understand the evolution of a speed process which begins with points given by a fixed rate such as \eqref{eq:def:sandwichfixed}.
In this section, we study an age-dependent critical branching process which is closely related with the actual speed process. In the first (or second) order approximation of the speed, it will turn out that its main term stays close to a certain critical branching process, and hence understanding the latter process will be crucial in our analysis.
For two parameters $t_0^{-}<t_0$, we define the age-dependent critical branching process starting from a configuration of points $\Pi_{0}[t_0^{-},t_0]$ in the interval $[t_0^{-},t_0]$. From now on, we use the following notation to emphasize the prescribed initial points: For $t\ge t_0$, we define
\begin{equation}\label{eq:def:Rb}
R(t) = \mathcal{R}_b(t;\Pi_0[t_0^{-},t_0],\alpha)
:=
\intop_{t_0^{-}}^{t_0} K_\alpha(t-x) d\Pi_0(x)+
\intop_{t_0}^{t} K_\alpha(t-x) d\Pi_R(x),
\end{equation}
and set $R(t)=\alpha$ for $t<t_0.$
In words, the process $R(t)$ can be described as follows. The points $\Pi_0[t_0^{-},t_0]$ are given initially, acting as roots of the branching process. Each particle at $x\in [t_0^{-},t_0]$ independently gives births to its children on $[t_0,\infty)$, at position $y\geq t_0$ at rate $K_\alpha(y-x)dy$. Each child of a root then gives birth to another generation with the same rule, and the branching continues.
Since $\int_0^\infty K_\alpha(x)dx=1$, all particles except the roots ($\Pi_0[t_0^-,t_0]$) perform critical branching process (i.e., average number of offspring is 1). Moreover, the distance between a child and a parent is determined by the density $K_\alpha$, and hence it is an age-dependent branching process.
Throughout the section (and the rest of the paper), $\theta>0$ denotes a large absolute constant, say, $\theta=10000$ and $C_\circ=50$ is a fixed constant that is significantly smaller than $\theta$. For given positive number $\alpha$ and $t_0$, we denote $$\beta:= \log(1/\alpha).$$ Further, let $t_0^-$ be an arbitrary fixed number satisfying
$$ t_0-2\alpha^{-2}\beta^\theta < t_0^- <t_0-\frac{1}{2}\alpha^{-2}\beta^\theta,$$
and set $ \acute{t}_0:= t_0 + 3\alpha^{-2} \beta^{\theta}$.
In what follows, we illustrate a way of interpreting $R(t)$ as a perturbation of the fixed rate process. Consider the process $R_0(t)$ defined as
\begin{equation}\label{eq:def:R0:ib}
R_0(t):= \begin{cases}
\alpha & \ \textnormal{ for } t_0^{-}\le t<t_0;\\
\intop_{t_0^{-}}^t K_\alpha(t-x) d\Pi_\alpha(x) & \ \textnormal{ for } t\ge t_0,
\end{cases}
\end{equation}
For $t\ge t_0$, observe that $R(t) = R_0(t)$ until we see the first point in $\Pi_{R_0 \triangle \alpha}$. This motivates us to define the following two processes $R_0^-(t)$ and $R_0^+(t)$ as follows: For all $t_0^{-} \le t < t_0$, we set $R_0^-(t)=R_0^+(t)=\alpha$. For $t\ge t_0$, let
\begin{equation}\label{eq:def:R:ib:aux}
\begin{split}
R_0^-(t)&:= \intop_{t_0^{-}}^t K_\alpha(t-x) d\Pi_{R_0^- - (R_0-\alpha)_-}(x);\\
R_0^+(t) &:= \intop_{t_0^{-}}^t K_\alpha(t-x) d\Pi_{R_0^+ + (R_0-\alpha)_+}(x),
\end{split}
\end{equation}
where $(s)_+ := s\vee 0$ and $(s)_-:= (-s) \vee 0$. Later in Proposition \ref{prop:branching:ib:sandwich}, we will see that $R_0^-$ (resp. $R_0^+$) lower (resp. upper) bounds $R(t)$.
To state the main result of this section, let
\begin{equation}\label{eq:def:R0plushat}
\hat{R}_0^+(t) := \left(R_0^+(t) + |R_0(t)-\alpha| \right) \vee \alpha.
\end{equation}
Recalling the definition of $\sigma_1(t;g)$ \eqref{eq:def:sig closest points}, define
\begin{equation}\label{eq:def:tau:branching:ib}
\begin{split}
&\tau_{{\textnormal{B}1}} = \tau_{{\textnormal{B}1}}(\alpha,t_0):= \inf \left\{t\ge t_0: \left(R (t)-\alpha \right) \notin \left( - \alpha^{\frac{3}{2}}\beta^{5\theta}, \, \alpha \beta^{5\theta} \sigma_1(t; \hat{R}_0^+)+ \alpha^{\frac{3}{2}}\beta^{5\theta}\right) \right\};\\
&\tau_{{\textnormal{B}2}}=\tau_{{\textnormal{B}2}}(\alpha,t_0):= \inf\left\{t\ge t_0: \left|\Pi_{\hat{R}_0^+}[(t-\alpha^{-1})\vee t_0, t] \right| \ge \beta^{C_\circ} \right\} ;
\\
&\tau_{\textnormal{B}}=\tau_{\textnormal{B}}(\alpha,t_0):=
\tau_{{\textnormal{B}1}} \wedge \tau_{{\textnormal{B}2}}.
\end{split}
\end{equation}
Then, we have the following result for $R(t)$.
\begin{thm}\label{thm:branching:ib}
Under the above setting, for all sufficiently small $\alpha>0$, we have
\begin{equation}
\mathbb{P} \left( \tau_{\textnormal{B}}(\alpha,t_0) >\acute{t}_0 \right) \ge 1-\exp\left(-\beta^2 \right).
\end{equation}
\end{thm}
The theorem will be essential later, serving as the induction base of our inductive argument in settling the regularity (Section \ref{sec:reg:intro}; Theorem \ref{thm:induction:base:main}).
Establishing the theorem consists of the four main steps as follows.
\begin{itemize}
\item Section \ref{subsec:branching:mg}: We develop further theory on representing the branching process using martingales. Not only will this be helpful in studying $\tau_{\textnormal{B}1}$, but also serve as a fundamental tool in the later sections.
\item Section \ref{subsec:branching:numberofpts}: We study the critical branching process that starts from a single initial particle. The analysis provides a useful tool to understand the gap between $R_0(t)$ and $R(t)$.
\item Section \ref{subsec:branching:ib}: We investigate the processes $R_0(t)$, $R_0^-(t)$ and $R_0^+(t)$, exploiting the martingale concentration lemmas from Section \ref{subsec:fixed:mgconcen}.
\item Section \ref{subsubsec:branching:ib:tau4}: Combining the above analysis, we deduce the bound on $\tau_{{\textnormal{B}1}}$ and $\tau_{{\textnormal{B}2}}$.
\end{itemize}
Before moving on, we record a simple lemma that will be used througout the rest of the paper.
\begin{lem}\label{lem:ind1:pi1int basicbd:basic}
Let $g:[0,h]\to (0,\infty)$ be a positive function which defines a (deterministic) set of points $\Pi_g[0,h]$, and set $n=|\Pi_g[0,h]|$. Recalling the definition $\pi_1(t;g)$ \eqref{eq:def:pi closest points}, there exists an absolute constant $C>0$ such that
\begin{equation}
\begin{split}
&\intop_0^h \frac{dt}{\pi_1(t;g)+1} \le (n+1)\log (h+1);\\
&\intop_0^h \frac{dt}{\sqrt{\pi_1(t;g)+1}} \le C\sqrt{(n+1)h}.
\end{split}
\end{equation}
\end{lem}
\begin{proof}
Let $\Pi_g[0,h]:= \{p_1<p_2<\ldots<p_n \}$ with $p_0=0, p_{n+1}=h.$
For the first one, we have
\begin{equation}
\intop_0^h \frac{dt}{\pi_1(t;g)+1} = \sum_{i=1}^{n+1}
\intop_{0}^{p_i-p_{i-1}} \frac{dx}{x+1} \le (n+1) \log(h+1).
\end{equation}
The second one follows by
\begin{equation}\label{eq:intbasicaux}
\intop_0^h \frac{dt}{\sqrt{\pi_1(t;g)+1}} =
\sum_{i=1}^{n+1}
\intop_0^{p_i-p_{i-1}} \frac{dx}{\sqrt{x+1}}
\le \sum_{i=1}^{n+1} 2\sqrt{ p_i-p_{i-1}}\le 2\sqrt{(n+1)h},
\end{equation}
where we used Cauchy-Schwarz inequality to obtain the last inequality.
\end{proof}
\subsection{Connection to martingales}\label{subsec:branching:mg}
Recall the definition $\mathcal{R}_b(t;\Pi_0[t_0^-,t_0],\alpha)$ \eqref{eq:def:Rb}. For a given point process $\Pi_Q$ with respect to $Q(t)>0$, we introduce a similar notation given by
\begin{equation}\label{eq:def:Qst}
\mathcal{R}_c(s,t;\Pi_Q[t_0^-,s],\alpha):=
\intop_{t_0^{-}}^s K_{\alpha_0}(t-x) d\Pi_Q(x) + \intop_{t_0^{-}}^{s} \intop_{s}^t K^*_{\alpha_0}(t-u)K_{\alpha_0}(u-x)du d\Pi_Q(x).
\end{equation}
In words, it is a conditional expectation of the rate at time $t$ of the critical branching process starting from the points $\Pi_Q[t_0^-,s]$, conditioned on $\Pi_Q[t_0^-,s]$. In particular,
$\mathcal{R}_c(t,t;\Pi_Q[t_0^-,t],\alpha) = \mathcal{R}_b(t;\Pi_Q[t_0^-,t],\alpha)$.
Abbreviating the notation by $Q_1(s,t):= \mathcal{R}_c(s,t;\Pi_Q[t_0^-,s],\alpha)$,
the main advantage of studying this process comes from the following observation, which is an analogue of \eqref{eq:Lt diff:basic}:
\begin{equation}\label{eq:integralform:branching:pre}
\begin{split}
\frac{\partial}{\partial s} Q_1(s,t) &= \left(K_{\alpha}(t-s) +\intop_s^t K_{\alpha}^*(t-u) K_{\alpha}(u-s) du \right) \frac{d\Pi_Q(s)}{ds} -\intop_{t_0^-}^s K_{\alpha}(s-x) d\Pi_Q(x) \\
&=K^*_{\alpha}(t-s) \left( \frac{d\Pi_Q(s)}{ds} - Q_1(s,s) \right) ,
\end{split}
\end{equation}
where the second line comes from the fact that $K^*_{\alpha_0}=K_{\alpha_0}+K_{\alpha_0}*K^*_{\alpha_0}$. This gives
\begin{equation}\label{eq:integralform:branching}
\begin{split}
Q_1(s_1,t)-Q_1(s_0,t) &= \intop_{s_0}^{s_1} K^*_\alpha (t-s) \big\{ d\Pi_Q(s)- Q_1(s,s)ds \big\}\\
&= \intop_{s_0}^{s_1} K^*_\alpha (t-s) \left\{ d\widetilde{\Pi}_Q(s)+ \big[Q(s)-Q_1(s,s) \big]ds \right\} ,
\end{split}
\end{equation}
where we wrote $d\widetilde{\Pi}_Q(s):= d\Pi_Q(s)-Q(s)ds$. This decomposes the identity to the ``martingale part'' and the ``drift part.''
In the definition of $\mathcal{R}_c$, the case $t=\infty$ will play important roles, and we specify this case with another notation:
\begin{equation}\label{eq:def:L}
\begin{split}
\mathcal{L}(t; \Pi_Q[t_0^{ -}, t],\alpha) :=
\intop_{t_0^{ -}}^{t} \intop_{t}^{\infty} K^*_\alpha \cdot K_\alpha(x-s) dx d\Pi_Q (s)
.
\end{split}
\end{equation}
where $K^*_\alpha = \lim_{t\to\infty} K^*_\alpha(t)= \frac{2\alpha^2}{1+2\alpha}$ (Lemma \ref{lem:estimat for K tilde:intro}). Note that the definition is an analogue of \eqref{eq:def:Lt:basic form}.
\subsection{The critical branching from a single particle}\label{subsec:branching:numberofpts}
In this subsection, we study the age-dependent critical branching process that starts from a single initial particle. The results in this subsection will be useful tools in the next subsection.
We let $\{r(t)\}_{t\ge 0}$ denote the rate of the critical branching process starting from a single point at the origin ($t=0$). That is, we define $r(t)$ by
\begin{equation}\label{eq:def:rt:branching singlept}
r(t) = r(t,\alpha) := K_\alpha(t)+ \intop_{0+}^t K_\alpha(t-x) d\Pi_r(x).
\end{equation}
Also, we set $\hat{h}:= 2\alpha^{-2}\beta^\theta$. Recall the definitions of $\pi_i(t;g)$ \eqref{eq:def:pi closest points} and $ \sigma_i(t;g)$ \eqref{eq:def:sig closest points}, and
consider the stopping times
\begin{equation}\label{eq:def:tauSB}
\begin{split}
\tau_{\textnormal{SB}1} &:= \inf\left\{t\ge 0: |\Pi_r[0,t]| \ge \beta^{\frac{7}{2}\theta} \right\};\\
\tau_{\textnormal{SB}2} &:=
\inf \left\{t\ge 0: r(t) \ge \alpha \beta^{4\theta} \sigma_1(t;r) + \alpha^2 \right\};\\
\tau_{\textnormal{SB}} &:= \tau_{\textnormal{SB}1} \wedge \tau_{\textnormal{SB}2}.
\end{split}
\end{equation}
Then, our main estimate can be described as follows.
\begin{prop}\label{prop:branching:sb:main}
Under the above setting, we have
\begin{equation}
\mathbb{P} \left(\tau_{\textnormal{SB}} \le \hat{h} \right) \le \exp\left(-\beta^4 \right).
\end{equation}
\end{prop}
Furthermore, we will deduce a bound on the probability that $\Pi_r$ contains more than certain number of points in a short interval. For instance, for any $t\in [k\alpha^{-\frac{3}{2}}, (k+1)\alpha^{-\frac{3}{2}} ]$ with $k\in \mathbb{N}$, we can write
\begin{equation}\label{eq:branching:sb:ge 1}
\mathbb{P} \left(|\Pi_r[t-\alpha^{-\frac{3}{2}}, t]| \ge 1 \right) \le \mathbb{E} |\Pi_r[t-\alpha^{-\frac{3}{2}},t]|
=
\intop_{t-\alpha^{-\frac{3}{2}}}^t K^*_\alpha(x) dx \le C \left\{k^{-\frac{1}{2}}\alpha^{\frac{1}{4}} \vee \alpha^{\frac{1}{2}} \right\},
\end{equation}
for some constant $C>0$. Note that the last inequality is from the estimate on $K_\alpha^*(x)$ (Lemma \ref{lem:estimat for K tilde:intro}). Our goal is to deduce a similar inequality as follows.
\begin{prop}\label{prop:branching:sb:ge3}
Under the above setting, for all $\alpha^{-\frac{3}{2}}\le t\le \hat{h}$ we have
\begin{equation}
\mathbb{P} \left(\big|\Pi_r[(t-\alpha^{-\frac{3}{2}})\wedge \tau_{\textnormal{SB}},t \wedge \tau_{\textnormal{SB}}] \big| \ge 3 \right) \le \alpha^{\frac{3}{4}} \beta^{19\theta}.
\end{equation}
\end{prop}
The reason why we are interested in the intervals of length $\alpha^{-3/2}$ will become more clear in the next subsection. To explain it briefly, we interpret $R(t)$ as a perturbation of the rate-$\alpha$ process, and it turns out that $\alpha^{-3/2}$ is the length of an interval where we start to see particles between $R(t)$ and $\alpha$. The proof of Proposition \ref{prop:branching:sb:ge3} follows as a consequence of Proposition \ref{prop:branching:sb:main}, and we discuss it at the end of this subsection.
Now we establish Proposition \ref{prop:branching:sb:main}, beginning with the control on $\tau_{\textnormal{SB}1}$.
\begin{lem}\label{lem:numberofpts:branching}
For any sufficiently small $\alpha>0$, we have $\mathbb{P} \left( \tau_{\textnormal{SB}1} \le \hat{h} \right) \le \exp\left(-\beta^{\frac{1}{200}\theta} \right)$. In other words,
\begin{equation}
\mathbb{P} \left( |\Pi_r[0,\hat{h}]| \ge \beta^{\frac{7}{2}\theta} \right) \le \exp\left(-\beta^{\frac{1}{200}\theta} \right).
\end{equation}
\end{lem}
To establish Lemma \ref{lem:numberofpts:branching}, we start with verifying a similar property for the critical Galton-Watson branching process with Poisson offspring.
\begin{lem}\label{lem:bound on branching}
Let $Z_n$ be a critical branching process with offspring distribution $\text{Poisson}(1)$ and $Z_0=1$. Then
\begin{equation}
\mathbb P \left( \sum _{k=1}^n Z_k\ge n^3 \right)\le e^{-cn}
\end{equation}
\end{lem}
\begin{proof}
Let $\varphi _n$ be the probability generating function of $X_n$. For any $u >0$ we have
\begin{equation}
\begin{split}
\varphi _n (u)=\mathbb E u ^{Z_n}= \mathbb E \left[ \mathbb E \left[ u ^{Z_n} |Z_{n-1}\right] \right]=\mathbb E \left[ (\mathbb E u ^{X_1} )^{X_{n-1}} \right]= \varphi _{n-1}(e^{u-1 }) = \psi^{(n)}(u),
\end{split}
\end{equation}
where $\psi(u):= e^{u-1}$ and where $\psi^{(n)}(u)$ is the $n$-fold composition of $\psi $ with itself.
Let $u :=1+ 1/5n$. We prove using induction on $k \le n$ that $\varphi _k(u ) \le 1+\frac{1}{5 n}+\frac{k}{16 n^2}$. In fact,
\begin{equation}
\varphi _{k+1}(u )=\exp (\varphi_{k}(u )-1)\le \varphi _k (u )+(\varphi _k (u )-1)^2 \le 1+\frac{1}{5n}+\frac{k+1}{16n^2}
\end{equation}
where in the last inequality we used that $\varphi _k(u )\le 1+1/4n$ which follows from the induction hypothesis when $k \le n$.
Finally, using Markov's inequality we get for any $k \le n$
\begin{equation}
\mathbb P (Z_k \ge n^2 )= \mathbb P \left( u ^{Z_k}\ge u ^{n^2 } \right) \le \varphi _k (u ) \cdot u ^{-n^2}\le e^{-cn}.
\end{equation}
The result now follows from a union bound.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:numberofpts:branching}]
Let $X_i$ be $i.i.d$ with density $K_\alpha$. We have $\mathbb P (X_1\le \alpha ^{-2})\le 1-c$ for some $c>0$. Thus
\begin{equation}\label{eq:bound on sum}
\mathbb P \left( \sum _{k=1}^{\beta^{\frac{11}{10}\theta} } X_k \le \hat{h}\right) \le \mathbb P \left( \sum _{k=1}^{\beta^{\frac{11}{10}\theta} } \mathds{1}_{\{ X_k \le \alpha ^{-2} \}} \ge \left(1-\frac{c}{2}\right) \beta^{\frac{11}{10}\theta} \right) \le \exp\left(-\beta ^{\frac{1}{100}\theta}\right),
\end{equation}
where in the last inequality we used Azuma. Now, if there are more than $\beta^{\frac{7}{2}\theta}$ particles in the branching before time $\hat{h}$ then either there are more than $\beta^{4\theta}$ particles up to the $\beta^{\frac{11}{10}\theta}$ generation or there are less particles than that up to the $\beta^{\frac{11}{10}\theta}$ generation but one of the particles in this generation came before time $\hat{h}$. The first event happens with very small probability by Lemma \ref{lem:bound on branching} and the second one happens with small probability by union bound and \eqref{eq:bound on sum}.
\end{proof}
We conlude the proof of Proposition \ref{prop:branching:sb:main}.
\begin{proof}[Proof of Proposition \ref{prop:branching:sb:main}]
For all $t\le \tau_{\textnormal{SB}1}$, observe that
\begin{equation}
\begin{split}
r(t) &=K_\alpha(t)+ \intop_{0+}^t K_\alpha(t-x) d\Pi_r(x) \\&\le K_\alpha(\pi_1(t;r)\wedge \alpha^{-2}\beta^\theta) \cdot \big| \Pi_r[0,t] \big| \le
\alpha\beta^{4\theta} \sigma_1(t;r) + \alpha^2,
\end{split}
\end{equation}
where the last inequality follows from the estimate on $K_\alpha(x)$ (Lemma \ref{lem:estimate on K:intro}). Thus, we obtain the desired result by combining with Lemma \ref{lem:numberofpts:branching}
\end{proof}
We conclude this subsection by verifying Proposition \ref{prop:branching:sb:ge3}.
\begin{proof}[Proof of Proposition \ref{prop:branching:sb:ge3}]
For any $\alpha^{-3/2}\le t\le \hat{h}$, denote $s_1 := t-\alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}}$ and $s_2:= t\wedge \tau_{\textnormal{SB}}$. First, based on the definition of $\tau_{\textnormal{SB}2}$, applying Lemma \ref{lem:ind1:pi1int basicbd:basic} with $n=\beta^{\frac{7}{2}\theta}$ gives that
\begin{equation}
\intop_{s_1}^{s_2} r(x)dx \le \alpha \beta^{4\theta} \cdot \alpha^{-\frac{3}{4}} \sqrt{n}\le \alpha^{\frac{1}{4}}\beta^{6\theta}.
\end{equation}
Thus, we can write
\begin{equation}
\begin{split}
\mathbb{P} \left( \left| \Pi_r[s_1,s_2] \right| \ge 3\right)
\le \intop_{s_1\le x<y<z\le s_2} r(x)r(y)r(z)dzdydx
\le \alpha^{\frac{3}{4}}\beta^{18\theta},
\end{split}
\end{equation}
noting that any addition of three extra points in $\Pi_r[s_1,s_2]$ does not change the bound given in the definition of $\tau_{\textnormal{SB}2}$. Thus, we conclude the proof by combining the above with Proposition \ref{prop:branching:sb:main}.
\end{proof}
\subsection{Understanding the fixed rate process and its perturbations}\label{subsec:branching:ib}
In this subsection, we study the processes $R_0(t), R_0^-(t)$, and $R_0^+(t)$ introduced in \eqref{eq:def:R0:ib}, \eqref{eq:def:R:ib:aux}. We begin with observing that $R_0^-$ (resp. $R_0^+$) lower (resp. upper) bounds both $R_0$ and $R$.
\begin{prop}\label{prop:branching:ib:sandwich}
Under the above setting, $R_0^-(t)\le R(t), R_0(t) \le R_0^+(t)$ for all $t\ge t_0$.
\end{prop}
\begin{proof}
Observe first that at $t = t_0$, $R^+(t_0)=R^-(t_0)=R(t_0)=R_0(t_0)$, and for $t\in[t_0^{-},t_0)$ they are all identical to $\alpha$. Recalling the definition of $R(t)$, which can be written as
\begin{equation}
R(t)= \intop_{t_0^{-}}^{t_0} K_{\alpha}(t-x)d\Pi_\alpha(x) + \intop_{t_0}^{t} K_{\alpha}(t-x)d\Pi_{R}(x),
\end{equation}
we observe that $R_0^-(t)\le R(t)\le R_0^+(t)$: They maintain the same value until time $t_0$, and then after that $R_0^+(t)$ picks up more points than $R(t)$, since it absorbs additional points from $(R_0-\alpha)_+$ in addition to all the points taken by $R (t)$. The same reasoning for opposite direction works for $R_0^-(t)$.
To see that the conclusion holds for $R_0$, we observe that $R_0^+(t) \ge R_0(t)$ for $t\ge t_0$ inductively: Initially, we have $R_0^+(t_0) = R_0(t_0)$. Furthermore, as long as we have $ R_0^+(s) \ge R_0(s)$ for all $s < t$, it holds that
\begin{equation}
R_0^+(t) + (R_0(t)-\alpha)_+
\ge
R_0(t) + (\alpha - R_0(t) ) \ge \alpha,
\end{equation}
meaning that the new points picked up by $R_0(t)$ will also be included in $R_0^+(t)$. Thus, it will continue to hold that $R_0^+(t) \ge R_0(t)$. The other inequality, $R_0^- \le R_0$, can be obtained analogously.
\end{proof}
We move on to the study of $R_0(t)$.
Define the notation
\begin{equation}\label{eq:def:Pi triangle}
\Pi_{R_0 \triangle \alpha} [ x_0,x_1]:=
\Pi_{R_0}[x_0,x_1] \triangle \Pi_{\alpha}[x_0,x_1] ,
\end{equation}
that is, the collection of points lying between $R_0$ and $\alpha$.
Then, define
\begin{equation}\label{eq:def:tauf}
\begin{split}
\tau_{\textnormal{f}1}&:= \inf\left\{t\ge t_0: (R_0(t)-\alpha) \notin \left(-\alpha^{\frac{3}{2}}\beta^{C_\circ},\, \alpha\beta^{C_\circ} \sigma_1(t;\alpha) + \alpha^{\frac{3}{2}}\beta^{C_\circ} \right) \right\};\\
\tau_{\textnormal{f}2}&:= \inf \left\{t\ge t_0^{-}: |\Pi_\alpha[(t-\alpha^{-1})\vee t_0^{-}, \, t ]| \ge \beta^{6} \right\};\\
\tau_{\textnormal{f}3}&:= \inf \left\{ t \ge t_0: \left| \Pi_{R_0 \triangle \alpha}[(t-\alpha^{-\frac{3}{2}})\vee t_0,\, t] \right| \ge \beta^{2C_\circ} \right\};\\
\tau_{\textnormal{f}4}&:=\inf \left\{t\ge t_0: R_0(t)\ge\alpha \beta^{C_\circ} \right\};\\
\tau_{\textnormal{f}}&:=
\tau_{\textnormal{f}1}\wedge \tau_{\textnormal{f}2}\wedge \tau_{\textnormal{f}3}\wedge \tau_{\textnormal{f}4}.
\end{split}
\end{equation}
The goal of this subsection is establishing the following lemma.
\begin{lem}\label{lem:branching:ib:R0 vs alpha}
We have $$\mathbb{P}(\tau_{\textnormal{f}} \le \acute{t}_0) \le \exp(-\beta^5).$$
\end{lem}
\begin{proof}
By an elementary estimate on Poisson processes, we have
\begin{equation}\label{eq:branching:ib:R0:med}
\mathbb{P}(\tau_{\textnormal{f}2} \le \acute{t}_0 ) \le \exp \left(-\beta^6 \right).
\end{equation}
To obtain appropriate control on $\tau_{\textnormal{f}1}$, we apply Corollary
\ref{lem:concentrationofint:continuity} to $R_0(t)$. Recalling the estimate Lemma \ref{lem:estimate on K:intro} on $K_\alpha$, we may set $M = \alpha^3\beta^2$. Writing $d\widetilde{\Pi}_\alpha(x) := d\Pi_\alpha(x) - \alpha dx$, this gives that
\begin{equation}\label{eq:branching:ib:R0:med1}
\begin{split}
\mathbb{P} \left( \intop_{t_0^{-}}^{t\wedge \tau_{\textnormal{f}2}} K_\alpha(t-x) d\widetilde{\Pi}_\alpha (x) \le \alpha \beta^{C_\circ}\sigma_1(t;\alpha) + \frac{1}{2}\alpha^{\frac{3}{2}}\beta^{C_\circ}, \ \forall t\in[t_0,\acute{t}_0] \right) \ge 1- \exp\left(-\beta^6\right);\\
\mathbb{P} \left( \intop_{t_0^{-}}^{t\wedge \tau_{\textnormal{f}2}} K_\alpha(t-x) d\widetilde{\Pi}_\alpha (x) \ge -\frac{1}{2}\alpha^{\frac{3}{2}}\beta^{C_\circ}, \ \forall t\in[t_0,\acute{t}_0] \right) \ge 1- \exp\left(-\beta^6\right).
\end{split}
\end{equation}
Moreover, we have for all $t\in [t_0,\acute{t}_0]$ that
\begin{equation}\label{eq:branching:ib:R0:med0}
\begin{split}
R_0(t)-\alpha &= \intop_{t_0^{-}}^t K_\alpha(t-x) d\widetilde{\Pi}_\alpha(x)
+
\intop_{t_0^{-}}^{t} K_\alpha(t-x) \alpha dx - \alpha\\
&= \intop_{t_0^{-}}^t K_\alpha(t-x) d\widetilde{\Pi}_\alpha(x) +O\left(\alpha^{100}\right),
\end{split}
\end{equation}
where the last identity follows again from the estimate of $K_\alpha(s)$ (Lemma \ref{lem:estimate on K:intro}) applied to $s\ge t_0-t_0^-$. Thus, combining this with \eqref{eq:branching:ib:R0:med} and \eqref{eq:branching:ib:R0:med1}, we obtain that
\begin{equation}\label{eq:branching:ib:R0:med2}
\mathbb{P} \left(\tau_{\textnormal{f}1} \le \acute{t}_0 \right) \le 2\exp\left(-\beta^6\right).
\end{equation}
Moving on to $\tau_{\textnormal{f}3}$, write $\tilde{\tau}_{\textnormal{f}}:= \tau_{\textnormal{f}1} \wedge \tau_{\textnormal{f}2}$, and observe that from applying Lemma \ref{lem:ind1:pi1int basicbd:basic} with $n=\alpha^{-\frac{1}{2}} \beta^{C_\circ}$, we get
\begin{equation}\label{eq:reg:branching:tau2:split0}
\intop_{(t-\alpha^{-\frac{3}{2}})\wedge \tilde{\tau}_{\textnormal{f}}}^{t\wedge \tilde{\tau}_{\textnormal{f}}} |R_0(t)-\alpha| \le 2\beta_0^{\frac{3}{2}C_\circ +1}.
\end{equation}
Thus, applying Corollary \ref{cor:concentration:numberofpts each interval} to $\Pi_{R_0 \triangle \alpha}$ gives that
\begin{equation}\label{eq:branching:ib:R0:med3}
\mathbb{P}\left({\tau}_{\textnormal{f}3} \le \acute{t}_0 \wedge \tilde{\tau}_{\textnormal{f}} \right) \le \exp\left(-\beta^6 \right).
\end{equation}
Lastly, we note that $\tau_{\textnormal{f}4} \ge \tau_{\textnormal{f}1}$ deterministically from their definitions.
Thus, along with \eqref{eq:branching:ib:R0:med}, \eqref{eq:branching:ib:R0:med2}, and \eqref{eq:branching:ib:R0:med3}, we obtain the desired conclusion.
\end{proof}
\subsection{Proximity to $\alpha$ for small $t$}\label{subsubsec:branching:ib:tau4}
In this subsection, we conclude the proof of Theorem \ref{thm:branching:ib}, combining the results from the previous subsections.
We begin with investigating $\tau_{{\textnormal{B}1}}$. Thanks to Proposition \ref{prop:branching:ib:sandwich}, we can write
\begin{equation}\label{eq:branching:ib:decomp}
|R(t)-\alpha| \le |R_0(t)-\alpha| +\left\{R_0^+(t)-R_0^-(t) \right\}.
\end{equation}
Since the bound on $|R_0(t)-\alpha|$ has already been given by $\tau_{\textnormal{f}1} $ in Lemma \ref{lem:branching:ib:R0 vs alpha}, we focus on the term $\left\{R_0^+(t)-R_0^-(t) \right\}$.
By the definition of $R_0^+$ and $R_0^-$ \eqref{eq:def:R:ib:aux}, the process $\delta R_0(t):= R_0^+(t)- R_0^-(t)$ satisfies
\begin{equation}\label{eq:def:delR}
\delta R_0(t) = \intop_{t_0}^t K_\alpha(t-x) d\Pi_{(R_0^++(R_0-\alpha)_+) \triangle (R_0^--(R_0-\alpha)_-)}(x),
\end{equation}
for all $t\ge t_0$, starting with $\delta R_0(t_0) = 0$. To lighten up our notation, we symbolically define
\begin{equation}\label{eq:def:branching:ib:delRhat}
\hat{\Delta} {R}_0 := (R_0^++(R_0-\alpha)_+) \triangle (R_0^--(R_0-\alpha)_-).
\end{equation}
\begin{remark}\label{rmk:branching:ib:delR}
In \eqref{eq:def:delR}, we can interpret $\delta R_0(t)$ in the following way: $\delta R_0$ starts as an empty process at time $t_0$. As time goes on, new particles are added to $\delta R_0$ at rate $|R_0-\alpha|$. We call these particles \textbf{the external additions}. Then, each particle from external addition performs the critical branching process.
\end{remark}
To study the process $\delta R_0(t)$, we introduce the following stopping time.
\begin{equation}
\begin{split}
\tau_{\delta \textnormal{B}1}= \tau_{\delta \textnormal{B}1}(\alpha, t_0) := \inf \left\{ t\ge t_0: \left|\Pi_{\hat{\Delta} {R}_0}[(t-\alpha^{-\frac{3}{2}})\vee t_0, t] \right| \ge \beta^{4\theta} \right\}.
\end{split}
\end{equation}
Our first goal is to show the following lemma on $\tau_{\delta \textnormal{B}1}$.
\begin{lem}\label{lem:branching:ib:taudelta}
Under the above setting, we have
\begin{equation}
\mathbb{P} \left(\tau_{\delta \textnormal{B}1} \le \acute{t}_0 \right) \le \exp\left(-\beta^3 \right).
\end{equation}
\end{lem}
\begin{proof}
We interpret $\delta R(t)$ as mentioned in Remark \ref{rmk:branching:ib:delR}, recalling the notion of \textit{external additions}. For each $j\in \mathbb{N}, \, 0\le j \le 2\alpha^{-\frac{1}{2}}\beta^\theta$, we write
$ \{ p_{j, 1}, \ldots, p_{j, n_j} \}$ to be the {external additions} at $\delta R_0(t)$ during time $[j\alpha^{-\frac{3}{2}}, (j+1)\alpha^{-\frac{3}{2}}]$.
For each point $p_{j,l}$ and the critical branching process $r_{j,l}(t)$ that starts from a single point at $p_{j,l}$, we define the stopping time $\tau_{\textnormal{SB}}(j,l)$ analogously as \eqref{eq:def:tauSB}:
\begin{equation}
\begin{split}
\tau_{\textnormal{SB}1}(j,l)&:= \inf \left\{ t\ge p_{j,l}: \big| \Pi'_{r_{j,l}}[0,t] \big| \ge \beta^{\frac{7}{2}\theta} \right\};\\
\tau_{\textnormal{SB}2}(j,l)&:=
\inf \left\{t\ge p_{j,l}: r_{j,l} \ge \alpha \beta^{4\theta}\sigma_1(t;r_{j,l}) +\alpha^2 \right\};\\
\tau_{\textnormal{SB}}(j,l)&:=
\tau_{\textnormal{SB}1}(j,l)\wedge \tau_{\textnormal{SB}2}(j,l);\\
\tau_{\textnormal{SB}} &:=
\tau_{\textnormal{f}} \wedge
\bigwedge_{j=0}^{2\alpha^{-\frac{1}{2}}\beta^\theta} \bigwedge_{l=1}^{n_j} \tau_{\textnormal{SB}}(j,l).
\end{split}
\end{equation}
(Recall the definition of $\tau_{\textnormal{f}}$ in \eqref{eq:def:tauf}) Moreover, for $k, j,l \in \mathbb{N}$ with $0\le k , j \le 2\alpha^{-\frac{1}{2}}\beta^\theta$ and $1\le l\le n_j$, we define
\begin{equation}
\begin{split}
N_{j,l}^{(k)} :=& \textnormal{ the number of points in } \left|\Pi_{\hat{\Delta} R_0}[k\alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}},\, (k+1) \alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}} ] \right| \\ &\textnormal{ that are descendents of } p_{j,l}.
\end{split}
\end{equation}
Note that for each $k$, $\{N^{(k)}_{j,l}\}_{j,l}$ forms a collection of independent random variables. Furthermore, due to the definition of $\tau_{\textnormal{f}3}$, we can consider $l\le \beta^{2C_\circ}$ only.
For each $k,j,l$ as above, define $\bar{N}^{(k)}_{j,l}$ to be the random variable as follows:
\begin{itemize}
\item $\bar{N}^{(k)}_{j,l} \equiv 0 $ for any $j>k$.
\item Otherwise, we define
\begin{equation}
\bar{N}^{(k)}_{j,l} = \begin{cases}
2, &\textnormal{ with probability } C\left\{ (k-j+1)^{-\frac{1}{2}}\alpha^{\frac{1}{4}} \vee \alpha^{\frac{1}{2}} \right\};\\
\beta^{\frac{7}{2}\theta},& \textnormal{ with probability } \alpha^{\frac{3}{4}} \beta^{19\theta};\\
0, & \textnormal{ otherwise.}
\end{cases}
\end{equation}
\item For each $k$, $\{\bar{N}_{j,l}^{(k)} \}_{j,l}$ is a collection of independent random variables.
\end{itemize}
Then, \eqref{eq:branching:sb:ge 1} and Proposition \ref{prop:branching:sb:ge3} tell us that $\{\bar{N}^{(k)}_{j,l} \}_{k,j,l}$ stochastically dominates $\{N^{(k)}_{j,l} \}_{k,j,l}$.
Now, we estimate
\begin{equation}
\left|\Pi_{\hat{\Delta} R_0}[k\alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}},\, (k+1) \alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}} ] \right| \preceq \sum_{j\le k} \sum_{l=1}^{n_j} \bar{N}^{(k)}_{j,l}.
\end{equation}
First, we note that each $\bar{N}_{j,l}^{(k)}$ is bounded by $\beta^{\frac{7}{2}\theta}$ and also observe that
\begin{equation}
\begin{split}
\sum_{j,l} \mathbb{E}\left[\bar{N}^{(k)}_{j,l} \right]&\le
\sum_{j=1}^{2\alpha^{-\frac{1}{2}}\beta^\theta} \sum_{l=1}^{\beta^{2C_\circ}} C \left\{
j^{-\frac{1}{2}}\alpha^{\frac{1}{4}} \vee \alpha^{\frac{1}{2}} \right\} \le \beta^{2\theta}
\\
\sum_{j,l} \mathbb{E}\left[\left(\bar{N}^{(k)}_{j,l} \right)^2 \right] &\le \sum_{j=1}^{2\alpha^{-\frac{1}{2}}\beta^\theta} \sum_{l=1}^{\beta^{2C_\circ}} C' \left\{
j^{-\frac{1}{2}}\alpha^{\frac{1}{4}} \vee \alpha^{\frac{1}{2}} \right\} \le \beta^{2\theta}.
\end{split}
\end{equation}
Thus, Bernstein's inequality (Lemma \ref{lem:branching:bernstein} below) tells us that for each $k$,
\begin{equation}
\begin{split}
\mathbb{P} \left( \left|\Pi_{\hat{\Delta} R_0}[k\alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}},\, (k+1) \alpha^{-\frac{3}{2}}\wedge \tau_{\textnormal{SB}} ] \right| \ge \beta^{4\theta} \right) \le \exp\left(-\beta^{\theta/3} \right).
\end{split}
\end{equation}
Thus, we can conclude the proof by taking a union bound over $k$, and by noting that
\begin{equation}
\mathbb{P} \left(\tau_{\textnormal{SB}} > \acute{t}_0 \right) \ge 1-\exp\left(-\beta^3 \right),
\end{equation}
which comes from Proposition \ref{prop:branching:sb:main} and Lemma \ref{lem:branching:ib:R0 vs alpha} followed by a union bound.
\end{proof}
The Bernstein's inequality used in the above proof can be stated as below.
\begin{lem}[Theorem 3.6,~\cite{chung2006concentration}]\label{lem:branching:bernstein}
Let $X_1, \ldots, X_n$ be independent random variables satisfying $|X_i|\le M$ almost surely for all $i$. Then, we have
\begin{equation}
\mathbb{P}\left( \sum_{i=1}^n (X_i-\mathbb{E}X_i) \ge x \right) \le \exp\left( -\frac{x^2}{2\sum_{i=1}^n \mathbb{E}[X_i^2] + \frac{2}{3} Mx } \right).
\end{equation}
\end{lem}
The following corollary translates Lemma \ref{lem:branching:ib:taudelta} into the form that we can apply to our analysis.
\begin{cor}\label{cor:branching:ib:taudelta}
Recall the definition of $\hat{\Delta} {R}_0(t)$ \eqref{eq:def:branching:ib:delRhat}, and
define the stopping times
\begin{equation}
\begin{split}
\tau_{\delta \textnormal{B}2}=\tau_{\delta \textnormal{B}2}&:=
\inf \left\{t\ge t_0 : \delta R_0(t) \le \alpha\beta^{4\theta+1}\sigma_1(t;\hat{\Delta} {R}_0) + \alpha^{\frac{3}{2}}\beta^{4\theta+1} \right\};\\
\tau_{\delta \textnormal{B}3}=\tau_{\delta \textnormal{B}3}&:=
\inf \left\{t\ge t_0 : \delta R_0(t) \le 2\alpha\beta^{4\theta+1}\right\}.
\end{split}
\end{equation}
Then, we have
\begin{equation}
\mathbb{P} \left( \tau_{\delta\textnormal{B}3} >\acute{t}_0 \right)\ge \mathbb{P} \left( \tau_{\delta\textnormal{B}2} >\acute{t}_0 \right) \ge 1-2\exp\left(-\beta^3 \right).
\end{equation}
\end{cor}
\begin{proof}
The first inequality is immediate from definition. To establish the second inequality, recall the definition of $\tau_{\textnormal{f}3}$ from \eqref{eq:def:tauf}. For $t \le \tau_{\textnormal{f}3}\wedge \tau_{\delta \textnormal{B}1}$, note that
\begin{equation}
\begin{split}
&\delta R_0(t) = \intop_{t_0}^t K_{\alpha_0}({t-x}) d\Pi_{\hat{\Delta} {R}_0}(x) \\
&\le
\big|\Pi_{\hat{\Delta} {R}_0}[t-\alpha^{-\frac{3}{2}}, t] \big| \cdot C\alpha \sigma_1(t;\hat{\Delta} {R}_0)
+
\sum_{k\ge 1} \big|\Pi_{\hat{\Delta} {R}_0}[t-(k+1)\alpha^{-\frac{3}{2}}, t-k\alpha^{-\frac{3}{2}}] \big| \cdot
\frac{C\alpha e^{-ck\alpha^{\frac{1}{2}}}}{\sqrt{k\alpha^{-\frac{3}{2}}}}\\
&\le \alpha\beta^{4\theta+1}\sigma_1(t;\hat{\Delta} {R}_0)+ \alpha^{\frac{3}{2}}\beta^{4\theta+1}.
\end{split}
\end{equation}
Thus, the proof follows from Lemmas \ref{lem:branching:ib:R0 vs alpha} and \ref{lem:branching:ib:taudelta}.
\end{proof}
We now are ready to establish Theorem \ref{thm:branching:ib}.
\begin{proof}[Proof of Theorem \ref{thm:branching:ib}]
Recall the decomposition \eqref{eq:branching:ib:decomp}, which can be rewritten as
\begin{equation}\label{eq:branching:ib:decomp:coupled}
R_0^-(t)-\alpha \le R(t)-\alpha \le (R_0(t)-\alpha )+ \delta R_0(t).
\end{equation}
From the upper bound, the results on $\tau_{\textnormal{f}3} $ (Lemma \ref{lem:branching:ib:R0 vs alpha}) and $\tau_{\delta \textnormal{B}2}$ give the control
$$R(t)-\alpha \le \alpha\beta^{5\theta}\sigma_1(t;\hat{R}_0^+) + \alpha^{\frac{3}{2}}\beta^{5\theta} . $$
For the lower bound, we study $R_0^-(t)$ instead. Let $t_0^\flat$ be another parameter such that $t_0^\flat := \frac{1}{2}(t_0^-+t_0). $ In particular, it satisfies
\begin{equation}
t_0-t_0^\flat = t_0^\flat - t_0^- \ge \frac{1}{4}\alpha^{-2}\beta^\theta.
\end{equation}
Noting that $R_0^-(t)=R_0(t)=\alpha$ for $t<t_0$, applying \eqref{eq:integralform:branching} to $R_0^-$ gives the following:
\begin{equation}\label{eq:branching:R0minus intform}
\begin{split}
R_0^-(t) - \mathcal{R}_c(t_0^\flat,t;\Pi_\alpha[t_0^-,t_0^\flat],\alpha )
=\intop_{t_0^\flat}^t K^*_\alpha (t-s) \big\{ \widetilde{\Pi}_{R_0^--(R_0-\alpha)_-}(s) - (R_0(s)-\alpha)ds \big\},
\end{split}
\end{equation}
The first integral can be estimated using Corollary \ref{lem:concentrationofint:continuity}: Since $R_0^-(t)-(R_0(t)-\alpha)_-\le R_0(t)$ for $t\ge t_0$, the definition of $\tau_{\textnormal{f}4}$ and Lemma \ref{lem:estimat for K tilde:intro} imply that
\begin{equation}
\begin{split}
\intop_{t_0^\flat}^{t_0\wedge \tau_{\textnormal{f}4}}
(K^*_\alpha(t-s) )^2 \alpha ds + \intop_{t_0}^{t\wedge \tau_{\textnormal{f}4}}
(K^*_\alpha(t-s) )^2 R_0(s) ds \\
\le \intop_{t_0^\flat}^{t} C\left(\frac{\alpha^2}{t-s+1} \vee \alpha^4 \right) \alpha\beta^{C_\circ} ds \le \alpha^3\beta^{\theta+2C_\circ},
\end{split}
\end{equation}
for all $t\le \acute{t}_0$. Thus, Corollary \ref{lem:concentrationofint:continuity} and Lemma \ref{lem:branching:ib:R0 vs alpha} tell us that
\begin{equation}\label{eq:branching:Rlbd:med1}
\mathbb{P} \left( \intop_{t_0^\flat}^t K^*_\alpha (t-s) \widetilde{\Pi}_{R_0^--(R_0-\alpha)_-}(s) \ge -\alpha^{\frac{3}{2}}\beta^\theta, \ \forall t\in [t_0,\acute{t}_0] \right) \ge 1- \exp\left(-\beta^5 \right).
\end{equation}
On the other hand, for $t\le \tau_{\textnormal{f}1}$ we can write (note that when $s\le t_0$, $R_0(s)-\alpha=0$)
\begin{equation}\label{eq:branching:Rlbd:med2}
\left|\intop_{t_0^\flat}^t K^*_\alpha (t-s) (R_0(s)-\alpha)ds\right|
\le \intop_{t_0}^t C\left(\frac{\alpha}{\sqrt{t-s+1}} \vee \alpha^2 \right) \left(\alpha\beta^{C_\circ} \sigma_1(s;\alpha) + \alpha^{\frac{3}{2}}\beta^{C_\circ} \right)ds.
\end{equation}
Then, the RHS can be bounded using Lemmas \ref{lem:ind1:pi1int basicbd:basic}, and \ref{lem:ind1:pi1int basicbd} below, by setting the parameters in Lemma \ref{lem:ind1:pi1int basicbd} as $\Delta_0=\Delta_1=\alpha^{-1}, K=\alpha^{-1}\beta^\theta$, and $ N=\beta^6.$ Whenever there is an empty interval of length $\Delta_1$, we can add a point to justify the choice of $\Delta_1$ which will only increase the value of the integral. This implies that the RHS is smaller than $\alpha^{\frac{3}{2}}\beta^{2\theta}.$
Lastly, we need to understand difference between the term $\mathcal{R}_c(t_0^\flat,t;\Pi_\alpha[t_0^-,t_0^\flat],\alpha)$ and $\alpha$. Recalling the definition of $\mathcal{L} $ \eqref{eq:def:L}, we first observe that
\begin{equation}\label{eq:branching:Rlbd:med3}
\begin{split}
&\big|\mathcal{R}_c(t_0^\flat,t;\Pi_\alpha[t_0^-, t_0^\flat],\alpha) - \mathcal{L}(t_0^\flat;\Pi_\alpha[t_0^-,t_0^\flat],\alpha) \big|
\\&\le
\intop_{t_0^-}^{t_0^\flat} K_\alpha(t-x) d\Pi_\alpha (x) +
\intop_{t_0^-}^{t_0^\flat} \intop_t^\infty K^*_\alpha \cdot K_\alpha(u-x) du d\Pi_\alpha(x) \\
&\quad +\intop_{t_0^-}^{t_0^\flat} \intop_{t_0^\flat}^t \big|K^*_\alpha - K^*_\alpha(t-u)\big| \cdot K_\alpha(u-x) du d\Pi_\alpha(x).
\end{split}
\end{equation}
For any $t\in [t_0, \tau_{\textnormal{f}2}]$, we see that the RHS is at most $\alpha^{100}$ since $t-t_0^\flat \ge \alpha^{-2} \beta^{C_\circ}$: The first two integrals are small due to the decay of $K_\alpha$ (Lemma \ref{lem:estimate on K:intro}), and in the last integral, either $|K^*_\alpha - K^*_\alpha(t-u)|$ or $K_\alpha (u-x)$ is small depending on the size of $u$ (see Lemma \ref{lem:estimat for K tilde:intro}). Moreover, letting $F(x) = \intop_x^\infty K_\alpha(y)dy$, we also have
\begin{equation}\label{eq:reg:ib:med1}
\begin{split}
\mathcal{L}(t_0^\flat;\Pi_\alpha[t_0^-,t_0^\flat],\alpha)& = K_\alpha^* \intop_{t_0^-}^{t_0^\flat} F(t_0^\flat-x) d\Pi_\alpha(x);\\
&=K_\alpha^* \intop_{t_0^-}^{t_0^\flat} F(t_0^\flat -x) d\widetilde{\Pi}_\alpha(x) + K_\alpha^* \intop_{t_0^-}^{t_0^\flat} F(t_0^\flat -x)\alpha dx .
\end{split}
\end{equation}
The first term in the second line can be bounded by Lemma \ref{lem:concentration of integral} using $|F(x)| \le1$, and the last term is dealt with integration by parts which is
\begin{equation}\label{eq:branching:Rlbd:med4}
\intop_{0}^{t_0^\flat-t_0^-} F(x)dx = \intop_{0}^{\infty} F(x)dx + O(\alpha^{50}) = \intop_0^\infty xK_\alpha(x)dx + O(\alpha^{50}) = (K_\alpha^*)^{-1} + O(\alpha^{50}).
\end{equation}
Thus, we get
\begin{equation}\label{eq:reg:ib:med2}
\big|\mathcal{L}(t_0^\flat;\Pi_\alpha[t_0^-,t_0^\flat],\alpha) - \alpha\big| \le \alpha^{\frac{3}{2}}\beta^{C_\circ},
\end{equation}
with probability at least $1-e^{-\beta^5}$.
Therefore, combining \eqref{eq:branching:R0minus intform}, \eqref{eq:branching:Rlbd:med1}, \eqref{eq:branching:Rlbd:med2}, \eqref{eq:branching:Rlbd:med3} and \eqref{eq:branching:Rlbd:med4}, we obtain
\begin{equation}\label{eq:branching:ib:tauB4}
\mathbb{P} \left( \tau_{\textnormal{B}1} >\acute{t}_0 \right) \ge 1-3\exp\left(-\beta^3\right).
\end{equation}
The remaining task is to control $\tau_{\textnormal{B}2}. $
First, for $t\le\tau_{\textnormal{f}}\wedge \tau_{\delta \textnormal{B}3}$, observe that
\begin{equation}
\begin{split}
\hat{R}_0^+(t)& := R_0^+(t) +|R_0(t)-\alpha| \le R_0(t) + \delta R_0(t) + |R_0(t)-\alpha|\\
&\le \alpha + 2|R_0(t)-\alpha| + \delta R_0(t)\le \alpha\beta^{C_\circ+1} + 2\alpha \beta^{4\theta+1} \le 3\alpha \beta^{4\theta+1},
\end{split}
\end{equation}
Thus, we have
\begin{equation}\label{eq:branching:ib:tauB6}
\begin{split}
&\tau_{ \textnormal{B}{3}} := \inf
\left\{
t\ge t_0: \big|\Pi_{\hat{R}_0^+}[(t-\alpha^{-1}) \vee t_0, t] \big| \ge \beta^{5\theta}
\right\};\\
&\mathbb{P} \left( \tau_{ \textnormal{B}{3}} >\acute{t}_0 \right) \ge 1-3\exp\left(-\beta^3 \right),
\end{split}
\end{equation}
from Corollary \ref{cor:concentration:numberofpts each interval},
based on Lemma \ref{lem:branching:ib:R0 vs alpha} and Corollary \ref{cor:branching:ib:taudelta}. We also note that $\tau_{\textnormal{B}3}$ is a weaker version of $\tau_{\textnormal{B}2}$.
The decomposition \eqref{eq:branching:ib:decomp:coupled} works similarly for $\hat{R}_0^+(t)$ as follows.
\begin{equation}
\begin{split}
|\hat{R}_0^+(t)-\alpha| \le \big| R_0^+(t) + |R_0(t)-\alpha| -\alpha \big| \le 2|R_0(t)-\alpha| + \delta R_0(t).
\end{split}
\end{equation}
Let ${\tau}_{\delta \textnormal{B}} := {\tau}_{\delta \textnormal{B}1}\wedge {\tau}_{\delta \textnormal{B}2} \wedge {\tau}_{ \textnormal{f}}\wedge \tau_{ \textnormal{B}{3}}$. By following similar computation as \eqref{eq:reg:branching:tau2:split0} and what follows afterwards,
\begin{equation}
\begin{split}
\intop_{(t-\alpha^{-1})\wedge{\tau}_{\delta \textnormal{B}}}^{t\wedge {\tau}_{\delta \textnormal{B}}}& \left| \hat{R}_0^+(s)-\alpha \right|ds
\le
\intop_{(t-\alpha^{-1})\wedge{\tau}_{\delta \textnormal{B}}}^{t\wedge {\tau}_{\delta \textnormal{B}}} 2\left| {R}_0(s)-\alpha \right|ds
+
\intop_{(t-\alpha^{-1})\wedge{\tau}_{\delta \textnormal{B}}}^{t\wedge {\tau}_{\delta \textnormal{B}}} \delta R_0(s) ds \\
&\le
\intop_{(t-\alpha^{-1})\wedge{\tau}_{\delta \textnormal{B}}}^{t\wedge {\tau}_{\delta \textnormal{B}}} \left\{2\alpha \beta^{C_\circ +\frac{1}{2}}\sigma_1(s;\hat{R}_0^+) +
\alpha \beta^{4\theta+1}\sigma_1(s;\hat{R}_0^+) + 2\alpha^{\frac{3}{2}} \beta^{4\theta+1}
\right\} ds \le 1,
\end{split}
\end{equation}
where the second inequality follows from the definitions of $\tau_{\textnormal{f}1}$ and $\tau_{\delta \textnormal{B}2}$, and the last one comes from Lemma \ref{lem:ind1:pi1int basicbd:basic} along with the bounds from $\tau_{ \textnormal{B}{3}}$. Then, applying Corollary \ref{cor:concentration:numberofpts each interval} implies that
\begin{equation}\label{eq:branching:ib:tauB5}
\mathbb{P} \left( \tau_{\textnormal{B}2} >\acute{t}_0 \right) \ge 1-7\exp\left(-\beta^3\right).
\end{equation}
Thus, we conclude the proof from \eqref{eq:branching:ib:tauB4} and \eqref{eq:branching:ib:tauB5}.
\end{proof}
To conclude the section, we record a technical but simple lemma used above, and it will be useful for the rest of the paper.
\begin{lem}\label{lem:ind1:pi1int basicbd}
Let $g:[0,h]\to (0,\infty)$ be a positive function which defines a (deterministic) set of points $\Pi_g[0,h]$.
Furthermore, assume additionally that $K, N_0, \Delta_0, \Delta_1 >0$ satisfy
\begin{itemize}
\item $\sup\{|\Pi_g[t-\Delta_0 , t] | : t\in[\Delta_0, h] \} \le N_0$;
\item $\Delta_1\ge \Delta_0$ and $\inf \{|\Pi_g[t-\Delta_1,t]| : t\in [\Delta_1,h] \} \ge 1$;
\item $h = K\Delta_0$.
\end{itemize}
Then, there exists an absolute constant $C>0$ such that the following hold true:
\begin{equation}
\begin{split}
&\intop_0^h \frac{dt}{\sqrt{(h-t+1)(\pi_1(t;g)+1) } } \le C \left( N_0 + \sqrt{\frac{K\Delta_1N_0}{\Delta_0}} \right) ;\\
&\intop_0^h
\frac{dt}{\sqrt{h-t+1}(\pi_1(t;g)+1)} \le
CN_0\left(1 + \frac{\sqrt{K} \log \Delta_1 }{\sqrt{\Delta_0}} \right);\\
&\intop_0^h \frac{dt}{(h-t+1)\sqrt{\pi_1(t;g)+1} } \le C \left( \frac{N_0 \log \Delta_1}{\sqrt{\pi_1(h;g)+1}} + \frac{N_0\sqrt{\Delta_1}}{{\pi_1(h;g)+1}} + \log K {\frac{\sqrt{\Delta_1N_0}}{\Delta_0}} \right)
;\\
&\intop_0^h \frac{dt}{(h-t+1)(\pi_1(t;g)+1) } \le CN_0 \log \Delta_1 \left( \frac{1}{\pi_1(h;g)+1} + \frac{\log K}{\Delta_0} \right).
\end{split}
\end{equation}
\end{lem}
\begin{proof}
Let $\Pi_g[0,h]:= \{p_1<p_2<\ldots<p_n \}$ with $p_0=0, p_{n+1}=h.$
To establish the first inequality, we let $\Gamma_k := \Pi_g[(K-k-1)\Delta_0, (K-k)\Delta_0 ]$. In the first integral, note that the integrals from $p_i$ to $p_{i+1}$ is bounded by an absolute constant, for $p_i \in \Gamma_0$. Hence,
\begin{equation}
\begin{split}
\intop_0^h \frac{dt}{\sqrt{(t-x)(\pi_1(x;g)+1) } } \le CN_0 + \sum_{k=1}^{K-1} \sum_{p_i \in \Gamma_{k+1}} \intop_{p_i}^{p_i+1} \frac{dx}{\sqrt{k\Delta_0 (x-p_i)} }\\
\le CN_0 + \sum_{k=1}^{K-1} \frac{\sqrt{\Delta_1N_0}}{\sqrt{k\Delta_0}}\le CN_0+ C \sqrt{\frac{K\Delta_1 N_0 }{\Delta_0}},
\end{split}
\end{equation}
where the second inequality came from Lemma \ref{lem:ind1:pi1int basicbd:basic}. The other inequalities can be obtained similarly, and we omit the details. For the last two inequalities, we just keep in mind that the contribution from the regime $[h-2\Delta_0, h]$ are dealt separately.
\end{proof}
\section{Regularity of the speed} \label{sec:reg:intro}
In this section, we formulate the notion of regularity, and build up an inductive analysis to study the regularity property of the speed. To motivate the works done in Sections \ref{sec:reg:intro}--\ref{sec:reg:next step}, we briefly review our future goal to explain why establishing the regularity is essential.
Recall the definition of $L$ given in \eqref{eq:def:Lt:basic form} (see also \eqref{eq:def:L}). The main goal of this paper is to establish the scaling limit of $L$ driven by the speed process. To this end, we need to study the mean and the variance of its increment at each interval, which can be conceptually described by
\begin{equation}\label{eq:Lincrement:conceptual}
\mathcal{L}(t_1;\Pi_S[t_1^-,t_1],\alpha_1) - \mathcal{L}(t_0;\Pi_S[t_0^-,t_0], \alpha_0),
\end{equation}
conditioned on the information up to time $t_0$. Here, $t_0^-<t_0<t_1^-<t_1$, and $\alpha_1, \alpha_0$ are the \textit{frame of reference} in each interval which are essentially the constants that approximates the speed. That is, on the interval $[t_0,t_1]$, the speed $S(t)$ behaves roughly like $\alpha_0$, with a smaller order fluctuation which will indeed turn out to be essentially $\alpha_0^{3/2} \textnormal{polylog}(\alpha_0)$.
The primary difficulty of computing the mean and variance of \eqref{eq:Lincrement:conceptual} stems from the complicated nature of the process $S(t)$. As we have seen in \eqref{eq:Lt:increment} (and \eqref{eq:integralform:branching}), the increment \eqref{eq:Lincrement:conceptual} can be described as an integral involving $d\Pi_S$, and if $S(t)$ does not behave nicely enough we may not be able to estimate such an integral accurately. This motivates us to demonstrate regularity properties of $S(t)$, which is going to tell us that $S(t)$ stays close to $\alpha_0$ during the interval $[t_0,t_1]$ in an appropriate sense, so that we can calculate the first and second moment of \eqref{eq:Lincrement:conceptual} conditioned on $\mathcal{F}_{t_0}$.
With this goal in mind, our argument is based on an inductive study of the speed. We will begin with defining the regularity property, which is a fairly complicated mixture of all the properties that $S(t)$ and $\Pi_S$ should satisfy. Then, we will show that
\begin{center}
If $S(t)$ is regular at time $t_0$ with respect to $\alpha_0$, \\
then $S(t)$ is regular at time $t_1$ with respect to $\alpha_1$, with high probability.
\end{center}
Before moving on to the definition of regularity, we briefly explain how we define the parameters used in the section. Throughout the rest of the paper, we fix $\epsilon=\frac{1}{10000}$ to be a small constant, and $C_\circ=50$, $\theta=10000$ be large constants (where $\theta$ needs to be larger depending on $C_\circ$) as in Sections \ref{sec:fixedrate} and \ref{sec:criticalbranching}. For given $\alpha_0>0$ and $t_0>0$, we set $\beta_0:= \log(1/\alpha_0)$, and the time parameters $t_0^{-}, t_0^+, \acute{t}_0, \textnormal{ and } \hat{t}_0$ are defined as follows:
\begin{itemize}
\item $t_0^-,t_0^+$ are arbitrary numbers satisfying
\begin{equation}\label{eq:def:t0t1}
t_0-2\alpha_0^{-2}\beta_0^\theta < t_0^- <t_0-\frac{1}{2}\alpha_0^{-2} \beta_0^{\theta}, \quad
t_0 + \frac{1}{2}\alpha_0^{-2}\beta_0^{\theta} < t_0^+ < t_0 + 2\alpha_0^{-2}\beta_0^{\theta}
.
\end{equation}
\item $\acute{t}_0, \hat{t}_0$ are fixed numbers set to be
\begin{equation}\label{eq:def:t0t11}
\acute{t}_0:= t_0 + 3\alpha_0^{-2} \beta_0^{\theta}, \quad
\hat{t}_0 := t_0 +2\alpha_0^{-2}\beta_0^{10\theta}.
\end{equation}
\end{itemize}
Note that the definitions of $t_0^-$ and $\acute{t}_0$ are consistent with those from Section \ref{sec:criticalbranching}. We also stress that $$(\acute{t}_0-t_0^+)\wedge (t_0^+-t_0) \wedge (t_0-t_0^-) \ge \frac{1}{2}\alpha_0^{-2}\beta_0^{\theta}.$$
For the speed $S(t)$, its \textit{first-order approximation} $S_1(t)$ \eqref{eq:def:S1:basic form} is set to be
\begin{equation}\label{eq:def:S1:reg}
S_1(t)= S_1(t;t_0^{-},\alpha_0):=
\intop_{t_0^{-}}^t K_{\alpha_0} (t-s) d\Pi_S(s).
\end{equation}
We also recall the definitions of $\pi_i(t;S)$ and $ \sigma_i(t;S)$ from \eqref{eq:def:pi closest points} and \eqref{eq:def:sig closest points}.
The rest of the section is organized as follows. We begin with giving a formal definition of regularity in Section \ref{subsec:regoverview:reg}, which will be studied exhaustively in Sections \ref{sec:reg:conti of reg} and \ref{sec:reg:next step}. In Section \ref{subsec:regoverview:overview}, we state the main theorem and give a more involved overview on the induction argument. We also discuss some important consequences of regularity in Section \ref{subsec:reg:conseq}.
Finally, in Section \ref{subsec:regoverview:history}, we introduce a preliminary analysis on the speed that makes it possible to ignore the history of the far past, enabling us to study $S'(t)$ \eqref{eq:def:Sprime:basic form} instead of $S(t)$.
\subsection{Formal definition of regularity}\label{subsec:regoverview:reg}
Let $\mathcal{F}_{t_0}$ be the $\sigma$-algebra generated by $(S(s))_{s\le t_0}$ and $\Pi_S[0,t_0]$. Since the process $(S(s))_{s\le t_0}$ can completely be recovered from the given configuration of points $\Pi_S[0,t_0]$, we can also identify $\mathcal{F}_{t_0}$ with $\Pi_S[0,t_0]$. Our idea is to define $\Pi_S[0,t_0]$ to be regular if
\begin{equation}
\mathbb{P} \left(\left. S \textnormal{ behaves }nicely \textnormal{ until time } \acute{t}_0\ \right|\ \Pi_S[0,{t_0}] \right) \approx 1.
\end{equation}
To introduce the \textit{nice} traits that $S$ should satisfy, we will investigate the process from the following perspectives.
\begin{enumerate}
\item Control on the size of the speed;
\item Control on the size of the aggregate;
\item Refined control of the first order approximation;
\item Regularity of the history until time $t_0$.
\end{enumerate}
In the following subsections, we detail the formal definitions of regularity for each category, and explain their purpose. Each criterion will be described in the language of stopping times, as we saw in Section \ref{sec:criticalbranching}. In the rest of the section and throughout Sections \ref{sec:reg:conti of reg} and \ref{sec:reg:next step}, $\kappa>0$ denotes a constant which is either $\frac{1}{2}$ or $2$. This constant is the parameter used to distinguish the ``stronger'' ($\kappa=\frac{1}{2}$) regularity from the ``weaker'' ($\kappa=2$) one: We show in Section \ref{sec:reg:conti of reg} that the initial ``weaker'' regularity leads to the ``stronger'' regularity at later times.
\subsubsection{Control on the size of the speed} \label{subsubsec:reg:Scontrol}
Maintaining an appropriate control on $S(t)$ is the primary property we want from the regularity. We begin with giving the formal definitions of the stopping times of interest. For simplicity, we write $\sigma_1\sigma_2(t):= \sigma_1(t;S)\sigma_2(t;S)$. Also, recall the definition of $S_1(t)$ from \eqref{eq:def:S1:reg}
\begin{equation}
\begin{split}
\tau_{1} (\alpha_0,t_0,\kappa )&:=
\inf\{t\geq t_0: S(t) \geq \kappa \alpha_0 \beta_0^{C_\circ} \};
\\
\tau_{2}(\alpha_0,t_0,\kappa)&:= \inf\{ t\geq t_0: S_1(t) \ge \kappa \alpha_0 \beta^{C_\circ}_0 \};
\\
\tau_{3}(\alpha_0,t_0) &:=
\inf\left\{t\ge t_0: S(t)-S_1(t)\notin \left(-\alpha_0^{\frac{3}{2}-\epsilon}\sigma_1(t) ,\alpha_0^{1-\epsilon} \sigma_1\sigma_2(t) \right) \right \};
\\
\tau_{4}(\alpha_0,t_0,\kappa)&:=
\inf\left\{t\ge t_0: \intop_{t_0}^{t} (S(s)-\alpha_0)^2ds \ge \kappa \alpha_0 \beta_0^{25\theta} \right\};
\\
\tau_{5}(\alpha_0,t_0,\kappa)&:=
\inf \left\{t\ge t_0: \intop_{t_0}^{t} (S(s)-\alpha_0)^2 S(s)ds \ge \kappa \alpha_0^2 \beta_0^{25\theta} \right\}.
\end{split}
\end{equation}
We briefly explain the purpose of each stopping time as follows:
\begin{itemize}
\item $\tau_{1}$ not only gives a fundamental understanding on $S$, but also useful in controlling other stopping times such as $\tau_{4}$ and $\tau_{5}$. In particular, it is heavily used in bounding the quadratic variations of various martingales, such as the second criterion in \eqref{eq:concentration:conditions:basic}.
\item $\tau_{3}$ provides a crucial estimate in bounding $|S(t)-\alpha_0|\leq |S(t)-S_1(t)|+|S_1(t)-\alpha_0|$, which is important in studying various different quantities including $\tau_{4}$ and $\tau_{5}$. Moreover, the lower bound on $S(t)-S_1(t)$ will lead us to obtaining the lower bound on $S(t)$: See Proposition~\ref{prop:reg:lbd of speed}
\item $\tau_{2}$, $\tau_{4}$ and $\tau_{5}$ verify the main assumptions of Proposition~\ref{prop:fixed perturbed:error}, enabling us to utilize the first- and second-order approximations.
\end{itemize}
\subsubsection{Size of the aggregate}\label{subsubsec:reg:aggsize}
Note that the number of points in $\Pi_S[0,t]$ describes the size of the aggregate, i.e., $X_t = |\Pi_S[0,t]|$. The stopping times that deals with the size of the aggregate have similar role as $\tau_{\textnormal{B}2}$ in Section \ref{sec:criticalbranching}, but we need several of them for different purposes.
\begin{equation}
\begin{split}
\tau_{6}(\alpha_0,t_0,\kappa)
&:=
\inf\left\{t\ge t_0: |\Pi_S[t_0^{-},\, t]| \le \frac{1}{100\kappa} \alpha_0 (t-t_0^{-}) \right\};
\\
\tau_{7}(\alpha_0,t_0,\kappa)
&:=
\inf\left\{ t\ge t_0:
|\Pi_S[(t-\alpha_0^{-1})\vee t_0,\, t]| \ge\kappa \beta_0^4
\right\};
\\
\tau_{8}(\alpha_0,t_0,\kappa)
&:=
\inf \left\{ t\ge t_0+\kappa \alpha_0^{-1}\beta_0^{C_\circ}:
|\Pi_S[t-\kappa\alpha_0^{-1}\beta_0^{C_\circ},\, t]| =0 \right\}.
\end{split}
\end{equation}
\begin{itemize}
\item $\tau_{6}$ provides useful estimates needed in Section \ref{subsec:regoverview:history}: If there are reasonably many particles in the interval $[t_0^{-},t]$, then it turns out we can eliminate the effect of information before $t_0^{-}$.
\item $\tau_{7}$ plays a similar role as $\tau_{\textnormal{B}2}$ from Section \ref{sec:criticalbranching}, such as helping to control $\tau_{2}$ and $\tau_{3}$.
\item $\tau_{8}$ makes it easier to control quantities involving $\sigma_1$ and $\sigma_2$, since the all neighboring points before $\tau_{8}$ cannot be too far from each other.
\end{itemize}
\subsubsection{Refined control of the first order approximation} \label{subsubsec:reg:refined1storder}
We introduce a stopping time which is a strengthened versions of $\tau_1$ and $\tau_{3}$. Instead of $\tau_1$, we describe a sharper control on the average of $|S(t)-\alpha_0|$. On the other hand, in $\tau_{3}$, the bound on $|S(t)-S_1(t)|$ has the term $\alpha_0^{-\epsilon}$ which is not strong enough in the later analysis (see Theorem \ref{thm:increment:formal}). Thus, we seek for a better bound, which is of $\beta_0^C$ rather than $\alpha_0^{-\epsilon}$. We define the stopping times $\tau_1^\sharp$ and $\tau_{3}^\sharp$ to be
\begin{equation}
\begin{split}
&\tau_1^\sharp = \tau_1^\sharp(\alpha_0,t_0):=
\inf \left\{ t\ge t_0: \intop_{t_0}^t |S(s)-\alpha_0| ds \ge \alpha_0^{-\frac{1}{2}-\epsilon} \right\};\\
&\tau_{3}^\sharp=\tau_{3}^\sharp(\alpha_0,t_0) := \inf \left\{ t\ge t_0: \, \intop_{t_0}^t |S(s)-S_1(s)| ds \ge \beta_0^{4\theta} \right\}.
\end{split}
\end{equation}
It is clear that $\tau_1$ does not imply $\tau_1^\sharp$, and also is not difficult to see that $\tau_3^\sharp $ is stronger than $\tau_3$; since integrating $|S(s)-S_1(s)|$ based on the bound in $\tau_3$ (along with $\tau_7, \tau_8$, and Lemma \ref{lem:ind1:pi1int basicbd:basic}) results in $\alpha_0^{-\epsilon}$, rather than $\beta_0^{4\theta}$-bound given in $\tau_3^\sharp$.
We remark that in $\tau_3^\sharp$, having a smaller exponent $4\theta$ of $\beta_0$ than $10\theta$ (which is the exponent in $\hat{t}_0-t_0$) plays a very important role in this analysis. See Section \ref{subsubsec:regoverview:overview:nextstep} for more discussion; details will be presented in Section \ref{sec:reg:next step}.
\subsubsection{Regularity of the history}\label{subsubsec:reg:history}
We introduce several additional events which are measurable with respect to $\mathcal{F}_{t_0}$. Recalling the definition of $\mathcal{L}$ \eqref{eq:def:L}, let
\begin{equation}\label{eq:def:alpha0prime}
\alpha_0'=\alpha_0'(\alpha_0,t_0,t_0^-):= \mathcal{L}(t_0;\Pi_S[t_0^{-},t_0],\alpha_0),
\end{equation}
where $\mathcal{L}$ is defined in \eqref{eq:def:L}. We also set $S(s) = \infty$ for $s<0$. Define
\begin{equation}\label{eq:def of the events A1 A2 A3}
\begin{split}
\mathcal{A}_1(\alpha_0,t_0)
&:=
\left\{ |\Pi_S[s, t_0^{-}]| \ge \sqrt{t_0^{-}-s+C_\circ} \log^2 (t_0^{-} -s +C_\circ) -\alpha_0^{-1} \beta_0^\theta, \ \ \forall 0\le s\le t_0^{-} \right\};
\\
\mathcal{A}_2(\alpha_0,t_0)&:= \left\{ \frac{1}{4} \le \frac{ |\Pi_S[t_0^{-},\, t_0]|}{\alpha_0^{-1}\beta_0^{\theta}} \le 4 \right\};
\\
\mathcal{A}_3(\alpha_0,t_0)&:=
\left\{|\alpha_0-\alpha_0'| \le \alpha_0^{\frac{3}{2}} \beta_0^{\theta} \right\}.
\end{split}
\end{equation}
\begin{itemize}
\item $\mathcal{A}_1$ is the key event used in Section \ref{subsec:regoverview:history}. By setting $|\Pi_S[s, t_0^{-}]|$ to be of strictly bigger order than $\sqrt{t_0^{-}-s}$ (which must actually be linear in $(t_0^{-}-s)$), the contribution from the history before $t_0^{-}$ becomes negligible from the representation \eqref{eq:speed:basic expression}, since the probability of the simple random walk to hit the aggregate before time $t_0^{-}$ is small enough.
\item $\mathcal{A}_2$ provides a basic estimate on the size of the aggregate in the past, and is used in investigating $\tau_{6}$.
\item $\mathcal{A}_3$, is introduced to ensure that the choice of $\alpha$ is appropriate. Later, our choice of $\alpha_1$ will satisfy $|\alpha_1-\alpha_0|=\alpha_0^{3/2} \beta_0^{O(1)}$, which is the same order as the bound in $\mathcal{A}_3$.
\end{itemize}
\subsubsection{Definition of regularity}\label{subsubsec:regoverview:regdef}
We give the formal definition of regularity combining the notions introduced in Sections \ref{subsubsec:reg:Scontrol}--\ref{subsubsec:reg:history}. We let
\begin{equation}\label{eq:def:tau:induction}
\begin{split}
&\tau(\alpha_0,t_0,\kappa ) := \hat{t}_0 \wedge \min\{\tau_i(\alpha_0,t_0,\kappa) : 1\le i \le 8 \},\\
&\tau^\sharp(\alpha_0,t_0,\kappa ):= \tau(\alpha_0,t_0,\kappa )\wedge \tau_1^\sharp(\alpha_0,t_0)\wedge \tau_3^\sharp(\alpha_0,t_0),
\end{split}
\end{equation}
where we define $\tau_3(\alpha_0,t_0,\kappa):= \tau_3(\alpha_0,t_0)$ which does not have dependence on $\kappa$. Note that $\tau$ is increasing in $\kappa$. Moreover, we set
\begin{equation}\label{eq:def:A:induction}
\begin{split}
\mathcal{A}(\alpha_0,t_0) := \mathcal{A}_1(\alpha_0,t_0) \cap \mathcal{A}_2(\alpha_0,t_0) \cap \mathcal{A}_3(\alpha_0,t_0) .
\end{split}
\end{equation}
Note that these events are $\mathcal{F}_{t_0}$-measurable. We now introduce the two notions of regularity.
\begin{remark}
In the following definitions and throughout the paper, the notation
\begin{equation}
\mathbb{P} \left( \ \ \cdot \ \ | \, \Pi_S(-\infty,t_0] \right)
\end{equation}
denotes the probability with respect to the law of the aggregate starting from time $t_0$ under the initial points given by $\Pi_S(-\infty,t_0]$, which was defined in Definition \ref{def:aggregate with initial condition}. Moreover, note that if $S(s)=\infty$ on $s< t^\flat$ for some $t^\flat$, then $\Pi_S(-\infty,t^\flat)$ is not well defined (for instance, $\overline{S}(t)$ from \eqref{eq:def:sandwich speed}). Even in this a case when there exists such an $t^\flat\le t_0$, we still use the notation $\Pi_S(-\infty,t_0]$ for convenience, which will actually refer to
\begin{equation}
\Pi_S(-\infty,t_0] := \textnormal{ the points in }[t^\flat,t_0] \textnormal{ are given by } \Pi_S[t^\flat,t_0], \textnormal{ and } S(s)=\infty \textnormal{ for all } s\le t^\flat.
\end{equation}
\end{remark}
\begin{definition}[Regularity]\label{def:reg} For $\alpha_0, t_0, t_0^-, r>0$, we denote $[t_0] := (t_0, t_0^-)$ and set $\acute{t}_0$ as \eqref{eq:def:t0t11}. We say $\Pi_0(-\infty,t_0]$ is $(\alpha_0,r; [t_0])$-\textit{regular} if it satisfies the following two conditions:
\begin{itemize}
\item $\mathbb{P} \left(\left.\tau (\alpha_0,t_0,\kappa=2) \le \acute{t}_0 \, \right| \, \Pi_S(-\infty,t_0] = \Pi_0(-\infty,t_0] \right) \le r;$
\item $\Pi_0(-\infty,t_0] \in \mathcal{A}(\alpha_0,t_0)$.
\end{itemize}
We also write $\Pi_0(-\infty,t_0]\in \mathfrak{R}(\alpha_0,r;[t_0])$ to denote that $\Pi_0(-\infty,t_0]$ is $(\alpha_0,r;[t_0])$-regular. Note that the dependence on $t_0^-$ comes from the definitions of $S_1$ and $\mathcal{A}$.
\end{definition}
\begin{definition}[Sharp regularity]\label{def:reg:sharp} For $\alpha_0, t_0,t_0^-,t_0^+, r>0$ with $[t_0]:=(t_0,t_0^-,t_0^+)$, and set $\acute{t}_0$ as \eqref{eq:def:t0t11}. We say $\Pi_0(-\infty,t_0]$ is $(\alpha_0,r;[t_0])$-\textit{sharp-regular} if it satisfies the following two conditions:
\begin{itemize}
\item $\mathbb{P} \left(\left.\tau^\sharp (\alpha_0,t_0,\kappa=2) \le \acute{t}_0 \, \right| \, \Pi_S(-\infty,t_0] = \Pi_0(-\infty,t_0] \right) \le r;$
\item $\Pi_0(-\infty,t_0] \in \mathcal{A}(\alpha_0,t_0)$.
\end{itemize}
We also write $\Pi_0(-\infty,t_0]\in \mathfrak{R}^\sharp(\alpha_0,r;[t_0])$ to denote that $\Pi_0(-\infty,t_0]$ is $(\alpha_0,r;[t_0])$-sharp-regular.
\end{definition}
The reason for separating two concepts of regularity will be clear when we state the main result in the next subsection (see also Remark \ref{rmk:regvsshreg}). There, we also briefly explain the main ideas of proof.
\subsection{Overview of the argument}\label{subsec:regoverview:overview}
Let $t_1$ be an arbitrary number that satisfies
\begin{equation}\label{eq:t1 regime}
t_0 + \alpha_0^{-2} \beta_0^{9\theta} \le t_1 \le t_0+\alpha_0^{-2} \beta_0^{10\theta} .
\end{equation}
We note that the upper bound of $t_1$ is substantially smaller than $\hat{t}_0$, namely, $t_1\le t_0 + \alpha_0^{-2}\beta_0^{10\theta} = \hat{t}_0-\alpha_0^{-2}\beta_0^{10\theta}.
$ Then, the goal of Sections \ref{sec:reg:intro}--\ref{sec:reg:next step} is to establish the following theorems.
\begin{thm}\label{thm:induction:main}
Let $\alpha_0, t_0, r>0$, let $t_0^-$ be any number satisfying \eqref{eq:def:t0t1}, set $\acute{t}_0$ as \eqref{eq:def:t0t11}, and let $t_1$ be an arbitrary fixed number selected within \eqref{eq:t1 regime}. Set $t_1^- := t_1- \alpha_0^{-2} \beta_0^\theta$, and define
\begin{equation}\label{eq:def:alpha1}
\alpha_1:= \mathcal{L}(t_1^-; \Pi_S[t_0^-,t_1^-],\alpha_0 ), \quad \beta_1:= \log(1/\alpha_1).
\end{equation}
Then, the following hold true for all sufficiently small $\alpha_0$ and for any $\Pi_S(-\infty,t_0]\in \mathfrak{R}(\alpha_0,r;[t_0])$:
\begin{enumerate}
\item $\mathbb{P} \left(\left.\Pi_S(-\infty,t_1] \notin \mathfrak{R}^\sharp(\alpha_1, e^{-\beta_1^{3/2}}; [t_1] ) \, \right| \, \Pi_S(-\infty,t_0] \right) \le r+ e^{-\beta_0^{3/2}}$;
\item $\mathbb{P} (| \alpha_1 - \alpha_0 |\ge 2\alpha_0^{3/2} \beta_0^{6\theta} \ | \ \Pi_S(-\infty,t_0] ) \le r+ e^{-\beta_0^{3/2}}$.
\end{enumerate}
\end{thm}
Note that the second statement directly implies
$$\mathbb{P} ( t_1^- \textnormal{ satisfies \eqref{eq:def:t0t1} in terms of } t_1, \alpha_1 \,| \, \Pi_S(-\infty,t_0] ) \ge 1-r-e^{-\beta_0^{3/2}},$$
which means that w.h.p., we can reapply the theorem for $\Pi_S(-\infty,t_1]$ and $[t_1]$ to obtain $\Pi_S(-\infty,t_2] \in \mathfrak{R}^\sharp(\alpha_2, e^{-\beta_2^{3/2}}; [t_2])$, and so on.
\begin{remark}
In the interval starting from $t_1$, the first-order approximation $S_1$ is defined as $S_1(t)=S_1(t;t_1^-,\alpha_1)$. Thus, by setting $\alpha_1$ to be $\mathcal{F}_{t_1^-}$-measurable as above, the points in $\Pi_S[t_1^-,t_1]$ before time $t_1$ do not affect the frame of reference $\alpha_1$ of $S_1(t)$.
\end{remark}
In addition to the above theorem,
we need the regularity of the fixed rate process obtained in Section \ref{sec:Sandwich}, which serves as the induction base of our argument. Recall the definitions of $\overline{Y}_{t_0, \alpha_0}$ and $\underline{Y}_{t_0, \alpha_0}$, and let $ \overline{S}(t)= \overline{S}_{t_0,\alpha_0}(t)$ and $\underline{S}(t)=\underline{S}_{t_0,\alpha_0}(t)$ be the corresponding speed processes \eqref{eq:def:sandwich speed}, respectively. For $t_0^-$ and $\acute{t}_0$ as above, we define
\begin{equation}
\begin{split}
\overline{S}_1(t) = \overline{S}_1(t;t_0^-,\alpha_0) := \intop_{t_0^-}^{t_0} K_{\alpha_0}(t-x) d\Pi_{\alpha_0}(x) + \intop_{t_0}^t K_{\alpha_0}(t-x) d\Pi_{\overline{S}}(x)
\end{split}
\end{equation}
Thus, using these processes, we can define the stopping times $\{\overline{\tau}_i(\alpha_0,t_0,\kappa) \}_{1\le i\le 8}$ (Sections \ref{subsubsec:reg:Scontrol}--\ref{subsubsec:reg:aggsize}), and the event $\overline{\mathcal{A}}(\alpha_0,t_0)$ \eqref{eq:def:A:induction}, analogously as before. Then, they define the regularity (Definition \ref{def:reg}) of $\overline{S}(t)$. Everything can be done in the same way for $\underline{S}(t)$. The main theorem on the regularity of $\overline{S}(t)$ and $\underline{S}(t)$ is stated as follows.
\begin{thm}\label{thm:induction:base:main}
Let $\alpha_0, t_0>0$, set $\acute{t}_0$ as \eqref{eq:def:t0t11}, and set $t_0^- := t_0-\alpha_0^{-2}\beta^\theta$. Then, under the above setting, we have
\begin{equation}
\mathbb{P}\left(\Pi_{\overline{S}}(-\infty,t_0], \, \Pi_{\underline{S}}(-\infty,t_0] \in \mathfrak{R}\left(\alpha_0, e^{-\beta_0^{3/2}}; [t_0]\right) \right) \ge 1-e^{-\beta_0^{3/2}}.
\end{equation}
\end{thm}
\begin{remark}\label{rmk:regvsshreg}
After some large constant time since the critical aggregate starts growing, we require the speed to be very small and the process to be sharp-regular. However, due to Theorem \ref{thm:induction:main}, a regular (does not have to be sharp-regular) interval beginning will result in sharp-regular intervals afterwards w.h.p., and hence Theorem \ref{thm:induction:base:main} will be enough for our purpose.
\end{remark}
The rest of the section and Sections \ref{sec:reg:conti of reg}--\ref{sec:reg:next step} are mostly devoted to the proof of Theorem \ref{thm:induction:main}, which consists of three major steps as follows.
\begin{enumerate}
\item Approximating $S(t)$ by $S'(t)$ \eqref{eq:def:Sprime:basic form} to validate the first and second order approximations;
\item Continuation of regularity (i.e., the stopping times) until time $\hat{t}_0$, which will be larger than $\acute{t}_1:=t_1+2\alpha_1^{-2}\beta_1^\theta$;
\item Regularity at time $t_1$, with new the frame of reference $\alpha_1$.
\end{enumerate}
In the remaining subsection, we briefly explain the main purposes and ideas of (1), (2) and (3).
\subsubsection{The first and second order approximation of the speed} \label{subsubsec:regoverview:overview:history}
To study $S(t)$, we attempt to use the first- and second-order approximations given in Section \ref{subsec:speed:speedagg}. However, the formulas \eqref{eq:speed:1storder main}, \eqref{eq:speed:2ndorder main} give expansions on $S'(t)$ defined in \eqref{eq:def:Sprime:basic form}, not $S(t)$. Thus, we need to justify that $S(t)$ is sufficiently close to $S'(t)$.
This analysis is conducted in the next subsection by studying the original formula \eqref{eq:speed:basic expression} of $S(t)$. We show that under the regularity condition, the hitting probability of the random walk affected by conditioning on $\mathcal{F}_{t_0^{-}}$ is negligible, since $t_0^{-}$ is far away from $t_0$.
\subsubsection{Continuation of regularity}\label{subsubsec:regoverview:overview:contiofreg}
The next step is to show that if $\tau(\alpha_0,t_0,\kappa=2)\ge \acute{t}_0$, then it is unlikely to be smaller than $\hat{t}_0$. The purpose of such an analysis is fairly obvious: In order to establish regularity at time ${t_1}$ in terms of $\alpha_1$, we would like to argue that (1) the process at time $t_1$ maintains the desired regularity properties in terms of $\alpha_0$, and (2) changing $\alpha_0$ to $\alpha_1$ does not have a significant effect. For (2), see the discussion in the following subsection.
To establish (1), we essentially argue that $\{\tau_i = \tau \}$ happens with small enough probability for all~$i$. In the following, we discuss several major ideas needed to process the argument.
\begin{itemize}
\item
We study $\tau_1$ via the original formula \eqref{eq:speed:basic expression}, by upper bounding $Y_t$ by an appropriate choice of $Y_t'$ that achieves the desired estimate. The construction of $Y_t'$ is done based on the control of $X_t$ which is given by $\tau_{7}$.
\item To obtain estimates on $\tau_{3}$, we appeal to the results we obtained in Section \ref{subsec:fixed:error} where $\tau_{1}, \tau_{2}, \tau_{4}$ and $\tau_{5}$ provide the appropriate assumptions we need for utilizing Proposition \ref{prop:fixed perturbed:error}.
\item To control the rest of the stopping times, we first express
\begin{equation}
|S(t)-\alpha_0|
\le
|S(t)-S_1(t)|+|S_1(t)-\alpha_0|.
\end{equation}
The first term in the RHS are controlled by $\tau_{3}$, and the remaining task is to control $|S_1(t)-\alpha_0|$. To this end, we observe that $S_1(t)=\mathcal{R}_c(t,t;\Pi_S[t_0^-,t],\alpha_0)$ (see \eqref{eq:def:Qst} for its definition) and interpret it via \eqref{eq:integralform:branching}.
After we understand the gap $|S(t)-\alpha|$, the results will follow mostly by applying the lemmas from Section \ref{subsec:fixed:mgconcen}.
\end{itemize}
\subsubsection{Regularity in the next time step}\label{subsubsec:regoverview:overview:nextstep}
The remaining step is to establish (2) in the first paragraph of Section \ref{subsubsec:regoverview:overview:contiofreg}. We note several major issues in achieving this goal:
\begin{itemize}
\item Our argument is mostly based on analyzing what happens when we change $\alpha_0$ to $\alpha_1$, and hence the control on $|\alpha_1-\alpha_0|$ is essential. This is done by expressing the difference in the integral form which is similar to \eqref{eq:integralform:branching} and studying it based on martingale techniques such as Lemma \ref{lem:concentration of integral} and the analytic properties of $K_\alpha(s)$ from Section \ref{sec:fourier and renewal}.
\item The most crucial task is to understand $\tau_{3}^\sharp(\alpha_1,t_1)$. We need to avoid having $\alpha^{-\epsilon}$ term, which is present in $\tau_{3}$. To this end, we need to study the gap between $S(t)$ and its second-order approximation $S_2(t)$ \eqref{eq:def:S2:basic form}, rather than $|S(t)-S_1(t)|$. Then, however, it turns out that the double integral term of $J_{t-s,t-u}$ in the formula of $S_2(t)$ cannot be studied in the same way as Section \ref{sec:fixedrate}, since the perturbation argument in Section \ref{subsec:fixed:perturbed} is too costly to guarantee a sharp control like $\tau_{3}^\sharp$.
To overcome this issue, we study this double integral rather directly, appealing to the proximity of $S(t)$ to $\alpha_0$ obtained from the previous time step.
One interesting issue we emphasize is that $\tau^\sharp_3 (\alpha_1,t_1)$ is typically much smaller than $t_2=t_1+\alpha_1^{-2} \beta^{10\theta}_1$: As $t$ gets closer to $t_2$, the variance accumulated from the double integral becomes so large that we need a larger exponent on $\beta_1$ for a correct bound.
\end{itemize}
\subsection{Consequences of regularity}\label{subsec:reg:conseq}
In this subsection, we highlight some properties that a regular aggregate must have. These will all be needed in Section \ref{sec:double int}, where we study the formal version of Theorem \ref{thm:speed:increment:informal}.
We begin with the control on the number of particles in $\Pi_S[t_0,t]$ when the process is regular. Its proof will be discussed in Section \ref{subsec:ind1:growth}.
\begin{prop}\label{prop:ind1:growth}
Let $\alpha_0,t_0,r>0$, set $\acute{t}_0, \hat{t}_0$ as \eqref{eq:def:t0t11}, and let $t_0^-$ be any number satisfying \eqref{eq:def:t0t1}. For any $\Pi_S(-\infty,t_0]\in \mathfrak{R}(\alpha_0,r;[t_0])$, we have
\begin{equation}
\begin{split}
\mathbb{P} \left( \left. \Big||\Pi_S[t_0, t]| - \alpha_0(t-t_0) \Big| \le \alpha_0^{-\frac{1}{2}-10\epsilon},\ \forall t\in[\acute{t}_0,\hat{t}_0] \, \right| \, \Pi_S(-\infty,t_0] \right) \ge 1-r-2\exp\left(-\beta_0^{1.9} \right).
\end{split}
\end{equation}
\end{prop}
Next, the following describes an important property of how the frame of reference changes.
\begin{prop}\label{prop:reg:newrates}
Let $\alpha_0,t_0,r>0$, and let $\acute{t}_0, t_0^-, t_1, t_1^-$ be as Theorem \ref{thm:induction:main}.
Furthermore, we define $\alpha_1':=\mathcal{L}(t_1;\Pi_{S}[t_1^{-},t_1], \alpha_1)$ analogously as \eqref{eq:def:alpha0prime}, and set
\begin{equation}
\tilde{\alpha}_1:= \mathcal{L}(t_1;\Pi_S[t_0^-,t_1],\alpha_0).
\end{equation}
Then, the following holds true for all sufficiently small $\alpha_0$ and for any $\Pi_S(-\infty,t_0]\in \mathfrak{R}(\alpha_0,r;[t_0])$:
\begin{equation}
\mathbb{P} \left(\left. |\alpha_1'-\tilde{\alpha}_1| \ge 2\alpha_0^{2} \beta_0^{8\theta+1} \, \right| \, \Pi_S(-\infty,t_0] \right) \le r+\exp\left(-\beta_0^2 \right).
\end{equation}
\end{prop}
In fact, we will later derive scaling limit of $\alpha'$ (see Sections \ref{sec:double int} and \ref{sec:scalinglimit}). To this end, we study the increment $\alpha_1'-\alpha_0'$ (defined in \eqref{eq:def:alpha0prime}) in Theorem \ref{thm:increment:formal}. Expressing that $$\alpha_1'-\alpha_0'=(\alpha_1'-\tilde{\alpha}_1) + (\tilde{\alpha}_1-\alpha_0'),$$
the estimate in the proposition gives an essentially sharp estimate on $(\alpha_1'-\tilde{\alpha}_1) $. Also, note that it is strictly stronger than the error bound in $\mathcal{A}_3$ (Section \ref{subsubsec:reg:history}). The proof of Proposition \ref{prop:reg:newrates} will be discussed in Section \ref{subsec:ind2:ratechange}.
Furthermore, we record estimates on the integrals of the errors such as $(S(t)-\alpha_0)$ in the following lemma.
\begin{lem}\label{lem:reg:errorint}
Let $\alpha_0,t_0,r>0$, and let $\acute{t}_0, t_0^-, t_1, t_1^-$ be as Theorem \ref{thm:induction:main}. Furthermore, let $S_1(t)=S_1(t;t_0^-,\alpha_0)$ and $S_2(t)=S_2(t;t_0^-,\alpha_0)$ be the first- and second-order approximations of the speed given by \eqref{eq:def:S1:basic form}, \eqref{eq:def:S2:basic form}. For any $\Pi_S(-\infty,t_0]\in \mathfrak{R}(\alpha_0,r;[t_0])$, we have
\begin{eqnarray}
\textnormal{(1)}\qquad \qquad & \mathbb{P} \left(\left. \intop_{\acute{t}_0}^{t_1} \big|S(t)-\alpha_0\big|dt \ge \alpha_0^{-\frac{1}{2}-\epsilon} \, \right| \, \Pi_S(-\infty,t_0] \right) \le r+ \exp\left(-\beta_0^{1.9} \right); &\qquad \qquad \\
\textnormal{(2)} \qquad \qquad & \mathbb{P} \left(\left. \intop_{\acute{t}_0}^{t_1} \big|S_1(t)-\alpha_0\big|dt \ge \alpha_0^{-\frac{1}{2}-\epsilon} \, \right| \, \Pi_S(-\infty,t_0] \right) \le r+ \exp\left(-\beta_0^{1.9} \right); & \qquad \qquad \\
\textnormal{(3)} \qquad \qquad & \mathbb{P} \left(\left.\intop_{{t}_0}^{t_1} \big|S(t)-S_1(t)\big|dt \ge \alpha_0^{-\epsilon} \, \right| \, \Pi_S(-\infty,t_0] \right) \le r+ \exp\left(-\beta_0^{1.9} \right); & \qquad \qquad \\
\textnormal{(4)} \qquad \qquad & \mathbb{P} \left(\left. \intop_{\acute{t}_0}^{t_1} \big|S(t)-S_2(t)\big|dt \ge \alpha_0^{\frac{1}{2}-\epsilon} \, \right| \, \Pi_S(-\infty,t_0] \right) \le r+ \exp\left(-\beta_0^{1.9} \right). & \qquad \qquad
\end{eqnarray}
\end{lem}
Note that all the regimes of the integrals are $[\acute{t}_0,t_1]$, except for (3) which is $[t_0,t_1]$. One can essentially prove the same for the integral over $[\acute{t}_0,t_1]$, but the above is enough for our purpose as well and is simpler to establish. The proof is discussed later:
\begin{itemize}
\item (1) and (2) are proven in Corollary \ref{cor:ind1:Sint S1int}
\item (3) is established in Lemma \ref{lem:ind1:intofSminS1}.
\item (4) is verified in Section \ref{subsec:ind1:1storder error}.
\end{itemize}
{ }Lastly, we state a lower bound on $S(t)$. Although the bounds given in $\tau_{1}$ implies that $S(t)$ can get pretty high compared to $\alpha_0$, it turns out that $S(t)$ cannot stay too low from $\alpha_0$.
\begin{prop}\label{prop:reg:lbd of speed}
Let $\alpha_0,t_0,r>0$, and let $\acute{t}_0, t_0^-,t_0^+, \hat{t}_0$ be as \eqref{eq:def:t0t1} and \eqref{eq:def:t0t11}.
Define the stopping time $\tau_{\textnormal{lo}}$ by
\begin{equation}
\tau_{\textnormal{lo}}=\tau_{\textnormal{lo}}(\alpha_0,t_0,t_0^+):= \inf \left\{ t\ge t_0^+: \ S(t) \le \alpha_0 - \alpha_0^{\frac{3}{2}-\epsilon} \right\}.
\end{equation}
Then, for any $\Pi_S(-\infty,t_0] \in \mathfrak{R}(\alpha_0,r;[t_0])$, we have $$\mathbb{P}(\tau_{\textnormal{lo}}\le \hat{t}_0 \, |\, \Pi_S(-\infty,t_0] )\le r+ \exp\left(-\beta_0^2 \right) . $$
\end{prop}
This proposition will be useful later in Section \ref{sec:double int}, and its proof is presented in Section \ref{subsec:ind1:1storder error}.
\subsection{Neglecting the far away history}\label{subsec:regoverview:history}
In this section, we resolve the issue addressed in Section \ref{subsubsec:regoverview:overview:history}. Recall the formula \eqref{eq:speed:basic expression}:
\begin{equation}\label{eq:def:S:in induction}
S(t) = \frac{1}{2}\mathbb{P} \left( W(s) \le Y_t(s), \ \forall s\ge 0 \ | \ Y_t \right).
\end{equation}
Our goal is to show that its dependence on $\mathcal{F}_{t_0^{-}} = \{Y_t(s)\}_{s\ge t-t_0^{-}}$ is negligible. To this end, we introduce an approximate version of speed and study its first and second order approximations. Let $\Pi'$ be the rate-1 Poisson point process on $\mathbb{R}^2$ independent from $\Pi$, and define $Y_t'(s)$ as
\begin{equation}
Y_t'(s):= \begin{cases}
Y_t(s), &\textnormal{ for } s\le t-t_0^{-};\\
Y_t(t_0^{-}) + |\Pi_{\alpha_0}'[t-s, t_{0}^{-}]|, &\textnormal{ for } s> t-t_0^{-}.
\end{cases}
\end{equation}
Then, we define the \textit{quenched} approximated speed $S_q'(t)$ as
\begin{equation}\label{eq:def:Sprime}
\begin{split}
S_q'(t) = S_q'(t;t_0^{-},\alpha_0) := \frac{1}{2} \mathbb{P}\left(W(s)\le Y_t'(s),\ \forall s\ge 0 \ | \ Y_t' \right) .
\end{split}
\end{equation}
We recall the previous definition of $S'(t)$ in \eqref{eq:def:Sprime:basic form}, and observe that
\begin{equation}
\mathbb{E}_{\Pi_{\alpha_0}'}[S_q'(t;t_0^{-},\alpha_0)] = S'(t;t_0^{-},\alpha_0).
\end{equation}
Recalling the definitions of $\tau_{6}$ and $\mathcal{A}$ from Sections \ref{subsubsec:reg:aggsize} and \ref{subsubsec:reg:history}, respectively, the main purpose of this subsection is to establish the following property.
\begin{prop}\label{prop:SvsSprime}
Suppose that we have $\Pi_S[t_0^-,t_0]\in \mathcal{A}(\alpha_0,t_0)$. Then, for all sufficiently small $\alpha_0>0$, we have $$|S(t)-S'(t;t_0^{-},\alpha_0)| \le \alpha_0^{100},$$ for all $t\in [t_0, \tau_{6}(\alpha_0,t_0,\kappa=2)]$. Furthermore, for any $t\in[\acute{t}_0,\tau_{6}\wedge\tau_{8}(\alpha_0,t_0,\kappa=2)]$, we have
\begin{equation}
|S(t)-S'(t;t_0,\alpha_0)| \le \alpha_0^{100}.
\end{equation}
\end{prop}
It turns out that the same proof as Proposition \ref{prop:SvsSprime} implies the following corollary, which will be used to study the induction base.
\begin{cor}\label{cor:SvsSprime}
Under the setting of Theorem \ref{thm:induction:base:main} and the discussions above, define $\overline{S}'(t;t_0^-,\alpha_0)$ analogously as above in terms of $\overline{S}(t)$, and suppose that $\Pi_{\overline{S}}(-\infty,t_0] \in \overline{\mathcal{A}}(\alpha_0,t_0)$. Then, for all sufficiently small $\alpha_0>0$, we have
\begin{equation}
|\overline{S}(t) - \overline{S}'(t;t_0^-,\alpha_0)| \le \alpha_0^{100},
\end{equation}
for all $t\in [t_0,\overline{\tau}_6(\alpha_0,t_0,2)].$ The same thing holds for $\overline{S}(t)$ and $\overline{S}'(t;t_0^-,\alpha_0)$.
\end{cor}
\begin{proof}[Proof of Proposition \ref{prop:SvsSprime}]
We prove the first statement of the proposition, and then the second one follows analogously. Let $t\in [t_0,\tau_{6}(\alpha_0,t_0,2)]$, and for simplicity we write $S'(t)=S'(t;t_0^{-},\alpha_0)$. Moreover, we write $\mathbb{P}(\ \cdot \ )$ to denote the probability over both the random walk $W$ and the independent point process $\Pi_{\alpha_0}'$ together. The main idea is to compare the formulas \eqref{eq:def:S:in induction} and \eqref{eq:def:Sprime} by coupling the random walk $W$ together. This gives that
\begin{equation}\label{eq:reg:history:SminusSprime}
\begin{split}
|S(t)- S'(t)|
&\leq
\mathbb{P}_{\Pi_{\alpha_0}'}\left(\left.W(s)\le Y_t(s) \ \forall s\le t-t_0^{-}, \textnormal{ and } \exists s'\ge t-t_0^{-} : W(s')\ge Y_t(s') \ \right| \ Y_t \right)
\\
&\ +\mathbb{P}_{\Pi_{\alpha_0}'}\left(\left.W(s)\le Y_t(s) \ \forall s\le t-t_0^{-}, \textnormal{ and } \exists s'\ge t-t_0^{-} : W(s')\ge Y_t'(s') \ \right| \ Y_t\right)\\
&=:P_1 + P_2.
\end{split}
\end{equation}
We begin with bounding $P_2$. Since $t\leq \tau_{6}(\alpha_0,t_0,2)$, \begin{equation}\label{eq:reg:SvsSprime:base}
Y_t(t_0^{-}) \ge \frac{\alpha_0}{200}(t-t_0^{-}).
\end{equation} Thus, we can write
\begin{equation}\label{eq:reg:history:P2}
\begin{split}
P_2 &\le \mathbb{P} \left(W(t-t_0^{-}) \ge \frac{\alpha_0}{400} (t-t_0^{-}) \right)\\
&+
\mathbb{P} \left( \exists s\ge t_0^{-}: \ W(s) \ge |\Pi'_{\alpha_0}[t-s, t-t_0^{-} ]| + \frac{\alpha_0}{400}(t-t_0^{-}) \right).
\end{split}
\end{equation}
Note that since $t-t_0^{-} \ge \alpha_0^{-2} \beta_0^\theta$,
\begin{equation}
\frac{\alpha_0}{400}(t-t_0^{-}) \ge
\sqrt{t-t_0^{-}} \log^{\theta/3}(t-t_0^{-}).
\end{equation}
Plugging this into the first term of \eqref{eq:reg:history:P2}, we can see that
\begin{equation}\label{eq:reg:history:P2:1}
\mathbb{P} \left(W(t-t_0^{-}) \ge \frac{\alpha_0}{400} (t-t_0^{-}) \right)\le \alpha_0^{101}.
\end{equation}
On the other hand, to study the second term of \eqref{eq:reg:history:P2}, we define
\begin{equation}
U'(s) = |\Pi_{\alpha_0}'[0,s] |-W(s) + \frac{\alpha_0}{400}(t-t_0^{-}),
\end{equation}
and define the stopping time $T':= \inf\{s>0: U'(s)\le0\}$. Then, we can write
\begin{equation}
\begin{split}
\mathbb{P} \left( \exists s\ge t_0^{-}: \ W(s) \ge |\Pi'_{\alpha_0}[t-s, t-t_0^{-} ]| + \frac{\alpha_0}{400}(t-t_0^{-}) \right)
=
\mathbb{P} \left(T'<\infty \right)\\
= \lim_{s\to \infty} \mathbb{E} \left[ \left(\frac{1}{1+2\alpha_0} \right)^{U'(s) \wedge T' } \right].
\end{split}
\end{equation}
As we saw in \eqref{eq:speed:Ut mg conv}, $(1+2\alpha_0)^{-U'(s)}$ is a martingale, and hence the Optimal Stopping Theorem tells us that
\begin{equation}\label{eq:reg:history:P2:2}
\mathbb{P}(T'<\infty) = \mathbb{E}\left[\left(\frac{1}{1+2\alpha_0} \right)^{U'(0)} \right] \le \left(\frac{1}{1+2\alpha_0} \right)^{\frac{\alpha_0^{-1}\beta_0^\theta}{400}} \le \exp\left( -\frac{\beta_0^\theta}{400}\right) \le \alpha_0^{101}.
\end{equation}
Combining \eqref{eq:reg:history:P2:1} and \eqref{eq:reg:history:P2:2} tells us that $P_2 \le 2\alpha_0^{101}$.
To control $P_1$, we first note that $\mathcal{A}_1(\alpha_0,t_0)$ and $t\le \tau_{6}(\alpha_0,t_0,2)$ implies the following for all $u\le t_0^{-}$:
\begin{equation}
\begin{split}
X_{t } - X_u
&=
X_t - X_{t_0^{-} }+ X_{t_0^{-} } - X_u\\
&\ge
\frac{\alpha_0}{200}(t-t_0^{-})+
\sqrt{t_0^{-}-u+C_\circ} \log^2 (t_0^{-}-u +C_\circ) -\alpha_0^{-1}\beta_0^{\theta/2}\\
&\ge
\frac{\alpha_0}{300}(t-t_0^{-})+\sqrt{t_0^{-}-u+C_\circ} \log^2 (t_0^{-}-u +C_\circ),
\end{split}
\end{equation}
where the last line is from $t-t_0^{-}\ge \alpha_0^{-2} \beta_0^\theta$. We can also see that
\begin{equation}
\frac{\alpha_0}{300}(t-t_0^{-}) \ge \sqrt{t-t_0^{-}+C_\circ} \log^2 (t-t_0^{-}+C_\circ),
\end{equation}
which gives
\begin{equation}
\begin{split}
X_{t } - X_u
&\ge
\sqrt{t-t_0^{-}+C_\circ} \log^2 (t-t_0^{-}+C_\circ)
+
\sqrt{t_0^{-}-u+C_\circ} \log^2 (t_0^{-}-u +C_\circ)\\
&\ge
\sqrt{t-u+C_\circ} \log^2 (t-u+C_\circ),
\end{split}
\end{equation}
where the last inequality follows from the fact that the function $\sqrt{x+20}\log^2(x+20)$ is increasing and concave. From this estimate, we can see that
\begin{equation}\label{eq:reg:history:P1}
P_1 \le
\mathbb{P}\left(\exists s'\ge t-t_0^{-}: W(s') \ge \sqrt{s'}\log^2(s') \right)
\le e^{-\beta_0^3
} \le \alpha_0^{101},
\end{equation}
where the second inequality is a standard estimate on a simple random walk. We conclude the proof by combining \eqref{eq:reg:history:SminusSprime}, \eqref{eq:reg:history:P2} and \eqref{eq:reg:history:P1}.
To obtain the second statement of the proposition, we apply the same argument with
\begin{equation}
Y_t(t_0) \ge \frac{\alpha_0}{2\beta_0^{C_\circ}} (t-t_0),
\end{equation}
instead of \eqref{eq:reg:SvsSprime:base}. This estimate is obtained from the definition of $\tau_8$.
\end{proof}
As a concluding remark of the section, we record a simple lemma on $\overline{S}(t)$ and $\underline{S}(t)$ as a step towards establishing Theorem \ref{thm:induction:base:main}.
\begin{lem}\label{lem:reg:base:A}
Under the setting of Theorem \ref{thm:induction:base:main} and the discussions above, we have
\begin{equation}
\mathbb{P} \left(\Pi_{\overline{S}}(-\infty,t_0] \in \overline{\mathcal{A}}(\alpha_0,t_0) \right) \ge 1- \exp \left(-\beta_0^{4} \right).
\end{equation}
The same holds for $\underline{S}(t)$.
\end{lem}
\begin{proof}
The estimate on the event $\{\Pi_{\overline{S}}(-\infty,t_0] \in \overline{\mathcal{A}}_1(\alpha_0,t_0) \cap \overline{\mathcal{A}}_2(\alpha_0,t_0) \}$ follows straight-forwardly from the properties of a fixed rate Poisson process, and we omit the details. The bound on the event $\overline{\mathcal{A}}_3(\alpha_0,t_0)$ comes from the same argument as \eqref{eq:reg:ib:med1} and \eqref{eq:reg:ib:med2}, replacing $t_0^\flat$ by $t_0$.
\end{proof}
\section{The inductive analysis on the speed: Part 1} \label{sec:reg:conti of reg}
The purpose of this section is to formulate the argument discussed in Section \ref{subsubsec:regoverview:overview:contiofreg}. Before we state the main theorem, we briefly explain a technical device required to understand the procedure of the induction argument. In the definition of $\tau$ from \eqref{eq:def:tau:induction}, recall that there was a parameter $\kappa \in \{\frac{1}{2},\, 2 \}$ that provides a constant multiplicative room in controlling the quantities inside $\tau_i$'s. One of the main observations in the induction argument is that even if we have $\{\tau(\alpha_0, t_0, \kappa) > \acute{t}_0\}$ with a weaker (i.e., larger) $\kappa$, the control on the regularity will bootstrap as the time passes by and at a larger time we still have $\{\tau(\alpha,t_0',\kappa') \ge \hat{t}_0 \}$ but with a stronger (i.e., smaller) $\kappa'$, with $t_0' > t_0$. Note that we can only have this bootstrapped estimate with $t_0'>t_0$, not with $t_0$, since the initial assumption is made under a weaker $\kappa$. By doing this bootstrapping, we acquire enough room to cover the error that comes from changing $\alpha_0$ to $\alpha_1$ in the next time step.
Having this idea in mind, we introduce another notation before stating the main objective. We define ${\tau}^+$ to be
\begin{equation}\label{eq:def:ind1:tauacute:basic}
{\tau}^+(\kappa ) := \tau(\alpha_0, t_0^+, \kappa),
\end{equation}
except that we use the same $S_1(t)=S_1(t;t_0^-,\alpha_0)$ when defining $\tau(\alpha_0, t_0^+, \kappa)$. That is, for instance, we define
\begin{equation}\label{eq:def:ind1:tauacute}
\begin{split}
{\tau}^+_2(\kappa) &:= \inf \{t\ge t_0^+: S_1(t;t_0^-,\alpha_0) \ge \kappa \alpha_0 \beta_0^{C_\circ} \};\\
{\tau}^+_4(\kappa)&:=
\inf\left\{t\ge {t}_0^+: \intop_{t_0^+}^{t} (S(s)-\alpha_0)^2ds \ge \kappa \alpha_0 \beta_0^{25\theta} \right\};\\
{\tau}^+_6(\kappa)
&:=
\inf\left\{t\ge {t}_0^+: |\Pi_S[t_0^{-},\, t]| \le \frac{1}{100\kappa} \alpha_0 (t-t_0^{-}) \right\},
\end{split}
\end{equation}
and corresponding analogue for all other ${\tau}^+_i(\kappa), \ i=1,3,5,7,8$, which are all defined in a straight-forward way unlike the three mentioned above. Then, set ${\tau}^+(\kappa)$ to be the minimum of ${\tau}^+_i$'s. The main result for this section can be stated as follows.
\begin{thm}\label{thm:reg:conti:main}
Recall the definitions of $\tau$ and $\mathcal{A}$ from \eqref{eq:def:tau:induction} and \eqref{eq:def:A:induction}, respectively. Write $\tau(\kappa)=\tau(\alpha_0,t_0,\kappa)$, $\mathcal{A}=\mathcal{A}(\alpha_0,t_0)$, and set ${\tau}^+(\kappa)$ as above. For all sufficiently small $\alpha_0>0$ and any $t_0>0$, we have
\begin{equation}\label{eq:ind1:main}
\mathbb{P} \left({\tau}^+(1/2 ) < \hat{t}_0 \ | \ \mathcal{A} \right)
\le
\mathbb{P} \left(\tau(2 ) \le \acute{t}_0 \ | \ \mathcal{A} \right)
+
\exp\left(-\beta_0^{1.9} \right).
\end{equation}
\end{thm}
\begin{remark}
The equations involving the notation $\mathbb{P} (\ \cdot \ | \, \mathcal{A} ) $ are understood as follows throughout Sections \ref{sec:reg:conti of reg} and \ref{sec:reg:next step}: The equation holds for any $\Pi_S(-\infty,t_0]$ such that $\Pi_S(-\infty,t_0]\in \mathcal{A}$.
\end{remark}
As done in Section \ref{sec:criticalbranching}, our idea is to show that each stopping time ${\tau}^+_i(1/2)$ is likely to be larger than $\tau(2)$ if $\tau(2)\ge \acute{t}_0$. In the following sections, we conduct this task through three major steps as follows.
\begin{enumerate}
\item In Section \ref{subsec:ind1:1storder error}, we study $|S(t)-S_1(t)|$ to control ${\tau}^+_3$, done by utilizing the result we saw in Section \ref{subsec:fixed:error}.
\item In Section \ref{subsec:ind1:S1vsalpha}, we estimate the gap $|S_1(t)-\alpha_0|$ based on ideas from Section \ref{subsec:branching:mg}.
\item In Section \ref{subsec:ind1:Svsalpha}, we control all stopping times except ${\tau}^+_3$, with the estimate
\begin{equation}\label{eq:ind1:Svsalpha:decomp:basic}
|S(t)-\alpha_0|
\le
|S(t)-S_1(t)|+|S_1(t)-\alpha_0|
\end{equation}
obtained from Sections \ref{subsec:ind1:1storder error}, \ref{subsec:ind1:S1vsalpha}.
\end{enumerate}
In each of these subsections, we add brief explanations on how the result is generalized to the case of induction base, Theorem \ref{thm:induction:base:main}. The only major difference compared to the proof of Theorem \ref{thm:reg:conti:main} is the availability of the assumption $\{\tau(2)>\acute{t}_0 \}$. Although we can only be interested in $t \ge t_0^+$ when establishing Theorem \ref{thm:reg:conti:main}, we need to consider all $t\ge t_0$ for Theorem \ref{thm:induction:base:main}. This is the place where Theorem \ref{thm:branching:ib} comes in to play.
As a consequence of the arguments used in Theorem \ref{thm:reg:conti:main}, we establish
\begin{itemize}
\item [(4)] Theorem \ref{thm:induction:base:main} in Section \ref{subsec:ind1:reg of fixed};
\item [(5)] Proposition \ref{prop:ind1:growth} in Section \ref{subsec:ind1:growth}.
\end{itemize}
For convenience, we restate Theorem \ref{thm:induction:base:main} to shape it into a more convenient form to work with. For its proof, we only discuss the case of $\overline{S}(t)$, since the corresponding result for $\underline{S}(t)$ follows analogously.
\begin{prop}\label{prop:ind1:base:main}
Assume the setting of Theorem \ref{thm:induction:base:main} and the discussions above, recall the definition of $\tau_{\textnormal{B}}=\tau_{\textnormal{B}}(\alpha_0,t_0)$ from \eqref{eq:def:tau:branching:ib} we define
\begin{equation}
\begin{split}
\overline{\tau}=\overline{\tau}(\alpha_0,t_0):=\tau_{\textnormal{B}} \wedge \min\left\{ \overline{\tau}_i(\alpha_0,t_0,\kappa=2) \,:\, 1\le i \le 8 \right\}.
\end{split}
\end{equation}
(Note that $\overline{\tau}_3(\alpha_0,t_0)$ does not depend on $\kappa$ although we wrote as above for convenience.)
Then,
\begin{equation}
\mathbb{P} \left( \overline{\tau}(\alpha_0,t_0) >\acute{t}_0, \ \Pi_{\overline{S}}(-\infty,t_0]\in \overline{\mathcal{A}}(\alpha_0,t_0) \right) \ge 1-\exp\left(-\beta_0^{1.9}\right).
\end{equation}
The same result holds for $\Pi_{\underline{S}}$ and $\underline{\tau}(\alpha_0,t_0).$
\end{prop}
Before moving on,
we establish the following lemma which is not only a useful tool in the later section, but also leads to the proof of (3) of Lemma \ref{lem:reg:errorint}.
\begin{lem}\label{lem:ind1:intofSminS1}
Under the setting of Theorem \ref{thm:reg:conti:main}, we have that
\begin{equation}
\begin{split}
\intop_{ t_0}^{\tau(2)} |S(t)-S_1(t)|dt \le \alpha_0^{-2\epsilon} .
\end{split}
\end{equation}
\end{lem}
\begin{proof}
From the definition of $\tau_{3}$ (Section \ref{subsubsec:reg:Scontrol}) and Lemma \ref{lem:ind1:pi1int basicbd:basic}, we have
\begin{equation}
\intop_{ t_0}^{\tau(2)} |S(t)-S_1(t)|dt \le \intop_{t_0}^{\tau(2)} \frac{\alpha_0^{1-\epsilon} dt }{\pi_1(t;S)+1} \le \alpha_0^{1-\epsilon}\beta_0^2 |\Pi_S[t_0,\tau(2)]| \le \alpha_0^{-2\epsilon} ,
\end{equation}
where we obtained the last inequality from the definition of $\tau_{7}$ (Section \ref{subsubsec:reg:aggsize}).
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:reg:errorint}-(3)]
Recall in Theorem \ref{thm:reg:conti:main} that ${\tau}^+(1/2) \le \tau(2)$ if $\tau(2)>\acute{t}_0$. Thus, combining Theorem \ref{thm:reg:conti:main} and Lemma \ref{lem:ind1:intofSminS1} directly implies (3) of Lemma \ref{lem:reg:errorint}.
\end{proof}
\subsection{The error from the first order approximation}\label{subsec:ind1:1storder error}
As the first step towards establishing Theorem \ref{thm:reg:conti:main}, we begin with studying $\tau_{3}$, the error of the first-order approximation.
\begin{lem}\label{lem:ind1:tau6}
Under the setting of Theorem \ref{thm:reg:conti:main}, we have
\begin{equation}
\mathbb{P} \left({\tau}^+_6 = \tau(2)\wedge \hat{t}_0,\ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
\noindent (Note that ${\tau}^+_6 \ge \tau(2)$ by definition.)
\begin{proof}
Let $t\in[\acute{t}_0, \hat{t}_0]$, and we begin with expressing that
\begin{equation}\label{eq:ind1:SminS1:decomp}
\begin{split}
|S(t)-S_1(t;t_0^-,\alpha_0)|\le& |S(t)-S'(t;t_0,\alpha_0)|+ |S'(t;t_0,\alpha_0)-S_1(t;t_0,\alpha_0)| \\
&+ |S_1(t;t_0^-,\alpha_0)-S_1(t;t_0,\alpha)|,
\end{split}
\end{equation}
and note that the first term in the RHS is bounded by $\alpha_0^{100}$ for any $\Pi_S(-\infty,t_0]\in \mathcal{A}$, which is from Proposition \ref{prop:SvsSprime}. Moreover, for any $t\in [\acute{t}_0, \tau(2)\wedge \hat{t}_0],$ Lemma \ref{lem:estimate on K:intro} tells us that
\begin{equation}\label{eq:ind1:S1timechange}
|S_1(t;t_0^-,\alpha_0)-S_1(t;t_0,\alpha_0)| = \intop_{t_0^-}^{t_0} K_{\alpha_0}(t-x)d\Pi_S(x) \le K_{\alpha_0}(\acute{t}_0-t_0) \cdot |\Pi_S[t_0^-,t_0]|\le \alpha_0^{100}.
\end{equation}
The rest of the proof follows as a direct consequence of Proposition \ref{prop:fixed perturbed:error}. In fact, by setting $\alpha$, $t^-$, $\hat{t}$ and $\tau$ in Proposition \ref{prop:fixed perturbed:error} to be
\begin{equation}
\alpha = \alpha_0, \ \ t^- = t_0, \ \ \hat{h} = \hat{t}_0, \ \ \tau = \tau(2),
\end{equation}
we see that the assumptions in the proposition are all satisfied. Thus, we obtain that
\begin{equation}\label{eq:ind1:tau6:1}
\begin{split}
\mathbb{P} \left( |S(t)-S_1(t;t_0^-,\alpha_0)| \le 4\alpha_0^{1-\epsilon}\sigma_1\sigma_2(t;S), \ \forall t\in[\acute{t}_0, \hat{t}_0\wedge \tau(2)] \, | \, \mathcal{A} \right) \\
\ge 1-4\exp\left(-\alpha_0^{-\frac{\epsilon}{3000}}\right) .
\end{split}
\end{equation}
The lower bound on $S(t)-S_1(t)$ is obtained analogously, from the expression
\begin{equation}
\begin{split}
S(t)-S_1(t;t_0^-,\alpha_0)\ge& (S'(t;t_0,\alpha_0)-S_1(t;t_0,\alpha_0)) -|S(t)-S'(t;t_0,\alpha_0)| \\
&- |S_1(t;t_0^-,\alpha_0)-S_1(t;t_0,\alpha)|,
\end{split}
\end{equation}
and Proposition \ref{prop:fixed perturbed:error}.
\end{proof}
Note that the corresponding analogue of \eqref{eq:ind1:tau6:1} holds the same for the second-order approximation $S_2(t)=S_2(t;t_0^-,\alpha_0)$ \eqref{eq:def:S2:basic form}, and we record this result in the following corollary.
\begin{cor}\label{cor:ind1:S2error}
Under the setting of Theorem \ref{thm:reg:conti:main}, there exists a constant $c_\epsilon>0$ such that
\begin{equation}
\begin{split}
\mathbb{P} \left(|S(t)-S_2(t)|\le 4\alpha_0^{1-\epsilon} \sigma_1\sigma_2\sigma_3(t;S) ,\ \forall t\in[\acute{t}_0, \hat{t}_0\wedge\tau(2)] \,| \,\mathcal{A} \right)\\
\ge 1-4\exp\left(-\alpha_0^{-c_\epsilon} \right) .
\end{split}
\end{equation}
\end{cor}
\begin{proof}
The only difference in the proof compared to the previous lemma is the way we control $|S_2(t;t_0,\alpha)-S_2(t;t_0^-,\alpha)|$. The difference is bounded by
\begin{equation}
\frac{1}{(1+2\alpha)^2} \intop_{t_0^-}^{t_0} K_{\alpha_0}(t-x) d\Pi_S(x) + \frac{\alpha}{1+2\alpha} \intop_{t_0^-}^t \intop_{t_0^-}^{t_0\wedge s} J_{t-s,t-u}^{(\alpha_0)} d\widehat{\Pi}_S(u)d\widehat{\Pi}_S(s),
\end{equation}
which is also smaller than $\alpha_0^{100}$ for all $t\in [\acute{t}_0,\tau(2)\wedge\hat{t}_0]$ due to the decay properties of $K$ and $J$ (Lemma \ref{lem:bound on deterministic J}).
\end{proof}
Along with Lemma \ref{lem:fixed perturbed:error int:perturbed}, this leads to the proof of (4) of Lemma \ref{lem:reg:errorint}:
\begin{proof}[Proof of Lemma \ref{lem:reg:errorint}-(4)]
The result is obtained by integrating the bound on $|S(t)-S_2(t)|$ in Corollary \ref{cor:ind1:S2error} using Lemma \ref{lem:fixed perturbed:error int:perturbed}, and relying on the estimate on $\tau(2) $ from Theorem \ref{thm:reg:conti:main}.
\end{proof}
We conclude this subsection by presenting the analogue of Lemma \ref{lem:ind1:tau6} for the induction base.
\begin{cor}\label{cor:ind1:tau6}
Under the setting of Proposition \ref{prop:ind1:base:main}, we have
\begin{equation}
\mathbb{P}\left( \overline{\tau}_3 = \overline{\tau}\wedge \hat{t}_0, \ \overline{\tau}>t_0 \right) \le \exp\left(-\beta_0^3 \right).
\end{equation}
\end{cor}
\begin{proof}
Relying on the same decomposition \eqref{eq:ind1:SminS1:decomp}, we can obtain the same estimate on $\overline{S}(t)-\overline{S}_1(t;t_0^-,\alpha_0)$ from Lemma \ref{lem:reg:base:A} and Corollary \ref{cor:SvsSprime} along with Lemma \ref{lem:ind1:tau6}.
\end{proof}
\subsection{Connection to the critical branching process}\label{subsec:ind1:S1vsalpha}
In this subsection, we estimate $|S_1(t)-\alpha_0|$. By understanding the size of this quantity, we will eventually be able to control $|S(t)-\alpha_0|$ in the next subsection. Define the stopping time $\tau_{\textnormal{b}}$ as
\begin{equation}\label{eq:def:taux}
\tau_{\textnormal{b}} = \tau_{\textnormal{b}}(\alpha_0,t_0,t_0^+) := \inf\left\{t\ge t_0^+:( S_1(t)-\alpha_0) \notin \left(-2\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} ,\alpha_0\beta_0^{C_\circ}\sigma_1(t;S) + 2\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} \right) \right\}.
\end{equation}
Our goal is to establish the following Lemma:
\begin{lem}\label{lem:ind1:taux}
Under the setting of Theorem \ref{thm:reg:conti:main}, we have
\begin{equation}
\mathbb{P} \left( \tau_{\textnormal{b}} <\hat{t}_0,\ \tau(2) >\acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
\end{lem}
To prove this lemma, we rely on the methods in Section \ref{subsec:branching:mg}. Recall the definition \eqref{eq:def:Qst}, and for $t\ge s \ge t_0$, define
\begin{equation}
\begin{split}
R(s,t)=\mathcal{R}_c(s,t;\Pi_{S}[t_0^{-},s],\alpha_0).
\end{split}
\end{equation}
We note that $R(t,t)=S_1(t)$ and for $t\ge t_0^+$ that
\begin{equation}
\begin{split}
|R(t_0,t)-\alpha_0'|& \le \intop_{t_0^-}^{t_0} K_{\alpha_0}(t-x) d\Pi_S(x) + \intop_{t_0^-}^{t_0} \intop_t^{\infty} K^*_{\alpha_0}\cdot K_{\alpha_0}(u-x) du d\Pi_S(x)\\
&\quad + \intop_{t_0^-}^{t_0} \intop_{t_0}^t |K^*_{\alpha_0}-K^*_{\alpha_0}(t-u) | K_{\alpha_0}(u-x) du d\Pi_S(x)\\
&\le \alpha_0^{100},
\end{split}
\end{equation}
for all $t\in[\acute{t}_0, \tau(2)\wedge\hat{t}_0]$, similarly as \eqref{eq:branching:Rlbd:med3}.
Then, \eqref{eq:integralform:branching} gives
\begin{equation}\label{eq:integralform:branching:ind1}
\begin{split}
S_1(t) - \alpha_0' + O(\alpha_0^{100}) &= R(t,t)-R(t_0,t) \\
&= \intop_{t_0}^t K^*_{\alpha_0}(t-x) \left\{ d\widetilde{ \Pi}_S(x) + (S(x)-S_1(x))dx\right\} ,
\end{split}
\end{equation}
where we wrote $d\widetilde{ \Pi}_S(x):= d\Pi_S(x) - S(x)dx$. This decomposes $S_1(t)-\alpha_0'$ into a martingale part and a drift part.
\begin{proof}[Proof of Lemma \ref{lem:ind1:taux}]
We begin with estimating the martingale part of $S_1(t)-\alpha_0'$ in \eqref{eq:integralform:branching:ind1}. We apply Corollary \ref{lem:concentrationofint:continuity} in the following setting:
\begin{itemize}
\item Set $\tau = \tau(2)$, $f_t(x) = K^*_{\alpha_0}(t-x)$, $D=1$ (such a choice of $D$ is justified by Corollary \ref{cor:bound on K'}).
\item From the definition of $\tau_{1}$ and the bound in Lemma \ref{lem:estimat for K tilde:intro}, we can set
\begin{equation}\label{eq:ind1:S1minusalpha QV}
M:=\alpha_0^3\beta_0^{12\theta} \ge \intop_{t_0}^{t\wedge \tau(2)} (K^*_{\alpha_0}(t-x))^2 S(x)dx.
\end{equation}
\item $\Delta = \alpha_0^{-1}$, $A = C\alpha_0^{\frac{3}{2}}$, $\eta = \alpha_0\beta_0^{C_\circ}$, $N= \beta_0^5$ and $\delta = \alpha_0^{10}$ (see definitions of $\tau_{1}$ and $\tau_{7}$).
\end{itemize}
Then Corollary \ref{lem:concentrationofint:continuity} gives
\begin{equation}
\begin{split}
\mathbb{P} \left( \left.\intop_{t_0}^{t\wedge \tau(2)} K_{\alpha_0}^*(t-x) d\widetilde{\Pi}_S(x ) \le \alpha_0\beta_0^{5}\sigma_1(t;S) +\alpha_0^{\frac{3}{2}}\beta_0^{6\theta}, \, \forall t\in[t_0, \hat{t}_0 ] \right| \, \mathcal{A} \right) \ge 1- e^{-\beta_0^{C_\circ}}; \\
\mathbb{P} \left( \left.\intop_{t_0}^{t\wedge \tau(2)} K_{\alpha_0}^*(t-x) d\widetilde{\Pi}_S(x ) \ge -\alpha_0^{\frac{3}{2}}\beta_0^{6\theta}, \, \forall t\in[t_0, \hat{t}_0 ] \right| \, \mathcal{A} \right) \ge 1- e^{-\beta_0^{C_\circ}}.
\end{split}
\end{equation}
To control the drift term of \eqref{eq:integralform:branching:ind1}, we use the bound given by $\tau_{3}$. Express that
\begin{equation}
\begin{split}
\intop_{t_0}^{t\wedge \tau(2)} K^*_{\alpha_0}(t-x) |S(x)-S_1(x)| dx \le \intop_{t_0}^{t\wedge \tau(2)} C\left(\frac{\alpha_0}{\sqrt{t-x}} \vee \alpha_0^2 \right) \frac{\alpha_0^{1-\epsilon}dx }{\pi_1(x;S)+1}
\le \alpha_0^{2-2\epsilon},
\end{split}
\end{equation}
where the last inequality is from Lemma \ref{lem:ind1:pi1int basicbd}, with parameters $\Delta_0 = \alpha_0^{-1}, \Delta_1= \alpha_0^{-1}\beta_0^{C_\circ}, K=\alpha_0^{-1}\beta_0^{11\theta}$ and $N_0 = \beta_0^{5}$. Thus, the conclusion follows by combining the above two bounds, along with the condition $|\alpha_0'-\alpha_0|\le \alpha_0^{\frac{3}{2}}\beta_0^{6\theta}$ from $\mathcal{A}_3(\alpha_0,t_0)$.
\end{proof}
We record a direct consequence of our analysis, for a future usage in Section \ref{subsec:increment:double int}.
\begin{cor}\label{cor:ind1:taux:fixedrate}
Under the setting of Theorem \ref{thm:reg:conti:main}, define the stopping time $\tilde{\tau}_{b}$ as
\begin{equation}
\tilde{\tau}_{\textnormal{b}}:= \inf\left\{ t\ge t_0^+: \left|\intop_{t_0^+}^t K^*_{\alpha_0}(t-x) d\widetilde{\Pi}_{\alpha_0}(x) \right| \ge \alpha_0\beta_0^{C_\circ} \sigma_1(t; \alpha_0) +2\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} \right\}.
\end{equation}
Then, we have $\mathbb{P}(\tilde{\tau}_{\textnormal{b}}<\hat{t}_0,\ \tau(2)>\acute{t}_0 \,| \, \mathcal{A} ) \le \exp(-\beta_0^5) .$
\end{cor}
Note that the exponent $6\theta$ of the error term $\alpha_0^{\frac{3}{2}}\beta_0^{6\theta}$ is related with the length of the interval $[\acute{t}_0, \hat{t}_0]$, as seen in \eqref{eq:ind1:S1minusalpha QV}. If we are interested in a shorter interval, we can obtain the following stronger bound, which will be essential in the analysis of $\tau_{3}^\sharp$ in Section \ref{subsec:ind2:bootstrappedJ}. Its proof is omitted since it is identical to that of Lemma \ref{lem:ind1:taux}.
\begin{cor}\label{cor:ind1:tauxprime}
Under the setting of Theorem \ref{thm:reg:conti:main}, define the stopping time
\begin{equation}\label{eq:def:tauxprime}
\tau_{\textnormal{b}}' = \tau_{\textnormal{b}}'(\alpha_0,t_0,t_0^+) := \inf\left\{t\ge t_0^+:| S_1(t)-\alpha_0'| \le \alpha_0\beta_0^{C_\circ}\sigma_1(t;S) + \alpha_0^{\frac{3}{2}}\beta_0^{\theta} \right\},
\end{equation}
with $\alpha_0' = \mathcal{L}(t_0;\Pi_S[t_0^-,t_0], \alpha_0)$ as before. Then, we have
\begin{equation}
\mathbb{P} \left( \tau_{\textnormal{b}}' \le \acute{t}_0, \ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
\end{cor}
We stress that the dependence on $t_0^-$ in the definition \eqref{eq:def:tauxprime} comes from $\alpha_0'$ and $S_1(t)=S_1(t;t_0^-,\alpha_0)$. We include $t_0^-$ in the notation (unlike the case of $\tau_{\textnormal{b}}$) to prevent confusion when it is used later in Section \ref{subsec:ind2:bootstrappedJ}.
We conclude this subsection by deriving an analogue for the process $\overline{S}(t)$ (and $\underline{S}(t)$). To be more concrete, we set $\overline{S}_1(t):= \intop_{t_0^-}^t K_{\alpha_0}(t-x) d\Pi_{\overline{S}}(x)$, and define the processes ${R}_0^+(t)$ and $\hat{R}_0^+(t)$ as in \eqref{eq:def:R:ib:aux} and \eqref{eq:def:R0plushat}, replacing $\alpha$ with $\alpha_0$. Define $\overline{S}^+(t):= \overline{S}(t)\vee \hat{R}_0^+(t)$, and let
\begin{equation}\label{eq:def:taux:init}
\overline{\tau}_{{\textnormal{b}}} = \overline{\tau}_{{\textnormal{b}}}(\alpha_0,t_0) := \inf\left\{t\ge t_0:( \overline{S}_1(t)-\alpha_0) \notin \left(-2\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} ,\alpha_0\beta_0^{5\theta}\sigma_1(t;\overline{S}^+) + 2\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} \right) \right\}.
\end{equation}
The major difference in the definition is that the stopping time reads the process starting from $t_0$, not from $t_0^+$, where we entail Theorem \ref{thm:branching:ib}.
Then, the following corollary holds true:
\begin{cor}
Under the setting of Proposition \ref{prop:ind1:base:main}, we have
\begin{equation}
\mathbb{P} \left( \overline{\tau}_{\textnormal{b}} = \overline{\tau} \wedge \hat{t}_0, \ \overline{\tau}>t_0 \right) \le \exp\left(-\beta_0^4 \right).
\end{equation}
\end{cor}
\begin{proof}
The proof is analogous as Lemma \ref{lem:ind1:taux}, using the decomposition \eqref{eq:integralform:branching:ind1} corresponding to $\overline{S}_1(t)$. This gives
\begin{equation}
\overline{S}_1(t)-\overline{R}(t_0,t) \in \left(-\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} , \, \alpha_0\beta_0^5 \sigma_1(t;\overline{S}) + \alpha_0^{\frac{3}{2}}\beta_0^{6\theta} \right)
\end{equation}
for all $t\in [t_0, \overline{\tau}(2)\wedge \hat{t}_0]$, which holds with high probability.
The only difference is that we are not guaranteed with the bound on $|\overline{R}(t_0,t)-\alpha_0|$ as before, since we need to cover $t\ge t_0$, not just $t\ge t_0^+$.
To study $|\overline{R}(t_0,t)-\alpha_0|$, consider the process
${R}(t) = \mathcal{R}_b(t;\Pi_{\alpha_0}[t_0^-,t_0],\alpha_0)$ \eqref{eq:def:Rb}. Then, $\tau_{\textnormal{B}1}$ from Theorem \ref{thm:branching:ib} gives the conclusion, since $\overline{R}(t_0,t)$ is an averaged version of $R(t)$.
\end{proof}
\subsection{Proximity of the speed from the base rate}\label{subsec:ind1:Svsalpha}
The goal of this subsection is to control all stopping times introduced in Sections \ref{subsubsec:reg:Scontrol} and \ref{subsubsec:reg:aggsize}, except $\tau_{3}$. Furthermore, we also establish Proposition \ref{prop:reg:lbd of speed}. To this end, we exploit the bound on $|S(t)-\alpha_0|$ based on \eqref{eq:ind1:Svsalpha:decomp:basic}.
Let $F(\alpha_0,t)$ be the following random function which is essentially the sum of error bounds from $\tau_{3}$ (Section \ref{subsubsec:reg:Scontrol}), and $\tau_{\textnormal{b}}$ \eqref{eq:def:taux}:
\begin{equation}
\begin{split}
F(\alpha_0,t):= 4\alpha_0^{1-\epsilon}\sigma_1\sigma_2(t;S) + \alpha_0\beta_0^{C_\circ+1} \sigma_1(t;S) + 3\alpha_0^{3/2} \beta_0^{6\theta}.
\end{split}
\end{equation}
Define
\begin{equation}\label{eq:def:tau12}
{\tau}_{9}^+:= \inf\{t \ge t_0^+: |S(t)-\alpha_0| \ge F(\alpha_0,t) \}.
\end{equation}
\begin{cor}\label{cor:ind1:tau12}
Under the setting of Theorem \ref{thm:reg:conti:main}, we have
\begin{equation}
\mathbb{P} \left({\tau}^+_{9} \le \tau(2),\ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le 3\exp\left(-\beta_0^2 \right).
\end{equation}
\end{cor}
\begin{proof}
This is an immediate consequence of Lemmas~\ref{lem:ind1:tau6} and \ref{lem:ind1:taux}.
\end{proof}
Recall the definition of ${\tau}^+_i(1/2)$ \eqref{eq:def:ind1:tauacute}. In the following subsections, we show that each stopping time ${\tau}^+_i$ is unlikely to be smaller than or equal to \begin{equation}\label{eq:def:tau2tilde}
\tilde{ \tau}(2):= \tau(2)\wedge {\tau}^+_{9}.
\end{equation}\begin{itemize}
\item In Section \ref{subsubsec:ind1:squareint}, we control ${\tau}^+_4(1/2), {\tau}^+_5(1/2)$, the stopping times related with the square integral of $|S(t)-\alpha_0|$.
\item In Section \ref{subsubsec:ind1:aggsize}, we study ${\tau}^+_6(1/2),{\tau}^+_{7}(1/2),{\tau}^+_{8}(1/2),$ which describe the size of the aggregate.
\item In Section \ref{subsubsec:ind1:Sbd}, we work with $\tau_{1}^+(1/2), \tau_{2}^+(1/2)$.
\item In Section \ref{subsubsec:ind1:base}, we show how the analysis from Sections \ref{subsubsec:ind1:squareint}--\ref{subsubsec:ind1:Sbd} can be done in the setting of Proposition \ref{prop:ind1:base:main}.
\item In Section \ref{subsubsec:ind1:Sminusalpha conseq}, we establish several consequences of our analysis which will be useful later in Section \ref{sec:double int}.
\end{itemize}
\subsubsection{The square integrals}\label{subsubsec:ind1:squareint}
The main goal of this subsection is to deduce the following lemma.
\begin{lem}\label{lem:ind1:tau78}
Under the setting of Theorem \ref{thm:reg:conti:main}, let $\tilde{\tau}(2)$ be as \eqref{eq:def:tau2tilde}. We have
\begin{equation}
\mathbb{P} \left({\tau}^+_{4}(1/2) \wedge {\tau}^+_5(1/2) \le \tilde{\tau}(2),\ \tau(2)> t_0^+ \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^3 \right).
\end{equation}
\end{lem}
\begin{proof}
The proof is based on estimating the integrals in the definition of $\tau_{4}^+,\tau_{5}^+$ directly from the bound we have from ${\tau}^+_{9}$. We first observe that
\begin{equation}\label{eq:ind1:squareint:splitinto4}
\begin{split}
\frac{1}{9}\intop_{t_0^+}^{ \tilde{ {\tau}}(2)} (S(t)-\alpha_0)^2 dt
\le
\intop_{ t_0^+}^{ \tilde{ {\tau}}(2)}
\left\{\frac{\alpha_0^{2-2\epsilon}}{(\pi_1(t;S)+1)(\pi_2(t;S)+1)} + \frac{\alpha_0^2\beta_0^{2C_\circ+2}}{\pi_1(t;S)+1}+ \alpha_0^3\beta_0^{12\theta}
\right\} \ dt.
\end{split}
\end{equation}
We control the integral of each term in the RHS separately.
The integral over the constant $\alpha_0^3\beta_0^{12\theta}$ turns out to give the leading order and satisfies
\begin{equation}\label{eq:ind1:tau78:term4}
\intop_{t_0^+}^{\hat{t}_0} \alpha_0^3 \beta_0^{12\theta} dt \le \alpha_0 \beta_0^{23\theta}.
\end{equation}
To control the first term in the RHS of \eqref{eq:ind1:squareint:splitinto4}, we write
\begin{equation}
\Pi_S [t_0^+, \tilde{ {\tau}}(2)] = \{p_1 < p_2 < \ldots < p_N \}.
\end{equation}
Setting $p_0=p_{-1}=t_0^+, p_{N+1}= \tilde{ {\tau}}(2)$, we express
\begin{equation}
\begin{split}
\intop_{{t}_0^+}^{ \tilde{ {\tau}}(2)}
\frac{\alpha_0^{2-2\epsilon}dt}{(\pi_1(t;S)+1)(\pi_2(t;S)+1)}&
\le 2\intop_{{t}_0^+}^{ \tilde{ {\tau}}(2)}
\frac{\alpha_0^{2-2\epsilon}dt}{(\pi_1(t;S)+1)(\pi_2(t;S)+2)}
\\
&\le 2\sum_{i=0}^{N} \int_{0}^{p_{i+1}-p_{i}} \frac{dx}{(x+1)(x+2+p_{i}-p_{i-1})}.
\end{split}
\end{equation}
We can bound the summand at $i=0$ by $1$, and the rest can be written as
\begin{equation}
\begin{split}
\sum_{i=1}^{N}& \int_{0}^{p_{i+1}-p_{i}} \frac{dx}{(x+1)(x+2+p_{i}-p_{i-1})}\\
&=
\sum_{i=1}^{N} \int_{0}^{p_{i+1}-p_{i}}
\frac{dx}{p_{i}-p_{i-1}+1} \left(\frac{1}{x+1}- \frac{1}{x+2+p_i-p_{i-1}} \right)\le \sum_{i=1}^N \frac{\beta_0^2}{p_i-p_{i-1}+1} ,
\end{split}
\end{equation}
where we used $\log(\hat{t}_0-{t}_0^+)\le \beta_0^2$ to obtain the last inequality. The following statement gives a control on the quantity in the RHS, and its proof is given after finishing the proof of Lemma \ref{lem:ind1:tau78}.
\begin{claim}\label{claim:ind1:pointsum:S}
Under the above setting,
\begin{equation}
\mathbb{P} \left( \left.
\sum_{i=1}^N \frac{1}{p_i-p_{i-1}+1} \ge \alpha_0^{-\frac{1}{2}-\epsilon} \ \right|
\ \mathcal{A} \right) \le \exp\left(-\beta_0^{5} \right).
\end{equation}
\end{claim}
Using the claim, we obtain
\begin{equation}\label{eq:ind1:tau78:term1}
\mathbb{P}\left( \left. \intop_{ {t}_0^+}^{ \tilde{ {\tau}}(2)}
\frac{\alpha_0^{2-2\epsilon}dt}{(\pi_1(t;S)+1)(\pi_2(t;S)+1)}
\ge \alpha_0^{\frac{3}{2}-4\epsilon} \ \right| \ \mathcal{A} \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
We move on to the second term in the RHS of \eqref{eq:ind1:squareint:splitinto4}. Using Lemma \ref{lem:ind1:pi1int basicbd:basic},
\begin{equation}\label{eq:ind1:tau78:term3}
\intop_{ {t}_0^+}^{ \tilde{ {\tau}}(2)}\frac{\alpha_0^2\beta_0^{2C_\circ+2}dt}{\pi_1(t;S)+1} \le \alpha_0^2\beta_0^{2C_\circ+4} \left|\Pi_{ S} [ {t}_0^+, \tilde{ \tau}(2)] \right| \le
\alpha_0 \beta_0^{11\theta},
\end{equation}
Note that we bounded $\log(\hat{t}_0-{t}_0^+)\le \beta_0^2.$
Thus, we obtain the conclusion for ${\tau}^+_4(1/2)$ by combining \eqref{eq:ind1:squareint:splitinto4}, \eqref{eq:ind1:tau78:term4}, \eqref{eq:ind1:tau78:term1}, and \eqref{eq:ind1:tau78:term3}, which gives that
\begin{equation}
\mathbb{P} \left(\left. \intop_{ {t}_0^+}^{ \tilde{ {\tau}}(2)} (S(t)-\alpha_0)^2 dt \ge \alpha_0 \beta_0^{24\theta} \ \right| \ \mathcal{A} \right) \le \exp\left(-\beta_0^3 \right).
\end{equation}
Note that the result for ${\tau}^+_5(1/2)$ follows as well, by writing
\begin{equation}
\intop_{{t}_0^+}^{ \tilde{ {\tau}}(2)} (S(t)-\alpha_0)^2 S(t)dt \le
2\alpha_0 \beta_0^{C_\circ} \intop_{{t}_0^+}^{ \tilde{ {\tau}}(2)} (S(t)-\alpha_0)^2 dt ,
\end{equation}
from the definition of $\tau_{1}(\kappa=2)$ (Section \ref{subsubsec:reg:Scontrol}).
\end{proof}
\begin{proof}[Proof of Claim \ref{claim:ind1:pointsum:S}]
Let $\hat{ \alpha}_0:= 2\alpha_0 \beta_0^{C_\circ }$, and let
\begin{equation}
\Pi_{ \hat{ \alpha}_0}[{t}_0^+,\hat{t}_0] = \{p_1'<p_2'<\ldots <p_{N'}' \}.
\end{equation}
Due to the definition of $\tau_{1}(\kappa = 2)$ (Section \ref{subsubsec:reg:Scontrol}), it suffices to establish the main inequality in terms of
\begin{equation}
\sum_{i=1}^N \frac{1}{p_i'-p_{i-1}'+1},
\end{equation}
where we set $p_0'={t}_0^+$ as before.
Our idea is to count the number of neighboring pairs of points with distance less than $\alpha_0^{-1/2}$. To this end, we make the following simple observation:
\begin{itemize}
\item If $p'_{i}-p'_{i-1} \le \alpha_0^{-1/2}$, then there exists $k\in \mathbb{Z}$ such that $p'_i, p'_{i-1}\in {t}_0^+ + \alpha_0^{-1/2}\big[k,k+2\big]$.
\end{itemize}
Note that for each $k\in \mathbb{Z}$,
\begin{equation}
\mathbb{P}\left(\left|\Pi_{ \hat{ \alpha} _0} \left[{t}_0^++k\alpha_0^{-\frac{1}{2}},\, {t}_0^++(k+2)\alpha_0^{-\frac{1}{2}}\right]\right| \ge 2 \right) \le C\alpha_0 \beta_0^{2C_\circ},
\end{equation}
where $C>0$ is an absolute constant. Further, for all even $k$, the above events are independent (and same for odd $k$). Thus, we apply a Chernoff bound for even $k$ and odd $k$ separately, and then use a union bound over the two to deduce that
\begin{equation}\label{eq:ind1:tau78:claim:numberofk}
\begin{split}
\mathbb{P}\bigg( \sharp \bigg\{ k\in \mathbb{Z}, \, k\le \alpha_0^{-3/2}\beta_0^{10\theta} : \Big|\Pi_{ \hat{ \alpha} _0} \big[{t}_0^++k\alpha_0^{-\frac{1}{2}},\, {t}_0^++(k+2)\alpha_0^{-\frac{1}{2}}\big]\Big| \ge 2 \bigg\}\,\ge \alpha_0^{-\frac{1}{2}}\beta_0^{11\theta} \bigg)\\
\le \exp\left(-\beta_0^\theta \right).
\end{split}
\end{equation}
Furthermore, in each interval $[{t}_0^++k'\alpha_0^{-1}, {t}_0^++(k'+1)\alpha_0^{-1}]$, there can be more than $2\beta_0^{C_\circ}$ particles with probability at most $\exp(-\beta_0^{C_\circ} )$. Thus, we take a union bound over $0\le k' \le \alpha_0^{-1}\beta_0^{10\theta}$ and obtain that
\begin{equation}
\mathbb{P} \left(\left|\Pi_{ \hat{ \alpha}_0} [t-\alpha_0^{-1},t] \right| \le 4\beta_0^{C_\circ}, \ \forall t\in [{t}_0^+,\hat{t}_0] \right) \ge 1- \exp\left(-\beta_0^{C_\circ}/2 \right).
\end{equation}
This implies that at each interval ${t}_0^++ [k\alpha_0^{-1/2}, (k+2)\alpha_0^{-1/2}]$ with at least two points, there cannot be more than $4\beta_0^{C_\circ}$ points. Also, we have that there are at most $\alpha_0^{-1} \beta_0^{11\theta}$ points in the entire interval.
Combining the above information, we have
\begin{equation}
\sum_{i=1}^{N'}\frac{1}{p_i'-p_{i-1}'+1} \le \alpha_0^{-\frac{1}{2}}\beta_0^{11\theta} \cdot 4\beta_0^{C_\circ} + \alpha_0^{\frac{1}{2}} \cdot \alpha_0^{-1}\beta_0^{11\theta} \le \alpha_0^{-\frac{1}{2}-\epsilon},
\end{equation}
with probability at least $1- \exp(-\beta_0^5)$.
\end{proof}
\subsubsection{The size of the aggregate}\label{subsubsec:ind1:aggsize}
The goal of this subsection is to study ${\tau}^+_6(1/2), {\tau}^+_{7}(1/2), {\tau}^+_{8}(1/2)$ (Section \ref{subsubsec:reg:aggsize}).
\begin{lem}\label{lem:ind1:tau9 10 11}
Under the setting of Theorem \ref{thm:reg:conti:main}, let $\tilde{\tau}(2)$ be as \eqref{eq:def:tau2tilde}, and set $\acute{ \tau}^+(\kappa)=\min\{{\tau}^+_i(\kappa):i=6,7,8 \}$. Then, we have
\begin{equation}
\mathbb{P} \left(\acute{\tau}^+(1/2) \le \tilde{\tau}(2),\ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
A key quantity in establishing the lemma is the following. For each $\Delta>\alpha_0^{-1}$, we define
\begin{equation}\label{eq:def:ind1:tau13}
{\tau}^+_{10}(\Delta):= \inf \left\{
t\ge t_0^+ : \left|\int_{(t- \Delta)\vee t_0^+}^t |S(s)-\alpha_0|ds \right| \ge \alpha_0^{\frac{3}{2}-7\epsilon} \Delta
\right\}.
\end{equation}
In fact, if we have a good control on ${\tau}^+_{10}$, then we can estimate the number of points in an interval of size $\Delta$ using Corollary \ref{cor:concentration:numberofpts each interval}. Based on Corollary \ref{cor:ind1:tau12}, we begin with showing the following result.
\begin{lem}\label{lem:ind1:tau13}
Under the setting of Theorem \ref{thm:reg:conti:main}, let $\tilde{\tau}(2)$ be as \eqref{eq:def:tau2tilde}. Then, we have for all $\Delta \geq \alpha_0^{-1}$ that
\begin{equation}
\mathbb{P} \left({\tau}^+_{10}(\Delta) \le \tilde{\tau}(2),\ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^3 \right).
\end{equation}
\end{lem}
\begin{proof}
Based on the definition of $\tilde{\tau}(2)$ and $\tau_{9}$ (\eqref{eq:def:tau12}, \eqref{eq:def:tau2tilde}), we have for $t\in[{t}_0^++\Delta,\tilde{ {\tau}}(2)]$ that
\begin{equation}\label{eq:ind1:Sminualpha:splitinto4}
\begin{split}
\intop_{ t-\Delta}^{t} |S(t)-\alpha_0| dt
\le
\intop_{t-\Delta}^{ t}
\left\{\frac{3\alpha_0^{1-\epsilon}}{\pi_1(t;S)+1}+ \frac{\alpha_0\beta_0^{5}}{\sqrt{\pi_1(t;S)+1}}+ 3\alpha_0^{3/2}\beta_0^{6\theta}
\right\} \ dt.
\end{split}
\end{equation}
Using Lemma \ref{lem:ind1:pi1int basicbd:basic}, the RHS is upper bounded by
\begin{equation}\label{eq:ind1:Sminualpha:splitbd}
\begin{split}
\alpha_0^{1-\epsilon} \beta_0^{2} \left|\Pi_{ S}[t-\Delta,t] \right| +
2\alpha_0 \beta_0^5 \sqrt{\Delta\left|\Pi_{ S}[t-\Delta,t] \right| } + 3\Delta \alpha_0^{3/2} \beta_0^{6\theta}.
\end{split}
\end{equation}
Among these terms, we control $\left|\Pi_{ S}[t-\Delta,t] \right|$ from Corollary \ref{cor:concentration:numberofpts each interval}. The definition of $\tau_{1}$ gives that
\begin{equation}
\intop_{ (t-\Delta)\wedge \tilde{ \tau}(2)}^{ t\wedge \tilde{ \tau}(2)} S(x)dx \le 2\Delta \alpha_0 \beta_0^{C_\circ},
\end{equation}
and the RHS is at least $\beta_0^{C_\circ}\ge 1$ since $\Delta\ge \alpha_0^{-1}$. Therefore, Corollary \ref{cor:concentration:numberofpts each interval} tells us that
\begin{equation}\label{eq:ind1:tau13:Sptsbd}
\mathbb{P} \left( \left|\Pi_{ S}[(t-\Delta)\wedge \tilde{ \tau}(2),\,t\wedge\tilde{ \tau}(2)] \right| \le 5\Delta \alpha_0 \beta_0^{C_\circ}, \ \forall t\in[{t}_0^++\Delta, \hat{t}_0] \right) \ge 1- \exp\left(-\beta_0^{C_\circ/3} \right).
\end{equation}
When the above event holds, \eqref{eq:ind1:Sminualpha:splitbd} is upper bounded by
\begin{equation}
\alpha_0^{2-2\epsilon} \Delta + \alpha_0^{1-\epsilon}\sqrt{\Delta} + \alpha_0^{\frac{3}{2}-\epsilon} \Delta \le \alpha_0^{\frac{3}{2}-\epsilon}\Delta,
\end{equation}
where the inequality followed from $\alpha_0^{1/2} \Delta \ge \sqrt{\Delta}$. Thus, combining the error probabilities in the three events, we deduce conclusion.
\end{proof}
We can conduct an analogous but simpler investigation on $|S(t)-\alpha_0|$ and $|S_1(t)-\alpha_0|$, leading to (1) and (2) of Lemma \ref{lem:reg:errorint}. We restate them in the following corollary without spelling out the details of the proof due to similarity.
\begin{cor}\label{cor:ind1:Sint S1int}
Under the setting of Theorem \ref{thm:reg:conti:main}, we have that
\begin{equation}
\begin{split}
\mathbb{P}\left(\left.\intop_{{t}_0^+}^t \big| S(s)-\alpha_0 \big| ds \le \alpha_0^{-\frac{1}{2}-\epsilon}, \ \forall t\in [{t}_0^+,\tau(2)] \, \right| \,\mathcal{A} \right) \le \exp\left(-\beta_0^{C_\circ/3} \right);\\
\mathbb{P}\left(\left.\intop_{{t}_0^+}^t \big| S_1(s)-\alpha_0\big| ds \le \alpha_0^{-\frac{1}{2}-\epsilon}, \ \forall t\in [{t}_0^+,\tau(2)] \, \right| \,\mathcal{A} \right) \le \exp\left(-\beta_0^{C_\circ/3} \right).
\end{split}
\end{equation}
\end{cor}
We conclude the subsection by establishing Lemma \ref{lem:ind1:tau9 10 11}.
\begin{proof}[Proof of Lemma \ref{lem:ind1:tau9 10 11}]
Set $\tilde{ \tau}'(\Delta):= \tilde{ \tau}(2)\wedge {\tau}^+_{10}(\Delta) = { \tau}(2)\wedge {\tau}^+_{9}\wedge {\tau}^+_{10}(\Delta)$.
Having Lemma \ref{lem:ind1:tau13} on hand, Lemma \ref{lem:ind1:tau9 10 11} follows from applying Corollary \ref{cor:concentration:numberofpts each interval} for different values of $\Delta$. Indeed, we can choose $\Delta=\alpha_0^{-1}$ for ${\tau}^+_{7}(1/2)$ and $\Delta=\alpha_0^{-1}\beta_0^{C_\circ}$ for ${\tau}^+_{8}(1/2)$ to see that
\begin{equation}\label{eq:ind1:tau1011bd}
\begin{split}
&\mathbb{P} \left( {\tau}^+_{7}(1/2) \le \tilde{ \tau}'(\alpha_0^{-1}), \ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^3 \right);\\
&\mathbb{P} \left( {\tau}^+_{8}(1/2) \le \tilde{ \tau}'(\alpha_0^{-1}\beta_0^{C_\circ}), \ \tau(2)>\acute{t}_0 \ | \ \mathcal{A} \right) \le \exp\left(-\beta_0^{C_\circ/3} \right).
\end{split}
\end{equation}
The remaining task is to study ${\tau}^+_6(1/2)$. On the event $\mathcal{A}_2 \supset \mathcal{A}$ (Section \ref{subsubsec:reg:history}), we have that
\begin{equation}\label{eq:ind1:tau9primevstau9}
\acute{\tau}^+_9:= \inf \left\{t\ge {t}_0^+: X_t-X_{t_0} \le \frac{\alpha_0}{50}(t-t_0) \right\} \le {\tau}^+_9(1/2).
\end{equation}
Letting $\Delta_0=\alpha_0^{-3/2}$, we have for all $t\in[{t}_0^+,\hat{t}_0]$, $t\le \tilde{\tau}'(\Delta_0)$ that
\begin{equation}
\intop_{t-\Delta_0 }^{t} S(t)dt \in \left[ \frac{\alpha_0\Delta_0}{2}, 2\alpha_0\Delta_0\right].
\end{equation}
Choose $M=2\Delta_0$ to apply Corollary \ref{cor:concentration of integral}, and union bound over all $t\in \mathcal{T}_{\Delta_0}$, where $\mathcal{T}_{ \Delta_0}:= \{t\in({t}_0^+,\hat{t}_0]: t={t}_0^++k\Delta_0,\, k\in\mathbb{Z}$. This gives that
\begin{equation}
\mathbb{P} \left(\left|\Pi_{ S}[t-\Delta_0,\, t] \right| \ge \frac{\alpha_0\Delta_0}{3},\,\forall t\in \mathcal{T}_{\Delta_0} , \, t\le \tilde{\tau}'(\Delta_0) \right) \ge 1- \exp\left(-\alpha_0^{-\epsilon} \right).
\end{equation}
We can also see that if this event holds, then $\acute{\tau}^+_9\ge \tilde{ {\tau}}'.$ Therefore, combining this with \eqref{eq:ind1:tau9primevstau9},
\begin{equation}\label{eq:ind1:tau9bd}
\mathbb{P} \left( \tau_{9}^+(1/2) \le \tilde{ \tau}'(\alpha_0^{-3/2}), \ \tau(2)> \acute{t}_0 \ | \ \mathcal{A} \right) \le\exp\left(-\alpha_0^{-\epsilon} \right)
\end{equation}
Thus, we obtain conclusion from Lemma \ref{lem:ind1:tau13}, equations \eqref{eq:ind1:tau1011bd} and \eqref{eq:ind1:tau9bd}.
\end{proof}
\subsubsection{The magnitude of the speed}\label{subsubsec:ind1:Sbd}
We conclude this section by establishing the estimates on $\tau_{1}$ and $\tau_{2}$, and their corresponding analogue for $\overline{S}(t)$. Further, we conclude the proof Theorem \ref{thm:reg:conti:main} by combining all results obtained throughtout this section. We begin with controlling $\tau_{2}$.
\begin{lem}\label{lem:ind1:tau5}
Under the setting of Theorem \ref{thm:reg:conti:main}, write $\tau_{2}^+(\kappa)=\tau_{2}(\alpha_0,t_0^+,\kappa).$ Conditioned on $\mathcal{A}$ and $\{\tau(2)> \acute{t}_0 \}$, we have $\tau_{2}^+(1/2) > \tau(2)$ almost surely.
\end{lem}
\begin{proof}
Recall the definition of $S_1(t)$ given by \eqref{eq:def:S1:basic form}:
\begin{equation}
S_1(t) = \intop_{ t_0^{-}}^t K_{\alpha_0} (t-x) d\Pi_S(x).
\end{equation}
Based on the definitions of $\tau_{7}$ (Section \ref{subsubsec:reg:aggsize}) and $\mathcal{A}_2$ (Section \ref{subsubsec:reg:history}), we express that
\begin{equation}
\begin{split}
\intop_{t_0^{-}}^{t} K_{\alpha_0}(t-x)d\Pi_S(x) &\leq \sum_{k:\,0\le k \alpha^{-1} \le t-t_0^{-}} \big|\Pi_S[t-(k+1)\alpha^{-1}, t-k\alpha_0^{-1} ]\big| K_{\alpha_0}(k\alpha_0^{-1})\\
&\le
\sum_{k:\, 0\le k \alpha_0^{-1} \le t'-t_0} \frac{C\alpha_0\beta_0^{C_\circ} }{\sqrt{k\alpha_0^{-1}+1} } e^{-c\alpha_0 k} + \alpha_0^{-1} \beta_0^{\theta+10} K_{\alpha_0}(t- t_0)
\leq \frac{1}{2}\alpha_0\beta_0^{C_\circ+1},
\end{split}
\end{equation}
where we used Lemma \ref{lem:estimate on K:intro} to obtain the second inequality.
\end{proof}
\begin{cor}\label{cor:ind1:tau5:base}
Under the setting of Proposition \ref{prop:ind1:base:main}, we have
\begin{equation}
\mathbb{P} \left( \overline{\tau}_2= \overline{\tau}\wedge\hat{t}_0, \ \overline{\tau}>t_0 \right) \le \exp\left(-\beta_0^{3} \right).
\end{equation}
\end{cor}
\begin{proof}
The proof is identical to the previous lemma, except that we are not conditioning on ${\mathcal{A}}_2$. Instead, we use Lemma \ref{lem:reg:base:A} to conclude the proof.
\end{proof}
\begin{lem}\label{lem:ind1:tau4}
Under the setting of Theorem \ref{thm:reg:conti:main}, write $\tau_{1}^+(\kappa)=\tau_{1}(\alpha_0,t_0^+,\kappa).$ Conditioned on $\mathcal{A}$ and $\{\tau(2) >\acute{t}_0 \}$, we have $\tau_{1}(1/2) > \tau(2)$ almost surely.
\end{lem}
We establish Lemma \ref{lem:ind1:tau4} by studying the original formula \eqref{eq:speed:basic expression} of $S(t)$. To this end, we first introduce the following lemma.
\begin{lem}\label{lem:ind1:tau4:aux}
Let $\alpha >0$ sufficiently small and $0<x \le \alpha ^{-1}$. Let $W(s)$ be a continuous time simple random walk. We have
\begin{equation}
\mathbb P (\forall s>0 , \ W(s) < x +\alpha s) \le 3 \alpha x
\end{equation}
\end{lem}
\begin{proof}
Let $\xi _\alpha $ be the unique solution in $(0,1)$ to the equation
\begin{equation}\label{eq:equation for theta}
\left(\frac{\xi +\xi ^{-1}}{2}-1\right)+\alpha \log \xi =0.
\end{equation}
We claim that $\xi _\alpha= 1-2 \alpha +O(\alpha ^2 )$. Indeed, if $\xi =1 -\delta $ then
\begin{equation}
\xi ^{-1}=1+\delta +\delta ^2 +O(\delta ^3 ), \quad \log \xi =-\delta +O(\delta ^2 ).
\end{equation}
Substituting these estimates in \eqref{eq:equation for theta} we get the equation $\frac{\delta ^2 }{2}+O(\delta ^3 )-\alpha \delta +O(\alpha \delta ^2 )=0$ and therefore $\delta =2 \alpha +O(\alpha ^2 )$. We have that
\begin{equation}
\mathbb E \xi ^{\alpha s-W(s)}=\xi ^{\alpha s} \sum _{k=1}^{\infty } \frac{e^{-s}s^k}{k!} \left( \frac{\xi +\xi ^{-1}}{2}\right)^k=\exp \left(\alpha s \log \xi +\left(\frac{\xi +\xi ^{-1}}{2}-1\right)s\right).
\end{equation}
Thus, by the definition of $\xi _{\alpha }$ the process $M_s:=\xi _\alpha ^{\alpha s-W(s)}$ is a martingale. Define the stopping time
\begin{equation}
T_0 := \inf \{s>0 : \ W(s)\ge x +\alpha s \}.
\end{equation}
Almost surely we have $M_{t \wedge T_0 } \to \mathds 1 _{T_0 <\infty }\xi ^{\alpha T_0 -W(T_0) } $. On $\{ T_0 <\infty \}$ we have $-x \le \alpha T_0 -W(T_0) \le -x+1 $ and therfore, by the bounded convergence theorem we have
\begin{equation}
1=\mathbb E M_0=\mathbb E \left[ \mathds 1 \{ T_0 <\infty \} \xi ^{\alpha T_0 -W(T_0) } \right] \le \theta ^{-x} \mathbb P (T_0 <\infty ).
\end{equation}
Thus $\mathbb P (T_0 <\infty ) \ge \xi ^{x}\ge (1-3\alpha)^{x} \ge 1-3 \alpha x$. This finishes the proof of the lemma.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:ind1:tau4}]
Throughout the proof, we condition on $\mathcal{A}$ and $\{\tau(2) > \acute{t}_0 \}$, and let $t\in[{t}_0^+, \tau(2)]$. Because of $\tau_{7} $ and $\mathcal{A}_2$, we have for all $s\in [t_0^{-},\, t]$ that
\begin{equation}\label{eq:ind1:tau4:linearbd on Y}
|\Pi_S[t-s,t]| \le 2\beta_0^4 + 2\alpha_0\beta_0^4 s =: Y'(s).
\end{equation}
From Lemma \ref{lem:ind1:tau4:aux}, we have
\begin{equation}
S(Y') \le 12\alpha_0\beta_0^8,
\end{equation}
where $S(Y')$ is the speed process \eqref{eq:def:speedproc} in terms of $Y'$. Let us define $Y''_t$ as
\begin{equation}
Y''_t(s) := \begin{cases}
|\Pi_S[t-s,t]| & \quad \textnormal{ for } s\le t-t_0^{-};\\
|\Pi_S[t_0^{-},\,t]| + \alpha_0\beta_0^4(s-t+t_0^{-}) & \quad \textnormal{ for } s \ge t-t_0^{-}.
\end{cases}
\end{equation}
Then, an analogous argument as Proposition \ref{prop:SvsSprime} implies that under the event $\mathcal{A}$,
\begin{equation}
|S(t) - S(Y_t'')| \le \alpha_0^{100},
\end{equation}
for any $t\ge t_0$.
Thus, since $S(Y') \ge S(Y_t'')$, we obtain $S(t) \le 12\alpha_0\beta_0^8+\alpha_0^{100} \le \frac{1}{2} \alpha_0\beta_0^{C_\circ}$ which holds deterministically. This tells us that $\tau_{1}^+(1/2) > \tau(2)$ almost surely.
\end{proof}
It turns out that the proof of Lemma \ref{lem:ind1:tau4} applies similarly to the case of $\overline{S}(t)$, and we record the result below.
\begin{cor}\label{cor:ind1:tau4:base}
Under the setting of Proposition \ref{prop:ind1:base:main}, we have
\begin{equation}
\mathbb{P} \left(\overline{\tau}_1=\overline{\tau}\wedge \hat{t}_0 \right) \le \exp\left(-\beta_0^{3}\right).
\end{equation}
\end{cor}
\begin{proof}
The proof of Lemma \ref{lem:ind1:tau4} applies analogously except for the following issues:
\begin{itemize}
\item The statement we have does not condition on $\overline{\mathcal{A}}$, in contrast to Lemma \ref{lem:ind1:tau4};
\item Proof of Lemma \ref{lem:ind1:tau4} deals only with $t\ge {t}_0^+$, whereas we need to cover all $t\ge t_0$ in our case.
\end{itemize}
The first item is resolved using Lemma \ref{lem:reg:base:A}. For the second one, observe that the reason we wanted $t\ge {t}_0^+$ in the previous proof was to deduce \eqref{eq:ind1:tau4:linearbd on Y}: since we did not have control on $|\Pi_S[t-\alpha_0^{-1},t]|$ for $t\le t_0$, we started looking at ${t}_0^+$ to ensure such a linear bound on $Y_t(s)$.
In our case, since $\overline{S}(t)=\alpha_0$ on $t\in[t_0^-,t_0)$, we do not have such an issue. In fact, we can obtain \eqref{eq:ind1:tau4:linearbd on Y} for $\overline{S}(t)$, $t\ge t_0$ from \eqref{eq:def:tauf} and Lemma \ref{lem:branching:ib:R0 vs alpha} along with $\overline{\tau}_9$. One advantage of this is that we also have
\begin{equation}\label{eq:ind1:tau4:base:t0}
\mathbb{P} \left(\overline{\tau}_1 = t_0 \right) \le \exp\left(-\beta_0^5 \right),
\end{equation}
which means that we do not need to include the event $\{\overline{\tau} >t_0 \} $ as before. Thus, having this discussion in mind, following the proof of Lemma \ref{lem:ind1:tau4} concludes the proof.
\end{proof}
We conclude this subsection by giving the proof of Theorem \ref{thm:reg:conti:main}.
\begin{proof}[Proof of Theorem \ref{thm:reg:conti:main}]
The proof follows by combining the lemmas discussed in this section. Namely,
\begin{itemize}
\item Control on ${\tau}^+_3$: Lemma \ref{lem:ind1:tau6}.
\item Control on ${\tau}^+_4(1/2), {\tau}^+_5(1/2)$: Corollary \ref{cor:ind1:tau12} and Lemma \ref{lem:ind1:tau78}.
\item Control on ${\tau}^+_6(1/2),{\tau}^+_{7}(1/2),{\tau}^+_{8}(1/2)$: Lemma \ref{lem:ind1:tau9 10 11} (with Corollary \ref{cor:ind1:tau12}).
\item Control on $\tau_{1}^+(1/2)$ and $\tau_{2}^+(1/2)$: Lemmas \ref{lem:ind1:tau4} and\ref{lem:ind1:tau5}.
\end{itemize}
Combining all the aforementioned results gives the conclusion.
\end{proof}
\subsubsection{Proximity to $\alpha$ of the induction base }\label{subsubsec:ind1:base}
In this section, we explain how to deduce the corresponding analogues of Lemmas \ref{lem:ind1:tau78} and \ref{lem:ind1:tau9 10 11} for $\overline{S}(t)$. The only difference is that the definition of $\overline{\tau}_{\textnormal{b}}$ \eqref{eq:def:taux:init} compared to $\tau_{\textnormal{b}}$ \eqref{eq:def:taux}, and hence the corresponding change needs to be made in $\tau_9^+$ \eqref{eq:def:tau12}.
Recall the definition of $\overline{S}^+$ \eqref{eq:def:taux:init}. Let us define
\begin{equation}
\overline{F} (\alpha_0,t):= 4\alpha_0^{1-\epsilon}\sigma_1\sigma_2(t;S) + \alpha_0\beta_0^{5\theta} \sigma_1(t; \overline{S}^+) + 3\alpha_0^{3/2} \beta_0^{6\theta}.
\end{equation}
Then, the analogue of ${\tau}^+_{9}$ is given by
\begin{equation}
\overline{\tau}_{9}:= \inf \left\{t\ge t_0: |\overline{S}(t)-\alpha_0| \ge \overline{F}(\alpha_0,t) \right\}.
\end{equation}
Furthermore, we set $\overline{\tau}_{10}(\Delta)$ to be the analogue of \eqref{eq:def:ind1:tau13}, which is
\begin{equation}
\overline{\tau}_{10}(\Delta):= \inf\left\{t\ge t_0: \left| \intop_{(t-\Delta)\vee t_0}^{t} \left\{\overline{S}(s)-\alpha_0 \right\}ds \right| \ge \alpha_0^{\frac{3}{2}-7\epsilon} \Delta \right\}.
\end{equation}
Note the important difference from the previous definitions: they read off the process starting from $t_0$, not ${t}_0^+$. Then, we have the following as corresponding counterparts of Corollary \ref{cor:ind1:tau12} and Lemma \ref{lem:ind1:tau13}:
\begin{equation}
\begin{split}
&\mathbb{P} \left(\overline{\tau}_{9} \le \overline{\tau}\wedge \hat{t}_0, \ \overline{\tau}>t_0 \right) \le \exp\left( -\beta_0^3\right);\\
&\mathbb{P} \left( \overline{\tau}_{10}(\Delta) \le \overline{\tau}\wedge \overline{\tau}_{9}\wedge \hat{t}_0, \ \overline{\tau}>t_0 \right) \le 5\exp \left(-\beta_0^3 \right),
\end{split}
\end{equation}
where the second one holds for all $\Delta \ge \alpha_0^{-1}$. In fact, the first one is immediate from the definitions of $\overline{\tau}_3$ and $\overline{\tau}_{\textnormal{b}}$, and the second one follows from the same proof as Lemma \ref{lem:ind1:tau13} relying on $\overline{\tau}_7$ and $\tau_{\textnormal{B}2}$ (Theorem \ref{thm:branching:ib}) when estimating $|\Pi_{\overline{S}^+}[t-\Delta,t]|$. Thus, the following corollary can be obtained analogously as Lemmas \ref{lem:ind1:tau78} and \ref{lem:ind1:tau9 10 11}, whose details are omitted thanks to similarity.
\begin{cor}\label{cor:ind1:tau7891011:base}
Under the setting of Proposition \ref{prop:ind1:base:main}, let $\overline{\tau}' = \min \{ \overline{\tau}_i: 4\le i \le 8 \}$. Then, we have
\begin{equation}
\mathbb{P} \left(\overline{\tau}' \le \overline{\tau}\wedge \hat{t}_0, \ \overline{\tau}>t_0 \right) \le 2\exp\left(-\beta_0^2 \right).
\end{equation}
\end{cor}
\subsubsection{The martingale generated by the gap $S-\alpha_0$}\label{subsubsec:ind1:Sminusalpha conseq}
In this subsection, we conduct further analysis based on ${\tau}^+_{10}(\Delta)$ \eqref{eq:def:ind1:tau13} which will be useful later in Section \ref{sec:double int}. Define the \textit{gap} process
\begin{equation}\label{eq:def:Strianglealpha}
\Pi_{S\triangle\alpha_0}[a,b] := \Pi_S[a,b] \triangle \Pi_{\alpha_0}[a,b],
\end{equation}
and write $d\Pi_{S-\alpha_0}(x):= d\Pi_S(x)-d\Pi_{\alpha_0}(x).$ We begin with stating a direct consequence of Theorem \ref{thm:reg:conti:main} and Lemma \ref{lem:ind1:tau13}
\begin{cor}\label{cor:ind1:gap num of pts}
Let $\alpha_0,t_0,r>0$, set $t_0^-, t_0^+, \hat{t}_0$ as before, and let $\Delta_0:= \alpha_0^{-\frac{3}{2}+7\epsilon}$. Define
\begin{equation}
\tau_{\textnormal{gap}}^{(1)}:= \inf \left\{t\ge t_0^++\Delta_0: \, |\Pi_{S\triangle \alpha_0}|[(t-\Delta_0), t]| \ge\beta_0^5 \right\}.
\end{equation}
If $\Pi_S(-\infty,t_0]\in \mathfrak{R}(\alpha_0,r;[t_0])$, then we have
\begin{equation}
\mathbb{P} \left(\left. \tau_{\textnormal{gap}}^{(1)} <\hat{t}_0 \, \right| \, \mathcal{F}_{t_0} \right) \le 2\exp\left(-\beta_0^2 \right)+r.
\end{equation}
\end{cor}
\begin{proof}
Proof follows directly from applying Corollary \ref{cor:concentration:numberofpts each interval} based on Lemma \ref{lem:ind1:tau13}, and then combining with Theorem \ref{thm:reg:conti:main}.
\end{proof}
The next object of interest is the martingale defined as
\begin{equation}
G(t):= \intop_{t_0^+}^t K^*_{\alpha_0}(t-x) \big\{d{\Pi}_{S-\alpha_0} (x) -(S(x)-\alpha_0 )dx \big\},
\end{equation}
which satisfies the following property.
\begin{cor}\label{cor:ind1:gap mg}
Define the stopping time
\begin{equation}
\tau_{\textnormal{gap}}^{(2)}:= \left\{
t\ge t_0^+: |G(t)| \ge {\alpha_0\beta_0^{C_\circ}}\sigma_1(t;S\triangle \alpha_0) + \alpha_0^{\frac{7}{4}-\epsilon}
\right\}.
\end{equation}
Then, under the setting of Corollary \ref{cor:ind1:gap num of pts}, we have
\begin{equation}
\mathbb{P} \left(\left.\tau_{\textnormal{gap}}^{(2)} <\hat{t}_0 \, \right| \, \mathcal{F}_{t_0} \right) \le \exp\left(-\beta_0^2 \right) +r
\end{equation}
\end{cor}
\begin{proof}
Recall the definition of $\tau(2)$ (Theorem \ref{thm:reg:conti:main}), $F(\alpha_0,t)$ and $\tau_9^+$ \eqref{eq:def:tau12} and write
\begin{equation}
\begin{split}
&\intop_{t_0^+}^{t\wedge \tau(2)\wedge\tau_9^+ } (K^*_{\alpha_0}(t-x))^2 |S(x)-\alpha_0|dx \\
&\le \intop_{t_0^+}^t C\left(\frac{\alpha_0^2}{t-x+1} \vee \alpha_0^4 \right) \left( \alpha_0^{1-\epsilon}\sigma_1(t;S)^2 + \alpha_0\beta_0^{C_\circ+1}\sigma_1(t;S)+\alpha_0^{\frac{3}{2}}\beta_0^{6\theta} \right) dx.
\end{split}
\end{equation}
Each terms in the integral can be estimated using Lemmas \ref{lem:ind1:pi1int basicbd:basic} and \ref{lem:ind1:pi1int basicbd}, with parameters $\Delta_0 = \alpha_0^{-1}, \Delta_1 = \alpha_0^{-1} \beta_0^{C_\circ}, K=\alpha_0^{-1-\epsilon}$ and $N_0 = \beta_0^5.$ This gives that for $t\le \hat{t}_0$,
\begin{equation}
\intop_{t_0^+}^t (K^*_{\alpha_0}(t-x))^2 |S(x)-\alpha_0|dx \le \alpha_0^{\frac{7}{2}-\epsilon},
\end{equation}
and hence we can apply Corollary \ref{lem:concentrationofint:continuity} with parameters
\begin{equation}
A^2=M= \alpha_0^{\frac{7}{2}-\epsilon}, \ \Delta = \alpha_0^{-\frac{3}{2}-7\epsilon}, \ D=1, \ \eta = \alpha_0\beta_0^{C_\circ},\ N= \beta_0^5,\ \delta = \alpha_0^{10}.
\end{equation}
Combining the result with Theorem \ref{thm:reg:conti:main} concludes the proof.
\end{proof}
The same proof applies to establishing the next corollary, whose details are omitted due to similarity.
\begin{cor}\label{cor:ind1:gap mg 2}
Let $\underline{\alpha}_0:= \alpha_0 - \alpha_0^{\frac{3}{2}-\epsilon},$ $d\widetilde{\Pi}_{\triangle \alpha_0} (x) := d\Pi_{\alpha_0}(x)-d\Pi_{\underline{\alpha}_0}(x) - (\alpha_0-\underline{\alpha}_0)dx$, and define the stopping time
\begin{equation}
\tau_{\textnormal{gap}}^{(3)} := \left\{t\ge t_0^+: \left|\intop_{t_0^+}^t K^*_{\alpha_0}(t-x) d\widetilde{\Pi}_{\triangle \alpha_0} (x) \right| \ge \alpha_0\beta_0^{C_\circ}\sigma_1(t; \triangle\alpha_0) + \alpha_0^{\frac{7}{4}-\epsilon} \right\}.
\end{equation} Under the setting of Corollary \ref{cor:ind1:gap num of pts}, we have
\begin{equation}
\mathbb{P} \left(\left. \tau_{\textnormal{gap}}^{(3)} <\hat{t}_0 \,\right|\,\mathcal{F}_{t_0} \right) \le \exp\left(-\beta_0^2 \right)+r.
\end{equation}
\end{cor}
\subsection{The induction base: regularity of the fixed rate process}\label{subsec:ind1:reg of fixed}
We conclude the proof of Proposition \ref{prop:ind1:base:main} and Theorem \ref{thm:induction:base:main}. Since we collected most of the ingredients from the previous subsections, we need the last piece of argument that tells us $\overline{\tau}$ is not likely to be trivial, i.e., $\overline{\tau}>t_0$.
\begin{lem}\label{lem:ind1:tau t0:base}
Under the setting of Proposition \ref{prop:ind1:base:main}, we have
\begin{equation}
\mathbb{P} \left(\overline{\tau}>t_0 \right) \ \ge 1-3\exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
\begin{proof}
From their definitions, it is clear that $\overline{\tau}_i >t_0$ for all $4\le i \le 8$. Furthermore, we saw from \eqref{eq:ind1:tau4:base:t0} in Corollary \ref{cor:ind1:tau4:base} that $\overline{\tau}_1>t_0$ w.h.p.. Thus, what remain to investigate are $ \overline{\tau}_2$ and $\overline{\tau}_3$.
For $\overline{\tau}_2$, recall the definition of $R_0(t)$ from \eqref{eq:def:R0:ib} and note that $R_0(t_0) = \overline{S}(t_0).$ Thus, Lemma \ref{lem:branching:ib:R0 vs alpha} implies the desired estimate on $\overline{\tau}_2$.
To understand $\overline{\tau}_3$, we write
\begin{equation}
\left| \big[ \overline{S}(t_0) - \overline{S}_1(t_0)\big] -\big[ \overline{S}'(t_0;t_0^-,\alpha_0) - \overline{S}_1(t_0)\big] \right| \le \big|\overline{S}(t_0) - \overline{S}'(t_0;t_0^-,\alpha_0)\big| .
\end{equation}
The RHS is $O(\alpha^{50})$ w.h.p. due to Corollary \ref{cor:SvsSprime} and Lemma \ref{lem:reg:base:A}, and the second term of the LHS can be controlled from Proposition \ref{prop:fixed perturbed:error}. Since we are only interested in estimating the above at point $t_0$ (not for entire $t$), it is not difficult to see that the third assumption of \eqref{eq:error:assumptions} can be weakened into
\begin{equation}
\sup_{t_0^- \le t <t_0} \overline{S}(t) \le \alpha_0^{1-\frac{\epsilon}{400}},\quad \overline{S}_1(t_0) \le \alpha_0^{1-\frac{\epsilon}{400}}.
\end{equation}
We also have the first two assumptions since $\overline{S}(t)=\alpha_0$ for $t\in[t_0^-,t_0)$. Combining all the above discussions concludes the proof.
\end{proof}
\begin{proof}[Proof of Proposition \ref{prop:ind1:base:main}]
We can obtain Proposition \ref{prop:ind1:base:main} by linking all the ingredients we observed so far: Building upon Lemmas \ref{lem:reg:base:A} and \ref{lem:ind1:tau t0:base}, the estimates on $\{\overline{\tau}_i\}_{1\le i \le 3} $ (Corollaries \ref{cor:ind1:tau6} \ref{cor:ind1:tau5:base}, \ref{cor:ind1:tau4:base}), and $\{\overline{\tau}_i \}_{4\le i \le 8}$ (Corollary \ref{cor:ind1:tau7891011:base}) deduce the desired result. The proof of the corresponding result for $\underline{S}(t)$ follows analogously.
\end{proof}
We conclude this subsection by verifying Theorem \ref{thm:induction:base:main} based on Proposition \ref{prop:ind1:base:main}.
\begin{proof}[Proof of Theorem \ref{thm:induction:base:main}]
From Markov's inequality, Proposition \ref{prop:ind1:base:main} implies that
\begin{equation}
\mathbb{P}\left[ \mathbb{P} \left( \left.\overline{\tau}\le \acute{t}_0 \, \right| \, \Pi_{\overline{S}}(-\infty,t_0] \right) \ge e^{-\beta_0^{3/2}} \right] \le e^{-\beta_0^{1.8}},
\end{equation}
and it also tells us that
\begin{equation}
\mathbb{P} \left(\Pi_{\overline{S}}(-\infty,t_0] \in \overline{\mathcal{A}}(\alpha_0,t_0) \right) \ge 1- \exp\left(-\beta_0^{1.9} \right).
\end{equation}
This concludes Theorem \ref{thm:induction:base:main} for $\Pi_{\overline{S}}(-\infty,t_0]$, and the result for $\Pi_{\underline{S}}(-\infty,t_0]$ follows analogously.
\end{proof}
\subsection{The growth of a regular aggregate}\label{subsec:ind1:growth}
This section, we verify Proposition \ref{prop:ind1:growth}, which follows as a consequence of Theorem \ref{thm:reg:conti:main} and Lemma \ref{lem:ind1:tau13}.
\begin{proof}[Proof of Proposition \ref{prop:ind1:growth}]
Recall the definition of $\tau(\kappa)$ and ${\tau}^+(\kappa)$ from Theorem \ref{thm:reg:conti:main}, and ${\tau}^+_{10}(\Delta)$ from \eqref{eq:def:ind1:tau13}. In the proof, we let $$\tilde{\tau}:= \tau(2)\wedge {\tau}^+(1/2) \wedge {\tau}^+_{10}(\alpha_0^{-1}).$$
Then, Theorem \ref{thm:reg:conti:main} and Lemma \ref{lem:ind1:tau13} tell us that
\begin{equation}\label{eq:ind1:growth:1}
\mathbb{P} \left( \left. \tilde{\tau} = \hat{t}_0 \, \right| \, \Pi_S(-\infty,t_0] \in \mathfrak{R}(\alpha_0,r;[t_0]) \right) \ge 1-r-\exp\left(-\beta_0^{1.9} \right).
\end{equation}
(Note that $\tau(2)\ge {\tau}^+(1/2)$ if $\tau(2) > \acute{t}_0$, and also that by definition $\tilde{\tau}^+ \le \hat{t}_0$.)
Now, we apply Lemma \ref{lem:concentration of integral} under the following setting: $$f\equiv 1, \ g(t) = S(t),\ \tau = \tilde{\tau}, \ M= \alpha_0^{-1}\beta_0^{12\theta},\ \lambda = M^{-\frac{1}{2}}, \ \textnormal{ and } a = \beta_0^{\theta}.$$
The definition of $\tau_{1}$ justifies the assumption \eqref{eq:concentration:conditions:basic}. Hence, from the lemma we obtain that
\begin{equation}\label{eq:ind1:growth:2}
\mathbb{P} \left( \left. \sup_{\acute{t}_0\le t \le \tilde{\tau}^+} \left| |\Pi_S[t_0,t]| - \intop_{t_0}^t S(s)ds \right| \ge \alpha^{-\frac{1}{2}}\beta_0^{7\theta} \, \right| \, \Pi_S(-\infty,t_0] \right)
\le \exp\left(-\frac{1}{2}\beta_0^\theta \right),
\end{equation}
and this holds for any given $\Pi_S(-\infty,t_0]$.
Lastly, we observe that the definition of ${\tau}^+_{10}(\alpha_0^{-1})$ gives that
\begin{equation}\label{eq:ind1:growth:3}
\sup_{\acute{t}_0 \le t \le {\tau}^+_{10}\wedge \hat{t}_0} \left|\intop_{t_0}^t S(s) ds - \alpha_0(t-t_0) \right| \le \alpha_0^{-\frac{1}{2}-8\epsilon}.
\end{equation}
Thus, we obtain the conclusion by combining \eqref{eq:ind1:growth:1}, \eqref{eq:ind1:growth:2}, and \eqref{eq:ind1:growth:3}.
\end{proof}
\section{The inductive analysis on the speed: Part 2} \label{sec:reg:next step}
In this section, we present the proof of Theorem \ref{thm:induction:main}, completing the inductive argument on regularity. Recall the notations $t_0^-, t_0^+,\acute{t}_0$ and $\hat{t}_0$ (\eqref{eq:def:t0t1} and \eqref{eq:def:t0t11}), and the definitions of $\tau^\sharp$, $\mathcal{A}$, and ${\tau}^+(\kappa)$ given in Section \ref{subsec:regoverview:reg} (\eqref{eq:def:tau:induction}, \eqref{eq:def:A:induction}) and in \eqref{eq:def:ind1:tauacute:basic}. To emphasize the time step and the frame of reference where ${\tau}^+(\kappa)$ is defined, we write ${\tau}^+(\alpha_0, t_0,\kappa) = {\tau}^+(\kappa)$ (Hence, ${\tau}^+(\alpha_0, t_0,\kappa)= \tau(\alpha_0,t_0^+,\kappa)$). We also set $t_1$ to be any number that satisfies \eqref{eq:t1 regime}, and let
\begin{equation}
t_1^- := t_1 - \alpha_0^{-2}\beta_0^\theta, \quad t_1^+:= t_1+\alpha_0^{-2} \beta_0^{\theta}, \quad t_1^{\sharp}:= t_1+4\alpha_0^{-2} \beta_0^\theta.
\end{equation}
Further, let $\alpha_1$ be given as \eqref{eq:def:alpha1} and let $\beta_1:= \log(1/\alpha_1)$.
\begin{thm}\label{thm:ind2:main}
Let $\alpha_0,t_0>0$, let $t_0^-, t_0^+$ be as above, and let
$t_1$ be any number satisfying \eqref{eq:t1 regime}. Under the above notations, let $\tau':= \tau(\alpha_0,t_0,2)\wedge {\tau}^+(\alpha_0, t_0,1/2)$.
Then, for all sufficiently small $\alpha_0$, the following hold true:
\begin{enumerate}
\item $\mathbb{P} (\tau^\sharp(\alpha_1,t_1,2) \le t_1^\sharp \ | \ \mathcal{A}(\alpha_0,t_0) )
\le
\mathbb{P} (\tau' < \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) )
+ 4\exp\left(-\beta_0^{2} \right);$
\item $\mathbb{P} (\mathcal{A}(\alpha_1,t_1)^c \ | \ \mathcal{A}(\alpha_0,t_0) )
\le
\mathbb{P} (\tau' < \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) )
+ \exp\left(-\beta_0^{2} \right).$
\end{enumerate}
\end{thm}
Our program is to follow the description given in Section \ref{subsubsec:regoverview:overview:nextstep}.
\begin{itemize}
\item In Section \ref{subsec:ind2:ratechange}, we study $|\alpha_1-\alpha_0|$, the change of the frame of reference. Moreover, we establish that $\mathcal{A}(\alpha_1,t_1)$ holds w.h.p. at time $t_1$, proving Theorem \ref{thm:ind2:main}-(2). As a byproduct of our analysis, we obtain control on $\tau_1^\sharp$ and establish Proposition \ref{prop:reg:newrates} as well.
\item In Section \ref{subsec:ind2:remaining reg}, we study all the stopping times except $\tau_1^\sharp$ and $\tau_{3}^\sharp$.
\item In Section \ref{subsec:ind2:bootstrappedJ}, we establish control on $\tau_{3}^\sharp $.
\item In Section \ref{subsec:ind2:fin}, we combine all the analysis done in Sections \ref{subsec:ind2:ratechange}--\ref{subsec:ind2:bootstrappedJ} and complete the proofs of Theorems \ref{thm:ind2:main} and \ref{thm:induction:main}.
\end{itemize}
Before moving on to the proofs, we clarify the definitions of $\alpha_0',\alpha_1,\alpha_1'$: given $\alpha_0,t_0$, we have
\begin{equation}\label{eq:def:alpha:ind2}
\begin{split}
\alpha_0' &= \mathcal{L}(t_0; \Pi_S[t_0^{-},t_0],\alpha_0)=\intop_{t_0^{-}}^{t_0} \intop_{ t_0}^{\infty} K^*_{\alpha_0} \cdot K_{\alpha_0}(u-x) du d\Pi_S(x) ;\\
\alpha_1&=\mathcal{L}(t_1^-;\Pi_S[t_0^{-},t_1^-],\alpha_0)
=
\intop_{t_0^{-}}^{t_1} \intop_{ t_1}^{\infty} K^*_{\alpha_0} \cdot K_{\alpha_0}(u-x) du d\Pi_S(x) ;\\
\alpha_1'&=\mathcal{L}(t_1;\Pi_S[t_1^{-},t_1],\alpha_1)
=
\intop_{t_1^{-}}^{t_1} \intop_{ t_1}^{\infty} K^*_{\alpha_1} \cdot K_{\alpha_1}(u-x) du d\Pi_S(x) ,
\end{split}
\end{equation}
where $K^*_\alpha = \frac{2\alpha^2}{1+2\alpha}$.
\subsection{Change of rates for critical branching}\label{subsec:ind2:ratechange}
Our goal is to establish that $\mathcal{A}(\alpha_1,t_1)$ \eqref{eq:def:A:induction} happens with high probability, conditioned on the regularity in the previous step. As a consequence of our analysis, we prove Proposition \ref{prop:reg:newrates}.
We begin with studying $\mathcal{A}_3(\alpha_1,t_1)$. The other events $\mathcal{A}_1$ and $\mathcal{A}_2$ are investigated in Sections \ref{subsubsec:ind2:A1}, \ref{subsubsec:ind2:A3} respectively.
\begin{lem}\label{lem:ind2:ratechange}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P}\left(\mathcal{A}_3(\alpha_1,t_1)^c,\ \tau'\ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 10\exp\left(-\beta_0^5 \right).
\end{equation}
\end{lem}
Since the base rates in the definition of $\alpha_1$ and $\alpha_1'$ are $\alpha_0$ and $\alpha_1$, respectively, we may expect that understanding the size of $|\alpha_1-\alpha_0|$ is important, which is also what we want in Theorem \ref{thm:induction:main}-(2). We define the following $\mathcal{F}_{t_1}$-measurable event
\begin{equation}\label{eq:def:A4}
\mathcal{A}_4=\mathcal{A}_4(\alpha_0,t_0):= \left\{|\alpha_1-\alpha_0|\le 2\alpha_0^{3/2}\beta_0^{6\theta} \right\},
\end{equation}
and begin with showing that $\mathcal{A}_4$ happens with high probability.
\begin{lem}\label{lem:ind2:alpha1vsalpha0}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P}\left(\mathcal{A}_4^c ,\ \tau' \ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 2\exp\left(-\beta_0^5 \right).
\end{equation}
\end{lem}
\begin{proof}
In the proof, set $\mathcal{A}=\mathcal{A}(\alpha_0,t_0)$ for convenience.
Since the event $\mathcal{A}_3(\alpha_0,t_0)\supset \mathcal{A}$ is given, we can focus on deriving $|\alpha_1-\alpha_0'|\le \alpha_0^{3/2}\beta_0^{6\theta}$. Define $S_1(t)= S_1(t;t_0^-,\alpha_0)$ as before. From equation \eqref{eq:integralform:branching}, we can write
\begin{equation}\label{eq:ind2:a1vsa0:split}
\begin{split}
\alpha_1-\alpha_0'
&= \mathcal{L}( t_1^-; \Pi_S[t_0^{-},t_1^-],\alpha_0)
-\mathcal{L}(t_0; \Pi_S[t_0^{-},t_0],\alpha_0)\\
&=
\intop_{t_0}^{t_1^-} K^*_{\alpha_0} d\widetilde{ \Pi}_S(x) + \intop_{t_0}^{t_1^-} K^*_{\alpha_0} (S(x)-S_1(x))dx ,
\end{split}
\end{equation}
where we wrote $d\widetilde{\Pi}_S(x)= d\Pi_S(x)-S(x)dx$.
To study the first term, we note that
\begin{equation}\label{eq:ind2:rate:a1vsa0:qv}
\intop_{t_0}^{t_1^-\wedge \tau'}( K^*_{\alpha_0})^2 S(x)dx \le \alpha_0^3 \beta_0^{10\theta+2C_\circ},
\end{equation}
due to the definition of $\tau_{1}$ (Section \ref{subsubsec:reg:Scontrol}). Since $K^*_{\alpha_0} \le C\alpha_0^2 \le \alpha_0^{3/2}$, we apply Corollary \ref{cor:concentration of integral} to deduce that
\begin{equation}\label{eq:ind2:rate:a1vsa0:mg}
\mathbb{P}\left(\left.\left| \intop_{t_0}^{t_1^-\wedge \tau'} K^*_{\alpha_0} d\widetilde{\Pi}_S(x)\right| \ge \alpha_0^{3/2}\beta_0^{5\theta+2C_\circ} \ \right| \ \mathcal{A} \right) \le \exp\left(-\beta_0^{C_\circ/2} \right).
\end{equation}
The second term of \eqref{eq:ind2:a1vsa0:split} follows directly from Lemma \ref{lem:ind1:intofSminS1} (note that $\tau'\le \tau(2)$ for $\tau(2)$ in Lemma \ref{lem:ind1:intofSminS1}):
\begin{equation}\label{eq:ind2:rate:a1vsa0:drift}
\mathbb{P} \left( \left. \left|\intop_{ t_0}^{t_1^-\wedge \tau'} K^*_{\alpha_0}(S(x)-S_1(x))dt \right| \ge \alpha_0^{2-3\epsilon} \ \right| \ \mathcal{A} \right) \le \exp\left(-\beta_0^{C_\circ/3} \right).
\end{equation}
\noindent Combining \eqref{eq:ind2:a1vsa0:split}, \eqref{eq:ind2:rate:a1vsa0:mg}, \eqref{eq:ind2:rate:a1vsa0:drift} and the condition from $\mathcal{A}_3(\alpha_0,t_0)$, we obtain conclusion.
\end{proof}
Before moving on, we establish the following lemma, which gives a desired control on the stopping time $\tau_1^\sharp$ (Section \ref{subsubsec:reg:refined1storder}.
\begin{lem}\label{lem:ind2:tau1sharp}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P} \left( \tau_1^\sharp(\alpha_1,t_1) \le t_1^\sharp,\ \tau'\ge \hat{t}_0 \,| \, \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^4 \right).
\end{equation}
\end{lem}
\begin{proof}
This comes as a direct consequence of Lemmas \ref{lem:ind1:tau13} and \ref{lem:ind2:alpha1vsalpha0}: For $t' = t_1^\sharp\wedge \tau_{10}^+(\alpha_0^{-1})\wedge \tau'$ and on the event $\mathcal{A}_4$, note that
\begin{equation}
\intop_{t_1}^{t'} |S(s)-\alpha_1| ds \le \intop_{t_1}^{t'} |S(s)-\alpha_0|ds + \intop_{t_1}^{t'} |\alpha_0-\alpha_1|ds \le \alpha_0^{-\frac{1}{2}-\epsilon}.
\end{equation}
\end{proof}
To establish Lemma \ref{lem:ind2:ratechange}, we set
\begin{equation}\label{eq:def:t1minmin}
t_1^{--}:= t_1^- - \alpha_0^{-2}\beta_0^\theta = t_1-2\alpha_0^{-2}\beta_0^\theta,
\end{equation}
and introduce auxiliary parameters $\alpha_1^{(1)}, \alpha_1^{(2)}$ and $\alpha_1^{(3)}$ defined as
\begin{equation}\label{eq:def:alphaprimeprime}
\begin{split}
\alpha_1^{(1)}& := \mathcal{L}(t_1;\Pi_S[t_1^{-},t_1],\alpha_0) = \intop_{ t_1^{-}}^{t_1} \intop_{t_1}^\infty K^*_{\alpha_0 } \cdot K_{\alpha_0}(u-x) du d\Pi_S(x);\\
\alpha_1^{(2)}& := \mathcal{L}(t_1;\Pi_S[t_1^{--},t_1],\alpha_0) = \intop_{ t_1^{--}}^{t_1} \intop_{t_1}^\infty K^*_{\alpha_0 } \cdot K_{\alpha_0}(u-x) du d\Pi_S(x);\\
\alpha_1^{(3)} &:= \mathcal{L}(t_1^-;\Pi_S[t_1^{--},t_1^-],\alpha_0) = \intop_{ t_1^{--}}^{t_1^-} \intop_{t_1^-}^\infty K^*_{\alpha_0 } \cdot K_{\alpha_0}(u-x) du d\Pi_S(x).
\end{split}
\end{equation}
From these notations, we write
\begin{equation}\label{eq:ind2:rate:a1vsa1prime:decomp}
|\alpha_1-\alpha_1'| \le |\alpha_1-\alpha_1^{(3)}|+|\alpha_1^{(2)}-\alpha_1^{(3)}| +|\alpha_1^{(1)}-\alpha^{(2)}|+|\alpha_1' -\alpha_1^{(1)}|.
\end{equation}
Recalling that $\alpha_1:= \mathcal{L}(t_1^-; \Pi_S[t_0^-,t_1^-], \alpha_0)$, we can see that $|\alpha_1^{(2)}-\alpha_1^{(3)}|$ can be investigated by the same method as Lemma \ref{lem:ind2:alpha1vsalpha0}, leading us to the following corollary.
\begin{cor}\label{cor:ind2:alpha1vsalphaaux3}
Under the setting of Theorem \ref{thm:ind2:main} and the above notations, we have
\begin{equation}
\mathbb{P} \left(\left. \left|\alpha_1^{(2)} -\alpha_1^{(3)} \right| \ge \frac{1}{2}\alpha_0^{\frac{3}{2}}\beta_0^{\theta}, \ \tau' \ge \hat{t}_0 \ \right|\, \mathcal{A}(\alpha_0,t_0) \right) \le 2\exp\left(-\beta_0^5\right).
\end{equation}
\end{cor}
\begin{proof}
From the proof of Lemma \ref{lem:ind2:alpha1vsalpha0}, our task is to control
\begin{equation}
\intop_{t_1^{-}}^{t_1} K^*_{\alpha_0} \big\{d\widetilde{\Pi}_S(x) + (S(x)-S_1(x;t_1^{--},\alpha_0))dx \big\}.
\end{equation}
The rest of the proof is analogous to Lemma \ref{lem:ind2:alpha1vsalpha0} except the following two things:
\begin{itemize}
\item Since the regime of integration $[t_1^-,t_1]$ is much shorter than $[t_0^-,t_1^-]$ from the previous lemma, the RHS \eqref{eq:ind2:rate:a1vsa0:qv} can be improved to $\alpha_0^3\beta_0^{\theta+2C_\circ}$, resulting in a corresponding enhancement in \eqref{eq:ind2:rate:a1vsa0:mg}.
\item For any $t_1^-\le t \le t_1\wedge \tau'$, we have
\begin{equation}
|S_1(t;t_0^-,\alpha_0)-S_1(t;t_1^{--},\alpha_0)| = \intop_{t_0^{-}}^{t_1^-} K_{\alpha_0}(t-x) d\Pi_S(x) \le \alpha_0^{100},
\end{equation}
due to the decay property of $K_{\alpha_0}(s)$. Thus, we can obtain \eqref{eq:ind2:rate:a1vsa0:drift} for our case as well.
\end{itemize}
\end{proof}
Next, we observe that the first and the third terms in the RHS are negligible, which comes as a simple consequence of regularity.
\begin{lem}\label{lem:ind2:alphavsalphaPP}
Under the setting of Theorem \ref{thm:ind2:main} and \eqref{eq:def:alphaprimeprime}, we have
\begin{equation}
\begin{split}
\mathbb{P}\left(|\alpha_1-\alpha_1^{(3)}|\ge \alpha_0^{100} ,\ \tau' \ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 2\exp\left(-\beta_0^5 \right);\\
\mathbb{P}\left(|\alpha_1^{(1)}-\alpha_1^{(2)}|\ge \alpha_0^{100} ,\ \tau' \ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 2\exp\left(-\beta_0^5 \right).
\end{split}
\end{equation}
\end{lem}
\begin{proof}
We start with investigating the first inequality. We observe that
\begin{equation}\label{eq:ind2:rate:a1vsa1PP}
\alpha_1-\alpha_1^{(3)}
= \intop_{ t_0^{-}}^{t_1^{--}} \intop_{t_1^{-}}^\infty K^*_{\alpha_0 } \cdot K_{\alpha_0}(u-x) du d\Pi_S(x)
\le Ce^{-c\beta_0^{\theta}} \left|\Pi_S [t_0^{-},t_1^{--}] \right|,
\end{equation}
where the last inequality came from the estimate Lemma \ref{lem:estimate on K:intro}, noting that $t_1^--t_1^{--} = \alpha_0^{-2}\beta_0^\theta$.
Moreover, from the definition of ${\tau}^+_{10}(\alpha_0, t_0,1/2)$ and $\mathcal{A}_2(\alpha_0,t_0)$ (Sections \ref{subsubsec:reg:aggsize}, \ref{subsubsec:reg:history}), we have $|\Pi_S[t_0^{-},\tau']|\le \alpha_0^{-1}\beta_0^{11\theta}$. Plugging this estimate to \eqref{eq:ind2:rate:a1vsa1PP}, we obtain the first estimate of Lemma \ref{lem:ind2:alphavsalphaPP}.
For the second one, observe that
\begin{equation}
\begin{split}
\alpha_1^{(2)}-\alpha_1^{(1)} = \intop_{t_1^{--}}^{t_1^-} \intop_{t_1}^\infty K^*_{\alpha_0}\cdot K_{\alpha_0}(u-x) du d\Pi_S(x)
\le
Ce^{-c\beta_0^\theta} |\Pi_S[t_1^{--},t_1^-]|,
\end{split}
\end{equation}
analogously as \eqref{eq:ind2:rate:a1vsa1PP}. Thus, the same reasoning as above gives the conclusion.
\end{proof}
In \eqref{eq:ind2:rate:a1vsa1prime:decomp}, what remains to understand is $|\alpha_1'-\alpha_1^{(1)}|$. The two terms differ only in the parameter $\alpha_1$ and $\alpha_0$ in the integrand, and hence we can expect to study the difference $|\alpha_1'-\alpha_1^{(1)}|$ based on analytic properties of $K^*_\alpha$ and $K_\alpha$.
Building upon such an intuition, we prove the following result.
\begin{lem}\label{lem:ind2:alphaPvsalphaPP}
Under the setting of Theorem \ref{thm:ind2:main} and \eqref{eq:def:alphaprimeprime}, we have
\begin{equation}
\mathbb{P}\left(|\alpha_1'-\alpha_1^{(1)} |\ge \alpha_0^2 \beta_0^{8\theta+1},\ \tau'\ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
\end{lem}
Proof of Lemma \ref{lem:ind2:alphaPvsalphaPP} is given in the following subsection. Before delving into the proof, we establish Lemma \ref{lem:ind2:ratechange}.
\begin{proof}[Proof of Lemma \ref{lem:ind2:ratechange}]
Lemma \ref{lem:ind2:ratechange} follows directly from a union bound over the four probabilities in Corollary \ref{cor:ind2:alpha1vsalphaaux3}, Lemmas \ref{lem:ind2:alphavsalphaPP} and \ref{lem:ind2:alphaPvsalphaPP}. Note that by the four estimates along with Lemma \ref{lem:ind2:alpha1vsalpha0}, we get
$$|\alpha_1-\alpha_1'|\le \frac{2}{3}\alpha_0^{\frac{3}{2}}\beta_0^\theta \le \alpha_1^{\frac{3}{2}}\beta_1^\theta, $$
with high probability, which is much stronger than $\mathcal{A}_3(\alpha_1,t_1)$ in terms of the exponent of $\beta_1$.
\end{proof}
Before moving on, we also conclude the proof of Proposition \ref{prop:reg:newrates}.
\begin{proof}[Proof of Proposition \ref{prop:reg:newrates}]
Recall the definition of $\tilde{\alpha_1}$, and we write
$$|\alpha_1'-\tilde{\alpha}_1|\le |\alpha_1'-\alpha_1^{(1)}| + |\tilde{\alpha}_1- \alpha_1^{(1)}|, $$
where $\alpha_1^{(1)}$ is given as \eqref{eq:def:alphaprimeprime}.
Note that we can write
\begin{equation}
\tilde{\alpha}_1-\alpha_1^{(1)} = \intop_{t_0^-}^{t_1^-} \intop_{t_1}^\infty K^*_{\alpha_0} \cdot K_{\alpha_0}(u-x)du d\Pi_S(x),
\end{equation}
which can be bounded in the same way as Lemma \ref{lem:ind2:alphavsalphaPP}. Thus, combining Theorem \ref{thm:reg:conti:main}, Lemmas \ref{lem:ind2:alphavsalphaPP} and \ref{lem:ind2:alphaPvsalphaPP} concludes the proof.
\end{proof}
\subsubsection{Proof of Lemma \ref{lem:ind2:alphaPvsalphaPP}: stability under change of rate}
Let $Y_\alpha (s)$ be a rate $\alpha $ Poisson process and let $W(s)$ be a continuous time random walk. Define the stopping time $T _\alpha : =\inf \{ t>0: \ W(s)>Y_\alpha (s) \}$ and the function $h_\alpha(t):=\mathbb P (t \le T _\alpha <\infty )$. Recall that $K_\alpha (t)=\alpha (1+2 \alpha ) h_\alpha(t)$ (Definition \ref{def:Kalpha:main}). In the following claim we bound the derivative of $h_\alpha (t)$ with respect to $\alpha $.
\begin{claim}\label{claim:derivative of h}
We have that
\begin{equation}
\left| \frac{d}{d \alpha } h_\alpha (t) \right| \le Ce^{-c \alpha ^2 t}
\end{equation}
\end{claim}
\begin{proof}
Throughout the proof we write $h_\alpha '(x)$ for the derivative of $h$ with respect to $x$ and write explicitly $\frac{d}{d \alpha }h_\alpha (x)$ when we differentiate with respect to $\alpha $. Let $\delta \le \alpha ^{10}$ and let $Y_\alpha ,Y_{\delta }$ be independent Poisson processes with rates $\alpha $ and $\delta $ respectively. Let $Y_{\alpha +\delta }:=Y_\alpha +Y _{\delta }$ and note that $T _\alpha \le T _{\alpha +\delta } $. Recall that $\mathbb P (T _\alpha <\infty ) =h_\alpha (0)=1/(1+2 \alpha )$. Integrating over the values $x$ that $T _\alpha $ can take we obtain
\begin{equation}
\begin{split}
&\left| h_{\alpha +\delta }(t)-h_{\alpha }(t) \right| \le \mathbb P (T _\alpha <t,\ t \le T _{\alpha +\delta } <\infty ) + \mathbb P (t \le T _\alpha <\infty , \ T _{\alpha +\delta } =\infty ) \\
& \le - \intop _0^t h_\alpha '(x) \mathbb P (Y_{\delta }(x)=1)h_{\alpha +\delta }(t-x) - \frac{2(\alpha +\delta ) }{1+2(\alpha+\delta ) }\intop _t^{\infty }h_\alpha '(x) \mathbb P (Y_{\delta }(x)=1)
- \intop _0^{\infty } h_\alpha '(x) \mathbb P (Y_\delta (x)\ge 2 ) \\
&\le C \intop _0^t x^{-\frac{3}{2}}e^{-c \alpha ^2 x} \cdot \delta x \cdot (t-x)^{-\frac{1}{2}} e^{-c \alpha ^2(t-x)}dx+ C\alpha \intop _t^{\infty } x^{-\frac{3}{2}} e^{-c \alpha ^2 x} \cdot \delta x dx +C\intop _0^\infty x^{-\frac{3}{2}} e^{-c\alpha ^2 x} \delta ^2 x^2 \\
&\le C\delta e^{-c \alpha ^2 t } \intop _0^t x^{-\frac{1}{2}} (t-x)^{-\frac{1}{2}} + C \delta \alpha e^{-c \alpha ^2 t } \intop _0^{\infty } x^{-\frac{1}{2}} e^{-c \alpha ^2 x} + C \alpha ^{-3 } \delta ^2 \le C \delta e^{-c \alpha ^2 t } +C_\alpha \delta ^2 ,
\end{split}
\end{equation}
where in the third inequality we used Corollary~\ref{cor:bound on K'} in order to bound $h_\alpha '$. The claim follows from the last bound.
\end{proof}
We claim that we can approximate the speed by a fixed value in the time interval $[t_1^-,t_1]$.
\begin{claim}\label{claim:speed is almost fixed}
Under the setting of Theorem \ref{thm:induction:main}, there exists a random variable $\alpha _2 \in \mathcal F _{t_1^-}$ such that
\begin{equation}
\mathbb{P} \left(\left. \intop _{t_1^{-}} ^{t_1} |S(t)-\alpha _2 | dt\ge \alpha_0 ^{-\frac{1}{2}} \beta_0^{2 \theta }, \, \tau' \ge \hat{t}_0 \, \right| \, \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
\end{claim}
\begin{proof}
Recalling the definition of $\alpha_1^{(3)}$ from \eqref{eq:def:alphaprimeprime}, we show that the choice $\alpha_2=\alpha_1^{(3)}$ is enough for our purpose. To this end, we split the integrand into
$$ |S(t)-\alpha_1| \le |S(t)-S_1(t;t_1^{--},\alpha_0)| + |S_1(t;t_1^{--},\alpha_0) - \alpha_1^{(3)}|, $$
where $t_1^{--}$ is given as \eqref{eq:def:t1minmin}.
Note that if $t_1^-\le t\le t_1\wedge \tau'$, then $|S_1(t;t_0^-,\alpha_0)-S_1(t;t_1^{--},\alpha_0)| \le \alpha_0^{100} $, by the same argument as \eqref{eq:ind1:S1timechange}. Thus, the definition of $\tau_3$ (Section \ref{subsubsec:reg:Scontrol}) gives
\begin{equation}\label{eq:ind2:SminS1timechange}
|S(t)-S_1(t;t_1^{--},\alpha_0)| \le 2\alpha_0^{1-\epsilon} \sigma_1\sigma_2(t;S).
\end{equation}
On the other hand, the same expression as \eqref{eq:integralform:branching:ind1} gives
\begin{equation}
S_1(t;t_1^{--},\alpha_0) - \alpha_1^{(3)} = \intop_{t_1^-}^{t} K_{\alpha_0}^*(t-x) \big\{ d\Pi_S(x)-S_1(x;t_1^{--},\alpha_0) dx \big\}.
\end{equation}
Using the bound \eqref{eq:ind2:SminS1timechange}, we can control this term analogously as the proof of Lemma \ref{lem:ind1:taux}, which gives that
\begin{equation}\label{eq:ind2:S1minusalphapp}
\begin{split}
\mathbb{P} \left( \left. \bigcap_{t\in[t_1^-,t_1]} \left\{ \big|S_1(t;t_1^{--},\alpha_0)-\alpha_1^{(3)} \big| \le \alpha_0\beta_0^{C_\circ}\sigma_1(t;S) + \alpha_0^{\frac{3}{2}}\beta_0^{\theta} \right\}, \ \tau'\ge\hat{t}_0 \ \right| \ \mathcal{A}(\alpha_0,t_0) \right)\\
\ge 1-\exp\left(-\beta_0^5 \right).
\end{split}
\end{equation}
Note that the exponent $\theta$ in the term $\alpha_0^{\frac{3}{2}}\beta_0^\theta$ is smaller than that of Lemma \ref{lem:ind1:taux} since the length of the interval we are looking at is of scale $t_1-t_1^-=\alpha_0^{-2}\beta_0^\theta$, not $\alpha_0^{-2}\beta_0^{10\theta}$.
When we have both \eqref{eq:ind2:SminS1timechange} and \eqref{eq:ind2:S1minusalphapp}, we can control the integral of $|S(t)-\alpha_1^{(3)}|$ by Lemma \ref{lem:ind1:pi1int basicbd:basic}. This tells us that
\begin{equation}
\intop_{t_1^-}^{t_1} |S(t)-\alpha_1^{(3)}|dt \le \intop_{t_1^-}^{t_1} \left(\frac{\alpha_0^{1-\epsilon}}{\pi_1(t;S)+1} + \frac{\alpha_0\beta_0^{C_\circ}}{\sqrt{\pi_1(t;S)+1}} + \alpha_0^{\frac{3}{2}}\beta_0^{\theta} \right)dt \le 2\alpha_0^{-\frac{1}{2}}\beta_0^{2\theta},
\end{equation}
concluding the proof.
\end{proof}
We are now ready to prove Lemma~\ref{lem:ind2:alphaPvsalphaPP}
\begin{proof}[Proof of Lemma~\ref{lem:ind2:alphaPvsalphaPP}]
Let $\alpha _2$ be the random variable from Claim~\ref{claim:speed is almost fixed}. Recall that
\begin{equation}
\alpha _1^{(1)} =\frac{2\alpha_0 ^2 }{1+2 \alpha_0 }\intop _{t_1^-}^{t_1} \intop _{t_1}^\infty K_{\alpha _0 }(y-x) dy \, d \Pi _S (x) =\frac{2\alpha_0 ^2 }{1+2 \alpha _0 }\intop _{t_1^-}^{t_1} \intop _{t_1-x}^\infty K_{\alpha _0 }(z) dz \, d \Pi _S (x)
\end{equation}
Next, we break the measure $d \Pi _S$ into $(d \Pi _S(x)- \alpha _2 dx)+\alpha _2 dx$ and start by estimating the integral with respect to the Lebesgue measure. We have
\begin{equation}
\begin{split}
\frac{2\alpha_0 ^2 \alpha _2 }{1+2 \alpha_0 } \intop _{t_1^-}^{t_1}\intop _{t_1-x}^\infty K_{\alpha _0}(z) dz \, dx &=\frac{2\alpha_0 ^2 \alpha _2 }{1+2 \alpha_0 } \intop _{0}^{t_1-t_1^-} \intop _{t_1-z}^{t_1} K_{\alpha _0}(z)dx\, dz+\frac{2\alpha_0 ^2 \alpha _2 }{1+2 \alpha_0 }\intop _{t_1-t_1^-}^{\infty } \intop _{t_1^-}^{t_1} K_{\alpha _0}(z) dx \, dz\\
=\frac{2\alpha_0 ^2 \alpha _2 }{1+2 \alpha_0 } &\intop _{0}^{t_1-t_1^-} z K_{\alpha _0} (z) dz+\frac{(t_1-t_1^-)2 \alpha _0 ^2 \alpha _2 }{1 +2 \alpha _0 } \intop _{t_1-t_1^-}^{\infty } K_{\alpha _0} (z)dz= \alpha _2 +O(\alpha _0^{10}),
\end{split}
\end{equation}
where in the last equality we used Corollary~\ref{cor:moments of K} and Lemma~\ref{lem:estimate on K:intro} and the fact that $t_1-t_1^- \ge \alpha _0^{-2 }\beta_0 ^2 $. Thus,
\begin{equation}
\alpha _1 '=O(\alpha_0 ^{10})+\alpha _2+\frac{2\alpha _0 ^2 }{1+2 \alpha_0 }\intop _{t_1^-}^{t_1} \intop _{t_1-x}^\infty K_{\alpha }(z) dz \, (d \Pi _S (x)-\alpha _2dx).
\end{equation}
Using the same arguments for $\alpha _1'$ we get
\begin{equation}\label{eq:integral of F}
\alpha _1'- \alpha _1^{(1)}=O(\alpha _0 ^{10})+\intop _{t_1^{-}}^{t_1} F(x) (d \Pi _S (x)-\alpha _2dx),
\end{equation}
where
\begin{equation}
F(x):=\intop _{t_1-x}^\infty \frac{2\alpha_1 ^2 }{1+2 \alpha_1 }K_{\alpha _1}(z) - \frac{2\alpha _0 ^2 }{1+2 \alpha _0} K_{\alpha _0 }(z) dz=2\intop _{t_1-x}^\infty \alpha_1 ^3 h_{\alpha _1}(z) - \alpha _0^3 h_{\alpha _0 }(z) dz.
\end{equation}
We turn to bound the function $F$. We have
\begin{equation}
\left| \frac{d}{d\alpha }(\alpha ^3 h_\alpha (z) ) \right| \le 3 \alpha ^2 h_\alpha (z)+\alpha ^3 \left| \frac{d}{d\alpha} h_\alpha (z) \right| \le C \alpha ^2 z^{-\frac{1}{2}}e^{-c \alpha ^2 z} +C \alpha ^3e^{-c \alpha ^2 z}\le C \alpha ^2 z^{-\frac{1}{2}}e^{-c \alpha ^2 z},
\end{equation}
where in the second inequality we used Lemma~\ref{lem:estimate on K:intro} and Claim~\ref{claim:derivative of h} and in the third inequality we changed the values of $C$ and $c$ slightly. Thus,
\begin{equation}\label{eq:bound on F in stability}
|F(x)|\le C |\alpha _1 -\alpha _0| \alpha _0^2 \intop _0^{\infty } z^{-\frac{1}{2}}e^{-c\alpha _0 ^2 z}dz = C \alpha _0 |\alpha _1 -\alpha _0| \intop _0^\infty y^{-\frac{1}{2}} e^{-cy} dy\le C \alpha _0 |\alpha _1 -\alpha _0|.
\end{equation}
Next, we use this bound in order to bound the integral on the right hand side of \eqref{eq:integral of F}. To this end, define
\begin{equation}
I_1:=\intop _{t_1^{-}}^{t_1} F(x) (d \Pi _S (x)-S(x) dx),\quad I_2:= \intop _{t_1^{-}}^{t_1} F(x) (S (x)-\alpha _2) dx.
\end{equation}
We start by bounding $I_1$ using Corollary~\ref{cor:concentration of integral}. We compute the quadratic variation
\begin{equation}
M:=\intop _{t_1^-} ^{t_1} F(x)^2 S(x) dx \le \alpha _0^2 |\alpha _1-\alpha _0|^2 \intop _{t_1^-} ^{t_1} S(x) dx \le C \alpha _0^4 \beta _0 ^{13 \theta +C_{\circ}} .
\end{equation}
where the last inequality holds on the event $\mathcal A _4 \cap \{ \tau'\ge \hat{t}_0 \}$ using the definition of $\tau_1$. Thus, by Lemma~\ref{lem:ind2:alpha1vsalpha0} and Corollary~\ref{cor:concentration of integral} we have
\begin{equation}\label{eq:bound on I _1 in the proof of stability}
\mathbb{P}\left( |I_1|\ge \alpha _0^{2}\beta_0^{6.5\theta+C_\circ },\ \tau'\ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
We turn to bound $I_2$. On the event in Claim~\ref{claim:speed is almost fixed} we have
\begin{equation}\label{eq: bound on I_2 in the proof of stability}
|I_2|\le C \alpha _0|\alpha _1-\alpha _0 | \intop _{t_1^-} ^{t_1} |S(x)-\alpha _2 | dx \le C \alpha _0^{2} \beta _0^{8 \theta } .
\end{equation}
Thus, using Claim~\ref{claim:speed is almost fixed} and substituting the bounds on $I_1$ and $I_2$ into \eqref{eq:integral of F} finishes the proof of the lemma.
\end{proof}
We also record a generalized version of Lemma \ref{lem:ind2:alphaPvsalphaPP} which comes from a simple union bound.
\begin{cor}\label{cor:ind2:Ldiff}
Under the setting of Theorem \ref{thm:ind2:main}, let $L(\alpha)$ denote
\begin{equation}
L(\alpha):= \mathcal{L}(t_1^-; \Pi_S[t_1^{--},t_1^-],\alpha).
\end{equation}
Then, we have
\begin{equation}
\mathbb{P} \left(\left. \sup_{\alpha : \, |\alpha-\alpha_0| \le \alpha_0^{\frac{3}{2}}\beta_0^{6\theta} } \left|L(\alpha)-L(\alpha_0) \right| \ge 2\alpha_0^2 \beta_0^{7\theta}, \ \tau' \ge \hat{t}_0 \, \right| \, \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left( -\beta_0^4\right).
\end{equation}
\end{cor}
\begin{proof}
Let $\delta = \alpha_0^{10}$, and define $\mathcal{D}_{\alpha_0}:= \left\{\alpha: \, \left|\alpha-\alpha_0 \right| \le \alpha_0^{\frac{3}{2}}\beta_0^{6\theta}, \, \alpha -\alpha_0 = k \delta \textnormal{ for some } k\in \mathbb{Z} \right\}$. Note that Lemma \ref{lem:ind2:alphaPvsalphaPP} controls the difference $|L(\alpha_1)-L(\alpha_0)|$, and we can apply a union bound over the choice of $\alpha_1$, extending the result into
\begin{equation}
\mathbb{P} \left(\left. \sup_{\alpha \in \mathcal{D}_{\alpha_0} } \left|L(\alpha)-L(\alpha_0) \right| \ge \alpha_0^2 \beta_0^{7\theta}, \ \tau' \ge \hat{t}_0 \, \right| \, \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left( -\beta_0^4\right).
\end{equation}
Then, note that if $|\alpha- \hat{\alpha}| \le \delta$ for some $\hat{\alpha } \in \mathcal{D}_{\alpha_0}$ and $t_1\le \tau'$, then $|L(\alpha)-L(\hat{\alpha})| \le \alpha_0^5$, by crudely bounding \eqref{eq:integral of F}, concluding the proof.
\end{proof}
\subsubsection{Estimating the event $\mathcal{A}_1$}\label{subsubsec:ind2:A1}
Recall the definition of $\mathcal{A}_1(\alpha_1,t_1)$ in Section \ref{subsubsec:reg:history}. Our goal in this subsection is to establish the following lemma.
\begin{lem}\label{lem:ind2:A1}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P}\left(\mathcal{A}_1(\alpha_1,t_1)^c,\ \tau'\ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le
\exp \left(-\beta_0^3 \right)
\end{equation}
\end{lem}
To establish this lemma, we study the size of $X_{t_1^{-}}-X_s$ in two separate regime: for $s>t_0^{-}$ and $s \le t_0^{-}$. To analyze the first case, we consider the following event $\mathcal{A}_{1,1}$:
\begin{equation}
\mathcal{A}_{1,1}=\mathcal{A}_{1,1}(\alpha_0,t_0) :=
\left\{\forall \Delta \in [\alpha_0^{-2}\beta_0^{\theta/2}, t_1^{-}-t_0^{-}], \ X_{t_1^{-}}-X_{t_1^{-}-\Delta} \ge \frac{\alpha_0\Delta}{3} \right\}.
\end{equation}
\begin{lem}\label{lem:ind2:A11}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P} \left(\mathcal{A}_{1,1}^c,\ \tau'\ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^4\right).
\end{equation}
\end{lem}
\begin{proof}
This comes as a consequence of Lemma \ref{lem:ind1:tau13} and Corollary \ref{cor:concentration of integral}, applied to
\begin{equation}
|\Pi_S[s,t_1^-]|-\intop_{s}^{t_1^-} S(x)dx.
\end{equation}
The argument goes analogously as the proof of Lemma \ref{lem:ind1:tau9 10 11}, and the details are omitted due to similarity.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:ind2:A1}]
Suppose that $s\in [t_1^{-}-\alpha_0^{-2}\beta_0^{\theta/2},t_1^{-} ] $. Then, given the event $\mathcal{A}_4$ from $\eqref{eq:def:A4}$, the condition for $\mathcal{A}_1(\alpha_1,t_1)$ is automatically satisfied, since
\begin{equation}\label{eq:ind2:A1:pf1}
\alpha_1^{-1} \beta_1^{\theta} \ge \frac{1}{2}\alpha_0^{-1}\beta_0^{\theta} \ge
\sqrt{\alpha_0^{-2}\beta_0^{\theta/2}} \log^2(\alpha_0^{-2}\beta_0^{\theta/2})\ge \sqrt{t_1^{-}-s+C_\circ}\log^2 (t_1^{-}-s+C_\circ).
\end{equation}
Now consider the case $s\in[t_0^{-},t_1^{-}-\alpha_0^{-2}\beta_0^{\theta/2}]$. Conditioned on the event $\mathcal{A}_{1,1}$ from Lemma \ref{lem:ind2:A11}, we have
\begin{equation}\label{eq:ind2:A1:pf2}
X_{t_1^{-}}-X_s \ge \frac{1}{3}{\alpha_0(t_1^{-}-s)} \ge \sqrt{t_1^{-}-s+C_\circ} \log^2 (t_1^{-}-s+C_\circ),
\end{equation}
which satisfies the criterion for $\mathcal{A}_1(\alpha_1,t_1)$.
Finally, for the case $s\le t_0^{-}$ conditioned on $\mathcal{A}(\alpha_0,t_0)$, we observe that
\begin{equation}\label{eq:ind2:A1:pf3}
\begin{split}
X_{t_1^{-}}-X_s &= X_{t_1^{-}}-X_{t_0^{-}} + X_{t_0^{-}} X_s\\
&\ge
\frac{\alpha_0}{3}(t_1^{-}-t_0^{-}) + \sqrt{t_0^{-}-s+C_\circ}\log^2 (t_0^{-}-s+C_\circ)-\alpha_0^{-1}\beta_0^{\theta/2}\\
&\ge
\frac{\alpha_0}{4}(t_1^{-}-t_0^{-})+ \sqrt{t_0^{-}-s+C_\circ}\log^2 (t_0^{-}-s+C_\circ)\\
&\ge
\sqrt{t_1^{-}-t_0^{-}+C_\circ}\log^2 (t_1^{-}-t_0^{-}+C_\circ)
+
\sqrt{t_0^{-}-s+C_\circ}\log^2 (t_0^{-}-s+C_\circ)\\
&\ge \sqrt{t_1^{-}-s+C_\circ}\log^2 (t_1^{-}-s+C_\circ),
\end{split}
\end{equation}
where the last step follows from the fact that the function $f(x)=\sqrt{x+50}\log^2(x+50)$ is positive, increasing and concave.
Thus, from \eqref{eq:ind2:A1:pf1}, \eqref{eq:ind2:A1:pf2} and \eqref{eq:ind2:A1:pf3}, we see that $\mathcal{A}_1(\alpha_1,t_1)$ holds if $\mathcal{A}(\alpha_0,t_0)$, $\mathcal{A}_{1,1}$ and $\mathcal{A}_4$ are given. Thus, due to Lemmas \ref{lem:ind2:alpha1vsalpha0} and \ref{lem:ind2:A11}, we obtain the conclusion.
\end{proof}
\subsubsection{Estimating the event $\mathcal{A}_2$}\label{subsubsec:ind2:A3}
Recall the definition of $\mathcal{A}_2(\alpha_1,t_1)$ in Section \ref{subsubsec:reg:history}. In this subsection, we show that $\mathcal{A}_2(\alpha_1,t_1)$ occurs with high probability.
\begin{lem}\label{lem:ind2:A3}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P}\left(\mathcal{A}_2(\alpha_1,t_1)^c,\ \tau'\ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^3 \right)
\end{equation}
\end{lem}
\begin{proof}
This is a direct consequence of Lemma \ref{lem:ind1:tau13} and Corollary \ref{cor:concentration of integral}, following the same argument discussed in Lemma \ref{lem:ind1:tau9 10 11} (see also Lemma \ref{lem:ind2:A11}). We omit its details due to similarity to the previous proofs.
\end{proof}
Wrapping up the discussion in Section \ref{subsec:ind2:ratechange}, we finish the proof of Theorem \ref{thm:ind2:main}-(2).
\begin{proof}[Proof of Theorem \ref{thm:ind2:main}-(2)]
Combining Lemmas \ref{lem:ind2:ratechange}, \ref{lem:ind2:A1} and \ref{lem:ind2:A3}, we obtain that
\begin{equation}
\mathbb{P}\left(\mathcal{A}(\alpha_1,t_1)^c, \ \tau'\ge\hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left(-\beta_0^2\right),
\end{equation}
which directly implies Theorem \ref{thm:ind2:main}.
\end{proof}
\subsection{Regularity of speed in the next step}\label{subsec:ind2:remaining reg}
Recall the definition of $\tau(\alpha_1,t_1,2)$, ${\tau}^+(\alpha_0,t_0,1/2)$, $\mathcal{A}(\alpha_1,t_1)$ and $\mathcal{A}(\alpha_0,t_0)$ given in \eqref{eq:def:tau:induction} and \eqref{eq:def:A:induction}. Our goal in this subsection is to deduce the desired control on \begin{equation}
\tilde{ \tau} := \tau(\alpha_1,t_1,2)
\end{equation}
(excluding $\tau_{3}^\sharp(\alpha_1,t_1)$), which is given by the following proposition.
\begin{prop}\label{prop:ind2:prereg}
Under the setting of Theorem \ref{thm:ind2:main}, the following holds true:
$$\mathbb{P}(\tilde{ {\tau}}\le \tau' ,\ \tau'\ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) ) \le \exp(-\beta_0^{2}) .$$
\end{prop}
Recall the event $\mathcal{A}_4$ in \eqref{eq:def:A4}. On the event $\{\tau' \ge \hat{t}_0 \}\cap \mathcal{A}_4$, observe that the following hold true with probability 1:
\begin{itemize}
\item $\tau_{1}(\alpha_1,t_1,2)\ge {\tau}^+_1(\alpha_0,t_0,1/2) \ge \tau'$, since $\frac{1}{2}\alpha_0 \beta_0^{C_\circ} \le 2\alpha_1 \beta_1^{C_\circ}$.
\item $\tau_{7}(\alpha_1,t_1,2)\ge {\tau}^+_{7}(\alpha_0,t_0,1/2)\ge \tau'$, since for any $t\le {\tau}^+_{7}(\alpha_0,t_0,1/2)$, \begin{equation}
X_t - X_{t-\alpha_1^{-1}} \le X_t - X_{t-2\alpha_0^{-1}} \le 2\cdot \frac{1}{2} \beta_0^4 \le 2\beta_1^4.
\end{equation}
\item $\tau_{8}(\alpha_1,t_1,2) \ge {\tau}^+_{8}(\alpha_0,t_0,1/2) \ge \tau'$, since for any $t\le {\tau}^+_{8}(\alpha_0,t_0,1/2)$, we have
\begin{equation}
X_t - X_{t-2\alpha_1^{-1} \beta_1^{C_\circ}}
\ge X_t - X_{t-\frac{1}{2}\alpha_0^{-1 } \beta_0^{C_\circ}} >0.
\end{equation}
\end{itemize}
Moreover, conditioned on $\{\tau' \ge \hat{t}_0 \}\cap \mathcal{A}_4$, with probability 1 we have
\begin{equation}
\tau_{4}(\alpha_1,t_1,2) \wedge \tau_{5} (\alpha_1,t_1,2) \ge {\tau}^+_4(\alpha_0,t_0,1/2) \wedge {\tau}^+_5 (\alpha_0,t_0,1/2)\ge \tau',
\end{equation}
since for any $ t\le {\tau}^+_1(\alpha_0,t_0,1/2) {\tau}^+_4(\alpha_0,t_0,1/2) \wedge {\tau}^+_5 (\alpha_0,t_0,1/2) $,
\begin{equation}
\begin{split}
\intop_{t_1}^t (S(t)-\alpha_1)^2 dt
&\le \intop_{t_1}^t 2(S(t)-\alpha_0)^2 dt + 2(\hat{t}_0-t_1) (\alpha_1-\alpha_0)^2
\\
&\le 2\cdot \frac{1}{2}\alpha_0 \beta_0^{25\theta} + 2\cdot 2\alpha_0^{-2}\beta_0^{10\theta}\cdot 4\alpha_0^3 \beta_0^{12\theta}
\le 2\alpha_1 \beta_1^{25\theta},
\end{split}
\end{equation}
where the last inequality followed from the definitions of ${\tau}^+_4(\alpha_0,{t_0},1/2)$ and $\mathcal{A}_4$.
The integral in the definition of $\tau_{5}(\alpha_1,t_1,2)$ follows similarly, using the definition of ${\tau}^+_1(\alpha_0,t_0,1/2)$ together. Wrapping up the above discussion, $\tilde{ {\tau}}^{(1)} := \tau_{1}\wedge\tau_{4}\wedge \tau_{5} \wedge \tau_{7}\wedge \tau_{8}(\alpha_1,t_1,2) $, and obtain that
\begin{equation}\label{eq:ind2:tau4781911}
\mathbb{P}\left(
\tilde{ \tau}^{(1)} \le \tau', \ \tau'\ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0)
\right) \le \exp\left(-\beta_0^4 \right).
\end{equation}
Thus, the remaining task is to control $\tau_{2}, \tau_{3}, $ and $\tau_{6}$. We conduct this separately in Lemmas \ref{lem:ind2:tau5}, \ref{lem:ind2:tau6} and \ref{lem:ind2:tau9} below. We begin with studying $\tau_{2}.$
\begin{lem}\label{lem:ind2:tau5}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P} \left(\tau_{2}(\alpha_1,t_1,2) \le \tau',\ \tau'\ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left( -\beta_0^4\right).
\end{equation}
\end{lem}
\begin{proof}
In the proof, we set
\begin{equation}
\begin{split}
S_1(t)=S_1(t,t_1^{-},\alpha_1) := \mathcal{R}_b (t;\Pi_S[t_1^{-},t],\alpha_1 ).
\end{split}
\end{equation}
We can bound this quantity very similarly as in Lemma \ref{lem:ind1:tau5} as follows. We use the definition of ${\tau}^+_{7}(\alpha_0,t_0,1/2)$ to see that for all $t\in[t_1,\tau']$,
\begin{equation}
\begin{split}
S_1(t)&= \intop_{t_1^{-}}^{t } K_{\alpha_1}(t-x) d\Pi_S(x)\\
&\le
\sum_{k: 0\le k\alpha_0^{-1} \le 2\alpha^{-2}\beta^{10\theta} }
\frac{\beta_0^4}{2} \cdot \frac{C\alpha_1}{\sqrt{k\alpha_0^{-1}+1}} e^{-c\alpha_1^2 (k\alpha_0^{-1})} \le \alpha_1 \beta_1^{C_\circ},
\end{split}
\end{equation}
where the last inequality holds if the event $\mathcal{A}_4$ is given. Thus, thanks to Lemma \ref{lem:ind2:alpha1vsalpha0}, we obtain the conclusion.
\end{proof}
\begin{lem}\label{lem:ind2:tau6}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P} \left(\tau_{3}(\alpha_1,t_1,2) \le \tau',\ \tau'\ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 2\exp\left( -\beta_0^2\right).
\end{equation}
\end{lem}
\begin{proof}
The proof is analogous as Lemma \ref{lem:ind1:tau6}, using the definitions of $\tau_{4}(\alpha_1,t_1,2)$ and $\tau_{5}(\alpha_1,t_1,2),$ and the estimate \eqref{eq:ind2:tau4781911}.
\end{proof}
\begin{lem}\label{lem:ind2:tau9}
Under the setting of Theorem \ref{thm:ind2:main}, we have
\begin{equation}
\mathbb{P} \left(\tau_{6}(\alpha_1,t_1,2) \le \tau',\ \tau'\ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \exp\left( -\beta_0^3\right).
\end{equation}
\end{lem}
\begin{proof}
This comes as a consequence of Lemma \ref{lem:ind1:tau13} and Corollary \ref{cor:concentration of integral}, applied to
\begin{equation}
|\Pi_S[t_1^-,t]| - \intop_{t_1^-}^t S(x)dx.
\end{equation}
We also rely on $\mathcal{A}_4$ to switch $\alpha_0$ to $\alpha_1$. The rest of the proof is analogous to that of Lemma \ref{lem:ind1:tau9 10 11}, and we omit the details due to similarity (see also Lemmas \ref{lem:ind2:A11} and \ref{lem:ind2:A3}).
\end{proof}
We conclude Section \ref{subsec:ind2:remaining reg} by proving Proposition \ref{prop:ind2:prereg}.
\begin{proof}[Proof of Proposition \ref{prop:ind2:prereg}]
It follows from combining \eqref{eq:ind2:tau4781911}, Lemmas \ref{lem:ind2:tau5}, \ref{lem:ind2:tau6} and \ref{lem:ind2:tau9}.
\end{proof}
\subsection{Refined control on the first order approximation}\label{subsec:ind2:bootstrappedJ}
In this subsection, we provide a refined analysis of the first order approximation, deducing an appropriate control on $\tau_{3}^\sharp$ (Section \ref{subsubsec:reg:refined1storder}). The main result we aim to establish is the following.
\begin{lem}\label{lem:ind2:tau6sharpreal}
Under the setting of Theorem \ref{thm:ind2:main} and Proposition \ref{prop:ind2:prereg}, we have
\begin{equation}
\mathbb{P} \left(\tau_{3}^\sharp(\alpha_1,t_1 ) \le t_1^\sharp, \ \tilde{\tau}\wedge \tau' \ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 10\exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
Having this on hand, we can deduce Theorem \ref{thm:ind2:main} by combining the results obtained so far.
\begin{proof}[Proof of Theorem \ref{thm:ind2:main}]
Theorem~\ref{thm:ind2:main}-(1) follows directly from combining Proposition \ref{prop:ind2:prereg}, Lemmas \ref{lem:ind2:tau1sharp} and \ref{lem:ind2:tau6sharp}. The second part of the theorem was already established in Section \ref{subsec:ind2:ratechange}.
\end{proof}
$\tau_{3}^\sharp$ studies the integral of $|S(t)-S_1(t)|$, and our approach is to first estimate $|S(t)-S_1(t)|$ itself. To this end, we define the stopping time $\tau_{{3,1}}^\sharp$ as
\begin{equation}
\tau_{{3,1}}^\sharp(\alpha_0,t_0)
:=
\inf \left\{t\ge t_0: |S(t)-S_1(t)| \ge \alpha_0 \beta_0^{3\theta} \sigma_1(t;S)^2 + \alpha_0^{1-\epsilon} \sigma_1\sigma_2\sigma_3(t;S) \right\}.
\end{equation}
\begin{lem}\label{lem:ind2:tau6sharp}
Under the setting of Theorem \ref{thm:ind2:main} and Proposition \ref{prop:ind2:prereg}, we have
\begin{equation}
\mathbb{P} \left(\tau_{{3,1}}^\sharp(\alpha_1,t_1 ) \le t_1^\sharp, \ \tilde{\tau}\wedge \tau' \ge \hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 10\exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
Assuming Lemma \ref{lem:ind2:tau6sharp}, we conclude the proof of Lemma \ref{lem:ind2:tau6sharpreal}.
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau6sharpreal}]
We set
$\tau_{\textnormal{int}}$ to be
\begin{equation}
\tau_{\textnormal{int}} := \inf\left\{ t\ge t_1: \intop_{t_1}^t \sigma_1\sigma_2\sigma_3(t;S) dt \ge \alpha^{-\frac{1}{2}-\epsilon} \right\}.
\end{equation}
We also recall the definition of $\tau'$ in Theorem \ref{thm:ind2:main}, and set
\begin{equation}
\tilde{\tau}_3^\sharp := \tau_{3,1}^\sharp\wedge \tau' \wedge \tau_{\textnormal{int}} .
\end{equation}
Then, Lemmas \ref{lem:fixed perturbed:error int:perturbed}, \ref{lem:ind2:tau6sharp} and Proposition \ref{prop:ind2:prereg} tell us that
\begin{equation}
\mathbb{P} \left( \tilde{\tau}_{3}^\sharp \le t_1^\sharp , \ \tau' \ge \hat{t}_0 \, | \,\mathcal{A}(\alpha_0,t_0) \right) \le \exp \left(-\beta_0^2 \right).
\end{equation}
The proof of the desired statement follows by observing that
\begin{equation}
\begin{split}
\intop_{t_1}^{\tilde{\tau}_3^\sharp} |S(t)-S_1(t)|dt \le \intop_{t_1}^{\tilde{\tau}_3^\sharp} \left\{\frac{\alpha_0\beta_0^{3\theta}}{\pi_1(t;S)+1} + \alpha_0^{1-\epsilon} \sigma_1\sigma_2\sigma_3(t;S) \right\} dt <\beta_1^{4\theta},
\end{split}
\end{equation}
where we used Lemma \ref{lem:ind1:pi1int basicbd:basic} and $\tau_7$ to deduce the last inequality, relying on $\mathcal{A}_4$ \eqref{eq:def:A4} to show that $\beta_0$ and $\beta_1$ are close.
\end{proof}
The rest of this subsection is devoted to the proof of Lemma \ref{lem:ind2:tau6sharp}. Unlike when we studied ${\tau}^+_3$ in Lemmas \ref{lem:ind1:tau6} and \ref{lem:ind2:tau6}, we can no longer work with the first-order approximation of the speed. Thus, the second-order approximation \eqref{eq:def:S2:basic form} is required. However, for technical reasons (which will be clarified in Remark \ref{rmk:ind2:tau6sh:frame shift}), we will read the process from time $t_1^{--}$ given by
\begin{equation}
t_1^{--}:= t_1 -2\alpha_0^{-2}\beta_0^\theta = t_1^- - \alpha_0^{-2}\beta_0^\theta,\quad t_1^\flat := t_1 - \frac{1}{3}\alpha_0^{-2}\beta_0^\theta,
\end{equation}
with respect to the same frame of reference $\alpha_1$.
Recall that $\alpha_1$ is measureable with respect to $\mathcal{F}_{t_1^-}.$ In what follows, we first introduce the setting we use in this subsection, which the framework in terms of $\alpha_1$ and $t_1^-$.
To begin with, the second-order approximation that we work with in this subsection is defined as
\begin{equation}\label{eq:def:S2:ind2}
S_2(t)=S_2(t;t_1^{--},\alpha_1) := \frac{2\alpha_1^2}{(1+2\alpha_1)^2} + \frac{S_1(t)}{(1+2\alpha_1)^2} + \frac{\alpha_1}{(1+2\alpha_1)} \intop_{t_1^-}^t \intop_{t_1^-}^{s} J_{t-s, t-u}^{(\alpha_1)} d\widehat{\Pi}_{S}(u) d\widehat{\Pi}_S(s),
\end{equation}
where
$S_1(t)$ in this subsection is given by
\begin{equation}\label{eq:def:S1:ind2}
\begin{split}
S_1(t)= S_1(t,t_1^{--},\alpha_1)
=
\intop_{ t_1^{--}}^t K_{\alpha_1}(t-x) d\Pi_S(x),
\end{split}\end{equation}
and $J_{u,s}^{(\alpha_1)}$ is as \eqref{eq:def:J:basic form} and \eqref{eq:def:J:expected val}. We also remark that $d\widehat{\Pi}_S(x)= d\Pi_S(x)-\alpha_1 dx$.
Recall that Lemma \ref{lem:ind2:alpha1vsalpha0} implies that the event
$\mathcal{A}_4:= \{|\alpha_1-\alpha_0|\le 2\alpha_0^{3/2}\beta_0^{6\theta}\}$
happens with high probability if $\Pi_S(-\infty,t_0]$ was regular, and it is $\mathcal{F}_{t_1^-}$-measurable. In such a case, $(t_1^-)^-:= t_1^{--}$, $(t_1^-)^+:= t_1^\flat$ satisfies \eqref{eq:def:t0t1} in terms of $t_1^-$ and $\alpha_1$, and thus we can redefine $\tau(\alpha_1, t_1^-,\kappa)$ \eqref{eq:def:tau:induction} and $\mathcal{A}(\alpha_1, t_1^-)$ \eqref{eq:def:A:induction} in terms of $(t_1^-)^- , t_1^-$ and $ S_1(t)$. Furthermore, the stopping time $\tau_{\textnormal{b}}'(\alpha_1,t_1^-,{t_1^\flat})$ from \eqref{eq:def:tauxprime} can be defined in the same way, and it will play an important role in this subsection. Since the choice of $t_1$ is flexible in Theorem \ref{thm:ind2:main}, the results from Theorem \ref{thm:ind2:main}-(2), Proposition \ref{prop:ind2:prereg} and Corollary \ref{cor:ind1:tauxprime} tell us the following.
\begin{cor}\label{cor:ind2:prereg:past}
Under the setting of Theorem \ref{thm:ind2:main}, let $\alpha_1 , $ $t_1^-, (t_1^-)^-=t_1^{--}, (t_1^-)^+ = t_1^\flat$, $\tau(\alpha_1,t_1^-,\kappa)$, $\tau_{\textnormal{b}}'(\alpha_1,t_1^-,t_1^\flat)$ and $\mathcal{A}(\alpha_1,t_1^-) $ be as above. Then, we have
\begin{itemize}
\item[(1)] $\mathbb{P} \left( \tau(\alpha_1,t_1^-,2) <\hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \mathbb{P}\left(\tau' <\hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) + 4 \exp (-\beta_0^2) $;
\item[(2)] $\mathbb{P} \left( \tau_{\textnormal{b}}'(\alpha_1,t_1^-,t_1^\flat) \le t_1^\sharp \ | \ \mathcal{A}(\alpha_0,t_0) \right) \le \mathbb{P}\left(\tau' <\hat{t}_0 \ | \ \mathcal{A}(\alpha_0,t_0) \right) + 4 \exp (-\beta_0^2) $;
\item[(3)] $\mathbb{P} \left(\mathcal{A}(\alpha_1,t_1^-)^c \, | \, \mathcal{A}(\alpha_0,t_0) \right) \le \mathbb{P} \left(\tau'<\hat{t}_0 \, | \, \mathcal{A}(\alpha_0,t_0) \right) + \exp (-\beta_0^2)$.
\end{itemize}
\end{cor}
Before moving on, we clarify the role of $\tau_{\textnormal{b}}'(\alpha_1,t_1^-, t_1^\flat )$. It tells us the bound on $|S_1(t)- \alpha_1''|$, where we denoted
$$\alpha_1'' := \mathcal{L}(t_1^-; \Pi_S[t_1^{--},t_1^-], \alpha_1 ). $$
On the event $\mathcal{A}_4$ where $|\alpha_1-\alpha_0|\le \alpha_0^{\frac{3}{2}}\beta_0^{6\theta}$, we have with high probability that
\begin{equation}
\begin{split}
|\alpha_1'' - \alpha_1| \le& |\mathcal{L}(t_1^-;\Pi_S[t_1^{--},t_1^-],\alpha_1) -\mathcal{L}(t_1^-;\Pi_S[t_1^{--},t_1^-],\alpha_0)|
\\
&+ |\mathcal{L}(t_1^-;\Pi_S[t_1^{--},t_1^-],\alpha_0) - \mathcal{L}(t_1^-;\Pi_S[t_0^{-},t_1^-],\alpha_0)|\\
\le & \alpha_0^2 \beta_0^{7\theta},
\end{split}
\end{equation}
where we bounded two terms using Lemma \ref{lem:ind2:alphavsalphaPP} and Corollary \ref{cor:ind2:Ldiff}. Note that Corollary \ref{cor:ind2:Ldiff} is necessary (instead of Lemma \ref{lem:ind2:alphaPvsalphaPP}) since $\alpha_1$ is not $\mathcal{F}_{t_1^{--}}$-measurable. Thus, combining with the estimates on $\tau_{\textnormal{b}}'(\alpha_1,t_1^-,t_1^{\pm})$, we have control on the stopping time $\tau_{\textnormal{b}}''$ given by
\begin{equation}\label{eq:def:tauxpprime}
\tau_{\textnormal{b}}'':=\inf \left\{ t
\ge t_1^{\flat} :\, |S_1(t)-\alpha_1| \le \alpha_0 \beta_0^{C_\circ} \sigma_1(t;S) + 2\alpha_0^{\frac{3}{2}}\beta_0^\theta \right\},
\end{equation}
in such a way that
\begin{equation}\label{eq:ind2:tauxpprime}
\mathbb{P} ( \tau_{\textnormal{b}}'' \le t_1^\sharp \, | \, \mathcal{A}(\alpha_0,t_0) ) \le \mathbb{P}(\tau' <\hat{t}_0 \,|\, \mathcal{A}(\alpha_0,t_0)) + \exp (-\beta_0^2) .
\end{equation}
Moving on to estimating $|S(t)-S_1(t;t_1^-,\alpha_1)|$ in $\tau_{{3,1}}^\sharp(\alpha_1,t_1)$, we write
\begin{equation}\label{eq:ind2:SminS1decomp}
\begin{split}
|S(t)-S_1(t;t_1^-,\alpha_1)|
\le& \,
|S(t)-S'(t;t_1^{--},\alpha_1)| +|S'(t;t_1^{--},\alpha_1)-S_2(t)|\\
& + |S_1(t)-S_2(t)| + |S_1(t;t_1^-,\alpha_1)-S_1(t)|
\end{split}
\end{equation}
If $\Pi_S(-\infty,t_1^-]\in\mathcal{A}(\alpha_1,t_1^-)$ (which happens with high probability due to (3) of Corollary \ref{cor:ind2:prereg:past}), Proposition \ref{prop:SvsSprime} tells us that
\begin{equation}\label{eq:ind2:SvsSprime}
|S(t)- S'(t;t_1^{--},\alpha_1)|\le \alpha_1^{100}.
\end{equation}
Also, if $\mathcal{A}_4$ is given, then for any $t_1^\flat \le t\le \tau'$ we have
\begin{equation}\label{eq:ind2:S1vsS1minus}
|S_1(t;t_1^-,\alpha_1)-S_1(t)| =\intop_{t_1^{--}}^{t_1^-} K_{\alpha_1}(t-x) d\Pi_S(x) \le e^{-c\beta_0^\theta} |\Pi_S[t_1^{--},t_1^-]| \le \alpha_1^{100},
\end{equation}
due to the decay property of $K_{\alpha}(s)$.
Moreover, note that the assumptions of Proposition \ref{prop:fixed perturbed:error} are satisfied with $t^- = t_1^{--}$, $\hat{t} = \hat{t}_0$, $\alpha=\alpha_1$, and $\tau = \tau(\alpha_1,t_1^-,2)$. Hence, it tells us that
\begin{equation}\label{eq:ind2:SprimevsS2}
\begin{split}
\mathbb{P} \left(\left. |S'(t;t_1^{--},\alpha_1)-S_2(t)| \le 2\alpha_1^{1-\frac{\epsilon}{2}}\sigma_1\sigma_2\sigma_3(t;S), \ \forall t\in[t_1^-,\, \hat{t}_0\wedge \tau(\alpha_1,t_1^-,2)] \,\right| \, \mathcal{A}(\alpha_0,t_0) \right)\\
\ge
1-\exp\left(- \beta_0^4 \right).
\end{split}
\end{equation}
Note that we again used $\mathcal{A}_4$ (Lemma \ref{lem:ind2:alpha1vsalpha0}) to claim that $\alpha_1$ is close enough to $\alpha_0$, enabling us to exchange $\alpha_1$ with $\alpha_0$ (in RHS).
Therefore, among the terms in \eqref{eq:ind2:SminS1decomp}, what remains is estimating $|S_2(t)-S_1(t)|$. It can written as
\begin{equation}\label{eq:ind2:S2vsS1}
S_2(t)-S_1(t)
=
\frac{2\alpha_1^2}{(1+2\alpha_1)^2} - \frac{4\alpha_1+4\alpha_1^2}{(1+2\alpha_1)^2} S_1(t)+ \frac{\alpha_1}{(1+2\alpha_1)} \intop_{t_1^-}^t \intop_{t_1^-}^{s} J_{t-s, t-u}^{(\alpha_1)} d\widehat{\Pi}_{S}(u) d\widehat{\Pi}_S(s).
\end{equation}
Here, the first two terms $\frac{2\alpha_1^2}{(1+2\alpha_1)^2}$ and $\frac{4\alpha_1+4\alpha_1^2}{(1+2\alpha_1)^2}S_1(t)$ are going to absorbed by $\alpha_1\beta_1^{3\theta} \sigma_1(t;S)^2$ in $\tau_{3,1}^\sharp$, due to the definitions of $\tau_{2}(\alpha_1,t_1^-,2)$, $\tau_{8}(\alpha_1,t_1^-,2)$, and $\mathcal{A}_4$ \eqref{eq:def:A4}. Thus, our main duty now is to investigate the double integral, the third term in RHS.
To this end, we recall that $t_1^\flat=t_1-\frac{1}{3}\alpha_0^{-2}\beta_0^\theta$, and we attempt to switch $t_1^-$ in the double integral to $t_1^\flat$, which will give us a significant technical advantage (see Remark \ref{rmk:ind2:tau6sh:frame shift}). We claim that if we have $ |\alpha_1-\alpha_0| \le 2\alpha_0^{\frac{3}{2}}\beta_0^{6\theta}$, then for any $t_1\le t\le \tau'$
\begin{equation}\label{eq:ind2:tau6sh:switch base}
\begin{split}
\left|\intop_{t_1^-}^t \intop_{t_1^-}^{s} J_{t-s, t-u}^{(\alpha_1)} d\widehat{\Pi}_{S}(u) d\widehat{\Pi}_S(s) - \intop_{t_1^\flat}^t \intop_{t_1^\flat}^{s} J_{t-s, t-u}^{(\alpha_1)} d\widehat{\Pi}_{S}(u) d\widehat{\Pi}_S(s)\right|\\
= \left|\intop_{t_1^-}^t \intop_{t_1^-}^{s\wedge t_1^\flat} J_{t-s, t-u}^{(\alpha_1)} d\widehat{\Pi}_{S}(u) d\widehat{\Pi}_S(s) \right| \le
\alpha_1^{50}.
\end{split}
\end{equation}
This is becuase for any $u,s,t$ such that $t_1^-\le u \le t_1^\flat$, and $t\ge t_1$, we have $t - u \ge \frac{1}{3}\alpha_0^{-2}\beta_0^{\theta}$, and hence the estimate on $J$ (Lemma \ref{lem:bound on deterministic J}) implies
\begin{equation}
\left| J_{t-s,t-u}^{(\alpha_1)} \right| \le \alpha_1^{100}.
\end{equation}
Combined with the definitions of $\tau_{1}, \tau_{7}$, we obtain \eqref{eq:ind2:tau6sh:switch base}.
Thus, it suffices to consider the following stopping time:
\begin{equation}
\begin{split}
\tau_{{3,2}}^\sharp &= \tau_{{3,2}}^\sharp(\alpha_1,t_1;\alpha_0)\\
&:=
\inf \left\{t\ge t_1: \left| \intop_{t_1^\flat}^t \intop_{t_1^\flat}^{s} J_{t-s, t-u}^{(\alpha_1)} d\widehat{\Pi}_{S}(u) d\widehat{\Pi}_S(s)\right| \ge \beta_0^{2\theta+4C_\circ} \sigma_1(t;S) \right\}.
\end{split}
\end{equation}
\begin{remark}\label{rmk:ind2:tau6sh:frame shift}
In the definition of $\tau_{{3,2}}^\sharp$, note that it reads the process starting from $t_1^\flat$, not $t_1^-$. In controlling the double integral in $\tau_{{3,2}}^\sharp$, we will rely on the bound on $|S_1(t)-\alpha_1|$, which works for $t\ge t_1^\flat$, due to the definition of $\tau_{\textnormal{b}}''$. This is the main reason why we shift the interval of interest to $[t_1^{\flat}, t]$ from $[t_1^-,t]$ and consider an estimate such as \eqref{eq:ind2:tau6sh:switch base}.
\end{remark}
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau6sharp}]
Summing up the discussions from \eqref{eq:ind2:SminS1decomp} to \eqref{eq:ind2:tau6sh:switch base}, the proof follows directly from Lemma \ref{lem:ind2:tau61sharp} below.
\end{proof}
\begin{lem}\label{lem:ind2:tau61sharp}
Under the setting of Theorem \ref{thm:ind2:main} and Proposition \ref{prop:ind2:prereg}, we have
\begin{equation}
\mathbb{P}\left(\tau_{{3,2}}^\sharp \le t_1^\sharp, \ \tilde{\tau}\wedge \tau' \ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 3 \exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
Proof of Lemma \ref{lem:ind2:tau61sharp} turns out to be more involved than the methods used in Section \ref{sec:fixedrate}, since we now do not allow the error to be of size $\alpha_1^{-\epsilon}$. We rather analyze the double integral directly, relying on the fact that the length scale of $t_1^\sharp-t_1^\flat $, which is roughly $\alpha_0^{-2}\beta_0^\theta$, is much smaller than $\alpha_0^{-2}\beta_0^{10\theta}$.
To begin with, we first study the inner integral. Define the stopping time $\tau_{{3,3}}^\sharp$ as
\begin{equation}\label{eq:def:tau63sharp}
\tau_{{3,3}}^\sharp:= \inf\left\{ t\ge t_1: \,\exists s\in[t_1^\flat,t] \textnormal{ s.t. } \left|\intop_{t_1^\flat}^s J^{(\alpha_1)}_{t-s,t-u} d\widehat{ \Pi}_S(u) \right| \ge \frac{\alpha_1^{\frac{1}{2}}\beta_1^{\theta+3 }}{\sqrt{t-s+1}} + \frac{\beta_1^{C_\circ +1} }{t-s+1} \right\}.
\end{equation}
\begin{lem}\label{lem:ind2:tau63sharp}
Under the setting of Theorem \ref{thm:ind2:main} and Proposition \ref{prop:ind2:prereg}, we have
\begin{equation}
\mathbb{P}\left(\tau_{{3,3}}^\sharp \le t_1^\sharp, \ \tilde{\tau}\wedge \tau'\ge \hat{t}_0\ | \ \mathcal{A}(\alpha_0,t_0) \right) \le 3\exp\left(-\beta_0^2 \right).
\end{equation}
\end{lem}
Recall the definitions $S_1(t)$ and $R(t)$.
In order to study the integrals over $d\widehat{\Pi}_S(x)$ which appear in Lemmas \ref{lem:ind2:tau61sharp} and \ref{lem:ind2:tau63sharp}, we decompose it as follows:
\begin{equation}\label{eq:ind2:tausharp:tilPiSdecomposition}
\begin{split}
d\widehat{\Pi}_S (x) = ~d \widetilde{\Pi}_S(x) + \big[S(x)-S_1(x) \big] + \big[S_1(x)-\alpha_1 \big],
\end{split}
\end{equation}
where we remind that $d\widetilde{\Pi}_S(x)=d\Pi_S(x)-S(x)dx$.
We will rely on this formula when establishing both Lemmas \ref{lem:ind2:tau61sharp} and \ref{lem:ind2:tau63sharp}.
In the remaining of Section \ref{subsec:ind2:bootstrappedJ}, we first prove Lemma \ref{lem:ind2:tau63sharp} in Section \ref{subsubsec:ind2:tau63sharp}, and then establish Lemma \ref{lem:ind2:tau61sharp} in Section \ref{subsubsec:ind2:tau61sharp}.
\subsubsection{Proof of Lemma \ref{lem:ind2:tau63sharp}}\label{subsubsec:ind2:tau63sharp}
We decompose the integral given in the definition of $\tau^\sharp_{3,3}$ into four parts using the decomposition \eqref{eq:ind2:tausharp:tilPiSdecomposition}, and then we study the following lemmas in order to show Lemma \ref{lem:ind2:tau63sharp}. Recall the definition of the event $\mathcal{A}_4$ given in \eqref{eq:def:A4}, and set
\begin{equation}\label{eq:def:ind2:Aprime}
\mathcal{A}':= \mathcal{A}(\alpha_0,t_0)\cap \mathcal{A}_4.
\end{equation}
Moreover, recall the definitions of $\tau_{\textnormal{b}}''$ \eqref{eq:def:tauxpprime} and set
\begin{equation}\label{eq:def:tauhat:ind2}
\hat{\tau}:= \tilde{\tau} \wedge \tau_{\textnormal{b}}'' .
\end{equation}
We note that Lemmas \ref{lem:ind2:tau63sharp:1} and \ref{lem:ind2:tau63sharp:2} below hold the same with the weaker stopping time $\tilde{\tau}$ instead of $\hat{\tau}$. However, we state them in terms of $\hat{\tau}$ to be consistent with Lemma \ref{lem:ind2:tau63sharp:4}.
\begin{lem}\label{lem:ind2:tau63sharp:1}
Assume the setting of Theorem \ref{thm:ind2:main}. For each $t\in [t_1,t_1^\sharp]$ and $s\in[t_1^\flat,t]$, define the event \begin{equation}
\mathcal{B}_1(s,t):= \left\{ \left| \intop_{t_1^\flat \wedge\hat{\tau}}^{s\wedge \hat{\tau}} J^{(\alpha_1)}_{t-s,t-u} d\widetilde{\Pi}_S(u) \right|
\le \left(\frac{\beta_0^{C_\circ+1}}{t-s+1} +\frac{\alpha_0^{\frac{1}{2}} \beta_0^{2C_\circ} }{\sqrt{t-s+1} }
\right) e^{-c\alpha_0^2(t-s)} \right\}.
\end{equation}
Then, we have
\begin{equation}
\mathbb{P}\left(\left. \bigcap_{t\in [t_1, t_1^\sharp] } \bigcap_{s\in [t_1^\flat,t] }\mathcal{B}_1(s,t) \ \right| \ \mathcal{A}' \right) \ge 1-\exp\left(-\beta_0^4 \right).
\end{equation}
\end{lem}
\begin{lem}\label{lem:ind2:tau63sharp:2}
Under the setting of Theorem \ref{thm:ind2:main} suppose that we condition on the event $\mathcal{A}_4$. Then, we have
\begin{equation}
\left| \intop_{t_1^{\flat}\wedge\hat{\tau}}^{u \wedge \hat{\tau}} J^{(\alpha_1)}_{t-s,t-u} (S(u)-S_1(u)) du\right| \le \frac{\alpha_0^{1-\epsilon}\beta_0^{C_\circ} }{\sqrt{t-s+1}}e^{-c\alpha_0^2(t-s)},
\end{equation}
for all $t\in[t_1,t_1^\sharp]$ and $s\in [t_1^\flat,t]$.
\end{lem}
\begin{lem}\label{lem:ind2:tau63sharp:4}
Under the setting of Theorem \ref{thm:ind2:main}, suppose that we condition on the event $\mathcal{A}_4$. For all $t\in [t_1,t_1^+]$ and $s\in[t_1^\flat,t]$, we have
\begin{equation}
\left| \intop_{t_1^\flat\wedge\hat{\tau}}^{s\wedge \hat{\tau}} J^{(\alpha_1)}_{t-s,t-u} (S_1(u)-\alpha_1)du \right|
\le \frac{\alpha_0^{\frac{1}{2}} \beta_0^{\theta+2}} {\sqrt{t-s+1} }e^{-c\alpha_1^2(t-s)}.
\end{equation}
\end{lem}
Before we delve into the details, we address a simple fact that is useful to keep in mind for the rest of the subsection. Note that \eqref{eq:ind2:tauxpprime} and Theorem \ref{thm:ind2:main}-(2) shown in Section \ref{subsec:ind2:ratechange} tells us that
\begin{equation}\label{eq:ind2:tau63sh:aux1}
\mathbb{P} \left( \left. \hat{\tau} \le t_1^\sharp, \ \tilde{\tau}\wedge\tau'\ge \hat{t}_0 \ \right| \ \mathcal{A}(\alpha_0,t_0) \right) \le 2\exp\left(-\beta_0^2 \right).
\end{equation}
Furthermore, we see from Lemma \ref{lem:ind2:alpha1vsalpha0} and Theorem \ref{thm:ind2:main}-(2) that
\begin{equation}\label{eq:ind2:tau63sh:aux2}
\begin{split}
\mathbb{P}\left( \left. (\mathcal{A}_4)^c \cup \mathcal{A}(\alpha_1,t_1^-)^c,\ \tilde{\tau} \wedge \tau' \ge \hat{t}_0\ \right| \ \mathcal{A}(\alpha_0,t_0) \right) \le 2 \exp\left(-\beta_0^5 \right).
\end{split}
\end{equation}
This tells us that along with the event $\{\tilde{\tau}\wedge\tau'\ge \hat{t}_0 \}$, conditioning on $\mathcal{A}'$ or on $\mathcal{A}(\alpha_1,t_1^-)$ instead of $\mathcal{A}(\alpha_0,t_0)$ produces only a small additional error probability. Moreover, we stress that $\mathcal{A}_4$ and $\mathcal{A}(\alpha_1,t_1^-)$ are $\mathcal{F}_{t_1^-}$-measurable.
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau63sharp}]
Having \eqref{eq:ind2:tau63sh:aux1} and \eqref{eq:ind2:tau63sh:aux2} combining the results of Lemmas \ref{lem:ind2:tau63sharp:1}--\ref{lem:ind2:tau63sharp:4}, we have with probability at least $1- 3\exp(-\beta_0^4)$ conditioned on $\mathcal{A}(\alpha_0,t_0)$ that
\begin{equation}
\left| \intop_{t_1^\flat \wedge\hat{\tau} }^{u\wedge\hat{\tau}} J^{(\alpha_1)}_{t-s,t-u} d\widehat{\Pi}_S(u)\right| \le \left(\frac{ \alpha_0^{\frac{1}{2}}\beta_0^{\theta+3} }{\sqrt{t-s+1}} + \frac{\beta_0^{C_\circ +1}}{t-s+1}\right) e^{-c\alpha_0^2(t-s)},
\end{equation}
and hence we obtain the conclusion.
\end{proof}
In the remaining of this subsection, we prove Lemmas \ref{lem:ind2:tau63sharp:1}--\ref{lem:ind2:tau63sharp:4}. In the proof, we repeatedly use Lemma \ref{lem:bound on deterministic J} which is
\begin{equation}\label{eq:ind2:Jbd}
J^{(\alpha_1)}_{x,y} \le \frac{Ce^{-2c y\alpha_0^{2} }}{\sqrt{(x+1)(y+1)}} \le \frac{Ce^{-c (x+y)\alpha_0^{2} }}{\sqrt{(x+1)(y+1)}},
\end{equation}
where the first inequality holds upon the event $\mathcal{A}_4.$
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau63sharp:1}]
Define the set
\begin{equation}\label{eq:ind2:tau63sh:1:discretizedset}
\mathcal{T}:= \left\{ \alpha_0^{10}k \in [t_1^\flat,t_1^\sharp] : k\in \mathbb{Z} \right\}.
\end{equation}
For any $t,u\in \mathcal{T}$ with $u\le t$, we first observe from \eqref{eq:ind2:Jbd} that on $\mathcal{A}_4$,
\begin{equation}\label{eq:ind2:tau61sh1:small s}
\begin{split}
\left|\intop_{(s-\alpha_0^{-1})\wedge \hat{\tau}}^{s\wedge \hat{\tau}} J^{(\alpha_1)}_{t-s,t-u} d\widetilde{\Pi}_S(u)\right|
&\le
\frac{\beta_0^{5}e^{-c\alpha_0^2(t-s)}}{t-s+1 } + \frac{Ce^{-c\alpha_0^2(t-s)}}{\sqrt{t-s+1}} \intop_{s-\alpha_0^{-1}}^s \frac{\alpha_0 \beta_0^{C_\circ}du}{\sqrt{t-u+1}} \\&\le \frac{2C\beta_0^{C_\circ}e^{-c\alpha_0^2(t-s)}}{t-s+1}.
\end{split}
\end{equation}
To take care of the integral from $t_1^\flat$ to $(s-\alpha_0^{-1})$, note that we have
\begin{equation}
\begin{split}
\intop_{t_1^\flat\wedge \hat{\tau}}^{s\wedge \hat{\tau}} \left\{J^{(\alpha_1)}_{t-s,t-u}\right\}^2 S(u)du
\le \frac{Ce^{-2c\alpha_1^2(t-s)}}{t-s+1} \intop_{t_1^{\flat}}^{s} \frac{\alpha_0 \beta_0^{C_\circ}du}{t-u+1}& \le
\frac{ \alpha_0 \beta_0^{C_\circ +2} }{t-s+1}e^{-2c\alpha_0^2(t-s)},\\
\sup \left\{\left|J^{(\alpha_1)}_{t-s,t-u} \right|: s\in[t_1^\flat,s-\alpha_0^{-1}] \right\} &\le \frac{C\alpha_0^{\frac{1}{2}}e^{-c\alpha_0^2(t-s)}}{\sqrt{t-s+1}}.
\end{split}
\end{equation}
Thus, we apply Lemma \ref{lem:concentration of integral} and use \eqref{eq:ind2:tau61sh1:small s} to obtain that
\begin{equation}
\mathbb{P} \left(\left.\left| \intop_{t_1^\flat\wedge \hat{\tau} }^{s\wedge \hat{\tau}}
J^{(\alpha_1)}_{t-s,t-u} d\widetilde{\Pi}_S(u)
\right| \ge \left(\frac{2C\beta_0^{C_\circ}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{C_\circ} }{\sqrt{t-s+1}} \right)e^{-c\alpha_0^2(t-s)} \ \right| \ \mathcal{A}' \right) \le \exp\left(-\beta_0^5 \right).
\end{equation}
Note that we can condition on $\mathcal{A}'$ when applying Lemma \ref{lem:concentration of integral} since it is $\mathcal{F}_{t_1^-}$-measurable. Moreover, the event $\mathcal{A}_4$ is used to ensure that $\alpha_1$ and $\alpha_0$ are close enough.
Now define
\begin{equation}
\mathcal{B}'_1(s,t):= \left\{ \left| \intop_{t_1^\flat\wedge \hat{\tau} }^{s\wedge \hat{\tau}}
J^{(\alpha_1)}_{t-s,t-u} d\widetilde{\Pi}_S(u)
\right| \le \left(\frac{2C\beta_0^{C_\circ}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{C_\circ} }{\sqrt{t-s+1}}\right)e^{-c\alpha_0^2(t-s)} \right\}.
\end{equation}
Note that its bound is slightly stronger than $\mathcal{B}_1(s,t)$, to leave some room to take a union bound and then conduct a continuity argument.
Then, we use a union bound to deduce that
\begin{equation}
\mathbb{P} \left(\left.\bigcap_{t\in \mathcal{T} } \bigcap_{s\in \mathcal{T}:\, s\le t} \mathcal{B}_1'(s,t)\ \right| \ \mathcal{A}' \right) \ge 1- \exp\left(-\beta_0^4 \right).
\end{equation}
The remaining duty is to improve this bound to work on general $t,s$ which are not necessarily in $\mathcal{T}$. This is done by the exact same argument presented in Lemmas \ref{lem:Qint 1st} and \ref{lem:concentrationofint:continuity}, and the details of the proof are left to the reader.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau63sharp:2}]
In the proof, we assume that we have $\mathcal{A}_4$.
Using \eqref{eq:ind2:Jbd} and the definition of $\tau_{3}(\alpha_1,t_1^-)$ (Section \ref{subsubsec:reg:Scontrol}), we express that
\begin{equation}\label{eq:ind2:tau63sh:2:main}
\begin{split}
\intop_{t_1^\flat\wedge \hat{\tau}}^{s\wedge\hat{\tau}} \left|J^{(\alpha_1)}_{t-s,t-u} \right| \left|S(u)-S_1(u) \right|du
\le
\intop_{t_1^\flat\wedge \hat{\tau}}^{s\wedge\hat{\tau}} \frac{Ce^{-c\alpha_0^2(t-s) }e^{-c\alpha_0^2(t-u)}}{\sqrt{(t-u+1)(t-s+1)}} \cdot\frac{\alpha_0^{1-\epsilon}}{\pi_1(u;S)+1}ds\\
= \frac{C\alpha_0^{1-\epsilon}e^{-c\alpha_0^2(t-s)}}{\sqrt{t-s+1}} \intop_{t_1^\flat\wedge \hat{\tau}}^{s\wedge\hat{\tau}} \frac{Ce^{-c\alpha_0^2(t-u)}}{\sqrt{t-u+1}} \cdot\frac{ds}{\pi_1(u;S)+1}.
\end{split}
\end{equation}
To simplify the RHS, we recall Lemma \ref{lem:ind1:pi1int basicbd}, with parameters $\Delta_0=\alpha_0^{-1}, \Delta_1= \alpha_0^{-1}\beta_0^{C_\circ}$ and $N_0 = \beta_0^5$. This gives
\begin{equation}
\intop_{t_1^\flat\wedge \hat{\tau}}^{s\wedge\hat{\tau}} \frac{Ce^{-c\alpha_0^2(t-u)}}{\sqrt{t-u+1}} \cdot\frac{ds}{\pi_1(u;S)+1} \le \beta_0^{C_\circ}.
\end{equation}
Note that although we need to set $K\ge \alpha_0^{-1}\beta_0^{10\theta}$ in Lemma \ref{lem:ind1:pi1int basicbd}, the contribution from $k\ge \alpha_0^{-1}\beta_0^2$ is negligible due to the extra exponential decay. This concludes the proof of Lemma \ref{lem:ind2:tau63sharp:2}.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau63sharp:4}]
Suppose that we are on the event $\mathcal{A}_4.$
Using the bound \eqref{eq:ind2:Jbd} and the definition of $\tau_{\textnormal{b}}''$ from \eqref{eq:def:tauxpprime}, we have
\begin{equation}
\intop_{t_1^\flat\wedge\hat{\tau}}^{s\wedge\hat{\tau}} \left| J^{(\alpha_1)}_{t-s,t-u} \right| |S_1(u)-\alpha_1| du
\le \frac{Ce^{-c\alpha_0^2(t-s)}}{\sqrt{t-s+1}}
\intop_{t_1^\flat\wedge\hat{\tau}}^{s\wedge\hat{\tau}}
\frac{ e^{-c\alpha_0^2(t-u)}}{\sqrt{t-u+1}} \left(\frac{\alpha_0\beta_0^{C_\circ}}{\sqrt{\pi_1(u;S)+1}} + \alpha_0^{\frac{3}{2}}\beta_0^\theta \right)ds.
\end{equation}
We first note that
\begin{equation}\label{eq:ind2:tau63sh:4:int1}
\intop_{t_1^\flat}^s \frac{\alpha_0^{\frac{3}{2}}\beta_0^\theta e^{-c\alpha_0^2(t-u)}}{\sqrt{t-u+1}} ds \le \alpha_0^{\frac{1}{2}}\beta_0^{\theta+1}.
\end{equation}
Furthermore, from Lemma \ref{lem:ind1:pi1int basicbd} with $\Delta_0 = \alpha_0^{-1}, \Delta_1= \alpha_0^{-1}\beta_0^{C_\circ}$, and $N_0 =\beta_0^5 $,
\begin{equation}\label{eq:ind2:tau63sh:4:int2}
\begin{split}
\intop_{t_1^\flat\wedge\hat{\tau}}^{s\wedge\hat{\tau}} \frac{\alpha_0\beta_0^{C_\circ}e^{-c\alpha_0^2(t-u)}du}{\sqrt{ (t-u+1)(\pi_1(u;S)+1) }}
\le
\alpha_0^{\frac{1}{2}} \beta_0^{2C_\circ}
.
\end{split}
\end{equation}
Note that the contribution from $k\ge \alpha_0^{-1}\beta_0^2$ is negligible similarly as in Lemma \ref{lem:ind2:tau63sharp:2}. Combining the two estimates, we obtain conclusion.
\end{proof}
\subsubsection{Proof of Lemma \ref{lem:ind2:tau61sharp}}\label{subsubsec:ind2:tau61sharp}
To finish the proof of Lemma \ref{lem:ind2:tau61sharp}, we define
\begin{equation}
Z(s,t):= \intop_{t_1^\flat}^s J^{(\alpha_1)}_{t-s,t-u} d\widehat{\Pi}_S(u),
\end{equation}
which we have studied in the previous subsection. Moreover, for $\hat{\tau}$ and $\tau^\sharp_{3,3}$ given in \eqref{eq:def:tauhat:ind2} and \eqref{eq:def:tau63sharp}, respectively, we set
\begin{equation}
\hat{\tau}' := \hat{\tau} \wedge \tau^\sharp_{3,3}.
\end{equation}
Splitting $d\widehat{\Pi}_S(s)$ into four parts as described in \eqref{eq:ind2:tausharp:tilPiSdecomposition} and recalling the definition of $\mathcal{A}'$ from \eqref{eq:def:ind2:Aprime}, we prove the following lemmas.
\begin{lem}\label{lem:ind2:tau61sharp:1}
Assume the setting of Theorem \ref{thm:ind2:main}. For each $t\in [t_1^\flat,t_1^\sharp]$, define the event
\begin{equation}
\mathcal{C}_1(t):= \left\{ \left| \intop_{t_1^\flat}^{t} Z(s,t) d\widetilde{\Pi}_S(s) \right|
\le \frac{ \beta_0^{2\theta+3C_\circ} }{ \pi_1(t;S)+1 }
\right\}.
\end{equation}
Then, we have
\begin{equation}
\mathbb{P}\left(\left. \bigcap_{t\in [t_1^\flat, t_1^\sharp] } \mathcal{C}_1(t) \ \right| \ \mathcal{A}' \right) \ge 1-\exp\left(-\beta_0^4 \right).
\end{equation}
\end{lem}
\begin{lem}\label{lem:ind2:tau61sharp:2}
Under the setting of Theorem \ref{thm:ind2:main}, suppose that we condition on $\mathcal{A}'$. Then, we have
\begin{equation} \left| \intop_{t_1^\flat\wedge\hat{\tau}'}^{t\wedge \hat{\tau}'} Z(s,t) (S(s)-S_1(s))ds \right|
\le \frac{ \alpha_0^{1-2\epsilon} }{ \pi_1(t;S)+1 } + \alpha_0^{\frac{3}{2}-2\epsilon}
,
\end{equation}
for all $t\in[t_1^\flat,t_1^\sharp]$
\end{lem}
\begin{lem}\label{lem:ind2:tau61sharp:4}
Under the setting of Theorem \ref{thm:ind2:main}, suppose that we condition on $\mathcal{A}'$. Then, we have
\begin{equation}
\left| \intop_{t_1^\flat\wedge\hat{\tau}'}^{t\wedge \hat{\tau}'} Z(s,t) (S_1(s)-\alpha_1)ds \right|
\le \alpha_0\beta_0^{2\theta+5}
,
\end{equation}
for all $t\in [t_1^\flat, t_1^\sharp].$
\end{lem}
Proofs of Lemmas \ref{lem:ind2:tau61sharp:1}--\ref{lem:ind2:tau61sharp:4} are similar to those of Lemmas \ref{lem:ind2:tau63sharp:1}--\ref{lem:ind2:tau63sharp:4}. Before delving into those, we first conclude the proof of Lemma \ref{lem:ind2:tau61sharp}.
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau61sharp}]
As in the proof of Lemma \ref{lem:ind2:tau63sharp}, \eqref{eq:ind2:tau63sh:aux1} and \eqref{eq:ind2:tau63sh:aux2} tell us that along with the event $\{\tilde{\tau}\wedge\tau'\ge \hat{t}_0 \}$, conditioning on $\mathcal{A}'$ or on $\mathcal{A}(\alpha_1,t_1^-)$ instead of $\mathcal{A}(\alpha_0,t_0)$ produces only a small additional error probability.
Thus, combining Lemmas \ref{lem:ind2:tau61sharp:1}--\ref{lem:ind2:tau61sharp:4}, we have with probability at least $1-3e^{-\beta_0^4}$ conditioned on $\mathcal{A}(\alpha_0,t_0)$ that
\begin{equation}
\begin{split}
\left| \intop_{t_1^\flat\wedge \hat{\tau}'}^{t\wedge \hat{\tau}'} Z(s,t) d\widehat{\Pi}_S(s) \right| \le \frac{2\beta_0^{2\theta +3C_\circ}}{\pi_1(t;S)+1}+2\alpha_0\beta_0^{2\theta+5}
\le
\frac{\beta_1^{2\theta+4C_\circ}}{\pi_1(t;S)+1},
\end{split}
\end{equation}
where the last inequality is from $ \pi_1(t;S) \le \alpha_0^{-1}\beta_0^{C_\circ}$ (see the definition of $\tau_{8}$ in Section \ref{subsubsec:reg:aggsize}). Then, we obtain the conclusion using Theorem \ref{thm:ind2:main}-(2) (proven in Section \ref{subsec:ind2:ratechange}) and Lemma \ref{lem:ind2:tau63sharp}.
\end{proof}
In the remaining of this subsection, we prove Lemmas \ref{lem:ind2:tau61sharp:1}--\ref{lem:ind2:tau61sharp:4}.
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau61sharp:1}]
The proof goes analogously as Lemma \ref{lem:ind2:tau63sharp:1}, by splitting the integral into two parts, from $t-\alpha_0^{-1}$ to $t$ and $t_1^\flat$ to $t-\alpha_0^{-1}$, which is to apply Lemma \ref{lem:concen of int:conti:forwardtime}. From the definition of $\tau_{{3,3}}^\sharp$, we see that for all $t\le \hat{\tau}'$,
\begin{equation}\label{eq:ind2:tau61sh:1:int}
\begin{split}
\left| \intop_{t-\alpha_0^{-1} }^{t} Z(s,t) d\widetilde{\Pi}_S(s)\right|
&\le
\frac{\beta_0^{2C_\circ}}{\pi_1(t;S)+1}+ \frac{\alpha_0^{\frac{1}{2}} \beta_0^{\theta+C_\circ+3}}{\sqrt{\pi_1(t ;S)+1}} +\alpha_0\beta_0^{\theta+C_\circ+4}
\\&\le
\frac{\beta_0^{\theta+2C_\circ}}{\pi_1(t ;S)+1},
\end{split}
\end{equation}
where the second inequality follows from $\pi_1(t\wedge \hat{\tau}' ;S)\le 2\alpha_0^{-1}\beta_0^{C_\circ}$.
The other integral can be estimated using Lemma \ref{lem:concen of int:conti:forwardtime}, by observing that
\begin{equation}
\begin{split}
& \intop_{t_1^\flat\wedge\hat{\tau}'}^{(t-\alpha_0^{-1})\wedge \hat{\tau}} Z(s,t)^2 S(s)ds \le \intop_{t_1^\flat}^{t} \left(\frac{2\beta_0^{2C_\circ+2}}{(t-s+1)^2} +\frac{2\alpha_0\beta_0^{2\theta+6} }{t-s+1} \right) 2\alpha_0 \beta_0^{C_\circ} ds \le \alpha_0^2 \beta_0^{2\theta+2C_\circ},\\
&\qquad \qquad \qquad \sup \left\{ |Z(s,t)|: s\in[t_1^{\flat}\wedge \hat{\tau}', (t-\alpha_0^{-1})\wedge \hat{\tau}'] \right\} \le 2\alpha_0 \beta_0^{\theta +3}.
\end{split}
\end{equation}
Thus, applying the lemma gives that
\begin{equation}
\mathbb{P} \left( \left.\left| \intop_{t_1^\flat\wedge\hat{\tau}'}^{(t-\alpha_0^{-1})\wedge \hat{\tau}'} Z(s,t)d\widehat{\Pi}_S(s) \right| \le \alpha_0\beta_0^{\theta+2C_\circ} , \ \forall t\in [t_1^\flat, t_1^\sharp ] \right| \ \mathcal{A}' \right) \ge 1- \exp\left(-\beta_0^{5} \right),
\end{equation}
and hence combining with \eqref{eq:ind2:tau61sh:1:int} concludes the proof, since $\hat{\tau}'\ge t_1^\sharp$ with high probability (Proposition \ref{prop:ind2:prereg}, equation \eqref{eq:ind2:tauxpprime} and Lemma \ref{lem:ind2:tau63sharp}).
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau61sharp:2}]
Recalling the definition of $\tau_{3}(\alpha_1,t_1^-)$ (Section \ref{subsubsec:reg:Scontrol}), we have
\begin{equation}\label{eq:ind2:tau61sh:2:int}
\begin{split}
\intop_{t_1^\flat\wedge\hat{\tau}'}^{t\wedge \hat{\tau}'} |Z(s,t) (S(s)-S_1(s))| ds
& \le
\intop_{t_1^\flat}^{t\wedge \hat{\tau}'} \left(\frac{\beta_0^{C_\circ+1}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{\theta+3} }{\sqrt{t-s+1}} \right) \frac{\alpha_0^{1-\epsilon}e^{-c\alpha_0^2(t-s)}}{\pi_1(s;S)+1}ds.
\end{split}
\end{equation}
Then, applying Lemma \ref{lem:ind1:pi1int basicbd} with $\Delta_0 = \alpha_0^{-1}, \Delta_1 = \alpha_0^{-1}\beta_0^{C_\circ}$, and $N_0 = \beta_0^5$ bounds the RHS by
$$\frac{\alpha_0^{1-2\epsilon}}{\pi_1(s;S)+1} + \alpha_0^{\frac{3}{2}-2\epsilon}, $$
recalling that the contribution from $k\ge \alpha_0^{-1}\beta_0^2$ is negligible similarly as in Lemma \ref{lem:ind2:tau63sharp:2}.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:ind2:tau61sharp:4}]
Having the definition of $\tau_{\textnormal{b}}''$ \eqref{eq:def:tauxpprime} in mind, we express that
\begin{equation}\label{eq:ind2:tau61sh:4:int}
\begin{split}
& \intop_{t_1^\flat\wedge \hat{\tau}'}^{t\wedge \hat{\tau}'}
\left|Z(s,t) (S_1(s)-\alpha_1) \right|ds\\
& \le
\intop_{t_1^\flat\wedge\hat{\tau}'}^{t\wedge \hat{\tau}'} e^{-c\alpha_0^2(t-s)} \left( \frac{\beta_0^{C_\circ+1}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{\theta+3}}{\sqrt{t-s+1}} \right) \left( \frac{\alpha_0 \beta_0^{C_\circ+1}}{\sqrt{\pi_1(u;S)+1}} + 2\alpha_0^{\frac{3}{2}}\beta_0^\theta \right)ds.
\end{split}
\end{equation}
We begin with observing that
\begin{equation}\label{eq:ind2:tau61sh:4:int2}
\intop_{t_1^\flat\wedge\hat{\tau}'}^{t\wedge \hat{\tau}'} e^{-c\alpha_0^2(t-s)} \left( \frac{\beta_0^{C_\circ+1}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{\theta+3}}{\sqrt{t-s+1}} \right) 2\alpha_0^{\frac{3}{2}}\beta_0^\theta ds
\le \alpha_0^{\frac{3}{2}} \beta_0^{\theta +C_\circ+3} + \alpha_0 \beta_0^{2\theta +4}.
\end{equation}
On the other hand, the remaining integral can be bounded using Lemma \ref{lem:ind1:pi1int basicbd} as in the proof of Lemma \ref{lem:ind2:tau61sharp:2}. This gives
\begin{equation}\label{eq:ind2:tau61sh:4:int3}
\begin{split}
\intop_{t_1^\flat\wedge\hat{\tau}'}^{t\wedge \hat{\tau}'} e^{-c\alpha_0^2(t-s)} \left( \frac{\beta_0^{C_\circ+1}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{\theta+3}}{\sqrt{t-s+1}} \right) \frac{\alpha_0 \beta_0^{C_\circ+1}ds}{\sqrt{\pi_1(s;S)+1}}\le \alpha_0 \beta_0^{2\theta},
\end{split}
\end{equation}
concluding the proof.
\end{proof}
\subsubsection{Some consequences of the integral calculation}\label{subsubsec:ind2:tausharpconseq}
Before moving on, we record several direct consequences of the integral calculations we have seen above. These results will be useful later in Section \ref{sec:double int} when we encounter similar formulas under a slightly different setting.
We begin with stating the analogue of Lemmas \ref{lem:ind2:tau63sharp:1} and \ref{lem:ind2:tau63sharp:2}, noting that the previous analysis works analogously in another interval instead of $[t_1^\flat, t_1^\sharp]$. In this subsection, $S_1(t)$ is defined to be $S_1(t)=S_1(t;t_0^-,\alpha_0).$
\begin{cor}\label{cor:ind2:tau63sharp:1}
Let $\alpha_0,t_0,r>0$, set $t_0^-, \acute{t}_0, \hat{t}_0$ as \eqref{eq:def:t0t1}, \eqref{eq:def:t0t11}, and let
\begin{equation}
t_0^\dagger := t_0 + \frac{5}{2}\alpha_0^{-2}\beta_0^{\theta} = \acute{t}_0 - \frac{1}{2}\alpha_0^{-2}\beta_0^\theta.
\end{equation}
Define the stopping times
\begin{equation}
\begin{split}
\tilde{\tau}_{3,3}^{(1)}&:= \inf \left\{
t\ge \acute{t}_0: \, \exists s\in[t_0^\dagger,t] \textnormal{ s.t. } \left|
\intop_{t_0^\dagger}^s J^{(\alpha_0)}_{t-s,t-u} d\widetilde{\Pi}_S(u)
\right|
\ge
\left(\frac{\beta_0^{C_\circ+1}}{t-s+1} + \frac{\alpha_0^{\frac{1}{2}}\beta_0^{2C_\circ}}{\sqrt{t-s+1}} \right) e^{-c\alpha_0^2(t-s)}
\right\}
;\\
\tilde{\tau}_{3,3}^{(2)}&:=
\inf \left\{
t\ge \acute{t}_0: \, \exists s\in[t_0^\dagger,t] \textnormal{ s.t. } \left|
\intop_{t_0^\dagger}^s J^{(\alpha_0)}_{t-s,t-u} (S(u)-S_1(u))du
\right|
\ge
\frac{\alpha_0^{1-\epsilon}\beta_0^{C_\circ} }{\sqrt{t-s+1}} e^{-c\alpha_0^2(t-s)}
\right\}
.
\end{split}
\end{equation}
Suppose that $\Pi_S(-\infty,t_0] \in \mathfrak{R}(\alpha_0, r; [t_0])$. Then, we have
\begin{equation}
\mathbb{P} \left(\left.\tilde{\tau}_{3,3}^{(1)}\wedge \tilde{\tau}_{3,3}^{(2)} <\hat{t}_0 \,\right|\,\mathcal{F}_{t_0} \right) \le \exp\left(-\beta_0^3 \right) +r.
\end{equation}
\end{cor}
To be precise, we remark that the conclusion can be obtained from combining Theorem \ref{thm:reg:conti:main} and the argument from Lemma \ref{lem:ind2:tau63sharp}.
We also derive a similar estimate on the integral of $K^*_{\alpha_0}$ instead of $J_{t-s,t-u}^{(\alpha_0)}.$
\begin{cor}\label{cor:ind2:tau63sharp:Kstar}
Under the setting of Corollary \ref{cor:ind2:tau63sharp:1}, define the stopping time
\begin{equation}
\tilde{\tau}_{3,3}^{(3)} :=
\inf \left\{
t\ge \acute{t}_0: \, \exists s\in[t_0^+,t] \textnormal{ s.t. } \left|\intop_{t_0^+}^s K^*_{\alpha_0}(s-u) (S(u)-S_1(u)) \right| \ge \alpha_0^{2-2\epsilon}
\right\}.
\end{equation}
Then, we have
\begin{equation}
\mathbb{P} \left(\left.\tilde{\tau}_{3,3}^{(3)} <\hat{t}_0\, \right|\, \mathcal{F}_{t_0} \right) \le \exp\left(-\beta_0^3 \right)+r.
\end{equation}
\end{cor}
\begin{proof}
The conclusion can be obtained from the same method as Lemma \ref{lem:ind2:tau63sharp:2}, since $K^*_{\alpha_0} (x) $ satisfies the estimate from Lemma \ref{lem:estimat for K tilde:intro} which is very similar to \ref{eq:ind2:Jbd}.
\end{proof}
The following analogue of Lemma \ref{lem:ind2:tau63sharp} can be obtained similarly.
\begin{cor}\label{cor:ind2:tau63sharp}
Under the setting of Corollary \ref{cor:ind2:tau63sharp:1}, define the stopping time
\begin{equation}\label{eq:def:tau63sharptil}
\tilde{\tau}_{{3,3}}^\sharp:= \inf\left\{ t\ge \acute{t}_0: \,\exists s\in[t_0^\dagger,t] \textnormal{ s.t. } \left|\intop_{t_0^\dagger}^s J^{(\alpha_0)}_{t-s,t-u} d\widehat{ \Pi}_S(u) \right| \ge \frac{\alpha_0^{\frac{1}{2}}\beta_0^{6\theta }}{\sqrt{t-s+1}} + \frac{\beta_0^{C_\circ +1} }{t-s+1} \right\}.
\end{equation}
Then, we have
\begin{equation}
\mathbb{P}\left(\left.\tilde{\tau}_{{3,3}}^\sharp < \hat{t}_0\ \right| \ \mathcal{F}_{t_0} \right) \le 3\exp\left(-\beta_0^2 \right) +r.
\end{equation}
\end{cor}
Note that in the term $\frac{\alpha_0^{1/2}\beta_0^{6\theta} }{\sqrt{t-s+1}}$, we have a larger exponent $6\theta$ than in $\tau_{3,3}^\sharp$, since we deal with a longer interval $[t_0, \hat{t}_0]$.
The other lemmas in the previous subsections can be extended similarly. Letting
$$Z_0(s,t):=\intop_{t_0^\dagger}^s J^{(\alpha_0)}_{t-s,t-u} d\widehat{\Pi}_S(u),$$
we have the analogue of Lemma \ref{lem:ind2:tau61sharp:2}, which can be deduced from using Corollary \ref{cor:ind2:tau63sharp} instead of Lemma \ref{lem:ind2:tau63sharp}.
\begin{cor}\label{cor:ind2:tau61sharp1}
Under the setting of Corollary \ref{cor:ind2:tau63sharp:1}, define the stopping time
\begin{equation}
\begin{split}
\tilde{\tau}_{3,1}^{(1)} := \inf \left\{
t\ge \acute{t}_0: \, \left|
\intop_{t_0^\dagger}^t Z_0(s,t) (S(s)-S_1(s))ds
\right|
\ge \frac{\alpha_0^{1-2\epsilon}}{\pi_1(t;S)+1} + \alpha_0^{\frac{3}{2}-2\epsilon}
\right\}
.
\end{split}
\end{equation}
Then, we have
\begin{equation}
\mathbb{P} \left(\left. \tilde{\tau}_{3,1}^{(1)} <\hat{t}_0 \, \right| \, \mathcal{F}_{t_0} \right) \le \exp\left( - \beta_0^3\right)+r.
\end{equation}
\end{cor}
Due to their similarity, proofs of Corollaries \ref{cor:ind2:tau63sharp:1}, \ref{cor:ind2:tau63sharp} and \ref{cor:ind2:tau61sharp1} are omitted.
\subsection{Proof of Theorem \ref{thm:induction:main}}\label{subsec:ind2:fin}
In this final section, we present the proof of Theorem \ref{thm:induction:main}. Throughout the proof, we assume that $\mathcal{F}_{t_0} \equiv \Pi_S(-\infty,t_0]$ is $(\alpha_0,r;[t_0])$-regular (see Definition \ref{def:reg}).
We begin with verifying the second item of Theorem \ref{thm:induction:main}. Let $\tau := \tau(\alpha_0,t_0,2)$ \eqref{eq:def:tau:induction}, ${\tau}^+:= {\tau}^+(\alpha_0,t_0,1/2)$ \eqref{eq:def:ind1:tauacute:basic}, and $\tau' := {\tau}^+\wedge \tau$. Observe that
\begin{equation}\label{eq:ind2:fin:main1}
\begin{split}
\mathbb{P} \left(\left.|\alpha_1-\alpha_0| \ge 2\alpha_0^{3/2}\beta_0^{6\theta} \ \right| \ \mathcal{F}_{t_0} \right)& \le
\mathbb{P} \left(\tau \le \acute{t}_0 \ | \ \mathcal{F}_{t_0} \right) + \mathbb{P} \left( \tau' < \hat{t}_0, \ \tau >\acute{t}_0 \ | \ \mathcal{F}_{t_0} \right)\\
&\quad +\mathbb{P}\left( |\alpha_1-\alpha_0| \ge 2\alpha_0^{3/2}\beta_0^{6\theta},\ \tau' \ge \hat{t}_0\ | \ \mathcal{F}_{t_0} \right)\\
&\le r+ \exp \left(-\beta_0^{3/2} \right),
\end{split}
\end{equation}
where the last line follows from the definition of regularity, Theorem \ref{thm:reg:conti:main} and Lemma \ref{lem:ind2:alpha1vsalpha0}.
To study the event $\{\mathcal{F}_{t_1} \textnormal{ is } (\alpha_1,e^{-\beta_1^{3/2}}; [t_1])\textnormal{-regular} \}$, we verify the two conditions of regularity separately. The second condition follows from Theorems \ref{thm:reg:conti:main} and \ref{thm:ind2:main}-(2). Namely, we write
\begin{equation}\label{eq:ind2:fin:main2}
\begin{split}
\mathbb{P} \left( \mathcal{A}{(\alpha_1,t_1)}^c \ | \ \mathcal{F}_{t_0} \right)
&\le
\mathbb{P} \left( \tau' <\hat{t}_0 \ | \ \mathcal{F}_{t_0} \right) +\exp \left(-\beta_0^2 \right)
\\
&\le
\mathbb{P} \left( \tau' < \hat{t}_0,\ \tau>\acute{t}_0\ | \ \mathcal{F}_{t_0} \right) + \mathbb{P} \left(\tau\le t_0^+ \ | \ \mathcal{F}_{t_0} \right)\\
&\le \exp\left(-\beta_0^2 \right)+ r+ \exp\left(-\beta_0^2\right) \le r+ \exp\left(-\beta_0^{3/2} \right).
\end{split}
\end{equation}
To verify the first condition of regularity, recall the definition of $\mathcal{A}_4$ \eqref{eq:def:A4}, and $t_1^\sharp = t_1+4\alpha_0^{-2}\beta_0^\theta$ which is larger than $\acute{t}_1=t_1+2\alpha_1^{-2}\beta_1^\theta$ on $\mathcal{A}_4$. Observe that
\begin{equation}\label{eq:ind2:fin:1}
\begin{split}
\mathbb{P} \left(\left. \mathbb{P}\left(\left.\tau^\sharp(\alpha_1,t_1,2) \le \acute{t}_1 \ \right| \ \mathcal{F}_{t_1} \right) > e^{-\beta_1^{3/2}}, \ \tau > \acute{t}_0, \ \mathcal{A}_4 \ \right| \ \mathcal{F}_{t_0} \right)\\
\le e^{2\beta_0^{3/2}} \mathbb{P} \left(\tau^\sharp (\alpha_1,t_1,2) \le t_1^\sharp,\ \tau>\acute{t}_0\ | \ \mathcal{F}_{t_0} \right),
\end{split}
\end{equation}
by Markov's inequality. Note that $\mathcal{A}_4$ ensures $\beta_1^{3/2}\le 2\beta_0^{3/2}$. Furthermore, since $$t_1^\sharp= t_1 + 4\alpha_0^{-2}\beta_0^\theta \le t_0 +\alpha_0^{-2}\beta^{10\theta} +4\alpha_0^{-2}\beta_0^\theta< t_0+2\alpha_0^{-2}\beta^{10\theta}= \hat{t}_0,$$
combining Theorem \ref{thm:reg:conti:main}, Proposition \ref{prop:ind2:prereg}, and Lemma \ref{lem:ind2:tau6sharp} gives that
\begin{equation}\label{eq:ind2:fin:2}
\mathbb{P} \left(\tau^\sharp (\alpha_1,t_1,2) \le t_1^\sharp,\ \tau>\acute{t}_0\ | \ \mathcal{F}_{t_0} \right) \le 3\exp\left(-\beta_0^2 \right).
\end{equation}
We also know from Theorem \ref{thm:reg:conti:main} and Lemma \ref{lem:ind2:alpha1vsalpha0} that
\begin{equation}\label{eq:ind2:fin:3}
\mathbb{P} \left(\mathcal{A}_4^c, \tau>\acute{t}_0 \ | \ \mathcal{F}_{t_0} \right) \le 2\exp\left(-\beta_0^2 \right).
\end{equation}
Thus, \eqref{eq:ind2:fin:1}, \eqref{eq:ind2:fin:2} and \eqref{eq:ind2:fin:3} tell us that
\begin{equation}\label{eq:ind2:fin:main3}
\begin{split}
\mathbb{P} \left(\left. \mathbb{P}\left(\left.\tau^\sharp(\alpha_1,t_1,2) \le \acute{t}_1 \ \right| \ \mathcal{F}_{t_1} \right) > e^{-\beta_1^{3/2}} \ \right| \ \mathcal{F}_{t_0} \right)&\le 3e^{2\beta_0^{3/2}-\beta_0^2} + 2e^{-\beta_0^2}+ r \\ &\le r+ e^{-\beta_0^{3/2}}.
\end{split}
\end{equation}
Thus, we conclude the proof of Theorem \ref{thm:induction:main} from \eqref{eq:ind2:fin:main1}, \eqref{eq:ind2:fin:main2} and \eqref{eq:ind2:fin:main3}. \qed
\section{The second order contributions and the moments of the increment}\label{sec:double int}
Building upon the analysis of regularity, the goal of this section is establishing the main estimate needed in deriving the scaling limit of the speed. We aim to state and proving the formal version of Theorem \ref{thm:speed:increment:informal}, which will be done in Section \ref{subsec:increment proof}. To this end, an essential step is to compute the expectation of the double integral term in $S_2(t)$ \eqref{eq:def:S2:basic form}, namely,
\begin{equation}
\mathcal{J}(t)=\mathcal{J}(t;\alpha,t_0^-):= \frac{\alpha}{1+2\alpha}\intop_{t_0^-}^{t} \intop_{t_0^-}^s J^{(\alpha)}_{t-s,t-u} d\widehat{\Pi}_S(u) d\widehat{\Pi}_S(v).
\end{equation}
The following theorem gives an appropriate control on the mean of $\mathcal{J}(t)$, and it is established in Section \ref{subsec:increment:double int}.
\begin{thm}\label{thm:double int:main}
Let $\alpha, t_0>0$, set $t_0^-, t_0^+, \acute{t}_0$ and $\hat{t}_0$ as \eqref{eq:def:t0t1} and \eqref{eq:def:t0t11}, and let $r=e^{-\beta^{3/2}}$ for $\beta:=\log(1/\alpha)$. Suppose that $\Pi_S(-\infty,t_0]$ is $(\alpha,r;[t_0])$-regular. Then, for all $t\in [\acute{t}_0,\hat{t}_0]$, we have
\begin{equation}
\mathbb{E}[\mathcal{J}(t)\, | \, \Pi_S(-\infty,t_0]] = 2\alpha^2 \left(1+ o(\alpha^{\frac{1}{5}}) \right).
\end{equation}
\end{thm}
We note that the error bound $\alpha^{\frac{1}{5}}$ is not essential: something better than $O(\beta^{-2})$ will suffice for our purpose.
\subsection{The contribution of the double integral}\label{subsec:increment:double int}
From the analysis in the previous sections, we can deduce that $\mathcal{J}(t)$ is typically of order $\alpha^{2-\epsilon}$. Then, the main difficulty in establishing Theorem \ref{thm:double int:main} is narrowing down its size to $2\alpha^2$ plus a smaller order error. Here, we require more refined tools to study the mean of $\mathcal{J}(t)$ accurately.
Let $t_0^-, t_0^+, \acute{t}_0$ and $\hat{t}_0$ be as in Theorem \ref{thm:double int:main}, we introduce another parameter $t_0^\dagger$ defined as
\begin{equation}
t_0^\dagger := t_0 + \frac{5}{2}\alpha^{-2}\beta^\theta.
\end{equation}
An important relation we stress here is that
\begin{equation}
(\acute{t}_0 - t_0^\dagger) \wedge (t_0^\dagger - t_0^+) \ge \frac{1}{2}\alpha^{-2}\beta^\theta.
\end{equation}
Since $\Pi_S(-\infty,t_0]$ is regular and we consider $t\ge \acute{t}_0$, we can ignore the contributions to the integral from the regime $[t_0^-,t_0^\dagger]$, analogously as we have seen in \eqref{eq:ind2:tau6sh:switch base}. Namely, with probability $1- e^{-\beta^5},$
\begin{equation}
\left|\mathcal{J}(t) - \intop_{t_0^\dagger}^t\intop_{t_0^\dagger}^s J_{t-s,t-u}^{(\alpha)} d\widehat{\Pi}_S(u )d\widehat{\Pi}_S(s) \right| \le \alpha_0^{50}, \quad \textnormal{for all } t\in[\acute{t}_0,\hat{t}_0].
\end{equation}
Hence, from now we will be interested in investigating the double integral starting from $t_0^\dagger$. The reason for our choice of such $t_0^\dagger$ is explained in Remark \ref{rmk:doubleint:dagger}. Controlling the contribution from the error event where the above does not hold will be discussed in the proof of Lemma \ref{lem:doubleint:out} below, and it will also be bounded by $O(\alpha^{50})$.
Writing $\mathcal{F}_t :=\Pi_S(-\infty,t_0]$ as before, we begin with observing that
\begin{equation}\label{eq:doubleint:mg:basic}
\mathbb{E} [\mathcal{J}(t) \, |\, \mathcal{F}_{t_0} ] = \frac{\alpha}{1+2\alpha} \mathbb{E} \left[\left. \intop_{t_0^{\dagger}}^t \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widehat{\Pi}_S(u) (S(s)-\alpha)ds \, \right| \, \mathcal{F}_{t_0} \right] + O(\alpha^{50}),
\end{equation}
since the outer integral with respect to $d\widetilde{\Pi}_S(s) = d\Pi_S(s) - S(s)ds$ is a martingale and thus has mean zero. Suppose that we can switch $(S(s)-\alpha)ds$ into $(S_1(s)-\alpha)ds$ with a negligible error, recalling the definition $S_1(s) = S_1(s;t_0^-,\alpha) = \mathcal{R}_c(s,s;\Pi_S[t_0^-,s],\alpha)$. Then, from \eqref{eq:integralform:branching}, we can write
\begin{equation}\label{eq:doubleint:S1minalpha:basic}
(S_1(s)-\alpha) = \intop_{t_0^+}^s K^*_\alpha(s-x) (d\Pi_S(x)-S_1(x)dx) +
(\mathcal{R}_c(t_0^+,s;\Pi_S[t_0^-,t_0^+],\alpha)-\alpha).
\end{equation}
\begin{remark}\label{rmk:doubleint:dagger}
Note that this integral starts from $t_0^+$, not $t_0^\dagger$, in order to keep our control on $\mathcal{R}_c(t_0^+,s;\Pi_S[t_0^-,t_0^+],\alpha)$ for all $s\in [t_0^\dagger,\hat{t}_0]$ (see \eqref{eq:doubleint:errorbasic} below). The integral \eqref{eq:doubleint:S1minalpha:basic} needs to be from $t_0^+$, not $t_0$, since we have control on $|S_1(u)-\alpha| $ only on $t\ge t_0^+$ (Lemma \ref{lem:ind1:taux}).
\end{remark}
From \eqref{eq:doubleint:S1minalpha:basic}, we again attempt to approximate $(S_1(s)-\alpha)ds$ by
\begin{equation}
(S_1(s)-\alpha)ds \approx \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_\alpha (x) + (\alpha''-\alpha)ds,
\end{equation}
with $\alpha'':= \mathcal{L}(t_0^+;\Pi_S[t_0^-,t_0^+],\alpha)$.
If we make these two steps of approximations rigorous and repeat a similar procedure to the inner integral, then we arrive at the following proposition.
\begin{prop}\label{prop:double int:main approx}
Define the integrals $I_1$ and $I_2$ by
\begin{equation}
\begin{split}
I_1(t)&:= \intop_{t_0^{\dagger}}^t \left[ \intop_{t_0^\dagger}^s J_{t-s,t-u}^{(\alpha)} d\widetilde{\Pi}_\alpha(u) \cdot \intop_{t_0^+}^s K^*_\alpha(s-u) d\widetilde{\Pi}_\alpha(u) \right] ds;
\\
I_2(t)&:= \intop_{t_0^\dagger}^t \left[ \intop_{t_0^\dagger}^s \intop_{t_0^+}^u J^{(\alpha)}_{t-s,t-u} K^*_\alpha(u-v) d\widetilde{\Pi}_\alpha(v) du \cdot \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_\alpha(x) \right]ds.
\end{split}
\end{equation}
Then, under the setting of Theorem \ref{thm:double int:main}, for all $\acute{t}_0\le t\le \hat{t}_0$ we have
\begin{equation}
\begin{split}
\mathbb{E}[\mathcal{J}(t)\,| \, \mathcal{F}_{t_0}] =
\frac{\alpha}{1+2 \alpha} \mathbb{E}\left[I_1(t)+I_2(t) \right]+ O\left(\alpha^{2+\frac{1}{5}} \right).
\end{split}
\end{equation}
Note that the integrals $I_1$ and $I_2$ are independent of $\mathcal{F}_{t_0}$.
\end{prop}
Before establishing the proposition rigorously, we first deduce Theorem \ref{thm:double int:main} from it.
\begin{proof}[Proof of Theorem \ref{thm:double int:main}]
Observe that for any deterministic functions $f,g$ and any numbers $a_1\le a_2$, $b_1\le b_2$, we have
\begin{equation}
\mathbb{E}\left[\intop_{a_1}^{a_2} f(x)d\widetilde{\Pi}_\alpha(x) \intop_{b_1}^{b_2} g(y)d\widetilde{\Pi}_\alpha (y) \right] = \alpha \intop_{a_1\vee b_1}^{a_2\wedge b_2} f(x)g(x ) dx,
\end{equation}
since the only nontrivial correlation comes from the cases when both $x$ and $y$ are at the same point in the point process. Thus, we obtain that
\begin{equation}
\begin{split}
\mathbb{E}[I_1(t)] &=
\alpha \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J_{t-s,t-u}^{(\alpha)} K_\alpha^*(s-u) duds = \alpha \intop_0^{t-t_0^\dagger} \intop_0^s J^{(\alpha)}_{u,s} K^*_\alpha(s-u) duds.
\end{split}
\end{equation}
Similarly, we can see that
\begin{equation}
\begin{split}
\mathbb{E}[I_2(t)] &= \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s\mathbb{E}\left[\intop_{t_0^+}^u J_{t-s,t-u}^{(\alpha)} K^*_\alpha(u-v) d\widetilde{\Pi}_\alpha(v) \intop_{t_0^+}^s K_\alpha^*(s-x) d\widetilde{\Pi}_\alpha(x) \right]duds \\
&=
\alpha \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s \intop_{t_0^+}^u J_{t-s,t-u}^{(\alpha)} K^*_\alpha(u-v) K^*_\alpha(s-v)dv du ds \\
&=\alpha \intop_0^{t-t_0^\dagger} \intop_0^s \intop_0^u J_{v,u}^{(\alpha)} K_\alpha^*(s-u) K^*_\alpha (s-v) dv du ds + O(\alpha^{100}),
\end{split}
\end{equation}
where in the third line, the integral over $[t_0^+,t_0^\dagger]$ of $dv$ has a negligible $O(\alpha^{100})$-order contribution due to the decay property of $J$, since $t\ge \acute{t}_0$.
The two deterministic integrals above are computed via a precise understanding of the quantities $K^*_\alpha$ and $J^{(\alpha)}$. This is done using analytical methods based on Fourier analysis, and can be found in Lemma \ref{lem:integral computation} in the appendix.
Applying the results from the lemma concludes the proof.
\end{proof}
The rest of the subsection is devoted to the proof of Proposition \ref{prop:double int:main approx}. The analysis is highly technical, and relies on a similar approach as that in Section \ref{subsec:ind2:bootstrappedJ} based on the decomposition
\begin{equation}
d\widehat{\Pi}_S(x) = d\widetilde{\Pi}_S(x ) + (S(x)-S_1(x)) dx + (S_1(x)- \alpha) dx.
\end{equation}
($d\widetilde{\Pi}_S(x) := d\Pi_S (x)-S(x)dx$ as before.)
\subsubsection{Reshaping the outer integral}
We begin with modifying the outer integral of $\mathcal{J}(t)$. Since it suffices to study the integral in the RHS of \eqref{eq:doubleint:mg:basic},
define
\begin{equation}
\begin{split}
&I_{\textnormal{out}}^{(1)}(t):=\intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J_{t-s,t-u}^{(\alpha)} d\widehat{\Pi}_S(u) (S(s)-S_1(s))ds;\\
&\tau_{\textnormal{out}}^{(1)}: = \inf \left\{
t\ge \acute{t}_0: \, \left| I_{\textnormal{out}}^{(1)}(t) \right| \ge \alpha^{\frac{3}{2}-\epsilon} + \frac{\alpha^{1-\epsilon}}{\pi_1(t;S)+1}
\right\}.
\end{split}
\end{equation}
Moreover, recalling \eqref{eq:doubleint:S1minalpha:basic}, the remaining outer integral is decomposed by the following formula:
\begin{equation}\label{eq:doubleint:S1minusalpha decomp}
\begin{split}
S_1(s)-\alpha = & \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_\alpha(x) + \intop_{t_0^+}^s K^*_\alpha ( s-x) d\widetilde{\Pi}_{S-\alpha} (x) \\
&+ \intop_{t_0^+}^s K^*_\alpha(s-x) (S(x)-S_1(x))dx + \big[\mathcal{R}_c(t_0^+,s;\Pi_S[t_0^-,t_0^+],\alpha) -\alpha'' \big] + \big[\alpha''-\alpha \big],
\end{split}
\end{equation}
where we defined $d\widetilde{\Pi}_{S-\alpha}(x) := d\Pi_S(x)-d\Pi_\alpha(x) -(S(x)-\alpha)dx$ and $\alpha'' = \mathcal{L}(t_0^+;\Pi_S[t_0^-,t_0^+],\alpha)$. We show that the contributions to the double integral coming from the terms other than the first and the last in the RHS are negligible. To this end, recall the notation $\Pi_{S\triangle \alpha}$ \eqref{eq:def:Strianglealpha} and define
\begin{equation}
\begin{split}
& I_{\textnormal{out}}^{(2)}(t):= \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J_{t-s,t-u}^{(\alpha)} d\widehat{\Pi}_S(u) \intop_{t_0^+}^s K^*_\alpha (s-x)d\widetilde{\Pi}_{S-\alpha}(x) ds ;\\
&\tau_{\textnormal{out}}^{(2)} := \inf \left\{
t\ge \acute{t}_0: \left| I_{\textnormal{out}}^{(2)}(t) \right| \ge \alpha^{1-\epsilon} \sigma_1(t;S\triangle \alpha) + \alpha^{\frac{1}{4}-\epsilon} \sigma_1(t;S\triangle\alpha)^2+ \alpha^{\frac{5}{4}-\epsilon}
\right\};\\
& I_{\textnormal{out}}^{(3)}(t):=
\intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J_{t-s,t-u}^{(\alpha)} d\widehat{\Pi}_S(u) \intop_{t_0^+}^s K^*_\alpha (s-x) (S(x)-S_1(x))dx ds;\\
& \tau_{\textnormal{out}}^{(3)} := \inf \left\{
t\ge \acute{t}_0: \left| I_{\textnormal{out}}^{(3)}(t) \right| \ge \alpha^{\frac{3}{2}-3\epsilon}
\right\}.
\end{split}
\end{equation}
We remark that the contribution from the fourth term is negligible due to Lemma \ref{lem:ind2:alphavsalphaPP}, which tells us that with probability at least $1-e^{-\beta^5}$,
\begin{equation}\label{eq:doubleint:errorbasic}
\big|\mathcal{R}_c(t_0^+,s;\Pi_S[t_0^-,t_0^+],\alpha) - \alpha'' \big| \le \alpha^{100},
\end{equation}
for all $t\in [t_0^\dagger,\hat{t}_0]$.
\begin{lem}\label{lem:doubleint:out1}
Under the setting of Theorem \ref{thm:double int:main}, we have
\begin{equation}
\mathbb{P} \left(\left. \tau_{\textnormal{out}}^{(1)} \wedge \tau_{\textnormal{out}}^{(2)} \wedge \tau_{\textnormal{out}}^{(3)} < \hat{t}_0 \ \right| \, \mathcal{F}_{t_0} \right) \le \exp \left( -\beta^2 \right) +r.
\end{equation}
\end{lem}
From this lemma, deduce the conclusion of this subsection which can be written as follows.
\begin{lem}\label{lem:doubleint:out}
Define the integral $I_{\textnormal{out}}(t)$ as
\begin{equation}
I_{\textnormal{out}}(t):= \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widehat{\Pi}_S(u) \cdot\left\{\intop_{t_0^+}^s K^*_\alpha(s-u) d\widetilde{\Pi}_\alpha(u) + \big[\alpha''-\alpha \big] \right\} ds.
\end{equation}
Under the setting of Theorem \ref{thm:double int:main}, we have for all $t\in[\acute{t}_0,\hat{t}_0]$ that
\begin{equation}
\mathbb{E}[\mathcal{J}(t) \, | \, \mathcal{F}_{t_0}] = \frac{\alpha}{1+2\alpha} \mathbb{E} [I_{\textnormal{out}}(t) \,|\, \mathcal{F}_{t_0} ] + o (\alpha^{2+\frac{1}{5}} ).
\end{equation}
\end{lem}
\begin{proof}
Let $\tau_{\textnormal{out}} := \tau_{\textnormal{out}}^{(1)}\wedge \tau_{\textnormal{out}}^{(2)}\wedge \tau_{\textnormal{out}}^{(3)}$.
We first show that
\begin{equation}\label{eq:doubleint:remainder}
\mathbb{E}[\mathcal{J}(t) \mathds{1}\{ \tau_{\textnormal{out}} <\hat{t}_0 \} \, | \, \mathcal{F}_{t_0} ] = o(\alpha^{40}) = \mathbb{E}[I_{\textnormal{out}}(t) \mathds{1}\{ \tau_{\textnormal{out}} <\hat{t}_0 \} \, | \, \mathcal{F}_{t_0} ].
\end{equation}
To establish the left estimate, we use the fact that $S(t)\le \frac{1}{2}$ and $|J_{u,s}^{(\alpha)} |\le 2$ from Proposition \ref{prop:speed:basicdef} and \eqref{eq:coupeling form of J}. Namely, we can express that
\begin{equation}\label{eq:doubleint:error:upd rough}
|\mathcal{J}(t)| \le \frac{\alpha}{1+2\alpha} \left[ |\Pi_{1/2}[t_0^\dagger,t]|^2 + \intop_{t_0^\dagger}^{t}\intop_{t_0^\dagger}^s 2\cdot \frac{1}{4} duds \right].
\end{equation}
Since the event $\{\tau_{\textnormal{out}} <\hat{t}_0 \}$ happens with probability less than $2e^{-\beta^{3/2}}$, we can couple this event with the tail events of the rate-$\frac{1}{2}$ Poisson process $|\Pi_{1/2}[t_0, \hat{t}_0]|$ and deduce the first equality of \eqref{eq:doubleint:remainder}. The second estimate can be obtained analogously.
What remains to show is that the contributions from $I_{\textnormal{out}}^{(1)}(t)$, $I_{\textnormal{out}}^{(2)}(t)$ and $I_{\textnormal{out}}^{(3)}(t)$ are small under the event $\{\tau_{\textnormal{out}} \ge \hat{t}_0 \}.$ Note that we already know that the contribution from $I_{\textnormal{out}}^{(3)}(t)$ is small, namely,
\begin{equation}\label{eq:doubleint:1}
\frac{\alpha}{1+2\alpha}\mathbb{E} \left[I_{\textnormal{out}}^{(3)}(t ) \mathds{1}\left\{\tau_{\textnormal{out}} \ge \hat{t}_0 \right\} \right] = o\left(\alpha^{2+\frac{1}{5}} \right).
\end{equation}
Moving on to the investigation of $I_{\textnormal{out}}^{(2)}(t)$, let $\delta_0 = \alpha^{-\frac{3}{2}+8\epsilon}$. From Lemma \ref{lem:ind1:tau13}, the event
\begin{equation}
\bigcap_{t\in[t_0^\dagger+\delta_0,\hat{t}_0]} \left\{ \intop_{t-\delta_0}^t |S(s)-\alpha|ds \le \alpha^\epsilon \right\}
\end{equation}
happens with probability at least $1- e^{-\beta^3}-r$. Conditioned on this event, the probability that $\pi_1(t;S\triangle \alpha)<\delta_0$ is smaller than $\alpha^\epsilon$ for each fixed $t$. Thus, we have
\begin{equation}\label{eq:doubleint:exp of closest pts}
\begin{split}
&\mathbb{E}\left[\left.\sigma_1(t;S\triangle \alpha)\mathds{1}\left\{\tau_{\textnormal{out}}\ge \hat{t}_0 \right\} \, \right| \,\mathcal{F}_{t_0} \right] \le \alpha^{\frac{3}{4}-5\epsilon};\\ &\mathbb{E}\left[\left.\sigma_1(t;S\triangle \alpha)^2\mathds{1}\left\{\tau_{\textnormal{out}}\ge \hat{t}_0 \right\} \, \right|\, \mathcal{F}_{t_0} \right] \le \alpha^{\frac{3}{2}-9\epsilon}.
\end{split}
\end{equation}
Combining with the definition of $\tau_{\textnormal{out}}^{(2)}$ and Lemma \ref{lem:doubleint:out1}, this gives that
\begin{equation}\label{eq:doubleint:2}
\frac{\alpha}{1+2\alpha}\mathbb{E} \left[\left.I_{\textnormal{out}}^{(2)}(t) \mathds{1}\left\{\tau_{\textnormal{out}}\ge\hat{t}_0 \right\} \, \right| \, \mathcal{F}_{t_0} \right] = o\left(\alpha^{2+\frac{1}{5}} \right).
\end{equation}
Similarly, we have
\begin{equation}
\mathbb{E}[\sigma_1(t;S)^2\mathds{1}\{\tau_{\textnormal{out}}\ge \hat{t}_0 \} \, | \, \mathcal{F}_{t_0} ] \le \alpha^{1-\epsilon},
\end{equation}
and hence from the definition of $\tau_{\textnormal{out}}^{(1)}$, we get
\begin{equation}\label{eq:doubleint:3}
\frac{\alpha}{1+2\alpha} \mathbb{E}\left[\left. I_{\textnormal{out}}^{(1)}(t) \mathds{1}\{\tau_{\textnormal{out}} \ge \hat{t}_0 \} \, \right| \, \mathcal{F}_{t_0} \right] = o \left(\alpha^{2+ \frac{1}{5}} \right).
\end{equation}
To conclude the proof, recall that
$$\mathcal{J}(t) = \frac{\alpha}{1+2\alpha} \big(I_{\textnormal{out}}(t) + I_{\textnormal{out}}^{(1)}(t)+ I_{\textnormal{out}}^{(2)}(t ) + I_{\textnormal{out}}^{(3)}(t )\big),$$
up to an $O(\alpha^{50})$-error that comes from \eqref{eq:doubleint:errorbasic}, whose contribution can be controlled analogously as \eqref{eq:doubleint:error:upd rough}, and combine \eqref{eq:doubleint:1}, \eqref{eq:doubleint:2} and \eqref{eq:doubleint:3}.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:doubleint:out1}]
We begin with noting that $\tau_{\textnormal{out}}^{(1)} \ge \hat{t}_0$ with high probability, due to Corollary \ref{cor:ind2:tau61sharp1}. To establish the control on $\tau_{\textnormal{out}}^{(2)}$ and $\tau_{\textnormal{out}}^{(3)}$, recall the definitions of $\tau_{\textnormal{gap}}^{(1)}$, $\tau_{\textnormal{gap}}^{(2)}$ (Corollaries \ref{cor:ind1:gap num of pts}, \ref{cor:ind1:gap mg}), $\tilde{\tau}_{3,3}^{(3)}$ and $\tilde{\tau}_{3,3}^\sharp$ (Corollaries \ref{cor:ind2:tau63sharp:Kstar}, \ref{cor:ind2:tau63sharp}). On the event $\{ \tau_{\textnormal{gap}}^{(1)}\wedge \tau_{\textnormal{gap}}^{(2)} \wedge \tilde{\tau}_{3,3}^{(3)} \wedge \tilde{\tau}_{3,3}^\sharp \ge \hat{t}_0 \}$, we can write
\begin{equation}
\begin{split}
I_{\textnormal{out}}^{(2)}(t)\le
\intop_{t_0^\dagger}^t \left( \frac{\alpha^{\frac{1}{2}}\beta^{6\theta}}{\sqrt{t-s+1}} + \frac{\beta^{C_\circ +1}}{t-s+1} \right) \left( \frac{\alpha\beta^{C_\circ}}{\sqrt{\pi_1(s;S\triangle \alpha)}} + \alpha^{\frac{7}{4}-\epsilon} \right)ds .
\end{split}
\end{equation}
Then, the RHS is bounded using Lemma \ref{lem:ind1:pi1int basicbd} with parameters $\Delta_0 = \alpha^{-\frac{3}{2}+7\epsilon}, N_0 = \beta_0^5, K = \alpha^{-\frac{1}{2}} $ and $\Delta_1 = \alpha^{-\frac{3}{2}}.$
This gives
\begin{equation}
\begin{split}
\intop_{t_0^\dagger}^t \left( \frac{\alpha^{\frac{1}{2}}\beta^{6\theta}}{\sqrt{t-s+1}} + \frac{\beta^{C_\circ +1}}{t-s+1} \right) \left( \frac{\alpha\beta^{C_\circ}}{\sqrt{\pi_1(s;S\triangle \alpha)}} + \alpha^{\frac{7}{4}-\epsilon} \right)ds \\
\le \frac{\alpha^{1-\epsilon}}{\sqrt{\pi_1(t;S\triangle \alpha)+1}} + \frac{\alpha^{\frac{1}{4}-\epsilon}}{\pi_1(s;S\triangle \alpha)+1} + \alpha^{\frac{5}{4}-2\epsilon} .
\end{split}
\end{equation}
Note that if there were intervals of length $\Delta_1$ without any points, then we can just artificially add points to justify the choice of $\Delta_1$; this operation will only increase the value of the whole integral.
Continuing our work on the same event as above, we express that
\begin{equation}
I_{\textnormal{out}}^{(3)}(t) \le \intop_{t_0^\dagger}^t \left( \frac{\alpha^{\frac{1}{2}}\beta^{6\theta}}{\sqrt{t-s+1}} + \frac{\beta^{C_\circ +1}}{t-s+1} \right) \alpha^{2-2\epsilon} ds \le \alpha^{\frac{3}{2}-3\epsilon},
\end{equation}
concluding the proof of Lemma \ref{lem:doubleint:out1}.
\end{proof}
\subsubsection{Reshaping the inner integral}\label{subsubsec:doubleint:inner}
Building upon Lemma \ref{lem:doubleint:out}, we investigate the inner integral of $I_{\textnormal{out}}(t)$ and establish Proposition \ref{prop:double int:main approx}. Define
\begin{equation}
\begin{split}
& F(s):= \intop_{t_0^+}^s K^*_\alpha(s-u)d\widetilde{\Pi}_\alpha(u) + \big[\alpha''-\alpha \big];\\
& \tau_{\textnormal{in}}^{(1)}:= \inf \left\{ s\ge t_0^+: |F(s)| \ge \alpha \beta^{C_\circ} \sigma_1(s;\alpha) + 3\alpha^{\frac{3}{2}}\beta^{6\theta} \right\}.
\end{split}
\end{equation}
Moreover, note that $\alpha''-\alpha = (\alpha''-\alpha' )+ (\alpha'-\alpha)$, and \eqref{eq:integralform:branching} gives
\begin{equation}
\alpha''-\alpha' =\intop_{t_0}^{t_0^+} K^*_{\alpha} \{d\Pi_S(x) - S_1(x)dx \}.
\end{equation}
Then, the same argument as Lemma \ref{lem:ind2:alpha1vsalpha0} tells us that
\begin{equation}
\mathbb{P} \left(\left. \left| \alpha''-\alpha' \right| \ge \alpha^{\frac{3}{2}}\beta^{2\theta} \,\right| \, \mathcal{F}_{t_0} \right) \le \exp \left(-\beta^5 \right).
\end{equation}
Thus, Theorem \ref{thm:reg:conti:main}, Corollary \ref{cor:ind1:taux:fixedrate}, and the definition of $\mathcal{A}_3(\alpha,t_0)$ (Section \ref{subsubsec:reg:history}) tell us that
\begin{equation}
\mathbb{P}\left(\left. \tau_{\textnormal{in}}^{(1)} <\hat{t}_0 \,\right|\, \mathcal{F}_{t_0} \right) \le \exp\left(-\beta^4 \right)+r.
\end{equation}
We also decompose $d\widehat{\Pi}_S(u)$ in $I_{\textnormal{out}}(t)$ as
\begin{equation}\label{eq:doubleint:innerint decomp}
\begin{split}
&d\widehat{\Pi}_S(u) = d\widetilde{\Pi}_\alpha(u) + d\widetilde{\Pi}_{S-\alpha}(u) + (S(u)-S_1(u))du + (S_1(u)-\alpha) du,
\end{split}
\end{equation}
and then writing $(S_1(u)-\alpha)$ as \eqref{eq:doubleint:S1minusalpha decomp}. Define the stopping times
\begin{equation}
\begin{split}
&\tau_{\textnormal{in}}^{(2)}:= \inf\left\{ t \ge \acute{t}_0: \left|\intop_{t_0^\dagger}^t\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} (S(u)-S_1(u)) F(s)duds \right| \ge \alpha^{\frac{3}{2}-\epsilon} \right\};\\
& \tau_{\textnormal{in}}^{(3)}:= \inf\left\{ t \ge \acute{t}_0: \left|\intop_{t_0^\dagger}^t\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} F(s) \intop_{t_0^+}^u K^*_\alpha(u-v)d\widetilde{\Pi}_{S-\alpha}(v) du ds \right| \ge \alpha^{\frac{5}{4}-\epsilon} \right\};\\
& \tau_{\textnormal{in}}^{(4)}:= \inf\left\{ t \ge \acute{t}_0: \left|\intop_{t_0^\dagger}^t\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} F(s) \intop_{t_0^+}^u K^*_\alpha(u-v) (S(u)-S_1(v))dvduds \right| \ge \alpha^{\frac{3}{2}-\epsilon} \right\}.
\end{split}
\end{equation}
We again stress that the integrals of $dv$ are over $[t_0^+,u]$, not $[t_0^\dagger,u]$.
Among the rest of the integrals rather than the ones stated in the above definitions, $d\widetilde{\Pi}_\alpha(u)$ and $\intop_{t_0^+}^u K^*_\alpha(u-v) d\widetilde{\Pi}_\alpha(v)du$ form the leading order $I_1(t)$ and $I_2(t)$. The other contributions coming from $d\widetilde{\Pi}_{S-\alpha}(u)$ and $(\alpha''-\alpha)du$ will be studied later. Note that in the above we consider $u \ge t_0^\dagger$, and hence the contribution from $[\mathcal{R}_c(t_0^+,u;\Pi_S[t_0^-,t_0^+],\alpha ) -\alpha''] du$ is negligible from \eqref{eq:doubleint:errorbasic} by the argument in Lemma \ref{lem:doubleint:out}.
\begin{lem}\label{lem:doubleint:in1ez}
Under the setting of Theorem \ref{thm:double int:main}, we have
\begin{equation}
\mathbb{P} \left(\left. \tau_{\textnormal{in}}^{(2)}\wedge \tau_{\textnormal{in}}^{(3)}\wedge \tau_{\textnormal{in}}^{(4)} < \hat{t}_0 \,\right|\, \mathcal{F}_{t_0} \right) \le \exp\left(-\beta_0^2 \right) +r.
\end{equation}
\end{lem}
\begin{proof}
The proof is very similar to that of Lemma \ref{lem:doubleint:out1}, and hence we concisely explain which lemmas from the previous sections are used to establish the conclusion.
For $\tau_{\textnormal{in}}^{(2)},$ we rely on Corollary \ref{cor:ind2:tau63sharp:1} to estimate the inner integral over $du$, and then combine with the bound from $\tau_{\textnormal{in}}^{(1)}$, using Lemma \ref{lem:ind1:pi1int basicbd} to perform integration over $ds$.
For $\tau_{\textnormal{in}}^{(3)}$, recall the definition of $\tau_{\textnormal{gap}}^{(1)}, \tau_{\textnormal{gap}}^{(2)}$ from Corollaries \ref{cor:ind1:gap num of pts}, \ref{cor:ind1:gap mg}, and note that on $\{\tau_{\textnormal{gap}}^{(1)}\wedge \tau_{\textnormal{gap}}^{(2)} \ge \hat{t}_0 \}$ we have
\begin{equation}
\begin{split}
&\left|\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} \intop_{t_0^+}^u K^*_\alpha(u-v)d\widetilde{\Pi}_{S-\alpha}(v) du\right|\\
&\le
\frac{e^{-c\alpha^2(t-s) }}{\sqrt{t-s+1}} \intop_{t_0^+}^s \frac{1}{\sqrt{t-u+1}} \left( \frac{\alpha\beta^{C_\circ}}{ \sqrt{\pi_1(u;S\triangle \alpha)}} + \alpha^{\frac{7}{4}-\epsilon} \right)du
\le \frac{\alpha^{\frac{3}{4}-9\epsilon} e^{-c\alpha^2(t-s) }}{\sqrt{t-s+1}} ,
\end{split}
\end{equation}
where the last inequality follows from applying Lemma \ref{lem:ind1:pi1int basicbd} as before in Lemma \ref{lem:doubleint:out1}. Then the control on $\tau_{\textnormal{in}}^{(3)}$ is obtained by performing the outer integral over $ds$, using the bound on $F(s)$ and Lemma \ref{lem:ind1:pi1int basicbd}.
Lastly, for $\tau_{\textnormal{in}}^{(4)}$, we rely on $\tilde{\tau}_{3,3}^{(3)}$ from Corollary \ref{cor:ind2:tau63sharp:Kstar}, where on $\{ \tilde{\tau}_{3,3}^{(3)} \ge \hat{t}_0 \}$ we have
\begin{equation}
\begin{split}
\left|\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} \intop_{t_0^+}^u K^*_\alpha(u-v) (S(u)-S_1(v))dvdu \right| \le \intop_{t_0^\dagger}^s \left|J^{(\alpha)}_{t-s,t-u}\right| \alpha^{2-2\epsilon} du \le \alpha^{2-3\epsilon}.
\end{split}
\end{equation}
Then, conducting the outer integral over $ds$ similarly as above gives the estimate on $\tau_{\textnormal{in}}^{(4)}$, concluding the proof.
\end{proof}
As mentioned right before Lemma \ref{lem:doubleint:in1ez}, we now study the remaining integrals, beginning with
\begin{equation}
\begin{split}
& I_{\textnormal{in}}^{(5)}(t):= \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widetilde{\Pi}_{S-\alpha}(u) \cdot \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_\alpha(x)ds ;\\
&I_{\textnormal{in}}^{(6)}(t):= \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widetilde{\Pi}_{S-\alpha}(u) \cdot \big[\alpha''-\alpha \big]ds.
\end{split}
\end{equation}
\begin{lem}\label{lem:doubleint:in2 mg}
Under the setting of Theorem \ref{thm:double int:main}, we have for all $t\in[\acute{t}_0,\hat{t}_0]$ that
\begin{equation}
\mathbb{E}\left[\left. I_{\textnormal{in}}^{(5)}(t)\, \right| \, \mathcal{F}_{t_0} \right] = O\left(\alpha^{\frac{5}{4}-\epsilon} \right)= \mathbb{E}\left[\left. I_{\textnormal{in}}^{(6)}(t)\, \right| \, \mathcal{F}_{t_0} \right] .
\end{equation}
\end{lem}
Although these integrals should have a negligible size compared to the leading order, one can see that the same argument as the previous analysis does not give the correct estimate on $I_{\textnormal{in}}^{(5)}$. To overcome this issue, we remind ourselves that our actual interest is to control the expected value of this quantity.
\begin{proof}[Proof of Lemma \ref{lem:doubleint:in2 mg}]
We begin with investigating $I_{\textnormal{in}}^{(5)}$. Define $\underline{\alpha}:=\alpha - \alpha^{\frac{3}{2}-\epsilon},$ and recall from Proposition \ref{prop:reg:lbd of speed} that $\tau_{\textnormal{lo}}(\alpha,t_0) \ge \hat{t}_0$ with high probability, that is, $S(t)$ is likely to stay above from $\underline{\alpha}.$ Thus, writing $\tau_{\textnormal{lo}}=\tau_{\textnormal{lo}}(\alpha,t_0)$, observe that
\begin{equation}
\mathbb{E}\left[\left.
\intop_{t_0^\dagger}^{t\wedge \tau_{\textnormal{lo}}} \left\{ \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widetilde{\Pi}_{S-\alpha}(u) \cdot \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_{\underline{ \alpha}}(x) \right\}ds
\,\right|\, \mathcal{F}_{t_0} \right] =0,
\end{equation}
since $d\widetilde{\Pi}_{S-\alpha}$ and $d\widetilde{\Pi}_{\underline{\alpha}}$ define martingales and depend on disjoint set of points. Thus, under extra conditioning on $\Pi_{\underline{\alpha}}[t_0,\hat{t}_0]$, we can see that the above will always be zero.
Therefore, relying on Proposition \ref{prop:reg:lbd of speed} and the argument from Lemma~\ref{lem:doubleint:out} gives that for all $t\in[\acute{t}_0,\hat{t}_0]$,
\begin{equation}
\mathbb{E}\left[\left.
\intop_{t_0^\dagger}^{t}\left\{ \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widetilde{\Pi}_{S-\alpha}(u) \cdot \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_{\underline{ \alpha}}(x)\right\} ds
\,\right|\, \mathcal{F}_{t_0} \right] =O(\alpha^{50}).
\end{equation}
Writing $d\widetilde{\Pi}_{\triangle \alpha} = d\widetilde{\Pi}_\alpha - d\widetilde{\Pi}_{\underline{\alpha}}$, define the stopping time
\begin{equation}
\tau_{\textnormal{in}}^{(5)} :=\inf \left\{
t \ge \acute{t}_0: \left| \intop_{t_0^\dagger}^t \left\{\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widetilde{\Pi}_{S-\alpha}(u) \cdot \intop_{t_0^+}^s K^*_\alpha(s-x) d\widetilde{\Pi}_{\triangle\alpha}(x)\right\}ds \right| \ge \alpha^{\frac{5}{4}-\epsilon}
\right\}.
\end{equation}
To conclude the proof, it suffices to show that
\begin{equation}
\mathbb{P} \left(\left. \tau_{\textnormal{in}}^{(5)} <\hat{t}_0 \, \right|\, \mathcal{F}_{t_0} \right) \le \exp\left(-\beta_0^2 \right) +r.
\end{equation}
Since this estimate follows similarly as the proof of Lemmas \ref{lem:doubleint:out1} and \ref{lem:doubleint:in1ez}, we give a brief explanation on its proof. Since $J^{(\alpha)}_{t-s,t-u}$ satisfies a similar bound as $K^*_\alpha$ (Lemma \ref{lem:estimat for K tilde:intro}), analogous argument as Corollary \ref{cor:ind1:gap mg} gives an estimate the first part of the integral, written as
\begin{equation}
\begin{split}
&\tau_{\textnormal{in}}^{(6)}:= \inf \left\{ t\ge \acute{t}_0: \, \exists s\in[t_0^\dagger,t] \textnormal{ s.t. } \left| \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} d\widetilde{\Pi}_{S-\alpha} (u)\right| \ge \frac{\beta_0^{C_\circ} }{t-s+1} + \frac{\alpha^{\frac{3}{4}-\epsilon}}{\sqrt{t-s+1}}\right\};\\
& \mathbb{P} \left(\left.\tau_{\textnormal{in}}^{(6)} < \hat{t}_0 \, \right|\, \mathcal{F}_{t_0} \right) \le \exp\left(-\beta^3 \right)+r.
\end{split}
\end{equation}
(Note that $\sigma_1(t;S\triangle \alpha)$ in Corollary \ref{cor:ind1:gap mg} becomes $(t-s+\pi_1(s)+1)^{-1/2} \le (t-s+1)^{-1/2}$ in this case.)
The second part of the integral is controlled by Corollary \ref{cor:ind1:gap mg 2}, and hence combining the two bounds with an application of Lemma \ref{lem:ind1:pi1int basicbd} concludes the proof, similarly as the previous lemmas.
Finally, we note that $I_{\textnormal{in}}^{(6)}(t)$ is a martingale (note that $\alpha''$ is $\mathcal{F}_{t_0^+}$-measurable while the integral starts from $t_0^\dagger$) and hence its expectation is zero.
\end{proof}
Now we are left with the two final integrals, which are
\begin{equation}\label{eq:doubleint:in:last piece}
\begin{split}
& I_{\textnormal{in}}^{(7)} (t):= \intop_{t_0^\dagger}^t \left\{\intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} du \cdot \intop_{t_0^+}^s K^*_\alpha(s-x)d\widetilde{\Pi}_\alpha(x)\right\} (\alpha''-\alpha) ds;\\
& I_{\textnormal{in}}^{(8)} (t):= \intop_{t_0^\dagger}^t \intop_{t_0^\dagger}^s J^{(\alpha)}_{t-s,t-u} (\alpha''-\alpha)^2du ds = \intop_0^{t-t_0^\dagger} \intop_0^s J_{u,s}^{(\alpha)} (\alpha''-\alpha)^2duds.
\end{split}
\end{equation}
Note that $I_{\textnormal{in}}^{(7)}(t)$ is a martingale, since $\alpha''$ is $\mathcal{F}_{t_0^+}$-measurable. Thus, we have $\mathbb{E} [I^{(7)}_{\textnormal{in}}(t) | \mathcal{F}_{t_0} ]=0$. On the other hand, $I_{\textnormal{in}}^{(8)}(t),$ which is a deterministic integral, is bounded from \eqref{eq:I_4}: $I_{\textnormal{in}}^{(8)}(t) = O(\alpha^{2-\epsilon}) $ for any $t\in[\acute{t}_0,\hat{t}_0]$.
Combining all the analysis done in this subsection, we conclude the proof of Proposition \ref{prop:double int:main approx}.
\begin{proof}[Proof of Proposition \ref{prop:double int:main approx}]
In \eqref{eq:doubleint:innerint decomp} and the discussions below, we have seen that $I_{\textnormal{out}}(t)$ in Lemma \ref{lem:doubleint:out} can be decomposed into
\begin{equation}
I_1(t), \, I_2(t), \,\, \left\{\textnormal{integrals in } \tau_{\textnormal{in}}^{(2)},\, \tau_{\textnormal{in}}^{(3)},\, \tau_{\textnormal{in}}^{(4)} \right\}, \ I_{\textnormal{in}}^{(5)}(t), \, I_{\textnormal{in}}^{(6)}(t), \, I_{\textnormal{in}}^{(7)}(t), \textnormal{ and } I_{\textnormal{in}}^{(8)}(t),
\end{equation}
up to an $O(\alpha^{50})$ error coming from \eqref{eq:doubleint:errorbasic}.
Then, Lemmas \ref{lem:doubleint:in1ez}, \ref{lem:doubleint:in2 mg} and \eqref{eq:doubleint:in:last piece} tell us that the conditional expectations given $\mathcal{F}_{t_0}$ of all of the above integrals except $I_1(t)$ and $I_2(t)$ are of order $O(\alpha^{\frac{5}{4}-\epsilon})$. This implies
\begin{equation}
\mathbb{E} \left[I_{\textnormal{out}}(t) \,|\,\mathcal{F}_{t_0} \right] = \mathbb{E} \left[I_1(t)+I_2(t) \,| \,\mathcal{F}_{t_0} \right] + O\left(\alpha^{\frac{5}{4}-\epsilon}\right),
\end{equation}
which concludes the proof combined with Lemma \ref{lem:doubleint:out}.
\end{proof}
\subsection{The analysis on the increment}\label{subsec:increment proof}
We give a formal statement and proof of Theorem \ref{thm:speed:increment:informal}, thus verifying the assumptions of Theorem 4.1 are indeed accurate.
\begin{thm}\label{thm:increment:formal}
Let $\alpha_0,t_0>0$, set $t_0^-, t_0^+,\acute{t}_0$ and $\hat{t}_0$ as \eqref{eq:def:t0t1} and \eqref{eq:def:t0t11} in terms of $\alpha_0$, and let $r=e^{-\beta_0^{3/2}}$. Also, let $t_1$ be any number satisfying \eqref{eq:t1 regime}, set $t_1^-:= t_1-\alpha_0^{-2}\beta_0^\theta$. Furthermore, define
\begin{equation}
\alpha_1 := \mathcal{L}(t_1^-; \Pi_S[t_0^-,t_1^-], \alpha_0),
\end{equation}
and $\alpha_0', \alpha_1'$ as \eqref{eq:def:alpha0prime}.
Then, for any $\Pi_S(-\infty,t_0]$ that is $(\alpha_0,r;[t_0])$-sharp-regular, we have
\begin{equation}
\begin{split}
\mathbb{E}\left[\alpha_1'-\alpha_0'\, | \, \Pi_S(-\infty,t_0] \right] &= o\left(\alpha_0^4\beta_0^{-3} \right) \cdot (t_1 -t_0);\\
\mathrm{Var} \left[ \alpha_1'-\alpha_0'\, | \, \Pi_S(-\infty,t_0] \right] &= (1+o(\alpha_0^{\frac{1}{3}})) 4\alpha_0^5(t_1-t_0).
\end{split}
\end{equation}
\end{thm}
\begin{proof}
Let $\tau =\tau(\alpha_0,t_0,2)$, $\tau^\sharp = \tau^\sharp(\alpha_0,t_0,2)$ be as \eqref{eq:def:tau:induction}, and recall the definition of $\tilde{\alpha}_1$ from Proposition \ref{prop:reg:newrates}. We define the event
\begin{equation}
\begin{split}
\mathcal{B}_1:=& \Big\{ \tau \ge \hat{t}_0 \Big\} \bigcap \Big\{\tau^\sharp \ge \acute{t}_0 \Big\} \bigcap \Big\{|\alpha_1'-\tilde{\alpha}_1| \le 2\alpha_0^2\beta_0^{8\theta+1} \Big\};\\
\mathcal{B}_2:=&\left\{\intop_{\acute{t}_0}^{t_1} S(t)dt = \alpha_0(t_1-\acute{t}_0) + O\left(\alpha_0^{-\frac{1}{2}-\epsilon}\right) =\intop_{\acute{t}_0}^{t_1} S_1(t)dt \right\};\\
\mathcal{B}_3:=&\left\{ \intop_{\acute{t}_0}^{t_1} |S(t)-S_1(t)| dt \le \alpha_0^{-2\epsilon} \right\} \bigcap \left\{ \intop_{\acute{t}_0}^{t_1} |S(t)-S_2(t)| dt \le \alpha_0^{\frac{1}{2}-2\epsilon} \right\} ;\\
\mathcal{B}_4:= & \left\{ \intop_{t_0}^{t_1} S(t)dt = \alpha_0(t_1-t_0) + O \left(\alpha_0^{-\frac{1}{2}-\epsilon} \right) \right\};\\
\mathcal{B}_5:=&
\left\{ \intop_{t_0}^{t_1} \sigma_1\sigma_2\sigma_3(t;S) dt \le \alpha_0^{-\frac{1}{2}-\epsilon} \right\} ;\\
\mathcal{B}:=& \mathcal{B}_1 \bigcap \mathcal{B}_2 \bigcap \mathcal{B}_3\bigcap \mathcal{B}_4 \bigcap\mathcal{B}_5,
\end{split}
\end{equation}
where $S_1(t)=S_1(t;t_0^-,\alpha_0)$, $S_2(t)=S_2(t;t_0^-,\alpha_0)$ are given as \eqref{eq:def:S1:basic form}, \eqref{eq:def:S2:basic form}.
Conditioned on $\Pi_S(-\infty,t_0]$, $\mathcal{B}$ happens with probability at least $1-2e^{-\beta_0^{3/2}}$:
\begin{itemize}
\item $\mathcal{B}_1$ is obtained from the definition of sharp-regularity, Proposition \ref{prop:reg:newrates}, and Theorem \ref{thm:reg:conti:main}.
\item $\mathcal{B}_2, \mathcal{B}_3$ come from Lemma \ref{lem:reg:errorint}, and the sharp-regularity along with $\mathcal{B}_2$ implies $\mathcal{B}_4$.
\item $ \mathcal{B}_5$ is from Lemma \ref{lem:fixed perturbed:error int:perturbed} (whose assumptions are satisfied by $\tau_{1}, \tau_{4}$, and $ \tau_{5}$).
\end{itemize}
Hence
the contributions to the mean and variance of increments on $\mathcal{B}^c$ are negligible (since $||S|| \le \frac{1}{2}$).
We write $\alpha_1' - \alpha_0' = (\alpha_1' - \tilde{\alpha}_1) + (\tilde{\alpha}_1-\alpha_0')$ and similarly as \eqref{eq:Lt diff:basic} and \eqref{eq:ind2:a1vsa0:split}, we express that
\begin{equation}\label{eq:increment:mean:0}
\tilde{\alpha}_1-\alpha_0' = \intop_{t_0}^{t_1} K^*_{\alpha_0} \{d\Pi_S (t) - S_1(t)dt\}.
\end{equation}
From these formulas, we begin with estimating the mean of the increment. Note that from Lemma \ref{lem:ind2:alphaPvsalphaPP},
\begin{equation}\label{eq:increment:mean:1}
\mathbb{E}[|\alpha_1'-\tilde{\alpha}_1|\,;\, \mathcal{B} \ | \, \mathcal{F}_{0}] = o(\alpha_0^4 \beta_0^{-3}) \cdot (t_1-t_0).
\end{equation}
Furthermore, we have
\begin{equation}\label{eq:increment:mean:2}
\begin{split}
\mathbb{E}[\tilde{\alpha}_1-\alpha_0' \, | \, \mathcal{F}_{0}]
= K^*_{\alpha_0} \cdot \mathbb{E}\left[\left. \intop_{t_0}^{t_1} \{S(t)-S_1(t) \} dt \, \right| \, \mathcal{F}_{0}\right]
\end{split}
\end{equation}
We decompose this integral into two parts, from $t_0$ to $\acute{t}_0$ and from $\acute{t}_0$ to $t_1$. For the first one, observe that the definition of $\mathcal{B}$ gives
\begin{equation}\label{eq:increment:mean:3}
\begin{split}
\left| \mathbb{E}\left[\left.\intop_{t_0}^{\acute{t}_0} \{S(t)-S_1(t) \}dt \,; \, \mathcal{B}\ \right| \, \mathcal{F}_{0}\right]\right|
\le
\beta_0^{4\theta} = o(\alpha_0^2\beta_0^{-3}) (t_1-t_0).
\end{split}
\end{equation}
To study the integral from $\acute{t}_0$ to $t_1$ of \eqref{eq:increment:mean:2}, we recall the formula \eqref{eq:def:S2:basic form} to see that
\begin{equation}
S(t) - S_1(t)= S(t)-S_2(t) + \frac{2\alpha_0^2}{(1+2\alpha_0)^2} - \frac{4\alpha_0+4\alpha_0^2}{(1+2\alpha_0)^2} S_1(t) + \mathcal{J}(t),
\end{equation}
where $\mathcal{J}(t)$ is defined in the beginning of this section.
Thus, we can see that for all $t\in[\acute{t}_0, t_1]$,
\begin{equation}
\mathbb{E} \left[ \left. \frac{2\alpha_0^2}{(1+2\alpha_0)^2} + \mathcal{J}(t) \, ; \, \mathcal{B}\, \right| \ \mathcal{F}_0 \right] = (1+ O(\alpha^{\epsilon})) 4\alpha_0^2.
\end{equation}
Moreover, the event $\mathcal{B}$ gives
\begin{equation}
\mathbb{E} \left[ \left. \intop_{\acute{t}_0}^{t_1} S_1(t)dt \, ; \, \mathcal{B}\, \right| \ \mathcal{F}_0 \right] = \alpha_0 (t_1-\acute{t}_0) + O\left(\alpha_0^{-\frac{1}{2}-\epsilon}\right).
\end{equation}
In addition, the integral of $|S(t)-S_2(t)|$ is bounded from the definition of $\mathcal{B}$, which is smaller than $\alpha_0^{\frac{1}{2}-2\epsilon}=o(\alpha_0^4\beta_0^{-1})\cdot (t_1-t_0)$. Combining these three estimates, we observe that the terms of order $\alpha_0^2(t_1-t_0)$ cancel out, and hence deduce that
\begin{equation}\label{eq:increment:mean:4}
\mathbb{E}\left[\left. \intop_{\acute{t}_0}^{t_1} \{S(t)-S_1(t) \} dt \,; \ \mathcal{B}\, \right| \ \mathcal{F}_0 \right] = o(\alpha_0^2 \beta_0^{-3})\cdot (t_1-t_0).
\end{equation}
Therefore, combining \eqref{eq:increment:mean:1}, \eqref{eq:increment:mean:2}, \eqref{eq:increment:mean:3} and \eqref{eq:increment:mean:4} concludes the proof of the first statement.
\vspace{2mm}
To control the variance, we first write $\alpha_1'-\alpha_0' = K_{\alpha_0}^*\mathcal{M}+\mathcal{D}$, where
\begin{equation}
\mathcal{M}:= \intop_{t_0}^{t_1} \{d\Pi_S(t)-S(t)dt\}, \quad \mathcal{D}:= (\alpha_1'-\tilde{\alpha}_1) + K^*_{\alpha_0}\cdot\intop_{t_0}^{t_1}\{S(t)-S_1(t)\} dt,
\end{equation}
and also define $\overline{\mathcal{D}}:=\mathbb{E} [ \mathcal{D} \, | \, \mathcal{F}_0].$
Then, we express that
\begin{equation}\label{eq:increment:var:1}
\begin{split}
\left| \mathrm{Var}\left[\left. \alpha_1'-\alpha_0' \, \right|\,\mathcal{F}_0 \right] -\mathrm{Var}\left[\left. K_{\alpha_0}^*\mathcal{M} \, \right|\,\mathcal{F}_0 \right] \right|
\le
\mathrm{Var}[\mathcal{D}\,|\,\mathcal{F}_0]+ 2K^*_{\alpha_0} \cdot \Big\{ \mathbb{E} [ \mathcal{M}^2\,|\, \mathcal{F}_0] \cdot \mathrm{Var}[\mathcal{D}\,|\,\mathcal{F}_0] \Big\}^{1/2},
\end{split}
\end{equation}
where we used Cauchy-Schwarz to deduce the inequality. Furthermore, since $\mathcal{M}$ is a martingale, we have
\begin{equation}\label{eq:increment:var:2}
\begin{split}
\mathbb{E}[\mathcal{M}^2\,|\, \mathcal{F}_0] &= \mathbb{E}\left[\left.\intop_{t_0}^{t_1} S(t)dt \, \right|\,\mathcal{F}_0 \right] = o(\alpha_0^{10}) + \mathbb{E}\left[\left.\intop_{t_0}^{t_1} S(t)dt \, ;\ \mathcal{B}\, \right|\,\mathcal{F}_0 \right] \\
&=
\alpha_0(t_1-t_0) + O\left(\alpha_0^{-\frac{1}{2}-\epsilon} \right) = \left(1+ o(\alpha_0^{\frac{1}{3}}) \right) \alpha_0(t_1-t_0).
\end{split}
\end{equation}
Moreover, on the event $\mathcal{B}$, we have $\mathcal{D} = O(\alpha_0^{2-2\epsilon})$. Thus, we see that $\mathrm{Var}[\mathcal{D}\,|\, \mathcal{F}_0] = O(\alpha_0^{4-4\epsilon})$, and hence
\begin{equation}
\mathrm{Var}[\mathcal{D} |\mathcal{F}_0 ] = o(\alpha_0^{1-5\epsilon}) (K_{\alpha_0}^*)^2 \mathbb{E} [\mathcal{M}^2 | \mathcal{F}_0 ] .
\end{equation}
Therefore, we combine the above computations \eqref{eq:increment:var:1} and \eqref{eq:increment:var:2} to deduce the second statement which finishes the proof.
\end{proof}
To conclude this section, we establish a variant of Theorem \ref{thm:increment:formal} which will be useful when applying to a slightly different setting in the next section.
\begin{cor}\label{cor:increment:formal:goodevent}
Under the setting of Theorem \ref{thm:increment:formal}, let $\mathcal{E}$ be an event that satisfies $\mathbb{P}(\mathcal{E} | \mathcal{F}_{t_0}) \ge 1- \alpha_0^{5}$, for any $\mathcal{F}_{t_0}=\Pi_S(-\infty,t_0]$ that is $(\alpha_0,r;[t_0])$-sharp-regular. Then, we have
\begin{equation}
\mathrm{Var}\left[\left.\left(\alpha_1'-\alpha_0'\right) \mathds{1}\{\mathcal{E} \} \, \right|\, \Pi_S(-\infty,t_0] \right] = (1+ o(\alpha_0^{\frac{1}{3}}))4\alpha_0^5(t_1-t_0).
\end{equation}
\end{cor}
\begin{proof}
For convenience, we write $\mathcal{F}_0= \mathcal{F}_{t_0}=\Pi_S(-\infty,t_0] $. Due to the analogous decomposition as \eqref{eq:increment:var:1}, it suffices to show that
\begin{equation}
\mathrm{Var} \left[ (\alpha_1'-\alpha_0')\mathds{1}\{\mathcal{E}^c \}\, | \, \mathcal{F}_0 \right] = o(\alpha_0^4),
\end{equation}
since we have $ \mathrm{Var} \left[ \alpha_1'-\alpha_0' | \mathcal{F}_0 \right]= \Theta(\alpha_0^3\beta_0^{10\theta})$. The proof goes analogously as that of Lemma \ref{lem:doubleint:out}, based on the same idea as \eqref{eq:doubleint:remainder}. Observe from \eqref{eq:ind2:a1vsa0:split} that
\begin{equation}
\begin{split}
\alpha_1-\alpha_0' &= \mathcal{L}(t_1^-;\Pi_S[t_0^-,t_1^-],\alpha_0) = K_{\alpha_0}^* \intop_{t_0^-}^{t_1^-} d\Pi_S(x) - S_1(x)dx\\
&= K_{\alpha_0}^*|\Pi_S[t_0^-,t_1^-]| - K_{\alpha_0}^* \intop_{t_0^-}^{t_1^-} \intop_{y}^{t_1^-} K_{\alpha_0}(x-y) dx d\Pi_S(y).
\end{split}
\end{equation}
Since $\intop_y^{t_1^-} K_{\alpha_0}(x-y)dx \in [0,1]$, we can see that $|\alpha_1-\alpha_0'|$ is stochastically dominated by
\begin{equation}
|\alpha_1-\alpha_0'| \preceq
K_{\alpha_0}^* |\Pi_{1/2} [t_0^-,t_1^-]|,
\end{equation}
due to the fact that $S\le \frac{1}{2}$ (Proposition \ref{prop:speed:basicdef}). Moreover, we also have
\begin{equation}
\begin{split}
\alpha_1' = K_{\alpha_1}^* \intop_{t_1}^\infty \intop_{t_1^-}^{t_1} K_{\alpha_1}(x-y) d\Pi_S(y) dx \preceq K_{\alpha_1}^* | \Pi_{1/2}[t_1^-,t_1]|.
\end{split}
\end{equation}
Thus, we can control the size of $\alpha_1$ and $\alpha_1'$ on the event $\mathcal{E}^c$ by coupling with the tail events of the Poisson process, which easily gives us the desired estimate. (Note that $\alpha_0'$ is already close to $\alpha_0$ since the speed is $(\alpha_0,r;[t_0])$-sharp-regular.)
\end{proof}
\section{Multi-scale analysis and the scaling limit}\label{sec:scalinglimit}
In this section we repeatedly use the induction step theorem in order to prove the main results of this paper. The proof goes as follows. Let $t>0$ be sufficiently large and recall that Theorem~\ref{thm:main theorem 2} says that $t^{-\frac{2}{3}}S(st)$ should behave roughly like the solution of the SDE \eqref{eq:SDE equation in main theorem}. The proof has the following parts:
\begin{enumerate}
\item
We show that in time $\approx \delta t$ the aggregate is regular and the speed is of order $\approx \delta ^{-\frac{1}{3}}t^{-\frac{1}{3}}$.
\item
We show that $t^{\frac{1}{3}}S(\delta t+st)$ is close to the solution of \eqref{eq:SDE equation in main theorem} with initial condition $Z(0)=\delta ^{-\frac{1}{3}}$.
\end{enumerate}
Theorem~\ref{thm:main theorem 2} follows from part (2) by taking $\delta \to 0$ and integrating the speed to obtain the aggregate size. The proof of the first part is given in Subsection~\ref{sec:stitching} and it includes analysis of $\log t$ many scales of the aggregate speed. We show that in each scale the aggregate speed is likely to decrease and that the time it takes for this to happen is not too long. The second part is given in Subsection~\ref{sec:convergence in distribution} and it is a direct application of a result by Helland \cite{helland1981minimal}. In \cite{helland1981minimal} it is shown that, under some assumptions, if the increments of a sequence of processes have the right conditional expectation and variance then the sequence of processes converges to the solution of the corresponding SDE.
Throughout this section it is convenient to work with a slightly different definition of regular. Recall the definition of $\mathfrak{R}^\sharp (\alpha_0,r,[t_0])$ in Section~\ref{subsec:regoverview:reg}. As usual, we identify a set of points $\Pi $ with the cumulative function $Y_t(s):=|\Pi [t-s,t]|$ and denote by $\Pi _Y$ the set of points corresponding to the step function $Y$. We also write $Y\in \mathfrak{R}^\sharp (\alpha_0,r,[t_0])$ when $\Pi _{Y}\in \mathfrak{R}^\sharp (\alpha_0,r,[t_0])$. For convenience, in the following definition and throughout this section we also switch the roles of $\alpha $ and $\alpha '$ from Section~\ref{sec:reg:intro} and equation \eqref{eq:def:alpha0prime}.
\begin{definition}\label{def:def of regularity prime}
Let $\alpha >0$ sufficiently small and $t>0$. We say that a deterministic step function $Y _0 :\mathbb R _+ \to \mathbb N \cup \{\infty \} $ satisfies $Y _0 \in \mathfrak{R}'(\alpha ,t)$ if there exists $t^{-}$ with
\begin{equation}
t -2 \alpha ^{-2} \log ^{\theta } (1/\alpha ) \le t^{-} \le t -\frac{1}{2} \alpha _0^{-2} \log ^{\theta } (1/\alpha )
\end{equation}
and $\alpha '>0$ such that $Y_0 \in \mathfrak{R} ^\sharp (\alpha ' ,\alpha ^6 ;[t])$ and
\begin{equation}
\alpha = \mathcal{L}(t; \Pi _{Y_0}[t^{-},t],\alpha '):= \frac{2 (\alpha') ^2 }{1+2 \alpha' }\intop _{t^-} ^{t} \intop _{t}^{\infty} K_{\alpha '} (z-x) dz \, d \Pi _{Y_0} (x).
\end{equation}
\end{definition}
Using this definition we state the main results of the induction step in the following theorem
\begin{thm}\label{thm:induction step for repeatd use}
Let $\alpha >0$ sufficiently small, $t >0$ and $\tilde{t}>0$ such that
\begin{equation}
\alpha ^{-2}\log ^{10\theta-4 } (1/\alpha ) \le \tilde{t}-t\le \alpha ^{-2}\log ^{10\theta } (1/\alpha ).
\end{equation}
Let $Y _{t} \in \mathfrak{R}'(\alpha ,t)$ and let $(Y_x ,S(x),X_x)$ be the aggregate with initial condition $(Y _{t},t)$. Then, there is a random variable $\tilde{\alpha } \in \mathcal F _{\tilde{t}}$ such that the following holds:
\begin{enumerate}
\item
\begin{equation}
| \tilde{ \alpha } -\alpha |\le \sqrt{\alpha ^5(\tilde{t}-t )} \log ^{2 \theta } (1/\alpha )
\end{equation}
\item
\begin{equation}
\mathbb P \left( Y_{\tilde{t}} \in \mathfrak{R}'(\tilde{\alpha } ,\tilde{t} ) \right)\ge 1- \alpha ^5
\end{equation}
\item
\begin{equation}
\left| \mathbb E \left( \tilde{\alpha } -\alpha \right) \right| \le \frac{\alpha ^4 }{\log ^3 (1/\alpha )}(\tilde{t}-t ).
\end{equation}
\item
\begin{equation}
\mathbb E \left( ( \tilde{ \alpha } -\alpha )^2 \right) = 4 \alpha ^5 (\tilde{t}-t )\left(1+O(\alpha ^{\frac{1}{3} } )\right).
\end{equation}
\item
\begin{equation}
\mathbb P \left( \left| X_{\tilde{t} }-X_{t}-\alpha (\tilde{t}-t) \right| \le \alpha ^{1.4}(\tilde{t}-t ) \right) \ge 1-\alpha ^5
\end{equation}
\end{enumerate}
\end{thm}
\begin{proof}
By Definition~\ref{def:def of regularity prime} there exists $\alpha '$ such that $Y_{t}\in \mathfrak{R} ^\sharp (\alpha ' ,\alpha ^6 ,[t])$ which in particular means by the definition of $\mathcal A _2$ in \eqref{eq:def of the events A1 A2 A3} that $|\alpha -\alpha '| \le (\alpha ') ^{\frac{3}{2}} \log ^{\theta } (1/\alpha ') \le 2 \alpha ^{\frac{3}{2}} \log ^{\theta } (1/\alpha ) $. Thus by Theorem~\ref{thm:induction:main} we have some $\tilde{\alpha }'$ such that
\begin{equation}
\mathbb P \big( Y_{t_1} \in \mathfrak{R} ^\sharp ( \tilde{\alpha } ' , ( \tilde{ \alpha }') ^7 ,[\tilde{t}]) \big) \ge 1-2\alpha ^6 \quad \text{and} \quad \mathbb P \big( |\alpha '- \tilde{ \alpha }'| \le 2(\alpha' ) ^{\frac{3}{2}} \log ^{6 \theta }(1/\alpha ') \big) \ge 1-2\alpha ^6 .
\end{equation}
Let
\begin{equation}
\tilde{\alpha } :=\mathcal{L}(\tilde{t}; \Pi _{Y_{\tilde{t}}}[\tilde{t}^{-},\tilde{t}], \tilde{\alpha }').
\end{equation}
Using the definition of $\mathcal A _2 $ again we have $|\tilde{ \alpha } -\tilde{ \alpha }'|\le (\tilde{ \alpha }')^{\frac{3}{2}} \log ^\theta (1/\tilde{\alpha }') \le 2\alpha _0^{\frac{3}{2}} \log ^{\theta } (1/\alpha )$ with probability at least $1-4 \alpha ^6 $. Thus we get
\begin{equation}
\mathbb P \big( |\alpha -\tilde{ \alpha } | \le \sqrt{\alpha ^5 (\tilde{t}-t)} \log ^{2 \theta } (1/\alpha \big) \ge 1-4 \alpha ^6.
\end{equation}
We also get that $\mathbb P \big( Y_{\tilde{t}} \in \mathfrak{R} ^\sharp ( \tilde{\alpha } ' , \tilde{ \alpha }^6 ,[\tilde{t}]) \big) \ge 1-6\alpha ^6$ and therefore by Definition~\ref{def:def of regularity prime}, $\mathbb P ( Y_{\tilde{t}} \in \mathfrak{R} ' ( \tilde{ \alpha } , \tilde{t}) ) \ge 1-6\alpha ^6$. This finishes the proof of the second part of the theorem.
The third and fourth part of the theorem follows from Theorem~\ref{thm:increment:formal}. The last part of the theorem follows from Proposition~\ref{prop:ind1:growth}.
This finishes the proof of all parts of the theorem except for part one which holds with probability $1-4 \alpha ^6$ instead of deterministically. In order to fix this issue we change the definition of $\tilde{\alpha} $ slightly. On the event where part (1) doesn't hold we change $\tilde{\alpha }$ to be $\alpha$. It is clear that parts (2),(3) and (5) still hold. Part (4) of the theorem still holds by Corollary~\ref{cor:increment:formal:goodevent}.
\end{proof}
\subsection{Stitching intervals}\label{sec:stitching}
In this section we use Theorem~\ref{thm:induction step for repeatd use} repeatedly to show that at time $t$ the aggregate speed is $\approx t^{-\frac{1}{3}}$. To this end, we stitch a lot of short intervals of time to obtain the right growth in a medium interval of time. Then we stitch a lot of medium intervals of time to obtain the right growth in a long interval of time. Finally, we stitch a lot of long intervals of time to obtain the right growth in the time interval $[0,t]$. See Subsections~\ref{sec:short},\ref{sec:medium} and \ref{sec:long}.
\subsubsection{Stitching short intervals}\label{sec:short}
In this section we build inductively a process that approximates the speed of the aggregate and study the time it takes for this process to get smaller or larger by a factor of $2$.
For the proof of the main lemma of this subsection we'll need the following martingale concentration result due to Freedman \cite{freedman1975tail}. See also Theorem 18 in \cite{chung2006concentration} for a short proof.
\begin{claim}\label{claim:bernstein}
Let $X_0,X_1,\dots ,X_n $ be a martingale with filtration $\mathcal F _0,\dots ,\mathcal F _n$. Suppose that almost surely for all $1\le i \le n$ we have:
\begin{equation}
\mathrm{Var} (X_i \ | \ \mathcal F _{i-1}) \le \sigma _i^2 \quad \text{and} \quad \left| X_i-X_{i-1} \right| \le M
\end{equation}
Then,
\begin{equation}
\mathbb P \left( |X_n-X_0|\ge \lambda \right) \le 2\exp \left(-\frac{\lambda ^2 }{\sum _{i=1}^n \sigma _i^2 +\frac{1}{3}M \lambda }\right).
\end{equation}
\end{claim}
Let $t_0>0$ and let $\delta _0>0$ sufficiently small. Let $\alpha _0>0$ such that $|\alpha _0-\delta _0| \le \delta _0^{1.1}$. Here $\delta _0$ should be understood as the scale of the speed and $\alpha _0$ as the initial speed. Let $Y _{t_0} \in \mathfrak{R}'(\alpha _0 ,t_0)$ and let $\Pi _S$ be the aggregate with initial condition $Y _{t_0}$. Let $t_1 > 0 $ such that
\begin{equation}
\delta _0 ^{-2}\log ^{10\theta -3}(1/\delta _0) \le t_1- t_0 \le \delta _0 ^{-2}\log ^{10\theta -1 }(1/\delta _0)
\end{equation}
and $t_i:=t_0+i(t_1-t_0)$ for any $i\ge 2$. We define the sequence $\alpha _i$ inductively. Suppose that $\alpha _1,\dots ,\alpha _i$ were already defined. Define the stopping times
\begin{equation}
\begin{split}
&\zeta _1:=\min \left\{t_i\ge t_0:\ \alpha _i < \delta _0/2 \right\}\\
&\zeta _2 :=\min \{t_i\ge t_0:\ \alpha _i > 2 \delta _0 \} \\
&\zeta _4:=\min \{t_i \ge t_0: Y_{t_i} \notin \mathfrak{R} '(\alpha _i,t_i) \}\\
& \zeta _5:=\min \{t_{i+1}\ge t_1: \left| X_{t_{i+1}}-X_{t_{i}}-\alpha _{i} (t_1-t_0) \right| \ge \alpha _i ^{1.4}(t_1-t_0) \},
\end{split}
\end{equation}
$\zeta _3:=\zeta _4 \wedge \zeta _5$ and $\zeta :=\zeta _1 \wedge \zeta _2 \wedge \zeta _3 $. We note that so far $\zeta $ is not well defined because $\alpha _j$ for $j>i$ was not defined but the event $\{\zeta >t_i\}$ is well defined. On the event $\{ \zeta >t_i \}$ we have that $Y_{t_i} \in \mathfrak{R} '(\alpha _i,t_i)$ and therefore we can define $\alpha _{i+1} \in \mathcal F _{t_i}$ to be the random variable $\tilde{\alpha }$ from Theorem~\ref{thm:induction step for repeatd use} with $t_i$ and $t_{i+1}$ as $t$ and $\tilde{t}$ respectively and with $\alpha _i$ as $\alpha $. On the event $\{\zeta \le t_i\}$ we just let $\alpha _{i+1}:=\alpha _i$. Using Theorem~\ref{thm:induction step for repeatd use} we have almost surely on the event $\{\zeta >t_{n}\}$ that $\mathbb P (\zeta _3=t_{n+1} \ | \ \mathcal F _{t_{n}})\le 2\alpha _{n}^5 \le C \delta _0^5$ and moreover
\begin{equation}
\left| \mathbb E ( \alpha _{n+1} -\alpha _n \ | \ \mathcal F _{t_{n}}) \right| \le \frac{(t_1-t_0)\alpha _{n}^4}{\log ^3 (1/\alpha _{n})} \le C\frac{(t_1-t_0)\delta _0^4}{\log ^3 (1/\delta _0)}.
\end{equation}
\begin{lem}\label{lem:stitching short intervals}
The stopping time $\zeta $ satisfies the following properties
\begin{enumerate}
\item
\begin{equation}
\mathbb P (\zeta =\zeta _3 ) \le C \delta _0 ^3
\end{equation}
\item
\begin{equation}
\mathbb P (\zeta =\zeta _1 )=\frac{2}{3}+O(\log ^{-1}(1/\delta _0)), \quad \mathbb P (\zeta =\zeta _2 )=\frac{1}{3}+O(\log ^{-1}(1/\delta _0))
\end{equation}
\item
For all $k \ge 1 $,
\begin{equation}
\mathbb P (\zeta \ge t_0 +k \delta _0^{-3} ) \le Ce^{-ck}
\end{equation}
\item
For all $ \epsilon <1$,
\begin{equation}
\mathbb P (\zeta \le t_0+ \epsilon \delta _0^{-3} )\le C\delta _0^4 +Ce^{-c \epsilon ^{-1}}
\end{equation}
\end{enumerate}
\end{lem}
\begin{proof}
We start by studying the Doob decomposition of $\alpha _i$. Consider the martingale
\begin{equation}
\beta _n:=\alpha _0+\sum _{i=1}^{n} \alpha_{i }- \mathbb E [\alpha _i \ \big| \ \mathcal F _{t_{i-1}}]= \alpha _n-\sum _{i=1}^n \mathbb E [\alpha _i \ | \ \mathcal F _{t_{i-1}}]- \alpha _{i-1 }.
\end{equation}
We show that the increments of $\beta _n$ are close to those of $\alpha _n$. We have
\begin{equation}\label{eq:increments are close}
\begin{split}
| (\beta _{n+1}-\beta _n) -(\alpha _{n+1 }-\alpha _{n }) | \le \mathds 1\{\zeta >t_n\} \left| \mathbb E [\alpha _{n+1}\ | \ \mathcal F {t_n}]-\alpha _n \right| \le C\mathds 1 \{\zeta >t_n\} \frac{\delta _0^4 (t_1-t_0)}{\log ^3 (1/\delta _0 )}.
\end{split}
\end{equation}
Thus, for any $n\le 2 \delta _0^{-3} \log ^2 (1/\delta _0)/(t_1-t_0)$ we have that
\begin{equation}\label{eq:beta is close to alpha}
|\beta _n-\alpha _{n }|\le C n\frac{ \delta _0^4 (t_1-t_0)}{\log ^3 (1/\delta _0)}\le C\frac{\delta _0 }{\log (1/\delta _0)}.
\end{equation}
and in particular $0.4 \delta _0<\beta _n \le 2.1 \delta _0$ for such $n$. By \eqref{eq:increments are close} we also have
\begin{equation}\label{eq:bound on increments of beta_n}
|\beta _{n+1}-\beta _n |\le |\alpha _{n+1 }-\alpha _{n }|+C\frac{\delta _0^4(t_1-t_0)}{\log (1/\delta _0)}\le C\sqrt{\delta _0^5(t_1-t_0)}\log ^{2 \theta } (1/\delta _0)
\end{equation}
and therefore, using \eqref{eq:increments are close} once again
\begin{equation}
\begin{split}
|(\beta _{n+1}-\beta _n)^2-(\alpha _{n+1 }-\alpha _{n })^2 |\le C\sqrt{\delta _0^5(t_1-t_0)}\log ^{2\theta } (1/\delta _0) | (\beta _{n+1}-\beta _n) -(\alpha _{n+1 }-\alpha _{n }) |\\
\le C \mathds 1 \{\zeta >t_n\}\delta _0^{\frac{13}{2}} \log ^{2 \theta } (1/\delta _0) (t_1-t_0)^{\frac{3}{2}}\le C\mathds 1 \{\zeta >t_n\} \delta _0^{\frac{7}{2}} \log ^{2 \theta }(1/ \delta _0).
\end{split}
\end{equation}
Thus
\begin{equation}\label{eq:variance of increments}
\begin{split}
\mathbb E \left[ (\beta _{n+1}-\beta _n)^2 \ | \ \mathcal F _{t_n} \right]&=\mathbb E \left[ (\alpha_{n+1 }- \alpha _{n })^2 | \ \mathcal F _{t_n} \right] +\mathds 1 \{\zeta >t_n\} O (\delta _0^{\frac{7}{2} }\log ^{2 \theta }(1/\delta _0 )) \\
&=\mathds 1 \{\zeta > t_n\} 4\alpha _n ^5(t_1-t_0) (1+O(\delta _0 ^{0.4} ))
\end{split}
\end{equation}
It follows that $\mathbb E \left[ (\beta_{n+1}-\beta _n )^2 \right]\ge 0.1 \delta _0^5 (t_1-t_0)\mathbb P (\zeta >t_n )$. Now, letting $n_0:=\lceil 100\delta _0^{-3} /(t_1-t_0) \rceil $ we have that
\begin{equation}
\begin{split}
5 \delta _0^2 \ge \mathbb E (\beta _{n_0+1}^2 )&\ge \mathrm{Var} (\beta _{n_0+1} )=\sum _{n=0}^{n_0} \mathbb E \left[ (\beta _{n+1}-\beta _n)^2 \right] \\
&\ge 0.1 n_0 \delta _0^5 (t_1-t_0) \mathbb P (\zeta \ge t_{n_0} ) \ge 10 \delta _0^2 \cdot \mathbb P (\zeta \ge t_{n_0}).
\end{split}
\end{equation}
We get that $ \mathbb P (\zeta > t_{n_0} )\le 1/2$. Let $k\ge 1$. By repeating the above arguments on the event $\{\zeta >t_{(k-1)n_0} \}$ with the initial condition $Y _{ t_{(k-1)n_0}}$ instead of $Y _{t_0}$ and the random variable $\alpha _{(k-1)n_0}$ instead of $\alpha _0$ we get that $\mathds 1 \{\zeta >t_{(k-1)n_0}\} \cdot \mathbb P (\zeta > t_{kn_0} \ | \ \mathcal F _{t_{(k-1)n_0}})\le 1/2$ almost surely (we note that we might not have $|\alpha _{kn_0}-\delta _0|\le \delta _0^{1.1}$ but we did not use the fact that $| \alpha _0-\delta _0|\le \delta _0^{1.1}$ in the proof of $ \mathbb P (\zeta > t_{n_0} )\le 1/2$). Thus, inductively we have that
\begin{equation}\label{eq:probability that the stopping time is larger than kn_0}
\begin{split}
\mathbb P (\zeta > t_{kn_0} )=\mathbb E \left[ \mathds 1 \{\zeta \ge t _{(k-1)n_0}\} \cdot \mathbb P (\zeta > t_{kn_0} \ | \ \mathcal F _{t_{(k-1)n_0}}) \right] \le \frac{1}{2}\mathbb P (\zeta > t_{(k-1)n_0} ) \le \cdots \le 2^{-k}.
\end{split}
\end{equation}
Finally, by the definition of $n_0$, for $c_1=0.001$ and $k$ sufficiently large we have that $t_{\lfloor c_1k \rfloor n_0}=t_0+ \lfloor c_1k \rfloor n_0 (t_1-t_0)\le t_0+k\delta _0^{-3} $ and therefore
\begin{equation}
\mathbb P (\zeta > t_0+k\delta _0^{-3} ) \le \mathbb P (\zeta > t_{ \lfloor c_1k \rfloor n_0} ) \le 2^{-\lfloor c_1 k \rfloor } \le C e^{-ck}.
\end{equation}
This finishes the proof of the third part.
We turn to prove the first part. For any $n\ge 1$
\begin{equation}\label{eq:zeta 3 happens fast}
\begin{split}
\mathbb P \left( \zeta _3 =\zeta \le t_n \right) &=\mathbb P \left( \bigcup _{i=1}^n \{\zeta _3 =t_i \} \cap \{\zeta \ge t_{i-1} \} \right) \le \sum _{i=1}^n \mathbb P \left( \zeta > t_{i-1} \text{ and } \zeta _3=i \right)\\
&=\sum _{i=1}^n \mathbb E \left[ \mathds 1 \{\zeta > t_{i-1} \} \cdot \mathbb P ( \zeta _3=t_i \ | \ \mathcal F _{t_{i-1}}) \right]\le C n \delta _0^5.\\
\end{split}
\end{equation}
Thus, if we let $n_1:=\lfloor \delta _0^{-1} \rfloor n_0 \le \delta _0^{-2}$ we get by \eqref{eq:probability that the stopping time is larger than kn_0} that
\begin{equation}
\mathbb P \left( \zeta _3 =\zeta \right)\le \mathbb P \left( \zeta _3 =\zeta \le t_{n_1} \right)+\mathbb P (\zeta >t_{n_1} )\le C\delta _0^3+Ce^{-c\delta _0^{-1}} \le C\delta _0^3.
\end{equation}
We turn to prove the fourth part of the lemma. Let $\log ^{-2} (1/\delta _0) \le \epsilon <1$ and let $n_3:= \lceil \epsilon \delta _0^{-3}/(t_1-t_0)\rceil \le \delta _0 ^{-1}$. Define
\begin{equation}
M:=C\sqrt{\delta _0^5 (t_1-t_0)}\log ^{2 \theta } (1/\delta _0) \le \delta _0^{\frac{3}{2}}\log ^{C_\theta } (1/\delta _0),\quad \sigma :=C\sqrt{\delta _0^5 (t_1-t_0)}
\end{equation}
The martingale $\beta _n$ satisfy the assumptions of Theorem~\ref{claim:bernstein} with this $M$ and $\sigma _i=\sigma $ by equations \eqref{eq:bound on increments of beta_n} and \eqref{eq:variance of increments}. Thus, taking $\lambda =\delta _0/4 $ we get
\begin{equation}
\mathbb P ( |\beta _{n_3}-\alpha _0| \ge \lambda ) \le 2 \exp \left( -\frac{\lambda ^2 }{n_3\sigma ^2 +\frac{1}{3}M \lambda} \right) \le 2 \exp \left( -\frac{c \delta _0^2 }{2 \epsilon \delta _0^2 +c\delta _0^{\frac{5}{2}}\log ^C(1/\delta _0)} \right)\le C e^{-c\epsilon ^{-1} }
\end{equation}
Thus, using that $|\alpha _0-\delta _0| \le \delta _0^{1.1}$ we get
\begin{equation}
\begin{split}
\mathbb P (\zeta \le t_0 +\epsilon \delta _0^{-3} ) &\le \mathbb P (\zeta \le t_{ n_3} )\le \mathbb P (\zeta _3=\zeta \le t_{ n_3} ) + \mathbb P (\zeta _1 \wedge \zeta _2=\zeta \le t_{ n_3} ) \\
&\le C n_3 \delta _0^5+ \mathbb P (|\alpha _{n_3}-\alpha _0| \ge 4 \delta _0/10) \\
&\le C \delta _0 ^4+ \mathbb P (|\alpha _{n_3}-\beta _{n_3}| > \delta _0/10)+\mathbb P (|\beta _{n_3}-\alpha _0| \ge \delta _0/4) \le C\delta _0^4 +Ce^{-c \epsilon ^{-1}},
\end{split}
\end{equation}
where in the third inequality we used \eqref{eq:zeta 3 happens fast} and in the fifth inequality we used \eqref{eq:beta is close to alpha}. This finishes the proof of the fourth part of the lemma when $\epsilon \ge \log ^{-2}(1/\delta _0)$. It is clear that the same inequality holds for $\epsilon <\log ^{-2}(1/\delta _0)$ as well (since the $e^{-c\epsilon ^{-1}}<\delta _0^4$ in this case).
Finally, we prove the second part of the lemma. Let $n_4:=\lceil \log ^2 (1/\delta _0) \delta _0^{-3}/(t_1-t_0) \rceil \le \delta _0^{-1}$ and
\begin{equation}
\mathcal A := \{\zeta =\zeta _1\le t_{n_4}\},\quad \mathcal B:=\{\zeta =\zeta _2\le t_{n_4}\} ,\quad \mathcal C :=\{ \zeta _1\wedge \zeta _2 >\zeta \text{ and } \zeta \le t_{n_4} \} \cup \{ \zeta >t_{n_4}\}.
\end{equation}
We have
\begin{equation}\label{eq:equality of expectations}
\delta _0+O(\delta _0^{1.1})=\alpha _0=\mathbb E [\beta _{n_4}]= \mathbb E [\mathds 1 _{\mathcal A}\beta _{n_4}] +\mathbb E [\mathds 1 _{\mathcal B}\beta _{n_4}]+\mathbb E [\mathds 1 _{\mathcal C}\beta _{n_4}].
\end{equation}
Next, we estimate each one of the terms separately. We have
\begin{equation}\label{eq:prob of C}
\mathbb P( \mathcal C ) \le \mathbb P (\zeta =\zeta _3 \le t_{n_4} )+ \mathbb P (\zeta \ge t_{n_4} ) \le C n_4 \delta _0^5+C e^{-c \log ^2(1/\delta _0)} \le C \delta _0^4,
\end{equation}
where in the second inequality we used \eqref{eq:zeta 3 happens fast} and \eqref{eq:probability that the stopping time is larger than kn_0}. Thus,
\begin{equation}\label{eq:expectation on C}
\mathbb E [\mathds 1 _{\mathcal C}\beta _{n_4}] \le C \delta _0^4.
\end{equation}
On the event $\mathcal A$ we have
\begin{equation}
|\beta _{n_4}- \delta _0/2 |\le | \beta _{n_4}-\alpha _{n_4}| +|\alpha _{n_4}-\delta _0/2| \le C \frac{\delta _0}{\log (1/\delta _0)} +C\delta _0^{\frac{3}{2}} \log ^C (1/\delta _0) \le C \frac{\delta _0}{\log (1/\delta _0)},
\end{equation}
where in the second inequality we bound the first term using \eqref{eq:beta is close to alpha} and bound the second term using the definition of $\zeta _1$ and the fact that the increments of $\alpha _n$ are small. Thus
\begin{equation}\label{eq:expectation on A}
\mathbb E [\mathds 1 _{\mathcal A}\beta _{n_4} ]=\frac{\delta _0}{2} \cdot \mathbb P (\mathcal A ) +O\left(\frac{\delta _0}{ \log (1/\delta _0)}\right)
\end{equation}
Using the same arguments we have
\begin{equation}\label{eq:expectation on B}
\begin{split}
\mathbb E [\mathds 1 _{\mathcal B}\beta _{n_4}]=2 \delta _0 \cdot \mathbb P (\mathcal B )+O\left(\frac{\delta _0}{ \log (1/\delta _0)}\right)
&= 2\delta _0 \left(1- \mathbb P (\mathcal A ) +O(\delta _0 ^4) \right)+O\left(\frac{\delta _0}{ \log (1/\delta _0)}\right) \\
& =2 \delta _0-2\delta _0 \cdot \mathbb P (\mathcal A ) +O\left(\frac{\delta _0}{ \log (1/\delta _0)}\right),
\end{split}
\end{equation}
where in the second equality we used \eqref{eq:prob of C}.
Substituting \eqref{eq:expectation on A},\eqref{eq:expectation on B} and \eqref{eq:expectation on C} into \eqref{eq:equality of expectations} we get
\begin{equation}
\delta _0= \frac{\delta _0}{2} \mathbb P (\mathcal A )+2 \delta _0-2 \delta _0 \cdot \mathbb P (\mathcal A )+O\left(\frac{\delta _0}{ \log (1/\delta _0)}\right)
\end{equation}
which means that
\begin{equation}
\mathbb P (\mathcal A )= \frac{2}{3} +O\left( \log ^{-1} (1/\delta _0)\right),\quad \mathbb P (\mathcal B )= \frac{1}{3} +O\left( \log ^{-1} (1/\delta _0)\right)
\end{equation}
The second part of the lemma follows as the event $\{\zeta =\zeta _1\} \setminus \mathcal A \subseteq \{\zeta >t_{n_4}\}$ has a small probability.
\end{proof}
\begin{remark}
On the event $\{\zeta <\zeta _3\}$ (which happens with high probability by Lemma~\ref{lem:stitching short intervals} ) we have that
\begin{equation}
\frac{\delta _0}{3} (\zeta-t_0) \le X_{\zeta }-X_{t_0} \le 3 \delta _0 (\zeta -t_0)
\end{equation}
\end{remark}
In the next subsection we are going to change $t_0$, $\alpha _0$ and $\delta _0$ and therefore we write more specifically $\zeta (t_0,\alpha _0, \delta _0)$ for the stopping time $\zeta $ and similarly with $\zeta _1,\zeta _2 ,\zeta _3$.
We define the random variable $\alpha '=\alpha '(t_0,\alpha _0, \delta _0)\in \mathcal F _{\zeta }$ as follows. on the event $\{ \zeta =t_i \}$, we let $\alpha ':=\alpha _i$.
\subsubsection{Stitching medium intervals}\label{sec:medium}
In this section we stitch the medium intervals of time together to create the large intervals of time in which the speed decreases with high probability. The idea is to repeatedly use the previous section to show that the speed behaves somewhat like a random walk with a negative drift on the dyadic scales. As in the previous section we let $t_0>0$, $\delta _0>0$ sufficiently small and $\alpha _0 >0$ with $|\alpha _0-\delta _0| \le \delta _0^{1.1}$. We also let $M >0$ with $2\delta _0\le M \le \sqrt{\delta _0}$. Finally, let $Y _{t_0} \in \mathfrak{R}'(\alpha _0,t_0)$ and let $\Pi _S$ be the aggregate with initial condition $Y _{t_0}$. We define the processes $\delta _i, \alpha _i$ and the stopping times $t_i$ inductively (note that the $t_i$ are defined not as in the previous subsection). Suppose we defined $\delta _1,\dots ,\delta _i,\alpha _1,\dots ,\alpha _i$ and $t_1,\dots ,t_i$ (note that the definition of $t_i$ will be different this time). Define the stopping times
\begin{equation}
\begin{split}
&\zeta _1':=\min \left\{t_i\ge t_0: \delta _i\le \delta _0/2 \right\}\\
&\zeta _2' :=\inf \{t_i\ge t_0:\delta_i > M \} \\
& \zeta _3' :=\inf \{t_i\ge t_1: \zeta (t_{i-1},\alpha _{i-1},\delta _{i-1})=\zeta _3(t_{i-1},\alpha _{i-1},\delta _{i-1})\}
\end{split}
\end{equation}
and $\zeta '=\zeta _1'\wedge \zeta _2 ' \wedge \zeta _3' $. Let $t_{i+1}:=\zeta (t_i,\alpha _i,\delta _i)$. We define $\delta _{i+1}\in \mathcal F _{t_{i+1}}$ as follows
\begin{equation}
\delta _{i+1}:=
\begin{cases}
\delta _{i}/2,\quad &\text{on the event }\{\zeta '>t_{i} \text{ and } \zeta '_3>t_{i+1} \text{ and } \zeta (t_i,\alpha _i,\delta _i)=\zeta _1(t_i,\alpha _i,\delta _i) \}\\
\ \delta _{i},\quad &\text{on the event }\{\zeta '\le t_{i} \text{ or } \zeta '_3=t_{i+1} \}\\
2\delta _{i},\quad &\text{on the event }\{\zeta '>t_{i} \text{ and } \zeta '_3>t_{i+1} \text{ and } \zeta (t_i,\alpha _i,\delta _i)=\zeta _2(t_i,\alpha _i,\delta _i) \}
\end{cases}
\end{equation}
We also let $\alpha _{i+1}:=\alpha '(t_i,\alpha _i,\delta _i)$ where $ \alpha '$ is defined in the end of the previous section. On the event $\{\zeta '> t_i\}$ we have that $\delta _i \le M$, and there exist a random variable $\alpha _i$ with $|\alpha _i -\delta _i|\le \delta _i^{1.1}$ so that $Y_{t_i} \in \mathfrak{R}'(\alpha _i,t_i)$. Thus, on this event, we can apply Lemma~\ref{lem:stitching short intervals} with $t_i, \delta _i$ and $\alpha _i$ as $t_0, \delta _0$ and $\alpha _0$ respectively. By the lemma, with probability at least $1-C \delta _i^3$ we have that $\zeta _3 '>t_{i+1}$. The reader should think of $\delta _i$ as the sequence of different dyadic scales that the speed traveled through.
\begin{lem}\label{lem:medium}\label{lem:stitching medium}
The stopping time $\zeta '$ satisfies the following properties:
\begin{enumerate}
\item
\begin{equation}
\mathbb P (\zeta ' =\zeta '_3 ) \le C \delta _0
\end{equation}
\item
\begin{equation}
\mathbb P (\zeta ' =\zeta '_2 ) \le \left( \frac{\delta _0}{M} \right)^c
\end{equation}
\item
For all $k\ge 1 $,
\begin{equation}
\mathbb P ( \zeta ' >t_0+ k \delta _0^3 ) \le C e^{-c\sqrt{k}}
\end{equation}
\item
For all $\epsilon <1$,
\begin{equation}
\mathbb P ( \zeta ' <t_0+ \epsilon \delta _0^3 ) \le C \delta _0^4+ C e^{-c\epsilon ^{-1}}
\end{equation}
\end{enumerate}
\end{lem}
\begin{proof}
Define $W_i:=\log _2 (\delta _i/\delta _0)$. Let $n\ge 1 $ and let $\zeta _1,\zeta _2 ,\zeta _3 ,\zeta $ be the stopping times from the previous subsction with $t_0,\alpha _0,\delta _0$ being $t_n,\alpha _n,\delta _n$ respectively. By Lemma~\ref{lem:stitching short intervals} we have that
\begin{equation}
\begin{split}
\mathbb E (W_{n+1}-W_n \ | \ \mathcal F _{t_n} )&=\mathds 1 \{ \zeta '>t_n \} \left( -\mathbb P ( \delta _{i+1}=2 \delta _i \ | \ \mathcal F _{t_n}) +\mathbb P ( \delta _{i+1}= \delta _i /2 \ | \ \mathcal F _{t_n}) \right) \\
&= \mathds 1 \{ \zeta '>t_n \} \left( -\frac{1}{3} +O\left(\frac{1}{\log (1/\delta _n)} \right) \right)\le -\frac{1}{4} \mathds 1 \{ \zeta '>t_n \},
\end{split}
\end{equation}
where the last inequality holds when $\delta _0$ is sufficiently small and as $\delta _n\le 2M \le 2 \sqrt{\delta _0}$. Thus, the process $M_n:=W_n+\frac{1}{4} \sum _{j=0}^{n-1 } \mathds 1 \{\zeta ' >t_j\}$ is a super-martingale with increments bounded in absolute value by $2$. We get, by Azuma inequality
\begin{equation}\label{eq:zeta' larger than t_n}
\mathbb P (\zeta '>t_n )=\mathbb P (W_n\ge 1, \zeta '>t_n )\le \mathbb P (M_n-M_0 \ge n/4 ) \le Ce^{-cn}.
\end{equation}
We can now prove the second part of the lemma. By \eqref{eq:zeta' larger than t_n} with $n_0:=\frac{1}{2} \log _2 (\frac{M}{\delta _0})$ we have
\begin{equation}
\mathbb P (\zeta '=\zeta _2' ) \le \mathbb P (\zeta '\ge t_{n_0} ) \le C e^{-cn_0} \le \left(\frac{\delta _0}{M}\right) ^c.
\end{equation}
We turn to prove the first part of the lemma. For all $n\ge 1 $, by Lemma~\ref{lem:stitching short intervals} we have that
\begin{equation}
\begin{split}
\mathbb P (\zeta '=\zeta _3'= t_n ) \le \mathbb E \left[ \mathds 1 \{\zeta '>t_{n-1} \} \cdot \mathbb P (\zeta _3' =t_n \ | \ \mathcal F _{t_{n-1}}) \right] \le \mathbb E \left[ \mathds 1 \{\zeta '>t_{n-1} \} C\delta _{n-1}^3 \right] \le C M^3
\end{split}
\end{equation}
and therefore $\mathbb P (\zeta '=\zeta _3' \le t_n ) \le CnM^3$. Thus, using also \eqref{eq:zeta' larger than t_n} with $n_1:=\lfloor \delta _0^{-\frac{1}{2}} \rfloor $ we get
\begin{equation}
\mathbb P (\zeta '=\zeta '_3 ) \le \mathbb P (\zeta '=\zeta '_3\le t_{n_1} )+\mathbb P (\zeta '\ge t_{n_1} ) \le C \delta _0.
\end{equation}
Next, we prove part (3). The lower bound for $\zeta '$ follows immediately from Lemma~\ref{lem:stitching short intervals}. Indeed, $\zeta '\ge \zeta (t_0,\alpha _0,\delta _0)$. We turn to prove the upper bound. By \eqref{eq:zeta' larger than t_n} we have that
\begin{equation}
\begin{split}
\mathbb P (\zeta ' \ge t_0+k \delta _0^{-3} ) &\le \mathbb P ( \zeta '\ge t_n )+\sum _{i=1}^n \mathbb P (\zeta '=t_i \ge t_0+k \delta _0^{-3} ) \\
&\le C e^{-cn} +\sum _{i=1}^n \sum _{j=1}^i \mathbb P \left( \zeta '=t_i, \ t_j-t_{j-1}>\frac{k}{n}\delta _0^{-3} \right)\\
&\le C e^{-cn} +\sum _{i=1}^n \sum _{j=1}^i \mathbb E \left[ \mathds 1 \{\zeta '>t_{j-1}\} \mathbb P \left( t_j-t_{j-1}>\frac{k}{n}\delta _0^{-3} \ | \ \mathcal F _{t_{j-1}} \right) \right]\\
&\le C e^{-cn } + \sum _{i=1}^n \sum _{j=1}^i e^{-ck/n} \le Ce^{-cn} +n^2e^{-ck/n},
\end{split}
\end{equation}
where the fourth inequality we used Lemma~\ref{lem:stitching short intervals}. The third part of the lemma follows from the last result by substituting $n=\sqrt{k}$.
\end{proof}
On the event $\{\zeta '<\zeta '_3\}$ we have that
\begin{equation}
\frac{\delta _0}{3} (\zeta '-t_0) \le X_{\zeta '} -X_{t_0} \le 2M(\zeta '-t_0).
\end{equation}
For $\epsilon _0>0$, we denote the good event by
\begin{equation}
\mathcal A =\mathcal A (t_0,\alpha _0,\delta _0,M,\epsilon ):=\{\zeta '<\zeta '_3 \wedge \zeta '_2\}\cap \{ t_0+\epsilon _0 \delta _0^{-3}\le \zeta ' \le t_0+ \epsilon _0 ^{-1} \delta _0^{-3}\}.
\end{equation}
By Lemma~\ref{lem:stitching medium} we have
\begin{equation}\label{eq:A is likely}
\mathbb P (\mathcal A ^c )\le C \delta _0 +C \left(\frac{\delta _0}{M}\right)^c +Ce ^{-c\epsilon ^{-\frac{1}{2}} } \le C \left(\frac{\delta _0}{M}\right)^c +Ce ^{-c\epsilon ^{-\frac{1}{2}} }
\end{equation}
Define $\alpha''= \alpha ''(t_0,\alpha _0,\delta _0,M):=\alpha '(t_{i-1},\alpha _{i-1},\delta _{i-1})$ on the event $\{ \zeta '=t_i \}$ for $i>0$ and $\alpha ''(t_,\alpha _0,\delta _0,M):=\alpha _0$ on $\{ \zeta '=t_0\}$. Note that, on $\{\zeta '<\zeta ' _2 \wedge \zeta ' _3\}$ we have $|\alpha ''-\delta _0/2|\le (\delta _0/2)^{1.1}$. Since all the parameters are going to change again in the next section we write more specifically $\zeta '(t_0,\alpha _0,\delta _0,M)$ instead of $\zeta '$. We do the same with the other stopping times $\zeta '_1,\zeta '_2$ and $\zeta '_3$.
\subsubsection{Stitching long intervals}\label{sec:long}
In this section we stitch together the long intervals of time in order to prove that aggregate has the $t^{\frac{2}{3}}$ growth.
\begin{thm}\label{thm:stopping time of getting to speed T to the minus one third}
Let $\epsilon , \alpha _0 >0$ such that $\alpha _0$ is sufficiently small depending on $\epsilon $. Let $Y_{t_0} \in \mathfrak{R} ' (\alpha _0, t_0)$ and let $ \Pi _S$ be the aggregate with initial condition $Y_{t_0}$. There exist $C_\epsilon >0$ such that the following holds. Let $t\ge t'(t_0, \alpha_0,\epsilon )$ and define the stopping time
\begin{equation}
\zeta _t:=\inf \{s>t_0 : \exists \alpha \text{ such that } |\alpha -t^{-\frac{1}{3}}| \le t^{-\frac{2}{5}} \text{ and } Y_s \in \mathfrak{R}' ( \alpha, s) \}.
\end{equation}
We have that
\begin{equation}\label{eq:event in theorem}
\mathbb P \left( \zeta _t \le C_\epsilon t \text{ and } X_{\zeta _t} \le C_\epsilon t^{\frac{2}{3}} \right) \ge 1-\epsilon
\end{equation}
\end{thm}
\begin{proof}
Let $m\ge 1$ sufficiently large independently of $t$ and let $n=n_t:=\lceil \log _2(\alpha _0^{-1} t^{\frac{1}{3}}) \rceil $. Define for all $0\le j \le n$
\begin{equation}
\delta _j:=\alpha _0 2^{-j}, \quad \quad M_j:=\begin{cases}
\ \sqrt{\delta _j} ,\quad &j \le \frac{1}{2} n \\
\delta _j 2^{m+\frac{1}{4}(n-j)}, \quad &j> \frac{1}{2} n
\end{cases}, \quad \quad \epsilon _j:=(m+n-j)^{-2}.
\end{equation}
It is easy to verify that $\frac{1}{2}t^{-\frac{1}{3}} \le \delta _{n}\le t^{-\frac{1}{3}} $ and that $\delta _j \le M_j \le \sqrt{\delta _j}$ for all $j$ as long as $t$ is sufficiently large.
Define the sequence of stopping times $t_i$ and the sequence of random variables $\alpha _i\in \mathcal F _{t_i}$ inductively by
\begin{equation}
t_{i+1}:=\zeta ' (t_i,\alpha _i,\delta _i, M_i),\quad \alpha_{i+1}:=\alpha '' (t_i,\alpha _i,\delta _i, M_i)
\end{equation}
Let $\mathcal A _i: =\mathcal A (t_i,\alpha _i,\delta _i,M_i,\epsilon _i)$ and $\mathcal B _j :=\bigcap _{i=1}^j \mathcal A_i$. We start by showing that $\mathbb P (\mathcal B _{n})$ is large. Using \eqref{eq:A is likely} we get
\begin{equation}\label{eq:prob of bad event}
\begin{split}
\mathbb P (\mathcal B _{n} ^c ) \le \mathbb P (\mathcal A _1^c) +\sum _{j=2}^{n} \mathbb P ( \mathcal B _{j-1 } \cap \mathcal A _j ^c )&= \mathbb P (\mathcal A _1^c)+\sum _{j=2}^{n} \mathbb E \left[ \mathds 1 _{\mathcal B _{j-1} }\mathbb P ( \mathcal A _j ^c \ | \ \mathcal F _{t_j-1}) \right] \\
&\le C \sum _{j=1}^{n} \left(\frac{\delta _j}{M_j}\right)^c +C \sum _{j=1}^{n} e^{-c \epsilon _j^{-\frac{1}{2}}}.
\end{split}
\end{equation}
Let $S_1,S_2$ be the corresponding first and second sums in the right hand side of \eqref{eq:prob of bad event}. We bound the second term using the definition of $\epsilon _j$ by
\begin{equation}
S_2\le C e^{-cm } \sum _{j=1}^{n} Ce^{-c(n-j)} \le C e^{-cm } \sum _{k=0}^{\infty }e^{-ck} \le Ce^{-cm}.
\end{equation}
The first term is bounded as follows
\begin{equation}
S_1 \le C \sum _{j =1}^{n/2} \delta _j ^c+Ce^{-cm} \sum _{j= \lceil n /2 \rceil } ^{n} e^{-c(n-j)} \le C \alpha _0^c+C e^{-cm }\sum _{k=1}^{\infty } e^{-ck} \le C \alpha _0^c +C e^{-cm}.
\end{equation}
Since $\alpha _0 $ is sufficiently small and $m$ is sufficiently large we get that $\mathbb P (\mathcal B _{n} ) \ge 1-\epsilon $.
Next, we show that on $\mathcal B _{n}$ the event in \eqref{eq:event in theorem} holds. To this end we bound the stopping time $t_{n}$ and the aggregate size $X_{t_{n}}$ and then show that, on $\mathcal B _{n}$ we have that $\zeta _t \le t_{n}$. We have
\begin{equation}
\begin{split}
t_{n}=t_0+\sum _{j=0}^{n-1} t_{j+1}-t_j \le t_0+\sum _{j=0}^{n} \epsilon _j^{-1} \delta_j^{-3} \le t_0+ \delta _{n}^{-3}\sum _{j=0}^{n} (m+n-j)^2 2^{3(n-j)}\\
\le t_0+C t \sum _{k=0}^{\infty } (2m+k)^2 2^{-3k} \le t_0+Cmt \le C_{t_0,m}t.
\end{split}
\end{equation}
We turn to bound the aggregate size at time $t_n$. We have
\begin{equation}
\begin{split}
X_{t_{n}} \le \sum _{j=0}^{n-1} X_{t_{j+1}}-X_{t_{j}} &\le C\sum _{j=0}^{n-1 } M_j(t_{j+1}-t_{j}) \le C\sum _{j=0}^{n} M_j\epsilon _j^{-1}\delta _j^{-3} \\
&\le C\sum _{j=1}^{n/2} \epsilon _j^{-1} \delta _j^{-3} +C_{m} \sum _{j=\lceil n/2 \rceil }^{n } (m+n-j)^2\delta _j^{-2} 2^{\frac{1}{4}(n-j)} \\
&\le C\epsilon _0^{-1} \delta _{\lfloor n/2 \rfloor }^{-3}+C_m \delta _{n}^{-2} \sum _{j=\lceil n/2 \rceil }^{n }(m+n-j)^2 e^{-c(n-j)} \le \\
&\le C_{m} n^2 \delta _{\lfloor n/2 \rfloor }^{-3} + C_m \delta _{n}^{-2} \sum _{k=0}^{\infty} (m+k)e^{-ck} \\
&\le C_{m} n^2 \delta _{\lfloor n/2 \rfloor }^{-3} + C_m \delta _{n}^{-2}
\le C_{m,\alpha _0} t^{\frac{1}{2}} \log ^2 t+C_{m,\alpha _0} t^{\frac{2}{3}} \le C_{m,\alpha _0} t^{\frac{2}{3}}.
\end{split}
\end{equation}
Thus, it suffices to prove that on $\mathcal B _{n}$ we have $\zeta _t \le t_n$. To this end, recall that the time interval $[t_0,t_{n}]$ is the union of many short intervals. On $\mathcal B _{n}$ the process is regular in the endpoints of each of these intervals with a parameter $\alpha $ that do not change by more than $\alpha ^{\frac{3}{2}} \log ^C (1/\alpha )$ between consecutive short intervals. Thus, by a discrete mean value theorem the process is regular at some time $t'<t_{n}$ with $\alpha '$ such that $| \alpha '- t^{-\frac{1}{3}}|\le t^{-\frac{2}{5}}$. It follows that $\zeta _t \le t_{n}$ on $\mathcal B _{n}$.
\end{proof}
\subsection{Convergence in distribution}\label{sec:convergence in distribution}
In this section we show how to deduce the convergence to the diffusion from the induction step theorem. To this end we'll use the following theorem due to Helland \cite{helland1981minimal}. The theorem we state here is weaker than the result in \cite{helland1981minimal} but suffices for what we need.
\subsubsection{The result of Helland}
We start with the basic settings and notations in the paper of Helland \cite{helland1981minimal} . Let $x_0>0$, $\mu :[0,\infty ]\to \mathbb R $ and $\sigma : [0,\infty ] \to [0,\infty ]$. Let $X(t)$ for $t>0$ be a solution to the stochastic differential equation
\begin{equation}
dX(t)= \mu (X(t)) dt +\sigma (X(t)) d B_t,\quad X(0)=x_0.
\end{equation}
such that almost all sample paths are continuous. Suppose that there are no accessible boundaries. That is $\mathbb P ( \forall t>0 ,\ 0<X(t)<\infty )=1$
For any $n\ge 1$, let $X^{(n)}(t)>0$ be a sequence of processes. Let $\Delta _n \to 0$ and for any $k \ge 0$ let $t_k=t_k^{(n)}:=k \Delta _n$. In what follows $\overset{p}{\longrightarrow }$ denotes convergence in probability. Suppose that:
\begin{enumerate}
\item
$X_n(0)\overset{p}{ \longrightarrow } x_0$
\item
$X_n(t)$ is fixed on the intervals $[t_k, t_{k+1})$
\item
For all $t,\epsilon >0$ we have as $n \to \infty $
\begin{equation}
\sum _{\substack{k: \\ t_k \le t}} \mathbb P \left( |X_n(t_{k+1})-X_n(t_k) | \ge \epsilon \ \big| \ \mathcal F _{t_k} \right) \overset{p}{\longrightarrow } 0
\end{equation}
\item
For all $t>0$ and a compact set $K \subseteq (0,\infty )$ we have as $n \to \infty $
\begin{equation}
\sum _{\substack{k: \\ t_k \le t}} \Big| \mathbb E \left( X_n(t_{k+1})-X_n(t_k) \ \big| \ \mathcal F _{t_k} \right)-\mu (X_n(t_k)) \cdot \Delta _n \Big| \cdot \mathds 1 \{X_n(t_k) \in K\} \overset{p}{\longrightarrow } 0
\end{equation}
\item
For all $t>0$ and a compact set $K \subseteq (0,\infty )$ we have as $n \to \infty $
\begin{equation}
\sum _{\substack{k: \\ t_k \le t}} \Big| \mathbb E \left( (X_n(t_{k+1})-X_n(t_k) )^2 \ \big| \ \mathcal F _{t_k} \right) -\sigma ^2 (X_n(t_k)) \Delta _n \Big| \cdot \mathds 1 \{X_n(t_k) \in K\} \overset{p}{\longrightarrow } 0
\end{equation}
\end{enumerate}
\begin{thm}[Helland]\label{thm:helland}
Suppose that the assumptions $(1)-(5)$ above hold. Then,
\begin{equation}
(X_n(t))_{t>0} \overset{d}{ \longrightarrow } (X(t))_{t>0}
\end{equation}
\end{thm}
\begin{remark}\label{remark:on convergence}
The convergence in Theorem~\ref{thm:helland} is in the sense of the Stone topology defined in \cite{stone1963weak} and discussed in \cite{helland1981minimal}. We will not define this topology. Instead, we just note that:
\begin{enumerate}
\item
when $X$ is almost surely continuous, the convergence is equivalent to the following two conditions
\begin{enumerate}
\item
The finite dimensional distributions of $X_n$ converge weakly to the finite dimensional distributions of $X$.
\item
For all $\epsilon ,T >0$ we have
\begin{equation}
\lim _{\delta \to 0} \limsup _{n\to \infty } \mathbb P \bigg( \sup _{\substack{ 0 \le s,t\le T \\ |s-t|\le \delta }} \left|X_n(t)-X_n(s) \right| \ge \epsilon \bigg) =0
\end{equation}
\end{enumerate}
\item
It can be shown, using Skorhod representation theorem, that if $X$ is almost surely continuous then there is a coupling of the processes $X_n$ and $X$ such that for all $T>0$ almost surely we have $\sup _{t\le T} |X_n(t)-X(t)| \to 0$.
\end{enumerate}
\end{remark}
\subsubsection{Proof of Theorem~\ref{thm:main theorem 2}}
Throughout this section $X_t$ is the size of the usual aggregate with no initial condition. Let $\epsilon >0$ sufficiently small, $\alpha _0 >0$ sufficiently small depending on $\epsilon $ and let $t_0:=\alpha _0^{-20}$. Recall the definition of $\overline{Y} _{t,\alpha }$ given in \eqref{eq:def:sandwichfixed}. Define the event
\begin{equation}\label{eq:def of A_0}
\mathcal A _0 := \{ X_{t_0} \le t_0 \} \cap \{\forall s >0,\ Y_{t_0}(s) \le \overline{Y} _{t_0,\alpha _0} (s) \} \cap \{ \overline{Y} _{t_0,\alpha _0} \in \mathfrak{R} (\alpha _0, \alpha _0^8 ;[t_0]) \}.
\end{equation}
The first event on the right hand side of \eqref{eq:def of A_0} holds with high probability as $S(t)\le \frac{1}{2}$ almost surely and therefore $X_{t_0}\precsim \text{Poisson}(t_0/2)$. The second and third events hold with high probability by Lemma~\ref{lem:aggregate is bounded} and Theorem~\ref{thm:induction:base:main} respectively. Thus $\mathbb P (\mathcal A _0) \ge 1-\epsilon $ as long as $\alpha _0$ is sufficiently small depending on $\epsilon $. Let $( \overline{Y}_t, \overline{S}(t) ,\overline{X} _t)$ be the aggregate with initial condition $\overline{Y} _{t_0,\alpha _0}$. We note that, according to Definition~\ref{def:aggregate with initial condition} we have that $\overline{X} _{t_0}=0$ and by Claim~\ref{claim:aggregate with initial condition} we have almost surely on $\mathcal A _0$ that $X_t \le t_0+\overline{X} _t$ for all $t>t_0$.
We turn to bootstrap the regularity by applying Theorem~\ref{subsec:regoverview:overview}. Let $t_1:=t_0+\alpha _0^{-2} \log ^{10 \theta }(1/\alpha _0)$. Define
\begin{equation}
\mathcal A_1 := \{\overline{X}_{t_1} \le t_0 \} \cap \{ \exists \alpha _1 \text{ with } \overline{Y} _{t_1} \in \mathfrak{R} '(\alpha _1 ,t_1) \}.
\end{equation}
Almost surely on $\mathcal A _0$ we have that $\mathbb P (\mathcal A _1 \ | \ \mathcal F _{t_0} )\ge 1- \epsilon $. Indeed, the first event holds with high probability by the same argument as above. For the second one, by Theorem~\ref{subsec:regoverview:overview}, there is a random variable $\alpha _1'\le 2 \alpha _0$ such that almost surely on $\mathcal A _0$ we have $\mathbb P ( Y_{t_1}\in \mathfrak{R} ^\sharp (\alpha _1', \alpha _1^8,[t_1] \ | \ \mathcal F _{t_0} )\ge \alpha _0^{10}$ and therefore, by Definition~\ref{def:def of regularity prime} there is some $\alpha _1 \le 3 \alpha _0$ such that almost surely on $\mathcal A _0$, $\mathbb P ( Y_{t_1}\in \mathfrak{R} ' (\alpha _1, t_1 ) \ | \ \mathcal F _{t_0})$. We get that
\begin{equation}
\mathbb P (\mathcal A _0 \cap \mathcal A _1) = \mathbb E (\mathds 1 _{\mathcal A _0} \mathbb P (\mathcal A _1 \ | \ \mathcal F _{t_0})) \ge (1-\epsilon ) \mathbb P (\mathcal A _0) \ge (1-\epsilon )^2 \ge 1-2 \epsilon .
\end{equation}
Next let $\delta >0$ sufficiently small such that $\delta^{\frac{2}{3}} C _\epsilon <\epsilon $ where $C_\epsilon $ is from Theorem~\ref{thm:stopping time of getting to speed T to the minus one third}. Finally, let $t>0$ sufficiently large (depending on all other parameters) such that $\delta t \ge t'(t_1, \alpha _1 ,\epsilon )$ where $t'$ is from Theorem~\ref{thm:stopping time of getting to speed T to the minus one third}. Let $t_2$ and $\alpha _2$ be the stopping time and random variable from Theorem~\ref{thm:stopping time of getting to speed T to the minus one third} with $\delta t$ instead of $t$. Define the events
\begin{equation}
\mathcal A _2 :=\{ t_2 \le \epsilon t \} \cap \{\overline{X}_{t_2} \le \epsilon t ^{\frac{2}{3}}\}.
\end{equation}
and $\mathcal A := \mathcal A _0 \cap \mathcal A _1 \cap \mathcal A _2$. The random variable $\alpha _2$ is defined on $\mathcal A _2 $ and satisfies $|\alpha _2 -(\delta t )^{-\frac{1}{3}}| \le C _\delta t^{-\frac{2}{5}}$. By Theorem~\ref{thm:stopping time of getting to speed T to the minus one third} and the choice of $\delta $ almost surely on the event $\mathcal A _0 \cap \mathcal A _1$ we have $\mathbb P (\mathcal A _2 \ | \ \mathcal F _{t_1}) \ge 1-\epsilon $. Thus we get $\mathbb P (\mathcal A )\ge 1-3 \epsilon $.
We let $ (\tilde{Y } ,\tilde{S}, \tilde{X})$ be the aggregate $(\overline{Y },\overline{S} , \overline{X })$ condition on the event $\mathcal A $. Starting with $t_2$ and $\alpha _2$, We define the sequence $t_i$ and the random variables $\alpha _i$ like in Section~\ref{sec:short}, with a minor difference. Instead of stopping the process when the speed changes by a factor of $2$, we stop the process when the speed changes by a factor of $\log (1/\alpha _2)$. More precisely, we let $t_3:=t_2+\alpha _2^{-2} \log ^{10 \theta -2 }(1/\alpha _2)$ and $t_{i+1}:=t_2+(i-1)(t_3-t_2)$. We define the sequence $\alpha _i$ inductively. Let
\begin{equation}
\begin{split}
&\zeta_1:=\min \left\{t_i\ge t_2:\ \alpha _i < \alpha _2 \log ^{-1} (1/\alpha _2) \right\} \\
&\zeta _2 :=\min \{t_i\ge t_2:\ \alpha _i > \alpha _2 \log (1/\alpha _2) \} \\
&\zeta _3:=\min \{t_i \ge t_2: \tilde{Y}_{t_i} \notin \mathfrak{R} '(\alpha _i,t_i) \}\\
& \zeta _4:=\min \{t_{i+1}\ge t_3: \left| X_{t_{i+1}}-X_{t_{i}}-\alpha_{i} (t_1-t_0) \right| \ge \alpha _{i} ^{1.4}(t_1-t_0) \},
\end{split}
\end{equation}
We let $\zeta := \zeta _1 \wedge \zeta _2 \wedge \zeta _3 \wedge \zeta _4$. On $\{\zeta >t_i\}$ we let $\alpha _{i+1}$ be the random variable $\alpha _1 $ from the Theorem~\ref{thm:induction step for repeatd use} and on $\{\zeta \le t_i\}$ we let $\alpha _{i+1}:=\alpha _i$.
\begin{claim}\label{claim:zeta is large}
We have that $\mathbb P (\zeta \le t_2+ \alpha _2^{-3} \log (1/\alpha _2))\le C \log ^{-c}(1/\alpha _2)$
\end{claim}
We'll not give all the details of the proof as it is similar to some other results in Sections~\ref{sec:short} and \ref{sec:medium}.
\begin{proof}[Sketch of Proof]
We define the sequence of scales $\delta _2,\delta _3,\dots $ as in Section~\ref{sec:medium} with $\delta _2:=\alpha _2$ and $\delta _i \in \{\alpha _2 2^k \ | \ k\in \mathbb Z \}$. Unlike Section~\ref{sec:medium} we do not change the steps of time every time we cross a scale. Let $W_i :=\log _2 (\delta _i/\alpha _2)$ as in the proof of Lemma~\ref{lem:stitching medium}. Since Lemma~\ref{lem:stitching short intervals} holds even when the steps of time are larger or smaller by a $\log $ factor, we still have that $W_i$ is roughly a random walk with a downward drift. We let $n_0:=\log \log (1/\alpha _2)$. By the martingale argument in the proof of Lemma~\ref{lem:stitching medium} the sequence $\delta _i$ will decrease to $4 \alpha _2/ \log (1/\alpha _2)$ before hitting $\zeta $. Once the speed is so small, the time it will take for the speed to change its scale by a factor of $2$ will be at least $\alpha _2^{-3} \log (1/\alpha _2)$ with high probability by Lemma~\ref{lem:stitching short intervals}.
\end{proof}
We let $A(t)$ be the continuation of $\alpha _i$ that is $A(t)$ is defined by $A(t):=\alpha _i$ for any $t_i\le t <t_{i+1}$. We also define the rescaled process
\begin{equation}
V_t(s):= t^{\frac{1}{3}} A(t_2+ s t)
\end{equation}
\begin{lem}\label{lem:speed to sde}
We have that
\begin{equation}
\left( V_t(s) \right) _{s>0} \overset{d}{\longrightarrow } \left(Z_\delta (s)\right)_{s>0}, \quad t \to \infty
\end{equation}
where $Z_\delta $ is the solution to $dZ=2Z^{\frac{5}{2}}dB_t$ with $Z_\delta (0)=\delta ^{-\frac{1}{3}}$. The convergence is in the sense of Remark~\ref{remark:on convergence}.
\end{lem}
\begin{proof}
We check all the conditions of Theorem~\ref{thm:helland} with $\mu =0$, $\sigma (x)=2x^{\frac{5}{2}}$ and $x_0:=\delta ^{-\frac{1}{3}}$. Let
\begin{equation}
\Delta _t:=\frac{(t_3-t_2)}{t} \le C_\delta t^{-\frac{1}{3}} \log ^Ct \to 0, \quad t \to \infty
\end{equation}
Finally, we let $s_k:=k \Delta _t$ so that $V_t(s_k)=A(t_{k+2})=\alpha _{k+2}$. The first condition holds since
\begin{equation}
V_t(0)=t^{\frac{1}{3}} A(t_2)= t^{\frac{1}{3}} \alpha _2 \to \delta ^{-\frac{1}{3}}.
\end{equation}
It is clear that the second condition holds. Next, we check the third condition. For sufficiently large $t$ we have that
\begin{equation}
\mathbb P ( | V_t(s_{k+1})-V_t(s_k) |\ge \epsilon ' \ | \ \mathcal F _{t_k}) \le \mathbb P( | \alpha _{k+3}-\alpha _{k+2} | \ge \epsilon ' t^{-\frac{1}{3}} \ | \ \mathcal F _{t_k} ) =0
\end{equation}
so in particular the third condition holds.
We turn to prove the fourth condition. Let $s>0$ and $K \subseteq (0,\infty )$ a compact set. We have that
\begin{equation}
\begin{split}
\sum _{\substack{k: \\ s_k \le s}} \left| \mathbb E \left[ V_t(s_{k+1})-V_t(s_{k}) \ \big| \ \mathcal F _{t_{k+1}} \right] \right| \mathds 1 \{V_t(t_k) \in K \} =t^{\frac{1}{3}} \sum _{\substack{k: \\ s_k \le s}} \left| \mathbb E \left[ \alpha _{k+3}-\alpha _{k+2} \ \big| \ \mathcal F _{t_{k+1}} \right] \right| \mathds 1 \{ t^{\frac{1}{3}}\alpha _{k+2} \in K\} \\
\le C t^{\frac{1}{3}} \sum _{\substack{k: \\ t_k \le st}} \frac{(t_{k+3}-t_{k+2}) \alpha _{k+2}^4}{\log ^3(1/\alpha _{k+2})} \mathds 1 \{t^{\frac{1}{3}} \alpha _{k+2} \in K\} \le C _{K,\delta } t^{\frac{1}{3}} \frac{\alpha _2 ^4 }{\log ^3 (1/\alpha _2 )} \sum _{\substack{k: \\ t_k \le st}} (t_{k+2}-t_{k+1}) \le \frac{C_{K,\delta }}{\log ^3 t} \to 0
\end{split}
\end{equation}
Lastly, we check the fifth condition
\begin{equation}
\begin{split}
&\sum _{\substack{k: \\ s_k \le s}} \left| \mathbb E \left( (V_t(s_{k+1})-V_t(s_k) )^2 \ \big| \ \mathcal F _{t_{k+1}} \right) -4 V_t(t_k) ^5 \Delta _t \right|\mathds 1 \{V_t(t_k) \in K \} \\
&=t^{\frac{2}{3}}\sum _{\substack{k: \\ s_k \le s}} \left| \mathbb E \left( (\alpha _{k+3}-\alpha _{k+2} )^2 \ \big| \ \mathcal F _{t_{k+2}} \right) - 4 \alpha _{k+2} ^5 (t_3 -t_2) \right| \mathds 1 \{V_t(t_k) \in K \} \\
&\le C t^{\frac{2}{3}}\sum _{\substack{k: \\ s_k \le s}} \alpha _{k+2}^{5+c} (t_1-t_0) \mathds 1 \{V_t(t_k) \in K, \ \zeta > t_{k+2} \} +Ct^{\frac{2}{3}}\sum _{\substack{k: \\ s_k \le s}} \alpha _{k+2} ^5 (t_3 -t_2) \mathds 1 \{V_t(t_k) \in K , \ \zeta \le t_{k+2}\}\\
&\le C t^{\frac{5}{3}}\alpha _2 ^{5+c} +Ct^{\frac{5}{3}}\alpha _1^{5} \mathds 1 \{ \zeta \le t_2 +\alpha _2^{-3} \log (1/\alpha _2) \} \overset{p}{\longrightarrow } 0.
\end{split}
\end{equation}
Where the convergence follows from Claim~\ref{claim:zeta is large}. The result now follows from Theorem~\ref{thm:helland}.
\end{proof}
\begin{cor}\label{cor:integrate both sides}
We have the following convergence in finite dimensional distributions
\begin{equation}
t^{-\frac{2}{3}} \left( \tilde{X}_{t_2+st}-\tilde{X}_{t_2} \right) \overset{d}{\longrightarrow } \intop _0^s Z_\delta (x)dx
\end{equation}
\end{cor}
\begin{proof}
On the event $\{\zeta > t_2 + \alpha _2^{-3} \log (1/\alpha _2)\}$ we have for all $k \ge 2 $ with $t_k \le st$ that
\begin{equation}
\left| \tilde{X} _{t_{k+1}}-\tilde{X}_{t_k} -\alpha _k (t_{k+1}-t_k) \right| \le \alpha _k^{1.4} (t_3-t_2) \le \alpha _2^{1.3}(t_3-t_2)\le t^{-\frac{2}{5}}(t_3-t_2).
\end{equation}
Thus
\begin{equation} \label{eq:aggregate to integral}
\left| \tilde{ X} _{t_2+st}-\tilde{X}_{t_2 } - \intop _{t_2}^{t_2+st} A(t')dt'\right| \le C_st^{\frac{3}{5}}
\end{equation}
Next, we have that
\begin{equation}\label{eq:integral to sde}
t^{-\frac{2}{3}}\intop _{t_2}^{t_2+st} A(t')dt'=t^{\frac{1}{3}} \intop _0^s A(t_2+xt)dx =\intop _0^s V_t(x)dx \overset{d}{\longrightarrow } \intop _0^s Z_\delta (x)dx,
\end{equation}
where the convergence is in finite dimensional distributions and it follows from Lemma~\ref{lem:speed to sde} and by part (2) of Remark~\ref{remark:on convergence}. Indeed, there is a coupling of the processes such that the convergence is almost sure and uniform on compact subsets of $(0,\infty )$ and therefore the integrals of the processes converge as well. The statement of the corollary follows from \eqref{eq:aggregate to integral} and \eqref{eq:integral to sde}.
\end{proof}
We need one last claim for the proof of Theorem~\ref{thm:main theorem 2}.
\begin{claim}\label{claim:convegence of integrals of sdes}
Let $Z$ and $Z_M$ for $M \ge 1$, be the solutions of $dZ(t)=2 Z(t)^{\frac{5}{2}}d B _t$ with $Z(0)=\infty $ and $Z_M(0)=M$. We have
\begin{equation}
\left(\int _0^s Z_M(x) dx\right)_{s\ge 0} \overset{d}{\longrightarrow } \left(\int _0^s Z(x) dx\right)_{s\ge 0},
\end{equation}
as $M \to \infty $, where the convergence is in finite dimensional distributions.
\end{claim}
\begin{proof}
Let $V_a (x)$ and $V(x)$ be a $\frac{8}{3}$-Bessel processes with initial condition $V_a (0)=a $ and $V(0)=0$ By Remark~\ref{remark:on bessel} it suffices to show that
\begin{equation}\label{eq:convergence of Bessel}
\left(\int _0^s V_a (x) dx\right)_{s\ge 0} \overset{d}{\longrightarrow } \left(\int _0^s V(x) dx\right)_{s\ge 0},
\end{equation}
as $a \to 0$. To this end we couple the two processes. Recall that $V_a $ and $V$ are solutions to the equation
\begin{equation}
dV(x)=dB_x +\frac{5}{6}\frac{dx}{V(x)}.
\end{equation}
Suppose that both $V_a$ and $V$ are driven by the same Brownian motion. The drift term is larger as long as the process is smaller and therefore, under this coupling, $V_a (x)-V(x)$ is positive and decreasing. Thus $|V_a (x)-V(x)|\le a $ for all $x>0$. Therefore, for all $s>0$
\begin{equation}
\left| \intop _0^s V_a(x)dx - \intop _0^s V(x)dx \right| \le \intop _0^s |V_a(x) -V(x)|dx\le as\to 0 ,\quad \text{a.s.},
\end{equation}
as $a\to 0$. Equation \eqref{eq:convergence of Bessel} follows from this.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main theorem 2}]
On the event $\mathcal A $ we have almost surely
\begin{equation}\label{eq:bound onX}
X_t \le X_{t_0}+\overline{X} _t \le t_0+\overline{X}_{t_2}+\overline{X}_{t_2+t}-\overline{X}_{t_2}\le 2 \epsilon t^{\frac{2}{3}}+ \overline{X}_{t_2+t}-\overline{X}_{t_2},
\end{equation}
where the last inequality holds for sufficiently large $t$. Thus, for all $z>0$ we have
\begin{equation}
\begin{split}
\liminf _{t\to \infty } \mathbb P (X_t \le zt^{\frac{2}{3}} ) &\ge \liminf _{t\to \infty } \mathbb P (X_t \le zt^{\frac{2}{3}} , \mathcal A ) \\
&\ge \liminf _{t\to \infty } \mathbb P (\overline{X}_{t_2+t}-\overline{X}_{t_2} \le (z-2 \epsilon )t^{\frac{2}{3}} , \mathcal A ) \\
&\ge \liminf _{t\to \infty } \mathbb P ( \mathcal A )\cdot \mathbb P (\overline{X}_{t_2+t}-\overline{X}_{t_2} \le (z-2 \epsilon ) t^{\frac{2}{3}} \ | \ \mathcal A )\\
&\ge (1-2 \epsilon ) \cdot \liminf _{t\to \infty } \mathbb P (\tilde{X}_{t_2+t}-\tilde{X}_{t_2} \le (z-2 \epsilon ) t^{\frac{2}{3}}) \\
&\ge (1-2 \epsilon ) \cdot \mathbb P \Big( \int _0^1 Z_\delta (x) dx \le (z-2\epsilon ) \Big) ,
\end{split}
\end{equation}
where in the last inequality we used Corollary~\ref{cor:integrate both sides}. Recall that the following inequality holds for all $\epsilon, \delta $ as long as $\delta $ is sufficiently small depending on $\epsilon $. Taking $\delta $ to zero and using Claim~\ref{claim:convegence of integrals of sdes} we get
\begin{equation}
\liminf _{t\to \infty } \mathbb P (X_t \le zt^{\frac{2}{3}} ) \ge (1-2 \epsilon ) \cdot \mathbb P \Big( \int _0^1 Z (x) dx \le (z-2\epsilon ) \Big) ,
\end{equation}
where $Z$ is the solution to the same SDE with $Z(0)=\infty$. Next, taking $\epsilon $ to zero we get
\begin{equation}
\liminf _{t\to \infty } \mathbb P (X_t \le zt^{\frac{2}{3}} ) \ge \mathbb P \Big( \int _0^1 Z (x) dx \le z \Big) .
\end{equation}
By the same arguments for all $z_1 ,\dots ,z_k>0$ we have
\begin{equation}
\liminf _{t\to \infty } \mathbb P (X_{s_1t} \le z_1t^{\frac{2}{3}}, \dots , X_{s_kt} \le z_kt^{\frac{2}{3}} ) \ge \mathbb P \Big( \int _0^{s_1} Z (x) \le z_1, \dots , \int _0^{s_k} Z (x) \le z_k \Big) .
\end{equation}
We turn to prove a matching upper bound on the last probability. As the arguments are similar we will not give all the details. To this end, we change the definition of $\mathcal A _0$ slightly. We take a sufficiently small $\alpha _0$ and $t_0=100 \alpha _0^{-2}$ and use the same definition as in \eqref{eq:def of A_0} with $\underline{Y}_{t_0,\alpha _0}$ instead of $\overline{Y}_{t_0,\alpha _0}$. We also let $( \underline{Y}_t, \underline{S}(t) ,\underline{X} _t)$ be the aggregate with initial condition $\underline{Y}_{t_0,\alpha _0}$ and work with $( \underline{Y}_t, \underline{S}(t),\underline{X} _t)$ instead of $( \overline{Y}_t ,\overline{S}(t),\overline{X} _t)$. The rest of the definitions such as $\mathcal A_1 , \mathcal A _2, \mathcal A $ and $\tilde{X}$ stay the same. As in \eqref{eq:bound onX} we have on $\mathcal A $ that
\begin{equation}
X_t\ge \underline{X} _t-\underline{X}_{t_2}\ge \underline{X}_{t_2+(1-2\epsilon )t}-\underline{X}_{t_2}.
\end{equation}
Thus
\begin{equation}
\begin{split}
\limsup _{t\to \infty } \mathbb P (X_t \le zt^{\frac{2}{3}} ) &\le \limsup _{t \to \infty } \mathbb P (\mathcal A^c) +\limsup _{t\to \infty } \mathbb P (\overline{X}_{t_2+(1-2\epsilon )t}-\overline{X}_{t_2} \le z t^{\frac{2}{3}} , \mathcal A ) \\
&\le 3 \epsilon + \limsup _{t\to \infty } \mathbb P (\tilde{X}_{t_2+(1-2\epsilon )t}-\tilde{X}_{t_2} \le z t^{\frac{2}{3}}) \\
&\le 3 \epsilon + \cdot \mathbb P \Big( \int _0^{1-2\epsilon } Z_\delta (x) dx \le z \Big) .
\end{split}
\end{equation}
Taking $\delta \to 0$ and then $\epsilon \to 0$ we get
\begin{equation}
\limsup _{t\to \infty } \mathbb P (X_t \le zt^{\frac{2}{3}} ) \le \mathbb P \Big( \int _0^1 Z (x) dx \le z \Big) .
\end{equation}
By the same arguments
\begin{equation}
\limsup _{t\to \infty } \mathbb P (X_{s_1t} \le z_1t^{\frac{2}{3}}, \dots , X_{s_kt} \le z_kt^{\frac{2}{3}} ) \le \mathbb P \Big( \int _0^{s_1} Z (x) \le z_1, \dots , \int _0^{s_k} Z (x) \le z_k \Big) .
\end{equation}
Thus
\begin{equation}
\lim _{t\to \infty } \mathbb P (X_{s_1t} \le z_1t^{\frac{2}{3}}, \dots , X_{s_kt} \le z_kt^{\frac{2}{3}} ) = \mathbb P \Big( \int _0^{s_1} Z (x) \le z_1, \dots , \int _0^{s_k} Z (x) \le z_k \Big) .
\end{equation}
This finishes the proof of Theorem~\ref{thm:main theorem 2}.
\end{proof}
\section*{Acknowledgements}
We are deeply grateful to Vladas Sidoravicius who, with great enthusiasm, introduced us to the MDLA model and wish we could have shared these results with him.
DE thanks Ryan Alweiss, Ofir Gorodetsky, Oren Yakir, Sang Woo Ryoo, Elad Zelingher and Jonathan Zung for useful discussions. DN is supported by a Samsung Scholarship. AS thanks Amir Dembo for his thoughtful discussions on the mildly supercritical case. AS is supported by NSF grants DMS-1352013 and DMS-1855527, Simons Investigator grant and a MacArthur Fellowship.
\bibliographystyle{amsplain}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction}
There has been a renaissance in the study of metastable
supersymmetry breaking vacua
in string theory and field theory in the past several years.
Several of the examples are stringy constructions which
involve brane dynamics
and/or fluxes in
a nontrivial gravitational background \cite{KachruMcGreevy, VafaLargeN,
KachruPearsonVerlinde,
ABSV, VerlindeMonopole, HeckmanVafa, Kutasov}. There are also
purely field-theoretic constructions,
like the Intriligator-Seiberg-Shih (ISS) models \cite{ISS}
where strong/weak coupling dualities \cite{SeibergDuality}
allow one to find such states, or the retrofitted models \cite{Retro}
where canonical theories like O'Raifeartaigh or Polonyi models can be
coupled to additional dynamics
in a way that naturally produces an exponentially small SUSY breaking scale.
These states have potential applications
both in understanding the existence and properties of stable non-supersymmetric
string compactifications \cite{BP, Silverstein, KKLT} (for recent reviews see
\cite{DouglasKachru,Lust}), and in building realistic
models of gauge-mediation \cite{GM} or sequestered, high-scale SUSY
breaking \cite{Seq}.
It is natural to think that via AdS/CFT duality \cite{juanAdS} or brane engineering
\cite{Hanany-Witten}, one can sometimes relate the stringy and field
theoretic constructions.
Indeed, many groups have recently
engineered various D-brane field theories which exhibit
dynamical SUSY breaking (DSB) and reduce to
known DSB field theories in the decoupling limit
\cite{Ooguri:2006in,Franco:2006es,OoguriOokouchi, FrancoUranga, MQCDSeibergShih,
Argurio:2006ny, TatarMeta, Kitano:2006xg, Wijnholt, Antebi}.
Here, we continue our investigation \cite{Argurio:2006ny} into the possible relations
between ISS-like states in field theory, and SUSY-breaking states where SUSY
is broken at the end of a ``warped throat," as in \cite{KachruPearsonVerlinde} (where
SUSY was broken by anti-D3 brane probes at the tip of a warped, deformed conifold
\cite{KS}). At a qualitative level, it is natural to think that SUSY breaking at
the end of a warped throat is AdS/CFT dual to dynamical supersymmetry breaking.
Then one should try to interpret the SUSY breaking states involving warped
antibranes, which can tunnel to supersymmetric states of the
same gravitational system \cite{KachruPearsonVerlinde}, as metastable states in the
dual SUSY field theory. While it is not necessary that such a correspondence should
hold (since non-supersymmetric vacua are not protected upon extrapolation in the
't Hooft coupling $g_sN$), it would be interesting and suggestive to find examples
where such states can be argued to exist both in the
gravitational system and the field theory dual.
In \cite{Argurio:2006ny}, we argued that the gauge/gravity duals derived by studying
fractional branes in simple quotients of the conifold are a natural place to find such
a correspondence. In a particular $\mathbb{Z}_2$ orbifold of the conifold, we were able to
realize a close relative of SUSY QCD where the small mass parameter of the ISS models is
dynamically generated, and where the dual gravitational system plausibly admits
metastable anti-brane states. However, that work left several open questions. Firstly,
the SQCD model that could be realized had $N_f = N_c$, while it is in the free
magnetic range $N_c + 1 \leq N_f < {3\over 2}N_c$ that one is really confident of the
existence of the metastable states in the field theory. Secondly, because we wish to
dynamically generate the small mass parameter of the ISS models, we must take care to
ensure that the dynamical masses are ${\it stabilized}$ against relaxation to
zero or infinity. In the models of \cite{Argurio:2006ny}, this delicate question
rested entirely on (largely unknown) properties of the K\"ahler potential.
In this paper, we argue that a simple variant of the model of \cite{Argurio:2006ny} solves
both of these problems. By considering other orbifolds of the conifold, we are able to
find models where we can reach the free magnetic regime necessary to prove
existence of the ISS vacua, and where we can argue that the superpotential itself helps
to stabilize the dynamical quark masses in the range where SUSY breaking occurs. More precisely,
the models we construct have $N_f=N_c+1$.
As an added bonus, our model shares a nice feature with the model of Kitano, Ooguri and
Ookouchi \cite{Kitano:2006xg}: the quiver superpotential naturally comes with terms that
break the R-symmetry of the original ISS models, which is problematic in model building
applications (as it forbids a gaugino mass).
In the rest of this section, we introduce our model. Its field theory dynamics is analyzed
in \S2, where we show how an effective massive SQCD arises in a given corner of its moduli
space. An important role is played by a stringy instanton
generated contribution to the
effective superpotential, whose origin we discuss in \S3. In \S4
we prove that the quark masses can be
dynamically stabilized, and in \S5
we estimate the lifetime of the metastable vacua. In \S6,
we briefly discuss the gravity dual IIB description (involving fluxes
and branes in the deformed geometry). We also
present a IIA T-dual of the IIB picture, where the gauge theory arises from a
configuration of NS 5 branes and D4 branes.
These descriptions allow one to visualize many (though not all) aspects
of the field theory dynamics, and in particular, make it obvious that the metastable
SUSY-breaking vacua are dual to models which contain anti-branes. We conclude in \S7.
\subsection{The structure of the model}
\label{model}
The models we would like to analyze are obtained by considering (fractional) D3 branes
at the tip of a non-chiral $\mathbb{Z}_N$ orbifold of the conifold, which is nothing but a
straightforward generalization of the system considered in \cite{Argurio:2006ny} for the case $N=2$.
The corresponding quiver gauge theory admits $2N$ gauge factors and $4N$ bifundamental chiral superfields $X_{ij}$
interacting via the following quartic superpotential
\begin{equation}
\label{wtree}
W = h \sum_{i=1}^{2N} (-1)^{i+1} X_{i,i+1}X_{i+1,i+2}X_{i+2,i+1}X_{i+1,i}~,
\end{equation}
where the index $i$ is understood $mod(2N)$.
Because the quiver is non-chiral, we can consistently assign arbitrary ranks to the quiver nodes, which
suggests that there should be $2N-1$ independent fractional branes one can define. This is indeed
mirrored in the geometric structure of the singularity, which admits $2N-1$ independent shrinking
2-cycles the branes can wrap. In a given basis, which will be relevant later, one obtains a natural classification
into $N$ deformation fractional branes and $N-1~$ ${\cal N}=2$ fractional branes, following the
definition proposed in \cite{seba}. We remind the reader that deformation
branes correspond to isolated nodes in the quiver (and hence gauge groups with no matter)
and lead to confinement, while ${\cal N}=2$
branes arise from occupying two connected nodes, which yields
a product of two SQCD theories with $N_f=N_c$, and hence have a moduli
space of vacua.
As we are going to show, for our present purposes it is enough to take $N=3$ (any
$\mathbb{Z}_{N}$ with $N>3$ naturally works in the same way). Therefore,
from now on we stick to this case, for simplicity.
This specific $\mathbb{Z}_{3}$ quotient admits five shrinking 2-cycles. Two of them
see, locally, a ${\mathbb{C}}^2/{\mathbb{Z}}_2$ singularity. The other three are dual (via geometric
transitions) to compact 3-cycles $A_i$ ($i=1,2,3$) which can be made finite by
a complex deformation.
The
corresponding quiver is shown in Figure
\ref{quiverg}.
\begin{figure}[ht]
\centering
\includegraphics[width=4.5cm]{quiverg.eps}
\caption{The quiver describing the field content of the gauge theory at the tip of the
non-chiral $\mathbb{Z}_3$ orbifold. The ranks of
the gauge groups can be chosen arbitrarily.}
\label{quiverg}
\end{figure}
We would like to consider the gauge theory arising from the following assignment of
ranks in the quiver
\begin{equation}
\label{5rank}
(N_c~,~N_c~,~N_c~,~1,~0~,~0)~.
\end{equation}
In terms of fractional branes, this may be viewed as $N_c$ ${\cal N}=2$ branes at nodes
1 and 2, $N_c$ deformation
branes at node 3 and another single deformation brane at node 4 (this definition is basis-dependent, of course).
The``$SU(1)$" fourth node is not actually a gauge group. Its interpretation is that $X_{34}$ and $X_{43}$ transform purely as
fundamental and antifundamental representations of node 3, respectively.
The corresponding quiver is depicted in Figure \ref{4nodes}.
\begin{figure}[ht]
\centering
\includegraphics[width=5.5cm]{4nodes.eps}
\caption{The quiver of the theory (\ref{5rank}).}
\label{4nodes}
\end{figure}
As we shall show, this system, whose dynamics we are going to study in detail, reduces exactly (in a region
of the moduli space to be specified below) to massive $SU(N_c)$ SQCD with $N_f=N_c+1$ massive (but light) flavors,
and therefore admits both the supersymmetric and the metastable non-supersymmetric vacua of that theory. Hence,
this system provides a string
embedding of an ISS model. Moreover, it has some additional virtues: the small flavor masses are dynamically generated
(and stabilized), it
is possible to give a simple gravity dual interpretation of the metastable non-supersymmetric vacua,
and R-symmetry is explicitly broken (which could be useful in any model-building applications).
\section{The dynamics of the model}
The quiver gauge theory we are going to analyze is the one depicted
in Figure \ref{4nodes}. This theory has a superpotential of the form
\begin{equation}
W = h( X_{12}X_{23}X_{32}X_{21} - X_{23}X_{34}X_{43}X_{32})
+ m X_{43} X_{34}~.
\label{wtree5}
\end{equation}
As already discussed, the two quartic terms follow from the conifold by standard orbifold techniques.
The mass term for $X_{34}$ and $X_{43}$ is generated by a stringy instanton. We postpone
discussion of the relevant instanton to \S3, and
we proceed to analyze the above superpotential.
The quartic coupling $h$ has the dimensions of an inverse mass, and
is inversely proportional to the UV scale generating the non-renormalizable interaction.
In this context, it is natural to take $h \sim 1/M_s^*$. Here, $M_s^*$ indicates the string
mass scale effectively warped down to a lower value due to the RG flow,
which manifests itself as a duality cascade. For the field theory interpretation
to be valid, we need $M_s^*$ to be bigger than any of the dynamical scales of the gauge groups
involved in the quiver.
To start analyzing the gauge theory, we will make some assumptions about the
scales of the gauge groups on every node.
Node 3 is the main node where the ISS-like SQCD dynamics takes place.
Node 2 acts as a subgroup of the flavor symmetry, broken
as $SU(N_c+1)\times SU(N_c+1) \supset SU(N_c+1) \supset SU(N_c)$.
Accordingly, we will take its dynamical scale $\Lambda_2$ to be (much) smaller
than the others, so that this gauge group
can be effectively considered as classical.
Node 1, which has a number of flavors which equals the number of
colors, undergoes confinement so that its effective dynamics is described
in terms of mesons and baryons. The mesons are going to supply mass terms
for some of the flavors of node 3, through the superpotential couplings
in (\ref{wtree5}), much as in \cite{Argurio:2006ny}. These masses are
subject to the constraint on the deformed moduli space of node 1.
In particular, we will need the scale $\Lambda_1$ to be such
that those masses are still lower than the scale of the SQCD node, $\Lambda_3$.
The additional mass $m$ will be larger than these $N_c$ masses, but still
smaller than $\Lambda_3$. We will explain later how this parameter
range can be obtained.
This model is quite similar to the one in
\cite{Kitano:2006xg}. One difference is that all the small parameters are generated
dynamically. In addition, this model arises naturally at a fairly simple Calabi-Yau
singularity.
Assuming now that node 1 is confining, the tree level superpotential
reads
\begin{equation}
\label{wlec}
W = h ( M_{22} X_{23}X_{32} - X_{23}X_{34}X_{43}X_{32})
+m X_{43}X_{34}~,
\end{equation}
where we have defined $M_{22}=X_{21}X_{12}$. This superpotential is not quite complete:
we should really implement the constraint relating the mesons and the
baryons of node 1 through the introduction of a
Lagrange multiplier. We delay that
to later on. Let us now assume that the interactions are such that
the mesonic and baryonic branches of node 1 decouple, and in particular that
when the meson matrix has maximal rank the baryons are required to vanish --
we will argue that this is the case in \S\ref{stabilization}.
For the time being, we assume that node
1 is on the mesonic branch, where the constraint
describing the quantum-deformed moduli space reads
\begin{equation}
\det M_{22} = \Lambda_1^{2N_c}~.
\label{wewant}
\end{equation}
This constraint is necessary for the generation of dynamical masses but does not
fully fix the eigenvalues of
$M_{22}$. In the non-supersymmetric vacua, their stabilization occurs at tree-level as we explain shortly.
At the stable point, the VEV of $M_{22}$ is proportional to
the identity matrix. Hence we see that in the effective SQCD theory at node 3,
we have $N_c$ flavors of mass $\sim h \Lambda_1^2$
and $1$ flavor of mass $m$. We will take the latter
to be the heavier one, so that $h\Lambda_1^2 < m$.
Therefore, along this branch the theory on node 3 with superpotential (\ref{wlec}) is
nothing but SQCD with $N_f = N_c+1 $ massive flavors, together with a quartic coupling
(which is irrelevant in the IR).
Integrating out the flavors, we obtain pure $SU(N_c)$ SYM characterized
by a dynamical scale
\begin{equation}
\Lambda_L^{3N_c} = \Lambda_3^{2N_c-1}h^{N_c} \det M_{22} ~m
= \Lambda_3^{2N_c-1}(h \Lambda_1^2)^{N_c} m~.
\end{equation}
Implementing the constraint on the deformed moduli space of node 1
with a Lagrange multiplier in an effective superpotential, it is easy to
see that we indeed have a moduli space of supersymmetric vacua where
$M_{22}$ has non zero VEV, while the baryons
are vanishing. When taking into account that node 2 is actually
gauged, we see that at low-energies the moduli space will be described
by $\mathbf{C}^{N_c-1}$ together with a residual $U(1)^{N_c-1}$ gauge symmetry.
We now move on and show that our theory also has meta-stable, SUSY breaking vacua.
Since node 3 has $N_f=N_c+1$, its low-energy dynamics is governed by a theory
of mesons and baryons. This case can actually be seen as a limiting
case of Seiberg duality, where the dual magnetic gauge group is a trivial
$SU(1)$, and the dual quarks are nothing but the baryons of the electric theory.
In the following, we will adopt this terminology.
The bifundamentals are combined
into effective mesons as $X_{i3}X_{3j}= \Lambda_3 \phi_{ij}$, and the
dual quarks are labeled $Y_{i3}$ and $Y_{3i}$. The superpotential is
\begin{equation}
W=(h\Lambda_3) (M_{22}\phi_{22} - \Lambda_3
\phi_{24}\phi_{42}) + m\Lambda_3 \phi_{44}
-\phi_{22}Y_{23}Y_{32} - \phi_{44}Y_{43}Y_{34} +\phi_{24}Y_{43}Y_{32}
+\phi_{42}Y_{23}Y_{34}~.
\label{wmag}
\end{equation}
Note that we have rescaled the mesons to canonical dimension using the scale
$\Lambda_3$. Accordingly, the cubic terms generated by the duality
have a coupling of $\mathcal{O}(1)$, which we set to one (shifting the undetermined
constant to the normalization of the canonical K\"ahler potential).
Strictly speaking, we should also add a term linear in the determinant
of the meson matrix, but it is highly irrelevant in the IR (and to our
considerations).\footnote{It does play a role if we want to recover the
SUSY vacua in the low-energy picture.} It is easy
to see that the above model of mesons and dual quarks would
have, in the absence of the $\phi_{24} \phi_{42}$ coupling, an accidental IR
$U(1)_R$ symmetry. The R charges would be 2 for the
mesons and 0 for the dual quarks, as in an O'Raifeartaigh model.
The quadratic coupling in the mesons which
arises naturally in the present model provides an explicit
breaking of this R-symmetry.
This theory is now amenable to an analysis very similar to
\cite{ISS,Kitano:2006xg}. There is supersymmetry breaking by the rank
condition. The F auxiliary field that vanishes is the one related to the
more massive flavor,
i.e. the F-term of $\phi_{44}$. On the other
hand, the F-components of $\phi_{22}$ are non-vanishing. As a consequence,
there is a tree level vacuum energy given by
\begin{equation}
V_{tree} = |h\Lambda_3|^2 \sum_{i=1}^{N_c} | M_i |^2 = N_c |h\Lambda_3
\Lambda_1^2 |^2~,
\end{equation}
where we obtain the final result by extremizing on the eigenvalues of the
matrix $M_{22}$ given the constraint on its determinant.
We are going to show later that indeed the constraint on the determinant
is not destabilized by baryonic VEVs.
A standard analysis of this model shows that it is
a sum of O'Raifeartaigh models with an additional coupling $m_{24}=h\Lambda_3^2$, which
is the one quadratic in the two mesons $\phi_{24}$ and $\phi_{42}$.
As we will discuss in \S\ref{stabilization}, $\phi_{22}$ gets a non-zero VEV
due to the one-loop potential (see also \cite{Kitano:2006xg}).
This VEV is directly related to the presence of
the quadratic meson coupling $m_{24}$, so that we can actually estimate it as
\begin{equation}
|\phi_{22}| \sim |h \Lambda_3^2|~.
\end{equation}
As noted in \cite{Kitano:2006xg}, the vacua analyzed here are unstable
if the size of $m_{24}$ exceeds the larger flavor mass.
Here this bound reads
$|h \Lambda_3^2|^2 < |m\Lambda_3|$, or in other words
(recall that $h=1/M_s^*$)
\begin{equation}
\left(\frac{\Lambda_3}{M_s^*}\right)^2 < \frac{m}{\Lambda_3}~.
\label{mzvsm}
\end{equation}
Note that all these relation must be taken with a grain of salt, since
there are factors of $\mathcal{O}(1)$ that we are not retaining (most
of which are non-calculable anyway).
All other pseudomoduli are lifted by the
one-loop potential, and acquire a non-tachyonic mass.
This shows that the present model has metastable supersymmetry
breaking vacua, provided we can show that there is no instability
towards turning on baryonic VEVs at node 1.
We demonstrate in \S\ref{stabilization} that this is the case, in an appropriate regime
of
parameters.\footnote{In the model that we presented
in \cite{Argurio:2006ny} there is potentially such an instability. In that model
the SQCD node is in a confining rather than IR free regime, so that
(non-calculable) corrections to the K\"ahler potential are present,
and make it difficult to determine what happens.
The gravity dual description provides another source of information; such
an instability is not readily apparent there, but it is a complicated
system which would benefit from further study.}
\section{A mass term generated by a stringy instanton}
\label{instanton}
Before analyzing in detail the stability of the model, we have to
explain how the mass term for the additional flavor is generated.
It turns out that
somewhat novel stringy instanton effects which have recently been
investigated in several other contexts \cite{FKMS,Cvetic,UrangaIbanez,
Haack,Bianchi:2007fx,cvetic,abflp}
contribute corrections to $W$ which depend on gauge invariants that
usually do not appear in the superconformal quiver superpotential.
Recall that these effects arise when
Euclidean D branes wrap cycles corresponding to quiver nodes which are
{\it not} occupied by space-filling branes. In this respect, they
are specific to set ups with fractional branes. We now show that
such an instanton generates the mass term $m$.
To understand the instanton contributions,
consider a D1 instanton (a Euclidean D1 or ED1 brane)
wrapping node 5 of the quiver.
It is BPS and preserves precisely 1/2 of the ${\cal N}=1$ supersymmetry;
acting on the instanton solution with the broken supercharges then produces
two fermion zero modes in the ED1 - ED1 open string sector. These are
the two fermion zero modes that are necessary to give rise to a contribution
to the space-time superpotential (we discuss the possibility of extra
``accidental'' zero modes at the end of this section).
Considering the extended quiver diagram including a node for the instanton, there are also fermionic
strings $\alpha, \beta$ stretching to node 4, in the $(+,\overline{N_4})$ and
$(-,N_4)$ representations of the $U(1) \times SU(N_4)$ gauge group (in the
present case we will have $N_4=1$). As follows from
the computations in \cite{FKMS,UrangaIbanez}, the fermionic spectrum in the extended quiver
is the same as it would be if
the instanton were actually a space-filling brane, except the fermions live in a different
dimension. By the simple argument of \cite{UrangaIbanez}, we also expect that there are
no bosonic zero modes in this sector. Bosons would arise from NS sector strings,
but the NS sector ground state energy receives a contribution from the number of
ND boundary conditions, which pushes the ground state energy above zero
in this configuration. The relevant part of the
extended quiver is reported in Figure \ref{inst}.
\begin{figure}[ht]
\centering
\includegraphics[width=6cm]{inst.eps}
\caption{The extended quiver describing the
interaction between the fractional brane system and a Euclidean
D1 brane on node 5, represented by a square. The relevant coupling is a quartic one, involving $\alpha,\beta, X_{34}$
and $X_{43}$, \eref{instac}.}
\label{inst}
\end{figure}
In the instanton action, we expect a gauge invariant coupling
\begin{equation}
\label{instac}
L = \alpha X_{43} X_{34} \beta~.
\end{equation}
In evaluating the instanton contribution to the 4d effective action, we should
integrate over the only two charged fermionic zero modes $\alpha, \beta$.
This yields a simple contribution to the superpotential
\begin{equation}
\label{wmass}
c ~X_{43} X_{34} ~e^{-{\rm Area}}.
\end{equation}
where $c$ is a dimensionful constant and the relevant
area is the area of the curve corresponding to node 5.
We thus identify the mass term as $m=c ~e^{-{\rm Area}}$. If we reasonably
take $c\sim M_s^*$, we see that it is not difficult to assume that
the area of the cycle wrapped by the instanton is such that $m<\Lambda_3$.
(Roughly that would amount to assume that $\Lambda_5\ll \Lambda_3$ if there was
a gauge group on node 5.)
Now, a similar instanton on node 6, with fermionic strings stretching
to node 1, produces another term in the superpotential.
The gauge invariant coupling in the instanton action is
\begin{equation}
\label{instac6}
L = \alpha X_{12} X_{21} \beta~.
\end{equation}
If we let $a,b$ denote gauge indices at node 1
and $f,g$ denote gauge indices at node 2, then the gauge contractions in
(\ref{instac}) give rise to $\alpha_{\bar a}X_{12}^{a\bar f} X_{21, f\bar b}
\beta^{b}$. So performing the integral over the $\alpha, \beta$ fermions,
which are now a set of $2N_c$ zero modes,
gives the contribution
\begin{equation}
\label{barmass}
c' B \tilde B ~e^{-{\rm Area}'}
\end{equation}
to the 4d effective theory, where $c'$ is similarly a dimensionful constant
and the area is now the one of the curve corresponding to node 6.
The coupling (\ref{barmass}) provides
a mass term for the baryons $B=\det X_{12}$ and $\tilde B=\det X_{21}$ of node 1.
We will see in the next section that this term does not however play
an important role in the stabilization of the baryons.
Let us end this section with a comment regarding a subtle point. With the above reasoning, the coefficients
$c$ and $c'$ have been determined up to a dimensionless number whose precise value
we cannot directly compute
in our geometric set-up. A crucial
ingredient for such coefficients not to
vanish involves
the number of uncharged fermionic zero modes on the ED1 brane.
Before taking into account the quiver branes back-reaction, there
are four, since
the ED1 is a 1/2 BPS state in the Calabi-Yau. While, as already discussed, two zero modes
are necessary to provide the chiral superspace integral for the superpotential term
(\ref{wmass}), the other two would provide a dangerous vanishing contribution.
Still, one should take into account the full back-reacted closed string
background, which includes non-trivial fluxes. This background preserves only 4
supercharges out of the 8 preserved by
the CY, so that the instanton has only 2 zero modes associated to
broken supersymmetries. Then, at least in many backgrounds,
it is reasonable to expect that the extra
zero modes get lifted by the interactions with other background fields.
This is an interesting problem in itself, but we leave it
for further research.
Instead, in order to provide a background where we can
explicitly identify an
object responsible for lifting the additional zero modes,
we can introduce orientifold planes in such a way that the instanton
wraps a cycle that is mapped to itself. In this way, half of the zero modes
are projected out from the start. One concrete
embedding of our model in an orientifold that accomplishes this
task, while not spoiling all other nice features of our model, is described in
Appendix \ref{appendix_orientifold}.
We therefore conclude
that the coefficient $c$ is non-vanishing in many suitable models.\footnote{
A more complete discussion of these issues, for backgrounds
where a simpler worldsheet CFT description is available, will appear in
\cite{abflp}. A general discussion on the introduction of O-planes
in generic Calabi-Yau geometries will appear in \cite{francoetal}.}
\section{Stabilizing dynamical masses}
\label{stabilization}
We have explained how our theory has metastable supersymmetry breaking vacua
under certain assumptions regarding the stability of the dynamically generated
masses. An important question is whether the dynamical masses relax to zero by turning on expectation values
for the baryonic fields at node 1. We show in this section that these
a priori dangerous directions are lifted because
$\langle \phi_{22} \rangle\neq 0$. New superpotential interactions
generated by the D1 instanton wrapping node 6 of the quiver also contribute to
stabilization, although they are not the dominant effect.
We thus first sketch how the one-loop potential gives
the crucial VEV to the field $\phi_{22}$. We start from (\ref{wmag}) and expand around the metastable vacuum. This is
characterized by VEVs for $Y_{34}, Y_{43}=\sqrt{m\Lambda_3}$ and, at tree
level, by an arbitrary $\phi_{22}$. The latter is also the field with
non-vanishing F-terms. The superpotential
for the fluctuations of the fields, expanded about the vacuum,
takes the form
\begin{equation}
W = m_{22}^2 \phi_{22} - Y_{32} \phi_{22} Y_{23}
+ m_{44} \phi_{24} Y_{32} + m_{44} \phi_{42} Y_{23} - m_{24} \phi_{24}\phi_{42}~,
\end{equation}
In writing this down, we have dropped several fields: $\phi_{44}$,
$\delta Y_{34}$
and $\delta Y_{43}$ do not feel SUSY breaking at this order so will cancel out
of the one-loop energy. The masses appearing above are given by
$m_{22}^2= h\Lambda_3 \Lambda_1^2$, $m_{44}^2= m \Lambda_3$ and $m_{24}=
h \Lambda_3^2$. All fields have a canonical K\"ahler potential.
It is straightforward to realize that the F-terms
set $\phi_{22}$ to a diagonal form. Then, we obtain
just a superpotential for $N_c$ decoupled O'Raifeartaigh
like models. Each such model has, besides the field with the non-zero
F-term, 4 other fields.
An important parameter is the coupling $m_{24}$ of the meson bilinear
$\phi_{24}\phi_{42}$.
The tree level vacuum energy is just:
\begin{equation}
E_{vac} = N_c \,|m_{22}|^4~.
\label{E_vac_0}
\end{equation}
We had already noted that the eigenvalues of $M_{22}$ are all trivially
stabilized at their common values \cite{Argurio:2006ny}.
To compute the one-loop vacuum energy of this model, we simply compute the boson/fermion masses as a
function of pseudo-moduli using
\begin{equation}
m_0^2=\left(\begin{array}{ccc} W^{\dagger ac}W_{cb} & \ & W^{\dagger abc}W_c \\
W_{abc}W^{\dagger c} & & W_{ac}W^{\dagger cb} \end{array} \right) \ \ \ , \ \ \
m_{1/2}^2=\left(\begin{array}{ccc} W^{\dagger ac}W_{cb}& \ & 0 \\ 0 & & W_{ac}W^{\dagger
cb} \end{array} \right)
\label{m_scalars_fermions}
\end{equation}
and plug them into the famous 1-loop Coleman-Weinberg result.
The eigenvalues of both the fermionic and the bosonic mass square matrices
can be computed analytically. In this model $\phi_{22}$ remains massless
at tree level and its Fermi partners do too. The other 8 eigenvalues
split in pairs as follows. The bosonic ones are given by
\begin{equation}
|m_{44}|^2 +\frac{1}{2}(|\phi_{22}|^2+|m_{24}|^2\pm |m_{22}|^2) \pm'
\frac{1}{2}\sqrt{(|\phi_{22}|^2-|m_{24}|^2\pm |m_{22}|^2)^2 +
4|m_{44}\phi_{22}^\dagger + m_{44}^\dagger m_{24}|^2}~,
\end{equation}
with the $\pm$ and $\pm'$ standing for two independent sign choices.
The fermionic ones are given by the same expression as the bosonic
ones, save that we formally set $m_{22}=0$ (the fermionic sector does not
talk directly to the F-term).
If we set $m_{24}=0$ we obtain the classic O'Raifeartaigh result
for the one-loop energy, i.e.
\begin{equation}
E_{1} = {N_c \over 32\pi^2}|m_{22}|^4 \left( y^{-2} (1+y)^2 {\rm log}(1+y) +
y^{-2} (1-y)^2 {\rm log} (1-y) + 2 {\rm log} ({|m_{44}|^2\over \Lambda^2})
\right)~,
\label{E1_zeroth_order}
\end{equation}
where we have defined $y = \vert{m_{22} \over m_{44}}\vert^2$ and $\Lambda$ is
the UV cutoff. In this case, $\phi_{22}$ stabilizes around zero.
When $m_{24}\neq 0$, the analytical form of the one-loop energy is not
very illuminating. However it is reasonable to expect that $\phi_{22}$
will pick up a tadpole around zero, so that its vacuum expectation
value is displaced to a non-zero value. Moreover, the size of the
VEV is directly
controlled by $m_{24}$, as they enter almost symmetrically in the
expressions for the eigenvalues. This is confirmed by a numerical analysis,
which also shows the existence of tachyons when $m_{24}$ is too close
to or larger than $m_{44}$. Indeed, one might have guessed that in this
range some dangerous mixings can occur.
Let us remark at this stage on a possibility which could have been
considered. We could have actually tried to generate the higher masses
like $m_{44}$ dynamically in the same way as the lower ones, $m_{22}$.
That would be simply implemented in a 5-node quiver with
ranks $N_f-N_c$ at the 4th and 5th node. The model would be very similar
to the above, except that every O'Raifeartaigh model would have now
$1+4(N_f-N_c)$ fields. If the masses were dynamical, one would have
a sum of $N_f-N_c$ contributions like (the generalization to $m_{24}\neq 0$ of)
eq. (\ref{E1_zeroth_order}). The latter potential attracts all
of the higher masses to smaller values. However one can see that the
deformed moduli space constraint is not sufficient
in this case to stabilize them. Indeed,
in the dominant contribution (the $\log |m_{44}|^2$ piece), the constraint
gives trivially a constant. The rest of the potential asymptotes to
a constant for very large $m_{44}$. Hence, it will always be favorable
to bring down some masses while sending the other(s) to infinity.
This is the reason why we cannot really access the full IR free region
$N_c+1 \leq N_f < {3\over 2} N_c$ of SQCD in this class of models.
\subsection{Stabilization of baryonic directions}
\label{section_stabilizing_masses}
At this stage, we have seen that as long as we can be on the mesonic branch
at node 1, we will successfully obtain a model of metastable
supersymmetry breaking with no small parameters added by hand.
An important question now arises, however.
Assuming we are working in the regime $h\Lambda_1^2 < m$, the energy of
the SUSY breaking vacuum is given by
\begin{equation}
V \sim N_{c}\,h^2 \Lambda_{3}^2 \,\Lambda_{1}^4 ~,
\label{vmeta}
\end{equation}
since it arises by summing the masses squared of the $N_c$ lightest flavors
of the $SU(N_c)$ gauge group at node 3 \cite{ISS}.
The quantum moduli space constraint for the mesons $M_{22}$ is really
\begin{equation}
\det M_{22} - B \tilde B = \Lambda_{1}^{2N_{c}}~.
\end{equation}
So, at least naively, it appears that by relaxing the mesonic VEVs and turning on
$B, \tilde B$, one can lower the vacuum energy to zero, destabilizing the
SUSY breaking vacuum. It is conceivable that the K\"ahler potential (which is
not computable) introduces a barrier that prevents such relaxation, but confidence
in the construction would be considerably enhanced if additional superpotential
terms were present which prevent
the baryons from `turning on' when one expands around the point (\ref{wewant}).
We now begin estimating the leading $B, \tilde B$ mass matrix by expanding
the potential around the would-be non-supersymmetric vacuum. To do so, we assume a canonical K\"ahler potential. We consider this is reasonable, since both potential instabilities and stabilizing effects arise under this assumption.
The leading off-diagonal term is
\begin{equation}
V_{,B \tilde{B}}=V_{,\tilde{B}B}=- \, h^2 \, \Lambda_3^2/\Lambda_1^{2N_c-4}~.
\label{off_diagonal_terms}
\end{equation}
This contribution appears at tree-level and favors the condensation of baryons
as discussed above.
However, there are several further terms in the (super)potential which
impart diagonal terms in the mass matrix, and overwhelm the tachyonic
contribution (\ref{off_diagonal_terms}) for reasonable choices of parameters.
One source of such a term is the tree-level coupling of $\phi_{22}$ in
(\ref{wmag}). This will in fact turn out to be the dominant effect
stabilizing $B, \tilde B$ at zero.
For completeness, we also include the sub-dominant effect caused by
the stringy instantons discussed in the previous section.
Putting (\ref{barmass}) together with (\ref{wmag}), we can check for stability of the mesonic
branch VEVs (\ref{wewant}) as follows. Assume the $M_{22}$ matrix is diagonal,
with equal eigenvalues given by $x$. $x$ is then determined by the deformed quantum
moduli space constraint of the node 1 gauge theory to be
\begin{equation}
x^{2N_c} = \Lambda_{1}^{2N_c} - B \tilde B~.
\end{equation}
We could impose this constraint by adding a Lagrange multiplier $\lambda_1$ to the
superpotential, multiplying the constraint equation. Then, subject to the constraint,
we should minimize the potential
\begin{equation}
\label{potis}
V = \Lambda_1^2 \, \vert h \, \Lambda_3 \, \phi_{22} + \lambda_{1} x^{N_c-1}\vert^2 + \Lambda_1^{2N_c-2}\,
\vert \lambda_1 B + c_1 B\vert^2 + \Lambda_1^{2N_c-2}\, \vert \lambda_1 \tilde B + c_1 \tilde B\vert^2~,
\end{equation}
where the first term arises from $|F_x|^2$, the second term from $|F_{\tilde B}|^2$,
and the third from $|F_{B}|^2$, and we have redefined $c_1 = c'~e^{-{\rm
Area}'}$ with respect to (\ref{barmass}).
Equation \eref{potis} only contributes to the diagonal
\begin{equation}
\label{baryonmass}
V_{,BB} = V_{,\tilde B \tilde B} = {2\over \Lambda_1^{2N_c+2}} ( c_1 \Lambda_{1}^{2N_c} -
h \Lambda_1^2 \Lambda_3 \phi_{22})^2 ~.
\end{equation}
From \eref{baryonmass} we can obtain the leading diagonal terms in the matrix of second derivatives of the potential. The leading
contribution is a non-zero expectation value $\phi_{22}\sim m_{24}=h \Lambda_3^2$. The net result is
\begin{equation}
V_{,BB}=V_{,\tilde{B}\tilde{B}}=h^4\Lambda_3^6/\Lambda_1^{2N_c-2}.
\label{diagonal_terms}
\end{equation}
Both $h$ and the instanton coefficient $c_1$ are suppressed by $M_s^*$, as $h\sim M_s^{*-1}$ and
$c_1\sim M_s^{*3-2N_c}$. (There is also a suppression by the volume of the
curve representing node 6 for the instanton, but since this effect plays
no role in our theory as the $\phi_{22}$ VEV already stabilizes the baryons of
node 1, we can take that volume to be anything
$\geq {\cal O}(1)$. For simplicity we've chosen ${\cal O}(1)$ here).
The only consistency requirement on the relevant scales is then
that $M_s^* > \Lambda_3$.
The eigenvalues of the matrix of second derivatives of the potential are $V_{BB}\pm V_{B\tilde{B}}$.
Then, we are free of tachyons provided that $V_{BB}\geq |V_{B\tilde{B}}|$. From \eref{off_diagonal_terms}
and \eref{diagonal_terms} we conclude the conditions for stability of the baryonic directions are
\begin{equation}
(\Lambda_1/\Lambda_3) < (\Lambda_3/M_s^*)
\label{stability_baryons}
\end{equation}
We see that we can always satisfy the above inequality, as well
as the ones coming from the hierarchy of mass scales in the low-energy model
(as in e.g. \cite{Kitano:2006xg}), by imposing the following hierarchy
\begin{equation}
\Lambda_1 \ll \Lambda_3 < M_s^* ~~\mbox{and}~~ m < \Lambda_3~.
\end{equation}
As discussed previously, we also need $m$ to satisfy the bound \eref{mzvsm}.
To summarize, we have checked that the potential baryonic instability
is cured. To do this, the one-loop generated VEV of $\phi_{22}$
is enough. On the other hand, to generate $m$, a crucial role was played by
an additional term in the superpotential, generated by a string instanton.
This makes our model a bona-fide version of SQCD with dynamically
generated exponentially small quark masses, and allows it to stably
display the related metastable vacua.
\section{Lifetime of the meta-stable vacuum}
In this section we study the possible decay channels for our non-supersymmetric ISS-like vacuum.
As reviewed in \S2, there is a SUSY vacuum where the mesons of node 3
acquire VEVs, which in turn are fixed by the VEVs of the mesons of node 1.
This is what we refer to as the mesonic branch, and is the usual SUSY
vacuum of SQCD, as considered e.g. in \cite{ISS}.
In addition to this SUSY vacuum, there is also another direction of possible
decay, which is precisely the one discussed in the previous section. Along
this direction, the baryons of node 1 acquire VEVs, and we are essentially
led to an SQCD at node 3 with $N_c$ massless and one massive flavors.
By standard arguments used when discussing
the deep IR of cascading quivers, one can see that
after a Seiberg duality on node 3,
the quiver reduces to SQCD with one flavor at node 2. The flavor acquires
a mass which is directly related to $m$.
We thus want to estimate the decay rate towards these two (classes of) vacua.
We do this by estimating the bounce action in the following form,
using the triangular approximation \cite{dj}
\begin{equation}
S \sim \frac{(\Delta \Phi)^4}{\Delta V},
\end{equation}
where $\Delta \Phi$ is roughly the width of the barrier while
$\Delta V$ is its height. We will see below that
we do not need to be more precise, since we are really interested
in lower bounds for the bounce action anyway. If we can tune these
lower bounds to be large enough, we can be confident that decay through
tunneling is suppressed and meta-stability is not affected.
Let us first consider decay towards the mesonic branch SUSY vacua.
Here $\Delta V$ is readily evaluated to be of the order of \eref{vmeta}, since
the energy of the metastable state and the peak only differ by a numerical
factor. To estimate $\Delta \Phi$, we first note that, as in \cite{ISS},
the fields which have the biggest variation are the mesons of node 3.
In the SUSY vacuum, their VEVs are given by
\begin{equation}
\phi_{22} = \Lambda_3 \left(\frac{m}{\Lambda_3}\right)^\frac{1}{N_c}, \qquad
\qquad \phi_{44} = \Lambda_3 \frac{h\Lambda_1^2}{m}
\left(\frac{m}{\Lambda_3}\right)^\frac{1}{N_c}.
\end{equation}
It is obvious from the above that $\phi_{22}\gg \phi_{44}$ for the
range of parameters discussed in the previous sections.
Moreover, recall that in the metastable state $\phi_{22}$ had a VEV
of the order of $h\Lambda_3^2$. This is however very small with respect
to the VEV it has in the SUSY vacuum, since we assume that $h\Lambda_3=
(\Lambda_3/M_s^*) \ll 1$. We can then identify $\Delta \Phi$
with the VEV of $\phi_{22}$ listed above.
Putting all together, we have the following estimate for
the bounce action towards the mesonic branch
\begin{equation}
S_\mathrm{mesonic} \sim \left(\frac{m}{\Lambda_3}\right)^\frac{4}{N_c}
\left(\frac{\Lambda_3}{\Lambda_1}\right)^4
\left(\frac{M_s^*}{\Lambda_3}\right)^2.
\end{equation}
Every factor in the expression above is (much) greater than one,
and we thus conclude that decay towards the mesonic branch is highly
suppressed.
As for the decay towards the baryonic branch of node 1, let us
provide the most conservative estimate. The field which varies the most
along the path is taken to be a representative baryon $B$. Its variation,
after the field has been canonically normalized, is taken to be
$\Delta \Phi \sim \Lambda_1$. Note that since $\Lambda_1$ is the smallest
scale in the game, this is really the most adverse situation.
As for $\Delta V$, we can take \eref{diagonal_terms} and plug in
the maximal VEV of the baryons $B \sim \Lambda_1^{N_c}$, so that
we get $\Delta V \sim h^4 \Lambda_3^6 \Lambda_1^2$. (Note that this
$\Delta V$ is much larger than the energy of the metastable vacuum.)
The bounce action is thus
\begin{equation}
S_\mathrm{baryonic} \sim \left(\frac{\Lambda_1}{\Lambda_3}\right)^2
\left(\frac{M_s^*}{\Lambda_3}\right)^4.
\end{equation}
Consistently with the inequality \eref{stability_baryons}, the above
bounce action can be made parametrically large, and thus also
the decay towards the baryonic branch is suppressed.
In this crude estimate, it seems that the latter
decay channel is the dominant one. To conclude that this is really
so would require a more serious investigation
of the potential and tunneling path. In any case we see that the
simplest estimates indicate that the apparent decay channels are
parametrically suppressed.
\section{The string dual description}
As we have seen, the quiver gauge theory we have analyzed in previous
sections admits a number of supersymmetric as well as metastable non-supersymmetric
vacua. A natural question is whether is it possible to provide a supergravity/string
dual description of such vacua. The well defined type IIB string embedding of
our model outlined in
\S \ref{model} makes this a realistic task.
\subsection{On the gravity dual}
\label{sugra}
In what follows we sketch the structure of only those vacua which are most relevant to our story:
the ISS-like vacua and their supersymmetric counterparts, i.e. the vacua
belonging to the mesonic branch. The discussion is very similar to the
one for the $\mathbb{Z}_2$ conifold quotient presented in our previous work \cite{Argurio:2006ny}, to which we refer
for details.
In the present construction a crucial role is
played by the presence of an additional fractional brane, that we have to
treat as a probe since its backreaction cannot be captured classically
(by definition, we cannot take the large $N$ limit for a single brane).
Moreover, a second equally crucial ingredient is that there is a
mass term constraining the position of this probe brane. Indeed, it can
be checked that for $m=0$ both classes of supersymmetric vacua become
runaway. Again, the mass term is the product of a stringy instanton
which is not expected to backreact on the classical geometry in any
simple manner.
Below, we will take the pragmatic point of view that, because of the
mass term, we can roughly integrate out the effect of the additional
probe brane. We are then left with the same gravity dual as discussed
in \cite{Argurio:2006ny}, albeit embedded in a higher singularity.
The effects of the additional fractional brane presumably show up as
(important) $1/N$ corrections to the geometry.
Our brane system can be embedded into a weakly curved gravity dual
background by adding (a large number of) regular D3 branes. One can easily
show that the resulting fractional/regular
brane system enjoys a duality cascade, i.e. a non-trivial RG-flow along which the effective
number of regular branes diminishes (in units of $N_c$, in this case). Hence,
choosing $N= k N_c$ regular D3 branes (with $k$ as large as we
wish), the IR field theory at the end of the cascade will be the quiver
field theory we have been studying. The number of cascade steps $k$ will
just control the final warp factor in the IR region of the gravity dual.
As discussed in \S2, the region of the moduli space where an effective massive
SQCD $SU(N_c)$ theory emerges is along the mesonic branch of node 1.
This corresponds to having the $N_c$ ${\cal N}=2$ branes at a distance
$(\det M_{22})^{1/N_c} \propto \Lambda_1^2$ along $\mathbb{C}$, the complex direction parametrizing the
VEV's of the adjoint scalar of the corresponding effective $SU(N_c)$ ${\cal N}=2$
vector superfield.
We will refer to the ${\cal N}=2$ fractional
branes as wrapped D5's, throughout this section.
The $\mathbb{Z}_3$ orbifold of the conifold\footnote{As already mentioned, we can
embed the same 4-node quiver in a $\mathbb{Z}_N$ orbifold with $N>3$. A reason to
go to larger $N$ might be to achieve the desired range of
scales, since there would be more geometrical quantities to tune.}
is described by the following equation in $\mathbb{C}^4$
\begin{equation}
\label{undefgeo}
x^3 y^3 = uv~.
\end{equation}
As already discussed in \S \ref{model}, this geometry supports three independent complex deformations, leading to
the completely smooth geometry
\begin{equation}
\prod_{i=1}^3 (xy-\epsilon_i) = uv~.
\end{equation}
Consequently, there are three non trivial 3-cycles $A_i$ whose minimal
size is given by $\epsilon_i$
\begin{equation}
\int_{A_i} \Omega = \epsilon_i~.
\end{equation}
In our particular case we would like to consider the case where
only two of the three 3-cycles are blown-up, and moreover they have the same size \cite{Argurio:2006ny}
\begin{equation}
(xy-\epsilon)^2 xy= uv~.
\end{equation}
This deformation is triggered by the $N_c$ deformation branes we have at node 3.
The geometry above has a $\mathbb{C}^2/\mathbb{Z}_2$ line of singularities
(also called $A_1$-singularities, not to be confused with the label
of the 3-cycles above) at the locus $xy=\epsilon$, $u=v=0$. Moreover,
it has an innocuous conifold singularity at
$x=y=u=v=0$.
We construct the geometrical dual to the supersymmetric vacua of our theory,
which were discussed at the beginning of \S2, in the following way.
After a geometric transition, we expect the brane at node 3 to transmute and
turn into flux,
\begin{equation}
\label{susyflux}
\int_{A} G_3 = N_c~~,~~ \int_{B} G_3 = \frac i g_s k
\end{equation}
where $A$ is the compact 3-cycle absorbing the RR flux of the original branes,
$B$ its non-compact dual, and $k$ is the
number of duality cascade steps (and can be naturally taken to be very
large).
As already noticed, the D5's wrapping the $\mathbb{C}^2/\mathbb{Z}_2$ singularity
are instead explicitly present in the dual geometry, lying somewhere
along the mesonic branch. Finally, the
single deformation brane at node 4 cannot transmute,
and remains as a probe at the remaining conical singularity,
which we expect to be slightly deformed
by the instanton discussed in \S\ref{instanton}.
It can be checked that the above set up has the
same charges and supersymmetric moduli space as our theory.
We now move to the description of the metastable state. As originally discussed in \cite{KachruPearsonVerlinde}, and
recently applied in similar contexts by many authors, a natural way to construct metastable non-supersymmetric vacua
is by adding anti-branes. The positive vacuum energy is proportional to the number of such branes.
The fact that the vacuum energy is exponentially small is due, in the gravity dual, to the
warping of the anti-brane tension.
In order for these configurations to describe states in the same gauge theory,
one should check that the supergravity charges, at
infinity, are unchanged. In the present context this can be achieved by adding $N_c$ anti-D3 branes and simultaneously
jumping the NS fluxes by one unit
\begin{equation}
\label{nsusyflux}
\int_{B} G_3 = \frac i g_s k \longrightarrow \int_{B} G_3 = \frac i g_s (k+1)~,
\end{equation}
so as to leave the full D3 brane charge untouched.\footnote{Let us remind the reader that the full D3 charge is
$Q_3 = \int H_3 \wedge F_3 + N - \overline{N}$, where $N$ and $\overline{N}$ are the net number
of D3 and anti-D3 branes, respectively.}
It is a nice check of our proposal that it is only by adding $N_c$
such branes (no more, no less), hence providing the correct energetics
for the ISS metastable vacua, that we can leave the global charges
at infinity untouched, and hence describe non-supersymmetric
states in the same gauge theory. Notice that as far as fluxes are concerned,
this shift corresponds to moving one step down in the cascade. That this is the case, will become
apparent when we discuss the type IIA T-dual description of this system in the
next subsection.
Due to the $F_5$ background in the gravity solution (dual to the large number of D3
branes present in the cascade), the anti-D3 branes are attracted to the tip of the geometry.
The metastable configuration presumably has all anti-D3s absorbed and dissolved as gauge
flux into the $N_c$ D5 branes
\begin{equation}
\int_{{\cal C}} {\cal F} = - N_c~,
\end{equation}
where ${\cal C}$ is the 2-cycle which the $N_c$ D5s wrap. This flux, via
the Chern-Simons coupling in
the DBI action of the D5 branes, accounts for the $N_c$ units of anti-brane charge.
Stability against decay through the Myers
effect can be argued as in \cite{Argurio:2006ny}, but of course a more detailed
study would be valuable.
A natural question is to ask whether one of the anti-branes can annihilate with the deformation brane associated to node 4,
which is sitting at the conifold singularity. In the next subsection we will provide a simple
argument as to why this is energetically disfavoured.
The supersymmetric vacua corresponding to the baryonic branch were
discussed in \cite{Argurio:2006ny}. It is not immediately clear how one
would directly relate them to the metastable states.
\subsection{Type IIA dual}
\label{IIadual}
In this section we study the Type IIA dual configurations describing our model. These constructions were first introduced in \cite{Uranga:1998vf}.
This approach provides a vivid picture of how the anti-branes arise in
the non-supersymmetric state.
\fref{IIA_5nodes_1} shows the type IIA T-dual brane configuration for our 4 node quiver for
$\langle {M}_{22}\rangle=m=0$.
\begin{figure}[ht]
\centering
\includegraphics[width=11cm]{IIA_5nodes_1.eps}
\caption{Type IIA configuration for the electric theory with
$\langle {M}_{22}\rangle=m=0$.
}
\label{IIA_5nodes_1}
\end{figure}
Performing a Seiberg duality on the middle node corresponds to moving the
second NS brane and second NS' brane across each other \cite{Elitzur:1997fh}. In the process,
some anti D4 branes are generated in the middle interval, but they are annihilated against
D4 branes sitting on top of them. The result is shown in \fref{IIA_5nodes_2}.
\begin{figure}[ht]
\centering
\includegraphics[width=11cm]{IIA_5nodes_2.eps}
\caption{Type IIA configuration for the magnetic theory.}
\label{IIA_5nodes_2}
\end{figure}
A non-zero VEV for $\langle {M}_{22}\rangle$ in the electric theory corresponds to moving the D4-branes
stretched between the first and second NS-brane in \fref{IIA_5nodes_1} in the $45$ directions.
Similarly, the mass $m$ is mapped to a displacement of the third NS brane together with the D4-branes
that stretch from it to the second NS'-branes in $89$. This is shown in \fref{IIA_5nodes_3}.
\begin{figure}[ht]
\centering
\psfrag{M22}{$\langle {M}_{22}\rangle$}
\includegraphics[width=11cm]{IIA_5nodes_3.eps}
\caption{Type IIA configuration for the electric theory with
non-zero $\langle {M}_{22}\rangle$ and $m$.}
\label{IIA_5nodes_3}
\end{figure}
If we now perform the Seiberg duality, we obtain the configuration in \fref{IIA_5nodes_4}.
The anti-D4's are not annihilated, since they are now displaced from the D4s due to the
meson VEVs (following the discussion in \S3, $\langle {M}_{22}\rangle$
is stabilized at tree level).
\begin{figure}[ht]
\centering
\psfrag{M22}{$\langle {M}_{22}\rangle$}
\includegraphics[width=11cm]{IIA_5nodes_4.eps}
\caption{Type IIA configuration for the magnetic theory with
non-zero $\langle {M}_{22}\rangle$ and $m$. Anti-D4 branes are indicated in red.}
\label{IIA_5nodes_4}
\end{figure}
\fref{IIA_5nodes_4} shows that the avatar of SUSY breaking in the ISS vacuum
is explicit un-annihilated anti D4 branes.
The failure to annihilate these branes is a direct consequence of the meson VEVs and $m$ (the dynamically generated masses).
Relative to the IIB story we described in the previous subsection, this
IIA configuration is an intermediate picture between the state with
explicit anti-D3 branes and the (final) state where they are dissolved within the $N_c$ D5s
at the $A_1$ singularity.
Collapsing D4 and anti D4-branes as much as possible, we
obtain \fref{IIA_5nodes_8} for the IIA picture of the ISS vacuum. We have labeled the NS branes to
simplify the discussion. We have annihilated the anti-D4's against D4's between NS1 and NS2 (and not NS2' and NS3) because,
since $|h\langle {M}_{22}\rangle| < m$, this clearly produces a lower energy configuration. Notice
that the objects the gauge flux (represented here by the anti-D4's) combines with are precisely D4 branes stretched between
the parallel NS1 and NS2. The resulting tilted D4 branes are dual to the D5/anti-D3 bound states
of the IIB configuration.
\begin{figure}[ht]
\centering
\includegraphics[width=11cm]{IIA_5nodes_8.eps}
\caption{The type IIA configuration once the anti-D4 branes has been dissolved into the D4's as gauge flux.}
\label{IIA_5nodes_8}
\end{figure}
In this type IIA setting it is also easy to see how
the configuration in \fref{IIA_5nodes_4} is related to the addition of anti D3-branes
and jumping fluxes in the gravity dual, as described in \S\ref{sugra}.
An anti-D3 brane maps to an anti-D4 brane
wrapping the entire $x_6$ circle in Type IIA. We can form such a complete anti-D4 by adding
D4/anti-D4 pairs to all intervals with the exception of the one between the second NS' and second NS. Grouping D4 branes in
each interval together, we are left with the configuration in \fref{IIA_5nodes_6}. It corresponds to $N_c$ full
anti-D4 branes (T-dual to $N_c$ anti-D3 branes) and the number of D4 branes in each interval corresponds
to moving up one step in the cascade from the magnetic theory (the last step). Increasing the cascade by one step is exactly how
increasing the NS flux by one unit manifests in this context. This matches nicely with our previous type IIB description
of the metastable non-supersymmetric vacua.
\begin{figure}[ht]
\centering
\includegraphics[width=11cm]{IIA_5nodes_6.eps}
\caption{The ISS vacuum configuration of \fref{IIA_5nodes_4} can be interpreted as adding $N_c$ anti-D3 branes
and moving down one step in the cascade.}
\label{IIA_5nodes_6}
\end{figure}
\section{Conclusions}
We have presented a D-brane construction that engineers metastable vacua closely related to those
of \cite{ISS}. The construction has some interesting conceptual features and some interesting
features for model building.
Conceptually, the most interesting thing about the construction is that it readily admits
a IIB gravity dual description (in the general framework of AdS/CFT). The metastable
states cannot be directly followed from weak 't Hooft coupling to strong 't Hooft
coupling, but they quite plausibly match on to strong-coupling analogues where the
SUSY breaking is well described by the presence of anti-D3 branes, and a
picture very similar to the one in \cite{KachruPearsonVerlinde}. In addition, the quivers that arise
in our construction are some of the simplest cases where the new string instanton effects
explored in many recent works make important contributions.
One may also wish to construct pseudo-realistic models of SUSY breaking
and/or direct mediation using quiver
gauge theories. In this case, our model has two virtues: the small dynamical masses
of the ISS model are explained naturally without any fine tuning of parameters
(as could also be done by retrofitting \cite{Retro}), and
the problematic R-symmetry which could
forbid gaugino masses is lifted by the extra
terms that automatically appear in our superpotential.
We note that because the R-symmetry is broken by an irrelevant operator
suppressed by the high scale $M_s^*$, and because for metastability it
is necessary to take $M_s^*$ somewhat higher than the SUSY-breaking scale,
it is likely that in real model-building applications, our theory would
produce low gaugino masses. Given the lower bounds on gaugino masses,
this would necessitate heavy squarks and sleptons, resulting in the need
for a (mild) tune to obtain a reasonable Higgs mass.
It
would be very interesting to study the gravity dual geometry in further detail.
While there aren't BPS or protected quantities that are guaranteed to match between weak
and strong coupling, one may find interesting patterns of qualitative agreement between
the two classes of non-supersymmetric states. Conversely, some quantities (e.g. lifetimes
or barrier heights) may change in a striking way upon extrapolation in $g_sN$.
\begin{center}
\bf{Acknowledgements}
\end{center}
\medskip
We would like to
thank O. Aharony, F. Bigazzi, M. Buican, G. Ferretti, B. Florea, A. Lerda,
N. Saulina, N. Seiberg and A. Uranga for helpful discussions.
R.A. and M.B. are partially supported by the European Commission FP6
Programme MRTN-CT-2004-005104, in which R.A is associated to V.U. Brussel.
R.A. is a Research Associate of the Fonds National de la Recherche
Scientifique (Belgium). The research of R.A. is also supported by IISN - Belgium
(convention 4.4505.86) and by the ``Interuniversity Attraction Poles Programme --Belgian Science Policy''.
M.B. is also supported by Italian MIUR under contract PRIN-2005023102 and by a
MIUR fellowship within the program ``Rientro dei Cervelli''.
S.F. is supported by the DOE under contract DE-FG02-91ER-40671.
The research of S.K. was supported in part by a David and Lucile Packard Foundation
Fellowship, the NSF under grant PHY-0244728, and the DOE under contract
DE-AC03-76SF00515. S. F. would like to thank the Galileo
Galilei Institute for Theoretical Physics
for hospitality while this work was being completed.
S.K. similarly acknowledges the kind hospitality of the International Centre
for Theoretical Physics.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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Installation - A Smart GitHub Gist Client With The TextExpander Feature.
You only need to configure the relevant Python environment variables to complete the installation.
Tips: please restart the software after setting, otherwise the normal software cannot be run.
If some of these commands are missing, please download and install them yourself.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,399
|
The Deadliest Dinosaurs of All Time – Watch out!
brain tease
By Gur Tirosh
On the 22nd of June, the movie "Jurassic World: Fallen Kingdom" will start showing in cinemas. As a dinosaur enthusiast, I'm sure that you cannot wait for this day. As you prepare yourself for this day, why not familiarize yourself with some of the most dangerous dinosaurs to ever exist?
Even though dinosaurs are known to be dangerous, there are some species which are considered to have been more dangerous than others. In no particular order, below is a list of ten of the most dangerous dinosaurs that existed.
1. The Tyrant Llizard
Tyrannosaurus, also known as the 'tyrant lizard' had a bite force that was so tremendous that it was considered to be the strongest among all dinosaurs and all the living terrestrial animals. Its bite force could reach up to 5,805 Kilograms. It had a mass of 5,800 – 15,000 kg and was about 4.6 – 6.1 m tall, half as tall as a telephone pole.
The Tyrannosaurus Rex skeleton known as Sue stands on display at Union Station June 7, 2000, in Washington D.C. (Photo by Mark Wilson/Newsmakers)
Considered to be the largest land-living carnivore of all time, Tyrannosaurus was capable of delivering bites that could crush stones. Its most massive tooth had a length of about 12 inches. It fed on plant-eating dinosaurs.
2. The Should be the 'King of Dinosaurs'
Spinosaurus aegyptiacus is considered as one of the largest carnivorous dinosaurs ever to live. This dinosaur can swim, and measured over 50 feet long, almost as tall as a three-story building. The Spinosaurus is one of the few dinosaurs with the ability to swim. It was so huge that many dinosaur lovers believe that it rivaled the tyrant lizard, commonly known as, Tyrannosaurus Rex or T Rex. It's also referred to as the 'spiny lizard' because of the large spined sail which ran down its back. This huge dinosaur's weight was about 7.4 tonnes, and its length was 46 feet.
An Extinct Spinosaurus – Spinosaurus Aegyptiacus. Spinosaurus is a genus of theropod dinosaur that lived in what now is North Africa, during the lower Albian to lower Cenomanian stages of the Cretaceous period, about 112 to 97 million years ago. (Photo by Roberto Machado Noa/LightRocket via Getty Images)
It had an elongated skull and jaws that resembled those of a crocodile that was about 5 feet long.
3. The Toothy Masiakasaurus Knopfleri
This one was so dangerous that there is no other predator dinosaur with the kind of teeth that this dinosaur possessed. It had front teeth which were projecting forward, and it is these teeth that it could use when slicing and then cut its captives into chunks that were chewy by using its rears that looked like blades.
Photo Credit: James St. John
Also known as a vicious lizard, this dinosaur with 6 feet body length, is believed to have feasted on lizards, fish and other critters about 65 – 70 million years ago in Madagascar, an island nation of Africa. It had a long neck and tail which made up most of its length.
4. Gryposaurus monumentensis/ the Hooked Beak Lizard
(Pronunciation: GRIP-oh-SORE-us MAHN-you-men-TEN-siss) This was a big-boned and duck-billed dinosaur that could have eaten any vegetation it stumbled across. It had a massive skull that was able to pack over 300 teeth that it used for cutting up fibrous greens. In addition to the 300 teeth, it had hundreds of more teeth for a replacement that rested on its jawbone for use anytime they were required.
Photo credit: Jens Lallensack
It lived about 75 million years ago in the Late Cretaceous. This plant-eating dinosaur may have been about 30 feet long as a fully grown dinosaur and had a head that was 3-foot long, almost as big as T Rex.
5. Allosaurus / Different Lizard
These existed about 155 million years ago. Its name translates to 'different lizard.' They are alleged to have had an enormous skull with numerous sharp teeth. They were 9 meters long although some studies suggest that they were about 12 meters long. They possessed a massive tail which offered great balance. They are said to have weighed 2.3 metric tons. Just imagine what would happen if they could come alive.
Allosaurus ('Other Lizard'), A Large, Fearsome Predatorial Dinosaur From The Late Jurassic Period. (Photo By Encyclopaedia Britannica/UIG Via Getty Images)
This massive carnivore found in Utah, weighed over 1,500 kilograms in its adulthood, with a height of 4.5 – 5 meters.
6. Troodon – Hunt in Packs
Although small in size, these dinosaurs are considered to have been very dangerous. This is because they could hunt in packs and they were also blessed with very large eyes which made it possible for them to hunt in the dark at night. It only weighed about 150 pounds, but trust me, a group of five of them was equivalent regarding danger, to one hungry and full grown T Rex. What made this dinosaur different from the rest was the size of its brain.
Illustration of Troodon catching young dinosaurs from the nest (Photo by De Agostini Picture Library/De Agostini/Getty Images)
Compared to the other carnivorous dinosaurs that lived in the Late Cretaceous North America, Troodon had a relatively big brain.
7. Majungasaurus / Cannibal Dinosaur
Previously known as Majungatholus, this dinosaur was given the name, 'cannibal dinosaur' by the media. They dubbed them cannibals because it is said that they preyed on the others of their kind when hunger pangs were too much to bear. There were tooth marks that were found on some of their bones which were tooth marks from the same species.
(Photo Credit Wikipedia)
They weighed about one ton, and they are alleged to have spent most their time terrorizing the smaller plant-eating dinosaurs that were found in late Cretaceous Africa.
8. Utahraptor – The Slasher
No raptor was as dangerous as the Utahraptor. This raptor weighed about a ton and lived approximately 125 million years ago, way before their descendants; the deinonychus and velociraptor. It had single claws that were curved. It had claws on its hind feet that were almost a foot long. These claws were used for slashing and disemboweling their prey.
Utahraptor dinosaur in the forest (Utahraptor ostrommaysorum) (Photo by De Agostini Picture Library/De Agostini/Getty Images)
Discovered in Utah in 1991, these raptors were almost entirely covered in feathers. It weighed over a ton and was 25 feet long. This means that it was and three car links long and could run speeds of up to 20 miles per hour. According to its weight, it was ten times heavier than the average raptor.
9. Giganotosaurus – One of the Biggest
This dinosaur found in South America was enormous. It weighed about eight to ten tons. Can you imagine? It had three fingers and is considered to be among the biggest dinosaurs to ever walk on the earth. It was one of the theropods that were capable of taking down a titanosaur that was fully grown or even a more manageable juvenile. It was larger than T Rex or Spinosaurus.
The Giganotosaurus disinterred from rocks, 1912. From a reconstruction drawing by A. Forrestier. (Photo by Universal History Archive/Getty Images)
An adult Giganotosaurus had a mass of 8,000 kilograms and a length of about 12 – 14 meters. It had a speed of 50 km/h.
10. Liopleurodon – The Swimmer
These are another species of dinosaurs that were able to swim. In fact, they were excellent swimmers because they had four strong limbs. They were capable of growing to an enormous length of 25 meters. They are believed to have existed in France and England. One of these species is alleged to have existed in Russia also.
Illustration of Liopleurodon catching fish (Photo by De Agostini Picture Library/De Agostini/Getty Images)
The structure of their body provided them with excellent acceleration and top speed. It weighed between 1,000 – 1,700 kg. It lived between 166.1 and 139.8 million years ago.
10 Interesting Facts About Crossword Puzzles
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 8,822
|
package org.kie.server.services.impl.marshal;
import java.util.Date;
import java.util.HashSet;
import java.util.Set;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import org.apache.commons.lang3.builder.EqualsBuilder;
import org.hamcrest.BaseMatcher;
import org.hamcrest.Description;
import org.json.JSONException;
import org.junit.Test;
import org.kie.server.api.marshalling.MarshallingFormat;
import org.kie.server.api.model.definition.ProcessInstanceQueryFilterSpec;
import org.kie.server.api.model.definition.QueryFilterSpec;
import org.kie.server.api.model.definition.TaskQueryFilterSpec;
import org.kie.server.api.util.ProcessInstanceQueryFilterSpecBuilder;
import org.kie.server.api.util.QueryFilterSpecBuilder;
import org.kie.server.api.util.TaskQueryFilterSpecBuilder;
import org.kie.server.services.api.KieServerRegistry;
import org.mockito.Mockito;
import org.mockito.internal.matchers.apachecommons.ReflectionEquals;
import org.skyscreamer.jsonassert.JSONAssert;
import org.xmlunit.matchers.CompareMatcher;
import static org.junit.Assert.assertEquals;
import static org.junit.Assert.assertThat;
public class MarshallerHelperTest {
@Test
public void testMarshallWithoutContainer() {
KieServerRegistry kieServerRegistryMock = Mockito.mock(KieServerRegistry.class);
MarshallerHelper helper = new MarshallerHelper(kieServerRegistryMock);
QueryFilterSpec queryFilterSpec = new QueryFilterSpecBuilder().get();
String marshalledQFS = helper.marshal(MarshallingFormat.JAXB.toString(), queryFilterSpec);
String expectedMarshalledQFS = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" + "<query-filter-spec>"
+ "<order-asc>false</order-asc>" + "</query-filter-spec>";
assertThat(marshalledQFS, CompareMatcher.isIdenticalTo(expectedMarshalledQFS).ignoreWhitespace());
}
@Test
public void testMarshallWithoutContainerWithExtraClasses() {
KieServerRegistry kieServerRegistryMock = Mockito.mock(KieServerRegistry.class);
Set<Class<?>> extraClasses = new HashSet<>();
extraClasses.add(TestExtraClass.class);
Mockito.when(kieServerRegistryMock.getExtraClasses()).thenReturn(extraClasses);
MarshallerHelper helper = new MarshallerHelper(kieServerRegistryMock);
TestExtraClass extraClass = new TestExtraClass();
extraClass.setBla("hallo");
String marshalledQFS = helper.marshal(MarshallingFormat.JAXB.toString(), extraClass);
String expectedMarshalledTEC = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" + "<test-extra-class>"
+ "<bla>hallo</bla>" + "</test-extra-class>";
assertThat(marshalledQFS, CompareMatcher.isIdenticalTo(expectedMarshalledTEC).ignoreWhitespace());
}
@Test
public void testJsonMarshallWithEmptyRegistry() throws Exception {
KieServerRegistry kieServerRegistryMock = Mockito.mock(KieServerRegistry.class);
Mockito.when(kieServerRegistryMock.getExtraClasses()).thenReturn(new HashSet<Class<?>>());
MarshallerHelper helper = new MarshallerHelper(kieServerRegistryMock);
QueryFilterSpec queryFilterSpec = new QueryFilterSpecBuilder().get();
String marshalledQFS = helper.marshal(MarshallingFormat.JSON.toString(), queryFilterSpec);
System.out.println(marshalledQFS);
String expectedMarshalledTEC = "{\"order-by\" : null, \"order-asc\" : false, \"query-params\" : null, \"result-column-mapping\" : null}";
JSONAssert.assertEquals(expectedMarshalledTEC, marshalledQFS, false);
}
@Test
public void testJsonMarshallWithNullRegistry() throws Exception {
MarshallerHelper helper = new MarshallerHelper(null);
QueryFilterSpec queryFilterSpec = new QueryFilterSpecBuilder().get();
String marshalledQFS = helper.marshal(MarshallingFormat.JSON.toString(), queryFilterSpec);
System.out.println(marshalledQFS);
String expectedMarshalledTEC = "{\"order-by\" : null, \"order-asc\" : false, \"query-params\" : null, \"result-column-mapping\" : null}";
JSONAssert.assertEquals(expectedMarshalledTEC, marshalledQFS, false);
}
/**
* Tests that MarshallerHelper can also be used when passing in a <code>null</code> KieServerRegistry.
*/
@Test
public void testMarshallWithNullRegistry() {
MarshallerHelper helper = new MarshallerHelper(null);
QueryFilterSpec queryFilterSpec = new QueryFilterSpecBuilder().get();
String marshalledQFS = helper.marshal(MarshallingFormat.JAXB.toString(), queryFilterSpec);
String expectedMarshalledQFS = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" + "<query-filter-spec>"
+ "<order-asc>false</order-asc>" + "</query-filter-spec>";
assertThat(marshalledQFS, CompareMatcher.isIdenticalTo(expectedMarshalledQFS).ignoreWhitespace());
}
@Test
public void testUnmarshallWithoutContainer() {
KieServerRegistry kieServerRegistryMock = Mockito.mock(KieServerRegistry.class);
MarshallerHelper helper = new MarshallerHelper(kieServerRegistryMock);
QueryFilterSpec expectedQueryFilterSpec = new QueryFilterSpecBuilder().get();
String marshalledQFS = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" + "<query-filter-spec>"
+ "<order-asc>false</order-asc>" + "</query-filter-spec>";
QueryFilterSpec unmarshalledQFS = helper.unmarshal(marshalledQFS, MarshallingFormat.JAXB.toString(), QueryFilterSpec.class);
// QueryFilterSpec does not implement equals method, so using Mockito ReflectionEquals.
assertThat(expectedQueryFilterSpec, new BaseMatcher<QueryFilterSpec>() {
@Override
public void describeTo(Description description) {
}
@Override
public boolean matches(Object item) {
return new ReflectionEquals(unmarshalledQFS).matches(item);
}
});
}
@Test
public void testUnmarshallWithoutContainerWithExtraClasses() {
KieServerRegistry kieServerRegistryMock = Mockito.mock(KieServerRegistry.class);
Set<Class<?>> extraClasses = new HashSet<>();
extraClasses.add(TestExtraClass.class);
Mockito.when(kieServerRegistryMock.getExtraClasses()).thenReturn(extraClasses);
MarshallerHelper helper = new MarshallerHelper(kieServerRegistryMock);
TestExtraClass expectedExtraClass = new TestExtraClass();
expectedExtraClass.setBla("hallo");
String marshalledTEC = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" + "<test-extra-class>" + "<bla>hallo</bla>"
+ "</test-extra-class>";
TestExtraClass unmarshalledTEC = helper.unmarshal(marshalledTEC, MarshallingFormat.JAXB.toString(), TestExtraClass.class);
assertEquals(expectedExtraClass, unmarshalledTEC);
}
/**
* Tests that MarshallerHelper can also be used when passing in a <code>null</code> KieServerRegistry.
*/
@Test
public void testUnmarshallWithoutNullRegistry() {
MarshallerHelper helper = new MarshallerHelper(null);
QueryFilterSpec expectedQueryFilterSpec = new QueryFilterSpecBuilder().get();
String marshalledQFS = "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" + "<query-filter-spec>"
+ "<order-asc>false</order-asc>" + "</query-filter-spec>";
QueryFilterSpec unmarshalledQFS = helper.unmarshal(marshalledQFS, MarshallingFormat.JAXB.toString(), QueryFilterSpec.class);
// QueryFilterSpec does not implement equals method, so using Mockito ReflectionEquals.
assertThat(expectedQueryFilterSpec, new BaseMatcher<QueryFilterSpec>() {
@Override
public void describeTo(Description description) {
}
@Override
public boolean matches(Object item) {
return new ReflectionEquals(unmarshalledQFS).matches(item);
}
});
}
@Test
public void testJsonUnmarshallQueryFilterSpec() {
MarshallerHelper helper = new MarshallerHelper(null);
String marshalledQFS = "{\"order-by\" : null, \"order-asc\" : false, \"query-params\" : null, \"result-column-mapping\" : null}";
QueryFilterSpec qfs = helper.unmarshal(marshalledQFS, MarshallingFormat.JSON.toString(), QueryFilterSpec.class);
//TODO: assert value.
}
@Test
public void testJsonUnmarshallProcessInstanceQueryFilterSpec() {
ProcessInstanceQueryFilterSpec expectedPiQfs = new ProcessInstanceQueryFilterSpecBuilder().get();
MarshallerHelper helper = new MarshallerHelper(null);
String marshalledQFS = "{\"order-by\" : null,\"order-asc\" : false,\"query-params\" : null}";
ProcessInstanceQueryFilterSpec unmarshalledPiQfs = helper.unmarshal(marshalledQFS, MarshallingFormat.JSON.toString(), ProcessInstanceQueryFilterSpec.class);
assertThat(expectedPiQfs, new BaseMatcher<QueryFilterSpec>() {
@Override
public void describeTo(Description description) {
}
@Override
public boolean matches(Object item) {
return new ReflectionEquals(unmarshalledPiQfs).matches(item);
}
});
}
@Test
public void testJsonUnmarshallNotNull() throws JSONException {
KieServerRegistry kieServerRegistryMock = Mockito.mock(KieServerRegistry.class);
Set<Class<?>> extraClasses = new HashSet<>();
// simulate server conditions
extraClasses.add(Date.class);
extraClasses.add(org.kie.server.api.model.type.JaxbByteArray.class);
Mockito.when(kieServerRegistryMock.getExtraClasses()).thenReturn(extraClasses);
MarshallerHelper helper = new MarshallerHelper(kieServerRegistryMock);
JSONAssert.assertEquals("{\"order-asc\" : false}", helper.marshal("application/json ; fields = not_null ", new QueryFilterSpecBuilder().get()), true);
// test reset
JSONAssert.assertEquals("{\"order-by\" : null, \"order-asc\" : false, \"query-params\" : null, \"result-column-mapping\" : null, \"order-by-clause\" : null}", helper.marshal("application/json", new QueryFilterSpecBuilder().get()),true);
}
@Test
public void testJsonUnmarshallTaskQueryFilterSpec() {
TaskQueryFilterSpec expectedTaskQfs = new TaskQueryFilterSpecBuilder().get();
MarshallerHelper helper = new MarshallerHelper(null);
String marshalledQFS = "{\"order-by\" : null, \"order-asc\" : false, \"query-params\" : null}";
TaskQueryFilterSpec unmarshalledTaskQfs = helper.unmarshal(marshalledQFS, MarshallingFormat.JSON.toString(), TaskQueryFilterSpec.class);
assertThat(expectedTaskQfs, new BaseMatcher<QueryFilterSpec>() {
@Override
public void describeTo(Description description) {
}
@Override
public boolean matches(Object item) {
return new ReflectionEquals(unmarshalledTaskQfs).matches(item);
}
});
}
@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "test-extra-class")
public static class TestExtraClass {
@XmlElement(name = "bla")
private String bla;
public String getBla() {
return bla;
}
public void setBla(String bla) {
this.bla = bla;
}
@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (obj == this) {
return true;
}
if (obj.getClass() != getClass()) {
return false;
}
TestExtraClass rhs = (TestExtraClass) obj;
return new EqualsBuilder().append(bla, rhs.bla).isEquals();
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,505
|
Who Will Ride the IT M&A Wave Following Oracle-Sun?
This one's been circulating literally for years. But the latest talk now is that IBM may be getting serious about making an offer that CEO Dan Warmenhoven, co-founder Dave Hitz and the NetApp Board of Directors cannot refuse. IBM saved up a bunch of cash for Sun; since that deal's kaput, why not put the money into a growing sector (storage) with a well-managed company that knows how to make and sell the next-generation products that Big Blue needs?
This is another oldie that is regularly resurrected. The synergies have always been there: West Coast, fairly conservative corporate cultures; longtime partnerships in several product areas; product lines that complement each other; plenty of customers that do not overlap, so there are new sales fields to mine for HP.
Seems as natural as running Photoshop on Mac OS.
There's no way this was going to happen a year ago, when talks first surfaced. But times have changed drastically since then. There's still no way their cultures will ever mesh. However, Microsoft CEO Steve Ballmer wants his company's consumer online business to stop losing money, and acquiring Yahoo is a fast way to turn that around.
Why eBay ever bought Skype has been a mystery for the ages. A company that really knows how to use Skype's capabilities needs to own it. One good move by eBay was getting Skype all that face time on "Oprah."
We've been all over this before and it still makes a lot of sense. VMware and its virtual machinery has been a cash cow for EMC. But the companies' cultures clash big time, and a lot of excellent VMware talent has moved on. Cisco wants to control virtualization for its network-centric data center strategy. EMC's strengths are in storing and protecting data.
EMC has built a hugely successful business by acquiring companies in and around the storage and data protection business (Iomega, RSA Security, Documentum, Avamar deduplication, Mozy online storage, etc.)It's always looking for the next conquest. How about somebody in the mobile computing sector, such as Motorola? Data has to be stored there, too, and more of it goes there every day.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,617
|
Parkside Primary Academy
Learning with Pride
About Parkside Primary Academy
Pioneer Academies Community Trust
Catch Up Funding 2020-2021
Learning and Curriculum
F1 Curriculum
Y1 Curriculum
Personal, Social, Health and Economic education (PHSE)
Discovery RE
SATS Information for Parents
Sex and Relationships Education
Term Time and Holiday Dates
ICT Blog
Foundation 1
Parkside Nurture
On May 25th 2018 the General Data Protection Regulation (GDPR) will come into effect across the EU. It is a new law that is being implemented in exactly the same way across all EU countries to create a level set of rules to ensure all data held and processed by organisations within the EU is secure and is processed lawfully.
You will no doubt have received communications from all kinds of businesses, including any you shop or bank with, so this may be something you are already aware of.
Public organisations, including schools, are required to comply with GDPR in the same way. At the Pioneer Academies Community Trust (PACT) we would like to reassure you that we have been working on our GDPR readiness over the last year.
PACT has always highly valued and protected all of its pupil, parent/carer and staff information - and will continue to do so as the GDPR legislation is enacted.
Schools across PACT already have strong Data Protection policies and protections in place, and these are now being updated and reviewed by the governing body in line with GDPR. This means that the school website will be updated shortly with improved policies and you will receive documents such as updated privacy notices, as you do usually each September.
Parent/Carers will begin to receive letters or updates from the schools regarding GDPR, and some of those may be about consent and some about updating your information with us.
It is really important that you read, and respond promptly, to any correspondence from school on this matter.
The Trust has a designated Data Protection Officer (DPO), who is helping to guide the schools and work to ensure full compliance with the new GDPR. The DPO can be contacted with any queries using the following email address or in writing to any of our schools:
DPO@pioneeract.org.uk
P Steadman
Chair of Trust
GDPR Data Protection Policy PACT 2018
Retention of Records Policy PACT 2018
Privacy Notice Pupil Information PACT 2018
Child-friendly Privacy Notice 2018
Privacy Notice School Workforce PACT 2018
Privacy Notice Recruitment
Dress to Express Fundraiser for Children's Mental Health Week 2021 21st January 2021
School Open 26 November 2020 25th November 2020
School Closure 25th November 2020
Newsletter 14/1/2022 14th January 2022
Newsletter 17/12/21 17th December 2021
Please note if you are invited into school for a meeting or your child's Reward Assembly you will be required to take a Lateral Flow test and provide a negative result.
All website content copyright © Parkside Primary Academy
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,884
|
ITS Technology Support Specialist
Go to our website »
Position Number: 2097940
Location: Hamilton, NY
Position Type: Computer and Information Technology
Requisition Number: 2020S055Posting
Full Time/Part Time: Full Time
Division: Dean of the Faculty
Department: Information Technology Services - Engagement and SupportColgate University is seeking an organized, motivated person for an IT Support Specialist position. This position provides the opportunity to coach a group of diverse students to grow a high-performance team. ITS endeavors to foster an inclusive environment that values diversity, professional development, creativity, and innovation to support the growth of individuals and the organization. Under the leadership of the CIO, ITS is embarking on a strategic planning process to align services and resources with several exciting new initiatives identified in Colgates Third Century Plan. The Engagement and Support team is the primary point of contact for our community and we strive to enhance the productivity of our user communities by empowering them through making information and tools accessible to them. In this role you will work closely with other IT Support Specialists and our hardware installation team to support our students, staff, and faculty and work with the rest of the IT teams as escalations and projects require. Wed love to hear from you even if you do not meet 100% of the preferred/desired qualifications.
Department Statement:
Colgate University is seeking an organized, motivated person for an IT Support Specialist position. This position provides the opportunity to coach a group of diverse students to grow a high-performance team.
ITS endeavors to foster an inclusive environment that values diversity, professional development, creativity, and innovation to support the growth of individuals and the organization. Under the leadership of the CIO, ITS is embarking on a strategic planning process to align services and resources with several exciting new initiatives identified in Colgates Third Century Plan.
The Engagement and Support team is the primary point of contact for our community and we strive to enhance the productivity of our user communities by empowering them through making information and tools accessible to them. In this role you will work closely with other IT Support Specialists and our hardware installation team to support our students, staff, and faculty and work with the rest of the IT teams as escalations and projects require.
Wed love to hear from you even if you do not meet 100% of the preferred/desired qualifications.
Accountabilities:
Under general direction, the ITS Technology Support Specialist provides broad scope system and service support to students, staff, and faculty of the Colgate community in furtherance of the strategic mission of the university. The ITS Technical Support Specialist may also participate in regularly scheduled or limited scope projects as assigned. Provide services as a member of the Service Desk staff including, but not limited to:
Serve as the first point of contact for customers seeking technical assistance over the phone, or email (and in-person interactions).
Perform remote troubleshooting through diagnostic techniques and pertinent questions.
Provide client follow up as necessary, reply using email, phone, and ensure proper recording, documentation, and closure for all issues.
Utilize customer service skills to meet business goals and objectives and end-user needs. Ensure a high level of client satisfaction through frequent and clear communication on all work orders.
Manage the technical assignments and activities of a diverse team of student workers and cooperative work/study employees, and provide training and leadership to them as necessary.
Create and maintain documentation on internal procedures and document various department activities as assigned.
Maintain general awareness of advances and current trends in technology. Research issues and resolutions to technical problems. Acquire and maintain technical knowledge through reading, self-paced training, computer-based training or video-based training, courses, seminars, and other methods as necessary.
Professional Experience/Qualifications
Professional experience in ITS/Information Technology Support roles.
Bachelors Degree or 3 years of experience in support/customer service role.
Job Open Date:
Job Close Date:
Special Instructions Summary
Colgate is committed to attracting and retaining a diverse faculty, staff, and student population.
We strive to be an inclusive community one that embraces and values diversity (in the broadest sense possible) in an environment of mutual respect, communication, and engagement. A variety of cultures and perspectives enriches the quality of campus life, and the opportunity to share different views and experiences is at the core of Colgates educational enterprise.
These differences can include but are not limited to: race; ethnicity; gender and gender expression; sexual orientation; socioeconomic status; geographic background; national origin; culture; age; mental, cognitive, and physical abilities; religious beliefs; and political beliefs.
As a result, we ask all candidates seeking consideration for the ITS Technology Support Specialist position to submit a diversity statement with their application materials.
You can upload the statement under the Diversity Statement document heading.
A diversity statement may be any length (e.g. a short paragraph, a page) explaining your experiences, contributions, and/or commitments related to diversity, equity, and inclusion. Examples of topics you may discuss include (but are not limited to): 1) how your experiences or competencies might serve to advance the universitys commitment to creating a diverse and inclusive campus, 2) your experiences working with diverse populations, or 3) how you will contribute to the universitys strategic initiatives in this area (see Colgates Diversity, Equity, and Inclusion Plan).
It is the policy of Colgate University not to discriminate against any employee or applicant for employment on the basis of their race, color, creed, religion, age, sex, pregnancy, national origin, marital status, disability, protected Veteran Status, sexual orientation and gender identity and expression, genetic information, victims of domestic violence and stalking, familial status, and all other categories covered by law. This policy shall apply to all employment actions, including but not limited to recruitment, hiring, upgrading, promotion, transfer, demotion, layoff, recall, termination, rates of pay or other forms of compensation and selection for training at all levels of employment. Colgate University will not discharge or discriminate against employees or applicants who inquire about, discuss, or disclose their own compensation or the compensation of another employee or applicant. Colgate University is an Equal Opportunity Employer. Minorities/Females/Persons with Disabilities/Protected Veterans are encouraged to apply.
CAMPUS CRIME REPORTING AND STATISTICS
The Campus Safety Department will provide upon request a copy of Colgates Annual Security and Fire Safety Report. This report includes statistics as reported to the United States Department of Education for the previous three years concerning reported: 1. crimes that occurred on-campus; in certain off-campus buildings or property owned or controlled by Colgate University; and on public property within, or immediately adjacent to and accessible from, the campus and 2. fires that occurred in student housing facilities. The report also includes institutional policies concerning campus security and fire safety, such as policies concerning sexual assault, life safety systems, and other related matters. To obtain a copy, contact the Campus Safety Compliance Manager via e-mail at cusafety@colgate.edu. You may also access the report from the Campus Safety web page at: www.colgate.edu/offices/support/campussafety.
To view the full job posting and apply for this position, go to: https://careers.colgate.edu/postings/3142
Copyright 2017 Jobelephant.com Inc. All rights reserved.
jeid-21539241685a7f47a892e6cdc171d507
IT Jobs in Higher Education specializes in helping match highly skilled tech workers and the higher education community. Typical jobs include: Application Programmer, Business Systems Analyst, CIS faculty, Computer Instructor, Computer Technician, Data Architect, Database Administrator, Director of Computer Systems and Services, e-Learning Specialist, Executive Director, Interactive Media Specialist, Instructional Technology Admin, Network Engineer, Programmer Analyst, Service Desk Administrator, Software Developer, Systems, and Webmaster.
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|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 6,922
|
All Products : Latitude 9 NeedleArts, The Beauty of Costa Rica at Your Fingertips!
We suggest this fabric for the Bamboo Orchid (L9-004) pattern.
We suggest this fabric for the Keel-Billed Toucan (L9-009) and Spot-Bellied Poison Dart Frog (L9-013) patterns.
We suggest this fabric for the Cattleya Orchid (L9-011) and Arenal Erupts at Night (L9-012) patterns.
We suggest this fabric for the Costa Rican Oxcart (L9-003) and Purple Passiflower (L9-008) patterns.
We suggest this fabric for the Red-eyed Tree Frog (L9-006) pattern.
We suggest this fabric for the Pink Hibiscus (L9-007) pattern.
We suggest this fabric for the Guaria Morada Orchid (L9-010) pattern.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 513
|
Q: Mongo groupby an object key but preserve the whole object/dict I am using mongodb and trying to construct a groupby query. A single document looks something like this
{
"_id" : "ca5ff110697611e89c2d9a0012d6e110",
"CreatedAt" : ISODate("2018-06-06T16:15:55.000Z,
"CreatedBy" : {
"Email" : "neo@orangescape.com",
"Name" : "Neo",
"_id" : "User001"
},
},
"Salary" : 200000
"Branch" : "C2"
}
and my groupby query looks something like this
[{'$project': {'Branch': 1, 'CreatedAt.year':{$year:"$CreatedAt"}, 'CreatedBy': 1, 'Salary': 1}},
{'$group': {'_id': {'Branch': '$Branch',
'Salary': '$Salary',
'CreatedAt.year': "$CreatedAt.year" ,
'CreatedBy': '$CreatedBy._id'},
'Salary': {'$sum': '$Salary'},
'CreatedBy': "$CreatedBy"
}},
]
As you see the CreatedBy is a nested field and grouping on its _id field results in just _id field but I want to preserve the name and email while grouping. Grouping by the whole CreatedBy because the user may change their name at some point and the old documents will have old name.
A: The $group stage needs an accumulator operator for each field. To keep the CreatedBy field, you can use $first for example:
the new query:
db.collection.aggregate([
{
"$project": {
"Branch": 1,
"CreatedAt.year": {
$year: "$CreatedAt"
},
"CreatedBy": 1,
"Salary": 1
}
},
{
"$group": {
"_id": {
"Branch": "$Branch",
"Salary": "$Salary",
"CreatedAtYear": "$CreatedAt.year",
"CreatedBy": "$CreatedBy._id"
},
"Salary": {
"$sum": "$Salary"
},
"CreatedBy": {
"$first": "$CreatedBy"
}
}
}
])
result:
[
{
"CreatedBy": {
"Email": "neo@orangescape.com",
"Name": "Neo",
"_id": "User001"
},
"Salary": 200000,
"_id": {
"Branch": "C2",
"CreatedAtYear": 2018,
"CreatedBy": "User001",
"Salary": 200000
}
}
]
you can try it online here: mongoplayground.net/p/cVNHRD5Rv21
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 108
|
{"url":"https:\/\/www.physicsforums.com\/threads\/torque-force-of-rockets-on-a-satellite.355178\/","text":"# Torque Force of Rockets On A Satellite\n\n1. Nov 15, 2009\n\n### Lancelot59\n\n1. The problem statement, all variables and given\/known data\n\nA satellite has a mass of 4000 kg, a radius of 4.9 m. 4 rockets tangentially mounted each add a mass of 220 kg, what is the required steady force of each rocket if the satellite is to reach 31 rpm in 5.1 min, starting from rest?\n\n2. Relevant equations\n\n$$\\Sigma$$T=I $$\\alpha$$\n\n3. The attempt at a solution\n\nSimple enough. I had already solved pretty much the same problem in my textbook.\n\nI got $$\\alpha$$ by taking the RPM as the delta Vtangental and then multiplying it by 2$$\\Pi$$\/60, and then dividing it by 306 seconds.\n\nThen for I I used (1\/2)Mr2 for the satellite body, and treated the rockets as point particles using mr2, and multiplying by 4.\n\nI = (1\/2)Mr2 + 4(mr2)\n\nI wound up with about 183 newtons per rocket, which the program says is wrong. This online system has a habit of telling you your stuff is wrong when in fact you're doing everything correctly. It uses some silly method of rounding off each step instead of just the final answer.\n\nAm I doing something wrong?\n\n2. Nov 16, 2009\n\n### tiny-tim\n\nHi Lancelot59!\n\n(have an alpha: \u03b1 and an omega: \u03c9 and a sigma: \u2211 and a pi: \u03c0 )\nerm why 2\u03c0\/60 ?\n\nv = r\u03c9\n\n3. Nov 16, 2009\n\n### Lancelot59\n\nWell the speed is in rotations per minute. So I changed it into radians per second.\n\n31RPM x (2pi radians\/1 rpm) x (1 minute\/60 seconds)\n\nI tried using v=r$$\\omega$$ which gave me twice the $$\\alpha$$ but the thing still says it's wrong. I calculated the moment of inertia to be 69148.8.\n\n4. Nov 16, 2009\n\n### tiny-tim\n\nah, I got confused by your Vtangential (still am, actually)!\n\nYour equations, and moment of inertia, look ok, except I don't understand where you're using v =r\u03c9.\n\nI have a feeling that your attempt to introduce V is somehow spoiling he result.\n\n5. Nov 16, 2009\n\n### Lancelot59\n\nWell then how could I find alpha?","date":"2018-01-20 11:51:52","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5361824631690979, \"perplexity\": 2115.000808595852}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-05\/segments\/1516084889567.48\/warc\/CC-MAIN-20180120102905-20180120122905-00306.warc.gz\"}"}
| null | null |
<?php
class Pages extends CI_Controller {
public function view($page = 'home')
{
if (! file_exists(APPPATH.'views/pages/'.$page.'.php'))
{
show_404();
}
$data['title'] = ucfirst($page); //bikin kapital huruf pertama
$this->load->view('templates/header', $data);
$this->load->view('pages/'.$page, $data);
$this->load->view('templates/footer', $data);
}
}
?>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,489
|
Q: realm BAD_ACCESS error I get error BAD_ACCESS on
bool read_only() const { return schema_mode == SchemaMode::ReadOnly; }
when this method is called:
- (void) writeItems: (NSArray *) rmItems {
dispatch_async(self.backgroundQueue, ^{
RLMRealm *realm = [[RLMRealm alloc] init];
[realm beginWriteTransaction];
[realm addObjects:rmItems];
[realm commitWriteTransaction];
});
}
Use Realm version: 2.4.3 with Objective-C project
realm was installed via CocoaPods
I use singleton class to work with realm:
@implementation RMDataManager
+ (id)sharedManager {
static RMDataManager *sharedManager_ = nil;
static dispatch_once_t onceToken;
dispatch_once(&onceToken, ^{
sharedManager_ = [[self alloc] init];
});
return sharedManager_;
}
- (instancetype)init {
self = [super init];
if(self) {
[self setVersionAndMigrations];
}
return self;
}
- (void)setVersionAndMigrations {
RLMRealmConfiguration *config = [RLMRealmConfiguration defaultConfiguration];
// Set the new schema version. This must be greater than the previously used
// version (if you've never set a schema version before, the version is 0).
config.schemaVersion = 1;
// Set the block which will be called automatically when opening a Realm with a
// schema version lower than the one set above
config.migrationBlock = ^(RLMMigration *migration, uint64_t oldSchemaVersion) {
// We haven't migrated anything yet, so oldSchemaVersion == 0
if (oldSchemaVersion < 1) {
// Nothing to do!
// Realm will automatically detect new properties and removed properties
// And will update the schema on disk automatically
}
};
// Tell Realm to use this new configuration object for the default Realm
[RLMRealmConfiguration setDefaultConfiguration:config];
// Now that we've told Realm how to handle the schema change, opening the file
// will automatically perform the migration
[RLMRealm defaultRealm];
}
- (dispatch_queue_t) backgroundQueue {
if(!_backgroundQueue) _backgroundQueue = dispatch_queue_create("RealmBackground", nil);
return _backgroundQueue;
}
- (void)clearAllData {
dispatch_async(self.backgroundQueue, ^{
RLMRealm *realm = [[RLMRealm alloc] init];
// Delete all objects from the realm
[realm beginWriteTransaction];
[realm deleteAllObjects];
[realm commitWriteTransaction];
});
}
@end
What do I do wrong? I've read realm docs and couldn't find any information which helps me.
A: You don't instantiate RLMRealm instances yourself.
You normally create them either using [RLMRealm defaultRealm] or [RLMRealm realmWithConfiguration], but directly creating them with [[RLMRealm alloc] init] isn't supported.
Assuming you're just working with the default Realm, you just need to change your code to this:
- (void) writeItems: (NSArray *) rmItems {
dispatch_async(self.backgroundQueue, ^{
RLMRealm *realm = [RLMRealm defaultRealm];
[realm beginWriteTransaction];
[realm addObjects:rmItems];
[realm commitWriteTransaction];
});
}
I definitely recommend you read the Realms section of the documentation to learn more about the RLMRealm class.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,338
|
\chapter{Amalgamation}\label{sec:amalgamation}
Amalgamation is a technique to build iterations which admit a homomorphism.
We need two types of amalgamations: using type-1 amalgamation, we make sure a stage of our iteration has an automorphism extending an isomorphism of two complete sub-algebras $B_0$, $B_1$ of the previous stage of the iteration. Using type-2 amalgamation, we take care that we can extend automorphisms of initial segments (e.g. those created by type-1 amalgamation).
The technique presented here differs substantially from that of \cite{shelah:amalgamation} (described also in \cite{jr:amalgamation})
in two important (and related) aspects: firstly, it has a ``full support'' flavour rather than a ``finite support'' flavour;
secondly, additional fine tuning was needed to allow for amalgamation to preserve stratification (most instances are discussed in detail below).
In \ref{modf}, we define the forcing $P^\Int_f$ which will be put to use when we define either type of amalgamation.
Before issuing this definition, we pause to analyze $P^\Int_f$ and find that it can be decomposed as a product after forcing with $B_0$ (section \ref{factor}; this will be put to use in lemmas \ref{newreal} and \ref{index:sequ} to show we can close off $\Gamma^0$ under automorphisms).
In section \ref{am} we define type-1 amalgamation (denoted by $\am$) and show it is a stratified extension (by finding a dense set where ``boolean values are stable''; two small details here are the use of ``reduced pairs of random reals'' in lemma \ref{Q:in:D}, and the stable meet operator in lemma \ref{q:cdot:p:in:D} to preserve that $Q$ is a strong sub-order),
and in section \ref{simpleram} we do the same for the simpler type-2 amalgamation (denoted by $\simpleram$).
In the last section, we construct a stable meet operator for amalgamation and discuss remote sub-orders.
There, we prove lemma \ref{remote:lemma} which helps to ensure ``coding areas'' don't get mixed up by the automorphisms (it will be put to use in lemmas \ref{index:sequ} and \ref{coding:survives}).
Also, we show that indeed there is a stable meet for amalgamation, which completes the inductive proof that there is a stable meet for each stage of the iteration.
\section{Basic amalgamation}\label{modf}
Let $P$ be a forcing, $Q$ a complete sub-order of $P$ such that $\pi\colon P \rightarrow Q$ is a strong projection (see \ref{stron:proj:equiv}, p.~\pageref{stron:proj:equiv} and the preceding discussion).
For $i \in \{0,1\}$, let $\dot B_i$ be a $Q$-name such that $\forces_Q \dot B_i$ is a complete sub-algebra of $\quot{P}{Q}$.
Moreover, say we have a $Q$-name $\dot f$ such that $\forces_Q \dot f\colon \dot B_0 \rightarrow \dot B_1$ is an isomorphism of Boolean algebras.
Our task is to find $P'$ containing $P$ as a complete sub-order, carrying an automorphism $\Phi \colon P' \rightarrow P'$ which extends the isomorphism of $\dot B_0$ and $\dot B_1$ (in the extension by $Q$) and which is trivial on $Q$.
Moreover, we want to preserve stratification:
say $(Q,P)$ is a stratified extension on $I = [\lambda_0, \kappa)\cap \mathbf{Reg}$, where we allow $\kappa = \infty$.
We want $\lambda_1<\kappa$ (possibly strictly) greater than $\lambda_0$ such that $(P',P)$ is a stratified extension above $\lambda_1$.
We shall assume for this purpose that $\kappa$ is a limit cardinal (or $\infty$).
We first make some observations:
Let $\ro(Q) * \dot B_i$ be denoted by $B_i$. This is a complete sub-algebra of $B=\ro(P)$, consisting $Q$-names (or if you prefer, $\ro(Q)$-names) $\dot b$ such that $1_Q\forces_Q\dot b\in\dot B_i$. Keep in mind that we can canonically identify the partial order $Q * (\dot B_i\setminus\{ 0 \})$
with the set of $b \in B_0$ such that $\pi_Q(b) \in Q$. Also, don't confuse this with the set of $b \in B_0$ such that $\pi_Q(b) = 1$---or, equivalently, $1_Q\forces_Q [b]_{\dot G} > 0$, which is called the term-forcing, usually denoted by $(\dot B_i\setminus\{ 0 \})^{Q}$.
Let $\pi_i$ denote the canonical projection from $P$ to $B_i$.
Then $\pi_i$ coincides with $\pi$ on $Q$ (by \ref{strong:vs:canonical:proj}).
Moreover, $\dot f$ can be viewed as an isomorphism $f$ of $B_0$ and $B_1$ (mapping names to names). We have
\begin{equation}\label{f:and:pi:commute}
\pi \circ f = f \circ \pi = \pi.
\end{equation}
In fact, for any pair of sub-algebras $B_0$, $B_1$ of $\ro(P)$ such that $Q\subseteq B_0\cap B_1$ and an isomorphism $f\colon B_0 \rightarrow B_1$, equation (\ref{f:and:pi:commute}) holds if and only if $f$ generates an isomorphism of the pair $\quot{B_i}{Q}$, $i\in\{0,1\}$ in any $Q$-generic extension.
Thus instead of starting with $\dot f$ and $\dot B_0$, $\dot B_1$ as in the first paragraph, we could also have started with $f$, $B_0$ and $B_1$ as above, satisfying (\ref{f:and:pi:commute}).
In a first step, we define $P^\Int_f$, the amalgamation of $P$ over $f$. $P^\Int_f$ contains $P$ as a complete sub-order and has an automorphism $\Phi$ extending $f$.
\begin{rem}
If we want to preserve stratification of $P$, we have to be more careful: we must carefully pick a dense subset $\D$ of $P$, such that $P'=\D^\Int_f$ is stratified. The partial order $\D^\Int_f$ is in general not equivalent to $P^\Int_f$, but solves the problem described in the first paragraph.
Finally, we will define a forcing $\am$ which is equivalent to $\D^\Int_f$, and moreover $(P, \am)$ is a stratified extension. Let's postpone these complications, and first look at $P^\Int_f$.
\end{rem}
Amalgamation is not a canonical operation. Firstly, if $D$ is a dense subset of $P$, we cannot infer that $D^\Int_f$ is dense in $P^\Int_f$.
This in combination with the fact that stratification is also not canonical is the main obstacle in this proof. Secondly, even the weaker statement fails: if $\ro(P)=\ro(R)$, we cannot conclude $\ro(P^\Int_f)=\ro(R^\Int_f)$.
Without precautions, we cannot even preclude $P^\Int_f=\{ 1_P \}$, although this pathology does not arise if we ask $B_0\cup B_1 \subseteq P$.
On the other hand, we cannot simply work with $\ro(P)$; for although $\ro(P)$ has a dense stratified subset (namely $P$),
this doesn't mean that $\ro(P)^\Int_f$ will have a dense stratified subset. Therefore, we want to stick as closely to $P$ as possible, but still have $B_0, B_1 \subseteq P$, so we define a ``hybrid'':
\begin{dfn}\label{def:blowup}
Consider the set $P \times B_0 \times B_1$, i.e. the set of triples $(p,\dot b^0,\dot b^1)$ where $p \in P$ and $\forces_Q \dot b^i \in \dot B_i$ for $i \leq 2$.
Order this set by $(p,\dot b^0,\dot b^1)\leq (q,\dot d^0,\dot d^0)$ \emph{if and only if} $p \leq q$ and $p \cdot \dot b^0 \cdot \dot b^1 \leq q \cdot \dot d^0 \cdot \dot d^1$ in $\ro(P)$.
This makes sense since we can canonically identify $\dot b^j, \dot d^j$ with elements of $B_j$.
We call $\blowup{P}=\blowup{P}(Q,f)$ the set of $(p,\dot b^0,\dot b^1) \in P \times B_0 \times B_1$ such that
\begin{equation}
\pi(p)\forces p \cdot \dot b^0 \cdot \dot b^1 \neq 0,
\end{equation}
or equivalently,
\begin{equation}\label{eq:blowup:def:alt}
\pi(p \cdot \dot b^0 \cdot \dot b^1)=\pi(p).
\end{equation}
For $\hat p \in \blowup{P}$, when we refer to the components of $\hat p$, we use the notation $\hat p = (\hat p^P, \hat p^0, \hat p^1)$.
When appropriate, we identify $\hat p$ with $\hat p^P\cdot\hat p^0\cdot\hat p^1$, i.e. the meet of the components in $\ro(P)$. In particular,
if $g$ is a function such that $\dom(g)=\ro(P)$, we write $g(\hat p)$ for $g(\hat p^P\cdot\hat p^0\cdot\hat p^1)$.
\end{dfn}
Clearly, $P$ is isomorphic to the subset of $\blowup{P}$ where the two latter components are equal to $1_{\ro(P)}$, and this set is in turn dense in $\blowup{P}$. So $P$ can be considered a dense subset of $\blowup{P}$.
Thus, the separative quotient of $\blowup{P}$ is the completion under $\cdot$ of $P\cup B_0 \cup B_1$ in $\ro(P)$ (leaving aside the $0$ element).
Observe, moreover, that if $D \subseteq P$ is dense in $P$, then $\{\hat p \in \blowup{P} \setdef \hat p^P \in D \}$ is the same as $\blowup{D}$, and
we shall often use this fact tacitly.
Lastly, observe that
\begin{equation}\label{blowup:order}
\hat p \leq \hat q \iff \big[ \text{ $\hat p^P \leq \hat q^P$ and $\pi_j(\hat p)\leq \pi_j(\hat q)$ for $j\in\{0,1\}$ }\big]
\end{equation}
and $\hat p \approx (\hat p^P,\pi_0(\hat p), \pi_1(\hat p))$.\footnote{We may regard $(\hat p^P,\pi_0(\hat p), \pi_1(\hat p))$ the canonical representative of $\hat p$ if $P$ is separative.} These two observations together would make for an equivalent, more strict definition of $\blowup{P}$, yielding separative $\blowup{P}$ provided $P$ is separative. Notwithstanding, we find the current definition more convenient---if less elegant.
In the following, we identify $P$ with $\{ \hat p \in \blowup{P} \setdef \hat p^0=\hat p^1=1 \}$.
Much of the following would work if we replace (\ref{eq:blowup:def:alt}) by the weaker $p \cdot \dot b_0 \cdot \dot b_1 \neq 0$.
The advantage of asking (\ref{eq:blowup:def:alt}) is that it makes the projection $\bar \pi\colon P^\Int_f \rightarrow P$ take a simple form.
\begin{dfn}
We define $P^\Int_f$ to consist of all sequences $\bar p: \Int \rightarrow \blowup{P}$ such that
for all but finitely many $k$ we have $\bar p(k)^P \leqlo^{\lambda_0} \pi(\bar p(k)^P)$ and for all $k$ we have
\begin{equation*}
f(\pi_0(\bar p(k+1)^P\cdot \bar p(k+1)^0 \cdot \bar p(k+1)^1))=\pi_1(\bar p(k)^P\cdot \bar p(k)^0 \cdot \bar p(k)^1),
\end{equation*}
or, simply
\begin{equation}\label{thinout}
f(\pi_0(\bar p(k+1)))=\pi_1(\bar p(k)).
\end{equation}
The ordering on $P^\Int_f$ is given by $\bar r \leq \bar p$ if and only if for all $k$, $\bar r(k) \leq \bar p(k)$ in $\blowup{P}$.
We define a map $\Phi \colon P^\Int_f \rightarrow P^\Int_f$ by:
\begin{equation*}
\Phi(\bar p)(i)= \bar p(i+1) \text{ for }i \in \Int.
\end{equation*}
Obviously, $\Phi$ is one-to-one and onto, and $\Phi(\bar p)\leq \Phi(\bar q) \iff \bar p \leq \bar q$.
\end{dfn}
\noindent
Observe that (\ref{f:and:pi:commute}) together with (\ref{thinout}) and (\ref{eq:blowup:def:alt}) imply that for all $i \in \Int$,
\begin{equation}\label{proj:am:to:Q}
\pi(\bar p(i))=\pi(\bar p(0))=\pi(\bar p(0)^P).
\end{equation}
Let $F \colon \blowup{P} \rightarrow B_1$ be defined by $F(x)=f(\pi_0(x))$ and let $G \colon \blowup{P} \rightarrow B_0$ be defined by $G(x)=f^{-1}(\pi_1(x))$.
It may seem more natural to replace (\ref{thinout}) by the weaker requirement that $f(\pi_0(p(k+1)))$ and $\pi_1(p(k))$ be compatible;
however, I'm not sure how to show $P$ is a complete sub-order in this case.
Moreover, (\ref{proj:am:to:Q}) simplifies the proof that $\D$ remains dense in a stratified extension, as it allows to build conditions in $\D$ in a coherent way, that is, without changing initial segments over and over again (see \eqref{ext:D:dense}).
We now define a complete embedding $e\colon\blowup{P} \rightarrow P^\Int_f$
and a strong projection $\bar \pi \colon P^\Int_f \rightarrow\blowup{P}$.
For $\hat u \in \blowup{P}$
define $e(\hat u)\colon \Int \rightarrow \blowup{P}$ by
\begin{equation*}
e(\hat u) (i) = \begin{cases}
(\pi(\hat u^P), G^i(\hat u),1) & \text{ for $i>0$,}\\
\hat u & \text{ for $i=0$,}\\
(\pi(\hat u^P),1,F^i(\hat u)) & \text{ for $i<0$.}
\end{cases}
\end{equation*}
\noindent
For $\bar p \in P^\Int_f$, define $\bar \pi(\bar p)\in \blowup{P}$ by $\bar \pi(\bar p) = \bar p(0)$.
\begin{lem}\label{strong:proj}
The map $\bar \pi$ is a strong projection, that is:
if $\hat w \leq \bar \pi(\bar q)$ in $\blowup{P}$, we may find $e(\hat w) \cdot \bar q \in P^\Int_f$.
\end{lem}
\begin{proof}
Let $\hat w \leq \bar \pi(\bar p)$. We define $\bar w$ by induction, as follows:
\[ \bar w(0)=\hat w \]
Assume $\bar w(i) \in \blowup{P}$ has already been defined. We know $\pi (\bar w(i))=\pi(\bar w(i)^P)$.
Assume by induction that $\pi(\bar w(i)^P)=\pi({\hat w}^P) $.
Also, assume by induction that $\bar w(i) \leq \bar p(i)$ and $\bar w(i) \leq e(\hat w)(i)$ in $\blowup{P}$.
To inductively define $\bar w$ on the positive integers, assume $i \geq 0$ and define:
\[ \bar w(i+1) = (\pi(\hat w)\cdot\bar p(i+1)^P,\bar p(i+1)^0,\bar p(i+1)^1 \cdot F(\bar w(i))). \]
The definition of $\bar w$ on the negative integers is also by induction. Assuming $i\leq 0$, we set:
\[ \bar w(i-1) = (\pi(\hat w)\cdot\bar p(i-1)^P,\bar p(i-1)^0 \cdot G(\bar w(i)),\bar p(i-1)^1) \]
For $i \geq 0$, as $\bar w(i) \leq \bar p(i)$, we have
\begin{multline*}
f(\pi_0(\bar w(i)))=F(\bar w(i)) \cdot f(\pi_0( \bar p(i))) \\
=F(\bar w(i)) \cdot \pi_1( \bar p (i+1))\\
=\pi_1( \pi({\hat w}^P)\cdot \bar p (i+1) \cdot F (\bar w(i))\ %
\end{multline*}
where the second equation holds as (\ref{thinout}) holds for $\bar p$,
and the last equation follows from $F(\bar w(i)) \leq \pi(\bar w(i))=\pi({\hat w}^P)$.
We conclude, by definition of $\bar w (i+1)$, that
\begin {equation}\label{e:complete:main}
f(\pi_0(\bar w(i)))=\pi_1( \bar w(i+1)).
\end{equation}
\noindent Applying $\pi$ to (\ref{e:complete:main}), we see $\pi(\bar w(i+1))=\pi(\bar w(i))$, and so
\begin{align*}
\pi(\bar w(i+1)) &= \pi({\hat w}^P) =\\%\label{e:complete:ih}
\pi (\pi({\hat w}^P)\cdot\bar p(i+1)^P) &= \pi( \bar w(i+1)^P),%
\end{align*}
where the first equation follows from the induction hypothesis and the second follows from
\[ \pi({\hat w}^P)\leq \pi(\bar p(0)^P)=\pi(\bar p(i+1)^P). \]
Thus, $\bar w(i+1) \in \blowup{P}$, $\pi(\bar w(i))=\pi({\hat w}^P)$ and by construction, both
$\bar w(i+1) \leq \bar p(i+1)$ and $\bar w(i+1) \leq e({\hat w}^P)(i+1)$ hold.
Replacing $F$ by $G$ in the above, we obtain a similar argument for the inductive step from $i\leq 0$ to $i-1$; we leave the details to the reader.
Finally we have that $\bar w(i) \in\blowup{P}$ and (\ref{e:complete:main}) holds for all $i \in \Int$, whence $\bar w \in P^\Int_f$. We have already shown $\bar w \leq \bar p$ and $\bar w \leq e(\hat w)$.
We now show $\bar w \geq e(\hat w)\cdot\bar p$:
Say $\bar r \in P^\Int_f$ such that $\bar r \leq e(\hat w)\cdot \bar p$. Clearly $\bar r(0) \leq \bar w(0)=w$.
Now assume by induction that $\bar r(i) \leq \bar w(i)$.
Then by (\ref{thinout}),
\[
\bar r(i+1) \leq \pi_1(\bar r(i+1)) \leq F(\bar w(i))%
\]
so as $\bar r(i+1) \leq \bar p(i+1)$, we have $\bar r(i+1) \leq \bar w(i+1)$.
A similar argument shows $\bar r(i-1)\leq\bar w(i-1)$, so we we've shown by induction that $\bar r \leq \bar w$.
So finally, $\bar w = e(\hat w)\cdot\bar p$.
\end{proof}
\noindent For $i \in \Int$, we write $e_i$ for $\Phi^i \circ e$ and $\bar \pi_i$ for $\bar \pi \circ \Phi^i$.
\begin{cor}\label{strong:proj:cor}
For each $i \in \Int$,
the map $e_i$ is a complete embedding of $\blowup{P}$ into $P^\Int_f$. It is well-defined and injective on the separative quotient of $\blowup{P}$. The map $\bar \pi_i\colon P^\Int_f\rightarrow \blowup{P}$ is a strong projection. The map $e_i \res P$ is a complete embedding of $P$ into $P^\Int_f$.
Letting $R = \{ \bar p \in P^\Int_f \setdef \bar p(i)^0=\bar p(i)^1=1 \}$, $R$ is dense in $P^\Int_f$, we have $e_i[P]\subseteq R$ and $\bar \pi_i\res R\colon R \rightarrow P$ is a strong projection.
\end{cor}
\begin{proof}
The first claim is an obvious corollary of the lemma. The rest follows straightforwardly from elementary properties of $e$ and $\bar \pi$.
\end{proof}
\noindent From now on, we identify $\blowup{P}$ with $e[\blowup{P}]$ and accordingly $P$ with $\{ e(p,1,1) \setdef p \in P \}$.
\begin{cor}
$\Phi$ is an automorphism of $P^\Int_f$ extending $f$.
\end{cor}
\begin{proof}
Let $b \in B_0$. We may assume $\pi(b)\in Q$ (this holds for a dense set of conditions in $B_0$). Thus $b \in \blowup{P}$ (to be precise, we should write $(\pi(b),b,1)$ instead of $b$).
Now as $F^n(f(b))=F^{n+1}(b)$ and $G^{n+1}(f(b))=G^n(b)$,
\begin{align*}
\Phi(e(b)) = \Phi((\hdots, G^2(b), G(b),\stackrel{\stackrel{0}{\downarrow }}{ b}, f(b), F^2(b)&, \hdots))=\\
(\hdots, G^2(b), G(b), b, \stackrel{\stackrel{0}{\downarrow }}{f( b)}, F^2(b), \hdots&) = e( f(b))
\end{align*}
So since $\Phi$ and $f$ agree on a dense set of conditions in $B_0$, they are equal on $B_0$.
\end{proof}
\section{Factoring the amalgamation}\label{factor}
\providecommand{\theorem}{theorem}
Interestingly, we can factor the amalgamation over a generic for $B_0$. We will put this to use when we investigate the tail $\am \colon P$.
In particular, it enables us to show that if $\dot r$ is a $P$-name which is unbounded over $V^Q$, $\Phi(\dot r)$ will be unbounded not just over $V^Q$ but over $V^P$.
This will play a crucial role in the proof of the main theorem, ensuring that when we make the set without the Baire property definable, the coding (ensuring its definability) doesn't conflict with the homogeneity afforded by the automorphisms. The main point of the present section is lemma \ref{newreal}; it is used in section \ref{sec:reals:areas} on p.~\pageref{sec:reals:areas}, to prove lemma \ref{index:sequ}. This is in turn used in section \ref{sec:preserving:coding} to prove the crucial lemma \ref{coding:survives}.
For an interval $I \subseteq \Int$, let $\restam{P}{I}$ be the set of $\bar p\colon I \rightarrow \blowup{P}$ such that whenever both $k\in I$ and $k+1 \in I$, (\ref{thinout}) holds.
In other words
\begin{gather*}
\restam{P}{I} =\{ \bar p \res I \setdef \bar p \in P^\Int_f \}.
\end{gather*}
It is clear that for each $k\in I$, the map $e^I_k\colon \blowup{P} \rightarrow\restam{P}{I}$, defined by $e^I_k(p)=e_k(p)\res I$ is a complete embedding. Similarly, there is a strong projection $\pi^I_k \colon\restam{P}{I}\rightarrow \blowup{P}$.
\begin{lem}\label{factor:am}
Let $G_0=G_Q * H_0$ be $Q* \dot{B_0}$-generic. Then in $V[G_0]$, there is a dense embedding of $P^\Int_f:G_0$ into
\[\big[\restam{P}{(-\infty,0]} : {e_0[G_Q * H_0]}\big] \times \big[\restam{P}{[1,\infty)} : {e_1[G_Q * f[H_0]]}\big]\]
and another one into
\[ \big[\restam{P}{(-\infty,-1]} : {e_{-1}[G_Q * H_0]}\big] \times \big[\restam{P}{[0,\infty)} : {e_{0}[G_Q * f[H_0]]}\big].\]
\end{lem}
\begin{proof}
We only show how to construct the first embedding; the second part of the proof is only different in notation.
Let $R_0$ denote $\loweram{P}{0}$ and $R_1$ denote $\upperam{P}{1}$, let $H_1 = f[H_0]$ and $G_1= G_Q*H_1$.
In $V$, let $S$ denote the obvious map $S\colon P^\Int_f \rightarrow R_0 \times R_1$: $S(\bar p)_0= \bar p \res (-\infty,0]$ and $S(\bar p)_1= \bar p\res[1,\infty) $.
Let $S^* = S \res ( P^\Int_f:G_0 )$. We show that the range of $S^*$ is dense in $(R_0:G_0) \times (R_1 : G_1)$. Since $S(\bar p) \leq S(\bar q) \iff \bar p \leq \bar q$, this implies that $S^*$ is injective on the separative quotient of its domain and thus is a dense embedding.
To show that $\ran(S^*)$ is dense, let $\bar p_0$, $\bar p_1$ be given such that $\bar p_i \in R_i:G_i$, for $i\in\{0,1\}$.
Fix $i\in\{0,1\}$ for the moment.
Without loss of generality, $\bar p_i(i) \in P$ (and not just in $\blowup{P}$).
Let $b_i = \pi_i(\bar p_i(i)) \in \dot B_i^{G_Q}$. Then as $\bar p_i \in R_i:G_i$, $b_i \in H_i$.
Find $q \in G_Q$ and a $Q$-name $\dot b$ such that $q \leq \pi_Q(\bar p_0),\pi_Q(\bar p_1)$ and $q$ forces that
\begin{gather}\label{dot b links p0 and p1}
\dot b = b_0\cdot f^{-1}(b_1) > 0 \quad\text{in $(\dot B_0)^{G_Q}$.}
\end{gather}
We have that $q \forces \dot b \cdot \bar p_0(0) \neq 0$ and $f(\dot b)\cdot \bar p_1(1) \neq 0$,
or in other words, $q \cdot \dot b \in \blowup{P}$,
$q \cdot \dot b \leq \bar \pi ( \bar p_0)$ and $q \cdot f(\dot b) \leq \bar \pi ( \bar p_1)$.
So we can define
$\bar p^*_0=e(q \cdot \dot b)\cdot \bar p_0$ and $\bar p^*_1=e_1(q \cdot f(\dot b))\cdot \bar p_1$.
As
\begin{equation}\label{in G:0}
q \in G_Q\text{ and }\dot b^{G_Q} \in H_0,
\end{equation}
we have $p^*_0 \in R_0 : G_0$ and $p^*_1 \in R_1 : G_1$. Define $\bar p^*$:
\begin{equation*}%
\bar p^* = ( \hdots, \bar p^*_0(-1),\bar p^*_0(0),\bar p^*_1(1),\bar p^*_1(2), \hdots )
\end{equation*}
Then $\pi_Q(\bar p^*)=q$ and by (\ref{dot b links p0 and p1}), $q$ forces that the following hold in $(\dot B_0)^{G_Q}$:
\begin{gather*}
f(\pi_0(\bar p^*_0(0)))=f(\pi_0(q\cdot\bar p_0(0)\cdot \dot b))=f(\dot b) \\
\pi_1(\bar p^*_1(1))=\pi_1(q\cdot\bar p_1(1)\cdot f(\dot b))=f(\dot b).
\end{gather*}
Thus $\bar p^* \in D^\Int_f$, and again by (\ref{in G:0}), $\bar p^* \in D^\Int_f:G_0$.
As $S^*(\bar p^*)=(\bar p_0^*, \bar p_1^*) \leq (\bar p_0,\bar p_1)$, we are done.
\end{proof}
\noindent
Let $\dot{R_i}$ be a $Q*\dot{B_0}$-name for $R_i$, for each $i \in\{0,1\}$.
We just showed that $Q*\dot{B_0}$ forces that there is a dense embedding from $\quot{P^\Int_f}{G_0}$ into $\dot R_0 \times \dot R_1$.
So there is a dense embedding of $P^\Int_f$ into $Q*\dot{B_0} * (\dot R_0 \times \dot R_1)$. Since the latter is equivalent to $\restam{P}{(-\infty,0]} * \dot R$ for some $\dot R$, we find that $\restam{P}{(-\infty,0]}$ is a complete sub-order of $P^\Int_f$. The same is true for
$\restam{P}{[0,\infty)}$ (or more generally, for $\restam{P}{I}$, where $I$ is any interval in $\Int$).
In fact, it's easy to show that the natural embedding and projection witness this.
The previous lemma affords insight concerning the action of the automorphism $\Phi$.
E.g. it enables us to show that if $\dot x$ is a $P$-name which is not in $V^{B_0}$ (and hence also not in $V^{B_1}$), then
for all $i \in \Int\setminus\{0\}$, $\Phi(\dot x) \not \in V^{P}$. In fact, for the proof of the main theorem, we shall need something a bit more specific:
\begin{lem}\label{newreal}
Assume that $\dot r_0, \dot r_1$ be $P$-names for reals random over $V^Q$, and
assume $\forces_Q \dot B_i = \gen{ \dot r_i}^{\quot{P}{Q}}$ (as is the case in our application).
If $\dot r$ is a $P$-name for a real such that $\dot r$ is unbounded over $V^Q$, then for any $i \in \Int\setminus\{0\}$, $\Phi^i(\dot r)$ unbounded over $ V^{P}$.
\end{lem}
\begin{proof}
Firstly, $\dot r$ is unbounded over $V^{B_i}$, for each $i\in\{0,1\}$, since the random algebra does not add unbounded reals.
For a start, let's assume $i = 1$.
Let $G_1=G_Q * f[H_0]$ be $Q* \dot{B_1}$-generic and work in $W=V[G_1]$.
We have that $\dot r$ is a $\quot{P}{ G_1 }$ name for a real which is unbounded over $W$ in the sense of definition \ref{generic:reals}---in any $\quot{P}{G_1}$-generic extension of $V[G_1]$, the interpretation of $\dot r$ will be unbounded over $V[G_1]$.
Let $R_0$, $R_1$ be defined as in the previous proof, i.e.
\begin{align*}
R_1&=\quot{ \restam{P}{[1,\infty)} }{G_1 },\\
R_0&=\restam{P}{(-\infty,0]} : {(G_Q * H_0)},
\end{align*}
let $\dot{R_i}$ be a $Q*\dot{B_0}$-name for $R_i$, for each $i \in\{0,1\}$, and let $I=[1,\infty)$.
As $\quot{P}{ G_1 }$ is a complete sub-order of $R_1$,
$e^I_1(\dot r)$ is an $R_1$-name which is unbounded over $W$.
By lemma \ref{unbounded}, viewing $e^I_1(\dot r)$ as a $R_0\times R_1$-name, it is unbounded over $W^{R_0}$.
As $G_1$ was arbitrary, $e^I_1(\dot r)$ is a $Q * \dot{B_0} * (\dot{R_1}\times \dot{R_0})$-name unbounded over $V^{Q * \dot{B_0} * \dot{R_0}}$.
By the previous theorem this means that $e_1(\dot r)$ is a $P^\Int_f$-name unbounded over $V^{Q * \dot{B_0} * \dot{R_0}}=V^{\restam{P}{(-\infty,0]}}$ and hence over $V^P$,
since $S \circ e_0=e^I_0$ shows that $P$ is a complete sub-order of $Q * \dot{B_0} * \dot{R_0}$.
For arbitrary $i \in \Int$ such that $i > 0$:
We just showed that $e_1(\dot r)$ is a $P^\Int_f$-name unbounded over $V^{\restam{P}{(-\infty,0]}}$.
Since $e^I_{-i+1}[P]$ is a complete sub-order of $\restam{P}{(-\infty,0]}$, we know $e_1(\dot r)$ is unbounded over $V^{e_{-i+1}[P]}$.
Apply $\Phi^{i-1}$ to see $\Phi^i(\dot r)$ is unbounded over $V^{e_0[P]}$, as $e_0= \Phi^{i-1} \circ e_{-i+1}$.
For $i < 0$, argue exactly as above but use the second dense embedding mentioned in lemma \ref{factor:am}.
\end{proof}
\section{Stratified type-1 amalgamation}\label{am}
We now turn to the matter of stratification.
Assume $(Q,P)$ is a stratified extension on $I=[\lambda_0,\kappa)\cap \mathbf{Reg}$, as witnessed by
$$\pss_Q=(\D_Q,c_Q, \leqlol_Q,\lequpl_Q,\Cl_Q)_{\lambda\in I}$$
and $\pss_P=(\D_P, c_P,\leqlol,\lequpl,\Cl)_{\lambda\in I}$.
As mentioned, we allow $\kappa$ to be $\infty$, as well.
We never need to mention $\leqlol_Q$, $\lequpl_Q$, $\Cl_Q$ and $\D_Q$ as we can always use the corresponding relation from $\pss_P$ (see the remark following definition \ref{pss:is}, p.~\pageref{pss:is}).
Moreover, assume $\forces_Q \card{\dot B_0} \leq \lambda_0$.
The main problem with stratification and amalgamation is quasi-closure:
Consider two sequences $(p_\xi)_{\xi<\rho}$ and $(q_\xi)_{\xi<\rho}$ such that
$p_\xi$ and $q_\xi$ are compatible for every $\xi<\rho$, with greatest lower bounds $p$ and $q$ respectively. In general, $p$ and $q$ don't have to be compatible. A similar problem occurs with regard to the defining equation (\ref{thinout}) of amalgamation: say we have a sequence of conditions $\bar p_\xi \in \am$ and for each $i \in \Int$, $\bar p(i)$ is a greatest lower bound of $(\bar p_\xi(i))_\xi$. Even though (\ref{thinout}) holds for every $\bar p_\xi$, it could fail for $\bar p$.
The solution to this problem is to thin out to a dense subset of $P$ where $\pi_i$ is stable with respect to ``direct extension'', before we amalgamate.
That is, on this dense subset, $\pi_i$ doesn't change (in a strong sense) when conditions are extended in the sense of $\leqlol$, for $\lambda\in I$.
\begin{dfn}\label{am:gen:D}
Let $\Dam=\Dam(Q,P,f,\lambda_0)$ be the set of $p \in P$ such that %
for all $q \in P$, if $q \leqlo^{\lambda_0} p$ we have
\begin{equation}\label{lower:part:frozen:nice}
\forall (b_0,b_1) \in B_0 \times B_1 \quad \big(\pi(q)\cdot p \cdot b_0 \cdot b_1 \neq 0\big) \Rightarrow \big(q\cdot b_0 \cdot b_1 \neq 0\big)
\end{equation}
\end{dfn}
\noindent
Observe that (\ref{lower:part:frozen:nice}) is equivalent to:
\begin{equation}
\forall j\in\{0,1\} \quad \pi(q)\forces_Q \forall b \in \dot B_{1-j} \quad \pi_j(q\cdot b)=\pi_j(p \cdot b),\label{lower:part:frozen}\\
\end{equation}
and also to the following:
\begin{equation}
\forall j\in\{0,1\}\quad \forall b \in B_{1-j} \quad \pi_j(q\cdot b) = \pi(q) \cdot \pi_j(p \cdot b).\label{lower:part:frozen:alt}
\end{equation}
\begin{lem}\label{lem:shrink}
$\Dam$ is open dense in $\langle P, \leqlo^{\lambda_0}\rangle$.
\end{lem}
\begin{proof}
Let $p_0$ be given. We inductively construct an adequate sequence of $p_\xi$, $0<\xi\leq{\lambda_0}$ with $p_{\lambda_0} \in \Dam$.
First fix $x$ such that the following definition is $\qcdefSeq$ in parameters from $x$. Fix $Q$-names $\dot b_j$ such that $\forces_Q \dot b_j\colon \lambda_0 \rightarrow \dot B_j$ is onto, for $j=0,1$,
and let $\xi \mapsto ( \alpha_\xi, \beta_\xi, \zeta_\xi)$ be a surjection from $\lambda_0$ onto $(\lambda_0)^3$.
For limit $\xi$, let $p_\xi$ be the greatest lower bound of the sequence constructed so far.
Say we have constructed $p_\xi$, we shall define $p_{\xi+1}$.
Let's first assume there are $p^*, \bar p$ such that
$\bar p \leqlo^{\lambda_0} p_\xi$, $\bar p\in\D(\lambda_0,x,p_\xi)$, $p^* \leq \bar p$ and
\begin{enumerate}
\item\label{descend} $\pi(p^*)\forces \bar p \cdot \dot b_0(\alpha_\xi)\cdot \dot b_1(\beta_\xi) = 0$,
\item\label{colour} $\zeta_\xi \in \C^{\lambda_0}(p^*)$.
\end{enumerate}
In this case pick $p_{\xi+1}$ such that $p_{\xi+1} \leqlo^{\lambda_0} \bar p$ and $p_{\xi+1}\lequp^{\lambda_0} p^*$ (using interpolation).
If, on the other hand, no such $\bar p, p^*$ exist, just pick $p_{\xi+1} \in\D(\lambda_0,x,p_\xi)$.
We now show (\ref{lower:part:frozen}) holds for the final condition $p_{\lambda_0}$: say, to the contrary, we can find $j \in \{0,1\}$ and $\dot b \in B_{1-j}$ together with $\bar q \leqlo^{\lambda_0} p_{\lambda_0}$ such that
\[ \pi(\bar q )\not\forces_Q \pi_j(\bar q\cdot \dot b)=\pi_j(p_{\lambda_0}\cdot \dot b). \]
Without loss of generality say $j=0$. We can find $q^* \leq \bar q$ such that for some $\alpha, \beta < \lambda_0$
\begin{enumerate}[(i)]
\item \label{compatible} $\pi(q^*)\forces \pi_0(p_{\lambda_0}\cdot \dot b) - \pi_0(\bar q\cdot \dot b) = \dot b_0(\alpha)\neq 0$,
\item $\pi(q^*)\forces \dot b = \dot b_1(\beta)$,\label{fix_b}
\item $q^* \in \dom(\C^{\lambda_0})$.
\end{enumerate}
Find $\xi<{\lambda_0}$ so that $\alpha=\alpha_\xi$, $\beta=\beta_\xi$ and $\zeta_\xi \in \C^{\lambda_0}(q^*)$.
By construction, at stage $\xi$ of our construction we had $\bar p$ and $p^*$ satisfying (\ref{descend}) and (\ref{colour}).
As $\C^{\lambda_0}(p^*)\cap \C^{\lambda_0}(q^*)\neq 0$ and $q^*\leq p_{\xi+1}\lequp^{\lambda_0} p^*$, we can find $w \leq p^*, q^*$.
But by \eqref{compatible} and \eqref{fix_b}, $\pi(w)\forces p_{\lambda_0} \cdot \dot b_0 (\alpha) \cdot \dot b_1(\beta) \neq 0$.
But since $w \leq p^*$,
$\pi(w)\forces \bar p \cdot \dot b_0 (\alpha) \cdot \dot b_1(\beta) = 0$ and so also
$\pi(w)\forces p_{\lambda_0} \cdot \dot b_0 (\alpha) \cdot \dot b_1(\beta) = 0$, contradiction.
Now we show $\Dam$ is open:
For any $r \leqlo^{\lambda_0} q$, $j=0,1$ and $\dot b \in B_{1-j}$, since $r \leqlo^{\lambda_0} p$, we have $\pi(r)\forces \pi_j (r \cdot \dot b) = \pi_j (p \cdot \dot b)$.
Since $\pi(q)\forces \pi_j (q \cdot \dot b) = \pi_j (p \cdot \dot b)$ and $r \leq q$,
$\pi(r)\forces \pi_j (r \cdot \dot b) = \pi_j (q \cdot \dot b)$. So $q \in \Dam$.
\end{proof}
Having $Q \subseteq \Dam$ helps in many circumstances, in particular we like to have $1_P \in \Dam$.
To this end we introduce the notion of $B_0, B_1$ being $\lambda_0$-reduced.
\begin{dfn}
We say the pair $B_0, B_1$ is $\lambda$-reduced over $Q$ if and only if whenever $p \in P$, $p \leqlo^{\lambda} q$ for some $q\in Q$ and $b\in B_j$ for $j=0$ or $j=1$, we have
\[ \pi_{1-j}(p\cdot b)=\pi(p)\cdot\pi(b). \]
\end{dfn}
Henceforth assume $B_0$, $B_1$ is a $\lambda_0$-reduced pair. We will later see that this is a very mild assumption, see lemmas \ref{reduced:equ} and \ref{reduce:a:pair}.
\begin{lem}\label{Q:in:D}
If $p \leqlo^{\lambda_0} q$ for some $q \in Q$ and $j\in\{0,1\}$ we have
\[
\pi(p)\forces \forall b \in \dot B_{1-j}\setminus \{ 0\} \quad\pi_j(p\cdot b)=1,
\]
and moreover, $p \in \Dam$. In particular, we have $Q \subseteq \Dam$.
\end{lem}
\begin{proof}
Fix $p$ as in the hypothesis. Say $r \in Q$, $r \leq \pi(p)$ and $b \in B_0$ such that $r \forces b \in \dot B_0\setminus\{0\}$. Then $r \leq \pi( b)$.
So as $B_0$, $B_1$ is $\lambda_0$-reduced, $r \leq \pi_1(p\cdot b)$, whence $r \forces \pi_1(p\cdot b)=1$. This proves the first statement for $j=1$, and in the other case the proof is the same.
We now show $p \in \Dam$: Say $p' \leqlo^{\lambda_0} p$. Since also $p' \leqlo^{\lambda_0} q$, we have
\[
\pi(p')\forces \forall b \in \dot B_{1-j}\setminus\{0\}\quad \pi_j(p'\cdot b) = 1 =\pi_j(p\cdot b),
\]
and thus $p \in \Dam$.
\end{proof}
\noindent
In fact, the first statement of lemma \ref{Q:in:D} is equivalent to $B_0$, $B_1$ being a reduced pair (this is really just a slight variation of lemma \ref{reduced:equ}).
The following provides a hint as to how we can assume that $B_0$, $B_1$ is $\lambda_0$-reduced:
\begin{fct}\label{reduced:equ}
Assume that $\dot r_0, \dot r_1$ are $P$-names for reals random over $V^Q$, and
assume $\forces_Q \dot B_i = \gen{ \dot r_i}^{\quot{P}{Q}}$ (as is the case in our application).
Say $j=0$ or $j=1$. The following are equivalent (interestingly, in (\ref{reduced:alt}), there is no mention of $j$):
\begin{enumerate}
\item\label{reduced:orig} Whenever $p \in P$, $p \leqlo^{\lambda} q$ for some $q\in Q$ and $b\in B_j$, we have
\[ \pi_{1-j}(p\cdot b)=\pi(p)\cdot\pi(b).\]
\item\label{reduced:alt}
Whenever $p \in P$, $p \leqlo^{\lambda} q$ for some $q\in Q$ and $b_0$, $b_1$ are $Q$-names for Borel sets such that for some $w \leq \pi(p)$,
$w \forces_Q$`` both $b_0$ and $b_1$ are not null'', there is $p' \leq p$ such that $p'\forces_P \dot r_0 \in b_0$ and $\dot r_1 \in b_1$.
\end{enumerate}
\end{fct}
\begin{proof}
First, assume (\ref{reduced:alt}). We carry out the proof for $j=0$ (the other case is exactly the same). Let $p \in P$ such that for some $q \in Q$, $p \leqlo^{\lambda_0} q$ and let $b_0 \in B_0$. As for any $r \in \ro(P)$, $r\leq\pi_j(r)\leq\pi(r)$ holds, we have
$\pi_j(p\cdot b_0) \leq \pi(p) \cdot \pi(b_0)$.
We now show $\pi_j(p\cdot b_0)\geq\pi(p)\cdot\pi(b_0)$. It suffices to show that whenever $b_1 \in B_1$ is compatible with $\pi(p)\cdot\pi(b_0)$, it is compatible with $p\cdot b_0$.
So fix $b_1\in B_1$.
We have $\pi(b_1)\cdot \pi(b_0)\cdot \pi(p) \neq 0$, so we may pick $w \leq \pi(b_1)\cdot \pi(b_0)\cdot \pi(p)$.
For $j=0,1$, let $\dot b_j$ be a $Q$-name for a Borel set such that $b_j = \bv{ \dot r_j \in \dot b_j}^{\ro(P)}$.
The last inequality means $w \forces \dot b_0$ and $\dot b_1$ are not null. So by assumption, we can find $p'$ forcing $\dot r_j \in \dot b_j$ for both $j=0,1$.
In other words, $p' \leq p \cdot b_0 \cdot b_1$, whence $b_1$ is compatible with $p\cdot b_0$.
For the other direction, assume (\ref{reduced:orig}) and again assume $j=0$, fix $p$ as above, and say $\dot b_0, \dot b_1$ are $Q$-names such that
$w \forces \dot b_0, \dot b_1 \in \Borelplus$ for some $w\leq \pi(p)$.
Let $b_j = \bv{ \dot r_j \in \dot b_j}^{\ro(P)}$. As $\pi(b_0)\cdot \pi(b_1)\cdot \pi(p)\neq0$,
$b_1$ is compatible with $\pi(b_0)\cdot \pi(p) = \pi_1(p\cdot b_0)$.
Thus $b_1$ is compatible with $p\cdot b_1$. So we may pick $p' \in P$, $p' \leq p\cdot b_0 \cdot b_1$.
\end{proof}
\begin{dfn}
Under the assumptions of the previous lemma, we also say the pair $\dot r_0$, $\dot r_1$ is $\lambda$-reduced.
\end{dfn}
We shall need the next lemma to show that $P$ completely embeds into $\am$ (see \ref{strong:proj:am}). Observe that the next lemma does not make the assumption that $B_0,B_1$ is a
$\lambda_0$-reduced pair obsolete, i.e. by itself the lemma does not imply $Q \subseteq \Dam$.
\begin{lem}\label{Q:cdot:D}
Assume that there exists a $Q$-stable meet operator $\qstm$ on $P$ with respect to $\pss$.
Then $Q \cdot \Dam \subseteq \Dam$.
More precisely, if $p \in \Dam$ and $q \in Q$ are such that $q \leq \pi(p)$, we have $q \cdot p \in \Dam$.
Moreover, if $(p,r)\in\dom(\qstm)$ and $p \in \Dam$, for any $j\in\{0,1\}$ and $b \in B_{1-j}$ we have $\pi_j((p\qstm r)\cdot b)=\pi_j(p\cdot b)$.
\end{lem}
\begin{proof}
Let $p \in P$, $q \in Q$ and $q\leq\pi(p)$. We check that $q \cdot p \in \Dam$.
So let
\begin{equation}\label{schnitzel:dom:var}
r \leqlo^{\lambda_0} q \cdot p,
\end{equation}
and fix $j \in \{0,1\}$ and $b \in B_{1-j}$.
To prove that $q \cdot p \in \Dam$, it suffices to show
\begin{equation}\label{q:cdot:p:in:D}
\pi_j(r\cdot b)=\pi(r)\cdot \pi_j(q\cdot p\cdot b).
\end{equation}
Observe that (\ref{schnitzel:dom:var}) implies that $r \leqlo^{\lambda_0} \pi(r) \cdot p$---for by \ref{pcs:is}(\ref{pi:mon}), $\pi(r)\leqlol \pi(q \cdot r) = q$; now use \ref{pss:is}(\ref{pss:is:lo:init}). Thus $(p,r) \in \dom(\qstm)$ and $p\qstm r \leqlo^{\lambda_0} p$.
Thus $p\qstm r \in \Dam$ and
\[\pi_j((p\qstm r )\cdot b)=\pi(p \qstm r) \cdot \pi_j(p\cdot b) = \pi_j(p\cdot b),\]
where the last equation holds because $\pi(p \qstm r)=\pi(p) \geq \pi_j(p\cdot b)$. Note in passing that this proves of the ``moreover'' clause of the lemma. We continue with the proof of the remaining part of the lemma.
By the previous, as $r=\pi(r)\cdot (p\qstm r)$,
\[ \pi_j(r \cdot b) = \pi(r)\cdot \pi_j( (p\qstm r) \cdot b)= \pi(r) \cdot \pi_j(p \cdot b)= \pi(r) \cdot \pi_j(q\cdot p \cdot b). \]
The last equation holds as $\pi(r)\leq q$.
This finishes the proof of the lemma.
\end{proof}
\noindent
From now on, assume we have a $Q$-stable meet $\qstm$ on $P$.
While it is true that $(\blowup{\Dam}, \Dam^\Int_f)$ is a stratified extension, this is not quite the partial order we use in the main theorem:
for this construction would require to repeatedly thin out to a dense set.
As a consequence, we would need the main iteration theorem \ref{thm:it:strat} not just for iterations but rather for sequences $(D_\xi)_{\xi<\theta}$ where $(D_\xi,D_{\xi+1})$ is a stratified extension, but we do not have strong projections from $D_{\bar\xi}$ to $D_{\xi}$ for $\xi<\bar \xi \leq \theta$. Moreover, we would need to prove that the limits in this directed system of partial orders are what we expect them to be (in particular, that each $D_\xi$ is embedded in this limit as a complete sub-order).\footnote{It seems plausible that theorem \ref{thm:it:strat} would go through in this broader case. This provided, it is possible, but lengthy to show that limits contain the $D_\xi$'s.} Instead, we have a much simpler solution.
\begin{dfn}[Type-1 amalgamation]
Let $\am=\am(Q,P,f,\lambda)$ be the set of $\bar p\colon\Int\rightarrow \blowup{P}$ such that the following conditions are met.
\begin{enumerate}
\item For all $i \in \Int$, $\pi(\bar p(i)^P) = \pi(\bar p(0)^P)$.
\item For all but finitely many $i\in\Int$, $\bar p(k)^P \leqlo^{\lambda_0} \pi(\bar p(k)^P)$.
\item For all $i \in \Int\setminus\{-1,0\}$, $f(\pi_0(\bar p(i)))=\pi_1(\bar p(i+1))$ --- that is, (\ref{thinout}) holds.
\item $\bar p(0) \in P$, i.e. $\bar p^0(0)=\bar p^1(0)=1$ and
\begin{align}
f(\pi_0(\bar p(-1)))& \geq \pi_1(\bar p(0)),\label{middle:l}\\
f(\pi_0(\bar p(0)) & \leq \pi_1(\bar p(1))\label{middle:r}.
\end{align}
\item For $i \in \Int\setminus\{0\}$, $\bar p(i)^P \in \Dam(Q,P,f,\lambda)$.
\end{enumerate}
\end{dfn}
\noindent
Observe we can replace (\ref{middle:l}) and (\ref{middle:r}) by
\begin{equation}\label{middle:both}
\bar p(0)^P \leq f(\pi_0(\bar p(-1))) \cdot f^{-1}(\pi_1(\bar p(1))).
\end{equation}
and obtain an equivalent definition.
Thus, $\bar p \in \am$ if and only if the following conditions are met:
\begin{enumerate}
\item $\bar p(0) \in P$,
\item $\bar p\res [1,\infty) \in \Dam^{[1,\infty)}_f$ and $\bar p\res (-\infty,-1] \in \Dam^{(-\infty,-1]}_f$
\item for both $j\in\{-1,1\}$ we have $\pi(\bar p(0))=\pi(\bar p(j))$ and (\ref{middle:both}) holds.
\end{enumerate}
Let $a\colon P \rightarrow \am$ be defined by $a(p)(0)=(p,1,1)$ and $a(p)(i)=(\pi(p),1,1)$ for all $i\in\Int\setminus\{0\}$.
As before, let $\bar\pi(\bar p)=\bar p(0)^P$ (we see no problem in using the same designation as for the projection from $\Dam^\Int_f$ to $\Dam$---see the remark after the next lemma).
\begin{lem}\label{strong:proj:am}
The map $a\colon P\rightarrow \am$ is a complete embedding and
\[\bar \pi \colon \am \rightarrow P
\] is a strong projection.
\end{lem}
\begin{proof}
Let $\bar p \in \am$, $w \in P$, and $w \leq \bar p(0)$. Define $\bar p'$ by
\[ \bar p' (i)= \begin{cases} w &\text{for $i=0$,}\\
(\pi(w)\cdot \bar p(i)^P,\bar p(i)^0,\bar p(i)^1) &\text{for $i \in\Int\setminus\{0\}$.} \end{cases} \]
Clearly $\bar p' \in \am$, $\bar p' \leq a(w)$ and $\bar p' \leq \bar p$.
Moreover, for arbitrary $\bar q \in \am$, if $\bar q\leq a(w)$ and $\bar q \leq \bar p$, clearly $\bar q \leq \bar p'$; so $\bar p' = a(w)\cdot \bar p$.
This shows that $\bar \pi$ is a strong projection and accordingly, $a$ is a complete embedding.
\end{proof}
\noindent
In what follows, we identify $P$ and $a[P]$---except when we feel this would hide the point of the argument.
Next we show that in fact, $\am$ and $\Dam^\Int_f$ are presentations of the same forcing.
\begin{lem}\label{common:dense:set}
The set $D^*=\{ \bar p \in \Dam^\Int_f \setdef \bar p(0)^0=\bar p(0)^1=1 \}$ is dense in both $\Dam^\Int_f$ and $\am$.
\end{lem}
\begin{proof}
First, we notice that $D^* \subseteq \am$ and that the ordering of $\Dam^\Int_f$ and that of $\am$ coincide on $D^*$.
Given $\bar p \in \Dam^\Int_f$, find $d \in \Dam$ such that $d \leq \bar p(0)^P\cdot p(0)^0\cdot p(0)^1$; clearly, $d \cdot \bar p \in D^*$.
Now let $\bar p \in \am$. We find $\bar w \leq \bar p$, such that $\bar w \in D^*$.
Find $d \in \Dam$ such that $d \leq \bar p(0)$.
First let $I=(-\infty,0]$ and construct $\bar w^{-}=\bar w\res I$.
Let $b_0=f(\pi_0(\bar p(-1)))$ and define $\bar p^{-} \in \Dam^{I}_f$ by
\[ \bar p^{-}=( \hdots, \bar p(i), \hdots, \bar p(-1), b_0 ), \]
where of course we identify $b_0$ and $(1_P, \pi(b_0),b_0)\in \blowup{\Dam}$.
Since $d \leq \bar p(0) \leq b_0$ and $b_0 = \bar \pi^I_0(\bar p^{-})$,
we can let $\bar w^{-}= d \cdot \bar p^{-} \in \Dam^{I}_f$.
Observe that $\bar \pi^I_0(\bar w^{-})=d$. %
Now let $I=[0,\infty)$. In an analogous fashion, define $\bar w^{+} \in \Dam^I_f$ such that $\bar w^{+} \leq \bar p\res I$ and $\pi^I_0(\bar w^{+})=d$. Letting
\[ \bar w(i)=\begin{cases} \bar w^{-}(i) &\text{for $i < 0$,}\\
\bar w^{+}(i) &\text{for $i \geq 0$,} \end{cases}
\]
we conclude $\bar w \in \am$.
Moreover, $\bar\pi(\bar w)=d \in \Dam$ whence $\bar w \in D^*$, and $\bar w \leq \bar p$ in $\am$.
\end{proof}
Thus, although $\Phi$ is not an automorphism of $\am$, since it is an automorphism of $\Dam^\Int_f$, it gives rise to an automorphism of the associated Boolean algebra.
We call $\Phi$ the \emph{ automorphism resulting from the amalgamation, } and we refer to $Q$ as the \emph{base of the amalgamation}\label{base} or, interchangeably, \emph{the base of $\Phi$}.
That $\ro(\am) = \ro(\Dam^\Int_f)$ justifies that we use the same notation for the strong projections $\bar \pi \colon \am \rightarrow P$ and $\bar \pi \colon \Dam^\Int_f \rightarrow \Dam$---as we know a strong projection coincides with the canonical projection on (the separative quotient of) its domain.
The next lemma clarifies the role of $\Dam$.
\begin{lem}\label{stay:in:am}
Let $\bar p \in \am$ and say $\bar q \colon \Int \rightarrow P\times B_0\times B_1$ satisfies the following conditions:
\begin{enumerate}
\item \label{stay:pi} for each $i \in \Int$, $\pi(\bar q(i)^P)=\pi(\bar q(0)^P)$.
\item $\bar q(0)^0=\bar q(0)^1=1$.
\item $\forall i \in \Int\setminus\{0\}\quad \bar q(i)^P \leqlo^{\lambda_0} \bar p(i)^P$.
\item \label{stay:pi:j} $\forall i \in \Int\setminus\{0\}\quad \pi_j(\bar q(i)) = \pi(\bar q(i)^P)\cdot \pi_j(\bar p(i))$
\end{enumerate}
Then $\bar q \in \am$.
\end{lem}
\begin{proof}
First, let $I=[1,\infty)$ and show $\bar q \res I \in \Dam^I_f$.
Let $i\in I$ be arbitrary.
By \ref{stay:pi:j} above, we have
\begin{equation}\label{q:p:proj}
\pi_j(\bar q(i))=\pi(\bar q(i)^P)\cdot \pi_j(\bar p(i))
\end{equation}
for $j \in \{0,1\}$.
Since by \ref{stay:pi} we have $\pi(\bar q(i)^P) = \pi(\bar q(0)^P) \leq \pi(\bar p(0)^P) = \pi(\bar p(i))$,
applying $\pi$ to (\ref{q:p:proj}) yields
\begin{equation}
\pi(\bar q(i)) = \pi(\bar q(i)^P)\cdot\pi(\bar p(i))= \pi(\bar q(i)^P),
\end{equation}
which means
\begin{equation}\label{in:blowup}
\bar q(i) \in \blowup{P}.
\end{equation}
Since $\bar p \in \am$ and since (\ref{q:p:proj}) holds, we have
\begin{equation*}
f(\pi_0(\bar q(i)))=\pi(\bar q(0))\cdot f(\pi_0(\bar p(i)))=\pi(\bar q(0)) \cdot \pi_1(\bar p(i+1)) =\pi_1(\bar q(i))
\end{equation*}
Thus $\bar q\res I \in \Dam^I_f$.
Repeat the argument above to show $\bar p\res (-\infty,-1] \in \Dam^{(-\infty,-1]}_f$.
As $\bar q(0)^0=\bar q(0)^1=1$ by assumption, (\ref{in:blowup}) holds for $i=0$.
Let $b = f(\pi_0(\bar p(-1))) \cdot f^{-1}(\pi_1(\bar p(1)))$.
As $\bar q(0) \leq \bar p(0) \leq b$, clearly
\[ \bar q(0) \leq \bar\pi(\bar q(0)) \cdot b = f(\pi_0(\bar q(-1))) \cdot f^{-1}(\pi_1(\bar q(1))). \]
Thus, finally $\bar q \in \am$.
\end{proof}
\noindent
Finally, we are ready to state and prove the main theorem of this section:
\begin{thm}\label{thm:am:s:ext}
$(P,\am)$ is a stratified extension on $J=[(\lambda_0)^+,\kappa)$.
\end{thm}
\begin{proof}
We proceed to define a stratification of $\am$. $\am$ is going to be stratified above $(\lambda_0)^+$,
but in general not above $\lambda_0$, which comes from the fact that possibly $B_0$ and $B_1$ conspire to yield antichains of size $(\lambda_0)^+$.
\footnote{This is the case if we amalgamate e.g. over copies of the Cohen algebra, as was shown by Shelah, which is why he invented sweetness. The same seems to occur for Random algebras, although I have no concrete example.}
For notational convenience, we define $\bar q \alol \bar p$ for arbitrary $\Int$-sequences $\bar q,\bar p \in {}^\Int(P\times B_0\times B_1)$ and for
$\lambda \geq \lambda_0$: $\bar q \alol \bar p$ exactly if for every $i\in\Int$, $\bar q(i)^P \leqlo^\lambda \bar p(i)^P$ and for every $i\in\Int\setminus\{0\}$ we have $\pi(\bar q(i)^P)\forces_Q \pi_j(\bar q(i)) = \pi_j(\bar p(i))$---or equivalently,
\begin{equation}\label{alol:bool}
\pi_j(\bar q(i)) = \pi(\bar q(i)^P)\cdot \pi_j(\bar p(i))
\end{equation}
for both $j\in\{0,1\}$.
\begin{cor}\label{stay:in:am:cor}
Using this notation we can state lemma \ref{stay:in:am} in the following way:
If for some regular $\lambda\geq\lambda_0$, $\bar p \in \am$ and $\bar q \colon \Int \rightarrow P\times B_0 \times B_1$ satisfy $\bar q \alol \bar p$ and moreover $\bar q(0) \in P$ and for all $i \in\Int$, $\pi(\bar q(i)^P)=\pi(\bar q(0)^P)$ holds, then $\bar q \in \am$.
\end{cor}
\begin{lem}\label{alo}
Observe that if $\bar q \colon \Int \rightarrow P\times B_0 \times B_1$ and $\bar p \in \am$ satisfy $\bar q(i)^P\leqlo^{\lambda} \bar p(i)^P$ for all $i \in \Int$ and $\bar q(i)^j=\bar p(i)^j$ for all $i \in\Int\setminus\{0\}$ and $j\in\{0,1\}$, then $\bar q \alol \bar p$.
\end{lem}
\begin{proof}
For $i \in \Int\setminus\{0\}$ and $j\in\{0,1\}$, we have
\begin{multline*}
\pi_j(\bar q(i))=\bar p(i)^j \cdot\pi_j(\bar q(i)^P\cdot \bar p(i)^{1-j}) \\
=\bar p(i)^j\cdot \pi(\bar q(i)^P)\cdot\pi_j(\bar p(i)^P\cdot \bar p(i)^{1-j})=\pi(\bar q(i)^P)\cdot\pi_j(\bar p(i)).
\end{multline*}
where the second line is equal to the first as $\bar p(i)^P \in \Dam$ and $\bar q(i)^P \leqlol \bar p(i)^P$.
Thus, $\bar q \alol \bar p$.
\end{proof}
\noindent
Now let $\bar p, \bar q \in \am$ and say $\lambda \in I$ and $\lambda>\lambda_0$.
Define
\[\bar q \in \aD(\lambda,x,\bar p)\iff \forall i \in \Int \quad\bar q(i)^P \in \D(\lambda,x,\bar p^P(i)).\]
We say $\bar q \aupl \bar p$ exactly if
\[ \forall i \in \Int\quad \bar q(i)^P \lequp^\lambda \bar p(i)^P.\]
Next we define $\aCl$.
Fix a name $\bij$ such that
\[ \forces_P \bij \colon \dot B_0 \cup \dot B_1\rightarrow \check{\lambda_0}\text{ is a bijection.}\]
Let $\dom(\aCl)$ be the set of all $\bar p \in \am$ such that for each $i \in \Int$, we have $\bar p(i)^P \in \dom(\Cl)$ and if $i\neq 0$, there is $\lambda'\in\lambda\cap I$ such that for $j \in\{0,1\}$ we have that $\bij(\pi_j(\bar p(i)))$ is $\lambda'$-chromatic below $\pi(\bar p(i)^P)$.
If $\bar p\in \dom(\aCl)$, we define $\aCl(\bar p)$ to be the set of all $(c(i),\lambda'(i),H^0(i),H^1(i))_{i\in\Int}$ such that
for all $i \in \Int$, $c(i)\in \Cl(\bar p(i))$ and for all $i \in \Int\setminus\{0\}$ and $j\in\{0,1\}$, $H^j(i)$ is a $\lambda'(i)$-spectrum of $\bij(\pi_j(\bar p(i)))$ below $\pi(\bar p(0)^P)$. Observe that $\lambda'(0)$, $H^0(0)$ and $H^1(0)$ can be chosen arbitrarily---they merely serve as place-holders to facilitate notation.
This finishes the definition of the stratification of $ \Dam^\Int_f$.
First we check that $\aD$ and $(\alol)_{\lambda\in I}$ give us a \pre closure system, see \ref{def:pcs}, p.~\pageref{def:pcs}.
That $\aD$ is $\qcdefF$ is immediate (without any further assumptions on the parameter $x$).
For the following, let $\bar p$, $\bar q$, $\bar r \in \am$, $\lambda \in I$ and $x$ be arbitrary.
It is clear that \eqref{qc:D} holds, for if $\bar q \leq \bar p \in \aD(\lambda, x, \bar r)$,
then $\bar q(i)^P \leq \bar p(i)^P \in \aD(\lambda, x, \bar r(i)^P)$ for each $i\in \Int$.
Thus by \eqref{qc:D} for $P$, $\bar q(i)^P \in \aD(\lambda, x, \bar r(i)^P)$ for each $i\in \Int$ and we are done.
For (\ref{qc:preorder}), we must prove transitivity, so say $\bar p \alol \bar q \alol \bar r$ and show $\bar p \alol \bar r$.
Fix $i\in \Int$ and $j\in\{0,1\}$.
Clearly, $\bar p(i)^P \leqlol \bar r(i)^P$.
As $\pi(\bar p(i)^P)\forces_Q\pi_j(\bar p(i))=\pi_j(\bar q(i))$ and $\pi_j(\bar q(i))=\pi_j(\bar r(i))$, we get $\pi(\bar p(i)^P)\forces_Q \pi_j(\bar p(i))=\pi_j(\bar r(i))$ and so as $i$, $j$ were arbitrary, $\bar p\alol \bar r$.
It remains to show that $\bar p \alol \bar q \Rightarrow \bar p \leq \bar q$.
So assume $\bar p \alol \bar q$ and fix $i\in\Int$. Firstly, $\bar p(i)^P\leq\bar q(i)^P$; moreover, (\ref{alol:bool}) implies $\pi_j(\bar p(i))\leq \pi_j(\bar q(i))$ for $j \in\{0,1\}$, and so as $i\in\Int$ was arbitrary and by (\ref{blowup:order}), we infer $\bar p\leq \bar q$.
(\ref{er}): Say $\bar p \leq \bar q \leq \bar r$ and $\bar p \alol \bar r$. Let $i \in \Int$ be arbitrary; clearly $\bar p(i)^P\leqlol\bar q(i)^P$.
Let $j \in \{0,1\}$ be arbitrary; as
\[ \pi_j(\bar p(i))\leq \pi(\bar p(i)^P)\cdot \pi_j(\bar q(i))\leq \pi(\bar p(i)^P)\cdot \pi_j(\bar r(i)) \]
and the terms on the sides of the equation are equal, we conclude $\bar p \alol \bar q$.
Condition (\ref{qc:leqlo:vert}) is trivial.
We continue by checking the remaining conditions of \ref{def:pss}, i.e. that we have a \pre stratification system on $\am$.
The conditions (\ref{up:extra}), (\ref{up}) and (\ref{s:lequp:vert}) are immediate by definition.
We prove (\ref{density}):
\begin{lem}\label{am:dom(C):dense}
Density holds; i.e
for $\lambda \in J$, $\bar p \in \am$ and $\lambda' \in [\lambda_0, \lambda)$ there is $\bar q \in \am$ such that
$\bar q \in \dom(\aCl)$ and $\bar q \alo^{\lambda'} \bar p$.
\end{lem}
\begin{proof}
First, look through the following definition and find a set of parameters $x$ such that it is $\qcdefSeq$ in parameters from $x$.
We define conditions $p^n_i \in P$ for $n \in \nat$ and $i\in\Int$ and $q^n \in Q$ for $n\in\nat$. We do so by induction on $n$, in each step using induction on $i$.
First, as $Q$ is stratified we can find $q^0 \in Q$ such that $q^0 \leqlo^{\lambda'} \pi(\bar p(0)^P)$ and for all $i \in \Int$ and both $j\in\{0,1\}$, $\pi_j(\bar p (i))$ is $\lambda'$-chromatic below $q^0$.
Set $p^0_i=\bar p(i)^P$, for $i\in\Int$.
Now say we have already defined a $\Int$-sequence $\bar p ^n = (p^n_i)_{i\in\Int}$ of conditions in $P$ and $q^n \in Q$.
We will define a stronger
$\Int$-sequence $\bar p^{n+1}= (p^{n+1}_i)_{i\in\Int}$ of conditions in $P$ and a $q^{n+1} \in Q$.
We first define $\bar p^{n+1}$ on the positive integers by induction, then on the negative ones.
At the end we find $q^{n+1} \in Q$.
So find $p^{n+1}_i \in P$ for $i\geq 0$, by induction on $i$.
Find $p^{n+1}_0 \leqlo^{\lambda'} q^n \cdot\bar p^n_0$ such that $p^{n+1}_0 \in \D(\lambda',x,q^n \cdot\bar p^n_0)$ and $p^{n+1}_0 \in \dom(\Cl)$.
Assume by induction that for all $i\in\Int$, $q^n \leqlo^{\lambda'} \pi(p^n_i)$, whence also $\pi(p^{n+1}_0) \leqlo^{\lambda'} q_n \leqlo^{\lambda'} \pi(p^n_i)$.
Continue by induction, choosing, for each $i \in \nat\setminus \{0\}$, a condition $p^{n+1}_i$ such that
\begin{equation}
\begin{aligned}\label{domC:dense:F:spaced}
p^{n+1}_i \leqlo^{\lambda'} &\pi(p^{n+1}_{i-1})\cdot p^n_i\\
p^{n+1}_i \in \D(\lambda', x, & \pi(p^{n+1}_{i-1})\cdot p^n_i)
\end{aligned}
\end{equation}
and $p^{n+1}_i \in \dom(\Cl)$.
By induction hypothesis, $\pi(p^{n+1}_{i-1}) \leqlo^{\lambda'} q_n \leqlo^{\lambda'} \pi(p^n_i)$, so $\pi(p^{n+1}_{i-1})\cdot p^n_i$ is a well defined condition in $P$ and $\pi(p^{n+1}_{i-1})\cdot p^n_i \leqlo^{\lambda'}p^n_i$.
Thus we have defined $p^{n+1}_i$ for $i \geq 0$.
Before we consider the case $i<0$, observe that for any $i \in \nat\setminus\{0\}$, by (\ref{domC:dense:F:spaced}), \ref{pcs:is}(\ref{pi:mon}) and (\ref{F:coh}), we have
\begin{equation*}
\begin{aligned}
\pi(p^{n+1}_i)\leqlo^{\lambda'} &\pi(p^{n+1}_{i-1})\\
\pi(p^{n+1}_i)\in \D(\lambda',x, &\pi(p^{n+1}_{i-1})).
\end{aligned}
\end{equation*}
We also use that by construction, $\pi(p^{n+1}_{i-1}) \leq q^{n+1}\leq \pi(p^n_i)$.
Thus, $(\pi(p^{n+1}_i))_{i\in \nat}$ is $(\lambda',x)$-strategic and we may assume by choice of $x$ and by lemma \ref{adequate} that it is $(\lambda',x)$-adequate.
Let $q^*$ be a greatest lower bound for $(\pi(p^{n+1}_i))_{i\in \nat}$.
Now we define $p^{n+1}_i$, by induction on $i$ for $i < 0$:
Find $p^{n+1}_{-1} \in P$ such that
\begin{align*}
p^{n+1}_{-1} \leqlo q^* \cdot p^n_{-1}\\
p^{n+1}_{-1} \in \D(\lambda',x,q^* \cdot p^n_{-1})
\end{align*}
and such that $p^{n+1}_{-1} \in \dom(\Cl)$.
Again, continue choosing for each $i\in \nat$, $i>1$ a condition $p^{n+1}_{-i} \in \dom(\Cl)$ such that the sequence $(\pi(p^{n+1}_{-i}))_{i\in \nat}$ is $(\lambda',x)$-adequate.
Finally, let $q^{n+1}$ be a greatest lower bound of $(\pi(p^{n+1}_{-i}))_{i\in \nat}$.
This finishes the inductive definition of $\bar p^{n}$.
For each $i \in \Int$, $(p^n_i)_{n\in\nat}$ is a $\lambda'$-adequate sequence and thus has a greatest lower bound which we call $\bar q(i)^P$.
By lemma \ref{lem:adeq:proj} and by choice of $x$, $\{ \pi(p^n_i)\}_{n\in\nat}$ is a $\lambda'$-adequate sequence in $Q$.
Since $(Q,P)$ is a quasi-closed extension, by \eqref{qc:glb} $\pi(\bar q(i)^P)$ is a greatest lower bound of this sequence.
As for each $n \in \nat$, $q^{n+1}\leqlo^{\lambda'}\pi(p^n_i)\leqlo^{\lambda'} q^n$, $(q_n)_{n \in \nat}$ also has greatest lower bound $\pi(\bar q(i)^P)$, whence for all $i \in \Int$, $\pi(\bar q(i)^P) = \pi(\bar q(0)^P)$.
Set $\bar q(i)^j=\bar p(i)^j$ for $j=0,1$ and observe that $\bar q \bar \leqlo^{\lambda'} \bar p$.
Thus as $\lambda' \geq \lambda_0$, we see $\bar q$ satisfies the hypothesis of lemma \ref{stay:in:am} and thus $\bar q \in \am$.
Lastly, as $\bar q(i)^P$ is a greatest lower bound of $\{ p^n_i \}_{n \in \nat}$, we conclude $\bar q(i)^P \in \dom(\Cl)$.
For each $i\in\Int$, fix $c(i) \in \Cl(\bar q(i)^P)$.
Fix $i \in \Int\setminus\{0\}$ and $j\in\{0\}$.
At the beginning, we chose $q_0$ such that $\pi_j(\bar p (i))$ is $\lambda'$-chromatic below $q_0$.
So we may fix a $\lambda'$-spectrum $H^j(i)$ of $\pi_j(\bar p (i))$ below $q_0$---and hence also below $\pi(\bar q(0)^P)\leq q_0$.
As $\bar q \alo^{\lambda'} \bar p$ we have $\pi_j(\bar q(i))=\pi(\bar q(i)^P) \cdot \pi_j(\bar p(i))$.
Thus, as $i\in\Int\setminus\{0\}$ and $j\in\{0,1\}$ were arbitrary,
\[(c(i), \lambda', H^0(i),H^1(i))_{i\in\Int} \in \aCl(\bar q). \]
\end{proof}
\noindent
Now we check that the \pre stratification system on $\am$ extends that of $P$.
Conditions (\ref{ext}), (\ref{pi:mon}) and (\ref{D:coh}) are immediate.
For (\ref{pss:is:cdot:lo}), it suffices to check (\ref{pss:is:lo:init}) and (\ref{pss:is:lo:tail!}).
(\ref{pss:is:lo:init}): Say $q \in P$ and $\bar p$ are such that $q \leqlol \bar \pi(\bar p)$. Let $i \in \Int\setminus\{0\}$.
By (\ref{pss:is:lo:init}) for $(Q,P)$ we have $\pi(q)\cdot \bar p(i)^P\leqlol \bar p(i)^P$.
Moreover, $\pi_j( (\bar p \cdot q) (i))=\pi(q)\cdot\pi_j( \bar p(i))$,
so as $(\bar p \cdot q)(0)=q\leqlol \bar p(0)^P$, we conclude $\bar p \cdot q \alol \bar p$.
(\ref{pss:is:lo:tail!}):
Say $q \in P$ and $\bar p$, $\bar r \in \am$ are such that $q \leq \bar \pi(\bar r)$ and
$\bar r \alol \bar p$. Let $i \in \Int\setminus\{0\}$. By (\ref{pss:is:lo:tail!}) for $(Q,P)$ we have
$\pi(q)\cdot \bar r(i)^P\leqlol \pi(q)\cdot \bar p(i)^P$. Moreover,
\begin{multline*}
\pi_j((\bar r \cdot q)(i))=\pi(q)\cdot\pi_j( \bar r(i))
\\=\pi(q)\cdot\pi(\bar p(0)^P)\cdot\pi_j( \bar p(i))=\pi(q)\cdot\pi_j( \bar p(i))=\pi_j((\bar p \cdot q)(i)),
\end{multline*}
so as $\bar r \cdot q(0)=q=\bar p \cdot q(0)$, we conclude $\bar r \cdot q \alol \bar p \cdot q$.
Conditions (\ref{pss:is:ext}), (\ref{pss:is:pi:mon}) and (\ref{pss:is:c}) are left to the reader.
Being cautious, we check (\ref{pss:is:cdot:up}). Say $w \in P$, $\bar d$, $\bar r \in \am$ and $w \leq \bar\pi(\bar d)$ while $\bar d \alol \bar r$.
By (\ref{pss:is:cdot:up}) for $(Q,P)$, we have $w \cdot \bar d(i)^P \lequpl w \cdot \bar r(i)^P$. As $\bar d \cdot w (0)=w=\bar r \cdot w(0)$, we conclude that $w \cdot \bar d \lequpl w \cdot \bar r$.
We check \eqref{qc:ext}, i.e. that $(P,\am)$ is a quasi-closed extension.
Start with \eqref{ext:D:dense}, that $\aD$ is coherently dense.
Observe that to find $\bar q \in \aD(\lambda, x, \bar p)$, with a given $\bar q(0)^P$, we need only make a direct extension $\bar q(i)^P$ of $\bar p(i)^P$ for every $i\in\Int\setminus\{0\}$
such that $\bar q(i)^P=\pi(\bar q(0)^P)$.
This is possible by \eqref{qc:ext} for $(Q,P)$.
We obtain $\bar q$ wich satisfies all the requirements of \ref{stay:in:am}.
Thus we can find such $\bar q \in \aD(\lambda, x, \bar p)$ without falling out of $\am$
and without changing $\bar q(0)^P$.
We prove (\ref{ext:glb}):
So say $\bar\lambda<\lambda$ and $\bar{\bar p}=(\bar p)_{\xi<\rho}$ is a $(\lambda^*,\bar\lambda,x)$-adequate sequence in $\am$ with a $\bar \pi$-bound $p \in P$.
Towards finding a greatest lower bound $\bar p$, set $\bar p(0) = p$.
Fix $i \in \Int\setminus \{0\}$.
By definition of $\aD$ and $\alol$, the sequence $\{\bar p_\xi(i)^P\}_{\xi<\rho}$ is $(\lambda^*,\bar\lambda,x)$-adequate in $P$.
Since $\{\pi(\bar p_\xi(0)^P)\}_{\xi<\rho}$ is the same as $\{\pi(\bar p_\xi(i)^P)\}_{\xi<\rho}$, the condition $\pi(\bar p(0))=\pi(p(i)^P) \in Q$ is a $\pi$-bound of $\{\pi(\bar p_\xi(i)^P)\}_{\xi<\rho}$.
Thus by (\ref{ext:glb}) for $(Q,P)$, the sequence $\{\bar p(i)^P\}_{\xi<\rho}$ has a greatest lower bound $p_i \in P$ such that for all $\xi <\rho$,
$p_i \leqlo^{\lambda^*} \bar p_\xi(i)^P$ and $\pi(p_i)=\pi(\bar p(0)^P)$.
Moreover, if $\lambda^*<\bar \lambda$, we have $p_i \leqlo^{\bar\lambda} \pi(p_i)$.
For each $i \in \Int$, let $\bar p(i)^P=p_i$ and for $j\in\{0,1\}$ let $\bar p(i)^j=\bar p_0(i)^j$.
By corollary \ref{stay:in:am:cor}, $\bar p \in \am$ and $\bar p \alo^{\lambda^*} \bar p_0$.
We must check that for all $\xi <\rho$, $\bar p \alo^{\lambda^*} \bar p_\xi$.
This is clear as for every $i\in \Int$ we have $\bar p(i)^P \leqlo^{\lambda^*} \bar p_\xi(i)^P$ by construction, and for every $i \in\Int\setminus\{0\}$ and $j\in\{0,1\}$ we have
\[ \pi_j(\bar p(i))=\pi(\bar p(i)^P)\cdot \pi_j(\bar p_0(i)) = \pi(\bar p(i)^P)\cdot \pi_j(\bar p_\xi(i)), \]
where the first equation holds since $\bar p \alo^{\lambda^*} \bar p_0$ second equation holds since
\[ \pi(\bar p(i)^P)=\pi(p)\leq \pi(\bar p_\xi(i)^P) \]
and $\bar p_\xi \alo^{\lambda^*} \bar p_0$ gives us
\[ \pi_j(\bar p_\xi(i))=\pi(\bar p_\xi(i)^P)\cdot \pi_j(\bar p_0(i)).\]
We check the remaining conditions of \ref{def:s:ext}, showing that $(P,\am)$ is a stratified extension on $J$.
(\ref{continuous}):
Say $\bar{\bar p}=(\bar p)_{\xi<\rho}$ and $\bar{\bar q}=(\bar q)_{\xi<\rho}$ are both $(\lambda^*,\bar\lambda,x)$-adequate for $\bar\lambda<\lambda$,
such that
\begin{equation}\label{sequence:colors}
\forall \xi<\rho \quad \aCl(\bar p_\xi)\cap \aCl(\bar q_\xi) \neq \emptyset.
\end{equation}
Say the sequence $\bar{\bar p}=(\bar p)_{\xi<\rho}$ has a greatest lower bound $\bar p$, the sequence $\bar{\bar q}=(\bar q)_{\xi<\rho}$ has a greatest lower bound $\bar q$.
We show
\begin{equation}\label{color}
\aCl(\bar p)\cap \aCl(\bar q) \neq \emptyset.
\end{equation}
First, observe that in each $P$-component we obtain a common colour for $\bar p$ and $\bar q$:
For each $i\in\Int$, as in the previous proof, $\{ \bar p_\xi(i)^P \}_{\xi<\rho}$ and $\{ \bar q_\xi(i)^P \}_{\xi<\rho}$ are $(\lambda^*,\bar\lambda,x)$-adequate and so by (\ref{continuous}) for $P$ we can find $c(i) \in \Cl(\bar p(i)^P)\cap\Cl(\bar q(i)^P)$.
To obtain the colours fo the boolean values, we can use the spectra of the first condition on either sequence, say $\bar q_0$:
fix $(c_0(i),\lambda_0'(i),H^0_0(i),H^1_0(i))_{i\in\Int} \in \aCl(\bar p_0)\cap \aCl(\bar q_0)$.
We shall now check that
\begin{equation}\label{color:am}
(c(i),\lambda_0'(i),H^0_0(i),H^1_0(i))_{i\in\Int} \in \aCl(\bar p)\cap \aCl(\bar q).
\end{equation}
This is clear by definition: fix $i \in \Int\setminus\{0\}$ and $j \in \{0,1\}$. Firstly, $\bar p \alol \bar p_0$ and so
\[ \pi_j(\bar p(i)) = \pi(\bar p(i)^P)\cdot \pi_j(\bar p_0(i)).\]
Moreover, by choice of $\lambda_0'(i)$ and $H^j_0(i)$ there is $b^j \in B_j$ such that we have
\[ \pi_j(\bar p_0(i)) = \pi(\bar p_0(i)^P)\cdot b^j. \]
and $H^j_0$ is a $\lambda'_0(i)$-spectrum for $b^j$ below $\pi(\bar p(i)^P)$.
The last two equations together yield
\[ \pi_j(\bar p(i))=\pi(\bar p(i)^P)\cdot b^j,\]
and so (\ref{color:am}) holds. This finishes the proof of (\ref{continuous}).
(\ref{s:ext:exp}), \emph{coherent expansion}: Assume $\bar q \aupl \bar p$ and $\bar p \alol \bar a(\bar p(0))$. Moreover, assume $\bar q(0) \leq \bar p(0)$. We show $\bar q \leq \bar p$. Let $i\in\Int\setminus\{0\}$ be arbitrary. As $\bar q(i)^P \lequpl \bar p(i)^P$ and
$\bar p(i)^P \leqlol a(\bar p(0)^P)(i)^P= \pi(p(i)^P)$, and as $\pi(\bar q(0)^P)\leq \pi(\bar p(0))$, we have $\bar p(i)^P\leq \bar q(i)^P$ by (\ref{s:ext:exp}) for $(Q,P)$.
Say $i\neq 0$ and $j \in \{0,1\}$. Then
\[ \pi_j(\bar p(i)) \leq \pi(\bar p(i)) = \pi(\bar p(0)^P) \leq \pi(\bar q(0)^P)=\pi(\bar q(i))=\pi_j(\bar q(i)), \]
where the last equality holds as $\bar q(i)^P \leqlol \pi(\bar q(0)^P) \in Q \subseteq \Dam$.
By (\ref{blowup:order}),
$\bar q \leq \bar p$, and we are done.
We show \emph{coherent interpolation} (\ref{s:ext:int}):
Let $\bar d, \bar r \in \am$ be such that $\bar d \alol \bar r$, and say $p \in P$ interpolates $\bar\pi(\bar d)$ and $\bar \pi(\bar r)$.
We find $\bar p \in \am$ such that $\bar p \alol \bar r$ and $\bar p \aupl \bar d$ and moreover $\bar \pi(\bar p)=p$.
For $i \in \Int\setminus\{0\}$, use coherent interpolation for $(Q,P)$ to find $p_i \in P$
such that $p_i \leqlol \bar r(i)^P$ and $p_i \lequpl \bar d(i)^P$ and moreover $\pi(p_i)=\pi(p)$.
Now we define a sequence $\bar p \colon \Int \rightarrow P\times B_0 \times B_1$.
Set $\bar p(0) = (p,1,1)$ and set $\bar p(i)=(p_i, \bar r(i)^0, \bar r(i)^1)$ for $i \in\Int\setminus\{0\}$.
Clearly, $\bar p \alol \bar r$ and so $\bar p \in \am$. By construction, $\bar p \aupl \bar d$ and $\bar \pi(\bar p)=\bar p(0)^P=p$.
It remains to demonstrate (\ref{s:ext:cent}):
\begin{lem}
Coherent centering holds: Say $\lambda \in J$, $\bar p \aupl \bar d$ and $\bar \Cl (\bar p) \cap \bar \Cl(\bar d)\neq\emptyset$.
Say further we have $w_0\in P$ such that $w_0 \leqlo^{<\lambda}\bar p(0)^P$ and $w_0 \leqlo^{<\lambda}\bar d(0)^P$.
Then there is $\bar w \in \am$ such that $\bar \pi(\bar w)=w_0$ and both $\bar w \alo^{<\lambda} \bar p$ and $\bar w \alo^{<\lambda} \bar d$.
\end{lem}
\begin{proof}
Fix $\bar p, \bar d$ and $w_0$ as in the hypothesis. Fix $i \in \Int\setminus\{0\}$ for the moment.
Observe we have
Since $\Cl(\bar p(i)^P)\cap \Cl(\bar d(i)^P)\neq\emptyset$, by coherent centering for $(Q,P)$ we can find $\bar w_i \in P$ such that $\pi(w_i)=\pi(w_0)$. If the additional assumption at the end of the lemma holds, we may assume $w_i \leqlo^{<\lambda} \bar p(i)^P$ and $w_i \leqlo^{<\lambda} \bar d(i)^P$.
For $i \in \Int$, set
\[ \bar w(i) = (w_i, \bar p(i)^0,\bar p(i)^1). \]
Since $\bar w \alo^{\lambda_0} \bar p$ and $\pi(\bar w(i)^P)=\pi(w_0)$ for each $i\in\Int$, by lemma \ref{stay:in:am}, $\bar w \in \am$.
Now say the additional assumption holds.
By construction, $\bar w(i)^P \leqlo^{<\lambda} \bar p$ for each $\lambda'\in[\lambda_0,\lambda)$.
Fix $i \in \Int$.
Since $\aCl(\bar p)\cap \aCl(\bar d)\neq\emptyset$, $\bar p(i)^j$ and $\bar d(i)^j$ have a common $\lambda$-spectrum below $\pi(w_0)$, and so
\begin{equation}\label{forces:same:bv}
\pi(w_0)\forces_Q \bar p(i)^j=\bar d(i)^j.
\end{equation}
Thus for each $i \in \Int$,
\[\bar w(i) = w(i)^P \cdot d(i)^0 \cdot d(i)^1 \leq \bar d(i)\]
whence $\bar w \leq \bar d$.
In fact, as $\bar w(i)^P \leqlo^{\lambda'} \bar d(i)^P$ and (\ref{forces:same:bv}) holds, $\bar w \alo^{\lambda'} \bar d$ for each $\lambda'\in[\lambda_0,\lambda)$.
\end{proof}
\end{proof}
\noindent
\section{Stratified type-2 amalgamation}\label{simpleram}
We now consider the simpler case when we want to extend an automorphism already defined on an initial segment of the iteration.
Let $P$ be a forcing, $Q$ a complete sub-order, $f$ an automorphism of $Q$ and $\pi\colon P \rightarrow Q$ a strong projection.
Assume $\lambda_0$ is regular and $(Q,P)$ is a stratified extension on $I= [\lambda_0, \kappa)$.
We denote the stratification on $Q$ by $\leqlol_Q$, $\lequpl_Q, \hdots$ and write $\leqlol$, $\lequpl, \hdots$ for the stratification of $P$.
Further we assume that for each regular $\lambda\in I$ and $q,r \in Q$,
\begin{enumerate}
\item $q\leqlol_Q r \iff f(q)\leqlol_Q f(r)$;
\item $q \lequpl_Q r \iff f(q)\lequpl_Q f(r)$;
\item $p\in \D_Q (\lambda,x,q))\iff f(p) \in \D(\lambda,x, f(q))$;
\item $\Cl_Q(q)\cap \Cl_Q(r) \neq \emptyset \iff \Cl_Q(f(q))\cap \Cl_Q(f(r)) \neq \emptyset$.
\end{enumerate}
We define the type-2 amalgamation $\simpleram(Q,P,f)$ (or just $\simpleram$ where the context allows) as the set of all $\bar p \colon \Int \rightarrow P$ such that for all but finitely many $i\in\Int$ we have $\bar p(i) \leqlo^{\lambda_0} \pi(\bar p(i))$ and for all $i \in \Int$,
\begin{equation}\label{s:thinout}
f(\pi(\bar p(i)))=\pi(\bar p(i+1)).
\end{equation}
The ordering is $\bar p \leq \bar q$ if and only if for each $i\in\Int$, $\bar p(i) \leq \bar q(i)$.
Define $\bar \pi\colon \simpleram\rightarrow P$ by $\bar \pi(\bar p)=\bar p(0)$. The map $\bar e\colon P\rightarrow\simpleram$ is defined by
$\bar e(p)(0)= p$ and $\bar e(p)(i)=\pi(p)$ for all $i \in\Int$, $i\neq 0$.
It is straightforward to check that $\bar e$ is a complete embedding and $\bar \pi$ is the restriction of the canonical projection
from $\ro(\simpleram)$ to $\ro(\bar e[P])$.
Moreover, if $q \in P$, $\bar p \in \simpleram$ and $q \leq \bar \pi(\bar p)$, then $q\cdot\bar p \in \simpleram$.
We now define the stratification of $\simpleram$, consisting of $\aCl$, $\aFl$, $\alol$, $\aupl$ for each regular $\lambda \in I$.
We say $\bar q \alol \bar p$ exactly if for every $i \in \Int$, $\bar q(i) \leqlol \bar p(i)$, and
$\bar q \aupl \bar p$ exactly if for every $i \in \Int$, $\bar q(i) \lequpl \bar p(i)$.
Similarly for $\aD(\lambda,x,\bar p(i))$.
For $\bar p$ such that for each $i \in \Int$, $\bar p(i) \in \dom(\Cl)$, we define $\aCl(\bar p)$ to be the set of all $c \colon \Int \rightarrow \lambda$ such that for each $i\in\Int$, $c(i) \in \Cl(\bar p(i))$.
\begin{lem}
$(P,\simpleram)$ is a stratified extension on $I$.
\end{lem}
\begin{proof}
The proof is a slight modification of the argument for type-1 amalgamation.
Therefore, we only touch the main points, and leave the rest to the reader.
\begin{lem}
$(P,\simpleram)$ is a quasi-closed extension on $I$.
\end{lem}
\begin{proof}
Let $\lambda \in I$.
Let $(\bar p_\xi)_{\xi<\theta}$ be $\lambda$-adequate. Fix $i \in \Int$ and let $\bar p(i)$ be the greatest lower bound of the $\lambda$-adequate sequence $(\bar p_\xi(i))_{\xi<\theta}$. By coherency, $(\pi(\bar p_\xi(i)))_{\xi<\theta}$ is also adequate and its greatest
lower bound is $\pi(\bar p(i))$.
As $f$ is an automorphism, for each $i\in\Int$, $f(\pi(\bar p(i)))$ is a greatest lower bound of $(q_\xi(i))_{\xi<\theta}$, where $\bar q_\xi(i) = f(\pi(\bar p_\xi(i)))$. As the latter is equal to $\pi(\bar p_\xi(i-1))$, we obtain (\ref{s:thinout}) for $\bar p$.
So $\bar p \in\simpleram$; it is straightforward to check it is a greatest lower bound of $(\bar p_\xi)_{\xi <\theta}$.
\end{proof}
\begin{lem}
Coherent interpolation holds, i.e whenever $\bar r,\bar d \in \simpleram$, $\bar d \leq \bar r$ and $p_0 \in P$ such that $p_0 \leqlol \bar \pi(\bar r)$ and $p_0 \lequpl \bar \pi(\bar d)$,
there is $\bar p \in \simpleram$ such that $\bar p \alol \bar r$, $\bar p \aupl \bar d$ and $\bar \pi(\bar p)=p_0$.
\end{lem}
\begin{proof}
Let $\lambda \in I$.
Given $\bar r, \bar d$ and $p_0$ as above, first set $\bar p(0)=p_0$.
As $\pi(\bar r(i))=\pi(\bar r(0))$ and $\pi(\bar d(i))=\pi(\bar d(0))$, $p_0 \lequpl \pi(\bar d(i))$ and $p_0 \leqlol \pi(\bar r(i))$, for all $i\in\Int$. Coherent interpolation for $(Q,P,\pi)$ allows us to find, for each $i\in\Int$, $i\neq 0$ a condition $\bar p(i) \in P$ such that $\pi(\bar p (i))=\pi(p_0)$, $p_0 \lequpl \bar d(i)$ and $p_0 \leqlol\bar r(i)$. As for each $i\in\Int$, $\pi(\bar p(i))=\pi(p_0)$, $\bar p \in \simpleram$.
\end{proof}
\begin{lem}
Coherent centering holds. That is: Say $\bar p \aupl \bar d$ and either of the following holds: $\bar \Cl (\bar p) \cap \bar \Cl(\bar d)\neq\emptyset$ or for some $q \in Q$, $\bar p \alol q$ or $\bar d \alol q$.
Say further $w_0\in\Dam$ such that for each regular $\lambda' \in [\lambda_0, \lambda)$, $w_0 \leqlo^{\lambda'} \bar \pi (\bar p)$ and $w_0 \leqlo^{\lambda'} \bar \pi (\bar d)$.
Then there is $\bar w \in \simpleram$ such that for each regular $\lambda' \in [\lambda_0, \lambda)$, $\bar w \alo^{\lambda'} \bar p$, $\bar w \alo^{\lambda'} \bar d$ and $\bar \pi(\bar w)=w_0$.
\end{lem}
\begin{lem}
If $\bar p \in \simpleram$ and $\lambda' , \lambda \in I$ and $\lambda' < \lambda$, there is $\bar q \in \simpleram$ such that
$\bar q \in \dom(\aCl)$ and $\bar q \alo^{\lambda'} \bar p$.
\end{lem}
\begin{proof}
We define a sequence $(q_i)_{i\in\Int}$ of conditions in $P$, by induction on $i$. As usual, read through the following definition and pick $x$ such that it is $\qcdefSeq$ with parameters in $x$.
First, find $q_0 \leqlo^{\lambda'} \bar p(0)^P$ such that $q_0 \in \dom(\Cl)$.
Continue by induction, choosing, for each $n \in \nat\setminus \{0\}$, a condition $q_n \leqlo^{\lambda'} \bar p(n)\cdot \pi(q_{n-1})$ such that
$\pi(q_n) \in \D_Q(\lambda,x,\pi(q_{n-1}))$
$q_n \in \dom(\Cl)$.
Let $q^*_1$ be a greatest lower bound for $(\pi(q_k))_{k\in \nat}$; it exists by quasi-closure for $Q$.
Find $q_1 \leqlo^{\lambda'} q^*_1 \cdot\bar p(1)$ such that $q_1 \in \dom(\Cl)$.
Again, continue by induction, choosing for each $n\in \nat\setminus \{0,1\}$, a condition $q_{-n} \in \dom(\Cl)$ ensuring that $(\pi(q_{-k}))_{k\in \nat}$ form an adequate sequence.
Finally, let $q$ be a greatest lower bound of $(\pi(q_{-k}))_{k\in \nat}$.
For each $i \in \Int$, $q \leqlo^{\lambda'} \pi(q_i)$, so $q\cdot q_i \leqlo^{\lambda'} q_i\leqlo^{\lambda'}\bar p(i)$.
Observe that by coherent stratification, $q\cdot q_i \in \dom(\Cl)$ for each $i\in\Int$.
Setting $\bar q(i) = q\cdot q_i$, we have $\pi(\bar q(i))=q$, for all $i\in\Int$.
Thus $\bar q \in \simpleram$, $\bar q \alo^{\lambda'} \bar p$ and $\bar q \in \dom(\aCl)$.
\end{proof}
We leave the rest of the proof that $(P,\simpleram)$ is a stratified extension on $I$ to the reader.
\end{proof}
\section{Remoteness and stable meets}\label{am:remote:stm}
The following lemma helps to ensure ``coding areas'' don't get mixed up by the automorphisms, as we shall see in lemmas \ref{index:sequ} and \ref{coding:survives}.
Also see the discussion at the beginning of section \ref{sec:remote}.
\begin{lem}\label{remote:lemma}
Say $C$ is remote in $P$ over $Q$ (up to some height $\kappa'$, where $\kappa'\leq\kappa$). Then $\Phi^k[C]$ is remote in $\am$ over $P$ (up to the same height) for any $k \in \Int\setminus\{0\}$.
\end{lem}
\begin{proof}
Let $D^* = \bar\pi^{-1}[\Dam] \subseteq \am\cap \Dam^\Int_f$ as in lemma \ref{common:dense:set}.
Let $j \in \{0,1\}$ arbitrary. If $\hat p \in \blowup{\Dam}$, $c \in C$ and $c \leq \pi_C(\hat p^P)$, by the definition of $\Dam$,
\[ \pi_Q(\hat p\cdot c)\forces\pi_j(\hat p\cdot c)=\pi_j(\hat p), \]
that is, $\pi_j(\hat p\cdot c)=\pi_j(\hat p) \cdot \pi_Q(\hat p^P\cdot c)$, so as $C$ is independent over $Q$ and thus $\pi_Q(\hat p^P\cdot c) = \pi_Q(\hat p^P)$, we have $\pi_j(\hat p\cdot c)=\pi_j(\hat p)$.
In fact, if we have $\hat p^P \in \Dam$, we have $c\cdot \hat p \in \blowup{\Dam}$.
Observe further that for any $c \in C$, $\pi_j(c)=1$, and moreover, $C\subseteq \Dam$. Thence, $C \subseteq D^* \subseteq \dom(\Phi^k)$.
Moreover, $\Phi^k(c) (0)= e_k(c) (0)=(1,1,1)$ and so $\Phi^k(c) \in D^* \subseteq \am$.
We now show $\Phi^k[C]=e_k[C]$ is independent in $\am$ over $P$:
Let $c\in C$, $\bar p \in \am$ and say $c \leq (\bar \pi_k\circ \pi_C)(\bar p)=\pi_C(\bar p(k))$.
Since $\pi_j (c \cdot \bar p(k))=\pi_j(\bar p(k))$, for every $i \in \Int$,
\begin{equation}\label{remote:equality:on:sides}
i \neq k \Rightarrow e_k(c) \cdot \bar p (i) = \bar p(i).
\end{equation}
Thus $e_k(c) \cdot \bar p \in \am$. This firstly shows that $\bar \pi_k\circ \pi_C$ is a strong projection from $\am$ to $C$.
Moreover $\bar \pi(\bar p \cdot e_k(c)) = \bar p(0) =\bar \pi (\bar p)$, and we are done with the proof that $\Phi^k[C]=e_k[C]$ is independent in $\am$ over $P$.
It follows that $\Phi^k[C]$ is remote in $\am$ over $P$, by definition of $\alol$: let $\lambda \in [\lambda_0, \kappa')$. Say $c \leq \bar \pi_k (\bar p)$.
Then $e_k(c) \cdot \bar p (k) = (\bar p(k)^P \cdot c, \bar p(k)^0,\bar p(k)^1)$ and $\bar p(k)^P \cdot c \leqlol p(k)^P$, by clause (\ref{remote:leqlo}) of definition \ref{remote}. So by (\ref{remote:equality:on:sides}), $e_k(c) \cdot \bar p \alol \bar p$, and we are done.
\end{proof}
The last lemma of this section is the counterpart of lemmas \ref{stable:meet:products} and \ref{stable:meet:comp}. Together these lemmas make sure that in the iteration used in our application, we have stable meet operators for every initial segment.
We assume $P$ is stratified on $J$.
\begin{lem}
There is a $P$-stable meet operator $\astm$ on $\am$.
\end{lem}
\begin{proof}
Of course we set
\[ \dom(\astm)=\{ (\bar p,\bar r) \setdef \exists \lambda\in J \quad\bar r \alol \bar \pi(\bar r) \cdot \bar p \}. \]
Say we have $\bar p$, $\bar r \in \am$ such that $(\bar p, \bar r) \in \dom(\astm)$.
This means we can fix a regular $\lambda \in J$ such that for each $i \in\Int\setminus\{0\}$, $\bar r(i)^P \leqlol \pi(\bar r(i)^P)\cdot\bar p(i)^P$.
Let $w_i = \bar p(i)^P \stm \bar r(i)^P$ for $i \in \Int\setminus\{0\}$ and
set
\[
\bar p \astm \bar r = \begin{cases} (w_i, \bar p(i)^0, \bar p(i)^1) &\text{ for $i \in \Int\setminus\{0\}$}\\
\bar p(0) &\text{ for $i=0$.}
\end{cases}
\]
Let $\bar p \astm \bar r$ be denoted by $\bar w$.
By lemma \ref{Q:cdot:D}, for $i\in\Int\setminus\{0\}$ and $j\in\{0,1\}$ we have
\begin{equation}\label{am:stm:proj:equal}
\begin{split}
\pi_j(\bar w(i))&=\bar p(i)^j\cdot \pi_j(w_i \cdot \bar p(i)^{1-j})\\
&=\bar p(i)^j\cdot \pi_j(\bar p(i)^P \cdot \bar p(i)^{1-j})=\pi_j(\bar p(i)). \\
\end{split}
\end{equation}
In particular, as $w_i \leqlol \bar p(i)^P$ and $i$ was arbitrary, %
we have
\begin{equation}\label{astm:leqlol}
\bar p \astm \bar r \alol \bar p.
\end{equation}
Moreover, $\pi(w_i)= \pi(\bar p(i)^P)=\pi(\bar p(0))=\pi(\bar w(0))$.
So $\bar w$ satisfies the hypothesis of lemma \ref{stay:in:am} and therefore $\bar w \in \am$.
Clearly, $\bar \pi(\bar p \astm \bar r)= \bar p(0)$.
It remains to see that $\bar \pi(\bar r) \cdot (\bar p \astm \bar r) \approx \bar r$;
we have
\[ \bar \pi(\bar r)\cdot \bar w = \begin{cases} (\pi(\bar r(0)^P)\cdot w_i, \bar p(i)^0, \bar p(i)^1) &\text{ for $i \in \Int\setminus\{0\}$}\\
\bar r(0) &\text{ for $i=0$.}
\end{cases}
\]
Write $\bar u = \bar \pi(\bar r)\cdot \bar w$ and write $\bar v = \bar\pi(\bar r)\cdot \bar p$. For arbitrary $i \in \Int \setminus\{0\}$ and $j \in \{0,1\}$ we have
\begin{align*}
\pi_j(\bar u(i))&=\pi(\bar r(0)^P)\cdot \pi_j(\bar w(i))= \pi(\bar r(0)^P)\cdot \pi_j(\bar p(i)) &&\text{by (\ref{am:stm:proj:equal}),}\\
&=\pi(\bar r(0)^P)\cdot \pi_j(\bar v(i)) = \pi_j(\bar r(i)) &&\text{as $\bar r \leqlol \bar v$.}
\end{align*}
Thus by (\ref{blowup:order}), $\bar u(i)\approx \bar r(i)$.
As $i \in \Int \setminus\{0\}$ was arbitrary and as $\bar u(0)= \bar r(0)$,
we conclude $\bar u \approx \bar r$, finishing the proof that $\astm$ is a $P$-stable meet on $\am$.
\end{proof}
\chapter{Easton Supported Jensen Coding}\label{sec:coding}
In this section we shall discuss Easton supported Jensen coding.
To avoid repetition, see the beginning of section \ref{sec:main} (definition \ref{def:it:succ:stage} and the preceding discussion) for a comprehensive motivation.
\section{A variant of Square}
The variant of square we discuss in this section is a technical prerequisite which we use to obtain a smooth transition from inaccessible to singular coding
at certain points in our construction, an approach we shall refer to as \emph{fake inaccessible coding}.
We will say more about this when we use it.
\begin{lem}
There is a class $(E_\alpha)_{\alpha\in\mathbf{Card}}$ such that for all $\alpha \in \mathbf{Card}$ which are not Mahlo, $E_\alpha$ is club in $\alpha$,
$E_\alpha \subseteq \mathbf{Sing}$ and
whenever $\beta$ is a limit point of $E_\alpha$ we have $E_\beta= E_\alpha \cap\beta$
and $E_\alpha \in J^{A\cap\alpha}_{\delta+2}$ whenever $J^{A\cap\alpha}_\delta\models$``$\alpha$ is not Mahlo.''
\end{lem}
\begin{proof}
Let $(C_\alpha)_{\alpha \in \mathbf{Sing}}$ be the standard global square on singulars, constructed as in \cite[11.63,~p.~228]{ralfbook}.
For $\alpha \in \mathbf{Sing}$, let $\eta^*(\alpha)$ be the maximal limit ordinal such that $\alpha$ is regular in $J^{A\cap\alpha}_{\eta^*(\alpha)}$ and let $M^*_\alpha = J^{A\cap\alpha}_{\eta^*(\alpha)}$.
Observe that if $\beta$ is a limit point of $C_\alpha$, there is
$\sigma\colon M^*_\beta \rightarrow M^*_\alpha$ which is (at least) $\Sigma_0$-elementary such that $\crit(\sigma)=\beta$ and $\sigma(\beta)=\alpha$.
Say $\alpha$ is not Mahlo.
Let $\eta$ be least such that $J^{A\cap\alpha}_{\eta}\models$``$\alpha$ not Mahlo'', let $m_\alpha = J^{A\cap\alpha}_{\eta}$, $\eta(\alpha)=\eta+1$ and let $M_\alpha = J^{A\cap\alpha}_{\eta(\alpha)}$.\footnote{
Observe we could write $\eta(\alpha)=\eta$ above.
Then $m_\alpha$ would be minimal with a definable club of definably singular cardinals.
This is still enough to make the rest of the argument go through, as our goal was to witness the non-Mahlo-ness in a way that is preserved with $\Sigma_0$-embeddings with large enough range.}
\begin{description}
\item[Case 1] If $\alpha \in \mathbf{Sing}^{M_\alpha}$ we let $E_\alpha = C_\alpha$, where the latter comes from the standard square.
\item[Case 2] Otherwise, if $\alpha$ is $\mathbf{\Sigma}_1$-singular in $M_\alpha$, to ensure coherency we define $E_\alpha$ to be the tail of $C_\alpha$ obtained by requiring that the maps witnessing that $\xi \in C_\alpha$ have $m_\alpha$ in their range.
In detail:
Note that $\rho_1(M^*_\alpha)=\alpha$.
Let $p(\alpha)$ be the 1st standard parameter and let
\[
W(\alpha)= \{ W^{\nu, p_1(M^*_\alpha)}_{M^*_\alpha}\setdef \nu \in p_1(M^*_\alpha) \}
\]
be the appropriate ``solidity witness'' (following the notation in \cite{hbst:fs}).
Observe that by construction, we can find a minimal $\theta(\alpha)<\alpha$ such that
$h_{\Sigma_1}^{M^*_\alpha}(\theta(\alpha)\cup\{p(\alpha)\})$ is unbounded in $\alpha$.\footnote{Pick a minimally definable function $F$ witnessing that $\alpha$ is singular;
the parameter of $F$, if any, can be absorbed into $\theta(\alpha)$ since $M_\alpha$ projects to $\alpha$;
now the Skolem hull is a fortiori unbounded.}
Let
$E^*_\alpha$ consist of those $\beta \in\alpha\cap\mathbf{Sing}$ such that there is a %
$\Sigma_0$-elementary
map $\sigma\colon J^{A\cap\beta}_{\bar \eta}\rightarrow M^*_\alpha$ such that $\{ \alpha, p(\alpha), m_\alpha \} \cup W(\alpha)\subseteq \ran(\sigma)$, $\crit(\sigma)=\beta$ and $\sigma(\beta)=\alpha$.
As in the proof of $\Box$, we show that if $E^*_\alpha$ is bounded in $\alpha$, $\cof(\alpha)=\omega$.
In this case, we
can set $E_\alpha = C_\alpha$.
\begin{fct}
If $E^*_\alpha$ is bounded in $\alpha$, $\cof(\alpha)=\omega$.
\end{fct}
\begin{proof}[Sketch of the proof.]
We need to find an embedding $\sigma$ witnessing $\beta \in E^*_\alpha$ for some large enough $\beta$.
Let $\xi <\alpha$ be given and let $M \prec M_\alpha$ be a countable elementary submodel such that $\{\xi, m_\alpha \} \subseteq M$ and let $\pi \colon\bar M \rightarrow M$ be the inverse of the collapsing map.
Let $E$ be the $(\crit(\pi), \beta)$-extender, where $\beta=\sup(\ran(\pi)\cap\alpha)$, derived from $\pi$.
Let $\sigma\colon \Ult(\bar M, E) \rightarrow M_\alpha$ be the factor map.
Check that $\sigma$ is as required, in particular it has critical point $\beta$ and $m_\alpha \in \ran(\sigma)$.
\end{proof}
If $E^*_\alpha$ is unbounded in $\alpha$, we define two sequences as follows.
Let $\gamma_0 = \min E^*_\alpha$.
Given $\gamma_\xi$, let $\delta_\xi$ be the least $\delta$ such that $h^{M^*_\alpha}_{\Sigma_{n(\alpha)}}(\delta\cup\{p(\alpha)\})\setminus (\gamma_\xi+1)\neq\emptyset$.
Let
\[
\gamma_{\xi+1} = \min [ E^*_\alpha \setminus h^{M^*_\alpha}_{\Sigma_{n(\alpha)}}(\delta_\xi\cup\{p(\alpha)\}) ].
\]
Here, assume that $h^{M_\alpha}_{\Sigma_{n(\alpha)}}$ to be defined such that its range grows by one element for every ordinal, slightly abusing notation.
At limit points $\lambda$, let $\gamma_\lambda= \bigcup_{\xi<\lambda} \gamma_\xi$, if
this yields a point below $\alpha$.
Otherwise set $\lambda=\bar\theta(\alpha)$ and stop the construction.
Observe that $\delta_\xi<\theta(\alpha)$ and the $\delta_\xi$ are increasing, so the ordertype $\bar \theta$ of the sequence constructed is at most $\theta(\alpha)<\alpha$.
Set
\[
E_\alpha = \{ \gamma_\xi \setdef \xi <\bar\theta(\alpha) \}\cap\mathbf{Card}.
\]
Note the only difference to $\Box$ is the requirement $m_\alpha \in \ran(\sigma)$ when $\eta^*(\alpha) = \eta(\alpha)$, i.e. $M_\alpha=M^*_\alpha$.
\item[Case 3] $\alpha$ is $\mathbf{\Sigma}_1$-regular over $M_\alpha$.
So
\[
E_\alpha = \{ \beta <\alpha \setdef h_{\Sigma_1}^{M_\alpha}(\beta\cup\{m_\alpha\})\cap\alpha=\beta \}
\]
defines a club.
Observe that for some $C\in m_\alpha$ , we have $m_\alpha\models C\subseteq \mathbf{Sing}$ and $C$ is club in $\alpha$;
moreover, $\{\alpha, C\} \subseteq h_{\Sigma_n}^{M_\alpha}(\emptyset\cup\{m_\alpha\})$ ($\alpha$ is the cardinity of $m_\alpha$).
So for every $n$ and $\beta \in E^n_\alpha$, we have $\beta \in C$ by elementarity, and thus $\beta \in \mathbf{Sing}$.\footnote{The same argument would work in case 1.}
\end{description}
It remains to see that $(E_\alpha)_{\alpha\in \mathbf{Sing}}$ is coherent.
So let $\alpha\in\mathbf{Sing}$ and $\beta$ be a limit of $E_\alpha$.
We must check that $\beta$ falls into the same case as $\alpha$.
Assume $\alpha$ falls into case 1.
We know that there is a $\Sigma_0$-elementary $\sigma\colon M^*_\beta\rightarrow M^*_\alpha$.
Since $\eta^*(\alpha) < \eta(\alpha)$, it must be the case that
$\eta^*(\beta) < \eta(\beta)$: otherwise, $m_\beta$ is a $J$-structure which is a model of ``$\beta$ is not Mahlo'' and by elementarity $\sigma(m_\beta)$ is a $J$-structure and $\sigma(m_\beta)\models$``$\alpha$ is not Mahlo,''
contradicting $\eta^*(\alpha) < \eta(\alpha)$.
Thus $\beta$ also falls into case 1 and coherency follows by the coherency of $\Box$.
Assume $\alpha$ falls into case 2.
It follows that $\eta^*(\alpha) = \eta(\alpha)$.
We know that there is $\sigma\colon J^{A\cap\beta}_{\bar \eta}\rightarrow M^*_\alpha$ as in the definition.
Standard fine structural arguments (e.g. see \cite{ralfbook}) show that $\eta^*(\beta)=\bar\eta$, i.e. $J^{A\cap\beta}_{\bar \eta} = M^*_\beta$ and $\beta$ is $\Sigma_1$-singular in $M^*_\beta$.
We have $m_\alpha \in \ran(\sigma)$, so
by $\Sigma_0$-elementarity $\sigma^{-1}(m_\alpha)$ must be a $J$-structure and $\sigma^{-1}(m_\alpha) = J^{A\cap\beta}_{\eta^*(\beta)-1}$, for otherwise the rudimentary closure of $m_\alpha\cup\{m_\alpha\}$ would have to exist in $M_\alpha$.
Thus $\eta^*(\beta)=\eta(\beta)$, for by a similar argument, $J^{A\cap\beta}_{\eta^*(\beta)-2}\models$``$\beta$ is Mahlo.''
Thus $\beta$ falls into case 1.
Now the same arguments as in the proof of $\Box$ show that $E_\beta=E_\alpha\cap\beta$ (maps factor because $W(\alpha) \subseteq \ran(\sigma)$).
Now assume $\alpha$ falls into case 3.
Let $\sigma\colon J^{A\cap\beta}_{\bar\eta}\rightarrow M_\alpha$ be the inverse of the collapsing map of $h_{\Sigma_1}^{M_\alpha}(\beta\cup\{m_\alpha\})$, and let
$\bar m =\sigma^{-1}(m_\alpha)$.
Clearly $\sigma(\beta)=\alpha$, $\crit(\sigma)=\beta$.
By elementarity, $\bar m$ is the first $J$-structure witnessing that $\beta$ is not Mahlo.
Thus, $\bar\eta = \eta(\beta)$.
Also, by elementarity and since $\beta$ is a limit point of $E_\alpha$, $\beta$ is $\mathbf{\Sigma}_1$-regular in $M_\beta = J^{A\cap\beta}_{\bar\eta}$.
Thus $\beta$ falls into case 2.
It is easy to check that $E_\beta = E_\alpha\cap\beta$.
\end{proof}
\section{Notation}\label{sec:full:setting}
\begin{dfn}[Notation]
In the following we shall use a convenient partition of the ordinals into $4$ components:
for $0\leq i\leq4$, write $[\On]_i$ for the set of $\xi$ which is equal to $i$ modulo $4$.
Write $(\xi)_i$ for the $\xi$-th element of $[\On]_i$ .
Given a set (or class) $B$,
write $[B]_i = \{ (\xi)_i \setdef \xi \in B \}$;
this is consistent with the notation $[\On]_i$ for the $i$-th component.
When we say \emph{$A$ codes $B$ on its $i$-th component},
we mean $A \cap [\On]_i = [B]_i$; that is $\xi \in B \iff (\xi)_i \in A$.
We also write $[A]^{-1}_i$ for $\{ \xi \setdef (\xi)_i \in A \}$,
so that $[A]^{-1}_i$ is the set coded by the $i$-th component of $A$.
We shall sometimes write $B\oplus C$ for $[B]_0 \cup [C]_1$.%
If $s$ and $t$ are strings of $0$'s and $1$'s, we shall write $s \conc t$ for the concatenation of $s$ and $t$.\footnote{That is, supposing $\dom(s) = \alpha$ and $\dom(t) =\beta$,
we have $\dom(s \conc t) = \alpha + \beta$, $(s \conc t )\res\alpha = s$ and $s \conc t (\alpha + \xi) = t(\xi)$ for $\xi \in [\alpha, \beta)$.}
If $X\subset \On$, we write $s\conc X$ for $s\conc\chi_X$, the concatenation of $s$ with the characteristic function of $X$.
Write $s \conc i$ for $s\conc t$ where $t$ is the string with length $1$ and value $i$.
Let $\langle .\,, .\rangle$ denote the G\"odel pairing function.
\end{dfn}
We work in the following settting, which captures the essence of the situation we shall find ourselves in at (some of the) successor stages in our iteration.
Again, see the beginning of section \ref{sec:main} (definition \ref{def:it:succ:stage} and the preceding discussion) for a comprehensive motivation.
Say $\kappa$ is the only Mahlo and $V=L[G^o][\bar B^-]$,
where, $G^o \subseteq \HSize(\kappa^+)$ and $\bar B^- \subseteq \HSize(\kappa^{++})$.
We also assume $\HSize(\kappa^{++})=L_{\kappa^{++}}[G^o]$ (intuitively, $\bar B^-$ contributes no subsets of $\kappa^+$ --- in fact, it comes from a $\kappa^{++}$-distributive forcing).
Lastly, for some reals $x_0$, $\mathbf{Card}^V=\mathbf{Card}^{L[x_0]}$.
Let $\bar T(\sigma, n, i)$ be a constructible, canonically definable sequence of $\kappa^{++}$-Suslin trees, indexed by $(\sigma, n, i) \in \kappa \times \omega\times 2$.
Further, assume that in $L[G^o][\bar B^-A]$, there is a set $I$ of size $\kappa$ and a real $r$ such that
$\bar B^-(\sigma,n,i)$ is a branch through $T(\sigma, n, i)$ if and only if $r(n)=i$ and $\sigma \in I$;
otherwise, we assume $\bar B^-(\sigma,n,i) = \emptyset$, and $T(\sigma,n,i)$ remains $\kappa^{++}$-Suslin in $L[G^o][\bar B^-]$.\footnote{The current proof could be formulated without this last requirement, but this would be cumbersome.
Also, to show both quasi-closure of our iteration $P_\kappa$ and that $r$ is projective in the generic extension via $P_\kappa$, we must work in a model where the requisite trees remain Suslin when we define Easton supported Jensen coding.}
Find $A_0\subseteq \kappa^{++}$ such that for each cardinal $\alpha \leq \kappa^{++}$,
we have $\HSize(\alpha) = L_\alpha[A_0]$.
We can also assume that $A_0\cap [\On]_0=[I]_0$,
$A_0 \cap [\On]_1= [x_0 \oplus \{ 2n + i \setdef r(n)=i \}]_1$ and
$A_0 \cap [\kappa^+, \kappa^{++}) \cap [\On]_1$ is the set
$\{ (\#(\sigma,n,i, t))_1 \setdef
r(n)=i, \sigma \in I, t\in \bar B(\sigma,n,i) \setminus \HSize(\kappa^+) \}$.
Thus we have $V=L[G^o][\bar B^-]=L[A_0]$ and for each cardinal $\alpha$,
we have $\HSize(\alpha) = L_\alpha[A_0]$.
Moreover, $\mathbf{Card}^V=\mathbf{Card}^{L[A_0\cap\omega]}$.
We shall write $s_{\kappa^+}$ for the characteristic function of $A_0 \cap [\kappa^+, \kappa^{++})$, as we will sometimes treat this set similar to conditions in our forcing (the top ``string'').
Our goal is to find a forcing $P$ which codes $L[A_0]$ into a subset of $\omega_1$, using David's Trick (or Killing Universes), with Easton support.
This means that if $G$ is $P$-generic over $L[A]$,
$L[G] = L[G\cap\omega_1]$ and $A_0$ is ``locally'' definable in $L[G]$.
It is then easy to code $L[G\cap\omega_1]$ into a real.
\section{Coding structures and coding apparatus}
We now define the building blocks of our forcing.
These consist of three types of almost-disjoint coding (for successor cardinals, for inaccessible cardinals, and for singular cardinals) using a specific \emph{coding apparatus}.
We will make use of our partition in the following way:
the coding of $A_0$ by $G$ uses $[\On]_0$, the ``successor coding'' of $G\cap[\alpha^+, \alpha^{++})$ uses $[\On]_1$,
the ``inaccessible limit coding'' (from $G\cap[\alpha, \alpha^+)$ to $G \cap \alpha$ for inaccessible $\alpha$) uses $[\On]_2$, except at the Mahlo, where $[\On]_3$ is used.
Finally, the singular limit coding (from $G\cap[\alpha, \alpha^+)$ to $G \cap \alpha$ for singular $\alpha$) uses $[\On]_3$.
The definition of the building blocks will be relative to a set $A$, which stands in lieu of $A_0$ above.
We will later not use $A_0$ itself,
but a slightly adapted
$A=A_p$, depending on $p \in P$ (and in notation from below, on $\alpha$).
This is due to the nature of the Mahlo coding from $\kappa^+$ into $\kappa$ and can be circumvented by doing the coding as a three step iteration.
\begin{dfn}[The locally regular case]\label{reg_defs}
Let $\alpha$ be an uncountable cardinal $\leq\kappa^{++}$ and let $A \subseteq \On$. Let $\langle \diamondsuit_\alpha; \alpha \in \On \rangle$ be the global $\diamondsuit$-sequence
of $L$ concentrating on the inaccessibles below $\kappa$ (i.e. the following holds in $L$: if $X \subseteq \kappa$, the set
$\{ \alpha \setdef \diamondsuit_\alpha = X\cap\alpha \}\cap\mathbf{Inacc}$ is stationary).
For a more uniform notation, write $E_\kappa = \emptyset$.
\begin{description}%
\item [Basic Strings]
Let $S_\alpha=S(A)_\alpha$ denote the set of $s\colon [\alpha, |s|) \rightarrow 2$, where $\alpha \leq |s| < \alpha^+$.
We abuse notation by writing $\emptyset_\alpha$ for the empty string at $\alpha$ and note $|\emptyset_\alpha| = \alpha$.
\item[Steering Ordinals]
For $\xi\in [\alpha, \alpha^+)$, define $\mu_0^\xi=\mu(A)_0^\xi$ %
and $\mu_0^{<\xi}$ simultaneously by induction:
Let $\mu_0^{<\alpha}$ be least $\mu$ such that $E_\alpha, \diamondsuit_\alpha \in L_\mu[A\cap\alpha]$\footnote{We want $E_\alpha$ in the coding structures of locally inaccessibles, whence $E_\alpha$ is hit and we can easily distinguish between conditions as they turn out in the distributivity proof and those as built to show extendibility.
The presence of $\diamondsuit_\alpha$ is important for a kind of $\kappa$-chain condition in \ref{it:reals:are:caught}, when we show that $\kappa$ is not collapsed in our iteration and stays Mahlo until the last stage.}
and for $\xi > \alpha$ let
\[
\mu_0^{<\xi} = \sup_{ \nu < \xi} \mu^{\nu}.
\]
Define $\sigma(\xi)$ to be least above $\mu_0^{<\xi}$ such that $L_{\sigma(s)}\vDash$``$\alpha$ is the greatest cardinal.''
Let $\mu_0^\xi= \sigma(\xi) + \alpha$.
Note that by induction $\xi \leq \mu_0^{<\xi}$.
If $\alpha \in \mathbf{Reg}$, we write $\mu^\xi$ and $\mu^{<\xi}$ for $\mu^\xi_0$ and $\mu_0^{<\xi}$.
We write $\mu_0^s$ and $\mu_0^{<s}$ for $\mu^{|s|}_0$ and $\mu_0^{<|s|}$; similarly for $\mu^s$ and $\mu^{<s}$.
\item [Coding structures]
We let $\mathcal{A}_0^s=\mathcal{A}(A)_0^s$ denote $L_{\mu_0^s}[A\cap\alpha,s]$.
If $\alpha \in \mathbf{Reg}$, we write $\mathcal{A}^s$ for $\mathcal{A}_0^s$ and $\mathcal{A}^\xi$ for $\mathcal{A}_0^\xi$.
\item [Coding apparatus]
For cardinals $\alpha < \kappa$, $s \in S_\alpha$ such that $\alpha$ is regular in $\mathcal{A}_0^s$, let
\[
H^s_i=H(A)^s_i = h^{\mathcal{A}_0^s}_{\Sigma_1}(i \cup \{ A\cap \alpha, D_\alpha, E_\alpha, s\})
\]
and let
$f^s(i)$ be the order type of $H^s_i \cap \On$.
If $\alpha=\beta^+$ for $\beta\in\mathbf{Card}$, let $B^s$ be %
$\{ i \in [\beta,\alpha) \setdef H^s_i \cap \alpha= i\}$ and
let $b^s = \{ \langle i, f^s(i)\rangle \setdef i \in B^s\}$ (where $\langle .\,, .\rangle$ denotes the G\"odel pairing function).
If $\alpha$ is inaccessible in $\mathcal{A}_0^s$, let
$B^s= \{ \beta^+ \setdef H^s_\beta \cap \alpha= \beta, \beta \in \mathbf{Card}\cap\alpha\}$
and let $b^s = \ran(f^s\res B^s)$.\footnote{For $\alpha < \kappa$ inaccessible in $\mathcal{A}_0^s$, we could also just let $B^s = \{ \beta^+ \setdef \beta \in \diamondsuit_\alpha \cap E_\alpha \}$ if this is unbounded in $\alpha$ and $B^s = \{ \beta^+ \setdef \beta \in E_\alpha \}$ otherwise. We could let $B^s = \mathbf{Succ}\cap\kappa$ for $s \in S_\kappa$.}
\end{description}
\end{dfn}
The intuition is that we want to code $s$ via almost-disjoint coding, using the almost disjoint family $\{ b^{s \res \nu}\setdef \nu \in [\alpha, |s|)\}$ (but in fact, $s$ will grow at the same time).
The particular form of the $b^s$ is very convenient when $\alpha\in\mathbf{Sing}$ (see the proof of extendibility \ref{ext1}).
For successor $\alpha$, using the pair $\langle i, \otp (H_i\cap\On)\rangle$ is vital to ensure that the $b^s$ are almost disjoint.
Observe that by definition,
for $s\in S_\alpha$ when
$\alpha < \kappa$ is locally inaccessible (that is, inaccessible in $\mathcal{A}^s_0$) we have $\lVert \xi \rVert^- \in E_\alpha$ for any all $ \xi\in b^s$.
For $s\in S_\alpha$ when
$\alpha$ is inaccessible
and $\diamondsuit_\alpha$ is club in $\alpha$, also $\lVert \xi \rVert^- \in \diamondsuit_\alpha$ for any all $ \xi\in b^s$.\footnote{As mentioned before, this is needed in the proof of \ref{it:reals:are:caught}.}
As we will see later, fake inaccessible coding makes it necessary that we be able to put all of $b^s$ into the support of a condition in $P$;
otherwise the coding structures will change as conditions are extended (see below).
Thus we want $b^s$ to be Easton.
This, together with coherency issues arising in the proof of \emph{extendibility} at singular $\alpha$ (see \ref{ext1} below), is why we use $E_\alpha$.
In said proof, we shall have no control over what happens at limits of $E_\alpha$, which is why we use successors for $B^s$.
Now we define coding structures for singular $\alpha$.
This is complicated by the fact that a discontinuity arises when $\alpha$ is not seen to be singular by a structure arising in the proof of distributivity.
In this proof, we will build conditions $p$ which inevitably use inaccessible coding at $\alpha$ which are really singular.
This is the phenomenon we call ``fake inaccessible coding.''
The idea behind my solution is that we make singular coding structures (say, at $\alpha$) large enough to
``see'' the fake inaccessible coding.
This requires that we can distinguish wether or not fake inaccessible coding has occurred at all.
Also the coding structures now depend on the condition \emph{below} $\alpha$ and we have to be careful they be stable with respect to extending $p$.
\begin{dfn}[The singular case]\label{sing_defs}
Let $\alpha$ be a limit cardinal now. Except for the first definition, we will always assume
$\alpha$ is singular.
The following definitions will be relative to some
(partial) function $p \colon \alpha \rightarrow 2$.
We extend the definition of support to such $p$ by $\supp(p)= \mathbf{Card} \cap \dom(p)$.
Given $p$ as above and a function $f$ with $\dom(f)\subseteq \mathbf{Card}$ such that $f \in \prod_{\beta \in \dom(f)} [\beta,\beta^+)$, we let $f_p$ be the partial function where
$f_p(\beta)$ is the least $\eta \in [\On]_3$ such that $f(\beta) < \eta$ and $p(\eta)=1$.
\begin{description}
\item [Regular decoding]
First, define $s(A,p)=s(p) \in S_\alpha$.
To this end, define a sequence $(s_\gamma)_\gamma < \gamma(p)$ of basic strings in $S_\alpha$.
Let $k=1$ if $\alpha \in \mathbf{Succ}\cap\kappa^{+}+1$ and $k=2$ if $\alpha \in \mathbf{Inacc}\cap\kappa+1$.
Let $s_0 = \emptyset$.
Given $s_\gamma$, first assume
$b^{s_\gamma} \subseteq \dom(p)$ and $\alpha$ is regular in $\mathcal{A}_0^{s_\gamma}$.
In this case let $s_{\gamma+1}$ be ${s_\gamma} \conc j$, where $j$ is such that $p((i)_k) =j$ for cofinally many $i \in b^{s_\gamma}$ (and ${s}\conc t$ denotes concatenation of strings of $0$'s and $1$'s).
This is well-defined as long as $\alpha$ is regular in $\mathcal{A}_0^{s_\gamma}$ ($b^s$ is not defined otherwise).
If $b^{s_\gamma} \not \subseteq \dom(p)$, or if $\mathcal{A}_0^{s_\gamma} \vDash \alpha$ is singular, let $s(p)= s_\gamma$ and $\gamma(p)=\gamma$.
Write
$\mu(p)$ for $\mu^{s_\gamma(p)}$.
\item [Steering Ordinals at singulars]
Inductively define $\mu(A,p)^\xi$, $\tilde \mu(A,p)^\xi$ and $\mu(A,p)^{<\xi}$ or more simply,
$\mu(p)^\xi$, $\tilde \mu(p)^\xi$ and $\mu(p)^{<\xi}$ by induction on $\xi \in [\alpha,\alpha^+)$:
If %
$p((\beta^+)_2)=1$ for a cofinal set of $\beta \in E_\alpha$, we let
$\mu(p)^{<\alpha} = \mu(p)$ and we say $p$ \emph{uses fake inaccessible coding}\footnote{Conditions with this property have to be dealt with when we prove quasi-closure.}.
Otherwise, let
$\mu(p)^{<\alpha}$ be the least $\mu$ such that $E_\alpha, \diamondsuit_\alpha \in L_{\mu}[A\cap\alpha]$, i.e.
$\mu(p)^{<\alpha} = \mu^{<\alpha}$.
We say in this case that $p$ \emph{immediately uses singular coding}.\footnote{such conditions are built in a straightforward manner in the proof of extendibility \ref{ext1}.}
For $\xi > \alpha$, let
\[
\mu(p)^{<\xi} = \sup_{\nu<\xi} \mu^\nu.
\]
and for $\xi \geq \alpha$ let, first let
$\sigma(p)^\xi > \mu^{<\xi}$ be least such that $L_{\sigma(\xi)}\vDash$``$ \alpha$ is the greatest cardinal and $\alpha\in \mathbf{Sing}$''.
Now let $\tilde \mu(p)^\xi = \sigma(p)^\xi + \alpha$ and $\mu(p)^\xi= \sigma(p)^s + \omega\cdot \alpha $.
Notice that by induction $\xi \leq\mu(p)^{<\xi}$ (equality may hold if $\xi$ is a limit)
and $\mu(p)^{\xi}\geq \mu_0^\xi$.
Again, write $\mu(p)^s, \tilde \mu(p)^s, \mu(p)^{<s}$ for $\mu(p)^{|s|}, \tilde \mu(p)^{|s|}, \mu(p)^{<|s|}$.
We say \emph{$p$ recognizes the singularity of $\alpha$} if and only if $\alpha$ is singular in $\mathcal{A}_0^{s(p)}$.
Observe this is never relevant when $p$ immediately uses singular coding.
\item[Coding structures at singulars]
We let
\begin{gather*}
\mathcal{B}(A,p)^{s}=\mathcal{B}(p)^{s} = L_{\mu(p)^s}[A\cap\alpha, s(p), s]\\
\mathcal{\tilde B}(p)^s = L_{\tilde\mu(p)^s}[A\cap\alpha, s(p), s],
\end{gather*}
where we set $s(p)=\emptyset$ if $p$ immediately uses singular coding.
Note that $\mathcal{B}(p)^{s}\models \alpha\in \mathbf{Sing}$ whenever $|s| > 0$ or $p$ immediately uses singular coding.
Again note that $E_\alpha \in \mathcal{B}(p)^{\emptyset_\alpha}$ and $\mathcal{A}_0^s \subseteq \mathcal{B}(p)^s$.
\item[Coding apparatus at singulars]
For $s \in S_\alpha$, let
\[
H(A,p)_i =H(p)_i = h^{B(p)^s}_{\Sigma_1}(i \cup \{ A\cap \alpha, s(p), s\})
\]
and let
$f(p)^s(i)$ be the order type of $H(p)_i \cap \On$.
Note again that $f(p)^{\emptyset_\alpha} = f^{\emptyset_\alpha}$.
\item [Singular coding with delays]
We define $t(A,p)=t(p)\in S_\alpha$.
To this end, we define a sequence $(t(p)_\xi)_{\xi<\delta(p)}$.
Let $t(p)_0 = \emptyset$.
For limit $\delta$, let $t(p)_\delta = \bigcup_{\delta'<\delta}t(p)_{\delta'}$.
Now suppose $t=t(p)_\delta$ is defined.
For $\gamma\in \mathbf{Card}$ let
$f^{t}_p(\gamma)$ be the least $\eta$ such that $f(p)^{t}(\gamma) \leq \eta < \gamma^+$ and $p((\eta)_3)=1$.
If $f^{t}_p(\beta^+)$ is undefined for cofinally many $\beta \in E_\alpha$, let $\delta(p)=\delta$ and $t(p) = t(p)_\delta$.
Otherwise, if possible define $X=X(p,\delta) \subseteq \alpha$ by $\eta \in X$ if and only if $p( (f^{s_\gamma}_p(\beta) + 1 + \eta )_3 ) = j$ for a tail of successor cardinals $\beta <\alpha$.
If $X(p,\delta)$ is undefined, again stop the construction at $\gamma$ and let $\delta(p)=\delta$ and $t(p)=t_\delta$\footnote{%
This will never occur.
Note that it follows from later definitions that the construction of $t(p)$ finishes only for one reason, namely that $f^{t}_p(\beta^+)$ is undefined for cofinally many $\beta \in E_\alpha$.
For since we require $p\res\alpha$ exactly codes $p_\alpha$, the construction cannot stop before we have $t(p)=p_\alpha$, where we
halt for the given reason, by the requirement that $p\res\alpha \in \mathcal{B}(p_{<\alpha})^{p_\alpha}$.}
If $[X(\xi)]^{-1}_0$ precodes a string $t \in S_\alpha$ which end-extends $t(p)_\delta$ and such that $f^{s_\gamma}_p \in \mathcal{B}(p)^{t}$, let
$t(p)_{\delta+1} = t$.
Otherwise let $t= {t(p)_\delta }^\frown [X]_3$.
If $f^{s_\gamma}_p \in \mathcal{B}(p)^{t}$, let
$t(p)_{\delta+1} = t$.
If not, again stop the construction at $\gamma$ and let $\delta(p)=\delta$ and $t(p)=t_\delta$ (we will later see this case never occurs).
We say \emph{ $p$ exactly codes $t$} if and only if $t = t(p)$.
\end{description}
\end{dfn}
\begin{dfn}[Decoding]\label{decoding_def}
We now describe the process of \emph{decoding}.
This will be run in the generic extension $L[A_0][G]$, the $P$-generic extension and there, $s_0 = \bigcup_{p \in G} p_{\omega_1}$ will yield $A_0$ via this process.
We shall also run this process ``locally'' in the transitive collapse $\bar M$ of certain (small) elementary submodels $M$.
In this case, the reader should anticipate that a condition of the forcing may be sufficiently generic over $M$ to ensure correct coding e.g. in the sense that if $\kappa^{++}\in M$, the decoding process relative to $M$ yields $A_0 \cap M$ --- whereas of course conditions will never code anything non-trivial over $L[A_0]$ as they are bounded.
If the following is run over $\bar M$, all definitions should be relativized to $\bar M$, so that $\kappa$ is replaced by the least Mahlo in $\bar M$, coding structures are as defined in $\bar M$ etc.
The following definition is by induction on cardinals $\beta \in [\beta_0, \kappa^+)$.
Assume $\beta_0\in \mathbf{Card}$, $A'\subseteq \beta_0$ and $p_{\beta_0} \colon [\beta_0,{\beta_0}^+) \rightarrow 2$, or $p_{\beta_0}\colon \beta_0 \rightarrow 2$ and $\beta_0 =\kappa$ (or, respectively, the least Mahlo in $\bar M$).
We inductively construct $A^*\subseteq \kappa$, setting $A^*\cap\beta = A'$.
If $\beta \in [\beta_0, \kappa)$ and we have constructec $A^*\cap \beta$ and $p_\beta \colon [\beta,{\beta}^+) \rightarrow 2$, define
$p_{{\beta}^+} \colon [{\beta}^+,{\beta}^{++}) \rightarrow 2$ via the decoding process at regulars described in definition \ref{sing_defs} from $p_{\beta}$, i.e. let $p_{{\beta}^+}= s(A^*\cap\beta, p_{\beta})$.
Also, let $\xi \in A^*\cap [\beta, \beta^{+3})\iff p_\beta( ( \langle \xi, \nu \rangle )_0)=1$ for $\beta^+$-many $\nu \in [\beta, \beta^+)$.
At limit $\beta < \kappa$, let $p_{<\beta} = \bigcup_{\beta' < \beta} s_{\beta'}$ and let
$s_\beta = s(A^*\cap\beta,p_{<\beta})$ when $\beta$ is inaccessible
and $s_\beta=t( A^*\cap\beta, p_{<\beta})$ when $\beta$ is singular.
When $\beta$ is the least Mahlo $\kappa$, first extract $A^*_0\cap \kappa$ and the generic for the Mahlo-coding from $A^*$: let
$A^*_0 \cap \kappa = [A^*\cap\kappa]^{-1}_0$, let $p_{<\kappa} = [A^*\cap\kappa]^{-1}_1$,
and let $p_\kappa = s(A^*_0 \cap \kappa, p_{<\kappa})$
(when $\beta = \beta_0 =\kappa$ and we are at the beginning of the induction, we just set $A^*_0\cap\kappa=A'$).
Let $\xi \in A^*_0\cap [\kappa, \kappa^{+})\iff p_\kappa( ( \langle \xi, \nu \rangle )_0)=1$ for $\kappa^+$-many $\nu \in [\kappa, \kappa^+)$.
For $\beta = \kappa^+$, we continue exactly as we did below $\kappa$, but with $A^*_0$ instead of $A^*$: let
$p_{\beta^{+}}= p_{\kappa^{++}} = s(A^*_0\cap\kappa^+, p_{\kappa^+})$
and let $A^*_0\cap [\kappa^+, \kappa^{++})$ be the set whose characteristic function is $p_{\kappa^{+}}$.
The set $A_0^* \subseteq \kappa^{++}$ is the outcome of the decoding procedure run up to $\kappa^{++}$ and we write $A^*(p_{\beta_0})=A^*(A', p_{\beta_0})$ for this set.
\end{dfn}
\begin{dfn}[Strings]\label{strings_dfs}
We now define $S^*_\alpha= S(A)^*_\alpha$.
Let $\Phi(r)$ be the statement ``$ r(n)=i \Rightarrow$ for cofinally many $\sigma <\bar\kappa$, $\bar T(\sigma,n,i)$ has a branch, where $\bar T$ is the canonical $\bar\kappa$-sequence of $\bar\kappa^{++L}$-Suslin trees and $\bar \kappa$ is the least Mahlo in $L$.''
We also say \emph{$r$ is coded by branches} for $\Phi(r)$.
We say $s \in S^*_\alpha$ if and only if $s \in S_\alpha$ and for all $\zeta \leq |s|$ and all $\eta, \bar \kappa$
such that $N= L_\eta[A\cap\alpha, s\res \zeta]\vDash$``$\zeta =\alpha^+$, $\bar \kappa$ is the least Mahlo in $L$, $\bar \kappa^{++}$ exists, $\mathbf{Card} = \mathbf{Card}^{L[A\cap\omega]}$ and $\ZF^-$ holds'',
we have that $L[A^*]\vDash \Phi(r)$, where $A^*=A^*(A\cap\alpha, s\res\zeta)$ is the predicate Jensen-coded by $s\res\zeta$ over $N$.
When $s \in S_\alpha$, we call $N$ as in the hypothesis a \emph{test model for $s$}.
Thus, $s \in S^*_\alpha$ if and only if for every test model $N$ for $s$, $L^N[A^*(s\res\alpha^{+N})]\models r$ is coded by branches.
\footnote{Observe that requiring $N \models \ZF^-\wedge\exists \bar\kappa^{++}$ ensures that $L^N$ thinks that the canonical $\bar\kappa$-sequence of $\bar\kappa^{++}$-Suslin trees exists, but this is not the only reason to make this requirement.
Rather, we will this requirement and $s\res\zeta$ to ensure $\zeta$ collapses to $\alpha$ ``quickly''.
In effect, this allows us to use $s$ to thin out the set of test models.
We could do the same without $\ZF^-$ but that would take an argument.}
Let $S_{<\kappa}$ denote the set of $s\colon [0,|s|) \rightarrow 2$, where $|s| < \kappa$.
We say $s \in S^*_{<\kappa}$ if and only if $s \in S_{<\kappa}$ and for all $\bar \kappa \in \mathbf{Card}\cap |s|+1$ and all $\eta$
such that $N= L_\eta[A_0\cap\bar\kappa, s\res \bar\kappa]\vDash$ ``$\bar \kappa$ is the least Mahlo in $L$, $\bar \kappa^{++}$ exists, $\mathbf{Card} = \mathbf{Card}^{L[A_0\cap\omega]}$ and $\ZF^-$ holds'', we have that $N\vDash \Phi(r)$.
Similarly to the above, when $s \in S_{<\kappa}$ we call $N$ as in the hypothesis a \emph{$<\kappa$-test model for $s$}.
\end{dfn}
\begin{dfn}[Building blocks]\label{reg_codings_dfs}
We now define three partial orders, each of which codes a set via almost disjoint coding using
the almost disjoint family $(b^t)_t$ and simultaneously localizes it.
These partial orders serve as building blocks for the final forcing.
\begin{description}
\item [Successor coding]
Let either $\alpha = \kappa^+$, $\beta=\kappa$ and $s=s_{\kappa^+}$ or let $\beta \in \mathbf{Card} \cap \kappa$, $\alpha= \beta^+$ and $s \in S^*_\alpha$ and let $A\subset\alpha$.
We define the partial order $R^s=R(A)^s$ to consist of conditions $p=(p_\beta,p^*)$ such that
$p \in S^*_\beta$ and $p^* \subseteq \{ b^{s\res\xi} \setdef \xi \in [\alpha, |s|) \} \cup |p|$ of size at most $\beta$.
It is ordered by: $(q,q^*)\leq (p,p^*)$ if and only if $q$ end-extends $p$, $p^* \subseteq q^*$
and
\begin{enumerate}
\item If $b^{s\res \xi} \in p^*$ and $s(\xi)=0$ then for any $\gamma \in (|p|,|q|)\cap b^{s\res\xi}$ we have $q((\gamma)_1)=0$.
\item If $\xi \in p^* \cap |s|$ and $\xi \in A$ then if $\gamma$ is such that $\langle \xi, \gamma \rangle \in (|p|,|q|)$, we have $q((\langle \xi, \gamma \rangle)_0)=0$.
\end{enumerate}
\item[Inaccessible coding]
Let $\alpha\leq \kappa$ be inaccessible, $s \in S^*_\alpha$.
We define the partial order $R^s=R(A)^s$ to consist of conditions $(p,p^*)$ such that
$p$ is a partial function $p \colon \alpha \rightarrow 2$ and $p^* \subseteq \{ b^{s\res\xi}\setminus\eta \setdef \xi \in [\alpha, |s|), \eta < \alpha \} \cup \alpha$ of size less than $\alpha$ and $\alpha\cap p^*$ is an ordinal.
We write $\rho(p^*)$ for this ordinal.
For $\alpha = \kappa$, we additionally demand that $p_{<\kappa} \in S^*_{<\kappa}$.
$R^s$ is ordered by: $(q,q^*)\leq (p,p^*)$ if and only if $q$ end extends $p$, $p^* \subseteq q^*$
and
\begin{enumerate}
\item If $b=b^{s\res\xi}\setminus\eta$, $b \in q^* \setminus p^*$ then
$$b \cap (\rho(p^*)\cup \sup\dom(p)) \subseteq \bigcup \{ b'\setdef b' \in p^*\}.$$
\item If $b^{s\res \xi} \setminus\eta\in p^*$ and $s(\xi)=0$ then for any $\gamma \in (|p|,|q|)\cap (b^{s\res\xi} \setminus \eta)$ we have $q((\gamma)_2)=0$.
\end{enumerate}
\end{description}
\end{dfn}
The additional construct $\rho(p^*)$ and its use in the definition of $\leq$ for inaccessible coding is necessary to preserve requirement \eqref{coding_areas_bounded} of the definition of $P(A_0)$ at the limit (see also definition \ref{D}, item \eqref{restraints_start_high}),
Also, together with the similar use of $\sup(\dom(p_{<\alpha}))$ this becomes important in the proof that $\kappa$ is not collapsed in the limit of our iteration (see \ref{it:reals:are:caught}).
With the intuition that our iteration can \emph{almost} be decomposed into an upper and a lower part, the idea is that this device ensures some independence of upper and lower parts in that argument, since restraints are the only possible interaction.
This also means that the part of the restraint below $ \rho(p^*)\cup \sup\dom(p_{<\alpha})$ as well as $\rho(p^*)$ itself must part of the value of the ``centering function'' $\Cl(p)$ (see \ref{sec_coding_strat}).
The use of $\sup(\dom(p_{<\alpha}))$ can be eliminated (but renders the argument slightly more elegant).
The use of $\sup(\dom(p_{<\alpha}))$ cannot obviate that of $\rho(p^*)$ because of the last clause in \eqref{qc:redundant}.
\section{Definition of the forcing $P(A_0)$}
\begin{dfn}[The conditions]\label{conditions}
Let $\mathbf{Card}'$ for now denote $(\mathbf{Card} \setminus \omega) \cup \emptyset \emptyset$.
A condition in $P=P(A_0)$ is a sequence
$$p = (p_{<\alpha}, p_\alpha, p^*_\alpha)_{\alpha \in \mathbf{Card}' \cap \kappa^{+}+1}.$$
For limit cardinals $\alpha< \kappa$, we demand $p_{<\alpha} = \bigcup_{\delta < \alpha} p_\delta$ (so $p_{<\alpha}$ is redundant unless $\alpha = \kappa$).
For $p$ as above to be a condition, we demand:
\begin{enumerate}
\item $p_{\kappa^+}=s_{\kappa^+}$ (where $s_{\kappa^+}$ is the characteristic function of $A\cap [\kappa^+, \kappa^{++}]$) and $p_\alpha \in S^*_\alpha$ for all $\alpha \leq \kappa$, while $p_{<\kappa} \in S^*_{<\kappa}$ (this is redundant by the definition of the Mahlo coding). We write $\supp(p) = \{ \alpha \setdef p_\alpha \neq \emptyset_\alpha \}$ and $\alpha(p) = \sup (\supp(p) \cap \kappa)$.
\item $\supp(p) \cap \kappa$ is an Easton subset of $|p_{<\kappa}|$.
For $\alpha \in \{ \kappa, \kappa^+\}$ we let $\mathcal{A}^{p_\alpha}$ be defined relative to $A_0$.
For $0<\alpha \leq |p_{<\kappa}|$, we let $\mathcal{A}^{p_\alpha}$ and $\mathcal{B}(p_{<\alpha})^{p_\alpha}$ be defined relative to
\begin{equation*}
A_p = A_0\oplus \{ \xi < \alpha \setdef p_{<\kappa}(\xi)=1\}.
\end{equation*}
Observe that we will order $P$ in such a way that coding structures never change when $p$ is extended, that is, the dependence on $p$ is ``static''.
In the following, $R^s$ is defined relative to these coding structures.
\item For all $\alpha \in \supp(p)$, $(p_\alpha, p^*_{\alpha^+}) \in R^{p_{\alpha^+}}$, where we let $R^{p_\omega}$ denote the standard almost disjoint coding of $p_\omega$ by a real relative to some convenient almost disjoint family in $L[A_0\cap\omega]$.
\item For all inaccessible $\alpha \leq\kappa$, $(p_{<\alpha}, p^*_{\alpha}) \in R^{p_\alpha}$,
remembering $p_{<\alpha}=\bigcup_{\alpha'<\alpha} p_\alpha$ for $\alpha < \kappa$.
\item For all singular $\alpha <\kappa$, $p\res \alpha \in \mathcal{B}(p_{<\alpha})^{p_\alpha}$, $t(p_{<\alpha})=p_\alpha$ and if $p_{<\alpha}$ is unbounded below $\alpha$ and uses fake inaccessible coding, then $p_{<\alpha}$ recognizes singularity of $\alpha$.\footnote{\label{why_stop_here}We let the fake inaccessible coding stop when the singularity of $\alpha$ is realized; without this natural stopping point, the coding structure for singulars will change when a condition is extended below $\alpha$, i.e. $\mathcal{B}(q_{<\alpha})^{\emptyset_\alpha} \neq \mathcal{B}(p_{<\alpha})^{\emptyset_\alpha}$ for $q\leq p$, because $q$ might carry new information in $t(q_{<\alpha})$.
Intuitively, $p$ hasn't exhausted the room for fake inaccessible coding.}
\item \label{coding_areas_bounded} For $\alpha\in\mathbf{Reg}$, there is a $\gamma < \alpha$ such that for all $\delta\in \mathbf{Inacc}\setminus \alpha+1$ and all $(\xi, \eta)\in p^*_\delta$ we have $b^{p_\delta\res\xi}\setminus \eta \cap \alpha \subseteq \gamma$.
\end{enumerate}
Note again we have $p_{\kappa^+}=s_\kappa$ for every condition $p \in P$.
We say $q \leq p$ if and only if for all $\alpha \in \supp(p)$, $(q_\alpha, q_{\alpha^+})\leq (p_\alpha, p_{\alpha^+})$ in $R^{p_{\alpha^+}}$ and for all inaccessible $\alpha \in \supp(p)$,
$(q_{<\alpha}, q_{\alpha})\leq (p_{<\alpha}, p_{\alpha})$ in $R^{p_\alpha}$.
\end{dfn}
Observe that the limit coding from $\kappa^+$ into $\kappa$ takes a slight detour, via $p_{<\kappa}$;
The reason for this is that the coding into $\kappa$ is easier if we use strings (mainly because of the proof that $\kappa$ isn't collapsed, \ref{it:reals:are:caught}).
In contrast, $p_{<\alpha}$ can be seen as a shorthand for $\bigcup_{\delta < \alpha} p_\delta$;
we use these ``broken strings'' to code into inaccessible $\alpha < \kappa$,
which is the natural choice.
We will need that being a condition, and in fact all of the definitions in \ref{reg_defs} and \ref{sing_defs} are absolute for $\Sigma^A_1$-correct models.
In fact, all these notions are Boolean combinations of $\Sigma^A_1$ statements.\footnote{more generally, it would suffice that they be provably $\Delta^A_2$ in a weak enough theory.}
For $p \in P$ and $\alpha \in \mathbf{Card}$,
let $B_p^\alpha\subseteq \alpha$ be defined by
\[
B^\alpha_p=\bigcup\{ b\cap\alpha \setdef b=b^s\setminus \eta \in p^*_\gamma, \gamma\in \mathbf{Inacc}\setminus\alpha+1\},
\]
and let $b_p \in \prod _{\alpha \in \mathbf{Card}\cap\kappa^{++}} \alpha$ be defined by
\begin{align*}
b_p(\alpha) =\sup B^\alpha_p
\end{align*}
That is, $B_p^\alpha$ is the set of coding ordinals used by inaccessibles above $\alpha$, and
$b_p$ locally bounds the height of this set.
\begin{rem}
Note that \ref{coding_areas_bounded} of \ref{conditions} (the definition of $P$), is equivalent to asking
$b_p(\alpha) < \alpha$ for every $\alpha \in \mathbf{Reg}$.
\end{rem}
\section{Doing the same in three steps.}
To aid the readers intuition, we describe the same forcing as a three-step iteration.
This illustrates further the special role of $p_{<\kappa}$ and the detour in the Mahlo coding,
but will not be needed for the rest of the proof.
\begin{enumerate}
\item The first step is to force with the successor coding $R^{A_0}$.
We will later see this forcing is fully stratified.
Let $G_1$ be the generic and let
\[
A_1 = \{ \xi \in [\kappa, \kappa^+) \setdef \exists (p, p^*) \in P\; p(\xi)=1\} \cup A_0 \cap \kappa.
\]
Observe that $A_1$ has the following property:
For any $N = L_\eta[A_1\res\zeta]$ such that $\zeta \geq \kappa$ and
\item The second step is to force with the Mahlo coding $R^{A_1}$ in $L[A_1]=L[A_0][G_1]$.
We will later see this too, preserves all cardinals and $\kappa$ is still Mahlo.
Let $G_2$ be the generic and let
\[
A_2 = \{ \xi \in \kappa \setdef \exists (p, p^*) \in P\; p(\xi)=1\}.
\]
\end{enumerate}
Observe that $A_2$ satisfies the following:
\begin{eqpar}
for all $\bar \kappa\in\mathbf{Card}$, $\eta \in \On$
such that $N= L_\eta[A_2\res \bar\kappa]\vDash$ ``$\bar \kappa$ is the least Mahlo, $\bar \kappa^{++}$ exists and $\ZF^-$ holds'', we have that $N\vDash \Phi(r)$.
\end{eqpar}
Lastly, we define $P(A_2)$ which codes $A_2$ by a subset of $\omega_1$.
For this sake, let $A$ denote $A_2$, and let all coding structures be defined relative to this $A$.
\begin{dfn}[The conditions for coding below $\kappa$]\label{conditions_3steps}
A condition in $P(A)$ is a function $p\colon \mathbf{Card}\cap \kappa\rightarrow V$, $p(\alpha)=(p_\alpha, p^*_\alpha)$ such that
\begin{enumerate}
\item $(p_{\kappa},p^*_{\kappa^+}) \in R^{s_{\kappa^+}}$.
\item $\supp(p) = \{ \alpha < \kappa \setdef p_\alpha \neq \emptyset \}$ is an Easton set.
\item For all $\alpha \in \supp(p)$, $(p_\alpha, p^*_{\alpha^+}) \in R^{p_{\alpha^+}}$.
\item For all inaccessible $\alpha \in \supp(p)$, $(p_{<\alpha}, p_{\alpha}) \in R^{p_\alpha}$,
where we define $p_{<\alpha}=\bigcup_{\alpha'<\alpha} p_\alpha$ for any $\alpha$.
\item For all singular $\alpha \in \supp(p)$, $p\res \alpha \in \mathcal{B}(p_{<\alpha})^{p_\alpha}$, $t(p_{<\alpha})=p_\alpha$ and if $p_{<\alpha}$ is unbounded below $\alpha$ and uses fake inaccessible coding, then $p_{<\alpha}$ recognizes singularity of $\alpha$.\footnote{See footnote \ref{why_stop_here}}
\item %
For $\alpha\in\mathbf{Reg}$, there is a $\gamma < \alpha$ such that for all $\delta\in \mathbf{Inacc}\setminus \alpha+1$ and all $(\xi, \eta)\in p^*_\delta$ we have $b^{p_\delta\res\xi}\setminus \eta \cap \alpha \subseteq \gamma$.
\end{enumerate}
We say $q \leq p$ if and only if for all $\alpha \in \supp(p)$, $(q_\alpha, q_{\alpha^+})\leq (p_\alpha, p_{\alpha^+})$ in $R^{p_{\alpha^+}}$ and for all inaccessible $\alpha \in \supp(p)$,
$(q_{<\alpha}, q_{\alpha})\leq (p_{<\alpha}, p_{\alpha})$ in $R^{p_\alpha}$.
\end{dfn}
This ends our description of $P$ as a three-step iteration.
We find it more convenient to talk about $P$ instead of this iteration,
and this is the forcing we work with in all of the following.
\section{Extendibility}
\begin{lem}[Extendibility for the Mahlo coding]\label{ext_mahlo}
Let $t \in S_\kappa$ or
$$t\colon [\kappa, \kappa^+) \rightarrow 2.$$
For any $\alpha< \kappa$ and any
$p \in R^{t}$ there is $q \in R^{t}$ such that $q \leq p$ and $|q_{<\kappa}| \geq \alpha$.
In addition, we can demand that $q_{<\kappa} (\nu)=0$ for $\nu \in [ |p_{<\kappa}|, |q_{<\kappa}|)\cap\mathbf{Card}$.
\end{lem}
\begin{proof}
Let $\alpha$ be the least counterexample.
If $\alpha \not \in \mathbf{Card}$ it suffices to extend $p_{<\kappa}$ to $p_{<\kappa}^1$ with
$|p_{<\kappa}^1| \geq \lVert \alpha \rVert$.
We can then further extend $p_{<\kappa}^1$ by appending 0s to obtain $q$.
If $\alpha \in \mathbf{Succ}$, the proof is similar (as no test model will think that $\alpha$ is Mahlo).
So assume $\alpha$ is a limit cardinal.
Let $C \subset \alpha$ be club in $\alpha$, $\otp C = \cof \alpha $, $C \subseteq \mathbf{Sing}$ and
let $(\beta_\xi)_{\xi \leq \rho}$ be the increasing enumeration of $C\cup\{\alpha\}$.
Moreover we demand that $\beta_0 > \rho$ if $\alpha$ is singular.
Let $p^0 \leq p$ such that $|p^0_{<\kappa}|=\beta_0$ and build a descending chain of conditions.
Assume you have $p^\xi$ such that $|p^\xi_{<\kappa}|=\beta_\xi$.
Extend to get $p'$ with $|p'| = \beta_{\xi+1}$.
Let $p^{\xi+1}$ be obtained from $p'$ by shifting values of $p'_{<\kappa}$ above $\beta_\xi$ away from the cardinals in a gentle manner, putting a 1 on $\beta_{\xi}$ and padding with 0s, as follows: let $p^{\xi+1}_{<\kappa} \res\beta_\xi = p'_{<\kappa} \res\beta_\xi = p^{\xi}_{<\kappa} \res\beta_\xi$ and for $\nu \in [ \beta_\xi, \beta_{\xi+1})$, let
\[
p^{\xi+1}_{<\kappa}(\nu)=\begin{cases}
p'_{<\kappa}(\delta + k)&\text{if $\nu = \delta + k +1 $ for some $\delta \in
\mathbf{Card}$, $k \in \omega$,}\\
1 &\text{if $\nu = \beta_{\xi}$,}\\
0 &\text{if $\nu \in \mathbf{Card}\cap (\beta_\xi, \beta_{\xi+1})$.}\\
p'_{<\kappa}(\nu) &\text{otherwise.}
\end{cases}
\]
Observe that $p^{\xi+1}\leq p^{\xi}$ since no restraints $b \in (p^\xi)^*$ are violated, as we have $\delta + k \not\in [b]_3$ for $\delta \in\mathbf{Card}$, $k \in \omega$.
We still have $p^{\xi+1}_{<\kappa} \in S^*_{<\kappa}$, as $L_\eta[A\cap\bar \kappa, p^{\xi+1}_{<\kappa}\res\bar\kappa] =L_\eta[A\cap\bar \kappa, p'_{<\kappa}\res\bar\kappa] $ for all relevant $\eta, \bar\kappa$.
At limit $\xi\leq \rho$, if $N = L_\eta[ p^{\xi}_{<\kappa}\res\bar\kappa]$,
note that $C\cap \beta_\xi \in N$ and so $N \models \bar \kappa$ is not Mahlo.
Thus $q = p^\rho$ is as desired.
\end{proof}
\begin{lem}[Extendibility for the successor coding.]
For each $p \in P$, $\xi \in \kappa^+$ there is $q\leq p$ such that $\xi \in \dom(q_{<\kappa^+})$.
Equivalently,
let $\alpha \in \mathbf{Card} \cap |p_{<\kappa}| \cup\{ \kappa\}$, $A\subseteq\On$ when $\alpha = \kappa$ and $A=A_q$ otherwise and let $s\in S(A)^*_{\alpha^+}$ or $s = s_{\kappa^+}$.
Any $p=(p_\alpha,p^*) \in R(A)^s$ can be extended to $q \leq p$ such that $|q| \geq \xi$, for any $\xi \in [\alpha,\alpha^+)\cap\mathbf{Card}$.
\end{lem}
\begin{proof}
The two statements are equivalent by lemma \ref{ext_mahlo}.
To avoid repetition, we leave the proof of this lemma as a by-product of lemma \ref{D}:
any $q \in \D^{L[A]}(0,p,\{\xi\})$ does the trick.
\end{proof}
The next lemma allows us to extend conditions at a singular cardinal $\alpha$ without violating that the extension be coded exactly below $\alpha$.
By the last item below, we may at the same time capture a set $X$ locally (in a sense);
this included for completeness and is not needed in the rest of the proof.
\begin{lem}[Extendibility at singulars]\label{ext1}
Let $p\in P$, $\alpha \in \mathbf{Sing}\cap |p_{<\kappa}|$ and $s \in S(A_p)_\alpha$ such that $p_\alpha \is s$, further say $X \subseteq \On$ and %
$ X\in \mathcal{B}(p_{<\alpha})^s$.
We can find $q \in P$, such that
\begin{itemize}
\item $q\res(\alpha, \infty] = p\res(\alpha, \infty]$,
\item $q_\alpha = s$,
\item for each each $\beta$ which is a limit point of $E_\alpha$ (in fact, for all $\beta=\delta^+$ for some $\delta \in E_\alpha$), $X\cap\beta \in \mathcal{B}(q_{<\alpha})^{q_\beta}$. \end{itemize}
\end{lem}
We write $A=A_p$ for the discussion of this lemma.
Before we proof this lemma, we make two technical observations:
\begin{lem}\label{codingapp_define}
Let $s,t \in S_\alpha$, $p$ be a condition such that $p\res \alpha \in \mathcal{B}(p_{<\alpha})^s$ and $s$ a proper initial segment of $t$. Then
$f(p_{<\alpha})^s \in \mathcal{B}(p_{<\alpha})^t$. In fact,
$f(p_{<\alpha})^s \in L_{\mu(p_{<\alpha})^s + \alpha +1}[A\cap\alpha,s(p),s]$.
\end{lem}
\begin{proof}[Proof of lemma \ref{codingapp_define}]
This is because the Skolem hulls are definable over $\mathcal{B}(p_{<\alpha})^s$,
and the transitive collapse of the Skolem hull of $\gamma < \alpha$ is constructible at most $f(p_{<\alpha})^s(\gamma) + 1$ steps above $\mathcal{B}(p_{<\alpha})^s$,
by the recursive definition of the transitive collapse.
Thus, $f(p_{<\alpha})^s$ is a definable subset of $L_{\nu(p_{<\alpha})^s + \alpha}[A\cap\alpha,s]$, and
$f(p_{<\alpha})^s \in L_{\nu(p_{<\alpha})^s + \alpha+1}[A\cap\alpha,s]$.
Since possibly, $\nu(p_{<\alpha})^s + \alpha = \tilde\nu(p_{<\alpha})^t$, it needn't be the case that $f(p_{<\alpha})^s \in \mathcal{\tilde B}(p_{<\alpha})^t$
but clearly, $f(p_{<\alpha})^s \in \mathcal{B}(p_{<\alpha})^t$.
\end{proof}
\noindent
Note that if the length of $\lh(t) > \lh(s)+1$, then we even have $f(p_{<\alpha})^s \in \mathcal{B}(p_{<\alpha})^{<t}$.
\begin{lem}\label{coding_app_stable}
If $\beta\in\mathbf{Card}\cap\alpha+1$, $t \in S_\beta$, $\mathcal{B}(p_{<\beta})^{t} = \mathcal{B}(q_{<\beta})^{t}$ and $f(q\res\beta)^t = f(p\res \beta)^t$.
\end{lem}
\begin{proof}
Fix $t$ and $\beta$ as in the first statement.
First, assume $p\res\beta$ does not use fake inaccessible coding.
Then neither does $q$.
In this case, $\mu(q_{<\beta})^{\emptyset_\beta}=\mu^{\emptyset_\beta}$ and $p\res\beta$ and $q\res\beta$ play no role at all in the definition of $f(q_{<\beta})^t$ and $f(p_{<\beta})^t$.
In the other case, when $p\res\beta$ uses inaccessible coding, by definition we have
$b^{\bar s} \subseteq \dom(p)$ for any $\bar s \is s(p\res\beta)$.
But then as $q \leq p$,
$s(p\res\beta)=s(q\res\beta)$ and again $\mu(q_{<\beta})^{\emptyset_\beta}=\mu(p_{<\beta})^{\emptyset_\beta}$ by definition.
Thus in this case as well, we have $f(q_{<\beta})^t = f(p_{<\beta})^t$.
\end{proof}
Observe that for limit cardinals $\beta < \alpha$, $[Y]^{-1}_0\cap\beta$ correspond to the G\"odel-numbers of true $\Sigma_1$ sentences of $\mathcal{\tilde B}(p_{<\alpha})^s$ with parameters from $\beta\cup\{A\cap\alpha, s\}$. Similarly for $[Y]^{-1}_1\cap\beta$.
Also observe that for large enough $\beta < \alpha$, we never have $H_\beta \cap\alpha = \beta$.
This means also that $Y\cap \beta$ does not precode a $t\in S_\beta$, for $[Y]^{-1}_0\cap\beta$ codes a model where $\beta^+$ exists.
\begin{proof}[Proof of lemma \ref{ext1}]
The proof is by induction on $\alpha$.
Pick $M = L_\mu[A\cap\alpha,s]$ so that $\mu$ is a limit ordinal, $\tilde \mu(p_{<\alpha})^s < \mu < \mu(p_{<\alpha})^s$
and $p, X, b^\alpha_p \in M$.
Let
\[
H_\beta = h^M_{\Sigma_1}(\beta\cup\{ A\cap\alpha, s\}),
\]
let $\pi_\beta\colon H_\beta \rightarrow \bar M_\beta$ be the transitive collapse.
Let $g$ be defined by $g(\beta)=(\beta^+)^{\bar M_\beta}$ for successor cardinals $\beta <\alpha$, noting that this is well defined since $\beta \in H_\beta$.
Pick $Y$ such that $[Y]^{-1}_0=\Th_{\Sigma_1}^{\mathcal{\tilde B}(p_{<\alpha})^s}(\alpha \cup \{ A\cap\alpha, s\})$,
while $[Y]^{-1}_1=\Th_{\Sigma_1}^{M}(\alpha \cup \{ A\cap\alpha, s\})$.
Observe that $Y$ pre-codes $s$.
Moreover, demand that $[Y]^{-1}_2 = X$.
The same construction works, wether or not $p_\alpha \neq \emptyset_\alpha$.
For $\beta \in E_\alpha$, let
$q_{\beta^+} = p_{\beta^+} \conc 0^{g(\beta)} \conc 1\conc ( [Y]_3\cap\beta)$ (where $ 0^{\nu}$ denotes a string of $0$'s of length $\nu$).
For $\beta$ which is a limit point of $E_\alpha$, let $q_\beta = p_\beta \conc ( [Y]_3\cap\beta)$.
Note that clearly, if $p$ didn't use fake inaccessible coding, neither does $q$.
We now show that this definition works for $\beta$ large enough.
First, note that
\[
H_\alpha = h^M_{\Sigma_1}(\alpha\cup\{ A\cap\alpha, s\}) = M,
\]
since $M$ can be characterized as the minimal model such that the $\Sigma_1$ statement
asserting the existence of $\mathcal{\tilde B}(p_{<\alpha})^s$ and certain statements describing the height of the $M$ via ordinal addition hold inside $M$. Thus, $M = \bigcup_{\beta < \alpha} H_\beta$.
\begin{lem}\label{exact_coding_collapse}
Assume $\beta$ is large enough so that $p \in H_\beta$, and also large enough so that letting
$\tilde \beta = \min H_\beta \cap [\beta, \alpha]$, we have $\tilde \beta \neq \alpha$.
Then $\pi_\beta(p_{\tilde \beta}) = p_\beta$.
\end{lem}
\begin{proof}
If $\tilde\beta=\beta$, there is nothing more to proove.
Otherwise, first note that $E_{\tilde\beta} \in H_\beta$ by elementarity and
$\pi_\beta(E_{\tilde\beta})=E_{\tilde\beta}\cap\beta=E_\beta$.
In fact, that $p\res\tilde\beta$ exactly codes $p_{\tilde\beta}$
is expressible as a $\Sigma_1$ statement inside $L_\alpha[A]$.
By elementarity, this statement also holds of $\pi_\beta(p \res\tilde\beta)$ and $\pi_\beta(p_{\tilde\beta})$ in $\bar M_\beta$ and is upwards absolute, so $\pi_\beta(p \res\tilde\beta)=p\res\beta$ exactly codes $\pi_\beta(p_{\tilde\beta})$.
But since by definition of $P$, $p\res\beta$ exactly codes $p_\beta$, we have $p_\beta = \pi_\beta(p_{\tilde\beta})$.
\end{proof}
\begin{lem}
The following hold:
\begin{enumerate}
\item
$g\res \beta$ eventually dominates $f(q\res\beta)^{p_{\beta}}$
for large enough $\beta \in\mathbf{Card}\cap\alpha$.
\item If $p_\alpha$ is a proper initial segment of $s$, $g$ eventually dominates $f(q\res\alpha)^{p_\alpha}$.
\item On the other hand $f(q_{<\alpha})^s$ eventually dominates $g$,
in fact $g\in \mathcal{B}(q_{<\alpha})^{s}$.
\end{enumerate}
\end{lem}
\begin{proof}
Let $\beta \leq \alpha$ be a limit cardinal large enough so that $p \in H_\beta$.
In case $\beta = \alpha$, note that $p_\alpha \in M$.
We may assume that either $\beta = \alpha$ or $H_\beta \cap \alpha \neq \beta$ (as $\alpha$ is singular in $M$).
Let $\tilde \beta = \min H_\beta \cap[\beta,\alpha]$.
For the case $\alpha = \beta$, we may assume $p_\alpha$ is a proper initial segment of $s$.
Then $f(p\res\alpha)^{p_{\tilde\beta}}\in M$ (by lemma \ref{codingapp_define} and the previous remark if $\beta = \alpha)$, and in fact it is the solution to a $\Sigma_1$-formula in $M$ with parameters $\tilde \beta$ and $p$ - or $p_\alpha$, in case $\beta = \alpha$.
Thus $f(p\res\alpha)^{p_{\tilde\beta}} \in H_\beta$.
Say $\gamma<\beta$ is a successor large enough so that $f(p\res\alpha)^{p_{\tilde\beta}} \in H_\gamma$.
Observe $\gamma \in H_\gamma$.
Thus by elementarity, $\bar M_\gamma \vDash$``$f(p\res\alpha)^{p_\beta}(\gamma)$ has size $\gamma$''.
Finally,
as $M \in \mathcal{B}(p_{<\alpha})^{s}$, clearly $g (\gamma) < \otp(\On \cap H_\gamma) < \otp (\On \cap H^s_\gamma)$ and so $g(\gamma) < f(p\res\alpha)^{s}(\gamma)$ for all $\gamma < \beta$.
Thus $g$ is slower than $f(p\res\alpha)^s$.
\end{proof}
We now show that for any limit $\beta \leq \alpha$, $q\res\beta \in \mathcal{B}(p_{<\alpha})^{q_\beta}$ and exactly codes $q_\beta$.
First, let $\beta=\alpha$.
Since $\mathcal{\tilde B}(p_{<\alpha})^{s} \in\bar M \in \mathcal{B}(p_{<\alpha})^{s}=\mathcal{B}(q_{<\alpha})^{q_\alpha}$, we have
$Y\in \mathcal{B}(q_{<\alpha})^{q_\alpha}$and $g \in \mathcal{B}(q_{<\alpha})^{q_\alpha}$ (because the height of the latter model is a limit mutiple of $\beta$, we can argue as in lemma \ref{codingapp_define}).
So also $q\res\alpha \in \mathcal{B}(q_{<\alpha})^{q_\alpha}$.
Say $p_\alpha=s=q_\alpha$.
Then we have seen that $f(p\res\alpha)^{s}$ dominates $g$, so $q\res\alpha$ exactly codes $p_\alpha$ just as $p\res\alpha$ does.
Now say $p_\alpha$ is a proper initial segment of $s=q_\alpha$.
Moreover, we have seen that $g$ eventually dominates $f(p\res\alpha)^{p_\beta}$.
So $f(p\res\alpha)^{p_\alpha}_{q\res\alpha}$ is defined,
the limit coding for $q\res\alpha$ goes on for one more step than for $p\res\alpha$ and we obtain $Y$ pre-coding $s=q_\alpha$.
Also, the argument of lemma \ref{codingapp_define} shows that $f(p\res\alpha)^{p_\beta}_{q\res\alpha} \in \mathcal{B}(p_{<\alpha})^{s}$.
Observe that also $\mathcal{B}(p_{<\alpha})^{s}= \mathcal{B}(q_{<\alpha})^{q_\alpha}$ by lemma \ref{coding_app_stable}.
Since also $Y \in \mathcal{B}(q_{<\alpha})^{q_\alpha}$, $q\res\alpha$ codes $q_\alpha$.
As $q\res\alpha \in \mathcal{B}(q_{<\alpha})^{q_\alpha}$, the coding stops there and $q\res\alpha$ exactly codes $q_\alpha$.
Now say $\beta < \alpha$.
Remember $H_\beta \cap\alpha\neq\beta$.
Clearly, $Y\cap\beta \in \mathcal{B}(q_{<\beta})^{q_\beta}$ as $q_\beta = p_\beta \conc [Y\cap\beta]_3$.
The $\Sigma_1$-theory of $\bar M_\beta$ allows us to reconstruct $g$ inside
$\mathcal{B}(q_{<\beta})^{q_\beta}$ using an argument like that of lemma \ref{codingapp_define}.
Thus, $q\res\beta \in \mathcal{B}(q_{<\beta})^{q_\beta}$.
Moreover, in this case we have also shown that $g\res\beta$ eventually dominates $f(p\res\alpha)^{p_\beta}$,
and as before we obtain $Y\cap\beta$ in the next step of the limit coding (after the exact coding of $p_\beta$ by $p_{<\beta}$, which could of course be trivial if $p_\beta=\emptyset_\beta$).
We have seen $Y\cap\beta$ does not pre-code an element of $S_\beta$,
so $q\res\beta$ exactly codes $q_\beta = p_\beta \conc [Y\cap\beta]_3$,
provided that $f(p_{<\beta})^{p_\beta}_{q\res\beta} \in \mathcal{B}(q_{<\beta})^{q_\beta}$.
This holds again by an argument similar to that of lemma \ref{codingapp_define}.
Again, as $q\res\beta \in \mathcal{B}(q_{<\beta})^{q_\beta}$, the coding stops there and $q\res\beta$ exactly codes $q_\beta$.
Finally, we obtain a $q$ such that the desired properties hold except for the capturing of $X$, which only holds for a tail of $\beta$;
we can use induction to get $q$ that works for all $\beta$.
Obviously, $q \leq p$ (we have never put 1s on any partition affected by restraints).
For the beginning of the induction, assume $\alpha$ is the least limit cardinal.
Argue as in the general limit case described above to get a condition $q$ that works on a tail below $\alpha$. Now make finitely many extensions to obtain a $q$ that works everywhere.
\end{proof}
Note the subtle role of the proof of \ref{ext1} in my choice to work with the $E_\beta$'s: it seems I could have easily done with the standard $C_\beta$ from $\Box$ instead.
Fake inaccessible coding requires an Easton set, but this is also no reason to go beyond $\Box$.
It is the mechanism of distinguishing between fake inaccessible coding and immediate singular coding
which requires that $E_\beta$ appear in the smallest relevant type of structure---where $\beta$ is seen to be non-Mahlo,
but not necessarily singular.
Without this distinction, we could choose to always do fake inaccessible coding---but then the proof of \ref{ext1} seems to fail.
\section{The main theorem for quasi-closure of $P(A_0)$}\label{jensen_qc}
We shall now define $\D$ and $\leqlol$ witnessing that $P$ is quasi-closed.
\begin{dfn}
We define $p \leqlol q$ just if $p\leq q$, for all $\delta \in \mathbf{Reg} \cap\lambda$ both $p_\delta = q_\delta$ and $p^*_{\delta^+} = q^*_{\delta^+}$. In accordance with the section \ref{sec:qc}, define $p \leqlo^{<\lambda} q$ just if $p\leq q$ and both $p_\delta = q_\delta$ and $p^*_{\delta} = q^*_{\delta}$ for all $\delta \in \mathbf{Reg} \cap\lambda$ when $\lambda$ is a limit cardinal.
\end{dfn}
The class $\D$ is defined in a more elementary way than might be expected from the literature (e.g. \cite{bjw:82} or \cite{friedman:codingbook}), ensuring a more basic kind of genericity over certain Skolem-hulls, owing to the fact that we've assumed $\mathbf{Card} = \mathbf{Card}^{L[A_0\cap\omega]}$.
The proof that this class is dense, i.e. \eqref{qc:redundant}, takes the form of an intricate inductive argument using that quasi-closure already holds over smaller (in the sense of the induction) coding structures.
The next theorem treats two situations simultaneously:
the first is a localized version of quasi-closure which will carry us through the singular limit step of the inductive argument we just mentioned (to show $\D$ is dense).
The second is the global version of quasi-closure.
The proof readily suggests such an aggregation.
We shall work under the following assumption.
\begin{ass}\label{M}
Let $\beta \in \mathbf{Card}\cap\kappa$ or $\beta=\infty$.
In the first case, let $q \in P$, $M=\mathcal{A}^{q_{\beta^+}}$;
also demand that $|q_{<\kappa}| \geq \beta^+$ (we need to ask this so $A=A_q$ is defined, since $\mathcal{A}^{q_{\beta^+}}=\mathcal{A}(A)^{q_{\beta^+}}$).
Define
\[
P(q)^{\beta^+} = \{ p \res\beta^+ \setdef p \in P\wedge p \leq q \wedge p^{\beta^+}=q^{\beta^+}\},
\]
and let $R=P(q)^{\beta^+}$.
In the second case let $M=L[A]$ and $R=P$.
\end{ass}
The definition of $P(q)^{\beta^+}$ was chosen so that for $p\in P(q)^{\beta^+}$, $p\cdot q \in P$ and $P(q)^{\beta^+}$ generically codes both $A\cap\beta$ and $q^{\beta^+}$ over $\mathcal{A}^{q^{\beta^+}}$.
In fact, $(p\cdot q) \res \beta^{++} = p$.
Equivalently, $p \in R$ if and only if
$p \leq q \res \beta^{++}$, $p^{\beta^+}=q^{\beta^+}$ and $p$ obeys all restraints from $q$ for inaccessibles $\delta > \beta$.
\begin{dfn}\label{D}
Given $p\in P$, $\lambda\in \mathbf{Reg}$, $\lambda < \beta\in\mathbf{Card}$, $x\in M$ arbitrary,
we now define
$\D^M_{[\lambda,\beta)}(p,\vec{x}) \subseteq P$.
For $\delta \in \On$, let
\begin{align*}
H_\delta &= H^M_\delta (p,\vec{x}) = h^M_{\Sigma_1}(\delta\cup\{\vec{x}\}),\\
H_{<\delta} &= H^M_{<\delta} (p,\vec{x}) = h^M_{\Sigma_1}(\sup(\supp(p)\cap \delta)\cup\{ \vec{x}\}).
\end{align*}
We define $\D^M_{[\lambda,\beta)}(p,\vec{x})$ as
the set of $q \in P$ such that if $\tau = \min( \supp(p)\cap[\lambda,\beta))$ exists then $q \leqlo^{\tau} p$ and
\begin{enumerate}[label=(D \arabic*), ref=D \arabic*]
\item\label{restraints_start_high} if $\tau > \lambda$ and $\tau\in\mathbf{Inacc}$ then $\rho(q^*_\tau)\geq \lambda$ (of course, this is vital to preserve \ref{coding_areas_bounded} in the definition of $P(A_0)$; see also the last clause of \eqref{qc:redundant});
\item\label{weak_genericity} for all $\delta\in [\tau,\beta)$ such that $\delta \in H_{<\delta} \cup \supp(p)$
\begin{enumerate}[label=(\alph*), ref=\ref{weak_genericity}\alph*]
\item\label{grow} $|q_\delta| > H_\delta \cap \delta^+$;
for $\delta=\kappa$ in addition, $|q_{<\kappa}| > \sup (H_{|p_{<\kappa}|}\cap\kappa)$.
\item\label{restraints} if $\nu \in H_{<\delta}\cap[\delta,\delta^+)$ then $b^{p_\delta\res\nu}\setminus \eta \in q^*_\delta$ for some $\eta \in [\lambda,\delta)$;
\item\label{A_coding} if $\xi \in H_\delta \cap [\delta,\delta^+)$ there is $\nu > |p_\delta|$ such that
$p_\delta((\langle \xi, \nu \rangle)_0) = 1$ if $\xi\in A$ and $p_\delta((\langle \xi, \nu \rangle)_0) = 0$ if $\xi\not\in A_p$;
\item \label{succ_coding} if $b^{p_{\delta^+}\res\nu}\in p^*_{\delta^+}$ there is $\zeta > |p_\delta|$ such that $q_\delta((\zeta)_1)=p_{\delta^+}(\nu)$;
\item \label{fake_inacc_coding} if $\delta\in\mathbf{Inacc}\cap\kappa$
there is $\beta \in E_\delta \setminus \sup(\supp(p)\cap\delta))$ such that
$q_{\beta^+}((\beta^+)_2)=1$;\footnote{This somewhat technical requirement makes it easy to distinguish the fake inaccessible coding from the singular coding. Conditions which immediately use singular coding, i.e. those constructed in the extendibility lemma will have $q_{\beta^+}((\beta^+)_2)=1$ on a tail of $\beta \in E_\delta$.}
\item \label{inacc_coding} if $\delta\in\mathbf{Inacc}\cap\kappa^+$ and $b^{p_\delta\res\nu}\setminus \eta \in p^*_\delta$ then
there is $\xi \in b^{p_\delta\res\nu}\setminus\eta$ such that $\xi > \sup(\supp(p)\cap\delta))$ and
$q_{<\delta}((\xi)_2)=p_\delta(\nu)$ (note that this clause elegantly covers both the Mahlo coding and the inaccessible, non-Mahlo coding);
\item\label{coding_areas_in_supp} if $\delta\in\mathbf{Inacc}$ and $b \in p^*_\delta \setminus\On$ then $b\cap\sup(\supp(p)\cap\delta) \subseteq \dom(q_{<\delta})$;
\end{enumerate}
\end{enumerate}
We also write
\[
U(p)=U^M(p, \vec{x})=\{ \delta \in \mathbf{Card}\cap\delta \setdef \delta \in H_{<\delta}\cup\supp(p)\}.
\]
Finally, for the proof of quasi-closure we set $\D(\lambda, \vec{x}, p) = \D^{L[A]}_{[\lambda, \infty)}(p,\vec{x})$.
\end{dfn}
\begin{rem} To show that $P$ is stratified, we also need to put something like the first clause into the definition of $\dom(\C^\lambda)$.
The easiest is to require $\lambda \in \supp(p)$.
\end{rem}
Note that we define $\D^M_{[\lambda,\beta)}(p,\vec{x})$ as a subset of $P$, and for $p \in P$ rather than as a subset of $R$ and for $p \in R$.
This is a notational convenience we will make use of when we show that these sets are non-emty.
To build sequences with greatest lower bounds, it is only the restriction to $R$ of $\D^M_{[\lambda,\beta)}(p,\vec{x})$ which is useful.
We also introduce the following terminology, which provides good intuition and will be useful when we show the least Mahlo is not collapsed in our iteration (see \ref{it:reals:are:caught}).
\begin{dfn}
Let $H$ be any set and let $p, q \in P$, $q \leq p$.
We say that $q$ is basic generic for $(H,p)$ at $\delta^+$ if and only if
\begin{enumerate}
\item\label{gen_grow} $|q_\delta| > H \cap \delta^+$;
\item\label{gen_restraints} %
if $\nu \in H\cap[\delta^+,\delta^{++})$ then $b^{p_{\delta^+}\res\nu} \in q^*_{\delta^+}$;
\item \label{gen_succ_coding} if $\nu \in H\cap[\delta^+,\delta^{++})$ then there is $\zeta > |p_\delta|$ such that $q_\delta((\zeta)_1)=p_{\delta^+}(\nu)$;
\item\label{gen_A_coding} if $\xi \in H \cap [\delta,\delta^+)$ there is $\nu > |p_\delta|$ such that
$q_\delta((\langle \xi, \nu \rangle)_0) = 1$ if $\xi\in A$ and $q_\delta((\langle \xi, \nu \rangle)_0) = 0$ if $\xi\not\in A_p$;
\end{enumerate}
When $\delta\in\mathbf{Inacc}$, we say that $q$ is basic generic for $(H,p)$ at $\delta$ if and only if
\begin{enumerate}
\item\label{gen_lim_grow} $|q_{<\delta}| \geq \sup (H\cap\delta)$.
\item\label{gen_lim_restraints} if $\nu \in H\cap[\delta,\delta^{+})$ then $b^{p_{\delta}\res\nu}\setminus \eta \in q^*_\delta$ for some $\eta$;
\item \label{gen_lim_mahlo_coding} if $\nu \in H\cap[\delta,\delta^{+})$ then
there is $\xi \in b^{p_\delta\res\nu}\setminus\eta$ such that $\xi \geq |p_{<\delta}|$ and
$q_{<\delta}((\xi)_2)=p_\delta(\nu)$;
\item\label{gen_lim_A_coding} if $\xi \in H \cap \delta$ there is $\nu > |p_\delta|$ such that
$q_{<\delta}((\langle \xi, \nu \rangle)_0) = 1$ if $\xi\in A$ and $q_{<\delta}((\langle \xi, \nu \rangle)_0) = 0$ if $\xi\not\in A_p$;
\end{enumerate}
The second definition we shall only use for $\delta = \kappa$ (in
\ref{it:reals:are:caught}).
\end{dfn}
\begin{thm}\label{jensen_qc_main}
Let $\rho \leq \lambda \leq \beta$, $\lambda\in \mathbf{Reg}$.
If $\beta<\infty$, let $(\beta_\xi)_{\xi < \rho} \in M$ be an increasing sequence of cardinals such that $\sup_{\xi < \rho} \beta_\xi = \beta$
and set $\beta_\xi = \infty$ for $\xi<\rho$ if $\beta=\infty$.
Say $(p^\xi)_{\xi < \rho} \in M$ is a sequence of conditions in $R$ which has $\bar w= (w^\xi)_{\xi < \rho} \in M$ s.t. $M\models\bar w$ is a $(\lambda, x)$-canonical witness.
Moreover say $(p^\xi)_{\xi < \rho}$ is strategic in the following sense:
\begin{enumerate}[label=(\Alph*), ref=\Alph* ]
\item \label{lambda} for any $\xi < \bar \xi < \rho$, $p^{\bar \xi} \leqlol p^\xi$,
\item \label{lambda_xi} for any $\xi < \rho$, there is $\lambda_\xi$ such that $p^\xi \in \D_{[\lambda_\xi, \beta_\xi)}(p^\xi, \{ \bar w\res\xi+1,x\})$ and
$p^{\xi +1} \leqlo^{\lambda_{\xi+1}} p^\xi$,
\item \label{dense_top} in case $\beta<\infty$ , letting $H = h^M_{\Sigma_1}(\beta_\xi \cup\{ \bar w\res\xi+1, x\})$,
$p^{\xi+1}$ is $(H,p^\xi)$-generic at $\beta^+$, which means in the present case:
\begin{enumerate}[label=(\arabic*), ref=\ref{dense_top}\arabic*]
\item $|p^{\xi+1}_\beta| > H \cap \beta^+$;
\item if $\nu \in H\cap[\beta^+, |s|)$ then $b^{s\res\nu} \in (p^{\xi+1})^*_\beta$;
\item if $\xi \in H \cap [\beta, \beta^+)$ there is $\nu > |p^\xi_\beta|$ such that
$p^{\xi+1}_\beta((\langle \xi, \nu \rangle)_0) = 1$ if $\xi\in A_p$ and $p^{\xi+1}_\beta((\langle \xi, \nu \rangle)_0) = 0$ if $\xi\not\in A_p$;
\item if $b^{s\res\nu}\in (p^\xi)^*_{\beta^+}$ there is $\zeta > |p^\xi_\beta|$ such that $p^{\xi+1}_\beta((\zeta)_1)=s(\nu)$;
\end{enumerate}
\end{enumerate}
Then $\bar p$ has a greatest lower bound $p^\rho \in R\cap M$.
\end{thm}
\begin{cor}\label{jensen_qc_main_cor}
Say $\bar p$ is a $(\lambda,x)$-adequate sequence.
Then $\bar p$ has a greatest lower bound.
\end{cor}
\begin{proof}[Proof of corollary]
In case $\beta = \infty$, strategic in the sense of the theorem means exactly that $\bar w$ is a strategic witness, except for one point: we must check is that the proof of the theorem goes through even if the hypothesis that $\bar p$ is adequate is only satisfied on a tail, so we shall pay some attention to this in the proof of the theorem.
\end{proof}
\begin{proof}[Proof of theorem]
The proof is split over several lemmas.
Fix a sequence $(p^\xi)_{\xi<\rho}$ and $\bar w$ as in the hypothesis.
For each $\xi < \rho$, pick $\lambda_\xi\in \mathbf{Reg}$ as in \eqref{lambda_xi}.
Note we may assume $\lambda_\xi \geq \lambda$, for we may replace $\lambda_\xi$ by $\lambda$ whenever this assumption fails:
firstly, \eqref{lambda} holds, and secondly,
$\D^M_{[\lambda_\xi, \beta_\xi)}(p^\xi, \{ x, \bar w \res\xi+1\}) \subseteq \D^M_{[\lambda, \beta_\xi)}(p^\xi, \{ x, \bar w \res\xi+1\})$ if $\lambda_\xi \leq \lambda$.
Let $p= p^\rho$ be the obvious candidate for a greatest lower bound, i.e.\ for $\delta \in \bigcup_{\xi<\rho} \supp(p^\xi)$,
\begin{equation}\label{obvious_candidate}
\begin{gathered}
(p_\delta, p^*_\delta)= ( \bigcup_{\xi<\rho} p^\xi_\delta, \bigcup_{\xi<\rho} (p^\xi)^*_\delta),\\
p_{<\kappa} = \bigcup_{\xi<\rho} p^\xi_{<\kappa}\text{, if $\kappa\leq\beta$.}
\end{gathered}
\end{equation}
Most of this proof will now be devoted to check that for each $\delta \in \supp(p)$ both
\begin{gather}
p_\delta \in S^*_\delta \label{is_string}\\
p\res\delta \in
\mathcal{B}(p_{<\delta})^{p_\delta} \text{ if }\delta \in \mathbf{Sing}.\label{growth}
%
\end{gather}
After that we conclude by checking that $p^*_{<\kappa} \in S^*_{<\kappa}$.
Let $\bar \lambda$ be minimal such that for an unbounded set of $\xi<\rho$, we have $\lambda_\xi \leq \bar \lambda$.
Obviously, if $\delta \in \mathbf{Card} \cap \bar \lambda$, $p^\rho_\delta = p^\xi_\delta$ for some $\xi$, and so \eqref{is_string} and \eqref{growth} hold.
So let $\delta \geq \bar \lambda$.
Let %
\[
X^\xi = \begin{cases}
h^M_{\Sigma_1}(\delta \cup \{x, \bar w\res\xi+1\}) &\text{for }\delta<\beta, \\
h^M_{\Sigma_1}(\beta_\xi \cup\{x, \bar w\res\xi+1\})&\text{for }\delta=\beta.
\end{cases}
\]
Let $X= \bigcup_{\xi<\rho} X^\xi$, let $\pi\colon \bar X \rightarrow X$ be the transitive collapse.
Let $j = \pi^{-1}$.
Write $\bar A = \bigcup_{\nu\in X} \pi(A\cap\nu)$ and (if $\beta<\infty$) $s= q_{\beta^+}$, $\bar s = \pi(s)$, so that
\[
\bar X = \begin{cases}
L_{\bar\mu}[\bar A]&\text{ in case }\beta=\infty, M=L[A]\\
L_{\bar\mu}[\bar A,\bar s]&\text{ when }\beta<\infty, M=\mathcal{A}^s.
\end{cases}
\]
We now embark on a series of lemmas which will be used several times when prooving \eqref{is_string} and \eqref{growth}
in various cases.
\begin{lem}[Definability]\label{def}
The sequence $(\pi(p^\xi))_{\xi<\rho}$ (or at least a tail) is a definable class in $\bar X$.
\end{lem}
\begin{proof}[Proof of lemma \ref{def}]
The notion of canonical witness was chosen precisely to ensure this definability, and in a sense, the lemma is trivial.
We proove the lemma under the weaker assumption that $\bar w$ is a $(\lambda,x)$-canonical witness for $(p_\xi)_{\xi<\rho}$ holds only for $\xi_0 < \xi < \rho$, as in the definition of quasi-closure \ref{def:qc}.
We show that there is a formula $\Theta$ such that $M\vDash \Theta(\xi,p^*) \iff p^*= p_\xi$ for $\xi \in [\xi_0, \rho)$.
Let $\Psi$ and $G$ be as in the definition of canonical witness, i.e. $\Psi$ is the formula defining $\bar w$ and $G$ is the $\Sigma^T_1(\lambda\cup\{x\})$ function s.t. $p^\xi=G(\bar w\res\xi+1,\vec{x})$ for $\xi<\rho$.
Let $\Gamma$ be a $\Sigma^T_1(\lambda\cup\{x\})$ formula representing $G$.
The formula $\Theta(\xi,p^*)$ is
\begin{multline*}
\exists\bar w^*= (w^*_\nu)_{\nu < \xi+1}\text{ s.t. }
\bar w^* \res\xi_0 = \bar w\res \xi_0 \\
\wedge \; [ \forall \nu \in [\xi_0, \xi +1] \quad \Psi(w^*_\nu, \bar w^*\res\nu,x) ]\\
\wedge \;\Gamma(p^*, (w^*_\nu)_{\nu < \xi+1}, x)
\end{multline*}
As $\rho\subseteq X \prec_{\Sigma_1} M$ and $\{ p_\xi , \bar w\res\xi+1\} \subseteq X$ for each $\xi < \rho$, the same formula defines (a tail of) $\bar p$ in $X$.
Now apply $\pi$.
\end{proof}
Observe that only $\delta$ such that %
\begin{equation}
\delta \in \supp(p^\rho) \text{ or } \delta = \bigcup_{\xi<\rho} \sup(\supp(p^\xi)\cap\delta) \label{delta_relevant}
\end{equation}
are relevant to the proof of \eqref{is_string} and \eqref{growth}.
Thus we restrict our attention to such $\delta$ in the following.
We shall now show that because of \eqref{lambda_xi}, $p^\rho$ enjoys a basic type of genericity over $X$.
First, note that $\mathbf{Card} \cap X \cap(\kappa^++1)\subseteq \supp(p^\rho)$, because of \eqref{grow} and the following:
\begin{lem}
If $\tilde \gamma \in \mathbf{Card} \cap X$, then
$\tilde \gamma \in U^M(p^\xi, \{ \bar w\res\xi+1, \vec{x}\})$ for large enough $\xi<\rho$.
\end{lem}
\begin{proof}
Fix $\xi$ large enough such that $\tilde\gamma\in X^\xi$.
If first alternative of \eqref{delta_relevant} obtains, we may assume $\xi$ is large enough such that $\delta \in \supp(p^\xi)$.
We can assume $\tilde\gamma > \delta$ as trivially,
$ \supp(p^\xi)\subseteq U^M(p^\xi, \{ \bar w\res\xi+1, \vec{x}\})$.
Then $\sup(\supp(p^\xi)\cap\tilde\gamma) \geq \delta$,
and so $\tilde\gamma \in H^M_{\tilde\gamma}(p^\xi, \{ \bar w\res\xi+1, \vec{x}\})$.
Now say the second alternative of \eqref{delta_relevant} obtains.
Then we may assume $\xi$ is large enough so that $\tilde\gamma \in h^M_{\Sigma_1}((\sup(\supp(p^\xi)\cap\delta) \cup\{ \bar w\res\xi+1, \vec{x}\})$.
Since $\delta\leq\tilde\gamma$, we immediately conclude $\tilde\gamma \in H^M_{\tilde\gamma}(p^\xi, \{ \bar w\res\xi+1, \vec{x}\})$.
\end{proof}
For $\gamma \in \mathbf{Card}^{\bar X}$, let
\[
\bar p_\gamma = \bigcup_{\xi<\rho}\pi(p^\xi)_{\gamma}.
\]
In fact, \ $j(\gamma)^+\cap X =\bigcup_{\xi < \rho} |p^\xi_\sigma(\gamma)|$, i.e.
\begin{equation}\label{total}
|\bar p_\gamma |= \gamma^{+L_{\bar \mu}},
\end{equation}
by the following:
\begin{lem}\label{nu_in_H}
If $\nu \in [\gamma, \gamma^{+\bar X})$ and either $\nu > \delta$ or the second alternative of \eqref{delta_relevant} obtains, then $j(\nu) \in H^M_{<j(\gamma)}(p^\xi, \{ \bar w\res\xi+1, \vec{x}\})$, for large enough $\xi<\rho$.
\end{lem}
\begin{proof}
Exactly as the previous lemma.
\end{proof}
\noindent
So clearly by \eqref{grow}, we have $\nu < |\bar p^{\xi+1}_\gamma |$ for $\xi, \nu$ as above, prooving \eqref{total}.
Similarly, \eqref{grow} makes sure $|\bar p^\xi_\gamma|$ strictly increases with $\xi < \rho$.
\begin{rem}
Observe for \eqref{total} we can assume $\nu > \delta$ in the hypothesis of the lemma, but the full generality of the lemma will be needed in \ref{local_coding}.
\end{rem}
Thus, letting
\[
\bar p = \bigcup_{\gamma \in \mathbf{Card}^{\bar X}} \bar p_\gamma
\]
we see $\bar p\colon\bar \mu \rightarrow 2$ is a total function.
We continue exploiting the basic genericity of $\bar p$ over $\bar X$, in the sense that $\bar p_\delta$ codes all of $\bar A$ and $\bar p$:
\begin
{lem}[Local coding]\label{local_coding}
Provided $\delta$ satisfies \eqref{delta_relevant} and letting
$$Y= L_{\bar\mu}[A\cap\delta, \bar p_\delta] $$
we have
$\bar X\subseteq Y$ and $\bar X$ is a definable class in $Y$.
In fact, for each $\nu < \bar \mu$,
$\bar A \cap \nu$ and $\bar p \res \nu$ are definable in
$(\mathcal{A}^{<\bar p_{\lVert \nu \rVert}\res\nu})^Y$ if $Y \models \lVert \nu \rVert \in \mathbf{Reg}$ and definable in $(\mathcal{B}(\bar p_{<\lVert \nu \rVert})^{<\bar p_{\lVert \nu \rVert}\res\nu})^Y$ if $Y \models \lVert \nu \rVert \in \mathbf{Sing}$.
\end{lem}
\begin{proof}[Proof of lemma \ref{local_coding}]
We show the definability of $\bar p$ and $\bar A$ by checking they can be reconstructed from $\bar A \cap \delta$ and $\bar p_\delta$ by the decoding procedure run over $\bar X$ (see the last item of definition \ref{sing_defs}).
We now describe this procedure, which is carried out in $L_{\bar \mu}[\bar A\cap \delta, \bar p_\delta]$, making three claims which we proove thereafter.\footnote{A lot of the technical complexity of the following stems from the fact that we do not know the cardinals of $L_{\bar \mu}[A\cap \delta, \bar p^\rho_\delta]$.
Thus we have to work in a mixture of models below.
If the sets $\D$ are defined to provide stronger genericity over $X$, the construction can be carried out entirely over the natural model, namely $L_{\bar \mu}[A\cap \delta, \bar p^\rho_\delta]$.
}
Observe that $\mathbf{Card}^{\bar X} = \mathbf{Card}^{L_{\bar\mu}}$.
\begin{enumerate}[label=(\roman*), ref=\roman*]
\item \label{local_sing_coding}
Say for $\gamma\in \mathbf{Sing}^{L_{\bar \mu}}$ we have already constructed $\bar A \cap \gamma$ and $\bar p\res\gamma$ inside $L_{\bar \mu}[\bar A\cap \delta, \bar p_\delta]$.
We claim that $\bar p\res\gamma$ codes $\bar p_\gamma$ via singular limit coding using as coding apparatus the functions $f^{\bar p_\gamma\res\nu}$ as defined in $L_{\bar \mu}[\bar A \cap \gamma, \bar p_\gamma\res\nu]$, for $\nu < \gamma^{+L_{\bar\mu}}$.
Observe this is well defined as $L_{\bar \mu}[\bar A \cap \gamma, \bar p_\gamma\res\nu] \subseteq \bar X$ and so $\gamma$ is a singular cardinal in that model (this needn't be the case for $L_{\bar \mu}[\bar A \cap \gamma, \bar p \res\gamma, \bar p_\gamma\res\nu]$).
\item \label{local_reg_coding}
Say for $\gamma \in \mathbf{Reg}^{L_{\bar \mu}}$ below the Mahlo of $L_{\bar \mu}$ (if there is one) and $\nu \in [\gamma,\gamma^{+L_{\bar \mu}})$ we have already constructed $\bar A \cap \gamma$ and $\bar p\res\nu$ inside $L_{\bar \mu}[\bar A\cap \delta, \bar p_\delta]$.
Letting $k=1$ if $\gamma\in \mathbf{Succ}^{L_{\bar \mu}}$ and $k=2$ if $\gamma$ is inaccessible in ${L_{\bar \mu}}$, for each $\nu \in [\gamma,\gamma^{+L_{\bar \mu}})$ we claim
\begin{equation*}
\begin{gathered}
\bar p_\gamma(\nu)=1 \iff \\
\{ \xi < \gamma \setdef \xi \in b^{\bar p_\gamma\res\nu}\wedge \bar p((\xi)_k) = 1\}\text{ is unbounded below }\gamma,
\end{gathered}
\end{equation*}
where by $b^{\bar p_\gamma\res\nu}$ we mean $(b^{\bar p_\gamma\res\nu})^{L_{\bar \mu}[\bar A \cap \gamma, \bar p_\gamma\res\nu]}$.
\item \label{local_mahlo_coding}
Say for $\gamma=\bar \kappa$, the Mahlo of $L_{\bar \mu}$, and $\nu \in [\bar \kappa,\bar \kappa^{+L_{\bar \mu}})$, say we have already constructed $\bar A \cap \bar \kappa$ inside $L_{\bar \mu}[\bar A\cap \delta, \bar p_\delta]$.
Let $\bar p_{<\bar \kappa} = [\bar A]_1$ and $\bar A_0 = [\bar A]_0$.
For each $\nu \in [\bar \kappa,\bar \kappa^{+L_{\bar \mu}})$ we claim
\begin{equation*}
\begin{gathered}
\bar p_{\bar \kappa}(\nu)=1 \iff \\
\{ \xi < \bar \kappa \setdef \xi \in b^{\bar p_{\bar \kappa}\res\nu}\wedge \bar p_{<\bar\kappa}((\xi)_3) = 1\}\text{ is unbounded below }\gamma,
\end{gathered}
\end{equation*}
where by $b^{\bar p_{\bar \kappa}\res\nu}$ we mean $(b^{\bar p_{\bar \kappa}\res\nu})^{L_{\bar \mu}[\bar A \cap \bar \kappa, \bar p_{\bar \kappa}\res\nu]}$.
\item \label{local_A_coding}
Say for $\gamma \in \mathbf{Card}^{L_{\bar \mu}}$ below the Mahlo of $L_{\bar \mu}$ (if there is one), we have already constructed $\bar A \cap \gamma$ and $\bar p\res\gamma$ inside $L_{\bar \mu}[\bar A\cap \delta, \bar p_\delta]$.
We claim:
\begin{gather*}
\nu \in \bar A \cap [\gamma, \gamma^{+L_{\bar \mu}})\iff \\
\{ \xi \in [\gamma, \gamma^{+L_{\bar \mu}}) \setdef \bar p_\gamma ((\langle \xi, \nu \rangle)_0)=1 \}\text{ is unbounded below }\gamma^{+L_{\bar \mu}}.
\end{gather*}
The same holds for $\bar A_0$ if $\gamma = \bar \kappa$, the Mahlo of $L_{\bar \mu}$.
\end{enumerate}
From these claims, the lemma follows by induction.
In a sense the proof of these claims is again trivial, by definition of $\D$, i.e. by strategicity or \eqref{lambda_xi}.
The claim in \eqref{local_sing_coding} is a straightforward consequence of elementarity:
since $\bar p_\gamma\res\nu \in \bar X$ for $\nu < | \bar p_\gamma |$,
$j(f^{\bar p_\gamma\res\nu})=f^{p^\xi_{j(\gamma)} }\res j(\nu)$ for large enough $\xi < \rho$,
and $X \vDash$``$p^\xi \res j(\gamma)$ exactly codes $p^\xi_{j(\gamma)}$.
By the obvious continuity of singular limit coding, the claim in \eqref{local_sing_coding} follows.
Now for the claim in \eqref{local_reg_coding}.
We restrict our attention to the exemplary case that $\gamma < \beta$ and is inaccessible in $L_{\bar \mu}$.
Observe for the claim we can assume $\delta < \gamma$, but note for later that the present proof also works if $\delta = \gamma$ and the second alternative of \eqref{delta_relevant} obtains.
Now say $\nu \in [\gamma, \gamma^{+L_{\bar\mu}})$ and show the claim in \eqref{local_reg_coding}.
We have seen for $\gamma \in U^M(p^\xi, \{\bar w\res\xi+1, \vec{x}\})$ and
$j(\nu) \in H_{j(\gamma)} (p^\xi, \{\bar w\res\xi+1, \vec{x}\})$
for large enough $\xi < \rho$
(we assume for now that $\nu > \delta$, but note for later reference
the present proof also goes through if the second alternative of \eqref{delta_relevant} obtains).
Thus by \eqref{restraints}, $b^{p_{j(\gamma)}^\xi \res j(\nu)} = b^{p_{j(\gamma)}^\rho \res j(\nu)} \in (p^\xi)^*_{j(\gamma)}$ for large enough $\xi < \rho$,
restraining extensions of $p^\xi$, i.e. making sure $p^\rho_{<j(\gamma)} ((\zeta)_1) = 0$ for a tail of $\zeta \in b^{p_{j(\gamma)}^\rho \res j(\nu)}$ if $p^\rho_{j(\gamma)}(j(\nu))=0$.
Moreover, \eqref{inacc_coding} makes sure
$p^\rho_{<j(\gamma)} ((\zeta)_1) = 1$ for a tail of $\zeta \in b^{p_{j(\gamma)}^\xi \res j(\nu)}$ if $p^\rho_{j(\gamma)}(j(\nu))=1$.
Together, for large enough $\xi < \rho$ we have
\begin{gather*}
p^\xi_{j(\gamma)}(j(\nu))=1 \iff X \models\text{``}p^\xi_{<j(\gamma)} ((\zeta)_1) = 1
\\\text{ for unboundedly many }\zeta \in b^{p_{j(\gamma)}^\xi \res j(\nu)}\text{''}
\end{gather*}
and since
\[
\sigma( b^{p_{j(\gamma)}^\xi \res j(\nu)} ) = (b^{\bar p_\gamma \res \nu})^{L_{\bar \mu}[\bar A\cap \gamma, \bar p \res \nu]}
\]
applying $\sigma$ proves the claim.
The other case of the claim in \eqref{local_reg_coding} is entirely analogous, substituting the use of \eqref{inacc_coding} by \eqref{succ_coding}
if $\gamma < \beta$ and using \eqref{dense_top} if $\gamma = \beta$.
Claim \eqref{local_mahlo_coding} is proved in the same manner, once more using \eqref{inacc_coding}.
The claim in \eqref{local_A_coding} is proved using using \eqref{A_coding}.
\end{proof}
\begin{lem}\label{catch_structure}
Provided $\delta$ satisfies \eqref{delta_relevant},
$\bar X \in \mathcal{A}_0^{\bar p_\delta}$
\end{lem}
\begin{proof}[Proof of lemma \ref{catch_structure}]
It suffices to observe that $L_{\bar \mu}[A\cap\delta]\vDash |\bar p_\delta| =\pi^{-1}(\delta^+)\in \mathbf{Card}$.
Thus by the definition of steering ordinal, $\bar \mu +\omega < \mu^{\bar p_\delta}$.
Since by lemma \ref{local_coding}, $\bar A$ and (in case $\beta<\infty$) $\bar s$ are definable over $L_{\bar \mu}[A\cap\delta, \bar p_\delta] \in L_{\mu^{\bar p_\delta}}[A\cap\delta, \bar p_\delta] = \mathcal{A}_0^{\bar p_\delta}$.
\end{proof}
\begin{lem}\label{collapsed_string}
If $\delta = \bigcup_{\xi<\rho} \sup(\supp(p^\xi)\cap\delta)$,
\begin{equation}
p\res\delta \in \mathcal{A}^{\bar p_\delta}_0\label{growth_collapse}.
\end{equation}
\end{lem}
\begin{proof}[Proof of lemma \ref{collapsed_string}]
As $\pi(p^\xi)\res\delta = p^\xi\res\delta$, lemmas \ref{def} and \ref{catch_structure} together yield
$p^\rho \res\delta \in \mathcal{A}^{\bar p_\delta}_0$.
\end{proof}
We proceed to show \eqref{is_string} and \eqref{growth}.
Let $\tilde\delta = \min \On\cap X \setminus \delta$, so that $\pi(\tilde\delta)=\delta$.
First, say $\delta < \tilde\delta$.
Then we must have $X \models \tilde\delta \in \mathbf{Inacc}$ ($\delta$ is the critical point of $\pi$).
Observe that for each $\xi < \rho$, by Easton support and since $\sup(\supp(p^\xi)\cap\tilde\delta)\in X$ we have $\sup(\supp(p^\xi)\cap\tilde\delta) < \delta$.
Thus \eqref{is_string} is trivially satisfied as $p^\rho_\delta = \emptyset_\delta$.
We may assume without loss of generality that
\begin{equation}\label{relevant_support_unbounded}
\delta = \sup_{\xi<\rho} (\supp(p^\xi)\cap\tilde\delta)
\end{equation}
hold, for otherwise \eqref{growth} is trivially satisfied and we are done.
This means $\delta \in \mathbf{Sing}$ ($\rho\leq\lambda$ and we must have $\lambda < \delta$ as the conditions grow below $\delta$).
\paragraph{Fake inaccessible coding:}
We show that $\bar p_\delta$ is coded using fake inaccessible coding, i.e.
\begin{equation}\label{fake_inaccessible_coding}
s(\bar p_{<\delta})=\bar p_\delta
\end{equation}
(of course $\bar p_{<\delta} = p_{<\delta}$).
We adapt the proof of claim \eqref{local_reg_coding}, lemma \ref{local_coding}.\footnote{At this point we need lemma \ref{nu_in_H} for the second alternative of \eqref{delta_relevant}.}
Because of \eqref{fake_inacc_coding}, $p_{<\delta}$ does not immediately use singular coding.
We remind the reader that the proof of the claim in \eqref{local_reg_coding}, lemma \ref{local_coding} also goes through if $\nu=\gamma= \delta$ because \eqref{relevant_support_unbounded} holds.
This means that for $\nu \in [\delta, \delta^{+L_{\bar\mu}})$ we have
\[
\bar p_\delta(\nu) = 1 \iff \bar p_{<\delta}((\zeta)_1)\text{ for a tail of } \zeta \in (b^{\bar p_\delta\res\nu})^{\bar X}.
\]
By \eqref{growth_collapse} $b^{\bar p_\delta}$ is eventually disjoint from $\bar p_{<\delta}$.
Moreover $p\res\delta\in \mathcal{A}_0^{\bar p_\delta}$ so the coding stops there.
It remains to show that %
\begin{equation}\label{restraints_in_dom}
\forall \nu \in [\delta, |\bar p_\delta| ) \quad b^{\bar p \res\nu}\subseteq^* \dom(\bar p_{<\delta}).
\end{equation}
Fixing $\nu$ as above, in the proof of lemma \ref{local_coding} we've seen $j(\delta) \in U^M(p^\xi, \{ \bar w\res\xi+1, \vec{x})$ and $j(\nu) \in H_{<\delta} ^M(p^\xi, \{ \bar w\res\xi+1, \vec{x})$ for large enough $\xi <\rho$, as the second alternative of \eqref{delta_relevant} obtains.
So by \eqref{coding_areas_in_supp} (which we haven't used up to now) and \eqref{relevant_support_unbounded} it follows that
\[
b^{\bar p_{j(\delta)}^\rho \res j(\nu)} \cap [\eta, \delta) \subseteq \dom(p^\rho_{<\delta})
\]
for some $\eta <\delta$.
Now use elementarity and apply $\sigma$ to get \eqref{restraints_in_dom}.
By \eqref{restraints_in_dom} and \eqref{fake_inaccessible_coding}, we have $\mathcal{A}_0^{\bar p_\delta} = \mathcal{B}(p^\rho_{<\delta})^{\emptyset_\delta}$, so \eqref{growth} follows from \eqref{growth_collapse} and we are done.
Now say $\delta \in X^\rho_\delta$.
Since $\bar p_\delta = p^\rho_{\tilde\delta}$ and $\mathcal{A}_0^{p_\delta} \subseteq \mathcal{B}(p_{<\delta})^{p_\delta}$ if $\delta \in \mathbf{Sing}$, \eqref{growth} immediately follows from \eqref{growth_collapse}. It remains to show \eqref{is_string} (definition \ref{strings_dfs}).
We only need to check the requirement given in definition \ref{strings_dfs} for $\zeta = |p^\rho_\delta|$, for when $\zeta < |p^\rho_\delta|$, the requirement is met as
$\zeta < |p^\xi_\delta|$ for some $\xi<\rho$ and
$p^\xi_\delta \in S^*_\delta$.
So let $N =L_\eta[A\cap\delta, p^\rho_\delta] $ be a test model, i.e. let $\eta, \bar \kappa$ be such that
\begin{gather}
N\vDash \ZF^-\text{and }|p^\rho_\delta| = \delta^+, \label{still_card} \\
N\vDash \mathbf{Card}=\mathbf{Card}^{L[A\cap\omega]}\label{cards_r_L_cards} \\
L_\eta\models \text{$\bar \kappa$ is the least Mahlo, }\bar\kappa^{++} \text{ exists.}\label{mahlo}
\end{gather}
We must show that $L_\eta[\bar A^*] \models r$ is coded by branches, where we let $\bar A^* = A^*(A\cap\delta, p_\delta)^N$, i.e. the set obtained when running the decoding procedure (see \ref{decoding_def}) relative to $N$.
By lemma \ref{def} and \ref{local_coding} and as $\pi(p^\xi_\delta)=p^\xi_\delta$ for $\xi < \rho$ the sequence $(|p^\xi_\delta|)_{\xi < \rho}$ is definable over $L_{\bar \mu}[A\cap\delta, p^\rho_\delta]$.
Thus \eqref{still_card} implies $\eta < \bar \mu = \bar X\cap \On$.
Let $\alpha = \mathbf{Card}^{L_{\bar \mu}} \cap\eta$.
Clearly $\eta \in [\alpha, \alpha^{+\bar X})$.
\paragraph*{Case 1:}
$\alpha \geq \bar\kappa^{+\bar X}$.
Observe $\eta > \bar \kappa^{+N} = \bar \kappa^{+\bar X}$, where the first inequality holds by \eqref{mahlo}.
Thus also $\bar X \vDash \bar \kappa$ is the least Mahlo and $j(\bar\kappa)=\kappa$.
It also easily follows by elemenarity of $j\colon \bar X \rightarrow M$ that $\beta=\kappa$ or $\beta=\infty$:
if $\beta<\kappa$, $M=\mathcal{A}^{q_{\beta^+}}\vDash$ there is no Mahlo cardinal.
By lemma \ref {local_coding} and by \eqref{cards_r_L_cards}, the decoding procedure of
$\bar p_\delta$ over $L_\eta[A\cap\delta]$ run up to $\kappa^{++N}$ yields
$\bar A_0 \cap\bar \kappa^{++N}$, where of course $\bar A_0=\bar A=\pi[A_0]$.
By elementarity, $L_{\bar\mu}[\bar A_0]\models \Phi(r)$ and also $L_\eta[\bar A_0\cap\bar\kappa^{++N}]\models \Phi(r)$ by the construction of $\bar T$.\footnote{In more detail: If If $\eta = \bar \mu$, this is obvious, so assume $\eta < \bar \mu$. Then $j(\bar A_0 \cap\bar \kappa^{++N}) = \bar B^- \res \tilde \kappa$, where $\tilde \kappa = j(\bar\kappa^{++N}) = \kappa^{++L_{j(\eta)}}$ and $L_{j(\eta)}$ is a model of $\ZF^-$.
Thus the trees defined in $L_{j(\eta)}$ are just initial segments of the real ones, and the respective components of $\bar B \res \tilde \kappa$ are branches through the local versions.}
\paragraph{Case 2:} Otherwise, assume $\alpha \leq\bar \kappa$ and if $\alpha = \bar \kappa$, then $\bar \kappa$ is Mahlo in $\bar X$.
Write $\bar \zeta$ for $\alpha^{+N}=\alpha^{+L_\eta}$.
Observe that as in the previous case, by the proof of lemma \ref{local_coding} and as $\mathbf{Card}^N=\mathbf{Card}^{L_\eta}$, the decoding procedure of $\bar p_\delta\res\bar\zeta$ over $L_\eta[A\cap\delta]$ run up to $\alpha$ yields $\bar A \cap\bar\zeta$ and $\bar p\res\bar\zeta$.
In the case $\alpha = \bar \kappa$, this uses the assumption that $\bar \kappa$ is Mahlo in $\bar X$ so that $N$ contains enough of the coding apparatus of $\bar X$ to run the proof of lemma \ref{local_coding} inside $N$.
As $\bar \zeta< \alpha^{+\bar X} = \sup_{\xi<\rho} |p^\xi_\alpha| $,
we can pick $\xi<\rho$ such that $\bar\zeta < |p^\xi_\alpha|$.
By elementarity,
\[
\tilde N= L_{j(\eta)}[A \cap j(\alpha), p^\xi_{j(\alpha)}\res j(\bar\zeta)],
\]
is a valid test-model for $\tilde s = p^\xi_{j(\alpha)}\res j(\bar\zeta)$ and as $\tilde s \in S^*_{j(\alpha)}$, this string codes a predicate $A^*$ over $L_{j(\eta)}[A\cap j(\alpha)]$ such that $L_{j(\eta)}[A^*]\models \Phi(r)$.
By the way the coding works and using elementarity of $\pi$, $\bar p_\delta$ also codes $\bar A^* = \pi(A^*)$ and $L_{\eta}[\bar A^*]\models \Phi(r)$.
\paragraph*{Case 3:}
It remains to deal with the case $\alpha = \bar \kappa$ when $\bar \kappa$ is not Mahlo in $\bar X$.
Let's assume for simplicity that $\kappa \in X$, or equivalently, $\beta \geq \kappa$.
The other case differs only in notation.
Again using the proof of lemma \ref{local_coding} and as $\mathbf{Card}^N=\mathbf{Card}^{L_\eta[A\cap\omega]}$, the decoding procedure of $p_\delta$ over $L_\eta[A\cap\delta]$ run up to
$\bar\kappa$ yields $\bar p_{<\pi(\kappa)} \res\bar \kappa$ (or if $\beta < \kappa$, the appropriate collapse of $p^0_{<\kappa} \res X$).
As $\bar \kappa< \pi(\kappa)$, we can find $\xi<\rho$ such that $|p^\xi_{<\kappa}| > j(\bar\kappa)$ (if $\beta<\kappa$, this holds for $\xi=0$). Since $N$ is a model of $\ZF^-$, the construction can be carried out in $N$.
By elementarity,
\[
\tilde N= L_{j(\eta)}[A \cap j(\bar\kappa), p^\xi_{<\kappa}\res j(\bar\kappa)],
\]
is a valid test-model for $\tilde s = p^\xi_{<\kappa}\res j(\bar\kappa)$ and as $\tilde s \in S^*_{<\kappa}$, this string codes a predicate $A^*$ over $L_{j(\eta)}[A \cap j(\bar\kappa)]$ such that $L_{j(\eta)}[A^*]\models \Phi(r)$.
By the way the coding works and using elementarity of $\pi$, $\bar p_\delta$ also codes $\bar A^* = \pi(A^*)$ and $L_{\eta}[\bar A^*]\models \Phi(r)$.
Analogously for $\beta < \kappa$.
Finally, we prove $p^\rho_{<\kappa} \in S^*_{<\kappa}$, i.e. the requirement in definition \ref{strings_dfs} is met
(to avoid trivialities, we assume $\lambda < \kappa$).
So let $N$ be a test-model and let $\bar \kappa$ be the least Mahlo in $N$.
We assume $\bar \kappa = |p^\rho_{<\kappa}|$, as otherwise (as before) there is nothing to prove.
Let
$$X = \bigcup_{\xi<\rho} H_{|p^\xi_{<\kappa}|},$$
let
$\pi\colon X \rightarrow \bar X$ be the the collapsing map, $j=\pi^{-1}$, $\mu = X \cap \On$ and $\bar\mu = \bar X \cap \On$.
By \eqref{grow}, we have $X = H_{\bar \kappa}$, $\bar\kappa = |p^\rho_{<\kappa}|$ and $j(\bar \kappa)=\kappa$ (so $\crit(j)=\bar\kappa$).
As before, let $\eta = N\cap\On$.
Since the sequence $\{ |p^\xi_{<\kappa}| \}_{\xi <\rho}$ is definable over $\bar X$, we can argue as previously that $\eta < \mu$.
Also, as before, $\bar p_{\bar\kappa} = \bigcup_{\xi < \rho} \pi(p^\rho_{\kappa})$
is decoded from $p^\rho_{<\kappa}$ relative to $\bar X$,
and $\bar p_{\bar\kappa}\res {\bar\kappa}^{+N}$ is decoded relative to $N$.
Writing $\tilde \eta = j(\eta)$ %
it follows that the model $\tilde N= L_{\tilde\eta}[A\cap \kappa, p^\rho_\kappa \res j({\bar\kappa}^{+N})]$
is a valid test-model for $\kappa$ and so we finish the argument once more using that $p^\rho_\kappa \in S^*_\kappa$.
\end{proof}
\section{The strategic class $\D$ is dense}\label{sec:D}
The next lemma shows density of $\D$, i.e. \eqref{qc:redundant}:
in a sort of bootstrapping process using theorem \ref{jensen_qc_main} in its local version, we shall see that
the auxiliary sets
$\D^M_{[\lambda,\gamma)}(p, \vec{x})\neq \emptyset$, for larger and larger $\gamma$.
To be able to meet \eqref{coding_areas_in_supp}, we have to strengthen our inductive hypothesis and assume that densely, any Easton set can be ``covered'' by the domain of a condition.
\begin{lem}[]
Let $\gamma\in \mathbf{Card}$. For any $q \in P$ and both $\beta\geq\gamma$ and $M$ %
as in assumption \ref{M}, the following holds:
For any $\lambda \in \mathbf{Reg}$ and $p \in P$,
\begin{enumerate}
\item\label{D_dense_ih} for any $\vec{x} \in M$, there is $p' \in P$ such that $p'\leqlol p$ and $p' \in \D^M_{[\lambda,\gamma)}(p, \vec{x})$
\item\label{ext2_ind} for any Easton set $B \subseteq [\lambda,\gamma)$ there is $p' \in P$ such that $p' \leqlol p$ and $B \subseteq \dom(p'_{<\gamma})$.
\end{enumerate}
\end{lem}
\noindent
In the proof, you will notice a somewhat unexpected role-reversal of $\gamma$ and $M$:
the induction is over $\gamma$ while $q$, $\beta$ and $M$ vary freely.
Also, note how $\D^M_{[\lambda,\gamma)}(p, \vec{x})$ has been decoupled from $R$; we talk about density in $P$ and $q$ is merely a parameter.
To make life a little easier, we could proove from \eqref{D_dense_ih} the seemingly stronger fact that for any $\bar \gamma \geq \gamma$ and $r\in P$ such that $r \leq p$,
$\D^M_{[\lambda,\gamma)}(p, \vec{x})$ is $\leqlol$-dense in $P(r)^{\bar\gamma^+}$ (but we only use it in this proof, so we leave it implicit).
\begin{cor}\label{cor_qc}
$P$ is quasi-closed:
Setting $\beta= \infty$, $M=L[A]$ and $\gamma = \kappa^{+3}$, it clearly follows that
for any $p \in P$, $\lambda \in \mathbf{Reg}$ and $\vec{x} \in L[A]$, there is $q\in \D(\lambda, p, \vec{x})$ such that $q \leqlol p$.
This complements theorem \ref{jensen_qc_main} in its global form (corollary \ref{jensen_qc_main_cor}).
\end{cor}
\begin{proof}[Proof of the lemma.]
The proof is by induction on $\gamma$, so say both statements are true for all cardinals $\gamma' <\gamma \in \mathbf{Card}$.
Fix $q$, $M$, $\lambda\in \mathbf{Reg}$, $\vec{x}\in M$ and an Easton set $B$ as in the hypothesis,
and let $p \in P$ be arbitrary.
We shall find $p' \leqlol p$ satisfying both $p' \in \D^M_{[\lambda,\gamma)}(p, \vec{x})$
and $B \subseteq \dom(p'_{<\gamma})$.
For the succcessor case, assume $\gamma=\delta^+$, for $\delta \in \mathbf{Card}$.
We need to take care of \eqref{restraints_start_high} and \eqref{grow} to \eqref{coding_areas_in_supp} (in definition \ref{D}, p.~\pageref{D}) for $\delta$, and then we can finish this case quickly by induction.
So let $\iota_0$ be least above %
$|p_\delta| + \delta$ such that
any pair $b, b' \in p^*_{\delta^+}$ is disjoint above $\iota_0$ and let
\[
\iota = \sup (H_\delta \cap \On ) \cup \sup_{b \in p^*_{\delta^+}} (\min (b\setminus\iota_0))_1 \cup \sup B
\]
Note that $\iota < \delta^+$ and let
\[
p^1_\delta = (p^0_\delta \conc E) \cup Z
\]
where $E \subseteq \delta$ codes the relation $\in\res\iota\times\iota$ in some recursive way and where
\begin{gather*}
Z \colon [ |p^0_\delta| + \delta, \iota) \rightarrow 2\\
Z(\zeta)=\begin{cases}
1 & \text{ if $\zeta = (\langle \xi, |p^0_\delta|\rangle )_0$ for some $\xi \in [\delta, |p^0_\delta|)$,}\\
1 & \text{ if $\zeta = (\xi)_1$ for some $\xi \in b\setminus \iota_0$ and $b \in p^*_{\delta+}$,}\\
0 &\text{ otherwise.}\\
\end{cases}
\end{gather*}
Moreover, let $\eta = \rho(p^*_\delta) \cup \sup(U\cap\delta)\cup \lVert p^*_\delta \rVert\cup\lambda$ and let
\[
(p^1)^*_\delta = (p^0)^*_\delta \cup \{ b^{p_\delta \res\nu} \setminus \eta \setdef \nu \in H_{<\delta}\} \cup \lambda.
\]
If $\delta\in \mathbf{Sing}$, we must find $p^1\res\delta$ so that it exactly codes $p^1_\delta$,
using the extendibility lemma \ref{ext1}.
Otherwise we can set $p^1\res\delta = p^0\res\delta$.
It is clear that $p^1$ satisfies \eqref{restraints} and also requirement \eqref{restraints_start_high} in definition of $\D$.
Also, by the choice of $Z$ and since $\iota$ was chosen large enough, $p^1$ satisfies \eqref{grow}, \eqref{A_coding} and \eqref{succ_coding}.
It is easy to see that $p^1_\delta \in S^*_\delta$:
by choice of $E$,
$p^1_\delta$ collapses the size of $\iota$ to $\delta$ over every $\ZF^-$ model.\footnote{Of course, much less than $\ZF^-$ is needed here.}
Thus any test model for $p^1$ is a test model for $p^0$ and we are done.
We must also arrange \eqref{coding_areas_in_supp} for $\delta$.
For this we made the inductive assumption that \eqref{ext2_ind} holds below $\delta$.
Let $\gamma'= \sup(\supp(p)\cap\delta)$ and let
\[
B'= \bigcup \{ b \cap \gamma' \setdef b\in p^*_{\delta}\setminus \On\}.
\]
By \eqref{ext2_ind} we can find $p^2 \leqlol p^1$ such that
$B' \cup (B \cap \delta) \subseteq \dom(p^2_{<\delta})$.
By induction hypothesis, we can find $p' \leqlol p^{2}$ such that
$$p' \in \D^M_{[\lambda,\delta)}(p^2,\vec{x})\subseteq \D^M_{[\lambda,\delta)}(p,\vec{x}).$$
It follows that $p' \in \D^M_{[\lambda,\gamma)}(p,\vec{x})$ and $B \subseteq \dom(p'_{<\gamma})$, finishing the successor case.
Let's say $\gamma$ is a limit ordinal.
We can assume $\gamma \in \mathbf{Sing}$, for otherwise $\supp(p)\cap\gamma$ and $B$ are both bounded below $\gamma$ and we can simply use the induction hypothesis.
So let $ (\beta'_\xi)_{\xi<\rho}$ be the increasing enumeration of a club in $\beta' = \gamma$, $\rho = \cof(\beta')$ and $\beta'_0 > \rho$.
We can assume without loss of generality that $|p_{<\kappa}| \geq \beta^+$ (by \ref{ext_mahlo}, extendibility for the Mahlo coding).
Let $R'= P(p)^{\beta'^+}$, $M' = \mathcal{A}^{p_{\beta'^+}}=\mathcal{A}(A_p)^{p_{\beta'^+}}$
and let $p^0=p\res\beta'^++1$, observing $p^0 \in R'$.
For $\xi < \rho$, let
$$D_\xi = \D^{M}_{[\lambda, \beta'_\xi)}(p,\vec{x})\cap L_{\beta'^+_\xi}[A]$$
and let
$\vec{x}' = \{ \lambda, p^0, (\beta'_\xi)_{\xi<\rho)}, (D_\xi)_{\xi<\rho}, B\}$, noting that $\vec{x}' \in M'$.
The point here is that for any $p' \in P$,
\begin{equation*}
p' \in \D^{M}_{[\lambda, \beta'_\xi)}(p,\vec{x}) \iff p'\res\beta'_\xi \in D_\xi,
\end{equation*}
so that we can talk about $\D^{M}_{[\lambda, \beta'_\xi)}(p,\vec{x})$ inside $M'$.
We shall now use theorem \ref{jensen_qc_main} for $M'$ and $R'$ to construct a sequence whose greatest lower bound will be the desired condition.
Let $X^0$ be least such that $X^0 \prec_1 M'$, $x'\in X^0$ and $X^0 \in M'$;
we can find such $X^0$ by the fact that $\cof(M'\cap\On) = \beta'^+$. %
We now find a sequence $\bar w = (w^\xi)_{\xi < \rho}$ such that for each $\xi < \rho$,
$w^\xi = (X^\xi, p^\xi, \bar w\res\xi)$
and
\begin{gather*}
w^\xi \in M'\\
X_{\xi} = h^{M'}_{\Sigma_1}(\{ p^\xi, \bar w\res\xi\}),
\end{gather*}
and if $\xi$ is limit, then $p^\xi$ is a greatest lower bound of $\bar p\res\xi+1$ in $R'$.
Note that this implies $(X^\nu)_{\nu<\xi}\in X_{\xi}$ (also note the sequence $(X_\xi)_{\xi<\rho}$ is not continuos).
To complete the definition of $\bar w$, we must specify how to construct $p^{\xi+1}$ in the successor case.
By induction, assume we have built such a sequence $\bar w\res\xi+1$.
Now let $p^{\xi+1} \in R'$ be least such that
\begin{itemize}
\item $p^{\xi+1} \res \beta'_\xi \in D_{\xi}$---equivalently, $p^{\xi+1}\in \D^{M}_{[\lambda, \beta'_\xi)}(p,\vec{x})$,
\item $p^{\xi+1} \in \D^{M'}_{[\lambda, {\beta'_\xi})}(p^{\xi},\{\vec{x}', \bar w \xi+1\})$ (note this is $\Pi^T_1(\{\vec{x}', \bar w \xi+1, p^\xi, \xi \})$ in $M'$)
\item $B \cap \beta'_\xi\subseteq \dom(p^{\xi+1})_{<\beta'^+_\xi}$,
\item the clauses in \eqref{dense_top} of theorem \ref{jensen_qc_main}
hold for $p^{\xi+1}$ with $\beta$, $\beta_\xi$, $M$, $\vec{x}$ in \eqref{dense_top} replaced by $\beta'$, $\beta'_\xi$, $M'$, and $\vec{x}'$ (note this is $\Pi^T_1(\{\vec{x}', \bar w \xi+1, p^\xi, \xi \})$ in $M'$ as well).
\end{itemize}
The last point can be arranged as in the first part of the argument in the successor case (using extendibility, \ref{ext1}).
The first point can be arranged as the induction hypothesis implies that $D_\xi$ is dense in $R'$.\footnote{Literally, the induction hypothesis talks about $P$, not $R'$. In fact, we have that $\D^{M}_{[\lambda, \gamma)}(p,\vec{x})$ is dense in $P$ below $p$ if and only if it has non-empty intersection with $P(q)^{\bar\gamma^+}$ for any $q \leq p$ and any $\bar\gamma \geq \gamma$.}
If in doubt, here is a pedestrian proof: since $p^\xi \in R'=P(p)^{\beta'^+}$, we know $p^\xi\cdot p\neq0$; so there is $p' \leqlol p^\xi\cdot p$ such that
$p' \in \D^{M}_{[\lambda, \beta'_\xi)}(p^\xi\cdot p,\vec{x})$;
But then
since trivially $\D^{M}_{[\lambda, \beta'_\xi)}(p^\xi\cdot p,\vec{x}) \subseteq \D^{M}_{[\lambda, \beta'_\xi)}(p,\vec{x})$,
$$p' \res\beta'^+_\xi \cup p^\xi \res [\beta'^+_\xi, \beta') \in \D^{M}_{[\lambda, \beta'_\xi)}(p,\vec{x}) \cap R'.$$
Similar arguments work for the remaining two points.
Note that the conjunction of theses items can be expressed by a $\Pi^T_1(\vec{x}'\cup\{ p^\xi, \xi \})$ formula (say $\Phi(p^{\xi+1}, \vec{x}',\bar w\res\xi+1)$) inside $M'$.
The sequence $\bar w$ is well defined since
by the following fact and theorem \ref{jensen_qc_main},
the sequence $\bar p \res \rho'$ has a greatest lower bound $p^{\rho'}$ when $\rho' \leq \rho$ is a limit ordinal.
Finally, we have that $p^\rho \in \bigcap_{\xi < \rho} D_\xi = \D^M_{[\lambda,\gamma)}(p,\vec{x})$ and $B \subseteq \dom(p^\rho_{<\gamma})$.
Since also $p^\rho \in R'= P(p)^{\gamma^+}$,
$p' = (p^\rho \cdot p) $ is the desired condition.
It remains to proove:
\begin{fct}
The sequence $\bar w = (w^\xi)_{\xi<\rho'} = (X^\xi, p^\xi)_{\xi<\rho'}$ is a $(\lambda, x')$-canonical witness for $\bar p \res\rho' = (p^\xi)_{\xi<\rho'}$ inside $M'$
\end{fct}
\begin{proof}[Sketch of proof.]
The idea is that $w^\xi$ is the unique $x=(X,p,\bar w')$ such that
\begin{equation}\label{canonical_witness_D}
\begin{gathered}
X = h^{M'}_{\Sigma_1}(\{ p, \bar w^*\})\text{ and}\\
X\models p \text{ is $\leq_{M'}$-least such that }\Phi(p,\vec{x}',\bar w)
\end{gathered}
\end{equation}
where $\Phi$ comes from the inductive definition of $\bar p$ discussed above,
and $\bar w^* = \bar w \res \xi$.
We leave it to the reader to check that \eqref{canonical_witness_D} can be expressed by a $\Pi^T_1(\vec{x}')$-formula $\Psi(X,p,\bar w')$.
Now it is easy to see the fact holds:
$x = w^\xi$ if and only if $x=(X,p,\bar w^*)$, $\Psi(X,p,\bar w^*)$ holds and $\bar w^*$ is a sequence of length $\xi$ such that for each $\nu < \xi$, we have $\Psi( \bar w^*(\nu)_0 ,\bar w^*(\nu)_1, \bar w^*\res\nu)$.
This is obviously $\Pi^T_1(\vec{x}')$ if $\Psi$ is.
See fact \ref{density reduction}, especially lemma \ref{adequate} for a similar (but harder) argument.
\end{proof}
\end{proof}
\section{Stratification of $P(A_0)$}\label{sec_coding_strat}
We now define the stratification system for $P(A_0)$.
Remember $\leqlol$ and $\D$ was defined in \ref{D}.
\begin{dfn}
\begin{enumerate}
\item Let $q \lequpl p$ just if $q\res [\lambda, \infty) \leq p \res [\lambda, \infty) $ in $P(A_0)$.
\item For any $p \in P(A_0)$, writing $\rho$ for $\rho(p^*_\lambda) \cup \sup\dom(p_{<\lambda})$, let
$\Cl(p)=\{ (p_{<\lambda}, \rho, B^{\rho}_p) \}$.
\end{enumerate}
\end{dfn}
We remind the reader that we take the singleton on the right-hand side in the above equation only to satisfy the abstract definition of stratification, which allows for a ``multifunction'' $\Cl$, as this is necessary for iterations.
A more straightforward definition would be $\Cl(p)=\{ p_{<\lambda} \}$
for all $p \in P(A_0)$.
This would work with the more traditional definition of $\leq$ in \cite{friedman:codingbook}.
The problem is that we had to introduce $\rho(p^*_\lambda)$ for inaccessible $\lambda$.
Recall that $\leq$ was defined so that ``new restraints'' (those in $q^*_\lambda$ for $q\leq p$) may only differ from old ones (those already in $p^*_\lambda$) \emph{above} $\rho(p^*_\lambda)$.
This means that with the more straightforward definition, in \eqref{centering}, when $\Cl(p)\cap\Cl(q)\neq\emptyset$ and $p \lequpl q$, and we want to form $q\cdot p$, the natural candidate (the point-wise union) might not actually lie below $p$ and $q$: not if any ``new'' restraints disagree with old restraints below $\rho(p^*_\lambda)$ (or $\rho(q^*_\lambda)$, respectively).
Thus, it is natural to make $\rho(p^*_\lambda)$ and the restraints below it part of $\Cl$.
Having checked quasi-closure, we invite the reader to check the rest of stratification, all of which is outright trivial.
\chapter{Projective measure without Baire}\label{sec:main}
We begin with the assumption $V=L$ and fix $\kappa$, the least Mahlo.
The first step is to force with $\bar T =\prod_{\xi<\kappa} T(\xi)$, the (full) product $\kappa$-many independent, $\kappa^+$-closed $\kappa^{++}$-Suslin trees.
In fact, $\bar T$ itself has the $\kappa^{++}$-cc, by construction.
This adds a sequence of branches $\bar B = (B(\xi))_{\xi<\kappa}$, where $B(\xi)$ denotes the branch through $T(\xi)$.
As a notational convenience, we often assume the sequence of trees (resp. branches) is indexed by elements of $$J=\kappa\times {}^{<\kappa}2 \times \omega\times 2 \times 2$$
rather than by ordinals in $\kappa$, that is as $B(\xi, s,n,i,j)$ and $T(\xi, s,n,i,j)$ for $\xi<\kappa$, $s \in{}^{<\kappa}2$, $n \in \nat$ and $i,j\in\{0,1\}$.
Since $\bar T$ is $\kappa^{++}$-cc, it is also $\kappa^{++}$-distributive, whence $W=L[\bar B]$ has the same cardinals and the same subsets of $\kappa^+$ as $L$ and the $\GCH$ still holds in $W$. In particular, $L$ and $W$ have the same reals.
We now define $(P_\xi)_{\xi\leq\kappa}$, by induction on $\xi$.
We start with $P_0 = \bar T$ and identify $\bar B$ with $G_0$.
Working in $W$, setting $P^*_0 = \emptyset$ and interpreting the following definition verbatim in $W$, we obtain $(P^*_\xi)_{\xi \leq\kappa}$.
The only formal difference being limit and amalgamation stages,
there is a very canonical equivalence of $\bar T * \dot P^*_\xi$ with $P_\xi$.
Note though that the proof shows $(P_{\xi+1}, P_\xi)$ to be a stratified extension on $[\lambda_\xi,\kappa]$, but the same \emph{is not directly obvious} for $(P^*_{\xi+1}, P^*_\xi)$ in $W$.
We shall call $G_\xi$ the $P_\xi$ generic over $L$ and $G^*_\xi$ the $(P^*_\xi)$-generic over $W$.
We construct this iteration to deal with the following tasks:
\begin{description}
\item[Task 1]\label{task:cohens} Add a set of reals $\Gamma^0$ such that $P_\kappa$ forces that the Baire-property fails for $\Gamma^0$;
\item[Task 2]\label{task:code} For each real $r$ added by $P_\kappa$, make sure that $P_\kappa$ forces $r \in \Gamma^0 \iff \Psi(r,0) \iff \neg \Psi(r,1)$,
where $\Psi(x,y)$ is $\Sigma^1_3$.
\item[Task 3]\label{task:amalgamation} Make sure every projective set of reals is Lebesgue-measurable in the extension by $P_\kappa$.
\item[Task 4]\label{task:collapse} To make the construction more uniform, we force with a Levy-collapse at certain stages.
\end{description}
We force with the Levy-collapse for two reasons: firstly, when we amalgamate, and at limits, whether we collapse the continuum depends on factors beyond our control. So we always make sure we collapse the current continuum at the next stage. Secondly, for the purpose of task 2 (which involves Jensen coding), we want to make sure CH holds all the time.
Task 2 requires the sophisticated technique of Jensen coding, which made its first appearance in \cite{bjw:82} and has since undergone a long development culminating in \cite{friedman:codingbook}.
We will make the real $r$ (along with information about its membership in $\Gamma$) definable by coding a subset of our set of branches $\bar B$ by a real $s$, where $s$ is generic for Jensen coding.
Say we have iterated for $\xi$ steps and are in $L[G_\xi]$.
For now, let's call the set of branches we ``code'' at the $\xi+1$-th step $\bar B^- =\{ B(\zeta) \setdef \zeta \in I \}$, where $I \subseteq \kappa$ of size $\kappa$.
Why do we use a subset of size $\kappa$? Since a real carries only a countable amount of information, one would think that a countable set of branches would suffice. The point here is that the automorphisms that arise from amalgamation (task 3) will make any such coding ``unreadable'' (see section \ref{sec:preserving:coding}). This is not surprising since by \cite{shelah:amalgamation}, a definable well-ordering of a set of reals of length $\omega_1$ yields a definable non-measurable set. In fact, if the present construction is altered so that each real is coded using a block of trees of length $\omega$, we must fail since this would add such a well-order (since the set of trees is of course well-ordered).
It is also easy to see how such a coding is made unreadable: if the trivial condition forces that the real $\dot r$ is coded using the branches indexed by the block $[\xi, \xi+\omega)$,
for any automorphism $\Phi$, also $\Phi(\dot r)$ would be coded on the same block. See section \ref{sec:preserving:coding} for the solution.
We shall pick a set $A_\xi \subseteq \kappa^+$ such that $(\Hhier_\alpha)^{L[A_\xi]}=L_\alpha[A_\xi]$ for every cardinal $\alpha \leq \kappa$ and $\{ B(\xi) \setdef \xi \in I\}$ is definable in some simple recursive fashion from $A_\xi$ (see below).
Note $L[A_\xi]$ will be a proper sub-model of $L[G_\xi]$ (since we don't want to code \emph{all of} $\bar B$).
We shall then force with $P(A_\xi)$ of section \ref{sec:coding} \emph{as defined in the model $L[A_\xi]$} to obtain a real $s$ such that $A_\xi \in L[s]$ and moreover the following is true in the extension:
\begin{eqpar}\label{david}
for all $\alpha, \beta<\kappa$ \emph{ if } $L_\beta[s]$ is a model of $\ZF^-$ and of
``$\alpha$ is the least Mahlo and $\alpha^{++}$ exists'' \emph{ then:}\\
$I\cap\alpha \in L_\beta$ and $L_\beta[s] \models$``$\forall \xi \in I \cap \alpha$ $\bar T^\beta(\xi)$ has a branch,''
\end{eqpar}
\noindent
where $ \bar T^\beta$ denotes the outcome of the construction of $\bar T$ carried out in $L_\beta$.
This is the vital use Jensen coding with localization (also called ``David's trick'' or \emph{killing universes}).
In order to make lemma \ref{it:reals:are:caught} below go through, we have modified it in several ways (the use of $\diamondsuit$, $\rho(p^*_\delta)$ and most notably, we use Easton support).
Note that forcing over $L[A_\xi]$, a proper sub-model of $L[G_\xi]$, means we shall have to argue that this step represents a stratified extension.
There are $4$ types of forcing involved, so we fix a simple and convenient partition $E^0, \hdots, E^3$ of $\kappa$: let $E^n$, for $0\leq n\leq 3$, denote the set of ordinals $\xi<\kappa$ such that for some limit ordinal $\eta$ and $k \in \omega$, $\xi=\eta+k$ and $k \equiv n \pmod 4$. For an ordinal $\xi<\kappa$, let $E^n(\xi)$ denote the $\xi$-th element of $E^n$.
Also fix, for each $\rho<\kappa$, an increasing sequence $\bar \alpha_\rho = (\alpha^\zeta_\rho)_{\zeta<\kappa}$ of ordinals $>\rho$ cofinal in $\kappa$: we let $\alpha^\zeta_\rho = G(\zeta,\rho)$, where $G$ is the G\"{o}del pairing function.
As we have to tackle certain tasks for every real of the extension, our definition will make use of two book-keeping devices,
$\bar s=(\dot s_\xi)_{\xi<\kappa}$ and $\bar r=(\bar\iota(\xi),\dot r^0_\xi, \dot r^1_\xi)_{\xi<\kappa}$.
We define $\bar s$ to list all reals which end up in the complement of $\Gamma^0$, in order to handle task 2 for each of these. To make sure all projective sets of reals are measurable (task 3) we ask that for each $\iota<\kappa$, the set of $\dot r^0_\xi, \dot r^1_\xi $ such that $\bar \iota(\xi)=\iota$ list all the pairs of reals in $L[G_\kappa]$ which are random over $L^{P_{\iota}}$.
We also ask that each pair $\dot r^0_\xi, \dot r^1_\xi $ be $\lambda$-reduced over $P_{\bar \iota(\xi)}$ for some large enough $\lambda$.
We shall first proceed with the definition of the iteration, and after that argue that a book-keeping with the requisite properties can be defined at the same time.
We define a sequence $\bar \lambda= (\lambda_\xi)_{\xi\leq\kappa}$ by induction, so that for each $\xi \leq \kappa$, $P_\xi$ will be stratified on $[\lambda_\xi,\kappa]$.
We let $\lambda_0 = \omega$. For limit $\xi$, let $\lambda_\xi$ be the minimum of $\mathbf{Reg} \setminus \bigcup_{\nu<\xi}\lambda_\nu$. For successors, define
\begin{equation}\label{def:lambda:xi}
\lambda_{\xi+1} =\begin{cases}
(\lambda_{\xi})^+& \text{if $\xi \in E^0$ or $\exists \rho$ s.t. $\xi=E^3(\alpha^0_\rho)$},\\
\lambda_{\xi}&\text{otherwise.}
\end{cases}
\end{equation}
For the readers orientation, be aware we will always have that $P_\xi$ collapses $\lambda_\xi$, except when $\xi$ is limit or $P_\xi$ is an amalgamation of type-1.
In those cases $P_\xi$ may or may not preserve $\lambda_\xi$.
The continuum of $L[G_\xi]$ is always at most $(\lambda_\xi)^+$
so $\CH$ always holds in $L[G_\xi]$, except when $P_\xi$ does not collapse $\lambda_\xi$.
When we define $P_\xi$, we will also define some auxiliary sets $D_\xi$. At stages where we do type-1 amalgamation, they are similar to $\Dam$ of section \ref{am}; at other stages $D_\xi=P_\xi$.
At all limit stages $\xi \leq \kappa$, we define $P_\xi$ to be the $\bar \lambda$-diagonal support limit of the iteration up to that point.
We generally write $B_\xi = \ro(P_\xi)$ for $\xi\leq \kappa$. In the inductive definition of the iteration, we also define
\begin{enumerate}
\item A sequence of sets $G^o_\xi$, for $\xi<\kappa$; these arise because at coding stages, the next forcing is not taken from the natural model $L[G_\xi]$ but from a smaller model $L[\bar B^-][G^o_\xi]$ (of course, since we do not want to code \emph{all of }$\bar B$).
The $G^o_\xi$ help to ``integrate out the $\bar T$-part'' and to show that the $(P_\xi,P_{\xi+1})$ is a stratified extension.
Note that we will have $\pset(\omega)^{L[G\res\xi]}= \pset(\omega)^{L[G^o_\xi]}$. %
\footnote{Moreover, $L[G^o_\xi]= L[G^*_\xi] \subsetneq L[\bar B][G^*_\xi] = L[G_\xi]$; we only introduce the sets $G^o_\xi$ because $L[G^o_\xi]$ and $L[\bar B^-, G^o_\xi]$ are less ambiguous than $L[G^*_\xi]$ and $L[\bar B^-, G^*_\xi]$.}
\item A sequence $\bar c=(\dot c_\xi)_{\xi<\kappa}$ of names for reals where each $\dot c_\xi$ is Cohen over $L[G\res E^0(\xi)]$.
\item A sequence $(C_\xi)_{\xi<\kappa}$ of so-called coding areas, where each $C_\xi \in {}^\kappa 2$ is generic over
$L[G\res E^1(\xi)]$ but has constructible initial segments.
\item Maps $\Phi^\zeta_\rho$, for $\rho,\zeta < \kappa$, where $\Phi^{\bar \zeta}_\rho$ extends $\Phi^\zeta_\rho$ for $\zeta < \bar \zeta$.
Finally, $\bigcup_{\zeta<\kappa}\Phi^\zeta_\rho$ uniquely determines an automorphism $\Phi_\rho$ of $\ro(P_\kappa)$ such that $\Phi_\rho (\dot r^0_\rho)=\dot r^1_\rho$ and $\Phi_\rho \res P_{\bar\iota(\rho)}$ is the identity. For a more uniform notation, we also write $\Phi^\kappa_\rho$ for $\Phi_\rho$.
We call any stage of the iteration $P_{\xi+1}$ such that $\xi=E^3(\alpha^\zeta_\rho)$ for some $\zeta < \kappa$ and thus such that $\Phi^\kappa_\rho$ extends $\Phi^\zeta_\rho$, \emph{ associated to $\Phi_\rho$}.
\end{enumerate}
\section{The successor stage of the iteration}\label{def:it:succ:stage}
For the successor stage, assume by induction that we have already defined $P_\xi$ for $\nu<\xi$ and $\dot r^0_\nu, \dot r^1_\nu$, $\dot s_\nu$ for $\nu \leq \xi$.
Fix $k$ and $\eta$ such that $\xi=E^k(\eta)$.
In any case except when $\xi \in E^3(\alpha^0_\rho)$, for some $\rho<\kappa$---that is, $\xi$ is a stage where we do type-1 amalgamation---we let $D_\xi=P_\xi$.
Let $G_\xi$ denote a generic for $P_\xi$.
We may assume we have already defined $G^o_\xi$, letting $G^o_0=\emptyset$.
\begin{description}
\item[$k=0$]
At this stage we collapse the continuum of $L[G_\xi]$ and make sure the GCH holds (task 4).
Let $P_{\xi+1}= P_\xi*\dot Q_\xi$,
where
\[ \forces_\xi \dot Q_\xi=\Coll(\omega,\lambda_{\xi+1}),\]
and let $G(\xi)$ be the $\Coll(\omega,\lambda_{\xi+1})$-generic.
Observe that $\lambda_{\xi+1} =(\lambda_\xi)^+\geq2^\omega$ in $L[G_\xi]$, so we can pick a $P_{\xi+1}$-name which is fully Cohen over $P_\xi$, and define $\dot c_\eta$ to be this name.
In $L[G\res\xi+1]$, simply let $G^o_{\xi+1} = G^o_\xi \times G(\xi)$.
\item[$k=1$] Let $P_{\xi+1}= P_\xi\times (\Add(\kappa))^L$.
We denote by $G(\xi)$ the generic and by $C_\eta$ the new subset of $\kappa$ it represents (and let $\dot C_\eta$ denote its canonical $P_{\xi+1}$-name). This will be the generic ``coding area'' used in the next step.
Again let $G^o_{\xi+1} = G^o_\xi \times G(\xi)$.
\item[$k=2$] We take care of task 2, making sure $\Psi(c,j)$, holds for some real $c$ given to us by book-keeping ($j=0,1$ indicates whether $c \in \Gamma^0$).
If $\eta$ is a limit or $\eta=0$, let $c$ denote $\dot c_\eta^{G_\xi}$ (the Cohen real defined at stage $E^0(\eta)$), and let $j=0$ (indicating that $c$ will be in $\Gamma^0$).
If $\eta$ is a successor, let $c$ denote $\dot s_{\eta-1}^{G_\xi}$, and let $j=1$ (indicating that $c$ will not be in $\Gamma^0$).
We wish to code a branch through $T(s,n,i,j)$ if and only if $s$ is an initial segment of $C=C_\eta$ (the coding area from the previous step) and $c(n)=i$. That is, we let
\[B(\eta, C,c,j)=\{ B(\eta, s,n,i,j) \setdef s \is C, c(n)=i \}\]
be the set of branches to code, and represent it in a $\Delta_1$ way as a subset of $\kappa^{++}$:
\[\bar B^-=\{\#(\eta, s,n,i,j,t) \setdef s \is C, c(n)=i, t \in B(s,n,i,j) \}\]
where $\#x$ denotes the order-type of $x$ in the well-ordering of $L[\bar B^-][G^o_\xi]$.
Working in $L[\bar B^-][G^o_\xi]$, define $A_\xi \subseteq \kappa^{++}$ as in \ref{sec:full:setting} so that
$L[\bar B^-][G^o_\xi] = L[A_\xi]$ and let $P(A_\xi)$ be the forcing discussed in section \ref{sec:coding}.
Finally, define
\[ Q_\xi = P(A_\xi)^{L[A_\xi]},\]
and let $\dot Q_\xi$ be a $P_\xi$-name for this forcing.
Observe that it takes an argument to show we get a stratified extension; $\dot Q_\xi^{G_\xi}$ isn't obviously stratified (on any interval) in $L[G_\xi]$.
Letting $G(\xi)$ denote the generic, let $G^o_{\xi+1} =G^o_{\xi} \times \{ p_{<\kappa^{++}} \setdef p \in G(\xi) \}$.
\item[$k=3$]
Say $\eta=\alpha^\zeta_\rho$.
We first treat the case where $\zeta=0$:
By induction the book-keeping device $\bar r$ gives us $\bar r(\rho)=(\bar\iota(\rho), \dot r^0_\rho, \dot r^1_\rho)$, where $\bar\iota(\rho)<\xi$ (in fact, $<\rho$) and the pair of names reals $\dot r^0_\rho,\dot r^1_\rho$ is fully random over $L^{P_{\bar\iota(\rho)}}$ and $\lambda_\xi$-reduced over $P_{\bar\iota(\rho)}$.
Let $f$ be the automorphism of the complete Boolean algebras generated by $\dot r^0_\rho$ and $\dot r^1_\rho$ in $B_\xi$ and let $P_{\xi+1}$ be the type-1 amalgamation of $P_\xi$ over $f$ and $P_{\bar\iota(\rho)}$:
\[ P_{\xi+1}= \am(P_{\bar\iota(\rho)},P_\xi,f,\lambda_\xi). \]
Set $D_\xi=\D(P_{\bar\iota(\rho)},P_\xi,f,\lambda_\xi)$.
The resulting automorphism of $P_{\xi+1}$ we denote by $\Phi^0_\rho$.
Observe that, in general, this automorphism need not extend to an automorphism of $B_\kappa$.
Also observe that by induction and theorem \ref{thm:am:s:ext} $(P_\xi,\am(P_{\bar\iota(\rho)},P_\xi,f,\lambda_\xi))$ will be a stratified extension above $\lambda_{\xi+1}$.
In the second case, when $\eta=\alpha^\zeta_\rho$ and $\zeta > 0$, we make sure $\Phi^0_\rho$ is extended by an automorphism of $P_{\xi+1}$. So we let
\[ P_{\xi+1}= \simpleram(\dom(\Phi),P_\xi, \Phi), \]
where $\Phi$ is (an extension of) $\Phi^0_\rho$, constructed at an earlier stage of the iteration:
If $\zeta$ is a successor ordinal, at a previous stage $E^3(\alpha_\rho^{\zeta-1})$, we defined $\Phi_\rho^{\zeta-1}$ extending $\Phi^0_\rho$. Set $\Phi=\Phi_\rho^{\zeta-1}$.
If $\zeta$ is a limit, we have a sequence $(\Phi_\rho^\nu)_{\nu<\zeta}$, forming an increasing chain, and all extending $\Phi^0_\rho$. Letting $\delta=\bigcup_{\nu<\zeta}\alpha_\rho^\nu<\xi$, there is a unique automorphism of $P_\delta$, extending each of them. Let $\Phi$ be this automorphism.
The resulting automorphism of $P_{\xi+1}$ we denote by $\Phi_\rho^\zeta$.
In both cases we say \emph{$\xi+1$ or $P_{\xi+1}$ is an amalgamation stage associated to $\Phi^\zeta_\rho$}.
$G^o_{\xi+1}$ is defined to be the $\Int$-sequence of the sets defined like $G^o_\xi$ in each of the $\Int$-many components of the amalgamation.
\end{description}
We see that in a very concrete sense, we can identify $G^*\res\xi$, the $\dot P_\xi^{\bar B}$-generic over $L[\bar B]$ with $G^o_\xi$.
\begin{lem}\label{iteration:prop}
\begin{enumerate}
\item\label{P:strat:ext} For all $\xi<\kappa$, $(P_\xi, P_{\xi+1})$ is a stratified extension on $[\lambda_{\xi+1},\kappa]$.
\item \label{it:strat} For all $\xi\leq\kappa$, $\bar P_\xi$ is stratified on $[\lambda_\xi, \kappa]$.
\item \label{P:cont:small} For all $\xi$, $ P_\xi \forces 2^\omega \leq (\lambda_\xi)^+$ over $W$.
\item \label{P:coll} If $\xi$ is not of the form $E^3(\alpha^0_\rho)$ for some $\rho$, i.e. if $P_{\xi+1}$ is not a type-1 amalgamation, we have that $ P_{\xi+1} \forces \card{\lambda_{\xi+1}} = \omega$ over $W$ and $P_{\xi+1}\forces\GCH$ over $W$. \label{it:CH}
\item \label{P:cent} In $W$, for each $\xi \leq \kappa$, $\dot P_\xi^{\bar B}$ is $\kappa^+$-centered.
\end{enumerate}
\end{lem}
Observe it follows that $P_\xi$ preserves all cardinals greater than $\lambda_\xi$, for $\xi<\kappa$.
We will show later that it also preserves Mahlo-ness of $\kappa$ and that $P_\kappa$ preserves $\kappa$.
\begin{proof}
When $\xi \not \in E^2$, the first item holds by induction and lemmas \ref{stratified:comp:implies:ext} (composition), \ref{products:ext} (products) and theorem \ref{thm:am:s:ext} (amalgamation);
the \pre stratification systems are the ones stemming from the constructions in lemmas \ref{stratified:comp:implies:ext}, \ref{products:ext} and theorem \ref{thm:am:s:ext} of course.
It remains to show $(P_\xi, P_{\xi+1})$ is a a stratified extension on $[\lambda_{\xi+1},\kappa]$ when $\xi$ is a coding stage.
We let $q \in \D_{\xi+1}(\lambda, p, x)$
if and only if $q\res\xi \in \D_\xi(\lambda, p\res\xi, x)$,
$q \res\xi \forces q(\xi) \in \dot \D_\xi(\lambda, p(\xi), x)^{L[A_\xi]}$
and both $p(\xi)_{<\kappa^+}$ and $p(\xi)^*_{\kappa^+}$ are $\kappa$-chromatic names (with $A_\xi$ defined above in the definition of forcing at coding stages);
observe this makes sense since we may view $p(\xi)_{<\kappa^+}$ and $p(\xi)^*_{\kappa^+}$ as functions with domain $\kappa$, and since by induction $P_\xi$ is stratified at $\kappa$.
The proof of the density property of $\D$ goes through verbatim (since we only have to look at $\lambda\leq \kappa$).
The rest $\Cl$, $\lequpl$, $\leqlol$ can be defined in a straightforward manner as in the proof for composition of stratified forcings \ref{stratified:composition}.
To see that this is a stratified extension, let $\bar p=p^\nu_{\nu<\rho}$ be a $(\lambda, \bar \lambda,x)$-adequate sequence of conditions in $P_{\xi+1}$ with $(\bar\lambda,x)$-canonical witness $\bar w$ and let $q\res \xi$ be a greatest lower bound of $(p^\nu\res\xi)_{\nu<\rho}$.
It suffices to show that $q$ forces that $(p^\nu(\xi))_{\nu<\rho}$ is $(\bar\lambda,x)$-adequate in $L[A_\xi]=L[\bar B^-, G^o_\xi]$ (where $\bar B^-$ is defined as above in the definition of forcing at coding stages).
That $\bar w$ is a strategic guide is clear by definition of $\D$ and the usual argument for composition; so we check that it is a canonical witness.
We check $\bar w$ is $\Pi^{A_\xi}_1(\{ x\} \cup\bar\lambda)$; this is clear (as in \ref{stratified:composition}), since it is $\Pi_1(\{ x\} \cup\bar\lambda)$ in $L$ and thus also in $L[A_\xi]$.
Since each $P_\xi$-name $q^\nu(\xi)$ is $\kappa$-reduced, it's interpretation can be found
inside $L[G^0_\xi]$ and hence in $L[A_\xi]$.
In fact, the (partial) function assigning to $q^\nu(\xi)$ its interpretation is $\Sigma_1^{A_\xi}(\{ x\} \cup\bar\lambda)$ provided $\kappa^{+++}$ is among the parameters in $x$---in fact, the existential quantifier can be bounded by $L_{\kappa^{+++}}[A_\xi]$ since we only have to find a $\kappa$-spectrum witnessing the interpretation.
Thus $q^\nu(\xi)^{G_\xi}$ can be obtained by application of a $\Sigma^{A_\xi}_1(x)$ (partial) function from the name $q^\nu(\xi)$, which can in turn be obtained by application of recursive function from $q^\nu$, which can be obtained by applying a ${\Sigma_1(\{ x\} \cup\bar\lambda)}^L$ (partial) function to $w^\nu$.
Thus, $q^\nu(\xi)^{G_\xi}$ can be obtained by applying a $\Sigma^{A_\xi}(\{ x\} \cup\bar\lambda)$-function to $w^\nu$ in $L[A_\xi]$.
This shows that $\bar w$ is a canonical witness.
The second item holds by theorem \ref{thm:it:strat}.
The third one is a corollary of the previous ones and lemma \ref{density reduction}and the next follows since we collapse $\lambda_\xi$ at the right stage.
Lastly, the centeredness follows since stratification at $\kappa$ allows us to define $\C^{\kappa^+}$;
If $(t,p), (s,q) \in \bar T * \dot P_\xi$ are such that $t$ and $q$ are compatible and $\C^{\kappa^+}(t,p) \cap \C^{\kappa^+}(s,q) \neq \emptyset$ then clearly $s\cdot t$ forces that $p \cdot q \neq 0$.
This implies that cardinals above $\kappa$ are preserved, which is by the way not essential.
\end{proof}
\begin{rem}
Note that we can also use $\Cl$ of the following explicit form.
The abstract definition with guessing systems serves the purpose of proving an abstract iteration theorem.
and may help to aid the intuition of the reader.
The two forms are equivalent for successor $\lambda$, modulo a recursive translation.
For $\lambda\in\mathbf{Inacc}$ this equivalence holds on a set which is dense in the same sense as $\dom(\Cl)$.
Let $\lambda \leq\kappa$ be regular and let $\sigma <\lambda$.
First, define $D^\sigma_{\lambda} \subseteq P_\kappa$, by induction on the length of a condition.
For the successor step, say $p \in P_{\nu+1}$. We let $p \in D^\sigma_{\lambda}$ if and only if $\pi_\nu(p) \in D^\sigma_{\lambda}$ and the following hold:
\begin{enumerate}
\item in case $\nu \in E^0$ (i.e. $\dot Q_\nu$ is $\Coll(\omega,\lambda_\nu)$), we require that
$\pi_\nu(p) \forces \ran(p(\nu))\subseteq \sigma$,
\item in case $\nu \in E^2$ (i.e. $\dot Q_\nu$ is Jensen coding of $L[A_\nu]$), we require that
$\pi_\nu(p) \forces$ for all $\sup\dom(p(\nu)_{<\lambda})=\sigma$ and $\rho(p(\nu)^*_\lambda)=\sigma$, if $\lambda\in\mathbf{Inacc}$.
\item in case $\nu \in E^3$ (i.e. $P_{\nu+1}$ is an amalgamation), we require that for all $i \in \Int\setminus\{0\}$ we have $p(i)^P \in D^\sigma_{\lambda}$ (it would be redundant to require this also for $i=0$).
\end{enumerate}
For $p \in P_\nu$ where $\nu \leq\kappa$ is a limit ordinal, let $p \in D^\sigma_{\lambda}$ if and only if for all $\nu'<\nu$, $\pi_{\nu'}(p)\in D^\sigma_{\lambda}$.
Finally, define $p \in D^\Sigma_\lambda$ if and only if there is $\sigma<\lambda$ such that $p \in D^\sigma_\lambda$; if $\lambda\in\mathbf{Inacc}$, we require that $\sigma\in\mathbf{Card}$.
Also, for any $p \in D^\Sigma_\alpha$, let $\sigma^2_\lambda(p)$ be the least $\sigma<\alpha$ such that $p \in D^\sigma_{\lambda}$ and $\supp^\lambda(p)\cap\lambda\subseteq \sigma$.
The sets $\Cl$ can now be defined in a similar fashion:
they are binary relations,
\[ \Cl \subseteq \{\text{ sequences of length }\leq \kappa \} \times P_\kappa. \]
that is, for any such sequence $H$, $\Cl(H) \subseteq P_\kappa$.
So let $\lambda\leq \kappa$ be regular.
The definition of $\Cl(H)$ is by induction on the length of conditions: for the successor step, assume we have already defined $\Cl$ on
\[ \{\text{ sequences of length }\leq \nu \} \times P_\nu. \]
Fix an arbitrary sequence $H$.
Assume $p \in P_{\nu+1}$ and let $p \in \Cl(H)$ if and only if $\pi_\nu(p) \in \Cl(H\res\nu)$ and either $p \leqlo^\alpha \pi_\nu(p)$ or the following hold:
\begin{enumerate}
\item $H=(H(\xi))_{\xi<\nu+1}$ is a sequence of length $\nu+1$,
\item $p \in D^\Sigma_\alpha$,
\item in case $\nu \in E^0$, we require that $p(\nu)$ (a collapsing condition) is $\beta$-chromatic below $\pi_\nu(p)$, for some $\beta <\alpha$, with spectrum $H(\nu)$,
\item in case $\nu \in E^2$, we require that $H(\nu)((\zeta)_0)$ is defined for each for each $\zeta < \delta= \sigma^2_\lambda(p(\nu))$ and we have $p(\nu)_{<\delta}(\zeta)$ is $\beta$-chromatic below $\pi_\nu(p)$, for some $\beta <\lambda$, with spectrum $H(\nu)(\zeta)$.
In a similar manner, for inaccessible $\lambda$, require $H(\nu)((\zeta)_1)$ is a spectrum for
the characteristic function of $B^\rho_p(\nu)$.
\item in case $\nu \in E^3$, we require that $H(\nu)=(\bar H_i^P, \bar H^0_i, \bar H^1_i)_{i\in\Int\setminus\{0\}}$ and for all $i \in \Int\setminus\{0\}$, $p(i)^P \in \Cl(\bar H^P_i)$ and for $j \in \{0,1\}$, $p(i)^j$ is $\beta$-chromatic with spectrum $\bar H^j_i$ below $\pi_\iota(p(0)^P)$ for some $\beta<\lambda$---where $\iota$ is chosen so that $P_\iota$ is the base of the amalgamation $P_{\nu+1}$ (see p.~\pageref{base} for the definition of \emph{base}).
\end{enumerate}
\end{rem}
\section{A word about book-keeping}\label{book-keeping}
We give a recipe for cooking up a definition of $\bar r=(\bar\iota(\rho),\dot r^0_\rho,\dot r^1_\rho)_{\rho<\kappa}$. %
The definition is given by induction ``on blocks''.
Assume $\bar r \res \xi$ has been defined.
We shall now define $\bar r$ and $\bar \iota$ on $[\xi,(\lambda_\xi)^+)$---the ``next block''.
First, consider some fixed $\iota <\xi$ (or, for the induction start, assume $\xi=\iota=0$).
As $\lambda_{\xi+1} \geq 2^\omega$ in $L[G_\xi]$,
we can enumerate all the reals in $L^{P_\xi}$ which are random over $L^{P_\iota}$ in order type $\lambda_{\xi+1}$.
In other words, find names $P_\xi$-names $(\dot x^\iota_\nu)_{\nu<\beta}$ such that
\[ P_\xi \forces \reals \setminus \dot N = \{ \dot x^\iota_\nu \}_{\nu<\beta},\]
where $\dot N$ is a name for the union of the Borel null sets with code in $L^{P_\iota}$ and $\beta=\lambda_{\xi+1}$.
By assumption, $\xi \in E^0$ is a limit ordinal or $0$, so the last forcing of $P_{\xi+1}$ collapses $\lambda_{\xi+1}$ to $\omega$.
For each $\nu,\nu' < \beta$, apply the lemma \ref{reduce:a:pair} below.
You obtain a set $Y=Y(\nu,\nu',\iota)$ of size $\beta$ consisting of pairs which are $\lambda_{\xi+1}$-reduced over $P_\xi$.
If there are no reals in $L^{P_\xi}$ which are random over $L^{P_\iota}$, let $Y$ be any set of pairs of random reals in $L^{P_{\xi+1}}$ which are $\lambda_{\xi+1}$-reduced over $P_\xi$ (such a set exists---if in doubt, look at the proof of lemma \ref{reduce:a:pair}).
Now define $\bar r \res \beta$ and $\bar\iota\res\beta$ (using a bijection of $\beta$ with $\xi \times \beta^3$) in such a way that all pairs obtained in this way are listed, i.e. for each $\iota < \xi$, each pair and $\nu,\nu' < \beta$ and each $y \in Y(\nu,\nu',\iota)$ there is $\rho \in [\xi,\beta)$ such that $\bar\iota(\rho)=\iota$ and $(\dot r^0_{\rho},\dot r^1_{\rho})=y$.
Note that by lemma \ref{it:reals:are:caught}, we catch our tail and $\bar r$ enumerates all the pairs of random reals of the final model $L[G_\kappa]$ (see lemma \ref{book-keeping:catches:all}).
\begin{lem}\label{reduce:a:pair}
Let $\iota<\xi$, where $\xi \in E^0$and say $1_{P_\xi}$ forces $\dot x^0, \dot x^1$ are $P_\xi$-names random over $L^{P_\iota}$.
Then there is a set $Y=\{ (\dot y^0_\nu,\dot y^1_\nu) \}_{\nu<\lambda_{\xi+1}}$such that
\[ 1\forces (\dot x^0,\dot x^1)\in \{ (\dot y^0_\nu,\dot y^1_\nu) \}_{\nu<\lambda_{\xi+1}} \]
and each pair in $Y$ is $\lambda_{\xi+1}$-reduced over $P_\xi$.
\end{lem}
\begin{proof}
Write $\beta = \lambda_{\xi+1}$.
Find $\{ \dot q_\zeta\}_{\zeta < \beta}$ such that $\forces_\xi \{ \dot q_\zeta\}_{\zeta < \beta}$ is a maximal antichain in $\dot Q_\xi = Coll(\omega,\beta)$.
Note that $\{ (1_{P_\xi}, \dot q_\zeta)\}_{\zeta < \beta}$ is maximal antichain in $P_{\xi+1}$.
Fix a map
\[ b\colon \zeta \mapsto (\dot b_0(\zeta),\dot b_1(\zeta))\]
such that $\forces_\xi b\colon \beta \rightarrow (\Borelplus)^2$ is onto, where $\Borelplus$ denotes the set of Borel sets with positive measure coded in $L[G_\xi]$.
For each $\zeta < \beta$ and $j=0,1$ pick $R^0_\zeta,R^1_\zeta$ such that $(1_{P_\xi}, \dot q_\zeta)$ forces $R^j_\zeta$ is random over $L^{P_\xi}$ and $R^j_\zeta \in \dot b_j(\zeta)$ for both $j=0,1$.
This is possible since $Coll(\omega,\beta)$ collapses the continuum of $L[G_\xi]$.
Fix $\nu < \theta$ for the moment, in order to define $\dot y^0_\nu,\dot y^1_\nu$:
for both $j=0,1$, pick $\dot y^j_\nu$ such that $(1_{P_\xi},\dot q_\nu) \forces \dot y^j_\nu=\dot x^j$ and for each $\zeta\in \beta\setminus\{ \nu\}$ we have $(1_{P_\xi}, \dot q_\zeta)\forces \dot y^j_\nu=\dot R^j_\zeta$.
As $\{ (1_{P_\xi}, \dot q_\zeta)\}_{\zeta <\beta}$ is maximal, $1_{P_\xi}$ forces $\dot r_j$ is random over $L^{P_\iota}$.
For each $\nu<\theta$, the pair $\dot y^0_\nu,\dot y^1_\nu$ is $\beta$-reduced over $P_\xi$:
Let $p \leqlo^{\beta} q \in P_\xi$, let $\dot b_0$, $\dot b_1$ be $P_\xi$-names and fix $w \leq \pi_\xi(p)$ such that $w\forces_\iota \dot b_0$ and $\dot b_1$ are codes for positive Borel sets.
Find $w' \in P_\xi$ and $\zeta<\theta$, such that $w'\leq w$, $\zeta\neq\nu$ and
\begin{equation}\label{r:fix:b:j}
w' \forces \dot b_j(\zeta) \subseteq \dot b_j\text{ for $j=0,1$.}
\end{equation}
We can ask $\zeta \neq \nu$ because we are content with $\subseteq$ instead of $=$ in (\ref{r:fix:b:j}).
As $w'\forces p(\xi)\dot \leqlo^{\beta}_\xi 1$, $w'\cdot p$ is compatible with $(1_{P_\xi}, \dot q_\zeta)$.
If $p' \leq w'\cdot p$ and $p' \leq (1_{P_\xi}, \dot q_\zeta)$, we have $p'\forces \dot y^j_\nu \in \dot b_j(\zeta)\subseteq\dot b_j$.
Clearly, this also shows that $\dot y^0_\nu,\dot y^1_\nu$ is $\beta$-reduced over $P_\iota$: any code for a positive Borel set in $L[G_\iota]$ is remains one in $L[G_\xi]$.
Lastly, as $\{ (1_{P_\xi}, \dot q_\zeta)\}_{\zeta < \theta}$ is maximal and $(1_{P_\xi}, \dot q_\nu)\forces \dot x^j=\dot y^j_\nu$,
\[ 1\forces (\dot x^0,\dot x^1)\in \{ (\dot y^0_\nu,\dot y^1_\nu) \}_{\nu<\lambda}. \]
\end{proof}
We now define $\Gamma^0_\xi$, an approximation of $\Gamma^0$ at stage $\xi<\kappa$ of the iteration.
Let $\Gamma^0_\xi$ be the smallest superset of $\{ \dot c_\eta \setdef E^0(\eta)<\xi \text{ and $\eta$ is limit or }\eta=0\}$ (for limit $\eta$ of course $E^0(\eta)=\eta$, but never mind) closed under all of the functions $F=\Phi^\zeta_\rho,(\Phi^\zeta_\rho)^{-1}$ such that $\dom F \subseteq P_\xi$, i.e. closed under functions in
\[
\{ \Phi^\zeta_\rho,(\Phi^\zeta_\rho)^{-1} \setdef E^3(\alpha^\zeta_\rho) \leq \xi \}
\]
Let
\[\dot \Gamma^0_\xi = \{(1_{P_\xi},\dot c)\setdef \dot c \in \Gamma^0_\xi \}, \]
that is, $\dot \Gamma^0_\xi$ is the canonical choice for a name whose interpretation consists of the interpretations of the elements of $\Gamma_\xi$.
When defining $\bar s$ at stage $\xi$, we need to make sure that all $P_\xi$-names for reals $\dot s$ which have the following property are listed (in the course of the iteration) by $\bar s$:
for any $\dot r\in \Gamma^0_\xi$, $\forces_{P_\xi} \dot r \neq \dot s$.
We can easily make sure this is the case using arguments as above.
As $P_\xi$ forces $\card{\reals}<\kappa$ (in fact $\leq \kappa$ would suffice), we can find $\dot f_\xi$ such that
\[ \forces_\xi \dot f_\xi\colon\kappa\rightarrow \reals\setminus \dot \Gamma_\xi\text{ is onto.} \]
We may assume (by induction hypothesis) we have such $\dot f_\nu$ for $\nu<\xi$.
Pick $\dot s_\xi$ such that for $\xi=G(\eta,\zeta)$, $\forces_\xi \dot s_\xi=\dot f_{\eta}(\zeta)$.
Later (see lemma \ref{disjoint}), we show that $\bar s$ lists exactly the reals of the final model $L[G_\kappa]$ which are not in $\Gamma^0$ (which we are about to define).
This concludes the definition of $(P_\xi)_{\xi\leq\kappa}$, $\bar c$, $(C_\xi)_{\xi<\kappa}$, $\Phi^\zeta_\rho$ for ${\zeta\leq\kappa}$ and ${\rho<\kappa}$, $\bar r$, and $\bar s$, as well as that of $\Gamma^0_\xi$ and $\dot \Gamma^0_\xi$.
\section{Cohen reals, coding areas and the set $\Gamma^0$}\label{sec:reals:areas}
Let $\Gamma^0$ be the least superset of
\[\{ \dot c_\xi \setdef \xi<\kappa, \text{ $\xi$ limit ordinal or }\xi=0\}.\]
closed under all functions $\Phi_\xi, (\Phi_\xi)^{-1}$, $\xi < \kappa$ and let $\dot \Gamma^0$ be the $P_\kappa$-name $\Gamma^0 \times \{ 1_{P_\kappa}\}$. Recalling $\Gamma^0_\xi$ from the previous subsection (defined in the discussion of the coding device $\bar s$), note that $\bigcup_{\xi<\kappa} \Gamma^0_\xi \subseteq \Gamma^0$; that the two sets are in fact equal, if not clear, follows from the next lemma.
The lemma also helps to see that $\Gamma^0$ and $\Gamma^1$ (defined below) give rise to disjoint sets in the extension, as intended.
Lastly, the lemma also is important to show that the coding areas $C_\nu$ behave in the same way as do the reals $c_\nu$, and this will be used in \ref{coding:survives} to show that the coding does not conflict with the automorphisms coming from amalgamation.
Let $\Gamma^1=\{ \dot s_\xi \setdef \xi<\kappa \}$ and let $\dot \Gamma^1$ be the $P_\kappa$-name $\Gamma^1 \times \{ 1_{P_\kappa}\}$.
Let $\dot x_\nu$ denote either $\dot c_\nu$ or $\dot C_\nu$.
Say $\dot y$ is of the following form:
\[
\dot y = (\Phi_{\xi_m})^{k_m} \circ \hdots \circ (\Phi_{\xi_1})^{k_1} (\dot x_\nu)
\]
where $\nu, \xi_1, \hdots, \xi_m < \kappa$ and $k_i \in \Int$ for $1\leq i\leq m$.
For $1\leq i\leq m$, write
\[
\dot y_i = (\Phi_{\xi_i})^{k_i} \circ \hdots \circ (\Phi_{\xi_1})^{k_1} (\dot x_\nu),
\]
and write $\dot y_0$ for $\dot x_\nu$.
Note that we can trivially assume that $\xi_{i+1} \neq \xi_i$, for $i$ such that $1\leq i< m$.
We can also assume $\not \forces_{P_\kappa} \dot y_{i+1}=\dot y_i$ for such $i$.
We then call $\nu, \xi_1, \hdots, \xi_m$,$k_1, \hdots, k_m$ an \textit{ index sequence of $\dot y$ }.
Observe that every $\dot y \in \Gamma^0$ can be written in the form above.
\begin{lem}\label{index:sequ}
There are $\rho_0, \hdots, \rho_m < \kappa$ such that
\begin{enumerate}
\item $\nu < \rho_0 < \hdots < \rho_m$,
\item if $1 \leq i \leq m$, $P_{\rho_i+1}$ is an amalgamation stage associated to $\Phi_{\xi_i}$,
\item if $0 \leq i \leq m$, $\dot y_i$ is a $P_{\rho_i+1}$-name not in $L^{P_{\rho_i}}$. Moreover, $\dot y_i$ is either unbounded over $L^{P_{\rho_i}}$ (if $\dot y_0 = \dot c_\nu$) or remote over $P_{\rho_i}$ up to height $\kappa$ (if $\dot y_0 = \dot C_\nu$).
\end{enumerate}
Moreover, for $\dot y, \dot y' \in \Gamma^0$, either $\forces_{P_\kappa} \dot y = \dot y'$ or $\forces_{P_\kappa} \dot y \neq \dot y'$.
If $\dot y$ and $\dot y'$ have different index sequences, the latter holds.
\end{lem}
\begin{proof}
By induction on $m$. For $m=0$, since $\dot y_0 = \dot x_\nu$, we pick $\rho_0$ so that $\dot Q_{\rho_0}$ adds $\dot x_\nu$ (over $L[G_{\rho_0}]$).
Then all of the above holds.
Now assume by induction the above holds for $i \leq m$.
Let $\rho_{m+1}$ be the least $\rho<\kappa$ such that $P_{\rho+1}$ is an amalgamation stage associated to $\Phi_{\xi_{m+1}}$, and $\dot y_m$ is a $P_{\rho+1}$-name.
Since by induction, $\dot y_m$ is forced to be different from any element of $L^{P_{\rho_m}}$, $\rho_m \leq \rho_{m+1}$. Moreover, $\rho_m < \rho_{m+1}$, for otherwise, $\xi_m = \xi_{m+1}$, contrary to assumption.
We have that either $P_{\rho_{m+1}+1}= \am(P_\zeta, D, \dot r_0, \dot r_1)$ (for some $\zeta$, $D$ a dense subset of $P_{\rho_{m+1}}$, and some $\dot r_0$, $\dot r_1$), or $P_{\rho_{m+1}+1}= \simpleram(P_\zeta, P_{\rho_{m+1}}, \Phi)$ (for some $\Phi$ and $\zeta$).
We prove the lemma assuming the first holds; very similar arguments work for the other alternative, which we leave to the reader.
Observe that $\dot y_m$ is not a $P_\zeta$-name, as otherwise, contrary to assumption,
\[\forces_{P_\kappa} \dot y_{m+1}=\Phi_{\xi_{m+1}}(\dot y_m)=\dot y_m.\]
We have to consider two cases: If $\dot x_\nu = \dot c_\nu$, we can assume by induction that $\dot y_m$ is unbounded over $L^{P_\zeta}$ and thus also over $L^{P_\zeta * \dot B (\dot r_i)}$ for $i=0,1$.
Thus by lemma \ref{newreal} applied for $P = D$, $\dot y_{m+1}$ is unbounded over $L^{P_{\rho_{m+1}}}$ and we are done.
If on the other hand, $\dot x_\nu = \dot C_\nu$, we can assume by induction that $\dot y_m$ is remote over $P_\zeta$ up to height $\kappa$ and $\kappa > \lambda_{\xi_{m+1}}$. So by lemma \ref{remote:lemma}, $\dot y_{m+1}$ is remote over $P_{\rho_{m+1}}$.
In either sub-case, we conclude that $\dot y_{m+1}$ is not in $L^{P_{\rho_{m+1}}}$ (for the second case, using lemma \ref{remote:lemma:not:in}).
Lastly, say $\nu, \xi_1, \hdots, \xi_m$,$k_1,\hdots,k_m$ is an index sequence of $\dot y$ and say $\nu', \xi_1', \hdots, \xi_m'$,$k'_1,\hdots,k'_m$ is an index sequence of $\dot y'$.
Assume $\not \forces_{P_\kappa} \dot y = \dot y'$; we show $\forces_{P_\kappa} \dot y \neq \dot y$.
Let $\rho_0, \hdots \rho_m$ and $\rho_0', \hdots \rho_m'$ be obtained as above for $\dot y$ and $\dot y'$ respectively.
Let $l$ be maximal such that $\xi_l \neq \xi'_l$ or $k_l \neq k'_l$, if such $l$ exists.
We may assume $\rho_l = \rho'_l$, for otherwise
\[ \forces_{P_\kappa} \dot y_l \neq \dot y_l',\]
and we can apply $(\Phi_{\xi_m})^{k_m} \circ \hdots (\Phi_{\xi_{l+1}})^{k_{l+1}}$ to this to obtain
\[\forces_{P_\kappa} \dot y \neq \dot y',\]
and we are done.
So we may also assume $\xi_l = \xi'_l$, as otherwise, also $\rho_l \neq \rho'_l$.
Thus $\xi_l = \xi'_l$ and $\rho_l = \rho'_l$ , but $k_l \neq k'_l$. From now on may we write $\xi$ and $\rho$ for $\xi_l = \xi'_l$ and $\rho_l = \rho'_l$, respectively.
Observe that $\dot y_{l-1} \in V^{P_{\rho_{l-1}+1}}$, and $\dot y'_{l-1} \in V^{P_{\rho'_{l-1}+1}}$
where both $\rho_{l-1}, \rho'_{l-1} < \rho$.
Also note that $\dot y_{l-1}$ is either remote or unbounded over $V^{P_{\rho_{l-1}}}$.
Moreover, if we let $P_\zeta$ be the base of $\Phi_\xi$, we have $\rho_{l-1} \geq \zeta$,
for otherwise again $\Phi^{k-l}(\dot y_{l-1} )= \dot y_{l-1}$.
Thus we have $\dot y_{l-1}$ is either remote or unbounded over $ V^{P_\zeta}$.
Now apply lemma \ref{newreal} to see that $\forces_{P_\kappa} \dot y'_{l-1} \neq (\Phi_{\xi_l})^{-k'_l+k_{l}} (\dot y_{l-1})$, noting again $\dot y'_{l-1}$ is in $V^{P_\rho}$.
Apply $(\Phi_{\xi_l})^{k_l}$ to see $\forces_{P_\kappa} \dot y_{l} \neq \dot y'_{l}$.
As above, apply $(\Phi_{\xi_m})^{k_m} \circ \hdots (\Phi_{\xi_{l+1}})^{k_{l+1}}$ to this to obtain
\[\forces_{P_\kappa} \dot y \neq \dot y'.\]
If no $l$ as above exists, the index sequences for $\dot y$ and $\dot y'$ are identical except possibly in the first coordinate.
Now observe that $\forces_{P_\kappa} \dot y_0 = \dot y'_0$ if $\nu = \nu'$ and $\forces_{P_\kappa} \dot y_0 \neq \dot y'_0$ if $\nu\neq\nu'$.
Apply $(\Phi_{\xi_m})^{k_m} \circ \hdots (\Phi_{\xi_{0}})^{k_{0}}$ and we're done.
\end{proof}
\begin{lem}\label{disjoint}
$\forces_{P_\kappa} \dot \Gamma^0=\reals \setminus \dot \Gamma^1$.
\end{lem}
It would be easier to show this in the following way: Show by induction on the number of applications of automorphisms that all
names $\dot c$ in $\Gamma^0 \setminus \Gamma^0_\xi$ have the following property: there is $\rho \geq \xi$ such that $\dot c$ is
in $L^{P_{\rho+1}}$ but not in $L^{P_\rho}$. This would make the slightly more complicated proof of lemma \ref{index:sequ} unnecessary.
\begin{proof}
First we show $\forces_{P_\kappa}\dot \Gamma^0\cup\dot \Gamma^1=\reals$. Let $r \not \in (\dot \Gamma^0)^G$. Find $\xi<\kappa$ such that $r\in L[G_\xi]$.
As $r \not \in (\dot\Gamma^0_\xi)^{G_\xi}$, $\bar s$ was defined to list a name for $\dot r$, so $r \in \dot (\Gamma^1)^G$.
Now let $\dot c \in \Gamma^0$ and $\dot s \in \Gamma^1$, and show $\forces_{P_\kappa}\dot s \neq \dot c$.
Fix $\xi <\kappa$ so that $\dot s$ is a $P_\xi$-name and $\forces_{P_\xi} \dot s\not \in \Gamma^0_\xi$.
Let $v, \rho_1, \hdots, \rho_n$ be obtained as in the previous lemma from an index sequence for $\dot c$ and write $\rho=\rho_n$.
By the last lemma $\dot c$ is a $P_{\rho+1}$ name not in $L^{P_\rho}$.
If $\rho+1 \leq \xi$, we are clearly done, for then $\dot c \in \Gamma^0_\xi$.
Otherwise, if $\rho \geq \xi$, $\dot c$ is not in $L^{P_\rho}\supseteq L^{P_\xi}$, so in any case, $\forces_{P_\kappa} \dot c \neq \dot s$.
\end{proof}
\section{The $\kappa$-stage of the iteration}\label{sec:stage kappa}
The next lemma shows that $\kappa$ remains a cardinal in the final model, $\kappa$ remains Mahlo at each earlier stage, and that all reals appear in some initial segment of the construction.
We write conditions of $P_\theta$ as $(t,p)$ in the following only to have a convenient way to refer to the $\bar T$ part and to aid intuition---we do not need to work in $\bar T * P^*_\theta$.
\begin{lem}\label{it:reals:are:caught}
Let $\theta \leq \kappa$, let $(\dot \alpha_\xi)_{\xi<\kappa}$ be a sequence of $P_\theta$-names for ordinals below $\kappa$ and let $(t,p) \in P_\theta$. Then for any $\beta_0<\kappa$ there is an inaccessible $\alpha \in (\beta_0,\kappa)$ and a condition $(t',p') \leq (t,p)$ such that for all $\xi <\alpha$,
$(t',p') \forces \dot \alpha_\xi <\alpha$. Moreover if $\theta = \kappa$, there is a sequence of $P_\alpha$-names $(\dot \alpha'_\xi)_{\xi<\alpha}$ such that for each $\xi<\alpha$,
$(t',p') \forces \dot \alpha_\xi = \dot \alpha'_\xi$.
\end{lem}
\noindent
The ``moreover'' clause is of course meaningless if $\theta < \kappa$.
Before we treat the lemma, we draw two corollaries.
\begin{cor}
\begin{enumerate}
\item If $r \in L[G_\kappa]$ is a real, there is $\alpha<\kappa$ such that $r\in L[G_\alpha]$. In particular, $\kappa$ remains uncountable in $L[G_\kappa]$ (i.e. $\kappa=\omega_1$ in the final model).
\item If $\theta <\kappa$, $\kappa$ remains Mahlo in $L[G_\theta]$.
\end{enumerate}
\end{cor}
\begin{proof}
For the first corollary, fix a real $r \in L[G_\kappa]$ and let $\dot \alpha_n$ be a $P_\kappa$-name for $r(n)$, for each $n \in \nat$.
The lemma shows we can find $(t',p') \in \bar G_\kappa$, $\alpha<\kappa$ and a sequence of $P_\alpha$-names $\{\dot \alpha'_n \setdef n\in\nat\}$, such that for each $n \in \nat$, $(t',p')\forces \dot \alpha_n=\dot \alpha'_n$. Obviously, $r \in L[G_\alpha]$.
For the second, say $\theta < \kappa$ and fix a $P_\theta$-name $\dot C$ for a closed unbounded subset of $\kappa$.
Let $\dot \alpha_\xi$ be a name for the least element of $\dot C$ above $\xi$.
By the lemma, we may find an inaccessible $\alpha \in (\lambda_\theta,\kappa)$ and $(t',p') \in G_\theta$ such that for each $\xi<\alpha$, $(t',p') \forces \dot \alpha_\xi<\alpha$.
Thus, $(t',p') \forces \check \alpha \in \dot C$.
Now observe that as $P_\theta$ is stratified above $\lambda_\theta$, $\alpha$ is inaccessible in $L[G_\theta]$.
\end{proof}
Before we begin the proof we'd like to remind the reader of the following terminology.
Let $\eta < \theta$ a coding stage and work in $L[\bar B^-, G^o_\eta] = L[A_\eta]$, where $\dot B^-$, $G^o$, $A_\eta$ are as defined in section \ref{def:it:succ:stage} (see p.~\pageref{def:it:succ:stage}; what we call $\eta$ here is called $\xi$ there unfortunately) and $A_\eta$ is the set to be coded at this stage, as defined in section \ref{sec:full:setting}, see p.~\pageref{sec:full:setting}.
Let $H$ be any set and let $p, q \in P$, $q \leq p$.
We say that $q \in P(A_\eta)$ is basic generic for $(H,p)$ at $\kappa^+$ if and only if
\begin{enumerate}
\item\label{grow_kappa_plus} $|q_\kappa| > H \cap \kappa^+$;
\item\label{restraints_kappa_plus}
if $\nu \in H\cap[\kappa^+,\kappa^{++})$ then $b^{p_{\kappa^+}\res\nu} \in q^*_{\kappa^+}$;
\item \label{succ_coding_kappa_plus} if $\nu \in H\cap[\kappa^+,\kappa^{++})$ then there is $\zeta > |p_\kappa|$ such that $q_\kappa((\zeta)_1)=p_{\kappa^+}(\nu)$;
\item\label{A_coding_kappa_plus} if $\xi \in H \cap [\kappa,\kappa^+)$ there is $\nu > |p_\kappa|$ such that
$q_\kappa((\langle \xi, \nu \rangle)_0) = 1$ if $\xi\in A_\eta$ and $q_\kappa((\langle \xi, \nu \rangle)_0) = 0$ if $\xi\not\in A_\eta$;
\end{enumerate}
We say that $q \in P(A_\eta)$ is basic generic for $(H,p)$ at $\kappa$ if and only if
\begin{enumerate}
\item\label{grow_kappa} $|q_{<\kappa}| \geq \sup (H\cap\kappa)$.
\item\label{restraints_kappa} if $\nu \in H\cap[\kappa,\kappa^{+})$ then $b^{p_{\kappa}\res\nu}\setminus \eta' \in q^*_\kappa$ for some $\eta'<\kappa$;
\item \label{mahlo_coding_kappa} if $\nu \in H\cap[\kappa,\kappa^{+})$ then
there is $\xi \in b^{p_\kappa\res\nu}\setminus\eta$ such that $\xi \geq |p_{<\kappa}|$ and
$q_{<\kappa}(\{ \xi \}_2)=p_\kappa(\nu)$;
\item\label{A_coding_kappa} if $\xi \in H \cap \kappa$ there is $\nu > |p_\delta|$ such that
$q_{<\kappa}((\langle \xi, \nu \rangle)_0) = 1$ if $\xi\in A_\eta$ and $q_{<\kappa}((\langle \xi, \nu \rangle)_0) = 0$ if $\xi\not\in A_\eta$;
\end{enumerate}
We shall use these notions as a replacement for $\D$ in the following proof, since this approach simplifies the argument showing that the sequence we build is appropriately definable (or ``canonical'').
We need to do this because we build the canonical witness (as usual, a sequence of models, this time of two different types) in advance, before we build the associated sequence of conditions.
\begin{proof}[Proof of lemma \ref{it:reals:are:caught}]
The proof is by induction on $\theta$, so assume lemma \ref{it:reals:are:caught} holds for all $\theta' < \theta$.
Let $\{ . \}_i$ for $i \in\{ 0,1\}$ be a pair of recursive functions such that $\xi \mapsto ( \{ \xi\}_0, \{ \xi\}_1 ) \in \On^2$ is surjective and each pair in $\alpha\times\alpha$ occurs $\alpha$-many times as $( \{ \xi\}_0, \{ \xi\}_1 )$ for $\xi <\alpha$, whenever $\alpha \in \mathbf{Card}$.
Also, we denote by $h\colon \On \rightarrow L$ the map enumerating the constructible universe according to its canonical well-ordering.
We assume without loss of generality that $(t,p) \in \dom(\C^\kappa)\cap\dom(\C^{\kappa^+})$.
Start by inductively choosing continuous $\in$-chains $N^{\xi}$ and $M^{\xi}$ for $\xi<\kappa$ as follows.
Set
\begin{gather*}
\vec{x}=\{\kappa^{+++}, (\dot\alpha)_{\xi<\kappa}, P_\theta, (t,p) \}\\
M^{-1}= \vec{x} \\
N^{-1}= \kappa\cup\{ \vec{x}\}.\\
\end{gather*}
Assuming we have $M^\xi \in N^\xi$, let $M^{\xi+1}\prec_{\Sigma_1} L$ be least (of size $<\kappa$) such that
$N^\xi \in M^{\xi+1}$ and $M^{\xi+1} \cap \kappa \in \kappa$, and choose
$N^{\xi+1}\prec_{\Sigma_1} L$ least (of size $\kappa$) such that
$M^{\xi+1} \in N^{\xi+1}$ and $N^{\xi+1} \cap\kappa \in \kappa^+$.
For limit $\rho <\kappa$, let
\begin{gather*}
M^\rho = \bigcup_{\xi<\rho} M^\xi\\
N^\rho = \bigcup_{\xi<\rho} N^\xi\\
\end{gather*}
and write $\kappa(\xi) = M^\xi \cap \On$. %
Observe each $M^\xi$ and $N^\xi$ are closed under $h$ and $h^{-1}$ and each $\kappa(\xi)$ is a strong limit cardinal.
Let $C = \{ \kappa(\xi) \setdef \xi < \kappa\}$ and find $\alpha\geq \beta_0$, an inaccessible limit point of $C$ such that
$\diamondsuit_\alpha = C\cap \alpha$ (remember we have chosen to denote by $(\diamondsuit_\xi)_\xi$ the canonical diamond sequence of $L$ concentrating on inaccessibles below $\kappa$).
Pick $C^* \subseteq \lim C\cap \mathbf{Sing}$ such that $C^*$ is club in $\alpha$ (where $\lim C$ of course denotes the set of limit point of $C$).
Now we construct a sequence of $(t^\xi, p^\xi) \in \bar P_\theta$ by induction on $\xi\leq\alpha$, starting with $(t^0,p^0)=(t,p)$.
\paragraph{The successor step:}
At successor stages, assume we have constructed $(t^{\xi}, p^{\xi}) \in M^{\xi+1}$ and when $\xi$ is a limit,
also assume that for each coding stage $\eta < \theta$,
$$(t^\xi, p^\xi\res\eta)\forces |p^\xi(\eta)_{<\kappa}|= \kappa(\xi).$$
First pick $(t', p') \mathbin{\leqlo^{<\kappa}} (t^{\xi}, p^{\xi})$ least such that
$(t', p') \in M^{ \xi+1}$ meeting the following requirements:
\begin{enumerate}
\item \label{Cstar}
If $\xi$ is a limit ordinal, then for each coding stage $\eta < \theta$ we have
$$
(t', p'\res\eta)\forces p'(\eta)_{<\kappa} (\kappa(\xi))=i,
$$
where $i = 0$ if $\kappa(\xi)\in C^*$ and $i=1$ otherwise.
\item \label{kappa:work}
If there is a condition $(s_0,r_0) \leq (t^{\xi}, p^{\xi})$
with $h(\{\xi\}_0) \in \C^{\kappa(\xi)}(s_0,r_0)$, we demand that for some $(s_1, r_1)\leq (s_0,r_0)$ such that $(s_1, r_1)\in \dom(\C^\kappa)\cap\dom(\C^{\kappa^+})$ which decides $\dot \alpha_{\{\xi\}_1}$, we have that
$$ t' \leq s_1,$$
for every stage $\eta \in E^1$ were we force with $\kappa$-Cohen $p'(\eta)\leq r^1(\eta)$
and
for every coding stage $\eta < \theta$, $(t',p'\res\eta)$ forces both that
$$
(p'(\eta)_{<\kappa}, p'(\eta)^*_{\kappa}) \leq (r_1(\eta)_{<\kappa}), r_1(\eta)^*_{\kappa}) \text{ in } P^{p'(\eta)_\kappa}
$$
and that
$$
(p'(\eta)_{\kappa}, p'(\eta)^*_{\kappa^+}) \leq (r_1(\eta)_{\kappa}), r_1(\eta)^*_{\kappa^+}) \text{ in } P^{p'(\eta)_{\kappa^+}}
$$
Moreover we demand $(t', p') \mathbin{\leqlo^{<\kappa}} (t^\xi, p^\xi)$.
Write $r^\xi_i = r_i$, where $i \in \{0,1\}$.
We also define $\alpha^\xi_\nu$ to be the ordinal such that $(s_1, r^\xi_1)\forces \dot \alpha_\nu = \check\alpha^\xi_\nu$.
\item\label{rho}
For every $\eta < \theta$, $(t', p' \res \eta) \forces \rho(p'(\eta)_{<\kappa})\geq \kappa(\xi)$.
\item\label{basic_generic}
For every $\eta < \theta$ which is a coding stage, we have that
\begin{eqpar}\label{kappa:strategic}
$(t', p'\res\eta)\forces$``$p'(\eta)_{<\kappa}$ is basic generic for $(M^{\xi}, p^\xi(\eta))$ at $\kappa$ and
$p'(\eta)_{\kappa}$ is basic generic for
$(N^{\xi}, p^\xi(\eta))$ at $\kappa^+$.''
\end{eqpar}
\item\label{generic}
Notice that in the previous item we have ensured basic genericity over subsets of $L$, while we want to capture some information about $L[A_\eta]$; so we ask the following.
For every $\eta < \theta$ and every $P_\eta$-name for an ordinal $\dot \alpha \in M^{\xi}$ there is a set $a$ of size $<\kappa$ such that
$(t', p'\res\eta) \forces \dot\alpha \in \check a$.
Also, for every $\eta < \theta$ and every $P_\eta$-name for an ordinal $\dot \beta \in N^{\xi}$ there is a set $b$ of size $\kappa$ such that
$(t', p'\res\eta) \forces \dot\beta \in \check b$.
\end{enumerate}
All of the above requirements can be expressed by a formula which is $\Sigma_1$ in parameters from $\vec{x}\cup\{ M^{\xi}, (t^\xi, p^\xi) \}$,
so if $(t',p')$ satisfying the above can be found at all then we can demand $(t',p') \in M^{\xi+1}$.
Requirements \ref{kappa:work} and \ref{rho} are trivial.
Requirement \ref{basic_generic} can be met by a density argument identical to the one showing the ``density'' of strategic class $\D$ in the limit of an iteration.
The difference is purely notational (we've treated the Mahlo coding as lower part at $\kappa$ and as upper part for $\lambda<\kappa$, now we are ``in-between'' these cases):
\begin{lem}
The set of all $(t', p'\res\eta)$ such that
for every $\eta < \theta$, \eqref{kappa:strategic} holds
is dense below $(t^\xi, p^\xi)$
\end{lem}
\begin{proof}[Proof of claim.]
Say we are given $(t', q) \leq (t^\xi, p^\xi)$.
Exactly as in \ref{thm:it:qc}, construct $p'(\eta)$ by induction on $\eta$.
At successor stages, let $p'(\eta)$ be a name for a condition such that \eqref{kappa:strategic} is met.
This is possible as $(t', q\res\eta)$ forces such a condition to exist, by corollary \ref{cor_qc} (see p.~\pageref{cor_qc}).
Observe that if $(t', q\res\eta) \forces q(\eta)_{<\kappa} = \emptyset$, then we can choose $p'(\eta) \leqlo^{\kappa} q(\eta)$.
Thus, the resulting condition $(t', p')$ has legal support.
\end{proof}
Lastly, we can satisfy \ref{generic}: for the first line, use induction hypothesis and $\lVert M^{\xi} \rVert < \kappa$, for the second use $\{\kappa\}$-stratification.
This shows we can find $(t',p')$ as above.
Find $(t^{\xi+1},p^{\xi+1}) \mathbin{\leqlo^{<\kappa}} (t',p')$ least such that $(t^{\xi+1},p^{\xi+1})\in \dom(\C^{\kappa^+})$ and $(t^{\xi+1},p^{\xi+1}) \in M^{\xi+1}$.
Now set $(t^{\xi+1},p^{\xi+1})$ to be $(t',p')$.
Note that by diagonal support, requirement \ref{generic} and since $M^{\xi+1}$ is $\Sigma_1$-elementary, for each $\eta<\theta$ we have
\begin{equation}\label{p_star_control}
(t^{\xi+1}, p^{\xi+1}(\eta)) \forces |p^{\xi+1}(\eta)_{<\kappa}| < \kappa(\xi+1).
\end{equation}
(and analogously, a similar equation holds for $|p^{\xi+1}(\eta)_\kappa|$).
By induction, we can infer
$$
(t^{\xi}, p^{\xi}(\eta)) \forces \{ \zeta \setdef p^{\xi}(\zeta)_{<\kappa}= 1 \}\cap \lim C = C^* \cap \kappa(\xi) ,
$$
for all limit $\xi$.
\paragraph{Taking greatest lower bound at the limit step:}
At limit $\rho \leq \alpha$, we take $(t^\rho, p^\rho)$ to be the greatest lower bound of $(t^\xi,p^\xi)_{\xi<\rho}$; this is well-defined:
we show by induction on $\eta <\theta$ that $(t^\rho, p^\rho\res\eta)$ is a condition.
Since we are always taking $\leqlo^{<\kappa}$-direct extensions,
at amalgamation stages we can simply use induction to take the point-wise limit in each of the $\Int$-many components and the resulting $\Int$-sequence will be a condition in the amalgamation.
By $\kappa$-closure of $\bar T$ and $\kappa$-Cohen forcing, we only have to treat coding stages.
So fix a coding stage $\eta$ and assume $w=(t^\xi, p^\xi\res\eta)$ is a condition.
We let $p^\xi(\eta)$ by the name for the obvious candidate for a lower bound of the sequence $(p^\nu(\eta))_{\nu<\eta}$ (see \eqref{obvious_candidate}, p.~\pageref{obvious_candidate}).
\begin{claim}\label{claim:generic}
The condition $w$ is $M^\rho$-generic, in the sense that
$$w \forces M^\rho[\dot G\res\eta]\cap \On = M^\rho \cap \On.$$
It is $N^\rho$-generic in the same sense.
\end{claim}
\begin{proof}
This follows from requirement \ref{generic} in the construction of the sequence and by $\Sigma_1$-elementarity.
\end{proof}
Observe a consequence of this is that we can treat $M^\rho[\dot G\res\eta]$ as a generic extension of $M^\rho$ by $P_\eta$; likewise for $N^\rho[\dot G\res\eta]$.
\begin{claim}
We have that $w \forces p^\rho(\eta)_{<\kappa} \in S^*_{<\kappa}$.
\end{claim}
\begin{proof}
Let $G\res\eta$ denote an arbitrary $P_\eta$-generic for the moment.
We work in $L[\bar B^-, G^o\res\eta] = L[A_\eta]$, as usual (see sections~\ref{sec:full:setting} and~\ref{def:it:succ:stage}).
Let $N = L_\gamma[A_\eta\cap\alpha, p^\xi(\eta)_{<\kappa}]$ be a $<\kappa$-test model such that $N \vDash \alpha$ is the least Mahlo.
Let $\bar M^\xi$ be the transitive collapse of $M^\xi$ for $\xi \leq \rho$ .
Letting $\bar \mu = \bar M^\rho \cap \On$, we show $\gamma < \bar \mu$;
for otherwise,
since $( \bar M^\xi)_{\xi<\rho}$ is definable over $\bar M^\rho = L_{\bar\mu}$ which is in turn definable in $N$,
we find $C\cap\alpha \in N$ and thus $\lim C^* \in N$, contradicting that $N \vDash \alpha$ is Mahlo.
Thus indeed, $\gamma < \bar \mu$. Now we can quote the last part of the proof theorem \ref{jensen_qc_main}, which easily shows that
by requirement
\ref{basic_generic} and by claim \ref{claim:generic} in the construction of the sequence and by elementarity,
for $A^*= A^*(p^\rho_{<\kappa}(\eta))$,
$L_\gamma[A^*] \vDash r^\eta$ is coded by branches.
\end{proof}
\begin{claim}
We have that $w \forces p^\xi(\eta)_\kappa \in S^*_\kappa$.
\end{claim}
\begin{proof}
This is completely analogous to the previous claim:
The proof for theorem \ref{jensen_qc_main} on page \pageref{jensen_qc_main} carries over to the present situation almost verbatim (in the case $\delta=\kappa$).
As in the previous claim, to show that the height of any test-model is less than that of the collapse of $N^\rho$, use that $(t^\xi,p^\xi)_{\xi<\rho}$ is appropriately definable over the transitive collapse of $N^\rho$ using $\vec{x}$ and $C^*$ as parameters.
The additional parameter is unproblematic since $C^* \in \Hhier(\kappa)$ and is therefore an element of each $N^\rho$.
As in the previous claim, we may directly use basic genericity (requirement \ref{basic_generic} in the construction) and claim \ref{claim:generic} instead of $\D$ as we did in the proof of theorem \ref{jensen_qc_main}.
\end{proof}
Observe when $\rho < \alpha$ is a limit ordinal,
have $(t^\rho, p^\rho) \in M^{ \rho +1}$:
this is because $(t^{\xi},p^{\xi})_{\xi<\rho}$ is definable over
$\langle M^{\rho}, C^*\cap\kappa(\rho)\rangle$ and $C^* \cap\kappa(\rho) \in M^{\rho +1}$ since
$\kappa(\bar\rho+1)$ is a strong limit cardinal.
This ends the construction of the sequence $(t^\xi, p^\xi)$, for $\xi \leq \alpha$.
Finally, we extend $(t^\alpha, p^\alpha)$ to $(t^\alpha, \bar p^\alpha)$ by making sure $\alpha$ is in the support at each coding stage, i.e.
\begin{equation}\label{alpha_in_supp}
\forall \eta < \theta\cap \alpha\cap E^2\quad (t^\alpha, \bar p^\alpha \res\eta) \forces \alpha \in \supp(p^\alpha(\eta))
\end{equation}
We show $(t^\alpha, \bar p^\alpha) \forces \dot \alpha_\nu < \alpha$, for each $\nu < \alpha$.
In fact, letting $\dot \alpha'_\nu$ be the name such that whenever $\{ \xi \}_1=\nu$ and $r^\xi_1$ is defined,
$r^\xi_1 \forces \dot \alpha'_\nu = \check \alpha^\xi_\nu$, we will show that
$(t^\alpha, \bar p^\alpha) \forces \dot \alpha_\nu = \dot \alpha'_\nu$.
Observe that this is a $P_\alpha$-name whenever $\alpha \leq \theta$, so this proves the theorem.
So let $\nu < \alpha$, $(s,r) \leq (t^\alpha, \bar p^\alpha)$ and assume $(s,r) \in \dom(\C^\alpha)$ and $(s,r)$ decides $\dot \alpha_\nu$.
Pick $\xi$ such that $h(\{ \xi \}_0)\in \C^\alpha(s, r)$, $\{ \xi \}_1=\nu$ and also,
$h(\{ \xi \}_0) \in \Hhier(\kappa(\xi))$.
The latter may be achieved by increasing $\xi$ if needed, without changing $\{ \xi \}_i$, $i\in \{0,1\}$.
Observe that it follows that $h(\{ \xi \}_0)\in \C^{\kappa(\xi)}(s, r)$, by definition of $\C$.
Since $\alpha$ is inaccessible, we have $\xi < \alpha$, and
$r$ witnesses that $r^\xi_1$ is defined.
We now show that $(s,r) \cdot (t^\alpha, r^\xi_1) \neq 0$.
This is clear for the $\bar T$-part, since $s \leq t^\alpha$.
We show that for each $\eta < \theta$, $(s,r \res \eta \cdot r^\xi_1\res\eta )\forces r(\eta) \cdot r^\xi_1(\eta) \neq 0$.
The only non-trivial cases are amalgamation and coding stages.
At amalgamation stages, for each $i \in \Int$, we can takes point wise meets by induction (or by running the present argument again).
Observe that since the outcome is a $\leqlo^{<\kappa}$-direct extension of $r^\xi_1\res\eta+1$,
the resulting sequence is a condition in the amalgamation, i.e. in $P_{\eta+1}$.
Now assume $\eta$ is a coding stage ($\eta \in E^2$).
Letting $w = (s, r \res \eta \cdot r^\xi\res\eta )$, we have that
$w$ forces that
$$(r(\eta)_{<\kappa}, r(\eta)^*_\kappa) \leq (r^\xi_1(\eta)_{<\kappa}, r^\xi_1(\eta)^*_{\kappa}) \text{ in }P^{r(\eta)_\kappa}$$ and
$$
(r(\eta)_{\kappa}, r(\eta)^*_{\kappa^+}) \leq (r^\xi_1(\eta)_{\kappa}, r^\xi_1(\eta)^*_{\kappa^+}) \text{ in }P^{r(\eta)_{\kappa^+}}.
$$
Moreover, $w$ forces that $r^\xi_1(\eta)$ is an extension of a condition $r^\xi_0(\eta)$ which has ``lower part $h(\eta)$'', i.e. $h(\eta) \in \C^{\kappa(\xi)}(r^\xi_0(\eta))$.
Since $w\forces h(\eta) \in \C^\kappa(\xi)(r(\eta))$, $w$ forces that
$r^\xi_0(\eta) \res\kappa(\xi) = r(\eta)\res\alpha \geq r^\xi_1(\eta)\res\alpha(\xi)$.
It follows immediately that that $w \forces r(\eta)\res [\alpha, \kappa^+] \cup r^\xi_1\res\alpha$ is a condition, call it $w'$.
Moreover, note that $w$ forces that
\begin{equation}\label{spaced_out}
\supp(r^\xi_1(\eta))\cap\kappa \subseteq \kappa(\xi+1).
\end{equation}
It remains to show that $w \forces w'=r(\eta) \cdot r^\xi_1(\eta) $.
It is clear that $w \forces w' \leq r^\xi_1$; in order to see $w \forces w' \leq r(\eta)$ we must check that making the extension from $r(\eta)\res\alpha$ to $r^\xi_1(\eta)$ below $\alpha$ did not violate any of the restraints in $r(\eta)$ at and above $\alpha$.
Since $w\forces \rho(p^\alpha (\eta)_\kappa) \geq \alpha$, $w$ also forces that
making the extension of $r(\eta) \res\alpha$ to $r^\xi_1(\eta) \res\alpha$ respects all restrains in $r(\eta)^*_\kappa$.
Since the restraints in $r(\eta)^*_\alpha$ are spaced by $\diamondsuit_\alpha$ (see the discussion following definition \ref{reg_defs} on page \pageref{reg_defs}) and by \eqref{spaced_out},
making this extension obeys all restraints in $r(\eta)^*_\alpha$.
Lastly, since \eqref{alpha_in_supp} makes sure that $w \forces \alpha \in \supp(r(\eta))$, no
restraints are violated by this extension.
In other words, we have shown that $w \forces w'=r(\eta) \cdot r^\xi_1 \neq 0$.
This finishes the proof that $(t^\alpha, r^\xi_1) \cdot (s,r) \neq 0$ and so
since $(s,r)$ decides $\dot \alpha_\nu$, we must have
$(s,r) \forces \dot \alpha_\nu = \check \alpha^\xi_\nu$.
This concludes the proof of lemma \ref{it:reals:are:caught}
\end{proof}
It is crucial that by lemma \ref{it:reals:are:caught}, the book-keeping devices $\bar r$ and $\bar s$ ``catch'' all the relevant reals in the
final extension by $P_\kappa$:
\begin{lem}\label{book-keeping:catches:all}
If $\iota<\kappa$, $\dot r^0, \dot r^1$ are $P_\kappa$-names for reals and $p \in P_\kappa$ forces $\dot r^0, \dot r^1$ are random over $L^{P_\iota}$, there is $q \leq p$ and $\nu<\kappa$ such that $\bar\iota(\nu)=\iota$
\[ q \forces \dot r^j=\dot r^j_\nu. \]
If $\dot s$ is a $P_\kappa$-name for a real and $p \in P_\kappa$, there is $q\leq p$ and $\xi<\kappa$ such that either
$q \forces \dot s=\dot s_\xi$, or $q\forces \dot s \in \Gamma_\xi$.
\end{lem}
\begin{proof}
By lemma \ref{it:reals:are:caught} there is $q\leq p$, $\xi<\kappa$ and $P_\xi$-names $\dot x^0, \dot x^1$ such that $q \forces \dot r^j=\dot x^j$.
As $P_\kappa$ collapses the continuum of any initial stage of the iteration, we may assume $1_{P_\kappa}$ forces $\dot x^j$ is random over $L^{P_\nu}$.
Using the notation from lemma \ref{reduce:a:pair}, find $\nu, \nu'$ and $q'\leq q$ such that $q'\forces \dot x^0 = \dot x_\nu$ and $q'\forces \dot x^1 = \dot x_{\nu'}$, and find $q'' \leq q'$ and $y \in Y(\nu,\nu')$ such that $q'' \forces (\dot x^0, \dot x^1)=y$.
Thus, by construction of $\bar r$, we may find $\nu$ such that $\bar\iota(\nu)=\iota$ and $y=(\dot r^0_\nu, \dot r^1_\nu)$, and we have
$q'' \forces \dot r^0=\dot r^0_\nu$ and $\dot r^1=\dot r^1_\nu$.
The second claim follows immediately from lemmas \ref{it:reals:are:caught}, \ref{disjoint} and the definition of $\bar s$.
\end{proof}
\section{Every projective set of reals is measurable}
Write $G=G_\kappa$.
\begin{lem}
For any $\nu<\kappa$, $\bigcup N^*_\nu$ is a null set, where
\[ N^*_\nu=\{N \in L[G_\nu] \setdef L[G_\nu]\models N \subset \reals \text{ has measure zero} \} \]
\end{lem}
\begin{proof}
Every null set $N \in L[G_\nu]$ is covered by a null Borel set whose Borel code is also in $L[G_\nu]$. The set $C^*$ of Borel codes for null sets in $L[G_\nu]$ is countable in $L[G]$, so $\bigcup N^*_\nu$, which is equal to the union of all the Borel sets with code in $C^*$, is a countable union of null sets in $L[G]$.
\end{proof}
The following, together with the last lemma, suffices to show that in the extension by $P_\kappa$, every projective set of reals is measurable.
\begin{lem}\label{auto}
Let $\nu<\kappa$. There is a name $\dot r_*$ which is fully random over $L^{P_\nu}$ such that the following hold:
\begin{enumerate}
\item
Let $\dot B(\dot r_*)$ be a $P_\nu$-name for the complete sub-algebra of $\quot{B_\kappa}{P_\nu}$ generated by $\dot r_*$ in $L[G_\nu]$ and let $B^0 = P_\nu * \dot B(\dot r_*)$.
For any $b \in B_\kappa \setminus B^0$, there is an automorphism $\Phi$ of $B_\kappa$ such that $\Phi(b)\neq b$ and $\Phi\res B^0 = \id$.
\label{lem:mix}
\item
For any $P_\kappa$-name $\dot r$ which is random over $L^{P_\nu}$ and any $p\in P_\kappa$ there is $q\leq p$ and an automorphism $\Phi$ of $B_\kappa$ such that $q \forces \dot r =\Phi(r_*)$ and $\pi_\nu \circ \Phi = \Phi \circ \pi_\nu = \pi_\nu$.\label{lem:homogeneity:for:random:reals}
\end{enumerate}
\end{lem}
\begin{proof}
For $\dot r_*$ we may use any $\dot r^0_\eta$ (from our list $\bar r$) such that $\bar\iota(\eta)\geq \nu$ (i.e. its fully random over $L^{P_\nu}$).
Let $\pi$ be the canonical projection $\pi \colon B_\kappa \rightarrow B^0$, where $B^0$ is as in the hypothesis of item \ref{lem:mix} of the lemma.
Pick $\xi<\kappa$ such that
\begin{enumerate}
\item $\pi_\xi(b) \not \in B^0$; this holds for large enough $\xi$ since $b \not\in B^0$;
\item $r^*$ is a $P_\xi$-name, i.e. $B^0$ is a complete sub-algebra of $B_\xi$.
\item $P_{\xi+1}=\am(P_\iota,D_\xi,r_*,r_*)$, where $\iota \geq \nu$.
\end{enumerate}
Let $b_0$ denote $\pi_\xi(b)$.
Clearly, there is $p \in D_\xi$, $p \leq b_0$ such that $\pi(p)\not\leq b_0$: for otherwise, the set
\[X =\{ d \in B^0 \setdef d\leq b_0\} \]
would be predense in $P_\xi$ below $b_0$, and thus $b_0=\sum X\in B^0$, contradiction.
So pick $p$ as above and let $q \in D_\xi$, $q \leq \pi(p)$ such that $q \cdot b_0=0$. Let $b_1 = \pi(q)$.
Look at the condition $\bar p \in (D_\xi)^\Int_f$ such that $\bar p(-1)=q$, $\bar p (0)= b_1\cdot p$ and for $i \in \Int\setminus\{-1,0\}$,
$\bar p(i)=b_1$. Then we have $\bar p \in (D_\xi)^\Int_f$, $\bar p \leq p \leq b_0$. Letting $\Phi$ denote the automorphism of $B_\kappa$ resulting from $P_{\xi+1}$, we have $\Phi(\bar p) \leq q$ whence $\Phi(\bar p) \cdot b_0 = 0$. So as $\bar p \leq \pi_\xi(b)$ and $\Phi(\bar p)\cdot b =0$, it follows that $\Phi(b)\neq b$; for otherwise since $\bar p \leq \pi_\xi(b)$, we have $\bar p \cdot b \neq 0$ but $\Phi(\bar p \cdot b) = \Phi(\bar p) \cdot b = 0$.
The second claim is clear from the construction, as $\Phi_\rho(\dot r^0_\rho)=\dot r^1_\rho$ for each $\rho<\kappa$.
\end{proof}
Finally, we show in $L[G]$:
\begin{lem}
Say $s \in [\On]^\omega$, $\phi$ a formula. If $X =\{r \in \reals \setdef \phi(r,s) \}$, $X$ is measurable.
\end{lem}
\begin{proof}
Let $X, s$ be as above, and say $s=\dot s^{G}$. Without loss of generality, $\dot s$ is a $P_\nu$-name, where $\nu<\kappa$ and $\forces_\nu \dot s \in [\On]^\omega$ (by lemma \ref{it:reals:are:caught}).
Fix $\dot r_*$ as in the previous lemma. Let $\dot B(\dot r_*)$ be a $P_\nu$-name for the complete sub-algebra of $\quot{B_\kappa}{P_\nu}$ generated by $\dot r_*$ in $L[G_\nu]$.
\begin{claim*}
$\bv{\phi(\dot r_*,\dot s)}^{B_\kappa} \in \ro(P_\nu)*\bg{\dot r_*}$.
\end{claim*}
\begin{proof}[Proof of Claim]
Write $B^0=P_\nu * \dot B(\dot r_*)$ and $b = \bv{\phi(\dot r_*,\dot s)}^{{B_\kappa}}$. Towards a contradiction, assume $b \not\in B^0$. By lemma \ref{auto}, (\ref{lem:mix}), there is an automorphism $\Phi$ of $B_\kappa$ such that $\Phi(b)\neq b$ while $\Phi(\dot s)=\dot s$ and $\Phi(\dot r)=\dot r$.
This is a contradiction, as we infer
\[ b = \bv{ \phi (\dot r_*,\dot s ) }^{B_\kappa}= \bv{ \phi (\Phi(\dot r_*),\Phi(\dot s) ) }^{B_\kappa} = \Phi(b). \]
\end{proof}
\noindent
Let $N^*$ denote
\[ \bigcup \{N \in L[G_\nu] \setdef L[G_\nu]\models N \text{ has measure zero}\},\]
and let $\dot N^*$ be a $P_\nu$-name for this set. $N^*$ is null in $L[G]$.
We find a Borel set $B$ such that for arbitrary $r \not \in N^*$, we have $r \in X \iff r \in B$. Then $X\setminus N^*=B\setminus N^*$ is measurable, finishing the proof. We may regard $\bg{\dot r_*}$ as identical to the Random algebra in $L[G_\nu]$, so we may write $\bv{\phi(\dot r_*,\dot s)}^{\quot{B_\kappa}{B_\nu}} = \ecn{B}$ for a Borel set $B$.
To show $B$ is the Borel set we were looking for, let $r \not \in N^*$ be arbitrary.
Find $\dot r$ and $p \in G$ such that $\dot r^G=r$ and $p \forces \dot r \not \in \dot N^*$, i.e. $p$ forces $\dot r$ is random over $L^{P_\nu}$. By \ref{lem:homogeneity:for:random:reals} of the previous lemma, there is an automorphism $\Phi$ of $P_\kappa$ and $q \in G$ such that $q\forces \Phi(\dot r_*)=\dot r$, and thus $\Phi(\dot r_*)^G={\dot r_*}^{\Phi^{-1}[G]}=\dot r^G$.
We also have $\pi_\nu \circ \Phi = \Phi \circ \pi_\nu = \pi_\nu$ and so
$\Phi(\dot s)^G={\dot s}^{\Phi^{-1}[G]}=\dot s^G$.
We have
\begin{multline*}
\phi(\dot r^G, \dot s^G) \iff \bv{\phi(\dot r, \dot s)}\in G \iff \\
\iff \bar \Phi^{-1}(\bv{\phi(\dot r, \dot s)}) \in \bar \Phi^{-1}[G] \iff \\
\iff \bv{\phi(\bar \Phi^{-1}(\dot r), \dot s)} \in \bar \Phi^{-1}[G]\iff \\
\iff \bv{\phi(\dot r_*, \dot s)} \in \bar \Phi^{-1}[G]\cap\bg{\dot r_*}\iff \dot r_*^{\bar \Phi^{-1}[G]}\in B
\end{multline*}
As $\dot r_*^{\bar \Phi^{-1}[G]}=\dot r^G$, we are done.
\end{proof}
\chapter{Introduction}
A central theme in descriptive set theory of the reals is that of the relation between definability and regularity. That is, one asks which sets in the projective hierarchy have regularity, in the sense of e.g. being Lebesgue measurable, having the Baire property or the perfect set property.
In recent years, regularity in this sense has been associated to the study of measurability with respect to an ideal on the real numbers and a forcing notion associated to this ideal, such as Sacks, Miller, Laver, Cohen, Random, Mathias and others.
For the purpose of this article, we will study only the examples of the null and the meager ideals, with the associated Cohen and Random forcing and the regularity properties of being Lebesgue measurable and the Baire property.
Nevertheless, the techniques developed here hold the promise of being applicable to the study of all these ideals.
The broad question we are concerned with in this article is: where in the projective hierarchy does the least irregular set appear?
Clearly this is independent of $ZFC$; under $V=L$, irregular sets appear on the lowest possible level, $\mathbf{\Delta}^1_2$, and in Solovay's model, all sets are regular.
It is also possible that the least irregular set appears at some higher level $n > 2$, in inner models of $\mathbf{\Pi}^1_{n-2}$-determinacy which have a $\mathbf{\Delta}^1_{n}$-good well-ordering of the reals (e.g. under sharps for $n=3$ and in the minimal canonical inner model with $n-3$ Woodin cardinals for $n>3$).
Observe that in all the above examples, the effects on different ideals are always similar, i.e. the least non-measurable set appears at the same level as the least one lacking the Baire property.
So a slightly more subtle question arises: can we manipulate the behavior independently, for distinct notions of regularity? E.g. Can we have a model where all projective sets are Lebesgue measurable but there is a projective set without the Baire property?
That some interaction is possible can be seen from \cite{b:meas=>baire}: if all $\mathbf{\Sigma}^1_2$ sets of reals are Lebesgue measurable, then also they have the Baire property.
A seminal result was Shelah's \cite{shelah:amalgamation}, where he finds two models of interest to our question:
In the first, all sets have the Baire property, but there is a set which is not Lebesgue measurable, starting from just the consistency of $ZFC$.
He proves that if all $\mathbf{\Sigma}^1_3$ sets are Lebesgue measurable, then $\omega_1$ must be inaccessible to reals; thus, in this model, there is a non-measurable, projective set.
In the second, all sets are Lebesgue measurable but there is a set which does not have the Baire property.
We improve this by making the counterexample projective:
\begin{thm}
Starting from ``$V=L$ and there exists a Mahlo cardinal'', we can force to obtain a model where all projective sets are Lebesgue measurable, but there is a (lightface) $\Delta^1_3$ set without the Baire property.
\end{thm}
The main tool developed in \cite{shelah:amalgamation} to selectively attack a specific ideal is \emph{amalgamation}, which allows to construct partial orders which admit certain automorphisms.
In our case we want a partial order where all partial isomorphisms of Random subalgebras extend to an automorphism of the whole forcing --- in fact, we want to build an iteration where this is true for every tail segment.
Thus, we obtain a model where all projective sets are measurable.\footnote{In $L({}^\omega \On)$ or in $L(\reals)$ of our extension, all sets will be measurable, as in Solovay's model, and there will be a projective set without the Baire property. Deliberately, we only consider the bigger model, where choice holds, and its projective sets.}
In \cite{david:singleton}, it is shown how iterations of Jensen-coding may be used to make a set of reals projective.
This is done by adding and then coding branches through large constructible Suslin trees, where the coding is projective, i.e. localized.
This localization is often referred to as ``Ren\'e David's trick'' or ``killing universes''.
Our iteration will be of length $\kappa$, where $\kappa$ is the least Mahlo in $L$.
In this iteration we collapse every cardinal below $\kappa$ to $\omega$.
Thus, we add many Cohen reals, which easily yield a set without the Baire property, which we call $\Gamma^0$.
For every real $s$ added by the iteration, we have to code a set of branches of $\kappa^{++}$-Suslin trees of $L$, projectively, into a real.
From this code it will possible to determine in a projective way whether $s \in \Gamma^0$.
This makes $\Gamma^0$ into a $\Delta^1_3$ set.
Also, for every pair of Random reals added by a tail of the iteration, we have to amalgamate.
Of course it is crucial to show that we can ``catch our tail'', i.e. that all reals in the final model appear in some initial segment of the iteration so that we have in fact treated all reals $s$ and all pairs of random reals of the final model.
To be able to code and amalgamate alternatingly in this iteration, we isolate the concept of iterations of stratified extensions of forcing, which is discussed in section \ref{sec:stratified forcing} and \ref{sec:ext} (the concept of stratified extension is needed since the components of our forcing need not be stratified---luckily, initial segment are stratified in a \emph{coherent way}).
This makes sure a tail of regular cardinals below $\kappa$ is preserved in every initial stage of the iteration.
In section \ref{sec:amalgamation}, we devise a version of amalgamation which is also a stratified extension, and therefore preserves stratification.
In section \ref{sec:main}, after having discussed stratified extension and amalgamation, we motivate and describe the iteration in detail.
We also prove that in the final model, every projective set of reals is measurable.
That $\kappa$ is not collapsed and thus becomes $\omega_1$ in the final model
does not follow directly from the theory of stratified forcing.
Instead, a kind of ``$\kappa$-properness'' is shown in a kind of $\Delta$-systems type argument involving some ideas from stratification, an archetypical use $\diamondsuit$, a special kind of $\Box$-sequence discussed in \ref{sec:coding} and the fact that enough cardinals are preserved in earlier stages of the iteration; this is the argument of \ref{sec:stage kappa}.
Here, we also show that the $\kappa$ stage of the iteration adds no new reals and that we indeed ``catch our tail''.
To allow this proof to go through we need a version of (localized) Jensen coding which uses \emph{Easton support}, and this is developed in section \ref{sec:coding}.
Finally in the last section we show that $\Gamma^0$ is indeed $\Delta^1_3$.
We advise the reader to skim through section \ref{sec:main}, especially the more detailed sketch of the iteration given at the beginning. Also, the beginning of each chapter contains a small introduction.
\chapter{Notation and preliminaries}
We first define \emph{strong projection} (from \cite{handbook_proper}), \emph{strong sub-order} and \emph{independence}.
These are practical in understanding how amalgamation is a stratified extension.
\begin{dfn}
We say \emph{$Q$ is a strong sub-order of $P$} if and only if $Q$ is a complete sub-order of $P$ and for every $p \in P$ and $q\in Q$ such that $q \leq \pi(p)$, we have $q\cdot p \in P$, where $\pi$ denotes the canonical projection from $\ro(P)$ to $\ro(Q)$.
\end{dfn}
\noindent
This is related to the notion of projection\footnote{\cite{handbook_proper} defines (ordinary) projection with
\ref{projection} replaced by the stronger: if $q \leq \pi(p)$, there is $\bar p\leq p$ such that $\pi(\bar p)=q$.} and strong projection:
We say $\pi \colon P \rightarrow Q$ is a \textit{projection} if and only if
\begin{enumerate}[1. ]
\item $p \leq p' \Rightarrow \pi(p)\leq \pi(p')$,
\item $\ran(\pi)=Q$,
\item \label{projection} if $q \in Q$ and $q \leq \pi(p)$, there is $\bar p \in P$ such that $\bar p \leq p$ and $\pi(\bar p) \leq q$.
\end{enumerate}
Observe this implies that $\pi(1_P)=1_Q$.
In \cite{handbook_proper}, $\pi$ is defined to be a \textit{strong projection} if and only if it satisfies the first two requirements above and the following strengthening of the third requirement:
\begin{enumerate}[1. ]
\item[$3'$.] If $q \leq \pi(p)$, there is $\bar p\leq p$ such that
\begin{enumerate}[a. ]
\item $\pi(\bar p)= q$,
\item for any $r \in P$, if $r \leq p$ and $\pi(r) \leq q$ then $r \leq \bar p$. \label{q.p}
\end{enumerate}
\end{enumerate}
This uniquely determines $\bar p$, and we denote it by $q \cdot p$.
If $\pi\colon P \rightarrow Q$ is a projection, $\pi[G]$ generates a $Q$-generic Filter whenever $G$ is a $P$-generic Filter, moreover in fact $\ro(Q)$ is a complete sub-algebra of $\ro(P)$.
If $\pi\colon P \rightarrow Q$ is a strong projection, the map $i$ sending $q \in Q$ to $i(q)=q \cdot 1_P$ is a complete embedding and we can assume that $Q$ is a subset of $P$. It follows from $3'$b. that $\forall p \in P \quad p \leq i(\pi(p))$.
When $Q$ is a complete sub-order of $P$, we say $q \in Q$ is a \emph{reduction (to $Q$) of} $p \in P$ if and only if for all $q' \in Q$, if $q' \leq q$ then $q'$ and $p$ are compatible.
\begin{lem}\label{strong:vs:canonical:proj}
Let $Q$ be a complete sub-order of $P$ and let $\pi$ be the canonical projection $\pi\colon \ro(P)\rightarrow \ro(Q)$.
Say $p \in P$ and $q \in Q$ is a reduction of $p$ such that $q \geq p$; then $q = \pi(p)$.
If $\bar \pi\colon P \rightarrow Q$ is a strong projection, then $\bar \pi$ coincides with the canonical projection on $P$.
\end{lem}
Observe that if $\pi\colon \ro(P) \rightarrow \ro(Q)$ is the canonical projection, then $\pi \res P$ is a strong projection if and only if
for every $p \in P$ and $q \in Q$ such that $q \leq \pi(p)$ we have $p\cdot q \in P$. All of the above gives us:
\begin{lem}\label{stron:proj:equiv}
The following are equivalent:
\begin{itemize}
\item $Q$ is a strong sub-order of $P$.
\item There is a strong projection $\pi\colon P \rightarrow Q$.
\item The restriction of the canonical projection $\pi\colon\ro(P)\rightarrow \ro(Q)$ to $P$ is the unique strong projection from $P$ to $Q$.
\end{itemize}
\end{lem}
Also observe that when $Q$ is a strong sub-order of $P$ which is a strong sub-order of $R$ with $\pi_P\colon R\rightarrow P$ a strong projection, then $1_Q$ forces $\pi_P\res\quot{P}{Q}$ is a strong projection from $\quot{P}{Q}$ to $\quot{R}{Q}$.
Imagine an iteration $R=(Q_0 \times Q_1 )* \dot Q_2$. Then in an extension by $Q_0$, the \pre order $Q_1$ is a complete sub-order of the tail $\quot{R}{Q_0}=Q_1*\dot Q_2$.
This special situation is captured well by the following:
\begin{dfn}\label{indie}
Let $Q$ and $C$ be sub-orders of $P$ with strong projections
$\pi_Q\colon P \rightarrow Q$ and
$\pi_C\colon P \rightarrow C$.
We say \emph{ $C$ is independent over $Q$ in $P$} if and only if for all $c \in C$ and $p \in P$ such that $c \leq \pi_C(p)$, we have
$\pi_Q(p \cdot c) = \pi_Q(p)$.
For a $P$-name $\dot C$, we say \emph{$\dot C$ is independent in $P$ over $Q$} if and only if $\dot C$ is a name for a generic of an independent complete sub-order of $P$; i.e. there is a complete sub-order $R_C$ of $P$ (with a strong projection $\pi_C\colon P \rightarrow R_C$) such that $R_C$ is a dense in $\gen{\dot C}^{\ro(P)}$ and $R_C$ is independent in $P$ over $Q$.
By $\gen{\dot C}^{\ro(P)}$ we mean the smallest Boolean subalgebra of $\ro(P)$ which contains
all the boolean values occurring in the name $\dot C$. Thus $\gen{\dot C}^{\ro(P)}$ is a ground model object.
\end{dfn}
We use the following in \ref{index:sequ} (p.~\pageref{index:sequ}), via the notion of ``remoteness'' (see also lemma \ref{remote:lemma:not:in}
and section \ref{sec:remote}).
\begin{lem}\label{indie:lemma:not:in}
If $C$ is independent over $Q$ in $P$, $1_Q$ forces that $C$ is a complete sub-order of $\quot{P}{Q}$ and $\pi_C$ is a strong projection.
If $\dot C$ is a $P$-name which is independent over $Q$, then $\dot C$ is not in $V^Q$.
\end{lem}
\paragraph*{Iterations.}
We say $\bar Q^\theta=(P_\iota,\dot Q_\iota)_{\iota<\theta}$ is an iteration if and only if
for each $\iota <\theta$,
\begin{enumerate}
\item $\forces_{P_\iota} \dot Q_\iota$ is a \pre order
\item \label{def:it:thread} $P_\iota$ consists of sequences $p$ such that $\dom(p) = \iota$ and for each $\nu <\iota$, $p(\nu)$ is a $P_\nu$-name such that
\begin{equation}\label{thread}
1_{P_\nu}\forces p(\nu) \in \dot Q_\nu.
\end{equation}
\item \label{def:it:order} The ordering of $P_\iota$ is given by:
\begin{equation}\label{threads:order}
r \leq p \iff \forall \nu<\iota \quad r \res \nu \forces_{P_\nu} r(\nu)\leq_{\dot Q_\nu} p(\nu).
\end{equation}
\end{enumerate}
\noindent
We state this as some would not agree with \eqref{thread}.
Fix an iteration $\bar Q^{\theta+1}$.
\begin{dfn}\label{it:terminology}
\begin{enumerate}
\item We call a sequence $p$ with $\dom(p) = \theta$ a \emph{thread through (or in) $\bar Q^\theta$} if and only if it satisfies (\ref{thread}). The set of threads through $\bar Q^\theta$ we shall sometimes denote by $\prod \bar Q^\theta$.\label{threads:def}
It is endowed with the ordering given by (\ref{threads:order}) (for $r$, $p \in \prod\bar Q^\theta$).
\item \label{threads:var:def} We also use the term \emph{thread} in a second, related sense: if $\bar p\in \prod_{\eta<\iota <\theta} P_\iota$ for some $\eta<\theta$---i.e $\bar p=(p_\iota)_{\iota\in(\eta,\theta)}$ and for each $\iota\in \dom(\bar p)$ we have $p_\iota \in P_\iota$---we say $\bar p$ forms or defines or simply \emph{is a thread (through $\bar Q^\theta$)} if and only if
\begin{equation}\label{threads:var}
\forall \iota, \bar\iota \in \dom(\bar p) \quad \iota\leq\bar\iota \Rightarrow \pi_\iota(p_{\bar\iota})=p_\iota.
\end{equation}
\end{enumerate}
\end{dfn}
\noindent
The point is that a thread in the first sense yields one in the second sense and vice versa.
\begin{dfn}
More generally we will also call sequence
$(P_\iota)_{\iota\leq\theta}$ an iteration when in comes strong projections $\pi^{\bar \iota}_\iota\colon P_{\bar \iota} \rightarrow P_\iota$, for each $\iota <\bar \iota \leq \theta$, and for each (limit) $\iota$, $P_\iota$ consists of threads in the second sense.
\end{dfn}
Clearly an iteration in the second sense can be written as an iteration in the first sense and vice versa.
\paragraph*{Ideals.}
If $c$ is a Borel-code, we write $B_c$ for the Borel set coded by $c$. Of course given two models of set theory, both containing a Borel code $c$, it may be that $c$ codes a different set in each model.
\begin{dfn}\label{generic:reals}
Say $\ro(Q)$ is a complete sub-algebra of $\ro(P)$.
Let $\dot G$ be is the canonical $P$-name for the $P$-generic over $V$ and $\pi\colon \ro(P) \rightarrow \ro(Q)$ the canonical projection.
Let $\dot I$ be a $P$-name for an ideal on $\reals$ in the extension via $P$.
For a $P$-name $\dot r$ and $p \in P$, we say is \textit{$p$ forces $\dot r$ is $\dot I$-generic over $V^Q$},
just if $p\forces_P \dot r \in \reals$ and for every $Q$-name for a Borel code $\dot c$,
\[ p \forces_P B_{\dot c} \in\dot I \Rightarrow \dot r \not \in B_{\dot c}.\]
We say $p$ forces $\dot r$ is \textit{fully} $\dot I$-generic over $V^Q$ if and only if $p$ forces $\dot r$ is $\dot I$-generic over $V^Q$ and in addition, for every $Q$-name $\dot c$ such that $\pi(p)\forces_Q \dot c$ is a Borel code,
\[p \forces_P \dot r \not\in B_{\dot c} \Rightarrow p \forces_P B_{\dot c} \in\dot I.\]
In other words, $p$ does not force anything non-trivial about $\dot r$.
We say \textit{$\dot r$ is (fully) $\dot I$-generic} just if $1_P$ forces $\dot r$ is (fully) $\dot I$-generic.
Instead of ``$\dot I$-generic'',
\begin{itemize}
\item If $\dot I$ is a name for the ideal of sets with measure zero, we say \textit{Random over $V^Q$}.
\item If $\dot I$ is a name for the ideal of meager sets, we say \textit{Cohen over $V^Q$}.
\item If $\dot I$ is a name for $\pset_{<\omega}(\reals)^{V[\pi(\dot G)]}$---or equivalently, for $\pset(\reals^{V[\pi(\dot G)]})$---we say $\dot r \not\in V^Q$ or \emph{$\dot r$ is not in $V^Q$}.
\item If $\dot I$ is a name for the ideal of sets which are bounded by a real in $V[\pi(\dot G)]$ in the sense of eventual domination, we say \textit{unbounded over $V^Q$}.
\end{itemize}
The terms \textit{$p$ forces $\dot r$ is fully Random over $V^Q$} and \textit{fully Cohen over $V^Q$} are to be understood analogously.
\end{dfn}
\begin{lem}\label{unbounded}
Let $P$ and $Q$ be arbitrary partial orders and let $\dot r$ be a $P$-name for a real.
If
$\dot r$ is unbounded over $V$, viewing $\dot r$ as a $P\times Q$ name via the natural embedding,
$\dot r$ is unbounded over $V^Q$.
\end{lem}
\noindent For a proof, see \cite[lemma 3.3, p.~392]{jr:amalgamation}.
\chapter{The set $\Gamma^0$ is $\Delta^1_3$}\label{sec:preserving:coding}
We now check that $\Gamma^0$ is in fact $\Delta^1_3$. By \cite{b:meas=>baire}, this is optimal, since under the assumption that all $\Sigma^1_2$ sets are Lebesgue-measurable, all $\Sigma^1_2$ sets do have the property of Baire.
Let $\Theta(r,s,\alpha,\beta)$ be the sentence
\begin{eqpar*}
$L_\beta[r,s]$ is a model of $\ZF^-$ and of
``$\alpha$ is the least Mahlo and $\alpha^{++}$ exists''.
\end{eqpar*}
\begin{dfn}
For an ordinal $\alpha$ and $C \in {}^\alpha 2$, write $\sigma \is C$ to express $\sigma$ is an initial segment of $C$, i.e. for some $\rho < \alpha$, $\sigma = C \res \rho$.
Let $\phi(x)$ be a formula.
When we write $\forall^* \sigma \is C \phi(\sigma)$, we mean there exists $\zeta<\alpha$ such that for all $\rho \in (\zeta, \alpha)$, $\phi(C\res\rho)$ holds. In other words, \emph{ for almost all initial segments $\sigma$ of $C$, $\phi(\sigma)$ holds}.
\end{dfn}
\noindent
For $j \in \{ 0,1\}$, let $\Psi(r ,j)$ denote the formula
\begin{eqpar*}
$\exists s \in {}^\omega 2 \quad \forall \alpha, \beta<\kappa$ \emph{ if } $\theta(r,s,\alpha,\beta)$ \emph{ then:}\\
$L_\beta[r,s] \models$``$\exists C \in {}^\alpha 2 \quad \forall^* \sigma \is C \quad \forall (n,i) \in \omega \times 2 \quad$ \\
$ r(n)=i \Rightarrow T^\alpha(\sigma,n,i,j)$ has a branch.''
\end{eqpar*}
\begin{lem}\label{coding:survives}
For $j\in\{0,1\}$ and any real $r$,
\[ r \in (\dot\Gamma^j)^G \iff \Psi(r,j).\]
\end{lem}
\begin{proof}
For $\xi\leq \kappa$, let $\mathcal{F_\xi}$ be the smallest set closed under (relational) composition and containing all functions $F=\Phi^\zeta_\rho,(\Phi^\zeta_\rho)^{-1}$ such that $\dom F \subseteq P_\xi$. In other words, $\mathcal{F_\xi}$ is the closure under relational composition of
\[
\{ \Phi^\zeta_\rho,(\Phi^\zeta_\rho)^{-1} \setdef E^3(\alpha^\zeta_\rho) < \xi \}.
\]
First assume $r \in (\dot\Gamma^j)^G$ and show $\Psi(r,j)$ holds. If $j=0$, by definition of $\dot \Gamma^0$ we can find $\eta < \kappa$ and $\Phi \in \mathcal{F}_\kappa$ such that $r= (\Phi(\dot c_\eta))^G$. If $j=1$, we fix $\eta<\kappa$ such that $r= (\dot s_{\eta-1})^G$.
Let $\dot r_0$ denote $\dot c_\eta$ if $j=0$ and let $\dot r_0$ denote $\dot s_{\eta-1}$ if $j=1$ and let $r_0=(\dot r_0)^G$ .
In either case, at stage $\xi=E^2(\eta)$ we force with Jensen coding, adding a real $s_0$ such that
\begin{eqpar*}
for all $\alpha,\beta <\kappa$, if $\theta(r_0,s_0,\alpha,\beta)$ then $C_\eta\res\alpha, r_0 \in L_\beta[s_0]$ and \\
$L_\beta[s_0]\models$ ``$\forall \sigma$ such that $\sigma \is C_\eta\res\alpha$ and for all $n,i$ such that $r_0(n)=i$, $T^\alpha(\sigma,n,i,j)$ has a branch''.
\end{eqpar*}
So
\[ 1_{P_\kappa}\forces \Psi(\dot r_0,j),\]
which completes the proof in case $j=1$. For $j=0$, apply $\Phi$ to get
\[ 1_{P_\kappa}\forces \Psi(\Phi(\dot r_0),j), \]
and we are done as $(\Phi(\dot r_0))^G=r$.
Now assume $\Psi(r,j)$ and show $r \in (\dot\Gamma^j)^G$:
Fix $s$ to witness $\Psi(r,j)$. It must be the case that
\begin{eqpar}\label{lots:of:branches}
$L[r,s] \models \exists C \in {}^\kappa 2 \quad \forall^* \sigma \is C \quad \forall (n,i) \in \omega \times 2 \quad$ \\
$ (r(n)=i) \Rightarrow T(\sigma,n,i,j)\text{ has a branch }$.
\end{eqpar}
\noindent
For let $L_\beta[r,s]$ be isomorphic to an elementary sub-model of $L_{\kappa^{+3}}[r,s]$ which contains $r$ and $s$, and let $\alpha$ be the least Mahlo in $L_\beta[r,s]$. Then as $\theta(r,s,\alpha,\beta)$ holds, by $\Psi(r,j)$,
\begin{eqpar*}
$L_\beta[r,s] \models \exists C \in {}^\alpha 2 \quad \forall^* \sigma \is C \quad \forall (n,i) \in \omega \times 2 \quad$ \\
$ (r(n)=i) \Rightarrow T^\alpha(\sigma,n,i,j)\text{ has a branch }$.
\end{eqpar*}
So by elementarity, (\ref{lots:of:branches}) holds.
Fix $\xi<\kappa$ such that $r,s \in W[G_\xi]$ and fix $C$ witnessing (\ref{lots:of:branches}). Pick $\zeta<\kappa$ such that
\begin{eqpar}\label{zeta:large:enough}
for $\Phi \in \mathcal{F}_\xi$ and $\nu, \nu' < \xi$ such that $\Phi\neq \id$ and $\nu\neq \nu'$ we have $\Phi(C_\nu)\res\zeta \neq C_{\nu'}\res\zeta$.
\end{eqpar}
This is possible by lemma \ref{index:sequ}.
As (\ref{lots:of:branches}) holds, we can also assume $\zeta$ to be large enough so that whenever $r(n)=i$, and $\rho \geq \zeta$,
$T(C\res\rho,n,i,j)$ has a branch in $L[r,s]$.
Since $r,s \in W[G_\xi]$, lemma \ref{no:unwanted:branches} below gives us: for any $n$ and $i$, if the tree $T(C\res\zeta,n,i,j)$ has a branch in $L[r,s]$
then there is $\Phi \in \mathcal{F}_\xi$ and $\eta <\xi$ such that $C\res\zeta \is \Phi(C_\eta)$.
By (\ref{zeta:large:enough}), $\Phi$ and $\eta$ are unique and do not depend on $n$ and $i$, so let $\Phi$ and $\eta$ be fixed.
By the way, it follows that $C=\Phi(C_\eta)$ (which we do not use in the following).
More importantly, whenever $r(n)=i$, $T(\Phi(C_\eta)\res\zeta,n,i,j)$ has a branch in $L[r,s]$.
Moreover, lemma \ref{no:unwanted:branches} yields that whenever $T(C\res\rho,n,i,j)$ has a branch in $L[r,s]$,
\begin{enumerate}
\item if $j=0$ then $\eta$ is a limit and $(\Phi(\dot c_\eta))^G(n)=i$
\item if $j=1$ then $\eta$ is a successor and $(\Phi(\dot s_{\eta-1}))^G(n)=i$.
\end{enumerate}
Thus, in the first case, $r=\Phi(c_\eta)$ and so $r \in (\dot \Gamma^0)^G$. In the second case, $r = (\Phi(\dot s_{\eta-1}))^G$. As $(\dot s_{\eta-1})^G \in (\dot \Gamma^1)^G$ and $(\dot \Gamma^1)^G$ is closed under all the automorphisms $\{\Phi^*_\rho \setdef \rho <\kappa \}$ (by lemma \ref{disjoint}), $r \in (\dot \Gamma^1)^G$.
\end{proof}
For $\xi \leq \kappa$, let $I_\xi$ be the set of triples $(\sigma,n,i,j)$ such that for some $\eta < \xi$ and $\Phi \in \mathcal{F}_\xi$, $\sigma \is \Phi(C_\eta)$ and
\begin{enumerate}
\item if $\eta$ is limit ordinal, $\Phi(c_\eta)(n)=i$ and $j=0$
\item if $\eta$ is a successor ordinal, $\Phi(s_{\eta-1})(n)=i$ and $j=1$.
\end{enumerate}
\begin{lem}\label{no:unwanted:branches}
Say $\xi_0<\kappa$ and let $u$ an arbitrary real in $L[G\res\xi\xi_0]$. Then $T(\eta,\sigma,n,i,j)$ has a branch in $L[u]$ only if $(\eta,\sigma,n,i,j) \in I_{\xi_0}$.
\end{lem}
\begin{proof}
Fix $\nu \in I$, $\xi_0<\kappa$ and $p_0 \in P_{\xi_0}$ such that $p_0 \forces \check \nu \not \in \dot I_{\xi_0}$ in $L[\bar B]$, where $\nu=(\eta,\sigma,n,i,j)$.
Let $\bar B^-$ denote $\{\bar B(\xi)\}_{\xi \in I\setminus\{\nu\}}$.
We will show in a moment that $P_{\xi_0}(\leq p_0)$ is equivalent to a forcing $P^*_{\xi_0} \in L[\bar B^-]$, whence $ \bar T * P_{\xi_0}(\leq p_0)$ is equivalent to
\[ \Big[ \big(\prod^{<\kappa}_{\zeta \in I\setminus\{\nu\}} T(\zeta)\big) * P^*_{\xi_0} (\leq p_0) \Big]\times T(\nu). \]
Assuming this for the moment, we can prove the lemma thus:
As $T(\nu)$ doesn't add reals, any real $u \in L[\bar B][G_{\xi_0}]$ is actually an element of $L[\bar B^-][G_{\xi_0}]$, and as $T(\nu)$ is Suslin in $L[\bar T^-]$ and $P^*_{\xi_0}$ is $\kappa^{+}$-centered,
$T(\nu)$ remains Suslin in $L[\bar B^-][G_{\xi_0}]$ and thus in $L[u]$.
It remains to see that $P_{\xi_0}(\leq p_0)$ is equivalent to a forcing which is an element of $L[\bar B^-]$. For the purpose of carrying out the inductive proof, we prove a stronger statement, in the claim below.
First, note that for $\xi < \bar \xi \leq \xi_0$, as $I_\xi \subseteq I_{\bar \xi}$,
\begin {equation*}\label{bool:val}
\bv{\nu\not\in \dot I_{\bar\xi}}\leq \bv{\nu\not\in \dot I_\xi},
\end{equation*}
and so
\begin {equation*}
\pi_\xi(\bv{\nu\not\in \dot I_{\bar\xi}})\leq \bv{\nu\not\in \dot I_\xi}
\end{equation*}
Let $b_\xi$ denote $\bv{\nu\not\in \dot I_\xi}^{B_\xi}$, for $\xi<\xi_0$.
\begin{claim}
For each $\xi \leq\xi_0$, there is an isomorphism\footnote{By isomorphism, we mean of course $j_\xi$ is injective on the separative quotient of its domain.}
\[ j_\xi \colon P_\xi(\leq b_\xi)\rightarrow P^*_\xi, \]
such that
\begin{eqpar}\label{fit:together}
for $\xi < \bar \xi \leq\xi_0$ and $p \in P_{\bar \xi}(\leq b_\xi)$,
$j_\xi(\pi_\xi(p))=\pi_\xi (j_{\bar \xi}(p))$.
\end{eqpar}
Moreover, $P^*_\xi\in L[\bar B^-]$ and $P^*_\xi = P_\xi(\leq b_\xi) \cap L_{\kappa^{++}}[B^-]$.
\end{claim}
There is no need to distinguish between $G_\xi$ and $j_\xi[G_\xi]$, we write $G_\xi$ for either one.
Observe $\bar T$ is $\kappa^{++}$-distributive,
so we have
\begin{equation}\label{H:kappa}
\Hhier(\kappa^{++})^{L[\bar B]}=\Hhier(\kappa^{++})^L
\end{equation}
At heart, the claim is a consequence of this simple observation:
\begin{fct}\label{H:kappa:names}
If $P$ has the $\kappa^{++}$-chain condition and $p\forces \dot x \in H(\kappa^{++})$, there is $\dot x'\in H(\kappa^{++})$ such that $p\forces \dot x=\dot x'$.
\end{fct}
\begin{proof}
Use nice names.
\end{proof}
The induction splits into cases.
For the successor case, assume $j_\xi$ is already defined and define $j_{\xi+1}$.
Observe that by induction, $\bar T * P_\xi$ is equivalent to
\[ \Big[\big(\prod^{<\kappa}_{\zeta \in I\setminus\{\nu\}}T(\zeta)\big) * P^*_\xi\Big]\times T(\nu), \]
and since $T(\nu)$ is $\kappa^{++}$-distributive in $L[\bar B^-][G_\xi]$ (because it is still Suslin in that model),
\begin{equation}\label{H:kappa:ext}
H(\kappa^{++})^{L[\bar B][G_\xi]}\subseteq L[\bar B^-][G_\xi],
\end{equation}
\paragraph*{Easiest Case:} As a warm up, assume $\xi \in E^1$. Thus $P_{\xi+1} = P_\xi \times Q$ for $Q\in L$.
Of course, $P^*_\xi \times Q \in L[\bar B^-]$.
We can set $j_{\xi+1}(p,q)=(j_\xi(p),q)$.
Observe that the claim asks for an isomorphism of $P^*_{\xi+1}$ with $P_\xi*\dot Q_\xi(\leq b_\xi)$, not $P_\xi(\leq b_\xi)*\dot Q_\xi$.
So to formally satisfy the claim -- and to make the induction work in the next step -- restrict $j_\xi$ to $P_{\xi+1}(\leq b_{\xi+1})$.
We should check $P^*_\xi \times Q (\leq b_{\xi+1}) \in L[\bar B^-]$, though:
Lightheartedly identify $P_{\xi+1}$ names and $P^*_\xi \times Q$-names and assume (by fact \ref{H:kappa:names} and (\ref{H:kappa})) that $\dot I_\xi \in L[\bar B^-]$.
Then for $p \in P^*_\xi \times Q$,
\[ p \forces \nu \in \dot I_{\xi+1} \]
is absolute between $L[\bar B]$ and $L[\bar B^-]$, so $P^*_\xi \times Q (\leq b_{\xi+1}) \in L[\bar B^-]$.
So set $P^*_{\xi+1}=P^*_\xi \times Q (\leq b_{\xi+1})$.
\paragraph*{Jensen Coding (and another easy case):}
Now, assume $\xi \in E^2$, i.e.
$P_{\xi+1}=P_\xi*\dot Q_\xi$
where $\dot Q_\xi$ is a name for Jensen coding.
Now it is crucial that we work below $b_\xi = \bv{\nu\not\in \dot I_\xi}^{B_\xi}$:
Work in $L[\bar B][G_\xi]$ for now, where $G_\xi$ is $P^*_\xi$-generic over $L[\bar B]$. Then $\nu \not \in (\dot I_\xi)^{G_\xi}$,
so the set of branches we code at this stage does not contain $B(\nu)$.
Thus $\dot Q_\xi^{G_\xi}$ is a subset of $H(\kappa^{++})$ (of the extension), which is definable over $\langle H(\kappa^{++}), \bar B^- \rangle$. By (\ref{H:kappa:ext}), $\dot Q_\xi^{G_\xi} \in L[\bar B^-][G_\xi]$.
This immediately implies that $P_{\xi+1}$ is equivalent to a forcing which is an element of $L[\bar B^-]$, but in order to carry out the inductive proof at limits, we need (\ref{fit:together}).
For this, let $\phi(x)$ be a formula defining membership in $\dot Q_\xi^{G_\xi}$ over $\langle H(\kappa^{++}), \bar B^- \rangle$ in $L[\bar B][G_\xi]$.
Set
\begin{multline}\label{def:in:H:kappa}
P^*_{\xi+1} =
\{ (p,\dot q) \setdef p \in P^*_\xi, \dot q \in H(\kappa^{++}) \text{ is a $P^*_\xi$-name, } P^*_\xi \forces \dot \phi(q)^{H(\kappa^{++})} \}
\end{multline}
As $P^*_\xi$ has the $\kappa^{++}$-chain condition, any $x \in H(\kappa^{++})^{L[\bar B][G_\xi]}$ has a $P^*_\xi$-name in $H(\kappa^{++})^{L[\bar B]}$.
Therefore, by (\ref{H:kappa}),
\[ P^*_\xi \forces \dot \phi(q)^{H(\kappa^{++})} \]
is absolute between $L[\bar B]$ and $L[\bar B^-]$ and thus (\ref{def:in:H:kappa}) witnesses that $P^*_{\xi+1} \in L[\bar B^-]$.
For $(p,\dot q) \in P_{\xi+1}$, we can now define $j_{\xi+1}(p,\dot q)$. Since
\[ P^*_\xi \forces j_\xi(\dot q) \in j_\xi(\dot Q_\xi)\subseteq H(\kappa^{++}) \]
using fact \ref{H:kappa:names}, we can find a $P^*_\xi$-name $\dot q' \in H(\kappa^{++})$ such that $P^*_\xi\forces j_\xi(\dot q)=\dot q'$. Let $j_{\xi+1}(p,\dot q)=(j_\xi(p),\dot q')$.
Clearly, $j_\xi(p,\dot q) \in P^*_{\xi+1}$. It is straightforward to check that $j_{\xi+1}$ preserves the ordering and is onto. It is injective on the separative quotient of $P_{\xi+1}$.
Again, as in the previous case, restrict $j_\xi$ to $P_{\xi+1}(\leq b_{\xi+1})$ to formally satisfy the claim.
The case $\xi \in E^0$ can be treated in an analogous -- but simpler -- way.
\begin{rem}
Observe, by the way that for any $P^*_\xi$ name $\dot Q$ such that $P^*_\xi \forces \dot Q = j_\xi(\dot Q_\xi)$,
\[ P^*_{\xi+1} = (P^*_\xi * \dot Q) \cap H(\kappa^{++}). \]
In particular, it follows by induction that $P^*_{\xi+1}=P_{\xi+1} \cap H(\kappa^{++})$.
Moreover, if $\dot Q \in L[\bar B^-]$ then
\[ P^*_{\xi+1} = ((P^*_\xi * \dot Q) \cap H(\kappa^{++}))^{L[\bar B^-]}. \]
If $\dot Q_\xi$ is chosen reasonably (e.g. in the most obvious way), in fact $j(\dot Q_\xi) \in L[\bar B^-]$.
\footnote{It would be tempting to define
$j_{\xi+1}(p,\dot q)=(j_\xi(p),j_\xi(\dot q))$,
but we do not know if $j_\xi(\dot q) \in L[\bar B^-]$.}
This means that we could find $P^*_\xi$ by interpreting the definition of $P_\xi$ \emph{in $L[B^-]$} if we commit to using only \emph{names which have size at most $\kappa^+$}.
\end{rem}
\paragraph*{Amalgamation:}
Now let $\xi \in E^3$ and say $\xi=E^3(\alpha^0_\rho)$, i.e. $P_{\xi+1}=\am(P_{\iota},P_\xi,f,\lambda_\xi)$, where $f$ is the isomorphism of the algebras generated by some $P_\xi$-names
$\dot r_0$ and $\dot r_1$, and let $\pi_i$ denote the canonical projection from $P_\xi$ to
the domain and range of $f$.
Let $\Phi$ be the resulting automorphism.
Let $R$ denote the set of $\bar p \in P_{\xi+1}$ such that for all $i \neq 0$, $\bar p(i)\in \blowup{D_\xi}(\leq b_\xi)$ and
$\bar p(0) \in P_\xi(\leq b_\xi)$.
We show $P_{\xi +1 }(\leq b_{\xi+1}) \subseteq R$ and that $R$ is in $L[\bar B^-]$.
For the first, it is crucial that $\dot I_{\xi+1}$ is closed under $\Phi$.
Say $\bar p \in P_{\xi +1 }(\leq b_{\xi+1})$, that is, $\bar p \forces \nu \not \in \dot I_{\xi+1}$.
By the definition of $\dot I_{\xi+1}$,
for each $i \in \Int$,
\begin{equation*}
\Phi^i(\bar p) \forces_{P_{\xi+1}} \nu \not\in\dot I_{\xi+1},
\end{equation*}
and so
\begin{equation*}%
\bar p(i) = \bar\pi(\Phi^i(\bar p)) \forces_{\blowup{P_\xi}} \nu \not\in\dot I_\xi,
\end{equation*}
where $\bar \pi$ denotes the canonical projection from $P_{\xi+1}$ to $\blowup{P_\xi}$. Thus, $\bar p \in R$.
By induction, $P_\xi(\leq b_\xi)$ is isomorphic to $P^*_\xi$.
A little care is needed to see $\blowup{D_\xi}(\leq b_\xi)$ is in $L[\bar B^-]$:
$\blowup{D_\xi}(\leq b_\xi)$ is not the same as $\blowup{D_\xi(\leq b_\xi)}$ in general.
The two orderings are equivalent, but once more, this doesn't mean that we can use them interchangeably in the definition of $R$.
At the same time, $(p, \dot b_0, \dot b_1) \in \blowup{D_\xi}(\leq b_\xi)$ \emph{does not imply} that $p \leq b_\xi$, and so $p$ needn't be in the domain of $j_\xi$.
So we have to check that in fact, $B^*_\xi = \ro(P^*_\xi) \in L[\bar B^-]$. This is because we may regard $B^*_\xi$ the collection of regular open cuts which are given by antichains in $P^*_\xi$.
As $P_\xi^*$ has the $\kappa^{++}$-chain condition all such antichains and hence all regular cuts are in $L[\bar B^-]$ (once more by (\ref{H:kappa})). So $B^*_\xi \in L[\bar B^-]$ and $j_\xi$ can be viewed as an isomorphism of $B_\xi(\leq b_\xi)$ with $B_\xi^*$.
Thus, as we have assumed $\dot r_0$, $\dot r_1$ are in $L[\bar B^-]$, we can define $\bg{ \dot r_i }^{B^*_\xi}$ and canonical projections from $B^*_\xi$ to $P^*_\iota * (\bc{ \dot r_i }{\quot{B^*_\xi}{P^*_\iota}})$ in $L[\bar B^-]$. In fact,
\[
P^*_\iota* (\bc{ \dot r_i }{\quot{B^*_\xi}{P^*_\iota}}) = j[C],
\]
where $C$ is the algebra obtained from $P_\iota *(\bc{ \dot r_i }{\quot{B_\xi}{P_\iota}})$ by factoring through the ideal of elements below $-b_\xi$.
Thus also
$D_\xi^*= j_\xi[ \blowup{D_\xi}(\leq b_\xi) ] \in L[\bar B^-]$ (it is a subset of $B^*_{\xi+1}$ with a sufficiently absolute definition).
We leave it to the reader to check that this suffices to find an isomorphic copy $R^*$ of $R$ in $L[\bar B^-]$.
Finally, let $P_{\xi+1}= R^*(\leq b_{\xi+1})$ and let $j_{\xi+1}$ be defined by $j_{\xi+1}(\bar p)(i) = j_\xi(\bar p(i))$.
A very similar but simpler argument works if $\xi=E^3(\alpha^\zeta_\rho)$ for $\zeta > 0$ and $P_{\xi+1} = \simpleram( \dom(\Phi), P_\xi,\Phi)$ for some $\Phi$.
This completes the successor cases.
For $\xi$ limit, check that the $\lambda_\xi$-diagonal limit is absolute between $L[\bar B^-]$ and $L[\bar B]$.
So let $P^*_\xi$ be the $\lambda_\xi$-diagonal limit of the sequence constructed so far, inside $L[\bar B^-]$, restricted to conditions below $b_\xi$.
By (\ref{fit:together}), the isomorphisms constructed at earlier stages can be glued together to form $j_\xi$.
This finishes the proof of the claim.
\end{proof}
\chapter{Extension and iteration}\label{sec:ext}
The proof of the main result makes it necessary to consider iterations $\bar Q^\theta$ such that each initial segment $P_\iota$ is stratified on $[\lambda_\iota,\kappa]$, but it is not forced that $\dot Q_\iota$ be stratified for all $\iota <\theta$---amalgamation is one example; also, I don't see a proof that $P(A)$ ``codes no unwanted branches'' if we don't use forcings from an inner model, and these are not quasi-closed in the model where we force with them.
We deal with this difficulty by introducing the concept of $(P_\iota,P_{\iota+1})$ being a \emph{stratified extension}.
With diagonal support, this ensures that the initial segments are sufficiently coherent so that we can conclude that $P_\theta$ is stratified.
This coherency provided by extension is vital: e.g. an iteration whose proper initial segments are all $\sigma$-strategically closed can add a real (see
\cite{kellner:sh:sacks}).
To further complicate things,
$\lambda_\iota$ is not the same fixed cardinal throughout the iteration.
We treat quasi-closed and stratified extension separately (sections \ref{sec:qc:ext} and \ref{sec:s:ext}). Each axiom of stratified (or quasi-closed) extension corresponds to an axiom of stratification (or quasi-closure)---in fact, interestingly, $P$ is stratified if and only if $(\{ 1_P\}, P)$ is a stratified extension.
To prove the iteration theorem, we also have to add some additional axioms concerning the interplay of the \pre stratification (\pre closure) systems on $P_\iota$ and $P_{\iota+1}$; see definitions \ref{pcs:is} and \ref{pss:is}.
In section \ref{sec:prod} we show products of stratified forcings are stratified extensions.
Finally, we introduce \emph{the stable meet operator} in section \ref{sec:stm} and \emph{remote sub-orders} in section \ref{sec:remote}. See the beginning of those sections for a motivating discussion.
\section{Quasi-closed extension and iteration}\label{sec:qc:ext}
In this section, we show that composition of quasi-closed forcing is a special case of quasi-closed extension.
We give a sufficient condition which makes sure that if $(P_0,P_1)$ is a quasi-closed extension, then $P_1$ is quasi-closed.
We prove that the relation of being a quasi-closed extension is transitive.
Finally, we formulate and prove an iteration theorem for quasi-closed forcing.
Let $P_0$ be a complete sub-order of $P_1$ and let $\pi\colon P_1 \rightarrow P_0$ be a strong projection.
Moreover, assume we have a system $\pcs_i=(\D_i, c_i, \leqlol_i)_{\lambda\in I}$ for $i \in \{0,1\}$ such that $\D_i \subseteq I \times V \times (P_i)^2$ is a class definable with parameter $c_i$
and for every $\lambda\in I$, $\leqlol_i$ is a binary relation on $P_i$.
\begin{dfn}\label{pcs:is}
We write $\pcs_0 \is \pcs_1$ to mean for every $x$ and every $\lambda \in I$,
\begin{enumerate}[label=(\pcsIsRef), ref=\pcsIsRef]
\item For all $p,q \in P_0$, $p\leqlol_0 q \Rightarrow p \leqlol_1 q$. \label{ext}
\item For all $p,q \in P_1$, $p \leqlol_1 q \Rightarrow \pi(p) \leqlol_0 \pi(q)$. \label{pi:mon}
\item For all $p,q \in P_1$, $p \in \D_1(\lambda,x,q) \Rightarrow \pi(p) \in \D_0(\lambda,x,\pi(q))$. \label{F:coh}\label{D:coh}
\end{enumerate}
\end{dfn}
\noindent
Observe that if $\pcs_0 \is \pcs_1$ we can drop the subscripts on $\leqlol_0$, $\leqlol_1$ and just write $\leqlol$ without causing confusion.
\begin{dfn}\label{qc:ext}
We say the pair $(P_0,P_1)$ is a \emph{quasi-closed extension on $I$, as witnessed by $(\pcs_0,\pcs_1)$}
if and only if $\pcs_0$ witnesses that $P_0$ is quasi-closed on $I$, $\pcs_1$ is a \pre closure system on $P_1$, $\pcs_0 \is \pcs_1$ and for $\lambda,\bar \lambda \in I$ such that
$\lambda \leq \bar \lambda$, the following conditions hold:
\begin{enumerate}[label=(\qcExtRef), ref=\qcExtRef]
\item
If $p\in P_1$ and $q\in P_0$ is such that $q \leqlol \pi(p)$ and $q\in \D(\lambda,x,\pi(p))$, there is $r \leqlol p$
such that $r \in \D(\lambda,x,p)$, $r \leqlol p$ and $\pi(r)=q$.
Moreover we can ask of $r$ that for any $\lambda' \in I$ such that $p \leqlo^{\lambda'}\pi(p)$ we also have $r \leqlo^{\lambda'} \pi(r)$.\label{ext:redundant}\label{ext:D:dense}
\item If $\bar p = (p_\xi)_{\xi<\rho}$ is a sequence of conditions in $P_1$ such that for some $q \in P_0$ and some $\bar w$
\begin{enumerate}
\item $q$ is a greatest lower bound of the sequence $(\pi(p_\xi))_{\xi<\rho}$ and for all $\xi <\rho$, $q \leqlol \pi(p_\xi)$,\label{ext:pi:bound}
\item $\bar w$ is a $(\lambda,x)$-strategic guide and a $(\bar \lambda, x)$-canonical witness for $\bar p$, \label{ext:glb:strat:def}
\item either $\lambda = \bar \lambda$ or $p_\xi \leqlo^{\bar\lambda}_1 \pi(p_\xi)$ for each $\xi<\rho$,\label{ext:glb:tail}
\end{enumerate}
then $\bar p$ has a greatest lower bound $p$ in $P_1$ such that for each $\xi < \rho$, $p \leqlol p_\xi$ and $\pi(p)=q$.
Moreover, if $p_\xi \leqlo^{\bar\lambda}_1 \pi(p_\xi)$ for each $\xi<\rho$, then also $p \leqlo_1^{\bar\lambda} \pi(p)$.\label{ext:glb}
\end{enumerate}
As before, if we say $(P_0,P_1)$ is a quasi-closed extension and don't mention either of $\pcs_0$, $\pcs_1$ or $I$,
that entity is either clear from the context
or we are claiming that one can find such an entity.
\end{dfn}
\noindent
We will grow tired of repeating all the conditions $\bar p$ has to satisfy in (\ref{ext:glb}), so we issue the following definition:
\begin{dfn}\label{adequate2}
We say $\bar p$ is $(\lambda,\bar\lambda,x)$-adequate if and only if $\lambda, \bar \lambda \in I$, $\lambda\leq\bar\lambda$ and $\bar p$ satisfies conditions (\ref{ext:glb:strat:def}) and (\ref{ext:glb:tail}) above. We say $q$ is a $\pi$-bound if and only if (\ref{ext:pi:bound}) holds.
\end{dfn}
\noindent
Of course, the obvious example for quasi-closed extension is provided by composition of forcing notions:
\begin{lem}\label{qc:comp:implies:ext}
If $P$ is quasi-closed on $I$ and $\forces_P \dot Q$ is quasi-closed on $I$, then $(P, P*\dot Q)$ is a quasi-closed extension on $I$.
To be more precise, let $\pcs_0$ denote the \pre quasi-closure system witnessing that $P$ is quasi-closed and let $\pcs_1=(\bar\D,\barc, \bleqlol)_{\lambda\in I}$ be the \pre quasi-closure system constructed as in the proof of \ref{stratified:composition}, where we showed that $P*\dot Q$ is stratified. Then $(\pcs_0,\pcs_1)$ witnesses that $(P, P*\dot Q)$ is a quasi-closed extension on $I$.
\end{lem}
\noindent
We give the proof after we prove the following simple lemma, which will be useful in several contexts.
\begin{lem}\label{bar:lambda:adeq}
Say $R$ carries a \pre closure system $\pcs$ on $I$ and $\bar p= (p_\xi)_{\xi<\rho}$ has a $(\lambda,x)$-strategic guide $\bar w$ which is also a $(\bar \lambda,x)$-canonical witness.
If for all $\xi < \rho$, $p_\xi \leqlo^{\bar\lambda} 1_R$, then $\bar p$ is in fact $(\bar \lambda,x)$-adequate.
\end{lem}
\begin{proof}[Proof of lemma \ref{bar:lambda:adeq}.]
For arbitrary $\xi<\bar\xi<\rho$,
by \ref{def:pcs}(\ref{er}), as $p_{\bar \xi} \leq p_\xi \leq 1_R$ and $p_{\bar \xi} \leqlo^{\bar\lambda} 1_R$, we have $p_{\bar\xi}\leqlo^{\bar\lambda} p_\xi$. Thus $\bar w$ is in fact a $(\bar \lambda,x)$-strategic guide for $\bar p$ and so $\bar p$ is $(\bar\lambda,x)$-adequate.
\end{proof}
\begin{proof}[Proof of lemma \ref{qc:comp:implies:ext}.]
Just by looking at the definition of $\bleqlol$ and $\bar \F$, it is immediate that $\pcs_0 \is \pcs_1$ and that $\pcs_1$ is a \pre closure system.
To check that $\pcs_1$ is a \pre closure system, observe \ref{def:pcs}(\ref{er}) has already been checked in the proof of theorem \ref{stratified:composition}. The other conditions we leave to the reader.
Condition \ref{qc:ext}(\ref{ext:redundant}) holds since $P$ forces \ref{def:qc}(\ref{qc:redundant}) for $\dot Q$:
say we have $(p,\dot p) \in P*\dot Q$ and $q \in P$ such that $q \in \D(\lambda,x,p)$.
By fullness we may find $\dot q$ such that $p \forces \dot q \in \D(\lambda,x,\dot q)$ and $p\forces \dot q \leqlol \dot p$.
We have $(q, \dot q) \leqlol (p,\dot p)$ and $(q,\dot q)\in \bar \D(\lambda,x,(p,\dot p))$.
Moreover, by the last clause of \ref{def:qc}(\ref{qc:redundant}), we can demand that
$p\forces$ for any $\lambda' \in I$, $\dot p \leqlo^{\lambda'} 1_{\dot Q} \Rightarrow \dot q \leqlo^{\lambda'} 1_{\dot Q}$. Thus, for any $\lambda' \in I$ such that $(p, \dot p) \bleqlo^{\lambda'} (p, 1_{\dot Q})$ we also have $(q, \dot q) \bleqlo^{\lambda'} (q, 1_{\dot Q})$, which proves the last statement of \ref{qc:ext}(\ref{ext:redundant}).
Now to the main point, that is \ref{qc:ext}(\ref{ext:glb}):\label{proof:of:def:qc:in:s:comp:impl:ext}
Say $\bar p = (p_\xi,\dot p_\xi)_{\xi<\rho}$ is sequence of conditions in $P_1$ which is $(\lambda,\bar\lambda,x)$-adequate. That is, we may fix $\bar w$ which is a $(\lambda,x)$-strategic guide and a $(\bar\lambda,x)$-canonical witness. We may also fix $q \in P_0$ which is a $\pi$-bound---i.e. (\ref{ext:pi:bound}) holds.
We have already seen that $q$ forces that
$\bar w$ is a $(\bar\lambda,x)$-canonical witness for $(\dot p_\xi)_{\xi<\rho}$ (this is the same argument as in the proof of theorem \ref{stratified:composition}).
It is easy to check that $q$ also forces that $\bar w$ is a $(\lambda,x)$-strategic guide for $(\dot p_\xi)_{\xi<\rho}$.
If $\lambda=\bar \lambda$, we conclude that $q$ forces that $(\dot p_\xi)_{\xi<\rho}$ is $\lambda$-adequate in $\dot Q$.
Thus we may pick a $P$-name $\dot q$ such that $q$ forces $\dot q \in \dot Q$ is the greatest lower bound of $(\dot p_\xi)_{\xi<\rho}$ in $\dot Q$.
Then $(q,\dot q)$ is the greatest lower bound of $\bar p$ and we are done with the proof of \ref{qc:ext}(\ref{ext:glb}) in this case.
If on the other hand, $\lambda<\bar\lambda$, we have that for all $\xi<\rho$, $(p_\xi,\dot p_\xi) \leqlo^{\bar\lambda}_1 (p_\xi,1_{\dot Q})$.
Thus by lemma \ref{bar:lambda:adeq} we have that
$q$ forces $(\dot p_\xi)_{\xi<\rho}$ is $\bar \lambda$-adequate.
Thus $q$ also forces that this sequence has a lower bound,
for which we may fix a name $\dot q$.
By quasi-closure for $\dot Q$ in the extension and since
for all $\xi<\rho$ we have
\[ q\forces\dot p_\xi \dot\leqlo^{\bar\lambda}1_{\dot Q},\]
we conclude that for any $\xi < \rho$ we have
\[ q \forces \dot q \dot \leqlo^{\bar \lambda} \dot q_\xi.\]
Thus $(q,\dot q)$ is a greatest lower bound of $\bar p$ and
\[ (q,\dot q)\bar\leqlo^{\bar\lambda}(q,1_{\dot Q}). \]
\end{proof}
\noindent
We now embark on a series of lemmas culminating in the insight that the second forcing of a quasi-closed extension $(P_0,P_1)$
is itself quasi-closed. Thus, we obtain a second proof that $P*\dot Q$ is quasi-closed (under the assumptions of the previous lemma).
This makes use of the fact that the projection map $\pi_0\colon P*\dot Q \rightarrow P$ is definable.
In general, we shall see that we have to assume that the strong projection map from $P_1$ to $P_0$ is sufficiently definable.
\begin{lem}\label{lem:strat:proj}
Assume for $i\in\{0,1\}$, $P_i$ carries a \pre closure system $\pcs_i$ on $I$ and $\pcs_0 \is \pcs_1$. If $\bar p=(p_\xi)_{\xi<\rho}$ is a sequence of conditions in $P_1$ and $\bar w$ is a $(\lambda,x)$-strategic guide with respect to $\pcs_1$, then $\bar w$ is also a $(\lambda,x)$-strategic guide for $(\pi(p_\xi))_{\xi<\rho}$ with respect to $\pcs_0$.
\end{lem}
\begin{proof}
Suppose we are given $\bar p$ and $\bar w$ as in the hypothesis.
If $\xi < \bar \xi < \rho$, since $p_\xi \leqlol_1 p_{\bar \xi}$, by \ref{pcs:is}(\ref{pi:mon}), $\pi(p_\xi)\leqlol_0\pi(p_{\bar\xi})$.
Let $\xi<\rho$ be arbitrary. Fix a regular $\lambda'$ such that
$p_{\xi+1} \leqlo^{\lambda'}_1 p_\xi$
and
$p_{\xi+1} \in \D_1(\lambda', (x, \bar w\res\xi+1), p_\xi)$
By \ref{pcs:is}(\ref{pi:mon}),
$\pi(p_{\xi+1})\leqlo^{\lambda'}_0 \pi(p_\xi)$
and by \ref{pcs:is}(\ref{F:coh}),
$\pi(p_{\xi+1}) \in \D_1(\lambda', (x, \bar w\res\xi+1), ,\pi(p_\xi))$,
finishing the proof.
\end{proof}
\begin{lem}\label{lem:adeq:proj}
Assume $(P_i, \pcs_i)$, $i\in\{0,1\}$ are as in lemma \ref{lem:strat:proj}.
Further, assume that the strong projection map $\pi\colon P_1\rightarrow P_0$ is $\qcdefG(\lambda\cup\{x\})$. %
If $\bar p=(p_\xi)_{\xi<\rho}$ is a sequence of conditions in $P_1$ which is $(\lambda,x)$-adequate with respect to $\pcs_1$, then $(\pi(p_\xi))_{\xi<\rho}$ is $(\lambda,x)$-adequate with respect to $\pcs_0$.
\end{lem}
\begin{proof}
Fix $\bar w$ which is both a $(\lambda,x)$-strategic guide and a $(\lambda,x)$-canonical witness for $\bar p$.
By the previous lemma, $\bar w$ is a $(\lambda,x)$-strategic guide for $(\pi(p_\xi))_{\xi<\rho}$.
We may find a $\qcdefG(\lambda\cup\{x\})$ function $G$ such that $p_\xi = G(\bar w\res\xi+1)$.
As $\pi \circ G$ is also $\qcdefG(\lambda\cup\{x\})$, $\bar w$ is also a
$(\lambda,x)$-canonical witness for $(\pi(p_\xi))_{\xi<\rho}$.
\end{proof}
\noindent
The following is useful e.g. when we show a condition has legal support. Here lies one of the reasons for asking (\ref{er}).
\begin{lem}\label{lem:support:simple}
Assume $(P_i,\pcs_i)$, for $i\in\{0,1\}$ are as in lemma \ref{lem:strat:proj}. For any $p\in P_1$ and any regular $\lambda\in I$ we have:
\begin{equation}\label{support:simple}
(\exists q \in P_0 \quad p \leqlol_1 q) \iff p \leqlol_1 \pi(p)
\end{equation}
\end{lem}
\begin{proof}
One direction is clear, so say $p \leqlol_1 q$ for some $q \in P_0$.
Apply \ref{def:pcs}(\ref{er}):
As $\pi$ is a strong projection, $p \leq \pi(p) \leq q$ and so $p \leqlol_1 \pi(p)$.
\end{proof}
The intuition behind definition \ref{qc:ext} is that $P_0$ and $P_1$ are both quasi-closed, not independently of each other, but in a very coherent way.
That $P_1$ is quasi-closed is almost implicit in definition \ref{qc:ext}---it depends on a further assumption about the definability of $\pi$ (this is responsible for the distinct flavor of quasi-closure, setting it apart from the other axioms of stratification):
\begin{lem}\label{lem:qc:succ}
If $(P_0,P_1)$ is a quasi-closed extension on $I$ and $\pi$ is $\qcdefG(\min I \cup\{c_1 \})$, then $P_1$ is quasi-closed on $I$.
\end{lem}
\noindent
Before we give the proof, note that this assumption on $\pi$ is not entirely trivial: in an iteration, the canonical projection $\pi\colon P_\theta \rightarrow P_\iota$ is $\Delta_0$ in the parameter $\iota$; it is not in general $\qcdefG(\min I)$.
Also we would like to note in passing that in fact $P$ is quasi-closed exactly if $(\{ 1_P\}, P)$ is a quasi-closed extension; the same will be true for stratified forcing.
\begin{proof}
First check \ref{def:qc}(\ref{qc:redundant}):
Say $p \in P_1$, $\lambda\in I$ and $x$ are given.
Use (\ref{qc:redundant}) for $P_0$ to find $q \in P_0$ such that $q \leqlol \pi(p)$ and $q \in \D_0(\lambda,x,\pi(p))$.
Now apply \eqref{ext:D:dense} to get $p' \leqlol p$ such that $p' \in \D_0(\lambda,x,p)$ and $\pi(p')=q$.
For the last clause of \ref{def:qc}(\ref{qc:redundant}), we can assume that $q$ has been chosen so that for any $\lambda'\in I$, if $\pi(p) \leqlo^{\lambda'}_0 1_{P_0}$, then $q \leqlo^{\lambda'}_0 1_{P_0}$.
We can also assume that $p'$ has been chosen so that for any $\lambda'\in I$, if $p \leqlo^{\lambda'}_1 \pi(p)$, then $p' \leqlo^{\lambda'}_1 \pi(p')$.
Thus, $p' \leqlo^{\lambda'}_1 \pi(p) \leqlo^{\lambda'}_1 1_{P_1}$.
It remains to check \ref{def:qc}(\ref{qc:glb}), so say $\bar w$ witnesses that $\bar p =(p_\xi)_{\xi<\rho}$ is $(\lambda,x)$-adequate for $P_1$.
Observe we assume that $x$ is a tuple with $c_1$ among its components.
By assumption, $\pi$ is $\qcdefSeq(\lambda\cup\{x\})$, so by lemma \ref{lem:adeq:proj}, $(\pi(p_\xi))_{\xi<\rho}$ is also $(\lambda,x)$-adequate.
Since $P_0$ is quasi-closed, $(\pi(p_\xi))_{\xi<\rho}$ has a greatest lower bound $q$.
Thus, applying \ref{def:qc}(\ref{qc:glb}) for $\bar\lambda=\lambda$, we conclude that $\bar p$ has a greatest lower bound.
\end{proof}
\noindent
The next lemma will be used in \ref{thm:it:qc} when we show that if the initial segments of an iterations form a chain of quasi-closed extensions, then the limit is itself a quasi-closed extension. It says that the relation of being a quasi-closed extension is transitive.
Let $P_0$, $P_1$ and $P_2$ be \pre orders such that for $i \in \{0,1\}$, $P_i$ is a strong sub-order of $P_{i+1}$.
\begin{lem}\label{qc:ext:transitive}
Say $\pi_1\colon P_2 \rightarrow P_1$ and $\pi_0\colon P_2 \rightarrow P_0$ are strong projection maps and $\pi_1$ is $\qcdefG(\min I \cup\{ c_2\})$.
If both $(P_0,P_1)$ and $(P_1,P_2)$ are quasi-closed extensions on $I$, then $(P_0,P_2)$ is also a quasi-closed extension on $I$.
\end{lem}
\begin{proof}
Let $(\pcs_0,\pcs_1)$ and $(\pcs_1,\pcs_2)$ witness that $(P_0,P_1)$ and $(P_1,P_2)$ are quasi-closed extensions.
We now check all the conditions of \ref{qc:ext} for $(P_0,P_2)$ and $(\pcs_0,\pcs_2)$.
That $\pcs_2$ is a \pre closure system holds by assumption, and that $\pcs_0 \is \pcs_2$ is obvious.
Observe that by \ref{pcs:is}(\ref{ext}), we don't need to distinguish between $\leqlol_0$, $\leqlol_1$ and $\leqlol_2$ and therefore we drop the subscripts in what follows.
We check \ref{qc:ext}(\ref{ext:redundant}):
Say $p \in P_2$ and $q \in \D_0(\lambda,x,\pi_0(p))$.
As in the previous proof, we can find $p' \in P_1$ such that $\pi_0(p')=q$, $p'\leqlol_1 \pi_0(p)$ and $p' \in \D_1(\lambda,x,\pi_1(p))$ and then $r$ such that $r\in\D_2(\lambda,x,p)$, with $\pi_1(r)=p'$ and $r\leqlol_2 p$.
Now if $\lambda'\in I$ and $p \leqlo^{\lambda'} \pi_0(p)$, we also have $p \leqlo^{\lambda'} \pi_1(p)$ by lemma \ref{lem:support:simple}.
This means we have both $p' \leqlo^{\lambda'} \pi_0(p')$ and $r \leqlo^{\lambda'} \pi_1(r)$.
As $\pi_1(r) = p'$, it follows that $r\leqlo^{\lambda'} \pi_0(r)$.
It remains to check \ref{qc:ext}(\ref{ext:glb}). So let $\bar p = (p_\xi)_{\xi<\rho}$ be a $(\lambda,\bar\lambda,x)$-adequate sequence of conditions in $P_2$ and let $q_0$ be a greatest lower bound of $(\pi_0(p_\xi))_{\xi<\rho}$ as in the hypothesis.
Fix $\bar w$ which is a $(\lambda,x)$-strategic guide and a $(\bar\lambda,x)$-canonical witness for $\bar p$.
Show exactly as in the proof of lemma \ref{lem:qc:succ} that $\bar w$ is both a $(\bar\lambda,x)$-canonical witness and a $(\lambda,x)$-strategic guide for $(\pi_1(p_\xi))_{\xi<\rho}$ in $P_1$.
Denote this sequence by $\bar q$.
Moreover, if it is the case that $\bar\lambda>\lambda$, then
\begin{equation}\label{glb:supp:long}
\forall \xi <\rho \quad p_\xi \leqlo^{\bar\lambda} \pi_0(p_\xi).
\end{equation}
By \ref{def:pcs}(\ref{pi:mon}), we have that
\begin{equation}\label{glb:supp:latter}
\forall \xi <\rho \quad\pi_1(p_\xi) \leqlo^{\bar\lambda} \pi_0(p_\xi).
\end{equation}
Thus $\bar q$ satisfies the hypothesis of \ref{qc:ext}(\ref{ext:glb}) for $(P_0,P_1)$ and we may find a greatest lower bound $q_1 \in P_1$ with $\pi_0(q_1)=q$.
Now use \ref{qc:ext}(\ref{ext:glb}) for $(P_1, P_2)$ to find a greatest lower bound $q$ of $\bar p$ such that $\pi_1(q)=q_1$ and so $\pi_0(q)=q_0$.
If $\lambda<\bar\lambda$, lemma \ref{lem:support:simple} and (\ref{glb:supp:long}) yield
\begin{equation}
\forall\xi < \rho \quad p_\xi \leqlo^{\bar\lambda} \pi_1(p_\xi).\label{glb:supp:former}
\end{equation}
So finally, as $q \leqlo^{\lambda} \pi_1(q)$ by (\ref{glb:supp:former}),
and $\pi_1(q)= q_1 \leqlo^{\bar\lambda} \pi_0(q_1)$ by (\ref{glb:supp:latter}),
we conclude
$q \leqlo^{\bar\lambda} \pi_0(q)$.
\end{proof}
\begin{dfn}\label{def:diagsupp}
Say $\theta$ is a limit ordinal, and $\bar Q^\theta$ is an iteration such that for each $\iota <\theta$, $P_\iota$ carries a \pre closure system $\pcs_\iota$ on $\mathbf{Reg}\cap[\lambda_\iota,\lambda^*)$, where the sequence $\bar \lambda=(\lambda_\iota)_{\iota<\theta}$ is a non-decreasing sequence of regulars.
All of the following definitions are relative to these \pre closure systems and to $\bar \lambda$.
\begin{enumerate}
\item For a thread $p$ through $\bar Q^\theta$, let
\[
\supp^\lambda(p) = \{ \iota < \theta \setdef \lambda_\iota\leq\lambda \text{ and } \pi_{\iota+1} \not \leqlol_{\iota+1} \pi_\iota(p) \},
\]
and let $\sigma^\lambda(p)$ be the least ordinal $\sigma$ such that $\supp^\lambda(p)\subseteq \sigma$.
\item Let $\lambda^*$ be regular such that $\lambda^*\geq\lambda_\iota$ for all $\iota<\theta$. We say $P_\theta$ is the \emph{$\lambda^*$-diagonal support limit of $\bar Q^\theta$} if and only if $P_\theta$ consists of all threads $p$ through $\bar Q^\theta$ such that
for each regular $\lambda\geq\lambda^*$, $\supp^\lambda(p)$ has size less than $\lambda$ and $\sigma^{\lambda^*}(p)<\theta$.
\item We define the \emph{natural system of relations on $P_\theta$} as follows
\begin{enumerate}
\item $p \leqlol_{\theta} q \iff \forall\iota < \theta \quad \pi_\iota(p) \leqlol_{\iota} \pi_\iota(q)$;
\item $p \in \D(\lambda,x,q) \iff \forall \iota<\theta \pi_\iota(q)\in\D(\lambda,x,\pi_\iota(q)$.
\item The parameter $c_\theta$ has as the first of its components a sequence $\bar c = ( c_\iota)_{\iota<\theta}$, where $c_\iota$ is the parameter from $\pcs_\iota$.
\end{enumerate}
We shall see in the proof of theorem \ref{thm:it:qc} that under natural assumptions the natural system of relations is a \pre closure system.
\item We say $\bar Q^\theta$ is a \emph{$\bar\lambda$-diagonal support iteration} if and only if for any limit $\iota<\theta$, $P_\iota$ is the $\lambda_\iota$-diagonal support limit of $\bar Q^\iota$.
\end{enumerate}
\end{dfn}
\begin{thm}\label{thm:it:qc}
Let $\bar Q^\theta$ be an iteration such that for each $\iota < \theta$, $P_\iota$ carries a \pre closure system $\pcs_\iota$ on $[\lambda_\iota,\kappa^*)$, where the sequence $\bar \lambda=(\lambda_\iota)_{\iota<\theta}$ is non-decreasing.
Moreover, let $\lambda_\theta=\min(\mathbf{Reg}\setminus \sup_{\iota<\theta}\lambda_\iota)$ and assume
\begin{enumerate}
\item For all $\iota<\theta$, $(P_\iota,P_{\iota+1})$ is a quasi-closed extension on $[\lambda_\iota,\kappa^*)$.
\item If $\bar \iota < \theta$ is limit, $\pcs_{\bar\iota}$ is the natural system of relations on $P_{\bar\iota}$ on $[\lambda_\iota,\kappa^*)$ and $\bar Q^\iota$ is a $\bar \lambda\res\iota$-diagonal support iteration.
\item We have $\cof(\theta)<\kappa$.
\end{enumerate}
Let $P_\theta$ be the $\lambda_\theta$-diagonal support limit of $\bar Q^\theta$. Then $P_\theta$ is quasi-closed on $I_\theta= [\lambda_\theta,\kappa^*)$, as witnessed by the natural system of relations $\pcs_\theta$.
\end{thm}
\noindent
In the proof of the theorem, we need the following lemmas \ref{supp:union}--\ref{supp:threads}, showing that the notion of $\lambda$-support behaves as we expect.
So fix an iteration $\bar Q^{\theta+1}$ and \pre closure systems as in the hypothesis of the theorem.
These lemmas are somewhat technical but straightforward to show.
\begin{lem}\label{supp:union}
For each $\lambda\in[\lambda_\theta,\kappa^*)$ and $p\in P_\theta$,
\[ \supp^\lambda(p)=\bigcup_{\iota<\theta} \supp^\lambda(\pi_\iota(p)).\]
\end{lem}
\begin{proof}
First, prove $\supseteq$:
Say $\xi$ is a member of the set on the right. Thus there is some $\iota<\theta$ such that
\begin{equation}\label{xi:not:in:support}
\pi_{\xi+1}(\pi_\iota(p))\not\leqlol \pi_\xi(\pi_\iota(p)).
\end{equation}
We consider two cases: first, assume $\iota \leq \xi$. Then $\pi_{\xi+1}(\pi_\iota(p))=\pi_\iota(p)=\pi_{\xi}(\pi_\iota(p))$,
and so as $\leqlol$ is a \pre order, (\ref{xi:not:in:support}) is false. Thus this case never occurs, and we can assume $\iota > \xi$.
Then $\pi_{\xi+1}(\pi_\iota(p))=\pi_{\xi+1}(p)$ and $\pi_{\xi}(\pi_\iota(p))=\pi_\xi(p)$, so (\ref{xi:not:in:support}) is equivalent to
$\pi_{\xi+1}(p) \not \leqlol \pi_\xi(p)$. We infer that $\xi \in \supp^\lambda(p)$.
All of the above inferences can be reversed, so $\subseteq$ holds as well.
\end{proof}
\begin{lem}\label{supp:mon}
If $\lambda,\bar\lambda$ are regular such that $\lambda_\theta \leq \lambda \leq \bar\lambda < \kappa^*$ and $p \leqlo^{\bar\lambda}q$, then $\supp^\lambda(p)\subseteq \supp^\lambda(q)$.
\end{lem}
\begin{proof}
Left to the reader.
\end{proof}
\noindent
Observe though that $\supseteq$ does not necessarily hold: in a two-step iteration $P*\dot Q$, we could have and $(p,1_{\dot Q}) \bleqlol (q,\dot q)$ but $q \not \forces_P \dot q \dleqlol 1_{\dot Q}$ (say e.g. $p \forces \dot q=1_{\dot Q}$).
In this example we have $\suppl(p,1_{\dot Q})=\{0\}\not \supseteq\suppl(q,\dot q)=\{0,1\}$.
\begin{lem}\label{supp:initial}
Fix $\bar \iota<\theta$.
If $p \in P_\theta$ and $\lambda,\bar\lambda$ are regular such that $\lambda_\theta \leq \lambda \leq \bar\lambda< \kappa^*$ and $p \leqlo^{\bar\lambda} \pi_{\bar\iota}(p)$, then
\[\supp^\lambda(p)=\supp^\lambda(\pi_{\bar\iota}(p)). \]
\end{lem}
\begin{proof}
A short proof: $\supseteq$ holds by lemma \ref{supp:union} and $\subseteq$ is a consequence of lemma \ref{supp:mon}.
We also give a direct proof:
If $\iota < \bar \iota$, $\pi_\iota(\pi_{\bar\iota}(p))=\pi_\iota(p)$, so for such $\iota$,
\[ \iota \in \supp^\lambda(p) \iff \iota \in \supp^\lambda(\pi_{\bar\iota}(p)). \]
If $\iota \geq \bar \iota$, we have $\pi_{\iota+1}(p)\leq \pi_\iota(p)\leq \pi_{\bar\iota}(p)$. By assumption and by \ref{pcs:is}(\ref{pi:mon}) for $(P_{\iota+1},P_\theta)$, we have $\pi_{\iota+1}(p)\leqlo^{\bar \lambda} \pi_{\bar\iota}(p)$.
So by lemma \ref{lem:support:simple}, $\pi_{\iota+1}(p)\leqlol\pi_\iota(p)$. We conclude that for $\iota \geq \bar \iota$, we have $\iota \not\in\supp^\lambda(p)$.
\end{proof}
\begin{lem}\label{supp:threads}
Let $\lambda_1 \in [\lambda_\theta,\kappa^*)$ such that $\lambda_1 \geq \cof(\theta)$.
Say $q = (q^\iota)_{\iota<\theta}$ is a thread through $\bar Q^\theta$ (see definition \ref{it:terminology}, p.~\pageref{it:terminology} for this terminology) and say there is $w \in P_\theta$ such that for all $\iota<\theta$, $q^\iota \leqlo^{\lambda_1}w$. Then $q$ has legal support, i.e. $q \in P_\theta$.
\end{lem}
\begin{proof}
Let $\lambda$ be regular.
First consider the case $\lambda_\theta \leq \lambda \leq \lambda_1$. As $q \leqlo^{\lambda_1}w$, by lemma \ref{supp:mon},
$\supp^\lambda(q)\subseteq\supp^\lambda(w)$, which satisfies the requirement of diagonal support by assumption.
Now say $\lambda>\lambda_1$ and fix a sequence $(\theta(\zeta))_{\zeta<\cof(\theta)}$ which is cofinal in $\theta$.
By lemma \ref{supp:union}
\[ \supp^\lambda(q)= \bigcup_{\zeta < \cof(\theta)} \supp^\lambda(q^{\theta(\zeta)}), \]
and by assumption the right hand side is a union over bounded subsets of $\lambda$. Thus $\supp^\lambda(q)$ is a bounded subset of $\lambda$.
\end{proof}
\noindent
\paragraph*{Proof of theorem \ref{thm:it:qc}:}
Observe we must make the assumption that $x$ includes the parameters $L_\mu[A]$ and $(\pi_\iota)_{\iota\leq\theta}$, where $\mu$ is a cardinal and and $\mu > \theta$, to make sure we can talk about the least sequence witnessing the cofinality of $\theta$ in a $\qcdefG(x)$ manner (alternatively, one could include this sequence in $c_\theta$; also note that in practice, we can replace $(\pi_\iota)_{\iota\leq\theta}$ as a parameter by just $\theta$ since it and $(P_\iota)_{\iota\leq\theta}$ will be recursive in $\theta$ and $L_\mu[A]$ and we can usually also drop $\theta$ since it will be definable from any condition $p \in P_\theta$).
We will show by induction on $\theta$ that for each pair $\iota < \bar \iota \leq \theta$, $(P_\iota, P_{\bar \iota})$ is a quasi-closed extension. Thus $(P_0,P_\theta)$ is a quasi-closed extension and so by lemma \ref{lem:qc:succ}, $P_\theta$ is quasi-closed.
The inductive hypothesis thus says that for each pair $\iota < \bar \iota <\theta$, $(P_\iota, P_{\bar \iota})$ is a quasi-closed extension
as witnessed by $(\pcs_\iota,\pcs_{\bar\iota})$.
We may assume $\theta$ is limit:
For if $\theta$ is a successor ordinal, $\pi^{\theta}_{\theta-1}$ is a $\Delta_1$-definable function and thus by induction hypothesis and lemma
\ref{qc:ext:transitive}, for any $\iota < \theta$, $(P_\iota, P_\theta)$ is a quasi-closed extension.
So assume $\theta$ is limit and let $\pcs_\theta$ be the natural system of relations on the diagonal support limit $P_\theta$.
Fix an arbitrary $\iota^*<\theta$. We show that $(P_{\iota^*},P_\theta)$ is a quasi-closed extension witnessed by $(\pcs_{\iota^*},\pcs_\theta)$.
By definition of $\pcs_\theta$, we have $\pcs_{\iota^*} \is \pcs_\theta$.
It is straightforward to show that $\pcs_\theta$ is a \pre closure system (as defined in \ref{def:pcs}, p.~\pageref{def:pcs}).
It is obvious that $\D_\theta$ is $\qcdefF(c_\theta)$:
Letting $\Phi$ be a universal $\Pi_1^A$ formula, by definition of the natural system of relations we can assume the first component $\bar c = ( c_\iota)_{\iota<\theta}$ of $c_\theta$ is such that for each $\iota<\theta$ and each $p, q \in P_\iota$,
\[ q\in \D_\iota(\lambda,x,p)\iff\Phi(c_\iota,q,\lambda,x,p).\]
Thus $\Phi$ and $\bar c$ witness that $\D_\theta$ is $\qcdefD(\{c_\theta\})$: for $q \in\D_\theta(\lambda,x,p)$ is equivalent to
\[ \forall \iota \in\dom(p) \quad\Phi(c_\iota,\pi_\iota(q),\lambda,x,\pi_\iota(p)).\]
We finish the proof that $\pcs_\theta$ is a \pre closure system by proving \ref{def:pcs}(\ref{er}), as the remaining conditions have similar proofs:
Say $p \leq_\theta q \leq_\theta r$ and $p \leqlol_\theta r$.
Fixing an arbitrary $\iota<\theta$, we have
$\pi_\iota(p) \leq_\iota \pi_\iota(q) \leq_\iota \pi_\iota(r)$ and $\pi_\iota(p) \leqlol_\iota \pi_\iota(r)$.
Thus, by \ref{def:pcs}(\ref{er}) for $P_\iota$, $\pi_\iota(p)\leqlol_\iota \pi_\iota(q)$. As $\iota<\theta$ was arbitrary, $p \leqlol_\theta q$ holds.
So as mentioned earlier, the natural system of relations is a \pre closure system.
We check \ref{qc:ext}(\ref{ext:redundant}).
Say $p \in P_\theta$ and $q^* \in P_{\iota^*}$ are such that $q^*\leqlol_{\iota^*} \pi_{\iota^*}(p)$ and $q^*\in \D_{\iota^*}(\lambda,x, \pi_{\iota^*}(p))$.
Assume first that $\cof(\theta)\in I$ and $\cof(\theta) > \lambda$.
Let $\sigma = \sigma^{\cof(\theta)}(p)$ and observe that $\sigma < \cof(\theta) \leq \theta$ by diagonal support.
If $\sigma \leq \iota^*$, set $r_0 = q^*$ and let $(\theta(\zeta))_{\zeta<\cof(\theta)}$ be the least normal sequence cofinal in $\theta$ such that $\theta(0)=\iota^*$.
Otherwise,
let $(\theta(\zeta))_{\zeta<\cof(\theta)}$ be the least normal sequence cofinal in $\theta$ such that $\theta(0)=\sigma$.
Use \ref{qc:ext}(\ref{ext:redundant}) for $(P_{\iota^*},P_\sigma)$ to get $r_0 \in P_\sigma$ such that $r_0 \in \D(\lambda, x, \pi_{\sigma}(p))$, $r_0 \leqlol_\sigma \pi_{\sigma}(p)$ and $\pi_{\iota^*}(r_0)=q^*$.
Observe that in either case, $p \leqlo^{\cof(\theta)} \pi_{\theta(0)} (p)$.
Let $(\theta(\zeta))_{\zeta<\cof(\theta)}$ be the least normal sequence cofinal in $\theta$ such that $\theta(0)=\sigma$.
Now construct, by induction on $\zeta < \cof(\theta)$, a thread $(r_\zeta)_{\zeta < \cof(\theta)}$: having $r_\zeta \in P_\zeta$, use \ref{qc:ext}(\ref{ext:redundant}) for $(P_{\theta(\zeta)},P_{\theta(\zeta+1)})$ to get $r_{\zeta+1} \in\D(\lambda, x, \pi_{\theta(\zeta+1)}(p))$ such that $r_{\zeta+1} \leqlol_{\theta(\zeta+1)} \pi_{\theta(\zeta+1)}(p))$,
$\pi_{\theta(\zeta)}(r_{\zeta+1})=r_\zeta$ and in addition, $r_{\zeta+1}\leqlo^{\cof(\theta)}_{\theta(\zeta+1)} r_\zeta$.
Let $r$ be the unique condition in $P_\theta$ defined by the thread $(r_\zeta)_{\zeta < \cof(\theta)}$.
Since $r \leqlo^{\cof(\theta)} r_0$, lemma \ref{supp:threads} allows us to conclude that $r$ has legal support. By construction, $r \in \D_\theta(\lambda,x,p)$, $r \leqlol_\theta p$ and $\pi_{\iota^*}(r)=q^*$.
If $\cof(\theta) \leq \lambda$, we can skip the first step in the above: we just let $r_0=q^*$.
At the end, we use $r \leqlol p$ and lemma \ref{supp:threads} allows us to conclude that $r$ has legal support.
We leave the rest of \ref{qc:ext}(\ref{ext:redundant}) to the reader, as it is similar to previous arguments.
\paragraph*{Main point of the argument:}
Now check \ref{qc:ext}(\ref{ext:glb}), the existence of greatest lower bounds.
Say $\bar p = (p_\xi)_{\xi<\rho}$ is a $(\lambda,\bar\lambda,x)$-adequate sequence of conditions in $P_\theta$; then $\lambda$ and $\bar \lambda$ are both regular, $\lambda_\theta\leq\lambda\leq\bar\lambda< \kappa^*$.
We may fix $\bar w$ which is both a $(\lambda,x)$-strategic guide and a $(\bar\lambda,x)$-canonical witness for $\bar p$.
Moreover, let $q^*$ be a greatest lower bound of the sequence $(\pi_{\iota^*}(p_\xi))_{\xi<\rho}$.
The construction of $q$ is by induction on $\zeta < \cof(\theta)$, and we shall use a sequence $(\theta(\zeta))_{\zeta<\cof(\theta)}$.
The construction is split in cases for the following reasons:
When $\cof(\theta) > \bar \lambda$, the definability of each $\pi_{\theta(\zeta)}$ poses a problem, and so we first have to find a $\pi_{\cof(\theta)}$-bound.
Moreover, the argument that supports are legal goes differently depending on wether $\cof(\theta) \leq\lambda$ or not.
There are several possibilities for distinguishing cases as the trick for dealing with $\cof(\theta)>\bar \lambda$ can be applied whenever $\cof(\theta) > \lambda_\theta$, but in my opinion this obscures the argument.
First, assume that $\cof(\theta)\leq \bar\lambda$.
Let $(\theta(\zeta))_{\zeta\leq\cof(\theta)}$ be the $\wo$-least increasing continuous sequence such that $\theta(0)=\iota^*$ and $\theta(\cof(\zeta))=\theta$.
By induction on $\zeta$, we now construct a lower bound $q^{\zeta}\in P_{\theta(\zeta)}$ of the sequence $(\pi_{\theta(\zeta)}(p_\xi))_{\xi<\rho}$ for each $\zeta \leq \cof(\theta)$.
Set $q^{0}=q^*$.
Now assume we have $q^{\zeta}$ and show how to find $q^{\zeta+1}$.
It is easily checked that $(\pi_{\theta(\zeta+1)}(p_\xi))_{\xi<\rho}$ is $(\lambda,\bar\lambda,x)$-adequate in $(P_{\theta(\zeta+1)},P_{\theta(\zeta)})$, because by the presence of $L_\mu[A]$ and $\theta$ in $x$ we only need bounded quantifiers (which we may assume come after the unbounded ones) to be able to talk about the sequence $(\theta(\zeta))_\zeta$.\footnote{In fact, again, it seems we could have done with a more restricted class of functions $G$ in definition \ref{canonical} (of canoncal witnesses).}
If $\theta$ is, as usual, equal to $\dom(p)$ for $p \in P_\theta$, we don't need $\theta$ in $x$.
By \ref{qc:ext}(\ref{ext:glb}) we obtain a greatest lower bound $q^{\theta(\zeta+1)}\in P_{\theta(\zeta)}$.
Now let $\zeta\leq\cof(\theta)$ be limit.
By construction and by (\ref{ext:glb}), the $q^{\zeta'}$, for $\zeta'<\zeta$ form a thread. Define $q^{\theta(\zeta)}$ to be this thread.
To show it has legal support, first assume that $\cof(\theta)\leq\lambda$.
By construction and by (\ref{ext:glb}), for each $\zeta$, we have $q^{\zeta}\leqlo^{\lambda} p_0$.
As $\lambda$ is greater than the maximum of $\lambda_\theta$ and $\cof(\theta)$, lemma \ref{supp:threads} allows us to infer that $q^{\zeta}$ has legal support, and thus is a condition in $P_{\theta(\zeta)}$ and a $\pi_{\theta(\zeta)}$-bound of $\bar p$ (in the sense of definition \ref{adequate2}).
The final condition $q^{\cof(\theta)}$ is a greatest lower bound of $(p_\xi)_{\xi<\rho}$ and for all $\xi<\rho$, $q^{\cof(\theta)} \leqlol p_\xi$.
To finish the case where $\cof(\theta) \leq\bar\lambda$, we have to consider the sub-case
where $\cof(\theta) \leq\lambda$ fails, i.e. we assume $\lambda < \bar \lambda$ and $\cof(\theta) \in (\lambda, \bar\lambda]$.
In this case we have $p_\xi\leqlo^{\bar\lambda} \pi_{\iota^*}(p_\xi)$ for all $\xi<\rho$.
By \ref{qc:ext}(\ref{ext:glb}) and by induction, we have $q^{\zeta} \leqlo^{\bar\lambda} q^{\iota^*}$ for each $\zeta<\cof(\theta)$, and so $q \leqlo^{\bar\lambda} q^{\iota^*}$.
Moreover, as $\bar\lambda$ is greater than both $\cof(\theta)$ and $\lambda_\theta$,
$q^\zeta$ has legal support by lemma \ref{supp:threads}.
Thus we are finished with the case $\cof(\theta) \leq \bar\lambda$.
Now assume $\cof(\theta) > \bar\lambda$.
We will now find $\iota'$ and $q'$ such that $q'$ is a $\pi_{\iota'}$-bound of $\bar p$ and for each $\xi < \rho$,
\begin{equation}\label{better_iota}
p_\xi \leqlo^{\cof(\theta)}\pi_{\iota'}(p_\xi).
\end{equation}
As $\rho\leq\lambda<\cof(\theta)$ and $\sigma^{\cof(\theta)}(p_\xi)<\cof(\theta)$ for each $\xi<\rho$, letting
\[ \iota' = \sup_{\xi<\rho} \sigma^{\cof(\theta)}(p_\xi),\]
we have $\iota' <\cof(\theta)$
and so $\iota' < \theta$.
Fix a $\nu < \rho$ and let $\sigma(\nu) = \sigma^{\cof(\theta)}(p_\nu)$.
Let $\bar q^\nu$ be the sequence defined by $q^\nu_\xi = \pi_{\sigma(\nu)}(p_\xi)$ for $\xi \geq \nu$. For $\xi < \nu$, the value of $q^\nu_\xi$ is arbitrary as long as $\bar q^\nu$ has $\bar w$ as a canonical witness (e.g. set $q^\nu_\xi = \pi_{1}(p_0)$).
It is straightforward to check that $\bar w$ is a $(\bar\lambda,x)$-canonical witness and a $(\lambda,x)$-strategic guide for each sequence $\bar q^\nu$, for $\nu < \xi$ --- this is why we only make demands on a tail in the definition of a strategic guide.
It is also clear that we can build a thread $(q_\nu)_{\nu<\rho}$ using (\ref{ext:glb}), where $q_\nu$ is a greatest lower bound of $\bar q^\nu$.
Let $q'$ be this thread.
We invite the reader to check that $q'$ is a greatest lower bound of the sequence $(\pi_{\iota'}(p_\xi))_{\xi<\rho}$ and that $q'\leqlol_{\iota'} \pi_{\iota'}(p_0)$.
Thus, $q'$ has legal support as $\lambda \geq \cof(\rho)$ and by lemma \ref{supp:threads}.
By choice of $\iota'$, \eqref{better_iota} holds.
Observe that in the case $\lambda<\bar \lambda$, (\ref{ext:glb}) also entails $q' \leqlol \pi_{\iota^*}(q')=q^*$.
Now we argue exactly as in the case where $\cof(\theta)\leq\lambda$, but this time
setting $q^0=q'$ and $\theta(0)=\iota'$. Again we build a sequence $(q^\zeta)_{\zeta<\cof(\theta)}$ by induction.
At each succesor step, $\{ \pi_{\theta(\zeta+1)}(p_\xi) \setdef \xi<\rho \}$ is $(\lambda,\cof(\theta),x)$-adequate in $(P_{\theta(\zeta)},P_{\theta(\zeta+1)})$.
By \ref{qc:ext}(\ref{ext:glb}) we obtain a greatest lower bound $q^{\zeta+1}\in P_{\theta(\zeta)}$ such that $q^{\zeta+1} \leqlo^{\cof(\theta)} q^{\zeta}$.
By induction, this entails $q^{\iota'}\leqlo^{\cof(\theta)} q^0 = q'$.
Thus at each limit stage $\zeta\leq\cof(\theta)$, $q^\zeta\leqlo^{\cof(\theta)} q'$
and lemma \ref{supp:threads} allows us to conclude that $q^{\zeta}$ has legal support.
Lastly, if $\lambda <\bar\lambda$, as $q^\zeta\leqlo^{\cof(\theta)} q' \leqlol q^*$, we also have
$q^\zeta\leqlol q^*=\pi_{\iota^*}(q^\zeta)$.
\qed
We conclude this section with an observation about the support of a greatest lower bound of an adequate sequence.
\begin{lem}\label{supp:glb}
Say $\bar p =(p_\xi)_{\xi<\rho}$ is a $(\lambda,x)$-adequate sequence with greatest lower bound $p$. Then for any regular $\bar \lambda$,
\[ \supp^{\bar\lambda}(p) \subseteq \bigcup_{\xi<\rho}\supp^{\bar\lambda}(p_\xi).\]
\end{lem}
\begin{proof}
Assume $\iota<\theta$ and $\iota \not \in \bigcup_{\xi<\rho}\supp^{\bar\lambda}(p_\xi)$. We may assume $\iota < \bar\lambda$ (since $p$ has diagonal support).
Then as $\pi_{\iota+1}$ is $\qcdefSeq(\bar\lambda)$, the sequence $(\pi_{\iota+1}(p_\xi))_{\xi<\rho}$ is $\qcdefSeq(\bar\lambda\cup\{x\})$-definable,
and for all $\xi<\rho$, $\pi_{\iota+1}(p_\xi)\leqlo^{\bar\lambda}\pi_{\iota}(p_\xi)$. Therefore we can apply \ref{qc:ext}(\ref{ext:glb}) for $(P_\iota,P_{\iota+1})$ (see \ref{qc:ext}, p.~\pageref{qc:ext}).
We conclude that $\pi_{\iota+1}(p)\leqlo^{\bar\lambda}\pi_\iota(p)$ and so $\iota \not\in \supp^{\bar\lambda}(p)$.
\end{proof}
\section{Stratified extension and iteration}\label{sec:s:ext}
In this section, we show that composition of stratified forcing is a special case of stratified extension.
We show that the second forcing in a stratified extension is stratified.
Finally we prove an iteration theorem for stratified forcing.
Let $P_0$ be a complete sub-order of $P_1$ and let $\pi\colon P_1 \rightarrow P_0$ be a strong projection and let $I$ be an intervall of regular cardinals.
Moreover, assume for $i \in \{0,1\}$, we have a system
\[\pss_i=(\D_i, c_i, \leqlol_i, \lequpl_i, \Cl_i)_{\lambda\in I}
\]
such that $\D\subseteq I \times V \times (P_i)^2 $ is a class which is definable with parameter $c_i$, and for every $\lambda\in I$,
$\leqlol_i$ and $\lequpl_i$ are binary relations on $P_i$ and $\Cl_i \subseteq P_i\times\lambda$.
\begin{dfn}\label{pss:is}
We write $\pss_0 \is \pss_1$ if and only if in addition to (\ref{ext}), (\ref{pi:mon}) and (\ref{F:coh}) (see \ref{pcs:is}, p.~\pageref{pcs:is}), the following hold:
\begin{enumerate}[label=(\pssIsRef), ref=\pssIsRef] %
\item If $q,q' \in P_0$ and $p,p' \in P_1$ are such that $q' \leqlol_0 q \leq \pi(p')$ and $p' \leqlol_1 p$, then $q'\cdot p' \leqlol_1 q\cdot p$. \label{pss:is:cdot:lo}
\item For all $p,q \in P_0$, $p\lequpl_0 q \Rightarrow p \lequpl_1 q$. \label{pss:is:ext}
\item For all $p,q \in P_1$, $p \lequpl_1 q \Rightarrow \pi(p) \lequpl_0 \pi(q)$. \label{pss:is:pi:mon}
\item If $w \leq \pi(d), \pi(r)$ and $d \lequpl r$ then $w \cdot d \lequpl w \cdot r$. \label{pss:is:cdot:up}
\item If $\Cl_1(p)\cap\Cl_1(q)\neq 0$ then $\Cl_0(\pi(p))\cap\Cl_0(\pi(q))\neq 0$. \label{pss:is:c}
\end{enumerate}
\end{dfn}
Observe that if $\pss_0 \is \pss_1$, we can drop the subscripts on $\lequpl_0$, $\lequpl_1$ and just write $\lequpl$ without causing confusion.
Observe also that by corollary \ref{lambda:is:big}, we can assume that $r \leqlo^{\card{P_1}} p$ holds exactly if $p=r$.
This implies\footnote{Interestingly, (\ref{suborder:and:leqlo}) also follows just from the assumption that for any $r,p\in P_1$, $r \lequp^{\card{P_1}} p$, together with (\ref{s:ext:exp}) \emph{coherent expansion}}
\begin{equation}\label{suborder:and:leqlo}
\forall p\in P_1 \big( p \leqlo^{\card{P_1}}\pi(p) \iff p \in P_0 \big).
\end{equation}
We could assume that $\Cl_0 = \Cl_1 \cap P_0\times \lambda$.
For if not, simply replace $\Cl_1$ by the following relation $\Cl_*$:
$s \in \Cl_*(p)$ if and only if $s \in {}^{\leq 2}\lambda$ such that $s(0) \in \Cl_0(\pi(p))$ and if $p \not \in P_0$ then $1 \in \dom(s)$ and $s(1) \in \Cl_1(p)$ (now in fact we get $\Cl_0(p) = \{ s\res 1 \setdef s \in \Cl_1(p) \}$ for $p \in P_0$).
To sum up, we could in principle completely eliminate any mention of $\pss_0$ from the definition of stratified extension.
Replacing (\ref{pss:is:cdot:lo})by the following two conditions yields an equivalent version of the above definition:
\begin{enumerate}[label=(\pssIsRe\Alph*), ref=\pssIsRe\Alph*]
\item $w \leqlol_0 \pi(p) \Rightarrow w \cdot p \leqlol_1 p$. \label{pss:is:lo:init}
\item If $w \leq \pi(p')$ and $p' \leqlol p$ then $w\cdot p' \leqlol w \cdot p$. \label{pss:is:lo:tail!}
\end{enumerate}
Sometimes it is more convenient to check both of these rather than (\ref{pss:is:cdot:lo}), which is concise but cumbersome to show.
Further notice that (\ref{pss:is:lo:tail!}) implies
\begin{enumerate}[label=(\pssIsRe b), ref=\pssIsRe b]
\item If $w \leq \pi(p)$ and $p \leqlol \pi(p)$ then $w\cdot p \leqlol w$. \label{pss:is:lo:tail}
\end{enumerate}
This is weaker than (\ref{pss:is:lo:tail!}). We note in passing that we could do entirely with (\ref{pss:is:lo:init}) and (\ref{pss:is:lo:tail}) and without (\ref{pss:is:lo:tail!}).
Neither condition (\ref{pss:is:cdot:lo}) nor any of its variants were included in \ref{def:pcs}, the definition of a \pre closure system simply because they are not needed to preserve quasi-closure in iterations---rather we need (\ref{pss:is:lo:tail}) to preserve coherent centering, and (\ref{pss:is:lo:init}) helps to preserve density at limits; see below.
We fix some convenient notation: If $d\leq r$, we say \emph{$p$ $\lambda$-interpolates $d$ and $r$} to mean that $p \lequpl d$ and $p\leqlol r$.
We say $p \leqlo^{<\lambda}q $ to mean that for all $\lambda'\in I \cap \lambda$,
we have $p \leqlo^{\lambda'} q$.
\begin{dfn}\label{def:s:ext}
Let $I$ be an intervall of regular cardinals.
We say the pair $(P_0,P_1)$ is a \emph{stratified extension on $I$, as witnessed by $(\pss_0,\pss_1)$}
if and only if $\pss_0$ witnesses that $P_0$ is stratified on $I$, $\pss_1$ is a \pre stratification system on $P_1$ and $\pss_0 \is \pss_1$;
Moreover, for all $\lambda \in I$ we have that (\ref{ext:redundant}), (\ref{ext:glb}) and all of the following conditions hold:
\begin{enumerate}[label=(\sExtRef), ref=\sExtRef]
\item \emph{Coherent Expansion}: For $p,d \in P_1$, if $p \lequpl d$, $d \leqlol \pi(d)$ and $\pi(p)\leq\pi(d)$, we have that $p \leq d$.\label{s:ext:exp}
\item \emph{Coherent Interpolation}: Given $d,r\in P_1$ such that $d \leq r$ and $p_0 \in P_0$ such that $p_0$ $\lambda$-interpolates $\pi(d)$ and $\pi(r)$ we can find $p \in P_1$ which $\lambda$-interpolates $d$ and $r$ such that $\pi(p)=p_0$. If moreover $d \leqlo^{<\bar\lambda}\pi(d)$, we can in addition assume $p \leqlo^{<\bar\lambda} \pi(p)\cdot r$.\label{s:ext:int}
\item \emph{Coherent Centering}: Say $d,p \in P_1$, $d \lequpl p$ and $\Cl_1(d)\cap \Cl_1(p) \neq\emptyset$.
Given $w_0 \in P_0$ such that both $w_0 \leqlo^{<\lambda} \pi(d)$ and $w_0 \leqlo^{<\lambda} \pi(p)$, we can find $w \in P_1$ such that $w\leqlo^{<\lambda}p,d$ and $\pi(w)=w_0$.\label{s:ext:cent}
\end{enumerate}
We find it relieving to notice that $P$ is stratified exactly if $(\{1_P\}, P)$ is a stratified extension.
Again, if we don't mention $\pss_0$, $\pss_1$ or $I$ we are either claiming that they can be appropriately defined or they can be inferred from the context.
\end{dfn}
\begin{lem}\label{stratified:comp:implies:ext}
If $P$ is stratified on $I$ and $\forces_P \dot Q$ is stratified on $I$, then $(P, P*\dot Q)$ \
is a stratified extension on $I$.
To be more precise, let $\pss_0$ denote the \pre stratification system witnessing that $P$ is stratified and let $\pss_1=(\bar\D,\barc,\bleqlol,\blequpl,\bCl)_{\lambda\in I}$ be the \pre stratification system constructed as in the proof of \ref{stratified:composition}, where we showed that $P*\dot Q$ is stratified. Then $(\pss_0,\pss_1)$ witnesses that $(P, P*\dot Q)$ is a stratified extension on $I$.
\end{lem}
\begin{proof}
We have already checked (\ref{ext}), (\ref{pi:mon}), (\ref{F:coh}), (\ref{ext:redundant}) and (\ref{ext:glb})---i.e. that $(P_, P*\dot Q)$ is a \emph{quasi-closed extension}---in lemma \ref{qc:comp:implies:ext}.
We showed that $\pss_1$ is a \pre stratification system when we proved theorem \ref{stratified:composition}.
It's technical but straightforward to check that $\pss_0 \is \pss_1$ (see definition \ref{pss:is}, p.~\pageref{pss:is}):
Fix $\bar p=(p,\dot p) \in P*\dot Q$ and $ w \in P$, $w \leq p$.
For (\ref{pss:is:lo:init}), say $w \leqlol p$. Then as $p \forces \dot p \dleqlol p$, we have $(w,p)\bleqlol \bar p$.
For (\ref{pss:is:lo:tail!}), fix another condition $\bar q=(q,\dot q) \in P*\dot Q$ such that $\bar p \bleqlol \bar q$. Then $p \forces \dot p \dleqlol \dot q$, whence $w\forces \dot p \dleqlol \dot q$ and so $(w,p) \bleqlol (w, 1_{\dot Q})$, done.
For (\ref{s:ext:exp}) and (\ref{pss:is:cdot:up}), let $\bar r=(r,\dot r) \in P*\dot Q$ and say $\bar p \blequpl \bar r$, i.e. $p \lequpl r$ and if $p\cdot r > 0$ then $p \cdot q \forces \dot p \dlequpl \dot r$.
To check (\ref{s:ext:exp}) \emph{coherent expansion}, assume $\bar r \bleqlol (r,1_{\dot Q})$ and $p \leq r$. Then $p \forces \dot p\dlequpl \dot r$. As $P$ forces \emph{expansion} for $\dot Q$, $p \forces \dot p \leq \dot q$ and we are done with (\ref{s:ext:exp}).
To check (\ref{pss:is:cdot:up}), say $w \leq p$. Then $w \cdot r \leq p \cdot r$, and so if $w\cdot r > 0$, it forces $\dot p \dlequpl \dot r$.
Since $w \lequpl w$, we infer that $(w,\dot p) \blequpl \bar r$.
The remaining (\ref{pss:is:ext}), (\ref{pss:is:pi:mon}) and (\ref{pss:is:c}) are immediate by the definition.
Now we check the conditions of \ref{def:s:ext} (see p.~\pageref{def:s:ext}).
For (\ref{s:ext:int}) \emph{coherent interpolation}, just look at how we found an interpolant in the proof of theorem \ref{stratified:composition}.
Do the same for (\ref{s:ext:cent}) \emph{coherent centering}.
\end{proof}
\noindent
The following is the analogue of \ref{lem:qc:succ} for quasi-closed extension:
\begin{lem}\label{lem:s:succ}
If $(P_0,P_1)$ is a stratified extension on $I$ and $\pi$ is $\qcdefG(\min I \cup\{c_1 \})$, then $P_1$ is stratified on $I$.
\end{lem}
\begin{proof}
The proof is a straightforward consequence of the definition and lemma \ref{lem:qc:succ}. We leave it to the reader.
\end{proof}
\begin{dfn}\label{natural:pss}
Say $\bar Q^\theta$ is an iteration such that each initial segment $P_\iota$ carries a \pre stratification system $\pss_\iota$ on $I_\iota = \mathbf{Reg}\cap[\lambda_\iota,\kappa)$, where the sequence $\bar \lambda=(\lambda_\iota)_{\iota\leq\theta}$ is a non-decreasing sequence of regulars.
Let $P_\theta$ be its $\lambda_\theta$-diagonal support limit.
We now add to the definition of the \emph{natural system of relations on $P_\theta$}.
Let $\lambda\in I_\theta$, where we set $I_\theta= [\lambda_\theta,\kappa)\cap \mathbf{Reg}$.
The relations $\leqlol$ and $\D$ are defined as in \ref{def:diagsupp}, p.~\pageref{def:diagsupp}. Let
\begin{enumerate}%
\item $p \lequpl_{\theta} q \iff \forall\iota < \theta \quad \pi_\iota(p) \lequpl_{\iota} \pi_\iota(q)$;
\item $p \in \dom(\Cl)$ if and only if for all $\iota <\sigma^\lambda(p)$, $\pi_\iota(p)\in\dom(\Cl_\iota)$;
\item $s \in \Cl_\theta(p)$ if and only if $s \colon \sigma^\lambda(p)\rightarrow\lambda$ and for all $\iota < \dom(s)$, we have $s(\iota)\in\Cl_\iota(\pi_\iota(p))$.
\end{enumerate}
\end{dfn}
\noindent
As before, the above yields a \pre stratification system under natural assumptions, as we shall see in the proof of theorem \ref{thm:it:strat}.
\begin{thm}\label{thm:it:strat}
Let $\bar Q^\theta$ be an iteration such that for each $\iota < \theta$, $P_\iota$ carries a \pre stratification system $\pss_\iota$ on $I_\iota = \mathbf{Reg}\cap[\lambda_\iota,\kappa)$, where the sequence $\bar \lambda=(\lambda_\iota)_{\iota<\theta}$ is a non-decreasing sequence of regulars.
Moreover, let $\lambda_\theta=\min(\mathbf{Reg}\setminus \sup_{\iota<\theta}\lambda_\iota)$, let $I_\theta= [\lambda_\theta, \kappa)$ and assume
\begin{enumerate}
\item For all $\iota<\theta$, $(P_\iota,P_{\iota+1})$ is a stratified extension on $I_\iota$.
\item If $\bar \iota < \theta$ is limit, $\pss_{\bar\iota}$ is the natural system of relations on $P_{\bar\iota}$
and $P_{\bar \iota}$ is the $\lambda_{\bar\iota}$-diagonal support limit of $\bar Q^{\bar \iota}$.
\item For each regular $\lambda\in[\lambda_\theta,\lambda^*)$ there is $\iota<\lambda^+$ such that for all $p \in P_\theta$ we have $\supp^\lambda(p) \subseteq \iota$.
\end{enumerate}
Let $P_\theta$ be the $\lambda_\theta$-diagonal support limit of $\bar Q^\theta$. Then $P_\theta$ is stratified on $I_\theta$.
\end{thm}
\begin{rem}
In our particular application we will have that for each regular $\lambda<\kappa$, there is $\iota<\lambda^+$ such that $\lambda<\lambda_\iota$.
Observe that by the definition of $\supp^\lambda(p)$, this implies that the last clause of the above is satisfied.
\end{rem}
\noindent
Of course, the following proof can be easily adapted to show that under the same hypothesis, for every $\iota<\theta$, $(P_\iota,P_\theta)$ is a stratified extension on $I_\theta$; while this approach facilitated the inductive proof in the case of quasi-closure, it would serve no purpose in the present context.
\begin{proof}[Proof of theorem \ref{thm:it:strat}.]
By lemma \ref{lem:s:succ}, we may assume $\theta$ is limit.
That $P_\theta$ is stratified on $I_\theta$ is witnessed by the natural system of relations $\pss_\theta$, as defined in \ref{natural:pss}.
The proof of the following lemma is a straightforward induction, which we leave to the reader:
\begin{lem}
For any $\iota <\bar\iota\leq\theta$, $\pss_\iota \is \pss_{\bar\iota}$.
\end{lem}
\noindent
Next, we check that $\pss_\theta$ is a \pre stratification system (see \ref{def:pss} p.~\pageref{def:pss}):
Conditions (\ref{up:extra}), (\ref{up}) and (\ref{s:lequp:vert}) are immediate by the definition of $\lequpl_\theta$ and the fact that for each $\iota<\theta$, $\pss_\iota$ is a \pre stratification system. The proofs resemble that of (\ref{exp}), see below.
The non-trivial condition is \ref{def:pss}(\ref{density}), \emph{Density}.\label{C:dense}
First we must check that $\ran(\Cl_\theta)$ has size at most $\lambda$: this is because by the last assumption of the theorem and by diagonal support,
$\supp^\lambda(p)\in[\iota]^{<\lambda}$ for some $\iota<\lambda^+$.
For the more interesting part of the argument, we use density and continuity for the initial segments $P_\iota$, $\iota<\theta$ together with quasi-closure.
Observe that by theorem \ref{thm:it:qc}, for any $\iota < \bar\iota \leq \theta$, $(P_\iota,P_\theta)$ is a quasi-closed extension on $I_\theta$.
Say we are given $p \in P_\theta$.
Let $\sigma = \sigma^\lambda(p)$. We may assume that $\sigma=\theta$, for otherwise we can use induction and \emph{Density} for $P_\sigma$ and are done.
Thus we can assume $\lambda_\theta<\lambda$, for otherwise, since $P_\theta$ is a diagonal support limit, $\supp^{\lambda}(p)$ is bounded below $\theta$.
So say we are given $\lambda' \in [\lambda_\theta,\lambda)$. We must find $q \leqlo^{\lambda'} p$ such that $q \in \dom(\Cl)$.
Let $\delta=\cof(\theta)$ and assume without loss of generality $\lambda'\geq\delta$ (otherwise we may increase $\lambda'$).
Fix a normal sequence $(\sigma(\xi))_{\xi\leq\delta}$ such that $\sigma(\delta)=\theta$.
We inductively construct a $\delta$-adequate sequence $(p_\xi)_{\xi<\delta}$ such that $p_0=p$ and for any $\nu,\xi$ such that $\nu < \xi < \delta$,
\begin{equation}\label{inDomainC}
\pi_{\sigma(\nu)}(p_{\xi}) \in \dom(\Cl_{\sigma(\nu)}).
\end{equation}
Fix appropriate $x$ so that the following sequence can be built in a $(\delta,x)$-adequate fashion and such that $(\sigma(\xi))_{\xi\leq\delta}$ is a component of $x$.
Let $p_0 = p$.
Assuming we have $p_\xi$ and $\bar w \res \xi+1$, find $p_{\xi+1}$ as follows.
We may find $q \in P_\theta$ such that $q \in \D_\theta(\lambda', (x,\bar w \res \xi+1) , p_\xi)$ and $q\leqlo^{\lambda'}p_\xi$.
Also, there is $q' \leqlo^{\lambda'} \pi_{\sigma(\xi)}(q)$ such that $q \in \dom(\Cl_{\sigma(\xi)})$.
Let $p_{\xi+1}=q'\cdot q$.
Since $\pss_{\sigma(\xi)}\is \pss_\theta$, by \ref{pss:is}(\ref{pss:is:lo:init}), $p_{\xi+1} \in \D_\theta(\lambda',x,p_\xi)$, and also $p_{\xi+1}\leqlo^{\lambda'} p_\xi$.
Moreover, by \ref{pss:is}(\ref{pss:is:c}), for any $\nu\leq\xi$,
\[
\pi_{\sigma(\nu)}(p_{\xi+1})\in\dom(\Cl_{\sigma(\nu)}).
\]
Apply the usual trick (see lemma \ref{adequate}) to find $w_{\xi+1}$ and $p_{\xi+1}$ as above so as to obtain a $(\delta,x)$-adequate sequence.
At limit stages $\bar \xi \leq \delta$, $p_{\bar \xi}$ is a greatest lower bound in $P_\theta$ of the sequence constructed so far.
It exists by quasi-closure for $P_\theta$.
We show
\begin{equation}\label{inDomainC:limit}
\pi_{\sigma(\bar\xi)}(p_{\bar \xi})\in\dom(\Cl_{\sigma(\bar\xi)}).
\end{equation}
Let $\nu < \bar \xi$ be arbitrary.
As $(P_{\sigma(\nu)},P_\theta)$ satisfies (\ref{qc:glb}),
\begin{equation}
\pi_{\sigma(\nu)}(p_{\bar \xi}) = \prod_{\xi\in(\nu,\bar\xi)} \pi_{\sigma(\nu)}(p_\xi).
\end{equation}
We want to apply (\ref{continuous}) for $P_{\sigma(\nu)}$.
By choice of $x$, $(\pi_{\sigma(\nu)}(p_\xi))_{\xi\in(\nu,\bar\xi)}$ is a $\lambda'$-adequate sequence;
and so we may use continuity for $P_{\sigma(\nu)}$.
Now by induction hypothesis, (\ref{inDomainC}) holds for all $\xi \in (\nu,\bar \xi)$ and so by (\ref{continuous}) we have $\pi_{\sigma(\nu)}(p_{\bar \xi}) \in \dom(\Cl_{\sigma(\nu)})$.
As $\nu<\bar \xi$ was arbitrary and by definition of $\Cl_{\sigma(\bar \xi)}$ we conclude that (\ref{inDomainC:limit}) holds.
In particular, for the last stage of our construction, we set $\bar\xi=\delta$ in (\ref{inDomainC:limit}) and conclude $p_\delta\in\dom(\Cl_{\theta})$, finishing the proof of \emph{Density}. So $\pss_\theta$ is a \pre stratification system.
\emph{Quasi-closure} was shown in lemma \ref{thm:it:qc}.
First we check conditions (\ref{exp})--(\ref{centering}) of \ref{stratified:main}, stratification (see p.~\pageref{stratified:main}).
\emph{Expansion} (\ref{exp}) is trivial:
If $d \lequpl_\theta r$ and $r \leqlol_\theta 1$, then for all $\iota<\theta$, $\pi_\iota(d) \lequpl_\iota \pi_\iota(r)$ and $\pi_\iota(r) \leqlol_\iota 1$. By induction, we may assume \emph{expansion} holds for each $P_\iota$, $\iota<\theta$. Thus $d \leq r$.
We show \emph{interpolation} (\ref{interpolation}) holds.
So fix $d,r \in P_\theta$ such that $d \leq r$ holds. We construct the interpolant $p$ by induction on its initial segments $p\res\iota$, for $\iota<\theta$.
Say we have already constructed $p\res\iota$.
Use coherent interpolation for $(P_\iota,P_{\iota+1})$ to obtain $p\res\iota+1$ interpolating $\pi_{\iota+1}(d)$ and $\pi_{\iota+1}(r)$:
demand that
\begin{equation}\label{int:above:supp:construction}
p\res\iota+1 \leqlo^{<\bar\lambda(\iota)} \pi_{\iota+1}(r)\cdot p\res\iota,
\end{equation}
where $\bar \lambda(\iota)$ is the maximal $\bar \lambda$ with the property that $\pi_{\iota+1}(d)\leqlo^{<\bar\lambda} \pi_\iota(d)$.\footnote{actually, it would suffice to demand this whenever $\iota \geq \sigma^\lambda(d)$}
We claim that for any $\gamma \in I_\theta$,
\begin{equation}\label{int:above:supp}
\iota \not \in \supp^\gamma(d)\cup\supp^\gamma(r)\Rightarrow p\res\iota+1 \leqlo^\gamma p\res\iota.
\end{equation}
So fix $\gamma\in\mathbf{Reg}$ and assume the hypothesis of (\ref{int:above:supp}). As $d\res\iota+1 \leqlo^\gamma d\res\iota$, by \ref{def:pcs}(\ref{qc:leqlo:vert}) and by definition of $\bar\lambda(\iota)$, we have $\gamma < \bar\lambda(\iota)$. Thus, (\ref{int:above:supp:construction}) yields
\begin{equation}\label{int:above:supp:p:pr}
p\res\iota+1 \leqlo^{\gamma} \pi_{\iota+1}(r)\cdot p\res \iota.
\end{equation}
Since $r\res\iota+1 \leqlo^\gamma r\res\iota$, by \ref{pss:is}(\ref{pss:is:lo:tail}) we infer
\begin{equation}\label{int:above:supp:pr:r}
\pi_{\iota+1} (r)\cdot p\res\iota \leqlo^{\gamma} p\res\iota.
\end{equation}
From (\ref{int:above:supp:p:pr}) and (\ref{int:above:supp:pr:r}) we get
$p\res\iota+1 \leqlo^\gamma p\res\iota$.
At limit stages $\bar\iota \leq \theta$ of the construction of the interpolant $p$, (\ref{int:above:supp}) holds for all $\iota<\bar\iota$, and so $p\res\bar\iota$ satisfies the support requirement.
This completes the proof of interpolation.
Now for \emph{centering} (\ref{centering}).
Say $p \lequpl d$ and fix $s \in \Cl_\theta (p) \cap \Cl_\theta(d)$.
Write $\sigma$ for $\dom(s)$.
By definition of $\Cl_\theta$, $\sigma = \sigma^\lambda(p)=\sigma^\lambda(d)$.
First, assume $\sigma=\theta$. In this case, we have $\lambda > \lambda_\theta$ by definition of diagonal support.
We construct $w$ by induction on its initial segments $w\res\iota$, for $\iota <\sigma$.
To start, use \emph{centering} for $P_1$ to obtain $w\res1$.
Assume we have $w\res\iota$; just use \emph{coherent centering} for $(P_\iota,P_{\iota+1})$ to obtain $w\res\iota+1$. At limits $\iota \leq \sigma$, use lemma \ref{supp:threads} and the fact that $\cof(\sigma)<\lambda$ and so $w_0 \res \iota \leqlo^{\cof(\sigma)} \pi_\iota(d)$.
Secondly, if $\sigma<\theta$, we can use \emph{centering} for $P_\sigma$ to obtain a lower bound $w_0$ of $\pi_{\sigma}(p)$ and $\pi_{\sigma}(d)$ with the desired properties.
We claim that $w=w_0 \cdot d$ is the desired condition, i.e. $w \leqlo^{<\lambda} p,d$.
The proof is of course by induction on $\iota \leq \theta$. For limit $\iota$, just use the induction hypothesis and the definition of $\leqlo^{<\gamma}_\iota$.
For the successor case, write
\begin{gather*}
d^*= \pi_{\iota+1}(d),\\
p^*= \pi_{\iota+1}(p),\\
w_0^*=w_0\cdot \pi_\iota(d) \\
\end{gather*}
and let $\pi$ denote $\pi_\iota$.
We may assume by induction that $w_0^* \leqlo^{<\lambda} \pi(d^*), \pi(p^*)$. In the following, use that $\pss_{\iota+1}$ is a \pre stratification system, $\pss_\iota \is \pss_{\iota+1}$ and \ref{pss:is}(\ref{s:ext:exp}), \emph{coherent expansion}.
Firstly, since $d^* \leqlol \pi(d^*)$ and $w^*_0 \leq \pi(d^*)$, by \ref{pss:is}(\ref{pss:is:lo:tail}),
we have
\begin{equation}\label{coh:cent:wd:small:supp}
w^*_0 \cdot d^* \leqlol w^*_0.
\end{equation}
In the same way, we can argue that
\begin{equation}\label{coh:centlower:triv:p}
w^*_0 \cdot p^* \leqlol w^*_0.
\end{equation}
Equation (\ref{coh:cent:wd:small:supp}) and $w^*_0 \leqlo^{<\lambda} \pi(d^*)$ give us
$w^*_0 \cdot d^* \leqlo^{<\lambda} \pi(d^*)$,
and together with $w^*_0 \cdot d^* \leq d^* \leq \pi(d^*)$ and \ref{def:pcs}(\ref{er}) we infer that $w^*_0 \cdot d^* \leqlo^{<\lambda} d^*$.
Since $d^* \lequpl p^*$ and $w^*_0 \leq \pi(d^*), \pi(p^*)$, we may conclude by $\pss_\iota \is \pss_{\iota+1}$ and \ref{pss:is}(\ref{pss:is:cdot:up}) that
\[ w^*_0 \cdot d^* \lequpl w^*_0 \cdot p^*.\]
This together with (\ref{coh:centlower:triv:p}), by \emph{coherent expansion} \ref{pss:is}(\ref{s:ext:exp}) yields
\[ w^*_0 \cdot d^* \leq w^*_0 \cdot p^*.\]
Thus $w^*_0 \cdot d^* \leq p^* \leq \pi(p^*)$ while at the same time $w^*_0 \cdot d^* \leqlol w^*_0 \leqlo^{<\lambda} \pi(p^*)$.
Another application of \ref{def:pcs}(\ref{er})
yields $w^*_0\cdot d^* \leqlo^{<\lambda} p^*$.
This ends the successor step of the inductive proof that $w \leqlo^{<\lambda} p,d$, and we are done with coherent centering.
Finally, check (\ref{continuous})\emph{Continuity}: Fix $\lambda^*, \lambda\in I_\theta$ such that $\lambda^*<\lambda$.
Say $\bar p$ and $\bar q$ are $(\lambda^*,x)$-adequate sequences of length $\rho$ with greatest lower bound $p$ and $q$ respectively, and for each $\xi<\rho$, $\Cl_\theta(p_\xi) \cap \Cl_\theta(q_\xi)\neq\emptyset$. We show that $p, q \in \dom(\Cl_\theta)$ and $\Cl_\theta(p)\cap\Cl_\theta(q)\neq\emptyset$.
For $\nu < \rho$, let $\sigma(\nu)= \sigma^\lambda(p_\nu)$.
Look at the sequence $\bar p^\nu = (p^\nu_\xi)_{\xi<\rho}$ of conditions in $P_{\sigma(\nu)}$, defined by
$p^\nu_\xi = 1_{P_{\sigma(\nu)}}$ for $\xi < \nu$ and $p^\nu_\xi = \pi_{\sigma(\nu)}(p_\xi)$ for $\xi\in [\nu,\rho)$.
As in the proof of theorem \ref{thm:it:qc}, it is easy to see $p^\nu_\xi = G(\bar w\res\xi+1)$ for some $\qcdefG(\lambda^*\cup\{x\})$ function $G$, where $\bar w$ is a canonical witness and strategic guide for $\bar p$.
Also as in the proof of theorem \ref{thm:it:qc}, $\bar p^\nu$ is $(\lambda^*,x)$-adequate. Its greatest lower bound is $\pi_{\sigma(\nu)}(p)$.
Thus by continuity for $P_{\sigma(\nu)}$, $\pi_{\sigma(\nu)}(p) \in \dom\Cl_{\sigma(\nu)}$.
Observe that by definition of $\Cl_\theta$, we have $\sigma^\lambda(p_\xi)=\sigma^\lambda(q_\xi)$ for all $\xi < \rho$.
Analogously, define adequate sequences $\bar q^\nu$ in $P_{\sigma(\nu)}$ with greatest lower bound $\pi_{\sigma(\nu)}(q)$.
By continuity for $P_{\sigma(\nu)}$, we infer
$\Cl_{\sigma(\nu)}(\pi_{\sigma(\nu)}(p))\cap\Cl_{\sigma(\nu)}(\pi_{\sigma(\nu)}(q))\neq\emptyset$ for each $\nu < \rho$.
Letting $\sigma = \sup_{\nu<\rho}\sigma(\nu)$, we infer
$\Cl_{\sigma}(\pi_{\sigma}(p))\cap\Cl_{\sigma}(\pi_{\sigma}(q))\neq\emptyset$, by the definition of $\Cl_{\sigma}$.
As $\sigma^\lambda(p), \sigma^\lambda(q) \leq \sigma$ by lemma \ref{supp:glb}, this means
$\Cl_{\theta}(p)\cap\Cl_{\theta}(q)\neq\emptyset$, by the definition of $\Cl_{\theta}$.
We are done with the proof of \emph{continuity}.
\end{proof}
\begin{cor}\label{cor:it:strat:comp}
Theorem \ref{thm:it:strat:comp} holds, that is, iterations with stratified components and diagonal support are stratified.
\end{cor}
\begin{proof}
By lemma \ref{stratified:comp:implies:ext}, composition is an example of stratified extension.
By theorem \ref{thm:it:strat}, since the iteration has diagonal support and its initial segments form a sequence of stratified extensions, the whole iteration is stratified.
\end{proof}
\section{Products}\label{sec:prod}
So far, stratified extension has only given us an overly complicated proof that iterations with stratified components are stratified. Here is a first non-trivial application: as a consequence of the next lemma, one can mix composition and products of stratified forcing freely in iterations with diagonal support, and the resulting iteration will be stratified.
\begin{lem}\label{products:ext}
If $P$ and $Q$ are stratified on $I$, $(P, P \times Q)$ is a stratified extension (on $I$).
\end{lem}
\begin{proof}
The proof is entirely as you expect.
Fix \pre stratification systems $\pss_P = (\D_P, c_P, \leqlol_P,\lequpl_P,\Cl_P)_{\lambda\in I}$ and $\pss_Q= (\D_Q, c_Q, \leqlol_Q,\lequpl_Q,\Cl_Q)_{\lambda\in I}$.
We now define a stratification system $\bar \pss = (\bar\D, \barc,\bleqlol,\blequpl,\bCl)_{\lambda\in I}$ on $P\times Q$ in the most natural way:
let $\bar \D(\lambda,x,(p,q))= \D_P(\lambda,x,p) \times \D_Q(\lambda,x,q)$ and let
\begin{align*}
(p,q) \bleqlol (\bar p,\bar q) \iff &p \leqlol_P \bar p \text{ and } q \leqlol_P \bar q \\
(p,q) \blequpl (\bar p,\bar q) \iff &p \lequpl_P \bar p \text{ and } q \lequpl_P \bar q \\
s \in \bCl(p,q) \iff & \Big[ s \in \Cl_P(p) \text{ and } q \leqlol 1_Q \Big] \text{ or } \Big[ s=(\chi,\zeta)\\
& \text{ where } \chi \in \Cl_P(p) \text{ and } \zeta \in \Cl_Q(q) \Big].\\
\end{align*}
That $\bar\pss$ is a \pre stratification system requires but a glance at the definitions (see \ref{def:pcs}, p.~\pageref{def:pcs} and \ref{def:pss}, p.~\pageref{def:pss}).
For example, \emph{Continuity}, (\ref{continuous}) is a straightforward application of continuity for both $P$ and $Q$.
The same holds for (\ref{ext}), (\ref{pi:mon}) and (\ref{F:coh}) (see p.~\pageref{pss:is} for the definition of $\pss_P \is \bar \pss$, and see p.~\pageref{pcs:is} for (\ref{ext}), (\ref{pi:mon}) and (\ref{F:coh}).
For the following, let $(p,q)\in P\times Q$, $w\in P$.
For your entertainment, we check \ref{pss:is}(\ref{pss:is:lo:init}) (see page~\pageref{pss:is:lo:init}).
Say $w \leqlol_p p$. Then clearly $(w,q) \bleqlol (p,q)$, done.
Now \ref{pss:is}(\ref{pss:is:lo:tail}):
say $w \leq p$ and $(p,q)\bleqlol(p,1_Q)$. This means $q \leqlol_Q 1_Q$ and so $(w,q)\bleqlol (w,1_Q)$, which is what we wanted to prove.
For the next two conditions, let $\bar d= (d,d^*) , \bar r=(r,r^*) \in P \times Q$ satisfy $\bar d \blequpl \bar r$.
We jump ahead and check (\ref{s:ext:exp}) of \ref{def:s:ext} (see p.~\pageref{def:s:ext}): say $d \leq r$ and $\bar r \bleqlol (r,1_Q)$. Then $r^* \leqlol_Q 1_Q$ and $d^* \lequpl_Q r^*$ by assumption, so by \ref{def:pss}(\ref{exp}) for $Q$, $d^* \leq r^*$ and thus $\bar d \leq \bar r$.
Let's check (\ref{pss:is:cdot:up})). Say $w \leq d$ and $w \leq r$.
By \ref{def:pss}(\ref{up:extra}) for $P$, $w \lequpl_P w$ and so $(w,d^*)\blequpl (w,r^*)$.
We omit the rest of \ref{pss:is} and conclude that $\pss_P \is \bar\pss$.
The most interesting part of the present proof is that of \emph{quasi-closed extension} (definition \ref{qc:ext}, see p.~\pageref{qc:glb}), of which we check (\ref{ext:glb}) , leaving (\ref{ext:redundant}) to the reader.
So say $(p_\xi,q_\xi)_{\xi<\rho}$ is $(\lambda,x)$-strategic and $\qcdefSeq(\bar\lambda\cup\{x\})$, and $(p_\xi)_{\xi<\rho}$ has a greatest lower bound $p$.
Firstly, since we can assume $x$ contains a parameter $X$ such that $P\subseteq X$, we conclude that $\bar q= (q_\xi)_{\xi<\rho}$ is $\qcdefSeq(\bar\lambda\cup\{x\})$ (by lemma \ref{lem:adeq:proj}). If $\lambda=\bar\lambda$, we are done as $\bar q$ is $\lambda$-adequate and $Q$ is quasi-closed.
If on the other hand, $\lambda<\bar\lambda$, we have that for all $\xi<\rho$, $q_\xi\leqlo^{\bar\lambda}_Q 1_Q$. By lemma \ref{bar:lambda:adeq},
$\bar q$ is $\bar\lambda$-adequate. Moreover, if $q$ is a greatest lower bound of $\bar q$, by quasi-closure for $Q$, we have $q \leqlo^{\bar\lambda}_Q 1_Q$. So $(p,q) \bar\leqlo^{\bar\lambda}(p,1_Q)$ and we are done.
To conclude that $(P, P\times Q)$ is a stratified extension, we check the remaining conditions of \ref{def:s:ext} (see p.~\pageref{def:s:ext}).
\emph{Coherent interpolation}, \ref{def:s:ext}(\ref{s:ext:int}) and \emph{Coherent centering}, \ref{def:s:ext}(\ref{s:ext:cent}) are identical to interpolation and centering for $Q$ in this context.
\end{proof}
\section{Stable meets for strong sub-orders}\label{sec:stm}
In the next section, we introduce the operation of amalgamation and show that the amalgamation of a stratified forcing $P$ is a stratified extension of $P$.
In that proof, we must show that a certain dense subset of $P$ is closed under taking meets with conditions from an ``initial segment'' (or rather, a strong sub-order) $Q$ (see lemma \ref{Q:cdot:D}, p.~\pageref{Q:cdot:D}). This will be facilitated by the so-called $Q$-stable meet operation $p \stm_Q r$, which we introduce in the present section.
In a standard iteration
this is a simple operation: $p \stm_Q r$ ``starts like $p$ on $Q$'' and then continues ``like $r$'' (as closely as possible) on $\quot{P}{Q}$.
What makes it useful is the following: if $r$ is ``a direct extension on the tail \quot{P}{Q}'' of a condition $p$, then $p\stm_Q r$ is a \emph{de-iure} direct extension of $p$, and moreover $r$ can be obtained straightforwardly from $p\stm_Q r$.
We now give a formal definition of such an operation, and then show that we can always define an operation $\stm$ on products and compositions.
Then we show how to define $\stm$ for infinite iterations. In the next section we shall see we also have a stable meet operator for amalgamation.
We take this formal, inductive approach (rather than defining $\stm$ directly on the iteration used in the main theorem) since amalgamation necessarily introduces an element of recursion into the definition of this operation.
Let $Q$ be a strong sub-order of $P$, and let
$\pi\colon P \rightarrow Q$ be the strong projection.
Say $\pss = (\hdots, \leqlol, \hdots)_{\lambda\in I}$ is a pre-stratification system on $P$.
\begin{dfn}\label{schnitzel}
We call $\stm$ a \emph{$Q$-stable meet operator on $P$} with respect to $\pss$ or a \emph{stable meet on $(Q,P)$} if and only if
\begin{enumerate}
\item $\stm\colon (p,r) \mapsto p \stm r$ is a function with $\dom(\stm)\subseteq P^2$ and $\ran(\stm)\subseteq P$.
\item $\dom(\stm)$ is the set of pairs $(p, r) \in P^2$ such that $r \leq p$ and
\begin{equation}\label{schnitzel:dom}
\exists \lambda \in I \quad r \leqlol \pi(r)\cdot p
\end{equation}
\item Whenever $r\leq p$ and $r \leqlol \pi(r)\cdot p$, the following hold:
\begin{gather}
p \stm r \leqlol p \label{schnitzel:leqlol}\\
\pi(p \stm r)=\pi(p)\label{schnitzel:proj}\\
\pi(r)\cdot (p \stm r) \approx r \label{schnitzel:meet}
\end{gather}
\end{enumerate}
As usual, we don't mention $\pss$ when context permits.
\end{dfn}
\noindent
A few remarks are in order to clarify this definition.
\begin{itemize}
\item We certainly don't have $p \stm r = r \stm p$.
\item The gist of (\ref{schnitzel:dom}) is that we try to express that $\pi(r)$ forces that in $\quot{P}{Q}$, the ``tail'' of $r$ is a direct extension (in the sense of $\leqlol$) of $p$; (\ref{schnitzel:dom}) captures the essence of this even when $\quot{P}{Q}$ is not stratified.
\item Observe that $r \leq p$ implies $\pi(r)\leq\pi(p)$ and so $\pi(r)\cdot p\in P$; thus (\ref{schnitzel:dom}) makes sense.
\item By $\pi(r)\cdot (p \stm r) \approx r $ we mean that $\pi(r)\cdot (p \stm r) \leq r $ and $\pi(r)\cdot (p \stm r) \geq r $.
Admittedly, we are very careful here. %
\item Observe that there could be more than one map $\stm$ satisfying the definition.
Intuitively, this is because (\ref{schnitzel:leqlol}) is not strong enough to fully determine $p \stm r$ on $\pi(p)-\pi(r)$.
If we add to the above the requirement that $-\pi(r) \cdot (p \stm r) = p$ hold in $\ro(P)$, this uniquely determines $\stm$. In fact this entails
\begin{equation}\label{schnitzel:var}
p \stm r= r + (p-\pi(r))
\end{equation}
in $\ro(P)$.
\footnote{In all the applications we have in mind, the natural definition of $\stm$ satisfies (\ref{schnitzel:var})---provided we work with the \emph{separative quotient} of $P$.}
For our purposes, this point is moot.
\end{itemize}
To understand the concept of stable meet operators, it is best to consider an instance of such an operator.
\begin{lem}\label{stable:meet:products}
Say $\bar P = Q_0 \times Q_1$, and say for each $\lambda\in I$, $\bleqlol$ is obtained from $\leqlol_0$ and $\leqlol_1$ as in the proof of \ref{products:ext} (where of course $\leqlol_i \subseteq (Q_i)^2$).
Then there is a stable meet operator on $(Q_0,\bar P)$ with respect to $\bleqlol$.
\end{lem}
\begin{proof}
Let $\pi$ denote the projection to the first coordinate.
Define $\dom(\stm)$ to be the set of pairs prescribed in definition \ref{schnitzel}.
Say $r=(r_0,r_1) \in Q_0\times Q_1$ and $p=(p_0,p_1)\in Q_0\times Q_1$ are such that $(r,p)\in\dom(\stm)$.
Define
\[(p_0,p_1) \stm (r_0,r_1) = (p_0,r_1).\]
As $(r,p) \in \dom(\stm)$, we can fix $\lambda$ such that $(r_0,r_1) \bleqlol \pi(r)\cdot p=(r_0,p_1)$, and so $r_1 \leqlo_1 p_1$.
Thus $(p_0, r_1) \bleqlol (p_0,p_1)$.
To check the other properties is left to the reader.
\end{proof}
\begin{lem}\label{stable:meet:comp}
Say $\bar P = Q * \dot R$, and say for each $\lambda\in I$, $\bleqlol$ is obtained from $\leqlol$ and $\dleqlol$ as in the proof of \ref{stratified:composition}.
Then there is a stable meet $\stm$ on $(Q_0,P)$ with respect to $\bleqlol$.
\end{lem}
\begin{proof}
Let $\pi$ denote the projection to the first coordinate.
Again, define $\dom(\stm)$ to be the set of pairs prescribed in definition \ref{schnitzel}.
Say $\bar r=(r,\dot r)$ and $\bar p=(p,\dot p)$ are such that $(\bar r,\bar p)\in\dom(\stm)$.
Define $\bar p \stm \bar r = (p,\dot r^*)$, where $\dot r^*$ is such that $r \forces \dot r^*=\dot r$ and $-r\forces \dot r^*=\dot p$.
Fixing a $\lambda$ witnessing that $(\bar r,\bar p)\in\dom(\stm)$, so that we have $(r,\dot r) \bleqlol \pi(\bar r)\cdot \bar p=(r,\dot p)$, and so $r\forces \dot r \dleqlol \dot p$.
Then $(p,\dot r^*) \bleqlol (p,\dot p)$, since $r \forces \dot r^*=\dot r \dleqlol \dot p$ and $p-r\forces \dot r^*= \dot p \dleqlol \dot p$.
To check the other properties is left to the reader.
\end{proof}
\noindent
The stable meet operator behaves very nicely in iterations:
\begin{lem}
Let $\bar Q^{\theta+1}$ be an iteration with diagonal support and say for each $\iota < \theta$, $P_\iota$ carries a \pre stratification system $\pss_\iota$ on $I$ and
\begin{enumerate}
\item For all $\iota<\theta$, we have $\pss_\iota \is \pss_{\iota+1}$.
\item If $\bar \iota \leq \theta$ is limit, $\pss_{\bar\iota}$ is the natural system of relations on $P_{\bar\iota}$.
\end{enumerate}
Moreover, say for each $\iota<\theta$, there is a stable meet operator $\stm^{\iota+1}_\iota$ on $(P_\iota,P_{\iota+1})$ with respect to $\pss_{\iota+1}$.
Then for each $\iota < \theta$ such that $\iota>0$ there is a $P_\iota$-stable meet operator on $P_\theta$.
\end{lem}
\begin{proof}
By induction on $\theta$, we show that for each pair $\iota, \eta$ such that $0<\iota<\eta \leq \theta$, there is a stable meet operator $\stm^\eta_\iota$ for $(P_{\iota},P_{\eta})$.
For $\iota, \eta$ as above and for $p$, $r \in P$ such that $r \leq p$ and (\ref{schnitzel:dom}) hold, define
\begin{equation}\label{schnitzel:it}
p \stm^\eta_\iota r = \prod_{\iota\leq\nu<\eta} \pi_{\nu+1}(p) \stm^{\nu+1}_\nu \pi_{\nu+1}(r).
\end{equation}
We prove by induction on $\theta$ that
\begin{enumerate}
\item For $\iota$, $\eta$ such that $0<\iota<\eta \leq \theta$ and for $(p,r)\in\dom(\stm^\theta_\iota)$,
\begin{equation}\label{schnitzel:comm}
\pi_\eta(p \stm^\theta_\iota r)=\pi_\eta(p) \stm^\eta_\iota \pi_\eta(r).
\end{equation}
The sequence of $\pi_\eta(p) \stm^\eta_\iota \pi_\nu(r)$, for $\eta \in (\iota,\theta]$ determines a thread in $P_\theta$, in the sense of definition \ref{it:terminology}.
\item For $\iota$ and $\eta$ as above, $\stm^\eta_\iota$ is a stable meet operator on $(P_\iota,P_\eta)$.
\end{enumerate}
Fix $\iota<\theta$.
Let $(p,r)\in\dom(\stm^\theta_\iota)$ be arbitrary and let $\lambda$ be an arbitrary witness to (\ref{schnitzel:dom}).
For the rest of the proof let $t^\eta_\iota$ denote $\pi_\eta(p) \stm^\eta_\iota \pi_\eta(r)$, for $0<\iota < \eta \leq \theta$.
First assume $\theta$ is limit. By induction hypothesis, $(t^\eta_\iota)_{\eta\in(\iota,\theta)}$ is a thread through $\bar Q^\theta$; by definition (\ref{schnitzel:it}), this thread is $p \stm^\theta_\iota r = t^\theta_\iota$.
We must show that $t^\theta_\iota$ has legal support.
It suffices to show that for each $\gamma \in I$, $\supp^\gamma (t^\theta_\iota) \subseteq \supp^\gamma (p) \cup \supp^\gamma (r)$.
So fix such a $\gamma$ and a $\xi <\theta$ such that we have
\begin{gather}
\pi_{\xi+1}(p)\leqlo^{\gamma} \pi_\xi(p),\label{xi:not:in:supp:p}\\
\pi_{\xi+1}(r)\leqlo^{\gamma} \pi_\xi(r).\label{xi:not:in:supp:r}
\end{gather}
We have $\pi_{\xi+1}(r)\leq \pi_\xi(r)\cdot \pi_{\xi+1}(p)\leq \pi_\xi(r)$ (simply because $r\leq p$) and so by (\ref{er}) and (\ref{xi:not:in:supp:r}), we have $\pi_{\xi+1}(r) \leqlo^\lambda \pi_\xi(r)\cdot \pi_{\xi+1}(p)$.
Since $\stm^{\xi+1}_\xi$ is a stable meet operator, and by (\ref{xi:not:in:supp:p}) we have
\[
\pi_{\xi+1}(p) \stm^{\xi+1}_\xi \pi_{\xi+1}(r)\leqlo^{\gamma}\pi_{\xi}(p).
\]
In other words,
$t^{\xi+1}_\xi \leqlo^\gamma \pi_\xi(p)$ and thus, taking the boolean meet with $t^\xi_\iota$ on both sides,
\[
t^{\xi+1}_\iota = t^{\xi+1}_\xi \cdot t^\xi_\iota \leqlo^\gamma \pi_\xi(p)\cdot t^\xi_\iota = t^\xi_\iota,
\]
where the last equation holds since $\stm^\xi_\iota$ is a stable meet operator by induction.
So we have $t^{\xi+1}_\iota \leqlo^\gamma t^\xi_\iota \in P_\xi$.
We conclude by lemma \ref{lem:support:simple} that $\xi \not \in \supp^\gamma (t^\theta_\iota)$, finishing the proof that $t^\theta_\iota$ has legal support.
It is straightforward to prove equations (\ref{schnitzel:leqlol}), (\ref{schnitzel:proj}) and (\ref{schnitzel:meet}) for $t^\theta_\iota=p \stm^\theta_\iota r$, assuming by induction that for each $\eta<\theta$, $\stm^\eta_\iota$ is a stable meet operator ($t^\theta_\iota$ is a thread whose initial segments satisfy these equations). We leave this to the reader.
Now let $\theta=\eta+1$. To see that $(t^\nu_\iota)_{\nu\leq\theta}$ is a thread, it suffices to show that $\pi_\eta(t^\theta_\iota)=t^\eta_\iota$. In order to show this, observe
\begin{equation*}
\pi_\eta(t^\theta_\iota)=t^\eta_\iota\cdot\pi_\eta(t^\theta_\eta)=t^\eta_\iota\cdot \pi_\eta(p)=t^\eta_\iota,
\end{equation*}
where the last equation holds since by induction, $t^\eta_\iota \leq \pi_\eta(p)$.
It follows by the induction hypothesis that $(t^\nu_\iota)_{\nu\leq\theta}$ is a thread.
It remains to show that $\stm^\theta_\iota$ is a $P_\iota$-stable meet on $P_\theta$, i.e we must show (\ref{schnitzel:leqlol}), (\ref{schnitzel:proj}) and (\ref{schnitzel:meet}).
Firstly, by induction,
\[ t^\eta_\iota \leqlol \pi_\eta(p),\]
and as $\stm^\theta_\eta$ is a $P_\eta$-stable meet on $P_\theta$,
\[\pi_\eta(p) = \pi_\eta(t^{\eta+1}_\eta).\]
By \ref{pss:is}(\ref{pss:is:lo:init}), this entails
\[ t^\eta_\iota \cdot t^{\eta+1}_\eta \leqlol t^{\eta+1}_\eta.\]
As $\stm^\theta_\eta$ is a $P_\eta$-stable meet on $P_\theta$, we have $t^{\eta+1}_\eta \leqlol p$, whence $t^\theta_\iota =t^\eta_\iota \cdot t^{\eta+1}_\eta \leqlol p$, proving (\ref{schnitzel:leqlol}).
Secondly,
\begin{equation*}\label{show:schnitzel:proj}
\pi_\iota(t^\theta_\iota) = \pi_\iota(t^\eta_\iota \cdot \pi_\eta(t^{\eta+1}_\eta))=\pi_\iota(t^\eta_\iota)=\pi_\iota(p).
\end{equation*}
The first equality here is trivial. The second holds since $t^\eta_\iota \leq \pi_\eta(p)$ by induction hypothesis and since by the assumption that $\stm^{\eta+1}_\eta$ is a $P_\eta$-stable meet, we have $\pi_\eta(p) = \pi_\eta(t^{\eta+1}_\eta)$.
The last equality of holds by induction.
Finally, we prove (\ref{schnitzel:meet}). We have
\[ \pi_\iota(r)\cdot t^\theta_\iota = \pi_\iota(r) \cdot t^\eta_\iota \cdot t^{\eta+1}_\eta = \pi_\eta(r) \cdot t^{\eta+1}_\eta = r,\]
where the first equation holds by definition, the second by induction hypothesis, and the last one since $\stm^{\eta+1}_\eta$ is a $P_\eta$-stable meet.
We are done with the successor case of the induction, and thus with the inductive proof of the lemma.
\end{proof}
\noindent
By the lemma, if $\bar Q^\theta$ is an iteration as in the hypothesis of the lemma and $\iota<\eta < \theta$, the map $\stm^\eta_\iota$
is the same as $\stm^\theta_\iota\res (P_\eta)^2$. So as we do for strong projections, we just write $\stm_\iota$ and we speak of the $P_\iota$-stable meet operator (without specifying the domain).
Moreover, we can formally set $p \stm_0 r = r$ and $p \stm_\iota r = p$ for $\iota \geq \theta$.
\section[Remoteness]{Remoteness: preserving strong sub-orders}\label{sec:remote}
Let $C, Q$ be complete sub-orders of $P$, and say $\pi_C\colon P \rightarrow C$ and
$\pi_Q\colon P \rightarrow C$ are strong projections.
We want to find a sufficient condition to ensure that $C$ is a complete sub-order of $\quot{P}{Q}$, after forcing with $Q$.
In our application $C$ will just be $\kappa$-Cohen forcing of $L$, for $\kappa$ the least Mahlo.
Our iteration will be of the form $P= Q * (\dot Q_0 \times C) * \dot Q_1$, so after forcing with $Q$, $C$ is a complete sub-order of $\quot{P}{Q} = (\dot Q_0 \times C) * \dot Q_1$.
We want the same to hold for $\Phi[C]$ (where $\Phi$ is a member of a particular family of automorphisms of $P$ which we construct using the technique of amalgamation); this helps to ensure ``coding areas'' don't get mixed up by the automorphisms, see lemma \ref{index:sequ} and lemma \ref{coding:survives}. So we have to introduce a property sufficient for $C$ to be a complete sub-order of $\quot{P}{Q}$, in such a way that this condition is inherited by $\Phi[C]$.
For this, we use of course the stratification of $P$.
This is necessary since forming $\quot{P}{Q}$ will not only ``take away an initial segment'' and leave $\Phi[C]$ in the tail in same obvious fashion as for $C$; instead forming $\quot{P}{Q}$ will also ``take away'' a small sub-algebra of $P$ (a copy of the random algebra).
Fix a \pre order $P$ which is stratified on $I$. The following definition is, as usual, relative to a particular \pre stratification system.
\begin{dfn}\label{remote}
We say $C$ is \emph{remote in $P$ over $Q$ (up to height $\kappa$)} if and only if for all $c \in C$ and $p \in P$ such that $c \leq \pi_C(p)$, we have
\begin{enumerate}
\item \label{remote:leqlo} $p\cdot c \leqlol p$ for every $\lambda \in I\cap\kappa)$;
\item $\pi_Q(p \cdot c) = \pi_Q(p)$.
\end{enumerate}
Observe that if we drop the first clause, this just says that $C$ is independent in $P$ over $Q$ (see definition \ref{indie}).
For a $P$-name $\dot C$, we say $\dot C$ is remote in $P$ over $Q$ if and only if it is a name for a generic of a remote complete sub-order of $P$; i.e. there is a complete sub-order $R_C$ of $P$ (with a strong projection $\pi_C\colon P \rightarrow R_C$) such that $R_C$ is dense in $\gen{\dot C}^{\ro(P)}$ and $R_C$ is remote in $P$ over $Q$.
\end{dfn}
\begin{lem}\label{remote:lemma:not:in}
If $\dot C$ is a $P$-name which is remote over $Q$, then $\dot C$ is not in $V^Q$.
\end{lem}
\begin{proof}
An immediate consequence of lemma \ref{indie:lemma:not:in}
\end{proof}
\chapter{Stratified Forcing}\label{sec:stratified forcing}
In this section we assume $V=L[A]$ for some class $A$.
We define \emph{stratified partial orders}, show such orders preserve cofinalities, give some examples and show that stratification is preserved under composition.
We also define diagonal support and state that iterations whose components are stratified are themselves stratified.\footnote{Most of these definitions are heavily inspired by \cite{sdf:iterated:cl}; see also \cite{friedman:codingbook}.}
The proof is left out, since we prove a slightly more general theorem in section \ref{sec:ext} where we deal with iterations with stratified initial segments but where the components aren't necessarily stratified.
We present the definition of stratification in two parts: the first we dub \emph{quasi-closure}.
We treat this first part separately from the remaining axioms of stratification for the following reasons: firstly, the proofs that each of these two groups of axioms is preserved in iterations are not only different but virtually independent of each other.
Secondly, we hope that the reader will agree that quasi-closure is interesting in its own right.
This view is in stark contrast to the fact that quasi-closure alone is not a very useful property---
in fact, every partial order is quasi-closed.
One should think of it as an incomplete notion, to which some other property has to be added in order to render it non-trivial.
Stratification is one example of this, closely connected to the notion of centered forcing.
There may be other examples, as well.
Before we define quasi-closure, we introduce pre-closure systems; analogously we will define pre-stratification systems.
We can reuse these notions when we define \emph{quasi-closed and stratified extension}; see section \ref{sec:ext} on page\pageref{sec:ext}.\\
\section{Quasi-Closure}\label{sec:qc}
Throughout, let $\langle R, \leq \rangle$ be a \pre order and let $I$ be an intervall of regular cardinals, that is $I = \mathbf{Reg} \cap [\lambda_0,\lambda_1)$ or $I = \mathbf{Reg} \cap[\lambda_0, \infty)$ for some regular $\lambda_0,\lambda_1$.
We introduce the notion of $R$ being \emph{stratified on $I$}, which implies that for any $\lambda \in I$, any ordinal of cofinality greater than $\lambda$ will remain so after forcing with $R$.
We have to use another property of $R$ to show cardinals greater $(\lambda_1)^+$ are preserved (in our application, $R$ will have an appropriate chain condition).
Moreover, we want to allow for $R$ to collapse some cardinals up to and including $\lambda_0$.
From now on, we write $\qcdefSeq(z)$ for the set of formulas provably equivalent to a $\Pi^A_1(z)$ formula in some weak version of $\ZFC$, $T$.
For concreteness, say $T$ states that the universe is closed under every function $F$ which is definable by a $\Sigma^A_1$ formula, say $\Phi$ such that $\ZFC \land V=L[A]$ proves ``$\forall x\exists y \Phi(x,y)$''.\footnote{This includes all rudimentary functions and the function $x \mapsto \operatorname{tcl}(x)$ assigning to $x$ its transitive closure; also, $T$ implies $\Delta_0^A$-separation.
Alternatively, we could take $T$ to be %
the theory of all $\Sigma^A_1$-elementary sub-models of $L[A]$.}
We only mention this to point put that unfortunately, these classes of formulas are not closed under bounded quantification.
In practice, the reader will see $T$ is of little relevance and we shall sometimes omit the superscript.
We now make a few convenient definitions that facilitate the treatment of quasi-closed partial orders, which we define afterward.
\begin{dfn}\label{def:pcs}
We say $\pcs=\pcsl$ is a \pre closure system for $R$ on $I$ if and only if $\D\subseteq \mathbf{Reg}\times V\times R^2$
is a $\qcdefD(c)$ class and for every $\lambda \in I$, $x \in V$, $p,q,r \in R$
\begin{enumerate}[label=(C \arabic*), ref=C~\arabic*]
\item if $p\leq q \in \D(\lambda, x, r)$ then $p \in \D(\lambda, x, r)$.\label{qc:D}%
\item The relation $\leqlol$ is a preorder on $R$ and $p\leqlol q \Rightarrow p \leq q$. \label{qc:preorder}
\item If $p \leq q \leq r$ and $p \leqlol r$ then $p \leqlol q$.\label{er}
\item If $\bar \lambda \in I$, $\bar \lambda \geq \lambda$ then $q\leqlo^{\bar\lambda}p \Rightarrow q \leqlol p$.\label{qc:leqlo:vert}
\end{enumerate}
\end{dfn}
As a notational convenience, define $\leqlo^0$ to mean $\leq_R$.
Clause (\ref{er}) can be dropped if one is not interested in iterations.
Observe that by (\ref{er}), $\leqlol$ is well-defined with respect to equivalence modulo $\approx$ (remember we say $p \approx q \iff p \leq q$ and $q \leq p$).
Think of each of the relations $\leqlol$ as a notion of \emph{direct extension}, as it is often called in the case of e.g. Prikry-like forcings. Intuitively, $p \leqlol q$ expresses that $p$ extends $q$ but some part ``below $\lambda$'' is left unchanged.
Think of $\D$ as providing a kind of strategy, as with strategically closed forcing. Together, this additional structure on $R$ allows us to express that certain sequences have lower bounds in $R$.
The missing ingredient and distinct flavor of quasi-closure is the condition that these sequences be \emph{definable} in a sense.
The main point is that the definability and the use of $\D$ are intertwined in that they are coordinated by a common object $\bar w$ which we shall call the \emph{strategic guide and canonical witness}.
For the next two definitions, fix a \pre closure system $\pcs$ for $R$ on $I$.
All the notions in the next two definitions have their meaning \emph{with respect to $\pcs$}.
\begin{dfn}\label{strategic}
Let $\bar p = (p_\xi)_{\xi<\rho}$ be a sequence of conditions in $R$ and $\rho \leq \lambda$.
We say $\bar w$ is a \emph{$(\lambda,x)$-strategic guide for $\bar p$} if and only if $\bar w= (w_\xi)_{\xi<\rho}$ is a sequence of the same length as $\bar p$ and for a tail of $\xi < \rho$
\begin{enumerate}
\item for some $\lambda' \in I$, $p_{\xi+1} \in \D(\lambda', (\{x,c\}, \bar w \res \xi+1) ,p_\xi)$
and $p_{\xi+1} \leqlo^{\lambda'} p_\xi$.
\item $p_{\xi+1}\leqlol p_\xi$.
\item if $\xi$ is a limit, $p_{\xi}$ is a greatest lower bound of $(p_\nu)_{\nu<\xi}$.
\end{enumerate}
\end{dfn}
\begin{dfn}\label{canonical}
\begin{enumerate}
\item We say a sequence $\bar w = (w_\xi)_{\xi<\rho}$ is $\qcdefSeq(z)$ if for some $\qcdefSeq(z)$ formula $\Psi$ we have $w = w_\xi \iff \Psi(w,\xi)$.
We say $G$ is a $\qcdefG(z)$ function if $G(x)=y$ is a $\qcdefG(z)$ formula and $G$ is a partial function.
\item Let $\bar p = (p_\xi)_{\xi<\rho}$ be a sequence of conditions in $R$ and $\rho \leq \lambda$.
We say $\bar w = (w_\xi)_{\xi<\rho}$ is a \emph{$(\lambda, x)$-canonical witness for $\bar p$}
if and only if $\bar w$ is $\qcdefSeq(\lambda\cup\{x, c\})$ and for some $\qcdefG(\lambda\cup\{(x, c)\})$ (partial) function $G$, we have $p_\xi = G(\bar w\res \xi +1 )$ for every $\xi <\rho$.
\item We say $\bar p$ is $(\lambda,x)$-adequate if and only if $\rho \leq \lambda$ and there is $\bar w$ which is both a strategic guide and a canonical witness for $\bar p$.
\item If $\bar p$ is $(\lambda,x)$-adequate for some $x$ we say $\bar p$ is \emph{$\lambda$-adequate}.
\end{enumerate}
\end{dfn}
\noindent
Note that a sequence $\bar p$ has a canonical witness simply if it is $\mathbf{\Sigma}^T_2$.
Also note there is some flexibility for with respect to the definability requirement we out in $G$; one could probably do with rudimentary $G$ or with one specific function.
\begin{dfn}\label{def:qc}
We say $\langle R, \pcs \rangle$ is \emph{quasi-closed on $I$} if and only if for any $x$ and $\lambda\in I$,
\begin{enumerate}[label=(C \Roman*), ref=C~\Roman*]
\item For any $p \in R$ there is $q \in \D(\lambda,x,p)$ such that $q \leqlol p$.
In addition we can demand that $q \leqlo^{\bar\lambda} 1_R$ for any $\bar \lambda \in I$ such that $p \leqlo^{\bar\lambda} 1_R$.
\label{qc:redundant}
\item Every $\lambda$-adequate sequence $\bar p = (p_\xi)_{\xi<\rho}$ in $R$ has a greatest lower bound $p$ in $R$
and for all $\xi<\rho$, $p \leqlol p_\xi$. If $\bar \lambda\in I$ is such that for each $\xi < \rho$, $p_\xi \leqlo^{\bar \lambda} 1_R$, then $p \leqlo^{\bar \lambda} 1_R$.\label{qc:glb}
\end{enumerate}
We also use the expression \emph{$R$ is quasi-closed as witnessed by $\pcs$}. If we omit $\pcs$ and no \pre closure system can be deduced from the context, we mean that there exists a \pre closure system $\pcs$ such that $\langle R, \pcs \rangle$ is quasi-closed.
When we say \emph{quasi-closed on $[\lambda_0, \lambda_1)$}, we mean of course
quasi-closed on $[\lambda_0, \lambda_1)\cap\mathbf{Reg}$ etc.
\end{dfn}
Clause (\ref{qc:redundant}) and the last sentence of clause (\ref{qc:glb}) are useful regarding infinite iterations of quasi-closed forcings.
\begin{rem}For arbitrary $R$, just define $p\leqlol q$ if and only if $p=q$ and $\D(\lambda,x,p)=\{p\}$ for all regular $\lambda \geq \lambda_0$ and all $x$.
Then $R$ is quasi-closed.
Quasi-closure becomes non-trivial under the additional hypothesis that certain questions about the generic extension can be decided by strengthening a condition in the sense of $\leqlol$, for some $\lambda$.
Stratified forcing satisfies such a hypothesis.
\end{rem}
\begin{rem}
Say $R$ is $\lambda^+$-closed; then $R$ trivially satisfies all the conditions of \ref{def:qc} for this one $\lambda$.
The same is true if $R$ is $\lambda^+$-strategic: for if $\sigma\colon R\rightarrow R$ is a strategy for $R$,
define $\D(\lambda,x,p)= \{ q \in R \setdef q \leq \sigma(p) \}$.
$\D$ is clearly $\qcdefSeq(\{\sigma \})$. We can define $\leqlol$ to be the same as $\leq$.
Of course every strategic and thus every adequate sequence has a greatest lower bound.
This is not vacuous, in that there are non-trivial adequate sequences. In fact, every sequence $\bar p$ of length less than $\lambda^+$ which adheres to $\sigma$ is $\lambda$-adequate:
For fix $\bar p$ of length less than $\lambda^+$. By re-indexing, assume the length of $\bar p$ is $\lambda$.
Since $\bar p$ is $\qcdefSeq(\{\bar p\})$, and since $\D$ does not depend on $x$ at all, $\bar p$ is $(\lambda,\{\bar p\})$-adequate.
These are our first examples of forcings which non-trivially satisfy the definition of quasi-closed (albeit for just one fixed $\lambda$), since any statement about the generic can be decided by extending in the sense of $\leqlol$.
\end{rem}
\section{A word about definability and set forcing}\label{rem:def:set}
The concept of quasi-closure is more natural in a class forcing context.
Since we only apply it for set forcing, we can make do with $\qcdefSeq(z)$ (as opposed taking into account $\Pi^T_n(z)$ for all $n > 1$).
We still have to use a form of $\Pi_1$-uniformisation, implicit in the construction of canonical witnesses (see \ref{adequate}).
In class forcing, this uniformisation can be achieved using the fact that conditions form a class, and the ``height'' of each condition in an adequate sequence effectively represents the canonical witness.
Using this technique, I do not know a proof that $\kappa$ is not collapsed.
See also \cite{friedman:codingbook}, chapter 8.2, p.~175.
Think of $x$ as a tuple of constants which can be used in in the definition of an adequate sequence $\bar p$.
From now on we shall always assume that $c$ is among the constants given by $x$.
We can and will assume that some large enough $L_\mu[A]$ is among the constants given by $x$, where $\mu$ is a cardinal and $R \in L_\mu[A]$ (i.e. assume that $L_\mu[A]$ is in or is recursive in $c$).
This allows us to bound quantifiers of certain statements and argue that they are $\qcdefSeq(\{x\}\cup\lambda)$.
Intuitively, this use of $c$ is analogous to the use of a large structure with predicates in the context of proper forcing.
In our application, it will suffice to set $c=\{ \kappa^{+++}\}$.
\section{Stratification}
\begin{dfn}\label{def:pss}
We say $\pss=\pssl$ is a \pre stratification system for $R$ on $I$ if and only if
$\pcsl$ is a \pre closure system for $R$ on $I$ and for every $\lambda \in I$ the following conditions are met:
\begin{enumerate}[label=(S \arabic*), ref=S~\arabic*]
\item \label{up:extra} The binary relation $\lequpl$ on $R$ satisfies $p \leq q\Rightarrow p \lequpl q$.
\item \label{up} If $p \leq q \lequpl r$ then $p \lequpl r$.\footnote{Note that we don't assume $\lequpl$ to be transitive, since this does not seem to be preserved by composition. If $\lequpl$ were transitive, condition (\ref{up}) would follow from (\ref{up:extra}).
We need (\ref{up}) for lemma \ref{density reduction}. We need that $\lequpl$ is reflexive (i.e. $p \lequpl p$ for all $p$) for \ref{pss:is}(\ref{pss:is:cdot:up}). In the context of (\ref{up}), reflexivity is the same as the last part of (\ref{up:extra}).}
\item \label{s:lequp:vert} If $\lambda \leq \bar \lambda$ and $\bar \lambda \in I$ then $p \lequpl q \Rightarrow p\lequp^{\bar\lambda} q$.
\item\label{density}
\textit{Density:} $\Cl \subseteq R \times \lambda$ is a binary relation such that $\dom(\C^\lambda)$ is dense in $R$. Moreover, for any
$\lambda' \in I \cap \lambda$ and $p \in R$, there is $q \leqlo^{\lambda'} p$ such that $q \in \dom(\C^\lambda)$.
%
\item\label{continuous}
\textit{Continuity}: If $\lambda'\in I\cap\lambda$ and $p$ is a greatest lower bound of the
$\lambda'$-adequate sequence $\bar p=(p_\xi)_{\xi<\rho}$ and for each $\xi<\rho$, $p_\xi \in \dom(\Cl)$, \emph{then} $ p \in \dom(\Cl)$.\footnote{In any application I know, we could ask this for all $\lambda'\in I$, not just those smaller than $\lambda$.}
If in addition $\bar q$ is another $\lambda'$-adequate sequence of length $\rho$ with greatest lower bound $q$ and for each $\xi <\rho$,
$\Cl(p_\xi)\cap\Cl(q_\xi)\neq\emptyset$, then
$\Cl(p)\cap\Cl(q)\neq\emptyset$.
\end{enumerate}
\end{dfn}
The last part of condition (\ref{up:extra}), all of (\ref{s:lequp:vert}) and and the ``moreover'' part of (\ref{density}) can be dropped if one is not interested in infinite iterations.
Don't think that $\lequpl$ is a \pre order or well-defined on the separative quotient of $R$, although (\ref{up}) guarantees some regularity with respect to $\approx$.
\begin{dfn}\label{stratified:main}
We say a \pre order $\langle R, \leq \rangle$ is \emph{stratified on $I$} as witnessed by $\pss=\pssl$ if and only if $\pss$ is a pre-stratification system on $I$, $\langle R, \leqlol, \D \rangle_{\lambda \in I} $ is quasi-closed, and for each $\lambda\in I$
the following conditions hold:
\begin{enumerate}[label=(S \Roman*), ref=S~\Roman*]
\item\label{exp} \emph{Expansion}: If $p \lequpl d$ and $d \leqlol 1_R$, then in fact $p \leq d$.
\item \label{interpolation}
\emph{Interpolation}: If $d \leq r$, there is $p \leqlo^\lambda r$ such that $p \lequp^\lambda d$.
In addition, whenever $\bar \lambda\in I$ and $d \leqlo^{\bar\lambda} 1_R$, then also $p \leqlo^{\bar\lambda} 1_R$.
\item \label{centering}
\emph{Centering}: If $p \lequpl d$ and and $\Cl(p)\cap\Cl(d)\neq\emptyset$ then $p$ and $d$ are compatible. In fact, there is $w$ such that for any
$\lambda' \in I \cap\lambda$, $w \leqlo^{\lambda'} p$ and $w \leqlo^{\lambda'} d$.
\end{enumerate}
If we omit $\pss$ and no \pre stratification system can be deduced from the context, we mean that there exists a \pre stratification system $\pss$ witnessing that $R$ is stratified.
When we say \emph{stratified on $[\lambda_0, \lambda_1)$}, we mean of course
stratified on $[\lambda_0, \lambda_1)\cap\mathbf{Reg}$ etc.
\end{dfn}
Conditions (\ref{continuous}) and (\ref{exp}) are important to preserve stratification in (infinite) iterations.
The second part of (\ref{centering}) was introduced to allow for amalgamation (see section \ref{sec:amalgamation}), but is also useful to control the diagonal support in iterations (see below).
We illustrate definition \ref{stratified:main} with some examples.
\begin{exm}\label{exm:centered}
A simple observation is that for any \pre order $R$, $R$ is stratified above $\card{R}$.
A little more generally, if $R$ is $\lambda_0$-centered, then $R$ is stratified on $[\lambda_0,\infty)$:
for if $\lambda \geq \lambda_0$, we can simply define $p \leqlol q$ just if $p=q$. Similarly, $\D(\lambda,x,p)=R$ for all $p \in R$. Thus, quasi-closure and continuity become vacuous.
Moreover, let $g:R\rightarrow \lambda_0$ be a a function such that if $g(p)=g(q)$ then $p$ and $q$ are compatible.
Set $\Cl(p)=\{ g(p) \}$ for any $p \in R$. Lastly, define $p \lequpl q$ to hold for \emph{ any } pair $p,q$.
Then the only non-vacuous condition in the definition of stratification is \emph{centering}, which holds for every $\lambda\geq \lambda_0$ since $g$ witnessed that $R$ was centered.
\end{exm}
\noindent
This example has a corollary:
\begin{cor}\label{lambda:is:big}
If a \pre ordered set $R$ is stratified, we can always assume that for $\lambda \geq \card{R}$,
$\D$, $\leqlol$, $\lequpl$ and $\Cl$ take the simple form discussed above in example \ref{exm:centered}.
\end{cor}
\noindent
Observe that (\ref{qc:leqlo:vert}) and (\ref{s:lequp:vert}) remain valid if we modify a given pre-stratification system in such a way as to ensure that the above assumption holds.
A more interesting example:
\begin{exm}\label{strat:ex:*}
Say $R=P*\dot Q$ where $P$ is $(\lambda_0)^+$-centered and $(\lambda_0)^+$-closed and $\forces_P \dot Q$ is $\lambda_0$-centered and $\lambda_0$-closed. Then $R$ is stratified on $\mathbf{Reg}$ --- ignoring (\ref{continuous}). If the centering functions for $P$ and $\dot Q_\xi$ in the extension are continuous in the sense of (\ref{continuous})---and it seems that for many centered forcings, this is the case---$R$ is actually stratified.
Define $\D$ as in the previous example.
For $\lambda < \lambda_0$, define $\leqlol$ to be identical to $\leq_R$;
define $p \lequpl q$ if and only if $p = q$ and let $\Cl(p)=\lambda$ for every $p\in R$. Then \emph{Interpolation} and \emph{centering} hold at $\lambda$ for trivial reasons, and quasi-closure at $\lambda$ expresses the fact that $R$ is closed under sequences of length at most $\lambda$.
For $\lambda = \lambda_0$, fix a name for a centering function $\dot g$; set $(p,\dot q) \leqlol (p', \dot q')$ if and only if $(p,\dot q) \leq (p', \dot q')$ and $\dot q = \dot q'$; set $(p,\dot q) \lequpl (p', \dot q')$ if and only if $p \leq_P p'$.
Let $\chi \in\Cl(p,\dot q)$ if and only if $p \forces \dot g(\dot q)=\check\chi$.
Lastly, $R$ has a subset $R'$ which is $(\lambda_0)^+$ centered and $\leqlo^{\lambda_0}$-dense. This allows us to define a stratification above $(\lambda_0)^+$, in a similar way to the previous example.
\end{exm}
Finally, we can discuss preservation of cofinalities and the $\GCH$.
\begin{dfn}\label{chromatic}
In the following, we fix a regular cardinal $\lambda\in I$ and drop the superscripts on $\C, \leqlo$ and $\lequp$.
\begin{enumerate}
\item Let $D, D^* \subseteq R$, and $r\in R$.
We say \textit{$r$ $\lambda$-reduces $D$ to $D^*$} (often, we don't mention the prefix $\lambda$) exactly if
\begin{enumerate}
\item $D^* \subseteq \dom(\C)$ and $\card{D^*}\leq\lambda$;
\item for each $d\in D^*$, $r \lequp d$;
\item for any $q \in \dom(\C)\cap D$, if $q\leq r$, there is $d \in D^*$ such that $\C(q)\cap\C(d)\neq 0$.
\end{enumerate}
\item \label{chromatic function} Let $\dot \alpha$ be a name for an element in the ground model $V$, and let $r \in R$.
We say $\dot \alpha$ is $\lambda$-chromatic below $r$ just if there is a function $H$ with $\dom H \subseteq\lambda$ such that if $q \leq r$ decides $\dot \alpha$ and $q \in \dom(\C)$,
then $\C(q)\cap \dom(H)\neq 0$ and for all $\chi \in \C(q)\cap \dom(H)$, $q \forces \dot \alpha = H(\chi)$ (to be pedantically precise, we mean the ``standard name'' for $H(\chi)$).
We call such $H$ a $\lambda$-\textit{spectrum (of $\dot \alpha$)}.
\item If $\dot s$ is a name and $p \forces \dot s \colon \lambda \rightarrow V$, then we say \emph{ $\dot s$ is $\lambda$-chromatic (with $\lambda$-spectrum $(H_\xi)_{\xi<\lambda}$) below $p$} if and only if for each $\xi<\lambda$, $\dot s(\check \xi)$ is chromatic with spectrum $H_\xi$ below $p$.
For notational convenience, we say $\dot x$ is $0$-chromatic below $p$ if for some $x$, $p \forces_R \dot x =\check{x}$.
\end{enumerate}
Observe that if for some ground model set $x$, $p \forces \dot \alpha = \check x$ (i.e. $\alpha$ is $0$-chromatic), then $\dot \alpha$ is in fact $\lambda$-chromatic for every regular $\lambda$,
and the function with domain $\lambda$ and constant value $x$ is a $\lambda$-spectrum.
\end{dfn}
To illustrate $\lambda$-reduction, observe that if $r$ $\lambda$-reduces $D$ to $D^*$, then $D^*$ is a subset of $D^*$ of size $\lambda$ which is predense below $r$:
for if $q\leq r$, we can assume that $q \in \dom(\Cl)$, so that there is $d \in D^*$ with $\Cl(d)\cap\Cl(q)\neq \emptyset$. By \eqref{up}, $q \lequpl d$ and so by \emph{Centering} \eqref{centering}, $q$ and $d$ are compatible.
Now say Say $R$ is stratified in $I$ and $\lambda\in I$.
\begin{thm}\label{single xdensity reduction}
Let $D$ be a dense subset of $R$ and $p \in D$. Then there is $r \leqlol p$ such that
$r$ $\lambda$-reduces $D$ to some $D^*$.
\end{thm}
\begin{proof}
We build an adequate sequence $\bar p = (p_0)_{\xi \leq \lambda}$, starting with $p_0=p$.
We shall first sketch a construction such that $r = p_\lambda$ is the desired condition, without specifying a canonical witness; then we argue how this construction can be carried out so as to obtain a canonical witness at the same time.
This will serve as a blueprint for later constructions, where we shall not explicitly carry out the construction of $\bar w$, as it is entirely analogous to the case at hand.
Let $x = (p_0, \leqlol, \lequpl, \Cl, R, \D, c)$, where $c$ contains any parameters in the definition of $\D$.
Say we have constructed $\bar w \res \nu$ and $\bar p \res \nu$. If $\nu \leq \lambda$ is a limit ordinal, assume by induction that $\bar p \res \nu$ is adequate and let $p_\nu$ be a greatest lower bound.
Now say $\nu = \xi +1$.
Choose, in a manner yet to be specified $w_{\xi+1}$, $p_{\xi+1}$, $p^*_\xi$ and $d_\xi$ which satisfy the following:
\begin{enumerate}
\item \label{build_ad_red_strategic}$p^*_\xi \in \D(\lambda,(x, \bar w\res\xi+1),p_\xi)$, and $p_{\xi+1}\leqlol p^*_\xi$.
\item \label{build_ad_red}
\begin{enumerate}
\item If there is no $d \leq p^*_\xi$ such that $d \in D$ and $\xi \in \Cl(d)$ we demand $p_{\xi+1}=p^*_\xi$;
\item \label{work} else $p_{\xi+1}$ and $d_\xi$ satisfy: $d_\xi \leq p^*_\xi$, $d_\xi \in D$ and $\xi \in \Cl(d_\xi)$; moreover $p_{\xi+1} \leqlo p^*_\xi$ and $p_{\xi+1} \lequpl d_\xi$.
\end{enumerate}
\end{enumerate}
Observe this does not even mention $w_{\xi+1}$; its role will be explained by lemma \ref{adequate}, below.
If the first alternative of item \ref{build_ad_red} obtains, the $d_\xi$ we choose is completely irrelevant for the rest of the construction.
It is clear that $\bar p$ will have $\bar w$ as a strategic guide:
since $p_{\xi+1} \leq p^*_\xi$, by item \ref{build_ad_red_strategic} and \eqref{qc:D}, $p_{\xi+1}\in\D(\lambda,x,p_\xi)$ and $p_{\xi+1}\leqlol p_\xi$.
Let $D^* = \{ d_\xi \setdef \xi < \rho\text{ and \ref{work} obtained at stage $\xi+1$ } \}$.
Clearly, $p_\lambda$ reduces $D$ to $D^*$.
For if $q \leq p_\lambda$, $q \in D$ and $\xi \in \Cl(q)$, then $q$ witnesses that at step $\xi+1$ of the construction of $\bar p$, \ref{work} obtained.
So we have $d_\xi \in D$ with $\xi \in \Cl(d_\xi)$.
Moreover, $q\leq p_\lambda \lequpl d_\xi$.
In order to conclude that this construction works, we need to show that for every $\nu \leq \lambda$, $\bar p \res \nu$ is $(\lambda,x)$-adequate. For this it is enough to explain how precisely we made our choices in the above construction:
\begin{lem}\label{adequate}
In the previous, $w_{\xi+1}$, $p_{\xi+1}$, $p^*_\xi$ and $d_\xi$ can be chosen so that $\bar w$ is a $(\lambda, x)$-canonical witness for $\bar p$.
\end{lem}
\begin{proof}
Observe that at successor stages, $p_{\xi+1}$, $p^*_\xi$ and $d_\xi$ are chosen so as to satisfy a property which is $\qcdefSeq$ in parameters $x$ and $\bar w\res\xi+1$.
That is, we may choose $\qcdefSeq(x)$ formula $\Phi_0$ such that $\Phi_0(p_{\xi+1},p^*_\xi,d_\xi, \bar w\res \xi+1)$ holds just if \ref{build_ad_red_strategic} and \ref{build_ad_red} hold.
In what follows, the choice of $\bar w$ will be such that for each $\xi < \lambda$,
$w_\xi$ is a quintuple, and $\bar p$ is obtained from $\bar w$ by projecting to the first coordinate.
In fact, we will have $w_{\xi+1} = (p_\xi, M, p^*_\xi, d_\xi, \bar w \res \xi+1)$ where $M$ is a model allowing us to uniformize $\Phi_0$ and the rest of the coordinates are as in the successor step of the above construction. At limits $\nu$ we shall have
$w_\nu = (p_\nu, M, p^*, d, \bar w \res \nu)$, where $p^*$ and $d$ are just dummies.
Using the initial segment $\bar w\res \xi$ as a last coordinate in $w_\xi$ makes $\bar w$ $\qcdefSeq(x)$ in the end (see below).
Let $\Phi_1(p,p^*,d,\tilde{w})$ be the formula expressing that if $\nu = \dom (\tilde{w})$ is a limit then $p$ is the greatest lower bound in $R$ of the sequence obtained from $\tilde{w}$ by projecting to the first coordinate (and for clarity, $p^*=d=\emptyset$ if you want); and if $\nu = \xi +1$ then $\Phi_0(p,p^*, d,\tilde{w})$ holds.
Now let $\Phi(w, \tilde{w})$ be the $\qcdefSeq(x)$ formula expressing: $w = (p,M,p^*,d,\tilde{w})$ is such that
\begin{enumerate}
\item $M$ is transitive, $M \prec_{\Sigma_1} L[A]$ and $p,p^*,d,\tilde{w} \in M$,
\item for any initial segment of $M$ containing $p,p^*,d,\bar w$ we have $N\not\prec_{\Sigma_1} M$,
\item $M \vDash (p,p^*,d)$ is $\leq_{L[A}$-least such that $\Phi_1(p,p^*,d, \tilde{w})$.
\end{enumerate}
We may choose $\bar w$ recursively such that for each $\xi < \lambda$, $\Phi(w_\xi, \bar w\res\xi)$ holds:
Observe that if such $w_\xi$ exists, it is unique.
For limit $\nu$, we may assume by induction that $\bar w\res \nu$ is a canonical witness for $\bar p\res\nu$, allowing us to infer $\bar p \res \nu$ is adequate and that $p_\nu$ and hence $w_\nu$ exists. At successor stages $\xi+1$, $w_{\xi+1}$ always exists, and so $\bar w$ is well-defined.
Lastly, we show $\bar w$ is $\qcdefSeq(x)$: this is because $w = w_\xi$ if and only if $w$ is a quintuple with last coordinate $\bar w$ such that $\Phi(w, \bar w)$ holds,
and $\bar w$ is a sequence such that for each $\xi \in \dom(\bar w)$, $\Phi(\bar w (\xi), \bar w \res\xi)$.
This finishes the proof of the lemma.
\end{proof}
Having shown that $\bar p$ may be chosen as a $(\lambda,x)$-adequate sequence, we are finished with the proof of the theorem.
\end{proof}
\begin{lem}\label{density reduction}
For each $\xi<\lambda$, let $D_\xi$ be an open dense subset of $R$. Let
\[
X = \{ q \in R \setdef \exists D^* \quad \forall \xi<\lambda \quad\text{q $\lambda$-reduces $D_\xi$ to $D^*$} \}.
\]
Then $X$ is dense in $\langle R, \leqlol \rangle$ and open in $\langle R, \leq \rangle$.
If $\dot s$ is a name such that $p \forces \dot s \colon \lambda \rightarrow V$, the set of $q$ such that $\dot s$ is $\lambda$-chromatic below $q$ is dense in $\langle R(\leq p), \leqlol \rangle$ and open.
\end{lem}
\begin{proof}
Let $\bar D=(D_\xi)_{\xi<\lambda}$ be a sequence of dense open subsets of $R$. Build a sequence as before:
let
\[
x = \langle p_0, \leqlol, \lequpl, \Cl, \Power(R), \bar D, y \rangle.
\]
At successor steps $\xi$, choose $p_\xi$ and $w_\xi$ such that $p_\xi \leqlol p_{\xi-1}$ and $p_\xi \in \D(\lambda,(x, \bar w\res\xi+1),p_{\xi-1})$ and such that we can pick $D^*_\xi$ such that $p_\xi$ reduces $D_\xi$ to $D^*_\xi$.
As in lemma \ref{adequate}, argue this can be done in such a way that the resulting sequence $\bar p$ is $(\lambda,x)$-adequate.
So a greatest lower bound $p_\lambda$ exists and for each $\xi<\lambda$, $p_\lambda$ reduces $D_\xi$ to $\bigcup_{\xi<\lambda} D^*_\xi$.
Now let $p \forces \dot s \colon \lambda \rightarrow V$. Let $D_\xi$ be the set of conditions $p \in R$ which decide $\dot f(\xi)$.
As above, find $q$ reducing all $D_\xi$ to $D^*$.
We now find a spectrum for $\dot s$: For $\chi < \lambda$, if $w \leq q$ decides $\dot s(\xi)$ and $\chi \in \Cl(w)$, there is also $d \in D^*$ which decides $\dot s(\xi)$ and such that $\chi \in \Cl(d)$.
Fix $z$ such that $d \forces f(\xi) = \check z$.
Then we may set $H_\xi(\chi)=z$.
It is easy to check that for each $\xi<\lambda$, $H_\xi$ is a spectrum for $\dot s(\xi)$ (and thus $(H_\xi)_{\xi<\lambda}$ is a spectrum for $\dot s$):
Say $w \leq q$ decides $\dot s(\xi)$ and fix some $\chi \in \Cl(w)$.
Then there is $d \in D^*$ with $\chi \in \Cl(d)$ such that $d \forces s(\xi) = H_\xi(\chi)$.
As $d \in D^*$, $q \lequpl d$. So as $\chi \in \Cl(w)\cap\Cl(d)$, $w$ and $d$ are compatible and thus $w \forces s(\xi) = H_\xi(\chi)$.
\end{proof}
\begin{cor}
Cofinalities greater than $\lambda$ remain greater than $\lambda$ after forcing with $R$ and $(2^\lambda)^V = (2^\lambda)^{V[G]}$ for any $R$-generic $V$.
\end{cor}
\section{Composition of stratified forcing}
In the main theorem of this section, theorem \ref{stratified:composition} below, we show stratification is preserved by composition. In the proof, we use ``guessing systems'', which we shall motivate now, before we state and prove the theorem.
Say $P$ is stratified and $\dot Q$ is forced by $P$ to be stratified on $I$, and let $\lambda\in I$ be fixed.
We know $P$ has a centering relation $\C$ and $\dot Q$ is forced to have a centering relation $\dot \C$ in the extension.
Similar to the proof that composition of centered forcing stays centered, we want to gain some control over $\dot \C$ in the ground model.
If we ignore the requirements \ref{def:pss}(\ref{density}) \emph{density} and \ref{stratified:main}(\ref{continuous}) \emph{continuity},
we could define $\bar \C$ on $P*\dot Q$ in the following way:
\[ (\chi,\xi) \in \bar \C(d,\dot d) \iff \chi \in \C(d)\text{ and }d \forces \check \xi \in \dot \C(\dot d) \]
Then $\dom(\bar\C)$ is dense and \ref{stratified:main}(\ref{centering}) \emph{centering} holds.
\noindent
The following definition also satisfies \ref{def:pss}(\ref{density}) \emph{density}: let
\begin{multline}\label{C:second:attempt}
(\chi,X) \in \bar \C(d,\dot d) \iff \Big( \chi \in \C(d)\text{ and for some $\dot \xi$ and }\lambda' \in\mathbf{Reg}\cap[\lambda_0,\lambda), \\
\text{ $X$ is a $\lambda'$-spectrum for $\dot \xi$ below $d$ and }
d \forces \dot \xi \in \dot \C (\dot d) \Big).
\end{multline}
Let's check \ref{def:pss}(\ref{density}) \emph{density} holds:
Given a condition $\bar p= (p,\dot p)$ and $\lambda'\in\mathbf{Reg}\cap[\lambda_0,\lambda)$, we can find $\bar d=(d,\dot d)$
such that $\bar d \alo^{\lambda'}\bar p$, $d \in \dom(\C)$ and $d \forces \dot d \in \dom(\dot \C)$. Moreover, we can assume that for some name $\dot \chi$, $d\forces \dot \chi \in \dot \C(\dot d)$ and $\dot \chi$ is $\lambda'$-chromatic. We have $\bar d \in \dom(\bar \C)$.
Let's also check that \ref{stratified:main}(\ref{centering}) \emph{centering} holds: say $\bar d \aupl \bar p$ and $(\chi,X)\in\bar\C(\bar p)\cap \bar\C(\bar d)$. First, observe that $p$ and $d$ are compatible. Fix $\dot \chi_0$, $\dot\chi_1$ such that both $p\forces\dot \chi_0 \in \dot \C(\dot p)$ and
$d\forces\dot \chi_1 \in \dot \C(\dot d)$
and $X$ is a spectrum for $\dot \chi_0$ below $p$ and for $\dot \chi_1$ below $d$.
As $p \cdot d \forces \dot \chi_0 = \dot \chi_1 \in \dot \C(\dot d) \cap \dot \C(\dot p)$, by \emph{centering} for $\dot Q$ in the extension,
$p \cdot d \forces \dot d$ and $\dot p$ are compatible, whence $\bar d$ and $\bar p$ are compatible.
To show stratification is preserved at limits, we will have to use \emph{Continuity} of the centering function;
Unfortunately, the approach described above does not yield a \emph{continuous} centering function in the sense of (\ref{continuous}).
For say $\bar d_\xi = (d_\xi, \dot d_\xi)$ form a $\lambda'$-adequate sequence of length $\rho$, and for each $\xi<\rho$,
$d_\xi\forces \dot \chi_\xi \in \dot \C(\dot d_\xi)$ and $X_\xi$ is a $\lambda_\xi$-spectrum for $\dot \chi_\xi$ below $d_\xi$.
By \emph{Continuity} for the components of the forcing,
if $(d,\dot d)$ is a greatest lower bound, we know $d \forces \dot d \in \dom(\dot\C)$;
but there is no reason to assume that there exists a $P$-name $\dot \gamma$, such that $d \forces \dot \gamma \in \dot\C(\dot d)$ and $\dot \gamma$ is $\lambda''$-chromatic for some $\lambda'' < \lambda$.
The solution to this problem is to allow a more general set of values for $\bar C(\bar d)$: in the situation described above, e.g. the sequence $(X_\xi)_{\xi<\rho}$ be used in much the same way as the single spectrum $X$.
This leads to the notion of a guessing system, which will be precisely defined in \ref{gs}.
\begin{thm}\label{stratified:composition}
Say $P$ is stratified on $I$ and $\dot Q$ is forced by $P$ to be stratified on $I$. Then $\bar P=P * \dot Q$ is stratified (on $I$).
\end{thm}
\begin{proof}
Say stratification of $P$ is witnessed by $\D, \Cl, \leqlol$, $\lequpl$ for each regular $\lambda\in I$, and we have class $\dot \D$, definable with parameter $\dot c$ and names $\dot\C^\lambda, \dot\leqlo^\lambda$, $\dlequpl$ for $\lambda$ regular which are forced by $P$ to witness the stratification of $\dot Q$. We now define $\bar \D, \bCl, \bleqlol$ and $\blequpl$ for regular $\lambda \geq \lambda_0$ to witness stratification of $P * \dot Q$.
\subsubsection*{The auxiliary orderings}
Let $\lambda \in I$ be regular.
We say $(p,\dot q)\bar \leqlo^\lambda(u,\dot v)$ if and only if $p \leqlo^\lambda u$ and $p\forces_P \dot q \dot \leqlo^\lambda \dot v$.
This defines a \pre order stronger than the natural ordering on $P*\dot Q$ (i.e. \ref{def:pcs}(\ref{qc:preorder}) holds).
Define $\bar p \blequpl \bar q$ if and only if $p \lequpl q$ and if $p \cdot q \neq 0$, $p \cdot q \forces_P \dot p \lequpl \dot q$.
\subsubsection*{The ordering axioms}
Let $\bar p=(p,\dot p)$, $\bar q=(q,\dot q)$ and $\bar r=(r,\dot r)$ be conditions in $\bar P$.
We check that \ref{def:pcs}(\ref{er}) holds: Say $\bar p\leq_{\bar P} \bar q \leq_{\bar P} \bar r$ and $\bar p \bleqlol \bar r$.
Then $p \leqlol r$ by \ref{def:pcs}(\ref{er}) for $P$. Moreover, $p$ forces \ref{def:pcs}(\ref{er}) for $\dot Q$ as well as $\dot p\leq\dot q \leq\dot r$ and $\dot p \dot\leqlo^\lambda \dot r$. So $p \forces_P \dot p \dot\leqlol \dot q$, and we conclude $\bar p \bleqlol \bar q$.
Check that \ref{def:pss}(\ref{up}) holds:
Say $\bar p \leq_{\bar P} \bar q \blequpl \bar r$. By (\ref{up}) for $P$, $p \lequpl r$. If $p \cdot r \neq 0$,
\[ p \cdot r \forces_P \dot p \leq_{\dot Q} \dot q \lequpl \dot r, \]
and so $p \cdot r\forces_P \dot p \lequpl \dot r$. Thus $\bar p \blequpl \bar r$.
Next, check \ref{def:pss}(\ref{exp}). Say $\bar p \blequpl \bar q$ and $\bar q \bleqlol 1_{\bar P}$. By (\ref{exp}) for $P$, $p \leq q$.
So $p \forces \dot p \dlequpl q \dleqlol 1_{\dot Q}$, so by (\ref{exp}) applied in the extension,
$p \forces \dot p \leq_{\dot Q} \dot q$, whence $\bar p \leq_{\bar P} \bar q$.
We leave it to the reader to check \ref{def:pcs}(\ref{qc:leqlo:vert}) and \ref{stratified:main}(\ref{s:lequp:vert}).
\subsubsection*{Quasi-Closure}
Define $(p, \dot p) \in \bar \D( \lambda, x, (q, \dot q))$ if and only if $p \in \D(\lambda,x,q)$ and $p\forces \dot p \in \dot \D( \lambda, x, \dot q)$.
It is straightforward to see that this definition is $\qcdefD( (c, \dot c))$.
Clearly, this defines a ``dense and open'' set, i.e. (\ref{qc:D}) and \eqref{qc:redundant} are satisfied.
Now say $(p_\xi, \dot q_\xi)_{\xi<\rho}$ is $(\lambda,x)$-adequate. We show this sequence has a greatest lower bound. Let $\bar w$ be a strategic guide and a canonical witness for $(p_\xi, \dot q_\xi)_{\xi<\rho}$.
We can immediately infer by the definition of $\bar \D$ that $\bar w$ is a strategic guide for $(p_\xi)_{\xi<\rho}$.
It is also clear that $\bar w$ is a canonical witness, since $p_\xi$ can be obtained from $(p_\xi, \dot q_\xi)$ by projecting to the first coordinate, and this map is $\Delta_0$.
Thus there is a greatest lower bound $p_\rho$ of $(p_\xi)_{\xi<\rho}$.
It is easy to see now that $p_\rho\forces_P$``$(\dot q_\xi)_{\xi<\rho}$ is $(\lambda,x)$-adequate'':
Fix a $\qcdefSeq(\lambda\cup\{x\})$ formula $\Phi(x,y)$ such that for $\xi<\rho$ we have
\[ \Phi(x,\xi) \iff x=w_\xi. \]
Then the relativization $\Phi(x,y)^{L[A]}$ witnesses that $1_P$ forces that $\bar w$ is $\Pi_1(\lambda\cup\{x\})$ in the extension by $P$, as well.
For a $\qcdefG(\lambda\cup\{x\})$-function $G$,
$(p_\xi, \dot q_\xi)=G(\bar w\res\xi+1)$ for each $\xi<\rho$, and so
$\dot q_\xi = \pi_1 (G (\bar w\res\xi+1))$, where $\pi_1$ is the projection to the second coordinate.
As $G(x)=y$ is $\qcdefG$, clearly the relativized formula $(\pi_1(G(x))=y)^{L[A]}$ is also $\qcdefG(\lambda\cup\{x\})$ in the extension by $P$.
So $\bar w$ is forced to be canonical witness.
Moreover, it is clear that $p_\rho$ forces that
$\bar w$ is a strategic guide for $(\dot q_\xi)_{\xi<\rho}$, by the definition of $\bar \D$.
So we can find $\dot q_\rho$ such that $p_\rho \forces_P$``$\dot q_\rho$ is a greatest lower bound of $(\dot q_\xi)_{\xi<\rho}$'', whence $(p_\rho, \dot q_\rho)$ is a greatest lower bound of the original sequence.
Leaving the last sentence of (\ref{qc:glb}) to the reader, we conclude that $\langle P*\dot Q, \bar \lequpl, (c, \dot c), \bar \D\rangle$ is $\lambda$-quasi-closed above on $I$.
To define $\bCl$, we first define the notion of a \emph{guessing system}.
Roughly speaking, a guessing system consists of conditions which are organized in levels; the conditions on the bottom have a $\bCl$-value in the sense of (\ref{C:second:attempt}). Conditions on higher levels are greatest lower bounds of conditions on the levels below, and we have some control over their $\dot\C^\lambda$-value by \emph{continuity} for $\dot Q$.
%
%
%
\begin{dfn}\label{gs}
Say $(p,\dot q)\in P*\dot Q$ and $\lambda$ is regular and uncountable.
A \emph{$\lambda$-guessing system for $\dot q$ below $p$} is a quadruple $(T_{\g},H_{\g}, \lambda_{\g}, q_{\g})$ such that
\begin{enumerate}[1.]
\item $T_{\g}$ is a tree, $T_{\g} \subseteq {}^{<\omega}\gamma$, where $\gamma=\width(T) <\lambda$ and $\is$ (initial segment) is reversely well founded on $T_{\g}$.
The root of $T_{\g}$ is $\emptyset$ (i.e. the empty sequence).
\item For $s \in T_{\g}$, $\rho_{\g}(s) = \{\xi \setdef s\conc\xi \in T_{\g} \}$
is an ordinal. Write $T_{\g}^0$ for the set of $\is$-maximal $s \in T_{\g}$, i.e.
$T_{\g}^0 = \{ s \in T_{\g} \setdef \rho_{\g}(s)=0 \}$.
\item $q_{\g}$ is a function from $T_{\g}$ into the set of $P$-names for conditions in $\dot Q$ and $\lambda_{\g}\colon T_{\g} \rightarrow \lambda\cap\mathbf{Reg}$.
\item\label{approximate} For $s \in T_{\g} \setminus T_{\g}^0$, $\{q_{\g}(s\conc\xi)\}_{\xi<\rho_{\g}(s)}$ is a $\lambda_{\g}(s)$-adequate sequence and $p$ forces that
$q_{\g}(s)$ is a greatest lower bound of $\{q_{\g}(s\conc\xi)\}_{\xi<\rho_{\g}(s)}$.
\item $\dom(H_{\g})=T_{\g}^0$.
\item \label{spectrum} For $s \in T_{\g}^0$, there is a $P$-name $\dot \chi$ such that
$p \forces_P \dot \chi \in \Cl(q_{\g}(s))$
and $H_{\g}(s)$ is a $\lambda_{\g}(s)$-spectrum of $\dot \chi$ below $p$.
\item $q_{\g}(\emptyset) = \dot q$.
\end{enumerate}
\end{dfn}
\noindent Now we are ready to define ${\bar \C}^{\lambda}$: let $s \in {\bar \C}^{\lambda} (p,\dot q)$ if and only if \emph{either}
\begin{enumerate}[(a)]
\item \label{short:center} $s \in \Cl(p)$ and $p \forces \dot q \dleqlol 1_{\dot Q}$ holds \emph{or else}
\item \label{long:center} if $\lambda>\min I$, $s=(\chi, T_{\g},H_{\g},\lambda_{\g})$ where $\chi \in \Cl(p)$ and for some $q_{\g}$, $(T_{\g},H_{\g},\lambda_{\g}, q_{\g})$ is a $\lambda$-guessing system for $\dot q$ below $p$.
\item if $\lambda=\min I$, $s=(\chi,\xi)$, where $\chi \in \Cl(p)$ and $p \forces_P \check \xi \in \dot \Cl(\dot q)$.
\end{enumerate}
It is straightforward to check that $\ran(\bCl)$ has size at most $\lambda$. Thus we may assume $\bCl \subseteq (P*\dot Q)\times \lambda$,
although this is not literally the case.
We have finally defined the stratification of $\bar P=P*\dot Q$. Let's check the remaining axioms.
\subsubsection*{Continuity}\label{proof:composition:cont}
Say $\lambda'\in I$, $\lambda'<\lambda)$, both $\bar p=(p_\xi,\dot p_\xi)_\xi$ and $\bar q=(q_\xi,\dot q_\xi)_\xi$ are $\lambda'$-adequate sequences of length $\rho$ and for each $\xi<\rho$, ${\bar \C}^{\lambda}(p_\xi,\dot p_\xi)\cap {\bar \C}^{\lambda}(q_\xi,\dot q_\xi)\neq\emptyset$.
Moreover, let $(p,\dot p)$ and $(q,\dot q)$ denote greatest lower bounds of $\bar p$ and $\bar q$, respectively.
First, by \emph{Continuity} for $P$, we can find $\chi \in \Cl(p)\cap \Cl(q)$.
For each $\xi<\rho$, fix $(T_{\g}^\xi,H_{\g}^\xi,\lambda_{\g}^\xi)$ such that for some $\chi'$,
\[ (\chi', T_{\g}^\xi,H_{\g}^\xi,\lambda_{\g}^\xi) \in \bCl(p_\xi, \dot p_\xi)\cap\Cl(q_\xi, \dot q_\xi). \]
and find $p_g^\xi$ such that $(T_{\g}^\xi,H_{\g}^\xi,\lambda_{\g}^\xi,p_g^\xi)$ is a guessing system for $\dot p_\xi$ below $p_\xi$.
Now construct a guessing system $(T_{\g},H_{\g},\lambda_{\g},p_g)$ for $\dot p$ below $p$, showing $(p,\dot p)\in\dom(\bCl)$.
It will be clear from the construction that $T_{\g},H_{\g}$ and $\lambda_{\g}$ do not depend on the sequence of $p_g^\xi$, $\xi<\rho$.
Let $s \in T_{\g}$ if and only if $s=\emptyset$ or $\xi\conc s \in T_{\g}^\xi$.
Let $\lambda_{\g}(\emptyset)=\lambda'$, and of course $p_g(\emptyset)= \dot p$.
Now let $s \in T_{\g}\setminus \{ \emptyset \}$ be given and define $\lambda_{\g}(s)$, $p_g(s)$ and, in the case that $s \in T_{\g}^0$, also define $H_{\g}(s)$.
Find $s'$ such that $s= \xi\conc s'$. Let $\lambda_{\g}(s) = \lambda_{\g}^\xi(s')$ and let $H_{\g}(s) = H_{\g}^\xi(s')$ if $s \in T_{\g}^0$ (or equivalently, if $s' \in (T_{\g}^\xi)^0$). Let $p_g(s)=p_g^\xi(s')$.
To check that $(T_{\g},H_{\g},\lambda_{\g},p_g)$ is a guessing system, first observe that $\is$ is reversely well-founded on $T_{\g}$.
Moreover, $\rho_{\g}(\emptyset) = \rho$ is an ordinal and $\lambda_{\g}(\emptyset)<\lambda$. Also, clause \ref{approximate} holds for $s=\emptyset$, by construction. The rest of the conditions are straightforward to check; they hold by construction and because for each $\xi<\lambda'$, $(T_{\g}^\xi,H_{\g}^\xi,\lambda_{\g}^\xi, p_g^\xi)$ is a guessing system.
If we carry out the same construction for $(q,\dot q)$, we obtain $q_g$ such that $(T_{\g},H_{\g},\lambda_{\g},q_g)$ is a guessing system for $\dot q$ below $q$. Thus,
\[ (\chi,T_{\g},H_{\g},\lambda_{\g}) \in {\bar \C}^{\lambda}(p,\dot p)\cap {\bar \C}^{\lambda}(q,\dot q). \]
\subsubsection*{Interpolation} Say $(d,\dot d)\leq_{\bar P} (r,\dot r)$. First find $p \in P$ such that $p \lequpl d$ and $p \leqlol r$.
If $p \cdot d \neq 0$, then $p \cdot d \forces_P \dot d \leq_{\dot Q} \dot r$, so we can find $\dot p$ such that $p \cdot d \forces_P \dot p \dlequpl \dot d$ and $p \forces_P \dot p \dleqlol \dot r$.
\subsubsection*{Centering} Say $\bar p\blequpl \bar d$, where $\bar p=(p,\dot p)$ and $\bar d=(d,\dot d)$,
and assume $\bCl(\bar p)\cap\bCl(\bar d)\neq\emptyset$.
First assume we can find $(\chi,T_{\g},\lambda_{\g},H_{\g}) \in \bCl(\bar p)\cap\bCl(\bar d) $ (i.e. \ref{long:center} holds in the definition of $\bCl$).
As $\chi \in \Cl(p)\cap\Cl(d)$, by \emph{centering} for $P$ there exists $w$ such that
for all regular $\lambda' \in 0\cup I \cap \lambda$, both $w\leqlo^{\lambda'}p$ and $p\leqlo^{\lambda'}d$.
Now fix $p_g$ and $d_g$ such that
$(T_{\g},\lambda_{\g},H_{\g},p_g)$ is a guessing system for $\dot p$ below $p$ and
$(T_{\g},\lambda_{\g},H_{\g},d_g)$ is a guessing system for $\dot d$ below $d$.
We show by induction on the rank of $s$ (in the sense of the reversed $\is$-order) that for each $s \in T_{\g}$,
\begin{equation}\label{share:colour:ext}
p\cdot d \forces_P \dot \Cl (p_g(s)) \cap \dot \Cl (d_g(s)) \neq 0.
\end{equation}
First, let $s \in T_{\g}^0$. By definition \ref{gs}, \ref{spectrum} we can find $P$-names $\dot \alpha$ and $\dot \beta$ such that both have
spectrum $H_{\g}(s)$ below $p$ and $d$, respectively, and moreover:
\[ p \forces_P \dot \alpha \in \dot \Cl(p_g(s)) \]
and
\[ d \forces_P \dot \beta \in \dot \Cl(d_g(s)). \]
Thus, as $\dot \alpha$ and $\dot \beta$ have a common spectrum below $p \cdot d$, (\ref{share:colour:ext}) holds.
For $s$ of greater rank, we may assume by induction that for each $\xi < \rho_{\g}(s)$,
\begin{equation}\label{share:colour:ext:ind}
p\cdot d \forces_P \dot \Cl (p_g(s\conc\xi)) \cap \dot \Cl (d_g(s\conc\xi)) \neq 0.
\end{equation}
As $p$ forces that
\begin{eqpar}
$\{p_g(s\conc\xi)\}_{\xi<\rho_{\g}(s)}$ is a $\lambda_{\g}(s)$-adequate sequence and
$p_g(s)$ is a greatest lower bound of $\{p_g(s\conc\xi)\}_{\xi<\rho_{\g}(s)}$,
\end{eqpar}
and as $d$ forces the corresponding statement for $d_g(s)$ and $\{d_g(s\conc\xi)\}_{\xi<\rho_{\g}(s)}$, \emph{Continuity} for $\dot Q$ in the extension allows us to infer (\ref{share:colour:ext}) for this $s$. This finishes the inductive proof on the rank of $s$.
Finally, (\ref{share:colour:ext}) holds for $s = \emptyset$, so as $p_g(\emptyset)=\dot p$ and $d_g(\emptyset)=\dot d$, by \emph{centering} for $\dot Q$ in the extension, $w \forces$ there exists $\dot w$ such that for all regular $\lambda' \in 0\cup I\cap\lambda$, both $\dot w\dot\leqlo^{\lambda'}\dot p$ and $\dot w\dot\leqlo^{\lambda'}\dot d$. Then $\bar w =(w,\dot w)$ is as desired.
Now secondly assume we have $\chi \in \bCl(\bar p)\cap\bCl(\bar d)$ and \ref{short:center} holds in the definition of $\bCl$.
In this case $\chi \in \Cl(p)\cap\Cl(d)$.
Let $w \in P$ such that $w \leqlo^{<\lambda} p$ and $w\leqlo^{<\lambda} d$.
By assumption, $w \forces \dot d \dleqlol 1_{\dot Q}$ and $\dot p \dlequpl \dot d$.
By \emph{expansion} (\ref{exp}) for $\dot Q$, we conclude $w \forces \dot p \leq \dot d$.
We claim $\bar w=(w,\dot p)$ is the desired lower bound:
$\bar w \bleqlo^{<\lambda} \bar p$ holds because $\dleqlol$ is a \pre order.
We show $\bar w \bleqlo^{<\lambda} \bar d$: we have $w \forces \dot p \leq \dot d \leq 1_{\dot Q}$ and by assumption $w \forces \dot p \dleqlol 1_{\dot Q}$. So by (\ref{er}), we conclude $w \forces \dot p \dleqlo^{<\lambda} \dot d$ and are done.
\subsubsection*{Density}
Let $(p_0,\dot q_0) \in \bar R$. First, assume $min I < \lambda$ and fix a regular $\lambda' \in I \cap \lambda$.
By \emph{Density} for $\dot Q$ in the extension, we can find $P$-names $\dot \chi$ and $\dot q_1$ such that $\forces_P$``$\dot q_1 \dot \leqlo^{\lambda'} \dot q_1$ and $\dot \chi \in \Cl(\dot q_1)$''.
By lemma \ref{density reduction}, we can find $p_1 \leqlo^{\lambda'} p_0$ such that $\dot \chi$ is $\lambda'$-chromatic below $p_1$,
and by \emph{Density} for $P$ we can find $p_2 \leqlo^{\lambda'} p_1$ and $\zeta$ such that $\zeta \in \Cl(p_2)$.
Let $T_{\g}\defeq\{ \emptyset \}$, $q_g(\emptyset)\defeq\dot q_1$, $\lambda_{\g}(\emptyset)\defeq \lambda'$ and let $H_{\g}(\emptyset)$ be a $\lambda'$-spectrum of $\dot \chi$ below $p_2$.
Thus $(T_{\g},\lambda_{\g},H_{\g},q_g)$ is a guessing system for $\dot q_1$ below $p_2$---the only non-trivial clause is (\ref{spectrum}),
which holds as $H_{\g}(\emptyset)$ is a $\lambda'$-spectrum of $\dot \chi$ below $p_1$ and $p_2 \leq_P p_1$.
So $(\zeta, T_{\g}, \lambda_{\g}, H_{\g}) \in \bCl(p_2,\dot q_1)$, and $(p_2, \dot q_1)\bar \leqlo^{\lambda'} (p_0, \dot q_0)$.
It remains to show $\dom(\Cl)$ is dense in the case that $min I =\lambda$. Find $(p_1,\dot q_1) \leq_{\bar R}(p_0,\dot q_0)$ such that for some ordinals $\zeta, \chi< \lambda$, $\zeta \in \Cl(p)$ and $p_1\forces_P \check{\chi}\in \dot \Cl(\dot q_1)$.
Then $(\zeta, \chi) \in \bCl(p_1,\dot q_1)$.
\end{proof}
\section{Stratified iteration and diagonal support}
We now proceed to show the notion of stratified forcing is iterable, if the right support is used. To this end, let's define stratified iteration with diagonal support.
To motivate this, imagine we want to take a product of forcings $P_\xi * \dot Q_\xi$, of the type of example $\ref{strat:ex:*}$.
The present approach to showing these forcings preserve cofinalities makes use of the fact that for large enough $\lambda_0$, $P_\xi$ is closed under sequences of length $\lambda_0$ while $\dot Q_\xi$ has a (strong form of) $(\lambda_0)^+$-chain condition.
If we want to preserve the latter, we should use support of size less than $\lambda_0$; if we want to preserve the first, our choice would be to use support of size $\lambda_0$.
This calls for a kind of mixed support:
i.e. define
\[ \Pi^{d}_{\xi<\lambda_0} ( P_\xi * \dot Q_\xi) \]
to be the set of all sequences $(p(\xi), \dot q(\xi))_{\xi<\lambda_0} \in \Pi_{\xi<\lambda_0} ( P_\xi * \dot Q_\xi)$
such that for all but less than $\lambda_0$ many $\xi$,
\begin{equation}\label{ex:support}
p(\xi)\forces_P \dot q(\xi)=\dot 1_\xi.
\end{equation}
Using the stratification of $P * \dot Q$, (\ref{ex:support})
may be written as
\[ (p(\xi), \dot q(\xi)) \leqlo^{\lambda_0} 1_{P_\xi*\dot Q_\xi}. \]
The use of the term ``diagonal'' is motivated by the intuition that we allow large support on $P_\xi$, which we regard as the ``upper'' part, and small support on $\dot Q_\xi$, which we regard as the ``lower'' part of the forcing $P_\xi * \dot Q_\xi$.
\begin{dfn}\label{def:it:strat:comp}
\begin{enumerate}
\item We say the iteration $\bar Q^\theta = \langle P_\nu; \dot Q_\nu, \leq_\nu , \dot 1_\nu\rangle_{\nu<\theta}$ \emph{has stratified components} if and only if for every $\nu<\theta$,
$\dot Q_\nu$ is a $P_\nu$-name and $P_\nu$ forces $\dot Q_\nu$ is a stratified partial order as witnessed by the system
\[
\bar \pss=(\dot\leqlol_\nu, \dot c_\nu, \dlequpl_\nu, \dot \D_\nu, \dot \C_\nu)_{\lambda\in \mathbf{Reg}, \nu<\theta}.
\]
(which is called its \emph{stratification}).
Formally, the reader may wish to replace $\dot \D_\nu$ in the above by a name for the G\"odel number of a formula defining the class $\dot \D_\nu$ with parameter $\dot c_\nu$.
Moreover, we demand that for all regular $\lambda$ there is $\bar\lambda<\lambda^+$ such that $\supp_\lambda(p) \subseteq \bar \lambda$, where $\supp_\lambda(p)$ is defined as
\[
\supp^\lambda(p)=\{ \xi \setdef p\res\xi \not \forces_\xi p(\xi) \dot \leqlo_\xi \dot 1_\xi\}.
\]
\item $P_\theta$ is the \textit{diagonal support limit } of the iteration with stratified components $\bar Q^\theta$ with stratification $\bar \pss$ if and only if
$P_\theta$ is the set of all threads though $\bar Q^\theta$ such that for each regular $\lambda$, $\supp_\lambda(p)$ has size less than $\lambda$.
\item We say $\bar Q_\theta$ is an iteration with diagonal support if for all limit $\nu < \theta$, $P_\nu$ is the diagonal support limit of $\bar Q\nu$.
\end{enumerate}
\end{dfn}
We omit the proof of the following theorem, since it will follow from theorem \ref{thm:it:strat} and lemma \ref{stratified:comp:implies:ext} as corollary \ref{cor:it:strat:comp}.
In the proof of the main theorem, we will need to use these stronger lemmas, theorem \ref{thm:it:strat:comp} does not suffice.
\begin{thm}\label{thm:it:strat:comp}
Say $\bar Q = \langle P_\nu, \dot Q_\nu \rangle_{\nu<\theta}$ is an iteration with stratified components and diagonal support.
Then $P_\theta$ is stratified.
\end{thm}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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Dr. Carolyn Warner
Co-Founder and Chairman, Corporate Education Consulting, Inc.
Honoree 2018
Carolyn Warner is Founder and Chairman of Corporate//Education Consulting, Inc., a consulting firm specializing in education, communications, and public private partnerships. She is nationally and internationally recognized as a fearless advocate for education and economic workforce issues.
Warner was a Congressional appointee to the National Skills Standards Board and was a delegate to the White House Conference on Small Business. She also served for 12 years as Arizona's elected State Superintendent of Public Instruction, the first non-educator to hold this post. She currently serves
as National Treasurer of Jobs for America's Graduates, the Nation's most successful school-to-work transition program.
In 1998, Warner received an honorary doctorate from Northern Arizona University in recognition of her service to education and the community.
She has received numerous awards, including Policy Leader of the Year by the National Association of State Boards of Education. She is also a bestselling author.
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
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## By Love Undone
## Suzanne Enoch
For the world's best agent,
Nancy Yost, who took a chance
and has yet to admit regretting it
For Micki Nuding,
my editor extraordinaire,
who likes the funny parts, too
And for the ladies of TLT bb—
you know who you are—
thanks for your support (LOL)
## Contents
Chapter 1
The yellow-red blossoms of the Lord Penzance rose-bush waved lazily...
Chapter 2
"He's here! He's here!"
Chapter 3
Maddie liked the word "obsequious." It sounded unpleasant and haughty...
Chapter 4
Maddie shook her head and glared at her employer. "I...
Chapter 5
After luncheon, the afternoon spiraled downward from merely painful to...
Chapter 6
"Wonderful," Maddie snarled, shutting her door with a thud and...
Chapter 7
Quin couldn't keep his eyes off her.
Chapter 8
"I do not need a dressmaker. I can sew my...
Chapter 9
On her first and only stay in London, Maddie had...
Chapter 10
Quin looked up as Edward Lumley hit the polished floor.
Chapter 11
"Good Lord, that's frightening." Maddie laughed, her voice and expression...
Chapter 12
"Quin, I have to admit, your little project is marvelous."
Chapter 13
Charles Dunfrey called on Maddie three times at Bancroft House...
Chapter 14
Maddie entered Willits House slowly, fighting the nagging idea that...
Chapter 15
The Duchess of Highbarrow sat in her private room, sewing.
Chapter 16
Quin wanted to strangle her.
Chapter 17
Quin rose early. A year ago—hell, six months ago—he would...
Chapter 18
Maddie gazed out the window of Charles Dunfrey's dilapidated coach.
About the Author
Other Books by Suzanne Enoch
Copyright
About the Publisher
## Chapter 1
The yellow-red blossoms of the Lord Penzance rose-bush waved lazily in the light breeze. Humming a counterpoint to the robins singing in the trees behind her, Madeleine Willits snipped three of the perfect blooms and gently dropped them into her basket. And then she pricked her finger.
"Ouch! Blast it."
At the same time, a stentorian bellow rumbled from the master bedchamber window like a clap of thunder, and she jumped.
"Miss Maddie!" the housekeeper called frantically.
"Good Lord," Maddie muttered. Dropping her clippers into the basket, she gathered up her skirt and ran for the kitchen entry.
Mrs. Hudson pulled open the door as she reached it, and Maddie shoved the basket into the housekeeper's plump arms. "What happened?" she called over her shoulder, running through the main hall toward the stairs to the second floor. Curious servants hurried into the hallway, creating obstacles for her to dodge.
"I don't know, Miss Maddie," came from behind her. "Garrett was in with him!"
"Garrett!" she called.
The butler appeared at the top of the stairs. Red-faced, he wiped at the thick brown trails of gravy running down the front of his black coat. "It was just the post!" he protested.
Bill Tomkins, closely followed by a tea saucer, exited the bedchamber at high speed behind Garrett. "He nearly killed me that time," the footman panted, leaning against the bannister.
"You shouldn't have been in there," Maddie said un-sympathetically, trying to regain her breath before she stepped into battle. She pulled the shawl from around her shoulders.
"What was I supposed to do, keep polishing the lamps while he's yowling like a Bedlamite? Scared the devil out of me," the footman exclaimed, shuddering.
The butler chuckled. "Then you should thank him."
Sending the servants a warning glance, Maddie fluttered the end of her shawl into the doorway. "We surrender, Mr. Bancroft. The household has been vanquished."
More rumbling issued from inside the room, followed by the thudding sound of a pillow hitting a wall. "Humph. Stop that nonsense and get your pretty face in here, girl," Malcolm Bancroft's irritated voice ordered.
Maddie entered the bedchamber. The remains of luncheon drippily decorated the near wall, while the pillows which had been propping Mr. Bancroft up in bed lay strewn about the floor, leaving her employer flat on his back amid a tangle of bed sheets.
"My, my. Such carnage." She clucked her tongue.
Awkwardly he lifted his head to pin her with a baleful, dark-eyed gaze. "Bah," he said, and lay flat again.
Stifling a grin, Maddie began gathering pillows in her arms. "Any interesting news in the post today?"
"I wouldn't be so blasted clever if I were you, Maddie. It's not news you'll relish, either. Damned stuffed shirts."
An edge of uneasiness ran through her as she levered him into a sitting position with the help of the pillows. "I see you've appropriated my favorite term for the nobility. The new king is coming to visit, I suppose. Shall I have the silver polished—or hidden? You know King George so much better than I."
As she expected, the mention of George IV distracted her employer from whatever it was that had upset him. "Mad King George, Fat King George. Who's next—Blind King George?"
Maddie chuckled, relieved as reluctant humor returned to his voice. "Royalty are blind to everything but their purses, anyway."
Mr. Bancroft snorted. "So they are." With his weakened left hand he gestured at a badly crumpled paper resting on a slice of toasted bread. "And that particular ailment infects most everyone in England who can lay claim to a title. Hand me that letter, my dear."
She complied, shaking crumbs off and resisting the urge to read it herself. He would tell her the news. He always did.
Awkwardly he flattened the paper against his chest. "Listen to this, Maddie. And brace yourself." Malcolm cleared his throat and lifted the wrinkled missive. "'Brother.'" He stopped and looked up at her, obviously waiting for the significance of that single word to sink in.
Maddie's insides jolted unpleasantly, and the last pillow slipped from her fingers onto the floor. "The Duke of Highbarrow has finally answered your letter," she muttered, sinking onto the comfortable chair beside his bed.
"It's been more than a fortnight since we wrote him. He was bound to answer eventually." He looked sideways at her. "I'd actually begun to wonder whether you'd burned the original letter."
Maddie straightened. "I told you I would send it," she said, wondering if he knew just how close she had come to 'accidently' misplacing the missive in her bedchamber fireplace.
"I know you did." Her employer smiled briefly, then returned his attention to the letter. "'Brother,'" he began again. "'I was away in York on business when news of your poorly timed illness arrived. I have sat to write you immediately upon my return to Highbarrow Castle.'"
"You were right," Maddie noted, as Mr. Bancroft paused to catch his breath. He tired so easily these days. "He always uses the word 'castle,' doesn't he?"
"At every opportunity. To continue, 'Victoria sends her wishes for a complete recovery, though as you know I really don't give a damn one way or the other.'"
"My word, he's awful."
"'I am planting my crop at Highbarrow Castle at the moment. Otherwise, despite your past errors of judgment, I would make an effort to call upon you at Langley Hall.'"
"Of course," Maddie and Mr. Bancroft agreed in skeptical unison.
All she knew of the duke were tales of his monumental stuffiness and arrogance, and Maddie let out her breath in a silent sigh of relief. He wasn't coming. "So that's that, then," she said, rising. "Hardly enough to warrant frightening me half to death, though. Shame on you."
"That's the good news, I'm afraid."
Slowly Maddie sat down again. "Oh."
"Now please remain calm."
She nodded. "Just as you did," she teased.
"Hush. 'However,'" he resumed, "'as getting the crop in at Langley is of paramount importance, I have spoken with Quinlan. He has agreed to journey to Somerset to oversee planting and to tend to the estate during your recovery. He follows immediately upon this letter, and should arrive at Langley on the fifteenth of the month. Yours, Lewis.'"
Maddie gazed out the window. The lovely spring morning, the first without rain in three days, had become a disaster. Worse than a disaster. She took a deep breath. "I assume His Grace is referring to Quinlan Ulysses Bancroft?"
Her employer nodded, a sympathetic grimace touching his gaunt face. "Afraid so. The Marquis of Warefield himself."
Maddie cleared her throat. "I see."
He reached out and squeezed her fingers. "I'm dreadfully sorry, my dear. You are acquainted with him, I suppose?"
She shook her head. "Thankfully not. I believe he was in Spain during my...visit to London—if you could call it that." Maddie frowned at the memory.
"It wasn't your fault, my girl," Malcolm soothed.
She eyed him fondly, wondering who was supposed to be comforting whom. "You're the only one who thinks so. None of them—not one of them—saw anything but that stupid kiss, and that stupid man trying to shove his hand down my dress. They didn't care that I wanted nothing to do with it, or with that awful scoundrel Spenser. And I want nothing to do with London society, ever again."
"Well, Quinlan wasn't there, so don't worry yourself. He wouldn't say anything, anyway. Wouldn't be polite, you know."
"I'm not worried." Maddie sat up straighter, pulling her fingers free from his comforting grip. "Nor am I the least bit faint of heart, Mr. Bancroft."
He chuckled. "I never said you were."
"It's merely that I'm...annoyed." Ready to throw a screaming fit would be closer to the truth, but she'd had the feeling lately that her peaceful days were numbered. Once the letter to Highbarrow Castle had gone out, someone had been bound to reply.
And even though she didn't know the Marquis of Warefield, she knew of him. Quinlan Ulysses Bancroft was the very pink of the ton, a favorite of the new king, the bluest of blue bloods, the epitome of propriety and dignity. She loathed him without ever having seen his pampered, spoiled, self-important visage. He was one of them.
"Nobility" might be what society called them, but from her experience, the word had nothing at all to do with their characters. "I thought we informed His Grace that you had someone tending to Langley during your illness."
"You didn't expect him to care about that, did you? He owns Langley Hall, my dear; I only manage it for him. And he will take whatever steps are necessary to preserve his considerable monetary well-being, with or without my consent. You know that."
She sighed. "Yes, I know that. Even so, he might have asked whether you wanted assistance before he foisted his son off on you."
Unexpectedly, Mr. Bancroft laughed again, rare color touching his pale cheeks. "I don't believe Quinlan allows himself to be 'foisted' on anyone."
"How noble he must be," Maddie said unenthusiastically.
Her employer narrowed his eyes, suspicion touching his expression. "Just remember, my dear, the less trouble you make for him, the shorter and less painful his visit is likely to be."
A flash of guilt ran through her. After all, this deuced marquis was Mr. Bancroft's nephew, and it had been at least four years since they had seen one another. Even though she might detest him and the rest of the damned aristocracy, she knew all too well how lonely Malcolm must feel being cut off from his family.
So, little as she liked Warefield's coming, she had no intention of stamping her feet and throwing a tantrum. Not in front of her employer, anyway. "I shall behave," she assured him.
He smiled. "I have no doubt that you will."
"As long as he does," Maddie added.
"He will. I already told you, he's the epitome of good manners."
"I am bereft of words at the very idea of setting eyes upon his illustrious personage."
"Maddie," Bancroft warned with a slight grin. He pulled himself into a more upright position, grunting with effort as his still legs hampered the movement. "Best send Mrs. Iddings down to the village and have her spread the word."
"So the local folk can flee into the hills, I suppose?"
"Our neighbors will never forgive me if I don't give them advance notice that the Marquis of Warefield is coming to Langley. An actual title appearing in this part of Somerset is rarer than a camel passing through the eye of a needle."
She sighed. "They will be beside themselves with excitement. I daresay I have no idea how I will contain my feelings, myself."
"Do try, won't you?"
Maddie smiled. "Of course. But only for you."
He looked at her fondly, with an understanding her own father had never possessed. "Thank you."
"You're welcome." She stood. "I'll bring you up some more tea."
"And peach tarts, if you don't mind. My luncheon seems to have met with an accident."
She glanced over her shoulder and chuckled in amusement. "Lucky we kept some sweets in reserve, isn't it?"
"Bah."
Maddie apprised Mrs. Iddings of the Marquis of Warefield's imminent arrival and then sent the cook down to Harthgrove to purchase vegetables and gossip away the afternoon. After bringing Mr. Bancroft his replacement luncheon, Maddie escaped to the garden potting shed, where she could bang about and curse without being overheard. Stupid, stupid noblemen, always showing up where they weren't wanted! Or needed.
"Madeleine?"
"Dash it," Maddie muttered, wiping her hands against her pelisse. "In here, Mrs. Fowler," she called.
She'd hoped to have until tomorrow before the neighbors came prying for information. Apparently Mrs. Iddings's gossip was even more efficient than Mr. Bancroft had anticipated. Smoothing the annoyed expression from her face, she stepped out of the shed.
"Oh, there you are, Maddie." Jane Fowler was wearing her favorite visiting dress; no doubt she intended to carry her news to every home along the lane once she'd pried it out of Maddie.
"Good afternoon."
"I should say so." Mrs. Fowler sighed happily, her rounded cheeks dimpling. She plucked a stray leaf from Maddie's hair. "I hear that we're to have an important guest in Somerset. I am quite beside myself."
"Oh, well, you—"
"My goodness," Mrs. Fowler continued, clapping her hands together, "a marquis." She leaned forward and lowered her voice, even though there was no one about to hear them except for the finches. "And I hear that he's very handsome, and that he has twenty thousand a year. Can you imagine? Twenty thousand pounds a year!"
Swallowing her annoyance at such awestruck pointlessness, Maddie nodded and started back toward the house at a brisk pace. Bad enough she had to host Warefield without having to talk about him as well. "You seem to know a great deal about him, Mrs. Fowler."
"Mrs. Beauchamp does. Her cousin is Baron Montesse, you know."
"Yes, I had heard that." Endlessly and repeatedly.
"How long will he be staying at Langley?"
"I really don't know. With the Season starting soon, I'm certain it can't be too long."
Mrs. Fowler sighed reverently. "Ah, yes, the Season."
The worshipful look on her face made Maddie want to laugh. "Have you told Lydia and Sally the news?"
"They were the ones who told me. Such good girls, they are. And Lydia has become quite proficient at the pianoforte, you know."
"Yes, I d—"
"Oh, I know Sally isn't quite out yet, but she is seventeen. Here in the country, so very far from London, Lord Warefield couldn't expect us to stand on such strict ceremony, don't you think?"
From what she'd heard of the marquis, he stood upon strictest ceremony at all times. "Of course not," she agreed, hiding her sly smile. If anything could encourage Warefield to shorten his stay at Langley, it would be the Fowler girls.
"Marvelous, marvelous." Mrs. Fowler continued on beside her, then stopped and lifted her handkerchief to her mouth, unsuccessfully stifling a rather giddy gale of giggles. "I have just thought of the very thing!"
Maddie reluctantly halted her escape. "Whatever might that be?"
"I will speak to Mr. Fowler, and we shall hold a country ball in honor of Lord Warefield. Won't that be spectacular? And I shall invite everyone—oh, everyone but the Dardinales. That Miss Dardinale is completely unacceptable."
By coincidence, Patricia Dardinale was also the prettiest young lady in the countryside. "I shouldn't think you would mind one more girl, Mrs. Fowler. I have heard Lydia's singing is much improved over last year. I believe if there is one thing to sway a gentleman's interest, it is a song well sung."
Mrs. Fowler clutched Maddie's arm. "Thank you, my dear. And you shall come as well, for I can't think that Mr. Bancroft would venture from Langley these days without you. Unless the marquis takes over his care, of course. How noble that would be. Oh, my, yes."
Maddie frowned. She hadn't considered that. It made sense that a meddling, busybody nobleman would think a woman incapable of her duties, however proficient she'd been with them over the past four years. "Yes, how noble, indeed."
For the forty-seventh time, Quinlan Ulysses Bancroft lost his place in Ivanhoe. He dropped the book onto the black leather seat beside him and, holding onto his hat with one hand, leaned his head out the window. "Really, Claymore, must we take a census of every wheel rut, rock, and puddle in Somerset?"
The groom's face appeared over the high corner of the coach. "Sorry, my lord," he said, and vanished again. "If you don't mind my saying," his voice drifted back, "it's my thinking that King George ain't traveled upon these roads lately."
Quin sat back and resumed reading, until another hard bump jolted him against the cushions. "Lucky George," he muttered.
Reluctantly he set the book aside again and stretched his long legs out to rest them on the opposite seat. With a sigh he settled back to watch southern Somerset County pass by outside. At least the weather had turned agreeable, and the green, tree-covered countryside smelled more of meadow grass than it did of cattle.
The marquis pulled his pocket watch out of his waistcoat and glanced down at it. By his estimation, another twenty minutes or so should finally put him at Langley Hall. Three damned days in a coach, with a perfectly good mount tethered behind. He might have left his luggage to follow and ridden on to Langley in half the time—except that the duke, his father, had written to indicate that he would arrive on the fifteenth.
Uncle Malcolm would undoubtedly take a hasty arrival as a threat against his management of Langley. And if there was one thing Quin did not wish to do, it was to further the antagonism between Lewis and Malcolm Bancroft. So he would not, under any circumstance, arrive before the fifteenth.
Little as he relished the idea of being his father's sacrificial lamb, the seven years of silence between the Bancroft brothers had long been a topic of jest and gossip in London's highest circles. Uncle Malcolm had always been his favorite relation, and even if it meant spending time with the rustics, he intended to do his damnedest to see that the wags were finally silenced. It looked very shabby and set a poor precedent before the rest of the nobility.
With any luck, he should be able to organize Langley's books and get the crop in the ground with little bother—which hopefully would make Uncle Malcolm look more favorably on making amends with his brother, which hopefully would leave His Grace feeling more amenable toward everyone in general.
And if events transpired as smoothly as he hoped, he would even have time to return to Warefield for a few weeks before the Season began. Lord knew, the coming summer would leave him little time for himself. Once he arrived in London his first task would be to make wedding arrangements, and the remainder of his social engagements would stem from that.
Quin stretched, yawning. Eloise had been dropping more hints than usual in her letters lately, and their understanding needed to be formalized. At least marriage looked to be fairly painless—so the sooner, the better. The duke's grumbles about grandchildren had grown into bellows—as if he needed another excuse to bellow.
"My lord," Claymore called from his perch, "Langley, I believe."
Quin shifted to look out the opposite window. Sprawling at the top of a slight rise and overlooking a quaint wildflower garden and a small forest glade, Langley Hall rose red and white into the cloud-patched noon sky. Barely more than a cottage by London standards, the estate did offer some of the best fishing in Somerset—small recompense though that was for the journey.
"I'll have stew of you, ye blasted beast!"
A gargantuan cream-colored pig squealed and ran full tilt across the road. A farmer followed by another man and a red-faced woman, all brandishing pitchforks and rakes, headed at full speed after it. The high-spirited coach horses skittered sideways, nearly dumping the lot of them into the spiny hedge bordering the road.
"Whoa, lads!" Claymore bellowed, while Quin slammed into the side of the coach and lost his hat to the floor. "Apologies, my lord!" the groom called. "Damned country folk. No manners at all!"
The marquis leaned down and retrieved his chapeau. "Splendid," he sighed, dusting off his hat and resettling it on his head. "Country folk. Bloody marvelous."
## Chapter 2
"He's here! He's here!"
Madeleine jumped at the kitchen maid's excited pronouncement of the bad news. The Marquis of Warefield had arrived, and exactly on time. No doubt he considered it gauche to arrive late—though she'd been hoping he'd be delayed.
She wanted to run to the nearest window and look for herself, if only to confirm that the nightmare had begun. But she'd seen hundreds of carriages before, and more than her share of English lords. And the good Lord knew they weren't worth gawking at; in fact, they weren't worth much of anything at all.
She doggedly finished stitching the brim of last year's yellow spring bonnet. With a little luck it should last her through the summer, anyway. Out of necessity her sewing skills had greatly improved over the last few years, but she was still surprised when she turned the hat to view it and found that the repairs actually looked quite satisfactory.
"He's here, Miss Maddie! Come quickly!" Mrs. Hodges exclaimed.
"I know, I know," she said, though she doubted Bill Tomkins or the housekeeper heard her as they hurried past the open morning room door. Blowing out her breath, she set the bonnet aside and went to join the others.
"Oh, look at that, Mrs. Hodges! What a fine carriage!" Tomkins said. He craned his tall frame and peered out the foyer window over the other servants' heads. "I'd wager the King himself has none finer."
Even the normally impassive Garrett was fidgeting. His gaze traveled from the window to the grandfather clock in the hallway and back again, as though he were trying to judge the precise moment he should open the front door to achieve the greatest effect.
"Don't worry, Garrett," Maddie said encouragingly. "I imagine the marquis will caterwaul from time to time, but I'm certain he won't bite."
Garrett glanced at her. "You may be confident of that, but you've never encountered the rest of the Bancrofts. I've no intention of making a false step in Lord Warefield's presence."
"Oh, please. The only difference between a noble and a pauper is that one can afford to be rude, and the other can't."
From the annoyed looks and disapproving comments sent in her direction, none of the other servants was particularly interested in hearing further ruminations on the topic from her. Maddie rolled her eyes and purposely stayed back from the excited crowd at the window. They'd see soon enough how little their overstuffed hero resembled their worshipful imaginings.
The rumble of hoofbeats neared, to the accompaniment of the creaking clatter of a large carriage. Garrett tugged once more at his coat, nodded at the assembled servants, and flung open the double front doors. The procession of Langley employees, pulling at neckcloths and straightening apron ribbons, streamed out the door and down the shallow steps to line either side of the front walk.
Following them, Maddie stepped onto the short, wide marble portico where she could watch without being obvious about it.
The coach that rolled up the crushed stone drive was huge and black, with the Warefield crest emblazoned in bright yellow and red on the door panel. A superb quartet of matched black geldings came to an impatient stop before the gawking servants, while a striking bay hunter pranced to a halt at the rear of the vehicle. Maddie sniffed. The pompous boor had even brought his own mount, as if the contents of Langley's stables wouldn't be good enough for him.
Before Bill Tomkins could take more than a step forward to open the coach's door, a liveried servant clambered down from beside the driver's perch to perform the service. With practiced efficiency he flipped down the steps tucked beneath the door, and then with a bow stepped backward.
An elegant leg, sheathed in a polished Hessian boot of finest black leather, emerged from the dark coach. The second limb followed, revealing muscular thighs molded into a pair of fawn-colored buckskin trousers. Maddie's skeptical gaze touched on an elegant gray and blue waistcoat, a coat of dark blue superfine over a broad chest, and a snowy white cravat snuggled between impeccably stiff shirt-points. White kid-gloved fingers handed the footman a polished mahogany cane tipped in ivory-inlaid ebony wood.
The marquis looked down as he stepped from the coach, and a blue beaver hat, set at a rakish angle on wavy hair the rich color of bees' honey, obscured her view of his face. "Buffoon," she muttered, unimpressed. Fencing clubs and boxing halls might keep him lean and athletic, but they couldn't improve a bulbous nose or crooked teeth. Or mask the lines of idle dissipation.
He finally looked up. Twin pools of jade, green as the forest after rain, took in the drive, the excited servants, and the red stone walls of Langley Hall. Maddie's eyes took in a finely chiseled nose completely lacking in any sort of deformity, and a strong, lean jaw. Lips that could disquiet a maiden's heart murmured something to the footman, who immediately handed the cane back and motioned at the anxious Langley footmen to begin unloading the gargantuan mound of luggage atop the coach.
Then Lord Warefield strolled forward to meet Langley's servants.
Mrs. Hodges favored him with a deep curtsey. "Welcome to Langley, my lord," she said, her plump cheeks red with excitement and nervousness. "I am Mrs. Hodges, the housekeeper."
"Good afternoon." With a slight nod he dismissed her, moving on down the line. The jade eyes barely gave each servant so much as a glance. "Good afternoon. Greetings. Pleased." Each followed the other in succession as he made his way to the shallow steps.
The marquis stepped onto the portico, and his aloof gaze passed over young Ruth, Mrs. Iddings, and then Maddie. For a second, he met her eyes and paused, his forward progress arrested. Swiftly she caught herself and curtseyed, lowering her eyes. When she dared look up again, she found him already past her, handing his cane and hat to the butler. She had anticipated being ignored, so she was surprised by the strength of her sudden annoyance at his compelling look and swift dismissal.
"And how are you after all this time, Garrett?" the marquis asked, dropping his gloves one by one into the hat, his attention on the decor of the hall. No doubt he found his uncle's taste completely vulgar and rustic.
"Quite well, my lord, thank you. Shall I show you to your chambers, or would you rath—"
"I'd prefer to see my uncle," Lord Warefield interrupted. "Kindly direct my luggage to my chambers. My valet follows a short distance behind with the rest of my things."
Maddie looked at him in disbelief. They'd already unloaded enough baggage to see him through the summer. If anything more arrived, she'd have to believe he meant to set up permanent residence.
"Very good, my lord."
Garrett glanced at Maddie, and with a start she stepped forward. "I will take you to see Mr. Bancroft, if you please," she said, reminding herself again that she'd promised to behave.
Warefield turned to look at her. A curved eyebrow arched slightly, and then he inclined his head. With a graceful gesture of his long fingers, the marquis motioned her to precede him. "I do please."
She moved past him into the entryway and down the wide hall, the quiet tapping of his boots against the hardwood floor following her to the curving staircase at the far end. Trying not to rush or trip and draw any more attention to herself than necessary, Maddie gripped the smooth mahogany railing and kept her eyes on the stairs before her. The less of a stir she made, the less notice the marquis was likely to pay to her.
Yet she hadn't expected him to be so aggressively handsome that she couldn't help wanting to look at him and touch him. It somehow made the entire visit even more irritating. Over the past few years, she had envisioned all English noblemen as fat, pig-eyed, pompous dandies.
The Marquis of Warefield was not remotely fat, nor pig-eyed, and though his attire was surely the very latest style, she certainly couldn't call him a dandy. Dandies were quite a bit less...capable looking. But judging by his haughty greeting, her memories of Londoners' inflated self-importance were still quite accurate. She held onto that thought as she continued to the second floor.
"Are you my uncle's nurse?"
"I am his companion," she corrected him, keeping her eyes to the front as they reached the top of the stairs. Silence followed her remark, and belatedly she realized what he must be waiting for. "My lord." She scowled at her stupid omission.
"And how did you come to be in this...position?"
Curiosity touched his cultured voice, and she clenched her jaw. "I applied for it, my lord."
"I see."
Maddie wanted to argue, for obviously he did not see. From his tone, he thought her Mr. Bancroft's mistress or some such scandalous thing, but she didn't wish to prolong the conversation by correcting his idiotic misapprehension. He had no right to be prying into her affairs, anyway.
The door of Mr. Bancroft's bedchamber stood before her. Gritting her teeth, she kept a tight rein on her fraying temper. She was nearly rid of the marquis for now.
"What am I to call you, then?" His deep voice sounded smug and amused.
Maddie hesitated, but since Warefield traveled in such elevated circles, he'd have no reason to recognize her name. "I am Miss Willits, my lord."
"You know, Miss Willits, it's not considered improper to face someone while conversing with them," he pointed out.
Maddie blinked. How dared he? Embarrassment, mortification, and fury shot through her all in a rush. She would drown him in politeness if that was what he required!
Stifling a furious growl beneath a smile, she whirled about in the doorway. "My apologies, my lord." She thrust her hand into the room. "Mr. Bancroft, my lord." As she glanced at him, his startled gaze caught hers. He opened his mouth to reply, but her legs swept her into swift, angry motion past him and back down the stairs. "If you'll excuse me, my lord."
Quin looked after his uncle's so-called companion as her slender form vanished down the stairs in a hurried blur of pink and white muslin. "Of course," he replied absently. Odd chit.
"Quinlan Ulysses Bancroft. Welcome to Langley Hall."
Quin shook himself and turned to face his uncle's bedchamber. "Uncle Malcolm, thank you."
With a smile he entered the room, taking in the colorful array of medicine bottles on the stand beside the bed, and the stack of books and cards and game pieces on the chest of drawers. Fresh flowers, set below the open window, waved gently in the soft spring breeze. Malcolm sat propped up by an enormous mound of pillows, his face pale and thin. Even so, his dark eyes twinkled as he smiled.
"Don't you look splendid, my boy."
Quin sketched a bow. "As do you. From Father's description I expected you to be already laid out in a coffin. You look quite well, I must say."
"No doubt my pending death was merely wishful thinking on Lewis's part." His uncle gestured at the chair beside the bed. "How was your journey?"
Refusing to take the bait and argue over his father's private ruminations, Quin instead took a seat. "Quite uneventful, thank you."
Malcolm shook his head and waved a finger energetically at his younger relation. "None of that, lad. You'll find I've become a regular gossip these days. You're to tell me who you passed on the road, what sort of weather you encountered, and how dismayed you were at being pulled away from Warefield before the Season."
For a moment Quin eyed his uncle. Malcolm had previously been known for his stubborn independence rather than for eccentricity, but as the duke had mentioned several times, there was no telling what an apoplexy might have done to his cognitive abilities. "Very well. I was not at all dismayed to come visit you, Uncle. It's been far too long, in fact. The sun shone periodically, though it rained once, and I passed two milk wagons, a mail stage, five farmers' wagons, and just a few minutes ago, a lone escaped pig with several angry persons in pursuit."
Malcolm slapped the quilted bed covering, the corners of his eyes crinkling as he chuckled. The action winded him, and it was a moment before he could speak. "That would be the infamous Miss Marguerite," he finally explained. "I'll have to tell Maddie."
"Maddie?"
"Miss Willits. Have you met her yet?"
Quin nodded slowly, still wondering at his initial, uncharacteristically heated reaction when he'd set eyes on her. "I have. She is—"
"Lovely, isn't she? A true lifesaver, she's been. I thought she was going to show you up here."
"She did." Apparently his assumption about Miss Willits's place in his uncle's household had been correct. "And yes, she's quite attractive. A bit...unique in her manners, perhaps, but I imagine none of you are exactly pleased to have me here."
Again his uncle grinned. "Some of us more than others."
Quin lifted an eyebrow, surprised at finding Malcolm in such good humor. "Well, thank you very much, Uncle."
"Excuse me, Mr. Bancroft, my lord."
Quinlan looked up. Miss Willits stood in the doorway. The high color in her unfashionably tanned cheeks and the tilt of her chin made it quite obvious that she had overheard his comment, and he stifled a scowl.
"Yes, Maddie?"
She stayed in the doorway, her light gray eyes averted from Quin. Irish blood, he decided admiringly, taking in her tall, slender figure and curling auburn hair, more slowly this time. By his guess, her unbound hair would hang to her waist, a far cry from the London fashion of daringly short curls. Nothing was so arousing in a woman as long, curling hair. His uncle had splendid taste in mistresses. Exquisite taste—as the rush of his pulse would indicate.
"Mr. Bancroft, Lord Warefield's luggage appears to be too substantial for the east room," she said stiffly. "If it's not too forward of me, perhaps his lordship might desire some of his things to be moved to the upstairs sitting room, so that he might utilize it as an office during his stay, especially if he should have need of reading or writing materials and space."
Her sensuous, full lips remained set in a stern, straight line, and his curiosity and interest rose another notch. Apparently she had a temper, this one. He'd corrected her behavior, and now she was angry at him for it.
"I've no wish to intrude," he offered easily, watching as her lovely eyes snapped in his direction and then away again. "I can read or write in the library."
"Nonsense. That's a splendid idea, my dear."
"I'll see to it at once." With a deep, formal curtsey, she left the doorway.
Quin returned his attention to Malcolm. "I want to assure you, Uncle, I am here only to be certain Langley runs smoothly until you are recovered. Father has no intention of turning you, or your staff," and he gestured in the direction Miss Willits had disappeared, "out into the wilds of Somerset."
Malcolm shifted the book lying across his lap. "Heavens, no. I never thought so—wouldn't look at all proper, you know. And believe me, I'm well aware that the titled Bancrofts' prime objective is to appear proper at all times."
Quin frowned. "That's hardly fair, Un—"
"Once I've hopped the twig, though, no doubt you or Rafe will be calling Langley your own."
"Rafe will be lucky to get a stick of furniture. And believe me, everything is still His Grace's." With a slight grin, Quin sat back in the chair. "He makes quite certain everyone knows it, and it's far too late to expect him to change. If I had my way, Langley would've been deeded over to you a long time ago."
His uncle glanced at him. "It's not much by your standards, I imagine, but it has its attractions."
"No doubt." Miss Maddie Willits seemed to be chief among them. He stood to leave. "I'd best get myself situated. Unless you have a different idea, I'd thought to begin on the books this afternoon, and take a tour of the fields in the morning. If the weather looks likely to hold, I don't see any reason not to get the wheat and barley put in immediately."
"No sense wasting time," his uncle agreed. "Maddie knows where everything is."
Quin nodded and turned for the door. "Very good."
"And Quinlan?"
He looked back again. "Yes?"
"Take...care with Maddie. She's more than an old buffoon like me deserves."
So he was being warned off already. It was easy to see why. In his incapacitated state, no doubt Malcolm hadn't been keeping her very occupied. If his brother had been the one sent to Langley, Rafael would likely have been more than willing to step into the void, as it were. Even for Quin, the thought was intriguing. "I shall take extreme care, Uncle."
Quin headed down the long portrait-lined hallway toward the east wing. Before the absurd quarrel between Lewis and Malcolm, he had spent several summers at Langley, and he had always regarded its generous wood beams and tall, wide windows with affection. The Hall seemed smaller than he remembered, but then he'd gained more than a foot in height and two decades in experience since his last visit.
He paused to gaze out one of the windows which overlooked the pond and the forest glade beyond. From his first brief view of Langley everything appeared to be fairly well organized, but he didn't look forward to delving into what must have become a chaos of bills and accounting over the past few weeks. Quin sighed.
"My lord?"
Starting, he turned around. Miss Willits and another servant stood in the hallway behind him, having apparently materialized out of thin air. "Yes?"
"Are you lost, my lord?"
"No, I'm not." He smiled. "But thank you for your concern."
Maddie nodded, then gestured at her companion. The older woman, though, flushed and backed away several steps. With an exasperated look, Miss Willits faced him again. "My most sincere apologies for disturbing you, my lord, but Mrs. Iddings, the cook, wishes to inquire whether you have any particular culinary preferences or dislikes. There was no time for us to inquire of your own cook at Warefield, my lord, and I believe the delicate constitutions of the nobility to be well documented."
"Oh." Quin nodded in what he hoped was an agreeable manner. "Of course. Thank you again." Now he was completely convinced; for some reason, Miss Willits strongly disliked him.
"Yes, my lord?"
Quin cleared his throat. "Well, I've no real love for blood pudding," he began, noting that Mrs. Iddings continued to keep her portly frame behind Miss Willits, while Maddie made no move at all. Not afraid of him either, then. He smiled in an attempt to placate the natives. "And I enjoy a nicely roasted pheasant, I suppose. All in all, I'm not very particular, despite any rumors to the contrary."
Maddie nodded coolly, no trace of a return smile in her gray eyes. "I am pleased to hear that, my lord." She turned to her companion. "Does that help you, Mrs. Iddings?' she asked, in a much warmer tone.
The cook curtsied. "Indeed it does. Thank you, Miss Maddie." She flushed again and bowed in the marquis's direction. "Thank—pleased—thank...God bless you, my lord." Mrs. Iddings hesitated, then hurried away down the hall.
"Thank you," he said to her back, though he doubted she heard him as she clumped down the back stairs. The servants at Langley seemed to be a rather odd bunch altogether. He looked back at Miss Willits, to find her glaring at him.
She swiftly wiped the look from her face and curtsied politely. "Thank you, my lord."
Before she could flee again, he stepped forward." My uncle said you would show me where the account ledgers are, Miss Willits," he suggested. "Do you have a moment to do so?"
"I thought, my lord, that you said you weren't lost."
"I know exactly where I am. I simply don't know where my uncle keeps his ledgers these days."
"Very well, my lord," she answered smoothly, turning with a swirl of her pink muslin skirt. She stopped at the head of the stairs, and pointedly faced him again. "If you please, this way, my lord."
Quin shook his head and followed her down the stairs. "How long have you been my uncle's companion, Miss Willits?"
She stopped so suddenly he nearly ran into her. Quin flung out a hand to catch his balance as she whirled on the steep stairs to face him. His fingers lightly grazed her cheek, and she flinched toward the railing. She took a quick breath and smoothed her skirt. "Four years, my lord."
Before he could apologize, or even open his mouth, she'd reversed direction and begun descending the stairs again. Torn between alarm and amusement, Quin continued after her. "Four years?" he repeated. "How old are you, Miss Willits?"
As she halted and spun to face him again, he grabbed onto the railing and stepped back. "I am twenty-three, my lord." Her polite tone didn't nearly hide the furious glint in her eyes.
Before she could begin spinning about again, Quin put a hand on her arm. "If you please, Miss—"
She blanched and jerked away. "Don't—"
Genuine dismay touched her eyes. Swiftly he lowered his hand, curiosity and surprise blazing through him as the color left her delicate, high-boned cheeks. Maddie Willits was primmer than any mistress he'd ever seen. "Of course. My apologies. I wished only to know whether you perform this unusual stairwell dance for all of Langley's guests."
She pinched her lips together before she lifted her chin in defiance again. "Only with those who order me to face them when I speak, my lord."
"I did not order you...well, perhaps I did, but I certainly didn't mean you should risk breaking your neck every time I ask you a question."
She eyed him expectantly for a moment. "Forgive me, my lord," she finally ventured, "but I am now at a loss as to how to proceed."
Quin stopped his frown before it could more than half form, plagued by the muddy feeling that he'd just lost some sort of odd contest. "Please proceed down the stairs, Miss Willits."
"Of course, my lord."
Quin followed his escort into a small, tidy office tucked into the far east corner of the manor. Miss Willits went to the desk beneath the double windows and pulled open a drawer, exposing a stack of ledgers.
"Thank you, Miss Willits. That will do," he said quickly, stopping her before she could disturb the order of the ledgers and paperwork and render his task even more difficult. "I can manage from here."
She froze, her hands tightening around the books. Abruptly she released them, dumping them back into the drawer with a loud thud. This time he was expecting it when she spun to face him, her face a polite mask and her eyes glinting. "Of course, my lord. How presumptuous of me. Pray forgive my ill behavior."
"No apology is necessary." As she walked away, he sat at the desk and pulled out the first of the books.
"Thank you, my lord."
"Have some tea sent in, if you please," he said, distracted, as he flipped to the first page. His uncle's hand scrawled haphazardly across the paper, a series of loops and slashes and serrated lines. Quin groaned inwardly. Deciphering the figures would be difficult enough without translating each line of writing, as well.
"As you wish, my lord."
He looked up. "Miss Willits?"
She was nearly out the door. "My lord?"
"Do you have a stutter?"
She furrowed her brow, the sight both enchanting and diverting despite the blatant hostility in the eyes beneath. "I don't believe so, my lord." She hesitated. "Why do you ask, my lord?"
"Generally one or two 'my lords' per conversation are enough to satisfy my pride," he said amiably, curious to see what her reaction would be. In most cases women didn't come to dislike him until after he'd explained that lovely as they were, he already had an understanding with Eloise Stokesley. "More than that begins to sound somewhat obsequious."
For the first time, her sensuous lips curved in a small smile as she curtsied and left the room. "Yes, my lord."
## Chapter 3
Maddie liked the word "obsequious." It sounded unpleasant and haughty at the same time, the very sort of arrow she meant to aim at the Marquis of Warefield, however handsome and charming he seemed to think himself.
With luck he would concentrate on merely being attractive and pleasant. Otherwise, he might actually take enough time with the ledgers to ruin the year's bookkeeping completely. As for the crops, if not for the last damned rain, she would already have had that business well begun. She could only pray Warefield would have as much aversion to dirtying his hands as the rest of the nobility, so she could organize the farm tenants before he deigned to set his lovely boots in a single field.
Once she had everything taken care of, Warefield would have absolutely no reason to hang about little Langley Hall in obscure Somerset County. He certainly wouldn't wish to prolong his stay just to visit, not with the rest of his kind beginning to prepare themselves for the ostentatious London Season.
"Twopence for your thoughts, Maddie."
Maddie shook herself and looked up at Mr. Bancroft. His amused expression deepened, and she wondered how long she'd been staring at her nicely roasted pheasant. Mrs. Iddings had certainly outdone herself, for she couldn't remember eating anything so delectable, damn it all. Not an undercooked or a burnt spot could be found. "Beg pardon?"
Malcolm chased a stray chunk of potato around his plate. "You had such a glare on your face. I merely wondered what you might be plotting—and against whom it might be aimed."
She narrowed her eyes. "I'm not plotting anything. I am merely wondering why your nephew decided to take his meal by himself in the formal dining room, rather than keeping company with his ailing uncle."
"Ah." He glanced up at her as she stabbed at her pheasant, imagining it to be a slice of Lord Warefield, well done. "Perhaps he's not comfortable with invalids. Many people aren't, you know."
She scowled. "I know that. But it's not as if you have the clap or something, for heaven's sake. A week ago you couldn't even move your hand, and now you're eating with it. You'll be out riding in a month, and the least he could do is—"
"Maddie," Mr. Bancroft interrupted.
At the same moment someone cleared his throat from the doorway. Flushing, Maddie added poor timing to Warefield's growing list of faults.
"Good evening, Quinlan," Malcolm said. "Have you dined?"
The marquis stepped into the room. He gave Maddie a quick, unreadable glance, then nodded at his uncle. "Indeed. Your cook is splendid, Uncle. That was quite possibly the most succulent pheasant I have ever tasted."
"I'll pass along the compliment. I'm certain Mrs. Iddings will be excessively pleased. Don't you think so, Maddie?"
Maddie kept her attention on her plate. "I'm certain she will be, Mr. Bancroft. She's quite delighted by his lordship's presence, as are the rest of us."
"Why, thank you, Miss Willits."
He hadn't even blinked at her considerable sarcasm. Perhaps she was too adept at it and he had thought her sincere. All the better, then; she could continue making a fool of him. "Not at all, my lord."
"I was wondering whom you might recommend as a guide when I tour the fields tomorrow," Lord Warefield continued, fiddling with the medicine bottles on the bed stand.
All of the dazzling bright colors were likely too tantalizing for him to resist, Maddie thought. Abruptly feeling cheery once more, she set aside her fork. "If I may be so bold, my lord," she said, "Sam Cardinal is quite knowledgeable. He's been a tenant here for more than fifty years, I believe."
The marquis nodded and smiled. "My thanks—"
"Nonsense, Maddie," Malcolm interrupted, with unusual obtuseness. "Sam Cardinal will talk poor Quinlan's ear off. Barley this, and barley that. I declare, the man's skull is stuffed with grain."
Lord Warefield chuckled, and glanced sideways at her again. "Do suggest someone else, then."
Maddie stifled her frown, wishing Mr. Bancroft would quit interfering. Then another, even more promising suggestion occurred to her, and she gleefully turned to the marquis. "Walter, the groom, grew up h—"
"Maddie," Mr. Bancroft scowled. "What in the blazes—"
"Actually, Miss Willits," Warefield interrupted, "I thought perhaps you might show me about, yourself."
"Me?'' She felt his cool green gaze on her face, and looked pleadingly at Mr. Bancroft. "Surely your uncle needs my—"
"I think Maddie would be the logical choice, yes," Malcolm agreed, nodding. "She's familiar with all the local farms and tenants, and with what's grown well, and where, over the last few seasons."
Maddie's jaw clenched. Traitor. It seemed she was completely on her own in this. Well, so be it: one marquis could hardly be that much of a challenge. "Surely Lord Warefield doesn't wish to follow a female about Langley."
"I have no objection," the marquis countered. "And I could hardly hope to find a more attractive guide anywhere in Somerset."
She met his gaze. He found her attractive, did he? My, my, how very flattering. And he had no objection to her presence. He must be even more obtuse than she'd imagined. Maddie kept the strained smile fixed on her face and tried not to grind her teeth. "I'm quite the early riser." She wished she could take just one minute and tell him exactly what she thought of him and his sod-headed, gossip-mongering peers.
"Splendid. I'd hoped to make an early start of it. Shall we say first thing in the morning, then?"
Apparently nothing short of pummeling the marquis would convince him to change his mind. "Whatever pleases your lordship."
He looked at her, an odd expression on his face. Perhaps her sarcasm had finally penetrated—though it might merely have been gas. With a slight smile, the marquis turned back to his uncle. "I'm a bit tired tonight—too much sitting about in the coach, no doubt—but I thought we might have time to chat together tomorrow evening at dinner. I imagine I can fill your ears with enough London gossip to amuse you for a while."
Mr. Bancroft looked sideways at Maddie, and she cursed herself for having pricked the marquis's pride, thereby ruining tomorrow evening for all of them. Surprisingly, though, her employer smiled. "That would be grand, Quinlan. Maddie and I were just about to begin a game of piquet, if you've a mind."
This time Maddie couldn't help her strangled exclamation. "Mr. Bancroft!"
Both men looked at her. "Is something wrong, Maddie?"
She madly tried to think up an excuse for her outburst. "His lordship said he was tired," she ventured hastily. "Of course he doesn't wish to play cards."
"Perhaps another time." Lord Warefield nodded, then sketched a shallow, elegant bow in Maddie's direction. "Good evening, then."
"Good evening, Quinlan."
"My lord."
His quiet footsteps retreated down the hallway. As soon as he had passed out of earshot, Maddie snorted and set the remains of her dinner aside. "Me, show him about Langley, indeed."
"You said you would," her employer pointed out.
"You said I would. I have things to do tomorrow. I've been calling on Mrs. Collins every day with fresh flowers and the post since she came down with the gout, and John Ramsey wanted me to look at his new sluice gate and see if you might be interested in using the same design for the northern field. I'm certain your nephew can find someone else to follow about."
For a moment Malcolm gazed at her, his serious look a little unsettling. "You know, Maddie, I've been thinking: it might do you good to associate with those of your own station from time to time. You've been rusticating in Somerset for four years now."
She stood, her cheeks flushing. "He is not of my station, and no doubt he would be quite insulted to hear you say such a thing." Maddie took a slow, deep breath to calm herself down. "And being in the country," she continued in a more moderate tone, "is called rusticating only when you wish to be somewhere else. Which I do not."
"Well, Miss Willits," he returned, the fine lines around his eyes deepening as he smiled, "I must say I'm very pleased to have you here." He gestured at the dresser. "Now, get the cards, if you don't mind. I need to regain some of my losses."
A reluctant grin touched Maddie's lips. "Yes, you owe me four million pounds, don't you?"
He scowled. "Only temporarily, girl."
Warefield had said they were to meet first thing in the morning. Therefore, when Maddie crept down to the stable just after dawn, she decided it would be his own fault for missing her. He should have been more specific about the time. She would be well on her way to Harthgrove before Lord Warefield even dreamed of rising.
Most of the household staff was still to bed as she made her way out the rear entrance through the kitchen. The spring morning air had a bite to it, and she shivered despite her warm gray riding habit. Cold or not, though, Maddie smiled in amusement as she headed down the path to the stable. Hopefully the marquis would be forced to spend the day with Sam Cardinal or Walter after all, while she could sample the new berry pastries at the bakery and chat the morning away with Squire John Ramsey and his sister.
She strolled around to the front of the stable—and stopped in confusion. Her mare, Blossom, stood saddled and waiting for her, Walter holding the reins. "Walter, what—"
"Good morning, Miss Willits."
She jumped and spun around. The marquis was mounted upon the huge bay hunter he'd brought with him from Warefield. Maddie refused to admire the beast, and instead kept her annoyed attention focused on Langley's unwelcome guest. He sat easily, looking down at her with a slight smile on his blasted handsome face. The early morning sun lightened the jade of his eyes to emerald. The marquis leaned forward with a creak of leather, taking several moments to peruse quite thoroughly her attire. No doubt he thought it cheap—though even her completely unflatterable self couldn't help but notice the twinkle in his gaze and the responding flutter along her nerves.
Feeling distinctly outmaneuvered, she smiled and curtsied. "Good morning, my lord."
"I'd begun to wonder whether you'd forgotten our appointment." He wheeled the bay about as it pranced impatiently. "We did say first thing in the morning, did we not?"
"Yes, we did, my lord," she conceded airily, yanking the mare's reins from Walter's startled hands and stalking over so the groom could hand her up into the saddle. "I daresay I shall be the envy of the entire countryside today, being in your lordship's company."
"I daresay that's far too generous, but thank you for the compliment."
"You are quite welcome, my lord." She nudged Blossom into a trot. "Shall we begin, my lord?"
For just a moment he glanced at her dubiously, but before he could comment, she was off and moving toward the nearest of the tenant farms. Since he'd demanded her company, he would simply have to accept her "my lords" as well.
A light breeze set the new elm leaves dancing in the treetops. Clouds tumbled slowly toward the rising sun, and she wondered whether they were in for another bout of rain. She certainly hoped not, for that would prolong the marquis's stay at Langley.
"You sit well, Miss Willits." The bay cantered up beside her, and again Warefield's eyes ran over her appraisingly.
Maddie flushed. He didn't need to sound so surprised, or to take so long gawking at her again. Once was quite enough to put her out of sorts. "You are far too kind, my lord," she returned, wanting to roll her eyes at his stuffiness. "I daresay I ride adequately for Somerset, but my goodness, I've never seen such a splendid mount as yours."
Predictably, his expression warmed at the faux praise. "Old Aristotle's a fine beast, indeed. I won him in a wager with my younger brother last year. If I don't take him everywhere with me, Rafe's been known to drop in at Warefield and attempt to make off with him."
"Only imagine, stealing a horse for amusement." Maddie chuckled and snapped the end of her reins against her thigh, harder than she intended. She flinched at the sting and blamed Warefield for that as well. "And to think that if common folk did such a thing, they could be hanged for it!"
He swung around to look at her, surprise and anger in his eyes. Perhaps he wasn't as obtuse as she'd thought—this time he'd actually noticed the insult. Before he could comment, though, a fortunate disaster caught her attention. She sat up straighter and pointed ahead of them.
"Oh, look, it's Miss Marguerite! Headed straight for Mrs. Whitmore's cabbage patch again, no doubt. We'd better head her off, or Langley will find itself in the middle of a tenant war."
The marquis regarded the pig trotting across the field in front of them. "She nearly sent my coach into a hedge yesterday. Several angry persons flailing farm implements were also involved."
Maddie grinned as she set Blossom into a canter behind the fleeing sow. "Yesterday? Miss Marguerite's being quite industrious. It's usually a week or more between escapes."
The marquis and Aristotle kept pace beside her. "Ah. This Miss Marguerite is a hardened criminal, then, I assume?"
"Most definitely." Abruptly, she remembered that she detested Warefield. "Though I daresay, my lord, that compared to the fascinating amusements of London, a marauding pig must be rather stale."
He laughed in response, an unexpectedly warm, infectious sound. "Not at all, Miss Willits. I don't spend all my time in London, you know. And we do have pigs at Warefield."
"Not like this one, my lord."
Granting him a smug glance, she took off across the field in pursuit. As she'd predicted, Miss Marguerite had angled along the stream bank, making straight for the Whitmores' small tenant farm. In a moment Warefield and Aristotle had flashed by her. He was a splendid rider, rust-colored coattails flying out behind him as he leaned low over the bay's back. However numerous his other faults, she had to admit that he looked magnificent. Maddie held Blossom back, curious to see how his lordship would handle this little adventure.
The marquis swept along the bank, swiftly closing the distance between himself and the pig. Miss Marguerite dodged sideways, and in response the marquis sent Aristotle forward in a sprint of speed, obviously trying to trap the pig between the horse and the stream. The maneuver would have been a sterling one, if not for the notorious mud at the edge of the water.
"Your lordship, be careful of...." Maddie trailed off, grinning delightedly. This should be interesting.
Aristotle reached the edge of the water. Feeling the ground give beneath his hooves, the bay balked and scrambled sideways to regain his footing. The marquis, his attention on the swerving pig in front of him, never saw it coming. With a startled yelp he went over the hunter's head, and, reins and hat flying, landed with a resounding splash in the stream.
"Damnation!"
The marquis swarmed to his feet, water cascading off his fine rust riding coat and filling his beautiful Hessian boots. Even with the spring melt, the water rose only as high as his hips, which she supposed was fortunate since he'd gone in head first. As he swept his wet blond hair out of his eyes, he issued several very colorful curses under his breath, which the morning breeze carried to Maddie's ears. They were quite imaginative, and he actually rose a notch or two in her estimation.
She took a deep breath, trying to stifle the laughter welling in her throat. "Oh, no, my lord! Are you un-hurt?" she asked belatedly, coaxing Blossom closer to the stream.
He spun around to glare at her. "Yes. Quite."
"How dreadful! I cannot imagine...Is the water very cold?"
"Yes." Slowly he turned in a circle, then glowered up at her again. "Frightfully. Where is my hat?"
"I believe I saw it...floating downstream, my lord." A chuckle erupted from her chest, and she quickly covered it with a cough. "Do you wish me to fetch assistance, my lord?"
"Absolutely not."
He eyed her blank face suspiciously, then shook water from his honey-colored hair and waded toward the bank. In the slippery mud he lost his footing and nearly went down again, and Maddie swiftly turned away, biting her lip to keep from laughing aloud.
"I shall fetch Aristotle for you, my lord," she said, and wheeled Blossom toward the stand of tall grass where the bay stood looking embarrassed by the whole affair.
As soon as Miss Willits turned away to fetch his blasted horse, Quin scrambled ungracefully through the slick mud and made his way back up onto the stream bank. Water and mud squished coldly in his boots, and he sloshed over to a clear, sunny spot of field and sat down to pull them off.
The deuced pig was out of sight, but Miss Willits seemed to know where the beast was going. He'd be damned if he'd let Miss Marguerite escape after this. In fact, he fully intended to dine on ham for luncheon. He dumped the first boot out and put it aside while he yanked off the second. Maddie approached with the horses behind him, cutting off any further cursing. "My thanks, Miss—"
"Oh, my goodness!"
At her shocked exclamation he froze. Miss Willits had clearly viewed his tumble into the stream with no anxiety and a great deal of amusement, and he couldn't believe that the removal of his boots would overset her. He turned his head.
Beside Maddie, with nearly identical expressions of astonishment on their pale faces, two young women, a brunette and a blonde, sat upon a pair of chestnut mares. Maddie's intelligent gray eyes gazed at him steadily for a moment, something very much like amused triumph, and very little like shocked dismay, in her gaze.
Quin had already begun to regard Miss Willits with some suspicion. Now, as her lips trembled with the effort of not breaking into out-and-out laughter, he was nearly ready to think her capable of actual sabotage.
Abruptly she blinked and straightened. "Oh, pray forgive my momentary upset, my lord. I had no idea you were en déshabille. May I introduce the Misses Lydia and Sally Fowler? Lydia, Sally, the Marquis of Warefield."
Quin shook more water out of his hair and swiftly climbed to his feet. "Ladies," he intoned, feeling completely ridiculous standing there in his stockings and with a sodden boot hanging from one hand, and even more distracted by the discovery that his uncle's companion spoke French. "Charmed."
"My...lord," the brunette returned, blushing bright red and thrusting one hand in his direction. "What an unexpected pleasure."
Stifling a grimace, Quin dropped his boot into the grass and stepped forward to grip her fingers.
"Lord Warefield, are you injured?" the blond asked, giggling nervously.
"Only my pride."
"Surely not, my lord," Maddie said warmly. "Not when the venture was as noble as this one, and not when your foe was so infamously and fiendishly clever."
There it was again: that definite sarcasm even her lovely innocent expression couldn't quite disguise. He'd been intrigued enough by her apparent dislike of him to suggest they spend the morning together. Obviously he'd underestimated the degree of her antipathy—no doubt because of the lust her mere appearance sparked in him. Even the very cold water had only managed to subdue his physical reaction to her.
"The venture was noble in thought, perhaps." He met her gaze squarely. "But I'm afraid that the poor cabbage patch may be done for by now."
Miss Willits's lips twitched, and she suddenly seemed to feel the need to examine the skyline over her shoulder. The sunlight accented the red highlights in her hair, and the gray riding habit, demure though it was, in no way disguised the curving lines of her body. Quin didn't realize how long he'd been staring at her until he heard the Misses Fowler whispering together.
"Lord Warefield," the older one said, "our home is just over the hill, if you'd care to dry your clothes."
"A splendid idea, Lydia," Miss Willits seconded.
An alarm bell immediately began ringing in Quin's skull. He was no fool, and it was rapidly becoming apparent that whatever Maddie favored had little to do with his own well-being. "My thanks, ladies, but Miss Willits and I have a great deal to do this morning. The sun is shining, and I daresay I'll dry soon enough."
Maddie looked disappointed, which convinced Quin that he'd made the correct decision. "Are you certain, my lord?" she pursued. "The Fowlers' cook makes a wonderful apple tart."
"Oh, yes!" Sally seconded, and actually reached down to put a hand on his arm. "Mama says Mrs. Plummer is the finest cook in Somerset. She's from Yorkshire, you know. It's because of Papa's sour stomach. He can't abide seasonings or spices of any kind, for they leave him with a terrible case of gas." She giggled.
Maddie made a sound in her throat, but when he glanced in her direction she had found a clump of grass to occupy her attention. "Are you well, Miss Willits?" he asked solicitously.
She started and glanced at him. "Quite, my lord. I was only thinking I should go see to Miss Marguerite before she completely decimates the cabbage crop."
She intended to abandon him to the Fowler sisters, then. "Yes, you're right," he said, hurriedly bending down to collect his boots and hobbling toward Aristotle. "We must be off."
"Are you certain you will not come to Renden Hall with us, my lord?" Lydia asked hopefully.
"My apologies, but I cannot." He pulled on one boot, his stockings squishing unpleasantly.
"Then you must call on us for tea tomorrow," Sally insisted.
Quin looked at her for a moment. Obviously she had no idea that only the head of the household was supposed to tender such an invitation, especially to a social superior. But he did not wish to appear as rude as she was. "Of course I shall," he answered, inwardly wincing. He glanced at Maddie and found sudden inspiration. "Would it be too forward of me to ask Miss Willits along as my guide? She does seem fond of Mrs. Plummer's tarts."
"What a grand idea!" Sally agreed. "Oh, Lydia, perhaps we could ask Squire John and his sister, and then we might play whist."
Quin watched Miss Willits from the corner of his eye. At the mention of this squire's name, her annoyed expression cleared.
"We shall see you tomorrow, then." Maddie clucked at her mare. "Come, Blossom, let's find Miss Marguerite."
Wondering if Uncle Malcolm was aware of Maddie's apparent fondness for the local squire, Quin quickly stomped into his other boot and grabbed Aristotle's dangling reins. "Ladies," he said absently, reaching up to touch the brim of his hat, only to remember that by now it must be on its way to Bristol Strait and the Atlantic Ocean. Drat it all, he'd been fond of that hat. "Until tomorrow."
"Until tomorrow, my lord."
"Yes, tomorrow, Lord Warefield."
A moment later he caught up to Miss Willits, who'd abandoned him without a backward glance. "What have I to look forward to with the Fowlers?"
"I could not say, my lord," she answered. The prickly annoyed female had returned, as though she hadn't found the previous encounter even the least bit amusing. "I don't believe they have ever entertained a member of the nobility, my lord."
Something in her tone, in her insistence on continually referring to him as "my lord," began to make him wonder whether her dislike was personal, or something more. "Have you ever entertained a member of the nobility?"
She glanced at him. "I have no household and no standing, my lord."
Neither had she answered the question. "Would they feel more comfortable if Uncle Malcolm accompanied us as well?"
"Mr. Bancroft cannot walk, my lord. Nor can he yet tolerate sitting upright for any length of time." Again she looked briefly in his direction, her expression unreadable but her eyes snapping. "I'm certain, though, that the Fowlers appreciate your concern over their comfort. And Mr. Bancroft as well, of course."
This time the cut wasn't even veiled. Although no one besides his brother had ever insulted him so bluntly before, he was far more curious than offended. "My, my, Miss Willits," he said mildly, "has your tongue ever caused anyone bodily harm?"
The muscles of her fine jaw clenched. "Not that I'm aware of, my lord. My most sincere apologies if I have in any way offended you."
"No apology necessary." A small farmhouse and a pig rooting through a cabbage patch came into view ahead of them. "I wish to ask, though, if I have in any way offended you."
She kicked out of the stirrup and with easy grace hopped to the ground. "Oh, my lord, do not tease," she said, with obviously exaggerated alarm.
For a brief moment Quin had the impression that if she'd been a man, they would have been selecting dueling pistols. "Do not think me a fool," he returned, dismounting and heading after Miss Marguerite before she could increase her cabbage carnage.
"Begging your lordship's pardon," Maddie said, moving swiftly to the far side of the cabbage patch to turn the determined pig back in his direction, "but why should you care what I think of you?"
"Why should I not?"
Maddie hesitated, an arrested expression on her face.
Just then the sow charged by him, squealing, and Quin nearly went down on his backside. "Dash it!"
Turning, he sprinted after the beast, abruptly angry at it—not for getting him tossed into the stream and nearly trampling him, but for interrupting the first genuine conversation he'd managed with Miss Willits.
Dirt and bits of grass and cabbage clung to Quin's damp breeches as he ran, but the blasted pig was not going to elude him again. He dodged after the animal, swearing under his breath. When Maddie called out behind him, she was quite a bit farther away than he'd realized.
"No, Mr. Whitmore!" she shouted. "Lord Warefield! Duck!"
At her alarmed tone, Quin threw himself forward into the muddy grass without hesitation. Before he'd hit the ground, he remembered her tendency to disregard his best interests, though, and cursed at his continued stupidity. An instant later, a musket thundered and a ball whistled over his head.
Breathing hard, Quin lurched to his feet and whipped around to see Miss Willits forcefully wrench a musket out of an older man's hands—one of the angry farmers he'd seen in pursuit of Miss Marguerite just the day before.
"Are you all right, Miss Willits?" he called, running back to the cabbage garden.
She quickly turned in his direction. "Yes, quite. And you, my lord?"
"No holes, thanks to you." At least she appeared not to want him dead, which was a relief. He stopped in front of the red-faced farmer. "You would be Mr. Whitmore, I presume?"
Maddie cleared her throat and returned the musket to its owner. "My lord, may I present Mr. Whitmore, one of your uncle's tenants? Mr. Whitmore, the Marquis of Warefield."
"The Mar—oh, good holy God, I'm sorry, my lord." Mr. Whitmore stumbled, blanching. "Terribly sorry."
He jabbed his free hand out at Quin, who lifted an eyebrow as he shook it. "Mr. Whitmore."
"I wasn't aiming for you, my lord, oh, no. It's just that blasted devil-spawned pig! That's the third time this month the beast has gotten into my vegetables!"
The marquis slapped at the mud and vegetation which continued to cling to him. "I'm none too fond of the animal, myself. What say we help ourselves to a side of pork, eh?"
The farmer grinned and hefted the musket. "Aye, my lord."
Miss Willits stepped forward and put her hand on the farmer's shoulder. "If I might make an alternate suggestion, my lord," she said hurriedly, "there will be piglets by May, and I'm certain no one would object if Mr. Whitmore chose the pick of Miss Marguerite's litter for himself."
The farmer scowled. "And what's to keep her from taking the rest of my crop before then?"
"The new fence Mr. Bancroft will see that the Hartleburys put up, to keep her where she belongs."
Mr. Whitmore eyed Maddie pleadingly, while Quin watched her with a great deal of interest. Langley didn't at all look like a holding in dire need of aid and repair, and he'd already begun to have a very strong suspicion why. Miss Willits spoke for his uncle readily enough, and the fanner accepted it as a matter of fact. And when, in frustration, Quin had skimmed ahead in the ledgers yesterday, the last few pages hadn't been in his uncle's indecipherable scrawl, but in a much neater, distinctly feminine hand. Maddie Willits was turning out to be quite an unusual mistress indeed.
"All right, Miss Maddie. I'll agree—if there ain't any more escapes."
Maddie smiled. "There won't be."
Considering that Miss Marguerite still wandered about in the wilds somewhere, it was an exceedingly bold statement. But the farmer bowed to Quin and turned away to examine the damage done to his cabbages.
She looked up at Quin. "Shall we continue your inspection, my lord?"
"Yes, of course. Though I'd like to return to Langley for a change of clothes first."
"I thought you might, my lord," she said mildly, perusing his water-, mud-, and grass-covered form with amusement, then turned to find Blossom.
Before she could ask the farmer for assistance in mounting, Quin stepped up behind her. Remembering her previous reaction to his touch, he cleared his throat. "If I may, Miss Willits."
She faced him, and then, with an exaggerated sigh of annoyance that didn't quite hide her discomfort, she nodded. "Thank you, my lord."
Quin slowly reached out to place his hands on her slender waist. Though his gloves were as dirty as the rest of him, for once Miss Willits had nothing disparaging to say. He looked down into her eyes and slowly lifted her up into the saddle. As she met his gaze evenly, he caught himself wishing she would smile just once at him.
Reluctantly he released her and retrieved Aristotle. As he swung up onto the hunter, he glanced in her direction. She was watching him, but swiftly turned her face away as he met her gaze.
"That was well done, Miss Willits," he complimented her, as they headed back down the road toward Langley. "You've granted Miss Marguerite a stay of execution."
"She is the Hartleburys' prize sow," she replied. "Without the income from her offspring, they'd not be able to make a go of it."
He waited for the omnipresent "my lord" to make an appearance. When it didn't, he smiled at her back. Apparently she'd made enough of a fool of him this morning that she felt them to be on more even ground—which Quin somehow didn't find insulting at all.
## Chapter 4
Maddie shook her head and glared at her employer. "I had nothing to do with it," she protested.
"He's been here for one day, Maddie," Mr. Bancroft returned, his expression equally exasperated. "Whatever you might think of the nobility, he is from a well-respected family. In addition, I happen to be rather fond of him. I cannot—I will not—have him drowned or shot! Is that clear?"
"How was I to know that Miss Marguerite had escaped again? Lord Warefield was the one who wanted to continue the pursuit, not me. And I didn't fire the musket, either!"
He eyed her. "I'd not wager on any of that, girl."
"I—"
"Under the circumstances, I won't ask you to apologize. But I do think you might exert yourself to be polite tonight at dinner and when you call on the Fowlers tomorrow!"
Maddie halted her sharp retort. In the four years she'd lived at Langley, this was the first time he'd vented anything more than good-humored frustration on her. And she was abruptly worried that he might bring another attack upon himself if he continued shouting at her, or that he might just decide he'd been charitable enough with her and ask her to leave. It had happened before, in other households, and for much less cause.
"I apologize to you, then, Mr. Bancroft," she said with as much dignity as she could muster, and turned to leave. She'd never liked losing, though, especially to a spoiled marquis, so she stopped in the doorway. "But I still think he was just trying to show off and ended up looking foolish. And that is certainly not my fault."
"Perhaps not," he sighed. "But do try not to kill him, won't you, my dear? Please?"
She turned and curtsied, immensely relieved at the reluctant amusement returning to his voice. "I shall try," she agreed solemnly, and headed downstairs.
Lord Warefield, accompanied by Walter, the groom, had left better than two hours ago to continue the Grand Tour of the surrounding fields. He hadn't asked her to continue as his guide, thank goodness, and in fact had barely said a word to her on the ride back to the manor, despite her solicitous concern. His jade eyes, the only part of him that didn't seem to be covered with grass and mud and water, had looked in her direction more than once, but she'd pretended not to notice. It really wasn't her fault, for heaven's sake—though she couldn't have asked for a more perfect way to reintroduce him to Langley.
Maddie chuckled. If she were the Marquis of Warefield, she would be packing by now to leave. After an afternoon spent with Walter, she doubted he'd even want to stay the night.
Mr. Bancroft typically napped in the early afternoon, so she slipped downstairs into the library to read for an hour or so. Far from being its usual quiet haven, though, the room reverberated with excited rattlings from the nearby kitchen and dining room. With the determination of a conquering army, Mrs. Iddings, Garrett, and the other servants went about preparing dinner and polishing the contents of the silver closet. They'd already done that, and only four days ago, but no doubt Warefield had found a spot on some utensil or other.
Maddie glared at the wall, then sighed and settled deeper into the comfortable chair. Entirely too much fuss and upheaval and upset, for no blasted reason. For heaven's sake, covered in mud, the Marquis of Warefield looked just as ridiculous as any commoner, if not more so.
Still, the rest of the staff insisted on looking upon him as though he were Apollo himself. If only he'd thrown some sort of tantrum, her morning would have been complete. But she supposed that becoming hysterical in front of social inferiors would be beneath his dignity. Perhaps he'd wept in private when he'd gone to change his clothes.
She smiled at the image, though given how calmly Warefield had accepted both the dunking and being shot at, she had to admit that his distress was more likely fantasy than fact. But if he thought he'd survived the worst Langley and Maddie Willits had to offer, he was in for a surprise. Several, in fact. As the American John Paul Jones had said some forty years ago during the war with the Colonies, she had not yet begun to fight.
"I'll take care of it myself."
Maddie started at the marquis's voice. Immediately following that, the library door opened, and the man himself strolled into the library.
"My thanks, Garrett," he continued, looking back over his shoulder. "And please let me know when the post comes."
"With pleasure, my lord."
Warefield closed the door and turned around. And froze. "Miss Willits," he said, clearly surprised to see her there. Jade eyes took her in as she sat curled up in Mr. Bancroft's favorite chair, the unopened book still sitting across her lap.
"My lord," she answered, straightening self-consciously and annoyed at the interruption. Of course he was surprised to see her. Servants did not drape themselves all over their employer's furniture. No doubt he expected her to be in the kitchen or assisting with the silver polishing, or mopping some floor before his boots could tread upon it.
He nodded and turned back for the door. "My apologies. I didn't realize the library was occupied."
Quickly she stood and set the book aside. "Oh, please, my lord. Do not let me keep you from your literary pursuits. Your uncle must be awake by now; I'll take him his tea."
"It's not necess—"
"Don't trouble yourself, my lord, I beg of you." Pasting another smile on her face, Maddie stepped past him. Dash it, she'd thought to be free of him until dinner, at least.
"Miss Willits, stop!"
Reaching for any remaining restraint, she stopped and turned to face him again. "Yes, my lord?" she ground out, though quite a few more colorful epithets came readily to mind.
"Do not leave the library on my account," he said slowly, enunciating each word. "I came only to find some paper. I neglected to bring any with me, and I need to catch up on my correspondence."
Of course he intended to report immediately to the Duke of Highbarrow on the poor state of Langley. "Oh, my lord!" she said, hurrying over to the small writing desk. "You should have told me! Allow me to fetch it for you. Pray do not exert yourself any further after the horror of this morning."
The marquis narrowed his eyes. "I would hardly describe being outwitted by a pig as a horror," he said dryly. "An embarrassment, perhaps." He paused. "And a tale my brother would be delighted to hear. I shall have to determine whom to bribe to prevent that from happening."
Of course, pompous peer that he was, he would think he could solve all his troubles—such as they were—with money. No doubt he always had. Maddie pulled a stack of paper from a drawer. As attractive as he looked, she knew what lay beneath his skin. And she would never like him—devilish captivating smile and pseudo-compassion or not.
"That was a joke, Miss Willits," he prompted.
Surprised, she looked up as she approached him with the paper. His eyes met hers, his expression amused.
"Yes, my lord."
"You may laugh if you wish, but I leave that wholly to your discretion."
"Thank you, my lord." So now he thought himself amusing and charming, as well. She might laugh at him, but she would never laugh with him. Ever. "I trust your tour of the fields went well?"
He lifted an eyebrow, his humor deepening. "Yes, quite. You know, I'd meant to ask you about that. My uncle's groom is...rather unique."
She nodded coolly. "Yes, we find him so, my lord."
"Quite knowledgeable about southern Somerset County, but I admit I hadn't anticipated his singular perspective on the nobility."
Maddie pinched her lips together and looked out the window. "Yes, my lord." She could feel his gaze on her, warm and discomfiting and unwelcome. Don't laugh, she reminded herself sternly. He's not amusing. "Apparently Walter was kicked in the head by a horse several years ago, and now believes himself to be a prince in hiding."
"So I gathered." The marquis took a step closer. "A son of the mad king, no less." He reached out and took the paper gently from her fingers. "And I thought the residents of Somerset were unused to nobles."
"Yes, we...." She trailed off. Now she was prolonging conversations with him for absolutely no good reason. "May I go now, my lord?"
"I said you needn't leave, Miss Willits."
"Yes, my lord, but I wish to go."
His searching gaze made her want to turn away, or better yet, to spit in his eye. Or kiss him.
"Then go," he said finally.
She blinked, stunned that she would think such a thing. "Thank you, my lord."
Quin checked his appearance in the mirror one last time, while Bernard put away his day coat. No trace of the morning's fiasco remained, other than the filthy pile of clothes Bernard had sent down to be washed, or failing that, destroyed. Even so, Quin couldn't rid himself of the feeling that something was out of place.
"Is there anything else, my lord?" Bernard asked, closing the portmanteau.
"Hm?" Quin shook himself and turned to find his valet. "Beg pardon?"
"Was there anything else you required, my lord?"
"No. Thank you." Quin scowled back at his reflection. "Bernard?"
The valet stopped his retreat. "My lord?"
"Do I have that beige coat with me? The one with the brown buttons?"
"No, my lord. My apologies. As your initial reaction to it was not entirely favorable, I did not realize you wished it packed. I shall send for it at once."
The marquis nodded absently. "Very good. And another pair of boots, while you're at it. I'm not convinced my riding boots are salvageable."
"Of course, my lord."
The valet slipped out of the room, and Quin took one last look at himself. Perhaps all the swimming and mucking about in the mud had addled his brain. All of his buttons were fastened correctly; he looked fine. Not too ostentatious, not too plain. Quin scowled at his reflection and turned away. He was acting like a damned dandy.
He headed for the west wing and his uncle's bedchamber. Voices came from the half-open door, and he paused in mid-knock, his attention snared.
"Mr. Bancroft," Maddie was saying, "I ordered it all the way from Surrey."
Quin leaned against the wail, unused to hearing the genuine and infectious amusement in Miss Willits's voice.
"It looks ridiculous."
"It does not. It's modern. If you just try it, you'll see how much easier a time you'll—"
"I'll break my neck!"
Maddie laughed. "I'll show you how easy it is."
Intrigued, Quin peered around the doorway. Miss Willits sat in a wheeled wooden chair, which she moved back and forward erratically by pulling two overly large wheels on either side.
"A wheelchair," he said, strolling into the room. "Mrs. Balfour uses one to get about now."
Miss Willits stopped rolling and shot to her feet. The good humor in her gray eyes vanished as she glared up at him and took several quick breaths, the bosom of her green muslin dress rising and falling enticingly in response. "Good evening, my lord."
"Uncle, Miss Willits." He nodded, disappointed. Maddie still hated him, apparently.
"Did someone die, my lord?" she asked politely.
He furrowed his brow. "Die?"
"Yes. Your coat, and...." She stopped and put one hand over her full, sensuous mouth. "Oh, my! My apologies. I have done it again, I fear. Black evening wear is quite the thing in London, no doubt."
Quin realized what had been nagging at him earlier. Miss Willits's sharp tongue. He glanced down at his garb and then met her eyes again. "It is, as a matter of fact."
"And you look splendid in it, my boy." Uncle Malcolm motioned for Quin to take one of the seats at the small table set beside the bed. "We may not be formal here, but there's no reason to ignore custom."
Quin caught his frown in time to force it into a smile. "Thank you, Uncle."
He made his way to the table. At that moment, however, Miss Willits found an urgent need to rearrange the vase of flowers by the window. Patiently he stood, waiting for her to take her seat first. With the same apparent patience, Maddie turned a red rose this way and that, pausing between adjustments to lean back and view her work. Out of the corner of her eye she sneaked a look at him, then continued with her arranging.
Then he realized he'd been correct all along. Maddie Willits was purposely taunting him, teasing him, and attempting to embarrass, frustrate, and humiliate him. And so far, she'd done a sterling job of it. Why, he had no idea, but he intended to find out.
With a glance at his uncle, who pretended to be absorbed in studying the plate of roast game hen set across his lap, Quin moved around the table and pulled out her chair. "Miss Willits?"
She turned around. "Oh, dear. Were you waiting for me, my lord?"
He smiled. "I was admiring two of the fairest blooms in Somerset," he corrected softly. "Waiting is my pleasure, I assure you."
For the space of several seconds she stared at him. "Thank you, my lord," she said slowly.
Quin sent another quick glance at his uncle, but Malcolm merely appeared amused at his mistress's antics. "Shall we?"
She reluctantly allowed him to seat her. Brushing her elbow with his hand, he took his own chair opposite her. By that time she'd rallied for another attack.
"Please forgive my being so forward as to have dinner served before your arrival, but there is so little room for footmen in here, and we did wait several minutes for you, my lord."
"Maddie," Malcolm grunted.
Quin took a sip of wine to cover his amusement. Sharp-witted, sharp-tongued, and sharp-mannered. It was a wonder he hadn't been killed already. "I'm afraid I must be the one to apologize for my tardiness. Unforgivably rude of me."
Miss Willits glared at him over her roast game hen. "Not at all, my lord," she said sweetly.
"Quinlan, you had some gossip for us, I believe?" his uncle broke in, making a bald attempt to turn the conversation.
"I do, though now that I've experienced the daily routine of Langley, I'm not certain anything I have to say could be as exciting."
"I'm just glad you didn't break your neck. You were damned lucky, my boy."
Miss Willits nodded at Malcolm. "Indeed. Thank heavens it's been raining, or Lord Warefield might have been seriously injured."
"And so might have Miss Marguerite," Quin added.
Maddie lifted an eyebrow, again the picture of beautiful disdain. "And the Hartleburys would have lost their best source of income over a muddied marquis." Even her chuckle sounded scornful. "What a jest."
"Or a joust," Quin muttered under his breath.
"Beg pardon, my lord?"
He looked up. "Hm? Oh, I was only saying that this morning's incident reminded me of my last encounter with the Earl of Westerly."
For the next hour, Quin regaled his audience with tales of his last sojourn to London. Uncle Malcolm seemed both relieved and amused at the change of subject, but then, he was acquainted with most of the participants. Miss Willits might as well have been regarding stones, for all the interest and enthusiasm she showed.
She pointedly retrieved a book from the chest of drawers, though Quin noted that either she was an exceptionally slow reader or she was paying more attention to his tales than to Walter Scott's. He had to consider that a small victory.
"Well, what do you think of the fields?" Malcolm asked finally.
Quin sipped the port Garrett had brought in earlier. "I noticed that the seed has already been prepared," he commented, glancing sideways at Miss Willits.
"No sense wasting valuable time," his uncle returned, "as you pointed out yesterday."
Quin was fairly sure Maddie had organized the crop. Everything was divided too fairly for the decision to have come from one of the tenant farmers. "I thought we might begin tilling tomorrow, on the three southernmost fields, anyway. They seem to be drier than the rest." Again he looked over at Maddie. "And I think Mr. Whitmore might appreciate a little preferential treatment after his cabbages were decimated this morning."
Malcolm nodded. "I agree."
"I assume you utilize all your tenants for joint planting and harvesting duties?"
"That is the fastest way to get the crop put in, my lord," Maddie said under her breath.
Quin had the distinct feeling he'd just been insulted again. "I'd hoped you would say that," he said, enjoying her disgruntled expression. "And perhaps you might assist me in assigning duties to the farmers, as you know their habits and character better than I?"
She narrowed her eyes. "I must see to Mr. Bancroft, my lord. And we have a luncheon which you engaged us to attend tomorrow."
"Maddie, Quin's quite correct. You might save us more than a day by doing the organizing." Malcolm held her indignant gaze. "And you will still have time for your luncheon with the Fowlers."
"Oh, very well." She stood, dropping her unread book on her chair. She leaned over to kiss Malcolm on the cheek, then coldly nodded at the marquis. "Good night, Mr. Bancroft, my lord."
"Good night, Maddie."
"Miss Willits."
Quin was surprised to see her leave, though he supposed he shouldn't be. In his unwell condition, Uncle Malcolm would probably have little use for her companionship. Even so, sharp-tempered and strong-willed as she was, he hadn't expected her to bother with making a show of going elsewhere to sleep. She certainly didn't seem interested in impressing him, at any rate—unless it was to kill him with kindness.
"Well, Quinlan, other than the stream and Miss Marguerite, how did you find Langley?"
"Quite exceptional, Uncle. Truth be told, it's hardly what I expected." Nothing here was—Miss Maddie Willits least of all.
"Hardly what Lewis expected, you mean. No burnt-out tenant houses, no floods or famine in Somerset without a Bancroft to oversee the land. I imagine he'll be quite disappointed."
"I shan't speculate." To avoid another family argument, Quin made a show of stretching. "Well, I'm off to bed myself. Shall I send your valet in for you?"
Malcolm shook his head. "Maddie will have seen to it. Good luck in the morning."
Quin grinned. "Thank you. I suspect I'll need it."
The Marquis of Warefield was up to something.
Maddie had watched him all morning: watched him chatting and being friendly with the awestruck farmers, and watched him actually pick a stone or two out of the soil and toss it out of the path of the plow.
She frowned at his lean, tanned profile and the curl of golden blond hair caressing his collar, and had to admit that he wasn't quite as dim as she had anticipated. Nobles didn't act nearly so amiable, she knew very well. Therefore, his affable demeanor toward the commoners was for some purpose. And she would find out what that was, before he could do any lasting damage to the good people of Somerset.
She'd instructed Bill Tomkins and the other footmen to erect a canopy at one edge of the Whitmores' field so that the marquis could take refuge from the sun when it should become too warm for him. Warefield had taken one look at the thing and the cushioned chair beneath, thanked her for her thoughtful attention to his health, and retreated to the far end of the field. There he'd stayed for the rest of the morning. She couldn't even hear him prattling on about whatever nonsense it was that had the farmers so interested.
"Coward," she muttered under her breath, glaring at him.
The workers' worshipful attention to him and his every command left her feeling practically useless, so she had to surmise that he'd insisted on her accompanying him merely to ensure himself an audience for his bloated, self-important conduct. Maddie plunked herself down on one of the shaded chairs and folded her arms. She was certainly not impressed by his efficiency. She could have done just as well—better, as a matter of fact—without him coming around at all.
"Miss Willits," he said, finally making his way around the plow horses toward her, "I believe we should return to Langley if we're not to be late to the Fowlers'."
Maddie stood and curtsied. "Of course, my lord. But who in the world will supervise while you're away?"
They walked toward Blossom and Aristotle, waiting in the shade of a stand of oaks. "I placed Sam Cardinal in charge. He seems to know a great deal about grain."
She concentrated on keeping the bland expression on her face. "I am aware of that, my lord."
"I know, Miss Willits. It was another attempt at humor. Pray forgive my ineptness."
He was making set-downs exceedingly difficult. It had to be deliberate. "Yes, my lord."
Unabashed at her lack of sympathy, he smiled and stepped toward her. "If you'll allow me," he murmured.
He slid his hands around her waist. She'd expected it, but even so, her breath caught at the warm strength of him. He hesitated, looking down into her eyes, and then lifted her easily up into the sidesaddle.
It was beastly awful, being a warm-blooded female helplessly attracted to a finely put together man, however pompous and useless she knew he must be. Maddie took a moment to fiddle with Blossom's reins before she looked at the marquis again and smiled stiffly. "My thanks, my lord."
"Again, a pleasure."
The marquis swung up onto Aristotle, and with a half wave at the farmers, who had already stopped their work to chatter about the noble in their midst, he led her back to Langley. Maddie glared at his back. There was no denying it any longer: he'd caught onto her game and was trying to make her feel guilty about making fun of him. That's why he was being nice to everyone. It wouldn't work, of course, because she knew all too well what his kind was really like. She'd have to modify her strategy a little.
Back at Langley she changed into her yellow and white sprig muslin gown. It had barely been fashionable two years ago and by now was hopelessly passé, but the Fowlers would hardly notice. And she certainly had no one else to impress.
Warefield had suggested they take his uncle's curricle. Not wanting to give him the opportunity to go lifting her yet again, Maddie hurried out to the stable, where Walter had the carriage waiting. He handed her up onto the narrow seat.
"I must say, Miss Maddie," the groom said, "it's terribly nice to have another blue blood in Somerset for a change. And a true gentleman, he is—promised right off not to let old Georgie know I was hereabouts."
Maddie nodded. "A true gentleman. No doubt he'll be the first to stand beside you if you decide to take your rightful place at court."
"My thanks for your vote of confidence, Miss Willits," the marquis's dry voice came from behind her. "I certainly shall."
Drat him for always sneaking up on her, and for the way his voice speeded her pulse. "I have great faith in...." Maddie trailed off as he climbed up beside her. She'd expected more overly formal black attire, and even had a comment ready for it. She had not expected a simple gray day jacket and buckskin breeches tucked into his mud-dimmed Hessian boots, nor the amused smile that warmed his face as he took the seat next to her.
"'Faith in....'" he prompted.
"The nobility," she finished.
He lifted an eyebrow as he accepted the reins from Walter. "That surprises me."
"And why is that?"
He snapped the reins, and the matching bays took off at a smart trot. "I have sensed a slight criticism in your tone from time to time."
"You are quite mistaken, my lord," she returned quickly, putting a shocked expression on her face. "I would never dream of such a thing, my lord, I assure you! Who am I to criticize the Marquis of Warefield?"
"Yes, Miss Willits, who are you?"
At first she thought he was agreeing with her faux humility, and opened her mouth to make an equally cutting remark. When she glared at him, though, his expression showed nothing but curiosity. "I am your uncle's companion," she said, amending the extremely insulting comment she'd been about to make.
"Yes, for four years, because you applied for the position. But what did you do before that?"
Maddie could only stare at him, disconcerted. "You remembered...."
"You do make something of an impression, you know," he returned dryly.
She swallowed, all of the insinuations and insults she'd planned for the ride vanishing in an instant. Blast him and his compliments. She neither needed nor wanted them. Maddie shook herself. She did know what to do with them: counterattack—immediately, before he realized he'd scored a hit. "Why do you bother with flattering me?"
"Is it flattery to ask a question?"
"It is flattery to feign interest for the sake of politeness, my lord."
"Ah," he nodded. "Then I am merely being polite?"
"Yes, of course."
"I see."
He made no other comment, and she dared to hope that she'd completely confused him with her dazzling logic. Pointedly ignoring him, she made a show of admiring the various wildflowers coming into bloom along the side of the road, and the robins and swallows building nests in the budding trees.
"You didn't answer my question."
Maddie shut her eyes for a moment. "Which question, my lord?"
"What did you do before you applied for the position with my uncle?"
"I...worked as a governess in several households," she answered slowly, wondering why she was so reluctant to lie to him. She owed him nothing. Yet she supposed she had no wish to be seen in the same light in which she saw him and his kind.
"Where?"
"Do you intend to check my references, my lord?"
He looked over at her again. "No. Of course not."
Maddie pointed down a rutted dirt track to the west. "Over there, my lord. About half a mile down."
Warefield turned the carriage in that direction. "You know," he said quietly, "I admire the job you've done here at Langley. I'm here only because my father wished it. I don't intend to turn you away."
She'd heard promises of integrity before. "Thank you for your assurances, my lord," Maddie said stiffly, "but they are completely unnecessary."
"And why is that?"
She turned to look directly at him. "You did not hire me, my lord."
He met her eyes, then pursed his lips and faced the road again. "True enough. Thank you for putting me in my place, Miss Willits."
Maddie pressed her advantage." You are quite wel—"
"Oh, good God," he muttered.
"What is it?"
As they came around the bordering hedge, she saw what had prompted his curse. The entire Fowler household, nearly as substantial as Langley's, stood lined up at full attention along the curving drive leading all the way up the steps to the front door. A chuckle tickled up Maddie's chest and burst out of her throat before she could stop it.
"Are you laughing, Miss Willits?"
She clapped a hand over her mouth and coughed. "No, my lord," she managed. "'Twould not be seemly."
He scowled. "No, it wouldn't."
Much more enthusiastic than she had been a moment earlier, Maddie smiled at the Fowlers' butler as he came forward to help her to the ground. "My thanks, Mason."
"Miss Maddie."
Warefield came around to her side of the curricle, and she led him forward to where the Fowler family stood waiting at the end of the line. "My lord, may I present Mr. Fowler and Mrs. Fowler? Mr. and Mrs. Fowler, the Marquis of Warefield."
Tall James Fowler bent himself almost double in a bow, his normally dour expression stretched into a rather alarming-looking smile. Beside him, Jane Fowler sank so far to the ground in her curtsey that Sally had to help her upright again. "My lord," they breathed, echoing one another.
"It is an honor to have you at Renden Hall," Mrs. Fowler continued reverently. "You have met our dear daughters Lydia and Sally, I believe."
"Yes, I have." The marquis stepped forward to shake Mr. Fowler's hand. "You've a lovely family, sir."
"Thank you, my lord." Mr. Fowler gestured toward the house. "Allow me to show you inside."
Quite delighted to be ignored, Maddie followed behind the Fowlers and Lord Warefield as they passed the retinue of servants. They had nearly reached the front door when the marquis made a show of turning around and coming back to collect her.
"Can't have you getting lost, Miss Willits, can we, now?" He tucked her hand around his arm and held it against him.
"I know my way around, my lord," she muttered, trying unsuccessfully to free her hand, and surprised at his iron strength.
"That is precisely what I am afraid of," he murmured, nodding at the wide-eyed procession of footmen and maids who swarmed up the steps and into the house behind them.
"This way to the dining room, my lord," Mrs. Fowler announced regally, scattering servants out of her way and thrusting her daughters ahead of her. "Mrs. Plummer has outdone herself today, if I do say so myself. You have inspired her."
"Glad to be of some use," he said, glancing about but keeping Maddie pinioned securely at his side.
They entered the dining room to find it stuffed practically to the rafters with fresh-baked bread, puddings, ham, and chicken. Maddie simply stared for a moment. She hadn't seen this much food even at the Fowlers' annual Christmas pageant, famous throughout southern Somerset. For a country luncheon, unless they intended to feed all the farmers out plowing the fields, it was absurdly overwrought.
"Do let me go," she whispered, tugging again at the marquis's hand warmly covering hers as they toured the repast on the table and the sideboards.
He looked down at her, and at the unexpected humor in his gaze, she stopped struggling. Damnation, he was ruining everything. He wasn't supposed to be amused at the Fowlers's toadying; he was supposed to accept it as his due.
"I've no intention of letting you out of my sight," he whispered back. "This was your idea."
"They wouldn't have invited me," she pointed out, more disconcerted by his attention to her than she cared to admit. "You are the Marquis of Warefield, my lord. This is all for you."
"And I'm supposed to be honored?"
She glared at him. Maddie wasn't particularly fond of the Fowlers, either, but there was no reason to be cruel. She yanked her hand free. "I'm certain you're used to much better, my lord," she retorted in a hushed voice, "but this is their ideal of highest elegance."
He looked at her for a long moment, his expression serious and unreadable. "I see."
Maddie intentionally seated herself between Lydia and Sally, keeping as far from the marquis as she could manage. He'd goaded her into being directly rude again, when she'd decided to stay with the more subtle approach of victory through flattery. From the occasional looks he sent in her direction he hadn't welcomed her last comment, but it served him right.
"Lord Warefield, please tell us about London," Sally begged, giggling. "Have you been to Almack's?"
"Of course he's been to Almack's," Lydia countered in an exasperated tone. "He's dined with the King."
"Is His Majesty as fat as they say?"
"Sally!" Mrs. Fowler fanned her napkin in front of her face. "For heaven's sake, mind your manners!"
"But Mama, King George is fat. Everyone says so."
"Sally! Silence!" Jane Fowler leaned across the table to capture the marquis's hand, nearly causing him to drop his forkful of sliced ham. "Pray forgive my daughter, my lord. She's a little high strung, perhaps, but quite proficient in all the gentler arts. We had tutors for both girls, all the way from Surrey."
Lord Warefield glanced at Maddie, then set his fork back on his plate. "I have no doubts on that count, Mrs. Fowler. Miss Sally, I do occasionally go hunting with His Majesty, and he is a rather...well-rounded individual. If you ever meet him, though, I suggest you not mention it. He's rather sensitive about the subject."
"You see?" Sally said gleefully, and took another biscuit from the bowl in the center of the table.
"Please, my lord, eat."
The marquis retrieved his utensil and obligingly took a bite. The entire family watched as he chewed and swallowed. Maddie decided she didn't feel the least bit sorry for him.
"It's quite good," he said after a moment, taking a sip of wine.
"And not a grain of salt used," Mr. Fowler said proudly. "No spices at all."
The butler stepped into the dining room and came to attention. "Mr. and Mrs. Fowler, Lord Warefield, I am instructed to inform you that Mrs. Beauchamp has arrived."
"What? What is she doing—"
Mrs. Beauchamp, wearing a low-cut gown two sizes too small for her ample bosom and two decades out of style, swept into the room. Warefield automatically came to his feet, though he looked somewhat startled. Mr. Fowler reluctantly rose a moment later, and Mrs. Beauchamp sank into a deep curtsey.
"My lord," she breathed, rising.
For several seconds Maddie sat where she was, dumbfounded at her tremendous luck, before she roused herself enough to stand. "Lord Warefield, another of Langley's esteemed neighbors, Mrs. Beauchamp."
The marquis nodded. "Mrs. Beauchamp. Charmed."
The lady came forward to clutch Lord Warefield's proffered hand and curtsied again. "I am delighted to meet you, my lord."
Mrs. Fowler leaned over to examine her neighbor's face. "Evelyn, whatever is that black spot on your cheek?"
"Oh, you silly thing," Mrs. Beauchamp tittered, fingering the small spot. "It's a patch. They're all the rage in London." She lifted her heavy jowled face toward the marquis's ear, still refusing to relinquish her grip on his hand. "My cousin is Baron Montesse," she whispered conspiratorially. "You may know him."
The marquis's lip twitched, and he cleared his throat. "Baron Montesse," he mused. "Of Berkshire?"
"Oh, no. Of Herefordshire, my lord."
"Ah. No, I don't believe we are acquainted, then. But I don't spend all that much time in London."
She frowned as Warefield finally managed to free his hand without pulling her over. "That's odd," she continued loudly, no doubt hoping the servants were listening. "I wrote that you were coming to Somerset, and he seemed familiar with you. Though he said something about your having a scar. But you don't, do you? Hmm. Perhaps he was mis—"
"A scar?" the marquis interrupted, and then smiled. "That explains it. He must be acquainted with my younger brother, Rafael. He was wounded at Waterloo. We do look something alike."
Mrs. Beauchamp's expression brightened. "Yes, that must be it, then." She gestured imperiously at one of the footmen to bring another chair forward. While Mrs. Fowler fumed, her neighbor sat on the far side of the marquis. "I knew we had mutual acquaintances. All of the nobility seems to know one another."
"Mother," Lydia hissed in protest, as Mrs. Beauchamp usurped her place of honor.
"Now, now, Lydia," her mother soothed, resuming her own seat. "You'll have your chance later, when you play for Lord Warefield."
Maddie cleared her throat and spread butter on her toasted bread. Again she felt the marquis's gaze on her face, but she refused to look. Let him suffer. At least she was enjoying herself...
Another footman brought a plate out for Mrs. Beauchamp, and she set to eating with her usual enthusiasm. Lord Warefield was barely able to touch his own plate, with all of the questions and absurd flattery sent in his direction. Maddie was amazed that the Fowlers and Mrs. Beauchamp were so eager to fawn over him. In all fairness, though, she'd grown up amid such nonsense and was used to it.
Sally leaned over the table to point her knife at Mrs. Beauchamp's soup. "I say, Mrs. Beauchamp, your patch has fallen off."
Laughter burst from Maddie's lips before she could stop it. Holding her napkin to her face and coughing to cover it as well as she could, she pushed to her feet. "Excuse me for a moment," she managed, and fled across the hall into the morning room. Pacing briskly, she tried to imagine something—anything—dull and somber.
"Miss Willits, are you quite all right?"
Still holding her hand over her mouth, Maddie whipped around. The marquis stood in the doorway gazing at her, while the dining room erupted in argument behind him. Attempting to cover her surprise, she leaned back against the couch. "Yes, I'm fine."
"I told them you had a bit of a cold, and that I wanted to make certain you were all right."
"Thank you. Though I don't need you to make my excu—" She stopped, her heart skittering with unexpected disquiet as he stepped into the room and shut the door behind him. An unbidden image of Benjamin Spenser entered her mind, but she sternly pushed it aside. She was not the stupid girl she once had been, and she knew quite well how to deal with men. "My lord?"
He took a deep breath and leaned back against the door. "Please, please, please tell me I may laugh at that," he muttered.
She stared at him and belatedly lowered the napkin from her lips. She'd half-expected an assault—not a conspiracy. "You...certainly don't need my permission, my lord."
Warefield folded his arms. "I certainly don't wish to give you another reason to scowl at me."
"I don't scowl at you," she shot back, trying to rally her anger again.
"Yes, you do." He pushed upright and took a step closer.
She backed away. "You awe and amaze me with your presence," she improvised weakly.
"Liar."
"Please, my lord, do not chastise me. I could not bear it."
"Yes, you could." He moved toward her again. "And how did you know the luncheon in there is not the height of elegance?"
Damnation. "I don't, my lord. I was simply judging by your reaction to—"
"Lord Warefield, are you going to return to the dining room?" Sally called, rapping at the door.
He didn't even glance behind him. "In a moment. Miss Willits is having difficulty catching her breath."
"I am not!" Maddie swerved around the couch and made for the door. "Don't use me to make excuses for your poor behavior, my lord. We should not be in here alone togeth—"
With unexpected swiftness the marquis dodged sideways and cut off her escape. "You said this luncheon was the Fowlers' ideal of elegance. Not yours."
She stopped, barely avoiding colliding with him. "I have been a governess in a variety of households, my lord. And your uncle is brother to the Duke of Highbarrow." She took a breath, reaching for her scattered wits and anger as she looked up at his heated expression. He was practically accosting her, after all. "Do you assume everyone you encounter in the country to be ignorant of the finer ways, Lord Warefield?"
He opened his mouth, then shut it again. "No. I do not." For a long moment he looked at her. "May I tell you something you will undoubtedly find shocking and annoying?"
Even more unsettled, Maddie swallowed. "Whatever pleases you, my lord."
"Sometimes—right at this moment, in fact—I have a rather strong urge to...kiss you."
Maddie flushed, her pulse suddenly pounding. "I—please restrain yourself, then, my lord," was the best she could manage. What she really wanted to do was flee back to Langley and lock all the doors before he could realize that she also had the recurring desire to fall upon him, damn him.
A slight, sensuous smile curved his mouth, and he nodded. "I shall attempt to do so." His long, elegant fingers gestured at the door. "If you please, Miss Willits."
Maddie smoothed her skirt, then approached. "Thank you, my lord."
"But I do intend to find out why you dislike me so much, Maddie."
She hesitated, then pulled open the door and hurried through. Sally eyed her with bald curiosity, but she pretended not to notice. Warefield followed close behind and resumed his seat and the luncheon conversation as though he'd never left the table.
Maddie glanced at him once. He was gazing directly at her, and she immediately looked away. The frustrated inquisitiveness in his eyes surprised and dismayed her. She'd expected anger—not this intense, extremely disconcerting curiosity. Nor had she expected him to be attracted to her, or her to him.
She was going to have to be more careful. Driving him away from Langley was one thing—but having the Marquis of Warefield discover her true identity was completely another.
## Chapter 5
After luncheon, the afternoon spiraled downward from merely painful to excruciating. They all crowded into the tiny drawing room, where Lydia Fowler sat at the pianoforte to regale them with a completely hideous interpretation of Beethoven's "Für Elise."
Quin found himself sandwiched between Mrs. Fowler and Mrs. Beauchamp, while Maddie sat in the corner gazing peacefully out the window. Closer to fidgeting than since he had been a boy in church, he lasted in his chair until the piece ended. Then, still applauding, he stood.
"Excuse me, ladies," he said pleasantly, "but the fire is a bit too warm for me."
Immediately Mrs. Fowler jumped to her feet and grabbed for the servants' bell. "I shall see to it at once, my lord!"
"No need. I'll just sit by the window for a bit." He turned to the elder Fowler daughter. "Miss Fowler, please play us another."
Mrs. Fowler beamed delightedly, and Lydia obligingly began another piece. He had no idea what she might be rendering this time, but she seemed enthusiastic enough. Quin strolled over to where Maddie sat and took the seat beside her before she could escape.
Something he'd said earlier—most likely his idiotic confession about wanting to kiss her—had her on the run, and he was intrigued enough to pursue his advantage. Since he'd arrived, he'd been defending himself against a myriad of attacks on his honor, his nobility, and his person, and he still had no real idea why.
"Enjoying yourself, Miss Willits?" he asked quietly.
She continued looking out the window. "Of course. I love Haydn."
So that's what it was. He wondered how she happened to know that. If she had been a governess, she was a very well-educated one. For a country landlord's mistress, she was simply extraordinary. His gaze lowered to the curving line of her throat, and the minute throbbing of her pulse beneath. Kissing her was only the beginning of what he wanted of Maddie Willits.
Quin took a breath. "You're aware that Mrs. Fowler has been planning a ball in my honor?"
Finally she faced forward again. "She mentioned it to me a few days ago, yes."
"It would be a grand opportunity for Uncle Malcolm to make an appearance in his wheeled chair, don't you think?"
"If he feels up to it, yes, I suppose," she answered grudgingly.
"And you shall join us then, I presume?"
She glanced at him, then away again. "I am Mr. Bancroft's companion. If he wishes me to join him, I shall."
Quin would personally see to it that Malcolm did wish it. She hadn't uttered one "my lord" yet, a good indication that she was still in full retreat. "And will you dance with me there?" he asked, pressing his advantage.
"You are better aware of propriety than I am, my lord."
He scowled before he could cover it. Miss Willits, it seemed, regrouped quickly.
"If you think it proper for a marquis to dance with his uncle's companion, then I will do as you say," she continued.
Quin looked at her. "I would hardly order you to dance."
"Thank you, my lord."
They were back on even ground, it seemed. "But I would like to."
She faced him again, her gray eyes lit by the afternoon sun through the window. "To what, my lord?"
"To dance with you." He glanced at Mr. Fowler and Sally, but they were busily chatting and ignoring Lydia's play. Quin edged closer to Maddie so that his knee brushed her skirt, and lowered his voice. "You have forbidden a kiss, and you seem to regard me as something of a monster, Miss Willits. I would like a chance to prove to you that I am nothing of the kind."
"I think nothing of you at all, my lord. It is not my place to do so."
He sighed. "Relentless, aren't you?" he muttered.
Her lips twitched, and she faced the pianoforte. "I'd like to think so, my lord."
She was amused at his expense—again. And somehow, though he'd never received such abuse in his life, he remained far more intrigued and diverted than angry. He'd always loved a good puzzle, and Maddie Willits was a virtual Sphinx's Riddle.
Yet he had no excuse whatsoever for his own very odd behavior. He was practically engaged to a perfectly lovely woman he'd known for twenty-three years, and at the same time he was lusting after a woman—his uncle's mistress, yet—he'd known for only three days.
As soon as he could politely do so, he made their excuses and they returned to Langley. Maddie was back to her previous hostile self, but Quin thought he detected more amusement in her demeanor than before. He hoped so, anyway. She vanished into the house as soon as he drove the curricle into the stable yard, so he really couldn't ask her.
Quin headed inside to change. He wanted to see how the plowing was progressing, and he needed to escape from Maddie's intoxicating, infuriating presence. Before he could do anything, though, Garrett intercepted him at the foot of the stairs.
"My lord, a letter arrived for you with the post." The butler held up for Quin's examination the silver tray containing the missive. "You said you wished to be notified."
"My thanks, Garrett." Quin took the letter upstairs with him. He immediately recognized the neat, ornate script of the address. Eloise Stokesley hadn't wasted any time in corresponding, but then she was always very prompt in such matters.
He summoned his valet and opened the one-page missive.
Dearest Quinlan,
I was heartbroken to read in your last letter that you've been banished to Somerset. I had hoped you would be able to visit us at Stafford Green, as we had planned.
"So had I," he muttered, though the stay at Langley had been much less of an annoyance than he had anticipated.
I know I shouldn't say so, but I miss you, Quinlan. I count the days until we shall be together in London, and even more the days until we shall be married. Please, tell me of your adventures in Somerset. I await your return letter with greatest anticipation.
Yours forever
Eloise
Bernard entered, and Quin set the letter on his dressing table. The best service he could do for Eloise was to get the planting and the ledgers finished and return to Warefield—hopefully by way of Stafford Green.
He had no intention of asking Maddie to accompany him out to the fields again. But if he allowed her to escape completely, she would have time to regroup and mount another attack.
If she was bored here at Langley, since her full services were no longer required by Malcolm, perhaps he had happened along just in time to capture her wandering attention. That theory didn't quite fit, since her actions were certainly not those of a woman trying to lure a man to her bed, but anything was worth a try.
Quin paused, frowning. Good God, now he was contemplating stealing another man's—his uncle's—lady. Perhaps it was Somerset itself making him mad. It seemed to have had that effect on several of the locals.
He went downstairs and out to the stable, stopping at the sound of voices coming from the garden.
"No, Bill, you're doing it all wrong."
"I am not, Miss Maddie."
"Yes, you are. You'll dump him on the ground."
"I will not!"
Quin leaned around the corner of the house. Bill Tomkins stood at the edge of the garden path. Malcolm, bundled in enough blankets to keep the entire Third Regiment warm in the middle of a Prussian winter, sat in the wheeled chair in front of the footman. Maddie stood beside them, scowling furiously.
"Maddie, it's all right."
"It is not, Mr. Bancroft. You need sunlight, not a dunking in the fish pond. Bill, leave off. I shall push him."
Malcolm chuckled. "You'd best do as she says, Bill."
The footman sighed heavily and relinquished his grip on the back of the chair. "Yes, Mr. Bancroft. Give me a yell when you want back up the stairs."
"Thank you, Bill." The footman strolled off back to the house, and Maddie took over the steering of the chair. "You should keep to the path, my dear."
"Nonsense. I want to show you the new roses. And did you bring the book?"
"The one thing that still works on me is my lap, Maddie."
She pushed him onto the soft grass, and they came to an abrupt halt. "Drat," she muttered.
"I think I'm sinking," Malcolm noted calmly.
Quin pushed away from the wall to perform a rescue.
Maddie shoved harder against the back of the chair, and with a sucking sound it came free of the mud. "See?" she said, moving him toward the roses at the far end of the garden. "Now, where were we?"
Malcolm produced a small book from the mountain of blankets surrounding him and opened it. "Let's see. Beatrice was telling Benedick that scratching up his face wouldn't make it look any worse."
She smiled. "Ah. And Benedick says, 'Well, you are a rare parrot-teacher.'" She released the chair and pranced in front of Malcolm, putting her hands on her hips. "And Beatrice rightly replies, 'A bird of my tongue is better than a beast of yours.'" As Malcolm awkwardly applauded, she turned about again and assumed a more masculine stance and voice. "'I would my horse had the speed of your tongue, and...'"
Quin cursed as she caught sight of him. Immediately she ceased her recitation and went back to pushing his uncle toward the rosebushes. Quin strode after them. "'I would my horse had the speed of your tongue, and so good a continuer,'" he finished.
"Ah, Quinlan. Good afternoon." Malcolm twisted around to greet his nephew.
"Miss Willits, I didn't know you enjoyed Shakespeare." He wanted to ask her where in blazes she'd learned to quote the bard by heart.
She turned to face him, her face flushed. "Of course you didn't know, my lord," she snapped, obviously embarrassed. "No doubt you also were unaware that your uncle enjoys it as well, or that he was quite looking forward to seeing you after four years—or you might have spent more than two hours speaking with him over the past three days."
"Maddie, that's enough," Malcolm said sharply. "I can fend for myself, thank you very much. You should not speak to Quinlan in that manner."
"Well, obviously no one else does." She turned to Quin. "Will you see your uncle safely back to his bedchamber?"
"Yes." He wondered why her words made him feel so completely...inadequate.
"Thank you. Mr. Bancroft, I'm going to see Squire John." With a flounce of her skirts, she stalked off around the side of the manor.
Quin looked down at Malcolm, who had an expression of mixed amusement and exasperation on his face. "I hadn't realized I'd been slighting you," he said quietly, crouching at his uncle's side.
Malcolm glanced at him, then smiled and patted his nephew on the cheek. "You didn't come here to be my nursemaid. You came to see that Langley remains a profitable holding."
"Yes, we must keep up appearances." He stood again and gripped the handles of the chair. "Do you wish to go inside, or see the roses?"
"I thought you'd be headed out to the fields to oversee the plowing."
"I changed my mind. Which direction?"
Malcolm's shoulders relaxed a little. "The roses, then, if you don't mind. I haven't been out of doors in nearly two months. And Maddie seems to have a unique understanding of the plants."
Quin resumed their trek. "Must be all the thorns."
His uncle laughed. "You may be right." He sobered and turned his head to look up at Quin. "Just remember how beautiful is the flower they protect."
"You don't mind her going off to visit this squire?"
"John Ramsey? No, he's a good fellow. I've known him and his sister since they were born." They stopped beside the nearest bush. "In fact, you used to play with him when you came to visit, as I recall."
Quin closed one eye, searching his memory. "John Ramsey...wasn't he the one who liked frogs?"
"No, that was Rafe. John was the one always building boats."
"Oh, I remember. Quiet little fellow."
"Yes, that's him. He's got a new irrigation system. He and some mathematician from Edinburgh came up with it. Maddie's been itching to try it on our north field."
"Why didn't she say anything to me about it?"
"I assume she didn't think you'd be interested. It'd add at least another week to your stay. We'll put it in next year. I should be up and about by then."
There it was again, that criticism, the assumption that he'd rather be elsewhere. The assumption that he was only doing his duty. "Perhaps I'll invite this squire up to the far field in the morning."
Malcolm smiled in pleased surprise. "Splendid."
In the morning, when Lord Warefield invited her to accompany him out to the fields again, Maddie wasn't certain she'd be able to keep her temper even for the short carriage ride to the crossroads. He must realize by now how much she detested him, yet it hadn't discouraged him one bit. That he'd sat with his uncle for most of the evening, playing piquet and being generally agreeable, didn't mean anything.
She hadn't mentioned his visit to Squire John or Lucy when she'd gone visiting, and though they must have known of his presence at Langley, they had been tactful enough not to bring it up. As she and the marquis came in view of the crossroads, she was therefore surprised to see the squire seated on his gelding, Dullard, apparently waiting for them.
"Good morning," Warefield said brightly, leaning from the curricle and holding out his hand. "John Ramsey. I believe I owe you an apology for sinking a boat several years ago."
"No apology necessary—though as I recall, it was nearly an entire fleet." The squire shook the marquis's hand. "Thank you for asking me out here. I was beginning to believe Maddie was the only one interested in modernizing hereabouts."
They rode over to John's holding to view the new irrigation system, and despite her skepticism, Maddie had to concede that the marquis's interest seemed genuine. Even more surprising, when she tried to make an escape into Harthgrove later, Lord Warefield volunteered to accompany her.
"I'd like to put in an order immediately for the planking we'll need to go about putting that system in at Langley," he said.
"But I'm only going to bring bread and vegetables to some of the tenants," she protested, as Walter helped her lift baskets into the back of the curricle.
"I'll go with you as my uncle's representative," he offered, loading another basket.
"I am your uncle's representative," she snapped.
He lifted his hands in surrender. "Very well, Miss Willits. May I accompany you to Harthgrove as no one at all?"
She clenched her jaw. "As you wish, my lord."
Maddie clambered onto the seat first and took the reins, clucking to the team as soon as she was settled. Still climbing up on the other side, the marquis nearly ended on his backside in the stable yard, but he managed to hang on and make his way onto the seat beside her. "Remind me not to turn my back on you, Maddie," he said.
She stifled an unexpected smile and concentrated on picking the most rutted part of the road all the way into Harthgrove. The vegetables no doubt got a little bruised, and so did her backside, but at least the marquis was too occupied with hanging on to try to converse.
When they reached the village, she pulled up the carriage. "The mercantile shop is over there," she said, pointing, "and the cottages I visit are further down the lane. Shall we meet here in an hour, my lord?"
He hopped to the ground and stepped around to offer her his hand. "I'm in no hurry, Miss Willits. And I'd like to meet more of Langley's tenants."
Reluctantly she gripped his fingers, and he helped her to the ground. "Why?"
Warefield shrugged. "I thought that by carrying a few baskets I might be of some use to you."
"Why would a marquis wish to be of use to me, of all people?" she asked, trying to ignore his tall, compelling warmth beside her as she gathered up baskets.
"Why wouldn't I?" he replied, taking the rest of the load himself.
She strode off down the lane, her green muslin skirts raising a light dust. A moment later, the marquis appeared beside her, his long legs taking the fast pace much more easily than her own. "Don't you have better things to do, my lord?" she suggested desperately. "The planting? The irrigation construction?"
He looked down at her and grinned. "Trying to get rid of me?"
Maddie sniffed and continued on her way. The other pedestrians stopped and stared as the Marquis of Warefield strolled down their dusty avenue. The bowing and curtseying that followed in their wake made Maddie feel as if she were leading some sort of parade, though of course everyone ignored her. At the Simmonds cottage she stopped and rapped on the door.
Usually one of the children rushed to the door and pulled it open. This morning, though, all she received was a muffled, "Come in, if ye please."
With a slight frown Maddie pulled the latch and pushed the door open—and stopped. With the exception of Mr. Simmonds, who was in Dorsetshire tending his sick mother, the entire Simmonds clan stood in a line along one wall of the cottage. The seven children, with Mrs. Simmonds in the middle, all bowed in ragged unison as the marquis stepped into the dark room beside her.
"My lord," they mumbled.
With renewed glee, Maddie set about introducing each of them to Lord Warefield as she set the basket on the small stone hearth and retrieved the emptied one from the week before. The children, especially the youngest of them, regarded the marquis with complete awe. He looked like a lean, tawny lion in a cage filled with squeaking mice as they leapt around him. Even after the performance was repeated twelve more times as they went from cottage to cottage, though, Warefield appeared only slightly embarrassed by the whole worshipful crowd, and not at all out of sorts.
"What shall I do with all the flowers they've given me?" he asked, as they returned to the curricle.
"I always make them up into a bouquet for your uncle's room," she said, dumping the empty baskets back into the carriage.
He examined the substantial handful of spring wildflowers for a moment, then met her eyes. "These are usually meant for you, then," he said, and held them out to her.
"Don't be silly," she said, embarrassed, and turned for the mercantile shop. No man had ever given her flowers before. Not even Charles. And she certainly had no intention of accepting a gift from the Marquis of Warefield. "We'd best put in an order for those planks."
"But I want you to have them," he insisted, not moving.
Maddie sighed heavily to cover her sudden discomfiture. Damn him for unsettling her so. Making a show of annoyance, she turned back around and took the bouquet from his fingers. "Thank you." She placed them in one of the baskets and faced him again. "May we go now?"
He smiled at her, though she couldn't see what in the world he was so pleased about. She'd only give them to Mr. Bancroft once they returned to Langley.
"By all means," he said, gesturing her to precede him. "Let's order those planks."
The next morning Malcolm had fresh roses in his bedchamber, which meant that Maddie either had kept the wildflowers for herself, or thrown them away. But she had accepted them from him, and without a negative comment.
Considering the opposition she'd been putting up, Quin felt a bit like Wellington at Waterloo. Perhaps the victory wasn't as definitive, nor as spectacular, but nevertheless he whistled as he rode out to the fields. She'd been in hiding all morning and had failed to appear for breakfast altogether. Most likely she was licking her wounds and readying for another attack—but Quin had been doing some battle planning of his own.
There was something about her—something he needed to pursue. He didn't quite know whether it was out of curiosity, or because he was, after all, a male and she was lovely. But the more skeptically she viewed him, the more determined he became to erase that look from her face.
He dismounted, leaving Aristotle to graze in the meadow while he headed for the group of farmers. Better yet, he wanted the opportunity to show Maddie Willits that not all of the nobility were as pompous as she apparently thought. And he knew just where he'd like to prove that to her, as well. In his bed, with her long auburn hair swept out across the white pillows, and....
"Look out, my lord!"
Quin blinked and stepped back just in time to avoid being run down by a very large plow horse. The nearest of the fanners eyed him, but immediately went back to clearing the field when he looked in their direction. He shook himself and bent down to clear a few of the last stones out of the plowed earth.
He was helping unload sacks of seed when he realized he'd forgotten about Eloise's correspondence. Rarely did he answer her letters on the same day he received them; he was often extremely busy, and besides, it seemed somewhat weak-kneed of him to do so. After all, he was not a dewy-eyed romantic, and he had had an understanding with Eloise since they were children. But forgetting completely was entirely uncharacteristic.
Remembering Maddie's harsh words about his concern over his uncle, Quin made a point of returning to the manor for luncheon with Malcolm. Again Maddie was nowhere in sight. "Where is Miss Willits this afternoon?" he asked offhandedly.
"Potting."
Quin looked up. "Beg pardon?"
"In the garden shed," Malcolm explained. "Maddie's roses have become quite popular in Somerset. They're in great demand in the spring, so she roots cuttings and sends them to the neighbors."
So she didn't hate everyone, then. Just marquises—or just the Marquis of Warefield. Quin pursed his lips, trying to decide how much he could ask Malcolm without giving away his own growing interest. "Malcolm, might I ask you a question?"
"Certainly."
"Why does Miss Willits seem to...dislike me so intensely? I haven't done anything to offend her, have I?"
Malcolm grinned. "You'll have to ask her. It's not for me to say."
Quin sighed and climbed to his feet. "You warned me to be careful with her. You might have warned me to bring a suit of armor along, as well."
His uncle only laughed.
Seeing Eloise's letter propped up on his dressing table reminded him once more that he hadn't written since he'd set out for Langley. With an impatient glance out the window toward the garden, he sat and pulled out a pen and some ink.
Dearest Eloise,
Uncle Malcolm is doing well. Unfortunately, it appears that I'll be staying here longer than we'd planned—in addition to the crops and accounts, a new irrigation system is needed at Langley. No real adventures to speak of....
Quin sat back. That last part wasn't exactly true, but he didn't wish to relate the near-drowning or near-shooting incidents, and he doubted Eloise would find his war with Miss Marguerite amusing. Nor was she likely to appreciate his odd battle of wits with Miss Willits.
He dipped the pen again.
...but Langley is rather rustic. I still plan to visit you at Stafford Green before the Season. Please give my regards to your father.
Yours,
Quinlan
It wasn't very long, but it would have to do for now. He'd give her more details in the next missive, when he knew how much longer he'd be staying. He sealed the letter, scrawled Eloise's address on the outside, and left it for Garrett to send out with the post.
He restlessly wandered about the house for a while, hoping Maddie would return from her seclusion before he had to return to the fields. Exasperated, he looked through the morning room window just in time to see her green skirt disappearing into the garden shed. Quin started outside, then stopped. If he appeared, she'd only accuse him of following her and neglecting his duty to Malcolm, and to Langley—and he'd damned well heard enough of that rubbish. So he'd have to be certain he wasn't neglecting anything.
Inspiration hit. "Aha," he muttered, grinning, and headed into the office at the far end of the hallway. Lifting the last ledger book out of its drawer, he flipped to the page where the handwriting changed.
With the book tucked under his arm, he marched into battle. "Miss Willits?" he called, making a show of looking about the grounds for her. "Miss Willits, are you here?"
For several moments she neglected to answer, but just as he was beginning to think he'd have to "accidentally" discover her in the potting shed, she emerged.
"Yes, my lord?" she said, brushing a stray lock of auburn hair back behind her ear.
"Ah, Miss Willits. I've a question for you."
Distrust entered her gray eyes as she watched him open the ledger. He stepped over next to her, holding the book so she could see it, too. She smelled of earth and lavender, and dirt smudged her fingers and one cheek. And the heat that began coursing along his veins had absolutely nothing to do with simple curiosity.
Attempting to return his attention to the accounts, Quin pointed at one of the last entries, dated only two days before. She'd been sneaking in and doing the accounts while he was out working—after he'd asked her to refrain from touching them. "What is this?"
She leaned a little closer to him to look at the page, then glanced up at his face. "How should I know, my lord?"
"Do you think me a complete idiot, Miss Willits?" He stifled a smile as she opened her mouth to respond. "No, don't answer that. Allow me to explain. Here," and he turned back several pages, "is my uncle's writing. The only Bancroft with worse writing is my brother." He pointed at another, much later entry. "This is my writing. Not much better, but at least you can tell the t's from the w's." He returned to the indicated page and its rows of neat entries. "And this writing, I believe, is yours."
She gave the page a sour look. "All right, my lord, I confess. I know how to do arithmetic." She pointed at one of the lines. "But if you'll note, my lord, nowhere did I allow my writing to touch yours."
Ignoring that, he lowered the book to look at her. "Why didn't you say you'd been tending Langley?"
"We wrote as much to your father." Maddie met his gaze. "You didn't seem to be interested."
Actually, he was more interested than he cared to admit. "Why don't we say that I was ignorant of the facts? You've done a great deal of good here, both for my uncle and for the Bancrofts."
"I do what I am employed to do," she answered shortly, and turned back to the shed. "Was that all you required, my lord?"
"I just wanted to thank you for tending my uncle."
She looked over her shoulder at him. "Someone needed to be here, my lord."
After yesterday, that same thought had occurred to him. If Maddie hadn't been Uncle Malcolm's companion, both the situation at Langley and Malcolm himself might have been in much worse condition than even the duke had surmised. He followed her, angry at the prick to his pride. "Ah. Now you are criticizing me for not coming to Langley sooner."
"It is not my place to criticize you for your shortcomings, my lord."
"Merely to point them out to me."
She curtsied. "As you wish, my lord."
That was too much. "Just a moment!" he growled, striding after her.
Maddie stopped and faced him. "Yes, my lord?" she asked, only the slightest hesitation showing in her eyes.
"Miss Willits, what in God's name...." He stopped, not because she lifted her chin at him defiantly, but because her hands shook before she moved them behind her back. "What," he began again, revising his choice of words, "have I done to distress you?"
She looked at him for a long moment. "Nothing, my lord," she said slowly. "And I wish to keep it that way."
Finally. "What if I were to assure you that I have nothing but the best intentions where you, my uncle, and Langley are concerned?" Quin put his hand over his heart. "I swear not to harm any of you. Including Miss Marguerite."
Maddie narrowed her eyes, her expression so clearly suspicious that Quin couldn't help smiling. She weighed what he'd said against whatever it was she thought she knew about him. For a moment he couldn't tell which side had won. Then she lifted her chin.
"If you had come here to help Mr. Bancroft," she said slowly, "I would have much less objection to you."
"I did come here to help him," Quin protested. "What do you think—"
She raised a dirt-smudged finger at him. "You came here to help your father. To see to the crops, and keep up the ledgers. And you didn't come for three weeks, and then arrived precisely on the fifteenth."
Quin took a deep breath. By God, this woman was infuriating. "I was at Warefield," he snapped. "I didn't know anything had happened to Malcolm until my father sent for me." He leaned toward her, closer to losing his temper than he'd been in years. "Why didn't you write sooner?" he accused hotly.
Maddie glared at him. "He wouldn't let me, at first. Mr. Bancroft doesn't like your side of the family." She folded her arms over her chest. "And now that I think about it, neither do I."
Quin leaned closer still, his face only inches from hers. And then his eyes focused on her lips. The anger in his veins became something else entirely—something equally hot and equally disturbing. "You don't know me," he murmured, meeting her gaze again.
She held his eyes, still fearless. "And you don't know me."
"I would like to," he said, in a low voice.
Her mouth opened, and then shut again. "You—I—" Maddie swallowed. "Bah," she finally snapped, and turned her back on him.
Quin watched her stalk back to the house. A slow smile touched his lips. This was becoming very interesting indeed.
## Chapter 6
"Wonderful," Maddie snarled, shutting her door with a thud and plunking herself down on her bed. "Wonderful. Now he really does want to seduce me."
She jumped to her feet again and strode to the window. He was still out there in the garden—she could see one leg and part of his arm if she craned her head against the cool glass. With all the insults she'd handed him, Lord Warefield still thought he could win her good favor simply by looking at her with those beguiling eyes and by talking nicely.
"Ha, ha, ha," she said to his elbow down below. "I've heard nice words before. Nicer than that." What she should have done—what she would have done, if he hadn't temporarily surprised her out of her wits—was tell him to go to Jericho and leave her in peace.
Maddie scowled. In point of fact, she'd had ample opportunity to tell him exactly that, and she hadn't done it. She'd gawked at him like some stupid, doe-eyed halfwit, and then even worse, had turned tail and run.
She banged her head against the glass. "Stupid, stupid girl," she muttered. "So what if he's handsome? So what if he can quote Shakespeare? I'm sure most everyone can. And so what if he doesn't mind chasing pigs through the mud and ruining a perfectly splendid set of clothes? And...."
Down in the garden, he turned around without warning and looked up at her.
"Drat!" Maddie ducked backward. She stayed hidden behind the curtains for a long moment, smacking her hands together in agitation. With the light outside, he might not have seen her gaping at him through the window. She counted to ten, took a breath, and then stepped forward again.
The marquis stood beneath her window, a white rose in one hand. With a smile, he held it up to her, then bowed with an absurdly grand flourish.
"Of all the nerve," she breathed. Her clever game had crumbled into a shambles, and he thought he had beaten her. The marquis had a surprise coming. She stuck out her tongue at him.
He blew her a kiss.
Her heart pounding, Maddie unlatched the window and shoved it open. "Don't you ruin my roses!" she shouted at him.
"'But soft,'" he called up to her with a deepening grin, "'what light through yonder window breaks? It is the east, and Juliet—'"
With a choked cry of outrage, she slammed the window shut again. No more speaking to him, or looking at him, or assisting him with anything. That would show him! She stomped to her door and threw it open. And she certainly wasn't going to the Fowlers' ball and dance with that arrogant, self-centered aristocrat.
She turned down the hallway and was nearly run over by Mr. Bancroft as he scooted around the corner in his wheeled chair. "Mr. Bancroft!" she said, trying to swallow her annoyance at his evil, seductive nephew. "How wonderful!"
"Perhaps this contraption isn't so bad, after all," he admitted. "Though I do seem to go in circles quite a bit."
"I told you that you'd like it." She cleared her throat. "Mr. Bancroft, I don't think I need to go—"
"And since I've been getting about so well," he interrupted happily, "I thought the Fowlers' ball on Friday would be a splendid opportunity for me to make my public debut."
Maddie closed her mouth again.
He reached out and took her hand. "Would you be my escort, Maddie?"
"I haven't anything to wear," she hedged, wondering if he could hear the reluctance in her voice.
"Nonsense. What about the gown you had made for the Dardinales' Christmas soiree? That was lovely."
"But it was for winter," she protested, "two years ago."
"No one here will notice."
Someone would notice, and she was unsettled to realize it was Lord Warefield's reaction she'd been thinking of. "I...suppose not."
"There you go. I'm not much for dancing at the moment, but I do quite a handy job at holding glasses of punch."
She laughed halfheartedly. "Yes, Mr. Bancroft. I would be delighted to accompany you."
But saying she would attend and actually preparing for the ball were two very different things. As soon as Warefield returned to his planting, she began rooting through her scanty wardrobe. When Maddie tried the gown on, it still fit, though the burgundy pelisse hardly seemed appropriate for a spring soiree, and the looser waist was hopelessly dated. And her shoes were the same black slippers she'd worn to every formal and semi-formal gathering since she'd come to Langley. Maddie frowned. She had three days. It was time to begin sewing.
"Miss Maddie?"
"Come in, Mrs. Hodges," she called, twisting in front of the mirror to eye her hemline skeptically. Using every ounce of skill and patience she possessed, she'd taken in the waist and adjusted the length a total of four times, and was still able to concede only that she looked passable. Barely.
"Oh, that's lovely," the housekeeper said approvingly, as she opened the door. "You'll have that Squire John mooning after you for certain."
Maddie smiled. "John Ramsey is a friend, Mrs. Hodges, nothing more."
"Hm. All the same, those Fowler girls'll be lucky to be noticed enough at their own party to keep from being bumped into." She laid a silver hair ribbon on the dressing table. "And I wager his lordship'll have no complaints, either."
A sudden nervous tremor shook Maddie's fingers, and she picked up the ribbon to cover it. Apparently she'd frightened him off, for he had barely spoken to her at all in the past three days. Of course, he would likely claim that he'd been occupied with the planting, but she knew better—the coward. "My thanks for the hair ribbon. I hadn't realized all of mine were so worn."
"That's because you never go into Harthgrove with us to look at the catalogs of London fashions."
"Who has time for London fashions so far from London?" Maddie returned, unable to keep an edge of disdain out of her voice.
"Well, at least I have a new hair ribbon to lend to other people who don't care about fashion," Mrs. Hodges sniffed with exaggerated haughtiness, turning for the door.
Maddie grinned. "Thank you very, very much for the loan, Mrs. Hodges."
The housekeeper returned to Maddie's side and kissed her on the cheek. "My pleasure, Miss Maddie."
Garrett clumsily rolled the wheeled chair down the stairs, while Bill Tomkins carried Mr. Bancroft to the ground floor. Maddie followed behind, still fiddling with her hair and muttering a prayer that an April snowstorm would sweep through Somerset before they reached the coach, and cause them to miss the soiree.
"You look ravishing, Maddie," Mr. Bancroft said appreciatively, smiling.
She looked back up the stairs, another nervous flutter running along her skin. "How long do you think we'll have to wait for Lord Warefield? If we arrive as fashionably late as they do in London, the ball will be over before we get there."
"I'm certain he'll be along before Michaelmas. And Maddie, try not to criticize his attire tonight, if you don't mind. You'll have the poor fellow weeping."
She chuckled reluctantly. "How else is he supposed to learn?"
Only a moment later, unhurried bootsteps sounded in the hall upstairs. Maddie leaned back against the wall and folded her arms. Affixing a look of bored disdain on her face, she kept her gaze on the grandfather clock. He might have caught her off guard once or twice, but the war was by no means over.
"Ah, good evening, Quinlan." Mr. Bancroft rolled to the foot of the stairs to greet his nephew.
"Good evening, Uncle. I trust I'm not too tardy?"
"Not at all."
Maddie waited for seven seconds to tick off the clock before she turned around. She'd intended to wait for ten, but she couldn't quite manage it. "Good evening, my lord," she said, curtseying. As she lifted her eyes to his, she nearly lost her balance.
"Miss Willits."
Quinlan stood looking down at her, a slight, amused expression on his lean, handsome face. She'd expected him to be wearing his black too-formal attire, but apparently he'd learned his lesson. Instead, a rich brown coat cut in a jaunty style cloaked his broad shoulders, while a cream-colored waistcoat and black breeches drew her helplessly attracted gaze once more to his well-muscled thighs.
For the first time in days, his Hessian boots were completely free of mud, and were shiny enough to reflect her own unwillingly mesmerized expression back at her. "My goodness, my lord. However will anyone be able to concentrate on dancing with such a splendid sight before them? I am quite ready to faint myself."
"Maddie," Mr. Bancroft warned.
"I think they'll be distracted enough by you to forget about me," the marquis said softly. "At least, all the men will be." He came forward and took her hand, lifting it to his lips. His gaze traveled down the length of her gown and back up again, pausing at her low-cut neckline. "You are lovely."
Maddie swallowed and swiftly retrieved her hand as a warm, pleasant flush crept from her toes all the way to her face. She took a quick breath, trying to gather her melting, scattering wits back into cohesion. "Not nearly as lovely as you, my lord."
Mr. Bancroft snorted. "Well, someone compliment me, so we can be on our way."
Immediately Maddie hurried over and kissed her employer on the cheek. "I can't tell you how happy I am to see you getting about so well." She smiled at him, taking his hand in hers. "Next month, you will be dancing."
When she straightened, Quinlan was looking from one to the other of them, his expression unreadable. "I don't think I can say it better than that," he murmured, his gaze stopping on Maddie. "Shall we be off?"
The wheeled chair was strapped to the back of Lord Warefield's coach, and the marquis lifted his uncle and placed him on the soft leather seat inside. As the coach started off, Maddie caught Quinlan looking at her from the opposite seat yet again, and far too smugly for her peace of mind.
"My lord," she began in her most deferential voice, "whyever didn't you wear this magnificent coat to dinner the other night? No one in the world could have found fault with such perfection."
The marquis glanced away for a moment, his expression distinctly uncomfortable. Barely able to keep from chortling gleefully, she leaned forward and gestured at his boots. "And your valet must truly be a marvel! It's taken him—what, eight days—to remove the last of the mud from those boots. How in the world did he manage it?"
"Aye," Mr. Bancroft agreed. "It would be a handy secret to have in Somerset, no doubt about that."
"Well...it will simply have to remain a secret," the marquis said rather brusquely, and looked out the window.
Maddie and Malcolm glanced at one another. "Excuse me, my lord," she said with carefully hidden amusement, "but do you mean your coat and boots are truly some sort of secret?"
He glared at her. "Yes. They are."
Before Maddie could pursue her interrogation any further, they arrived at the Fowlers' residence. Light shone from every window and from the lanterns scattered along the drive, which was already crowded with carriages and wagons. Apparently Mrs. Fowler had been as good as her word and had invited every landowner in the area to attend the marquis's grand unveiling.
A footman arrived to help Maddie to the ground and then assist the marquis with untying the chair. Once they had Mr. Bancroft settled, Maddie stepped behind the chair and took the handles.
"I'll do that, Miss Willits," the marquis said.
"Oh, I wouldn't hear of it, my lord!" she gasped with mock horror. "However will everyone be able to shake your hand if you are pushing Mr. Bancroft about?"
"Miss Willits, it is not seemly for you—"
"I know perfectly well what is seemly and what is not," she countered, unable to keep the abrupt anger out of her voice. "Far better—"
"Maddie," Mr. Bancroft said quietly.
She stopped. "Mr. Bancroft is my employer," she continued more evenly. "Allow me to do my duty by him."
His eyes studying hers, Quinlan slowly nodded and stepped back. "Of course."
The uneven drive didn't make things any easier, but stubborn determination could do wonders, and Maddie managed to maneuver the chair up to the stairs. Two footmen who'd obviously been primed regarding their duties then took over, lifting Mr. Bancroft and the contraption into the manor house, and then all the way upstairs to the second floor. The marquis fell into step beside her as she followed, and she could feel his gaze on her again.
"1 didn't intend to offend you, Maddie," he said.
"It would be an honor to be offended by such a gentleman as yourself," she replied coolly, hoping the trio of trumpeters she spied at the entrance to the ballroom was there to announce the marquis's arrival.
"Good God," he muttered as he, too, noticed them. "Did you have something to do with this, Miss Willits?"
She put a hand to her breast. "Me, my lord? I would never presume."
Quin's eyes followed the gesture, then returned to her face. He slowly reached out to straighten her sleeve, his fingers brushing her bare arm. "What a shame."
Maddie narrowed her eyes. "What—"
Before she could complete the sentence, the trilling fanfare began. Quinlan looked completely appalled, and Maddie was forced to clap a hand over her mouth to contain her amusement.
Apparently they'd been expected to arrive late. The entire assemblage, dressed in the finest attire Maddie had ever seen them wear, stood lining either side of the doorway. As their party entered, the guests, the footmen, and the musicians in the back of the room bowed almost in unison. Mrs. Fowler came forward, her arms outstretched in a gracious greeting, while her husband followed behind.
"My lord," she breathed, curtseying deeply. "You honor us again with your presence."
He smiled dazzlingly as he took her hand. "Thank you, Mrs. Fowler. I'm happy to be here."
"Please, my lord, allow me to introduce you."
With that the crowd swept forward, surrounding them and making a rather alarming racket. Maddie leaned forward over the back of Malcolm's chair as they waited, abandoned, in the entryway. "Would you care for some punch, Mr. Bancroft?"
"Thank you, my dear, yes, I would."
She wheeled him to the refreshment table. "You'd think they could at least be bothered to welcome you," she commented, glancing back at the multitude. "You've done more for them than Warefield would ever dream of."
"Perhaps, but Quinlan is a novelty. I'm merely an antique hereabouts."
"You're quite a bit more than that." She glanced at the marquis again to see him showing off his warm smile and jade eyes to anyone bold enough to speak to him. He played the role of gentleman marquis with absolute perfection. But no one was that nice—especially not a titled nobleman. "My, he does make a magnificent centerpiece, though, doesn't he?"
Mr. Bancroft accepted a glass of punch. "Are you certain, my dear, that this little game you've concocted is going as you think?"
She looked at him. "What game?"
"Come, Madeleine, we've known one another for four years. Do you think I can't tell that you're attempting to kill him with kindness?"
Maddie put a hand over her heart, aware that she'd already spent an inordinate amount of time this evening proclaiming her innocence. "I assure you, I have no idea—"
"Does he look like he's ready to be driven away, Maddie?" he said quietly.
She looked once more at the marquis. He was gazing over Mr. Fitzroy's head, directly at her. And then he grinned.
"Oh, damnation!" she hissed, turning away and feeling warmth creep up her cheeks again. And all he'd done this time was show her his teeth, for heaven's sake.
"Quinlan's used to getting his way, but he's no fool. What's he supposed to think, with you attacking him at every turn for no good reason?"
"I have a very good reason," she snapped. "And I'm certainly not trying to attract his interest."
"Perhaps you'd best tell him that."
She folded her arms indignantly, unable to slow the fast beating of her heart. "I'll be happy to."
The orchestra struck up a country dance, and Maddie jumped. The marquis was escorting Miss Fowler onto the narrow, polished dance floor, while her younger sister scowled and Jane Fowler simply glowed. Realizing how tensely she'd been holding herself, Maddie let her shoulders relax a little. Of course he wouldn't dance with her. There might be no other nobility about, but there were daughters of propertied gentlemen. She was only a companion.
"Maddie, may I have the honor?"
She looked up at Squire John Ramsey as he stopped before her. "Of course, John." Before she took his hand, though, she turned to Mr. Bancroft. "Do you wish me to stay?"
"Heavens, no. Go dance, girl."
Luckily the only open spot on the floor was halfway across the room from the marquis, so she wouldn't have to dance with him for more than a few seconds as they passed one another. She smiled at John, grateful that at least one person hadn't ignored her this evening.
"Lord Warefield seems to be enjoying himself," he said, as they stepped around one another.
"Even better, Langley will be adopting your watering system," she said. "Mr. Bancroft is quite pleased."
"I'm gratified," John admitted. "When Warefield asked me to meet him the other morning, I half thought he meant to tell me to mind my own business and let the Bancrofts take care of their own."
"You don't like the marquis, then?"
John grinned. "I don't know him well enough to say, either way."
"But you knew one another as children, didn't you?"
He shrugged and took her hand to step forward. "He visited Malcolm a few times during the summer, years ago. We played together, I suppose, though it mostly seemed to consist of him and his brother ambushing and sinking the toy boats I used to make."
She sniffed. "How typical."
"I haven't seen him since I was eight, Maddie. I doubt he stones frigates out on the Thames." He stopped speaking as they circled past Sally Fowler and James Preston, then took her hand again. "I take it you don't share the community's delight over our guest?"
"He's a bit stuffy for me."
"Well, he did come to help Malcolm."
Some help. "Yes, I suppose he did," she said reluctantly.
John moved past her, and Maddie wound around James Preston, Mr. Fowler, Mr. Dardinale, and then Lord Warefield. He kept hold of her fingers a moment longer than he should have. "You know John Ramsey quite well, don't you?" he murmured.
She looked straight at him and pulled her fingers free. "Yes."
It was completely ridiculous, but now that even Mr. Bancroft had noticed Quinlan's apparent interest, the marquis did sound almost jealous. Perhaps, though, he was only chastising her for dancing with one of her betters. That made more sense than anything else.
When he took Sally out for a quadrille, and James Preston danced with her, she decided she must have been right. She had only offended his overdeveloped sense of propriety, and he had been unable to resist pointing out her faux pas to her.
The butler announced that dinner was ready, and unmindful of any propriety at all, the entire female contingent present, minus one, herded around the marquis, undoubtedly hoping to be the one he chose to escort into the dining room. The chattering, giggling din was deafening. Lord Warefield didn't forget his own manners, though, and deftly he picked Mrs. Fowler out of the crowd, wrapped her arm around his, and led the way out of the ballroom.
With great ceremony Mr. Fowler pushed Mr. Bancroft's chair to the foot of the table. The marquis, of course, sat at the head. Maddie rolled her eyes and took the seat next to her employer. "I've lost my appetite, I think," she muttered.
He smiled, but didn't say anything. His face had become rather pale, only a few shades darker than his starched cravat. Immediately forgetting her annoyance, Maddie leaned close to him.
"Are you well?" she whispered.
"I'm fine," he returned. "Just a bit tired."
"You shouldn't have exerted yourself so soon. We'll go." She started to her feet, but he shook his head and put his hand over hers.
"No worries, my dear. I may fall asleep in my chair, but I shall survive the evening." He smiled. "I promise."
A warm hand slid down Maddie's shoulder to rest on her arm. "Uncle?"
Startled, Maddie looked up at Quinlan. He leaned over her shoulder, gazing at his uncle with the same concern in his eyes that she felt herself.
"You two are making me feel old," Malcolm grumbled. "Go back and sit down, boy, before you begin a riot."
Quinlan glanced down at her. "Keep an eye on him," he murmured.
She lifted her chin. "I always do."
He paused, his eyes holding hers. "I know."
He returned to his seat, and after innumerable toasts and speeches in his honor, the footmen finally brought out the food. Maddie did keep a close watch on Mr. Bancroft, but his appetite hadn't diminished, and she decided that he'd been telling the truth when he said he was only tired. All the same, she'd been so consumed with disdaining Warefield that she'd nearly forgotten her duties.
"Ladies and gentlemen?"
Quinlan stood at the head of the table, a glass of wine in his hand, and Maddie groaned. She'd already drunk a thousand toasts this evening, and now Warefield had to think of something clever to make everyone else look shabby.
"If I may," the marquis continued, as every eye looked at him, "I know there've been quite a few toasts already this evening, but I would feel remiss if I didn't add one more."
"Please do, my lord," Mrs. Fowler begged.
"This is a double toast, actually." He raised his glass. "To my uncle, Malcolm Bancroft, for his courage and strength and for his unflagging concern for the well-being of the people of Somerset."
"To Malcolm Bancroft," everyone echoed. For once Maddie was pleased to join in, and she smiled down at her employer.
"And to Madeleine Willits, for the great care she has taken with my uncle, and for her tolerance in putting up with a very annoying interloper at Langley." He grinned at her.
"To Maddie," came the second echo. Quinlan tipped his glass, his eyes still holding hers as he drank.
"Oh, dear," Maddie whispered, heat sliding along her veins.
Apparently Mr. Bancroft had been correct, after all. She'd never thought that the son of the Duke of Highbarrow Castle would take her antagonism to mean interest. If she'd encouraged him, it hadn't been done intentionally. She didn't think so, anyway—but from the way her body continued to react to his every look and expression, anything was possible.
Maddie looked at Quin, wondering when precisely she'd ceased hating him. And what precisely she was going to do about it now.
## Chapter 7
Quin couldn't keep his eyes off her.
Somerset featured a pleasant enough selection of eligible young ladies, he supposed; daughters of squires and knights and second sons of second sons of barons. Some of them wore the latest fashions of London and Paris and actually looked quite pretty in them. Bobbed and curled haircuts in the style encouraged by Beau Brummell's followers seemed to be the order of the day, even here.
And then there was Maddie Willits: long auburn hair with wispy tendrils escaping from silver ribbon and a dark burgundy dress easily two years out of fashion—yet which brought out the gray of her eyes. The elegant, practiced ease with which she danced made him yearn to take her in his arms. She fit in with these rustics as well as a plow horse would fit into Highbarrow Castle's stables. As well as he fit in at Langley. Or at least as well as she wanted him to think he fit in at Langley.
By the time dinner ended, he was becoming quite tired of everyone pointing out the graceful tilt of his hand as he brought a fork to his mouth, and the cultured turn of his wrist when he took a sip of wine. Being dissected in a physician school's anatomy class would have been less trying. At least he would have been dead, and wouldn't have had to listen to the ridiculous commentary.
After dinner he stood for a quadrille with Patricia Dardinale, mainly because Mrs. Fowler had been attempting to keep the two of them apart all evening. "You dance quite well," he said approvingly. The Fowler daughters had already assured him of bruised ankles by morning, and he'd had his toes stepped on twice.
Blue eyes beneath dark, curling lashes looked up at him, and she smiled. "Thank you, my lord. My governess came directly from London."
"You do her good credit."
He looked about for his reluctant house mate, and finally spied her in one of the other groups of dancers. Maddie hadn't lacked for a partner all evening and had always managed to be either in a different set, or at the far end of the line from him. The quadrille was her second dance with John Ramsey.
"How long do you plan to stay in Somerset, my lord?" Miss Dardinale asked, as he approached her again.
"I had planned to leave at the end of the week, but I may stay a bit longer, to see the new irrigation system finished."
"Oh, yes," she nodded. "Papa, Squire John, and Maddie have been trying to find a way to bring water to our east pasture for a year now. I think they've finally figured it out."
He glanced over at the blasted annoying female again. "Miss Willits seems quite adept at mathematics." Apparently she was the Leonardo da Vinci of Somerset.
"Mama tried to hire her away from Mr. Bancroft to be my governess," Patricia admitted. "She wouldn't go, but she has been coming over twice a week to teach me Latin." The alabaster brow wrinkled for a moment, then smoothed itself out again. "It's very difficult." She smiled. "I prefer French. It seems much more romantic, don't you think?"
Quin looked at her absently. "Yes, quite."
So his uncle's companion spoke French and wrote in Latin, knew Shakespeare well enough to quote the bard from memory, and could both keep estate account ledgers and engineer irrigation plans. "Do they ever play waltzes in Somerset?" he asked his partner.
"Oh, yes." She glanced about, then leaned a little closer as they linked elbows. "I doubt Mrs. Fowler will request one for tonight. Lydia's terrible at the waltz." She giggled.
He didn't plan to waltz with Lydia.
As soon as the quadrille ended, he strolled over to his hostess. "Mrs. Fowler, might I make a request of the orchestra?"
"Of course, my lord. They know all of the latest tunes and dances. We long to hear something that's popular right now in London."
"Splendid." He turned to face the dozen musicians. "Might we have a waltz?"
The violinist nodded. "Our pleasure, my lord. Any waltz in particular?"
"No. Anything at all." Quin turned around again as a score of females began heading in his direction. "Miss Willits?" he called, hoping she hadn't heard the request and bolted.
After a moment she came out from behind Uncle Malcolm's chair. "Yes, my lord?"
"You promised to show me a waltz, as I recall," he lied, not feeling the least bit guilty about it. "Will you do so now?"
She glared at him with thinly veiled annoyance, clearly realizing that if she argued, the crowd would turn against her faster than the villagers had against Frankenstein's monster. "Of course, my lord. It would be my greatest honor."
"Thank you."
He strolled up and took her hand. Behind her annoyance he sensed confusion and uncertainty, which was better than outright hostility. Beneath his thumb, the pulse at her wrist beat fast and hard, the one measure of her feelings she was unable to control. The music began, and he led her out to the middle of the floor.
"Shall we?" he murmured.
"I hate you," she whispered back, fitting her hand into his and allowing him to slide his arm about her waist.
He smiled. "And why is that?"
They glided into the waltz. As he had suspected, she danced superbly—which, added to her other accomplishments and abilities, made her the most talented, as well as most lovely, governess, companion, and mistress he could ever remember encountering. She glanced about the room, and following her gaze, he belatedly realized that they were the only couple on the dance floor. Undoubtedly the other guests had taken his request for a waltz to be a royal command. Well, that was perfectly fine with him.
"You shouldn't be dancing with me, my lord," she said, avoiding his gaze.
Quin wondered how far he could push her before she renewed her attempt to do him bodily harm. "I can dance with whomever I wish," he returned. "I'm the Marquis of Warefield."
She narrowed her eyes. "Not by any accomplishment of your own. Do you expect me to be impressed simply because you can afford to wear splendid clothes and drive fine carriages?"
He wished she would stop mentioning his damned suit. If she ever found out he'd actually sent to Warefield for it, he'd never hear the end of it. "No."
"And don't think I haven't heard that you sent all the way to Warefield for your magnificent attire. A four-day journey for a coat and a pair of boots."
Damnation. "If you weren't so hard to please, I wouldn't have had to do it," he countered.
"I am not hard to please. And you did it to please your own vanity."
"I did it out of a sense of self-preservation." Distracted and on the attack, Maddie seemed to forget just how many people were watching them. Shamelessly taking advantage, he pulled her lithe body closer. "So, Maddie, why do you hate me?"
She looked down at his cravat. "You needn't concern yourself with my feelings, my lord."
"So you keep repeating. But tell me, anyway."
"Because you are the Marquis of Warefield, I suppose," she said finally, in a voice so quiet and reluctant he could barely make it out, even with only inches between them.
"But you've already said I can't claim responsibility for the fact of my birth," he said softly. "If that's true, how can you blame me for it?"
He thought he'd cornered her, but she lifted her chin and met his gaze squarely. "Because I choose to."
"Now, that's hardly fair. I've been trying to play by your rules, but you keep changing them. Makes it rather difficult, you know."
Maddie hesitated. "Makes what rather difficult?"
Quin let his eyes drift to where they'd been wanting to go all evening. He focused on her soft, full lips. "Winning you over," he murmured. She jerked her hand, but he held her fingers and kept her close to him. "We couldn't possibly have met before, since you have only been employed as a governess, and I generally don't visit houses with young children. So is it my family?" He shook his head before she could answer. "No, because you work for my uncle."
"Don't let it disturb you, my lord. No doubt your mind is used to contemplating far loftier issues."
"Sweet Lucifer," he swore softly, wondering where he had gained such a great tolerance for insolence. He'd never had it before. "What do I have to do to earn a civil response from you?"
"I have been quite civil, I think."
Quin looked down at her, dancing calmly and gracefully in his arms while she flayed him alive with her tongue. And he wanted to kiss her only a little more than he wanted to wring her neck. "Miss Willits, I surrender. You are the victor. I am helpless before you. Have pity."
Her lips twitched. "No."
"How about a bargain?" he pursued. Behind him, at the edges of the dance floor, he could hear murmurs of conversation, but he dismissed them. Tonight, he was dancing with Maddie. And enjoying himself more than he could remember in a long time. "I will pretend I am not the Marquis of Warefield, and you will pretend you don't hate me."
"I don't..." She stopped. "Why do you insist on my liking you?" Maddie revised, her eyes meeting his.
"Because I like you, Maddie. My uncle regards you very highly. Your opinion is listened to and respected. And yet this beautiful, forthright woman," he continued, trying very hard not to kiss her right in the middle of the waltz, "apparently hates me. I just want to know what it is that I've done to you. Whatever it was, believe me, it was not intentional."
For a long moment she held his gaze. Finally she sighed, a little unsteadily. "All right. A truce. Until you leave. Not one second longer."
Ah, victory. Of a son, anyway. "So I no longer need to check my bed sheets every evening for thorns or poisonous spiders?"
Unexpectedly, she chuckled. "I hadn't thought of that."
He liked her laugh. "Thank God."
By the time the evening ended, Quin felt as tired as Malcolm looked. Maddie stayed quiet during the carriage ride back to Langley, and even when he intentionally left her several good openings for an insult, she didn't take the bait. Apparently she meant to honor the truce. He glanced at Malcolm. Honor had little to do with any of this: he was supposed to be aiding his uncle, and all he could think about was how he would go about maneuvering Maddie into his bed. It was pure madness—and he had never thought he'd enjoy madness quite so much.
It just didn't make sense that the one titled gentleman she'd spoken to in four years would be the one man able to look beyond his title, the one noble who could simply be...nice. Even so, Maddie was willing to concede that perhaps she'd been a little hard on Quinlan. If Charles Dunfrey had fallen into some stream, he probably would have drowned rather than surface to face ridicule.
Maddie paused in mid-snip and regarded the white rose before her. It had been a long time since she'd thought about Charles Dunfrey without either flinching or wanting to hit something badly. Good. As a fiancé he had been handsome enough, but he'd been severely lacking in the qualities of trust and loyalty. He'd also lacked the vision to see beyond the obvious, as had her own supposed friends and acquaintances. She'd assumed that every other person of his station was therefore the same. Apparently, she hadn't been entirely correct.
"Maddie?"
She jumped and turned around. Quinlan strolled through the garden toward her, his coat missing and his shirtsleeves rolled halfway up to his elbows. He looked like a Grecian statue of a mythical hero. "Yes, my lord?"
"Blasted hot this morning, isn't it?" he said, stopping before her. "I just returned from Harthgrove. It looks as though the last load of lumber may be in by this afternoon." He stepped closer, leaning down to lift one of the roses out of her arm basket. "Exquisite," he murmured, running a finger along the delicate edge of the white petals.
Maddie swallowed and continued choosing her bouquet. "That's good news. You'll be finished here by the end of the week." And that, to her surprise, didn't please her very much at all.
Quinlan grinned. "So now that you're bound to a truce, you only want me gone, hm?"
She met his gaze, hoping she looked more composed than she felt. But however amusing he might think himself, however handsome he might be, she'd agreed to the truce because what he'd said last evening had made sense. Not because he'd convinced her to surrender. Just one hint that he intended to act like—well, like a noble, and she would renew her attack. "I only wanted you gone before." She snipped another bloom.
"And you weren't exactly subtle about it." For a moment he was quiet, and then sun-warmed petals brushed against her cheek.
Unsettled, she stepped over to the next bush, crimson buds waving in the warm breeze. "Have you told Mr. Bancroft the news?"
He followed her. "Not yet."
"You should let him know. He's been anxious about it."
The rose and then his fingertips brushed across the back of her neck. "I will."
She shivered. "Stop it."
"How is it that a lady as lovely and intelligent as you is still unmarried?" he asked softly, ignoring her demand.
Maddie shut her eyes for a moment and tried to slow her breathing. "Because I choose to be," she lied.
"You know," he continued in the same quiet voice, "I think you never really disliked me at all." His fingers trailed down her arm to her wrist, and slowly he pulled her around to face him.
"Yes, I did."
Jade eyes caught and held hers. "Oh, I think you wanted to," he conceded, only his soft murmur separating his mouth from hers.
Quinlan was right. He was right, and this was wrong—and Maddie leaned up toward him and closed her eyes. His lips ever so gently touched hers. He tasted of tea and honey, and warm spring mornings and everything that had ever made her smile.
In helpless response she lifted her arms around his neck and pressed herself closer against him. Quinlan made a sound in his throat as he deepened the embrace of their mouths, and she trembled in answer. She hadn't been kissed in so long, and the last time....
"Quinlan!"
White-hot mortification shot through her at the sound of Mr. Bancroft's furious bellow. Gasping in horror, Maddie tore her mouth from the marquis's, and without looking at him or at her employer, she bolted around the back of the house.
"Oh, my God, oh, my God," she sobbed, holding her hands over her face and weeping as she slammed open the door to the servants' stairs and hurried up to her bedchamber.
She'd done it again. Even worse, this time she had known perfectly well what Quinlan's intention was, and she'd let him kiss her anyway. She'd even encouraged it! Everyone in London was right. She was stupid, fast, and loose.
Yelling began in the office downstairs, the words muffled, but the emotion behind them clear. First came Mr. Bancroft's low, angry rumble, and then Quinlan's sharper-voiced response. Maddie wiped her eyes and returned to the door. Everything had slipped out of control without anyone realizing it until it was too late. It had been an accident.
She took a deep breath and opened the door. An accident. Mr. Bancroft needed his nephew right now more than he needed her, and she would just explain that she'd been the stupid one and was completely at fault. She was ruined anyway, so it didn't really matter.
Quin paced angrily before the window of his uncle's office. "Look," he snapped, "I'll apologize for overstepping my bounds, if you want, but I won't have you bellowing at me as if I were some idiotic schoolboy!"
Malcolm kept the wheelchair moving to face his nephew. "I'll bellow at you in whatever manner I damned well please," he growled. "By God, Quinlan, I thought better of you than that!"
Attempting to rein in his temper, Quin took a deep breath. "It was just a bloody kiss," he grated, not mentioning that he'd been wanting to kiss her for days, or that he had hoped the kiss would be a prelude to something much more intimate. "And she didn't exactly try to rum me away."
"Quinlan—"
The marquis flung out his arm, furious and frustrated, half his thoughts still on how very good it had felt to have her in his arms, until his damned uncle had appeared and ruined everything. "You're no good to her now, anyway. Why not let someone else have a go at her?"
"What? You has—"
"Excuse me."
Quin whipped around to face the doorway. Maddie stood there, white-faced, tears trailing down her cheeks. He blanched, hoping she hadn't heard what he'd just said. God, he was an idiot. "Maddie, I didn't—"
"I just wanted to say that it was a misunderstanding and an accident," she said in a subdued voice, avoiding Quin's gaze. "Lord Warefield is not to blame. I'm sorry, Mr. Bancroft. You deserve better."
Malcolm, his face paling, wheeled forward. "Maddie, don't—"
She turned around and disappeared.
"Damnation! Now you've done it, boy!" The resemblance between Malcolm and the Duke of Highbarrow suddenly became more obvious.
"I have not done anything. It was a kiss, Uncle."
Malcolm glared at him for a long moment. "Close the door," he finally commanded, in a more controlled voice.
Quinlan complied, but refused to take the seat his uncle indicated. "Now what?" he demanded, crossing his arms over his chest.
"Just who do you think she is?"
"What do you mean, who do I think she—"
"You think she's my bedamned mistress, don't you, Quinlan?"
Quin narrowed his eyes. Something was going on. "Well, what else was I supposed to think? A beautiful, intelligent woman, out here in the middle of Somerset, tending...you?"
"Tending an old cripple, you mean?"
"No."
"Madeleine Willits is the oldest daughter of Viscount Halverston," Malcolm said, obviously reluctant to utter the words, "and she is not my mistress. Nor is she anyone else's."
Quin sat down. All the questions, all the intriguing hints he'd picked up about Maddie, and he'd never suspected she might be nobility. "What in Lucifer's name is she doing here with you?"
"She was engaged, five years ago. Apparently one of her betrothed's friends got drunk and kissed her, among other things. The wrong person saw it, and she was ruined."
"Over a...." Quin sat back. "Over a kiss," he said, half to himself. No wonder she'd looked so horrified.
"Yes. Maddie's a bit...spirited, and according to her, she left London and her family rather than listen to their stupid accusations when she hadn't done anything wrong."
Quin gazed at his uncle for a moment. "And so, five years later, she's become self-sufficient and found employment completely without references or assistance from her family or friends."
"Yes."
He shook his head. "Bloody remarkable."
Malcolm sighed. "She is a remarkable young woman."
"Why didn't you tell me?"
"It wasn't my story to tell. I thought I knew who she was, but it still took her three years to tell me. And I'm not titled. Thank God."
"So. What would you have me do, Uncle?"
The door opened. Maddie entered again, this time looking much more composed. And laden with two large valises.
Quin stood quickly, dismay tightening his chest. "Miss Willits."
"Excuse me again. I only wanted to say good-bye to Mr. Bancroft."
"I'd have you do what's right," Malcolm snapped, glaring at Quin.
"Do what's...." Quin closed his mouth, stunned out of any remaining composure. "You mean, marry her?"
"Absolutely not!" Maddie dumped her bags onto the floor, her face a mask of hurt and wounded fury. "Don't be ridiculous!"
"Now, Maddie, that's—"
"I'm already ruined, Mr. Bancroft," she interrupted hotly. "It doesn't matter."
"Then why are you leaving?" he barked at her.
She faltered, looking at her employer. Quin studied her face, fascinated at the play of emotions across her sensitive features. There was more to her than he'd begun to imagine. If not for Eloise—or his father—the idea of marrying Madeleine Willits wouldn't have been all that preposterous. Or, surprisingly, all that unwelcome.
"Maddie," he said softly, and her eyes darted in his direction, "it was my fault. Not yours." He hesitated, holding her gaze. "And I'm engaged already. Or just about. Otherwise...."
"I can take responsibility for my own stupidity, thank you very much," she said stiffly. "And you're a noble already. You don't need to pretend to be possessed of the quality."
Quin narrowed his eyes. Marriage to the spitfire might not have been preposterous, but it would have been dangerous. "I don't believe you have the right to question my nobil—"
"Please!" Malcolm bellowed.
Quin started and looked in his uncle's direction. He'd forgotten the older man's presence. From Maddie's reaction, she had as well.
"Thank you," Malcolm resumed, in a more even tone. "I am quite aware of your...arrangement with Eloise, Quinlan. I had something else in mind."
"Something else? What?" Maddie asked suspiciously.
"I've actually been considering this for several days now." Malcolm faced his nephew. "If you and the rest of the titled Bancrofts were to reintroduce Maddie to society, it could—"
"No!" Maddie gasped, paling.
"—It could undo the harm done to her reputation and enable her to secure a husband," he continued, undaunted. He looked over at her again. "It would set your life back the way it was before the scandal, my dear."
"Absolutely not!" she returned at high volume. "I am never going back to London. And certainly not with him!"
Quin smiled wryly. Apparently he'd broken the truce. "You liked me for a moment, I believe."
"You agree, then, Quinlan? Your ill behavior could turn this into something positive."
"It was my ill behavior, blast it!" Maddie argued. "Don't try to solve my problems. Please! Just let me leave in peace."
Quin frowned. His Grace would be beyond furious, but Malcolm was correct. Whatever Maddie might think, and whatever insanity had overcome him in the garden—and since he'd set eyes on her—he considered himself to be a man of honor. "I agree."
She turned on him. "It is not your decision."
He lifted an eyebrow. "I believe it is."
Maddie stomped her foot. "This is absurd! I am leaving!"
Quin strode forward and lifted her luggage before she could. "Yes, you are. I'll have to inform my father. We need to leave for Highbarrow Castle immediately." He turned to his uncle, plans and strategy already forming in his mind, and surprising elation running through him. Apparently, he and Maddie Willits weren't quite finished with one another yet, after all. "I'll go see John Ramsey and arrange to have him supervise the remainder of the irrigation work. The planting will be finished today."
Maddie grabbed for the bags, but he evaded her easily. "Give those to me at once!" she shouted.
"Maddie, listen to Quinlan. It's for the best."
"Do you always solve your problems by running away?" Quin said, taunting her. "I hadn't thought you a coward."
"I am not a coward!"
Malcolm lifted a hand to his forehead and sank back in his chair. Concerned, Quin dropped the valises and came forward. "Uncle?"
Maddie pushed him out of the way and knelt in front of her employer. "I'm sorry," she murmured, putting her hands on his knee and looking earnestly at Malcolm's face. "It's all right. Just take a deep breath."
"Stop arguing. Please," Malcolm muttered, rubbing at his temple.
"We have. Shh. You must be calm."
Maddie lowered her head, and Malcolm caught Quin's eye. Then he winked. Quin gaped at him for a moment, torn between astonishment and amusement at the old man's duplicity, and then he bent to take Maddie's shoulders. "We'll do as he says," he murmured. "It will be all right."
His uncle put his fingers under Maddie's chin so she had to look up at him. "Make me a promise, my dear. Do as Quinlan and his family say, just until you can be presented again at Almack's. If they accept you there, you will have no troubles anywhere in London."
"Mr. Bancroft," she pleaded, tears welling again in her gray eyes.
"After that, if you still don't wish to remain with your family and your friends, you may return to Langley."
She looked over her shoulder at Quin. Attempting to ignore the queer mix of anticipation and compassion she seemed to be stirring in him, he kept a solemn expression on his face and nodded. "I would like the chance to redeem myself. And to help you, if I may."
Maddie shut her eyes for a long moment. "All right. Just until Almack's."
"I just don't understand how you could simply hire someone from Harthgrove and expect them to be able to care for your uncle," Maddie snapped.
"I did not 'just hire' someone. Both Malcolm and your squire highly recommended him."
"John Ramsey is not my squire. And I don't care who recommended that man. It's my duty to care for Mr. Bancroft."
Maddie sat back in the carriage, attempting to ignore both the pretty wooded country outside and the handsome, annoying man seated opposite her. She should never have given in—and in any case, she should never have agreed to travel to Highbarrow Castle alone with him. Well, alone except for a second coach carrying their luggage, two drivers, his valet, and two footmen.
He'd called her a coward again, though, and then he'd flung her argument back in her face when she'd protested. If she was already ruined, what did it matter how she got to Highbarrow? Now, three days later, she could answer that it mattered a great deal, because she couldn't stop thinking about the stupid kiss, and how it had melted like fire along her veins.
"Miss Willits, for the eight thousand, nine hundred and thirty-second time, Uncle Malcolm will do quite well without you. He said so himself. Please, let it be. Whining about it will certainly not make me turn the coach around and take you back, or believe me, I would have done it already."
She folded her arms across her chest. "I am not whining."
He glanced out the window, the fourth time he'd done so over the last ten minutes, and then looked at her again. "You know, if I wasn't in dire fear of the consequences, I'd say I liked it better when you were fawning."
Maddie sniffed. "No doubt you did. I'm surprised you even noticed anything was out of the ordinary."
"You are hardly of the ordinary," he returned.
He'd been doing that to her for the past three days, giving her offhand compliments that could just as easily be taken as insults. He hadn't tried to kiss her again, and in fact had made it clear that he was doing what he saw as his duty, to compensate her for an unfortunate mistake. She tried to see it the same way, but dismissing the embrace—and her reaction to it—as a simple mistaken moment of madness took more effort than she expected.
When he glanced outside yet again, the butterflies which had begun dancing in her stomach turned into very large crows. Quinlan cleared his throat. "Well, have a look."
Taking a steadying breath, Maddie leaned forward. Immediately she saw why Highbarrow Castle was always referred to by its full title. She'd grown up at Halverston Hall, but it was nothing like this. Gray spires rose into the blue sky from an immense estate sprawled in the center of a vast clearing. A birch and oak forest bordered the grounds on three sides, with a glassy lake behind.
"It's...very nice," she offered, swallowing her sudden nervousness.
Unexpectedly, the marquis chuckled. "Don't let His Grace hear you say that. He wouldn't appreciate a four-hundred-and-thirty-eight-year-old symbol of Saxon resilience being called 'nice.'"
"Oh, I know," she said absently, continuing to gaze at the gray stones of Highbarrow Castle. It was beyond magnificent, by its very design meant to be overpowering and intimidating. But she did not intend to be intimidated. "Mr. Bancroft told me all about the duke."
He looked sideways at her. "Wonderful."
It took twenty more minutes for the coach to pass through the wooded glade, up the gradual slope, around the winding way, and across the moat bridge up to the front drive of Highbarrow Castle. Fleetingly she wondered what lay in the dark, still waters that flowed from the lake in a ring around the grounds.
Though she'd never seen him so much as ruffled before, she thought Quinlan seemed rather edgy. For once she couldn't blame him. She'd heard enough about the Duke of Highbarrow to know that he would not take the news of her arrival well. Not that she felt sorry for the marquis. She'd tried to leave, and he and Mr. Bancroft had insisted she stay. This was because of their stubbornness, not hers.
The coach rocked to a stop. Quinlan sat for a moment, looking at her. "I won't tell you to behave, because I know you wouldn't do as I tell you to save your life," he said.
"I might, to save my life," Maddie countered. "I'm not an idiot. But this has nothing to do with that."
"It has to do with your honor. Isn't that the same thing?"
She returned his curious gaze. "I used to think so."
The latch turned, and a meticulously dressed footman pulled open the carriage door. "Lord Warefield, welcome," he intoned, bowing.
The marquis gestured for her to precede him. "Thank you."
The footman helped Maddie to the ground, glancing at her curiously, and then moved back to let Quinlan step down on his own. She looked up. The huge dwelling looked even grander up close, with endless rows of windows gazing out imperiously over lush Suffolk County.
"Is His Grace home?"
"Yes, my lord. The duke and the duchess are taking tea in the south drawing room."
"Splendid."
The butler stood holding the front door and also greeted the marquis with the deference. Quinlan handed over Maddie's shawl and his hat and gloves, then took her by the elbow to lead her down a very long hallway with a high, arched ceiling. Portrait upon portrait lined one wall from floor to ceiling, the subjects ranging from men and women in current fashion to fierce-eyed, armor-clad Saxon chieftains.
Where Langley had been open and warm, Highbarrow seemed entirely designed to make Maddie more nervous than she already was. Servants appeared and disappeared through doorways, silent except for the quiet "Good day, my lord" s they murmured in Lord Warefield's direction.
At the end of the hallway, the marquis stopped. He released Maddie's elbow and looked down at her. "Do you wish me to explain you first, or would you rather accompany me into the lion's den?"
"Are you asking me for my sake, or for your own, my lord?" she returned coolly, somewhat bolstered by his tense demeanor.
He gave a brief grin. "You'd dance on my grave, too, wouldn't you?" He knocked on the door.
She shrugged. "If I ever visited it."
"Enter," a soft female voice called.
If she'd had any doubts about Mr. Bancroft's description of his older brother, they vanished as the marquis ushered her into the south drawing room. The Duke of Highbarrow Castle sat beneath the window, the afternoon sun silvering the gray in his dark hair. Cool brown eyes beneath straight black brows lifted from the London Times to rest on his son, and a moment later flicked over to assess her.
Maddie was suddenly acutely aware of the cheap fabric of her traveling gown, and of the thrice-mended yellow bonnet on her head. And she had no intention of letting any of them know it. With a slight lift of her chin she stopped beside Quinlan, turning her gaze to take in the rest of the room. Each piece of silver, from the candlesticks to the spoon sitting on the tea tray, shone brighter than starlight. Not one particle of dust showed on anything, and the polish of the mahogany furniture gave the smooth red wood an almost mirror-like appearance.
"Back so soon?" a low, cold voice rumbled. Maddie's eyes returned to the duke.
"Pleased to see you again as well, Father," Quinlan returned in the same cool tone. Maddie glanced at him curiously, because she never heard him sound so much, well, like a noble before.
"Welcome, Quin," a much warmer female voice said, and a small woman rose from one of the small chairs by the fireplace to grasp the marquis's hands. He smiled and kissed her on the cheek.
Silvery blond hair was coiled on top of her head, and her slender figure was draped in a beautiful muslin gown of green and white. The duchess's eyes were the same jade color as her son's, though their warmth cooled considerably as she turned to view Maddie.
"And who have we here?" she asked, only the slightest surprise entering her voice.
"Allow me to introduce Miss Willits. Maddie, the Duke and Duchess of Highbarrow."
"Your Graces," Maddie said, curtseying and bending her head, keenly aware of how very far she was from Somerset and any friends, or even acquaintances. She glanced at Quinlan. He was the closest to an ally she had here, and she could hardly rely on him. She wouldn't allow herself to rely on him, Maddie amended silently. No one but herself, ever again.
"Where did she come from?" The duke remained seated, and in fact, far from coming to his feet to greet either her or his son, crossed his ankles and flipped to the next page.
"Langley." Quinlan smiled at her, his eyes warning her to behave. "She was Uncle Malcolm's companion."
"I am Mr. Bancroft's companion," Maddie corrected politely, trying not to glare at the duke for his rudeness. After all, farfetched as the idea was, he might simply be shy. She'd give him the benefit of the doubt, for the sake of Mr. Bancroft. She'd given her word to make a go of this stupidity, and so she would—as long as the rest of the illustrious Bancrofts kept up their side of the agreement.
The duke snapped the paper and resumed reading. "Malcolm's whore, you mean."
Maddie flushed, while the marquis stirred beside her. "No, his companion," she corrected evenly, before Quinlan could do so. "And he's feeling much better. The doctor has even said he doesn't believe the paralysis will be permanent. Thank you for your concern."
The surprised expression the duchess wore deepened, and the London Times abruptly folded over and dropped to the floor. "Insolent thing, aren't you?"
The duke stood. Quinlan was the taller of the two, but Lewis Bancroft looked broader than his lean son. Maddie shifted a little closer to the marquis.
"What's she doing here, Quinlan?"
The marquis hesitated for just a moment, obviously searching for the least inflammatory words he could find. "She is the eldest daughter of Viscount Halverston. You may recall some—"
"You're the one Charles Dunfrey cast off when he found you lifting your heels for one of his friends." Highbarrow sneered. "Stupid mistake. And now you've had to settle for poor, crippled Malcolm." He glanced at Quinlan. "Or is it my son you've got your claws into now?"
"Father," Quin said sharply, his annoyed, wary expression deepening.
This was what Maddie was used to from the nobility. And somehow, it made her feel more comfortable to know that some of her memories and suppositions had been correct all along. She began to fume. "I do not, Your Grace."
"Actually, I am the one at fault," Quinlan offered in a calmer voice. "I had no idea who Miss Willits was, and I...." He looked at her again. "I behaved improperly toward her. Malcolm suggested I might be able to right two wrongs by bringing her here."
"She's not carrying your brat, is she? Good God, Quinlan! A shrew-mouthed whore, and three months before you're to marry Eloise."
"Lord Warefield kissed me," Maddie said sharply. "That is all. And I did not wish to come here. That was all his and Mr. Bancroft's idea. I would just as soon leave immediately."
"Good."
"Insufferable snob," Maddie muttered, and gathered her skirts. "Good day, Your Graces." Not even looking at the marquis, she turned and headed out the door and back down the long hall.
Quin strode after her and grabbed her arm to turn her about to face him. "For God's sake, give me a moment to explain," he whispered.
"In another few moments I will be forced to call your father out and shoot him," Maddie hissed back. "He's far worse than you are."
His lips twitched. Quin nodded, his fingers still hard and warm and tight around her arm. "Yes, he is. But I made a promise. Give me another damned minute, Maddie."
She jabbed a finger at him and wrenched her arm free. "One."
Quin took a deep breath and ushered her back inside. The duke had already seated himself again. The duchess, though, stood by the door and watched them approach. "Father," the marquis began over again, "I did wrong by Miss Willits. She is a properly bred young lady, falsely accused of wrongdoing. I wished to make amends. Uncle Malcolm thought that with the help of our family, she might be reintroduced into society. I agreed."
"Oh. You agreed."
"Yes. I did. And she stays, until we can all repair to London. I can't very well take her to Warefield without completing the damage, so she will stay here, as our guest."
The duke stood again. "You know, I might have expected this nonsense from Rafael, but until this morning I was not aware that you were an idiot. Pay her off and send her away." Highbarrow strode forward, stopping a few feet short of Maddie so that she had to look up at him. "What will it take, Miss Willits? Ten pounds? A hundred? Name the price it will take you to keep from wagging your tongue about my son's indiscretions, and be gone."
Maddie glanced over at the mantel clock. Fifty-eight seconds, fifty-nine, one minute was up. She'd kept her word. "Your Grace," she began, so angry her voice shook, "if I chose to wag my tongue about your son's misbehavior, every penny you own wouldn't be enough to keep me silent."
"Then wha—"
"I didn't come here for money," she interrupted. "I came here because Mr. Bancroft felt I had been wronged, and he took this silliness as a chance to set things right for me. I will tell you what I told him: I am perfectly happy with the way my life is. And I have no desire to spend another moment in your arrogant, self-centered, pompous company. Good day."
"Quin, did you agree to this?" the duchess asked, putting a hand on Maddie's arm before she could depart, ruining another chance at an effectively dramatic exit.
He nodded. "When I arrived at Langley, Miss Willits had done such a fine job of estate managing that I scarcely had anything to do. If nothing else, I intend to help her out of gratitude."
"No, you won't," the duke snarled, his face flushed. "I want her out of this house! Now!"
"Just a moment, Lewis," the duchess countered. For a second she looked at Maddie, then returned her gaze to her husband. "Quin gave his word. I won't have him break it because you feel inconvenienced."
"Victoria! I will not—"
"It's settled, Lewis," Lady Highbarrow said firmly.
Her fingers twitched on Maddie's arm, but she continued to eye the duke coolly.
Highbarrow clenched his fist as though he wanted to hit one of them. Abruptly he turned on his heel. "Bah. Do what you will. I'm going to London. You may join me when this...girl is gone."
"Lewis!"
"Enough, wife!" The duke's roaring retreated with him down the hallway and abruptly cut off as a door distantly slammed.
Quinlan looked after his vanished father. "That went well," he muttered.
"He was planning on leaving for London at the end of the week anyway. Some new trade agreement." The duchess removed her hand from Maddie's arm and placed it on her son's. "Excuse us for a moment, Miss Willits," she said, and walked with Quin to the door.
"Don't go anywhere," the marquis instructed Maddie, looking back at her.
Maddie tried to manufacture a scowl, but had to settle for nodding. He'd stood up for her when she hadn't needed or wanted him to—or so she thought, until he briefly grinned at her, and she suddenly felt as though everything would be all right. And Maddie wondered when, exactly, she'd begun to think of Quin Bancroft as an ally.
Upstairs the duke continued to bellow and slam things about as he stirred up the Highbarrow household in preparation for his departure. Quin listened, mostly to be certain His Grace didn't go back to the south drawing room and continue his argument with Maddie. She'd made his father genuinely angry, something people didn't dare do very often. And she'd done it in a rather spectacular manner. Quin couldn't recall anyone ever calling his father pompous before—certainly not to his face. Likely the duke couldn't recall it, either.
"I received a letter from Rafael a few days ago," the duchess said, stopping before one of the tall windows overlooking the gardens.
Quin leaned against the wall beside her. "And what's the scoundrel up to this time?"
"Apparently he's spent the past six months in Africa, as a special envoy for King George."
"Africa?" Quin repeated, surprised. "He's supposed to be guarding the Tower or something, isn't he?"
"I believe he volunteered for Africa, just as he did for Wellington's regiment at Waterloo. Anyway, he's been granted a leave, finally. He hopes to join us in London before the end of the Season, unless the local tribesmen begin their rebellion again."
"Do you think he's ready to sell out his commission?" Quin was very aware that his mother's attention was on the girl in the drawing room, but she'd get around to asking about Maddie when she was ready.
"Perhaps. He really didn't say." She smiled. "He did ask after Aristotle."
"Yes, well, I've a few things to tell him about the damned animal, too."
"Why did you kiss this Miss Willits?" She turned from the view to look at him. "And why did you feel the need to make amends for it?"
He shrugged. "I thought Malcolm might suffer another apoplexy if I didn't agree to do something. I practically expected him to horsewhip me."
"Does he care for her?"
"He's very fond of her. She's more like a daughter than anything else, though, I think."
"Why did you kiss her?"
Quin looked down for a moment, unable and unwilling to explain the turmoil of emotions Maddie Willits had awakened in him from the moment he'd first viewed her. "You know, I'm not really certain. She...she has a very bad opinion of the nobility, and she damned well says what's on her mind. And I suppose I wanted to prove to her afterward that not everyone is like this Charles Dunfrey who turned her away, or the bastard who ruined her before."
"Which I believe he did by kissing her when he had no business being anywhere near her."
"Yes, Mother, I see the parallel, thank you very much," he returned dryly.
She looked at him for a moment. "Did you argue with Eloise?"
"No, of course not. Why?"
"You haven't been known for dallying before, Quin. At least you've never felt the need to inform your father or me about it if you have." She turned back to the garden view. "Though I'd assume if you'd dallied, the ladies would have been of more solid standing in society than Miss Willits—or of no standing at all."
That had occurred to him as well. "I know." He started back for the drawing room. "At any rate, I gave my word. And as I said, I couldn't very well take her to Warefield and still expect her to have any chance of returning to society. So I thought you...might be persuaded to assist me."
"Your father doesn't want anything to do with it."
"Yes, but he won't be here." He strolled back to her side and took her hand in his. "Will you help me redeem myself, Mother?"
Her eyes twinkled. "She called Lewis arrogant and self-centered."
"And pompous."
She chuckled. "If you feel it's that important, of course I'll help." The duchess's smile faded. "But only to a point, Quin. I won't allow the Bancroft name to be sullied any more than your father would. If society continues to frown on her when she returns, she must be sent away. Agreed?"
He took a deep breath, far more relieved than he expected. He had Maddie for another few weeks, anyway. "Agreed."
## Chapter 8
"I do not need a dressmaker. I can sew my own clothes."
Maddie glared at Quinlan, who stood in the doorway of her borrowed bedchamber and glared back at her. The maid standing just inside the door looked as though she desperately wanted to flee, but she couldn't do so without going past the marquis.
"Making you presentable to London society is part of my promise to Uncle Malcolm," Quin returned sternly. "In two weeks' time you will not be able to sew enough clothes to last you more than a day or two in London. The dressmaker will be here this afternoon."
"No."
"Yes."
She wanted to throw something at him, but instead plunked herself down on the soft bed. "I refuse to end up owing you anything." She'd decided that from the beginning: being in his debt would be worse than being ruined in the first place. That's why she'd left home. She, and no one else, would be responsible for her welfare and her well-being. And if she was being stubborn and impractical, it was her right.
"Is that why you insisted on paying for you own room at the inns on the way here?" he asked. Far from becoming angry, he looked only curious. Apparently he was almost constantly curious about her, for he was always asking questions she'd rather not answer. And he had an oddly compelling way of making her want to answer at the same time.
After a moment he seemed to notice the maid, and absently gestured her to leave. The girl scurried out so fast, her apron might have been on fire.
"Yes. That is exactly why I paid my own way."
He strolled over to lean against the bed's tall footboard. "Maddie, I am extremely wealthy. When I become Duke of Highbarrow, I will be obscenely so. You can't match me."
"I'm very aware of that, my lord," she said stiffly. "You don't need to point it out."
Quin shook his head. "No, no, no. What I mean is, my uncle's idea was for you to be reintroduced in such a manner that no one could gainsay you. I can afford to do that without even noticing the loss. You've worked hard for what you've earned. Save it for something...for yourself."
She looked up at him, trying to summon the anger at him that had been absent since the night of the Fowler ball. Without the anger there, she kept noticing the slight, amused smiles that touched his mouth, and the lean line of his jaw, and the way the sunlight turned his honey-colored hair to gold. "If you cared about what I wanted," she answered finally, "you would never have dragged me away from Langley."
"Uncle Malcolm cares about what you want. And despite having nearly been drowned and shot in your presence, I do as well."
Maddie looked down at her hands. "I gave my word to go with you only for your uncle's sake. So please don't expect me to go to the gallows with a smile on my face."
To her surprise, he sat on the bed beside her. "The gallows? I can't say I've ever heard London referred to in quite that way before."
She smiled briefly, trying not to smell his light cologne or notice that a lock of hair had fallen across his forehead. "It was certainly the scene of my social execution."
"Don't you miss it, even a little?"
She shook her head vehemently. "No."
He fiddled with the edge of her skirt, the cheap muslin rustling against her legs and making her nerves tingle pleasantly. Good Lord, now she was thinking about kissing him again.
"But—"
"You have no idea what it's like, do you?" she interrupted, trying to rally her indignation again. "No one would dare cut you, whatever you did. Both you and your father are too wealthy and too powerful for anyone even to consider it. I'm only the daughter of a second generation viscount." She stopped, but he continued looking at her with his intense jade eyes, and she found herself continuing when she had meant not to.
"I was invited everywhere, especially once I became engaged. And after that...stupid, stupid night, not even my so-called friends would visit me, or even look at me. My parents locked me in my room for three days. I think they intended to send me to a convent. Ha! Can you imagine? Me, in a convent?"
"No, I can't." He lifted his hand and tucked a straying strand of hair behind her ear. "How did you get away?"
An unexpected shiver ran down Maddie's spine at his gentle touch. "I waited for bad weather, then packed a valise, threw it out my window, and climbed down the rose trellis. I walked to Charing Cross Road, and then took the stage to Brighton. I intended to set sail for America, but I didn't have enough money."
"By damn," he murmured, studying her face closely.
His scrutiny unsettled her, but he didn't seem to be laughing at her, so she shrugged. "So I hired on as a governess in Brighton. I lasted a fortnight, until the news over the scandal broke there, and my employer figured out who I must be. He gave me four shillings and set me out onto the street." She scowled. "After he offered to keep my tale quiet in exchange for certain...favors." Maddie flushed. Spenser pawing at her had been bad enough.
"Who was it?" Quin asked.
"It doesn't matter. They're all the same."
"No, we're not."
No, they didn't seem to be, and that was somehow hard to accept. "You kissed me," she pointed out, more to remind herself than him. "Was that just because you thought I was Mr. Bancroft's mistress, and of no account?"
He shot to his feet. "No! Absolutely not." Agitated, he strode to her window and then turned around again. "That kiss was...something else entirely."
"What, then?" She wanted to know. And not simply to confirm that it hadn't meant anything to him.
"A mistake. Of sorts."
She lowered her eyes, hurt. "Of what sort?"
"Of the sort that I really can't regret, but wouldn't dare to repeat."
"No?"
He held her gaze for a moment. "No," he said softly, then took a quick breath, as though he had only just realized they were alone in her bedchamber. "Mrs.... the dressmaker will be here at two. Don't—"
"You don't remember the poor woman's name?" she teased.
"Damnation. It rhymes with sunflower." His lips quirked.
"Hm. That's something, anyway."
"She's my mother's dressmaker. Not mine."
Before she could summon an insulting response to that, he was gone, whistling down the hallway. Maddie gazed after him for a long time.
She felt at a distinct disadvantage at Highbarrow. At Langley she'd been comfortable, on good terms with all the servants and the neighbors, and familiar with the routines and minute details.
Except for a few hours spent in the Marquis of Tewksbury's ballroom five years before, she'd never experienced such pomp and circumstance and wealth as she saw at Highbarrow Castle. It was unnerving—yet still nothing compared to what she would be going through the moment she set foot in London.
The duke thankfully departed with his two coaches and a retinue of servants before noon. She wouldn't have minded arguing with him some more, but as she was woefully short of allies, she didn't want to risk angering the duchess over something as foolish as His Grace.
Maddie took luncheon alone, sitting at a huge, polished oak dining table that could easily have seated the entire household staff at Langley. Quinlan had ridden off to visit some neighbors, and apparently the duchess, despite her earlier support, wasn't ready actually to socialize with the interloper.
Mrs. Neubauer arrived at two in the afternoon. The dressmaker was tall and thin, with an impossibly pointed chin that Maddie couldn't help staring at—especially after the woman spent a full minute walking around her, fingering her muslin gown and sniffing.
"No wonder the duchess wanted new clothes for you," she muttered, examining the hem of Maddie's sleeve. "Well below my standards, that's for certain. But then, my standards are why Her Grace sent for me."
"How fortunate for me." Maddie tried to decide whether she was annoyed or amused.
"Hm." Mrs. Neubauer finally stopped her circling and crossed her arms over her chest. "What am I to measure you for, then?"
Maddie folded her own arms, leaning decidedly toward annoyance. "I have no idea, I'm sure."
"Gowns, for morning, afternoon, and evening." The Duchess of Highbarrow glided into Maddie's bedchamber, one of her maids in tow. "Suitable for London society."
The maid pulled out the dressing table chair, and the duchess seated herself. Maddie looked at her for a moment, more uncomfortable than she had ever been in Quin's presence, then belatedly curtsied. "Your Grace."
"Quin says you have no manners. I see you do remember something of your upbringing."
Maddie clenched her jaw. "More than I care to, my lady," she answered as politely as she could.
The duchess looked at her for a moment, then sat back and waved her hand at the dressmaker. "Get on with it, Mrs. Neubauer."
"Of course, Your Grace."
After a thorough measuring session, Maddie had to stand and watch as Her Grace and the dressmaker decided on color and fabric and style. Neither of them asked her opinion, though they did spend some moments debating how best to showcase her bosom.
"I don't wish to be showcased," Maddie said stiffly. She'd been stared at enough the night of the disaster. Just the idea of going through something like that again left her feeling queasy. "And I won't wear blue, for heaven's sake. It makes me look tallow-faced."
The duchess glanced at her, then continued conversing with her dressmaker. "Substitute a gray and green silk for the blue. With gray slippers."
"Thank you, Your Grace," Maddie said, offering a slight smile.
"We certainly don't want you to look tallow-faced," the duchess said dryly.
Finally Mrs. Neubauer gathered her things together and left. Her Grace, though, remained seated in Maddie's chair.
"Do you have anything nicer than what you're wearing now, so you may dress for dinner?"
Again Maddie kept a rein on her flashing temper. If it had been Quinlan asking the question, she would have given him a sound set-down for it. But this haughty woman had stood up for her. She would take an insult or two in return. "A little nicer," she admitted. "We are—were—less formal at Langley."
"No doubt." The duchess stood with an elegant swirl of lavender. "We are more formal here. I expect you to comply with that." She headed out the door.
"If you didn't want me here, then why did you speak up for me?" Maddie said to her back.
The duchess stopped and turned around. "I spoke up for my son. We have all learned that the best way to maintain peace in the family is to concede to my husband's wishes. This time Quin chose not to do so." Lady Highbarrow spent another moment looking at Maddie, her expression the speculative one Maddie had seen Quinlan wear. "And I really can't think of a good reason why he should risk his father's temper over a foul-tempered flirt of inconsequential family." She shrugged and walked away down the hall. "We shall see."
Victoria Bancroft paused at the downstairs landing to listen. The girl's door had closed quietly, without any of the outburst or angry hysterics she'd half expected to hear. She waited a moment longer, then continued down to the first floor.
The whole affair was extremely odd. Quin chasing another female so close to his own engagement was not all that surprising, poorly as it must be regarded. Given his general levelheadedness and keen measure of common sense, his bringing that same woman to his parents' home and practically demanding that she be taken in and cared for was surprising in the extreme.
His Grace had, of course, chosen to view the entire incident as an affront to his dignity and stomped off to London, leaving her to sort out the absurd mess before the Bancrofts became the topic of the new Season's gossip. Just a whisper that the Marquis of Warefield had taken up with his estranged uncle's castoff would be enough to set the town ablaze.
Quin unexpectedly came in the front door as she started down the hallway. Since he rarely got the chance to amuse himself, when he went fishing with Jack Dunsmoore he always stayed out until well after nightfall. And it was barely past teatime. Victoria stopped and waited for him to catch up. "How was Lord Dunsmoore?"
"Quite well. I left him fishing. Not a damned thing biting this afternoon. How did the fitting go?" He slapped at the thin layer of dust covering his buckskin breeches.
"Miss Willits will have suitable attire beginning the day after tomorrow."
"She didn't try to throw anyone out a window, or go about stabbing old what's-her-name with pins?" The marquis chuckled.
Victoria stopped and faced her elder son. "Do you find it amusing that a supposedly well-bred young lady would throw a tantrum every few moments?"
Quin leaned back against the wall. "She's not some rabid wolf, Mother. She's merely been on her own for quite a—"
"She's merely been living off the good graces of your uncle, you mean," she interrupted.
Quin's smile faded. "I wasn't joking when I said she'd been tending to Langley, you know. And as well as any estate manager I've come across. Better than some."
"And?"
"And as for living off Malcolm's good graces, she purchased him a wheeled chair so he could begin to get about. I looked through the ledgers, and there was no notation of it. She finally admitted yesterday that the blunt had come out of her own salary—and she hadn't told Malcolm. She wanted it to be a gift."
"So she bought him a chair. It wasn't a diamond watch fob. You're being ridiculous, and it's not like you."
The marquis gave her another look and then straightened. "We didn't get him anything," he said quietly, and turned down the hallway. "Father even decided not to notice Malcolm was ill until there became a danger that Langley's crop wouldn't get put down in time."
"We sent you," she reminded him, but he'd already turned the corner. Victoria looked after him until the sound of his boots against the marble floor faded away, and then she continued on to the west drawing room. Two things were becoming clear. This kiss apparently hadn't been as much of an accident as Quin had claimed—and Lewis should never have sent him to Somerset in the first place.
"I thought perhaps you intended a hunger strike," Quin said mildly, watching Maddie take her seat at the dining table. He seated himself and gestured to the footmen to begin serving dinner.
"I thought perhaps I'd be dining alone again," she said demurely, folding her hands in her lap. "Is your mother going to join us, my lord?" She smiled at one of the servants as he offered her a selection from the platter of roast chicken.
"Uh-oh. What've I done this time?" Quin asked, noting that the pretty smile she gave the footman vanished as she met his gaze.
"Nothing, my lord. Why do you ask?"
"I'm being 'my lord'ed again. In your vocabulary I believe that to be an insult." He lifted an eyebrow. "Or do I err?"
"It is the proper way to address you, Quin," the duchess said from the doorway. "Don't fault her for it."
He stood again as Lady Highbarrow entered the dining room. Her customary place to the right of the duke's chair had been set, but disregarding that, she took the seat beside Maddie. As the head footman scurried to move her utensils, alarm bells began going off in Quin's head. Victoria Bancroft, though she was far more levelheaded than her husband, had as deep a sense of pride about the Bancroft line and standing as did the duke. Perhaps she wasn't as volatile as Maddie, but he'd seen her scar more than one upstart with her sharp tongue.
She'd made it fairly obvious that she had strong reservations about Maddie, and she rarely amended her opinion once it had been given. Quin was surprised to realize that he wanted his mother to like their reluctant houseguest—now eyeing him from beyond the duchess with accusing fury in her eyes—and that he didn't want to see Maddie turned away by his family, as she had been by her own.
"I'm not faulting Miss Willits for anything," he corrected innocently. "I am merely curious as to whether she is enjoying being at Highbarrow."
"How could I not enjoy it, my lord?" Maddie asked sweetly, her teeth clenched.
Quin stifled a grin as he locked eyes with her, abruptly deciding he'd best begin checking his bed sheets for poisonous spiders again. "My thought exactly."
The duchess leaned forward for her glass of wine, blocking Maddie from his view. Quin blinked.
"I meant to ask you, Quin," she said. "Do you have any word from Eloise?"
Eloise. Damnation, he'd forgotten to write her again. She thought him still at Langley. He shook his head. "I doubt her correspondence has had time to catch up to me," he hedged. "The last I heard from her, she was doing well and was looking forward to seeing you in London."
"Do you still plan to wait until autumn for the wedding? As I've said before, you'll have a much better turnout if you marry in June or July."
"Oh, I agree, Your Grace," Maddie said brightly, as Victoria sipped her wine.
The duchess lifted an eyebrow. "You do?"
"Most definitely. Once hunting season has begun, bringing everyone back together, even for such a prestigious event as the Marquis of Warefield's wedding, will be a monstrous headache."
Quin looked at her suspiciously. Maddie at her most solicitous was invariably Maddie at her most devious. "Why so helpful now?"
"Now, my lord?" she repeated, gazing at him quizzically. "Have I been unhelpful to you previously? I can't recall."
"No," he returned slowly, his deep suspicion growing. "I don't suppose you have been."
Lady Highbarrow continued to regard Maddie with cool green eyes. "You are in favor of this marriage, then?"
Maddie smiled engagingly. "I could hardly oppose it, even were it my place to do so. I barely know one of the participants, and I am not acquainted with the other at all."
The duchess looked at her for another moment. "Do you often kiss men with whom you are barely acquainted, then?"
Maddie compressed her lips, the only outward sign she gave of being angry. "I suppose that would depend on whose gossip you listen to, Your Grace."
"But you did kiss him," Lady Highbarrow pursued.
"It was a morning Byron would have admired," Quin interrupted. "Quite overwhelmingly romantic. And as it gives me the opportunity to repay Miss Willits for her kindness to Malcolm, I can't help but look upon the kiss as a fortunate...accident."
Behind his mother, Maddie glared at Quin. He returned her gaze coolly, wondering that this clever, witty woman had ever fooled him for even one second with her dim sycophantic veneer.
Maddie lifted her fork. "I have to admit," she said smoothly, "I have wondered why it is my poor character everyone is concerned with, when Lord Warefield keeps insisting he was the one at fault."
He rested his chin on his hand and regarded her. "Because you liked it?" he suggested.
Immediately he regretted the jibe, for, clearly embarrassed, Maddie paled and slammed her fork back down onto the table. "You big, arrogant—"
"Quin!" his mother snapped, even as he sought an apology. "Whatever feelings were involved, if you insist on reminding Miss Willits of her indiscretion, she will have no chance of redeeming her character."
"My indiscretion," Maddie repeated. "His lips, but my indiscretion." She looked at Quin. "I see now why you prize your nobility so highly. Apparently it automatically absolves you from any hint of wrongdoing, at the expense of the nearest social inferior." She stood. "Excuse me. I've lost my appetite."
"Maddie," Quin muttered, scowling.
Lady Highbarrow caught her hand before she could escape. "Miss Willits, at the risk of being blunt, Quin is the future Duke of Highbarrow. In comparison, you are a social inferior."
"My lady, I have never been more proud to be called so."
The duchess's patronizing smile froze in place.
Quin realized his jaw had dropped, and he snapped it shut before the surprised chuckle that began deep in his chest could make itself heard. He cleared his throat and shot to his feet.
"Mother, if you'll excuse me for just a moment," he said hurriedly, striding around the table to grab Maddie's hand away. "Apparently there are several things I did not make clear to our guest." He yanked Maddie toward the door. "Miss Willits, if you please," he continued sternly.
As soon as they were out of earshot, Maddie pulled her hand free from his. "I will not be dragged about like a mewling infant," she hissed, her gray eyes snapping with fury. "Next you'll be trying to take me over your knee!"
The image her words instantly conjured likely had little in common with what she was describing. This woman had just soundly insulted his mother, and he had no business daydreaming about having her seated naked on his lap, her long auburn hair tumbling down her shoulders past her bare breasts—
"Lord Warefield!" Maddie was growling, "I said, I am leaving!"
Quin grabbed hold of her arm again and spun her back around to face him. "No, you're not!" The sudden anger blazing through his veins surprised him—not because she hadn't said enough to make him angry, but because he absolutely did not want her to go.
"No one but you wants me here!" she snapped, coiling her delicate hand into a fist. "You pompous ass!"
He ducked backward as she swung at him. "Don't think I'm so refined that I wouldn't set you on your pretty ass if you hit me," he snarled, shaking her by the arm. "You're not making me break my word to Uncle Malcolm, and you are keeping your promise as well. Is that clear?"
For a long moment she glared at him, her bosom heaving with her fast, furious breathing. "I hate you, you bully," she muttered, wrenching her arm free again.
"Is that clear?" he repeated.
"Yes. Very clear."
Quin watched her stomp upstairs to her bedchamber. When her door finally slammed, he let out the breath he'd been holding and leaned back against the wall. Whatever it was he felt toward Madeleine Willits, it damned well wasn't hatred. And that scared him more than the blackest fit of anger ever could.
By the time Quin declared their absurd little group ready for London, Maddie possessed more gowns in more fabrics and colors than she'd ever owned in her life. She'd learned every waltz, country dance, and quadrille invented over the past five years, and been tested on all the ones in style before that time and since the beginning of history. Most painful of all, she'd been forced to read back issues of the London Times to re-familiarize herself with who had been married, buried, and welcomed into society's highest circles.
After her argument with the marquis, she had made every possible attempt to avoid him, and except for the annoying lessons and instructions, he had seemed to do the same. In a house as huge as Highbarrow Castle, it wasn't all that difficult. From time to time she actually felt bad about saying she hated him, but he'd deserved it, shaking her and ordering her about like that when she'd begun to think of him as an ally—and as a friend, if one could call a man one thought about kissing and touching and holding all the time as merely a friend.
She desperately wanted to avoid going to London, but she didn't want to prolong her stay at Highbarrow, either. She hadn't felt so trapped since her parents had locked her in her bedchamber five years ago, and she endured it only because she would be able to leave it all behind her again after Almack's.
They'd even hired a maid for her, and Maddie watched, her arms crossed, as poor Mary finished stuffing another portmanteau full of ballgowns. "We could always leave one behind by accident," she suggested with a smile.
Mary wiped her hand across her forehead. "It would be the one that Her Grace was especially counting on your wearing, Miss Maddie."
"No doubt. Are you certain you don't want my help?" At least Mary had a sense of humor, and she wondered fleetingly whether Quin had hired her, or whether he or the duchess had assigned the task of finding a maid to the head housekeeper.
"Oh, no, ma'am. It wouldn't be seemly, you know."
Maddie sighed. "Yes, I know."
A throat cleared from the open doorway. Immediately recognizing the sound, Maddie stiffened and turned around. "My lord," she acknowledged, echoing Mary's curtsey.
"Nearly packed?" Quin asked smoothly.
"Yes, thank you," she answered politely. He'd avoided speaking to her for almost a fortnight, so his seeking her out now couldn't bode well.
"Splendid. We're all set to leave in the morning, then."
"Splendid," she echoed unenthusiastically. He stayed in the doorway, and after a moment Maddie looked over at him again. "Was there something else, my lord?"
"Yes. Do you have a minute?"
Immediately Mary ducked her head and scurried for the doorway. Maddie put out a hand to stop the maid's retreat. "It's quite all right, Mary. My legs work as well as yours."
"Yes, Miss Maddie."
Quin straightened and opened his mouth. "Miss Wil—"
"My lord, shall we?" Maddie interrupted, and stepped past him into the hallway.
He followed her. "Why do you insist on the servants calling you Miss Maddie?"
She lifted her chin. "I don't. I ask them to call me Maddie, and then we compromise."
"It's not your proper address. You're a viscount's eldest daughter. Once we get to London, you will be addressed as Miss Willits."
Talking about her family still had the ability to upset Maddie. She shook her head and started back to her bedchamber. "You may wish to consult with my parents about that. I believe I may have been disowned."
He stood behind her, silent, for a long moment. "Maddie?"
She whirled back around. "Oh, my apologies, my lord. I'm supposed to face you when I speak to you. I'd forgotten." The words sounded brittle to her, but she tucked her arms behind her back defiantly, daring him to comment.
"Why didn't you tell me about your parents?" he asked instead.
"Does this change your mind? Should I leave now?"
He frowned. "No, of course not. It would have been helpful to know, though. I might have written Lord Halverston and—"
"No!" She strode back to him, dismay and dread tightening her throat. "You will not write my family about anything!"
"What do you suggest I do, then? We can't very well pretend you're someone else. You will be recognized, you know." He stepped closer, his jade eyes serious. "And it's you I promised to restore to society—not some mystery lady with no past."
Maddie turned away. "As I've told you all along, my lord, none of this is necessary. Nor is it going to be as simple and easy as you seem to think."
"Do you have any idea what I think, Maddie?"
She had no intention of being intimidated by his supreme kindness, or whatever it was he thought he'd bestowed upon her, and she looked up at him again. "I think that you kissed me to see what I would do, and once you discovered I wasn't a whore and wouldn't be your mistress, you were so embarrassed that you trapped yourself into going to ridiculous lengths to ease your own mind. Or do I err, my lord?"
Eyes glinting, he glared at her for a long moment. Slowly, though, and to her growing consternation, his expression eased. "Don't let your anger at a few idiots color the way you see the rest of the world, Miss Willits." He reached out and softly ran his ringer along her cheek. "Perhaps I kissed you because I was attracted to you. And perhaps you kissed me back because you were attracted to me."
Her pulse skittering at his caress, Maddie pulled away before he could realize he'd scored a hit. Only a conceited buffoon would throw her own unfortunate weakness back in her face. "You are in error, my lord," she said stiffly. "The only thing I've enjoyed where you are concerned was seeing you face down in the mud." Before he could reply, she hurried back to her bedchamber and slammed the door.
"That arrogant, pompous...." she muttered.
"Excuse me, Miss Maddie?" Mary straightened from stuffing a mountain of undergarments into a trunk.
"Oh, nothing." Scowling, Maddie sat at her dressing table and wrote Mr. Bancroft another very nice letter about how well everything was going and how well she and the titled Bancrofts were getting along, and how much she was looking forward to seeing London again. She wondered if he'd believe a word of it.
"Well, what does it say?"
Malcolm looked up from Maddie's letter. Chin in one hand, Squire John Ramsey sat glaring at him from the far side of their chess game. A leaf sailed down from the garden tree they sat beneath, and Malcolm brushed it off the board.
"Lewis—my brother—fled to London after five minutes with her, and she's apparently declared war on the rest of the family. I wouldn't be surprised if they had to clap her in irons to get her inside the coach to London."
"What's so amusing about that?" John challenged. "She must be absolutely miserable."
Malcolm couldn't explain that he was able to judge Maddie's spirits just from the mild tone of her letter. Langley seemed more quiet and calm than it had been for years, since the defiant beauty had first arrived and dared him to hire her. He missed her terribly, but from the moment he'd taken her up on the challenge, he'd known she wouldn't stay forever. He shook his head at his companion.
"Maddie's a fighter. She needs a challenge—something to push against. If my illustrious relations had greeted her with honey and cake, they'd never have been able to drag her to London, because she'd have them twisted around her little finger by now."
"As she does every male in Somerset," John sighed.
Malcolm looked at the letter once more, then set it aside to resume the game. "Yes, she does." And Quinlan had better be taking proper care of her, or there would be hell to pay.
"Maddie, please come down from the carriage," Quin pleaded soothingly, while he attempted to ignore the curious gawking of the butler and the scores of footmen needed to unload the Bancroft party's luggage.
"No," came her tense voice from inside the darkened coach.
"What nonsense." The Duchess of Highbarrow rolled her eyes, snapped her fan shut, and headed up the front steps into Bancroft House amid a sea of bowing servants.
Quin leaned against the open door of the coach. He should have ridden the last few miles with her—but then his mother or her maid would have had to join them, and he'd never have been able to talk to her. Not that they'd done much talking the last few times he'd made the attempt. Whenever he saw her, he immediately became seized with the desire either to bellow at her or kiss her. It had become quite irritating.
"Maddie, Bancroft House is surrounded by a very healthy border of oak trees, with a hedge of blooming pink rhododendrums beneath. In addition to its being quite picturesque, I assure you that the drive cannot be seen from the street."
"I want to go home," she stated.
The loneliness in her voice made him pause. "And where would that be, precisely?" he prompted quietly.
Given her keen sense of the practical, he thought that would get her attention. And indeed, a moment later her hand emerged from the dark. Swiftly he clasped it in his own. She was shaking, and he realized how unnerved she must be by the whole experience. Even before they'd entered the suburbs of London, she'd pulled the curtains shut in the carriage's small windows. From Aristotle's back he'd tried repeatedly to lure her to peek outside, but she wouldn't even answer him.
Slowly he drew her out of the coach. Her eyes were shut tight, and she stopped when her feet touched the drive. "Eventually you will run into something that way," he murmured, amused and sympathetic at the same time.
"I know," she said through clenched teeth. "Just give me a moment."
"Take several."
She continued to clasp his fingers tightly. Apparently she detested the rest of London more than she disliked him. Quin hadn't anticipated being elevated from enemy to ally, but the circumstance wasn't unwelcome. He gazed at her wan face. Good God, she was beautiful.
Finally, with a slow, deep breath, she opened her vulnerable gray eyes. She took in the huge house, the drive, the scattering of curious servants, and then Quin. "It's lovely," she said woodenly.
"Hm. I'll consider that high praise, coming from you. Shall we?" He gestured toward the open front door.
Maddie didn't budge, or loosen her grip on his hand. "Will you be staying here?"
Quin hadn't intended to. During the Season he typically stayed at Whiting House on Grosvenor Street, which had at one time belonged to his grandmother's family. Spending the entire summer at Bancroft House—with his parents—was a torture he hadn't had to endure since he turned eighteen and was admitted to Oxford. "Of course I'm staying here. Until you're settled, anyway."
The poisonous look Maddie shot at him was easy to read—she would never be settled in London.
"I am to be married sometime this summer, you know," he said in answer. "I can't very well have Eloise living here as well."
"Then perhaps you shouldn't have found me attractive," she said smoothly, a hint of color returning to her cheeks. "Though I suppose it's not uncommon for someone of your rank to promise yourself to someone and then throw yourself at someone else."
Apparently she'd recovered from her fit of nerves. "I did not throw myself at you. I believe it was a mutual collision."
A swiftly stifled grin touched her lips. "Don't flatter yourself," she said haughtily, as she freed her fingers from his and flounced past Beeks, the butler, and into the house.
"How can I possibly flatter myself, with you about?" Quin muttered at her back, before he followed her inside.
## Chapter 9
On her first and only stay in London, Maddie had been ecstatic. She had finally been able to see the famous places like Hyde Park, Bond Street, and the dark Tower of London—places she'd only heard about. Fabulous balls had been full of exciting, famous people who had treated her as an equal and claimed to be pleased to meet her.
And she had no desire to see any of those places or any of those people ever again.
"Miss Maddie, do you wish to change for luncheon?"
Maddie let the bedchamber curtains slide shut through her fingers, closing her off from the quiet view of elegant King Street. "I suppose I should."
She was still unused to having someone to help her dress and do her hair, but neither did she want to refuse Mary's help and cause the poor girl to be let go—no doubt exactly what Lord Warefield had anticipated. When she'd donned her new green and yellow silk gown, she glanced at the mantel clock and then reluctantly emerged from her bedchamber. Half a dozen servants nodded politely at her as she made her way downstairs to the dining room—where she stopped short.
The Duke of Highbarrow looked up from slicing a peach. "You're still about?" he asked gruffly, and returned to his luncheon.
"Good afternoon, Your Grace."
A footman hurried forward to pull out a chair, and rather than have the duke think her a coward, Maddie sat. His Grace rudely continued to ignore her, and she glanced about the room impatiently. Quin had said repeatedly that while in town the Bancrofts sat for luncheon precisely at one. So here she was, precisely at five minutes after one, when everyone else should have been there.
Another footman offered her a platter of fresh fruit, and with a grateful smile she selected a peach. Like everything else she'd seen in the house, it was perfect, round and golden. Maddie narrowed her eyes, imagining Quin's perfect smile and his handsome and very late backside, and sliced the fruit in two.
She glanced sideways at Lewis Bancroft again. Now that he wasn't bellowing at her and insulting her, she noticed that he was more heavyset than Malcolm, and that his dark brown hair was more generously tinged with silver about the temples. His complexion was ruddier, though Mr. Bancroft had been so pale over the past few weeks she'd been at Langley that she tended to think of that pallor as his natural coloring. And though she admitted that she might be prejudiced, she thought the duke's expression much less kind than Malcolm's.
"What are you staring at, girl?"
Maddie blinked. "I was looking for the resemblance between you and your brother, Your Grace."
"Bah. Malcolm's fortunate I still claim him as kin."
"Perhaps, Your Grace, it is you who is fortun—"
Quin skidded into the doorway. "Good afternoon, Father, Miss Willits," he said hurriedly, straightening his cravat and taking the seat opposite Maddie. "Apologies. I was catching up on some correspondence and lost track of the time."
The duke pinned him with annoyed brown eyes. "You're staying here now, as well? What in damnation's wrong with Whiting House?"
The marquis motioned for a cup of tea. "Nothing at all. I've merely decided to stay here for a few weeks."
His Grace lowered his brow. "Why?"
"He's keeping his word, as he was raised to do." The duchess glided into the room to sit opposite her husband. "He can't very well have Miss Willits at Whiting House. She needs a chaperon. And that would be me."
"Absolute nonsense. She's ruined already."
Well, that was enough of that. "I did not—"
"Perhaps so," the marquis said mildly, glancing warningly at Maddie, "but I will proceed, with or without your assistance."
"Without, I assure you." The duke pushed away from the table and stood. "At the first sign of trouble, it will be without your mother's assistance as well. And I don't want the girl getting in my way. With you here, Quinlan, it's too damned crowded already."
You could fit the entire Fifth Regiment in this house and still have room for a cannon. Seething, Maddie smiled brightly. "I will avoid you at every opportunity, Your Grace. You may be assured of that."
Lord Highbarrow paused on his way out the door. "Absolute nonsense," he repeated, and continued on his way.
"Please try to avoid antagonizing him," Quin asked, looking at Maddie.
"He antagonizes me," she protested.
"Still, it would be much easier if he was on our side, don't you think?"
"Why should he be, my lord? There is nothing in this to benefit him. Restoring me to society gains him absolutely nothing."
"Let's not begin this argument again, Maddie," Quin grumbled.
"I agree," the duchess contributed unexpectedly. "Lewis is not known for his patience. We must begin plans for your return to society without delay." She drummed her fingers on the table. "Nothing formal to start with, of course," she mused, eyeing Maddie with an uncomfortable intensity. "You should first be seen with me, so your coincidental connection to Quin doesn't become the gossips' primary focus."
"I have no connection with Quin—Lord Warefield," Maddie countered, the color rising in her cheeks.
"Shopping, I think," the duchess continued, as though she hadn't heard Maddie's protest. "Very good. Yes. Bond Street, tomorrow morning."
"But...I don't need anything." A wave of anguished nervousness suddenly made her fingers shake. People would see her. People who knew her.
"'Need' is not the reason one goes shopping on Bond Street. Being seen is the point of shopping on Bond Street. And so you shall be."
"But—"
"My mother is correct," Quin cut in. "This has to begin somewhere." He reached for a slice of fresh bread. "Besides, everyone will be addressing their conversation to Her Grace, with this being her first time out in public since her return to London. I doubt you'll have to say a word." He glanced up at her, his green eyes dancing. "That may be the most difficult thing for you."
"Oh, ha, ha," Maddie smirked, nevertheless bolstered by his comments. He was undoubtedly correct, after all, for he knew much more about snobbery and etiquette than she ever cared to learn. "And what important task will be occupying your day then, my lord?"
"I have to see that Whiting House is opened."
"My," she said, opening her eyes wide in awe. "Really?"
He sternly pointed a finger at her. "Yes, it means I shall stand about all day ordering servants hither and thither. No doubt I shall be quite exhausted by evening."
Lady Highbarrow cleared her throat delicately. "Not so exhausted that you can't attend dinner with us at Lady Finch's, I hope."
The duchess eyed her son, then flicked her gaze back over to Maddie, who quickly wiped the look of horror from her face. "Oh, my. Dinner?"
"I wrote Evelyn last week and asked her to put together an intimate gathering for a few select friends."
"Well, thank you, Mother," Quin said, his tone surprised. A moment later he kicked Maddie under the table.
She jumped. "Yes, thank you, Your Grace," she echoed, and kicked him back.
The Duke of Highbarrow skipped dinner that night, and instead went off to White's to smoke cigars and play cards. Actually, he could probably stand to miss a few dinners, for recently he'd become rather gouty, which left him even more ill-tempered than usual.
Quin would have liked to visit one of his clubs as well, but Maddie, of course, couldn't go anywhere, which also trapped his mother inside. So instead, the three of them played whist for several hours. Maddie had a natural cutthroat instinct for cards, which wasn't surprising in the least. The real surprise of the evening was the duchess smiling—not once, but twice—at things Maddie said.
In the morning, Quin rode off to Whiting House as he'd said he would. Once there, he instructed Baker, his butler, to open the house, with the explanation that he would be staying there from time to time, and would undoubtedly make use of it later in the Season. That accomplished, he swung back up on Aristotle and turned east for Bond Street.
His mother had been correct; if Maddie's purportedly wronged and injured character was to be redeemed, it would never do for him—or any man—to be seen with her on her first day back among the London ton. But nothing said he couldn't hang about in the shadows and make certain everything went well.
He left Aristotle and ten pence in the care of a young street urchin, and strolled up the crowded street in search of a new walking cane, which seemed the most logical thing for him to be looking for. It took nearly half an hour of aimless wandering before he spied the two women. The duchess emerged from a store, followed in succession by Maddie, four clerks carrying boxes, and Lady DeReese and Mrs. Oster. Quin dodged behind a parked barouche and peered over the top at them.
As he'd suspected, Maddie seemed to be of little interest to the two ladies in comparison with the esteemed Duchess of Highbarrow. Miss Willits stood a little to one side, clearly trying to look interested in the conversation, though just as clearly not. With her auburn hair glinting red in the sunlight, and her yellow silk gown showing off her lithe, slender figure, she was easily the most attractive lady on the street, if not in all of London.
He couldn't believe her parents would attempt to lock such a sprite away, much less consider sending her to a convent! What a wasted life that would have been for such a vibrant creature. Nor did he think she could have been completely happy at Langley. As much as Uncle Malcolm thought of her and she of him, he would never have been able to convince her to leave if she hadn't somehow truly wished it.
"Warefield!"
Quin started and looked up the street. "Danson," he replied, nodding. "Didn't know you were in London, yet."
"Yes, well my creditors think I'm still in Cornwall," Thomas Danson answered, clasping Quin's hand. "It's a bit early in the Season for you as well, isn't it?"
Quin shrugged, hoping the ladies across the busy lane hadn't noticed him or heard the conversation. "A bit. Had some business to attend to."
Danson turned away to toy with his dark hair in the reflection of the bakery window. "I say, why don't you buy me some luncheon at the Navy Club?"
With difficulty Quin kept from looking in Maddie's direction. "Why not?" he said, hoping his reluctance didn't show in his voice. He linked his arm through Thomas's, keeping his companion between him and the ladies. He was far too old to be acting like a schoolboy, and far too close to being engaged to be mooning after Maddie Willits. That was what had gotten him into such trouble in the first place.
"When is Eloise due back in town?" Danson asked.
Quin risked one glance at Maddie, to find her staring at him, barely contained fury in her gray eyes. Quickly he looked away again, cursing to himself. "Tomorrow, I should think," he answered, wondering how in the world he was going to explain himself this time.
Maddie would accuse him of spying, and she would be right. And he couldn't very well say that he'd only been gawking at her, not spying, because then she would accuse him of finding her attractive again. Next she would throw his near-engagement to Eloise back in his face as a coup de grâce. "Eloise said her father intended to be here by the twelfth."
"Have you made your declaration official yet?" Danson chuckled. "No, I don't suppose you have. Hasn't been a full-page announcement in the London Times yet."
"Please, nothing so tasteless as that. Half a page, at most."
Belatedly it occurred to him that he owed someone else an explanation for his rather ramshackle behavior. Eloise Stokesley probably wouldn't mind a stray kiss with a pretty, ruined girl in Somerset. Not nearly so much, anyway, as she would mind whatever madness had prompted him to take it upon himself to reintroduce Madeleine Willits into society. And he wasn't certain he could explain it adequately anyway.
Quin sighed. As much damned trouble as Maddie was causing him, he supposed he shouldn't be enjoying the entire debacle nearly as much as he was.
Maddie stood as close to the Duchess of Highbarrow as she could without tromping on the older woman's gown, and looked about Lady Finch's drawing room. Her Grace had declared their shopping expedition a success, leaving Maddie no excuses or protests to avoid the evening's dinner soiree.
Actually, walking about and shopping had been easy compared with tonight. She'd never been privileged to travel in such high circles. Tonight she would be expected to behave like a meek, demure young lady who would never have done such a thing as allow a gentleman to kiss her or fondle her front in public. And yet, she'd made that same mistake twice, now. Maddie smiled politely as Lady Finch nodded at her.
The first time truly hadn't been her fault, for she'd been caught by surprise by that snake Spenser. Slimy, wet, and cold—her first thought, before she realized she'd just been ruined, had been that she'd rather be kissed by a fish.
It was the second kiss that was becoming more troublesome by the moment, particularly as Quin refused to abandon her. She'd known that morning exactly what was going to happen—blast it all, she'd even encouraged it, wanted it, and relished it.
Maddie glanced across the room. Tall and handsome, Quinlan stood chatting with a few friends. He looked completely at ease, completely in his element, charming and witty and not a bit self-conscious. She hadn't been so close to hating him since he'd first appeared at Langley.
All afternoon she'd been looking for a chance to confront him on following her about like a spy. But all afternoon, obviously knowing he was in for a severe tongue lashing, he'd avoided her. And she couldn't very well have brought it up during the coach ride to Lady Finch's—not with the duchess there, reciting what she should and shouldn't do during the soiree.
Staring at Quin was one of the things she'd been directly told to avoid, but she was completely unable to resist several scowls and a glower. Shouting and hitting would have been much more satisfying. Almost as satisfying as kissing him again.
"Come, my dear," the duchess said, in the warm tone she seemed able to adopt the instant they were in public. "I wish to sit by the fire, where it's warmer."
She offered her arm, and Maddie hurriedly took it. "Of course, Your Grace."
Even before they'd settled themselves in, they were surrounded by a dozen other ladies, all inquiring after Lady Highbarrow's health, and that of her husband and two sons. No one inquired about Malcolm Bancroft's health, or about Maddie's, but at least it meant she didn't have to answer any questions. Instead, Maddie smiled and nodded agreement to the duchess's conversation at all the proper times, and offered no independent opinion or commentary—much as some of the ladies' silly gossip deserved it.
The duchess had warned her that while everyone would be looking at her and judging her comportment, no one was likely to speak to her at her first formal gathering. According to Her Grace, no one would approach her until she'd been deemed harmless. Maddie thought it more likely that no one wanted to be the first to acknowledge her.
"Mother, Miss Willits, may I fetch you a glass of Madeira?" Quin asked, stopping before them.
"Yes, thank you," the duchess answered, and nudged Maddie in the ribs.
"If you please, my lord," she blurted, glancing up at him and then away.
He vanished, then reappeared shortly with their drinks. As he handed Maddie hers, he leaned closer. "How are you doing?"
"I very much wish to spit at you, but I'm attempting to behave," she whispered back. "Go away."
He bowed, a smile tugging at his lips. "Yes, my lady."
"Quin, go away," the duchess repeated, glaring at him imperiously.
"I am, I am," he chuckled and wandered off.
"Miss Willits?"
Startled, Maddie looked up at the small, white-haired lady standing beside the fireplace. "Yes?" she said hesitantly, uncertain whether she should be ready to fight or to flee.
"Anne," the duchess said warmly, as she turned to look as well. "I didn't expect you in London yet."
"Neither did I." The lady smiled. "Ashton insisted."
"Anne, may I present Miss Willits? Maddie, Lady Ashton."
Then Maddie remembered her. "You were at the Tewksbury ball," she stated.
"Yes, I—"
"You called Spenser a drunken lout."
Lady Ashton nodded. "I should have said it louder. Would you and Her Grace join me for tea on Thursday?"
"We'd be delighted," the duchess answered.
"Oh, yes," Maddie seconded, smiling. Perhaps Quin was right, after all—some warmth and decency did survive in London.
Finally the duchess declared that they might depart, and Maddie practically bolted for the door and the coach. Quin joined them a moment later.
"That went quite well, don't you think?" he asked, leaning back in the seat opposite her.
"The soiree isn't what counts," Maddie said succinctly, gazing out the window until they passed Curzon Street—where Willits House stood. The conversation inside the coach suddenly became much more interesting.
"It's not? Then what, pray tell, were we doing there? And why did I spend twenty minutes conversing with that rattlebrain Lord Avery?"
She eyed him, amused by his pretend exasperation—though of course she'd never let him know it. "It's what everyone will say about me now that we're gone that counts. People rarely insult you to your face, as you have just demonstrated regarding Lord Avery."
"Now, Maddie—"
"She's correct, Quin," the duchess interrupted. "And you weren't helping things, hovering about like a footman."
"I was not hovering," he protested indignantly. "I was being a dutiful son and host."
"Well, do it less obtrusively next time, won't you, dear?"
Quin folded his arms. "I'll certainly try. Do I still accompany you to the opera tomorrow, or have you managed to recruit Father?"
"The opera?" Maddie gasped, her heart pounding in dismay. "Oh, no. Not yet."
"Yes, you will escort us," the duchess answered, ignoring their guest's protest. Unexpectedly, she reached over and took Maddie's hand. "Whatever they may mutter among themselves, they would do it whether you were there or not. Whatever they would say to you in my presence, they had best be polite about it."
"If I'm accepted only in your company, Your Grace, there seems little point in any of this," Maddie said shakily, thankful nevertheless for the duchess's unexpected support. Everyone went to the opera at the beginning of the Season, with no grand balls or soirees organized yet. Everyone who was in town would be there, not just the select acquaintances of the duchess.
"It's a beginning, Maddie," Quin said. "One step at a time."
"That's easy for you to say, Lord Warefield. You're not the one on the edge of the abyss."
"Neither are you."
"And what am I to do if I come across Charles Dunfrey?" She swallowed. "Or my parents?"
"Your parents are not in town yet," Quin answered calmly. "I already inquired. As for Dunfrey, my friend Danson tells me he sold his box last Season. I very much doubt he'll be at the opera tomorrow night."
"Yes, but what about—"
"Maddie," he interrupted. "Don't worry. I will keep my word to you. Whatever else happens, you'll end the Season well."
The duchess looked from one to the other, and settled back in her seat. Quin had told her the two of them fought every time they saw each other. To her, it didn't look nearly as much like fighting as it did flirting. And she wondered what would happen when they realized that as well.
The duke claimed a meeting, while the duchess and Maddie, not quite so unwilling as yesterday, undertook another preliminary shopping excursion. If Napoleon had planned his campaign as well as the Duchess of Highbarrow had planned Maddie's, he wouldn't be rotting on Saint Helena.
Quin, grateful for a few hours' reprieve, spent most of the morning pacing about the Bancroft library. Eloise would be in London by the afternoon, and what he hadn't been able to disclose in his correspondence he felt even less able to tell her in person—at least without making it sound as though he had some ulterior motive for bringing Maddie Willits to London. Which he didn't, of course.
"Like hell I don't," he muttered aloud, dropping the book he carried onto a chair. "Well, Eloise," he began, "I felt sorry for the girl, stuck rusticating in Somerset with my stodgy old uncle." He paused by the library window, then shook his head and began circling the room again, feeling rather like a bird searching for a safe roost before two very lovely nightingales could peck his eyes out.
He cleared his throat and tried again. "You see, my dear, Uncle Malcolm pleaded with me to assist Miss Willits, and sickly as he was, I could hardly refuse." Quin rubbed at his temples. "Ahem." He dropped into a chair. "Good God, I'm a dreadful liar."
"Might I suggest 'I tried tumbling the chit, and she threatened to cry rape unless I took her to London and foisted her off on my parents'?"
Quin looked up, scowling, as his father stopped in the doorway. "I did not try to tumble her," he snapped, wishing he'd remembered to close the door. For God's sake, Maddie might have heard him. He'd never have lived it down.
Lord Highbarrow scowled and folded his arms. "Oh, really?" he said, cynicism dripping from his voice.
"Yes, really. And I practically had to drag her by her hair to get her to come to London. And please keep your voice down, Your Grace. Someone might hear you."
"No one who matters. I always thought Rafael was my idiot son. She's abusing your generosity. You don't actually think she intends to storm back to Malcolm if she can drag the future Duke of Highbarrow before the archbishop, do you? Sweet Lucifer, boy, stop using your balls for brains!"
Quin shot to his feet, anger tearing through him. "I am marrying Eloise, as you require. I—"
"I didn't ask for a recital of the obvious, Quinlan. That's what my accountants are for. You'd best not forget that you are marrying Eloise, if you want to remain the Marquis of Warefield. It's my damned title, yours to use by my leave only!"
"I know that. I haven't done anything except feel pity for a—"
"That isn't pity I see in your eyes when you look at her, boy," the duke cut in. "If you want to rut with her, that's fine. But get your damned whore out from under my roof!"
His face flushing, Quin clenched his fist. "Maddie Willits is not a damned—"
"Bah!"
His Grace stomped out the door and down the hall, bellowing at the butler for a glass of port as he went. Incensed both at the accusation and at how close some of it was to the truth, Quin grabbed the brandy decanter and hurled it into the fireplace. It shattered against the hot bricks, the brandy exploding in a satisfying hiss of blue flames. "Bloody, pompous—"
"Should I pack, then?" Maddie's voice came from the doorway.
Quin whipped around, paling. "Damn...I didn't know you'd returned. Excuse my language, Maddie."
Shaking her head, she backed up into the hallway. "No need, my lord. And you don't have to be so polite, you know." She brushed at her eyes as a single tear ran down her cheek. "It's what they're all saying, I'm certain."
Quin followed her into the hall and grabbed her hand. "Wait," he said, pulling her back into the library and closing the door. "None of that was meant for you to hear."
She looked away, her lower lip trembling. Her slim wrist, clenched tightly in his fingers, shook a little. "It doesn't matter."
"Of course it does. His Grace just likes to roar and intimidate the rest of the pack. He's nothing but wind."
"That was...quite rude of him," she said unsteadily, obviously very hurt and making a heroic effort to stop crying. "No wonder Mr. Bancroft doesn't like him. I don't, either."
"Neither do I, at the moment," he said. "Please don't cry."
"I'm not crying. I'm very angry."
Slowly Quin drew her closer. "I'm sorry," he murmured, wanting to hold her in his arms. "And don't think you need to leave. I'll speak to him again—in a more civil tone. I promise."
He was rather surprised he'd made the offer: begging to His Grace on bended knee was not something he did on a frequent basis. In fact, he couldn't recall either one of them ever backing down after an argument. But if he didn't apologize, Maddie would leave. And he didn't want her to leave.
"This is ridiculous, anyway. If my parents or...or Charles—if they should see me, everything would be ruined. Especially me."
Quin reached down and lifted her chin with his fingers. "And what makes you think I don't enjoy a little bit of the ridiculous every now and then?"
He wanted to kiss her. He wanted to feel her lips against his—he wanted to feel her body against his.
Maddie met his gaze. "Oh, no," she whispered. "Not again."
"Should I not?"
"Yes. No. Oh, blast." She lifted up on her toes, and twined her arms about his neck. As she pulled herself up against him, he leaned down and brushed his lips against hers. The touch was electric. Unable to help himself, he kissed her again, more roughly, sliding his hands down her hips and pulling her against him.
"Quin?" the duchess called. "Quin, I need to talk to you."
With a strangled sound, Maddie wrenched away from him. "Stop it," she said sharply, shoving at his chest. "Stop it!"
He stared at her for a moment, stunned at his own reaction to her, and exceedingly frustrated at having been interrupted. "You started it. And don't go anywhere," he ordered, taking a last look at her and then slipping through the library door.
Maddie sighed and plunked herself down in one of the chairs. "I started it? Oh, I suppose I did. Drat." Slowly she reached up and traced her lips with her fingertips. Only a kiss, and it had gone through her like lightning—worse than before, and it left her with a raw, aching yearning for him.
A few moments later Quin came back into the room. His face was somber, and her heart began pounding again, this time with dread. "I have to leave, don't I?" And the duke would probably see to it that she couldn't go back to Langley now. Which narrowed down her choices to none. "Don't I?"
"No, you don't." He cleared his throat. "Unfortunately, though, there has been a complication."
She rallied enough to lift an eyebrow. "Only one?"
"So to speak. His Grace has forbidden my mother to assist you. Her compromise was that you be allowed to remain here until arrangements can be made to send you back to Langley. And she will still accompany us to the opera tonight, as she gave her word. My family is very big on honoring their word."
"I've noticed." She wondered for a wrenching moment what would have happened between them if she'd been able to stay. "It's over, then."
"No, it's not. Tonight will go a long way toward repairing the damage. And I have a few ideas."
"Just let me go back, my lord. You've done more than your part."
He tilted his head at her. "Call me Quin."
"I don't want to."
"Why not? We've kissed twice now."
She couldn't tell him that it meant there was some sort of connection between them, that she had a difficult enough time with distancing herself from him already, that over the past few weeks she had begun to regard him with a great deal more affection than she thought possible. "It's not proper."
The marquis actually laughed. "Call me Quin," he repeated.
Maddie took a deep breath. "Just let me leave, Quin."
"No."
She didn't know precisely why he was continuing to argue, for he was nearly engaged. Still, for a bare, exhilarating moment she was relieved that he insisted on pursuing this stubborn course of action. For whatever reason, he wanted her there.
"Now, if you'll excuse me, I need to go change."
Maddie blinked, reluctantly floating down from her daydream. "Why?"
Quin grimaced, edging for the door. "I'm meeting someone this afternoon."
At the sight of his sheepish expression, she immediately realized to whom he was referring. "Of course. Lady Stokesley arrives in London today." The remains of her fantasy sank with a thud into a very deep mud puddle.
"Yes." He paused. "Which could be quite opportune for you."
She looked at him skeptically. "For me?"
He strode forward to clasp her hands in his. "Of course. Father might be able to forbid my mother from assisting us, but he can't very well order Eloise around. She's only a second cousin."
Humiliation flushed Maddie's cheeks. "No, Quin. That's a terrible idea. I'm sure she'll want nothing to do with me. How would you ever explain me to her, anyway?"
His smile faltered a little. "Eloise is very understanding."
Maddie nodded, pulling her hands free. "Even so," she returned, trying to sound cynical and amused instead of heartbroken, "if I were you, I wouldn't mention that you kissed me. Twice."
"Maddie," he began, closing the distance between them again.
She backed away. "Don't try to explain. We'll just put it to the general weakness of men."
His eyes searched hers. "I'd rather you didn't," he said, "but I don't think it would be wise to pursue an explanation any further at the moment."
Looking at the slight smile curving his mouth, and the jade eyes studying her face, Maddie abruptly agreed with him. This was becoming extremely complicated. "Well? You'd best be off, then."
"Don't go anywhere while I'm away," he warned.
She put a hand to her chest. "Me?"
"Miss Willits, don't make me lock you—"
"All right, all right," she surrendered. "I won't go anywhere while you're off visiting."
He nodded. "Good."
"Quin!" Eloise, Lady Stokesley, hurried downstairs in a fashionable swirl of blue silk.
She'd cut her hair since last autumn, her long blond tresses now a short, daring cascade of curls framing her perfect alabaster features. Her blue eyes reflected the fine material of her gown as she stopped before him. They'd known one another for so long that Quin sometimes forgot how lovely she was—until he set eyes on her after a few months of being apart. He took her hands and drew them to his lips.
"Eloise," he said with a smile. She stood a few inches above Maddie's height—though he'd never thought of her as being overly tall before. "You are a vision, my dear, as always."
She curled up a delicate fist and hit him on the shoulder. "I'm very angry at you."
There was no heat in either the blow or her voice, and he lifted an eyebrow. "Whatever for?"
"I've had a terrible time keeping track of you, you beast. First you're at Warefield, and then you go off to Somerset, of all places. You hardly correspond with me, and then, when I finally think my letters have caught up to you, your uncle writes to inform me you're at Highbarrow Castle. Then you go off to London without a word. And without coming to Stafford Green first."
Quin led her into the Stokesley House morning room. "Yes, well, that's one of the reasons I wanted to see you right away. I've had...something of an adventure."
Eloise seated herself on the couch, beckoning him to join her. "Do tell me what's kept you so occupied."
He heard the slight censure in her voice but ignored it. He had been less than communicative over the past few weeks. "It's a rather long story."
"I should imagine so. Tell me."
Quin settled back against the cushions. "Well, you know Uncle Malcolm had an apoplexy, and that Father sent me to Langley to help with the planting."
She nodded. "Yes, you did manage to write me about that, and about how inconvenient it would be to your schedule."
It had been inconvenient only until he'd set eyes on Maddie. "It turned out that Uncle Malcolm already had an assistant who was quite proficient at estate management," he continued.
"Then why didn't you take your leave early and visit me, as we'd planned?"
"I'm getting to that. This assistant of my uncle's was a female."
Eloise's eyes widened, and she put a hand over her mouth. "Oh, my," she said slyly. "How wicked of your uncle. And such a shame that the best gossip is always within one's family, so one can't spread it about."
Quin frowned, then wiped the expression from his face when she looked at him curiously. Funny, he'd never taken notice of gossip before. But then, he'd never seen the effects of it—until Maddie. "I don't think it was like that at all," he countered, a little stiffly. "She was more like a daughter to him, I believe. And it turned out that she was the eldest daughter of Viscount Halverston."
"The...we came out the same Season. She's the one who ran off after half the ton discovered her lifting her skirts for that awful Benjamin Spenser, isn't she?"
"I'm not convinced that was her fault."
Eloise looked up at his sharp tone. "Oh, really?"
Charging to Maddie's defense would only make things worse. And Maddie certainly wouldn't appreciate it. "As you said," he continued, "Spenser doesn't have the best reputation as a gentleman. At any rate, my uncle believed her to be innocent of any wrongdoing. He...he asked me to reintroduce her to London society."
"He did what?" Eloise stood. "You had that...lightskirt with you when you traveled to Highbarrow?"
"Eloise, please," he cut in, before she could say something worse about Maddie. "My mother has been helping her. She's very sweet."
Her eyes narrowed suspiciously. "Sweet?"
It wasn't exactly the perfect word to describe Maddie, but he couldn't very well tell Eloise that he was half addle-brained over the wild-hearted chit. "Yes. Except that now Father's on a rampage about how she'll somehow tarnish the Bancroft name, and you know Mother won't directly defy him, so...so I need your help, Eloise."
"My help?"
Quin shrugged. "Well, yes. I promised Malcolm I'd see her able to marry any gentleman in London. I can't very well chaperon her myself."
For a long moment she looked at him. Finally, she sat again and took his hand in hers. "Of course I'll help, my marquis." Eloise moved closer, leaning her head against his shoulder. "Poor dear, she'd be all alone if not for us."
"Well, she's not exactly helpless. In fact, I believe her to be quite capable."
"Capable?" Eloise chuckled. "Oh, my. You make her sound like a milkmaid. I can't wait to meet her."
"You said you came out together."
She shook her head. "No, I said we came out the same Season. I imagine we attended a few of the same soirees, but Halverston's a very small holding, you know."
He knew what she meant—that Maddie hadn't been privileged to travel in the same high circles into which he and Eloise had been born. "Yes, I know. Mother's last official duty is to chaperon her to the opera tonight. I'll bring her by tomorrow, if that's all right."
"Oh, yes! I'm quite looking forward to helping her now. We shall be like sisters."
Quin smiled, relieved, and kissed Eloise on the cheek. "Thank you."
"Come by at one, and we'll have luncheon in the garden."
With a nod, Quin collected his hat and went outside to retrieve Aristotle. That hadn't been nearly as sticky as he'd anticipated. Perhaps there was a chance he would escape the Season in one piece.
With Mary's assistance, Maddie dressed in her dark green and gray gown with a scooped neck and short puffy sleeves. The flowing silk was easily the most lovely thing she'd worn in years, if not ever. It was a gown fit for a prime box at the opera, and as she twisted in front of the full-length mirror, she was terrified. "This is so stupid."
"You look beautiful, Miss Maddie," Mary protested, reaching out to adjust one last out-of-place curl.
"Well, thank you, but that's not quite what I meant." She'd made Quin promise four different times that her parents weren't in town yet, and another three times that Charles Dunfrey wouldn't be attending The Magic Flute. Which left only a thousand other people to stare at her and laugh that she would think of returning to London society—even in the company of the Bancrofts.
Her door rattled with Quin's confident knock. "Ready, Maddie?"
"I really don't think we should upset your father any further," she told the door. "You go. I'll begin packing for Langley."
The bedchamber door opened. "No excuses...." the marquis began, then closed his mouth as he ran his gaze over her. "'My prime request,'" he said softly, "'Which I do last pronounce, is, oh, you wonder, if you be maid or no?'"
She couldn't help the smile that touched her lips. "'No wonder, sir, but certainly a maid.'"
He stepped closer, apparently oblivious of Mary's presence. "Then you know The Tempest, as well."
"I've recently had a great deal of time to read." Belatedly she backed away from him. "And you shouldn't be in here."
He lifted an eyebrow. "Don't want you escaping out the window, you know. The coach is waiting. Shall we?"
Maddie shook her head. "No, I don't think so."
"After tonight it will be much easier. Come on."
He was entirely too high-spirited, which made her even more nervous than she had been a few minutes earlier. Reluctantly she followed him downstairs. "You've never been cut, have you, my lord?"
He looked sideways at her.
"Quin," she corrected.
"No, I haven't. Except by you, of course."
"Well, as I imagine you'll find out tonight," she said, while the butler helped her on with her shawl, "I don't matter, Quin. And thank you, Beeks."
"You could be a little more positive about this, Maddie." He gestured her to precede him out the door.
Reluctantly, Maddie complied.
Shrugging, she stepped up into the coach to sit beside the duchess. He didn't understand—and he wouldn't understand, because it had never happened to him.
Once they exited the coach at the front of the opera house, however, she thought perhaps he'd been right, after all. She received several startled second looks and heard the murmured commentary going about the huge lobby, but no one actually turned his back on her. She held tightly onto Quin's arm, grateful for his tall, strong presence, and tried to look relaxed. The marquis, with Her Grace on his other arm, smiled and greeted their friends and acquaintances as though nothing out of the ordinary was occurring.
"See?" Quin murmured at her, as they started up the long, winding staircase leading to the balcony and the exclusive boxes.
"They can't cut me without cutting you and the duchess," she muttered back, through teeth clenched in a determinedly amused smile.
"It was five years ago, Maddie. Probably no one so much as remembers the specifics. If you comport yourself well, they'll have no reason to cut you."
She had her doubts about that. Even so, as she stepped through the curtains at the back of the Bancroft box and took the forward seat beside the duchess, as she'd been instructed, she couldn't help a moment of optimism. So far, the nightmare she'd imagined hadn't materialized. Perhaps Quin was right, and no one remembered her at all.
A few pairs of opera glasses, then more, turned in her direction at the beginning of the first act, but she thought they might just as easily have been aimed at the illustrious Bancrofts. She hadn't been to the opera in a long time, and once everyone settled down to watch, she allowed herself to be absorbed in the comedy unfolding on the stage below.
She hardly realized intermission had arrived until the massive curtains swung shut and the house lights brightened. A wave of uneasiness ran through her again, even stronger than before. Everyone mingled at intermission.
"Shall we?" Quin said, rising and stretching.
"Polite nods and one- or two-word answers. For anything more, defer to me—is that clear, Maddie?" the duchess said, motioning her toward the back of the box.
She nodded, hoping no one could hear her knees knocking together. "Don't worry about that."
It was like remembering, like something she had done before, but not exactly in the same way. The French phrase for it—déjà vu—that was what it felt like, to emerge from a theater box into a crowd of glittering nobility, but at the same time to know with absolute certainty that she did not belong.
"Your Grace. How splendid to see you back in London." A short, very broad woman whose neck sparkled with diamonds appeared from one side of the wide hallway. She clasped the duchess's hand.
"Lady Hatton," the duchess returned. "I was so sorry to hear of your cousin's death. My condolences to you and your family."
"Oh, thank you, Your Grace. One would have thought India more civilized by now." The white-haired woman looked at Quin and curtsied, then turned her attention to Maddie. "I don't believe I've met your companion, Your Grace."
"Ah, of course. Lady Hatton, I should like to introduce Miss Willits, an old family friend. Madeleine, Lady Hatton of Staffordshire."
Maddie nodded politely, self-consciously taking Quin's arm again. "Pleased to meet you." It was actually four words, but she couldn't think how to limit it to less and not sound like a halfwit.
"And you as well, my dear." Lady Hatton smiled. "What a lovely girl you are."
"Thank you, my lady."
A moment later three other grand ladies had approached, all of them thankfully more interested in learning the state of the duke's health and wealth than in being introduced to Maddie. Relieved, she turned away—only to find herself being stared at intently by a very beautiful, tall woman with a profusion of short blond ringlets framing her face.
She looked familiar, and as she parted from her companions and came forward, Maddie remembered who she was. Her heart and her courage sank.
"Eloise," Quin said warmly, and Maddie belatedly released her grip on his arm. She'd likely left bruises, but he didn't give any indication of being in dire pain. Instead, he smiled again. "Why didn't you say you would be attending this evening?"
"It was something of a last minute decision," Lady Stokesley answered, and stopped before Maddie. "You must be Miss Willits. Your face is somewhat familiar, but I don't think we've ever been formally introduced."
"Then allow me," Quin said. "Eloise, Miss Willits. Maddie, Lady Stokesley."
"How do you do?" Maddie said, trying not to stare. This was the woman who held Quin's heart, and she wasn't certain she was ready to like her at all.
"Quin's told me all about you," Eloise replied. "I do hope we can be friends."
Maddie forced a smile. "So do I."
Lady Stokesley wrapped her arm around Quin's. "May I borrow Lord Warefield for just a moment?"
"Of course."
They stepped away, and Maddie abruptly realized that she was alone. Very alone. She turned to find the duchess, but Her Grace was nowhere in sight. And then neither was Quin, nor Lady Stokesley.
"Miss Willits?"
With a start, Maddie turned again. The stocky gentleman gazing at her looked somewhat familiar, but so did half of the London nobility, and she couldn't place him.
"I say, it is you. Maddie, isn't it?"
He came closer and took her hand, bringing it to his lips.
"Have we met?" she asked stiffly, trying to free her hand while he kissed her knuckles again.
"We have mutual friends, I think. I'm Edward Lumley."
"I'm afraid I don't—"
"By gum, the time away from London hasn't damaged you a spot," he said admiringly, taking in her gown, and particularly its low neckline.
"Thank you. If you'll excuse—"
"I'd wager you're a real goer," he continued, stepping still closer. "What'd it take to steal you away from Warefield?" He grinned, running his fingers up her arm. "He makes a bit of a stodgy ride, no doubt. I've been to the Spice Islands, you know. Learned all sorts of techniques."
For a moment Maddie stared at him, hardly following what he was saying, and even less able to believe that he would dare say it to her face. "I believe," she said coldly, narrowing her eyes as his gaze dropped to her bosom again, "you are mistaken about me."
"What, you don't belong to the duke, do you?"
Maddie belted him.
## Chapter 10
Quin looked up as Edward Lumley hit the polished floor.
"My goodness!" Eloise exclaimed.
Lumley scrambled back to his feet. Red-faced, he strode up to the young woman eyeing him with wounded fury, her fists clenched before her.
Quin swiftly reached out to grab Lumley's arm and wrenched him around. "Terribly sorry there, Lumley," he said, brushing at the idiot's coat with his free hand. "Didn't see you standing there. Not a pleasant thing, to get an elbow in the eye like that."
Lumley glared at him, the scarlet of his face deepening to crimson. He jutted a finger at Maddie. "She—"
"Yes, she is lovely, but I think you'd best recover yourself a little before I introduce you to my mother's companion, Lumley. Perhaps a breath of fresh air is what you need."
The iron grip on his arm and the angry glint in Quin's eyes was apparently enough to convince Mr. Lumley to make himself scarce, and with a parting sneer at Maddie, he collected his dented hat and wounded pride and departed.
Quin shrugged as he looked after the gentleman, and a polite round of chuckles answered the motion. Silently he counted to five, then turned back to Maddie, who stood staring at him, white-faced.
"I believe Her Grace is looking for you, Miss Willits." He smiled, forcibly unbending her clenched fingers and tucking them over his arm.
"Thank you, Lord Warefield," she returned gamely, smiling despite the hurt and anger still in her eyes.
They walked to the fringes of the crowd, with Maddie clinging ferociously to his forearm. When they finally reached the top of the stairs, he stopped and made a show of checking the elbow of his coat for damage. "What in damnation happened?" he murmured, looking at her sideways.
"He—that—that Edward Lumley person wanted me to be his mistress!" she sputtered, clearly still furious.
"Calm down," he said, glancing back at the noisy crowd.
"Calm down?" she repeated. "Calm down? He said you were too dull, and that he'd learned all sorts of things in the Spice Islands!"
"I'm too dull, am I?" Quin narrowed his eyes, abruptly liking Lumley even less now. "Maddie, whatever he said, you can't go about hitting people when they insult you. It's not at all comme il faut."
"No?" she replied indignantly, her color beginning to return. "This was your fault, Warefield."
"And why is that, pray tell?"
"You're a Bancroft: the great, grand Marquis of Warefield. Your name would protect me from all insult and innuendo, remember? Wasn't that what you said?"
"Maddie, I—"
"No one's forgotten what happened," she cut in. "They all think I'm some sort of whore. What am I supposed...supposed to do when someone says that to me?" Her defiance melted and tears began to form in her eyes.
"Give them a set-down," he returned more quietly, wanting to kiss the tears from her eyes. "Just don't set them down on the floor."
"That's easy for you to say. No one would dare utter something like that in your presence. You don't have to deal with it."
Quin sighed. She was right—and he had been too arrogant in his assumption about the protection his name would provide her. "That's still no excuse."
"I don't care. I want to leave."
"We can't leave after that little show of yours. We have to stay and act as if nothing happened. Come on, we'd better find my mother."
She put her hands on her hips and he tensed, ready for another of her uniquely spectacular scenes. "What about my honor?"
"What do you mean, 'your honor'?"
"That man insulted me, and all you're worried about is whether anyone noticed. You're exactly what I thought you were."
Quin didn't know exactly what that was, but he didn't like the implication. Especially since she was probably right. "I'm sorry," he murmured. "I should not have left you alone. I won't abandon you again, Maddie. I promise."
If he hadn't been so relieved that her first meeting with Eloise had gone so smoothly, he would have realized how idiotic it had been to leave her side. Quin touched her chin, tilting her face up, then hurriedly lowered his hand as Lady Granville passed by. "I'm sorry," he repeated.
She held his gaze for a long moment, then nodded.
"All right?"
She nodded again, and then abruptly grasped his arm again, holding him tightly.
He headed them back toward the box. "My, my, Maddie Willits, speechless," he teased softly. Something he'd said had finally been the right thing, though he had no idea which something it had been.
The duchess was already seated and waiting expectantly for them, but when he shook his head she refrained from inquiring about what had happened. As the opera began again, Quin sat back and studied Maddie's profile. Obviously she was nowhere near ready to be set loose into society, if she allowed herself to be so hurt by any comment about her past. And volatile and emotional as she was, convincing her to ignore the insults would be next to impossible.
He needed to come up with some other way to help her get past the innuendos and propositions of the less-principled members of society—though he could hardly blame them for desiring her companionship. The sooner he and Eloise could find her some nice, quiet, unassuming gentleman, the better off he'd be. Quin frowned. The better off Maddie would be.
Eloise Stokesley sat beside her mother, Lady Stafford, and looked down at the opera glasses in her lap. She and Quin Bancroft had been promised to one another since her birth. The idea had never bothered her. Quite the contrary, actually—the Marquis of Warefield was highly esteemed, very wealthy, and handsome as a Greek sun god. Her friends all knew she was to marry him, and they envied and admired her for it.
So when he'd asked last year if she would mind one year's delay before he made an official declaration, she had agreed. The Bancroft properties were so vast, a few months would be necessary just for his father and their horde of solicitors to decide which additional lands and funds should come to him upon his marriage. And it gave her another Season to flirt with her myriad male admirers and gloat about her coming nuptials.
She looked up, across the orchestra and the throng of the less wealthy below. The Bancroft box was nearly opposite the Stokesleys', and in the near darkness she could just make out the three of them sitting there. The duchess, Quin, and her. Eloise lifted the opera glasses.
Madeleine Willits. Eloise remembered her as a determined flirt with a smile for anyone who amused her, no matter how base they might have been. No wonder she'd earned such a fast reputation. And pretty as she was, no wonder she had snared Quin's attention. Eloise turned the glasses to view her nearly betrothed. She narrowed her eyes.
He sat in the shadows, his attention not on the stage, but on the woman seated in front of him. His expression as he studied her was amused, but it was also keenly interested. The moment Quin had approached her to ask her assistance in introducing Miss Willits back into society, she'd suspected his motive was more than simple compassion. Now she had no doubt of it.
But women had been unsuccessfully throwing themselves at the Marquis of Warefield for years. She knew he'd taken several mistresses in the past, but they hadn't been of any quality to threaten her own status, and he'd certainly never been overly attached to any of them. This one was different. He'd never taken any of the others to the opera, or bothered having them chaperoned by his mother.
Eloise lowered the glasses again before anyone could notice her discomfiture. Something would have to be done to set things back into their proper order. As the future Duchess of Highbarrow, she had a stake in the doings of the Bancroft family. And Maddie Willits did not belong there.
Maddie looked up the Bancroft House drive, and with a curse, ducked behind the duchess's roses. Quin and Aristotle rode by, returning to the stable from their morning ride in Hyde Park. Quin had barely spoken to her after the opera, but from the way he'd kept looking at her, he had something in mind. She just didn't want to hear it. They'd tried and failed.
She'd been away from this nonsense for too long, and she couldn't adapt to it again. She didn't want to adapt to it again. Only now, even the idea of returning to Langley Hall had its drawbacks: Quin wouldn't be there. Last night, when he'd promised not to abandon her, she had wanted so much to throw her arms around him and kiss him, even though he couldn't possibly have known how much those words meant to her.
She bent down to yank another weed out of the ground, shredding the offending plant into the bucket hanging off her arm, and grateful that the duchess had let her putter about in the garden. At least it kept her from feeling completely useless.
A terrible commotion erupted from the direction of the stables, and she whipped around, wondering with some alarm whether Quin had been trampled by his independent-minded mount.
Hefting the heavy bucket, she hurried down the carriage path toward the noise. As she rounded the hedge, a man all in black flashed out of the stable doors, mounted on Aristotle. Quin emerged right behind them, covered in straw and with a pitchfork in one hand.
"Stop him!" he bellowed.
The rider dodged around one of the stable hands and thundered down the path toward Maddie and the street beyond. Reacting instinctively, she swung the wooden bucket up as hard as she could. It caught the rider in the shoulder, and with a grunt he canted sideways out of the saddle and tumbled to the ground.
He rolled a short distance, then immediately returned to his feet to shake himself and stride angrily in her direction. Alarmed, Maddie backed away and raised the bucket menacingly.
"Blast it, that hurt!" he said, rubbing his left shoulder.
"Don't come any closer unless you want worse," she warned. She heard Quin coming up behind her and shifted sideways, hoping he was still armed.
"Where's my damned horse?" the marquis asked with surprising composure, leaning on the pitchfork and breathing hard.
"My damned horse, you mean," the other replied, and put two fingers to his mouth. The sharp two-toned whistle surprised Maddie. Even more surprising was the sight of Aristotle trotting back up the drive and coming to a stop beside his abductor.
"Show-off," Quin muttered.
The lean, sandy-haired man patted the gelding's neck and received a nuzzle in the shoulder in return. "Who's your trained assassin, Warefield?" he asked, looking at Maddie and grinning. A long, narrow white scar ran from high on his left cheekbone almost to his jaw, giving him a vaguely piratical air.
"You're Rafael," Maddie whispered, blanching. Now she'd nearly killed the marquis's brother. Good Lord, the Bancrofts would be lucky if she didn't do them all in.
He swept a bow, his light green eyes dancing. "I see my reputation as a horse thief precedes me. But it is my animal."
"Rafe, Miss Willits." Quin supplied with a reluctant grin. "Maddie, my idiot brother Rafael."
Rafael Bancroft did look a great deal like his older brother, though his face was thinner than Quin's, and darkened by much time spent out-of-doors. He lacked an inch or so of the marquis's height, but they shared the appearance of lean, contained strength.
She set down the bucket and stuck out her hand. "Pleased to make your acquaintance."
He shook her hand, his grip firm and friendly. "You're lethal, you know," he chuckled, rubbing at his shoulder again.
"I'm sorry. I heard Quin yell, and—"
"No need to explain," he said, glancing at his brother. "And where did you meet Quin?"
She heard his emphasis on the name and blushed. "It's a very long story," she offered, hesitating to explain her presence to another of the unpredictable Bancrofts.
"She was staying at Langley as Malcolm's companion," Quin said, reaching around his brother to grab Aristotle's bridle. "Help me put my damned horse up, and I'll tell you the whole story."
"My damned horse, Warefield."
Maddie watched the two of them stroll back to the stable, knowing she was being excluded, and wondering what sort of tale the marquis intended to spin about her presence. He'd best fill her in later if there was a lie involved, so she wouldn't trip over it by accident.
"She what?" Rafe asked, leaning over the stall door.
"She hit him," Quin repeated, hanging up Aristotle's bridle. "Laid him out flat."
"That's extraordinary."
"It's a deuced lot of trouble." The marquis glanced up at his younger brother and shook his head. "She's a deuced lot of trouble."
"Then you probably shouldn't have kissed her." Rafe grinned. "Not that I blame you. She's a diamond of the first water."
"Which doesn't help me any."
"From what you've said, she seems to know what she wants, Quin. Are you so certain Langley's not the best place for her?"
He'd asked himself the same question over and over. And he kept coming up with the same answer. Or at least, the answer he was willing to voice aloud. "She was wronged. If she wants to go back, that's fine. But she should have a choice about it. I won't have some loudmouthed jackanapes driving her off."
"Fair enough, but if she goes about hitting every rake who insults her, she won't last long here, anyway."
The beginning of an idea tickled at Quin's mind. He needed to think it through a little further, though, before he dared bring it up with Maddie. "How was Africa?"
"Full of angry Dutch settlers." Rafe shrugged. "And hot as Hades." He backed up as his brother exited the stall. "Is she the reason you're not at Whiting? I laid in wait for two hours there this morning, until I finally accosted your groom and he said you were here."
"I'm trying to be the buffer between Maddie and Father. Does he know you're back yet?"
"No." Rafael paused, letting Quin leave the stable ahead of him. "Warefield, you haven't set a date for you and Eloise yet, have you? Because although it's not really my place to say, have you considered another reason you might have Maddie staying h—"
"You're right," Quin said sharply. "It really isn't your place to say. And no, I haven't spoken to Eloise about our marriage yet."
"That's a fine welcome home, brother," Rafe said, unruffled. "I imagine I'll receive more of the same from Father."
"He wants you to resign your commission," Quin informed him, grateful they'd returned to a safer topic. "The idea of your being accepted into the Coldstream Guards was that you'd be stationed in London—not in Africa."
"That was Prinny's idea. I think he got tired of my being more carefully protected than he was."
"Nonsense," Quin countered. "You volunteered. You as much as said you wanted to go to Africa."
"His Grace didn't want me in Belgium, either. I got a nice, shiny medal for that."
"You nearly got yourself killed." Quin gestured at his brother's face. "And His Grace doesn't want you risking your hide."
"What would I do here?"
"I don't know," Quin answered, slowing as Maddie reappeared from behind the line of rosebushes. His heartbeat quickened at the sight of her, as it had every time since he'd first set eyes on her. "Something less dangerous."
"Yes, I can see myself in the priesthood," Rafe returned. "Miss Willits, surely you've not been assigned the task of gardening in return for your room and board at Bancroft House?"
"I like to garden," she said defensively.
"You are a rose among toadstools," he said grandly, lifting the bucket from her arm.
"Thank you." She smiled.
"Ignore him," Quin instructed her, frowning. "He's a terrible flirt."
"It's nice to be flirted with," she returned coyly, batting her eyes at him, which made him feel uncomfortable. "It doesn't happen to me very often anymore."
Rafael chuckled. "Then you must be constantly surrounded by idiots. I'm certain by the end of the Season you'll be wishing everyone would leave off so you can garden."
"I actually have an idea about that," Quin returned.
Her expression went from surprise to dismay to distrust so quickly he couldn't be certain he'd actually seen all of them. "About gardening, or about being left alone?"
"I'll explain later."
Rafe raised his hands. "Don't let me stop you. I'll go give the lion something else to roar at." With a jaunty salute he tossed the bucket to a stable hand and strolled toward the house.
Maddie turned to face Quin and folded her arms over her chest. "Well?"
"The art of the insult," he pronounced with satisfaction.
"Beg pardon?"
"The insult. You let Lumley insult you, and you let him get away with it."
"I hit him," she argued, flushing.
"That doesn't count. You can't keep going about doing that, you may run across someone who will hit you back. Not to mention how uncivilized it is. You have to fight them the same way they fight you: with words."
"What am I supposed to say when some...man calls me a whore and asks me to be his mistress?"
"That's exactly what we need to figure out."
"You're completely insane," she stated, and turned her back.
Quin grinned. "Now you're getting it."
With a heave of her shoulders, she faced him again. "What?"
"You need to have a reply for anything—any insult anyone might choose to fling at you."
"And how am I going to do that, pray tell?"
"We are going to practice. We'll call it...anti-rake training." He tried not to laugh at her suspicious expression. "We'll make it into a contest."
"Anti—you are insane."
Actually, the more he thought about it, the better it sounded. "No, I'm not. It's brilliant. Once we've come up with an appropriate reply for every inappropriate comment, you'll be invincible—and uninsultable."
"That's not even a word. Leave me alone, Warefield."
Despite her words, he heard the reluctant humor in her tone. "Why should I be the only man in London to do so?" he asked.
Her fist caught him in the shoulder as he dodged. "Stop it!"
Quin grabbed her hand and pulled her up against him. "No hitting allowed," he warned her. "Wound me with your wit, chit."
She pulled free. "And what good will that do?"
"As you know, word travels swiftly here in London. Once the rakes and disreputables learn that they are the ones who end up looking like fools if they bother you, they'll stop insulting you. And they'll begin respecting you. Or at least fearing you."
She narrowed her eyes. "Are you certain of this?"
"Absolutely."
"All right. But I still think it's ridiculous."
It might turn out to be, but at least it gave her something to think about besides how hurt she'd been last night. Quin smiled after her as she headed back inside. If he'd learned one thing about Maddie, it was that she loved a good fight. And hopefully he'd just given her one.
"What a stupid idea," Maddie muttered.
She sat in the drawing room with the duchess, waiting for the Bancroft men to finish their after-dinner discussion and join them. The conversation continued to grow in volume, which would seem to signal either that it was nearly over, or that bloodshed was about to ensue.
"What's a stupid idea, my dear?" Her Grace lowered her book and eyed Maddie over the top of it.
Maddie flushed. "Nothing, Your Grace. I'm sorry; I was talking to myself."
"Not a proper habit for a young lady to have."
"I don't think I have any proper habits at all," Maddie agreed.
The duchess closed her book. "How was Malcolm when you left him? I should have asked sooner."
Maddie didn't quite know how to take Lady Highbarrow's unexpected solicitude, but it seemed the duchess was beginning to tolerate her a little. "He could move his left arm quite well, and he claimed his legs were beginning to tingle. In his last letter, he said he'd actually taken three steps before he fell on Bill Tomkins."
"Who is Bill Tomkins?"
"Mr. Bancroft's footman."
The duchess nodded. "Do you write him? Malcolm, I mean."
Maddie set aside her embroidery, grateful to have someone to talk to. "Yes. Three times a week."
"And what do you tell him?"
"About the weather, how well everything is going, and how much fun I'm having being back in London."
Faint curiosity touched the duchess's dark green eyes, and she moved over to sit on the couch beside Maddie. "So you lie to him."
"Not exactly," Maddie said hesitantly. "He wanted this for me, though, and I don't wish him to worry."
"Wish who to worry?" Quin said from the doorway.
"Your uncle," Her Grace answered, before Maddie could. "Where's Lewis?"
Rafael dropped onto the couch beside his mother. "He said he was going up to his office to write a letter to King George, asking him to have me dismissed from the army before some filthy African native eats me."
The duchess lifted an eyebrow. "Lovely thought, dear."
"His Grace was appalled as well. Until Quin begets a male heir, I am the spare, after all. Wouldn't do for me to end up in some Zulu's belly."
Maddie wrinkled her nose, torn between alarm and amusement. "Ooh. Do stop that."
Rafe grinned. "Apologies."
Quin ignored the joking, looking at Maddie until she lowered her eyes and pretended to be distracted by Rafe and the duchess. He always looked at her that way: as though he was trying to see inside her. It didn't exactly make her uneasy, but it unsettled her—because she liked it.
"What are your plans, then, Madeleine?" the duchess asked.
Quin looked as startled as Maddie at the first genuine expression of interest Lady Highbarrow had shown. "I'm not certain, Your Grace. I think I've imposed on your kindness for too long."
The marquis frowned at her, but Her Grace smiled. "As I recall, the imposition was not entirely your idea."
"Nearly an abduction, from what I've heard," Rafael seconded.
"Even so, without a sponsor I have no reason to remain in London." Maddie glanced at Quin, and then away again.
"You have a sponsor," he said, bringing his brother a glass of port. "I spoke with Eloise yesterday. She was thrilled to offer her assistance. We were to meet her today for luncheon, but she sent a note begging off until tomorrow."
"Eloise?" Rafe broke in, raising both eyebrows.
"Yes, Eloise. Aren't you tired after your long journey?"
"From Bristol? Not really." Rafael looked at his brother, then stood, offering the duchess his arm. "Catch me up on the gossip, my dear," he said, as she rose. "I hate to laugh at the wrong people."
They exited the room, though the duchess pointedly left the door open. Quin looked after them for a moment, then took a sip of his port and turned to face her. "Well?"
"Well what?"
"You let Rafe get away without a comment on his blatant snobbery, while I can't look at Lord Barton's fourth footman without you expecting me to know his name and whether he has children."
"I'm not that bad," she protested, wondering why it hadn't occurred to her to be angry at Rafael.
"Yes, you are."
"I am not." She stood. "And I'm going to bed. Good night."
He was silent for a moment, looking toward the window. "'Bed'?" he finally repeated. "I thought the night would just be beginning for you."
Maddie looked at him sideways, her heart skittering again. "What do you mean?"
"Just that I thought you would be receiving most of your callers late in the evening." Obviously reading her puzzled look, he took a step closer. "Your gentlemen callers, that is," he added.
Abruptly she understood. "Stop it, Quin. I've thought about this, and it's a stupid idea."
"I don't think you really have a choice." He reached up and touched her cheek. "I hope I may be your first caller of the evening."
Maddie shivered at his touch, then slapped his hand away. "I mean it, Quin. Leave me alone."
He grabbed her hand and tugged her up against him. "I warned you before about hitting me," he growled. "Insult me instead."
She took a quick breath, angry and exhilarated at the same time. "I wouldn't have taken you for such a Jack-a-dandy, my lord. Apparently I was mistaken in thinking you had scruples."
Quin pursed his lips. "It's a start, I suppose. But you'll have to do better."
He was right; if insult was the only weapon she was allowed to use to defend herself, she would have to come up with something stronger. "I shall attempt to regroup, my lord," she said. "But I am tired tonight."
Quin studied her expression, then sketched a bow. "As you wish. But Maddie, we're not through here. I'm not about to give up."
As she headed upstairs to her bedchamber, Maddie wondered if Quin had been referring to her predicament, or to her—and she didn't know which to hope for.
The Season might have begun badly for Maddie Willits, but it had been even worse for Charles Dunfrey.
Mr. Wheating from the Bank of England had called at Dunfrey House twice, the second time not even bothering to be polite. Dunfrey had the satisfaction of putting the banker out on his backside, but he knew it was a futile gesture of defiance. Unless his luck at the table or in commerce improved, and soon, the bank would own every piece of his property that wasn't entailed.
And so he read the note sent over from Eloise Stokesley with great interest. It didn't explain much, but the simple fact that the daughter of the Earl of Stafford had written him snared his attention. What the "subject of mutual interest" might be he had no idea, but he had every intention of finding out.
He had been planning to call on Lord Walling, to try to convince the old fool to forgive at least part of the thousand quid gambling note he'd held for the past year. Instead, he donned a conservative gray jacket and waited impatiently for Lady Stokesley's arrival.
Twenty minutes later, the housekeeper showed Eloise Stokesley into his shabby drawing room. He stood and took her hand. "My lady, I must say, this is an unexpected pleasure."
She pulled her fingers free and sat on the end of his couch farthest from where he stood. "Hardly a pleasure for me, Mr. Dunfrey, I assure you."
He leaned one arm on the mantel and eyed his guest speculatively. "Ahh. How may I help you, then, my lady?"
She removed her gloves and folded them neatly on her lap. "I shall be blunt with you, Mr. Dunfrey."
Dunfrey nodded. "Please."
"You at one time were betrothed to a Madeleine Willits."
He frowned, truly startled. "Yes, I was."
"Then you are partly to blame for this fiasco."
"For which fiasco, if I may be so bold?"
"For leaving her to wander about, stealing other people's men and their fortunes."
Dunfrey left the fireplace and sat beside Eloise. "Begging your pardon, my lady, but what in damnation are you talking about? Madeleine Willits is gone, or dead. And good riddance."
"An interesting way to speak of your intended." Eloise fiddled with her gloves, then set them back down again. "She is neither gone nor dead. In fact, she is living at Bancroft House, with the permission of the Duke of Highbarrow."
Dunfrey stared at her, stunned dismay running coldly into his gut. "Highbarrow? Friendly, naive little Maddie climbed as high as that? By Lucifer, life is unfair."
She nodded. "The Marquis of Warefield himself is seeing her reintroduced into society."
Things suddenly began to make sense. "I thought Warefield was to marry you, my lady."
"He is. And he will." She sat back, curving her fine neck to regard him. "It has come to my attention that being a widower is disagreeable to you."
"In what way?"
"To be more precise, you have run through your late wife's money and are now heavily in debt." She looked pointedly at the carpet, threadbare in at least a half dozen spots.
Dunfrey's back stiffened. "I don't believe that's any of your bloody business, Lady Stokesley."
"Mm. A sensitive subject?" she purred. "I assure you, Mr. Dunfrey; you may speak in complete confidence to me."
Dunfrey looked at her assessingly. Business might not be his forte, but he had a good eye for opportunity. And he sensed that something in this unexpected tête-´-tête would come out to his benefit. "I am perhaps a little short of ready blunt this spring," he admitted. "But what does that have to do with Madeleine Willits and your Marquis of Warefield?"
"Everything. Quinlan's assisting her as a point of honor. He wants her to be able to marry well, as though she had never been ruined five years ago—or so he says. I'm even hosting the darling for luncheon this afternoon." She sighed distastefully. "If you publicly forgive her for her misbehavior, you will make the eventuality of a good marriage possible for her."
"And?"
"And I will pay you one thousand pounds for your efforts."
Slowly he smiled. "My, you do want Warefield badly. But isn't he worth more than a thousand quid?"
"Yes, but she isn't."
Dunfrey sat back and crossed his arms. "She was."
Eloise's perfect brow furrowed. "Beg pardon?"
"Her father, Halverston, was ready to give me three thousand quid to take her off five years ago."
"Seems to me that he would have given you even more to marry her after she was ruined."
"My thought exactly. Until the damned chit fled into the night."
She looked at him for a long moment, a slow smile spreading across her face. "You did it on purpose."
Dunfrey shrugged. "Odd bird like her, why not squeeze her father for a little more blunt to compensate me for the embarrassment?"
"Except it didn't work," Eloise surmised. "She ran off before you could suggest the solution to Halverston."
"No great loss. Patricia Giles's father gave me a property in York to make her my bride."
"You sold it off eight months ago, as I recall."
She'd obviously been looking into his personal finances. Which meant she knew that a thousand pounds would hold off the hawks for only a month or so, damn her. He needed a good five or six thousand just to make it through the year and give him any sort of foothold on recovering his financial standing. Dunfrey stood and walked over to the window. "I wonder," he mused, half to himself.
"Wonder what?"
"Are her parents in London?"
Eloise shook her head. "I believe they're expected in a week or so. Why—oh, my," she breathed. She pulled her gloves on again and stood. "Whatever you do with her, I don't care. I simply want her away from Warefield. And if you do that, you'll have my gratitude."
"And your thousand quid," he reminded her.
"And my thousand quid. Good hunting, Mr. Dunfrey."
"The same to you, Lady Stokesley."
"I don't want to go."
Quin stood in the doorway of Maddie's bedchamber, where he seemed to be spending a great deal of his time lately. "We were invited."
"You're her betrothed. I'm already having tea this afternoon with Lady Ashton. You go. I'll stay and have luncheon with Rafe."
The marquis frowned. "I'm not her betrothed—yet. And Rafe's going with us." Whether he wants to or not. "Be downstairs in five minutes, Maddie, or I'll carry you down."
Quin strode down the hallway to the billiards room, where Rafael was playing a game against himself. The marquis paused watching him for a moment. "Back in London after a year, and you stay in your parents' house and play billiards? Alone? You?"
Rafe glanced up at him, then made another superb shot. "I'm just daft. Ignore me."
"What's wrong?"
"Nothing. Things were just...getting a bit sticky in Africa. I needed a rest."
Quin leaned his hands against the billiards table. "What's wrong, Rafael?" he repeated.
His brother shrugged. "Nothing. Really. And I think you have enough to worry about without my adding to the confusion."
"And just what do I have to worry about?"
"Like what excuse you're going to think up this year to delay marrying Eloise."
"I am going to marry Eloise this year."
Rafe eyed him distastefully. "Why?"
"Because I gave my word. And because she'll make a fine wife." Quin took a stroll about, the room. "And it wasn't an excuse before; the properties in Cornwall needed to be signed over, and I had to be there."
"Mm-hm."
"What?"
Rafael nudged a ball across the table. "You had the right idea last year," he muttered, then looked up again. "Quin—"
"Oh, shut up." Quin frowned at him. "And go change. You're going with us to luncheon."
"I most certainly am not."
"Yes, you are."
"My memories of Africa are becoming fonder by the moment, you know."
"Look. Maddie's after any excuse to avoid going back into society. You're her latest. She says you're lonely, and she wants to stay and have luncheon with you. So I told her you were coming with us."
Rafe leaned on his cue stick, his expression brightening considerably. "She wants to spend time with me? I'm flattered. Perhaps I'll take her on a picnic."
"No!" Quin said, too sharply. "Just go change, will you?"
His brother tossed the stick onto the table and strolled out the door. "Fine. I'll attend for Maddie's sake—not for yours."
Quin looked after him. He and Rafael had always gotten along exceedingly well, probably because not much bothered the easygoing younger Bancroft. At the moment, though, he wished his brother back in Africa—anywhere, in fact, except where Rafe and Maddie could become acquainted.
By the time his two reluctant companions came downstairs, he was ready to wish himself somewhere else as well. While Maddie looked lovely in a dark gray gown, she glowered daggers at him. Rafe, on the other hand, had donned his dress uniform, gold-trimmed red and black, and far too formal for anything less than a grand ball.
"Rafe," he complained.
"You look lovely," Maddie told his brother, obviously recognizing a fellow rebellious spirit.
"Why, thank you, Maddie. I think it brings out my eyes." He fluttered his lashes seductively at her as they went outside.
She laughed. "Oh, definitely. You'll be the belle of the luncheon."
"Get in the damned carriage," Quin growled.
"You don't need to order me about." Maddie glared at him once more, then plunked herself down on the barouche seat.
"Don't expect me to feel guilty," Quin retorted, stepping in front of Rafe to take the seat beside her. "This is for your well-being, not mine."
"I remain unconvinced." She leaned forward and tapped Rafe's polished black Wellington boot with one finger. "You don't actually go into battle looking this splendid, do you?"
"Heavens, no." He sat back in the seat. "The dress uniform is only for surrenders, victories, and parties."
"Then why are you wearing it now?" Quin asked, cutting in on Maddie's pointed admiration.
Rafael shrugged. "I figure I'll be declaring either the first or the second before the afternoon is over."
Quin frowned, disgruntled. "You could at least make an attempt to get along with Eloise. I don't know why you decided to dislike her, anyway."
"I believe the feeling is mutual."
"Why don't you like her?" Maddie whispered.
"Oh, no. None of that," Quin protested. "I'm working hard enough to stop the gossip about you without your participating in spreading groundless rumors about other people."
Maddie folded her arms and scooted as far away from him as she could. "You are a big bully."
Quin didn't much like that, but he wasn't certain what to do about it. He realized sitting and conversing with Eloise couldn't be easy for Maddie, and he didn't want her becoming defensive before they'd even arrived.
"We've arrived, my lord," Claymore said, from the high driver's perch.
He took a deep breath. "All right, then."
This time Rafe was quicker, and he helped Maddie to the ground while Quin stewed behind them. He caught up and took Maddie's other arm. "Might I have a word with you?"
"What are you going to warn me about now?" she asked, but released Rafe's arm and stopped to look at him.
"Nothing. I...." He reached out and straightened a strand of her auburn hair. "I just want you to like her," he said quietly.
She met his gaze. "Why do you care?"
"Because I do."
The front doors of Stokesley House opened. The butler emerged, followed by Eloise in a patterned green gown that admirably showed off her tall, slender figure.
"Quin," she exclaimed, and held out her hands to him.
He took them. "Eloise, you remember Miss Willits?"
"Of course." Eloise shook Maddie's hand warmly. "I'm so pleased to be able to help you. And I can't believe Mr. Lumley would dare speak to you that way."
"Thank you," Maddie said, her expression noncommittal.
Quin pointed at his brother, who stood observing an oak tree with apparent fascination. "And my brother, of course."
Eloise glided past him to lead the way into the house. "Rafael," she said smoothly, barely glancing at him. "You look as though you're dressed for a funeral."
Rafael gave a lazy salute and followed her inside. "You know me, ever hopeful."
Maddie's lips twitched, and she leaned closer to Quin. "Why don't they like one another?" she whispered, her breath soft and warm against his cheek.
"I'm really not certain," he returned in the same tone, relieved and grateful that she seemed to have forgotten her anger at him. "Rafe claims to have reason, but he's never explained it to me. Eloise won't talk about him. Whatever it was, it apparently happened shortly after Rafe returned from Waterloo." He shrugged. "Politics, perhaps."
She looked sideways at him, but at least she didn't disagree with him aloud. In truth, he hadn't a clue. He had merely been hoping that whatever it was, it would go away before the wedding.
"Miss Willits," Eloise said as their party headed through the library and out to the small garden, "did you have any particular friends we might induce to come calling on you, now that you've returned?"
Maddie shook her head. "No."
"Oh, come now," Eloise coaxed with a smile. "Not one?"
"There is no one whose acquaintance I would care to renew," she said flatly. "Or to recall."
Eloise looked at her for a moment. "My goodness." She turned to Quin. "This makes things rather difficult, don't you think?"
Maddie's expression shifted from defiance to humiliation, and Quin stifled an unexpected spark of anger at Eloise. She had to know Maddie might be sensitive about this, and she was generally more tactful than that. "Not really," he answered her, turning from regarding Maddie. "I wouldn't want friends that fickle, either."
Eloise looked as though she wanted to say something, but instead she gestured them to sit at the table settled in the shade of an elm. "We'll need one more place set," she informed a footman, who hurried off.
"You must tell me all about your adventures after you left London," Eloise urged Maddie, as she took her own seat.
"I don't really consider them adventures, Lady Stokesley. I—"
"Oh, please call me Eloise. I feel as though we are practically family."
Maddie looked skeptical, but smiled. "All right, Eloise."
"I'd be happy to tell you all about my adventures in Africa," Rafe broke in, helping himself to a glass of Madeira.
Eloise looked over at him coldly. "Yes, Rafael, how many native girls did you bugg—"
"You know, Maddie," Quin interrupted hurriedly, surprised at the venom in Eloise's voice, "there's no reason we have to do this all at once. We'd be wiser to feel our way slowly, I think."
Maddie looked at him quizzically. "I wish you'd said that before you threw me to the wolves at the opera."
"I didn't thr—"
"You're right, Quin," Eloise agreed. "I thought we might begin with a luncheon, the day after tomorrow. Just a few of my particular friends would be invited. And then a picnic in the country, you know, with a few more friends, yours and mine."
"Yes, that would be excellent," Quin agreed, ignoring Maddie's glare at being excluded from the plans. "I think my mother was being a bit ambitious with the opera last evening."
"The waters had to be tested." Eloise motioned impatiently at the waiting footmen to serve lunch.
"And they were full of sharks," Maddie muttered.
Rafe chuckled, lifting an eyebrow at Eloise when she sniffed distastefully. Whatever antagonism was between them had gotten decidedly worse, and Quin had every intention of finding out what was going on.
Then he looked at Maddie, uncharacteristically quiet as she watched Eloise out of the corner of her eye. He might be curious about Rafe and Eloise, but Maddie came first, he decided—not caring to question why her predicament, and her happiness, had become so important to him.
## Chapter 11
"Good Lord, that's frightening." Maddie laughed, her voice and expression delighted.
Quin looked up. Rafael leaned around the corner of the stable, a native African mask pulled down over his face, striking in contrast with his blue coat and black breeches.
Rafe took it off. "I believe it's supposed to be the Zulu god of rain. Perhaps he's meant to scare the droplets out of the sky. Quite effective, I imagine." Rafe strolled into the stable and hung the mask on a bridle peg. "You're going riding?"
Quin swung up onto Aristotle before the gelding could escape to greet his former owner. "Yes, we are," he said shortly, tired of his brother's uncanny ability to sense whenever he wanted to spend time alone with Maddie. "See you in a bit."
Maddie looked at him curiously. Not wanting to give away his thoughts, he turned his attention to adjusting Aristotle's reins. She'd been subdued all yesterday afternoon, which had caused him to spend the evening wondering whether he'd done the right thing in turning her over to Eloise. With his mother forbidden from assisting, though, his cousin was the only other choice.
His almost-betrothed certainly knew the correct people, as did he—but once he thought about it, not many of them actually seemed like anyone Maddie would want to be acquainted with. And yesterday Eloise had been sharp-tongued and out of sorts. Perhaps Rafe's presence had rattled her usually calm demeanor, but the whole venture was becoming a damned nuisance. Things had been much simpler before he'd traveled to Langley Hall—before he'd encountered Maddie Willits.
He glanced at her slender figure as his brother helped her into the sidesaddle. "How do you like Honey?" he asked, indicating the spirited chestnut mare.
Maddie smiled "She's wonderful. I'm going to teach her to come when I whistle."
"You're not supposed to whistle," he pointed out, pleased that she approved of the mare.
"Mind if I tag along?" Rafe asked.
Quin, trying to hide his annoyance, turned the restless Aristotle in a circle. "Yes."
"My thanks, brother." Rafe motioned at the head groom. "Wedders? Saddle me a beast, will you? Unless you'll let me ride Aristotle, Quin," he suggested with a sly smile at his brother.
"Absolutely not."
The three of them headed out to Hyde Park, which sparkled with dewdrops in the cool morning sun. Rafael flanked Maddie on one side, while Quin commandeered the other. No doubt they looked ridiculous, like dogs after the same bone. But at this time of day there were few others about who would notice.
After a few moments of silence, Quin cleared his throat. "This early in the morning, I doubt anyone would see if the three of us found a hollow and became acquainted."
Maddie looked startled, then rolled her eyes. "Oh, not that nonsense again."
"What?" Rafe roared. "Are you completely jolter-headed? Apologize!"
Quin kept his attention on Maddie. "No."
"Leave me alone," Maddie snapped. "I'm trying to enjoy myself."
"The two of us together would be much more enjoyable."
"Quin!" Rafe bellowed, surprisingly sounding a great deal like their father.
Quin glanced sideways at him. Just about any lady was fair game for Rafe, and it was unlike him to be so protective.
"Don't make me clobber you," Maddie warned, looking as though she very much wanted to.
"You can't go about pummeling everyone who insults you." He leaned closer. "And it will happen again—unless you'd rather surrender to the Edward Lumleys of the world," he pressed, knowing she wouldn't be able to ignore the challenge.
"Of course not. But I doubt he'll be insulting me again, anyway."
Rafe looked from one to the other of them. "What the devil is going on?"
"Anti-rake training," Maddie informed him.
"Anti—are you mad, Quin?"
"Oh, shut up," Quin grumbled, attempting to ignore his unhelpful sibling. "Of course Lumley will continue insulting you. Not to your face, perhaps, but any other time he can. And that will hurt you far more than if you'd disposed of him in the proper manner."
"With a pistol?" she suggested.
Rafael burst into laughter, and she grinned back at him. Quin didn't find it amusing at all.
"Rafe, go away," he suggested through clenched teeth.
"Oh, all right," his brother sighed, no doubt sensing Quin was ready to knock him out of the saddle. He kicked his gray gelding into a trot, heading across the park toward Rotten Row. "Let me know if I need to avenge your honor, Maddie."
She continued to glare at Quin. "That was mean, Warefield."
"Maddie, this is serious. I want you to be able to hold your head up here."
Her expression hurt, she looked away. "I can hold my head up here. I didn't do anything wrong."
He reached out and caught Honey's bridle. "I know that."
When he'd planned this anti-rake training, the difficulty had not been in coming up with suggestive situations. Rather it had been in finding ways to word them so as not to give away his feelings toward her. And it wasn't getting any easier.
"You don't have much of a choice about it, darling. You're going to hear stupid, insulting things. You're ruined, remember? So answer me in kind."
She scowled, her eyes glinting. "All right, Warefield. If I were as glib as you seem to think yourself, I would certainly be able to come up with something much more clever than that with which to insult me—darling."
Quin nodded. "A passable riposte."
She looked at him sideways, her expression still dark and angry. "I wasn't joking." Maddie wrenched the bridle free of his grip. "And I'd still rather lay you out as flat as Lumley."
"I might let you, if you'd join me there."
She closed her eyes. "I doubt there'd be room for two, what with your swelled head, my lord."
He stifled a grin. "It's not my head that's swelling."
Maddie blushed, then lifted her chin. "You have a better chance of getting acquainted with your horse than you do with me."
She wasn't shy—that was for damned certain. "You can't respond by saying something more suggestive than what I said to you."
"Oh, so now there are rules?"
"Of course there are—"
"Warefield?"
Lord Avery rode toward them, a smile on his doughy face. Unwilling to have poor, dull-witted Peter face to face with Maddie at her most spirited, he wheeled Aristotle around. "I'll be right back. Don't go anywhere."
"'Don't go anywhere,'" Maddie mimicked imperiously. "As if a lady would wait about for further insults." Immediately she turned her mare to look for Rafael. She spied him after a moment, surrounded by at least a dozen ladies in carriages, and hesitated. "I don't think so," she said to herself, nervous at the crowd, and headed instead for the nearly deserted Ladies' Mile.
Admittedly, their last exchange had been somewhat amusing, but sometimes she absolutely hated Quin Bancroft. He always believed he knew what was best for her, whether she agreed or not. And unbearably self-righteous, he never had anything pleasant, or comforting, or sweet, or romantic to say to her.
Maddie blinked and drew Honey up short. Romantic? Where in the world had that come from? Even if she did like him, even if she happened to be desperately fond of him, he would never consider marriage with someone like her. A ruined chit—that was what he'd called her—and that was precisely what she was. But Quin....
Despite all her efforts, and even though he was stupidly stubborn and probably took in stray cats and dogs just because he felt sorry for them, all of her dreams and imaginings seemed to center around him. Not even attractive, easygoing, unattached Rafael stirred her pulse and made her heart pound like Quin did.
Maddie looked down at her hands. It was completely absurd, for her to fall for the Marquis of Warefield simply because he happened to be the first young, handsome gentleman of her own social status who'd been kind to her, both before and after he'd discovered her identity. And when they kissed, the attraction was certainly mutual. But then again, perhaps he was only being polite. If he was one thing, Quinlan Ulysses Bancroft was unfailingly polite.
She sent the chestnut along the quiet track, enjoying the sensation of actually being alone for once. It had been a long time since she'd been able to do much of anything without Quin barking at her heels.
"My...my God!"
Maddie yanked hard on the reins, dragging the mare to a halt. All the blood drained from her face, and suddenly she couldn't breathe. She knew that voice—and she'd hoped never to hear it again. Her eyes closed. She couldn't even look.
"Maddie? Maddie Willits? Is it you?"
At the sound of a horse approaching, Maddie took an uneven breath and opened her eyes again. "Charles," she faltered.
Charles Dunfrey looked the same as she remembered—tall, dark-haired, and exceedingly handsome. His brown eyes gazed at her in obvious astonishment, his square, chiseled jaw hanging open. "It is you. I can hardly believe it!"
Neither could she. "Ex...excuse me," she managed, and yanked the chestnut's reins around with shaking fingers.
"Don't go. Please. Please."
Hesitating, Maddie turned to look at him again, at his hopeful, earnest expression, and tried to ignore the tide of emotions tumbling her about. He had turned her away five years ago. He hadn't even wanted to hear her explain. She should be angry—not ill and lightheaded with nervousness. "What do you want, Mr. Dunfrey?"
"I thought I'd never see you again." Charles guided his mount a few slow steps closer, as though he were afraid she might bolt.
"That was what you intended, I believe," she said stiffly, groping for anger, indignation, bitterness—anything to bolster her flagging courage.
He shook his head. "No. I was angry—furious. But when you...left, I...." Charles looked down, then met her gaze again. "I had a lot of time to think about things, Maddie."
"So did I."
"I...." he began again, then trailed off. "Good Lord, I'm just so surprised to see you, I don't know what to say. Please, tell me you're not still angry. Might...might I call on you tomorrow? Are you staying with your parents?"
"No. They...I'm staying at Bancroft House, as a guest of the Duchess of Highbarrow. My parents don't know I'm here."
"At Bancroft House?" He reached out as though he wanted to touch her hand where she tightly clutched the reins. At the last moment he stopped himself. "Might I call on you there?"
Again she hesitated, completely unnerved. "Yes. Yes, if you wish."
"Thank you." With a last glance at her, he turned and rode away.
Maddie couldn't stop shaking. She'd dreaded that meeting for so long, and it had been nothing at all like she had imagined. Nothing.
"What in damnation did he want?"
Quin looked like a knight ready to charge into battle for his distressed damsel. His green eyes glinting and narrowed, he glared at Charles Dunfrey's retreating back.
Maddie shook herself. "Nothing."
Quin looked sideways at her, his jaw tight and angry. "'Nothing?'" he repeated. "You spent a long time discussing nothing, then."
"I think he wanted to apologize."
"Apo—" He snapped his mouth shut and looked after Charles again. "And you let him just apologize? After what he did to you? To your reputation?"
A small thrill ran down her spine. He was jealous—over her. "It would make things much easier for me—don't you think?—if Charles and I were to reconcile?"
"Yes...I suppose it would," he agreed, with supreme reluctance.
She nodded. "He's going to call on me tomorrow, at Bancroft House."
He glanced at her again, then away, and she could fairly hear his teeth grinding. "Fine. Splendid." Quin wrenched Aristotle around. "Let's go. Where's my damned brother?"
Quin knew his mood had deepened beyond foul when, less than five minutes after their return from Hyde Park, both Maddie and Rafe deserted him to find the duchess and challenge her to a game of piquet.
Damn Charles Dunfrey, anyway. And damn Maddie, for of course being right about a reconciliation between them. He could toil all summer in an attempt to repair the damage to her reputation, yet Dunfrey could smile at her once in public and do the same.
There was no use denying it any longer, though the very idea made him want to smash some very expensive breakables. He didn't just want Maddie to be restored to society; he wanted to be the one to do it. He wanted her to be grateful to him. He wanted her to need him—and to love him as much as he did her.
His heart pounding, Quin leaned back against the wall and stared at the closed door to the drawing room where they sat. Where she sat.
Sweet Lucifer, he loved her. Of all the idiotic things he'd ever thought or done in his entire life, this was the worst. Even if she hadn't been rained, Madeleine Willits was no one with whom he could consider anything more serious than an affair. And at the moment, he would have been happy—ecstatic—to have that.
He could hear the three of them in the drawing room, laughing and chatting while they played cards. Even the duchess had warmed to Maddie. Yesterday she'd accompanied Miss Willits to Lady Ashton's, practically daring His Grace to comment. And she'd convinced the duke to delay sending word to Malcolm that they were returning his disgraceful companion to Langley posthaste.
"What are you moping about now?"
Quin jumped, straightening. "I'm not moping," he said stiffly, as the duke emerged from his office, a fistful of papers in one hand. "I'm deciding."
"Deciding what?" His Grace asked skeptically.
Whether to tell Maddie how I feel about her. "Whether I should plan for a summer wedding or an autumn one," he said instead, remembering Eloise and their twenty-three-year agreement with a kind of detached horror. He looked at his father, seeing the swiftly masked surprise on his stern face.
The duke regarded him levelly. "Why not tomorrow, if you're suddenly so eager?"
"Fine by me," Quin snapped, furious and unsettled and trapped. "I'll send a note over to Eloise."
"Don't try to bluff me, Quinlan," Lord Highbarrow warned.
"I'm not," he said shortly, and turned on his heel. "You'd best send a messenger to King George and tell him we'll be needing Westminster Abbey in the morn ing," he continued over his shoulder. "I don't imagine that will be a problem."
"So you intend to have the entire peerage thinking you've got Eloise with child and I forced a quick marriage? I should say not!" the duke roared, his expression darkening. "You haven't been that much a fool, have you?"
Quin faced his father again. "I thought I was rutting with Miss Willits," he snarled, white-faced. "Make up your mind, Your Grace."
"Don't you dare speak that way to me, boy, or I'll see you as well liked in London as that red-haired whore!"
That was enough of that. "You will not speak about Maddie in that manner, you pomp—"
"I'm not the one—"
The drawing room door opened. "Lewis," the duchess interrupted in a low voice. No doubt the three of them had heard the entire exchange from the drawing room. Quin winced.
"Victoria, stay out—"
"Please," she interrupted. "Lady Finch and Lady DeReese will arrive here any moment. Calm down."
"Calm down? You tell me what to do, wife? Bah! I'm going to White's!" His Grace stalked down the hallway. "You'll wed Eloise this summer, Quinlan, or when it comes time for a new Duke of Highbarrow, it won't be you! Is that clear?"
Quin didn't answer; he wasn't expected to. His father had given a direct order, and it would be followed. End of argument, end of conversation. He met his mother's concerned, searching gaze, then nodded stiffly at her and turned on his heel.
Aristotle looked annoyed at being taken out twice in one day. Under the circumstances, Quin had little sympathy for him. He rode over to Queen Street and asked if Eloise was in. She wasn't, but the Stokesley butler gave him the direction of the acquaintance she'd gone to visit.
He felt ridiculous chasing Eloise about London. They'd known one another for so long that he could fairly well predict what her reaction would be if he appeared on the Countess Devane's doorstep, looking for her. She was lovely and intelligent and had been groomed from birth to be the future Duchess of Highbarrow—just as he had been schooled to be the future duke. But he wanted to know something that had abruptly become very important for him to discover. He wanted to know what he felt when he was with her.
He knew what he felt when he and Maddie were in the same room: frustrated, antagonized, and exhilarated. In truth, whether he felt anything toward Eloise, it didn't matter. He had always known he would marry her, and so he would. But he continued to Devane's home anyway, climbed the shallow steps, and rapped on the door.
"Lord Warefield." The butler bowed as Quin handed over his calling card and his request. "If you would care to wait in the foyer."
Only a few moments later Eloise appeared from the direction of the upstairs drawing room. "Quin, is something wrong?" she asked, descending the steps toward him.
"No," he said, taking her hand. "I just...wanted to make certain I wasn't imposing on you the other day, when I asked you to help me with Maddie."
She smiled warmly. "Of course not. In fact, I was just arranging for Miss Harriet DuChamps and Lady Devane to join us for luncheon tomorrow."
"Good. I appreciate your assistance."
"I'm happy to help." She looked at him for a moment, her perfect brow furrowing just a little. "Was there something else?"
"No. No, of course not." He started to turn away, then stopped again. He had to know. "Eloise, might I...make a request of you?"
"Anything, Quin."
Quin glanced up and down the hallway, which was thankfully deserted, and cleared his throat. "Might I kiss you?"
The brief look of puzzlement passed from her face, and she smiled again. "I would like that."
Taking a short breath, Quin stepped closer. He lowered his head as she lifted hers, and he brushed his lips against her soft mouth. For a long moment he lingered there, tasting her mouth, hearing her soft sigh.
Finally he stepped back again. "Thank you."
"Well," she prompted, smiling faintly, "how was it?"
He returned her smile. "Wonderful, Eloise. I just realized I had never kissed you in all this time. I'll see you tomorrow." He bowed and turned away.
"Quin?"
The marquis stopped. "Yes?"
"We need to decide on a date. If we delay much longer, no one will be around to celebrate with us."
"Yes, I know. I'm...working out the schedule with His Grace. Soon, though. It will be soon."
He made his way outside, and back to Aristotle. There he stopped, one hand on the gelding's bridle. She'd given him the perfect opportunity to declare himself, and he had done nothing. Well, nothing except admit to himself that he really had no interest at all in his wife-to-be—and a great deal of interest in a woman who could never be his.
"Hm, and what did Lord Warefield want?" Joanna, Lady Devane, curled a strand of her blond hair around and around her finger.
Eloise smiled and resumed her seat. "To kiss me," she murmured, and sipped her tea.
Harriet DuChamps sat forward. "To what?"
"To kiss you?" Joanna repeated skeptically. "He came all this way just for that?"
"We are to be married, you know," Eloise pointed out. "And he does dote on me."
"Seems to me he dotes on someone else these days," Lady Devane suggested.
"Quin's always been kind and generous. The poor little ruined bitch had nowhere else to go." She set aside her tea and leaned forward. "And our task, ladies, is to find her somewhere else to go. Posthaste."
Harriet giggled. "They drown unwanted puppies, don't they?"
Joanna and Eloise laughed, and Eloise resumed nibbling at her teacake. "I'll set a large punch bowl at luncheon tomorrow, just in case."
Charles Dunfrey sighed as his coach rattled to a halt. What a blasted nuisance, having to leave London in the middle of the Season. And for a trip to Devonshire, of all places, where there would be absolutely nothing of interest to do, and no one of interest to see.
Half surprised the vehicle had made the journey intact, Dunfrey settled his hat on his head and stood as the door opened. "Good evening, Hoskins," he said, stepping to the ground. "Would the viscount be in this evening?"
The butler stared at him open-mouthed for a moment, astonishment in every line of his thin, dignified countenance. "Mr.... Mr. Dunfrey. Yes, he...he is. This way, sir."
Hoskins showed him into the drawing room, and in his hurry to leave and inform his employer of their visitor, slammed the door behind him. Dunfrey smiled briefly, then wandered about, looking at the old familiar porcelain miniatures and collection of crystal vases. Little had changed in five years. He turned around as the door opened again.
The tall, silver-haired gentleman standing in the doorway looked poised between shock and dismay as he looked at his houseguest. Dunfrey could little blame him. He'd never thought to set eyes on the viscount again, himself.
Dunfrey bowed. "Good evening, Lord Halverston. I apologize for not sending advance word that I was coming, but I didn't know myself, until this morning." He gave an apologetic, slightly embarrassed smile. "Might I trouble you for a glass of port? I'm...a little unsettled."
The viscount nodded warily and motioned at the butler lurking in the hallway behind him. "Hoskins, port." He stepped into the drawing room and closed the door.
For a moment Dunfrey wished he'd asked for Lady Halverston, as she would be easier to deal with, but he didn't want either of them to go running off until he'd had a chance to explain things properly.
"Forgive my directness, Charles," the viscount said, in his dry voice, "but what brings you to Halverston? We did not part well, last time we spoke."
Shaking his head, Dunfrey sat at one end of the couch. "No, we did not," he said earnestly. "And I wish to apologize for that, as well. I...well, heat of the moment, you know."
The viscount nodded.
Dunfrey shifted, genuine nervousness augmenting his intentional appearance of agitation. If things went badly this evening, he wouldn't be willing to wager over his ability to avoid debtors' prison. "Well. I don't quite know how to say this. Ah, I—this morning, I saw...I saw Madeleine."
Lord Halverston's face went white. "Madeleine? You saw Maddie? My daughter, Maddie?"
Dunfrey hurried to his feet and helped Viscount Halverston into a chair before his knees could buckle, while his own mood continued to lift. Given the circumstances, Robert's continued interest in his daughter's whereabouts could only bode well for him—he hoped. "Yes. Actually, I spoke to her."
"Where is she?" Robert Willits asked, gripping the arms of his chair.
This would be the difficult part. He needed to make himself essential to all this. If Halverston thought himself able to go around outside assistance to get to his daughter, everything would be lost. "In London."
"Lon—where in London?"
"My lord, she seemed none too eager to speak to me, or to speak of you, other than to say that you didn't know she was there. Forgive my curiosity, but I...assume that you have not reconciled with her?"
"We haven't been able to find her to do so," Lord Halverston admitted, deep reluctance edging his voice. "Is she well?"
"She is beautiful," he answered truthfully. "Even more so than she was at eighteen." In fact, it had been almost disappointing to see her looking so well. She hadn't pined over him a bit, no doubt.
"Did she say where she's been? Is she—"
"Please, my lord." Dunfrey offered Robert another embarrassed smile. "I spoke to her only briefly. I...didn't want to press her tolerance. I have a great deal to make amends for, where she is concerned."
The viscount looked at him assessingly. "You wish to make amends, then?"
Dunfrey nearly smiled at Halverston's hopeful tone. "Yes. Yes, I do. You know that I married after Maddie disappeared. My wife...Patricia was dear to me, but she has been gone for over a year now. And when I saw Maddie this morning—well, I realized that she has stolen back into my heart, Robert. Time heals all wounds, they say."
"So they do, Charles."
"I am to call on her tomorrow. I thought, though, that you would want to hear my news immediately. And I also wanted your permission to proceed."
"You still want to marry her, then? Even though you know nothing of her whereabouts for the past five years?"
This was almost too easy. "My lord, of course I know I'm taking a risk with my reputation. I may very well be censured for my actions if I renew my offer to take Maddie as my wife. And the Lord knows, since Patricia's death, things have not been easy." He shrugged. "But I have come to believe that I owe Maddie another chance."
The viscount sat forward, his color returning. "You are a good and understanding man, Charles. I have always thought so."
"Thank you, Lord Halverston. It is my dearest hope that I shall be able to convince Maddie of the same."
"Where is she, then?" the viscount repeated.
Dunfrey had hoped he would forget the question. That, he supposed, would have been too much to hope for. "I will tell you, but might I suggest a plan of action first? No one wants her to flee again, I'm certain."
"No, of course not," the viscount agreed hastily. "What's your idea, Dunfrey?"
"Well, she has apparently won the favor of the Duchess of Highbarrow."
"Highbarrow? My goodness!" Lord Halverston looked stunned for the third time that evening. "The Duchess of Highbarrow?"
"Yes. She admitted to me—reluctantly—that she was staying in the Bancroft household. That is where I'm to call on her tomorrow afternoon."
For the second time, Viscount Halverston looked hopeful. "That's an exceedingly good sign, I would say. Please proceed, Charles."
"Of course," Dunfrey agreed, stifling a triumphant smile. He had Halverston now. "I think you should repair to London, in the—"
"Yes, at once," the viscount said eagerly.
"No, no...she would know that I had betrayed her trust. You must delay a week or two, and then come to London on some pretext or other. Then we can carefully arrange for you to come across her as if by accident...as I did."
The viscount was nodding. "I agree. We don't want her upsetting the duchess and taking flight again."
"Absolutely not. I don't wish to risk losing her again." Or her dowry. That, though, could be negotiated later, once he had her safely in hand.
Halverston took a breath. "Nor do I."
Dunfrey stood. "Splendid. I should get back to London posthaste. It would never do for me to miss calling on her."
"You have my deepest gratitude, Charles."
This time Dunfrey's smile was genuine. "Thank you, Robert."
Maddie looked about at her newfound acquaintances. Quin continually reminded her that she was of noble birth, and that she had as much right to hold her head up as anyone else. Rare and appreciated as the compliment had been, he really had no idea what he was talking about.
Even before she'd been ruined, she'd never moved in circles this golden. Daughters, wives, and sisters of this duke, that marquis, and a twelfth-generation viscount surrounded her, gossiping and nibbling daintily on pastries. She'd seen most of them during the short course of her debut Season but had never imagined actually being invited to luncheon with them. She stifled a grimace as a crumb fell from her peach pastry onto the floor. Even less had she thought to be the reason for such a luncheon.
"He actually fell over?"
Lady Margaret Penwide covered her mouth with her hand as she chuckled. "Oh, no. Mrs. Grady stopped his fall."
Eloise, seated beside Maddie, smiled at her and briefly squeezed her hand. The gesture was no doubt meant to be encouraging, but it caused another piece of pastry to break free. This one landed on Maddie's foot.
"She didn't!"
"I suppose it wasn't her fault, for given her girth, she undoubtedly couldn't get out of the way. But she ended up standing there in the middle of Hyde Park with Francis Henning hanging onto her bosom with both hands."
The rest of luncheon went like that, with someone revealing an embarrassing piece of gossip about a mutual acquaintance, and everyone else laughing over it. For Maddie the barbs seemed a little too familiar to be amusing. At the same time, five years ago she hadn't been all that different from Eloise and her friends. She had been a fool.
"Miss Willits?"
She looked up, surprised, as a dark-haired lady sat beside her: Beatrice Densen, she vaguely remembered, a refined lady several years older than Maddie, with a reputation—at least five years ago—for giving elegant salon parties. "Miss Densen," she replied.
"Excuse me a moment, my dear," Eloise said from her other side, and stood. "I need to see to the desserts."
"Of course."
"Miss Willits, if I may be so bold, I have always thought society treated you very cruelly," Beatrice said, taking Maddie's hand.
The abrupt intimacy left her feeling rather uneasy, but she smiled. At least someone was bothering to talk to her instead of simply staring, or worse yet, watching out of the corner of their eye. "Thank you."
"My brother, Gaylord, and I were planning a quiet evening tonight. Could I entice you to join us? Gaylord is a fair whist player."
Maddie smiled. Quin probably wouldn't approve, because it wasn't part of his carefully laid-out plan. "I would love to," she answered.
Beatrice smiled back at her. "I will come around for you myself, at seven, then."
"Thank you."
At half past two, the Warefield coach clattered onto the Stokesleys' short drive to bring her back to Bancroft House. Maddie looked at the red and yellow crest emblazoned on the carriage door in disgust. For someone trying to repair her reputation, Quin certainly had an odd way of going about it.
She'd heard the speculation, carefully out of Eloise's hearing, over why the marquis might be staying with his parents rather than his own perfectly lovely Whiting House. And then she heard her own name as the possible reason.
Still, she decided as she climbed into Quin's coach, at least they had all been kind and polite to her face. She hadn't expected even that much courtesy from them. And she'd handled herself rather well. She hadn't spilled any more than a few crumbs, which she had managed to hide beneath her skirts, and she'd been invited to a dinner. Altogether, she supposed she'd won some sort of victory.
When the coach entered the Bancroft House drive, though, she changed her mind. Charles Dunfrey's carriage stood there already, waiting. A flutter of nervousness quaked through her. Luncheon had been the easy part.
"Miss Willits, you have a caller," the butler informed her.
"Yes. Thank you, Beeks," she said, her fingers shaking as she removed her hat and shawl, handing them over to his care.
The butler nodded. "You will find him in the drawing room." He hesitated. "Best of luck, Miss Maddie."
She looked at him, surprised. "Thank you."
Her heart pounding, Maddie slowly climbed the stairs. With each step she tried to convince herself that whatever Charles said or thought about her didn't matter. He'd broken off their engagement, and she'd made a life for herself completely independent of him and her parents.
At the top of the stairs, she stopped. Quin stood in the doorway of the library, an open book resting in his hands. He glanced up at her, jade flashing beneath long, black lashes, and then went back to his reading.
"How was luncheon?" he said to the book as she passed by.
"No one called me any terrible names," she returned.
"Did anyone speak to you?"
"It is none of your affair."
"I think it is," he said with more heat, lifting his head again to look at her.
"Then you are wrong."
Before he could reply to that, she opened the drawing room door and stepped inside.
Charles stood as soon as she entered. He looked as uncomfortable as she felt, which actually left her a little more at ease. "Mr. Dunfrey," she said, in as calm a voice as she could manage. "Good afternoon."
"Must we begin with such formality?" he asked. "Please call me Charles."
She nodded. "Very well...Charles. Shall I ring for tea?"
He looked at her for another moment, then visibly shook himself and motioned for her to take a seat. "Yes, please."
They were both silent as a footman entered with a tea tray, then vanished again through the open door. Maddie dearly hoped that Quin was not still standing in the library doorway, where he would be able to hear clearly everything that was said. But even if he was, she didn't dare close the door. Here, she most especially needed to behave with propriety. Unnecessary and pointless as she had decided Charles's apology would be, she wanted to prove to him that he had been wrong about her.
"I cannot get over how beautiful you've become," he said into the silence, and she jumped. "And you were a rose among thorns before."
"Thank you. You haven't changed, I don't think."
He chuckled. "You are very kind, my dear."
The mantel clock softly chimed the hour, while Maddie sipped her too hot tea and tried desperately to think of something to say. "I heard that you married," she finally ventured.
"Yes, yes. Patricia Giles. She was several years older than you, I believe. From a good family, though."
"I was sorry to learn of her passing."
Charles nodded. "Thank you. You have a good heart, Maddie. I don't know if I could be so generous, were our positions reversed." He bowed his dark head for a moment. "Maddie, I broke with Spenser the night you...I...I saw the two of you. He—"
"Charles, I—"
"No, please," he cut across her interruption. "Let me say this. He wrote me several months ago, confessing that he'd been drinking and that his attentions to you had been unwelcome."
Maddie looked at him for a long moment, a thousand thoughts tumbling through her mind. "So you know the truth."
"Yes. Actually, I think I realized it quite a long time ago. When I first saw the two of you, I was so angry...jealous and hurt, I think. I wanted to be the only man you'd ever kissed."
An image of Quin jumped into her head. His warm lips, the light in his eyes when he looked into hers..."I wanted that as well, Charles. But that is an impossibility, and I will not dwell on it."
He sat forward, taking the teacup and saucer from her hands, and grasping her fingers. "I do not want you to dwell on it," he said earnestly, holding her eyes. "You have suffered, away from your family and friends, for five years. And not because of your own actions, but because of mine."
"Charles...."
He knelt at her feet. "Maddie, do you think, with your generous heart, that you might perhaps—not right away, of course—but do you think eventually you might be able to forgive me?"
She'd dreamed of this, in the first few months after she'd fled London—dreamed of everyone who'd been so awful to her, coming and begging on their hands and knees for her forgiveness. And even five years later, it still felt quite...satisfying. "Yes, Charles. I think I might be able to forgive you."
He smiled. "Thank you, Maddie."
"I—damn it!"
Maddie jumped again, pulling her hands free while Charles swiftly stood. Rafael stumbled into the drawing room with none of his usual grace.
"Rafe, what—"
"Beg pardon, Maddie. Tripped, or something." He turned his attention to Charles. "I say, you're Dunfrey, aren't you?" Rafe strode forward and clasped her former betrothed's hand. "Rafael Bancroft."
Charles looked at him somewhat warily as he retrieved his hand. "I'm pleased to finally meet you, Captain. I've heard a great deal about you."
"Only the good parts are true." Rafe grinned and winked at Maddie. "That your barouche out front?"
"Yes, yes it is. I hope—"
"Fine pair of bays you have there. Wouldn't be interested in selling 'em, would you?"
"My—well—I really hadn't thought about it."
Rafe clapped him on the back, leading him toward the door. "Well, think about it, Dunfrey. I might be willing to part with as much as a hundred quid for the pair, if they're sound."
"A hundred...." Charles looked over his shoulder at Maddie, who sat watching the men's departure with a mixture of relief, disbelief, and amusement. "Maddie—Miss Willits—might I call on you again?"
"Yes, you may."
They disappeared down the hallway. Maddie took a deep breath, and sinking back in the well-cushioned chair, slowly let it out again and closed her eyes. Charles Dunfrey still liked her. Handsome, witty Charles Dunfrey had apologized, and he wished to call on her again.
"The rat's gone, is he?"
Maddie opened one eye to regard the tall, lean figure in the doorway. "As if you didn't know."
"And what do you mean by that, pray tell?" Quin folded his arms across his chest.
The other eye opened as well. "You practically threw poor Rafe in here on his head."
"I did no such thing."
"Well, I really don't care, one way or the other. Miss Densen invited me to dinner with her and her brother." She stood.
"Having dinner with the Densens is not part of the plan," he said, straightening to block her exit from the room.
"You're just angry because perhaps you're not quite so necessary anymore, Warefield," she shot back at him, shoving against his hard chest with her palm and stalking past him.
Quin looked after her, his eyes narrowed. "'Not quite so necessary,'" he mimicked darkly. "Ungrateful chit."
## Chapter 12
"Quin, I have to admit, your little project is marvelous." Eloise hid the words behind her fine ivory fan. "There's nothing like a diversion to liven up the Season. And you were right, of course: Maddie is actually quite nice, if a bit quiet."
"'Quiet?'" Grateful for the darkened opera house, Quin lifted an eyebrow in keen curiosity. "How so?"
"Well, perhaps 'quiet' isn't precisely the right word. But you really can't blame her for being reserved. I would be a bit timid myself, not knowing how anyone would react to my presence. I practically had to tie Lady Anne Jeffries to a chair to convince her not to cut poor Madeleine and actually leave the luncheon."
"Maddie seemed to think it went fairly well," Quin said in a low voice. At least the opera below was fairly energetic, so no one was likely to overhear the conversation. "She spoke of a dinner invitation this evening." Actually, she'd thrown it in his face, but he didn't care to go into that.
"Yes. I advised her against it, but I think she was just grateful to have been invited."
Quin straightened. "What do you mean, you 'advised her against it'? Miss Densen is a good friend of yours, is she not?"
"Beatrice is, yes—if a little...eccentric. But I would not vouch for Gaylord and his cronies. They—"
"'They'?" Quin repeated sharply, suddenly and absurdly alarmed. "Maddie said it was to be a private dinner, with just the Densens."
Eloise rapped him on the arm with her fan. "You need to spend more time in London."
"So enlighten me."
She sighed. "Gaylord has been holding mixed-gender card parties at his home for better than a year. They began quite modestly—I even attended one myself. But lately—well, very few virtuous ladies attend any longer." She shrugged. "As I said, I tried discreetly to warn her. But Maddie's...obstinate nature is what got her into such trouble before, no doubt."
Densen's mansion was only ten minutes or so from the opera house. Quin stood. "I should go get her."
"Don't you dare leave me here to go chasing after Maddie Willits," Eloise protested. "You've already gone far beyond settling any debt. And I have listened to enough rumors about why the Bancrofts have been so helpful and generous to a little social insignificant like her."
Slowly Quin retook his seat. "I beg your pardon?" he murmured angrily, even though she was clearly correct.
She reached out to put a hand on his arm. "It's what everyone is saying, Quin. I wanted you to hear it. Don't be blinded by your wish to do a good deed."
"I'm not blinded by anything," he returned firmly, if not absolutely truthfully.
She sat back and looked at him. "Very well. I am only concerned. Your first duty is to your family."
Angrier still at her censure, he took a breath, flexing his shoulders to try to release some of the tension. "I am aware of that, Eloise."
"We must remain friends, Quin," Eloise said. "I know you are fond of Maddie—you've always taken pity on poor, lost creatures. I only ask that you keep your obligation to help her in perspective."
She was right—again—and he still didn't like hearing it. He still wanted to rush off and rescue Maddie from her own poor judgment. As Eloise had said, though, Maddie's obstinacy in going her own way had likely caused all her troubles in the first place. And he had his own troubles—obligations—to take care of.
"Eloise, might I escort you to Bond Street tomorrow?" he said, by way of answer. "I believe we have something of mutual interest to discuss."
She smiled. "It would be my pleasure, Quin."
Assuming a stolidly stone-faced expression, Maddie stepped past Beeks into Bancroft House's main hallway. They knew only that she'd gone to dinner with the Densens. There was no need for anyone to hear anything further, nor any explanation as to why it had taken her so long to return to her temporary haven. Especially Quin: she'd never hear the end of it if he learned that she'd been to a raucous card party and practically had to bribe the butler to find her a hack so she could leave.
"How was dinner?"
Maddie jumped and whipped around to see Quin exiting the darkened morning room, his expression tense and angry.
"What were you doing in there?"
"Reading," he said shortly. "How was dinner?"
No doubt sensing trouble, Beeks flashed her a sympathetic look and fled downstairs into the kitchen. Maddie put her hands on her hips. "Reading, in the dark? You were spying on me, waiting for me to come back."
"Do you really think I have nothing better to do than sit around and wait for you?"
"Apparently not." Maddie flounced past him, heading up the stairs to her bedchamber.
He followed right behind her. "At least tell me whether you enjoyed Gaylord Densen's company."
A flush reddened her cheeks. "He's very amusing."
A hand snaked around her waist and jerked her sideways with surprising strength. Trying to keep from falling over, Maddie grabbed onto Quin's shoulder and arm as he half-dragged her into the drawing room.
"What in the world are you doing?" she demanded, staggering away from him.
Quin closed and locked the door, then turned to face her. He leaned back against the sturdy oak frame. "We are going to have a little chat."
"More of your stupid rules? How can you possibly have thought up still more of them?"
"Practice."
"I'm very tired. I'd like to go to bed."
"Without telling me about Gaylord Densen's little card party?"
Maddie snapped her mouth shut, any thought of confessing that she'd been misled about the nature of the evening's engagement vanishing. He had no right to act so superior. "What do you wish to know about it, then, my lord?"
He stepped closer to her. "Why didn't you listen to Eloise's advice about going there? She warned you about it."
"She did no such...." Maddie looked at him, then walked over to the writing desk under the window to give herself a moment to think. She couldn't possibly tell Quin; he'd never believe her. "I don't know," she said instead.
"You don't know?"
"Oh, leave me alone."
She tried to push him out of the way, but he didn't budge. Instead, he grabbed her hands and pulled her up against his chest. "You are too bright for me to believe that you don't know why you would attend a card party with a herd of disrespectables. Do you really want that badly to leave here? Or was everyone correct about you encouraging Spenser?"
Her heart wrenched. "How dare you?" she spat, jerking free of his strong grip and stalking back across the room.
"A little less witty tonight, aren't you, dear? Exhausted, no doubt."
When she turned to glare at him, his jade eyes glittered with anger. But fury wasn't the only expression that touched his face. He desired her. He wanted her—and whatever tremors of need and want that triggered in her own soul, it abruptly made several things quite clear. "That's why you dragged me to London in the first place, isn't it?"
His expression darkened further. "What are you talking about?"
"When you kissed me at Langley—you as much as said you wanted me to be your mistress. You still do. You think I really did encourage Spenser—so why would I have any objection to the great, grand Marquis of Warefield's attentions?"
He strode toward her, then with obvious effort stopped himself, clenching his fists. "That is absolutely not true, and you know it, Maddie."
"Then why am I here?"
Quin glared at her. "Because Uncle Malcolm asked me to bring you here—and because I behaved poorly and wanted to make amends for it."
She narrowed her eyes. "Oh, really?"
"Yes, really." He jabbed a finger at her. "The least damned thing you could do is cooperate a little. For God's sake, would it kill you to admit that you're grateful to me?"
She pounced on the word. "Grateful? Grateful? For what? For being able to have stupid, drunken men try to grab my breasts and make idiotic jokes about me?"
"What in damnation are they supposed to think, when you attend gatherings like that? If you behave like a slut, they'll treat you like one!"
The vase was in Maddie's hands almost before she realized she'd snatched it up. With a furious hiss, she dashed the contents into his face. "Self-righteous ass!"
Water dripping down his finely chiseled nose, Quin grabbed her arm. Maddie, more furious than she could ever remember being, wrenched free of his grip. With a loud rip her delicate sleeve tore off in his hand. He glared at it in shock, then flung it to the ground and advanced on her again. "Lightskirt!"
"Oh, now you've hurt me," she taunted, and kicked him in the knee. "Tearing a dress you paid for—you beast!"
"Ow, damn it! That's right—I own almost everything you have on." He yanked off her other sleeve as she gasped in outrage. "And I get nothing in return. You probably gave more to Gaylord and his cronies tonight for fun!"
She grabbed up a porcelain miniature and hurled it at him. "Bastard! You said you didn't want anything in return!"
The diminutive Caesar hit his shoulder and fell to the carpet in a hundred pieces. Quin grabbed the room's second vase, and a cascade of cold water and daisies doused Maddie. "I don't anymore!"
She shrieked and flung pillows from the long couch at him. "Liar! I can't even imagine how dull your life must be—no wonder you keep me about!"
"That little error will be remedied tomorrow. And my life is perfectly happy without you in it!" he yelled, throwing a pillow into her face.
She hurled it back at him. "Ha! So that's why I'm all you talk about with your mother and brother—because you're so exciting."
"Hoyden!"
He grabbed for her again and she spun away, but her skirt came to a sudden stop without her. Ripping free of its stitches, it tangled around her legs and sent her tumbling against the writing desk. Her hair, drenched and coming down from its clips, hung in her face. She swiped it out of the way and spied the brass letter opener engraved with His Grace's initials. "You arrogant, stuffy—" She lifted the letter opener and swung her arm at him.
One of his waistcoat buttons flew off, the threads neatly slashed in two.
"That is why you want me about," she panted, her heart beating so furiously she thought it must explode. He backed away warily, looking for an opportunity to grab her weapon away from her. "Because you are so very dull."
"I am not dull."
Slash went another button to the floor. "Dull!"
He stopped when his back came up against a bookcase.
"Dull!" Through her fury, an odd, fluttering tingle began along her nerves, making her hand shake.
His eyes met hers, his anger deepening into something else entirely. "Damned nuisance," he growled.
The last button came free and rolled beneath the couch. "Dull," she breathed.
Quin grabbed her chin in his fingers and tilted her face up; his mouth closed roughly over hers. He pried the letter opener from her hand and flung it into the corner.
Maddie's pulse raced as her frustrated outrage swept into an equally fierce, wanting desire. She pressed herself against him, pulling his coat and then the buttonless waistcoat from his shoulders. She twined her fingers into his damp honey-colored hair and kissed him hungrily, matching his angry passion with her own.
He took the few threads of dress remaining around her shoulders and ripped them in two. His strength was a little frightening, yet wonderfully exhilarating. "Maddie," he murmured hotly, turning them so that she was the one pinned hard against the bookcase.
She couldn't stop kissing him. She didn't want to stop kissing him, and touching him. Her dress was nothing but a tattered rag hanging about her waist until he tore it free and dropped it to the carpet, leaving only her thin, flimsy shift covering her.
He pulled his fine lawn shirt free of his breeches, trailing his mouth along her jawline. What they were doing, she knew distantly in the part of her brain that remained sane, was very, very wrong. And she didn't care. All that mattered was that he didn't stop.
Maddie moved his hands out of the way and tugged the wet shirt over his head. She moaned as his lips found the hollow of her throat, and her fast-beating pulse. Her heart fluttered wildly as he pushed the one remaining strap of the shift from her shoulders, and kissed the bare skin it revealed.
"Oh, God," she murmured, gasping for another unsteady breath as the shift slid down her body to the floor.
He pushed her harder against the bookcase, as though trying to bring them still closer together. Quin covered her mouth with his again, as if to stifle any protest she might make. His hands slipped from her shoulders and down to her breasts, and she drew another ragged breath at his intimate touch.
Maddie ran her hands down his hard, smooth chest, and the muscles jumped beneath his skin. He captured one of her hands and lowered it to the fastenings of his breeches. She could feel his hard, growing arousal, and with rumbling, unsteady fingers undid the fastenings and freed him from confinement.
Twisting, Quin half pushed and half carried her down to the floor. His hot, hungry mouth immediately sought her breasts, and she tangled her hands in his hair and arched against him as her nipples hardened at the caress of his tongue. He must be insane, to want her—and to want her in his parents' own drawing room—and she must be equally mad to encourage him.
His hands swept down her flat belly to her rounded hips, squeezing and kneading her buttocks and pulling her against him. Quin's mouth claimed hers again as he stretched the long, lean length of his body atop hers. His skin against hers was warm and intimate, and when his knee nudged hers apart, she arched her hips again, feeling his throbbing manhood brushing against her thigh.
Moaning as his mouth teased hers open, their tongues caressing, Maddie slid her arms around his strong, muscular shoulders, holding him against her. "Quin," she whispered breathlessly.
He entered her with a growling moan of possession. At her gasp of pain and surprise and wonder, he froze. "Jesus," he muttered raggedly. Burying his face against her shoulder, he held his body very still except for the tremors of tension along his arms. He lifted his head again, his dark green eyes burning into hers. "Jesus," he repeated. "You were a virgin."
Before she could gather her wits enough to say that of course she was a virgin, he kissed her again, deeply and roughly. Slowly his hips began to move again, forward and back against her, inside her, and she gasped again. "Oh, Quin, that feels so good," she groaned, lifting her hips to meet his increasing rhythm. "You feel so good."
"God, so do you," he answered, shifting his arms so he could lean down and nibble at her ear. "Hold onto me, Maddie. It gets even better."
Waves of ecstasy made her whole body tremble as he made love to her. She wasn't certain she would live through anything better than this. Her ankles lifted around his hips, as though her body already knew what to do. She threw back her head, half-closing her eyes at the erotic, intimate feeling of him, of Quin, moving inside her.
Another shiver of tension began in her most secret core, and then grew until it exploded into a pulsing, shivering jumble of indescribable delight. Maddie cried out, her entire body arching. Above her Quin thrust deep and fast into her a few more times, and then convulsed against her. Slowly he collapsed on top of her, his weight hard and muscled and welcome.
All of the fight driven out of her, Maddie concentrated on regaining her breath and her senses, and absolutely not on contemplating what she and Quin had just done. And what she already wished he would do again.
Quin closed his eyes, savoring the warm, soft feeling of her body beneath his. Having just made possibly the greatest mistake of his thirty years, he probably should feel considerably worse than he did. The most he could manage, though, was a dim sense that his complicated life had just become much more confused.
The wild pounding of his heart gradually slowed, and with a deep breath he lifted his head to look down at her. "Why didn't you tell me?"
Puzzlement showed briefly in her eyes. "Tell you what?"
"That even after Spenser, in all your five years away from London, you'd managed to remain...pure?"
Maddie scowled, then shoved at his shoulders with surprising strength. "Get off me, you big oaf."
Reluctantly he shifted off of her and sat up. Unable to help himself, he let his gaze drift down to her soft, full breasts, heaving with her angry breath. Maddie had a spectacular bosom. "Was it such an idiotic presumption?"
"Only if you know nothing about me," she snapped. "You really did believe all those rumors, didn't you?"
"No, of course not."
"Then why did you presume I was a whore?"
He blinked, wishing his brain would hurry up and catch up with his body. That had never been a problem before. "I never did such a thing. It's just that...five years is a long time, Maddie." And he'd wanted to make love to her for so long, and then she'd made him so damned angry...sweet Lucifer, he was an idiot.
Her gray eyes studied his for a long moment, while the excited flush of her cheeks slowly faded. "So you decided that since I was already ruined, you might as well take advantage."
He shifted closer to her, noting that despite her words, she couldn't help dropping her eyes below his waist for a moment. "Now, just a damned minute—I don't recall your trying to stop anything. Just the opposite, I think."
"Well," she began, blushing again, and belatedly covering her lovely chest with her arms, "I simply...got a little carried away."
At that his lips twitched in a small grin. "I'll say."
"You stop that!"
"Well, now we're in a spot."
"No, we're not. Go fetch me another dress, and we'll go upstairs to bed. The end."
"To bed together?" he suggested hopefully, caressing her cheek with the palm of his hand.
For just a moment she shut her eyes, leaning into his embrace. "This isn't funny, you know."
He chuckled. "I know. It's a damned tragedy." And one he was rather enjoying. There was a powerful connection between them, and he affected her just as she affected him. And for once, where Maddie was concerned, he actually felt in command of the moment. Still touching her cheek, he rose to his knees and leaned forward to kiss her lightly, at the same time removing her one remaining hairclip and letting her damp auburn hair tumble down her shoulders. "And in my own, excessively dull way of thinking, I do have a solution in mind."
Slender fingers hesitantly lifted to trace the muscles of his abdomen. "And what might that be?"
She still desired him, whatever better sense she had. He looked at her face, at the yearning and passion even her renewed anger hadn't been able to drive away. He wondered if she saw a reflection of the same in his own gaze. "Marry me."
Pure surprise dilated her eyes. "What?"
He smiled. "I said, marry—"
"Why?" she interrupted. "Now that I'm truly ruined, why not just keep me as your mistress or something, so that you can still marry Eloise Stokesley like you're supposed to?"
"Don't be ridiculous. At Langley I said I would do right by you, Maddie. I may have...stumbled somewhat, but those weren't just words. I meant them."
"Oh, please."
Maddie stood and grabbed her shift, the only intact piece of clothing left to her. Looking at the carnage, Quin was surprised he hadn't left bruises on her. For Lucifer's sake, his breeches were still wrapped around one ankle, and he still wore his boots. He'd had mistresses before, but never—never—had he been so out of control.
"I can hardly wait for your next 'stumble,'" she continued, pulling the thin garment on over her head and hopelessly disheveled hair. "What will it be? Rendering me naked in the middle of Almack's assembly?"
"Maddie," he protested, standing and yanking his breeches back into place. "I am serious. I'll speak to His Grace in the morning. And we will be married. As soon as I can arrange it."
"No, you won't. And no, we won't. It's stupid to throw away two lives, Quin." She hesitated, looking up at him. "Whether this had happened or not, three quarters of London thinks I've done it with someone. Nothing has changed."
Now he was offended. "Something damned well has changed."
"Just listen, won't you?" she said with more heat. "My chances of tricking some gentleman into thinking me respectable and marrying me are zero."
She shrugged, loneliness touching her eyes again, making him want to hold her—even though he half thought she would seriously damage him if he tried.
"They are not," he said.
"They are...and they always were. I came here for the sake of your uncle. And once we complete the spectacle at Almack's, I will go back to Langley. I never thought otherwise. I can't believe you did, either."
He stared at her, righteous indignation warring with his desire to take her again, right there on the floor. He'd known she was bright, but he'd had no idea someone so...impassioned...could be so wise. Too damned wise for her own good. For his own good. "But, I—" He stopped the words just in time, just before he said "I love you." If she'd thought him an oaf before, that statement would drop him into irretrievable imbecility.
Maddie bent down to snatch up the tattered remains of her lovely gown. With the bits and pieces gathered in her arms, she unlocked the drawing room door. For a moment she leaned her forehead against the cool, smooth wood, then turned to face him again. "Quin, marry Eloise. Do what you're supposed to do. You don't need me in your life."
Silently she slipped out into the hallway, and a moment later her quiet tread climbed the stairs toward her bedchamber. Quin slowly went about dressing again, and cleaning up the remains of flowers and pillows and broken glass on the carpet. "You're wrong," he whispered, lifting a mangled rose and inhaling the light scent of its broken petals. "I do need you in my life."
"Rafe, I think you're just making this up."
The younger Bancroft brother finished a graceful series of turns about the huge ballroom and ended by Maddie's side. "I am not," he protested, his voice echoing in the empty mirrored room. "It's all the rage in Paris, and I have it on the best authority that Lady Beaufort loves Paris. She'll be sure to have at least one or two of the latest waltzes, and you don't want to be left out, do you?"
She sighed. "Actually, yes."
He chuckled. "Coward."
Maddie jumped as footsteps echoed into the room behind her, but it was only the duchess. Quin had been absent all morning, and she had to wonder whether he was avoiding her, or if—even worse—he'd gone to set the date for his wedding with Eloise. "Your Grace," she curtsied.
Lady Highbarrow nodded at Maddie. "My husband is in Parliament this morning," she announced, and took the seat at the pianoforte. "And you sound as though you could use some music to cover up the sound of all that awful stomping Rafael is doing."
"You know how to play a waltz, Mother?" Rafe asked in mock amazement.
She lifted an eyebrow at her son. "What your father doesn't know won't hurt us."
"I hope not." He turned and held his hands out to Maddie. "Come, my dear, let me teach you to waltz."
"I know how to waltz."
"Maddie, we've been through this already. Do cooperate a little."
She grinned. "Oh, very well."
The duchess began to play, and Rafe swept an arm about Maddie's waist to swing her into the dance. Like most things from Paris, this waltz seemed more scandalous than its British counterpart. Rafe held her so close to him, they were practically....
She blushed, turning her face so he wouldn't see her sudden discomfiture. She would never forget last night, for it seemed she'd been waiting for Quin to hold her for a lifetime.
"Oh, so now you'll dance with anyone, will you?"
Maddie stiffened at the sound of Quin's voice. Luckily, Rafael held her closely enough that she could regain her composure without stumbling. He looked down at her curiously, but she only smiled. "Anyone but you, my lord."
Rafe nodded approvingly. "Well said. Turned the insult right around, eh, Warefield? And didn't wallop anyone."
"Yes, she did," the marquis admitted grudgingly. At least the admission seemed reluctant to Maddie, for he stayed planted in the doorway, watching her and Rafe twirl about the floor.
Quin looked as impeccably dressed and calm as always, until she glanced at his face. Unless she was mistaken, he'd gotten even less sleep than she had last night.
It would have been so easy, as he held her in his arms. It would have been so simple, to tell him that she loved him. But it wouldn't have changed anything.
Whether he returned her affections, or had simply been guided by animal lust, he had been slated to marry Eloise Stokesley for twenty-three years—since Maddie's birth, and since his seventh year. And her presence could not and would not be allowed to change such an arrangement—not between two families as powerful as the Bancrofts and the Stokesleys.
"Penny for your thoughts, Maddie," Rafe murmured, glancing over her head at his brother. "Quin hasn't frightened you, has he?"
"Why in the world would you say that?"
"Thought I heard the two of you arguing last night."
"We always argue." Maddie blushed again, and Rafe's gaze sharpened a little.
"Yes, you do. I'm surprised Quin hasn't had an apoplexy of his own by now. I thought I was the only one who dared argue with the Marquis of Warefield. Except for His Grace, of course."
"Why do you call Lord Highbarrow 'His Grace'? Both you and Quin do it," she asked, to turn the subject.
Rafe shrugged. "He likes it better than being called 'Father.' I heard him once, bellowing at Quin: 'Any damned ass can be a father. I'm a duke!'"
Quin strolled over to sit beside his mother at the pianoforte, luckily still out of earshot.
"Might I ask you a question?" she continued carefully.
"Of course, my lady."
"Why do you and Eloise Stokesley not...deal well together?"
His expression tightened a little as he shook his head. "It's personal."
"It doesn't concern you that your brother is going to marry her?"
"You sound concerned," he replied promptly, obviously trying to put her on the defensive. "Why, do you believe her to be some sort of maniac?"
Maddie forced a smile. "Of course not."
"Rafe, might I have a go at that?" Quin asked, rising again.
"It's a bit modern for you, Quin, don't you think?"
"Very funny," Quin said dryly. "Hand Maddie over, if you please."
She didn't like the way that sounded, as though she was a piece of property. Rafael seemed to sense that, for he hurriedly relinquished her and strolled over to chat with his mother.
"Good morning," Quin said, studying her eyes as he took her in his arms.
"You might have asked me if I wished to dance with you," she snapped, trying to keep her attention on her anger rather than on the way her body just wanted to melt into his. It was deuced difficult, being hopelessly attracted to someone and having absolutely no hope of any future with him.
"Grumpy, aren't you? Am I to assume you didn't sleep well?" he continued mildly, far too calm about the whole mess, as far as she was concerned.
"No, I did not."
"Mm," he nodded. "Neither did I.I kept thinking of you."
Despite his lowered voice, Maddie couldn't help glancing at the duchess and Rafe. "Be quiet."
"Aha!" Quin grinned down at her.
"What is it, dear?" Lady Highbarrow asked.
"Nothing, Mother," he answered easily, still gazing at Maddie with an expression of idiotic triumph on his lean, handsome face.
"What 'aha'?" she muttered, trying not to scowl.
"You wish our liaison to remain a secret."
"Oh, so now it's a liaison? Last night it was a stumble, I believe," she said dryly, wishing he would find an easier topic.
"A tumble, at least," he agreed. "And an exceedingly pleasurable one."
Before she could kick him, he turned her in his arms and propelled her toward the door leading into the garden.
"We'll be back in a moment," he said over his shoulder, as he exited behind her. "Maddie's feeling faint."
"I am not feeling faint," she hissed, regaining her balance and backing away from him. "What are you doing?"
He pursued her, not stopping until he had her trapped between himself and a trio of exceedingly thorny rose bushes. "I didn't want any witnesses if you intended to become violent," he answered, and reached past her shoulder to pick a barely blooming red rose, still more a bud than a flower. Slowly he brushed the soft petals along her cheek. "You see, Maddie, if you truly thought you were ruined beyond hope of redemption, you wouldn't care if I shouted my conquest to the treetops. But you do think there's hope, don't you? Even now."
"After last night, my lord, I can't believe you're asking me the question. Our...our actions did ruin me beyond hope of redemption."
"'Our actions'? We made love, Maddie," he said softly, brushing the rose along the low neckline of her morning dress, the petals leaving a light, sweet scent on her skin. "Didn't you enjoy it? You said you did."
She shivered at the soft caress of the rose, and of his voice. "It doesn't matter whether I enjoyed it or not."
"Yes, it does." He leaned closer, replacing the rose petals with his lips in a feather-soft touch of his mouth to hers. "Did you enjoy being with me, Maddie?"
She drew a ragged breath, wanting nothing more than to fall to the ground with him and repeat exactly what they had done last night. "Yes."
Quin smiled. "So did I. Very much. Though the next time, I'd like to take more time, to...be more thorough."
"The next time?" she repeated, hoping the sudden heat running beneath her skin didn't show on her face. "There will not be a next time. You know that as well as I do."
"I'm fairly stubborn for a dull gentleman, wouldn't you say, my dear?"
"I...I didn't mean that, Quin," she said reluctantly. "You made me very angry."
"Even so," he answered softly, gently kissing her once more, and making her pulse begin to flutter all over again, "your point was taken."
Maddie leaned up for another kiss first, in case they should begin another argument, then narrowed her eyes. "What do you mean?"
"Everyone else has been running my life since I can remember. I've put up with it because I considered it to be my duty, and because it really wasn't all that difficult to tolerate. Until now."
"Oh, no you don't," she warned, slipping past his shoulder and backing away from him in alarm. "You will not use me as an excuse to rebel against your family. Don't be stupid. You have far too much to lose."
She couldn't tell if he was listening or not. He kept nodding, but his expression seemed anything but compliant. Rather, he looked like the idea of making love to her in the garden's soft grass appealed to him as much as it did to her.
"I'm going inside," she stated, holding up a hand to ward him off. "Charles has invited me to accompany him on a picnic. I need to change."
He stopped, his expression darkening. "Very well," he said stiffly. "Go. I'm to meet Eloise, anyway."
"Very well," she repeated, realizing that she didn't like Eloise Stokesley at all. "Please be sure to tell her I thank her for her advice about the Densens. I shall be more cautious next time." More cautious about Eloise and her friends, at any rate.
"And so she should be more cautious," Eloise agreed. She glanced over her shoulder to see Lord and Lady Pembroke and their daughter Lady Froston walking behind them, and she boldly wrapped her arm through Quin's. "For heaven's sake, I don't know how much more clear I could have been."
The marquis nodded at her. "Maddie is in agreement."
Again, though, his tone was rather absent, as though his mind was elsewhere. Eloise could guess where. "Good. If she expects me to see her reintroduced into society, she must at least cooperate with me."
His lips twitched. "She's not very good at that, I'm afraid."
"Honestly, she isn't good at much, it seems to me."
His arm tensed beneath hers. "I would appreciate if you didn't repeat that sentiment. We're here to quash rumors," he said quietly, steel beneath the soft tones, "not spread them."
So she'd gone too far and insulted the little mopsie. He didn't seem to care nearly as much about his future wife's feelings. "Oh, there's Darby's," she cooed, pretending not to hear his censure. "Buy me a new hat, will you?"
"Another one?"
"I don't believe there's supposed to be a limit, my dear," she chuckled, trying to coax him out of his doldrums. "Come on, I'll let you choose. You do have excellent taste, for a man."
"Hm. My thanks."
She let him choose a pretty green, if rather plain, bonnet, and instructed that it be sent to her address. Something still distracted him, and whatever it was, so far this morning it had kept him from asking that one particular question she'd been waiting to hear for five years—since she'd turned eighteen.
"Is something troubling you, Quin?" she finally asked, out of patience.
He shook himself. "No. My apologies. I suppose my mind's been elsewhere today."
"Where has it been?" she breathed, leaning against his arm as they strolled along crowded Bond Street. "On a certain question you said you needed to ask me today, I suppose?"
Quin stopped, looking down at her. "Yes, as a matter of fact."
Not liking the hesitant look in his eyes, Eloise forced a smile. "Asking it seems so silly. After all, we've known since...forever that we're to be married. If it makes things any easier on you, your father stopped by to see me yesterday."
"He did?"
"Yes. And he informed me that you'd set July the seventeenth as the date for our wedding. I think that's a lovely date, and splendid timing, as well." She put both hands on his arm. "So, what I'm trying to say, dear Quin, is that you really needn't even ask me. Just know that I say yes."
For a long moment he looked down at her, then slowly shook his head. "You are better than I deserve."
She chuckled, relieved. "Of course I am. Shall we go ask the duchess to begin compiling a guest list?"
Quin pulled his arm free of her ringers. "Eloise, I can't marry you."
Eloise froze, her relief turning to disbelieving horror. "What?"
"Not right now. I need a little time...to think."
"About what? Don't be ridiculous, Quin. Your father will cut you off if you delay this wedding for another year." She took a step back. "And so will I. I'm twenty-three, Quin. Most of my friends are already married. Some of them have children. I won't be made a laughingstock."
"That is not my intention," he said stiffly.
Hurriedly Eloise stepped forward again, putting a hand on his arm. "You are a very kind man, Quin; you always have been. If you need to take time to think, then do so. But know that I am here, and know that we both have an obligation to our families." She leaned closer, lowering her voice. "Do you think I haven't met anyone I couldn't fancy myself in love with? But I haven't allowed it to happen. There is too much at stake. You must do the same."
"Don't you think I know that, Eloise?" He took a deep breath. "Just give me a few days. A week. And then I will ask you, properly." Quin smiled a little. "On bended knee."
"A week," she agreed, returning his smile. "Now, take me home. I must choose a gown to wear to the Beauforts' tonight."
He bowed. "Yes, my lady."
Quin strolled ahead to signal his driver, and Eloise stopped to look at her reflection in a shop window. He wanted a week. It might as well have been forever. And something had to be done—before his little slut could ruin the fortunes and futures of two very important families. Apparently Charles Dunfrey wasn't having any effect. She would have to see to that. Immediately.
Maddie had called him dull. In a sense, she knew what she was talking about. Quin watched her waltzing with Rafe, her natural grace and exuberance rendering the rest of the females at the Beaufort ball pale and awkward by comparison.
Dull. The epithet hit closer than he felt comfortable acknowledging. Well, perhaps he wasn't exactly dull, but he'd certainly taken a great many things for granted. He'd never had to worry about—or even think about—an income, his place in society, or whom he would marry. It had all been taken care of by the time he knew enough to wonder about it.
Maddie laughed at something Rafe said, and a stab of jealousy wrenched Quin's insides unpleasantly. Because of her, because of what had happened to her and what she had accomplished all on her own, he could no longer take anything for granted.
He'd never thought to fall in love with Eloise; they'd known one another so long that a mutual fondness seemed adequate. But since he'd returned to London, even fondness seemed too strong a word.
Certainly Eloise had offered her assistance—or rather, she'd agreed to help him in order to get Maddie out of Bancroft House. To assume otherwise would be absurd. Quin glanced across the room at her, seated between her mother and his own, a charming, poised smile on her lovely face. As always she looked beautiful, her blond hair framing her face with delicate curling tendrils, and the sapphire of her dress matching the blue of her eyes. She looked like a marchioness, and she looked like a stunning future duchess.
And then there was Maddie—who looked like a beautiful wood sprite, captured for a moment before she made her escape into the morning mist. A smile touched his lips. She was certainly an auburn-haired element of nature, and half the gentry present wouldn't even speak to her. They all looked, though, especially the men. There was no doubt about that.
Perhaps she was right about him, after all. With her, he never knew what might happen next. With her, he felt...alive. And without her, the past years of his life seemed so lifeless.
"Ah, Warefield."
Quin blinked and turned around. "Mr. Dunfrey." The jealous twinge he'd felt at Rafe expanded into something much darker and angrier.
"Maddie—Miss Willits—has told me of your extraordinary generosity toward her. As her initial...predicament was in part my fault, I am exceedingly grateful to you, my lord."
"Don't trouble yourself," Quin said shortly, wishing Dunfrey would go away. "It has nothing to do with you. She was wronged, and I am setting it right."
Dunfrey nodded at him. "Just so. And I hope that I might take a final step in that direction, myself."
So Charles Dunfrey did want her again, after all. If Quin had even the least little right to do so, he would have posted a No Trespassing sign right then. Maddie was his. But even Dunfrey had more right to Maddie than he did.
And even half ready to strangle the ass for presuming to take away his very favorite person in the world, he knew the choice had to be Maddie's. Even if it killed him. "None of my affair, Dunfrey," he said stiffly. With a nod he turned to find Eloise for the next dance.
As far as his initial—and stupidly naive—plan was concerned, everything was coming together splendidly. Baron Grafford had escorted Maddie out for the quadrille, and her dance card was three-quarters filled. She'd already handed out two perfectly worded set-downs to a pair of overeager gentlemen, and had done no discernible damage to herself—or to them—in the process.
Almack's next assembly was in ten days, and Charles Dunfrey was ready to forgive all and take Maddie as his bride. His debt of honor would be satisfied, and he could go on to marry Eloise a month later, just as the duke had envisioned.
Only one thing was wrong. He'd never expected to fall in love with Maddie Willits—and now that he had, he wasn't certain he could give her up.
## Chapter 13
Charles Dunfrey called on Maddie three times at Bancroft House over the next four days, and he also took her on a picnic and horseback riding at Hyde Park. During that time, with the exception of a few unreadable looks, the Marquis of Warefield seemed to be avoiding her again. She began to think that perhaps Quin had given up on his absurd notion of marrying her, and the idea left her stupidly disappointed and brokenhearted.
After all, the whole idea of his marrying her to restore her honor was completely ridiculous. Even if she had been a paragon of virtue, the future Duke of Highbarrow would have set his gaze much higher than where she stood. It made no sense that she lay awake every night, imagining being with him again, and dreaming of what it would be like actually to many the man with whom—despite all her intentions to the contrary—she had fallen in love.
Apparently, though, the marquis hadn't given up entirely. The next morning Charles had an appointment, and Quin appeared before she'd finished eating breakfast. "Good morning," he said amiably, taking the seat opposite her.
"Good morning," she answered, wary of his seeming good humor. He was up to something.
He motioned for the platter of fruit, and one of the footmen hurried to bring it to him. "What are your plans for today?"
"I need to write Mr. Bancroft."
Quin sat back and gazed at her. "And tell him what?"
Maddie blushed. As if she could write him about what was actually happening. "I haven't written him yet that Charles Dunfrey has been calling on me, and that he's been quite nice."
"Ah. Considering that his bellowing is what ruined you, I should hope he would be quite nice."
"You don't need to be so hostile," she said testily.
"I know, I know," he muttered, half to himself, and sighed. "May I make amends by buying you a new hat?"
"I don't need a new hat."
He paused for a moment, as though she'd said something unexpected. "How about a new dress, then?" His eyes met hers lazily, and a warm, responding tremor went down her spine.
She wished he would quit referring to that night, exciting and intoxicating as it had been. Being reminded of his intimate touch and his passion only reminded her that she would never have that with him again. "A new hat will be more than adequate."
"Splendid. I'll have the phaeton hitched up." With a grin, he rose and snatched a peach from the platter. "Oh, by the way," he continued, "I thought you might want to know: His Grace is taking his breakfast at home this morning." He slipped out the door.
"Egad." Maddie hastily crammed the remaining biscuit in her mouth and washed it down with a gulp of tea. She'd become nearly as adept as the rest of the Bancrofts at evading the duke, thanks to a great deal of luck and a good measure of what Rafe termed her "uncanny nose for trouble."
With a quick and garbled word of thanks to the half dozen footmen awaiting the family's pleasure in the breakfast room, Maddie fled through the kitchen and up the back stairs to her bedchamber. By now even the lofty Bancroft servants were used to her untraditional ways, and her escape warranted a mere nod from the head cook.
After she snatched up her bonnet and gloves from her dressing table, she hurried downstairs by the same route, exiting into the stable yard through the kitchen door.
Quin sat in the driver's seat of the phaeton, waiting for her. "Finished breakfast already?" he asked, offering his hand as she nimbly clambered up beside him.
"You might have warned me earlier," she answered, tying the bonnet under her chin.
"Here, Maddie, let me do that."
She slapped his hand away and turned her shoulder to him. "I can manage, thank you very much."
"Don't you like me to touch you?" he murmured.
Maddie swallowed. "Yes. So don't."
Quin glanced at her, and then snapped the reins. The phaeton started smartly down the short path. They turned out onto the street, and he leaned closer. "You keep telling me you're already ruined. What's the harm—"
She'd been asking herself the same question—repeatedly. "I am not some actress or opera singer, Warefield. I suggest you find one of them to satisfy your baser needs."
For a moment his expression darkened. "You're fitting back into polite society quite well, my dear."
"You make it sound like an insult. My fitting in is what you wanted, my dear."
"Perhaps. But I wasn't speaking of my 'baser needs,' as you call them. I was talking about desire, Maddie."
She glanced at him, blushing. Everything was so much easier when they were arguing. "Well, stop that, too."
Unexpectedly, he chuckled. "I can stop talking about it."
She smiled reluctantly. "Sometimes—just sometimes—I'm glad I rescued you from Mr. Whitmore and Miss Marguerite."
"You rescued me from that damned pig?"
"I should say so. If not for me, you—"
"Maddie!"
Startled, she whipped her head around. Standing outside a clothier's shop were Lord and Lady Halverston, gaping at the phaeton as it passed by them.
Quin took one look at her face and pulled up the phaeton. "Who are they?" he snapped.
"My...my parents," she choked out.
"Sweet...." he growled, and grabbed hold of her arm before she could jump from the carriage and make an escape.
He needn't have worried. She couldn't move. She couldn't speak. She couldn't utter a word. All she could do was stare at her parents, staring back at her.
"Oh, good heavens," her mother continued, hiking up her skirt and running forward. "It is you. It is you! Maddie!"
The hack driver behind them whistled his annoyance, and Quin maneuvered the phaeton over to the side of the street—even though Maddie would much rather have grabbed the reins and fled. When he gently placed his hand over her clenched one, she jumped again.
"Go say hello," he whispered.
She shook her head tightly. "I can't. Just go."
"Go where?"
"Anywhere."
"I'm here, Maddie," he said quietly, stroking his fingers over hers. "I promised, remember? Nothing will happen to you."
His typical, self-confident arrogance brought her back to herself. "Where were you five years ago?" she muttered, and stood.
Hurriedly Quin tied down the reins and jumped to the ground. Before either of her parents reached them, he had moved to her side of the phaeton and reached up to take her hand. Reading the silent encouragement in his eyes, she grasped his fingers tightly and stepped down to face her parents.
"Mama, Papa," she said, her voice miraculously steady. "You both look well."
They stopped a few feet away, as though afraid she might bolt again if they came nearer. Her mother fluttered a handkerchief. "We look well? Where in the world have you been? Do you have any idea how worried we were when you vanished? You simply have no—"
Lord Halverston put a hand on his wife's shoulder. "Please, Julia. There will be time for explanations later. Is this your husband, Maddie?"
Flabbergasted again, Maddie looked at Quin. With a slight grin he pulled free of her grip and stepped forward to hold out his hand. "Quin Bancroft," he said amiably, glancing sideways at Maddie. "A friend."
Lord Halverston shook his hand vigorously. "You are too modest, my lord. Julia, this is the Marquis of Warefield."
Lady Halverston curtsied, her expression stunned and astonished, and her face nearly as white as Maddie's. "My lord."
Quin asked some innocuous question about when the family had arrived in London and Maddie glanced at him, grateful for the reprieve. She stood close beside him while he played the kind, pleasant marquis, and she tried not to shake. Whatever she claimed about being able to stand on her own, she was very glad he was there.
Silently she studied her parents. Except for a little more gray peppering his temples, Robert Willits looked almost unchanged. When he made an effort to be pleasant, as he was doing now for Quin, the viscount could be very charming. What she remembered best about him, though, was the constant barrage of harsh, disapproving words he had for her stubbornness and lack of propriety, and the even worse words he'd bellowed about how she had forever disgraced herself and her family.
Her mother always seconded what her father said, mean and unfair as it frequently seemed. Today, though, Lady Halverston had eyes only for her daughter.
"How long have you been in London?" her mother asked.
She shrugged. "A few weeks."
"A few weeks? Why didn't you write? Why didn't you let us know where you were?"
"I didn't want you to know."
"But you don't mind that everyone else in London knows of your return?" Lord Halverston scowled at her.
She was familiar with the expression. "Coming here was not my idea."
Quin stepped forward, taking her hand again and placing it over his arm. "My mother and my cousin are assisting with Miss Willits's return to society. She was very kind to our family, and we are attempting to repay the favor."
Again Maddie was grateful to the marquis, this time for keeping her past whereabouts a secret. "Your mother will be expecting us," she lied, looking up at him hopefully.
"Yes," he nodded. "I beg your pardon, Halverston, but the duchess hates to be kept waiting."
"Of course," her father hurriedly agreed.
"Perhaps you might wish to call on us this afternoon, at Bancroft House," Quin continued.
"Oh, yes. Bancroft House. We'd be delighted." The viscount nodded at Quin and shook his hand again.
"We'll see you at two, then."
Quin turned and helped Maddie back up into the phaeton. When he joined her on the seat she elbowed him hard in the ribs, unable to rein herself in any longer.
"Ouch. What was that for?"
"Traitor," she muttered at him, trying not to stare at her parents.
"Coward," he returned, whispering the word in her ear as he collected the reins. With her mother waving after them, he sent the rig back out into the street.
Maddie sat with her arms crossed over her chest, refusing to look in his direction again. No one could possibly be as annoying as he was and so kind and compassionate at the same time.
"That wasn't so bad, was it?"
"You had no right to invite them to call on me," she snapped. "And I am not a coward."
"I didn't invite them to call on you," he corrected, fleeting humor touching his eyes. "I invited them to call on me."
"Oh, how gracious. I thought you were...on my side," she said, searching for the right words. "But you weren't, were you? You were just worried I'd cause a scene and embarrass you."
"No, I—"
"Stop the phaeton. I'm getting out."
"No, you're not." Before she could react, Quin grabbed her arm and yanked her closer. "Maddie, you're upset. That's all right. But please, don't be angry at me for it. I am on your side. I'm trying to help, in my own dull, pompous way."
For a moment she let herself lean against his strong, warm shoulder and closed her eyes. It was so absurd that she could be mad enough to spit at him, and at the same time want nothing more than to just melt into his arms.
With a glance at the crowded walkways, she straightened. Melting in the middle of Mayfair would be decidedly unwise. "None of this mess was part of your bargain with Mr. Bancroft, you know."
He grinned. "I didn't bargain for a great deal of this, truth be told, Maddie. But I can't say I'm sorry for any of it."
"Well, that's one of us," she said, with as much sarcasm as she could muster.
"Oh, come now," he chided, obviously not fooled a bit. "If you could reconcile with your parents, wouldn't you want to? Your mother certainly seemed pleased to see you."
So she had. "Don't tell them where I've been, please."
"Your secret is safe with me," he said. Quin drew a long breath. "Maddie?"
She looked at him, watching her with serious green eyes. "Yes?"
He held her gaze for a long moment, then shook himself and faced forward again. "Nothing."
As soon as they returned to Bancroft House, Maddie fled upstairs to change. Just in case, Quin instructed the gardener to let him know if she tried to make an escape out her window. She seemed resigned to speaking with her parents again, but her temperament could be rather mercurial. And he didn't want to risk losing her now. Not until he'd figured things out.
He made his way up to the morning room to inform his mother of the Willitses' impending visit.
"Mama, Maddie's parents are in London," he said, pushing the door open and strolling into the room. "I've invited them...." Belatedly he noticed his mother's guest. "Eloise? I thought you would be visiting Lady Landrey this morning."
Eloise sipped her tea, her blue eyes regarding him warmly. "The poor thing canceled the brunch. Seems her son's been sent down from Cambridge in disgrace."
"I'm surprised Lester was tolerated as long as he was." Quin sat beside her and motioned to a footman for another cup.
"Yes, it's amazing what a healthy endowment will do for one's patience," she smiled. "Sugar?"
"No, thank you."
"What were you saying about Maddie's parents?" The duchess set aside her embroidery and regarded her son.
"We ran across them this morning."
His mother sat forward. "How did they react?"
Quin stifled a smile. Much as she tried to remain aloof, the Duchess of Highbarrow had completely fallen for Maddie's considerable charms. "I'm not certain. Her mother seemed relieved, but her father was apparently more interested in meeting me."
"Can you blame him?" Eloise chuckled. "A ruined bit or the future Duke of Highbarrow?"
"Yes, but the 'ruined bit,' as you call her, is his daughter—whom he hasn't seen in five years." Quin glanced at his second cousin, annoyed. She didn't sound very much like a willing confederate.
"You said you invited them somewhere," his mother broke in. "Here, I presume?"
"Yes. At two this afternoon. I explained that she had done our family a favor and we were repaying her by chaperoning her return to society."
Eloise eyed him coolly. "You didn't mention that you'd kissed her?"
So she'd found out about that. From the duke, no doubt. But he had done a great deal more than kiss Maddie. Quin gazed calmly at Lady Stokesley. "I didn't think it very wise, no. Is something bothering you, Eloise?"
"Only that you haven't kissed me more often, Quin. An oversight I hope you intend to correct soon." She held her cup up, and a footman hurried to refill it. A drop of hot tea splashed on her finger, and she gasped and threw the contents of the cup at the servant's chest. "You idiot! Are you trying to scar me?"
He bowed, wiping frantically at the hot liquid soaking his waistcoat. "No, my lady. Please accept my apologies. I'm terribly sorry. I—"
"Franklin, get out," the duchess ordered.
He bowed again. "Yes, Your Grace. Thank you, Your Grace."
Still bowing, Franklin backed out of the room. His place was immediately taken by another servant, who swiftly cleaned up the mess and provided Eloise with a new cup. Quin watched the incident, disturbed, while his mother glanced at him and calmly added another spoonful of sugar to her tea.
"Eloise tells me you've agreed on July the seventeenth," she said. "Your father will be pleased. In fact, I believe he intended on meeting with the archbishop this morning, to secure Westminster Cathedral." She sipped her tea again, then lifted a finger and set the cup aside. "Oh, and we need to send out invitations immediately. Otherwise, the whole gala will appear to be hastily planned."
"As if you could hastily plan something over twenty-three years in the making," Quin said. Of course His Grace would be pleased. He'd been the one to choose the date.
"There's no need for sarcasm," Eloise returned, obviously out of countenance with him today.
He really couldn't blame her. In all likelihood, the sooner everything was settled, the better for everyone concerned. Except for him—and except, perhaps, for Maddie. "No. It just seems a great deal of fuss over something everyone's known about for a quarter of a century."
Eloise stood. "Well, I think it's wretched of you," she snapped. "You didn't used to be so cruel and unfeeling."
"Oh, damnation." Reluctantly he stood and walked to the door before her. "My apologies, Eloise. I did not intend to be cruel," he said, feeling distinctly as though he'd enacted the same scene before, and would do so again. Endlessly.
Eloise stopped, looking up at him with her much-praised blue eyes. "I know. Take me riding tomorrow. And buy me something pretty."
Quin forced a smile. "With pleasure."
He escorted her outside and handed her up into her father's carriage. "Until tomorrow, Eloise," he said, kissing her knuckles.
Back in the morning room, his mother had summoned the head cook and was discussing a luncheon menu. He leaned in the doorway, waiting until she finished and dismissed the servant. "For Lord and Lady Halverston, I presume? Thank you, Mother."
"Guests are guests," she said, rising. "And Eloise has actually been quite patient and understanding—for Eloise. Quin, I know Maddie is charming. But—"
He raised a hand. "I know exactly what Maddie is. I don't need everyone between here and Yorkshire reminding me."
She looked at him for a moment. "Good. Now go tell Cook I've decided on the chicken rather than the ham."
More than ready to make his escape before he could hear another lecture about familial duty and obligation, he excused himself. On the way to the kitchen it struck him that his mother hadn't done all that much lecturing lately. Still trying to figure that out, he headed down the back stairs.
In the kitchen doorway, he stopped. A dozen servants gathered around the huge central preparation table, while Franklin, shirtless and grimacing, perched on one edge of it. Maddie stood before him, applying a clean, white bandage to his heat-reddened skin.
"It's not too bad," she comforted, wrapping the bandage about his chest, "though I imagine it hurts quite a bit."
"Just be glad Lady Amiable didn't shoot a bit lower, mate." John, another footman, chuckled.
"Hush now," Cook admonished. "There's a lady present."
John blushed. "Apologies, Miss Maddie."
Quin stifled a smile. Damn it all, now she had his parents' servants calling her that.
"No worries, John. Did you bring Franklin a dry shirt?"
"Aye. Just as you said."
She nodded, smiling up at Franklin. "All right. That should do, then. Take a look in a day or two. If the salve's worked, the redness should be almost gone by then."
The footman hopped down from the table. "My thanks, Miss Maddie."
"My pleasure. And for heaven's sake, duck next time."
He laughed. "I will."
As she turned for the door, Quin swiftly stepped back around the corner. She passed him, and he grabbed her by the arm. Before she could utter a word, he pushed her up against the wall and bent down to close his mouth over hers. After a moment of stunned surprise, she threw her arms around his shoulders and leaned up into his embrace, kissing him back with rough, hungry passion.
"Quin, stop it," she whispered breathlessly, running her mouth along the line of his jaw. "Someone will see."
"No, they won't." He captured her lips again, teasing her mouth open and kissing her desperately. Heart-pounding arousal ran through him, and he was hard pressed not to lift her skirt right there in the servants' hallway. Good God, he'd completely lost his mind. Finally he pulled one of her hands free and led her toward the back stairs. "Come on," he murmured.
"No," she said, attempting to straighten her disheveled hair and nibbling at his chin at the same time. "I'm mad at you."
"You are?" He kissed her again, both frustrated and amused. "Whatever for?"
She couldn't quite conceal her smile. "For inviting my parents here, of course. I told you that already."
With a determined breath he straightened and took her hands in his. She had such delicate hands. "You're making amends with the rest of London. Do you really want to leave your own mother and father out?"
"It's not that simple, you know," she said quietly, stepping into the circle of his arms.
His heart leaped. Strong as her character was, she'd never seemed to need anyone before—him least of all. Gently he slipped his arms around her shoulders and her slender waist. "I do know," he answered into her auburn hair. "But at least if you make peace with them, you can return to Langley knowing that."
She lifted her head, her gray eyes holding his. "Then you've accepted that I will return to Langley?"
He shook his head. "No. I've accepted only that you have the right to return to Langley, if you still wish to—after your debut at Almack's."
She sighed, still leaning against him. "Will His Grace be here this afternoon?"
"He's at a meeting righting for the dwindling rights of the nobility, I believe. Why?"
"For once I wish he was about. I'd like to see my father put in his place." Slowly she reached up and traced his lips with her fingers. "Quin," she said softly, "if you could have lived your life any way you chose, with no promises or obligations or debts, what would you have done?"
"I've never really considered it," he mused. "I suppose I would have liked to have been a professor of literature."
Maddie lifted an eyebrow. "Really?"
"Well, yes. You said anything."
"I know. Go on. You surprise me."
"And you constantly surprise me," he whispered, kissing her once more. "And of course, I would be married to you."
She pulled free, scowling again. "No, you wouldn't, because we would never have met."
He kept hold of one hand. "Considering the outlandish way we did encounter one another, how can you imagine it could be more unlikely for it to happen in another lifetime?"
A bell rang in the kitchen, corresponding to the pull in one of the rooms upstairs. With a curse Maddie fled past him up the stairs, just as John emerged to answer the bell. "My lord, do you require something?" he asked, obviously startled to see the marquis lurking below stairs.
Quin blinked. "Hm? Oh, just taking a walk. Don't mind me," he muttered, and turned to follow Maddie. Then he remembered about the chicken. "Damn."
Lord and Lady Halverston arrived promptly at two. Accompanying them were two girls. "I hope you don't mind, my lord," the viscount apologized, gesturing at the young ladies. "They insisted on coming along to see their sister."
Quin nodded. The little minx hadn't told him she had sisters, though he could immediately see the likeness in their light brown hair and high cheekbones. "Of course. Miss Willits is in the drawing room, with the duchess."
The Willits family hurried upstairs behind Beeks, while Franklin took the butler's place at the front door. "Franklin," Quin said, leaving the crowd to go ahead, "are you well?"
"Oh, yes, my lord. No harm done. My own fault, for being so clumsy." He bowed. "My apologies to you, my lord."
"No need. You weren't burned?"
The footman flushed. "I'm quite all right, my lord."
Not wanting to torture the poor fellow further, Quin turned for the stairs. "Very well."
The man wouldn't even admit that he'd been hurt, yet he—or one of the other servants—had actually gone to Maddie for help. Quin paused on the landing as excited giggles and laughter floated down from the drawing room. A year ago, he wouldn't have thought to ask after the footman. He would simply have accepted Eloise's tantrum as a matter of course. It wasn't her first. Nor, he was certain, would it be her last.
He stopped in the doorway. A cacophony of competing voices assaulted his ears, as everyone tried to be heard at once. Uncharacteristically, Maddie was the only one not talking.
Instead, she stood at the far end of the room, one hand gripped by each of her sisters, turning from one to the other as they regaled her with some story. The viscountess stood watching her three daughters with misty eyes, while Lord Halverston was profusely thanking the duchess for her extreme kindness to his willful daughter.
Maddie glanced up. As she saw him, a smile touched her lips, and for the first time a secret passion touched her eyes. Passion for him. "My lord," she said, curtseying.
"Maddie, introduce me to your family," he asked, stepping into the room to join her, and barely able to keep from swinging her in the air and laughing in delight. Sometime over the past few days, he and Maddie, without really even realizing it, seemed to have become a "we."
"This is Polly," she said, lifting the hand of the younger girl, who looked to be twelve or thirteen, with a smattering of freckles across the bridge of her nose. "And this is Claire."
The older of the two curtsied to him politely. She was pretty enough, though not striking as Maddie, her eyes more green than gray, and her face a little rounder. She looked to be sixteen or seventeen, no doubt on the verge of her own debut in society.
"Pleased to make your acquaintance," he said, smiling and taking her hand.
"I've arranged for luncheon," the duchess said, having to raise her voice to be heard over the viscount's continuing protestations of thanks. "This way, if you please."
They trooped into the dining room, where the noise continued unabated throughout the meal. Quin looked on in amazement, wondering how Maddie could have become as independent and witty as she was, in a household of such...obtuse silliness.
"Maddie, do you have much to pack?" the viscountess asked.
The comment immediately snared Quin's attention. "To pack for what purpose?"
The room quieted, though his ears were still ringing from the rebounding sound. The viscount cleared his throat. "Willits House is open now, my lord. It wouldn't be seemly for our unmarried daughter to be staying under someone else's roof, elegant as it is."
"My mother is chaperoning her," he replied testily. "There is no impropriety."
"Oh, of course not, my lord," Lord Halverston agreed. "But, well, people will talk, you know."
"They will talk anyway," Maddie said.
"And the more we can minimize it, the better."
Quin looked at Lord Halverston, what remained of his good humor sliding away. She was not leaving. "You might have taken that into consideration five years ago."
"Quin," his mother said sharply. "I believe the decision should be up to Maddie."
Maddie looked about the table, spending the longest moments gazing at Quin and at her father. Finally she turned to the duchess. "Your Grace, I think perhaps I should return to Willits House. Though—"
"No!" Quin snapped, rising.
She swallowed, refusing to meet his furious gaze. "Though I would be extremely grateful if I might continue to call on you—from time to time."
"Of course, my dear. Nothing would please me more."
A great many things would have pleased Quin more, including his never having set eyes on the rest of the damned Willits family. He swallowed the angry retort that came to his lips, and instead nodded and dropped his napkin into his chair. "Very well. I'll summon Mary and have her begin packing your things."
He strode out into the hallway, where he came to a halt, his breathing ragged and hard. He knew precisely what she was doing: because he refused to admit that they would never suit, she was attempting to take the issue out of his hands. Except that he wasn't about to give up. Not yet. Not ever.
## Chapter 14
Maddie entered Willits House slowly, fighting the nagging idea that somehow nothing would have changed, even after five years. Everett, the butler, certainly looked the same, despite his expression of stunned surprise.
"Good afternoon, Everett." She smiled, wishing she could feel as easy as she was attempting to act.
"Miss Willits," he stammered, bowing. "Welcome home."
"Thank you." Impulsively she held out her hand. After a startled moment, he shook it. "It's good to see you again."
"And you." A reluctant, admiring smile touched his lips. "You were missed."
The furnishings downstairs hadn't changed, and neither had the paintings nor the burgundy carpeting she'd always detested in the drawing room. Her sisters trailed along behind her, excitedly chatting about their five years of adventures, while she slowly climbed the stairs and tried not to remember the last time she had fled to her bedchamber. She hesitated at the half-closed door, but before she could push it open, Claire stepped forward and barred the way.
"This is my room, now," she said. "Papa said someone might as well have use of it, and you know I never liked the morning sun."
"Claire," her mother called from below, "Maddie may have whichever room she wants."
"Mama," the pretty, brown-haired girl protested, then sighed heavily. "Oh, all right."
"No, Claire, keep it," Maddie replied, turning down the hallway. "I don't especially like my old room, anyway." It had ceased to be a refuge; it had been only a prison, with her locked inside.
Her sisters milled about in the room she chose, then went back downstairs to suggest a trip to the horse auctions to purchase a new mount for Maddie, and horses for each of them. Quin had sent Mary along to Willits House with her, and once she and the luggage arrived upstairs, Maddie was only too happy to stay and help her maid unpack.
"I can manage, Miss Maddie," Mary said, pulling open the mahogany wardrobe three footmen had carried up to the room. "You should be with your family."
"I'm giving them time to adjust." She grimaced. "I'm giving myself time to adjust."
For a few moments, she had thought Quin wouldn't let her leave Bancroft House at all. He'd kept himself in check, but his anger showed in every line of his tensed muscles and his tightly clenched jaw. She really hadn't wanted to go, but if she'd stayed she knew she would have given in to him eventually. He was too compelling, his presence too intoxicating. And he was still going to marry Eloise Stokesley in a little over a month.
"Maddie?" her mother said through the half-open door.
"Come in." Self-consciously Maddie brushed at her skirt as she straightened.
The door swung open. "Might I have a word with you?"
Mary curtsied. "Excuse me, Miss Maddie," she said, stepping around Maddie and hurrying out the door.
"'Miss Maddie?'" Lady Halverston repeated. "Have you given up your place in our family?"
"I thought it was you who gave me up," she said, without heat. "And I've become used to being called Maddie." She sat on the edge of her bed. This was the moment she had dreaded, when her mother would ask where she'd been, and she would have to decide how much to tell her, and how much of an escape route she wanted to leave herself.
"We began to think you were dead, you know," the viscountess said, sitting at the dressing table. "Being angry and upset is one thing, Madeleine, but you disappeared for five years."
"I wanted to make my own way."
Her mother looked at her. "You say that as though it's nothing," she finally commented. "Your father would have forgiven you eventually, if you'd stayed. You know that."
Maddie kept her temper in check. "I did nothing wrong. I didn't—and I don't—need his forgiveness. And you should know, I am not going to stay. I...promised someone I'd remain in London until my debut at Almack's—my second debut at Almack's. I will do so. After that, though, there's no reason for me to be here any longer."
"I see. And to whom did you make this promise?"
"A friend."
"And what about your family?"
"Father told me quite clearly what a burden I was, and how unworthy I was of being a Willits. I haven't forgotten that." She looked down at her hands. "I don't think I ever shall."
"Maddie, the two of you have never been able to go a week without arguing. You never left before."
"Mama, after that, how could I have stayed?"
Lady Hal version looked down for a moment. "Then why did you return?"
"I've already imposed too much on the kindness of the Bancrofts," she said cautiously, not daring to mention anything of the tempest of emotions whirling between her and Quin.
"Lord Warefield seemed quite fond of you," the viscountess noted, examining Maddie's hairbrush and avoiding her gaze.
"Lord Warefield takes his family's obligations very seriously. And he expects people to do as he says. I..." She hesitated, trying to avoid saying too much. It would never do if anyone discovered how desperately she loved him. "I don't always agree with him."
"You disagree with the Marquis of Warefield? That seems rather unwise."
Maddie shrugged. "Someone needs to do it."
The viscountess looked at her speculatively. "Maddie—"
"Mama, nothing can be as it was before. I've been on my own for five years, and I liked it—for the most part." Honesty forced her to add the last. "If you wish me to go, I will. But I won't sit about and have Father yell at me, as he did before."
The viscountess stood. "You are still our daughter, however independent you believe yourself to be. The Bancrofts apparently had some hope for you, so we can have no less. But you cannot be allowed to embarrass this family again. Claire has her own debut next year, and I know you would not want to see her chances for a good marriage ruined simply because you have declared yourself independent of everyone and everything around you."
Maddie nodded as her mother left the room. "Very well."
She'd known it wouldn't be easy, coming back, and she had been right. Her mother, however grateful she was to have her daughter again, would bow to her husband's wishes. Maddie sighed. Five years had changed her so much, though she was certain her parents wouldn't think it was for the good. Maddie sank back on the bed, fighting a sudden attack of loneliness and abruptly wondering what Quinlan was up to, now that she was gone.
"What do you mean, 'she's gone'?" Rafe demanded, setting his billiard cue down hard enough to make the balls on the table jump. "And why the hell didn't you tell me when you first came in here?"
Quin glanced at his brother, then went back to chalking his stick. "I didn't feel like it. And I meant just what I said. Her parents came to see her, and she packed up her things and left with them. They all looked quite ecstatic at being together again."
"That's idiotic. They're the reason she left London in the first place. You shouldn't have let her go anywhere."
"Oh, really? And what was I supposed to do? Lock her in her bedchamber? It was my understanding that she didn't like that very much."
"She doesn't belong there," Rafael continued stubbornly.
The marquis eyed his younger brother. He could no longer tell whether he viewed every other male in London through a jealous gaze, but Rafe was certainly upset about something. "They are her family. We are not."
Rafe stalked to the sideboard and poured himself a snifter of brandy. "Yes, they are. The same ones who threw her in Charles Dunfrey's direction before. And now that he's gone and apologized to her, they'll likely do it again."
"And what's wrong with that?" Quin asked, mostly to have another opinion besides his own.
"Remember when you shoved me into the drawing room and nearly broke my neck? He took me up on my offer."
"What offer?"
"To buy his pair of bays for a hundred quid. I had to backpedal like a madman to keep from getting saddled with 'em."
"And?"
"Well, aside from the fact that I don't need a pair of coach horses in Africa, they were worth twice that, easily."
Quin set down his cue, keenly interested now. "Forgive me, Rafe, for not being as brilliant about shady dealings as you, but exactly what about this concerns you?"
His brother shrugged, rolling a billiard ball absently about the table. "It just seems to me that if Dunfrey wanted to sell his bays, he could have gotten a lot more for them than what he was willing to accept."
Finally Quin began to catch on. "Then you think he wasn't really interested in selling them."
Rafe nodded. "Precisely. He was interested in—"
"The money."
The marquis hefted his cue and returned it to its proper slot along the wall. "Excuse me, Rafe, I have an appointment."
"With whom?"
Quin turned for the door. "I don't know yet."
The manager of the Bank of England was quite flustered to see the Marquis of Warefield stroll into the bustling building unaccompanied by accountants or lawyers, and even more so when Quin requested a private audience.
"What may I do for you, Lord Warefield?" he asked solicitously, folding and unfolding his fingers on top of his scratched oak desk.
"I have a rather unusual request to make of you." Quin wondered why he didn't feel a single pang of guilt over what he was about to do.
"Anything, my lord. The Bancroft family's finances are beyond reproach."
"Thank you, Mr. Wheating. That's good to know."
The bank manager loosened his cravat a little. "I meant no offense, my lord. Oh, heavens, no."
"None taken. I don't require a loan, however. I require a little information."
Mr. Wheating's tufted eyebrows furrowed. "Information, my lord? What sort of information?"
Quin tapped his chin. "I'm contemplating something of a business venture with one of my fellows. I'm not terribly well acquainted with him, though, and I wished to know a bit more about his financial stability."
"Oh. Um, well—you know, my lord, information about all of our clients is, well, privileged."
"Of course. I wouldn't want you to break any rules. And I don't need any specifics." Quin leaned forward, smiling confidently, attempting to ignore the tickle in his mind that knew exactly what Maddie would say about his throwing his title around. "Just a general overview. I would be extremely grateful."
Mr. Wheating glanced about his empty office. "Who might this fellow be, my lord?"
"Mr. Charles Dunfrey." Quin sat back expectantly.
"Charles Dun—Dunfrey, you say?" Wheating's ruddy features paled. "Oh. Oh, my."
"Could you elaborate?"
"Well, my lord, speaking generally, I would have to say...." Even with the door closed and no one else in the tiny room, he leaned forward across the desk and lowered his voice. "I would have to say that in general, Mr. Dunfrey's finances are a bit shaky."
Quin raised an eyebrow. "A bit shaky?"
The manager cleared his throat. "Quite shaky."
"Ah."
"Yes. Into negative figures, one might say."
"Oh, dear," Quin said in mock distress, disliking Charles Dunfrey more with every passing moment, "this is troubling. I cannot thank you enough, Mr. Wheating." He stood and strolled to the door of the tiny office. "You have saved the Bancrofts a great deal of embarrassment."
Mr. Wheating climbed to his feet and bowed grandly. "My pleasure, my lord, of course."
Quin rode Aristotle back to Grosvenor Square by way of Curzon Street. The avenue was well out of his way, and he knew damned well why he was going that route—the Willitses lived on Curzon Street. He paused outside the wrought-iron gates barring him from Maddie, staring at the curtained windows until the gelding began to fidget.
He contemplated calling on her to inform her of Dunfrey's shaky finances, but for God's sake, she'd been gone from Bancroft House for only three hours. He'd look exactly like what he was—a complete fool, so in love with a ruined chit that he couldn't stand being away from her for more than five minutes.
Besides, just because Dunfrey had called on her once or twice didn't mean either of them was seriously considering marriage again. With Dunfrey's money troubles, it was entirely possible he wouldn't want to be saddled with volatile Maddie Willits for a wife. Marrying into an older, more respected title could do him more good than a few ready quid, if Viscount Halverston even had the kind of blunt that would satisfy him.
Feeling a little better, he kicked Aristotle into a trot and headed toward Bancroft House. They'd already planned to attend the Garrington ball tomorrow evening, and he would be able to see her and dance with her—and perhaps some miracle would occur and he would actually think of a way to get them out of this bloody big hole they'd fallen into. If not, he could always kidnap her and make off to the Orient. No doubt she would be furious, but at least she wouldn't think him dull.
Maddie had barely finished breakfast when Everett entered the room to announce that she had a caller. Her heart leaped. "Who is it?" she asked, trying to hide her excitement and knowing she must be doing a miserable job of it. He'd come to see her, after all!
"Mr. Charles Dunfrey, my lady."
The delight faded from her heart. "Oh."
"My, whatever can Charles want?" her mother asked, looking curiously at her father.
"No idea, I'm sure," he mumbled around his toasted bread.
When he glanced at Maddie, she quickly fixed a smile on her lips. They'd barely spoken since yesterday afternoon, and she had no intention of giving him an excuse to bellow at her again. "I'll go see, I suppose."
Charles turned away from the window as she entered the morning room. "Maddie. I'm so pleased you've returned home."
"Yes, so am I, Charles. Thank you."
"It seems everything has been set right again." He took her hand and brought it to his lips. "Well, almost everything, anyway. Maddie, I need to ask you something. You know I'm not one for speeches, but this has been weighing on me for some time now, and I can't deny it any longer."
Maddie sat in the chair Charles indicated. She had a fair idea of what he wanted to ask, and her own less than pleased reaction didn't surprise her. He'd prefaced his first proposal to her in nearly the same way—and she'd accused Quin of being dull.
Back then, she'd been excited and nervous and thrilled, barely able to keep from throwing her arms about him when he'd finally asked the question. And then he'd kissed her, and she had thrown her arms about his neck. For a brief two weeks, she'd thought fairy tales really did come true—until she'd been proved very, very wrong.
Charles took her hands and knelt before her. "Maddie, we have been apart for five years, but I believe we were meant to be together. Will you do me the very great honor of becoming my wife?"
For a long time she looked at him, waiting for the thrill, the jangle of nerves, that had accompanied this moment five years earlier. Nothing but a tremor of uneasy nervousness ran through her. Perhaps she was trying too hard—or perhaps it was just that he was no longer the one she dreamed of spending her life with. "May I have some time to consider, Charles?" she asked. "A great deal has changed for me over the past few weeks."
"Of course." He smiled and stood. "But at least allow me one liberty." Slowly he leaned forward and brushed his lips against hers.
Maddie smiled at him, even less moved, if possible, than she had been a moment ago. "Thank you for your patience, Charles. I will give you my answer tomorrow."
He kissed her knuckles again. "I do love you, Maddie. I always have." With a last look, he left the room.
Maddie sat back. Marrying Charles would solve all her problems. It didn't matter that she didn't feel anything toward him. Nothing close to what she felt when Quin merely looked in her direction. But he was marrying someone else. She would never hear Quin laugh again when she insulted him, and she would never feel his arms around her again, holding her, and never—
"Maddie?" Her father stepped into the room. "Where's Charles?"
"He left."
"He—what did he want?"
"To marry me."
"That's splendid!" For a moment he was silent, looking at her expectantly. "Then why did he go?"
She looked up at him. "I told him I would give him my answer tomorrow."
The viscount opened and closed his mouth. "What precisely did you do that for?"
She heard the anger in his voice and tried to answer in a reasonable tone, however tense and uncertain and lonely she might feel. "I wanted a few hours to think about things, Papa."
He folded his arms, his expression darkening even further. "To think about what things? He was good enough for you before. And being gone God-knows-where and doing Lucifer-knows-what for five years has hardly elevated your social standing." Lord Halverston narrowed his eyes. "Or is it that you think you're too good for all of us, now that the grand Duchess of Highbarrow has shown you some charity?"
"No! Of course not. Just give me until tomorrow to answer him, Papa. That's all I ask."
"Just so long as you give the correct answer, Madeleine."
When he'd left and closed the door behind him, Maddie shut her eyes. Everything had been so much easier at Langley Hall, where she could be Miss Maddie and spend her evenings playing whist or word puzzles with Mr. Bancroft and Squire John. But she couldn't deny that she'd been lonely there, too—nor that when John Ramsey asked her to marry him, as she'd sensed he eventually would, she would have said no.
"Miss Willits?" Everett scratched politely at the door.
"Yes?" she asked halfheartedly, signing.
"Miss, a Mr. Rafael Bancroft is here to see you."
Unexpected tears welled up in her eyes. Perhaps there was still some hope. She wiped at them hurriedly. "Show him in, please."
A moment later die door opened, and Rafe strolled in past Everett. With his usual jaunty grin he bowed, pulling a bright bouquet of flowers from behind his back. "My lady."
She mustered a smile, fighting more tears. "Hello, Rafe."
He looked at her for a moment, and then thrust the bouquet at the butler. "Put these in water, will you?" he asked, and closed the door in Everett's face. "Whatever is wrong? You look like a damned watering pot, Maddie." He dropped into the chair beside her.
"Oh, I don't know," she muttered irritably, wiping at her eyes again. "I'm just glad to see you."
"If I'm so popular with you, then you shouldn't have left Bancroft House or my illustrious company," he commented, reaching out to pluck a hard candy from the dish on the table.
"I had to."
"Mm-hm," he said around the candy, nodding. "Well, you can tell me all your troubles if you like, but it won't do you a bit of good. I'm not the one you need to talk to."
She looked at him sideways. "I don't need to talk to anyone."
Rafe sighed heavily. "Suit yourself, Maddie. I am absolutely not going to get pulled into the middle of this mess. I have enough problems of my own."
"Like what?" she asked innocently. Something had been bothering him from the moment he'd arrived in London, but as far as she knew, he hadn't confided in anyone.
"Like something I have no intention of telling you about," he answered easily. "But I will tell you this. My brother has had his entire life planned out for him, and he's been perfectly happy with it—until now. He's never had his head twisted around before, Maddie, and you can't expect him to be anything more than a complete idiot about it." He patted her on the hand, and stood. "And that is all I intend to say on the subject."
She looked at him, amused. "Is that why you came by? To inform me that you weren't going to say anything?"
"Actually, it was to invite you to go riding with me in Hyde Park tomorrow morning. I believe Quin was supposed to go with you this morning, but he's a bit...preoccupied."
"So you're fulfilling a familial obligation by offering to spend time with me?" she asked, hurt.
"I'm taking advantage of his stupidity." Rate winked at her. "I'll come by at seven. Do you have a mount here?"
"No." As she was beginning to realize, she had nothing here. Nothing that meant anything, anyway. Not anymore.
"I'll bring Sunny, or whatever her name is."
Maddie grinned. "Honey."
"Honey," he repeated, half to himself. "Sounds like something fat old Prinny would own."
That caught her attention. "What?"
He squinted one eye. "Nothing."
"Rafe," she warned, chuckling. "What did you say?"
The younger Bancroft leaned back against the door. "Well, apparently my daft brother searched all over London for the perfect mount for you—you know how he is—and Prinny—drat, I mean King Georgie—had the exact one Quin wanted."
"So Quin bought Honey from King George, for me?"
"Well, not precisely. Prinny's been after some architect to design a palace somewhere, and—"
"Brighton," she supplied, becoming more intrigued with every disjointed sentence.
"Oh, then you know the story."
"Rafe!"
"All right, all right. Prinny's got this architect at Brighton, but he couldn't get Parliament to put up enough blunt to keep him on the job. Quin agreed to make up the difference."
Maddie sat and looked at him in disbelief, a delighted grin tugging at her lips. "Quin helped King George keep John Nash on salary to renovate Brighton Pavilion, so 1 could have a horse to ride in London?"
Rafe nodded. "Mm-hm."
A peal of delighted laughter tumbled from her throat. "Oh, good grief! No wonder he didn't say anything about it to me."
"I say, that's right. I'm not supposed to tell you, you know." He winked again. "Tricked it out of me, you did. Honey and I will be by at seven."
She stood and came forward to rise up and kiss him on the cheek. Before she could complete the gesture, he turned his head and touched his lips to hers. Startled, Maddie rocked back on her heels. "Rafe?"
"I'm not some castrate, you know," he muttered, "and you're quite impressive." He pulled open the door. "Good God, he's an idiot."
"Rafe, this morning Charles Dunfrey asked me to marry him," she blurted, flushing.
He closed the door again. "And?" he asked slowly, his light green eyes sharpening perceptibly.
That was what she liked so much about Rafe: he wasn't nearly as daft as he liked to pretend. She wondered how it must be for him, to be a second son and have the Duke of Highbarrow for a father. "I'm to give him my answer in the morning."
He drummed his fingers against the door for several moments. "You'll be at the Garrington ball tonight, won't you?" he asked finally.
She nodded.
His eyes held hers. "I'll see you there, then."
"Yes, I'll see you there."
After he left, the room seemed quiet and gloomy, and Maddie sat wondering why, precisely, she'd bothered to tell him about Charles. She sighed. Because he would tell Quin, of course. And because no matter what she'd said about wanting the marquis to leave her alone and do his duty by Eloise, she was still in love with him. "Oh, drat it all."
Eloise sat in her coach and watched Rafael Bancroft retrieve his horse and ride away from Willits House. The damned interfering rat couldn't seem to stay out of her affairs. No doubt he'd spent the entire visit with Maddie, trying to convince her to return to Bancroft House before Quin forgot about her.
Well, Maddie was not going to return to Bancroft House. Dunfrey had timed it perfectly, having her parents arrive in London before Quin's stupid sense of honor could ruin everything. How he could possibly think pity was a respectable reason to marry a completely unsatisfactory person she had no idea. But something of that sort had been on his mind; she could see it in his eyes when he looked at Maddie. And she didn't see it in his eyes when he looked at her. That didn't matter so much, though, as long as she ended being the one wearing his ring, and his title.
From his brief note of this morning, Dunfrey's plan was working so far—but there were some things she didn't dare leave completely to chance. Not with her future at stake. With a deep breath she lifted her umbrella and rapped on the roof of her coach. The driver started the team and turned into Willits House's short drive. Another coachman jumped down from his perch to open the door and help her to the ground.
"Wait here," she instructed, climbing the shallow steps.
The door swung open just as she reached it. "I am Lady Stokesley," she announced, before the butler could inquire. "I am here to see Miss Willits." She handed over a gilded calling card.
The butler, who had the ill manners to look flustered, showed her into the foyer. "If you'll wait here a moment, my lady."
She had barely enough time to note the inferior artwork lining the hallway before Maddie, accompanied by a plump woman who must have been her mother, appeared. "Maddie," she said warmly, coming forward to take the smaller woman's hands, "how pleased I am to find you here, back with your family. I never expected it."
"We are pleased to have her here," the older woman said. "I am Lady Halverston."
"Oh, yes," Maddie said belatedly, blushing. "Mama, Lady Stokesley. Eloise, my mother, Lady Halverston."
"Charmed," Eloise cooed, gripping the viscountess's fingers. "Your daughter resembles you."
Lady Halverston chuckled with unbecoming amusement. "Thank you for the compliment, Lady Stokesley. Do come in." She led the way into the drab morning room.
It seemed Maddie was a better match for Dunfrey than she'd realized, Eloise thought. "I cannot stay," she said hastily, contemplating with horror the idea of actually taking tea with the woman. "I thought Maddie might wish to accompany me on a picnic."
Maddie looked at her, something almost suspicious touching her vapidly innocent gaze for a moment. "Well, thank you, Eloise, but I really don't—"
"Hush, Maddie," Lady Halverston interrupted. She put her hand on Eloise's gloved one. "We've been having a little difficulty adjusting to Maddie's return," she confided with a smile. "I think a bit of fresh air with some friends will be just the thing to restore her spirits."
Not if I have anything to do with it. Lady Stokesley smiled warmly. "Say no more. Come with me, Maddie, my dear."
The girl hesitated again, glancing at her mother, then shrugged. "I'll get my bonnet," she said, and hurried out the door.
"Thank you for your kindness to my daughter," the viscountess said. "We never expected to see her again, and certainly not under such pleasant circumstances."
"Yes," Eloise agreed. "I have already come to think of her as one of my dearest friends. And Quinlan—that is, Lord Warefield, my betrothed—speaks very highly of her."
"Lord Warefield does seem fond of Maddie. I think he was none too pleased when she left."
"We're all sorry to see her leave."
Maddie hurried back into the morning room, her gloves and a pink bonnet clutched in one hand.
"Ah, there you are, my dear. Let's be off, shall we?" a291/>Eloise smiled as she led the way out to her coach. With the extremely helpful friends she had selected to join them, this was going to be so easy, it was almost pitiful. Almost.
## Chapter 15
The Duchess of Highbarrow sat in her private room, sewing.
Her favorite chair had been placed before the large window which overlooked the quiet street in front of the mansion, but she had no desire to look outside. She knew very well what was going on out there. A rather annoying clattering and clanging, which had begun below some forty minutes earlier, gave way to a rattling, clopping sound, and then slowly faded away into silence.
Victoria's hands stilled in her lap, and she sighed. She also knew what the absence of sound meant: her sons were gone again.
It made sense. Quin had obviously stayed to keep an eye on Maddie, and Rafael had stayed because of Quin—and Maddie. Once she left, Quin spent an hour stomping about and pretending he wasn't in a black temper. By teatime he had sent for his footmen at Whiting House and moved his things back home. And as he had over the past few years, Rafe went to stay with his brother.
"Victoria?" The bellow echoed up from the hallway, the duke's method of avoiding the necessity of asking the servants for the whereabouts of his wife.
She didn't answer. She didn't need to, for Beeks would immediately inform his employer that she'd spent the afternoon in her private rooms and had asked not to be disturbed.
A moment later the door opened. "Victoria?"
"Yes?" She picked out her last row of stitches and began them over again. Apparently she hadn't been paying very much attention to her work. She had paid attention to several other interesting things in the Bancroft household, though, and for the duke's sake, he had better realize them as well—and soon.
"Where're the randy idiots and their mopsie?"
"If you are referring to our sons and Miss Willits, they are gone."
He closed the door and went to look out the window. "Gone where?"
"Maddie's parents came to see her, and she left with them. Quinlan and Rafael went to Whiting House. You just missed them."
For a moment the duke said nothing as he gazed outside. "Good," he muttered finally.
The duchess set aside her sewing and looked up at her husband. "And why is that good?"
He glanced back at her. "They were too damned noisy. It was like having a flock of geese about."
"And it's much better now—so quiet you can hear the minute hand of the grandfather clock on the landing?"
Slowly His Grace turned around. "Did you like all that nonsense?"
"I liked having my sons home. We don't see them very often, in case you hadn't noticed."
"We're busy folk."
Victoria shook her head. "Not that busy. They don't like coming around, now that they don't have to be here."
"I suppose you're going to blame that on me. Well, I expect guests under my roof to abide by my laws. Always have."
"I know," she said quietly.
"And always will. If that's too much for them, then they have no business being here. Glad they're gone." He nodded, as though attempting to convince himself of his own sincerity, and stalked out of the room again.
Occasionally Victoria wondered what would happen if she pushed against his "laws" to the point of open defiance. She'd come close several times, usually with regard to Rafe and his high spirits, but somehow the duke had always managed to deflect or ignore the attack. Lately it had begun to occur to her that his avoidance was no accident—and in a way, that was comforting. He wanted her there, even if the only way he could show it was by ignoring her direct questions and pronouncements.
The duchess picked up her sewing again. Such an obtuse method of rule couldn't last forever, and whether Lewis Bancroft realized it or not, his kingdom had already begun to crumble at the edges. A bright, fiery sprite had entered their lives, and nothing would ever be the same. She glanced out the window toward the light blue, cloud-patched sky. Her sons would certainly never be the same. One of them in particular.
The mutterings began well before Maddie arrived. Quin pretended not to hear them, while carefully tracing their source.
By now they'd spread across the Garrington ballroom, invading nearly every nook and cranny, but the center of the disturbance seemed to be a large group of his most intimate acquaintances. He stopped beside a potted plant and watched them for a moment as they chatted and laughed and managed to exclude all social inferiors by not even noting their presence.
As usual, the main attraction of the group seemed to be his second cousin, and as she leaned sideways to whisper into another intimate's ear, Quin decided he had several rather pointed questions to ask her.
"Eloise?" he said, strolling out from his hiding place and stopping beside her. "I hadn't expected to see you so early in the evening. You look lovely, as always."
She held out her hand for him to take. "It looked to be a sad crush, and I didn't want to have to wade in through the mud and horseshit." Her faithful circle of companions laughed, and she snapped her fan playfully. "Well, it's true, you know."
Quin smiled, unamused, and tucked her hand around his arm. "Might I have a word with you? And a waltz, of course, if you've still one unclaimed."
"I always leave one for you. Excuse me, ladies. My future husband would like to speak to me—in private."
The two of them strolled toward the wide doors that opened out onto the balcony, and with a glance into the half darkness, they stepped outside.
"Ah," Eloise murmured. "Alone at last." With another look around them and down at the darkened garden below, she slipped her hands up on either side of his face, and leaned up to kiss him slowly and deeply.
It was the first time she'd exhibited any kind of passion toward him, and at the moment he wasn't particularly interested. Not in her, anyway. "What was that for?" he asked as they parted.
"Just to remind you that our marriage will be more than a union of names and wealth. I think you'd forgotten that."
Recently his views of what a marriage should be had changed. "When did you decide that?" he asked.
She reached up to touch his cheek again, apparently undaunted by his cool tone. "Oh, Quin, we've known one another for so long. Sometimes I think it would have been better if our parents had kept us apart until it was time for us to marry."
Quin nodded. "You favor an element of mystery, I suppose?"
"No, not really. But sometimes I almost believe you think of me as a sister, or something equally awful."
"I don't, Eloise. But I do think of you as a friend." Or rather, he had, until the last few days. Quin lifted her hand away from his face and held it. "And as a friend, I'd like an explanation."
Her delicate brow furrowed. "An explanation of what?"
Quin looked down at her for a moment, wondering when, precisely, he'd ceased to think of her as a potential mate. Probably the moment he had set eyes on Madeleine Willits. "Did you go somewhere with Maddie today?"
She yanked her hand free. "I'm trying to seduce you, and you still ask me about her?"
"Eloise, she's here in London because of me," he returned flatly. "She's my responsibility. I have an obligation to look—"
"She is not your responsibility. She is her own responsibility. You didn't ruin her, Quin. You had nothing to do with it."
That wasn't exactly true anymore, but as he didn't want to begin a shouting match, he nodded. "All right. But tell me what happened today."
"Nothing happened. I invited her on a picnic, as we'd discussed, and—"
"We discussed taking her on a picnic," Quin agreed. "With mutual friends attending."
"Oh, Quin, don't you see? You already spend nearly every day with her. You're not helping her by accompanying her everywhere. Besides, everything was fine. She did very well."
He continued to watch her, looking for any sign that she could actually be as devious as he had begun to suspect. "Not according to what I've been hearing tonight."
"What have you been hearing?" she asked, meeting his gaze evenly.
She could be telling the truth, he supposed, and truly knew nothing of the widely circulating rumors. But for the first time, he doubted her word. "I heard that she suggested you and your female friends leave so she could enjoy the company of Lord Bramell and Lionel Humphries in private."
Eloise clapped her hand over her mouth, but the expression in her eyes wasn't all that surprised. "Nothing of the sort happened! John and Lionel were there, of course, because you know they always attend such things, but—I mean—when Lady Catherine Prentice arrived, we all went to see her new setter puppy, but Maddie was alone with John and Lionel for only a moment. Two at most."
"You shouldn't have left her alone."
"She wanted to stay behind, Quin. I couldn't drag her across the park, for heaven's sake."
"Damnation," he swore softly. Maddie knew better. Anything she did—anything—whether innocent or not, would be viewed in the worst light possible by her fellows. To stay behind, alone with two single gentlemen, was worse than stupid. And Maddie wasn't stupid. Far from it.
But neither was Eloise. He looked at her speculatively. If his suspicions were correct, Eloise had a great deal of explaining to do. In all fairness, though, his mind didn't exactly work to perfection where Miss Willits was concerned, and he had no proof. Blind in love with Maddie or not, he couldn't accuse Lady Stokesley until he knew for certain that she was guilty of sabotage.
"Just remember, Quin, in a month's time you won't be able to claim poor Maddie as your responsibility any longer." She leaned up against him, her short blond curls tickling his cheek. "That will be me."
"I remember." He wondered why he hadn't always found her so self-centered and cloying. "We'd best go back inside, or we'll be starting some rumors of our own."
He escorted her back to their group of friends and spent the next hour dividing his attention between polite conversation and keeping an eye on the ballroom doorway. Rafe had said Maddie would be attending. He'd also said a few other things, at a rather high volume, and Quin intended to take care of those issues as soon as Miss Willits arrived.
Finally, late enough that she'd likely fought against coming, Maddie and her parents arrived. Quin's breath caught at the sight of her, glorious in green and gray. He watched her take stock of the room and the other guests, and he knew precisely when she decided she didn't want to be there.
"Excuse me for a moment," he said, to whomever happened to be listening, and started across the room toward her. He couldn't help himself. He craved her like he breathed air.
Rafe, obviously making use of his military skills, damn him, reached her first. "Good evening, my dear." He took her hand. "So pleased you could join us this evening."
Quin made a valiant attempt not to break into a full-on charge. It would never do for the Bancroft brothers to begin a tug-of-war over her in the middle of the ballroom. He stopped beside her. "Miss Willits." He smiled, stealing her hand from Rafael's grip and lifting it to his lips. "You look...stunning."
"Thank you, my lord," she said, meeting his eyes and then looking quickly away. "Do let go of my hand."
He complied reluctantly. It seemed like days, rather than hours, since he had seen her last, and he wanted—needed—to touch her. "May I have a waltz with you?"
"I don't think you should," she said, still gazing determinedly at the punch bowl on the refreshment table.
"I do," he answered.
"No."
"Yes." As usual when he argued with her, Quin began to feel as if he was beating his head against a brick wall.
"Better do as he says, Maddie," Rafe put in, for once helpful. "But save one for me as well."
She smiled and looked at him. "Of course I will."
Quin didn't like that. Blast it, now he wanted to pummel people insensible just because she was smiling at them. Somehow, somewhere, he had completely lost control—and the oddest part was, he didn't mind it all that much.
Before he could ask her what in God's name had happened at the picnic, the orchestra began playing again, and he had to excuse himself to dance with his designated partner. For a moment he thought Maddie would have to remain alone beside her parents at the end of the room, but the Duchess of Highbarrow appeared from nowhere and led the Willits family off for a chat.
Whatever orders His Grace had given his wife regarding her assisting Maddie, she seemed to be ignoring them. He would have to call on her tomorrow and thank her: she'd just seen to it that Maddie would have partners for any dance she wished.
And so he danced a quadrille with that young lady, and a country dance with this one, and the entire time he kept his attention on Maddie. When Rafe claimed her hand for the first waltz of the evening, he barely managed to wipe the scowl from his face before he went to fetch Eloise.
"Rafael seems quite fond of Maddie," Eloise purred, as they circled grandly about the crowded room. "Do you think he might offer for her?"
"No," he answered sharply, glancing at the smiling couple again.
"No, I suppose not," she agreed smoothly. "Whatever Rafael's standards might be, your father would never allow such a poor match."
That caught his attention again. "What do you know of Rafe's standards?" he asked.
"Oh, just speculation," she returned. "I have to admit, though, they do look rather good together."
Yes, they did. Tall and muscular, with slightly tousled hair the color of ripened wheat and an easy grin made a little lopsided by the scar on his cheek, Rafe would look good with anyone. And Maddie, tonight wearing Quin's favorite gown because it brought out the gray of her eyes, her auburn hair piled high with curling wisps framing her face, was absolutely mesmerizing.
"So you think her a poor candidate for marriage to a peer?" he pursued, wondering how she would reply.
"Despite your commendable efforts, my dear, how could I think anything else?"
Quin nodded, remembering Eloise throwing tea at a servant, and Maddie slipping down to the kitchen to patch him up. "Then you were being kind to her only for my sake?" he continued.
"I like her, of course," Eloise retorted, her expression exasperated. "I'd like her more if you didn't seem so fond of her."
"Or if she was a social equal," he added.
"You make that sound like a bad thing. We all have standards to uphold, Quin. Especially you."
Quin nodded, wondering if she could sense his grow ing disgust. "Yes, I do. Thank you for reminding me," he said softly.
She smiled coyly at him. "You're welcome, my love."
By God, Eloise could be a snob. He realized how very much he disliked her, though he wondered how he might have reacted to Maddie's unexpected return to London, if he hadn't become acquainted with her first. Quin stifled a completely inappropriate grin. It was more accurate to admit that she'd pummeled him to his senses.
If Eloise had simply stated that she didn't like Maddie, and that she felt threatened by her presence, he could have accepted it. In fact, her honesty would have made him feel an absolute cad. But she'd been devious, and lied, and apparently had set Maddie up for several scandalous episodes. He couldn't help falling in love with Maddie, but he could help how he dealt with it. And Eloise was making the choice a surprisingly easy one.
"May I call on you in the morning?" he asked smoothly, as the music stopped.
"I look forward to it."
Finally he claimed Maddie for the last waltz of the evening, not particularly caring whether she'd already promised it to someone else or not. "I've missed you," he said, sweeping her out onto the dance floor.
"I've been gone for only two days. You will have to get used my absence, you know. I'm not your pet hunting dog."
However boldly and carelessly she spoke, Quin felt the tension in her lithe body, and the way her hand shook ever so slightly in his. Though he disliked causing her pain, he couldn't help feeling encouraged: at least she still felt an attraction toward him. Whether it was anything close to the torrent he felt for her, he could only hope.
"You're closer to a wild fox," he agreed, and took a deep breath. "Are you going to marry Charles Dunfrey?"
Gray eyes met his. "Rafe told you?"
"Of course Rafe told me," he snapped. "You knew he would."
"How could I possibly—"
"Are you going to marry him?" he interrupted.
For a long moment she looked up at him, her eyes searching his. "Who did you plan to have me marry when you dragged me back to London?"
"Damnation, Maddie, why won't you ever just answer a blasted question?"
Her lips twitched, humor replacing the somber look in her eyes for just a moment. "Why don't you?"
Quin wanted to be angry at her, but he was keenly aware that his time to converse with her had become severely limited. "All right. I surrender. I...hadn't thought that far ahead when I dragged you to London," he admitted. "I merely had a vague idea of saving you."
She actually smiled. "My white knight," she murmured. "Well, I suppose I'm saved."
"Don't marry him, Maddie."
For a moment she looked up at him. "So serious, now. Why are you so stubborn about this?"
Because I love you. "Because I am," he said instead. "I thought that was a character trait you'd admire."
"Quin, I don't want to marry him, really. But—"
"That's all I need to know."
"You are very annoying," Maddie stated. "First you drag me here to marry me off so you won't have to bother with me any longer, and now you tell me not to marry the only man who's asked. I don't—"
"Dunfrey is not the only man who's asked," he reminded her softly. "And he is not the man you're going to marry, I am."
Maddie looked down at his cravat. "I wish..." she said, in a very small voice.
He wanted to take her wan face in his hands and kiss her. "You wish what?"
She met his eyes again, as the music crashed to a halt. "I just wish," she repeated, and stepped back from him amid the applause of the other guests.
"Sometimes wishes come true," he whispered, and escorted her back to her parents.
Charles Dunfrey was there, waiting to take her out onto the floor for the next dance, and Quin's mood immediately soured. He nodded stiffly at his rival, and with a last glance at Maddie, strolled away. Tomorrow he would straighten things out.
Malcolm Bancroft sighed as the coach rattled to a stop. He sat where he was for a few moments, still half wondering what the hell he was doing.
A footman pulled the coach door open, and looked in at him expectantly. "Will you be leaving the coach, sir?" he asked politely.
"Yes, I suppose I will. You'd best call for assistance, though. You're a bit small, and I'm a bit ungainly."
After a moment the footman evidently believed he was serious, for he stepped back and whistled. Another liveried servant appeared, and Malcolm thrust a sturdy cane at each of them. Scooting as close to the door as he could, he heaved himself upright.
"Either catch me, or start running," he warned, and let his weight shift forward.
The footmen grabbed his arms as he half fell out of the coach, and among the three of them they managed to land him feet first on the ground. Taking his canes back, Malcolm began hobbling toward the front steps.
His left leg remained numb to the knee, and he had to keep a constant watch on it to make sure it didn't go wandering off without him. According to his physician, he still tired far too easily and should remain in bed for at least another fortnight. After reading Maddie's last letter, though, he'd decided he couldn't wait that long.
Beeks pulled open the front door as he reached the top step. The butler gaped at him for a moment, then hurried forward to offer his assistance. "Mr. Bancroft. We did not expect you."
Malcolm gave a slight grin. "No, I wouldn't imagine so. Is the old windbag about?"
"His Grace is just preparing to leave for the House of Lords," the butler replied.
"What about Quinlan?"
"Lord Warefield is residing at Whiting House."
Not surprised in the least to hear that, Malcolm headed into the morning room and carefully seated himself on the couch. "Please tell His Eminence I'm here, will you, Beeks?"
The butler nodded. "Of course, Mr. Bancroft."
He'd figured Lewis would make him wait fifteen minutes before he made his appearance, but it was only ten. His brother must have been curious in the extreme to forgo his usual performance.
"Who's watching over Langley?" the duke barked, strolling into the morning room.
"Squire John Ramsey. Who's watching over Highbarrow?"
"My estate manager." Lewis glanced at his younger brother, then headed to the window, pulling aside the curtain to look out at the drive. "You've brought luggage. I hope you don't think to stay here; I just cleared that other crowd of fools out."
"I wouldn't stay here if it was the only house left standing in England," Malcolm replied calmly. "I'll stay at Whiting House, with Quinlan."
"Good. What do you want here?"
Mr. Bancroft eyed His Grace for a moment. "To see if you were keeping up your end of the bargain. I can see you're not, though I'm not really surprised."
Lewis faced him. "What bargain?"
"Seeing Maddie Willits back into society."
"That wasn't my damned bargain."
"It is a matter of Bancroft honor."
The duke dropped into the chair opposite his brother. "It's a matter of you and Quin doffing the same girl, and then feeling guilty about it. She doesn't signify, and she's certainly not worth all this trouble."
"She's worth considerably more trouble than you are. You used to have at least a few remnants of usefulness about you. Now you're just loud."
Lewis glared at him. "Get out of my house."
"With pleasure." Using a cane and an arm of the couch for leverage, Malcolm hauled himself to his feet. For a brief moment something uncertain entered the duke's eyes, but it was just as swiftly banished. "I'll be at Whiting House, should you wish to stop by and apologize."
"Not while I'm breathing."
Malcolm smiled. "I can wait."
Quin was on his way out the front door when he saw the coach. He immediately recognized the pair of bays pulling it and couldn't help grinning in relief. Reinforcements had arrived, and it was about damned time.
"What in the world are you doing here?" he asked, striding forward and yanking the coach door open. "And where is your chair?"
"I am walking, thank you very much," Uncle Malcolm returned. "Or a reasonable impression of walking, anyway."
Quin helped his uncle down to the ground and into the foyer. "Did you stop by Bancroft House? Have you seen Maddie? Did you—"
"Quinlan, what a chatterbox you've become," Malcolm chastised with a grin. "Something on your mind, lad?"
He sat down in the drawing room, while Quin stalked the floor in a circle around him. "You know bloody well something's on my mind," the marquis stated. "It's your fault."
"My fault? What's my fault?"
"You knew I'd fall in love with her. And I thought you were sending her to London for her own sake." Quin glared at Malcolm. His uncle's smile didn't leave Quin feeling any less agitated; in fact, if he'd been Maddie, he would have been pummeling someone by now. "Well, say something!"
"This was for Maddie's sake," Malcolm offered, obviously considering his words. "As soon as I read that it was you Lewis was sending, I knew this would be my last chance to see her restored to her proper place in society. I also knew I'd likely lose her. Sending her away with you was not easy, you know."
"Well, you seemed to forget one rather important fact," Quin snapped at him. "I'm engaged."
Malcolm shrugged. "Obviously this was something I had to throw together rather quickly. You couldn't expect me to take care of all the details."
"The details?" Quin asked skeptically, angered at the idea that he'd somehow been manipulated into this whole disaster, whatever the ultimate reward might turn out to be.
"Yes. As for the part about you falling in love with her, I admit, it crossed my mind, but it wasn't my primary goal. Truly, Quin. I liked you as a youngster—very much—but for all I knew, you'd grown into your father. In which case, Maddie would have drowned you, instead of merely teasing you with the idea."
"I don't like this," Quin said flatly.
"That's why I hadn't intended to tell you any of it."
Quin sat opposite his uncle. "What changed your mind?"
"This." Malcolm pulled a much-folded piece of parchment out of his pocket and smoothed it open. "Maddie's latest letter."
"Let me see it," Quin demanded.
"Most of it's none of your affair," his uncle replied. "This part, though, I thought you should hear." He lifted the letter again. "'Charles Dunfrey has been calling on me, and while I doubt he could possibly be as smitten as he once claimed, at least he is polite and respectful. Con—'"
"The bastard," Quin growled, and Malcolm looked up at him.
"Hush. 'Considering that Quin has obligations of his own, and that I don't wish to cause any more trouble for him than I already have, perhaps setting things back the way they were would be the wisest choice, after all.'" He looked up. "It goes on from there, but I found that section particularly alarming."
Shooting to his feet, Quin began pacing again. He'd known she was thinking that, but to hear it worded that way.... She sounded so sad, and he had the absurd desire to go to her and snatch her up in his arms. "She won't marry him," he stated.
"What's to stop her?"
Quin looked down at his uncle. "I am." He headed for the door. "I hope you're not too tired to go visiting with me," he said over his shoulder. "I know someone who'd like to see you."
Maddie felt very much in need of an ally. She looked at her father, trying not to glare, trying to be reasonable, and trying very, very hard not to run out the front door and keep running until she reached Quin and Whiting House.
"Are you listening to me?"
"Yes, Father, I'm listening."
"So you will agree to Charles's proposal when he comes by today."
"I haven't decided yet," she said, her calmness edging toward anger. Whatever she'd been through over the past five years, it didn't seem to have changed her father's thinking in the least. And even if she did decide to marry Charles, it would be for her own reasons, and not because of the viscount's bullying and badgering.
"You're too good for him, now?" he asked with scathing sarcasm, folding his arms over his chest and tapping a boot against the floor.
She shrugged. "I don't love him."
"You don't—what in damnation does that matter? Do you think I love your mother?"
Maddie hoped Julia Willits wasn't listening. "I would have hoped so," she said, her voice beginning to shake a little. "She is a good-hearted and kind woman."
"And far too soft on you, obviously."
Maddie's indignant response would no doubt have seen her kicked out of Willits House, but before she could more than open her mouth, Everett scratched at the library door.
"What is it?" the viscount grumbled, obviously displeased at having his tirade interrupted.
Everett opened the door and stuck his head in. "Two callers for Miss Willits," he said.
"Who is it?" Lord Halverston continued to glower.
"A Mr. Bancroft, and the Marquis of Ware—"
"Mr. Bancroft?" Maddie cried, jumping to her feet and pushing past the surprised butler. "Sorry, Everett."
"Quite all right, Miss."
Both men stood in the foyer, and that in itself stopped her. "Mr. Bancroft, you're standing?"
"More or less," he grinned. "How are you, my dear?"
Maddie hurried forward and flung her arms about her former employer. "I'm so glad," she whispered, tears running down her face. "I'm so glad."
"Now, that is a proper greeting," Malcolm said, hugging her back.
"I'd settle for one of those, myself."
Maddie looked up at Quin, who gazed back at her. She smiled tearfully. He had come to see her. "Thank you."
"You're welcome," he answered, "but he brought himself to London. I think he's worried you're going to marry the wrong man."
"Ah, Lord Warefield," her father said grandly, emerging from the library and showing no sign at all of his former ill temper. "So kind of you to stop by to see my daughter."
"A pleasure," Quin replied, shaking the viscount's hand. "May I present my uncle, Mr. Bancroft? Malcolm, Viscount Halverston."
Maddie reluctantly relinquished her hold on Mr. Bancroft, and he shook hands with her father as well. "I've heard a great deal about you," Malcolm said noncommittally.
"I wish I could say the same," the viscount answered, looking at his daughter.
Maddie could guess what he was thinking—that she'd had some sort of sordid relationship with Mr. Bancroft. He seemed to think that about every gentleman she mentioned. And she had long ago reached the point that she really didn't care what he thought. She looked up at Quin again.
"How are you today, Miss Willits?" he asked politely.
"Very well, thank you, my lord."
"Might I have a word with you in private?" he continued. "My mother wished me to convey a message to you."
"Of course," she said, trying to cover her sudden excitement, and motioned him to join her in the library.
With her father standing right there in the hallway, they couldn't exactly close the door, but Quin took her hand and led her to a far corner of the room, beneath the high windows. "How are you getting along?" he asked quietly, running his fingers along her cheek.
Maddie closed her eyes as his lips touched hers. She had missed him so much. "As though I never left."
He smiled. "That poorly?"
"Yes." She sighed. "So you came to inquire after my health?"
"Not exactly. I never asked you," he murmured. "What is your dream of an ideal life?"
Unsettled, Maddie turned away. Quin slowly slid his arms about her waist, pulling her back against his chest. "I don't dream," she said. He was making doing the right thing supremely difficult—and he knew it, the bastard.
"Tell me anyway."
She shook her head. "You know." Little by little she let herself relax against him. "It would have been nice."
Quin rested his cheek against her hair. "It will be nice," he corrected.
It was too easy—too easy simply to lose herself in the moment, to pretend that it would last forever. Maddie straightened and turned to face him. "Quin, stop—"
His jade eyes held hers, warm and compassionate, looking deeper inside her than anyone had or ever would. And in that moment she knew: she was not going to many Charles Dunfrey. Thanks to Quin, she knew what it was to love someone. Whatever else happened, she would not marry for anything less.
Slowly he smiled. "What are you thinking?"
Maddie leaned up on her tiptoes and kissed him. "It is nice." Her father's voice echoed in from the hallway, and she jumped, taking a step back. "So. Have you been to call on Eloise yet?"
He scowled and shook his head. "I was on my way when Malcolm appeared. I'll go as soon as I see him back to Whiting House. Has Dunfrey been to call on you yet?"
"No."
Quin swallowed, his expression becoming uncertain. "I need to tell you something."
Now she was uneasy. She didn't need another lesson to know that nothing was simple where they were concerned. "I'm listening."
"Even if you decide for some unfathomable reason that you don't want to marry me, there's something you should know about Charles Dunfrey."
"Not spreading rumors, are you?" she asked, only half teasing. He wouldn't dare to stoop so low as to lie about Charles just to convince her not to marry him. Not now.
"This is a fact. Maddie, I discovered something the other day, and considering the circumstances, I don't think I should keep it from you."
"Stop stalling about and tell me, Quin."
"It's Dunfrey's finances. He's—"
"He's what, Warefield?" This was the sort of backbiting behavior she expected from the rest of the nobility; she hadn't expected it of Quin. "He's not as wealthy as you?" she suggested. "I suppose not. But then again, who is?"
"Maddie, you're taking it all wrong. This is not about my pompous snobbery, or your lack thereof."
She put her hands on her hips. "Why don't you explain it to me, then, my lord?"
"I'm trying to explain it, damn it," he snapped. "Dunfrey's one step ahead of the bloody moneylenders, Maddie. Without your dowry, he'll be done for, probably by the end of the Season. I'm worried that—"
"That he's marrying me only for my money? Or for my parents' money, rather?" She shrugged, furious and hurt. "What did you expect? I suppose he couldn't possibly just happen to be poor and simply wish to marry me because he loves me. For heaven's sake, who could be that abysmally stupid?"
"Maddie—"
"Thank you, my lord, you've been a great help. Now, go marry that wretched Eloise, and leave me alone." Tears danced in her eyes, and she lowered her gaze to his chest.
He opened and shut his mouth several times. "Damnation," he cursed. "You are impossible."
"That's what I've been trying to tell you, if you'll recall. Good day, my lord."
She turned on her heel and left the room, pausing only to nod at Malcolm before going upstairs to her bedchamber.
## Chapter 16
Quin wanted to strangle her.
He also wanted to kiss the tears from her eyes, and to kiss her sweet, soft lips, and to hold her in his arms again. The thought that he would never do so again wrenched something hard and painful loose in his chest. "Devil a bit," he muttered, stalking out into the hallway. "Let's go, Uncle," he snapped.
"My lord," the viscount said, touching his shoulder, "if my daughter has offended you, please let me apologize. She has no manners, and—"
Quin jabbed a finger at him. "Don't," he snarled, and strode outside.
His dramatic exit was somewhat ruined when it took Malcolm another four minutes to make his way down the stairs and out to the coach, so he sat in the half dark of the vehicle's interior and stewed.
"Another argument?" his uncle grunted, tossing both canes onto the floor.
Quin stood and helped haul him up into the coach and down into a seat. "This is ridiculous. She never listens to me, she misinterprets everything I say, and she is so damned stubborn, I just want to wring her neck."
Malcolm lifted an eyebrow. "And?"
"This is all your fault."
"All I did was take her in and have an apoplexy. You're the one who fell in love," his uncle pointed out. "Don't blame me."
"Why am I doing this?" Quin demanded. "Why am I putting myself through this? No one will appreciate it. I am the bloody future Duke of Highbarrow, for God's sake."
Malcolm just looked at him.
Quin glared right back at him. "Oh, shut up," he finally mumbled. "Love is highly overrated."
"I wouldn't know, Quin."
"Lucky you."
"Do you really mean that, son?"
The marquis sat back and folded his arms across his chest. "No."
Rafael was ecstatic to see Malcolm when they returned to Whiting House. The two of them toddled off together to the morning room, no doubt to gossip the day away about the idiot Marquis of Warefield and his infatuation with a stubborn, impossible...lovely, high-spirited, intelligent sprite who was absolutely nothing but trouble. And quite the best thing that had happened to him in his entire life.
By the time he remembered that he'd been heading over to see Eloise, she'd gone out shopping with some friends, and had left word that he was a rude, uncaring beast, and she would see him tomorrow.
"Just as well," Quin informed the Stokesley butler, as he turned on his heel. "I don't feel like another bludgeoning today, anyway."
"Very good, my lord."
He swung back up on Aristotle, intending to return home, until a paralyzing thought occurred to him: Dunfrey was still supposed to call on Maddie today. If he did, and if she was angry enough, there was no telling what she'd do. He kicked Aristotle into a gallop, upsetting the more conservative members of the gentry as they went about their early afternoon visiting.
When he stormed into his morning room, Rafael and Malcolm were playing chess. "Rafe, go visit Maddie," he ordered, ripping off his gloves and tossing them at his brother. "Aristotle's outside."
"Just a damned minute," Rafe said, deftly catching the gloves. "I am a captain in His Majesty's Coldstream Guards. I do not go and visit women on command. Nor am I your errand boy." He threw the gloves back. "Go visit her yourself. And don't try to bribe me with my own horse."
"I thought you liked her."
"I do like her. Enough that I'm not going to participate in this silliness any longer."
Quin narrowed his eyes. "What silliness, pray tell?"
Rafe stood. "I don't know what the devil's going on between you," he snapped, all humor for once missing from his light green eyes, "but I do know if you don't take care of it soon, I will." He stomped toward the door.
"What is that supposed to mean?" Quin asked coolly, seething.
Rafe glanced over his shoulder. "Whatever the hell you want it to." He stalked out the front door. A moment later, Quin heard him whistle sharply for a hack.
The marquis took a deep breath and sat in his brother's vacated chair. "Wonderful. Now everyone's angry at me."
"Why did you want Rafe to go see her?" Malcolm asked, calmly removing an opposing knight from the board and setting it with his other captives.
"Dunfrey's going to call on her today. She's supposed to decide whether she's going to marry him or not."
"Ah, that would explain several things." Still looking as though he hadn't a care in the world, Malcolm shifted one of his own pieces sideways and hauled himself to his feet.
"Which things?" Quin asked, following behind him.
"Maddie's being in tears when she saw you this morn—"
"You mean when she saw you," Quin amended.
"—When she saw you this morning, and all the colorful names Rafe has been calling you since we returned." Malcolm paused and looked sideways at him. "He was going to see her today, anyway." He smiled. "I think he's half in love with her, too. She has quite a talent for that."
Quin stopped. "For what?"
"For being irresistible to every man who's not terrified of her. Now, don't bother me until dinner is ready. I'm damned tired of limping about."
Rafe didn't return, but sent word that he was spending the night at Bancroft House, along with a heart-lifting "She said no" scrawled on a piece of paper. Staying voluntarily under the same roof as His Grace was a measure of how furious he must have been at his brother.
Actually, even with the good news, Quin was glad Rafael had stayed away—the desire to strangle his brother grew stronger every minute he thought about what Malcolm had said, about Rafe being half in love with Maddie. On the other hand, with Rafe gone, he had no way of knowing why Maddie had turned Dunfrey down.
He spent the night pacing before his bedchamber fireplace, fighting the urge every few minutes to ride off to Willits House and demand to know what had happened. The only thing that stopped him was that he still hadn't broken with Eloise, and that Maddie would know that when she looked at him.
He knew that ultimately she didn't believe he would keep his word to her, or that he would dare brook four hundred years of Bancroft history to marry a social outcast—that he would risk losing his fortune and his title just to be able to wake up each morning and argue with an auburn-haired wood sprite.
So Quin paced until two in the morning, and then he went downstairs to the drawing room and got very, very drunk. Then he thought of something. "Damn," he said, dropping into a chair and toasting himself with another snifter of brandy. "I'm not such an imbecile, after all."
Eloise had clearly decided to make use of deceit and subterfuge to secure his hand. He might as well do the same to prevent it. Quin smiled. With Eloise busy fawning over him, she wouldn't have time to make trouble for Maddie. And after Almack's, all bets were off.
Maddie sat on the edge of her bed, staring at the ivory dress laid out on the quilt. Ten days had gone by so quickly, she could scarcely believe it. And after Almack's tonight, all bets were off.
Rafe had called on her every day for the past week, as had Mr. Bancroft, and, surprisingly enough, the Duchess of Highbarrow. In addition to bolstering her flagging spirits, the Bancrofts' presence inspired her father to allow her to remain at Willits House despite her abysmal stubbornness regarding marriage to Charles Dunfrey. Of Quin Bancroft, though, there was no sign.
She didn't sleep, and she couldn't eat. If she'd been the woman her father wanted her to be, she would have put her feelings and affections aside and graciously informed Charles Dunfrey that she'd changed her mind, that of course she would be his wife.
But marrying for anything but love, and marrying anyone but Quin Bancroft, was unthinkable. After their last argument, though, he'd apparently decided to ignore her existence, instead turning to Eloise and happily carrying on with his stuffy, stupid, maddening engagement. It rendered even daydreaming about their own union completely ridiculous—which didn't stop Maddie from imagining it endlessly. She tried not to listen, but it seemed everyone she encountered had seen Quin just five minutes earlier, and always in the company of Eloise Stokesley.
Rafe hadn't tried to explain his brother's thinking, other than with a cool "Must always keep up appearances, you know." In fact, she knew that he and Quin hadn't been speaking and that Rafe had moved back into Bancroft House, where his mood was so foul, even the duke stayed clear of him.
She hated causing so much pain and anguish. But after tonight it would be over. The Bancrofts would have no more reason to be pleasant to her, and she would have carried out her promise to Mr. Bancroft. Best and worst of all, if she wanted, she would be able to return with him to Langley Hall. The thought wouldn't be so difficult if she could stop thinking of Quin's kisses, and Quin's laugh, and Quin's touch, every moment of every day.
Mary scratched at the door. "Miss Maddie, I need to start getting you ready," she said softly.
"Come in," Maddie said, making an effort to erase the tense, nervous expression from her face. "I may as well get this blasted nuisance over with."
When she came downstairs an hour later, everyone said she looked superb—except for her father, who hadn't said a word to her since she had sent Charles away. She wished she'd realized before how little he cared for her. It would have made the past five years much easier: she could have written her mother or Claire, and told them where she was, and at least they might have corresponded. She wouldn't make the same mistake again, but neither would she stay.
"Are you ready, Maddie?" her mother asked. "Her Grace is to meet us at Almack's."
"Yes, I'm ready. More than ready."
Despite her brave pronouncement, the carriage ride went far too quickly. She would have chosen even the company of her father, cold and stone-faced beside her, over addressing the patronesses of Almack's. It didn't help that she already hated them. How any dozen stodgy women had come to wield such power over their fellows she had no idea, but it couldn't possibly be fair.
Not only was the duchess there waiting for them; so were Rafe and Malcolm. As she glanced across the room at what seemed like hundreds of guests, all ready to second the judgment of the patronesses, her pulse leaped. Quin had come, too. He was dancing with Eloise Stokesley, but at least he had come. She would be able to see him, and perhaps even speak to him, one last time.
She knew she should be angry at him, but the pounding of her heart, the heat in her cheeks, and the way she wanted to kick off her shoes and run across the room to hug and kiss him made one thing very clear: whatever she'd said to him, and however loud she'd yelled and stomped her feet and tried to dislike him, she loved Quin Bancroft.
"Maddie," the duchess murmured, "stop looking at my son."
She jumped. "Was I?" she asked shakily, turning her gaze at once to the white- and ivory-clad girls moving in a shuffling, nervous line before the row of patronesses. Of course, every single one of the stuffy, conceited women seemed to have decided to attend tonight's assembly.
"Yes, you were. I fear, though, that now he seems to be staring at you as well. He used to know better. You've been a poor influence on him."
Maddie looked over at her sponsor. Despite her words, she didn't look angry. In fact, she seemed to be rather amused. "My apologies, Your Grace."
"Hmm. Well, come on, we'd best get you past the gauntlet."
"Your Grace?" Maddie said hesitantly, as she took the older woman's arm. "Thank you for everything you've done."
Lady Highbarrow chuckled. "I think you'd be wise to save your thanks. The evening's not over yet."
Quin watched. He couldn't help it, because he couldn't keep his eyes off her. He didn't know when he'd stopped looking at this moment as Maddie's triumph and begun looking at it as a death knell to the moments he could spend with her. But as his mother introduced her to each of the Almack patronesses, and as she curtsied politely and received the all-important nod of the head, the tension gnawing at the pit of his stomach grew into a knot of dread and anguish that made even the act of breathing difficult.
After tonight, she could leave. If he made one misstep, if he gave her one more split second to distrust him, he would lose her. And that, he was certain, would kill him.
"Quin, I thought you were over her," Eloise murmured, running her fingers along his sleeve.
"I am," he replied easily, turning back to face his betrothed. "Her success is something of a reflection on the Bancrofts, though, don't you think?"
Eloise glanced at Maddie. "I suppose, but only because you've made it so. You could easily have distanced yourself, and your family, from her anytime you chose."
"Wouldn't have been much honor in that." He risked another glance as Maddie made it to the end of the gauntlet and emerged triumphant. While he had the overpowering urge to applaud, he also wanted to race over and grab hold of her before she could flee into the night.
The orchestra struck up a waltz. In his dreams, almost from the beginning, he had been the one to dance with her. Instead, Rafe bowed and took her hand to lead her onto the floor.
"He doesn't miss an opportunity to stand up with her, does he?" Eloise noted. "Will you dance with me, my love?"
"Of course," he said.
They danced the waltz, and then he took his mother out for a quadrille. Another entire set followed until the music for another waltz finally began. Seeing several gentlemen heading in Maddie's direction, now that she was accepted again, he quickly excused himself from his circle of cronies.
"This one is mine, I hope," he murmured, as he came up behind her.
She jumped and turned to look up at him. "As you wish, my lord." Her cheeks were flushed from the heat of the room, and her eyes sparkled as he swept her into the dance.
"Congratulations," he said, smiling. "You are a triumph."
"How could I be anything less, with the Duchess of Highbarrow leading the introductions? You should have heard her. She practically threatened those women into accepting me." She chuckled. "It was quite wonderful. Rafe said when Her Grace is determined about something, she's more frightening than a herd of stampeding water buffalo."
"You see Rafe quite a bit these days, don't you?" he asked, unable to keep the jealous edge out of his voice.
Her smile faded. "I see those who come to visit me," she answered. "You've been busy elsewhere, apparently." Maddie glanced pointedly at Eloise, dancing with Thomas Danson.
He narrowed his eyes. "Yes, I have, thank you very much."
"Eloise is not my fault," she snapped. "You're the one who keeps promising to marry everyone in sight."
"I do not!" he protested.
She looked up at him. "I just wish you would tell me, straight out. You're not hurting anyone, Quin, by doing the right thing and marrying Eloise."
"The right thing is not to marry Eloise," he countered. "Which is exactly why I've been keeping her occupied for the past week."
Maddie's suspicious expression made him want to laugh. "What do you mean, keeping her occupied'? She's your betrothed, you big oaf."
"I've recently found that idea very unappealing," he said softly, holding her as closely as he dared in the blasted conservative assembly. "But I decided that while she spends time with me, she can't be making things more difficult for you."
"Oh, so now you're being gallant again?" she said skeptically.
"I thought so."
"Ah. And you'll gallantly stand beside her at Westminster Abbey in two weeks as well, I suppose?"
He heard it then—the jealousy in her voice. Quin smiled. "I don't want to marry Eloise," he whispered, very conscious of the other dancers circling them and how much damage could still be done to her reputation. "Nor do I intend to do so. I love you, Maddie."
She actually stumbled, tripping over his foot, and with a grin Quin pulled her against his chest until she could regain her balance.
"Are you all right?" he asked, feeling considerably more confident than he had a moment before.
"What...what did you say?" she asked almost soundlessly, staring up at him white-faced.
"I said that I love you. Is that such a surprise?"
"Well, yes. Quin, you can't love me—you're still betrothed to Eloise."
"Only until tomorrow morning. She can't hurt you now, and I'm not certain I can take another day with her. I'm finding that several of my acquaintances are rather unpleasant to be around—very arrogant, they are."
Maddie grinned, her eyes lighting. "You're an idiot," she whispered.
"I still love you. I'm beginning to think you're impossible not to love."
"Yours is a minority opinion." Maddie took a deep breath. "You shouldn't have told me, anyway—you're only making things more difficult. You're bound to come to your senses soon."
"Be with me tonight, Maddie," he murmured. "We'll figure everything else out tomorrow."
"But Eloise—"
"To hell with Eloise. I want you."
"That's very dishonorable," she said weakly.
He heard the reluctant desire in her voice. It was all he needed. "You're already ruined," he pressed, "thanks to me. Please, Maddie, whatever else happens, I want—"
"Shh," she said. "How in the world do you intend to be with me, Lord Marquis? Are we going to steal into the library and lock the door? Oh, dear me, we've already done that, haven't we?"
He chuckled, delighted. He'd won—at least, for the moment. "That was the drawing room. And when the dance is finished, follow me."
As soon as the waltz ended, people crowded onto the floor for a country dance. Quin ducked backward, Maddie trailing behind him. As the music started, he slipped out onto the dark balcony. Maddie peeked around the comer, and he pulled her onto the stones. He leaned down to kiss her, but she sidestepped, shoving him away.
"What is it?" he asked, pursuing her toward the railing.
Maddie raised her fist in his direction. "This is what happened before," she hissed, "with that blasted Spenser. I don't want anyone to see you trying to kiss me. Not after I just became respectable again."
"I'll check first, then," he said, and moved past her into the darker shadows. Under the circumstances, he was lucky she hadn't hit him; the only excuse he could give for being so obtuse was that he was having difficulty thinking of anything but feeling the warm, naked slide of her body against his. "All clear," he informed her, returning to her side. "Now will you kiss me?"
With another hesitant look around, she lifted up on her toes. Leaning her hands on his chest, she touched her lips to his. Quin shut his eyes at the soft touch, slipping his arms around her slender waist and wondering that she'd ever come this far in trusting him. If he'd been the one betrayed and reviled by his peers, he wasn't certain he'd have been able to do the same.
She sighed. "Oh, my, that's nice." After a moment, she relaxed against him. "What now?" she whispered against his mouth, placing feather-light kisses on his lips and along his jaw and cheek.
White-hot desire blazed through him. "We'd best do something soon," he murmured, "because I am becoming extremely uncomfortable."
"Out here?" she asked skeptically, her breathing uneven.
Quin stepped to the railing and looked down into the garden. A thick trellis of vines crept up the wall beside the stone abutment. They could climb down, but once there, only the uncertain shadows of the foliage would shield them from curious eyes. And there were plenty of those about. "Devil it." Banging his fist on the railing in growing frustration, he leaned out further and looked up.
"That way." He grinned, pointing to the dark window twenty feet above their heads. "The attic."
"You are completely mad," she declared, unable to stifle a nervous chuckle. "Quin, you're not serious. Almack's attic?"
"Yes, I am. Shall I go first?"
"Quin, I'm not climbing up there. I'll tear my dress."
"Then I suppose we'll have to make love right here." He smiled softly, running his finger along her cheek. "It wouldn't be the first time we've torn one of your lovely gowns, Maddie."
"Oh, damn," she swore, swallowing hard. "If I wasn't already ruined, you would take care of it. Climb the blasted trellis."
He made it up fairly easily. Luckily, the window was unlatched, and he pushed it open and swung his legs over the sill. When he looked down, Maddie had the hem of her gown tucked into her neckline, and she was looking ahead determinedly as she climbed.
"Don't you like heights?" he asked softly, as he guided her into the tall, narrow room.
"I don't believe I'm doing this," she panted, freeing her hem. "I've completely lost my mind. This is so stupid."
Quin lowered his lips over hers, stopping her complaints with a rough, deep kiss. He couldn't believe she'd done it, either, and he had no intention of letting her take herself back down the trellis right away.
Everything was covered with dust, but fortunately the spare furniture was draped with sheets. He released Maddie long enough to uncover an ornate serving table, a long gash across its otherwise smooth surface. Lifting Maddie around the waist, he set her down on the polished oak.
"Quin, we can't do this," she managed, clutching his shoulders and lifting her chin as he trailed his mouth down her soft throat.
"Stop saying that. I want to be with you."
"And I want to be with you. But—"
He backed away just enough to look her in the eye. "You are not marrying anyone but me, and I am most certainly not marrying Eloise. Is that clear?"
She scowled. "You can't order me—"
"And I'm not marrying Eloise Stokesley," he repeated, before she could manage to turn this into another battle. They didn't have a great deal of time.
Her mouth opened and then closed. "You're not? Truly?"
"Truly." He reached down to grasp her ankles, then slowly slid his hands up along her legs, lifting her skirt as he went.
"Just like that?"
Quin leaned down to brush his lips across her exposed thigh, and the muscles jumped beneath her skin. "Just like that."
"And what about your family?" she pursued raggedly.
"I'll tell them tomorrow." Hungrily he sought her mouth again.
She moaned as his gentle touch found the secret place between her thighs. "But what about—"
"Shh," he murmured. "Don't ask me anything else."
Whether she intended to ask anything or not, she became occupied with nibbling at his lip and then his ear. His heart pounding, Quin tugged her legs around his hips, pulling her close against him. It seemed impossible that such a fiery, passionate woman could have had no lovers before him, but he knew she hadn't. He was her first, and if everything went as he'd planned, as he'd dreamed, he'd be her only lover.
Maddie's hands tugged at his waist, pulling his shirt-tails free of his trousers. With a breathless chuckle, she kissed him, her gray eyes dancing with heat and passion. Her hands fumbled with his fastenings, then freed him from his breeches. Quin moaned as she folded her legs around his hips. He entered her slowly, relishing the feel of her warm, tight flesh around him.
She threw her head back as he pushed into her, twining her hands together behind his neck and holding herself hard against him. Quin grasped her buttocks, pulling her to him with every thrust of his hips. He wanted to remember everything—the darkness of the night with the half moon rising just over the rooftops, the muffled sound of the country dance below, the lavender scent of Maddie's skin, and the sparkle of her eyes as he looked into them.
They climaxed together, and he buried his face against her shoulder as he shuddered and spilled his seed inside her. Maddie threw her arms about his shoulders, holding tightly to him. After a long time, he lifted his head and gently kissed her again.
"Well, you've ruined me again," she said, still out of breath.
"I seem to be making a habit out of it," he agreed. "I can't help myself. You are irresistible. But one of these days, you and I are going to share an actual bed, with an entire night of nothing but the two of us."
She smiled at him and slowly lifted her hand to stroke his cheek. "That would be very nice," she whispered.
He couldn't help grinning. Maddie hadn't said she loved him, but she did care for him. She'd made that much obvious by the degree of trust she'd shown. Whether or not she would ever be able to tell him so, he had no idea. But he could hope. And he would wait.
"I love you, Maddie. I have since the moment we met."
Her expression sobered. "What do we do now?"
He grinned ruefully. "We climb back down the trellis. Then you and I will go home, to our separate houses. In the morning I will call on my parents, I will call on Eloise, and then I will call on you. You are my heart's desire, Maddie—you make me feel so alive. Nothing else matters but you and me."
"Not your family, or your honor, or your title? What if your father disinherits you?"
He reached up to clasp her hands and brought them around to hold against his chest, over his heart. "I will call on you in the morning," he repeated firmly. "Trust me."
"Do you promise?"
"I promise, my love."
With a last, lingering kiss, Quin helped her down from the table. She wanted him to keep holding her, but a loud laugh from the ballroom below reminded her that they couldn't very well stay hidden in Almack's attic forever—and that was a pity.
He tucked his shirt back into his breeches and attempted to put his hair back in some sort of style, for she'd tousled it rather badly. Maddie self-consciously straightened her skirt and her underthings, feeling rather tousled herself. Pure insanity. That was her only explanation for her behavior since she'd met Quinlan Ulysses Bancroft.
She had heard of men who, once they began, couldn't stop drinking liquor. Day in and day out they craved the stuff, paying attention to nothing else, until finally they drank themselves to death. For the first time she understood the attraction.
She craved Quin with every breath, with every beat of her heart. No one knew her as he did, and certainly no one cared for her as he did. They were completely wrong for one another, but nothing made as much sense as being with him. In the moonlight his hair looked white and silver, and the green of his eyes darkened almost to black.
"What are you looking at?" he asked, as he glanced over at her.
"You," she answered. "I can't figure you out."
He chuckled. "I thought I was being rather obvious, myself."
"Not that." Maddie flushed, though she had to wonder what in the world she had to be embarrassed about. She certainly had no secrets left from him.
Quin flung the old sheet back over the furniture. "What, then?"
"Since you've met me, you've nearly been drowned, shot, attacked by a mad sow, bellowed at by your father, kept—"
"My father bellows all the time," he interrupted. "It has nothing to do with you."
"I just don't understand how, after all that, you could possibly decide that you love me."
Quin looked at her for a long moment, then slowly came forward to fix a straying strand of her hair. "I'm not dull," he said quietly.
"I know that," she agreed. "I was only mad at you before."
Gently he put a finger over her lips. "That's not my point. I'm not dull, but my life is. As far back as I can remember, I've known I'd be the Duke of Highbarrow one day. I've known who I am to regard as a friend, and who is an enemy—not because I ever met them, but because of who their great-great ancestors were. I've known whom I'm to marry practically since she was born. You are...unexpected. And that's very rare and precious, where I come from." He grinned. "No one's actually ever attempted to drown me, before you."
She searched his eyes, but all she saw was a warmth and passion that matched her own. "What happens when I become dull and ordinary to you?"
Quin laughed, until she put her hand over his mouth before they heard him downstairs. "Hush," she ordered.
He removed her hand, holding it in his. "I don't think you could become dull if you wanted to, my dear." He chuckled softly, his eyes dancing, and leaned down to kiss her again.
That led to more kissing, until the music stopped below. She looked toward the window. "Oh, no."
"It's just the set, Maddie," he said, pulling her close against him again, his arms wrapped around her shoulders as if he never intended to let her go. She wouldn't have minded that at all. "We'll wait until the next one begins, and then we should go."
She looked up at him dubiously. "I don't suppose you mean we should sneak out the attic and back downstairs that way."
"We go down the way we came up," he said calmly.
"But I am considerably less motivated to climb down the trellis than I was to climb up," Maddie complained, only half joking. "That's a damned long way down, Quin."
"I'll go first, so you may fall on me if you feel the need to do so."
Releasing her, he strolled to the window, peering out carefully. "The garden appears to be deserted," he informed her, and leaned out to look down at the balcony. "Oops." He ducked back inside. "Apparently we're not the only amorous couple this evening. I must say, Almacks's standards are falling abominably."
Now that she wasn't quite so...involved with Quin, the night air coming in through the window felt chilly against her bare arms. She hugged herself. "I guess we have to go out the other way, then."
"No, we don't. We can wait another few minutes." He glanced down and then looked at her again. "Don't you want to know who it is?"
"Absolutely not. Whoever they are, no doubt they want privacy, or they wouldn't be out there."
Quin straightened and turned back to her. "Do you want my coat?" He started to shrug out of it.
"No, I don't want your coat, Sir Galahad," she retorted, yanking it back up onto his shoulders. "You'll have to put it on again in two minutes, anyway." He did feel nice and warm, though, so she slipped her arms around his lean waist, under his coat.
He had told her that he loved her, and she wished she could say the words back to him. She felt them, so much that it almost hurt to hold them in, but when she tried, they simply became stuck. Tomorrow, she would tell him. After he told Eloise and his parents that he intended to marry ruined little Maddie Willits.
Once he'd done that, she had the feeling that reality would come crashing down on his head, and he would regret having becoming temporarily mad and said all those wonderful things to her. Until then, she would let everything be a dream. A very pleasant, comforting dream.
"Why did you turn Dunfrey down?" he murmured into her hair.
She buried her face against his chest. "I thought about what you said. Charles claimed he loved me, but he sounded just as sincere when he called me a whore in front of all my friends. You were right. I think he just wanted my dowry."
"Maddie," he said softly.
"It's all right." The music began again, and she started at the sudden noise.
Quin leaned backward and glanced down again. "Hm. Apparently they weren't as amorous as we were. They're gone. In all fairness, though, they didn't remove any clothes."
She chuckled against his hard, well-muscled chest. "Neither did we."
"We rearranged some," he protested. "If you'd like me to be more thorough, I'm quite willing." He shifted. "Exceedingly willing."
She could tell. And if she didn't let go of him now, she never would. "Oh, no you don't," she said, pulling free of his arms. "Get going."
"Minx," he said, turning for the window.
"Oaf."
"Lightskirt." Hopping up onto the sill, he swung his legs outside.
"Blackguard."
"Sprite." Quin disappeared from view.
"Dullard."
His head reappeared. "I say. That last one I handed you was a compliment."
"Oh. Um—hero."
He grinned. "Much better. Come along, my sweet. And you'd best be quick about it." Quin vanished downward again.
He was right. If no one had already discovered their absence, they were luckier than they deserved. Frowning nervously, she hiked her skirts up to her knees and swung her legs out over the garden. Grasping one side of the trellis, she awkwardly stepped onto it.
"Very nice," he murmured from very close below her. "I should have let you lead before."
"Shut up," she snapped quietly, making a game effort to stomp on his head with her slippered foot. When she'd first seen him, she'd never have thought the Marquis of Warefield could be so very funny and witty and passionate and warm. Thank goodness he'd seen through her anger before she had.
Finally he gripped her about the waist and set her down on the balcony's hard stone. "Do you want to go first, or shall I?" he whispered.
"I will." Someone had left a half-empty glass of Madeira on the railing, and she picked it up, pasted a bored expression on her face, and slowly strolled back into the ballroom.
No one turned immediately to stare at her, and she took that as a good sign. When her mother grabbed her arm, she jumped and nearly spilled the glass down her front.
"Where have you been?" Lady Halverston hissed, her face flushed.
"Getting some air," she replied. "I've been rather nervous tonight."
"Even so, with your reputation, you know better than to go wandering off. People would be more than willing to believe you were up to something. And then all of Lady Highbarrow's efforts would have been for nothing. I could never have explained that to Her Grace."
Maddie tugged her arm free. "Don't worry, Mama. I know what I'm doing. I shan't embarrass you again."
She turned away and caught sight of Rafael. He was in his dress uniform again, splendid and dangerous and handsome. Even the scar across the left side of his face only served to make him look more rakish. He leaned against the wall, a glass of port in each hand, and looked at her. After a long moment, he straightened and made his way around the edge of the ballroom to deliver one of the glasses to his brother.
Maddie took a breath. Rafe, at least, knew something had happened between her and Quin. All she could do was pray that no one else did, and that Quin's optimism about tomorrow would be true. For both their sakes.
## Chapter 17
Quin rose early. A year ago—hell, six months ago—he would never have imagined a day like this. And he certainly would never have been looking forward to it. Lately his well-buried adventurous spirit seemed to have emerged, and he knew exactly whom he could thank for it. In fact, he intended to thank her for it as frequently as possible.
Early as it was, Malcolm was up before him, lurking in the upstairs hallway. Today he used only one cane, and unless Quin was mistaken, he looked as though he wanted to wallop his nephew with it.
"Good morning, Uncle." Quin smiled, near enough to whistling that he could easily believe he'd veered off into madness.
"I thought the nobility only rose before noon when residing in the country."
"You sound like Maddie."
"Speaking of whom," Malcolm put in, allowing Quin to help him navigate the stairway, "have you forgotten about her?"
"Forgotten about Maddie? I could as easily forget to breathe."
"Ah. And that is why, I suppose, you've spent practically every waking moment over the past week with Eloise Stokesley."
Quin grinned at him. "Precisely."
Malcolm eyed him for a moment. "Care to explain that?"
"I do not. I'm going out for a bit. If you wish to go anywhere, have Claymore drive you."
"Quinlan."
He turned around in the doorway. "Yes, Uncle?"
"What about Maddie?"
"I'm working on it." Until everything had been settled, he intended to disclose as little and to as few people as possible. Even to Maddie's staunchest supporters.
Aristotle glared at him when he went out to the stable, as the damned horse had done since Rafe had left again. Quin had him saddled anyway and rode west to Bancroft House. And then step number one of his carefully laid plan went awry.
"What do you mean, His Grace went out early?" he demanded, frowning at Beeks. "I sent a note yesterday, asking for an audience this morning."
The butler nodded. "I delivered the note into his hand myself, my lord. As far as I know, he did read it."
Quin swore under his breath. "Did he say where he was going?"
"No, my lord. He did say, however, that it wouldn't take long, if you'd care to wait."
"Blast." Little as he liked the idea of sitting about, which seemed decidedly unheroic, it was the most logical choice. His Grace could be anywhere in London. "Oh, very well. Is the duchess in?"
"No, my lord. Today is her charitable works day."
Quin frowned. "Rafe?" he asked, though he doubted he and his brother would have much to say to one another.
"Out riding, my lord."
"Fine, fine. I'll be in the morning room."
"Ah, my lord?" Beeks said hesitantly.
"What is it?"
"Lady Stokesley is already waiting in the morning room."
Quin looked at the butler for a moment. "Waiting for my father, I presume?"
"That is what she said, my lord."
Narrowing his eyes, Quin gazed down the hallway. That was certainly interesting. "Thank you, Beeks."
The marquis strolled down the long hallway and paused outside the half-open morning room door. The proper thing to do would have been to speak to his father first, but he wouldn't put it past His Grace to have figured out why he had demanded an audience and fled in order to avoid it. And Lucifer knew he was looking forward to a little chat with Eloise, anyway.
He smiled darkly and pushed open the door. "Eloise, good morning! I never would have expected to see you out and about so early."
She jumped, quickly rising. "I could say the same thing about you, Quin. What brings you to Bancroft House?"
Quin waved his hand. "Nothing much. Have you had tea?"
"Well, yes, I—"
"Beeks," he called, leaning out the door again, "have some tea brought in, will you?"
"Right away, my lord."
Studying her face for any sign of what she might be up to this time, Quin took the seat next to her. The London Times sat on the end table, but the entire front section was missing, and after a moment he set it aside.
Franklin brought in the tea, and flinched as he caught sight of Eloise. She barely favored him with a glance, obviously not even remembering that she'd scalded the servant with hot tea only a few days before. Quin remembered, though—quite well. Just as he remembered Maddie, below stairs in the kitchen, patching Franklin up again. "Close the door, will you, Franklin?" he asked, as the footman departed.
"Yes, my lord."
As the door shut, Eloise looked at him curiously, then leaned forward to pour them each a cup of tea. "My, my Quin, the two of us, alone?"
"I'd meant to call on you," he said. "You've saved me a trip."
"You have me curious, my love. Please, tell me what is on your mind."
For a moment he sat back, watching as she sipped her tea, a perfect porcelain figurine of impeccable manners and dress. "Eloise, do you believe in love?"
"What?" she asked, lifting an eyebrow. "Is that what you wanted to speak to me about? Of course I do."
He nodded. "Good."
Eloise smiled. "Why is that good?"
"Because it means you'll understand why I'm breaking off our betrothal."
"What?" she gasped. The cup of tea fell from her fingers and spilled on the expensive Persian carpet.
"I cannot marry you," he explained calmly.
"Quin, you can't mean that. Not after all this time! We're to be married in a fortnight, for heaven's sake. The invitations have gone out, and the announcement is to be made in the London Times tomorrow!"
He shook his head ruefully. "I know. Very poor timing on my part, I suppose."
"'Poor timing?' Is that all you have to say about it?"
"Well, that's up to you." Quin let the threads of anger that had been pulling at him for the past few weeks begin to twine together. "If you'll get up and leave now, I'm willing to end it at that. If you'd like me to elaborate, believe me, I'll be more than happy to do so."
Eloise pushed to her feet in a flurry of blue silk. "It's her, isn't it? That little shrew!"
"No, it's not. Yes, I'm in love with her, but the—"
"It's your damned brother, then!" she shrieked. "I'll kill him for this."
Quin looked at her intently. "What does Rafael have to do with this?"
"Nothing!" she snapped, wild-eyed. "Why, then? Why?"
"The fact is, Eloise, I find you to be a conceited, two-faced, malignant liar, and I really don't want to marry you—regardless of whether anyone else is involved or not."
Her face went white. "How dare you speak to me that way?" she hissed. "If it wasn't for her, you would be marrying me."
Quin stood. "Don't think," he said, in a controlled, quiet voice, "that because I have been polite to this point, I am some sort of fool. For a long time—for too long, I see now—I was willing to go along with this nonsense because I felt it was my duty to do so."
"It still is your duty."
"I have watched you, though," he continued, as if she hadn't spoken. "I have seen you be unfailingly petty and cruel whenever the chance arose, and I have seen you belittle those you thought you could because of the privilege of your rank."
"What about the privilege of your rank? You can't marry her—she's nothing!"
"Eloise, this is about you and me. Leave Maddie out of it."
"My God, Quin. I can't believe...have you told your father?"
"Not yet. I will as soon as he returns."
She looked at him for a moment, then took a breath and bent to pick up her teacup and set it back on the tray. "Well, that was nice of you, to tell me first. No one else knows. Nothing is lost." Eloise glanced toward the window, then back at him again. "Listen to me, Quin. I care for you, and I understand. Maddie is the poor, orphaned lamb you've worked very hard to save, and—"
"We were discussing your character," he interrupted.
"—And now you can't let her go. But for God's sake, don't marry her! Make her your mistress. As long as you're discreet, I don't care. Just do something—anything—to get her out of your system, and come to your senses before it's too late!"
"I suggest you never speak of Maddie in that tone again, Eloise. Now get out, before I throw you out."
With great dignity, her hands shaking with suppressed fury, Lady Stokesley turned for the door. "Don't you understand?" she said, as she pulled it open. "Your father will disown you when he hears of this. You will have nothing. Nothing. And then I won't want you."
He looked into her eyes, fighting the sensation that if she'd had a knife, she would have put it into his back by now. "I will have her."
Eloise grabbed her shawl from the butler and stomped down the front steps. Outside her carriage she paused. "No, you won't have her," she vowed, and reentered the house.
The duchess always kept pen, parchment, and ink in the front room, and it only took a moment to scribble out the note. She slipped back out again, and handed the paper to one of her footmen. "Take this to Dunfrey House, and deliver it into Mr. Dunfrey's hand. At once, if you wish to remain in my father's employ."
"Yes, my lady." He doffed his hat and ran off.
The driver helped her up into the coach. She closed the door and sat back. "That should take care of that." She smiled as the coach rocked into motion.
Fifteen minutes later, as her coach passed Hyde Park—nearly deserted at this time of morning—the door wrenched open.
"Hello, cousin," Rafael Bancroft said with a smile. "Keep going," he barked at the driver, and slipped off his hunter to step inside. His damned horse continued to keep pace with the coach as he slammed the door shut.
"Get out of here," she snapped, kicking at him.
He sat beside her, pushing her body against the wall of the coach. Grabbing her hands, he wrenched her around to face him. "Who was that note for?" he asked, hatred in his light green eyes.
"I don't know what you're talking about. Let go of me and get out, or I'll make certain Quin knows what you've done!"
"I saw you hand that note off," he snarled, shaking her. "I've kept my silence, Eloise, to keep my brother. But he doesn't want you anymore, does he? So I can confess our little indiscretion any time I like."
"I'll tell everyone you raped me."
"And will you tell them the same thing about Patrick Oatley? Considering which part of his body your mouth was attached to, I'm not certain anyone would believe it."
"I don't know what you're talking about."
"You did six years ago, when I mentioned that I'd happened to see the two of you together." He smiled, his eyes glinting. "And then, as I recall, you pounced on me, too, to—what was that you said afterward? Oh, yes. To keep me quiet. You're a spirited lover, Eloise, I'll give you that. With lots of practice, I presume. But I'm not going to keep quiet any longer."
She tried to wrench free. "I never wanted you, you pig!"
Rafael grinned. "Liar." He yanked her up against him. "Now, what did that note say, and to whom did you send it? If you don't tell me, Eloise, I swear, I'll remove every stitch of your clothing and throw you out onto the street."
He meant it. Eloise could see it in his eyes. "I hate you."
"The feeling is mutual, believe me," he answered, in the same tone. "Talk."
She stared at him, her mind racing. Dunfrey should have read the note by now, and if he had any sense, would have moved to act on it. If she could delay Rafael a few more minutes, it would be too late. "It was merely a business proposal," she spat out, fighting against his hard grip.
Shifting so that he held both of her hands pinned beneath one arm, Rafael leaned down and grabbed her leg. Eloise shrieked as he pulled one of her shoes off. He tossed it out the curtained window. "A proposal to whom?" he asked coolly.
"To Charles Dunfrey. Now, leave me alone!"
"Dunfrey?" he repeated, scowling. "What did it say?"
She snapped her jaw shut defiantly, until her other shoe followed the first. "You wouldn't dare."
"Wouldn't I?" Pushing her suddenly forward, he ripped open the back of her expensive dress all the way to her waist. "What did it say, damn it?"
Eloise stifled a furious, frightened sob. "I'll kill you for this!"
"You may try." With another wrench her dress came off completely, and he wadded it around one arm. "You're running out of wardrobe, dear."
"It said.... "She took a quick breath, trying to decide just how much more defiance he would stand for. As he started to stuff the dress out the window, she shrieked, "It said that I would give him five thousand pounds if he would make Maddie Willits disappear! Now go away, you snake!"
He shoved her away. "You cold-blooded bitch," he growled. "She did nothing to you."
"She took Quin!"
"You lost him. Six years ago, when you decided you could shut me up about Oatley by climbing into my bed."
"I did no—"
"Damn it, Eloise, why do you think I took an early leave? If you'd shown the slightest bit of genuine feeling for Quin, I—"
She lunged at him, her nails bared, but he was apparently expecting it, and he shoved her away again. Rafael looked at her coldly for another minute, then jerked her legs out from under her, sending her to the floor of the coach. While she struggled with him to get upright, her shift ripped off in his hands, leaving her in only her stockings.
"You had no right to hurt Maddie," he snarled.
"Then perhaps you should have confessed your sins to your brother before now."
"Bitch." He stood, looking her up and down while she flushed furiously and belatedly tried to cover herself. "Don't bother—I've seen it."
Abruptly mortified that he would throw her out into the street naked, she let her hands drop. "Wouldn't you like to see it again?" she suggested, swiping her disheveled hair out of her face.
He laughed, though his eyes glinted. "This is one snake who's not going near that hole again, dearest."
Shoving the coach door open, he whistled. A moment later, the horse appeared. Dropping her clothes out onto the street, he stepped into the stirrup and swung back into the saddle. "Good-bye, Eloise," he said jauntily, and wrenched the bay around. "And thank you for a lovely time. Again."
Eloise gasped and lurched forward to grab the door shut, but not before several very curious passersby glimpsed her inside.
"My lady?" the driver called, slowing.
"Take me home!" she screamed. "Now!"
Maddie looked at Everett in disbelief. "Who wants to see me?" she asked, setting her napkin down on the breakfast table, nervous flutters running through her stomach.
"The Duke of Highbarrow, Miss Willits."
Her father, the only other member of the family who'd already risen this morning, pushed away from the table. "Well, don't keep him waiting. Let's go."
"My lord, His Grace requested to see Miss Willits. Alone," the butler stated, and cleared his throat.
"Oh," Viscount Halverston said, and retook his seat. "Go, Maddie. For God's sake."
With a deep breath, Maddie went to find the Duke of Highbarrow in her morning room.
"Shoddy," he noted, turning around.
"Thank you, Your Grace," she answered, grateful he'd begun the conversation—if that's what this was—with an insult.
"How much will it take to convince you to leave London?" he asked, standing by the window and looking at her.
"I believe we've had this conversation before. I will not be bribed."
"What about ten thousand pounds? Is that enough to tear you away from my son?"
She gaped at him. Ten thousand pounds could keep her independent, and in style, for the rest of her life. "Out of consideration for your son, Your Grace," she said stiffly, "I will not repeat this conversation. Now, will you kindly leave?"
"Insolent chit." He flung a folded copy of the London Times onto the table in front of her. "You won't get any more out of me."
She glanced down at the page as it slowly fell open—and felt the blood drain from her face. In bold letters half an inch high, a full-page advertisement announced the wedding of the Marquis of Warefield and Lady Stokesley, to be held on Saturday, July the seventeenth. It named the illustrious parents of the illustrious pair, and the time and location of the ceremony. Numbly she noted that the duke had managed to secure Westminster Abbey, after all.
"That," the duke said, jabbing a finger at the paper, "is my son's future. You aren't fit to stand in his shadow, and your continued presence will be nothing but poison to him and to the entire Bancroft family. You are a ruined, inconsequential nothing, and while I might admire your courage at reaching so far beyond your grasp, Quinlan is, after all, beyond your grasp."
He stared at her while she continued to look helplessly at the bold, black words on the page. All she could think was that she couldn't have him. His Grace was right. Quin belonged to someone else, and if he tried to change that now, the scandal would be a hundred times worse than what Spenser had done to her. Slowly she sat down on the couch, her legs wobbly and numb.
"Listen, girl," he said in a quieter voice. "All you need to do is call on Bancroft House—at the servants' door. If I see you with packed bags, I'll give you ten thousand pounds, in currency. My offer stands until sunset. After that, you get nothing. Is that clear?"
Maddie didn't answer. After a moment, he stalked out the door. She heard his carriage creak into motion, but she couldn't look away from the announcement. She must have really believed Quin when he'd told her he loved her and that he meant to marry her. She must have believed it, or she wouldn't be feeling as though her heart had been ripped from her breast.
But it didn't matter. Nothing would happen now, except that the duke would have his way, after all. Slowly she stood. She couldn't go to Langley any longer, because with his pride pricked again, Quin might look for her. That much was obvious. Anywhere else in the world would do, so long as she never had to see him again.
From past experience she knew she didn't need much, and at least she had her savings from her employment with Mr. Bancroft this time. Silently she slipped upstairs, hardly noticing the tears wetting her cheeks, and threw a few things into her old, patched valise. She stopped at her dressing table, and quickly wrote out a note to anyone who should care to read it. And for Quin, if he should come to Willits House looking for her.
Hurrying back downstairs before her family could appear, she left the note on top of the newspaper, and her bag just inside the morning room door. "Everett," she said, stepping into the hallway.
"Miss Willits?"
"Would you mind terribly if I asked you to look for my riding gloves? I think I left them in the drawing room yesterday."
The butler smiled and nodded. "My pleasure, Miss Willits." He headed upstairs.
Maddie grabbed her valise and silently slipped out the front door. The nearest stage stop was only a few blocks away, and she set off out down the street at a fast pace.
"Maddie? It is you. I was just coming to see you."
She jumped. A coach slowed beside her, and Charles leaned out the half-open door. "I'm sorry, Charles, I'm in something of a hurry," she blurted.
"Randolph, stop," he called to his driver, and hopped to the ground. "Is something wrong?"
"It's...a long story, Charles. But no, nothing is wrong. I'm simply on my way to visit someone."
"On foot?"
"I need the fresh air."
She started off again, but he put a hand on her arm to stop her. "Maddie," he said quietly, stepping in front of her. "After what I did to you before, I shouldn't have expected you to agree to marry me." He tilted her chin up with his gloved fingers, his brown eyes holding hers. "But I am yours to command. I owe you at least that. May I take you somewhere?"
She looked up and down the street. People were beginning to stir from their homes, and with every moment the chance that someone would see her, and remember where she'd gone, grew. "Will you take me to the stage?" she asked quickly, before she could change her mind.
"I'll do better than that," he answered, taking her valise and motioning her into the coach. "Where do you want to go?"
With a quick breath she stepped up into the coach. "Anywhere. Dover."
Charles smiled and knocked his cane against the roof. "That's easy enough," he said, as the coach rocked into motion.
"Did you see that?" Polly said, turning away from her sister's bedchamber window.
"See what?" Claire asked sleepily, sitting up in bed and stretching.
"Maddie. She got into a coach and drove off."
"Don't be silly, Polly. She wouldn't do that without telling anyone. Not after the last time."
"But she did. I saw her. I think it was Mr. Dunfrey."
Claire smiled wisely. Polly was such a child sometimes. "Maddie would never get into a carriage with Charles Dunfrey. Never ever."
"Well, you look, then."
Scowling, Claire stood and, pulling on her dressing gown, made her way over to the window. "I don't see—" She stood on her tiptoes. "Oh, it's Rafael Bancroft." She breathed, watching as the gentleman swung down from a magnificent bay horse and ran toward their front door. "Hurry up and help me get dressed."
"What for?"
"Because I want to say good morning to him."
"Do you like him?" Polly asked.
"You're such a baby," Claire chastised. "Everyone likes Rafael Bancroft. He's handsome. And he's a Bancroft."
They could hear him downstairs, talking rather sharply to Papa, so Claire had to settle for combing out her hair and putting on her good slippers before she and Polly hurried down into the breakfast room.
"What do you mean, my father was here?" Rafael snapped, then turned as he saw them enter. "Ladies," he acknowledged, and turned back to the viscount.
"What's going on, Papa?"
"Not now, Clake. Go back to your rooms and get dressed, for heaven's sake."
"But Polly saw Maddie leave," she said, not wanting to miss anything.
"You did?" Rafael asked, turning quickly to Polly. "Where did she go?"
"She didn't go anywhere," Lord Halverston insisted. "She's in the morning room with the Duke of Highbarrow."
Scowling, Rafael turned on his heel and strode out into the hallway. The morning room door was open, and he went inside without asking. At the end table he stopped and picked up a scrap of paper and a section of the morning paper. A moment later he threw them down again, cursing.
"Miss Polly?" he said urgently. "Did you see where Maddie went?"
"She got into a coach," Polly repeated. "I think it was Charles Dunfrey's."
"I told you, that's ridiculous," Claire repeated. "She told him she didn't want to get married. She wouldn't go anywhere with him."
"Which way were they headed?" Rafael pursued.
Polly pointed. "That way."
"North. Gretna, no doubt." He leaned down and kissed her swiftly on the cheek. "My thanks, my lady," he said, and ran past them and back out the door.
Claire glared at her sister. "You should have let me tell him," she snapped. "That was my kiss."
"Oh, be quiet, girl. What in God's name is going on?" their father grumbled. He looked at her again. "And go get dressed!"
"I was about to go looking for you," Quin said, as his father stepped into the hallway.
The duke glanced at him for a brief moment, then turned to walk toward the stairs. "What do you want?"
"I sent over a note yesterday, remember? I wanted to see you this morning."
"Had something to take care of."
Quin followed him upstairs to his private office, uneasiness pulling at him. "Do you have a moment now?"
"Not really."
The marquis shut the door and leaned back against it. "This will only take a minute."
His Grace turned around to face him. "Don't bother. Do you think I don't know what kind of nonsense you're planning?"
"I hardly consider it nonsense," Quin said, immediately on the defensive, and still trying to maintain a reasonable tone.
"Quinlan Ulysses Bancroft," his father said, in an unexpectedly quiet voice. "You will be the twelfth Duke of Highbarrow. Twelve generations, Quin. Don't you think any of our ancestors ever fancied an unacceptable person? Do you think they married them?"
"I don't give a damn, Father," Quin said shortly. "This generation is in love with Maddie Willits. And I will marry her, if she'll have me."
"Hm. And do you know what that would look like? You'd be an embarrassment to the entire family."
Quin folded his arms. "What do you think it looks like for you and Malcolm to be practically spitting at one another in public?"
"That's none of your affair."
He nodded. "And this is none of yours."
"I made an agreement with the Earl of Stafford."
"I didn't." Further argument would likely result only in higher volume, so Quin straightened and turned away. "I just wanted you to know my intentions."
"It doesn't matter, anyway."
Quin stopped and turned around. "What do you mean, it doesn't matter?"
"I've taken care of it."
Sudden alarm tightened the muscles across Quin's back. There had to be a very good reason why his father was so calm about all this. "Just where did you go this morning, Your Grace?"
"You're going to marry Eloise Stokesley. It's settled." The duke sat at his desk and pulled out a stack of ledgers, his usual method of signaling dismissal.
Quin stared at his back. "Sweet Lucifer," he hissed, turning already to grab the door handle and yank it open. "If you've done anything—anything—to hurt Maddie, I'll—"
"You'll what, Quinlan?" the duke asked, not bothering to look up.
"I'll show you what a spectacle I can make, Father. In spades."
"Quinlan! Don't you dare go after that mopsie!"
Not bothering to respond, Quin strode out the door and down the stairs. The duke had been to see Maddie—he'd wager good blunt on it. And mercurial as her temper was, there was no telling what she might have done in response.
Outside, he stopped, looking around. "Wedders, where is my horse?" he snapped at the groom.
"Begging your lordship's pardon," the groom said, backing away, "but Master Rafe has 'im."
"What?"
"Aye, my lord. He took off with old Aristotle right after Lady Stokesley left, my lord. Looked madder 'n piss, if you'll forgive the expression."
"Damnation," he snarled. "Of all the stupid, poorly timed...Saddle me another horse. Now!"
"Aye, my lord."
When he burst into the Willitses' front room some moments later, Claire was standing in there, wearing her dressing gown and crying, and there was no sign of Maddie. Lord Halverston sat on the couch, a newspaper in one hand, and shook his head.
"She's ruined it," Claire sobbed. "She's ruined it again! Papa, it's not fair!"
The viscount stood as Quin, immediately sensing that they were discussing Maddie, strode into the room. "Shut up, Claire. Good morning, Lord Warefield. And congratulations."
Quin frowned. "Congratulations for what?"
Maddie's father handed him the front section of the London Times. "For this, of course, though in truth we did already know. I think everyone does."
Quin snatched the paper out of his hand and perused it quickly. "Blast," he swore. "Damn, damn, damn." His father had seen fit to forestall any argument simply by placing the announcement in the paper a day early. And of course Maddie had seen it. "Where's Maddie?" he demanded, ripping the paper in half and throwing it to the floor.
"I...There is some question about that, my lord." The viscount produced a smaller piece of paper.
Quin glanced down at it. "'Don't look for me,'" he read aloud. He looked up at Halverston again. "What in God's name is going on?"
"I wish I knew, my lord. Today has been completely...nonsensical. First His Grace your father very kindly stops by, and then your brother, and now we can't find Maddie, and who knows where—"
"My brother stopped by?" Quin repeated very slowly.
"Just for a moment," the viscount clarified. "And I beg your pardon, my lord, but he wasn't very polite. Demanded to know where my daughter was, and then off he went, without even a 'Good day.'"
"And now you can't find Maddie," Quin said quietly, something very black and angry stirring in his chest.
"Well, she left first, in a coach," Claire said, wiping at her eyes. "At least we think so. Rafael was on a horse."
He nodded. "Yes. He was on my horse. Do you by any chance know where they might have been heading?"
"I don't know," she mumbled. "He said something about Gretna."
"Gretna Green, perhaps?" Quin asked calmly, fury tearing through him. Rafe was a dead man. His brother's message was clear: he'd taken Aristotle, and he'd taken Maddie. Obviously he didn't intend on coming back. He'd practically threatened to elope with her once already.
"Perhaps. He really didn't say."
Quin turned for the door. "So that's how he wants it," he growled. "All right, Rafe. Let's play."
## Chapter 18
Maddie gazed out the window of Charles Dunfrey's dilapidated coach. Green meadows, stands of trees, and a scattered cottage now and again swept into view and then away again beneath the overcast sky. They were finally out of London, and she tried to relax a little.
She would never see Quin again. He would do his duty and marry Eloise Stokesley, and they would have children, and she would read something about him now and again in a newspaper, and that would be all.
The ache that had begun with the duke's announcement this morning deepened into a hole so black she knew she would never laugh or smile again. Quin would say she was a coward, and she probably was. But finally and ultimately, she hoped he would realize that she'd done it because she loved him—so very much that she would let Eloise marry him, and so much that she wouldn't be able to stand seeing him in the company of his new bride. Ever. Leaving London wasn't a choice, but a necessity.
Maddie shook herself. "Shouldn't we be able to see the coast by now?" she asked, glancing at Charles.
He'd been quiet for the more than two hours they'd been traveling, and he stirred as though he'd been daydreaming. "Soon, I'm sure," he said.
"This is really very kind of you," she continued, hoping that talking to him would at least keep her thoughts away from Quin for more than a heartbeat. "I'm sure you must have had other plans for today."
"It's my pleasure, Maddie. Do you mind my asking what happened? You do have a valise with you, after all."
"My aunt is suddenly ill," she improvised. If she could help it, there would be no scandal this time. "In Spain. I need to go tend her, right away."
"Your father's sister?"
She shook her head. "No, my mother's."
The sun broke through the cloud covering, its light shining in her eyes, and she glanced out the window again and frowned.
Just as quickly she wiped the expression from her face. Her heart began to beat at twice its normal rate, and she took a deep breath, trying to calm herself before Charles noticed her discomfiture. As far as she knew, his offer of assistance was completely legitimate. It was merely that the sun wasn't quite where it was supposed to be at this time of morning. "Isn't that odd?" she said, as casually as she could. "I thought the sun would be in front of us."
Charles nodded, yawning. "We're heading a little north, now. In a few minutes we'll turn east again."
"Of course," she agreed, growing more suspicious and uneasy by the moment. "But isn't Dover actually a little south of London?"
He chuckled. "Maddie, I have never doubted your wit and wisdom. Remind me, though, never to have you read a map for me."
Maddie smiled stiffly. "Actually, cartography is something of a hobby of mine."
He looked at her, his gaze sharpening a little. "Is that so?"
"Yes, as a—"
"Hold up there!"
Maddie jumped at the stentorian bellow coming from behind them. "Rafael?" Suddenly she was thankful her escape hadn't gone quite as smoothly as she'd envisioned.
"I say, hold up there, coachman!" came from much closer.
She furrowed her brow and gave Charles her best look of bewildered confusion. "What in the world could Rafael Bancroft want with us?"
Charles leaned forward and rapped his cane against the roof. "I'm sure I have no idea. Keep going, Randolph! We're in a hurry!"
The coach immediately accelerated, rocking precariously on the rutted road.
"I'm not asking again! Stop!"
"Mr. Dunfrey?" The coachman's voice sounded extremely nervous. "Sir, he has a pistol."
"A pistol?" Maddie gasped. Apparently, her hunch had been correct after all, though the realization was not very comforting. "Stop the coach. Something is terribly wrong, I'm sure."
"I thought you needed to be with your aunt as soon as possible," Charles commented, sitting back again. "Ignore him."
"But it must be important!" she insisted, trying to decide if she actually wanted to risk leaping from the carriage while the horses were at a full gallop.
Charles eyed her, clearly annoyed, but she couldn't read anything more than that in his expression. "Oh, very well. Randolph, stop!"
The coach lurched to a halt, nearly sending Maddie to the floor. She grabbed onto the window frame and hauled herself back onto the seat.
"Maddie!"
Charles pulled the curved handle free from his cane, revealing a very sharp-looking rapier. "Ask him to come in," he said, reaching out to rest the tip of the blade against her throat.
Torn between fury and fright, Maddie clenched her fists and scooted as far back in her seat as she could. The blade followed her. "Rafe? Come in, if you please."
The door opened, followed by a very deadly-looking pistol, and a winded, angry-looking Rafael. "Maddie, come out of—" Rafael began, then swore as he saw the sword. "Sweet Lucifer, Dunfrey. Put it down."
"I believe that's my line," Dunfrey said. "Turn it around slowly, and hand it to me. Randolph! Is he alone?"
"Aye, Mr. Dunfrey."
"Are you all right, Maddie?" Rafe asked, the pistol and his eyes still unwaveringly on Charles.
"Yes, I'm fine, for the moment. What are you doing here?"
"I came to tell you that you're being kidnaped."
"Hmm," Dunfrey sighed, far too calmly. "Dear Lady Stokesley, I suppose?"
"Yes."
"Well, now you're being kidnaped as well, Bancroft. Give me the pistol and sit down."
Still gazing at Dunfrey, Rafe backed off just a little. He whistled sharply.
"What in damnation was that for?" Dunfrey snapped.
Rafael continued to look at him coolly. "Just talking to my horse."
"Well, stop it, and get in here. Slowly."
With a scowl, Rafe turned the weapon and handed it, butt first, to Charles. "Watch out there," he said, taking the seat beside Maddie. "It's loaded."
"I should hope so. Randolph, go!"
The coach rattled to a start again. Maddie glanced sideways at Rafael, but his attention remained on Charles. "How did you know where I was?" she asked him quietly.
"Convinced Eloise to let me in on the secret," he muttered back, smiling at their captor. "She can be very cooperative, given the correct incentive."
"What does Eloise know about this?"
"Apparently she's been working from the beginning to keep you and Quin apart. I suppose—"
"That's enough of that," Charles interrupted. "You know, Bancroft, you've made this whole thing quite a bit stickier. You won't vanish nearly as easily as Madeleine, I'm afraid."
"No one's vanishing, Dunfrey. Except you, when they send you to the gallows."
"Charles," Maddie put in, trying to keep Dunfrey distracted so that he wouldn't shoot one or the other of them, "why would you do this? You have nothing to gain."
"Except five thousand quid." Rafe folded his arms over his chest and closed his eyes.
Maddie wished she could look so calm. "I don't have five thousand quid."
"Eloise does," Rafe said, already looking half asleep.
Charles smiled and, one-handed, slid the rapier back into the cane. "Actually, it's more than that. By my reckoning, Lord Halverston should be willing to part with at least that much again to get you married respectably."
Abruptly a great many things began to make sense. "You only wanted to marry me for the money?" she asked, anger beginning to edge out her fright. She'd suspected, of course, but this was too much.
"It's the way of the world, Maddie. And be grateful for it. If not for the extra blunt from your father, I would likely throw both you and Bancroft down a well. Now I've only one to worry about."
"I wouldn't count on that," Rafe murmured, so quietly Maddie could barely hear him.
She turned back to the window so Charles wouldn't see the sudden anticipation in her eyes. And she hoped desperately that she would see Quin at least one more time.
When Lady Highbarrow returned home for afternoon tea, a note awaited her. Beeks, looking even more stoic than usual, bowed as he handed it to her.
"From Lord Warefield, Your Grace. A messenger delivered it several hours ago."
"Thank you." Now her sons were reduced to conversing with her via messenger. At the butler's continued dour expression, she paused. "Is something wrong, Beeks?"
"I couldn't say, my lady."
"I see." Curious, she headed up to her private room, where a fresh pot of tea awaited her. Pouring herself a cup, she unfolded the missive—and rose so quickly, she tipped the entire tea tray onto the floor. "Lewis!"
The duke appeared a moment later. From his expression, her uncharacteristic shout had completely unsettled him.
"What is it, Victoria?"
"What did you do?" she demanded, stalking up to him, the note clenched in one hand.
He assumed his normal stubborn, imperturbable expression. "I set things to rights."
"Oh, really? Then tell me what you make of this." She unfolded the note again and read it.
Rafe and Maddie on their way to Gretna Green. Am following.
Q.
The duchess looked up at her husband. "So I repeat, Lewis, what did you do?"
"That damned fool!" the duke exploded. "Both of them! We'll be the laughingstock of London. Two Bancrofts chasing after a whore!"
"What concerns me, husband," Victoria said, in a quiet and controlled voice, "is what will happen when Quin catches up to them. 'Set everything to rights,' indeed. They'll kill one another."
The Duke of Highbarrow stared at her for a moment, the color slowly draining from his stern face. "Good God," he hissed, and turned on his heel. "Damned, damned fools."
If Maddie had left in a carriage, and Rafael on Aristotle, then logically they intended to meet somewhere along the way. If Quin had been thinking clearly, he would have asked Claire whether the coach had any identifying markings, but he hadn't seemed to be able to do much but ride at top speed along the north road and curse his brother in half a dozen languages. It was easier to focus on Rafael, who had taken her away, rather than to admit that Maddie had left him.
He had always prided himself on being reasonable and fair in his dealings, on being in control of his emotions, and on honoring the responsibilities of his title. As he traveled the busy road, dodging hay wagons and shepherds and stopping every closed carriage he passed, he didn't give a damn about any of that—or about the ruckus he was causing. Rafael had taken Maddie away, and Rafael would give her back.
It was past noon when he came upon the first clue. Just off the road, a group of young boys surrounded a horse and unsuccessfully tried to grab hold of its dragging reins. Quin looked at the animal more closely, and then sharply pulled up his own mount.
Aristotle dodged nimbly around his would-be captors, at the same time staying within the same small clearing rather than running off, where they would have had no chance of catching him. Quin kneed his gelding toward the group, stopping at the fringe of the trees.
"Aristotle," he called, though the horse had never listened to him before.
To his surprise, the gelding whinnied and trotted up to him. Quin leaned down and picked up the reins.
"He wouldn't leave you behind," he told the horse, his mind racing in a hundred different directions. "Not now. He ordered you to stay here, didn't he? Why would he do that?"
"Hey, milord, that horse yours?" one of the boys called.
"My brother's," he answered. "Have you seen him?"
"That beastie's been here for over an hour. Never seen nobody."
Angry as Quin was, the presence of Aristotle actually made him stop and think for a moment. And when he did, the idea of Rafe and Maddie running off together and leaving Aristotle behind to mark their trail made absolutely no sense at all. He looped the bay's reins around the cantle of his saddle and turned north again. Whatever was going on, he was bloody well going to find out what it was.
"You're going to kill that fine pair of horses of yours, if you insist on running them like this," Rafael noted calmly.
"Shut up," Charles snapped.
He'd become increasingly short-tempered all afternoon, and as satisfying as tormenting their captor was, Maddie wished Rafael would let up on him a little. Her own temper was becoming very fragile, and her bottom and legs were cramped from sitting in the ill-sprung coach all day.
"If you're going to kill me," Rafe began again amiably, "you might as well tell me where we're going."
"Rafe," she whispered, looking sideways at him, "do quit reminding him about that."
"No, he's quite right, my dear," Charles countered. "You'll figure it out eventually, anyway. We are going to Gretna Green, so that Maddie and I can be married."
She stared at him. "I am not going to marry you, in Scotland or anywhere else. So you may as well stop the coach right now, and let us—"
"Maddie, Maddie, Maddie," he chastised, shaking his head. "Please understand. I receive five thousand quid for taking you out of London. An additional sum will be mine when you are mine. If you make that idea too unpleasant, I will settle for the initial payment, and I'll bury you in the same hole as Bancroft here."
"Kidnaping is one thing," Maddie said, trying to keep her voice steady. "Murder is quite another. I hope you realize that. You're setting a price on the worth of your own life, as well as ours."
"Thank you for your unasked for bits of wisdom, my dear, but allow me some credit." As he had been for the past twenty minutes, he glanced toward the window, pushing the curtains aside with his free hand. "Randolph!" he called in a louder voice. "The eastern road, if you please."
"Aye, Mr. Dunfrey. I see it."
Dunfrey sat back again, the pistol still aimed at Maddie. She supposed that was to discourage Rafe from attempting any sort of rescue or escape, but she wished Charles would stop looking at her as though the idea of shooting her didn't trouble him in the least.
"Once you turned down my proposal, Maddie, I planned this little contingency. Of course, I didn't expect you to run out your front door and into my carriage with your bag all packed, but you have to admit, it did make things a bit easier on me."
The coach lurched as the road beneath them became steadily more rutted. Finally they rocked to a halt, and Randolph jumped down from his perch to pull the door open. Dunfrey gestured with the pistol. "Please follow my coachman, Bancroft. Maddie, you're to stay right behind him."
With a last, angry glance at their captor, Rafe jumped to the ground. Maddie followed, her long skirt catching on the carriage steps and nearly tripping her. The sun was already behind the tall elms to the west, and in both directions the muddy, rutted road was empty of other travelers. Directly before them stood a small inn, a single lantern hanging above a bench by the dark, scarred door.
They seemed to have the inn completely to themselves. The coachman led them into the deserted common room, which at least had a fire going in the stone fireplace. Obviously someone had lit the fire, thank goodness, and Maddie looked about for a friendly innkeeper—or at least, one who could be bribed.
The man who walked in through the kitchen door, though, with a tray of bread and fruit in his arms, didn't look the least bit friendly. He also looked extremely familiar. Maddie blanched, stopping in her tracks, and Dunfrey ran into her from behind, the muzzle of the pistol bruising her spine.
"Ouch! That hurt."
"Sit down," he grumbled.
"But—"
"Sit down in the chair there, Maddie, before I find a more accommodating position for you," Charles said in a darker voice, and pushed her toward the chair set before the fireplace.
Maddie did as he said, her eyes on the tall gentleman setting the food down on the table. He turned to face her and smiled.
"Good evening, Maddie. Haven't seen you for a while. You look more lovely than ever."
"Spenser, that's right, you know Maddie," Charles said more amiably, sitting at the long wooden table, "and this, unfortunately, is Rafael Bancroft. Don't worry, we'll kill him before we move on."
Benjamin Spenser eyed Rafael as the coachman dragged another chair over beside Maddie's. "Bancroft, as in the Duke of Highbarrow's kin?"
"Pleased to meet you," Rafael said, and held out his hand. "You're Benjamin Spenser, I presume? The ass who ruined Maddie?"
"Sit down," he ordered, picking up a coil of rope from the bench. "I've no objection to killing anyone, Dunfrey, but you think splitting a thousand quid is worth the risk of murdering a Bancroft?"
Charles glanced up at him. "It's twice that now."
Rafael snorted. Dunfrey rose and hit him hard across the face with the pistol. Rafael grunted and fell backward into the chair. Charles leaned over him. "I'd kill you for nothing, Bancroft."
"Charles, stop it!" Maddie protested, shooting to her feet. He shoved her back down into the chair.
She looked from her former betrothed to the man who had ruined her. Now that she saw them together, and now that she'd realized how highly Charles valued her dowry, quite a few things made sense.
"Why so sour-faced, Maddie?" Charles cajoled, while Randolph and Spenser tied Rafe to the chair before he could regain his senses.
"You never cared for me at all, did you?" she said quietly, unable to keep the bitterness from her voice. "All you wanted was money for marrying me. As much of it as you could get."
"Why else would I want to marry you?" Dunfrey asked, finally dropping the pistol in one of his coat pockets. "Though I have to admit, if I'd known what you'd end up looking like, I might have been willing to settle for slightly less currency."
"It's a bit late to try flattering me, you ape," she retorted.
Spenser moved behind her with another stout section of rope, and Maddie tensed again. Balling her fist, she surged to her feet and slugged Charles Dunfrey in the chin as hard as she could.
Not expecting the blow, Dunfrey rocked backward and lost his balance. He gripped the edge of the table, blinking. Attempting to take advantage of his momentary surprise, Maddie crashed into him, and they both fell to the floor.
She grabbed for his pocket, trying to recapture the pistol, but he threw her off. She landed hard on her back, the breath knocked out of her. With a curse Dunfrey pounced on her, pinning her by the shoulders with his hands and the weight of his body on top of her.
"This gives me an idea," he snarled, blood welling from a cut lip. Shoving his knee between her legs, he leaned down and kissed her wetly.
"Dunfrey!" Rafael roared, pulling against the ropes that bound him securely to the chair. The coachman gagged him with a rag.
"I warned you not to push me, Maddie," Charles continued. Laying his body harder against her, he kissed her again.
It was foul, wet, and disgusting. And, even worse, she could feel his growing arousal between their bodies. "Get off me," she demanded frantically.
Spenser knelt at her head and grabbed her flailing hands. He grinned down at her. "Share and share alike, I always say," he leered, pinning her arms above her shoulders.
The last of her anger slid into pure fear as Charles, his hands free now, ripped at the front of her dress. "Future husbands first," he said, licking her neck.
The door burst open. "That would be me," Quin snarled, white-faced and disheveled.
"Quin!" Maddie sobbed, relieved.
Quin leaped at Dunfrey. Twisting, Maddie grabbed Spenser's ankle as he scrambled to his feet, sending him sprawling. Dunfrey toppled off of her as Quin plowed into him with a furious growl.
The coachman standing behind Rafe looked as though he didn't know what to do, so Maddie yanked off one of her shoes and hurled it at him. It struck him in the shoulder and he jumped, then broke and ran for the door.
She tried to grab Spenser again, but he regained his feet and dived into the fight. Realizing she wouldn't be of much assistance to Quin against the two big men, Maddie scrambled over to Rafe to untie him. One wrist was already bloody, and the knots were slick and tight. "Stop pulling, or I'll never get you loose," she snapped, and he relaxed his arms a little.
Finally she had him free. He yanked the gag off and slammed into Spenser, knocking him away from Quin and Dunfrey.
Trying to recover her breath, Maddie staggered to her feet. As she watched in horror, Dunfrey scrambled away from Quin and dug into his coat pocket for the pistol. Her frantic gaze lit on the discarded cane, and she snatched it up.
Dunfrey stood and leveled the pistol at Quin. With a shriek, Maddie pulled the rapier free and stabbed it into Dunfrey's back. "No!"
Charles swung around and hit her in the face with the pistol, knocking her hard to the floor. Blurrily, she saw Quin grab the weapon and shove Dunfrey away from her. And then the Duke of Highbarrow, together with a dozen footmen, burst into the room, weapons drawn.
Maddie shut her eyes as the room spun drunkenly. Then someone knelt beside her and lifted her into his arms. "Maddie," Quin breathed, his voice shaking. "Maddie, can you hear me? Open your eyes."
She looked up into his beautiful jade gaze. Breathing a sigh of relief, he pulled her tightly against his chest. Maddie wrapped her arms around his neck, buried her face in his shoulder, and began to sob. "Quin," she said, over and over again. "Oh, Quin."
"Shh," he murmured into her hair. "It's all right. You're all right, Maddie."
"She's not hurt, is she?"
The muscles across Quin's back tensed, and she looked up to see Rafael squatting down beside them, though there seemed to be two or three of him. "No, I'm fine, Rafael. Really."
"Excuse us," Quin said brusquely, and lifted her in his arms. With a warning glance at the Duke of Highbarrow, who actually stepped aside, he carried her outside into the moonlit darkness. He sat on the bench beneath the lantern, and cradled her like a babe. "Why did you leave, Maddie?" he asked quietly. "You said you would wait for me."
She tried to focus her eyes on his lean face. "I've been enough trouble, Quin. Don't you understand, you and Eloise—"
"Eloise and I are nothing," he interrupted fiercely. "I have already told her my intentions. What about you and Rafael?"
She furrowed her brow. "What about us?"
"You were going with him to Gretna Green."
The muddiness in her head cleared a little. "No, we weren't. Charles was kidnaping me for my dowry. Rafe found out, and came to rescue me."
Quin glanced back toward the open doorway. "Some rescue," he said grudgingly. He looked back at her, holding her gaze for a long time as he stroked her cheek with gentle fingers. "Do you love me, Maddie?" he asked softly.
"Quin, I—"
He shook his head. "Do you love me?"
A tear ran down her cheek. "Of course I love you," she whispered.
Quin closed his eyes for just a moment. "Then marry me.
"I can't. I'm ruined. Twice now."
"At least," he smiled, and leaned down to kiss her softly on the lips. "Marry me."
"You came after me," she said, for the first time realizing exactly what had happened. "You came after me!" The dark, lonely knot in her chest finally broke apart and melted away.
"Of course I did. I love you."
"No, that's not what I mean," she argued, wishing the dull ache in her head would go away so she could speak coherently. "No one came after me the last time." She started crying again. "But you came."
He looked at her for a long time, his expression unreadable, then gathered her up and stood again. "That settles that, Miss Willits."
"Settles what?" she asked, twining her hands in his lapels and wondering that he could lift her so easily.
Quin walked back into the inn, Maddie still in his arms. "Your Grace," he said, and the duke turned from glowering at his captives to eye his son.
"What is it now, boy?"
"Maddie and I are continuing on to Gretna Green."
His Grace's face reddened. "You are not—"
"Do whatever the hell you want with your titles and your land and your heirs," Quin interrupted, and turned on his heel. "Tell mother we'll see her in London next week."
"Quin," Maddie said, "you've gone mad! Put me down!"
"Want company?" Rafe asked, hopping down from the table where he'd been perched.
"No."
"Do you realize what a scandal there'll be?" the duke bellowed, striding after them. "Your wedding to Eloise has already been announced. King George is going to attend!"
Quin stopped and turned around. "Father, I leave it to you to do what you will. In case I haven't made it clear, I don't care. I've been respectable my entire life, and I've discovered something about it."
"And what might that be?" the duke asked, his skeptical expression melting into concern as he realized his son wasn't bluffing.
"It's very dull. I'm tired of it." With a last glance at Rafael, Quin turned and headed them out the door again. "I'll see you in a week," he called over his shoulder.
For once, Maddie didn't know what to say. Quin Bancroft had always been a good-humored, reasonable man—but at the moment, she wouldn't have been surprised if he had decided to travel to Gretna Green on foot, carrying her the entire way. "Quin?" she said quietly.
"Hush. No more arguments. You're far too stubborn, so I'm simply not going to listen to you." He lifted her into Charles's coach. A moment later her valise followed her. A moment after that, Quin himself stepped in, and closed the door.
"Might I ask if anyone is going to drive us, or are we to sit here in the yard all night?" she ventured.
He sat beside her, tugging her close against him so she could rest her aching head on his shoulder. "I recruited Franklin. It seems he's rather fond of you."
"Yes, he's very nice," she agreed, closing her eyes as he wrapped a warm arm about her shoulders. "Remind me to check his bandages tomorrow."
"Yes, love." The coach jolted into motion, and Quin cursed. "You came all this way in this hell-sprung hack?"
"My bottom is sore," she confessed drowsily.
"I'll take care of it at the next town," he murmured into her hair.
"My bottom?"
Quin chuckled. "The coach. And your bottom, if I have anything to say about it."
With a supreme effort, Maddie managed to open one eye. "You can't marry me, Quin."
"I told you that I am not discussing that subject with you," he retorted. "Go to sleep."
"But I'll be an embarrassment," she protested. "You're going to be the Duke of Highbarrow, for heaven's sake."
"I wouldn't wager money on that," he returned, his voice amused. "Maddie, you are more precious to me than anything on this earth, including my title. If I have to be Quin Bancroft to marry you, then I will happily become him. We could raise pigs."
"You can't do that. Everything is planned out for you. You have everything."
"I want only you." He tilted her chin up and kissed her, a feather-light touch of his lips to hers. "Madeleine," he murmured. "Will you marry me?"
"I suppose I have to now," she answered, closing her eyes again, and unable to keep a smile from touching her lips. "This is the third time you've ruined me. Or the fourth. I can't remember."
For a long moment he was silent. "Are you certain it's not Rate you would rather have here?" he finally asked quietly.
"Rafe?" she asked, surprised, and lifted her head to look up at him. "Why Rafe?"
He shrugged, looking away from her out the window into the darkness. "You seem to get along well."
Maddie relaxed again, comforted by his jealous tone. "We get along too well. I could never argue with him. It would be very dull."
Quin made a sound in his chest that exploded into laughter. "You think," he managed finally, "that Rafe is dull?"
They purchased a new, considerably better-sprung carriage in Nottingham, and from there arrived in Gretna Green two days later. Quin kept a close eye on Maddie, worried that once her head cleared she would take to the hills and vanish, but when they entered the quaint little chapel and stood before the extremely surprised priest, she was still beside him.
And five days later, as the coach turned onto King Street in Mayfair, she was seated next to him, though she looked considerably less happy. "You saw the London Times yesterday, the same as I did," she said, eyeing him.
"Yes, and I told you not to look at it." Quin grinned and took her hand, stroking her fingers.
"Not looking at it doesn't change anything. And I know you have to care, at least a little bit. Highbarrow Heir Elopes to Scotland with Unknown Femme.' Really, Quin. I know you're proud, and I know you didn't like seeing it."
"I am happy," he stated, tugging her across the coach to sit on his lap. "No one can take you away from me now." He touched his lips to hers, thrilling in her quick, passionate response.
She sighed, running her fingers along the line of his jaw. "You may come to regret that. And very soon."
Quin narrowed his eyes. Grabbing her hands in his, he yanked her around to look at him. "Don't say that, Maddie. Ever. You are my life. Without you, I am incomplete. Do you understand?"
Maddie nodded, tears gathering in her eyes. "You say very nice things, you know."
"I mean them," he whispered, kissing her again.
It had been that way since their marriage. He couldn't seem to stop touching her and holding her and kissing her, as though, even with his ring on her finger, he wasn't certain she was real. And in bed, she matched him passion for passion. He'd never delighted in making love as much as he delighted in making love to Maddie. His wild wood sprite would not vanish again.
"Quin?"
"Yes, my sweet?"
"I love you."
Quin kissed each of her fingers, while she smiled happily at him. "I love you, too."
"My lord, my lady, Bancroft House," Franklin called.
"Oh, blast," she grumbled, scooting off his lap. "Are you certain we couldn't just sail off to Spain or something?"
The carriage rolled to a stop. "Actually, that idea had occurred to me," he confessed. "But I would at least like to say good-bye."
She wrinkled her nose, obviously attempting to put on a brave front. "You're just trying to make me look bad."
He chuckled as Franklin pulled open the door. "As if I could."
Seeks stood at the top of the stairs, waiting for them. "My Lord and Lady Warefield," he nodded. "Welcome."
"Thank you, Beeks."
Maddie smiled at him, and the butler took a step closer. "They've a gauntlet set up for you, I'm afraid," he said in a low voice.
She looked up at Quin. "I told you that we should have arrived a day early," she whispered. "But no, you still insist on being punctual."
"It's good manners. Who's here, Beeks?"
"The duke and duchess, Miss Maddie's parents, Mr. Bancroft, and Mr. Rafael. They've been in the drawing room since breakfast, waiting for you to arrive."
For the moment Quin refrained from pointing out that Maddie was no longer a miss, by any stretch of the imagination. Instead, he held out his hand to the butler. After a startled moment, Beeks smiled and shook it. "My thanks, Beeks."
"Good luck to both of you, my lord."
Quin looked down at Maddie, and took her hand in his. "Ready, my love?"
"May I bring a sword?"
A brief grin touched his lips. "No, you may not."
She sighed. "Then I suppose I'm ready."
The butler preceded them upstairs to the drawing room. He stepped inside the half open door. "Lord and Lady Warefield have arrived," he announced in a very stout voice.
Quin leaned forward. "If you ever lack for employment, Beeks," he murmured, "please see me."
"Yes, my lord."
With Maddie clutching his hand tightly, they stepped past the butler and into the drawing room. Even if he hadn't already known, it was easy to tell the enemies from the allies. Particularly when there seemed to be only one enemy. He'd rehearsed a hundred speeches, but none of them appealed to him as they all stood eyeing one another like dogs trying to decide whether to attack or join together in a pack.
Typically, it was Rafe who moved first.
"You look wonderful, Maddie." He grinned, and strolled forward to kiss her on the cheek. "Even with this old sod. Congratulations to both of you." To Quin's surprise, his younger brother kissed him on the cheek as well.
The Duke of Highbarrow sat by the window and hadn't even glanced in their direction. Rafe shrugged at Quin's look and captured Maddie's hand to drag her over to the room's other occupants. Quin glanced after her, but once Lady Halverston threw her arms around her daughter's neck and began sobbing joyfully, Maddie seemed to recover herself.
"Your Grace," he said quietly, walking over to stand by the window, "shall I begin moving my things out of Warefield?"
"So you did it?" die duke responded, glancing up at his son. "You went and married her?"
"Yes, I did."
"Eloise and her parents went home to Stafford Green. She hasn't been seen in public since you ruined everything. I'll be lucky if Stafford ever speaks to me again."
"I'm sorry for the inconvenience," Quin admitted, sitting in the deep sill. "But I'm not sorry for marrying the woman I love."
"I didn't expect you would apologize."
"So what's it to be, Father?"
Finally the duke turned to look at Maddie, who was standing beside the duchess and watching them. "Bah."
Malcolm released Maddie's hand and hobbled over to his older brother. "Lewis, don't be an ass," he said. "She's ten times the lady Eloise could ever be, and you know it."
"The marriage to Eloise was arranged."
"Because you wanted Stafford's vote in the House. I told you to let it go years ago."
"And I told you to go to hell."
"Well, Langley was as far as I got," Malcolm said calmly.
Quin looked from one to the other. "Just a moment. Do you mean that this not speaking to one another nonsense was because of Eloise and me?"
Malcolm shrugged, a slight grin touching his face. "More or less."
"Good God. You devious bastard," Quin stared at his uncle. "I should—"
"I didn't lie to you, Quin," Malcolm interrupted. "I didn't drink of you and Maddie until Lewis's letter."
A warm hand touched Quin's shoulder and slid down his arm to grasp his fingers. "You should have said something, Mr. Bancroft," Maddie chided. "I would have been sure to kill him, then."
His Grace finally stood. "So you think you've won, do you?" he snapped at Maddie.
She looked at him coolly, then released Quin's hand and stepped directly in front of the duke. Quin thought she meant to hit him, and he tensed, ready to leap forward and prevent bloodshed. Instead, though, she raised up on her tiptoes, put a hand on the duke's shoulder, and kissed him on the cheek.
"We began badly, Your Grace. If you wish to continue that way, believe me, I love a good argument. But I would like us to be friends, and I would like my children to know their grandfather."
Maddie was brilliant. She'd hit on Highbarrow's one weakness, and she knew it. Quin slid his arms about her waist and pulled her back against him. "All of your grandchildren," he seconded, kissing Maddie's ear, and fire burning through him at her nearness. "Scores of them."
The duke eyed her for a long moment, then nodded grudgingly. "There'd better be...Madeleine."
She nodded. "I shall make a gallant attempt, Your Grace."
"Hurrah!" Rafael shouted, and slapped Quin soundly on the back. "Justice triumphs. May I have my horse back, now, Warefield?"
Quin turned Maddie around and leaned down to kiss her softly. "I love you," he murmured.
"I love you, too," she whispered back, returning his kiss with one of her own. "Now, give Rafe his horse back."
"Anything you wish, my love." He grinned at her. "Stubborn chit."
## About the Author
SUZANNE ENOCH once dreamed of becoming a zoologist and writing books about her adventures in Africa. But those dreams were crushed after she viewed a National Geographic special on the world's most poisonous snakes—of which 99% seemed to be native to Africa. She decided to turn to the much less dangerous activity of writing fiction.
Now a New York Times and USA Today bestselling author of historical and contemporary romance, the most hazardous wildlife Suzanne encounters are dust bunnies under the sofa.
To see pictures of those dust bunnies, please visit www.suzanneenoch.com.
Visit www.AuthorTracker.com for exclusive information on your favorite HarperCollins author.
## [Avon Books by
Suzanne Enoch](contents.xhtml#adc_01)
Historical Titles
BEFORE THE SCANDAL
AFTER THE KISS • TWICE THE TEMPTATION
SINS OF A DUKE • SOMETHING SINFUL
AN INVITATION TO SIN • SIN AND SENSIBILITY
ENGLAND'S PERFECT HERO
LONDON'S PERFECT SCOUNDREL
THE RAKE • A MATTER OF SCANDAL
MEET ME AT MIDNIGHT • REFORMING A RAKE
TAMING RAFE • BY LOVE UNDONE
STOLEN KISSES • LADY ROGUE
Contemporary Titles
A TOUCH OF MINX • BILLIONAIRES PREFER BLONDES
DON'T LOOK DOWN • FLIRTING WITH DANGER
## Copyright
This book is a work of fiction. The characters, incidents, and dialogue are drawn from the author's imagination and are not to be construed as real. Any resemblance to actual events or persons, living or dead, is entirely coincidental.
BY LOVE UNDONE. Copyright © 1998 by Suzanne Enoch. All rights reserved under International and Pan-American Copyright Conventions. By payment of the required fees, you have been granted the non-exclusive, non-transferable right to access and read the text of this e-book on-screen. No part of this text may be reproduced, transmitted, down-loaded, decompiled, reverse engineered, or stored in or introduced into any information storage and retrieval system, in any form or by any means, whether electronic or mechanical, now known or hereinafter invented, without the express written permission of HarperCollins e-books.
EPub © Edition NOVEMBER 2008 ISBN: 9780061979354
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\section{Introduction}
Let $\mathscr{H}$ be the Ariki-Koike algebra over an commutative
integral domain $R$ with parameters $q, q^{-1}, Q_{1}, \ldots , Q_{r}
\in R$, associated to the complex reflection group $W_{n, r} = G(r,1,n)$.
In \cite{DJMa} Dipper, James and Mathas introduced the cyclotomic
$q$-Schur algebras $\mathcal{S}(\Lambda)$ with weight poset $\Lambda$
as a tool for studying the representations of the Ariki-Koike algebra
$\mathscr{H}$.
It is an important
problem to determine the decomposition matrix of $\mathcal{S}(\Lambda)$.
In the case where $r =1$, the cyclotomic $q$-Schur algebra
coincides with the $q$-Schur algebra of \cite{DJ}.
In that case, under the condition that
$R = \mathbb{C}$ and $q$ is a root of unity,
Varagnolo and Vasserot proved in \cite{VV}
the decomposition conjecture due to Leclerc and Thibon \cite{LT}, which
provides us an algorithm of computing the decomposition matrix
in connection with the canonical basis of the level 1 Fock space of type $A$.
\par
It is an open problem to determine the decomposition matrix for
$\mathcal{S}(\Lambda)$ in the case where $r \ge 2$.
It is known by \cite{A} (see also \cite{DM}) that the determination of the
decomposition matrix of $\mathcal{S}(\Lambda)$ is reduced to the case
where $r = 1$ if the parameters satisfy the separation condition
\par\medskip\noindent
(S) \quad $q^{2k}Q_i - Q_j$ are invertible in $R$ for $|k| < n$,
$i \ne j$.
\par\medskip\noindent
In \cite{U}, Ugolov constructed the canonical basis of the
level $r$ Fock space of type $A$, and gave an algorithm of computing them.
Assume that $R = \mathbb C$, and that $Q_i = q^{s_i}$ with
$s_{i+1} - s_i \ge n$ for a root of unity $q \in \mathbb C$.
In that case, Yvonne \cite{Y} formulated a conjecture
which makes it possible to determine the decomposition matrix of
$\mathcal{S}(\Lambda)$ by means of the algorithm of
Ugolov as in the case of $r = 1$. He showed in \cite{Y} that the
Jantzen sum
formula for $\mathcal{S}(\Lambda)$ also holds for the Fock space side
in an appropriate sense, which supports the conjecture.
\par
In this paper, we introduce a certain subalgebra
$\mathcal{S}^0(\Lambda)$ of $\mathcal{S}(\Lambda)$, motivated
in \cite{SawS}, and show that
$\mathcal{S}^0(\Lambda)$ of $\mathcal{S}(\Lambda)$ is a standardly
based algebra in the sense of Du and Rui \cite{DR1}.
$\mathcal{S}^0(\Lambda)$ has a natural quotient
$\overline{\mathcal{S}^{0}}(\Lambda)$, which turns out to be
a cellular algebra. We discuss a relationship among the
representations of $\mathcal{S}(\Lambda)$, $\mathcal{S}^0(\Lambda)$ and
$\overline{\mathcal{S}^{0}}(\Lambda)$. In particular we show, in the
case
where $R$ is a field, that
a certain important part of the decomposition matrix of
$\mathcal{S}(\Lambda)$ coincides with a part of the
decomposition matrix of $\overline{\mathcal{S}^{0}}(\Lambda)$.
This reduces a computation of some decomposition numbers of
$\mathcal{S}(\Lambda)$ to that of
$\overline{\mathcal{S}^{0}}(\Lambda)$ which seems to have a simpler
structure than $\mathcal{S}(\Lambda)$.
\par
The modified Ariki-Koike algebra
$\mathscr{H}^{\flat}$ over $R$ was discussed in \cite{SawS},
which has the same parameter set as $\mathscr{H}$, but under the condition that
\par\medskip\noindent
$(\ast)$ \quad $Q_{i} - Q_{j}$ are invertible in $R$ for any $i\neq j$.
\par\medskip\noindent
Note that the condition $(\ast)$ is weaker than the separation condition (S).
It is known in [S] that $\mathscr{H}^{\flat}$ is isomorphic to
$\mathscr{H}$ when $R$ is a field, and the parameters
satisfies the separation condition.
Let $\bs{m} = ( m_1, \ldots , m_r )$ be a tuple of non-negative
integers, and
$\widetilde{\mathcal{P}}_{n,r} =
\widetilde{\mathcal{P}}_{n,r}(\bs m)$ the set of all
$r$-compositions
$\lambda = ( \lambda^{( 1 )}, \ldots , \lambda^{( r )} )$
such that $\lambda^{(i)} \in \Bbb Z^{m_i}_{\ge 0}$.
A representation of $\mathscr{H}^{\flat}$
on a tensor space $V^{\otimes n}$ was constructed in [S], where
$V = \bigoplus_{i = 1}^{r} V_{i}$ and $V_{i}$ is a free
$R$-module of rank $m_{i}$ satisfying the condition that
\par\medskip\noindent
$(\ast\ast)$ \quad $m_{i} \geq n$ for all $i$.
\par\medskip
In \cite{SawS} we have defined the
cyclotomic $q$-Schur
algebra $\mathcal{S}^{\flat}(\bs{m}, n)$ with the poset
$\Lambda = \widetilde{\mathcal{P}}_{n,r}$ as the
endomorphism algebra $\operatorname{End}_{\mathscr{H}^{\flat}}V^{\otimes n}$.
Moreover, we have constructed a certain subalgebra
$\mathcal{S}^{0}(\bs{m}, n)^{\flat}$ of $\mathcal{S}(\Lambda)$ which has a
surjective algebra homomorphism
$\widehat{f} : \mathcal{S}^{0}(\bs{m},n)^{\flat}
\rightarrow \mathcal{S}^{\flat}(\bs{m}, n)$.
Note that $\mathcal{S}^{0}(\bs{m},n)^{\flat}$
(resp. $\mathcal{S}^{\flat}(\bs{m}, n)$)
coincides with the previous $\mathcal{S}^{0}(\Lambda)$
(resp. $\overline{\mathcal{S}^{0}}(\Lambda)$)
whenever the conditions $(\ast)$ and $(\ast\ast)$ are satisfied.
\par
By a result of \cite{SawS}, the decomposition matrix of
$\mathcal{S}^{\flat}(\bs{m}, n)$ can be described in terms of
the decomposition
matrices for $q$-Schur algebras, which are computable by
the theorem of Varagnolo-Vasserot.
Thus our results determine a part of the decomposition matrix
for $\mathcal{S}(\Lambda)$ in the case where $R = \mathbb C$ and
the parameters $Q_i = q^{s_i}$ are all distinct for a root of unity
$q \in \mathbb C$.
\par
The details of the paper are as follows. In Section 1
we recall the notations and basic results on the representation
theory
of the Ariki-Koike algebras and the cyclotomic $q$-Schur
algebras. In Section 2 we investigates the subalgebra
$\mathcal{S}^{0}(\Lambda)$ of $\mathcal{S}(\Lambda)$. We show
that $\mathcal{S}^{0}(\Lambda)$ is a
standardly based algebra in the sense of Du and Rui \cite{DR1}, which
is a natural extension of the
cellular algebra. Section 3 discusses
the relationship between $\mathcal{S}^{0}(\Lambda)$ and
$\mathcal{S}(\Lambda)$, and their Weyl modules $Z^{( \lambda, 0)}$ and
$W^{\lambda}$ for a multipartition (or $r$-partition) $\lambda$ in
$\Lambda$,
respectively. More precisely, we show that $Z^{( \lambda, 0)}
\otimes_{\mathcal{S}^{0}(\Lambda)} \mathcal{S}(\Lambda) \simeq
W^{\lambda}$. Even though $\mathcal{S}^{0}(\Lambda)$ is not cellular,
the Weyl module $Z^{( \lambda, 0)}$ has the radical and
we can discuss almost similar to the cellular theory, i.e, we can
define irreducible modules $L_{0}^{\lambda}$ of
$\mathcal{S}^{0}(\Lambda)$ in a similar way as defining of irreducible
modules
$L^{\lambda}$ of $\mathcal{S}(\Lambda)$. In turn, in Section 4 we
study the relationship between $\mathcal{S}^{\flat}(\bs{m}, n)$ (or
$\overline{\mathcal{S}^{0}}(\Lambda)$) and
$\mathcal{S}^{0}(\Lambda)$ and their Weyl modules. For a
multipartition $\lambda \in \Lambda$, let
$\overline{Z}^{\lambda}$ be the Weyl module of
$\overline{\mathcal{S}^{0}}(\Lambda)$. Since
$\overline{\mathcal{S}^{0}}(\Lambda)$ is a cellular algebra, we can
also define irreducible modules $\overline{L}^{\lambda}$. Because
$\overline{\mathcal{S}^{0}}(\Lambda)$ is a quotient of
$\mathcal{S}^{0}(\Lambda)$ by $\mathcal{S}^{00}(\Lambda)$, we can
regard $\overline{Z}^{\lambda}$ and $\overline{L}^{\lambda}$ as
$\mathcal{S}^{0}(\Lambda)$-modules. Under this setting, we can verify
that the decomposition number $[ \overline{Z}^{\lambda} :
\overline{L}^{\mu} ]$ is equal to the decomposition number $[
Z^{( \lambda, 0)} : L_{0}^{\mu} ]$ for any multipartitions $\lambda,
\mu$ in $\Lambda$ when $R$ is a field. In section 5, we estimate the
decomposition number $[ W^{\lambda} : Z^{\mu} ]$ by working use of the
induction of $Z^{( \lambda, 0)}$ and the restriction of
$W^{\lambda}$. Suppose that $R$ is a
field. Our main result asserts that
$[
\overline{Z}^{\lambda} : \overline{L}^{\mu} ] =
[ Z^{(\lambda, 0)} : L_{0}^{\mu} ] = [ W^{\lambda} : L^{\mu} ]$ for
all multipartitions $\lambda, \mu$ in $\Lambda$ such that $\lambda =
(\lambda^{( 1 )}, \ldots , \lambda^{( r )} )$, $\mu = (\mu^{( 1 )},
\ldots ,
\mu^{( r )} )$ with $| \lambda^{( i )} | = | \mu^{( i )} |$. Finally,
we denote by $[W^{\lambda^{( i )}} :
L^{\mu^{( i )}} ] ~ ( 1 \le i \le r )$ the decomposition number of
$L^{\mu^{( i )}}$ in $W^{\lambda^{( i )}}$. Again suppose that $R$
is a
field. Moreover, assume that $\Lambda =
\widetilde{\mathcal{P}}_{n,r}(\bs{m})$ and the previous conditions
$(\ast)$, $(\ast \ast)$. Then, as a corollary, we have that $[
W^{\lambda} :
L^{\mu} ] = \prod_{i = 1}^{r} [ W^{\lambda^{( i )}} :
L^{\mu^{( i )}} ]$ for all multipartitions $\lambda, \mu$ in
$\widetilde{\mathcal{P}}_{n,r}(\bs{m})$ as above.
\par\bigskip\noindent
\section*{Table of contents}
\par\medskip
0. Introduction
\par
1. Preliminaries on Ariki-Koike algebras and cyclotomic $q$-Schur
algebras
\par
2. The standard basis for $\mathcal{S}^{0}(\Lambda)$
\par
3. A relationship between $\mathcal{S}^{0}(\Lambda)$ and
$\mathcal{S}(\Lambda)$
\par
4. A relationship between $\mathcal{S}^{\flat}(\bs{m}, n)$ and
$\mathcal{S}^{0}(\Lambda)$
\par
5. An estimate for decomposition numbers
\par\medskip
\section{Preliminaries on Ariki-Koike algebras and Cyclotomic
$q$-Schur algebras}
\subsection{}\label{def:the definition of
Ariki-Koike algebras}
Fix positive integers $r$ and $n$ and let $\mathfrak{S}_{n}$ be the
symmetric group of degree $n$. Let $R$ be an integral
domain with 1 and $q, Q_1,
\ldots , Q_r$ be elements in $R$,
with invertible $q$. The Ariki-Koike algebra associated to the complex
reflection group $W_{n, r} = G( r, 1, n )$, is the associative unital
algebra
$\mathscr{H} = \mathscr{H}_{n, r}$ over $R$ with generators $T_1,
\ldots, T_n$ subject to the following conditions,
\begin{equation*}
\begin{array}{rll}
( T_{1} - Q_{1} ) \cdots ( T_{1} - Q_{r} ) &= 0, & \\
( T_{i} -q )( T_{i} + q^{-1} ) &= 0 &( i \geq 2), \\
T_{1} T_{2} T_{1} T_{2} &= T_{2} T_{1} T_{2} T_{1},& \\
T_{i} T_{j} &= T_{j} T_{i} & ( | i - j | \geq 2), \\
T_{i} T_{i+1} T_{i} &=T_{i+1} T_{i} T_{i+1} & ( 2 \le i \le n-1 ).
\end{array}
\end{equation*}
It is known that $\mathscr{H}$ is a free $R$-module of rank
$n!r^{n}$. The subalgebra $\mathscr{H}( \mathfrak{S}_{n} )$ of
$\mathscr{H}$ generated by $T_{2}, \ldots, T_{n}$ is isomorphic to the
Iwahori-Hecke algebra $\mathscr{H}_{n}$ of the symmetric group
$\mathfrak{S}_{n}$.
For $i = 2, \ldots , n$ let $s_i$ be the transposition $( i-1, i )$ in
$\mathfrak{S}_{n}$.Then $\{ s_2, \ldots, s_n \}$ generate
$\mathfrak{S}_{n}$. For $w \in \mathfrak{S}_{n}$, we set $T_{w} =
T_{{i}_{1}} \cdots T_{{i}_{k}}$ where $w = s_{i_1} \cdots s_{i_k}$ is
a reduced expression. Then
$T_{w}$ is independent of the choice of a reduced expression. We also
put $L_{k} = T_{k} \cdots T_{2}T_{1}T_{2} \cdots T_{k}$ for $k = 1, 2,
\ldots , n$. Note that all $L_1, \ldots , L_n$ commutes. Moreover,
these
elements produce a basis of $\mathscr{H}$.
\addtocounter{thm}{1}
\begin{thm}[{\cite[Theorem 3.10]{AK}}]
The Ariki-Koike algebra $\mathscr{H}$ is free as an $R$-module with
basis $\{ L_{1}^{a_{1}} \cdots L_{n}^{a_{n}} T_{w} \mid w \in
\mathfrak{S}_{n},~ 0 \le a_{i} < r \text{ for } 1 \le i \le n \}$.
\end{thm}
Recall that a composition of $n$ is sequence $\sigma = ( \sigma_{1},
\sigma_{2}, \ldots )$ of non-negative integers such that $| \sigma | =
\sum_{i} \sigma_{i} = n$. $\sigma$ is a partition if in addition
$\sigma_{1} \geq \sigma_{2} \geq \cdots $. If $\sigma_{i} = 0$ for all
$i > k$ then we write $\sigma = ( \sigma_{1}, \ldots , \sigma_{k} )$.
An $r$-composition (or multicomposition) of $n$ is an $r$-tuple
$\lambda = ( \lambda^{( 1 )}, \ldots, \lambda^{( r )} )$ of
compositions with $\lambda^{( i )} = ( \lambda^{( i )}_1, \lambda^{( i
)}_2, \ldots )$ such that $| \lambda^{( 1 )}
| + \cdots + | \lambda^{( r )} | = n$. An $r$-composition $\lambda$ is
an $r$-partition if each $\lambda^{( i )}$ is a partition. If
$\lambda$ is an $r$-partition of $n$ then we write $\lambda \vdash
n$. The diagram $[ \lambda ]$ of the $r$-composition $\lambda$ is the
set $[ \lambda ] = \{ ( i, j, s ) \mid 1 \le i \le \lambda_{j}^{( s )},
1 \le s \le r \}$. The elements of $[ \lambda ]$ are called nodes.
The set of $r$-compositions of $n$ is partially ordered by dominance,
i.e, if $\lambda$ and $\mu$ are two $r$-compositions then
$\lambda$ dominates $\mu$, and we write $\lambda \unrhd \mu$, if
\begin{equation*}
\underset{c = 1}{\overset{s-1}{\sum}} | \lambda^{( c )}| + \underset{j
= 1}{\overset{i}{\sum}} | \lambda_{j}^{( s )}| \geq \underset{c =
1}{\overset{s-1}{\sum}} | \mu^{( c )}| + \underset{j
= 1}{\overset{i}{\sum}} | \mu_{j}^{( s )}|
\end{equation*}
for $1 \le s \le r$ and for all $i \geq 1$. If $\lambda \unrhd \mu$
and $\lambda \neq \mu$ then we write $\lambda \rhd \mu$.
If $\lambda$ is an $r$-composition let $\mathfrak{S}_{\lambda}
=\mathfrak{S}_{\lambda^{( 1 )}} \times \cdots \times
\mathfrak{S}_{\lambda^{( r )}}$ be the corresponding Young subgroup of
$\mathfrak{S}_{n}$. Set
\begin{equation*}
x_{\lambda} = \underset{w \in \mathfrak{S}_{\lambda}}{\sum} q^{l( w )}
T_{w}, \qquad u^{+}_{\lambda} = \underset{s = 2}{\overset{r}{\prod}}
\underset{k = 1}{\overset{a_{s}}{\prod}} ( L_{k} - Q_{s} ),
\end{equation*}
where $a_s = | \lambda^{( 1 )} | + \cdots + | \lambda^{( s-1 )} |$ for
$2 \le s \le r$. If $s = 1$ then we set $a_{s} = 0$. Set $m_{\lambda} =
x_{\lambda} u^{+}_{\lambda} = u^{+}_{\lambda} x_{\lambda}$ and define
$M^{\lambda}$ to be the right ideal $M^{\lambda} = m_{\lambda}
\mathscr{H}$ of $\mathscr{H}$.
For any $r$-composition $\mu$, a $\mu$-tableau $\mathfrak{t} =(
\mathfrak{t}^{( 1 )}, \ldots , \mathfrak{t}^{( r )} )$ is a bijection
$t : [ \mu ] \rightarrow \{ 1, 2, \ldots , n \}$, where
$\mathfrak{t}^{( i )}$ is a tableau of $\text{Shape}( \mathfrak{t}^{(
i )} ) = \mu^{( i )}$. We write $\text{Shape}(
\mathfrak{t} ) = \mu $ if $\mathfrak{t}$ is a $\mu$-tableau. A
$\mu$-tableau $\mathfrak{t}$ is called standard
(resp. row standard) if all $\mathfrak{t}^{( i )}$ are standard
(resp. row standard).
Let
$\text{Std}(\lambda)$ be the set of standard $\lambda$-tableaux.
For each $r$-composition $\mu$, let $\mathfrak{t}^{\mu}$ be the
$\mu$-tableau with
the numbers $1, 2, \ldots , n$ attached in order from left to right
along its rows and from top to bottom, and from $\mu^{( 1 )}$ to
$\mu^{( r )}$. If $\mathfrak{t}$ is any row standard $\mu$-tableau
let $d( \mathfrak{t} ) \in \mathfrak{S}_{n}$ be the unique permutation
such that $\mathfrak{t} = \mathfrak{t}^{\mu} d( \mathfrak{t}
)$. Furthermore, let $* : \mathscr{H} \rightarrow \mathscr{H}$ be the
anti-isomorphism given by $T_{i}^{*} = T_{i}$ for $i = 1, 2, \ldots ,
n$, and set $m_{\mathfrak{s} \mathfrak{t}} = T_{d( \mathfrak{s} )}^{*}
m_{\lambda} T_{d( \mathfrak{t} )}$.
\begin{thm}[{\cite[Theorem 3.26; Murphy bases]{DJMa}}]
The Ariki-Koike algebra $\mathscr{H}$ is free as an $R$-module with
cellular basis $\{ m_{\mathfrak{s} \mathfrak{t}} \mid \mathfrak{s},
\mathfrak{t} \in {\rm{Std}}(\lambda) \text{ for some } \lambda
\vdash n \}$.
\end{thm}
Here, and below, whenever we write expressions involving a pair of
tableaux (such as $m_{\mathfrak{s} \mathfrak{t}}$ or $\varphi_{ST}$
and so on), we implicitly assume that the two tableaux are of the same
shape.
\addtocounter{subsection}{2}
\subsection{}\label{sub:the definition of
H^lambda}
For each $r$-partition $\lambda$ let $\mathscr{H}^{\lambda}$ be the
$R$-submodule of $\mathscr{H}$ with basis $\{ m_{\mathfrak{u}
\mathfrak{b}} \mid \mathfrak{u}, \mathfrak{b} \in
\text{Std}(\mu) \text{ for some } \mu \rhd \lambda \}$. Then
$\mathscr{H}^{\lambda}$ is a two-sided ideal of $\mathscr{H}$.
Let $S^{\lambda}$ be the Specht module (or cell module)
corresponding to the $r$-partition $\lambda$. Thus, $S^{\lambda}
= ( m_{\lambda} + \mathscr{H}^{\lambda} ) \mathscr{H}$, a submodule of
$\mathscr{H} / \mathscr{H}^{\lambda}$. For each $\mathfrak{t} \in
\text{Std}(\lambda)$ let $m_{\mathfrak{t}} = m_{\mathfrak{t}^{\lambda}
\mathfrak{t}} + \mathscr{H}^{\lambda}$. Then $S^{\lambda}$ is free
as an $R$-module with basis $\{ m_{\mathfrak{t}} \mid \mathfrak{t} \in
\text{Std}(\lambda) \}$.
Furthermore, there is an associative symmetric bilinear form on
$S^{\lambda}$ which is determined by
\begin{equation*}
m_{\mathfrak{t}^{\lambda} \mathfrak{s}} m_{\mathfrak{t}
\mathfrak{t}^{\lambda}} \equiv
\langle m_{\mathfrak{s}}, m_{\mathfrak{t}} \rangle
m_{\mathfrak{t}^{\lambda} \mathfrak{t}^{\lambda}} \mod
\mathscr{H}^{\lambda}
\end{equation*}
for all $\mathfrak{s}, \mathfrak{t} \in
\text{Std}(\lambda)$. The radical $\text{rad}{S^{\lambda}}$ of this
form is again an $\mathscr{H}$-module, so $D^{\lambda} = S^{\lambda} /
\text{rad}{S^{\lambda}}$ is an $\mathscr{H}$-module. When $R$ is a
field, $D^{\lambda}$ is either $0$ or absolutely irreducible and all
the simple $\mathscr{H}$-modules up to isomorphisms arise uniquely in
this way.
We can now give a definition of the cyclotomic $q$-Schur algebras. A
set $\Lambda$ of $r$-compositions of $n$ is saturated if $\Lambda$ is
finite and whenever $\lambda$ is an $r$-partition such that $\lambda
\unrhd \mu$ for some $\mu \in \Lambda$ then $\lambda \in \Lambda$. If
$\Lambda$ is a saturated set of $r$-compositions, we denote by
$\Lambda^{+}$ be
the set of $r$-partitions in $\Lambda$.
\addtocounter{thm}{1}
\begin{defi}
Suppose that $\Lambda$ is a saturated set of multicompositions of
$n$. The cyclotomic $q$-Schur algebra with weight poset $\Lambda$ is
the endomorphism algebra
\begin{equation*}
\mathcal{S}( \Lambda ) = \operatorname{End}_{\mathscr{H}}( M( \Lambda ) ), \qquad
\text{ where } M( \Lambda ) = \underset{ \lambda \in
\Lambda}{\bigoplus} M^{\lambda}.
\end{equation*}
\end{defi}
Let $\lambda$ be an $r$-partition and $\mu$ an $r$-composition. A
$\lambda$-Tableau of type $\mu$ is a map $T : [ \lambda ] \rightarrow
\{ ( i, s ) \mid i \geq 1, ~ 1 \le s \le r \}$ such that $\mu^{( s
)}_{i} = \sharp \{ x \in [ \lambda ] \mid T( x ) = ( i, s ) \}$ for
all
$i \geq 1$ and $1 \le s \le r$. We regard $T$ as an $r$-tuple
$T = ( T^{( 1 )}, \ldots , T^{( r )} )$, where $T^{( s )}$ is the
$\lambda^{( s )}$-tableau with $T^{( s )}(i, j) = T(i, j, s)$ for all
$(i, j, s) \in [ \lambda ]$. In this way we identify the standard
tableaux above with the Tableaux of type $w = ( ( 0 ), \ldots , (
1^{n} ) )$. If $T$ is a Tableau of type $\mu$ then we write
$\text{Type}( T )
= \mu$.
Given two pairs $( i, s )$ and $( j, t )$ write $( i, s ) \preceq ( j,
t )$ if either $s < t$, or $s = t$ and $i \le j$.
\begin{defi}
A Tableau $T$ is $($row$)$ semistandard if, for $1 \le t \le r$, the
entries in $T^{( t )}$ are
$({\rm{i}})$ weakly increasing along the rows with respect to
$\preceq$,
$({\rm{ii}})$ strictly increasing down columns,
$({\rm{iii}})$ $( i, s )$ appears in $T^{( t )}$ only if $ s \geq
t$.
\end{defi}
Let $\mathcal{T}_0(\lambda, \mu)$ be the set of semistandard
$\lambda$-Tableaux of type $\mu$ and let
$\mathcal{T}_0(\lambda) = \mathcal{T}^{\Lambda}_0(\lambda) =
\bigcup_{\mu \in \Lambda}
\mathcal{T}_0(\lambda, \mu)$.
Notice that if $\mathcal{T}_0(\lambda, \mu)$ is non-empty, then
$\lambda \unrhd \mu$.
Suppose that $\mathfrak{t}$ is a standard $\lambda$-tableau and let
$\mu$ be an $r$-composition. Let $\mu( \mathfrak{t} )$ be the Tableau
obtained from $\mathfrak{t}$ by replacing each entry $j$ with $( i, k
)$ if $j$ appears in row $i$ of $( \mathfrak{t}^{\mu} )^{( k )}$. The
tableau $\mu( \mathfrak{t} )$ is a $\lambda$-Tableau of type $\mu$. It
is not necessarily semistandard.
If $S$ and $T$ are semistandard $\lambda$-Tableaux of type $\mu$ and
$\nu$ respectively,
let
\begin{equation*}
\qquad m_{ST}
=
\underset{
\substack{
\mathfrak{s}, \mathfrak{t}\in \text{Std}(\lambda) \\
\mu( \mathfrak{s} ) = S, ~ \nu( \mathfrak{t} ) = T
}
}{\sum}
q^{l( d(\mathfrak{s}) ) + l( d(\mathfrak{t}) )} m_{\mathfrak{s}
\mathfrak{t}}.
\end{equation*}
For $S$ and $T$ as above we define a map $\varphi_{ST}$ on
$M(\Lambda)$
by $\varphi_{ST}( m_{\alpha} h ) =
\delta_{\alpha \nu} m_{ST} h$, for all $h \in \mathscr{H}$ and all
$\alpha \in \Lambda$. Here $\delta_{\alpha \nu}$ is the Kronecker
delta, i.e, $\delta_{\alpha \nu} = 1$ if $\alpha = \nu$ and it is
zero otherwise. Then $\varphi_{ST}$ is well-defined, and it belongs
to $\mathcal{S}( \Lambda
)$. Moreover,
\begin{thm}[{\cite[Theorem 6.6]{DJMa}}]
The cyclotomic $q$-Schur algebra $\mathcal{S}( \Lambda )$ is free as
an $R$-module with cellular basis $\mathcal{C}( \Lambda ) = \{
\varphi_{ST} \mid S, T \in \mathcal{T}^{\Lambda}_0(\lambda) \text{ for
some } \lambda \in \Lambda^{+} \}$.
\end{thm}
The basis $\{ \varphi_{ST} \}$ is called a semistandard basis of
$\mathcal{S}( \Lambda ) $. Since this basis is cellular, the map $* :
\mathcal{S}( \Lambda ) \rightarrow \mathcal{S}( \Lambda )$ which is
determined by $\varphi_{ST}^{*} = \varphi_{TS}$ is an
anti-automorphism
of $\mathcal{S}( \Lambda )$. This involution is closely related to the
$*$-involution on $\mathscr{H}$. Explicitly, if $\varphi : M^{\nu}
\rightarrow M^{\mu}$ is an $\mathscr{H}$-module homomorphism then
$\varphi^{*} : M^{\mu} \rightarrow M^{\nu}$ is the homomorphism given
by $\varphi^{*}( m_{\mu} h ) = ( \varphi( m_{\nu} ) )^{*} h$, for all
$h \in \mathscr{H}$.
For each $r$-partition $\lambda \in \Lambda^{+}$, we define
$\mathcal{S}^{\vee \lambda} = \mathcal{S}^{\vee}( \Lambda )^{\lambda}$
as the $R$-span of $\varphi_{ST}$ such that $S, T \in
\mathcal{T}^{\Lambda}_0(\alpha)$ with $\alpha \rhd \lambda$, which is
a two-sided ideal of
$\mathcal{S}( \Lambda )$. We define the Weyl module $W^{\lambda}$ by
the right $\mathcal{S}( \Lambda )$-submodule of $\mathcal{S}( \Lambda
) / \mathcal{S}^{\vee}( \Lambda )^{\lambda}$ generated by the image
$\varphi_{\lambda} = \varphi_{T^{\lambda} T^{\lambda}} \in
\mathcal{S}( \Lambda )$ where $T^{\lambda} = \lambda(
\mathfrak{t}^{\lambda} )$. For each $T \in
\mathcal{T}^{\Lambda}_0(\lambda)$, let
$\varphi_{T}$ be the image of $\varphi_{T^{\lambda} T}$ in
$W^{\lambda}$. Then the Weyl module $W^{\lambda}$ is $R$-free with
basis $\{ \varphi_{T} \mid T \in \mathcal{T}^{\Lambda}_0(\lambda)
\}$. As in the case of Specht modules there is an inner product on
$W^{\lambda}$ which is determined by
\begin{equation*}
\varphi_{T^{\lambda}
S} \varphi_{T T^{\lambda}} \equiv \langle \varphi_{S}, \varphi_{T}
\rangle \varphi_{T^{\lambda} T^{\lambda}} \mod \mathcal{S}^{\vee
\lambda}.
\end{equation*}
Let $\text{rad}{W^{\lambda}} = \{ x \in W^{\lambda} \mid \langle x, y
\rangle = 0 \text{ for all } y \in W^{\lambda} \}$. The quotient
module $L^{\lambda} = W^{\lambda} /
\text{rad}{W^{\lambda}}$ is absolutely irreducible and $\{ L^{\lambda}
\mid \lambda \in \Lambda^{+} \}$ is a complete set of non-isomorphic
irreducible $\mathcal{S}( \Lambda )$-modules.
\addtocounter{subsection}{3}
\subsection{}
For an $r$-composition $\mu$, we define the type $\alpha = \alpha( \mu
)$ of $\mu$ by $\alpha = ( n_1, \ldots , n_r )$ with $n_{i} = | \mu^{(
i )}|$, and the size of $\mu$ by $n = \sum_{i = 1}^{r} n_{i}$. We
also define a sequence $\bold{a} = \bold{a}( \mu ) = ( a_1, \ldots ,
a_r
)$. (Recall that $a_{i} = \sum_{k = 1}^{i - 1} | \mu^{( k )} | =
\sum_{k = 1}^{i - 1} n_{k} $.)
We define a partial order $\geq$ on the set $\mathbb{Z}^{r}_{\geq 0}$
by
$\bold{a} \geq {\bold{a}}'$ for $\bold{a} = ( a_1, \ldots , a_r )$,
${\bold{a}}' = ( {a}'_1, \ldots , {a}'_r ) \in \mathbb{Z}^{r}_{\geq
0}$
if $a_{i} \geq {a}'_{i}$ for any $i$. We write $\bold{a} >
{\bold{a}}'$ if
$\bold{a} \geq {\bold{a}}'$ and $\bold{a} \neq {\bold{a}}'$. It is
clear that
\begin{equation}\label{eq:A certain fact of r-compositions}
\text{If } \lambda \unrhd \mu, \text{ then } \bold{a}( \lambda ) \geq
\bold{a}( \mu ) \text{ for } r\text{-compositions } \lambda, \mu.
\end{equation}
Hence if $\mathcal{T}_0(\lambda, \mu)$ is non-empty, then
$\lambda
\unrhd \mu$, and so we have $\bold{a}( \lambda ) \geq \bold{a}( \mu
)$.
For any $r$-partition $\lambda$ and $r$-composition $\mu$, we define a
subset $\mathcal{T}^{+}_0(\lambda, \mu)$ of $\mathcal{T}_0(\lambda,
\mu)$ by
\begin{equation*}
\mathcal{T}^{+}_0(\lambda, \mu) = \{ S \in \mathcal{T}_0(\lambda,
\mu) \mid \bold{a}( \lambda ) = \bold{a}( \mu ) \}.
\end{equation*}
Note that the condition $\bold{a}( \lambda ) = \bold{a}( \mu )$ is
equivalent to $\alpha( \lambda ) = \alpha( \mu )$. Take $S \in
\mathcal{T}^{+}_0(\lambda, \mu)$. Then one can check that $S \in
\mathcal{T}^{+}_0(\lambda, \mu)$ if and only if each entry of $S^{( k
)}$ is of the form $( i, k )$ for some $i$. Hence in this case $S^{(
k )}$ can be identified with a semistandard $\lambda^{( k )}$-Tableau
of type $\mu^{( k )}$ under the usual definition of the semistandard
Tableaux for $1$-partitions $\lambda^{( k )}$ and $1$-compositions
$\mu^{( k )}$. It follows that we have a bijection
\begin{equation*}
\mathcal{T}^{+}_0(\lambda, \mu) \simeq \mathcal{T}_0(\lambda^{( 1 )},
\mu^{( 1 )} ) \times \cdots \times \mathcal{T}_0(\lambda^{( r )},
\mu^{( r )} )
\end{equation*}
via $S \leftrightarrow ( S^{( 1 )}, \ldots , S^{( r )} )$. Moreover,
if $\mathfrak{s} \in \text{Std}( \lambda )$ is such that $\mu(
\mathfrak{s} ) = S$ with $S \in \mathcal{T}^{+}_0(\lambda, \mu)$, then
the entries of $i$-th component of $\mathfrak{s}$ consist of numbers
$a_{i} + 1, \ldots , a_{i + 1}$ for $\bold{a}( \lambda ) = ( a_{1},
\ldots , a_{r} )$. In particular, $d( \mathfrak{s} ) \in
\mathfrak{S}_{\alpha}$ for $\alpha = \alpha( \lambda )$.
Fix an $r$-tuple $\bs{m} = (
m_1, \ldots , m_r )$ of non-negative integers.
Then, an $r$-composition $\mu = ( \mu^{( 1 )}, \ldots , \mu^{( r )} )$
with $\mu^{( i )} =( \mu^{( i )}_{1}, \ldots , \mu^{( i )}_{m_{i}} )
\in
\mathbb{Z}_{\geq 0}^{m_{i}}$ is called an $(r,
\bs{m})$-composition, and $(r, \bs{m})$-partition is defined
similarly. We denote by
$\widetilde{\mathcal{P}}_{n,r} = \widetilde{\mathcal{P}}_{n,r}( \bs{m}
)$ (resp. $\mathcal{P}_{n,r} =\mathcal{P}_{n,r}( \bs{m} )$) the set
of $(r, \bs{m})$-compositions (resp. $(r, \bs{m})$-partitions) of
size $n$. (Note that
$\mathcal{P}_{n,r}( \bs{m} )$ are naturally identified with each other
for any $\bs{m}$ such that $m_{i} \geq n$. However,
$\widetilde{\mathcal{P}}_{n,r}$ depends on the choice of $\bs{m}$.)
Finally, let
\begin{equation*}
\begin{array}{ll}
\mathcal{C}^{0}(\Lambda) = \underset{
\mu, \nu \in \Lambda, ~ \lambda \in \Lambda^{+}
}{\bigcup}
\{ \varphi_{ST} \in \mathcal{C}(\Lambda)
\mid
&S \in \mathcal{T}_0(\lambda, \mu), ~ T \in \mathcal{T}_0(\lambda,
\nu),\\
&\bold{a}( \lambda ) > \bold{a}( \mu ) \text{ if } \alpha( \mu ) \neq
\alpha( \nu ) \}
\end{array}
\end{equation*}
and we define $\mathcal{S}^{0}(\Lambda)$ as the $R$-submodule of
$\mathcal{S}(\Lambda)$ with basis
$\mathcal{C}^{0}(\Lambda)$.
\medskip
\section{The standard basis for $\mathcal{S}^{0}(\Lambda)$}\label{The
standard basis for S^0(Lambda)}
\subsection{}
In \cite{SawS}, the $R$-submodule
$\mathcal{S}^{0}(\Lambda)$ was constructed as an $R$-subalgebra of
$\mathcal{S}(\Lambda)$ when $\Lambda = \widetilde{\mathcal{P}}_{n,r}$
subject to the conditions $(\ast)$, $(\ast \ast)$ in Introduction.
But, as a fact, $\mathcal{S}^{0}(\Lambda)$ becomes an $R$-subalgebra
with no condition and it become
a standardly based algebra in the sense of \cite{DR1} which is an
extension of the cellular algebra. The standardly based algebra has a
basis, namely
standard basis, corresponding to the cellular basis in the case of the
cellular algebra. The aim of this section is to prove that
$\mathcal{S}^{0}(\Lambda)$ is an $R$-subalgebra of
$\mathcal{S}(\Lambda)$ and $\mathcal{C}^{0}(\Lambda)$ is a standard
basis for
$\mathcal{S}^{0}(\Lambda)$.
First, we recall the definition of standardly based algebras in
\cite{DR1}.
\addtocounter{thm}{1}
\begin{defi}[{\cite{DR1}}]\label{def:standardly based}
Assume that $R$ is a commutative ring with 1. Let $A$ be an
$R$-algebra and $(\Lambda,\ge)$ a poset. $A$ is called a
\emph{standardly
based algebra on} $\Lambda$ $($or \emph{standardly based}$)$ if the
following conditions hold.
$(a)$ For any $\lambda \in \Lambda$, there are index sets
$I(\lambda)$,$J(\lambda)$ and subsets
\begin{equation*}
\mathscr{A}^{\lambda}=\{a^{\lambda}_{i,j} \mid (i,j) \in I(\lambda)
\times J(\lambda) \}
\end{equation*}
of $A$ such that the union $\mathscr{A} = \underset{{\lambda \in
\Lambda}}{\bigcup} \mathscr{A}^{\lambda}$ is disjoint and forms an
$R$-basis for $A$.
$(b)$ For any $a \in A$, $a^{\lambda}_{i,j} \in \mathscr{A}$, we
have
\begin{equation*}
\begin{array}{c}
a\cdot a^{\lambda}_{i,j}\equiv \underset{i' \in
I(\lambda)}{\sum}f_{i',\lambda}(a,i)~a^{\lambda}_{i',j}\mod
A(>\lambda)\\
a^{\lambda}_{i,j} \cdot a \equiv \underset{j' \in J(\lambda)}{\sum}
f_{\lambda,j'}(j,a) ~ a^{\lambda}_{i,j'} \mod A(> \lambda),
\end{array}
\end{equation*}
where $A( > \lambda )$ is the $R$-submodule of $A$ spanned by
$\mathscr{A}^{\mu}$ with $\mu > \lambda$, and $f_{i',\lambda}(a,i)
$, $f_{\lambda,j'}(j,a) \in R$ are
independent of $j$ and $i$, respectively. Such a basis $\mathscr{A}$ is
called a \emph{standard basis} for the algebra $A$.
\end{defi}
Note that the cellular algebra is a special case of the standardly
based algebras.
We prepare some notation.\\
Let
\begin{equation*}
\varOmega= ( \Lambda^{+} \times \{0, 1\} ) \setminus \{ (\lambda, 1)
\mid
\mathcal{T}_0(\lambda,\mu) = \emptyset
\text{ for any } \mu \in \Lambda \text{ such that }
\mathbf a(\lambda) > \mathbf a(\mu)\}
\end{equation*}
and we define a partial order
$(\lambda_1,\varepsilon_1)\geq(\lambda_2,\varepsilon_2)$ on
$\varOmega$ by
$(\lambda_1,\varepsilon_1)>(\lambda_2,\varepsilon_2)$ if
$\lambda_1\rhd\lambda_2$, or $\lambda_1=\lambda_2$ and
$\varepsilon_1>\varepsilon_2$.
For a $( \lambda, \varepsilon ) \in \varOmega$, we define index sets
$I(\lambda,\varepsilon)$, $J(\lambda,\varepsilon)$ by
\begin{equation*}
\begin{array}{ll}
I(\lambda,\varepsilon)=
\left\{\begin{array}{ll}
\mathcal{T}_0^+(\lambda) & \qquad \text{ if } \varepsilon=0, \\[3mm]
\underset{
\mu \in \Lambda, ~
\bold{a}(\lambda)>
\bold{a}(\mu)
}{\bigcup}
\mathcal{T}_0(\lambda,\mu) & \qquad \text{ if } \varepsilon=1,
\end{array}\right.
\\[10mm]
J(\lambda,\varepsilon)=
\left\{\begin{array}{ll}
\mathcal{T}_0^+(\lambda) & \qquad \text{ if } \varepsilon=0, \\[3mm]
\mathcal{T}_0(\lambda) & \qquad \text{ if } \varepsilon=1,
\end{array}\right.
\end{array}
\end{equation*}
where $\mathcal{T}_0^+(\lambda) = \bigcup_{\mu \in \Lambda}
\mathcal{T}_0^{+}(\lambda, \mu )$. Then $I(\lambda,\varepsilon)$ and
$J(\lambda, \varepsilon)$ are not
empty for all $( \lambda, \varepsilon ) \in \varOmega$. Assume that
$( \lambda, \varepsilon ) \in \varOmega$. We define a subset $
\mathcal{C}^0(\lambda,\varepsilon)$ of $ \mathcal{S}^0(\Lambda)$ by
\begin{equation*}
\mathcal{C}^0(\lambda,\varepsilon)=\{ \varphi_{ST} \mid
(S,T) \in I(\lambda,\varepsilon)\times J(\lambda,\varepsilon) \}.
\end{equation*}
It is easy to see that
\begin{equation}\label{eq:the property of C^0(m,n)}
\text{the union} \underset{{(\lambda,\varepsilon) \in
\varOmega}}{\bigcup} \mathcal{C}^0(\lambda,\varepsilon) ~ \text{is
disjoint and is equal to the set} ~ \mathcal{C}^0(\Lambda).\\
\end{equation}
First, we need the following Lemma.
\begin{lem}[{\cite{M2}}]\label{lem:multiplication formula}
Suppose that $\lambda_i \in \Lambda^{+}$ and $\mu_i, \nu_i \in
\Lambda$ $(i=1,2)$. Assume that
$\varphi_{S_1T_1}$, $\varphi_{S_2T_2} \in
\mathcal{C}(\Lambda)$ where $S_i \in
\mathcal{T}_0(\lambda_i,\mu_i)$, $T_i \in
\mathcal{T}_0(\lambda_i,\nu_i)$. Then
\begin{equation*}
\varphi_{S_1T_1} \cdot \varphi_{S_2T_2} = \delta_{\nu_1
\mu_2} \cdot \underset{\substack{
\lambda \in \Lambda^{+}, ~ \lambda \unrhd \lambda_1 \text{and}
\lambda_2, \\
S \in \mathcal{T}_0(\lambda,\mu_1), ~ T
\in\mathcal{T}_0(\lambda,\nu_2)}}{\sum} r_{ST} \cdot \varphi_{ST}
\end{equation*}
where $r_{ST} \in R$, and $\delta_{\nu_1 \mu_2}$ is such that
$\delta_{\nu_1
\mu_2}=1$ if $\nu_1=\mu_2$ and it is zero otherwise.
\end{lem}
\begin{proof}
This lemma was shown by Mathas \cite[(2.8)]{M2}. However, since this
fact
itself is important for later
discussions, we give the proof here.
It is sufficient to consider the case where $\nu_1 = \mu_2$. For all
$\mu \in \Lambda$, by definition of
$\varphi_{ST}$, we may suppose that
$\varphi_{S_2T_2}(m_{\mu}) = \delta_{\mu
\nu_2}m_{S_2T_2}= \delta_{\mu \nu_2}
m_{\mu_2}h$ with some $h \in \mathscr{H}$. Since
$m_{S_iT_i} \in M^{\nu_i \ast} \cap
M^{\mu_i}$ $(i=1,2)$, we have
$(\varphi_{S_1T_1} \cdot
\varphi_{S_2T_2})(m_{\mu}) = \delta_{\mu
\nu_2}m_{S_1T_1}h \in M^{\nu_2 \ast} \cap
M^{\mu_1}$. By \cite[Proposition 6.3]{DJMa}, we deduce that
\begin{equation*}
m_{S_1T_1}h = \underset{\substack{
\lambda \in \Lambda^{+},\\
S \in \mathcal{T}_0(\lambda,\mu_1), ~ T \in
\mathcal{T}_0(\lambda,\nu_2)}}{\sum} r_{ST} \cdot m_{ST}
\end{equation*}
where $r_{ST} \in R$. Therefore,
\begin{equation*}
\varphi_{S_1T_1} \cdot \varphi_{S_2T_2}
=\underset{\substack{\lambda \in \Lambda^{+},\\
S \in \mathcal{T}_0(\lambda,\mu_1), ~ T \in
\mathcal{T}_0(\lambda,\nu_2)}}{\sum} r_{ST} \cdot \varphi_{ST}.
\end{equation*}
Note that the set $\{\varphi_{ST} \}$ is the cellular basis
for $\mathcal{S}(\Lambda)$ by \cite[Theorem 6.6]{DJMa}. Then by
the property of
cellular basis, the last sum is over $\lambda \in \Lambda^{+}$
with $\lambda \unrhd \lambda_1$ and $\lambda \unrhd \lambda_2$, $S
\in
\mathcal{T}_0(\lambda,\mu_1)$ and $T \in
\mathcal{T}_0(\lambda,\nu_2)$. This proves the lemma.
\end{proof}
For all $\lambda \in \Lambda^{+}$, $\mu, \nu \in
\Lambda$ with $\mathcal{T}_0(\lambda, \mu),
\mathcal{T}_0(\lambda, \nu )\neq
\emptyset$ and every $S \in \mathcal{T}_0(\lambda, \mu)$, $T \in
\mathcal{T}_0(\lambda, \nu)$, since definitions of $\varphi_{ST}$ and
$m_{ST}$
we
find that
\begin{equation*}
\begin{array}{ll}
(\varphi_{ST^{\lambda}}
\varphi_{T^{\lambda}T})(m_{{\mu}'})
& = {\delta}_{{\mu}' \nu} \cdot
\varphi_{ST^{\lambda}}(m_{T^{\lambda} T})
= {\delta}_{{\mu}' \nu} \cdot
\varphi_{ST^{\lambda}}(m_{\lambda})
\underset{
\substack{
\mathfrak{t} \in \text{Std}(\lambda)\\
{\nu}(\mathfrak{t}) = T
}
}{\sum} T_{d(\mathfrak{t})} \\
&= {\delta}_{{\mu}' \nu} \cdot m_{S T^{\lambda}}\underset{
\substack{
\mathfrak{t} \in \text{Std}(\lambda)\\
{\nu}(\mathfrak{t}) = T
}
}
{\sum} T_{d(\mathfrak{t})}
= {\delta}_{{\mu}' \nu} \cdot m_{ST}
\end{array}
\end{equation*}
where ${\mu}' \in \Lambda$. Therefore,
\begin{equation}\label{eq:decomposition of varphiST}
\varphi_{ST^{\lambda}} \varphi_{T^{\lambda}T} =
\varphi_{ST} \quad
(\text{for }
^{\forall}S \in \mathcal{T}_0(\lambda, \mu),
^{\forall}T \in \mathcal{T}_0(\lambda, \nu)).
\end{equation}
The next lemma is a sharper version of Lemma \ref{lem:multiplication
formula}.
\begin{lem}\label{lem:sharper multiplication formulae}
Suppose that $\lambda_i \in \Lambda^{+}$ and $\mu_i, \nu_i \in
\Lambda$ $(i=1,2)$, and let $ \nu_1 = \mu_2$. Assume that
$\varphi_{S_1T_1}$, $\varphi_{S_2T_2} \in
\mathcal{C}^0(\Lambda)$ where $S_i \in
\mathcal{T}_0(\lambda_i,\mu_i)$, $T_i \in
\mathcal{T}_0(\lambda_i,\nu_i)$. Then
\begin{equation*}\label{eq:sharper multiplication formulae}
\varphi_{S_1T_1} \cdot \varphi_{S_2T_2}
=
\begin{cases}
\underset{\varphi_{ST} \in
\mathcal{C}^0(\lambda_1,0)}{\sum}
r_{ST} \cdot\varphi_{ST}
+
\underset{
\lambda \rhd \lambda_1 ~
}{\sum}
\underset{\varphi_{ST} \in
\mathcal{C}^0(\lambda)}{\sum}
& r_{ST} \cdot
\varphi_{ST}, \\
& \text{if} ~ \varphi_{S_1T_1} \in
\mathcal{C}^0(\lambda_1,0),\\
& \\
\underset{
\lambda \unrhd \lambda_1 ~ }{\sum}
\underset{
\varphi_{ST} \in
\mathcal{C}^0(\lambda,1)}{\sum} r_{ST} \cdot
\varphi_{ST}, &
\text{if} ~ \varphi_{S_1T_1} \in
\mathcal{C}^0(\lambda_1,1),\\
\underset{
\lambda \unrhd \lambda_2 ~ }{\sum}
\underset{
\varphi_{ST} \in
\mathcal{C}^0(\lambda)}{\sum} r_{ST} \cdot
\varphi_{ST}, &
\text{if} ~ \varphi_{S_2T_2} \in
\mathcal{C}^0(\lambda_2,0),\\
\underset{
\lambda \unrhd \lambda_2 ~ }{\sum}
\underset{
\varphi_{ST} \in
\mathcal{C}^0(\lambda,1)}{\sum} r_{ST} \cdot
\varphi_{ST}, &
\text{if} ~ \varphi_{S_2T_2} \in
\mathcal{C}^0(\lambda_2,1), \\
\end{cases}
\end{equation*}
where $r_{ST} \in R$ and $\mathcal{C}^0(\lambda) =
\mathcal{C}^0(\lambda, 0) \cup \mathcal{C}^0(\lambda, 1)$ and all
$\lambda, S, T$ occurring in the above formulas are such that $\lambda
\in
\Lambda^{+}$,
and the
semistandard Tableaux $S, T$ with ${\rm{Type}}( S ) = \mu_{1}$ and
${\rm{Type}}( T ) = \nu_{2}$,
respectively.
Hence the $\mathcal{S}^0(\Lambda)$ is a subalgebra of
$\mathcal{S}(\Lambda)$.
\end{lem}
\begin{proof}
By Lemma \ref{lem:multiplication formula}, $\varphi_{S_1T_1} \cdot
\varphi_{S_2T_2}$ is a linear combination of
$\varphi_{ST}$, where $S \in
\mathcal{T}_0(\lambda,\mu_1)$, $T \in\mathcal{T}_0(\lambda,\nu_2)$
with $\lambda \in
\Lambda^{+}$, $\lambda \unrhd \lambda_1$ and $\lambda_2$.
First assume that $\varphi_{S_1T_1} \in
\mathcal{C}^0(\lambda_1,0)$. Then for
the above $\varphi_{ST}$, we have $\bold{a}(\lambda) > \bold{a}(\mu)$
if $\alpha(\mu) \neq \alpha(\nu_{2})$ since $\varphi_{S_2T_2} \in
\mathcal{C}^0(\Lambda)$ and $\lambda
\unrhd \lambda_2$ (Note that $\mu \unrhd \nu \Rightarrow
\bold{a}(\mu) \geq \bold{a}(\nu)$ for $\mu, \nu \in
\Lambda$. cf. \eqref{eq:A certain fact of r-compositions}). So
$\varphi_{ST} \in
\mathcal{C}^0(\Lambda)$, hence $\varphi_{ST} \in
\mathcal{C}^0(\lambda,0) \cup \mathcal{C}^0(\lambda,1)$. Moreover, the
case $\varphi_{ST} \in
\mathcal{C}^0(\lambda_1,1)$ cannot happen. In fact, if it happens,
then
$\alpha(\lambda_1) \neq \alpha(\mu_1)$ which contradicts our
assumption that $\varphi_{S_1T_1} \in
\mathcal{C}^0(\lambda_1,0)$. Therefore, the first equality holds.
The second equality is easy. In fact, assume that $\varphi_{S_1T_1} \in
\mathcal{C}^0(\lambda_1,1)$. Then $\varphi_{ST}$, as in the previous
case, are elements in the $\mathcal{C}^0(\lambda,1)$ since
$\lambda \unrhd \lambda_1$ and $\bold{a}(\lambda_1) >
\bold{a}(\mu_1)$. Hence we obtain the second one.
Next assume that $\varphi_{S_2T_2} \in
\mathcal{C}^0(\lambda_2,0)$. If $\alpha(\mu_1) \neq \alpha(\nu_2)$,
then the definition of
$\mathcal{C}^0(\lambda_2,0)$ and our assumption $\mu_2 =\nu_1$ implies
that $\alpha(\mu_1) \neq \alpha(\nu_1)$. It follows that
$\bold{a}(\lambda_1)
> \bold{a}(\mu_1)$ since $\varphi_{S_1T_1} \in
\mathcal{C}^0(\Lambda)$. Thus we have $\bold{a}(\lambda) >
\bold{a}(\mu_1)$ by $\lambda \unrhd \lambda_1$. This shows that
$\varphi_{ST}$ is in
$\mathcal{C}^0(\Lambda)$. Therefore $\varphi_{ST} \in
\mathcal{C}^0(\lambda,0) \cup \mathcal{C}^0(\lambda,1)$ and the third
equality holds.
Finally, suppose that $\varphi_{S_2T_2} \in
\mathcal{C}^0(\lambda_2,1)$. If
$\alpha(\mu_1) = \alpha(\nu_1)$ then $\bold{a}(\lambda_2)
> \bold{a}(\mu_1)$ since $\varphi_{S_2T_2} \in
\mathcal{C}^0(\lambda_2,1)$ and $\mu_2 = \nu_1$ by our
assumption. Hence $\bold{a}(\lambda) > \bold{a}(\mu_1)$, since
$\lambda \unrhd \lambda_2$. On the other hand, if $\alpha(\mu_1) \neq
\alpha(\nu_1)$ then also $\bold{a}(\lambda) > \bold{a}(\mu_1)$ and
$\varphi_{S_1T_1} \in
\mathcal{C}^0(\Lambda)$ and $\lambda \unrhd \lambda_1$. This argument
means that $\varphi_{ST} \in \mathcal{C}^0(\lambda,1)$ and
hence the fourth equality holds. The lemma is proved.
\end{proof}
\addtocounter{subsection}{3}
\subsection{}\label{standardly based subsection}
Let $\mu \in \Lambda$. We define $\varphi_{\mu} \in
\mathcal{S}(\Lambda)$ as the identity map on $M^{\mu}$
and zero map on $M^{\kappa}$ with $\kappa \neq
\mu$. Moreover, let $1_{\mathcal{S}(\Lambda)}$ be the
identity element of $\mathcal{S}(\Lambda)$. Then, from
the definition, we can
write $1_{\mathcal{S}(\Lambda)} = \sum_{\mu \in
\Lambda} \varphi_{\mu}$. On the
other hand, since $\varphi_{\mu}(m_{\mu}) =
m_{\mu} \in M^{\mu \ast} \cap
M^{\mu}$, we have
\begin{equation*}
\varphi_{\mu}(m_{\mu}) = \underset{\substack{
S,T \in \mathcal{T}_0(\lambda,\mu)\\
\lambda \in \Lambda^{+}}}{\sum} r_{S,T} \cdot m_{ST}
\qquad (r_{S,T} \in R)
\end{equation*}
by \cite[Proposition 6.3]{DJMa}. This shows that
\begin{equation}\label{eq:identity map on M^mu_natural}
\varphi_{\mu} = \underset{\substack{
S,T \in \mathcal{T}_0(\lambda,\mu)\\
\lambda \in \Lambda^{+}}}{\sum} r_{S,T} \cdot
\varphi_{ST} \qquad (\text{ for any } \mu \in
\Lambda ).
\end{equation}
Thus all the $\varphi_{ST}$ in the right hand side are contained in
$\mathcal{C}^0(\Lambda)$. Hence we have
\begin{equation}\label{id:identity element of S^0}
1_{\mathcal{S}(\Lambda)} = \underset{ \mu \in \Lambda}{\sum} \quad
\underset{\substack{
S,T \in \mathcal{T}_0(\lambda,\mu)\\
\lambda \in \Lambda^{+}}}{\sum} r_{S, T} \cdot
\varphi_{ST} \in \mathcal{S}^0(\Lambda).
\end{equation}
For any $(\lambda, \varepsilon) \in \varOmega$, we define by
$\mathcal{S}_{0}^{\vee (\lambda, \varepsilon)} =
\mathcal{S}^0(\Lambda)(>(\lambda, \varepsilon))$ the $R$-submodule of
$\mathcal{S}^0(\Lambda)$ spanned by
$\varphi_{UV}$ where $(U,V) \in I({\lambda}',{\varepsilon}')
\times J({\lambda}',{\varepsilon}')$ for some
$({\lambda}',{\varepsilon}') \in \varOmega$ with
$({\lambda}',{\varepsilon}') >
(\lambda, \varepsilon)$. Note that $\mathcal{S}^0(\Lambda) \cap
\mathcal{S}^{\vee \lambda} =
\mathcal{S}_{0}^{\vee (\lambda,1)}$ for every $\lambda \in
\Lambda^{+}$. Similarly, we define
$\mathcal{S}^0(\Lambda)(\geq (\lambda, \varepsilon))$ as the
$R$-submodule spanned by $\varphi_{UV}$ with
$({\lambda}',{\varepsilon}') \geq (\lambda, \varepsilon)$.
We can now state.
\addtocounter{thm}{1}
\begin{thm}\label{thm:standardly based}
The subalgebra $\mathcal{S}^0(\Lambda)$ is standardly based on
$(\varOmega, \geq)$ with standard basis $\mathcal{C}^0(\Lambda)$, that
is,
$({\rm{i}})$ The union $\underset{{(\lambda,\varepsilon) \in
\varOmega}}{\bigcup} \mathcal{C}^0(\lambda,\varepsilon) =
\mathcal{C}^0(\Lambda)$ is
disjoint and forms an $R$-basis for $\mathcal{S}^0(\Lambda)$.
$({\rm{ii}})$ For any $\varphi \in \mathcal{S}^0(\Lambda)$,
$\varphi_{ST} \in \mathcal{C}^0(\lambda,\varepsilon)$, we have
\begin{equation}\label{eq:multiplication formulae of S^0(m,n)}
\begin{array}{c}
\varphi \cdot \varphi_{ST} \equiv \underset{S'
\in
I(\lambda,\varepsilon)}{\sum}f_{S',(\lambda,\varepsilon)}(\varphi,S)
\cdot \varphi_{S'T} \mod
\mathcal{S}_{0}^{\vee (\lambda, \varepsilon)}\\
\varphi_{ST} \cdot \varphi \equiv \underset{T'
\in
J(\lambda,\varepsilon)}{\sum}f_{(\lambda,\varepsilon),T'}(T,\varphi)
\cdot \varphi_{ST'} \mod \mathcal{S}_{0}^{\vee (\lambda, \varepsilon)},
\end{array}
\end{equation}
where $\varphi_{S'T}, \varphi_{ST'} \in
\mathcal{C}^0(\Lambda)$ and $f_{S',(\lambda,\varepsilon)}(\varphi,S)$,
$f_{(\lambda,\varepsilon),T'}(T,\varphi) \in R$ are
independent of $T$ and $S$, respectively.
\end{thm}
\begin{proof}
The first condition (i) is immediate from \eqref{eq:the property of
C^0(m,n)}. We
show (ii). Take $\varphi_{ST} \in
\mathcal{C}^0(\lambda,\varepsilon)$ and $\varphi \in
\mathcal{S}^0(\Lambda)$. Note that $\varphi_{ST}$ is an element in
$\mathcal{C}(\Lambda)$ which is the cellular basis for
$\mathcal{S}(\Lambda)$ and $\varphi$ is an element in
$\mathcal{S}(\Lambda)$. Hence, by the property of
cellular
basis, $\varphi_{ST} \cdot \varphi$ can be written as
\begin{equation*}
\varphi_{ST} \cdot \varphi = \underset{T' \in
\mathcal{T}_0(\lambda) }{\sum}r_{T'} \cdot \varphi_{ST'} +
\underset{\substack{{\lambda}' \in \Lambda^{+}, ~ {\lambda}' \rhd
\lambda\\
U,V \in \mathcal{T}_0({\lambda}')}}{\sum}r_{UV} \cdot \varphi_{UV}
\end{equation*}
with $r_{T'}$, $r_{UV} \in R$, where $r_{T'}$ does not depend on
$S$. Then by rewriting $\varphi$ as
a linear combination of the basis elements in
$\mathcal{C}^0(\Lambda)$ and by combining the formulas in Lemma
\ref{lem:sharper multiplication formulae}, we obtain the second
equality. By a similar argument, applying the third and fourth
formulas in Lemma \ref{lem:sharper multiplication formulae} instead,
the
first equality also holds. The theorem follows.
\end{proof}
\addtocounter{subsection}{1}
\subsection{}(cf. {\cite{DR1}}).\label{def:the definition of
standardly (full-)based algebra}
Let $A$ be a standardly based algebra on $\Lambda$ as given in
\ref{def:standardly based}. For any
$\lambda \in \Lambda$, let $f_{\lambda}: J(\lambda) \times I(\lambda)
\rightarrow R$ be a function, whose value $f_{\lambda}(j,i')$ at
$(j,i') \in J(\lambda) \times I(\lambda)$ is defined by
\begin{equation*}
a_{ij}^{\lambda} a_{i'j'}^{\lambda} \equiv f_{\lambda}(j,i')
a_{ij'}^{\lambda} \mod A(>\lambda).
\end{equation*}
The function $f_{\lambda}$ induces a bilinear form ${\beta}_{\lambda}:
A^{\lambda} \times A^{\lambda} \rightarrow R$ such that
${\beta}_{\lambda}(a_{ij}^{\lambda},a_{i'j'}^{\lambda}) =
f_{\lambda}(j',i)$, where $A^{\lambda}$ is the free $R$-submodule of
$A$ spanned by $a_{ij}^{\lambda}$ for all $(i,j) \in I(\lambda) \times
J(\lambda)$. We say $A$ is a standardly \emph{full}-based algebra if
$ \text{Im}({\beta}_{\lambda})=R$ for all $\lambda \in \Lambda$.
The following result has been proved by Du and Rui
\cite[(3.2.1), (4.2.7)]{DR2}.
\addtocounter{thm}{1}
\begin{thm}[{\cite{DR2}}]\label{th:quasi-hereditary}
Suppose that $A$ is a standardly based algebra.
$(\rm{i})$ Let $R$ be a commutative Noetherian ring. If $A$ is a
standardly
full-based algebra, then $A$ is a quasi-hereditary algebra over $R$ in
the sense of \cite{CPS}.
$(\rm{ii})$ If $R$ is a commutative local Noetherian ring, then $A$ is
split
quasi-hereditary if and only if $A$ is a standardly full-based
algebra.
\end{thm}
\addtocounter{subsection}{1}
\subsection{}\label{sub:dagger tableau}
We shall show that the $\mathcal{S}^0(\Lambda)$ turns out to be a
standardly full-based algebra under a certain condition. For
$(\lambda,1)
\in \varOmega$, put $\lambda = (\lambda^{(1)}, \ldots
,\lambda^{(r)})$, $ \lambda^{(i)} = (\lambda^{(i)}_1, \ldots
,\lambda^{(i)}_{n_i})$ and take the smallest positive integer $l$ such
that $| \lambda^{(l)}| \neq 0 $ $(1 \leq l \leq r)$. Note that $l \neq
r$ since $(\lambda, 1) \in \varOmega$. Then, one can
define an $r$-partition $\lambda^{\dagger}$ associated to $\lambda$
as follows,
\begin{equation*}
\begin{array}{l}
\lambda^{\dagger} = (\lambda^{\dagger (1)}, \ldots ,\lambda^{\dagger
(r)}), \\[3mm]
\lambda^{\dagger (l)} = \underset{|\lambda^{(l)}|-1 ~
\text{times}}{(\underbrace{1,1, \ldots ,1})}, ~~ \lambda^{\dagger
(l+1)} = \underset{|\lambda^{(l+1)}|+1 ~
\text{times}}{(\underbrace{1,1, \ldots ,1})}, \\[10mm]
\lambda^{\dagger (j)} = \underset{|\lambda^{(j)}| ~
\text{times}}{(\underbrace{1,1, \ldots ,1})} \quad (j \neq l,
l+1). \qquad \\[10mm]
(\text{If } |\lambda^{(j)}| = 0 ~~ \text{then }
\lambda^{\dagger
(j)} ~ \text{is the empty partition.})
\end{array}
\end{equation*}
We remark that $\lambda^{\dagger}$ is defined only when $(\lambda, 1)
\in
\varOmega$, and it satisfies the property
\begin{equation}\label{geq:the property of lambda dagger}
\bold{a}(\lambda)> \bold{a}(\lambda^{\dagger}).
\end{equation}
Moreover, let $T^{\lambda \dagger}$ be a semistandard
$\lambda$-Tableau of type $\lambda^{\dagger}$ in which the entries are
laid in increasing order from ``left-upper'' to right along the
rows. (The ``left-upper'' means that the left has priority over the
upper.) For example, if
$\lambda=((3,2,1),(2,2))$ then $\lambda^{\dagger} = ( ( 1^5 ), ( 1^5 )
)$ and
\begin{center}
$T^{\lambda \dagger} = \biggl($
\begin{tabular}{ccc}
\begin{minipage}{13mm}
\setlength{\unitlength}{1.2mm}
\begin{picture}(21,6)
\put(0,0){\line(1,0){5}}
\put(0,2){\line(1,0){10}}
\put(0,4){\line(1,0){15}}
\put(0,6){\line(1,0){15}}
\put(0,0){\line(0,1){6}}
\put(5,0){\line(0,1){6}}
\put(10,2){\line(0,1){4}}
\put(15,4){\line(0,1){2}}
\put(0.5,3){\makebox(4,4){$\scriptscriptstyle{(1,1)}$}}
\put(5.5,3){\makebox(4,4){$\scriptscriptstyle{(2,1)}$}}
\put(10.5,3){\makebox(4,4){$\scriptscriptstyle{(3,1)}$}}
\put(0.5,1){\makebox(4,4){$\scriptscriptstyle{(4,1)}$}}
\put(5.5,1){\makebox(4,4){$\scriptscriptstyle{(5,1)}$}}
\put(0.5,-1){\makebox(4,4){$\scriptscriptstyle{(1,2)}$}}
\end{picture}
\end{minipage}
& \quad , &
\begin{minipage}{13mm}
\setlength{\unitlength}{1.2mm}
\begin{picture}(21,6)
\put(0,2){\line(1,0){10}}
\put(0,4){\line(1,0){10}}
\put(0,6){\line(1,0){10}}
\put(0,2){\line(0,1){4}}
\put(5,2){\line(0,1){4}}
\put(10,2){\line(0,1){4}}
\put(0.5,3){\makebox(4,4){$\scriptscriptstyle{(2,2)}$}}
\put(5.5,3){\makebox(4,4){$\scriptscriptstyle{(3,2)}$}}
\put(0.5,1){\makebox(4,4){$\scriptscriptstyle{(4,2)}$}}
\put(5.5,1){\makebox(4,4){$\scriptscriptstyle{(5,2)}$}}
\end{picture}
\end{minipage}
\end{tabular}
$\biggl)$.
\end{center}
Thus, if $\mathfrak{s} \in \text{Std}(\lambda)$ and
$\lambda^{\dagger}(\mathfrak{s}) = T^{\lambda \dagger}$ then
$\mathfrak{s} = \mathfrak{t}^{\lambda}$.
Finally, let
\begin{equation*}
P_n(q,Q_1, \ldots ,Q_r) = \prod_{i=1}^{n} (1+q+ \cdots +q^{i-1}) \cdot
\prod_{j=1}^{r-1} \prod_{-n < k < n} (q^{2k} Q_j - Q_{j+1}).
\end{equation*}
We have the following result.
\addtocounter{thm}{1}
\begin{prop}
Assume that $R$ is a commutative Noetherian ring. Suppose that
$P_n(q,Q_1,
\ldots ,Q_r) \in R$ is
invertible, and $\lambda^{\dagger} \in \Lambda^{+}$ for any $(
\lambda, 1 ) \in \varOmega$. Then $\mathcal{S}^0(\Lambda)$
is the standardly full-based, and
hence is quasi-hereditary over $R$.
\end{prop}
\begin{proof}
For all $(\lambda, \varepsilon) \in \varOmega$, it is enough to show
that
there exists some $T \in I(\lambda, \varepsilon) \cap
J(\lambda, \varepsilon)$ such that $(\varphi_{TT})^2 \equiv
r \cdot \varphi_{TT} \mod
\mathcal{S}_{0}^{\vee (\lambda,\varepsilon)}$ with $r \in R$
invertible. If $(\lambda, 0) \in \varOmega$ then one can take $T =
T^{\lambda}$. Since
$\varphi_{T^{\lambda}T^{\lambda}}$ is the identity map on
$M^{\lambda}$, this case is immediate.
Assume that $(\lambda, 1) \in \varOmega$. First, we claim that
\begin{equation}\label{eq:applying L to m}
m_{\lambda} \cdot (L_{|\lambda^{(l)}|} - Q_{l+1})
\equiv (q^{2k}Q_{l} - Q_{l+1}) \cdot m_{\lambda}
\mod \mathscr{H}^{\lambda}
\end{equation}
where $l$ is the integer attached to $\lambda$ as in \ref{sub:dagger
tableau}, and $k$ is
some integer such that $|k| < n$ and $\mathscr{H}^{\lambda}$ is as in
\ref{sub:the definition of H^lambda}. In fact,
applying James-Mathas' result \cite[Proposition 3.7]{JM} for
$\mathfrak{t}^{\lambda}$ and $|\lambda^{(l)}|$, we see that
$m_{\lambda} \cdot L_{|\lambda^{(l)}|} \equiv q^{2k}Q_{l}
\cdot m_{\lambda} \mod \mathscr{H}^{\lambda}$ and hence
\eqref{eq:applying L to m} holds.
Now, by \eqref{geq:the property
of lambda dagger} and by our assumption, one can choose the element
$\varphi_{T^{\lambda \dagger} T^{\lambda \dagger}} \in
\mathcal{C}^0(\lambda, 1)$. For any ${\mu}' \in
\Lambda$, we have $\varphi_{T^{\lambda \dagger} T^{\lambda
\dagger}}(m_{{\mu}'}) =
{\delta}_{{\mu}' {\lambda}^{\dagger}}
\cdot m_{{\mathfrak{t}}^{\lambda} {\mathfrak{t}}^{\lambda}}
={\delta}_{{\mu}' {\lambda}^{\dagger}} \cdot m_{\lambda}$. Hence
\begin{equation}\label{eq:calculation 1}
\begin{array}{ll}
(\varphi_{T^{\lambda \dagger} T^{\lambda \dagger}})^2(m_{{\mu}'})
&= {\delta}_{{\mu}'
{\lambda}^{\dagger}} (\varphi_{T^{\lambda \dagger}
T^{\lambda \dagger}})(m_{\lambda})
={\delta}_{{\mu}' {\lambda}^{\dagger}}
(\varphi_{T^{\lambda \dagger}
T^{\lambda \dagger}})(u^{+}_{\lambda}
x_{\lambda} )\\
&= {\delta}_{{\mu}' {\lambda}^{\dagger}}
(\varphi_{T^{\lambda \dagger} T^{\lambda
\dagger}})(m_{{\lambda}^{\dagger}} (L_{|{\lambda}^{(l)}|} -
Q_{l+1}) x_{\lambda}) \\
&= {\delta}_{{\mu}' {\lambda}^{\dagger}} \cdot m_{\lambda}
(L_{|{\lambda}^{(l)}|} - Q_{l+1}) x_{\lambda}
\end{array}
\end{equation}
where the second and the third equality follows from the definition of
$m_{\lambda}$ and $m_{{\lambda}^{\dagger}}$,
respectively. Note that $ m_{\lambda} x_{\lambda} =
P(q) m_{\lambda}$ where $P(q)$
is some products of Poincar\'e
polynomials. Therefore, by \eqref{eq:applying L to m}, we have
\begin{equation*}
(\varphi_{T^{\lambda \dagger} T^{\lambda
\dagger}})^2(m_{{\mu}'})=
\left\{\begin{array}{ll}
P(q) \cdot (q^{2k}Q_{l} - Q_{l+1}) m_{\lambda} + h & \text{ if
}{\mu}' = {\lambda}^{\dagger} \\
0 & \text{ otherwise },\\
\end{array}\right.
\end{equation*}
where $h \in \mathscr{H}^{\lambda}$.
Now, by definition, $({\varphi}_{T^{\lambda \dagger}
T^{\lambda \dagger}})^2 (m_{{\lambda}^{\dagger}}) \in
M^{{\lambda}^{\dagger} \ast} \cap
M^{{\lambda}^{\dagger}}$.
So, by \cite[Proposition 6.3]{DJMa},
$({\varphi}_{T^{\lambda \dagger} T^{\lambda \dagger}})^2
(m_{{\lambda}^{\dagger}})
= \sum r_{UV} m_{UV}$ where $r_{UV} \in R$ and the sum is over
$U, V \in \mathcal{T}_0(\alpha, {\lambda}^{\dagger})$ for some $\alpha
\in \Lambda^{+}$. Hence, by \eqref{eq:calculation 1} and by a property
of cellular basis, we deduce that
\begin{equation*}
\begin{array}{ll}
P(q) \cdot (q^{2k}Q_{l} - Q_{l+1}) m_{\lambda}
& \equiv ({\varphi}_{T^{\lambda \dagger} T^{\lambda \dagger}})^2
(m_{{\lambda}^{\dagger}}) \mod
\mathscr{H}^{\lambda} \\
&= m_{\lambda} (L_{|{\lambda}^{(l)}|} -
Q_{l+1}) x_{\lambda}
= m_{T^{\lambda \dagger} T^{\lambda \dagger}} (L_{|{\lambda}^{(l)}|} -
Q_{l+1}) x_{\lambda}\\
&= \sum_{T' \in \mathcal{T}_0(\lambda,
{\lambda}^{\dagger})} r_{T'} m_{T^{\lambda \dagger} T'} +
\sum_{U', V'} r_{U'V'}
m_{U'V'}
\end{array}
\end{equation*}
where $r_{T'}, r_{U'V'} \in R$ and the last sum is over $U', V' \in
\mathcal{T}_0(\alpha, {\lambda}^{\dagger})$ for $\alpha \in
\Lambda^{+}$ such that $\alpha \rhd \lambda$. Comparing the
coefficient of $m_{T^{\lambda \dagger} T'}$ on both sides reveals
that
$r_{T^{\lambda \dagger}} = P(q) \cdot (q^{2k}Q_{l} - Q_{l+1})$
and $r_{T'} = 0$ unless $T' = T^{\lambda \dagger}$. Thus,
$({\varphi}_{T^{\lambda \dagger} T^{\lambda \dagger}})^2
(m_{{\lambda}^{\dagger}}) = P(q) \cdot (q^{2k}Q_{l} -
Q_{l+1}) ({\varphi}_{T^{\lambda \dagger} T^{\lambda \dagger}})
(m_{{\lambda}^{\dagger}}) + \sum_{U', V'} r_{U'V'}
{\varphi}_{U'V'} (m_{{\lambda}^{\dagger}})$. Consequently,
$(\varphi_{T^{\lambda \dagger} T^{\lambda \dagger}})^2
\equiv P(q) \cdot
(q^{2k}Q_{l} - Q_{l+1}) \cdot
\varphi_{T^{\lambda \dagger} T^{\lambda \dagger}} \mod
\mathcal{S}_{0}^{\vee (\lambda, 1)}$. By our assumption, $P(q)
\cdot (q^{2k}Q_{l} - Q_{l+1})$ is invertible and the proposition
follows.
\end{proof}
\medskip
\section{A relationship between $\mathcal{S}^0(\Lambda)$ and
$\mathcal{S}(\Lambda)$}
\subsection{}
We shall describe a relationship between the original cyclotomic
$q$-Schur
algebra $\mathcal{S}(\Lambda)$ and its $R$-subalgebra
$\mathcal{S}^0(\Lambda)$.
Because $\mathcal{S}(\Lambda)$ is equipped with the involution $*$,
we can define subsets $\mathcal{S}^0(\Lambda)^*$,
$\mathcal{C}^0(\Lambda)^*$ of
$\mathcal{S}(\Lambda)$ by
\begin{equation*}
\mathcal{S}^0(\Lambda)^* = \{ \varphi^* \mid \varphi \in
\mathcal{S}^0(\Lambda) \}, \quad \mathcal{C}^0(\Lambda)^* = \{
\varphi^{*}_{ST} \mid \varphi_{ST} \in \mathcal{C}^0(\Lambda) \},
\end{equation*}
respectively. Then $\mathcal{S}^0(\Lambda)^*$ is an $R$-subalgebra of
$\mathcal{S}(\Lambda)$. Moreover, observe that
$\mathcal{S}^0(\Lambda)^*$ turns out to be a standardly based
algebra on $\mathcal{C}^0(\Lambda)^*$ by applying the involution
$*$ to Theorem \ref{thm:standardly based}.
\addtocounter{thm}{1}
\begin{prop}
$
\mathcal{S}(\Lambda) = \mathcal{S}^0(\Lambda) \cdot
\mathcal{S}^0(\Lambda)^* = \sum_{\lambda \in \Lambda^{+}}
\mathcal{S}^0(\Lambda) \varphi_{T^{\lambda} T^{\lambda}}
\mathcal{S}^0(\Lambda)^*
$.
\end{prop}
\begin{proof}
For all $\lambda \in \Lambda^{+}$, $\mu, \nu \in
\Lambda$ with $\mathcal{T}_0(\lambda, \mu)$, $\mathcal{T}_0(\lambda,
\nu) \neq \emptyset$ and for any $S \in \mathcal{T}_0(\lambda, \mu)$, $T
\in
\mathcal{T}_0(\lambda, \nu)$, we certainly have
$\varphi_{ST^{\lambda}} \in
\mathcal{C}^0(\Lambda)$ and hence $(\varphi_{TT^{\lambda}})^* =
\varphi_{T^{\lambda}T} \in
{\mathcal{C}^0(\Lambda)}^*$. Moreover, by using
\eqref{eq:decomposition of varphiST}, we see that all the basis
$\mathcal{C}(\Lambda)$ for
$\mathcal{S}(\Lambda)$ is contained in $\mathcal{S}^0(\Lambda) \cdot
\mathcal{S}^0(\Lambda)^*$. Hence $\mathcal{S}(\Lambda) =
\mathcal{S}^0(\Lambda) \cdot
\mathcal{S}^0(\Lambda)^*$. Finally, for any $S \in
\mathcal{T}_0(\lambda, \mu)$, $T \in \mathcal{T}_0(\lambda, \nu )$, we
see that $\varphi_{ST^{\lambda}}
\varphi_{T^{\lambda}T} = \varphi_{ST^{\lambda}}
\varphi_{T^{\lambda} T^{\lambda}}
\varphi_{T^{\lambda}T} \in \mathcal{S}^0(\Lambda)
\varphi_{T^{\lambda} T^{\lambda}}
\mathcal{S}^0(\Lambda)^*$, so the last equality follows.
\end{proof}
Next we introduce the Weyl module for
$\mathcal{S}^0(\Lambda)$. By \eqref{eq:multiplication
formulae of S^0(m,n)} in Theorem
\ref{thm:standardly based}, it is easy to see that $R$-modules
$\mathcal{S}^0(\Lambda)(\geq (\lambda, \varepsilon))$ and
$\mathcal{S}_{0}^{\vee (\lambda, \varepsilon)} =
\mathcal{S}^0(\Lambda)(> (\lambda, \varepsilon))$ are two-sided
ideals of $\mathcal{S}^0(\Lambda)$. Fix a $(\lambda, \varepsilon) \in
\varOmega$. For $S \in I(\lambda, \varepsilon)$, we define the Weyl
module $Z^{(\lambda, \varepsilon)}_{S}$ for
$\mathcal{S}^0(\Lambda)$ by the $R$-submodule of $\{
\mathcal{S}^0(\Lambda)(\geq (\lambda, \varepsilon)) \} / \{
\mathcal{S}^0(\Lambda)(> (\lambda, \varepsilon)) \}$ with basis $\{
\varphi_{ST} + \mathcal{S}_{0}^{\vee (\lambda, \varepsilon)} \mid T
\in J(\lambda, \varepsilon)\}$. Moreover, by \eqref{eq:multiplication
formulae of S^0(m,n)}, we see that $Z^{(\lambda, \varepsilon)}_{S}$
is the right
$\mathcal{S}^0(\Lambda)$-module and the action of
$\mathcal{S}^0(\Lambda)$ on $Z^{(\lambda, \varepsilon)}_{S}$ is
independent of the choice of $S$, i.e, $Z^{(\lambda,
\varepsilon)}_{S_1} \simeq Z^{(\lambda,\varepsilon)}_{S_2}$ for
all $S_1, S_2 \in I(\lambda, \varepsilon)$.
However, since $T^{\lambda}$ is not an element in
$I(\lambda, 1)$ for $(\lambda, 1) \in \varOmega$, one should
pay attention that
there is no ``canonical''-Weyl module for the case $(\lambda,
1)$. (That is, we can not define $Z^{(\lambda, 1)}_{T^{\lambda}}$.)
For the convenience sake let $Z^{(\lambda,
0)} = Z^{(\lambda,
0)}_{T^{\lambda}}$ and put
$\varphi^{0}_{T} = \varphi_{T^{\lambda} T} +
\mathcal{S}_{0}^{\vee (\lambda, \varepsilon)}$ for any $T \in
J(\lambda, 0) = \mathcal{T}_0^{+}(\lambda)$. We can now prove the
following.
\begin{thm}\label{thm:tensor theorem}
Let $\lambda \in \Lambda^{+}$. Then there exists an isomorphism of
$\mathcal{S}(\Lambda)$-modules
\begin{equation*}
Z^{(\lambda, 0)} \otimes_{\mathcal{S}^0(\Lambda)}
\mathcal{S}(\Lambda) \simeq W^{\lambda}
\end{equation*}
which maps $\varphi^{0}_{T^{\lambda}} \psi \otimes \varphi$ to
$\varphi_{T^{\lambda}} \psi \varphi$.
\end{thm}
\begin{proof}
First we note that $Z^{(\lambda, 0)} = {\varphi}^{0}_{T^{\lambda}}
{\mathcal{S}}^0(\Lambda)$. In fact, since $\mathcal{S}_{0}^{\vee
(\lambda, 0)}$ is a two-sided ideal of $\mathcal{S}(\Lambda)$, for
any $T \in
\mathcal{T}^{+}_0(\lambda, \mu)$ with $\mu \in
\Lambda$, we have ${\varphi}^{0}_{T^{\lambda}} \cdot
\varphi_{T^{\lambda} T} = {\varphi}^{0}_{T}$ by
\eqref{eq:decomposition of varphiST}. Now we can write
$Z^{(\lambda, 0)} \otimes_{\mathcal{S}^0(\Lambda)}
\mathcal{S}(\Lambda) = ({\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1) \cdot
\mathcal{S}(\Lambda)$. On the other hand, we can also
write $W^{\lambda} =
{\varphi}_{T^{\lambda}} \cdot
\mathcal{S}(\Lambda)$ (note that
${\varphi}_{T^{\lambda}} =
{\varphi}_{T^{\lambda} T^{\lambda}} +
\mathcal{S}^{\vee \lambda}$). Hence it is enough
to find two $\mathcal{S}(\Lambda)$-homomorphisms $f:
{\varphi}_{T^{\lambda}} \cdot
\mathcal{S}(\Lambda) \rightarrow ({\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1) \cdot
\mathcal{S}(\Lambda)$ and $g: ({\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1) \cdot
\mathcal{S}(\Lambda) \rightarrow
{\varphi}_{T^{\lambda}} \cdot
\mathcal{S}(\Lambda)$ such that
$f({\varphi}_{T^{\lambda}}) = {\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1$ and
$g({\varphi}^{0}_{T^{\lambda}} \otimes_{\mathcal{S}^0(\Lambda)} 1) =
{\varphi}_{T^{\lambda}}$.
We first define $f$. Consider the
$\mathcal{S}(\Lambda)$-homomorphism $\widetilde{f}:
{\varphi}_{T^{\lambda} T^{\lambda}} \cdot
\mathcal{S}(\Lambda) \rightarrow ({\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1) \cdot
\mathcal{S}(\Lambda)$ satisfying
$\widetilde{f}({\varphi}_{T^{\lambda} T^{\lambda}}) =
{\varphi}^{0}_{T^{\lambda}} \otimes_{\mathcal{S}^0(\Lambda)}
1$. Note that
$\widetilde{f}$ is a well defined homomorphism. To see this we take
an element ${\varphi} \in
\mathcal{S}(\Lambda)$ such that
${\varphi}_{T^{\lambda} T^{\lambda}} {\varphi}
= 0$. Write ${\varphi} = \sum r_{{\lambda}', ST} \cdot
{\varphi}^{{\lambda}'}_{ST}$ for some $r_{{\lambda}', ST} \in
R$ where ${\varphi}^{{\lambda}'}_{ST} =
{\varphi}_{ST}$ with $S,T \in \mathcal{T}_0({\lambda}')$
and the sum is taken over $r$-partitions ${\lambda}' \in
\Lambda^{+}$ and semistandard tableaux $S,T \in
\mathcal{T}_0({\lambda}')$. Then, since
${\varphi}_{T^{\lambda} T^{\lambda}}$ is the identity
map on $M^{\lambda}$ and is zero on
$M^{\kappa}$ for $\lambda \neq \kappa \in
\Lambda$ and $\{ {\varphi}^{{\lambda}'}_{ST}
\}$ is the basis for $\mathcal{S}(\Lambda)$,
$r_{{\lambda}', ST} = 0$ whenever
$S \in \mathcal{T}_0({\lambda}', \lambda), T \in
\mathcal{T}_0({\lambda}', \nu)$ with ${\lambda}' \in
\Lambda^{+}$ and $\nu \in
\Lambda$. It follows that ${\varphi}^{\lambda'}_{S T^{\lambda'}} \in
\mathcal{S}^{0}(\Lambda)$ for all $S \in \mathcal{T}_0({\lambda}')$
such
that $r_{\lambda', S T} \neq 0$. Hence, by using
\eqref{eq:decomposition
of varphiST}, we have
\begin{equation*}
\begin{array}{ll}
{\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi}
&= {\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} ( \sum
r_{{\lambda}', ST} \cdot {\varphi}^{{\lambda}'}_{ST} ) \\[3mm]
&={\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} ( \sum
r_{{\lambda}', ST} \cdot {\varphi}^{{\lambda}'}_{S
T^{{\lambda}'}} {\varphi}^{{\lambda}'}_{T^{{\lambda}'} T} ) \\[3mm]
&= \sum
r_{{\lambda}', ST} ( {\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi}^{{\lambda}'}_{S T^{{\lambda}'}} \cdot
{\varphi}^{{\lambda}'}_{T^{{\lambda}'} T} ) \\[3mm]
&=\sum
r_{{\lambda}', ST} ( {\varphi}^{0}_{T^{\lambda}}
\cdot {\varphi}^{{\lambda}'}_{S T^{{\lambda}'}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi}^{{\lambda}'}_{T^{{\lambda}'} T} )
\end{array}
\end{equation*}
where all sums are taken over ${\lambda}' \in
\Lambda^{+}$ and $S,T \in
\mathcal{T}_0({\lambda}')$. Now we have ${\varphi}^{0}_{T^{\lambda}}
{\varphi}^{{\lambda}'}_{S T^{{\lambda}'}} = 0$ unless $S$ is the
tableaux of type $\lambda$, and $r_{{\lambda}', ST} = 0$ if $S$ is
of type $\lambda$. It follows that ${\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi} = 0$ and so $\widetilde{f}$ is well-defined.
Take ${\varphi}^{\alpha}_{ST} \in
\mathcal{S}(\Lambda)$ where $S, T \in
\mathcal{T}_0(\alpha)$ with $\alpha \in \Lambda^{+}$ and
suppose that $0 \neq {\varphi}_{T^{\lambda} T^{\lambda}}
{\varphi}^{\alpha}_{ST} \in
\mathcal{S}^{\vee \lambda}$. Then $S \in
\mathcal{T}_0(\alpha, \lambda)$ and so ${\varphi}^{\alpha}_{ST} =
{\varphi}_{T^{\lambda} T^{\lambda}}
{\varphi}^{\alpha}_{ST}
\in
\mathcal{S}^{\vee \lambda}$ since
${\varphi}_{T^{\lambda} T^{\lambda}}$ is the identity map
on $M^{\lambda}$. Thus, $\alpha \rhd \lambda$.
Furthermore, again by using \eqref{eq:decomposition
of varphiST} and the property that ${\varphi}_{S T^{\alpha}} \in
\mathcal{S}^0(\Lambda)$, we have ${\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi}^{\alpha}_{ST}
= {\varphi}^{0}_{T^{\lambda}} \cdot {\varphi}^{\alpha}_{S T^{\alpha}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi}^{\alpha}_{T^{\alpha} T}$. Since
${\varphi}^{\alpha}_{S
T^{\alpha}}$ is an element in the standard basis
$\mathcal{C}^0(\Lambda)$ for
$\mathcal{S}^0(\Lambda)$ and $\alpha \rhd \lambda$ we
have ${\varphi}^{}_{T^{\lambda} T^{\lambda}}
{\varphi}^{\alpha}_{S T^{\alpha}} \in \mathcal{S}_{0}^{\vee
(\lambda, 1)}$. Thus, ${\varphi}^{0}_{T^{\lambda}} \cdot
{\varphi}^{\alpha}_{S T^{\alpha}} = 0$ and then
${\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)}
{\varphi}^{\alpha}_{ST} =0$.
It follows that $\widetilde{f}(
\mathcal{S}^{\vee \lambda} \cap
{\varphi}_{T^{\lambda} T^{\lambda}} \cdot
\mathcal{S}(\Lambda) ) = 0$. So
$\widetilde{f}$ induces an
$\mathcal{S}(\Lambda)$-homomorphism $f$ from
${\varphi}_{T^{\lambda}} \cdot
\mathcal{S}(\Lambda)$ to $({\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1) \cdot
\mathcal{S}(\Lambda)$ such that
$f({\varphi}_{T^{\lambda}}) = {\varphi}^{0}_{T^{\lambda}}
\otimes_{\mathcal{S}^0(\Lambda)} 1$.
Next we define $g$. To do this we define the
$\mathcal{S}^0(\Lambda)$-balanced map $\widetilde{g}$ from the
direct
product module ${\varphi}^{0}_{T^{\lambda}}
\mathcal{S}^0(\Lambda) \times \mathcal{S}(\Lambda)$
(for the right $\mathcal{S}^0(\Lambda)$-module
${\varphi}^{0}_{T^{\lambda}}
\mathcal{S}^0(\Lambda)$ and the left $\mathcal{S}^0(\Lambda)$-module
$\mathcal{S}(\Lambda)$) to ${\varphi}_{T^{\lambda}}
\mathcal{S}(\Lambda)$ such that
$\widetilde{g}( ( {\varphi}^{0}_{T^{\lambda}}
{\varphi}', {\varphi} ) ) =
({\varphi}_{T^{\lambda}} {\varphi}' )
{\varphi}$ for any ${\varphi}' \in
\mathcal{S}^0(\Lambda)$ and ${\varphi} \in
\mathcal{S}(\Lambda)$. Since the first two formulas in Lemma
\ref{lem:sharper multiplication formulae} assert that
${\varphi}_{T^{\lambda} T^{\lambda}}
{\varphi}' \in
\mathcal{S}^{\vee \lambda}$ if ${\varphi}_{T^{\lambda} T^{\lambda}}
{\varphi}' \in \mathcal{S}_{0}^{\vee (\lambda, 0)}$, we see that
$\widetilde{g}$ is a well-defined $\mathcal{S}^0(\Lambda)$-balance
map. Then by using the universality property of the tensor product,
$g$ is defined.
\end{proof}
The following lemma is an analogy of Lemma 6.7 (i) in \cite{SawS}.
\begin{lem}\label{lem:injective homomorphism f_lambda from
Z^(lambda,0) to W^lambda_natural}
Suppose that $\lambda \in \Lambda^{+}$. We regard
$W^{\lambda}$ as right $\mathcal{S}^0(\Lambda)$-module via
the restriction. Then there exists an injective
$\mathcal{S}^0(\Lambda)$-module homomorphism $f_{\lambda} :
Z^{(\lambda, 0)} \rightarrow W^{\lambda}$ such that
$f_{\lambda}( {\varphi}^{0}_{T}) =
{\varphi}_{T}$ for $T \in J(\lambda, 0) =
\mathcal{T}^{+}_0(\lambda)$.
\end{lem}
\begin{proof}
First, We can write $Z^{(\lambda, 0)} = \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) / \{ \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}_{0}^{\vee
(\lambda, 0)} \}$ and $W^{\lambda} = \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}(\Lambda) / \{ \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} \}$ and, moreover, there exists an inclusion map
\begin{equation*}
\varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) / \{ \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} \} \hookrightarrow \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}(\Lambda) / \{ \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} \} = W^{\lambda}
\end{equation*}
since $\mathcal{S}^{0}(\Lambda) \subset
\mathcal{S}(\Lambda)$. Then we have a natural surjective
$\mathcal{S}^{0}(\Lambda)$-module
homomorphism
\begin{equation*}
f : \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) / \{ \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}_{0}^{\vee
(\lambda, 0)} \}
\rightarrow \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) / \{ \varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} \}
\end{equation*}
since $\mathcal{S}_{0}^{\vee
(\lambda, 0)} \subset \mathcal{S}^{\vee
\lambda}$. But since $\mathcal{S}^{0}(\Lambda) \cap
\mathcal{S}^{\vee
\lambda} = \mathcal{S}_{0}^{\vee (\lambda, 1)} \subset
\mathcal{S}_{0}^{\vee (\lambda, 0)}$, we have $(
\varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} ) \subset \mathcal{S}_{0}^{\vee (\lambda, 0)}$. Hence $(
\varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} ) \subset (
\varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}_{0}^{\vee
(\lambda, 0)} )$ and, therefore, $(
\varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}^{\vee
\lambda} ) = (
\varphi_{T^{\lambda}
T^{\lambda}} \mathcal{S}^{0}(\Lambda) \cap \mathcal{S}_{0}^{\vee
(\lambda, 0)} )$. Consequently, $f$ is an isomorphism. The statement
for the basis now clear.
\end{proof}
Take a $(\lambda, 0) \in \varOmega$. Then $I(\lambda, 0) =
J(\lambda, 0) = \mathcal{T}^{+}_0(\lambda)$. Hence, by
\eqref{eq:multiplication formulae of S^0(m,n)} in Theorem
\ref{thm:standardly based}, we have the following.
\addtocounter{subsection}{3}
\subsection{}
Suppose that $S, T \in \mathcal{T}^{+}_0(\lambda)$. Then there exists
an element $r_{ST} \in R$ such that for any $U, V \in
\mathcal{T}^{+}_0(\lambda)$
\begin{equation*}
{\varphi}_{US} \cdot {\varphi}_{TV} \equiv
r_{ST} \cdot {\varphi}_{UV} \mod \mathcal{S}_{0}^{\vee (\lambda, 0)}.
\end{equation*}
We define a bilinear form $\langle ~ ,
~
\rangle_{0} : Z^{(\lambda, 0)} \times Z^{(\lambda, 0)} \rightarrow R$
by $\langle {\varphi}^{0}_{S}, {\varphi}^{0}_{T} \rangle_{0} =
r_{ST}$. Hence we have
\begin{equation}\label{eq:the bilinear form on Z^(lambda,0)}
\langle {\varphi}^{0}_{S}, {\varphi}^{0}_{T} \rangle_{0} \cdot
{\varphi}_{UV} \equiv
{\varphi}_{US} \cdot {\varphi}_{TV} \mod
\mathcal{S}_{0}^{\vee (\lambda, 0)},
\end{equation}
where $U$ and $V$ are any elements of $\mathcal{T}^{+}_0(\lambda)$. It
is easy to see that
\begin{equation}\label{eq:the relationship between <,>_0 to
<,>_{natural}}
\langle {\varphi}^{0}_{S}, {\varphi}^{0}_{T} \rangle_{0} = \langle
{\varphi}_{S},
{\varphi}_{T} \rangle \qquad \text{ for every }
S, T \in \mathcal{T}^{+}_0(\lambda).
\end{equation}
Let $\text{rad}Z^{(\lambda, 0)} = \{ x \in Z^{(\lambda, 0)} \mid
\langle x, y \rangle_{0} = 0 \text{ for all } y \in Z^{(\lambda, 0)}\}$.
\addtocounter{thm}{1}
\begin{lem}
${\rm{rad}}Z^{(\lambda, 0)}$ is an
$\mathcal{S}^{0}(\Lambda)$-submodule of $Z^{(\lambda, 0)}$.
\end{lem}
\begin{proof}
Take $x \in \text{rad}Z^{(\lambda, 0)}$ and $y \in Z^{(\lambda, 0)}$,
and write $x = \sum_{T_1 \in
\mathcal{T}^{+}_0(\lambda)} r_{T_1}^{( x )} \cdot
{\varphi}^{0}_{T_1}$ and $y = \sum_{T_2 \in
\mathcal{T}^{+}_0(\lambda)} r_{T_2}^{( y )} \cdot
{\varphi}^{0}_{T_2} \in Z^{(\lambda, 0)}$ with $r_{T_1}^{( x )}$,
$r_{T_2}^{( y )} \in
R$. Then, for any ${\varphi} \in
\mathcal{S}^{0}(\Lambda)$, we have $\langle x {\varphi}, y
\rangle_{0} = \sum_{T_1, T_2 \in \mathcal{T}^{+}_0(\lambda)}
r_{T_1}^{( x )} r_{T_2}^{( y )} \langle {\varphi}^{0}_{T_1}
{\varphi}, {\varphi}^{0}_{T_2} \rangle_{0}$. We
claim that
\begin{equation}\label{eq:claim 1}
\langle {\varphi}^{0}_{T_1}
{\varphi}, {\varphi}^{0}_{T_2} \rangle_{0} \cdot
{\varphi}_{S_1 S_2} \equiv ( {\varphi}_{S_1
T_1} {\varphi} ) {\varphi}_{T_2 S_2} \mod
\mathcal{S}_{0}^{\vee (\lambda, 0)}
\end{equation}
for any $S_1, S_2 \in
\mathcal{T}^{+}_0(\lambda)$. In fact, since
${\varphi}_{S_1 T_1} \in \mathcal{C}^0(\lambda, 0)$ and
${\varphi} \in \mathcal{S}^{0}(\Lambda)$, Theorem
\ref{thm:standardly based} implies that
\begin{equation*}
{\varphi}_{S_1
T_1} \cdot {\varphi} \equiv \sum_{T \in \mathcal{T}^{+}_0(\lambda)}
f_{(\lambda, 0), T}(T_1,
{\varphi} ) \cdot {\varphi}_{S_1 T} \mod
\mathcal{S}_{0}^{\vee (\lambda, 0)}
\end{equation*}
where $f_{(\lambda, 0), T}(T_1,
{\varphi} ) \in R$. Hence,
\begin{equation*}
\begin{array}{ll}
( {\varphi}_{S_1
T_1} \cdot {\varphi} ) \cdot {\varphi}_{T_2
S_2}
&\equiv \underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum} f_{(\lambda,
0), T}(T_1,
{\varphi} ) \cdot {\varphi}_{S_1 T}\cdot
{\varphi}_{T_2 S_2} \\[6mm]
&\equiv \underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum} f_{(\lambda,
0), T}(T_1,
{\varphi} ) \cdot \langle {\varphi}^{0}_{T},
{\varphi}^{0}_{T_2} \rangle_{0} \cdot
{\varphi}_{S_1 S_2} \\[6mm]
&\equiv \langle \underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum}
f_{(\lambda, 0), T}(T_1, {\varphi} ) \cdot
{\varphi}^{0}_{T}, {\varphi}^{0}_{T_2} \rangle_{0}
\cdot {\varphi}_{S_1 S_2}.
\end{array}
\end{equation*}
On the other hand, by the definition of ${\varphi}^{0}_{T_1}$
and by Theorem \ref{thm:standardly based}, we have
\begin{equation*}
\begin{array}{ll}
{\varphi}^{0}_{T_1} \cdot {\varphi}
&= ( {\varphi}_{T^{\lambda} T_1} +
\mathcal{S}_{0}^{\vee (\lambda, 0)} ) \cdot {\varphi}
= {\varphi}_{T^{\lambda} T_1} \cdot {\varphi} +
\mathcal{S}_{0}^{\vee (\lambda, 0)} \\[3mm]
&=\underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum} f_{(\lambda, 0),
T}(T_1, {\varphi} ) \cdot
{\varphi}_{T^{\lambda} T} +
\mathcal{S}_{0}^{\vee (\lambda, 0)}
= \underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum} f_{(\lambda, 0),
T}(T_1, {\varphi} ) \cdot {\varphi}^{0}_{T}.
\end{array}
\end{equation*}
Note that
$\mathcal{S}_{0}^{\vee (\lambda, 0)}$ is a two-sided ideal of
$\mathcal{S}^{0}(\Lambda)$ in the second equality. \eqref{eq:claim 1}
follows from this. By Theorem \ref{thm:standardly based}, one can
write $\varphi \cdot \varphi_{T_2 S_2} = \sum_{S \in
\mathcal{T}^{+}_0(\lambda)} f_{S, (\lambda, 0)}(\varphi, T_2)
\varphi_{S S_2}$. Then by \eqref{eq:claim 1}, we have, for $T_2$, $S_2
\in \mathcal{T}^{+}_0(\lambda)$,
\begin{equation*}
\begin{array}{ll}
\langle {\varphi}^{0}_{T_1}
{\varphi}, {\varphi}^{0}_{T_2} \rangle_{0} \cdot
{\varphi}_{S_1 S_2}
&\equiv ( {\varphi}_{S_1
T_1} \cdot {\varphi} ) \cdot {\varphi}_{T_2
S_2}
= {\varphi}_{S_1
T_1} \cdot ( {\varphi} \cdot {\varphi}_{T_2
S_2} ) \\
&\equiv {\varphi}_{S_1
T_1} \cdot ( \underset{S \in \mathcal{T}^{+}_0(\lambda)}{\sum}
f_{S, (\lambda, 0) } ( {\varphi}, T_2 ) \cdot
{\varphi}_{S S_2} ) \\
&\equiv \underset{S \in \mathcal{T}^{+}_0(\lambda)}{\sum}
f_{S, (\lambda, 0) } ( {\varphi}, T_2 ) \cdot \langle
{\varphi}^{0}_{T_1}, {\varphi}^{0}_{S} \rangle_{0}
\cdot {\varphi}_{S_1 S_2}
\end{array}
\end{equation*}
The last equality follows from \eqref{eq:the bilinear form on
Z^(lambda,0)}. Since ${\varphi}_{S_1 S_2}$ are free $R$-basis of
$\mathcal{S}^{0}(\Lambda)$ and ${\varphi}_{S_1 S_2} \not\in
\mathcal{S}_{0}^{\vee (\lambda, 0)}$, we have
\begin{equation*}
\langle {\varphi}^{0}_{T_1}
{\varphi}, {\varphi}^{0}_{T_2} \rangle_{0} =
\sum_{S \in \mathcal{T}^{+}_0(\lambda)} f_{S, (\lambda, 0) } (
{\varphi}, T_2 ) \cdot \langle {\varphi}^{0}_{T_1},
{\varphi}^{0}_{S} \rangle_{0}.
\end{equation*}
Hence
\begin{equation*}
\begin{array}{ll}
\langle x {\varphi}, y \rangle_{0}
&= \underset{T_1, T_2 \in \mathcal{T}^{+}_0(\lambda)}{\sum} r_{T_1}^{(
x )} \cdot r_{T_2}^{( y )} \cdot \langle {\varphi}^{0}_{T_1}
{\varphi}, {\varphi}^{0}_{T_2}
\rangle_{0} \\[6mm]
&= \underset{T_1, T_2 \in \mathcal{T}^{+}_0(\lambda)}{\sum} r_{T_1}^{(
x )} r_{T_2}^{( y )} \cdot ( \underset{S \in
\mathcal{T}^{+}_0(\lambda)}{\sum} f_{S,
(\lambda, 0) } ( {\varphi}, T_2 ) \cdot \langle
{\varphi}^{0}_{T_1}, {\varphi}^{0}_{S} \rangle_{0} )
\\[6mm]
&= \underset{T_2 \in \mathcal{T}^{+}_0(\lambda)}{\sum} r_{T_2}^{( y )}
\cdot ( \underset{S \in \mathcal{T}^{+}_0(\lambda)}{\sum} f_{S,
(\lambda, 0) } ( {\varphi}, T_2 ) \cdot ( \underset{T_1
\in \mathcal{T}^{+}_0(\lambda)}{\sum} r_{T_1}^{( x )} \cdot \langle
{\varphi}^{0}_{T_1}, {\varphi}^{0}_{S} \rangle_{0} )
~ ) \\[6mm]
&= \underset{T_2 \in \mathcal{T}^{+}_0(\lambda)}{\sum} r_{T_2}^{( y )}
\cdot ( \underset{S \in \mathcal{T}^{+}_0(\lambda)}{\sum} f_{S,
(\lambda, 0) } ( {\varphi}, T_2 ) \cdot \langle x,
{\varphi}^{0}_{S}
\rangle_{0} ) \\[6mm]
&= \langle x, \underset{T_2 \in \mathcal{T}^{+}_0(\lambda)}{\sum}
r_{T_2}^{( y )} \cdot ( \underset{S \in
\mathcal{T}^{+}_0(\lambda)}{\sum} f_{S,
(\lambda, 0) } ( {\varphi}, T_2 ) \cdot
{\varphi}^{0}_{S} ) \rangle_{0}.
\end{array}
\end{equation*}
Since $\underset{T_2 \in \mathcal{T}^{+}_0(\lambda)}{\sum}
r_{T_2}^{( y )} \cdot ( \underset{S \in
\mathcal{T}^{+}_0(\lambda)}{\sum} f_{S,
(\lambda, 0) } ( {\varphi}, T_2 ) \cdot
{\varphi}^{0}_{S} ) \in Z^{(\lambda, 0)}$ and $x \in
\text{rad}Z^{(\lambda, 0)}$, the last formula is equal to
$0$. This shows that $x {\varphi} \in \text{rad}Z^{(\lambda,
0)}$. It is immediate that if $x_1, x_2 \in \text{rad}Z^{(\lambda,
0)}$ then $x_1 + x_2 \in \text{rad}Z^{(\lambda, 0)}$. Thus, the
lemma follows.
\end{proof}
We put $L_{0}^{\lambda} = Z^{(\lambda, 0)} /
\text{rad}Z^{(\lambda, 0)}$. Then we have the following.
\begin{prop}\label{prop:the property of L^lambda and radZ^(lambda,0)}
Suppose that $R$ is a field, and $\lambda \in \Lambda^{+}$. Then
$({\rm{i}})$ $L_{0}^{\lambda} \neq 0$ and
$({\rm{ii}})$ ${\rm{rad}}Z^{(\lambda, 0)}$ is the unique maximal
submodule of $Z^{(\lambda, 0)}$ and $L_{0}^{\lambda}$ is absolutely
irreducible. Moreover, the
Jacobson radical of $Z^{(\lambda, 0)}$ is equal to
$\text{rad}Z^{(\lambda, 0)}$.
\end{prop}
\begin{proof}
For any $\lambda \in \Lambda^{+}$, we have
${\varphi}^{0}_{T^{\lambda}} \in
Z^{(\lambda, 0)}$ and ${\varphi}^{0}_{T^{\lambda} T^{\lambda}}
{\varphi}^{0}_{T^{\lambda} T^{\lambda}} ={\varphi}^{0}_{T^{\lambda}
T^{\lambda}}$. Hence $\langle {\varphi}^{0}_{T^{\lambda}},
{\varphi}^{0}_{T^{\lambda}} \rangle_{0} = 1 \neq 0$. Thus,
${\varphi}^{0}_{T^{\lambda}} \not\in \text{rad}Z^{(\lambda, 0)}$,
and we have $L_{0}^{\lambda}
\neq 0$. This proves (i). Let $x \in Z^{(\lambda, 0)} \setminus
\text{rad}Z^{(\lambda,0)}$. Then $\langle x, y \rangle_{0} \neq 0$
for some $y \in Z^{(\lambda, 0)}$. We write $y = \sum_{S \in
\mathcal{T}^{+}_0(\lambda)} r_S \cdot {\varphi}^{0}_{S}$. For each
$T \in \mathcal{T}^{+}_0(\lambda)$ set $y_{T} = \sum_{S \in
\mathcal{T}^{+}_0(\lambda)} r_S \cdot {\varphi}_{S T}$,
an element of $\mathcal{S}^{0}(\Lambda)$. Moreover, write $x =
\sum_{U \in
\mathcal{T}^{+}_0(\lambda)} r'_U \cdot {\varphi}^{0}_{U}$. We note
that ${\varphi}^{0}_{U} \cdot {\varphi}_{ST} = \langle
{\varphi}^{0}_{U}, {\varphi}^{0}_{S} \rangle_{0} \cdot
{\varphi}^{0}_{T}$
in $Z^{(\lambda, 0)}$ by \eqref{eq:the bilinear form on
Z^(lambda,0)}. Thus we
have
\begin{equation*}
\begin{array}{ll}
x y_{T}
&= \underset{S, U \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r_{S} r'_{U} \cdot
{\varphi}^{0}_{U}
{\varphi}_{ST}
= \underset{S, U \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r_{S} r'_{U} \cdot \langle
{\varphi}^{0}_{U}, {\varphi}^{0}_{S} \rangle_{0} \cdot
{\varphi}^{0}_{T} \\[6mm]
&= \langle \underset{U \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r'_{U} {\varphi}^{0}_{U}, ~
\underset{S \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r_{S} {\varphi}^{0}_{S}
\rangle_{0}
\cdot {\varphi}^{0}_{T} = \langle x, y
\rangle_{0} \cdot {\varphi}^{0}_{T}.
\end{array}
\end{equation*}
Now $R$ is a field and
$\langle x, y \rangle_{0} \neq 0$. Hence we have $x \cdot
( 1 / \langle x,
y \rangle_{0} ) \cdot
y_{T} = {\varphi}^{0}_{T}$ for any $T \in
\mathcal{T}^{+}_0(\lambda)$. Consequently, $x$ generates
$Z^{(\lambda, 0)}$ as an $\mathcal{S}^{0}(\Lambda)$-module. This
argument can be applied to any
element of $Z^{(\lambda, 0)}$ which does not belong to the
radical. It follows that $\text{rad}Z^{(\lambda, 0)}$ is the unique
maximal proper submodule
of $Z^{(\lambda, 0)}$ and $L_{0}^{\lambda}$ is irreducible. The same
argument shows that $L_{0}^{\lambda}$ is irreducible for any
extension field of $R$, so $L_{0}^{\lambda}$ is absolutely
irreducible. This proves (ii).
\end{proof}
\medskip
\section{A relationship between $\mathcal{S}^{\flat}(\bs{m},n)$ and
$\mathcal{S}^0(\Lambda)$}\label{sec:A relationship between
S^flat(m,n) and S^0(Lambda)}
First, we recall the definition of modified Ariki-Koike algebras and
their
cyclotomic $q$-Schur algebras (\cite{SawS}).
\subsection{}\label{sec:modified Ariki-Koike algebra}
From now on, throughout this paper, we consider the following
condition on parameters $Q_1, \ldots ,
Q_{r}$ in $R$ whenever we consider the modified Ariki-Koike algebras
(and their cyclotomic $q$-Schur algebras).
\begin{equation}\label{eq:The condition of the Modified Ariki-Koike}
Q_{i} - Q_{j} \text{ are invertible in } R \text{ for any } i \neq
j. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
\end{equation}
Let $A$ be a square matrix of degree $r$ whose $i$-$j$ entry is given
by $Q_{j}^{i-1}$ for $1 \le i, j \le r$. Thus $A$ is the Vandermonde
matrix, and $\Delta = \det{A} = \prod_{i > j} ( Q_i -Q_j )$ is
invertible by \eqref{eq:The condition of the Modified Ariki-Koike}. We
express the inverse of $A$ as $A^{-1} = \Delta^{-1} B$ with $B = (
h_{ij} )$, and define a polynomial $F_{i}( X ) \in R[ X ]$, for $1 \le
i \le r$, by $F_{i}( X ) =\sum_{ 1 \le j \le r } h_{ij} X^{j-1}$.
The modified Ariki-Koike algebra $\mathscr{H}^{\flat} =
\mathscr{H}^{\flat}_{n, r}$ is an associative algebra over $R$ with
generators $T_2, \cdots , T_n$ and $\xi_{1}, \ldots , \xi_{n}$ and
relations
\begin{equation}\label{eq:Defining relations of H^flat}
\begin{array}{ll}
( T_i - q )( T_i + q^{-1} )
= 0
& ( 2 \le i \le n), \\
(\xi_{i} - Q_{1} ) \cdots ( \xi_{i} - Q_{r} )
= 0
& ( 1 \le i \le n), \\
T_{i} T_{i+1} T_{i}
= T_{i+1} T_i T_{i+1}
& ( 2 \le i \le n), \\
T_{i} T_{j}
= T_{j} T_{i}
& ( | i - j | \geq 2 ), \\
\xi_{i} \xi_{j}
= \xi_{j} \xi_{i}
& ( 1 \le i, j \le n ), \\
T_{j} \xi_{j}
= \xi_{j-1} T_j + \Delta^{-2}
\underset{c_1 < c_2}{\sum}
(Q_{c_2} - Q_{c_1} )( q - q^{-1} ) F_{c_1}( \xi_{j-1} ) F_{c_2}(
\xi_{j} ),
& \\
T_{j} \xi_{j-1}
= \xi_{j} T_{j} - \Delta^{-2}
\underset{c_1 < c_2}{\sum}
(Q_{c_2} - Q_{c_1} )( q - q^{-1} ) F_{c_1}( \xi_{j-1} ) F_{c_2}(
\xi_{j} ),
& \\
T_{j} \xi_{k}
=\xi_{j} T_{j}
& ( k \neq j-1, j ).
\end{array}
\end{equation}
It is known that if $R = \mathbb{Q}( \overline{q}, \overline{Q}_{1},
\ldots
, \overline{Q}_{r} )$, the field of rational functions with variables
$\overline{q}, \overline{Q}_{1}, \ldots
, \overline{Q}_{r}$, $\mathscr{H}^{\flat}$ is isomorphic to
$\mathscr{H}$, and it gives an alternate presentation of $\mathscr{H}$
apart from \ref{def:the definition of Ariki-Koike algebras}.
The subalgebra $\mathscr{H}^{\flat}( \mathfrak{S}_{n} )$ of
$\mathscr{H}^{\flat}$ generated by $T_2, \ldots , T_n$ is isomorphic
to $\mathscr{H}_{n}$, hence it can be naturally identified with the
corresponding subalgebra $\mathscr{H}( \mathfrak{S}_{n} )$ of
$\mathscr{H}$. Moreover, it is known by \cite{S} that the set $\{
\xi_{1}^{c_1} \cdots \xi_{n}^{c_n} T_{w} \mid w \in
\mathfrak{S}_{n}, ~ 0 \le c_{i} < r \text{ for } 1 \le i \le n \}$
gives rise to a basis of $\mathscr{H}^{\flat}$.
Let $V = \bigoplus_{i = 1}^{r} V_{i}$ be a free $R$-module, with rank
$V_{i} = m_{i}$. We put $m = \sum m_{i}$. It is known by \cite{SakS}
that we can define a right $\mathscr{H}$-module structure on
$V^{\otimes n}$. We denote this representation by ${\rho} :
\mathscr{H} \rightarrow \operatorname{End}{V^{\otimes n}}$. Note that this
construction works without the condition \eqref{eq:The
condition of the Modified Ariki-Koike}. Also it is shown in
\cite{S} that, under the assumption \eqref{eq:The
condition of the Modified Ariki-Koike}, a right action
of
$\mathscr{H}^{\flat}$ on $V^{\otimes n}$ can be defined. We denote
this
representation by ${\rho}^{\flat} : \mathscr{H}^{\flat} \rightarrow
\operatorname{End}{V^{\otimes n}}$. By \cite[Lemma 3.5]{S}, we know that
$\text{Im}{\rho} \subset \text{Im}{{\rho}^{\flat}}$.
We consider the condition
\begin{equation}\label{con:m_i>=n}
m_{i} \geq n \text{ ~ for ~ } i = 1, \cdots, r. \qquad \qquad \qquad
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
\end{equation}
\addtocounter{thm}{1}
\begin{lem}[{\cite[Lemma 1.5]{SawS}}]
Under the conditions \eqref{eq:The condition of the Modified
Ariki-Koike}, \eqref{con:m_i>=n}, there exists
an $R$-algebra homomorphism $\rho_{0} : \mathscr{H} \rightarrow
\mathscr{H}^{\flat}$ such that $\rho_{0}$ induces the identity on
$\mathscr{H}_{n}$. $($Here we regard $\mathscr{H}_{n} \subset
\mathscr{H}$, $\mathscr{H}_{n} \subset \mathscr{H}^{\flat}$ under the
previous identifications.$)$ If ${\rm{Im}}{{\rho}^{\flat}} =
{\rm{Im}}{\rho}$ and $R$ is a field, then $\mathscr{H} \simeq
\mathscr{H}^{\flat}$.
\end{lem}
From now on, throughout the paper, we fix an $r$-tuple $\bs{m} = (
m_1, \ldots , m_r )$ of non-negative integers and always assume the
condition \eqref{con:m_i>=n} whenever we consider
$\mathscr{H}^{\flat}$.
Any $\mu \in \widetilde{\mathcal{P}}_{n,r}( \bs{m} )$ may be regarded
as an
element in $\mathcal{P}_{n,1}$ (i.e, $1$-composition) of $n$ by
arranging the entries of $\mu = ( \mu_{j}^{( i )} )$ in order
\begin{equation*}
\mu_{1}^{( 1 )}, \ldots , \mu_{m_1}^{( 1 )}, \mu_{1}^{( 2 )}, \ldots ,
\mu_{m_2}^{( 2 )}, \ldots , \mu_{1}^{( r )}, \ldots , \mu_{m_r}^{( r
)},
\end{equation*}
which we denote by $\{ \mu \}$.
For $\alpha = ( n_1, \ldots , n_r ) \in \mathbb{Z}_{\geq 0}$ such that
$\sum n_i = n$, we define $c( \alpha )$ by
\begin{equation*}
c( \alpha ) = ( \underset{n_1\text{-times}}{\underbrace{r, \ldots ,
r}},
\underset{n_2\text{-times}}{\underbrace{r-1, \ldots , r-1}}, \ldots ,
\underset{n_r\text{-times}}{\underbrace{1, \ldots , 1}} )
\end{equation*}
and let $c( \alpha ) = ( c_1, \ldots , c_n)$. We define $F_{\alpha}
\in
\mathscr{H}^{\flat}$ by $F_{\alpha} = \Delta^{-n} F_{c_1}( \xi_{1} )
F_{c_2}( \xi_{2} ) \cdots
F_{c_n}( \xi_{n} )$.
For any $\mu \in \widetilde{\mathcal{P}}_{n,r}$, put $m^{\flat}_{\mu}
=
F_{\alpha( \mu )} \cdot m_{ \{ \mu \} }$ where $m_{\{ \mu \}} =
\sum_{ w \in \mathfrak{S}_{ \{ \mu \} } } q^{l( w )} T_{w} ~ ( =
x_{\mu}) \in \mathscr{H}_{n}$.
We define
an $R$-linear anti-automorphism $h \rightarrow h^{*}$ on
$\mathscr{H}^{\flat}$ by the condition that $*$ fixes the
generators $T_{i} ~ ( 2 \le i \le n )$ and $\xi_{j} ~ ( 1 \le j \le n
)$. As discussed in \cite[2.7]{SawS}, this condition induces a
well-defined
anti-automorphism on $\mathscr{H}^{\flat}$. Moreover, by Lemma 2.9 in
\cite{SawS}, we know that $(
m^{\flat}_{\mu} )^{*} = m^{\flat}_{\mu}$.
For $\mathfrak{s}, \mathfrak{t} \in \text{Std}( \lambda )$ with
$\lambda \in \mathcal{P}_{n,r}$, we define an element
$m^{\flat}_{\mathfrak{s} \mathfrak{t}} \in \mathscr{H}^{\flat}$ by
$m^{\flat}_{\mathfrak{s} \mathfrak{t}} = T^{*}_{d( \mathfrak{s} )}
m^{\flat}_{\mu} T_{d( \mathfrak{t} )}$. By the above fact, we have
$( m^{\flat}_{\mathfrak{s} \mathfrak{t}} )^{*} =
m^{\flat}_{\mathfrak{t} \mathfrak{s}}$.
\begin{thm}[{\cite[Theorem 2.18]{SawS}}]
The modified Ariki-Koike algebra $\mathscr{H}^{\flat}$ is free as an
$R$-module with cellular basis $\{ m^{\flat}_{\mathfrak{s}
\mathfrak{t}} \mid \mathfrak{s}, \mathfrak{t} \in \text{Std}(
\lambda ) \text{ for some } \lambda \in \mathcal{P}_{n,r} \}$.
\end{thm}
Put $M_{\flat}^{\mu} = m^{\flat}_{\mu} \mathscr{H}^{\flat}$ for $\mu
\in \widetilde{\mathcal{P}}_{n,r}$. We define a cyclotomic $q$-Schur
algebra
$\mathcal{S}^{\flat}(\bs{m},n)$ as follows.
\begin{defi}
The cyclotomic $q$-Schur algebra for $\mathscr{H}^{\flat}$
with weight poset
$\widetilde{\mathcal{P}}_{n,r}$ is the
endomorphism algebra
\begin{equation*}
\mathcal{S}^{\flat}(\bs{m},n) = \operatorname{End}_{\mathscr{H}^{\flat}}( M^{\flat}(
\widetilde{\mathcal{P}}_{n,r} ) ), \qquad \text{ where } M^{\flat}(
\widetilde{\mathcal{P}}_{n,r} ) = \underset{\mu \in
\widetilde{\mathcal{P}}_{n,r}}{\bigoplus} M_{\flat}^{\mu}.
\end{equation*}
\end{defi}
For an $r$-tuples $\alpha \in \widetilde{\mathcal{P}}_{n,1}$, let
$M_{\flat}^{\alpha} = \bigoplus_{\mu ; \alpha( \mu ) = \alpha}
M_{\flat}^{\mu}$. Then by Proposition 5.2 (i) in \cite{SawS}, we
have $\mathcal{S}^{\flat}(\bs{m},n) \simeq \bigoplus_{\alpha \in
\widetilde{\mathcal{P}}_{n,1}} \operatorname{End}_{\mathscr{H}^{\flat}}
M_{\flat}^{\alpha}$ as $R$-algebras.
\begin{thm}[{\cite[Theorem 5.5]{SawS}}]\label{weak separation
condition}
Let $\mathcal{S}^{\flat}(\bs{m},n)$ be the cyclotomic $q$-Schur
algebra associated to the modified Ariki-Koike algebra
$\mathscr{H}^{\flat}$ and $\mathcal{S}( m_{i}, n_{i} )$ be the
$q$-Schur algebra associated to the Iwahori-Hecke algebra
$\mathscr{H}_{n_{i}}$. Then there
exists an isomorphism of $R$-algebras
\begin{equation*}
\mathcal{S}^{\flat}(\bs{m},n) \simeq
\underset{
\substack{
( n_1, \ldots , n_r ) \\
n = n_1 + \cdots + n_{r}
}
}{\bigoplus}
\mathcal{S}( m_{1}, n_{1} ) \otimes \cdots \otimes
\mathcal{S}( m_{r}, n_{r} ).
\end{equation*}
\end{thm}
Let $\mu, \nu \in \widetilde{\mathcal{P}}_{n,r}$ and $\lambda \in
\mathcal{P}_{n,r}$. We assume that $\alpha( \mu ) = \alpha( \nu ) =
\alpha( \lambda )$. For $S \in \mathcal{T}^{+}_0(\lambda, \mu)$ and $T
\in \mathcal{T}^{+}_0(\lambda, \nu)$, put
\begin{equation*}
m^{\flat}_{ST} = \underset{
\substack{
\mathfrak{s}, \mathfrak{t} \in \text{Std}( \lambda ) \\
\mu( \mathfrak{s} ) = S, ~ \nu( \mathfrak{t} ) = T
}
}
{\sum}
q^{l( d( \mathfrak{s} ) ) + l( d( \mathfrak{t} ) )}
m^{\flat}_{\mathfrak{s}
\mathfrak{t}}.
\end{equation*}
Moreover, for $S \in \mathcal{T}^{+}_0(\lambda, \mu)$ and $T \in
\mathcal{T}^{+}_0(\lambda, \nu)$, one can define
$\varphi^{\flat}_{ST} \in \mathcal{S}^{\flat}(\bs{m},n)$ by
$\varphi^{\flat}_{ST}( m^{\flat}_{\alpha} h ) = \delta_{\alpha \nu}
m^{\flat}_{ST} h$, for all $h \in \mathscr{H}^{\flat}$ and all $\alpha
\in \widetilde{\mathcal{P}}_{n,r}$.
\begin{thm}[{\cite[Theorem 5.9]{SawS}}]
The cyclotomic $q$-Schur algebra $\mathcal{S}^{\flat}(\bs{m},n)$ is
free as an $R$-module with cellular basis
$\mathcal{C}^{\flat}(\bs{m},n) = \{ \varphi^{\flat}_{ST} \mid S, T \in
\mathcal{T}^{+}_0(\lambda), \text{ for some } \lambda \in
\mathcal{P}_{n,r} \}$.
\end{thm}
\addtocounter{subsection}{5}
\subsection{}
Let $\mathcal{S}^0(\Lambda)$ be as in Section \ref{The standard basis
for S^0(Lambda)}. We describe a relationship between the algebra
$\mathcal{S}^0(\Lambda)$ and the cyclotomic $q$-Schur algebra
$\mathcal{S}^{\flat}(\bs{m},n)$ in the case where $\Lambda =
\widetilde{\mathcal{P}}_{n,r}$. But in the moment, we shall consider
an arbitrary $\Lambda$ as in Section \ref{The standard basis
for S^0(Lambda)}.
First, let $\mathcal{C}^{00}(\Lambda) = \{
{\varphi}_{ST} \mid (S, T) \in I(\lambda, 1) \times
J(\lambda, 1), ~ \lambda \in \Lambda^{+} \} \subset
\mathcal{C}^{0}(\Lambda)$ and $\mathcal{S}^{00}(\Lambda)$ be the
$R$-span of ${\varphi}_{ST} \in
\mathcal{C}^{00}(\Lambda)$, which is an $R$-submodule of
$\mathcal{S}^{0}(\Lambda)$. We note that, $\mathcal{S}^{00}(\Lambda)$
is a two-sided ideal of $\mathcal{S}^{0}(\Lambda)$ by the second
and fourth formula in Lemma \ref{lem:sharper multiplication formulae}.
Thus one can define the quotient algebra
$\overline{\mathcal{S}^{0}}(\Lambda) = \mathcal{S}^0(\Lambda) /
\mathcal{S}^{00}(\Lambda)$. We write $\overline{x} = x +
\mathcal{S}^{00}(\Lambda)$ $( x \in \mathcal{S}^0(\Lambda) )$. It is
easy to see that $\overline{\mathcal{S}^{0}}(\Lambda)$ has a free
$R$-basis $\{ \overline{{\varphi}}_{ST} \mid S \in
I(\lambda, 0), ~ T \in J(\lambda, 0), ~ \lambda \in \Lambda^{+}
\}$. Note that the condition $(S,T) \in I(\lambda, 0) \times
J(\lambda,
0)$ is nothing but $S,T \in \mathcal{T}^{+}_0(\lambda)$. For
$\lambda \in \Lambda^{+}$, let $\overline{\mathcal{S}_{0}}^{\vee
\lambda} = \overline{\mathcal{S}_{0}}^{\vee}(\Lambda)^{\lambda}$ be
the $R$-submodule of
$\overline{\mathcal{S}^{0}}(\Lambda)$ spanned by
$\overline{{\varphi}}_{ST}$ with $S, T \in
\mathcal{T}^{+}_0(\alpha)$ for various $\alpha \in \Lambda^{+}$
such that $\alpha \rhd \lambda$. We show the following.
\addtocounter{thm}{1}
\begin{thm}
The algebra $\overline{\mathcal{S}^{0}}(\Lambda)$ has a free basis
\begin{equation*}
\overline{\mathcal{C}^{0}}(\Lambda) = \{
\overline{{\varphi}}_{ST} \mid S,T \in
\mathcal{T}^{+}_0(\lambda), ~ \lambda \in \Lambda^{+} \}
\end{equation*}
satisfying the following properties.
$({\rm{i}})$ The $R$-linear map $\ast :
\overline{\mathcal{S}^{0}}(\Lambda) \rightarrow
\overline{\mathcal{S}^{0}}(\Lambda)$ determined by
$\overline{{\varphi}}^{\ast}_{ST} =
\overline{{\varphi}}_{TS}$, for all $S,T \in
\mathcal{T}^{+}_0(\lambda)$ and all $\lambda \in \Lambda^{+}$,
is an anti-automorphism of $\overline{\mathcal{S}^{0}}(\Lambda)$.
$({\rm{ii}})$ Let $T \in \mathcal{T}^{+}_0(\lambda)$. Then for all
$\overline{{\varphi}} \in
\overline{\mathcal{S}^{0}}(\Lambda)$, and any $V \in
\mathcal{T}^{+}_0(\lambda)$, there exists $r_{V} \in
R$ such that
\begin{equation*}
\overline{{\varphi}}_{ST} \cdot
\overline{{\varphi}} \equiv \underset{V \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r_{V}
\overline{{\varphi}}_{SV} \mod \mathcal{S}_{\overline0}^{\vee
\lambda}
\end{equation*}
for any $S \in \mathcal{T}^{+}_0(\lambda)$, where $r_{V}$ is
independent of the choice of $T$.
In particular, $\overline{\mathcal{C}^{0}}(\Lambda)$ is a cellular
basis of $\overline{\mathcal{S}^{0}}(\Lambda)$.
\end{thm}
\begin{proof}
To show (i) we first take a small detour. Let $f :
\mathcal{S}^{0}(\Lambda) \rightarrow
\overline{\mathcal{S}^{0}}(\Lambda)$ be the $R$-linear
map given by
\begin{equation*}
f({\varphi}_{ST})=
\left\{\begin{array}{ll}
\overline{{\varphi}}_{TS} & \text{ if }
{\varphi}_{ST} \in \mathcal{C}^0(\lambda, 0)
\text{ for some } \lambda \in \Lambda^{+},\\
0 & \text{ otherwise. } \\
\end{array}\right.
\end{equation*}
We claim that $f$ is a surjective algebra
anti-homomorphism. It is clear from the definition that $f$ is
surjective.
Then, to prove the claim it suffices to show the equations
\begin{equation}\label{eq:(i)}
f({\varphi}_{S_1T_1} {\varphi}_{S_2T_2})
= f({\varphi}_{S_2T_2}) \cdot
f({\varphi}_{S_1T_1})
\end{equation}
for all $S_i, T_i \in \mathcal{T}^{+}_0(\lambda_i)$, $\lambda_i \in
\Lambda^{+}$ ($i=1,2$) and
\begin{equation}\label{eq:(ii)}
f(1_{\mathcal{S}^{0}(\Lambda)})
= 1_{\overline{\mathcal{S}^{0}}(\Lambda)}.
\end{equation}
We show \eqref{eq:(ii)}. By
applying $f$ to
\eqref{id:identity element of S^0}, we have
\begin{equation*}
f( 1_{\mathcal{S}^{0}(\Lambda)} )
= \sum_{\lambda \in \Lambda^{+}}
\sum_{S, T \in \mathcal{T}^{+}_0(\lambda)}
r_{T, S} \cdot \overline{{\varphi}_{ST}}
= \overline{ (
1_{\mathcal{S}^{0}(\Lambda)} )^{*} }
= \overline{1}_{\mathcal{S}^{0}(\Lambda)}
= 1_{\overline{\mathcal{S}^{0}}(\Lambda)}
\end{equation*}
where $*$ is the involution of $\mathcal{S}(\Lambda)$. Hence we obtain
\eqref{eq:(ii)}. We show \eqref{eq:(i)}. There are three cases to
consider. First, if ${\varphi}_{S_1T_1} \in
\mathscr{C}^0(\lambda, 1)$ with some $\lambda \in \Lambda^{+}$
then ${\varphi}_{S_1T_1} \cdot {\varphi}_{S_2T_2} \in
\mathcal{S}^{00}(\Lambda)$ by Lemma \ref{lem:sharper multiplication
formulae} and hence $f({\varphi}_{S_1T_1}
{\varphi}_{S_2T_2}) = 0 =
f({\varphi}_{S_1T_1})$, so we are done in this
case. Similarly, it holds if ${\varphi}_{S_2T_2} \in
\mathcal{C}^0(\lambda, 1)$
with some
$\lambda \in \Lambda^{+}$. So, we only need to check the case where
${\varphi}_{S_iT_i} \in \mathcal{C}^0(\lambda_i, 0)$ with
some $\lambda_i \in \Lambda^{+} ~ (i=1,2)$. Let $\mu_i, \nu_i
\in \Lambda$ with
$\alpha(\lambda_i) = \alpha(\mu_i) = \alpha(\nu_i)$ and $S_i \in
\mathcal{T}_0(\lambda_i,\mu_i), T_i \in
\mathcal{T}_0(\lambda_i,\nu_i) ~ (i=1,2)$. If $\nu_1 \neq \mu_2$ then
${\varphi}_{S_1 T_1}
\cdot {\varphi}_{S_2 T_2} = 0 = {\varphi}_{T_2S_2} \cdot
{\varphi}_{T_1S_1}$. Hence we have
\begin{equation*}
f({\varphi}_{S_1T_1} \cdot {\varphi}_{S_2T_2})
= 0 = \overline{{\varphi}_{T_2S_2} \cdot
{\varphi}_{T_1S_1}} =
\overline{{\varphi}}_{T_2S_2} \cdot
\overline{{\varphi}}_{T_1S_1} =
f({\varphi}_{S_2T_2}) \cdot f({\varphi}_{S_1T_1}).
\end{equation*}
So we may assume that $\nu_1 = \mu_2$. It is
easy to see from our assumptions that $\bold{a}(\mu_{1})
=\bold{a}(\mu_{2})$. Hence, for $\lambda \in \Lambda^{+}$ and $S
\in \mathcal{T}_0(\lambda, \mu_1)$, $T \in \mathcal{T}_0(\lambda,
\nu_2)$, if ${\varphi}_{ST} \in \mathcal{C}^0(\lambda, 1)$
then ${\varphi}_{TS} \in
\mathcal{C}^0(\lambda, 1)$. Therefore, by the first formula in
Lemma \ref{lem:sharper multiplication formulae}, we have
\begin{equation}\label{eq:(1)}
{\varphi}_{S_1T_1} \cdot {\varphi}_{S_2T_2} =
\underset{\substack{
\lambda \unrhd \lambda_1, ~ S, T\\
{\varphi}_{ST} \in \mathcal{C}^0(\lambda, 0)}}{\sum} r_{ST}
\cdot
\varphi_{ST} + \underset{\substack{
\lambda \unrhd \lambda_1, ~ S, T\\
{\varphi}_{ST} \in \mathcal{C}^0(\lambda, 1) }}{\sum} r_{ST} \cdot
\varphi_{ST}
\end{equation}
where $ r_{ST} \in R$ and both sums are taken over $r$-partitions
$\lambda
\unrhd \lambda_1$ and semistandard Tableaux $S \in
\mathcal{T}_0(\lambda,
\mu_1)$, $T \in \mathcal{T}_0(\lambda,
\nu_2)$. This gives us two formulas. First, we have
\begin{equation}\label{eq:(2)}
f({\varphi}_{S_1T_1} \cdot
{\varphi}_{S_2T_2}) = \underset{\substack{
\lambda \unrhd \lambda_1, ~ S, T ~ \\
{\varphi}_{ST} \in \mathcal{C}^0(\lambda,0)}}
{\sum} r_{ST} \cdot
\overline{{\varphi}}_{TS}.
\end{equation}
Secondly, by applying
the involution $\ast$ of $\mathcal{S}(\Lambda)$ on both
sides,
\begin{equation}\label{eq:(3)}
\overline{{\varphi}_{T_2S_2} \cdot
{\varphi}_{T_1S_1}} = \underset{\substack{
\lambda \unrhd \lambda_1, ~ S, T ~ \\
{\varphi}_{ST} \in \mathcal{C}^0(\lambda, 0)}}
{\sum} r_{ST} \cdot
\overline{{\varphi}}_{TS}
\end{equation}
since in the second sum in \eqref{eq:(1)}, we have
${\varphi}_{ST}^{\ast} =
{\varphi}_{TS} \in \mathcal{C}^0(\lambda, 1)$. Comparing
\eqref{eq:(2)} and \eqref{eq:(3)}, we have
\begin{equation*}
f({\varphi}_{S_1T_1}
\cdot
{\varphi}_{S_2T_2}) = \overline{{\varphi}_{T_2S_2} \cdot
{\varphi}_{T_1S_1}} =
\overline{{\varphi}}_{T_2S_2} \cdot
\overline{{\varphi}}_{T_1S_1} =
f({\varphi}_{S_2T_2}) \cdot
f({\varphi}_{S_1T_1}).
\end{equation*}
This proves our claim. By definition of
$f$, we have $\ker f = \mathcal{S}^{00}(\Lambda)$. Hence $f$ induces
an
algebra anti-automorphism on $\mathcal{S}^{\overline0}(\Lambda) =
\mathcal{S}^{0}(\Lambda) / \mathcal{S}^{00}(\Lambda)$, which maps
$\overline{{\varphi}}_{ST}$ to
$\overline{{\varphi}}_{TS}$ for any $S,T \in
\mathcal{T}^{+}_0(\lambda)$ with $\lambda \in \Lambda^{+}$. Thus
(i) holds. Then (ii) follows from the second formula of
\eqref{eq:multiplication formulae of S^0(m,n)} by applying the natural
map $\mathcal{S}^{0}(\Lambda) \rightarrow
\overline{\mathcal{S}^{0}}(\Lambda)$.
\end{proof}
In the case where $\mathcal{S}^{\flat}(\bs{m},n)$ is defined,
$\overline{\mathcal{S}^{0}}(\Lambda)$ can be identified with
$\mathcal{S}^{\flat}(\bs{m},n)$, i.e, we have the following
proposition.
\begin{prop}\label{prop:the relationship between S^overline0 and S^0}
Let $\Lambda = \widetilde{\mathcal{P}}_{n,r}$ and assume that
\eqref{eq:The condition of the Modified Ariki-Koike} and
\eqref{con:m_i>=n} holds. Then there exists an algebra isomorphism
${\flat}:
\overline{\mathcal{S}^{0}}(\Lambda) \rightarrow
\mathcal{S}^{\flat}(\bs{m},n)$
satisfying the following. For $\overline{{\varphi}}_{ST}
\in \overline{\mathcal{C}^{0}}(\Lambda)$ such that $S, T \in
\mathcal{T}^{+}_0(\lambda)$ and $\lambda \in \Lambda^{+}$, we have $(
\overline{{\varphi}}_{ST} )^{\flat} = {\varphi}^{\flat}_{ST}$.
\end{prop}
\begin{proof}
By assumption, $\mathcal{S}^{\flat}(\bs{m},n)$ is defined. Now
\cite[Proposition 6.4 (ii)]{SawS} says that there exists a surjective
algebra
homomorphism $\widehat{f} : \mathcal{S}^{0}(\Lambda) \rightarrow
\mathcal{S}^{\flat}(\bs{m},n)$ such that
\begin{equation*}
\widehat{f}({\varphi}_{ST}) =
\left
\{
\begin{array}{ll}
{\varphi}^{\flat}_{ST}
&
\text{ if } S,T \in \mathcal{T}^{+}_0(\lambda), \\
0
&
\text{ otherwise } \\
\end{array}\right.
\end{equation*}
for ${\varphi}_{ST} \in \mathcal{C}^{0}(\Lambda)$. Then $\ker
\widehat{f} =
\mathcal{S}^{00}(\Lambda)$ and so $\widehat{f}$ induces an algebra
isomorphism $\overline{\mathcal{S}^{0}}(\Lambda) \rightarrow
\mathcal{S}^{\flat}(\bs{m},n)$, which maps
$\overline{{\varphi}}_{ST}$ to ${\varphi}^{\flat}_{ST}$ for any
$S,T \in \mathcal{T}^{+}_0(\lambda)$ and $\lambda \in
\Lambda^{+} = \mathcal{P}_{n,r}$. We obtain the result.
\end{proof}
We now return to the general setting, and consider
$\overline{\mathcal{S}^{0}}(\Lambda)$ for arbitrary $\Lambda$. The
above proposition says that the $\overline{\mathcal{S}^{0}}(\Lambda)$
is
a natural "cover" of the $\mathcal{S}^{\flat}(\bs{m},n)$.
For $\lambda \in \Lambda^{+}$,
$\overline{{\varphi}}_{\lambda} =
\overline{{\varphi}}_{T^{\lambda} T^{\lambda}}$ is an
element in $\overline{\mathcal{S}^{0}}(\Lambda)$. Hence, by the
cellular theory \cite{GL}, one can define a Weyl module
$\overline{Z}^{\lambda}$ of $\overline{\mathcal{S}^{0}}(\Lambda)$ as
the right $\overline{\mathcal{S}^{0}}(\Lambda)$-submodule of
$\overline{\mathcal{S}^{0}}(\Lambda) /
\overline{\mathcal{S}_{0}}^{\vee
\lambda}$ spanned by the image of
$\overline{{\varphi}}_{\lambda}$. We denote by
$\overline{{\varphi}}_{T}$ the image of
$\overline{{\varphi}}_{T^{\lambda} T}$ in
$\overline{\mathcal{S}^{0}}(\Lambda) /
\overline{\mathcal{S}_{0}}^{\vee
\lambda}$. Then the set $\{
\overline{{\varphi}}_{T} \mid T \in
\mathcal{T}^{+}_0(\lambda) \}$ is a free $R$-basis of
$\overline{Z}^{\lambda}$. Define a bilinear form $\langle ~ , ~
\rangle_{\overline{0}}$ on $\overline{Z}^{\lambda}$ by requiring that
\begin{equation*}
\overline{{\varphi}}_{T^{\lambda}
S}\overline{{\varphi}}_{T T^{\lambda}} \equiv \langle
\overline{{\varphi}}_{S},
\overline{{\varphi}}_{T} \rangle_{\overline{0}}
\cdot \overline{{\varphi}}_{\lambda} \mod
\overline{\mathcal{S}_{0}}^{\vee \lambda}
\end{equation*}
for all $S,T \in \mathcal{T}^{+}_0(\lambda)$. Let
$\overline{L}^{\lambda} = \overline{Z}^{\lambda} / \text{rad}
\overline{Z}^{\lambda}$, where $\text{rad}
\overline{Z}^{\lambda} = \{x\in \overline{Z}^{\lambda} \mid \langle x,
y
\rangle_{\overline{0}} = 0
\text{ for all } y \in \overline{Z}^{\lambda} \}$. In the case where
$R$ is a field, by a general theory
of cellular algebras, the set $\{
\overline{L}^{\lambda} \mid \lambda \in \Lambda^{+},
~ \overline{L}^{\lambda} \neq 0 \}$ gives a complete set of
non-isomorphic irreducible
$\overline{\mathcal{S}^{0}}(\Lambda)$-modules. Furthermore, we have
the following result.
\begin{prop}\label{prop:complete set of non-isomorphic irreducible
S^overline0(m,n) modules}
Suppose that $R$ is a field. Then $\overline{L}^{\lambda} \neq 0$ for
any $\lambda \in \Lambda^{+}$. Hence, $\{
\overline{L}^{\lambda} \mid \lambda \in \Lambda^{+} \}$ is a
complete set of non-isomorphic irreducible
$\overline{\mathcal{S}^{0}}(\Lambda)$-modules. Therefore,
$\overline{\mathcal{S}^{0}}(\Lambda)$ is quasi-hereditary.
\end{prop}
\begin{proof}
We have $\overline{{\varphi}}_{T^{\lambda}
T^{\lambda}}\overline{{\varphi}}_{T^{\lambda} T^{\lambda}} \equiv
\langle
\overline{{\varphi}}_{T^{\lambda}},
\overline{{\varphi}}_{T^{\lambda}} \rangle_{\overline{0}}
\cdot \overline{{\varphi}}_{\lambda} \mod
\overline{\mathcal{S}_{0}}^{\vee \lambda}$. But since
$\overline{{\varphi}}_{T^{\lambda}
T^{\lambda}}\overline{{\varphi}}_{T^{\lambda}
T^{\lambda}} = \overline{{\varphi}}_{T^{\lambda}
T^{\lambda}}$, we see that $\langle
\overline{{\varphi}}_{T^{\lambda}},
\overline{{\varphi}}_{T^{\lambda}} \rangle_{\overline{0}}
= 1$, and so $\overline{L}^{\lambda}$ is non-zero. In particular,
$\overline{\mathcal{S}^{0}}(\Lambda)$ is a standardly full-based
algebra (see Definition in \ref{def:the definition of
standardly (full-)based algebra}). Hence Theorem
\ref{th:quasi-hereditary} gives us that
$\overline{\mathcal{S}^{0}}(\Lambda)$ is quasi-hereditary. The
proposition is proved.
\end{proof}
\begin{lem}\label{lem:Z(lambda,0) is isomorphic to overlineZlambda}
Suppose that $\lambda \in \Lambda^{+}$. Then there exists an
$\mathcal{S}^{0}(\Lambda)$-module isomorphism $h_{\lambda} :
Z^{(\lambda,0)} \rightarrow \overline{Z}^{\lambda}$ such that
$h_{\lambda}({\varphi}^{0}_{T}) =
\overline{{\varphi}}_{T}$ for any $T \in
\mathcal{T}^{+}_0(\lambda)$, where $\overline{Z}^{\lambda}$ is
regarded as an $\mathcal{S}^{0}(\Lambda)$-module via the canonical map
$\pi : \mathcal{S}^{0}(\Lambda)
\rightarrow \overline{\mathcal{S}^{0}}(\Lambda)$.
\end{lem}
\begin{proof}
Let $\pi : \mathcal{S}^{0}(\Lambda) \rightarrow
\overline{\mathcal{S}^{0}}(\Lambda)$ be the natural surjection. Then
$\pi$ maps $\mathcal{S}_{0}^{\vee \lambda}$ to
$\overline{\mathcal{S}_{0}}^{\vee
\lambda}$, hence it induces an $\mathcal{S}^{0}(\Lambda)$-module
surjective
homomorphism $\overline{\pi} : \mathcal{S}^{0}(\Lambda) /
\mathcal{S}_{0}^{\vee \lambda}
\rightarrow \overline{\mathcal{S}_{0}}(\Lambda) /
\overline{\mathcal{S}_{0}}^{\vee
\lambda}$. Here $Z^{(\lambda, 0)} = {\varphi}^{0}_{T^{\lambda}}
\mathcal{S}^{0}(\Lambda)$ and $\overline{Z}^{\lambda} =
\overline{{\varphi}}_{\lambda} \overline{\mathcal{S}_{0}}(\Lambda) =
\overline{{\varphi}}_{\lambda} \mathcal{S}_{0}(\Lambda)$, and
$\overline{\pi}( {\varphi}^{0}_{T^{\lambda}} ) =
\overline{{\varphi}}_{\lambda}$. Hence $\overline{\pi}$ induces a
surjective map $h_{\lambda} : Z^{(\lambda, 0)} \rightarrow
\overline{Z}^{\lambda}$. Since it is easy to check that
$\overline{\pi}( {\varphi}^{0}_{T} ) = \overline{{\varphi}}_{T}$ for
any $T \in \mathcal{T}^{+}_0(\lambda)$, the map $h_{\lambda}$ is an
isomorphism.
\end{proof}
Combining Lemma \ref{lem:Z(lambda,0) is isomorphic to overlineZlambda}
and Lemma \ref{lem:injective homomorphism
f_lambda from
Z^(lambda,0) to W^lambda_natural}, we have the following.
\begin{lem}
We regard $\overline{Z}^{\lambda}$ and $W^{\lambda}$ as
right $\mathcal{S}^{0}(\Lambda)$-modules via the above and via the
restriction, respectively. Then the map $\overline{f}_{\lambda} =
f_{\lambda} \circ h_{\lambda}^{-1} : \overline{Z}^{\lambda}
\rightarrow W^{\lambda}$, which maps
$\overline{{\varphi}}_{T}$ to ${\varphi}_{T}$
for any $T \in \mathcal{T}^{+}_0(\lambda)$, is an injective
$\mathcal{S}^{0}(\Lambda)$-module homomorphism.
\end{lem}
We need two easy lemmas.
\begin{lem}\label{lem:irreducible S^0(m,n)-modules for composition
factors of Z^(lambda,0)}
Suppose that $R$ is a field. For every $\lambda \in
\Lambda^{+}$, we regard $\overline{L}^{\lambda}$ as right
$\mathcal{S}^{0}(\Lambda)$-modules via the map $\pi$. Then
$\mathcal{S}^{0}(\Lambda)$-module
$\overline{L}^{\lambda}$ is irreducible. Furthermore, $\{
\overline{L}^{\alpha} \mid \alpha \in \Lambda^{+}, ~ \lambda \unrhd
\alpha \}$ is a
complete set of pairwise inequivalent irreducible
$\mathcal{S}^{0}(\Lambda)$-modules occurring in the composition factors
of $Z^{(\lambda, 0)}$.
\end{lem}
\begin{proof}
By Proposition \ref{prop:complete set of non-isomorphic irreducible
S^overline0(m,n) modules}, we have $\overline{L}^{\lambda} \neq
0$. Since $\overline{L}^{\lambda}$ is irreducible as
$\overline{\mathcal{S}^{0}}(\Lambda)$-module, it is irreducible as
$\mathcal{S}^{0}(\Lambda)$-module. It is clear that $\{
\overline{L}^{\alpha} \mid \alpha \in \Lambda^{+}, ~ \lambda
\unrhd \alpha \}$ is a complete set of irreducible
$\mathcal{S}^{0}(\Lambda)$-modules occurring in $Z^{(\lambda, 0)}$.
\end{proof}
\begin{lem}\label{lem:L^lambda isomorphic to overlineL^lambda}
Suppose that $R$ is a field. Then, for all $\lambda \in
\Lambda^{+}$, we have $L_{0}^{\lambda} \simeq \overline{L}^{\lambda}$
as $\mathcal{S}^{0}(\Lambda)$-modules.
\end{lem}
\begin{proof}
(The following proof is almost similar to \cite[Lemma 6.8]{SawS}.)
Recall the bilinear form $\langle ~ , ~ \rangle_{0}$ (resp. $\langle ~
, ~ \rangle_{\overline{0}}$) on $Z^{(\lambda, 0)}$ (resp. on
$\overline{Z}^{\lambda}$) and the $\mathcal{S}^{0}(\Lambda)$-module
isomorphism $h_{\lambda}$ in Lemma
\ref{lem:Z(lambda,0) is isomorphic to overlineZlambda}. We note that
$h_{\lambda}^{-1} : \overline{Z}^{\lambda} \rightarrow Z^{(\lambda,
0)}$ maps $\text{rad}\overline{Z}^{\lambda}$ onto
$\text{rad}Z^{(\lambda, 0)}$. In fact, it is easy to see that $\langle
{\varphi}^{0}_{S}, {\varphi}^{0}_{T} \rangle_{0} =
\langle \overline{\varphi}_{S},
\overline{\varphi}_{T} \rangle_{\overline{0}}$ for any $S,
T \in \mathcal{T}^{+}_0(\lambda)$. Hence
$h_{\lambda}^{-1}(\text{rad}\overline{Z}^{\lambda}) =
\text{rad}Z^{(\lambda, 0)}$, and $h_{\lambda}^{-1}$ induces an
$\mathcal{S}^{0}(\Lambda)$-module isomorphism $\overline{L}^{\lambda}
\rightarrow L_{0}^{\lambda}$.
\end{proof}
The following result connects the decomposition numbers in
$\overline Z^{\lambda}$ and in $Z^{(\lambda,0)}$.
\begin{thm}\label{thm:the property of L^lambda and
[overlineZ^lambda:overlineL^mu]=[Z^(lambda,0):L_0^mu]}
Suppose that $R$ is a field. Then
\begin{enumerate}
\item
$\{ L_{0}^{\alpha} \mid \alpha \in
\Lambda^{+}, ~ \lambda \unrhd \alpha \}$ is a complete set of pairwise
inequivalent
irreducible $\mathcal{S}^{0}(\Lambda)$-modules occurring in the
composition
factors of the
$\mathcal{S}^0(\Lambda)$-module $Z^{(\lambda,0)}$.
\medskip
\item
For $\lambda$, $\mu \in \Lambda^+$, we have
\begin{equation*}
[\overline Z^{\lambda} : \overline L^{\mu}] = [ Z^{(\lambda, 0)} :
L_{0}^{\mu} ].
\end{equation*}
\item
For $\lambda$, $\mu \in \Lambda^+$ such that
$\alpha(\lambda) \ne \alpha(\mu)$, we have
\begin{equation*}
[\overline Z^{\lambda} : \overline L^{\mu}] = 0.
\end{equation*}
\end{enumerate}
\end{thm}
\begin{proof}
The first and the second statement follows from Lemma
\ref{lem:irreducible
S^0(m,n)-modules for composition
factors of Z^(lambda,0)} and Lemma \ref{lem:L^lambda isomorphic to
overlineL^lambda}.
We show the third statement.
Take $\lambda \in \Lambda^{+}$ and assume that $\alpha(\lambda) \neq
\alpha(\mu)$ for $\mu \in \Lambda^{+}$. Let
\begin{equation*}
0 = M_{1} \subsetneqq M_{2} \subsetneqq \cdots \subsetneqq M_{k} =
\overline{Z}^{\lambda}
\end{equation*}
be a composition series of $\overline{Z}^{\lambda}$ and assume that
$M_{i+1} / M_{i} \simeq \overline{L}^{\mu}$ for some integer $i$ with
$1 \le i \le k-1$. Note that, as in the proof of Proposition
\ref{prop:complete set of non-isomorphic irreducible
S^overline0(m,n) modules},
$\overline{\varphi}_{T^{\mu}} + \text{rad}\overline{Z}^{\mu}$ is a
non-zero element of $\overline{L}^{\mu}$. Then, there exists $c_{\mu}
\in M_{i+1} / M_{i}$ corresponding to $\overline{\varphi}_{T^{\mu}} +
\text{rad}\overline{Z}^{\mu}$. Take a representative
$\widetilde{c}_{\mu} \in M_{i+1} \subset \overline{Z}^{\lambda}$ of
$c_{\mu}$. Take ${\varphi}_{\mu} \in
\mathcal{S}^{0}(\Lambda)$ (cf. \eqref{eq:identity map on
M^mu_natural}). Since $M_{i+1}$ is an
$\overline{\mathcal{S}^{0}}(\Lambda)$-module of
$\overline{Z}^{\lambda}$, we have $\widetilde{c}_{\mu} \cdot
\overline{{\varphi}_{\mu}}
\in M_{i+1}$. Moreover, $\widetilde{c}_{\mu} \cdot
\overline{{\varphi}_{\mu}}$
corresponds to
\begin{equation*}
( \overline{\varphi}_{T^{\mu}} + \text{rad}\overline{Z}^{\mu} ) \cdot
\overline{{\varphi}_{\mu}} = \overline{\varphi}_{T^{\mu}} +
\text{rad}\overline{Z}^{\mu} \neq 0
\end{equation*}
Therefore, $\widetilde{c}_{\mu} \cdot \overline{{\varphi}_{\mu}} \neq
0$. On the
other
hand, for any $T \in \mathcal{T}^{+}_0(\lambda)$, the type of $T$ is
not equal to $\mu$ since
$\alpha(\lambda) \neq
\alpha(\mu)$. Hence, by the definition
of ${\varphi}_{\mu}$, we have
${\varphi}_{T^{\lambda} T} \cdot {\varphi}_{\mu} = 0$
for any $T \in \mathcal{T}^{+}_0(\lambda)$. This implies that $x \cdot
\overline{{\varphi}_{\mu}} = 0$ for any $x \in
\overline{Z}^{\lambda}$, a contradiction. The theorem follows.
\end{proof}
\medskip
\section{An estimate for decomposition numbers}
We are now ready to estimate the decomposition numbers for the
cyclotomic $q$-Schur algebras.
\subsection{}
We keep the notation in Section \ref{sec:A relationship between
S^flat(m,n) and S^0(Lambda)}, and consider the general $\Lambda$. By
abuse of the notation, $\mathcal{S}^{0},
\overline{\mathcal{S}^{0}}, \mathcal{S}$ denotes
$\mathcal{S}^{0}(\Lambda),
\overline{\mathcal{S}^{0}}(\Lambda), \mathcal{S}(\Lambda)$
respectively, and $M \otimes \mathcal{S}$ denotes $M
\otimes_{\mathcal{S}^{0}} \mathcal{S}$ for any right
$\mathcal{S}^{0}$-module $M$. Suppose that $\lambda \in
\Lambda^{+}$ and $\psi \in \mathcal{S}^{0},
{\varphi} \in \mathcal{S}$. Let $g_{\lambda} =
g : Z^{(\lambda,0)} \otimes \mathcal{S} \rightarrow
W^{\lambda}$ be the isomorphism given in Theorem \ref{thm:tensor
theorem}. Then for any $T \in
\mathcal{T}^{+}_0(\lambda)$, we have
\begin{equation*}
{\varphi}^{0}_{T} \otimes {\varphi} =
{\varphi}^{0}_{T^{\lambda}} \cdot
{\varphi}_{T^{\lambda} T} \otimes {\varphi} =
{\varphi}^{0}_{T^{\lambda}} \otimes
{\varphi}_{T^{\lambda} T} {\varphi}
\overset{g}{\longmapsto}
{\varphi}_{T^{\lambda}}{\varphi}_{T^{\lambda} T}
{\varphi} = {\varphi}_{T} {\varphi}
\end{equation*}
where the second equality follows from ${\varphi}_{T^{\lambda} T} \in
\mathcal{S}^{0}$ since $T
\in \mathcal{T}^{+}_0(\lambda)$. Thus, we see that
\addtocounter{thm}{1}
\begin{rem}\label{rem:Remark of the tensor theorem}
The $\mathcal{S}$-module isomorphism
\begin{equation*}
g_{\lambda} : Z^{(\lambda,0)} \otimes \mathcal{S}
\rightarrow W^{\lambda}
\end{equation*}
in Theorem \ref{thm:tensor theorem} maps ${\varphi}^{0}_{T}
\otimes {\varphi}$ to
${\varphi}_{T} {\varphi}$ for any $T \in
\mathcal{T}^{+}_0(\lambda)$.
\end{rem}
\begin{lem}\label{lem:the key Lemma1}
Suppose that $\lambda \in \Lambda$. Let $M_1, M_2$ be non-zero
$\mathcal{S}^{0}$-submodules of $Z^{(\lambda, 0)}$ with $M_1
\subsetneqq M_2$, and let $\iota : M_1 \hookrightarrow
M_2$,~~~ ${\iota}_1 : M_1 \hookrightarrow Z^{(\lambda, 0)}$,~~~
${\iota}_2 : M_2
\hookrightarrow Z^{(\lambda, 0)}$ be inclusion maps. Then
$({\rm{i}})$ the $\mathcal{S}$-module homomorphism ~ ~ $\iota
\otimes
{\rm{id}}_{\mathcal{S}} : M_1 \otimes
\mathcal{S} \rightarrow M_2 \otimes \mathcal{S}$ is a non-zero map.
$({\rm{ii}})$ $(\iota_{1} \otimes {\rm{id}}_{\mathcal{S}})(M_1 \otimes
\mathcal{S})$ is a proper $\mathcal{S}$-submodule of $(\iota_{2}
\otimes
{\rm{id}}_{\mathcal{S}})(M_2 \otimes
\mathcal{S})$.
\end{lem}
\begin{proof}
Put $I = \iota \otimes
\text{id}_{\mathcal{S}}$, $I_i = \iota_{i} \otimes
\text{id}_{\mathcal{S}}$ $(i=1,2)$. Take $0 \neq x \in
M_1$. Then, by $M_1 \subset Z^{(\lambda, 0)}$, we may write $x =
\sum_{T \in \mathcal{T}^{+}_0(\lambda)} r_{T}
{\varphi}^{0}_{T}$ with $r_{T} \in R$. Then we have $(g \circ I ) ( x
\otimes 1) = \sum_{T \in
\mathcal{T}^{+}_0(\lambda)} r_{T} {\varphi}_{T}$ by Remark
\ref{rem:Remark of the tensor theorem}. Now suppose that $I( x \otimes
1) = 0$. Then $( g_{\lambda} \circ I )( x \otimes 1 ) = 0$ and so
$r_{T} =0$ for any $T \in \mathcal{T}^{+}_0(\lambda)$ since $\{
{\varphi}_{T} \mid T \in \mathcal{T}^{+}_0(\lambda)\}$ is a
subset of the basis of $W^{\lambda}$. So $x
= 0$, which contradicts the choice of $x$. Hence $I$ is non-zero. This
proves (i).
Next we show (ii). Clearly, $I_1(M_1 \otimes \mathcal{S}) \subset
I_2(M_2 \otimes \mathcal{S})$ so it suffices to see that $I_1(M_1
\otimes \mathcal{S}) \neq I_2(M_2 \otimes \mathcal{S})$. By our
assumption,
there exists a non-zero element $x \in M_2 \setminus M_1$. Consider $x
\otimes 1 \in M_2 \otimes \mathcal{S}$ and, by way
of contradiction, suppose that $I_2(x \otimes 1) \in I_1(M_1
\otimes \mathcal{S})$. Hence there exists $y_i \in
M_1$ and ${\varphi}_{i} \in \mathcal{S}$ such
that
\begin{equation}\label{eq:the image of x tensor 1}
(g_{\lambda} \circ I_2)(x \otimes 1) = (g_{\lambda} \circ
I_1)(\sum_{i} y_i \otimes {\varphi}_{i}).
\end{equation}
Since $x, y_i \in Z^{(\lambda, 0)}$, one can write $x = \sum_{T \in
\mathcal{T}^{+}_0(\lambda)} r_{T} {\varphi}^{0}_{T}$
and $y_i = \sum_{T \in \mathcal{T}^{+}_0(\lambda)} r_{T}^{i}
{\varphi}^{0}_{T}$ with $r_{T}, r_{T}^{i} \in
R$. By substituting them into \eqref{eq:the image of x tensor 1}, we
have
\begin{equation}\label{eq:substituting formula (1)}
\sum_{T \in
\mathcal{T}^{+}_0(\lambda)} r_{T} {\varphi}_{T} =
\sum_{i} \sum_{T \in \mathcal{T}^{+}_0(\lambda)} r_{T}^{i}
{\varphi}_{T} {\varphi}_{i}.
\end{equation}
Hence, by using the definition of ${\varphi}_{T}$, we have
\begin{equation}\label{eq:substituting formula (2)}
\sum_{T \in
\mathcal{T}^{+}_0(\lambda)} r_{T} {\varphi}_{T^{\lambda}
T} - \sum_{i} \sum_{T \in \mathcal{T}^{+}_0(\lambda)} r_{T}^{i}
{\varphi}_{T^{\lambda} T} {\varphi}_{i} =
\psi
\end{equation}
for some $\psi \in
\mathcal{S}^{\vee \lambda}$. Take $\nu \in
\Lambda$ with $\alpha(\nu) = \alpha(\lambda)$, and multiply
${\varphi}_{\nu}$ on the both side of \eqref{eq:substituting formula
(2)} from the
right, we obtain
\begin{equation}\label{eq:multiply varphi_nu from the right}
\underset{T \in \mathcal{T}_0(\lambda, \nu)}{\sum} r_{T}
{\varphi}_{T^{\lambda} T} - \underset{i}{\sum} \underset{T \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r_{T}^{i}
{\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu} = \psi {\varphi}_{\nu}.
\end{equation}
(Recall that ${\varphi}_{\nu}$ is the identity map on
$M^{\nu}$ and zero on $M^{\kappa}$ with $\nu
\neq \kappa \in \Lambda^{+}$ see \ref{standardly based subsection}.)
Now, by using Lemma
\ref{lem:multiplication formula} and \eqref{eq:identity map on
M^mu_natural}, we see that ${\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu}$ is an $R$-linear combination of elements
of the form ${\varphi}_{UV}$ where $U \in
\mathcal{T}_0({\lambda}', \lambda)$ and $V \in
\mathcal{T}_0({\lambda}', \nu)$
with ${\lambda}' \in \Lambda^{+}$. Since $\alpha(\nu) =
\alpha(\lambda)$, this implies that ${\varphi}_{T^{\lambda} T}
{\varphi}_{i}
{\varphi}_{\nu} \in \mathcal{S}^{0}$. In addition, clearly
${\varphi}_{T^{\lambda} T} \in \mathcal{S}^{0}$ for any $T \in
\mathcal{T}_0(\lambda, \nu)$. Hence, the left side of
\eqref{eq:multiply varphi_nu from the right} is contained in
$\mathcal{S}^{0}$. On the other hand, since
$\psi \in
\mathcal{S}^{\vee \lambda}$ and
$\mathcal{S}^{\vee \lambda}$ is a two-sided
ideal of $\mathcal{S}$, $\psi
{\varphi}_{\nu}$ is an element in
$\mathcal{S}^{\vee \lambda}$. Comparing both
sides of \eqref{eq:multiply varphi_nu from the right}, we have $\psi
{\varphi}_{\nu} \in \mathcal{S}^{0} \cap
\mathcal{S}^{\vee \lambda} =
\mathcal{S}_{0}^{\vee (\lambda, 1)} \subset
\mathcal{S}_{0}^{\vee (\lambda, 0)}$. Therefore, we have
\begin{equation}\label{eq:multiply varphi_nu and working modulo
S_0^vee(lambda,0)}
\underset{T \in \mathcal{T}_0(\lambda, \nu)}{\sum} r_{T}
{\varphi}_{T^{\lambda} T}
\equiv \underset{i}{\sum} \underset{T \in
\mathcal{T}^{+}_0(\lambda)}{\sum} r_{T}^{i}
{\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu} \mod \mathcal{S}_{0}^{\vee (\lambda,0)}
\end{equation}
for any $\nu \in \Lambda$ with $\alpha(\nu) =
\alpha(\lambda)$. We note that ${\varphi}_{T^{\lambda} T}
{\varphi}_{i}
{\varphi}_{\nu} =
{\varphi}_{T^{\lambda} T^{\lambda}} \cdot
{\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu}$ with ${\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu} \in \mathcal{S}^{0}$. Then \eqref{eq:multiply
varphi_nu and working modulo
S_0^vee(lambda,0)} implies that
\begin{equation}
\sum_{T \in
\mathcal{T}_0(\lambda, \nu)} r_{T}
{\varphi}^{0}_{T} = \sum_{i} \sum_{T \in \mathcal{T}^{+}_0(\lambda)}
r_{T}^{i} {\varphi}^{0}_{T^{\lambda}} \cdot
{\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu}.
\end{equation}
It follows that
\begin{equation}\label{eq:summing up formulas}
\underset{T \in \mathcal{T}_0^{+}(\lambda)}{\sum} r_{T}
{\varphi}^{0}_{T} =
\underset{
\substack{
\nu \in \Lambda\\
\alpha(\nu) = \alpha(\lambda)
}
}
{\sum}
\underset{T \in \mathcal{T}_0(\lambda, \nu)}{\sum}
r_{T} {\varphi}^{0}_{T}
=
\underset{
\substack{
\nu \in \Lambda\\
\alpha(\nu) = \alpha(\lambda)
}
}
{\sum}
\underset{i}{\sum}
\underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum}
r_{T}^{i} {\varphi}^{0}_{T^{\lambda}}
\cdot {\varphi}_{T^{\lambda} T} {\varphi}_{i}
{\varphi}_{\nu}.
\end{equation}
Put ${\varphi}_{\Sigma} = \underset{\mu \in
\Lambda, ~~ \alpha(\mu) = \alpha(\lambda)}{\sum}
{\varphi}_{\mu}$. Then ${\varphi}_{T^{\lambda}
T} = {\varphi}_{T^{\lambda}
T} {\varphi}_{\Sigma}$ for all $T \in
\mathcal{T}^{+}_0(\lambda)$ and, by using Lemma
\ref{lem:multiplication formula} again,
${\varphi}_{\Sigma} {\varphi}_{i}
{\varphi}_{\nu}$ is an $R$-linear combination of elements
of the form ${\varphi}_{UV}$ where $U \in
\mathcal{T}_{0}({\lambda}', \mu)$ and $V \in
\mathcal{T}_{0}({\lambda}', \nu)$ with $\alpha(\mu) =
\alpha(\lambda) =\alpha(\nu)$ for various ${\lambda}' \in
\Lambda^{+}$ and $\mu \in
\Lambda$. Therefore
${\varphi}_{\Sigma} {\varphi}_{i}
{\varphi}_{\nu} \in \mathcal{S}^{0}$. We also know that
${\varphi}_{T^{\lambda} T} \in \mathcal{S}^{0}$ for $T
\in \mathcal{T}^{+}_0(\lambda)$. Combining these facts we can
compute
\begin{equation*}
{\varphi}^{0}_{T^{\lambda}}
\cdot {\varphi}_{T^{\lambda} T}
{\varphi}_{i}
{\varphi}_{\nu}
={\varphi}^{0}_{T^{\lambda}}
\cdot {\varphi}_{T^{\lambda} T}
{\varphi}_{\Sigma} {\varphi}_{i}
{\varphi}_{\nu}
=( {\varphi}^{0}_{T^{\lambda}}
\cdot {\varphi}_{T^{\lambda} T} )
\cdot {\varphi}_{\Sigma}
{\varphi}_{i}
{\varphi}_{\nu}
={\varphi}^{0}_{T}
\cdot {\varphi}_{\Sigma}
{\varphi}_{i}
{\varphi}_{\nu}.
\end{equation*}
Substituting this into \eqref{eq:summing up formulas}, we have
\begin{equation}
\begin{array}{l}
x
= \underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum} r_{T}
{\varphi}^{0}_{T} = \underset{
\substack{
\nu \in \Lambda \\
\alpha(\nu) =\alpha(\lambda)
}
}{\sum}
\underset{i}{\sum}
\underset{T \in \mathcal{T}^{+}_0(\lambda)}{\sum}
r_{T}^{i} {\varphi}^{0}_{T}
\cdot {\varphi}_{\Sigma}
{\varphi}_{i}
{\varphi}_{\nu}
= \underset{
\substack{
\nu \in \Lambda \\
\alpha(\nu) =\alpha(\lambda)
}
}{\sum}
\underset{i}{\sum}
y_{i} \cdot {\varphi}_{\Sigma}
{\varphi}_{i}
{\varphi}_{\nu}.
\end{array}
\end{equation}
Since ${\varphi}_{\Sigma}
{\varphi}_{i}
{\varphi}_{\nu} \in \mathcal{S}^{0}$ and $y_{i} \in M_1$ we have
$y_{i} \cdot {\varphi}_{\Sigma}
{\varphi}_{i}
{\varphi}_{\nu} \in M_1$. Hence $x \in M_1$, a
contradiction. Therefore $I_2( x {\otimes} 1 ) \not\in I_1( M_1
{\otimes} \mathcal{S} )$, and this proves (ii). The
Lemma is proved.
\end{proof}
\begin{lem}\label{lem:the unique maximal proper right
widetildeS-submodule of L^lambda tensor widetildeS}
Suppose that $R$ is a field. For every $\lambda \in
\Lambda^{+}$, let ~ ${\iota} :
{\rm{rad}}Z^{(\lambda, 0)} \rightarrow Z^{(\lambda, 0)}$ be the
inclusion map. Then there exists an $\mathcal{S}$-submodule
$N^{\lambda}$ of
$L_{0}^{\lambda} \otimes_{\mathcal{S}^{0}} \mathcal{S}$ such
that
\begin{equation}\label{eq:unique maximal submodule of L_0^lambda
otimes S}
( L_{0}^{\lambda} \otimes_{\mathcal{S}^{0}} \mathcal{S} ) /
N^{\lambda} \simeq L^{\lambda}
\end{equation}
as $\mathcal{S}$-modules. Moreover,
$N^{\lambda}$ is the unique maximal submodule of $L_{0}^{\lambda}
\otimes_{\mathcal{S}^{0}} \mathcal{S}$.
\end{lem}
\begin{proof}
By the definition of $L_{0}^{\lambda}$, for each $\lambda \in
\Lambda^{+}$, there exists a short exact sequence of
$\mathcal{S}^{0}$-modules
\begin{equation*}
\begin{CD}
0 @>>> \text{rad}Z^{(\lambda, 0)} @>>> Z^{(\lambda,
0)} @>>> L_{0}^{\lambda} @>>> 0.
\end{CD}
\end{equation*}
This sequence induces an exact sequence of right
$\mathcal{S}$-module
\begin{equation*}
\begin{CD}
\text{rad}Z^{(\lambda, 0)} \otimes \mathcal{S}
@>I>> Z^{(\lambda,
0)} \otimes \mathcal{S} @>>> L_{0}^{\lambda} \otimes
\mathcal{S} @>>> 0,
\end{CD}
\end{equation*}
with $I = \iota \otimes \text{id}_{\mathcal{S}}$. Further, note that
$\text{rad}W^{\lambda}$ is the unique
maximal submodule of
$W^{\lambda}$ and $L^{\lambda} \neq 0$ for every
$\lambda \in \Lambda^{+}$. Hence under the isomorphism in
Theorem \ref{thm:tensor theorem}, there exists an
$\mathcal{S}$-submodule $M \simeq
\text{rad}W^{\lambda}$ of $Z^{(\lambda, 0)} \otimes
\mathcal{S}$ such that $\text{Im}( I ) \subset M \subsetneqq
Z^{(\lambda,
0)} \otimes S$.
Then we have an isomorphism $f : ( Z^{(\lambda, 0)} \otimes S ) /
\text{Im}( I ) \xrightarrow{\sim} L_{0}^{\lambda} \otimes
\mathcal{S}$. If we put $N^{\lambda}
= f( M / \text{Im}( I ) )$, we obtain \eqref{eq:unique maximal
submodule of L_0^lambda
otimes S}.
The uniqueness of
$N^{\lambda}$ follows from the uniqueness of
$\text{rad}W^{\lambda}$ and we are done.
\end{proof}
\begin{lem}\label{lem:the key Lemma2}
Suppose that $R$ is a field. For any $\lambda \in
\Lambda^{+}$, we regard $L^{\lambda}$ as an $\mathcal{S}^{0}$-module
via the restriction. Then
$\mathcal{S}^{0}$-module $L^{\lambda}$ contains
$L_{0}^{\lambda}$ as an irreducible submodule.
\end{lem}
\begin{proof}
This Lemma can be proved in a similar way as \cite[Lemma
6.8]{SawS}. However, this fact itself is important for later
discussions, we give here the proof.
Recall the bilinear form $\langle ~ , ~ \rangle$ (resp. $\langle ~ , ~
\rangle_{0}$) on $W^{\lambda}$ (resp. on $Z^{(\lambda, 0)}$) and the
injective $\mathcal{S}^{0}$-module homomorphism $f_{\lambda} :
Z^{(\lambda, 0)} \rightarrow W^{\lambda}$ in Lemma
\ref{lem:injective homomorphism f_lambda from
Z^(lambda,0) to W^lambda_natural}. We note that $f_{\lambda}$ maps
$\text{rad}Z^{(\lambda, 0)}$ into
$\text{rad}W^{\lambda}$. In fact, by \eqref{eq:the
relationship between <,>_0 to <,>_{natural}}, $\langle
{\varphi}_{S}, {\varphi}_{T} \rangle
= \langle
{\varphi}^{0}_{S}, {\varphi}^{0}_{T} \rangle_{0}$
for any $S, T \in \mathcal{T}^{+}_0(\lambda)$. On the other hand, if
$S \in \mathcal{T}^{+}_0(\lambda, \mu), T \in \mathcal{T}_0(\lambda,
\nu) \setminus \mathcal{T}^{+}_0(\lambda, \nu)$, then $\alpha(\mu)
\neq \alpha(\nu)$, and so $\mu \neq \nu$. It follows that
${\varphi}_{T^{\lambda} S} {\varphi}_{T
T^{\lambda}} = 0$, and $\langle {\varphi}_{S},
{\varphi}_{T} \rangle = 0$. Hence
$f_{\lambda}(\text{rad}Z^{(\lambda, 0)}) \subset
\text{rad}W^{\lambda}$. Now $f_{\lambda}$ induces an
$\mathcal{S}^{0}$-module homomorphism $L_{0}^{\lambda} \rightarrow
L^{\lambda}$, which is clearly non-trivial. Since $L_{0}^{\lambda}$
is an irreducible $\mathcal{S}^{0}$-module, $L^{\lambda}$ contains an
irreducible module isomorphic to $L_{0}^{\lambda}$.
\end{proof}
We can now state our main result.
\begin{thm}\label{thm:On a composition multiplicity for the cyclotomic
q-Schur algebra 1}
Suppose that $R$ is a field. Then, for all $\lambda, \mu \in
\Lambda^{+}$, we have
\begin{equation*}
[ \overline{Z}^{\lambda} : \overline{L}^{\mu} ] = [ Z^{(\lambda, 0)} :
L_{0}^{\mu} ] \le [ W^{\lambda} : L^{\mu} ].
\end{equation*}
\end{thm}
\begin{proof}
Since the first equality was shown in Theorem \ref{thm:the property of
L^lambda and
[overlineZ^lambda:overlineL^mu]=[Z^(lambda,0):L_0^mu]}, it is
enough to see that $[
Z^{(\lambda, 0)} : L_{0}^{\mu} ] \le [ W^{\lambda} :
L^{\mu} ]$. Let
\begin{equation*}
0 = M_0 \varsubsetneqq M_1 \varsubsetneqq \cdots \varsubsetneqq M_{l-1}
\varsubsetneqq M_{l} = Z^{(\lambda, 0)}
\end{equation*}
be a composition series of $Z^{(\lambda, 0)}$, and for $k = 0, \ldots
, l-1$ assume that $M_{k+1} / M_{k} \simeq
L_{0}^{\mu_{k}}$ with some $\mu_{k} \in
\Lambda^{+}$. Let
${\iota}_{k} : M_{k} \hookrightarrow M_{k+1}$ be the inclusion map and
put $I_{k} = {\iota}_{k} \otimes \text{id}_{\mathcal{S}} :
M_{k} \otimes \mathcal{S} \rightarrow M_{k+1} \otimes
\mathcal{S}$. Note that
$I_{k}$ is an $\mathcal{S}$-module
homomorphism. Then we have an exact sequence $0 \rightarrow M_{k}
\rightarrow M_{k+1}
\rightarrow L_{0}^{\mu_{k}} \rightarrow 0$ of
$\mathcal{S}^{0}$-modules. Then, we obtain an $\mathcal{S}$-module
exact sequence
\begin{equation}\label{eq:S-module exact sequence}
\begin{CD}
M_{k} \otimes \mathcal{S} @>{I_{k}}>> M_{k+1}
\otimes \mathcal{S} @>>>
L_{0}^{\mu_{k}} \otimes \mathcal{S} @>>> 0.
\end{CD}
\end{equation}
Let ${\iota}'_{k} : M_{k}
\hookrightarrow Z^{(\lambda, 0)}$ be the inclusion map and put $M_{k}
= ( {\iota}'_{k} \otimes \text{id}_{\mathcal{S}} )( M_{k} \otimes
\mathcal{S}
)$. Then
by Lemma \ref{lem:the
key Lemma1} (i), (ii), we have a filtration of $\mathcal{S}$-module
in $W^{\lambda}$
\begin{equation*}
0 = \mathcal{M}_{0} \varsubsetneqq
\mathcal{M}_{1} \varsubsetneqq \mathcal{M}_{2}
\varsubsetneqq \cdots \varsubsetneqq \mathcal{M}_{l-1}
\varsubsetneqq \mathcal{M}_{l} = W^{\lambda}.
\end{equation*}
We shall compute $\mathcal{M}_{k+1} / \mathcal{M}_{k}$ for $0 \le
k < l$. Let $f : M_{k+1} \otimes \mathcal{S}
\rightarrow ( {\iota}'_{k+1} \otimes
\text{id}_{\mathcal{S}} )( M_{k+1} \otimes
\mathcal{S} ) / ( {\iota}'_{k} \otimes
\text{id}_{\mathcal{S}} )( M_{k} \otimes
\mathcal{S} ) \simeq \mathcal{M}_{k+1} /
\mathcal{M}_{k}$ be the natural map. Clearly, $f$ is a surjective
$\mathcal{S}$-module homomorphism. Moreover, it is easy to see that
$\ker{f} \supset I_{k}(
M_{k} \otimes \mathcal{S} )$. Hence $f$ induces an
$\mathcal{S}$-module surjective map $\widetilde{f} : ( M_{k+1} \otimes
\mathcal{S} ) /
I_{k}( M_{k} \otimes \mathcal{S} ) \rightarrow
\mathcal{M}_{k+1} / \mathcal{M}_{k}$. Since
$\mathcal{M}_{k} \varsubsetneqq
\mathcal{M}_{k+1}$, $\widetilde{f}$ is a non-zero
map. Note that $( M_{k+1} \otimes \mathcal{S} ) / I_{k}(
M_{k} \otimes \mathcal{S} ) \simeq L_{0}^{{\mu}_{k}} \otimes
\mathcal{S}$ by \eqref{eq:S-module exact sequence}. Hence, there
exists a certain $\mathcal{S}$-submodule $\mathcal{N}_{k}$ of
$L_{0}^{{\mu}_{k}} \otimes \mathcal{S}$ such that $(
L_{0}^{{\mu}_{k}} \otimes \mathcal{S} ) / \mathcal{N}_{k}
\simeq \mathcal{M}_{k+1} / \mathcal{M}_{k}$. However, by Lemma
\ref{lem:the unique maximal proper right widetildeS-submodule of
L^lambda tensor widetildeS}, there exists a unique maximal submodule
$N^{{\mu}_{k}}$ of
$L_{0}^{{\mu}_{k}} \otimes \mathcal{S}$ such that $(
L_{0}^{{\mu}_{k}} \otimes \mathcal{S} ) /
N^{{\mu}_{k}} \simeq L^{{\mu}_{k}}$. Since $N^{{\mu}_{k}} \supset
\mathcal{N}_{k}$, we have a surjective map
\begin{equation*}
\mathcal{M}_{k+1} / \mathcal{M}_{k} \simeq ( L_{0}^{{\mu}_{k}} \otimes
\mathcal{S} ) /
\mathcal{N}_{k} \rightarrow ( L_{0}^{{\mu}_{k}}
\otimes
\mathcal{S} ) / N^{{\mu}_{k}} \simeq L^{{\mu}_{k}}.
\end{equation*}
This proves that $[ Z^{(\lambda, 0)} : L_{0}^{\mu} ] \le [ W^{\lambda}
: L^{\mu} ]$.
\end{proof}
In the following special case, we have a more precise formula.
\begin{thm}\label{thm:On a composition multiplicity for the cyclotomic
q-Schur algebra 2}
Suppose that $R$ is a field. Then, for all $\lambda, \mu \in
\Lambda^{+}$ with $\alpha( \lambda ) = \alpha( \mu )$,
\begin{equation*}
[ \overline{Z}^{\lambda} : \overline{L}^{\mu} ] = [ Z^{(\lambda, 0)} :
L_{0}^{\mu} ] = [ W^{\lambda} : L^{\mu} ].
\end{equation*}
\end{thm}
\begin{proof}
Assume that $\alpha(\lambda) = \alpha(\mu)$ for $\lambda$, $\mu \in
\Lambda^{+}$. In view of Theorem \ref{thm:On a composition
multiplicity for the cyclotomic
q-Schur algebra 1}, it is enough to show that $[
W^{\lambda} :
L^{\mu} ] \le [ Z^{(\lambda, 0)} :
L_{0}^{\mu} ]$. Take $\lambda \in \Lambda^{+}$ and let
\begin{equation*}
0 = W_{0}
\varsubsetneqq W_{1} \varsubsetneqq \cdots \varsubsetneqq
W_{c} = W^{\lambda}
\end{equation*}
be a composition series of
$W^{\lambda}$ and assume that
$W_{i+1} / W_{i} \simeq
L^{{\mu}_{i}}$. We regard $W_{i}$, $L^{{\mu}_{j}}$ as
$\mathcal{S}^{0}$-modules, via the
restriction. By Lemma \ref{lem:the key Lemma2}, there exists a
submodule $W'_{i}$ of $W_{i+1}$ containing $W_{i}$ such that $W'_{i} /
W_{i} \simeq L_{0}^{{\mu}_{i}}$. Moreover, by Proposition
\ref{prop:the
property of L^lambda and radZ^(lambda,0)}, $L_{0}^{\mu} \neq 0$ for
all
$\mu \in \Lambda^{+}$. Therefore, $W_{i}
\varsubsetneqq W'_{i} \subset W_{i+1}$.
Recall the injective $\mathcal{S}^{0}$-module map $f_{\lambda} :
Z^{(\lambda, 0)}
\rightarrow W^{\lambda}$ given in Lemma \ref{lem:injective
homomorphism f_lambda from Z^(lambda,0) to W^lambda_natural} and put
$M_{i} = f_{\lambda}( Z^{(\lambda, 0)} ) \cap W_{i}$,
$M'_{i} = f_{\lambda}( Z^{(\lambda, 0)} ) \cap W'_{i}$. Then we
obtain a filtration of $\mathcal{S}^{0}$-modules
\begin{equation*}
0 = M_{0} \subset M'_{0} \subset M_{1} \subset M'_{1} \subset \cdots
\subset M_{c-1} \subset M'_{c-1} \subset M_{c} = f_{\lambda}(
Z^{(\lambda, 0)} )
\end{equation*}
of $f_{\lambda}( Z^{(\lambda, 0)} )$. Since $f_{\lambda}$ is an
injection from $Z^{(\lambda, 0)}$ to $W^{\lambda}$, we can regard the
above filtration as
a filtration of $Z^{(\lambda, 0)}$. We claim that
\begin{equation}\label{eq:M_i neq M'_i if alpha(mu_i)= alpha(lambda)}
M_{i} \neq M'_{i} ~ \text{ if } ~ \alpha( \mu_{i} ) = \alpha(
\lambda).
\end{equation}
We show \eqref{eq:M_i neq M'_i if alpha(mu_i)= alpha(lambda)}. Assume
that $\alpha( \mu_{i} ) = \alpha(
\lambda )$. Note that, as in the proof of Proposition \ref{prop:the
property of L^lambda and radZ^(lambda,0)},
$\varphi^{0}_{T^{{\mu}_{i}}} + \text{rad}{Z^{({\mu}_{i}, 0)}}$ is a
non-zero element of $L_{0}^{{\mu}_{i}}$. Then, there exists
$x_{{\mu}_{i}}
\in W'_{i} / W_{i}$ corresponding to $\varphi^{0}_{T^{{\mu}_{i}}} +
\text{rad}{Z^{({\mu}_{i}, 0)}}$. Take a representative
$\widetilde{x}_{{\mu}_{i}} \in W'_{i} \setminus W_{i}$ of
$x_{{\mu}_{i}}$. Take $\varphi_{{\mu}_{i}} \in \mathcal{S}^{0}$
(cf. \eqref{eq:identity map
on M^mu_natural}). Since $\widetilde{x}_{{\mu}_{i}} \in
W'_{i}$ and $W'_{i}$ is an $\mathcal{S}^{0}$-submodule of
$W^{\lambda}$, we have $\widetilde{x}_{{\mu}_{i}} \varphi_{{\mu}_{i}}
\in
W'_{i}$. Moreover, $\widetilde{x}_{{\mu}_{i}} \varphi_{{\mu}_{i}}$
corresponds to
\begin{equation*}
(\varphi^{0}_{T^{{\mu}_{i}}} + \text{rad}{Z^{({\mu}_{i}, 0)}} ) \cdot
\varphi_{{\mu}_{i}} = \varphi^{0}_{T^{{\mu}_{i}}} +
\text{rad}{Z^{({\mu}_{i}, 0)}} \neq 0
\end{equation*}
Therefore, $\widetilde{x}_{{\mu}_{i}}
\varphi_{{\mu}_{i}} \in W'_{i} \setminus W_{i}$.
On the other hand, since $\widetilde{x}_{{\mu}_{i}} \in
W_{i+1} \subset W^{\lambda}$, we can write $\widetilde{x}_{{\mu}_{i}}
= \sum_{T \in \mathcal{T}_{0}(\lambda)} r_{T} \varphi_{T} ~ ( r_{T}
\in R )$ and hence $\widetilde{x}_{{\mu}_{i}}
\varphi_{{\mu}_{i}} = \sum_{T \in \mathcal{T}_{0}(\lambda,
{\mu}_{i})} r_{T} \varphi_{T}$ since $\varphi_{{\mu}_{i}}$ is an
identity map on $M^{{\mu}_{i}}$ and is zero otherwise. By noticing
$\alpha({\mu}_{i}) = \alpha(\lambda)$, we define an element
$y_{{\mu}_{i}} \in Z^{(\lambda, 0)}$ by $y_{{\mu}_{i}} = \sum_{T \in
\mathcal{T}_{0}(\lambda, {\mu}_{i})}
r_{T} \varphi^{0}_{T}$. Then we have $f_{\lambda}( y_{{\mu}_{i}} ) =
\sum_{T \in
\mathcal{T}_{0}(\lambda, {\mu}_{i})}
r_{T} \varphi_{T} = \widetilde{x}_{{\mu}_{i}}
\varphi_{{\mu}_{i}}$. It follows that $\widetilde{x}_{{\mu}_{i}}
\varphi_{{\mu}_{i}} \in \{ f_{\lambda}( Z^{(\lambda, 0)} ) \cap
W'_{i} \} \setminus \{
f_{\lambda}( Z^{(\lambda, 0)} ) \cap W_{i} \}$.
Therefore, $M_{i} \neq
M'_{i}$ and \eqref{eq:M_i neq M'_i if alpha(mu_i)= alpha(lambda)}
holds.
By the claim \eqref{eq:M_i neq M'_i if alpha(mu_i)= alpha(lambda)},
the quotient $M'_{i} / M_{i}$ is a non-zero
$\mathcal{S}^{0}$-submodule of $L_{0}^{{\mu}_{i}}$. Hence $M'_{i} /
M_{i} \simeq L_{0}^{{\mu}_{i}}$. This proves that $[ W^{\lambda} :
L^{\mu} ] \le [ Z^{(\lambda, 0)} : L_{0}^{\mu} ]$ for $\lambda$, $\mu
\in \Lambda^{+}$ such that $\alpha(\lambda) = \alpha(\mu)$. The
theorem is proved.
\end{proof}
\addtocounter{subsection}{6}
\subsection{}
We return to the setting in \ref{sec:modified Ariki-Koike
algebra}. Let $\Lambda = \widetilde{\mathcal{P}}_{n,r}$ under the
condition \eqref{eq:The condition of the Modified Ariki-Koike} and
\eqref{con:m_i>=n}. For an
$r$-partition $\lambda \in \mathcal{P}_{n,r}$, we denote by
$\mathcal{S}_{\flat}^{\vee \lambda}$ the $R$-submodule of
$\mathcal{S}^{\flat}(\bs{m}, n)$ spanned by $\varphi^{\flat}_{ST}$
such that $S, T \in \mathcal{T}^{+}_{0}(\alpha)$ with $\alpha \rhd
\lambda$. Moreover, for an
$r$-partition $\lambda \in \mathcal{P}_{n,r}$, $T^{\lambda} \in
\mathcal{T}^{+}_{0}(\lambda, \lambda)$, and in fact $T^{\lambda}$ is
the unique semistandard $\lambda$-Tableau of type $\lambda$. Moreover,
$\mathfrak{t} = \mathfrak{t}^{\lambda}$ is the unique element in
$\text{Std}(\lambda)$ such that $\lambda( \mathfrak{t} ) =
T^{\lambda}$. Thus, $m^{\flat}_{T^{\lambda} T^{\lambda}} =
m^{\flat}_{\mathfrak{t}^{\lambda} \mathfrak{t}^{\lambda}} =
m^{\flat}_{\lambda}$, and $\varphi^{\flat}_{\lambda} =
\varphi^{\flat}_{T^{\lambda} T^{\lambda}}$ is the identity map on
$M_{\flat}^{\lambda}$. We define the Weyl
module $W_{\flat}^{\lambda}$ as the right $\mathcal{S}^{\flat}(\bs{m},
n)$-submodule of $\mathcal{S}^{\flat}(\bs{m},
n) / \mathcal{S}_{\flat}^{\vee \lambda}$ spanned by the image of
$\varphi^{\flat}_{\lambda}$. For each $T \in
\mathcal{T}^{+}_{0}(\lambda, \mu)$, we denote by $\varphi^{\flat}_{T}$
the image of $\varphi^{\flat}_{T^{\lambda} T}$ in
$\mathcal{S}^{\flat}(\bs{m},
n) / \mathcal{S}_{\flat}^{\vee \lambda}$. Then we know that the Weyl
module $W_{\flat}^{\lambda}$ is $R$-free with basis $\{
\varphi^{\flat}_{T} \mid T \in \mathcal{T}^{+}_{0}(\lambda) \}$.
The Weyl module $W_{\flat}^{\lambda}$ enjoys an associative symmetric
bilinear form, defined by the equation
\begin{equation*}
\varphi^{\flat}_{T^{\lambda} S} \varphi^{\flat}_{T T^{\lambda}} \equiv
\langle \varphi^{\flat}_{S}, \varphi^{\flat}_{T} \rangle_{\flat}
\cdot \varphi^{\flat}_{\lambda} \mod \mathcal{S}_{\flat}^{\vee \lambda}
\end{equation*}
for all $S, T \in \mathcal{T}^{+}_{0}(\lambda)$. Let
$L_{\flat}^{\lambda} = W_{\flat} / \text{rad}{W_{\flat}^{\lambda}}$,
where $\text{rad}{W_{\flat}^{\lambda}} = \{ x \in W_{\flat}^{\lambda}
\mid \langle x, y \rangle_{\flat} = 0 \text{ for all } y \in
W_{\flat}^{\lambda} \}$. By \cite[Proposition 5.11]{SawS}, we
know that, for all $r$-partition $\lambda \in \mathcal{P}_{n,r}$,
$L_{\flat}^{\lambda}$ is an absolutely irreducible and $\{
L_{\flat}^{\lambda} \mid
\lambda \in \mathcal{P}_{n,r} \}$ is a complete set of non-isomorphic
irreducible $\mathcal{S}^{\flat}(\bs{m}, n)$-modules. Furthermore, for
$\lambda,
\mu \in \mathcal{P}_{n,r}$, we denote by $[
W_{\flat}^{\lambda} : L_{\flat}^{\mu} ]$ the composition
multiplicity of $L_{\flat}^{\mu}$ in $W_{\flat}^{\lambda}$.
Note that the above definition of the Weyl module
$W_{\flat}^{\lambda}$ coincides with the
definition of the Weyl module $\overline{Z}^{\lambda}$ when
$\mathcal{S}^{\flat}(\bs{m}, n)$ is isomorphic to
$\mathcal{S}^{\overline0}(\Lambda)$ under the isomorphism $\flat$ in
Proposition \ref{prop:the relationship between S^overline0 and
S^0}. Consequently, under the isomorphism $\flat$, we have $[
W_{\flat}^{\lambda} : L_{\flat}^{\mu} ] = [
\overline{Z}^{\lambda} : \overline{L}^{\mu} ]$ for every $\lambda,
\mu \in \mathcal{P}_{n,r}$.
On the other hand, note that in the case where $r = 1$, the notation
for
$\mathcal{S}^{\flat}(\bs{m}, n)$ coincides with the standard notation
for $q$-Schur algebras discussed as in \cite[Chapter 4]{M1}. So,
we use freely such a notation. For $\lambda, \mu \in
\mathcal{P}_{n,r}$, we denote by $[ W^{\lambda^{( i )}} : L^{\mu^{( i
)}} ] ~ ( 1 \le i \le r )$ is defined as the composition
multiplicity of $L^{\mu^{( i )}}$ in $W^{\mu^{( i )}}$ for $\lambda =
( \lambda^{( 1 )}, \ldots , \lambda^{( r )} )$ and $\mu = ( \mu^{( i
)}, \ldots , \mu^{( r )} )$.
\addtocounter{thm}{1}
\begin{prop}[{\cite[Proposition 5.14]{SawS}}]\label{prop:on
decomposition number}
Let $\Lambda = \widetilde{\mathcal{P}}_{n,r}$. Suppose that $R$ is a
field, and that \eqref{eq:The condition of the Modified Ariki-Koike}
and \eqref{con:m_i>=n} are satisfied. Let $\lambda$, $\mu \in
\mathcal{P}_{n,r}$. Then under the isomorphism in Theorem \ref{weak
separation condition}, we have
\begin{equation*}
[ W^{\lambda} : L^{\mu} ] =
\begin{cases}
\prod_{i = 1}^{r} [ W^{{\lambda}^{( i )}} : L^{{\mu}^{( i )}} ] &
\text{ if } \alpha(\lambda) = \alpha(\mu), \\
0 & \text{ otherwise}.
\end{cases}
\end{equation*}
\end{prop}
Then, combining Theorem \ref{thm:On a composition multiplicity
for the cyclotomic q-Schur algebra 2} and Proposition \ref{prop:the
relationship
between S^overline0 and S^0} and Proposition \ref{prop:on
decomposition number}, we have the following.
\begin{cor}
Let $\Lambda = \widetilde{\mathcal{P}}_{n,r}$. Suppose that $R$ is a
field, and that \eqref{eq:The condition of the Modified Ariki-Koike}
and \eqref{con:m_i>=n} are satisfied. Then, for
all $\lambda, \mu \in
\mathcal{P}_{n,r}$ with $\alpha( \lambda ) = \alpha( \mu )$, we have
\begin{equation*}
[ W^{\lambda} : L^{\mu} ] =
\overset{r}{\underset{i=1}{\prod}} [ W^{\lambda^{( i )}} : L^{\mu^{( i
)}} ].
\end{equation*}
\end{cor}
\bigskip
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 3,204
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Death Cab for Cutie Announce Livestream for Plans 15th Anniversary
By Lexi Lane | September 1, 2020 | 11:20am
Photo by Eliot Lee Hazel Music News Death Cab for Cutie
Death Cab for Cutie will host a live broadcast on Sept. 5 on YouTube of Directions: The Plans Video Album at 5 p.m. ET.
The livestream is in honor of Death Cab's Plans turning 15 in August. Directions was originally released as a visual accompaniment to the album, highlighting 11 short films from different directors for each song on the album.
"15 years ago we made our major label debut with our fifth studio album Plans," said bassist Nick Harmer. "We told ourselves that it was just another album, that now being on Atlantic Records was no big deal, but it was and we knew it. I think that's why we chose the barn studio at Long View Farms in Massachusetts, during winter, to isolate ourselves from the pressure and our fears and just be able to focus on music. We weren't expecting getting snowed in and that gave the experience a heaping tablespoon of The Shining but in the end, we were able to shut out the world and make album that came from our hearts, a promise to ourselves that no matter what came at us next, we would always remember our music comes first."
The band have also announced "We Have The Facts And We're Voting 2020," a fundraiser contest to maintain voter rights in the upcoming election. Participants will be entered for anything from limited edition merch, a private acoustic concert and autographed test pressings. Proceeds raised from the raffle will go to Fair Fight Action and Future Now.
Listen to Death Cab for Cutie's 2008 Daytrotter session below.
More from Death Cab for Cutie
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Hear Death Cab for Cutie Play Songs From Transatlanticism & More on This Day in 2008By Ellen JohnsonApril 28, 2020 | 1:06pm
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10 New Albums to Stream TodayBy Paste StaffSeptember 6, 2019 | 11:18am
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,613
|
Persone
Giorgio Faletti – scrittore, attore e cantautore italiano
Guglielmo Faletti – vescovo italiano
Altro
Faletti's Hotel – storico albergo di Lahore (Pakistan)
Pagine correlate
Falletti
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,666
|
\section{Introduction}
Massive Open Online Courses (MOOCs), as the name suggests, are online courses open to anyone and aimed at teaching large audiences for free. The only entrance barrier to most of these courses is having access to a computer or a smart device with an internet connection. This, among other factors, causes them to have become increasingly popular over the last few years and attract hundreds or thousands of users in some cases. However, the number of students that effectively complete the courses is significantly smaller, as the dropout rate of many MOOCs is above 90\% \cite{khalil2014moocs}. Having significant amounts of data collected on the users within a course lends itself well for quantitative statistical analysis with respect to the factors indicative of completion or dropout.
This paper focuses on applying feature extraction and machine learning classification techniques on Maastricht University's MOOC on Problem Based Learning in order to better assess dropout behavior of learners. The contribution of this paper is threefold: First, a methodology of extracting features that describe user behavior within a MOOC (approach is applied to a specific dataset but can easily be extended to others as well) is described. Second, a machine learning framework (comparing three different algorithms, namely Logistic Regression, Random Forest and AdaBoost) to predict dropout within a temporal context (i.e. which is the exact time that a user will dropout) is presented. Finally, we conduct an analysis on which features are strong indicators of dropout (or persistence) from a MOOC. Novelty of this paper lies in the combination of static and temporal features while providing a straightforward machine learning framework for both predicting and highlighting indicators for dropouts.
The rest of the paper is organized as follows. Section \ref{sect:related} presents recent works relevant to dropout prediction. The dataset used in this paper is presented in Section \ref{sect:dataset}, followed by the methodology description in Section \ref{sect:tools}. Experimental setup and results are presented in Section \ref{sect:exp} and finally Section \ref{sect:discussion} concludes the paper.
\section{Related work}
\label{sect:related}
There have been several approaches tackling user dropout in MOOCs, but only a few take into account the temporal aspect. In \cite{balakrishnan2013predicting} authors utilized Hidden Markov Models to predict the persistence of a user in a MOOC, however their accuracy is not satisfying. Onah et. al \cite{onah2014dropout} delve into finding predictors of dropout by analyzing the behaviour and profile data of users. They conclude that the presence of interactions between users and tutors is a strong indicator of persistence, but their conclusions are not supported by any machine learning algorithm. Halawa et. al \cite{halawa2014dropout} explore many time related features, in order to find strong indicators of dropout, however their classification algorithm is very simple and could be improved. All these works attempt to predict one week ahead, except for \cite{ramesh2014learning} where they predict at three different time points during the course and \cite{taylor2014likely} where they provide predictions for all lags. Current work goes beyond \cite{taylor2014likely} by including profile information of learners so that prediction is also available when the course starts. Moreover, we compare three classification algorithms and come up with a way to measure feature importance.
\section{Dataset}
\label{sect:dataset}
As mentioned previously, the dataset used in this paper is from Maastricht University's MOOC on Problem Based Learning that was offered from October till December of 2015. In total, after removing duplicates, there were 2769 users registered for the course, but only 358 of those completed it. This represents a dropout rate of 87\%, which corresponds to the rates found in similar literature \cite{taylor2014likely}. It is interesting to note that 75\% of the users failed to submit any assignment, and of those that submitted at least one assignment, 51\% completed the course. Furthermore, it out of the 37\% of users that left their profile empty, less than 1\% completed the course.Figure \ref{fig:dropouts} shows the number of users that dropped out and those that remained per week.This graph shows a peak of dropouts in week 0, corresponding to users that did not start, and slightly decreasing trend for the rests of the weeks. Furthermore, it shows that for most weeks the minority class, dropping out, is a rare event and rationalizes the use of specific techniques and metrics described in Sections \ref{sect:tools} and \ref{sect:exp} for dealing with such scarceness.
The dataset is composed of profile data, course activity data, forum data, video data and team data.
All the features have to be dated as the classifiers should not use data posterior to the predicted week, hence team data and video data are omitted due to the lack of temporal information.
\begin{figure}[h]
\centering
\scalebox{0.9}{
\includegraphics[width = \linewidth]{Dropoutsperweek}}
\caption{Number of users dropped out and remaining per week, where R represents the remaining users and D the dropped out ones}
\label{fig:dropouts}
\end{figure}
\section{Methodology}
\label{sect:tools}
The dropout prediction framework consists of two main modules: the feature extraction and the prediction/classification algorithm. These sections describe in detail the process of constructing features from the data and the three classification algorithms tested.
\subsection{Feature extraction}
\begin{table}[]
\centering
\caption{Profile Data features}
\label{pdfeatures}
\scalebox{0.9}{
\begin{tabular}{|l|l|}
\hline
\textbf{Label} & \textbf{Definition {[}levels{]}} \\ \hline
\textit{pd\_a} & \textit{Country of residence {[}113{]}} \\ \hline
\textit{pd\_b} & \textit{Primary language {[}62{]}} \\ \hline
pd\_c & Gender {[}4{]} \\ \hline
pd\_d & Biography \\ \hline
pd\_e* & Track chosen {[}3{]} \\ \hline
\textbf{pd\_f} & \textbf{Range of age of the user {[}5{]}} \\ \hline
pd\_g* & Why choose this course? {[}4{]} \\ \hline
pd\_h* & Role in education {[}5{]} \\ \hline
\textbf{pd\_i} & \textbf{Education experience {[}3{]}} \\ \hline
\textbf{pd\_j} & \textbf{PBL experience {[}3{]}} \\ \hline
pd\_k* & Areas of interest {[}6{]} \\ \hline
pd\_l* & Work schedule preference {[}3{]} \\ \hline
pd\_m & Time-zone {[}25{]} \\ \hline
\textbf{pd\_n} & \textbf{Anxious to discover the content? {[}5{]}} \\ \hline
\textbf{pd\_o} & \textbf{Determination to finish the MOOC {[}5{]}} \\ \hline
pd\_p & Learning objectives \\ \hline
pd\_q & First MOOC? {[}2{]} \\ \hline
pd\_r* & Medium for finding this MOOC? {[}5{]} \\ \hline
\end{tabular}}
\end{table}
The features extracted can be grouped into two different sets. The first set contains the Profile Data features, which are not fixed in time and thus the same ones are used for all the predictions, regardless of the week predicted. The second set consists of the Forum data features, the Course Activity data features and the Google Hangout features. These are all specific to each week, which means that to predict the dropouts for a certain week, only features of that week and of the weeks before it can be used. This set of features will be referred to as the temporal features. \\
The Profile data features shown in Table \ref{pdfeatures} are all categorical except for biography and learning objectives where the values represent the standardized
length of the text written by the user. The features in \textbf{bold} are ordinal, which means that they are ordered, and conversely the rest of the categorical features which cannot be ordered, are said to be nominal. The predictors followed by an * need some disambiguation:
\begin{itemize}
\item pd\_e: the user has to choose between one of the following tracks:
\begin{itemize}
\item T1: Role of tutor in PBL
\item T2: Designing PBL problems and courses
\item T3: Assessment/organizational aspects of PBL
\end{itemize}
The track chosen will dictate which assignments must be handed in.
\item pd\_g: The user can choose one of the following motivations:
\begin{itemize}
\item Personal interest
\item Expand professional network
\item Increase career opportunities
\item other
\end{itemize}
\item pd\_h: The user can choose one of the following roles in education:
\begin{itemize}
\item Curriculum manager
\item Not involved
\item Teacher
\item Educational adviser
\item other
\end{itemize}
\item pd\_k: The user can choose one of the following areas of interest:
\begin{itemize}
\item Arts - literature - philosophy
\item Economics - Business - Trade
\item Healthy body and healthy mind
\item International relations - politics - law
\item Science - Technology
\item None of these - No difference
\end{itemize}
\item pd\_l: The user can choose of the the following options for his preferred work schedule:
\begin{itemize}
\item Synchronously
\item Asynchronously
\item No preference
\end{itemize}
\item pd\_r: The user can choose one of the following mediums for finding the course:
\begin{itemize}
\item Professional network
\item Social media
\item Maastricht University's website
\item NovoEd website
\item other
\end{itemize}
\end{itemize}
The features in \textit{italics} have been dropped during early experimental stages as it was observed that they did not positively impact the predictions and that they slowed down the classifiers significantly. Furthermore, it is important to note that each categorical feature has one more level which corresponds to Not Communicated (NC) which means that the users left it blank.
\begin{table}[]
\centering
\caption{Temporal features}
\label{temporalfeatures}
\scalebox{0.9}{
\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{\textbf{ID}} & \multicolumn{1}{c|}{\textbf{DESCRIPTION}} \\ \hline
\textbf{f} & \textbf{Interaction with the forum} \\ \hline
fp & Number of forum posts \\ \hline
fp\_l & Average forum post length \\ \hline
fr\_ba & Number of replies received \\ \hline
\textit{fr\_ba\_l} & \textit{Average length of replies received} \\ \hline
\textit{fr\_ba\_u} & \textit{Number of distinct user that replied} \\ \hline
fr\_ab & Number of replies given \\ \hline
fr\_ab\_l & Average length of replies given \\ \hline
\textit{fr\_ab\_p} & \textit{Number of distinct posts replied to} \\ \hline
fc\_ba & Number of comments received \\ \hline
\textit{fc\_ba\_l} & \textit{Average length of comments received} \\ \hline
\textit{fc\_ba\_u} & \textit{Number of distinct users that commented} \\ \hline
fc\_ab & Number of comments given \\ \hline
fc\_ab\_l & Average length of comments given \\ \hline
fc\_ab\_p & Number of distinct posts commented on \\ \hline
\textit{{[}...{]}\_i*} & \textit{\begin{tabular}[c]{@{}l@{}}All of the above containing "\_ba" but\\ where the user replying is an instructor\end{tabular}} \\ \hline
\textbf{a} & \textbf{Assignment submitted} \\ \hline
\textbf{ar} & \textbf{Assignment review submitted} \\ \hline
\textbf{gh} & \textbf{Participated at Google Hangout session} \\ \hline
\end{tabular}}
\end{table}
Table \ref{temporalfeatures} shows the temporal features. All are numerical and standardized, except for the ones in \textbf{bold} which are Boolean. The features in \textit{italics} have been dropped for the same reasons as in Table \ref{pdfeatures}.
Furthermore, "\textit{[...]\_i}" represents the 6 features containing "\_ab" but where the user replying is an instructor. From the literature in Section \ref{sect:related}, it was expected that these features would perform well. However, the instances in which these features take positive values are very rare, which leads to them decreasing the performance of the classifiers, hence they have been dropped.
\subsection{Logistic Regression}
Logistic Regression is a classification technique
widely used for predicting the outcome of binary classes \cite{cox1958regression}.
Given a training set of $N$ instances, let $\mathbf{x_i}$ be the feature vector of instance $i = 1,2,...,N$ and be of length $K +1$ for the $K$ features and $x_{i0} = 1$ as a dummy variable for $\beta_0$
Let $\mathbf{y}$ be the column vector of length $N$ representing the binary class of each instance.$\bm{\beta}$ is the column vector of coefficients of the predictors computed by the logistic regression. The probability that a given instance is predicted as a success is calculated by the logistic function defined as follows:
\begin{equation}
\pi_i = Pr(y_i = 1|\bm{x_i}) = \frac{1}{1+e^{-(\bm{\beta x_i})}}
\end{equation}
Hence, the probability that an outcome is a failure is equal to $1-\pi_i$.
\subsubsection{Training}
Maximum Likelihood Estimation (MLE) is used to iteratively train the logistic regression. It estimates the values of $\bm{\beta}$ such as to maximize the log-likelihood function defined as follows:
\begin{equation} \label{log-likelihood}
\ell(\bm{\beta}) = \sum_{i=1}^N y_i \textrm{log}(\pi_i)+\sum_{i=1}^N (1-y_i) \textrm{log}(1-\pi_i)
\end{equation}
To estimate $\bm{\beta}$, it initiates with a set of random coefficients, then at each iteration, it uses Newton's method to find the steepest gradient between the current predictions and the actual classes, and updates the coefficients of the features accordingly. It repeats the process until convergence of the coefficients \cite{menard2002applied}
\subsubsection{Feature selection}
The significant number of features extracted justifies the implementation of a feature selection method to improve the model.
The most important incentives for using such techniques are decreasing the variance of the coefficients, decreasing over-fitting, and increasing model interpretability \cite{janecek2008relationship}.
The feature selection method applied to the Logistic Regression is the Elastic-net regularization. It is a weighted combination of two other regularization techniques, namely the Lasso and Ridge regressions. The idea is to minimize the penalized negative log-likelihood ($PNLL$) function which is defined as follows:
\begin{equation} \label{PNLL}
PNLL = -\ell(\bm{\beta}) + \lambda J(\bm{\beta})
\end{equation}
Where $-\ell(\bm{\beta})$ is obtained from Eq. \ref{log-likelihood}, $J(\bm{\beta})$ is the shrinkage penalty and $\lambda \geq 0$ is a tuning parameter which controls how much weight is given to each term. The Elastic-net shrinkage penalty is defined as follows:
\begin{equation}
J^{EN}(\bm{\beta}) = (1-\alpha) \sum_{k=0}^K \frac{\beta_k^2}{2} + \alpha \sum_{k=0}^K |\beta_k|
\end{equation}
Where $\alpha \in [0:1]$ is a tuning parameter controlling the weight applied to each norm. When $\alpha = 0$, a ridge regularization is performed, whereas when $\alpha = 1$ a lasso regularization is performed. H. Zou et. al\cite{zou2005regularization}, who first proposed the elastic-net approach, argue that it inherits the benefits of the Lasso method, while at the same time being able to group strongly correlated predictors. Both $\lambda$ and $\alpha$ are tuned through cross-validation, as will be explained in Section \ref{section:trainingValidation}.
\subsection{Random Forest}
Random Forest is an ensemble learning method
often used for classification and that has shown to perform well \cite{svetnik2003random}. The main idea of this classifier is to create many decision trees independently, through a technique called bagging, and combining their outputs in order to make predictions. To select the best predictors, the decision trees use the Gini index defined as follows:
\begin{equation}
I_G(f) = 1- \sum_{i=1}^J f_i^2
\end{equation}
Where $f_i$ is the fraction of instances of class $i$ and $J$ is the number of classes. The lower the Gini index of a predictor, the better it is at splitting the data.
\subsubsection{Bagging}
In order to overcome the fact that decision trees are prone to overfitting the data, Random Forest uses a specific implementation of a technique called bootstrap aggregating (bagging) to grow many trees independently and combine their results \cite{pal2005random}. To do so, it samples with replacement $B$ times from the training set to generate $B$ new samples of equal size to the original one, and trains a classification tree on each sample using $\sqrt{p}$ randomly selected predictors of the total $p$ predictors available. Once all the trees have been trained, the Random Forest will be able to make predictions for unseen data by taking the mode of the predictions of each tree .\\
\subsection{AdaBoost}
AdaBoost, akin to Random Forest, is an ensemble learning algorithm that uses decision trees as weak learners in order to obtain a strong learner. The main idea of this classifier is to build decision trees sequentially through a technique called boosting, such that each tree improves on the previous one. AdaBoost is considered to be the first successful implementation of a boosting algorithm, and is still considered to be a very good classifier for binary classes \cite{freund1995desicion}. The decision trees used in this classifier are similar to those used in Random Forest, the only difference being that they generally only have a depth of a few nodes.
\subsubsection{Boosting}
Boosting, in the context of decision trees, consists of growing a tree on the training dataset where all the samples have the same weights, then re-weighting the dataset such that the weights of misclassified samples increase and repeating the process for an arbitrary number of trees. Once all the trees have been built, the final classifier combines the predictions through a weighted vote approach, where each weight is a function of the individual tree's performance measure.
\subsubsection{Training}
Let $\bm{Z}$ be the training set such that $\bm{z_n}=(\bm{x_n},y_n) \textrm{ for } n = 1,...,N$ where $\bm{x_n}$ is the predictor vector and $y_n \in \{-1,+1\}$ is the class label of instance $\bm{z_n}$. Then, let $\bm{W}$ be the vector of weights associated to each training instance and initiated such that $\bm{W}(\bm{z_n}) = \frac{1}{N} , \forall n \in N$. Now, for $t = 1,...,T$ with $T$ being an arbitrary number of trees, let $\bm{S_t}$ be the training set obtained by sampling with replacement $N$ times from $\bm{Z}$ with respect to the weights $\bm{W}$ and let $h_t(\bm{S_t})$ be the tree trained on that dataset. The error of the weak learner is calculated as follows:
\begin{equation}
\epsilon_t = \sum_{i:h_t(x_s)\neq y_s} \bm{W}(\bm{z_s}), \quad \forall \bm{z_s} \in \bm{S_t}
\end{equation}
The weights of the misclassified instances are then increased as follows:
\begin{equation}
\bm{W}(\bm{z_s}) = \bm{W}(\bm{z_s}) \times e^{\alpha_t \times I(y_s \neq h_t(\bm{x_s}))}, \quad \forall \bm{z_s} \in \bm{S_t}
\end{equation}
Where $I(y_s \neq h_t(\bm{x_s}))$ is an indicator function that takes the value 1 if the instance was misclassified and 0 otherwise. And the parameter $\alpha_t$ is defined as such:
\begin{equation}
\alpha_t = \frac{1}{2}\bigg(\frac{1-\epsilon_t}{\epsilon_t}\bigg)
\end{equation}
Then, $\bm{W}$ is normalized in order to represent a distribution function, and the steps of sampling, training and re-weighting are repeated until $T$ decisions trees have been generated. Once all the trees have been trained, the AdaBoost classifier is obtained by weighted majority voting defined as follows:
\begin{equation}
AdaBoost(\bm{x}) = \textrm{sign}\bigg(\sum_{t=1}^T \alpha_t h_t(\bm{x})\bigg)
\end{equation}
\section{Experimental setup}
\label{sect:exp}
\subsection{Training-validation}\label{section:trainingValidation}
\subsubsection{k-fold cross validation}
K-fold cross validation (k-CV) is used in order to train and validate the models. To do so, it partitions the data into $k$ equal sized samples (folds), then $k-1$ of those are used to train the model and the remaining one is used to test the model. This process of training and testing is repeated $k$ times, using a different testing sample every time. The performance metrics of the $k$ iterations are then averaged and used as an estimate of the performance of the model. It has been shown that validating through k-CV when one of the classes is relatively rare is generally one of the best options \cite{kohavi1995study}. The classifiers discussed in this paper include hyperparameters that are tuned through k-CV. To avoid ambiguity, the k-CV for evaluating the models will be referred to as the outer loop and the one for tuning the hyperparameters as the inner loop. This naming reflects the fact that the inner loop k-CV is applied to each training set created in the outer loop.
\subsubsection{SMOTE}
Synthetic Minority Over-sampling Technique (SMOTE) is a method regularly used to improve the performance of classifiers on datasets with rare events \cite{he2009learning}. To do so it synthesizes new samples of the minority class, dropouts in our case, using k-Nearest-Neighbours. SMOTE is applied to every training set created in the outer loop of the nested k-CV, but never to the test sets, as these must reflect the real distribution of the data.
\subsection{Evaluation metrics}
The choice of performance metrics is important as different metrics assess different aspects of the models, and using an inadequate one can be misleading. For instance, accuracy, one of the most commonly used metrics for binary classification tasks, is shown to be ineffective at representing the performance of a model when the data set is imbalanced \cite{weiss2004mining}.
Table \ref{table:confusion} represents a confusion matrix, which compares the predicted classes to the reference values.
\begin{table}[]
\centering
\caption{Confusion Matrix}
\label{table:confusion}
\begin{tabular}{c|c|c|c|}
\cline{3-4}
\multicolumn{2}{c|}{\multirow{2}{*}{}} & \multicolumn{2}{c|}{Predicted} \\ \cline{3-4}
\multicolumn{2}{c|}{} & Pos & Neg \\ \hline
\multicolumn{1}{|c|}{\multirow{2}{*}{\begin{tabular}[c]{@{}c@{}}Actual\\ Class\end{tabular}}} & Pos (P) & TP & FN \\ \cline{2-4}
\multicolumn{1}{|c|}{} & Neg (N) & FP & TN \\ \hline
\end{tabular}
\end{table}
\subsubsection{AUROC}
The Area Under the Receiver Operating Characteristic curve (AUROC) is a newer metric than the aforementioned accuracy and has become the norm in measuring the performance of binary classifiers \cite{huang2005using}. It takes advantage of the fact that most classifiers output probabilities instead of binary classes. To do so, it calculates the relationship between the True Positive Rate (TPR or Sensitivity or Recall) and the False Positive Rate (FPR or Fall-out) while sweeping through threshold values for the output probabilities. The TPR and FPR are calculated as follows:
\begin{equation}
TPR = \frac{TP}{TP+FN}
\end{equation}
\begin{equation}
FPR = \frac{FP}{FP+TN}
\end{equation}
The ROC curve is then obtained by plotting the values calculated with the different thresholds. Using this curve, the AUROC is derived by calculating the area underneath it. This one number metric summarizes the goodness of fit of a classifier. J. Davis et. al\cite{davis2006relationship} argue that the AUROC can be overly optimistic if the dataset is highly imbalanced, and propose to use other metrics such as the Area Under the Precision-Recall Curve (AUPRC). However, in our scenario, the cost of missing dropouts is considerably higher than the cost of predicting too many dropouts. Hence, AUROC is considered to be a more suitable metric.
\subsubsection{F2 measure}
The AUROC measures the performance of the classifiers over all the thresholds of the outcome probabilities, which is a good measure for comparing different models within the experiments. But in order to measure how well the classifiers would perform in a real life scenario where hard class labels have to be predicted, we use the F2 measure. This measure is a variant of the $F_\beta$ measure, which is defined as follows:
\begin{equation}
F_\beta = (1 +\beta^2) \frac{\textrm{precision}\times \textrm{recall}}{(\beta^2 \times \textrm{precision})+\textrm{recall}}
\end{equation}
where $\beta = 2$, and precision is calculated as follows:
\begin{equation}
\textrm{precision} = \frac{\sum TP}{\sum TP + FP}
\end{equation}
\subsection{Experimental Results}
Throughout the experiments we hold a few parameters fixed in order to be able to compare the results . Each experiment uses the k-CV approach where $k= 10$ for the outer loop and $k=3$ for the inner loop. The reason for choosing a relatively small $k$ value for the inner loop is because using greater ones would result in having the test sets of the inner folds contain too few positive cases, especially when predicting the exact week of dropout.
\subsubsection{State prediction}
The first experiments are aimed at predicting the state of the users, which comes down to predicting which users will still participate from a certain week on. Throughout this paper we described the dataset as being unbalanced such that the dropouts are the minority class, however this is reversed when predicting the states. Indeed, all users that have dropped out already are still used for future predictions, which means that for week 8 the remaining users represent 13\% of the data. Figures \ref{fig:lr_notExactWeek_auroc} and \ref{fig:lr_notExactWeek_f2} show the AUROC and F2 measures of the Logistic Regression trained and tested on every lag/week combination. We omitted the graphs corresponding to the same experiments using Random Forest and AdaBoost because the values differed at most by 0.01 for a few lag/week combinations. We can see that both the AUROC and the F2 measures are high throughout the combinations, except for the predictions using only the Profile features and for those of week 0 with a lag of 0. This shows that all three algorithms perform very well for predicting whether a user has dropped out by a certain week.
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"lr_notExactWeek_auroc"}
\caption{AUROC measures for state prediction with Logistic Regression}
\label{fig:lr_notExactWeek_auroc}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"lr_notExactWeek_f2"}
\caption{F2 measures for state prediction with Logistic Regression}
\label{fig:lr_notExactWeek_f2}
\end{figure}
\subsubsection{Dropouts per week}\
The second set of experiments focuses on predicting dropouts for each specific week. In this scenario, the minority class becomes the dropouts, except for week 0 as can be seen in figure \ref{fig:dropouts}.
Furthermore, in order to overcome the fact that the events of the minority class appear only rarely in most of the weeks, we use the SMOTE technique to over-sample the dropouts in the training sets. The Figures \ref{fig:lr_exactWeek_auroc},\ref{fig:gbm_exactWeek_auroc},\ref{fig:lr_exactWeek_f2},\ref{fig:rf_exactWeek_f2} and \ref{fig:gbm_exactWeek_f2} show the AUROC and the F2 measures for each of the classifiers. We directly see that these values are lower than for the previous experiments, which was expected. Week 7 has no values because no user dropped out during that week. Furthermore, we notice an important difference between the two metrics. This is mainly due to the fact that the AUROC barely penalizes for predicting too many dropouts, whereas the F2 measure highly penalizes this. Hence, the low F2 measures reflect low precision values, but the recall values are still relatively high. In other words, these algorithms tend to predict too many dropouts which drastically lowers the F2 score, however their performance at identifying those that will actually drop out, which is captured in the recall, is still quite good. Furthermore, the values for week 0 are still very high, this is because many learners drop out that week, hence the algorithms do not considerably overestimate the number of dropouts. We can see that overall, AdaBoost slightly outperforms the two other classifiers with respect to AUROC, while Logistic Regression slightly outperforms the two others with respect to the F2 measure.
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"lr_exactWeek_auroc"}
\caption{AUROC measures for dropout week prediction with Logistic Regression}
\label{fig:lr_exactWeek_auroc}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"rf_exactWeek_auroc"}
\caption{AUROC measures for dropout week prediction with Random Forest}
\label{fig:rf_exactWeek_auroc}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"gbm_exactWeek_auroc"}
\caption{AUROC measures for dropout week prediction with AdaBoost}
\label{fig:gbm_exactWeek_auroc}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"lr_exactWeek_f2"}
\caption{F2 measures for dropout week prediction with Logistic Regression}
\label{fig:lr_exactWeek_f2}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"rf_exactWeek_f2"}
\caption{F2 measures for dropout week prediction with Random Forest}
\label{fig:rf_exactWeek_f2}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 6cm]{"gbm_exactWeek_f2"}
\caption{F2 measures for dropout week prediction with AdaBoost}
\label{fig:gbm_exactWeek_f2}
\end{figure}
\subsubsection{Feature importance}
To be able to judge of the importance of the different features, we set up an experiment similar to the previous one but we remove the users that did not fill their Profile. The reason for this is that having left the Profile empty is a very strong indicator of dropout and pushes the coefficients of the Profile features down. As mentioned previously, the number of temporal features increases as the lag increases, therefore, we chose to combine the values of different weeks of each feature while still taking into account the temporal quality that they possess. Figure \ref{fig:featureImportance} shows the relative importance (RI) of each feature on a logarithmic scale. These values are obtained from AdaBoost and are based on the mean decrease in the Gini Index. No pre-processing is done for the Profile Data features, however for the temporal features we sum the relative importance for the different values of each feature for each prediction. For example, if the relative importance for $a_0,a_1,a_2$ for week 2 are $0,15$ and $65$ respectively, then $a = 80$ for that prediction. We can see that 'a' is almost ten times more important than any other feature. Furthermore, we see that 'pd\_d','pd\_k','pd\_p' and 'ar' are relatively high compared to the other features.
To reflect how spread the weights of a certain temporal feature are over time, we define a new metric, the temporal weight, which is computed as follows:
\begin{equation}
\textrm{TW}(x) = \sum_{i=0}^{lag} \frac{\textrm{RI}(x_i)}{\textrm{RI}(x)} \times (lag-i)
\end{equation}
Using the same values as the previous example, we would obtain: $\textrm{TW}(a) = \frac{0}{80} \times 2+\frac{15}{80} \times 1 +\frac{65}{80} \times 0 = \frac{15}{80}$. We then average these weights for each feature over all the predictions to obtain the box plots shown in Figure \ref{fig:temporalWeightFeatures}. We see that 'a', which has been shown to be the most important feature, has its values close to zero and only one outlier which implies that it mainly gives weight to the latest features. This can be explained by the fact that users either stop submitting assignments or persist until the end,but seldomly skip an assignment, hence looking more than one week back does not provide new information. The rest of these features denote activities which are not required to pass the course, which makes them interesting because they reflect a latent variable which can be defined as having an active attitude towards the MOOC. We see that all of them have many outliers and their range is significant, especially for 'fr\_ab','fp' and 'fc\_ba'. This implies that their values are often composed of the features of several weeks prior to the prediction. By analyzing the results of our experiments, we found that all these features are indicators of persistence rather than dropout. This knowledge combined with the results shown in Figure \ref{fig:temporalWeightFeatures} corroborates the hypothesis that Forum activity denotes an active attitude towards the course. Furthermore, the significant range of values suggests that the importance of these features barely decays over time, which means that having shown this active attitude towards the course at some point in time is still a strong indicator of persistence several weeks later.
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 10cm]{"featureImportance"}
\caption{Box plots of the relative importance of each Profile Data feature }
\label{fig:featureImportance}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width = \linewidth, height = 7cm]{"temporalWeightFeatures"}
\caption{Box plots of the relative importance of each Profile Data feature }
\label{fig:temporalWeightFeatures}
\end{figure}
\section{Conclusion}
\label{sect:discussion}
In this paper, a framework for extracting features from a MOOC course was presented and then these features were used as input to a classification system which is able to predict whether a user will dropout or not, taking into account the temporal dimension. In more detail, contributions of this paper are as follows: Firstly, predicting which users will drop out at some point is done with high accuracy for all three classifiers, even when using only the Profile Data. This is not to be translated that filling in the Profile leads to completing the course, but it does demonstrate that asking users to do so is important for teachers interested in predicting who will finish the MOOC.
Secondly, predicting the exact week that a user will drop out is considerably more difficult, and even when the lag is equal to the predicted week results obtained are not always looking good.
This can be explained by the fact that there are few dropouts per week which makes it difficult for the classifiers to train and test, even when using k-CV. Furthermore, we believe that if the Video Data and the Team Data had temporal information, which would allow us to use them, better results could be obtained.
Finally, we found some interesting indicators of dropout and persistence.
Although it is not a feature, we found that the users who leave their Profile empty have a high probability of dropping out, and this can be considered the strongest indicator of dropout. The strongest indicator of persistence is the assignment feature, which is the only feature that really denotes whether a user is actively involved in the course or not. Following this, we have a few features that reflect a hidden variable which could be to have an active attitude towards the MOOC. These features are the length of the Biography and of the Course objectives. Lastly, the temporal weights' graph showed us that the features which denote an active participation of the user on the forum are relatively important. Given a dataset with more active forum users, it would be interesting to dig deeper into the relationship between Forum activity and user persistence.
Despite this work brought interesting insights into the dropout behaviour of users in a MOOC, there is still more work to be done. First of all, aligning all features so that they can be correctly projected on a temporal axis will boost the timely (per-week) prediction accuracy. Furthermore, current features can be combined with usage (log) statistics on the actual website (clicks, active time, etc.). Finally, we are looking forward to integrating the implemented approach with a real MOOC in order to check real-time performance and whether instructors are assisted in improving learners' experience.
\section*{Acknowledgment}
The authors would like to thank Daniëlle Verstegen (School of Health Professions Education, Maastricht University) for providing the dataset described in Section \ref{sect:dataset}.
\bibliographystyle{IEEEtran}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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Your definitive football source. U.S. edition.
How to watch Coupe de la Ligue final in India?
The teenage forward has played sparingly this season, but has shown his vast potential in recent appearances Juventus have made a new contract for teenage striker Moise Kean a priority, Goal understands.
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{
"redpajama_set_name": "RedPajamaC4"
}
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\section{Introduction}
According to General Relativity, if the universe is filled with the particles
of the Standard Model of particle physics, gravity should lead to a
deceleration of the expansion of the universe. However, in 1998 two
independent evidences of present accelerated expansion were presented
\cite{Riess1,Perlmutter1} and later confirmed by different
observations \cite{KOBE, WMAP,LSS}. On the other hand, measurements of large
scale structure \cite{infLSS} and CMB anisotropy \cite{KOBE} also
indicate that the universe evolved through a period of early accelerated
expansion (inflation).
There is no compelling explanation for any of these cosmic accelerations,
but many intriguing ideas are being explored. In the case of inflation,
the origin of the accelerated expansion can be either a modification of
gravity at small scales \cite{hemg} or a coupling of the expansion of
the universe to the progress of phase transitions \cite{inflation}.
In the case of the present accelerated expansion these ideas can be classified
into three main groups: new exotic sources of the gravitational field with
large negative pressure \cite{DE} (Dark Energy), modifications of
gravity at large scales \cite{lemg} and rejection of the spatial
homogeneity as a good approximation in the description of the present
universe \cite{DEwoDE}.
Different models (none of them compelling) of the source responsible
for each of the two periods of accelerated expansion have been
considered. Einstein equations admit a cosmological constant
$\Lambda$, which can be realized as the stress-energy tensor of empty
space. This $\Lambda$, together with Cold Dark Matter, Standard Model
particles and General Relativity, form the current cosmological model
($\Lambda$CDM). However, quantum field theory predicts a value of
$\Lambda$ which is 120 orders of magnitude higher than observed.
Supersymmetry can lower this value 60 orders of magnitude, which is
still ridiculous \cite{Weinberg}.
In order to solve this paradox, dynamical Dark Energy models have been proposed.
This has also lead to explore the possibility that cosmic acceleration arises
from new gravitational physics. Also here several alternative
modifications of the Einstein-Hilbert action at large and small
curvatures \cite{fR, Carroll1, Soti, HS, Odin, Odin2, Odin8}, or even higher
dimensional models \cite{Deffayet, DGP}, producing an accelerated
expansion have been identified.
All these analysis include an \textit{ad hoc} restriction to actions involving simple
functions of the scalar curvature and/or the Gauss-Bonnet tensor.
This discussion is sufficient to establish the point that cosmic acceleration can
be made compatible with a standard source for the gravitational field, but it is
convenient to consider a more general framework in order to make a systematic
analysis of the cosmological effects of a modification of general relativity.
In this paper we parametrize the evolution of the universe (considered isotropic and
homogeneous) with the Hubble parameter $H$. One finds that it is possible to restrict
the domain of $H$ to a bounded interval. This restriction naturally
produces an accelerated expansion when the Hubble expansion rate
approaches any of the two edges of the interval.
Therefore, we find a new way to incorporate two periods
of cosmic acceleration produced by a modification of general
relativity. But the dependence on the two limiting values
of $H$ can be chosen independently. One could even consider a Hubble
expansion rate constrained to a semi-infinite interval with a unique
period of accelerated expansion. In this sense the aim to
have a unified explanation of both periods of accelerated expansion is
only partially achieved. This simple phenomenological approach to the problem
of accelerated expansion in cosmology proves to be equivalent at
the homogeneous level to other descriptions based on modifications
of the Einstein-Hilbert action or the introduction of exotic components
in the matter Lagrangian.
In the next section we will review how a general modification of the gravitational
action leads to a generalized first Friedman equation at the homogeneous level. In
the third section we will present a specific model (ACM) based on the simplest
way to implement a bounded interval of $H$. In the fourth section we will contrast
the predictions of ACM for the present acceleration with astrophysical observations.
In the fifth section we will show that it is always possible to find modified
gravitational actions which lead to a given generalized first Friedman equation, and
present some simple examples. In the sixth section we will show that it is also always
possible to find Dark Energy models which are equivalent to a given generalized
first Friedman equation, and present some simple examples. The last section is devoted
to summary and conclusions.
\section{Action of the cosmological standard model extension}
The spatial homogeneity and isotropy allow to reduce the
gravitational system to a mechanical system with two
variables $a(t)$, $N(t)$ which parametrize the Robertson-Walker
geometry
\begin{equation}
ds^2 \,=\, N^2(t) dt^2 - a^2(t) \Delta_{ij} dx^i dx^j
\end{equation}
\begin{equation}
\Delta_{ij} \,=\, \delta_{ij} + \frac{k x_i x_j}{1 - k{\bf x}^2} \, .
\end{equation}
Invariance under parameterizations of the time variable
imply that the invariant time differential $N(t)dt$ must be used. Also
a rescaling of the spatial variables
$x^i\rightarrow\lambda x^i$ together with $a(t)\rightarrow\lambda^{-1}
a(t)$ and $k\rightarrow\lambda^{-2} k$ is a symmetry that must be kept
in the Lagrangian. The action of the reduced homogeneous gravitational
system can be then written as
\begin{equation}
I_g \,=\, \int dt N \, L\left(\frac{k}{a^2}, H,
\frac{1}{N}\frac{dH}{dt},
\frac{1}{N}\frac{d}{dt}\left(\frac{1}{N}\frac{dH}{dt}\right),
...\right) \, ,
\label{Ig}
\end{equation}
with $H \,=\, \frac{1}{aN}\frac{da}{dt}$.
If we keep the standard definition of the gravitational coupling and
the density ($\rho$) and pressure ($p$) of a cosmological homogeneous
and isotropic fluid as a source of the gravitational field, we have
the equations of the reduced system
\begin{equation}
\left(\frac{8\pi G_N}{3}\right) \rho \,=\, - \frac{1}{a^3} \frac{\delta
I_g}{\delta N(t)}
\end{equation}
\begin{equation}
- \left(8\pi G_N\right) p \,=\, - \frac{1}{N a^2} \frac{\delta
I_g}{\delta a(t)}\, .
\end{equation}
It is possible to choose a new time coordinate $t^{'}$ such that
\begin{equation}
\frac{dt^{'}}{dt} \,=\, N(t)\, .
\end{equation}
This is equivalent to set $N=1$ in the action (\ref{Ig}) of the gravitational
system and the evolution equations of the cosmological model reduce to a
set of equations for the scale factor $a(t)$. If we introduce the
notation
\begin{equation}
H^{(i)} \,=\, \left(\frac{d}{dt}\right)^i \, H
\end{equation}
\begin{equation}
\left(\delta_i L\right)^{(j)} \,=\, \left(\frac{d}{dt}\right)^j
\left[a^3 \partial_i L\right]\, ,
\end{equation}
with $\partial_i L$ denoting the partial derivatives of the Lagrangian
as a function of the variables $(k/a^2, H, H^{(1)}, H^{(2)}, ...)$ we
have
\begin{equation}
\left(\frac{8\pi G_N}{3}\right) \rho \,=\, - L +
\sum_{i=2}^{\infty} \sum_{j=0}^{i-2} \frac{ (-1)^{i-j}}{a^3} H^{(j)} \left(\delta_i
L\right)^{(i-j-2)}
\label{rho}
\end{equation}
\begin{equation}
- \left(8\pi G_N\right) p \,=\, - 3 L + \frac{2 k}{a^2} \partial_1 L +
\sum_{i=2}^{\infty} \frac{(-1)^i}{a^3} \left(\delta_i
L\right)^{(i-1)} \, .
\label{p}
\end{equation}
In the homogeneous and isotropic approximation, the vanishing of the
covariant divergence of the energy-momentum tensor leads to the
continuity equation
\begin{equation}
\frac{d}{dt} \left(\rho a^3\right) \,=\, - p \frac{d}{dt} a^3
\label{cont_eq}\, .
\end{equation}
In the radiation dominated era one has
\begin{equation}
\rho \,=\ 3 p \,=\, \frac{\sigma}{a^4}
\label{rad}\, ,
\end{equation}
where $\sigma$ is a constant parameterizing the general
solution of the continuity equation. In a period dominated by matter
one has a pressureless fluid and then
\begin{equation}
p \,=\, 0 \, ,{\hskip 2cm} \rho \,=\, \frac{\eta}{a^3}\, ,
\end{equation}
with constant $\eta$. When these
expressions for the energy density and pressure are plugged in
(\ref{rho}-\ref{p}), one ends up with two compatible differential equations for
the scale factor $a(t)$ which describe the evolution of the
universe. From now on we will use the more common notation
$\frac{dH}{dt}\equiv\dot{H}$,...
\section{The Asymptotic Cosmological Model}
Let us assume that the Lagrangian L of the gravitational system is
such that the evolution equations (\ref{rho}-\ref{p}) admit a solution
such that $a(t)>0$ (absence of singularities), $\dot{a} >0$ (perpetual
expansion) and $\dot{H}<0$. In that case, one has a different value of
the scale factor $a$ and the Hubble rate $H$ at each time and then one
has a one to one correspondence between the scale factor and the
Hubble rate. Since the continuity equation (together with the equation
of state) gives a relation between the scale factor and the density,
one can describe a solution of the evolution equations of the
generalized cosmological model through a relation between the energy
density and the Hubble rate, i.e. through a generalized first Friedman
equation
\begin{equation}
\left(\frac{8\pi G_N}{3}\right) \rho \,=\, g(H)\, ,
\label{rho_g}
\end{equation}
with $g$ a smooth function which parametrizes the different algebraic
relations corresponding to different solutions of different
cosmological models. Each choice for the function $g(H)$ defines a
phenomenological description of a cosmological model. Then one
can take it as a starting point trying to translate any observation
into a partial information on the function $g(H)$ which parametrizes
the cosmological model.
Eq. (\ref{rho_g}) is all one needs in order to reconstruct the
evolution of the universe at the homogeneous level. The generalized
second Friedman equation is
obtained by using the continuity equation (\ref{cont_eq}) and the
expression for $\rho$ as a function of $H$. One has
\begin{equation}
- \left(8\pi G_N\right) p \,=\, 3 g(H) + \frac{g^{'}(H)}{H} \dot{H}
\label{pressure}\, .
\end{equation}
In the matter dominated era one has
\begin{equation}
\frac{\dot{H}}{H^2} \,=\, - 3 \frac{g(H)}{H g^{'}(H)}
\label{matter}
\end{equation}
and then the assumed properties of the solution for the evolution
equations require the consistency conditions
\begin{equation}
g(H) > 0 \, ,{\hskip 1cm} g^{'}(H) >0\, .
\label{gcons}
\end{equation}
In the period dominated by radiation one has
\begin{equation}
\frac{\dot{H}}{H^2} \,=\, - 4 \frac{g(H)}{H g^{'}(H)}
\label{radiation}
\end{equation}
instead of (\ref{matter}) and the same consistency conditions
(\ref{gcons}) for the function $g(H)$ which defines the generalized
first Friedman equation.
We introduce now a phenomenological cosmological model
defined by the condition that the Hubble rate has an upper bound $H_+$
and a lower bound $H_-$. This can be implemented through a function
$g(H)$ going to infinity when $H$ approaches $H_+$ and going to
zero when $H$ approaches $H_-$. We will also assume that there is an interval
of $H$ in which the behavior of the energy density with the Hubble parameter is,
to a good approximation, scale-free i.e. $g(H)\propto H^2$. The source
of the gravitational field will be a homogeneous and isotropic fluid
composed of relativistic and non-relativistic particles.
The Cosmological Standard Model without curvature is recovered in the
limit $H_-/H\rightarrow 0$ and $H_+/H\rightarrow\infty$ which is a
good approximation for the period of decelerated expansion.
Notice that this interpretation is independent of the underlying
theory of gravitation.
The Hubble parameter can be used to parametrize the history of
universe as long as $\dot{H}\neq0\,\forall \,t$. The total density can be
thus expressed as a function of $H$. If $H$ is bounded, then $\rho(H)$
will have a pole at $H=H_+$ and a zero at $H=H_-$. Far from these scales,
the behavior of $\rho(H)$ can be assumed to be approximately scale-free.
Under these conditions, we can parametrize the dependence of the
cosmological model on the lower bound $H_-$ by
\begin{equation}
g(H) \,=\, H^2 h_{-}\left(\frac{H_-^2}{H^2}\right)
\label{g-}
\end{equation}
and similarly for the dependence on the upper bound $H_+$
\begin{equation}
g(H) \,=\, \frac{H^2}{
h_{+}\left(\frac{H^2}{H_+^2}\right)}
\label{g+}\, ,
\end{equation}
where the two functions $h_{\pm}$ satisfy the conditions
\begin{equation}
\lim_{x\to 0} h_{\pm} (x) \,=\, \beta^{\pm 1}\, , {\hskip 1cm}
\lim_{x\to 1} h_{\pm} (x) \,=\, 0\, .
\end{equation}
$\beta$ is a constant allowed in principle by dimensional
arguments. If $\beta \neq 1$ then it can be moved to the lhs of the
Friedman equation, turning $G_N\rightarrow\beta G_N$, and can be
interpreted as the ratio between an effective cosmological value of
the gravitational coupling and the value measured with local
tests. But $\beta\neq 1$ would be in conflict with Nucleosynthesis,
through the relic abundances of $^4 He$ and other heavy elements (for
3 neutrino species) \cite{Mukh}, so we set $\beta= 1$ . Therefore
\begin{equation}
\lim_{x\to 0} h_{\pm} (x) \,=\, 1 \, ,{\hskip 1cm}
\lim_{x\to 1} h_{\pm} (x) \,=\, 0\, .
\label{hpm}
\end{equation}
The consistency conditions (\ref{gcons}) result in
\begin{equation}
h_{\pm} (x) > 0 \, ,{\hskip 1cm} h_{\pm} (x) > x h_{\pm}^{'}(x)
\label{hcons}
\end{equation}
for the two functions $h_{\pm}$ defined in the interval $0<x<1$. Thus,
we can divide the cosmic evolution history into three periods. In the
earliest, relativistic particles dominate the energy density of the
universe and the generalized first Friedman equation shows a
dependence on the upper bound $H_+$. There is also a transition period
in which the effect of the bounds can be neglected and the rhs of the
first Friedman equation is scale-free; this period includes the
transition from a radiation dominated universe to a matter dominated
universe. In the third present period, non-relativistic particles
dominate the energy density of the universe but the dependence on the
lower bound $H_-$ must be accounted for in the generalized first
Friedman equation.
\indent From the definition of the Hubble parameter, one has
\begin{equation}
\frac{\dot{H}}{H^2} \,=\, -1 + \frac{a \ddot{a}}{\dot{a}^2}\, .
\end{equation}
Then, in order to see if there is an accelerated or decelerated
expansion, one has to determine whether $\dot{H}/H^2$ is greater or smaller
than $-1$.
In the period dominated by radiation one has
\begin{equation}
\frac{\dot{H}}{H^2} \,=\, - 2 \, \frac{1}{1 - \frac{x h_+^{'}(x)}{
h_+(x)}}\, ,
\end{equation}
with $x=H^2/H_{+}^2$, where we have used (\ref{g+}) assuming that only
the dependence on the upper bound ($H_{+}$) of the Hubble parameter is
relevant. A very simple choice for this dependence is given by
\begin{equation}
g(H) \,=\, \frac{H^2}{\left(1 - \frac{H^2}{H_{+}^2}\right)^{\alpha_+}}
\label{alpha+}\, ,
\end{equation}
with $\alpha_+$ a (positive) exponent which parametrizes the departure
from the standard cosmological model when the Hubble rate approaches
its upper bound. With this simple choice one has a transition from an
accelerated expansion for $H^2 > H_+^2/(1+\alpha_+)$ into a decelerated
expansion when $H^2 < H_+^2/(1+\alpha_+)$, which includes the domain of
validity of the standard cosmological model ($H\ll H_+$).
In the period dominated by matter (which corresponds to lower values
of the Hubble rate) we assume that only the dependence on the lower
bound ($H_-$) of the Hubble rate is relevant. Then one has
\begin{equation}
\frac{\dot{H}}{H^2} \,=\, - \frac{3}{2} \, \frac{1}{1 - \frac{x
h_-^{'}(x)}{h_-(x)}}\, ,
\end{equation}
with $x=H_{-}^2/H^2$. We can also consider a dependence on $H_-$
parametrized simply by an exponent $\alpha_-$
\begin{equation}
g(H) \,=\, H^2 \, \left(1 - \frac{H_-^2}{H^2}\right)^{\alpha_-}
\label{alpha-}\, .
\end{equation}
With this choice one has a transition from a decelerated expansion
for $H^2 > H_{-}^2(1+\alpha_-/2)$ leaving the domain of validity of
the standard cosmological model and entering into an accelerated
expansion when the Hubble rate approaches its lower bound for $H^2 <
H_{-}^2(1+\alpha_-/2)$.
The possibility of describing $\rho$ as a function of $H$ is independent
of the existence of spatial curvature $k$. However, in the kinematics of
observables in the expanding universe we do need to specify the value of
$k$. In the rest of the paper we will assume that the universe is flat
($k=0$), although the same analysis could be done for arbitrary $k$.
The properties of the expansion obtained in this simple example (a
period of decelerated expansion separating two periods of accelerated
expansion) are general to the class of phenomenological models with a
generalized first Friedman equation (\ref{rho_g}) with $g(H)$
satisfying the consistency conditions (\ref{gcons}) and a Hubble rate
constrained to a finite interval. The specific part of the example
defined by (\ref{alpha+},\ref{alpha-}) is the simple dependence on the
Hubble rate bounds and the values of the Hubble rate at the
transitions between the three periods of expansion. From now on we
will name this description the Asymptotic Cosmological Model (ACM).
With respect to the late accelerated expansion, ACM can be seen as a
generalization of $\Lambda$CDM, which can be recovered by setting
$\alpha_-=1$. It also includes an early period of exponential
expansion which can be seen as a phenomenological description of the
evolution of the universe at inflation in the homogeneous approximation.
\subsection{Horizon problem}
One can see that the horizon of a radiation-dominated universe can be
made arbitrarily large as a consequence of an upper bound on the
Hubble parameter and in this way one can understand the observed
isotropy of the cosmic microwave background at large angular scales.
Let us consider the effect of the modification of the cosmological
model on the calculation of the distance $d_h(t_f,t_i)$ of a source of
a light signal emitted at time $t_i$ and observed at time $t_f$
\begin{equation}
d_h(t_f, t_i) \,=\, c \, a(t_f) \int_{t_i}^{t_f} \frac{dt}{a(t)}\, .
\end{equation}
We have
\begin{equation}
\frac{dt}{a} \,=\, \frac{da}{a^2 H}\,=\, \left(\frac{8 \pi G_N
\sigma}{3}\right)^{-1/4} \frac{g^{'}(H) dH}{4H g(H)^{3/4}} \, ,
\end{equation}
where in the first step we have used the definition of the Hubble
expansion rate $H$ and in the second step we have used the relation
between the scale factor $a$ and $H$ as given by
(\ref{rad}-\ref{rho_g}). We are considering both times $t_i$ and $t_f$
in the radiation dominated period.
The distance $d_h$ is then given by
\begin{equation}
d_h(H_f, H_i) \,=\, \frac 1{4 g(H_f)^{1/4}} \int_{H_f}^{H_i} \frac{dH
g^{'}(H)}{H g(H)^{3/4}} \label{dist}\, .
\end{equation}
If $H_i$ is very close to $H_+$ (i.e. if we choose the time $t_i$ when
the light signal is emitted well inside the period of accelerated
expansion) then the integral is dominated by the
region around $H_i$ which is very close to $H_+$. Then one can
approximate in the integrand
\begin{equation}
g(H) \approx \frac{H_+^2}{\left(1-\frac{H^2}{H_+^2}\right)^{\alpha_+}} \, .
\end{equation}
On the other hand if the observation is made at a time $t_f$ within the
domain of validity of the cosmological standard model ($H_-^2\ll
H_f^2\ll H_+^2$) then the factor $g(H_f)^{1/2}$ in
front of the integral can be approximated by $H_f$ and then one has
\begin{equation}
d_h(H_f, H_i) \approx \frac{\alpha_+}{4 \sqrt{H_f H_i}}
B_{\frac{H^2_i}{H^2_+}}(1/2,-\alpha/4) \, ,
\end{equation}
where $B_z(m,n)$ is the incomplete Beta function, and it can be made
arbitrarily large by choosing $H_i$ sufficiently close
to $H_+$. In this way we see that a cosmological model with a finite
interval of variation for $H$ solves the horizon problem.
\section{Constraints of ACM by Observations}
In this section we will carry out a more technical analysis about how
the astrophysical observations constrain the parameter space of ACM in
the matter dominance period. This analysis is based on the use of
(assumed) standard candles, basically Type Ia Supernovae \cite{Riess2}
and CMB \cite{WMAP3y}. These observations constrain the parameter
space to confidence regions in which the combination
$\Omega_m\equiv(1-H_-^2/H_0^2)^{\alpha_-}$ is constrained to be around
one quarter.
The consideration of both Type Ia SNe and CMB together favor
$\alpha_->1.5$. The results of this analysis can be seen in
figures (FIG. 1-3).
A reader not interested in technical details might well skip this section.
We center our discussion of the experimental tests of the Asymptotic
Cosmological Model
in the late accelerated expansion produced when $H$ approaches its
lower bound $H_-$. The vast amount of supernovae data collected by
\cite{Riess2} and \cite{Astier}, the data from the SDSS Baryon
Acoustic Oscillation \cite{Eisen}, the mismatch between total energy
density and total matter energy density seen at CMB anisotropies
\cite{WMAP3y} and the measurements of present local mass density by
2dF and SDSS \cite{LSS,Cole} compared with the measurements of $H_0$
from the HST Cepheids \cite{Cepheids} show that the universe undergoes
a surprising accelerated expansion at the present time.
We will firstly confront the model with the Supernovae Ia data from
Riess \textit{et al.} and SNLS collaboration.
The usefulness of the Supernovae data as a test of Dark Energy models
relies on the assumption that Type Ia SNe behave as standard candles,
i.e., they have a well defined environment-independent luminosity
$\mathcal{L}$ and spectrum. Therefore, we can use measured bolometric
flux $\mathcal{F}=\frac{\mathcal{L}}{4\pi d_L^2}$ and frequency to
determine luminosity distance $d_L$ and redshift $z$.
The luminosity distance is given now by
\begin{equation}
\begin{array}{c}
d_L(z)=c(1+z)\int_0^z \frac{dz'}{H(z')}\\
=\frac{c(1+z)}{3}\int^{H(z)}_{H_0}\frac{g'(H)dH}{H g^{1/3}(H_0)g^{2/3}(H)}\, .
\end{array}
\end{equation}
The computed value must be compared with the one obtained
experimentally from the measured extinction-corrected
distance moduli ($\mu_0=5 log_{10}(\frac{d_L}{1 Mpc})+25$) for each
SN. The SNe data have been compiled in references \cite{Riess2}, and
we have limited the lowest redshift at $cz<7000 km/s$ in order to
avoid a possible ``Hubble Bubble'' \cite{Jha,Zehavi}. Therefore our
sample consists on 182 SNe. We will determine the likelihood of the
parameters from a $\chi^2$ statistic,
\begin{equation}
\chi^2(H_0,H_-,\alpha_-)=
\sum_i\frac{(\mu_{p,i}(z_i;H_0,H_-,\alpha_-)-\mu_{0,i})^2}{\sigma^2_{\mu_{0,i}}+\sigma^2_v}\,,
\end{equation}
where $\sigma_v$ is the dispersion in supernova redshift due to
peculiar velocities (we adopt $\parallel\mathbf{v_p}\parallel = 400
km/s$ in units of distance moduli), $\sigma_{\mu_{0,i}}$
is the uncertainty in the individual measured distance moduli
$\mu_{0,i}$, and $\mu_{p,i}$ is the value of $\mu_0$ at $z_i$
computed with a certain value of the set of parameters
${H_0,H_-,\alpha_-}$. This $\chi^2$ has been marginalized over
the nuisance parameter $H_0$ using the adaptive method in reference
\cite{Wang}. The resulting likelihood
distribution function $e^{-\chi^2/2}$ has been explored using Monte
Carlo Markov Chains. We get a best fit
of ACM at $\alpha_-=0.36$ and $\frac{H_-}{H_0}=0.95$, for which
$\chi^2=157.7$. In contrast, fixing $\alpha_-=1$, we get
$\Omega_\Lambda\equiv\frac{H_-^2}{H_0^2}=0.66$ and
$\chi^2=159.1$ for the best fit $\Lambda$CDM. The confidence regions
are shown in Fig. 1 (top).
\begin{figure}
\centerline{\includegraphics[scale=0.6,
width=9cm]{sinrestriccion.png}}
\end{figure}
\begin{figure}
\centerline{\includegraphics[scale=0.6,
width=9cm]{restricciondos.png}}
\label{confidence}
\caption{Confidence regions in parameter space of the Asymptotic
Cosmological Model (ACM) with (bottom) and without (top) priors from
CMB and estimations of present matter energy density at $1\sigma$
(green), $2\sigma$ (red) and $3\sigma$ (blue). The $\Lambda$CDM is
inside the 1$\sigma$ region if we consider no priors, but is moved
outside the $2\sigma$ region when we confront the SNe data with CMB
and estimations from present matter energy density. In this case a
value $\alpha_-> 1.5$ is favored.}
\end{figure}
We can add new constraints for the model coming from measurements of
the present local matter energy density from the combination of 2dF
and SDSS with HST Cepheids, rendering $\Omega_m = 0.28 \pm 0.03$; and
the distance to the last scattering surface from WMAP, which leads to
$r_{CMB}\equiv\sqrt{\Omega_m}\int^{1089}_0 \frac{H_0
dz'}{H(z')}=1.70\pm0.03$ \cite{Riess2}.
Including these priors we get a best fit of ACM at $\alpha_-=3.54$ and
$\frac{H_-}{H_0}=0.56$ (our simulation explored the region with
$\alpha_-<3.6$), for which $\chi^2=164.1$. In contrast, we get
$\Omega_\Lambda=0.73$ and $\chi^2=169.5$ for the best fit
$\Lambda$CDM, which is outside the $2\sigma$ confidence region shown
in Fig. 1 (bottom). The fits of the best fit $\Lambda$CDM and ACM
taking into account the priors to the SNe data are compared in Fig. 2.
The information which can be extracted from the data is limited. This
can be seen in Fig. 3, in which it is explicit that the data
constrain mainly the value of the present matter energy density,
$\Omega_m\equiv\frac{\rho_m}{\rho_C}=(1-\frac{H_-^2}{H_0^2})^{\alpha_-}$.
\begin{figure}
\centerline{\includegraphics[scale=0.6, width=9cm]{bestfitACM.png}}
\label{bestfit}
\caption{Distance Moduli vs. Redshift comparison between the best
fits of ACM ($H_0 = 66 \, Km/s MPc$, red) and $\Lambda$CDM ($H_0 =
65 \, Km/s MPc$, blue) to the SNe data with priors. ACM fits clearly
better the medium redshift SNe. }
\end{figure}
\begin{figure}
\centerline{\includegraphics[scale=0.6, width=9cm]{omega.png}}
\label{omegabestfit}
\caption{$\Omega_m$ vs. $\alpha$ confidence regions with the CMB prior
only. The confidence regions show $\Omega_m = 0.25 \pm 0.03$ and
$\alpha > 1.15$ at $2\sigma$ level. }
\end{figure}
\section{Generalized First Friedman Equation and $f(R)$ gravity}
The extension of the cosmological model considered here could be
compared with recent works on a modification of gravity at large
or very short distances \cite{fR, Carroll1, Odin2, Soti, HS, Odin, Odin8}.
It has been shown that, by considering a correction to the
Einstein Hilbert action including positive and negative powers of the
scalar curvature, it is possible to reproduce an accelerated expansion
at large and small values of the curvature in the cosmological
model. Some difficulties to make these modifications of the
gravitational action compatible with the solar system tests of general
relativity have lead to consider a more general gravitational action,
including the possible scalars that one can construct with the
Riemann curvature tensor \cite{RR}, although the Gauss-Bonnet scalar is the only
combination which is free from ghosts and other pathologies. In fact, there
is no clear reason to restrict the extension of general relativity
in this way. Once one goes beyond the derivative expansion, one should
consider scalars that can be constructed with more than
two derivatives of the metric and then one does not have a good
justification to restrict in this way the modification of the
gravitational theory. It does not seem difficult to find an
appropriate function of the scalar curvature or the Gauss-Bonnet
scalar, which leads to a cosmology with a bounded Hubble expansion rate.
One may ask if a set of metric f(R) theories which include ACM as an
homogeneous and isotropic solution exists.
The answer is that a bi-parametric family of f(R) actions which lead to
an ACM solution exists. In the following section we will derive them
and we will discuss some examples. The derivation follows the same
steps of modified f(R)-gravity reconstruction from any FRW cosmology
\cite{Odin4}.
We start from the action
\begin{equation}
S=\frac 1{16 \pi G_N}\int d^4 x\sqrt{-g}[f(R) + 16\pi G L_m]\, ,
\end{equation}
which leads to the ``generalized first Friedman equation''
\begin{eqnarray}
\begin{array}{c}
-18(H \ddot H +4 H^2 \dot H)f''(R)\\
-3(\dot H +H^2)f'(R)
-\frac 12 f(R)=\left( 8\pi G_N \right) \rho
\end{array}\\
R=-6(\dot{H}+2H^2)\, .
\end{eqnarray}
If the energy density of the universe is mainly due to matter (as in
the late accelerated expansion), we can use (\ref{matter}) and
(\ref{rho_g}) to express $\rho$, $R$, $\dot{H}$ and $\ddot{H}$ as functions of
$H$. Thus we get
\begin{eqnarray}
\begin{array}{c}
[H^3 g g'^2-3H^2 g^2 g'+3H^3 g^2 g'']f''(R)\\
+\frac{[3H g g'^2-H^2 g'^3]}{18}f'(R)-\frac{g'^3}{108}f(R)=\frac{g g'^3}{18}
\end{array}
\label{difR}\\
R(H) = 18 H g(H)/g'(H)-12 H^2 \label{RH}\, .
\end{eqnarray}
$H(R)$ can be obtained from (\ref{RH}) and set into (\ref{difR}); then
we get an inhomogeneous second order linear differential equation with
non-constant coefficients. Therefore, there will always be a
bi-parametric family of $f(R)$ actions which present ACM as their
homogeneous and isotropic solution. The difference between these
actions will appear in the behavior of perturbations, which is not
fixed by (\ref{rho_g}). Some of these actions are particularly easy to
solve. If $g(H)=H^2$ then $R=-3H^2$ and the differential equation
becomes
\begin{equation}
6 R^2 f''(R) -R f'(R)-f(R) +2 R=0\, .
\end{equation}
Its general solution is
\begin{equation}
f(R)=R+c_1 R^{\frac 1{12}(7-\sqrt{73})}+c_2 R^{\frac 1{12}(7+\sqrt{73})}\, ,
\end{equation}
which will give (\ref{rho_g}) as First Friedman equation as long as
the radiation energy density can be neglected. In general the
differential equation (\ref{difR}) will not be solvable
analytically, and only approximate solutions can be found as power
series around a certain singular point $R_0$ (a value of $R$ such that
$H(R)$ cancels out the coefficient of $f''(R)$ in (\ref{difR})). These
solutions will be of the form $f(R)=f_p(R) +c_1 f_1(R) +c_2 f_2(R)$
with
\begin{equation}
\frac{f_i(R)}{R_0}=\sum_{m=0}^\infty a_m^{(i)} (\frac{R-R_0}{R_0})^{s_i+m} \, ,
\end{equation}
where $i=p,1,2$, $a_0^{(1)}=a_0^{(2)}=1$ and the series will converge
inside a certain radius of convergence. An interesting choice of $R_0$
is $R_0 = -12 H_-^2 \equiv R_-$, which is the value of $R$ at
$H_-$. One can also find the approximate solution of the differential
equation for $\mid R\mid\gg\mid R_-\mid$, which will be of the form
\begin{equation}
\frac{f_i(R)}{R}=\sum_{m=0}^\infty a_m^{(i)} (\frac{R_-}{R})^{s_i+m} \label{fR-}\, .
\end{equation}
One can in principle assume that the action (\ref{fR-}) could be
considered as valid also in the region in which radiation begins to
dominate, but this action reproduces (\ref{rho_g}) only if matter
dominates. However, some of the new terms appearing in (\ref{fR-})
will be negligible against $R$ when radiation begins to dominate. The
others can be canceled out by setting to zero the appropriate integration
constant and therefore the Cosmological Standard Model will be
recovered as a good approximation when radiation begins to dominate.
$f(R)$-theories are not the only modified gravity theories studied in
the literature; $f(G)$-theories \cite{GB} are also a popular field of research.
In these theories, the Einstein-Hilbert action is supplemented by a
function of the Gauss-Bonnet scalar
$G=R^2-4R_{\mu\nu}R^{\mu\nu}+R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$
(not to be confused with the gravitational coupling $G_N$).The same
approach can be used to answer what $f(G)$ actions are able to
reproduce ACM homogeneous evolution, once more in analogy with the
modified f(G)-gravity reconstruction of a FRW cosmology
\cite{Odin5}. The form of the action will be
\begin{equation}
S=\frac 1{16 \pi G_N}\int d^4 x\sqrt{-g}[R+f(G) + 16\pi G_N L_m]\, ,
\end{equation}
from which we derive the modified Friedman equation
\begin{equation}
\left(8\pi G_N\right) \rho = 3 H^2-\frac 12 G f'(G)+\frac 12
f(G)+12f''(G)\dot{G}H^3 \, .
\end{equation}
Following the same procedure as in the case of $f(R)$-theories we
arrive to the differential equation
\begin{eqnarray}
\begin{array}{c}
3 g(H) = 3 H^2-\frac 12 G f'(G)+\frac 12 f(G)\\
+\frac{864 H^6 g}{g'^3}[9 g g'-H g'^2-3 H g g''] f''(G)
\end{array}
\label{difG}\\
G=24 H^2(H^2-3H g/g')\label{GH}\, .
\end{eqnarray}
For a given $g(H)$, one can use (\ref{GH}) to get $H(G)$, then set it
into (\ref{difG}) and solve the second order linear differential
equation. As in the previous case, there will be a bi-parametric family
of solutions for $f(G)$ which will have (\ref{rho_g}) as their
homogeneous isotropic solution as long as matter dominates. The
parameters will need to be fixed in order to make the contribution of
undesired terms in the Friedman equation to be negligible when
radiation dominates. Again $g(H)=H^2$ is an example which can be
solved analytically. The differential equation turns to be
\begin{equation}
12 G^2 f''(G)-G f'(G)+f(G)=0\, ,
\end{equation}
which has the solution
\begin{equation}
f(G)=c_1 G + c_2 G^{1/12}\, .
\end{equation}
In most cases this procedure will not admit an analytical solution
and an approximate solution will need to be found numerically.
A similar discussion can be made to explain the early accelerated
expansion (inflation) as a result of an $f(R)$ or $f(G)$ action,
taking into account that in this case the dominant contribution to the
energy density is radiation instead of matter. The analogue to
(\ref{fR-}) will be now a solution of the form
\begin{equation}
\frac{f_i(R)}{R}=\sum_{n=0}^\infty a_n^{(i)} (\frac{R}{R_+})^{r_i+n} \label{fR+}\, ,
\end{equation}
for $\vert R \vert \ll \vert R_+\vert$ with $R_+ = -12 H_+^2$.
Terms in the action which dominate over $R$ when $R$ becomes small
enough should be eliminated in order to recover the Standard
Cosmological model before matter starts to dominate.
Both accelerated expansions can be described together in the
homogeneous limit by an $f(R)$ action with terms
$(\frac{R_-}{R})^{s_i+m}$ with $s_i>0$ coming from (\ref{fR-}) and
terms $(\frac{R}{R_+})^{r_i+n}$ with $r_i>0$ coming from
(\ref{fR+}). The action will contain a term, the Einstein-Hilbert
action $R$, which will be dominant for $H_-\ll H\ll H_+$ including the
period when matter and radiation have comparable energy densities.
\section{Alternative Descriptions of the ACM}
In general, by virtue of the gravitational field equations, it is
always possible to convert a modification in the gravitational term of
the action to a modification in the matter content of the universe. In
particular, at the homogeneous level, it is possible to convert a
generalized first Friedman equation of the type (\ref{rho_g}) to an
equation in which, apart from the usual matter term, there is a dark
energy component with an unusual equation of state $p=p(\rho)$ and in
which General Relativity is not modified.
The trivial procedure is the following. Assume that the source of the
gravitational field in the modified gravity theory behaves as
$p_d=\omega \rho_d$ (matter or radiation). This is the case when the
modification of gravity is relevant. We can then define a dark energy
or effective gravitational energy density as
\begin{equation}
\rho_g =\frac{3}{8\pi G_N} \left( H^2-g(H)\right) \label{dedensity}\, .
\end{equation}
The continuity equation (\ref{cont_eq}) allows to define a pressure
for this dark energy component as $p_g=-\rho_g-\dot{\rho}_g/3 H$. The
equation of state of the dominant content of the universe leads to
\begin{equation}
\dot{H}=-3(1+\omega)\frac{H g(H)}{g'(H)} \label{dH}\, ,
\end{equation}
which can be used to express $\dot{\rho}_g$ as a function of $H$, and
one gets the final expression for $p_g$,
\begin{equation}
p_g=\frac{-3}{8\pi G_N}\left(
H^2-2(1+\omega)H\frac{g(H)}{g'(H)}+\omega g(H)\right)
\label{depressure}\, .
\end{equation}
Then, for any given $g(H)$ we can use (\ref{dedensity}) to get
$H=H(\rho_g)$ and substitute in (\ref{depressure}) to get an
expression of $p_g(\rho_g;H_\pm,\omega)$ which can be interpreted as
the equation of state of a dark energy component. In the simple case
of a matter dominated universe with ACM, $\alpha_-=1$ gives obviously
a dark energy component verifying $p_g=-\rho_g$. Another simple
example is $\alpha_-=2$, for which
\begin{equation}
\left(\frac{8 \pi G_N}{3 H_-^2} \right) p_g=\frac{2}{\frac{8 \pi
G_N}{3 H_-^2}\rho_g -3}\, .
\end{equation}
The inverse procedure is also straightforward. Suppose we have a
universe filled with a standard component $p_d=\omega\rho_d$ and a
dark energy fluid $p_g=p_g(\rho_g)$. Using the continuity equation of
both fluids one can express their energy densities as a function of
the scale factor $a$ and then use this relation to express $\rho_g$ as
a function of $\rho_d$. Then the Friedman equation reads
\begin{equation}
H^2=\frac{8\pi G_N}{3}(\rho_d+\rho_g(\rho_d))\, ,
\end{equation}
which, solving for $\rho_d$, is trivially equivalent to (\ref{rho_g}).
This argument could be applied to reformulate any cosmological model
based on a modification of the equation of state of the dark energy
component \cite{Odin3} as a generalized first Friedman equation
(\ref{rho_g}).
Until now we have considered descriptions in which there is an exotic
constituent of the universe besides a standard component (pressureless
matter or radiation). In these cases, pressureless matter includes both
baryons and Dark Matter. However, there are also descriptions in which
Dark Matter is unified with Dark Energy in a single constituent of the
universe. One of these examples is the Chaplygin gas $p_g=-1/\rho_g$
\cite{Chaplygin, Fabris}. The use of the continuity equation leads to
\begin{equation}
\rho_g=\sqrt{A+B a^{-6}}\, ,
\end{equation}
where $A$ and $B$ are integration constants. The previous method can
be used to find the generalized first Friedman equation for the baryon
density in this model,
\begin{equation}
\frac{8\pi
G_N}{3}\rho_b=\frac{H^2}{k}\left(\sqrt{1+k(1-\frac{H_-^4}{H^4}})-1
\right) \label{Chap}\, ,
\end{equation}
where $H_-=\sqrt{\frac{8\pi G}{3}}A^{1/4}$, $k=B \rho_{b0}^{-2}-1$,
and $\rho_{b0}$ is the present value of $\rho_b$. In the $H\gg H_-$
limit this model can be interpreted as a universe filled with baryons
and dark matter or as a universe filled with baryons and with a higher
effective value of $G_N$. Equation (\ref{Chap}) does not fulfill
(\ref{hpm}) because it describes the behavior of just baryon
density. In the $H\gtrsim H_-$ limit, the model can be interpreted as
a universe filled with baryons and a cosmological constant or as an
ACM model with $\alpha_-=1$ and filled only with baryons.
A similar example in the early period of accelerated expansion would
be a universe filled with a fluid which behaves as
a fluid of ultra-relativistic particles if the energy density is low
enough but whose density has an upper bound
\begin{equation}
\rho_X=\frac\sigma{a^4+C}\, .
\end{equation}
This dependence for the energy density in the scale factor follows
from the equation of state
\begin{equation}
p_X=\frac 13 \rho_X-\frac {4 C}{3 \sigma}\rho_X^2 \, .
\end{equation}
This model turns out to be equivalent to a universe filled with
radiation following eq. (\ref{alpha+}) with $\alpha_+=1$ and
$H_+^2=(\frac{8 \pi G_N}{3})\frac{\sigma}{C}$.
Another equivalent description would be to consider that the universe
is also filled with some self interacting scalar field $\varphi$ which
accounts for the discrepancy between the standard energy-momentum and
GR Einstein tensors. Given an arbitrary modified Friedman equation
(\ref{rho_g}) a potential $V(\varphi)$ can be found such that the
cosmologies described by both models are the same. The procedure is
similar to the one used to reconstruct a potential from a
given cosmology \cite{Odin7}. From the point of
view of the scalar field, the cosmology is defined by a set of three
coupled differential equations
\begin{eqnarray}
\ddot{\varphi}+3 H \dot{\varphi}+V'(\varphi)&=0 \label{klein}\\
\frac{8\pi G_N}{3}(\rho_d+\frac{\dot{\varphi}^2}2 +V(\varphi))&=H^2
\label{quint}\\ -3(1+\omega)H\rho_d&=\dot{\rho_d} \label{contw}\, .
\end{eqnarray}
The solution of these equations for a certain $V(\varphi)$ will give
$H(t)$ and $\rho_d(t)$, and therefore $\rho_d(H)$ which is $g(H)$ up
to a factor $\frac{8 \pi G_N}{3}$. In this way one finds the
generalized first Friedman equation associated with the introduction
of a self interacting scalar field.
Alternatively, given a function $g(H)$ in a generalized first Friedman
equation (\ref{rho_g}) for the density $\rho_d$, one can find a scalar
field theory leading to the same cosmology in the homogeneous
limit. By considering the time derivative of (\ref{quint}) and using
(\ref{klein}) and (\ref{contw}) one gets
\begin{equation}
\dot{\varphi}^2=-(1+\omega)\rho_d-\frac{\dot{H}}{4\pi G_N} \label{dtvarphi}\, ,
\end{equation}
where we can use (\ref{dH}) and (\ref{rho_g}) to get $\dot{\varphi}^2$
as a function of $H$. Setting this on (\ref{quint}) we get
$V(\varphi)$ as a function of $H$. On the other hand,
$\varphi'(H)=\dot{\varphi}/\dot{H}$, so
\begin{equation}
\frac{8\pi G_N}{3}(\varphi'(H))^2=\frac{(2H
-g'(H))g'(H)}{9(1+\omega) H^2 g(H)}
\label{phi}
\end{equation}
and
\begin{equation}
\frac{8\pi G_N}{3}V(\varphi(H))=H^2-g(H)+\frac{(1+\omega)(2H-g'(H))g(H)}{2
g'(H)}\, .
\label{V(phi)}
\end{equation}
If $g(H)$ is such that the rhs of (\ref{phi}) is positive definite, it
can be solved and the solution $\varphi(H)$ inverted and substituted
into (\ref{V(phi)}) in order to get $V(\varphi)$. This is the case of
a function of the type (\ref{alpha-}) with $\alpha_->1$. For the
particular case of an ACM expansion with $\alpha_-=1$, the solution is
a flat potential $V=V_0$ and a constant value of $\varphi=\varphi_0$.
If $g(H)$ is such that the rhs of (\ref{phi}) is negative definite, as
it happens in (\ref{alpha-}) with $\alpha_-<1$ or in (\ref{alpha+}),
the problem can be solved by changing the sign of the kinetic term in
(\ref{quint}). The result is a phantom quintessence in the case of
(\ref{alpha-}) with $\alpha_-<1$. The case (\ref{alpha+}) is more
complicated because the energy density of the associated inflaton
turns out to be negative. Moreover, it is of the same order as the
energy density of ultra-relativistic particles during the whole period
of accelerated expansion. Therefore, it seems that this model is
inequivalent to other inflation scenarios previously studied.
In summary, there are many equivalent ways to describe the discrepancy
between observed matter content of the universe and Einstein's General
Relativity. At the homogeneous level, it is trivial to find relations
among them. The possibility to establish the equivalence of different
descriptions is not a peculiarity of the description of this discrepancy
in terms of a generalized first Friedman equation (\ref{rho_g}). The
same relations can be found among generalized equations of state,
scalar-tensor theories and f(R) modified gravity \cite{Odin6}.
\section{Summary and discussion}
It may be interesting to go beyond $\Lambda$CDM in the description of
the history of the universe in order to identify the origin of the
two periods of accelerated expansion. We have proposed to use the
expression of the energy density as a function of the Hubble parameter
as the best candidate to describe the history of the universe. In this
context the late time period of accelerated expansion and the early
time inflation period can be easily parametrized.
We have considered a simple modification of the cosmological equations
characterized by the appearance of an upper and a lower bound on the
Hubble expansion rate. A better fit of the experimental data can be obtained
with this extended cosmological model as compared with the
$\Lambda$CDM fit. Once more precise data are available, it will be
possible to identify the behavior of the energy density as a function
of the Hubble parameter and then look for a theoretical derivation of
such behavior.
We plan to continue with a systematic analysis of different
alternatives incorporating the main features of the example considered
in this work. Although the details of the departures from the standard
cosmology can change, we expect a general pattern of the effects induced
by the presence of the two bounds on $H$. We also plan to go further,
considering the evolution of inhomogeneities looking for new consequences
of the bounds on $H$.
The discussion presented in this work, which is based on a new description
of the periods of accelerated expansion of the universe, can open a new way
to explore either modifications of the theory of gravity or new components
in the universe homogeneous fluid. Lacking theoretical criteria to select
among the possible ways to go beyond $\Lambda$CDM, we think that the
phenomenological approach proposed in this work is justified.
We are grateful to Paola Arias and Justo L\'opez-Sarri\'on for
discussions in the first stages of this work. We also acknowledge
discussions with Julio Fabris, Antonio Segu\'{\i}, Roberto Empar\'an,
Sergei Odintsov, Aurelio Grillo and Fernando M\'endez.
This work has been partially supported by CICYT (grant
FPA2006-02315) and DGIID-DGA (grant2008-E24/2). J.I. acknowledges a
FPU grant from MEC.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,828
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\section{Introduction}
\label{introduction}
In this note we study existence of a solution of a system of reflected backward stochastic differential equations (BSDEs) with inter-connected obstacles. Letting $T>0$ and $t\in [0,T]$, the problem is to find $m$ trebles of $({\mathcal F}_s)_{s\in [t,T]}$-adapted processes $(Y^i,Z^i,K^i)_{i\in \Gamma}$, where $\Gamma:=\{1,\dots,m\}$, $Y^i,\,K^i\in\mathbb{R}$ and $Z^i\in\mathbb{R}^d$, $d\ge1$, such that for any $i\in \Gamma$ we have: $\forall s\in [t,T]$,
\begin{equation}\label{main-systemintro}
\left\{
\begin{array}{l}
{Y}^{i}_s=h_i(X^{t,x}_T)+\int^T_sf_i(r,X^{t,x}_r,({Y}_r^{k})_{k\in \Gamma},({Z}^{k}_r)_{k\in \Gamma})dr+{K}^{i}_T-{K}^{i}_s-\int^T_s{Z}^{i}_rdB_r\,\\[+4pt]
{Y}^{i}_s\geq \max\limits_{j\neq i}\{{Y}^{j}_s-{g}_{ij}(s,X^{t,x}_s)\}\\[+4pt]
\int_t^T( {Y}^{i}_s-\max\limits_{j\neq i}\{{Y}^{j}_s-{g}_{ij}(s,X^{t,x}_s)\})dK^i_s=0
\end{array}
\right.
\end{equation}
where:
\begin{itemize}
\item[ i)] $B$ is a $d$-dimensional Brownian motion and we denote $Z^i=(Z^{i1},Z^{i2}\ldots Z^{id})$ and ${Z}^{i}dB:=\sum_{j=1}^dZ^{ij}dB^j$;
\item[ii)] for any $i,j\in \Gamma$, the functions $h_i$, $f_i$ and $g_{ij}$ are deterministic;
\item[iii)] for any $(t,x)\in [0,T]\times\mathbb{R}^k$, the process $X^{t,x}$ is solution of the following SDE:
\[
X^{t,x}_{s}=x+\int_t^{s}b(r,X^{t,x}_r)dr+\int_t^{s}\sigma(r,X^{t,x}_r)dB_r,\quad t\le s\leq T.
\]
\end{itemize}
Since randomness in \eqref{main-systemintro} stems from the Markov process $X^{t,x}$, we say that the system \eqref{main-systemintro} is Markovian.
If for $i=1,...,m$, $f_i$ does not depend on $(y^i)_{i=1,m}$ and $(z^i)_{i=1,m}$, the solution of \eqref{main-systemintro} is linked to an optimal switching problem. The latter is a problem in which a decision maker (or controller) controls a (stochastic) system which may operate in different modes (e.g., a power plant). The aim of the controller is to maximise some performance criterion by optimally choosing controls of the form $\delta:=(\tau_n,\zeta_n)_{n\geq 0}$. Here $(\tau_n)_{n\geq 0}$ denotes an increasing sequence of (stopping) times at which the controller switches the system across different operating modes. Moreover, $(\zeta_n)_{n\ge 0}$ is a sequence of random variables taking their values in $\{1,...,m\}$. Each $\zeta_n$ represents the system's new operating mode after a switch has occurred at time $\tau_n$.
In this setting it is well known (see e.g. \cite{cek, dhp,hamadene-morlais,hutang}, etc.) that $Y^{i}_t$ is the \emph{value} of an optimal switching strategy, i.e., given $\tau_0=t$ and $\zeta_0=i$, it holds
\begin{equation}\label{carac:intro}
Y^{i}_t=\esssup_{\delta:=(\tau_n,\zeta_n)_{n\geq 0}}\mathsf{E}\Big[\int_t^T f_{a_s}(s, X^{t,x}_s)ds-A^\delta_T+h_{a_T}(X^{t,x}_T)\Big|{\mathcal F}_t\Big]
\end{equation}
where the process $a:=(a_s)_{s\leq T}$ is indicating the mode of the
system at time $s$, $A^\delta_T$ stands for the total switching cost
when the strategy $\delta$ is implemented and, finally,
$h_{a_T}(X^{t,x}_T)$ is the terminal payoff. It is also known that
the solution of \eqref{main-systemintro} enables to construct an
optimal strategy as well.
It is important to remark that a characterization as in
\eqref{carac:intro} also holds in non-Markovian frameworks and
we mention that switching problems often arise in economics, finance
and power system management, amongst many other applied fields (see
e.g. \cite{[BO], [BO1], [BS],[CL],
[DX],[DP],[DZ2],porchet,shi,tri1,tri} and the references therein).
Problems like \eqref{main-systemintro} have been studied at a
theoretical level in the case when, for any $i=1,...,m$, the
function $f_i$ depends only on the state variable $z^i$ and possibly
on $(y^i)_{i\in \Gamma}$ (see, e.g., \cite{cek,hamadene-morlais}). In
that setting existence (and uniqueness) results were provided (also
for the non-Markovian case) by using comparison principles for solutions of BSDEs. Such comparisons do not hold in our framework since $f_i$ depends on $(z^i)_{i\in \Gamma}$, hence we must rely on different methods.
The main objective of this paper is indeed to consider systems in which, for $i=1,...,m$, functions $f_i$ not only depend on the state variable $z^i\in\mathbb{R}^d$ but on all components of the state variable $z:=(z^i)_{i\in\Gamma}$.
In particular we show that if $\sigma \sigma^\top$ is bounded and uniformly elliptic, then \eqref{main-systemintro} has a solution, provided that the switching costs $(g_{ij})_{i,j\in \Gamma}$ are sufficiently regular. We adopt a usual penalization scheme (see \eqref{eq:penalisedBSDE} below) to handle the reflection constraints and rely deeply on essentially three facts: i) the representation of solutions of BSDEs as deterministic functions of $t$ and $X$; ii) smoothness of $g_{ij}$, which enables fundamental bounds in the penalisation scheme; iii) existence of a transition density of $X^{t,x}_s$ for any $s>t$, which satisfies a so-called \emph{domination condition}.
Our work is a first step towards the solution of \eqref{main-systemintro} in general non-Markovian setup. The paper is organized as follows. In Section \ref{sec:problemformulation} we set out the notations and make standing assumptions that hold throughout the paper. In Section \ref{sec:mainresult}, we prove our existence result in a number of steps. First we introduce the penalization scheme associated with \eqref{main-systemintro} and study its properties (in particular we show in Proposition \ref{prop:penalisation} that the time derivative of the penalizing term is uniformly bounded). Then we use an argument based on weak convergence and the aforementioned domination condition (see also \cite{HLP}) to obtain a convergent subsequence of solutions to the penalized problems. We finally show that the limit of such subsequence solves \eqref{main-systemintro} and provide a representation of $(Y^i,Z^i)_{i\in\Gamma}$ as deterministic functions of $(t,X)$. We leave for future investigation questions of uniqueness of the solution and its links to optimal switching problems. The latter will inevitably feature a more general structure than \eqref{carac:intro}.
\section{Setting and problem formulation}
\label{sec:problemformulation}
\subsection{Setting}
\label{sec:setting}
Let $T$ be a fixed positive real constant, and let $(\Omega,{\mathcal F},\mathsf{P})$ be a probability space on which we define a $d$-dimensional standard Brownian motion $B:=(B_t)_{t\in [0,T]}$. For $t\leq T$, we set ${\mathcal F}^\circ_t:=\sigma\{B_s, s\le t\}$, the $\sigma$-algebra generated by $B$ up to time $t$, and we denote by $({\mathcal F}_t)_{t\leq T}$ the completion of $({\mathcal F}^\circ_t)_{t\leq T}$ with the $\mathsf{P}$-null sets of ${\mathcal F}$. For arbitrary integer numbers $d\ge1$ and $m\ge 1$, we denote by $|\cdot|_{d}$ and $|\cdot|_{m\times d}$ the Euclidean norms in $\mathbb{R}^d$ and $\mathbb{R}^{m\times d}$, respectively. Occasionally, when no confusion may arise, we will simplify our notation using $|\cdot|$ for either $|\cdot|_{d}$ or $|\cdot|_{m\times d}$.
Next, we introduce the following sets:
\begin{itemize}
\item[(i)] ${\mathcal P}$ is the $\sigma$-algebra of ${\mathcal F}_t$-progressively measurable sets of $\Omega\times[0,T]$;
\item[(ii)] ${\mathcal B}(\mathbb{R}^d)$ is the Borel $\sigma$-algebra on $\mathbb{R}^d$, $d \geq 1$;
\item[(iii)] $\textbf{H}^2_T(\mathbb{R}^d):= \{\zeta:=(\zeta_t)_{t\leq T}$ is a $\mathbb{R}^d$-valued, ${{\mathcal P}}$-measurable process such that $\mathsf{E}\big[\int^T_0{\vert {\zeta_t}\vert}^2 dt\big]<\infty\}$;
\item[(iv)]$\textbf{S}^2_T(\mathbb{R}):= \{\xi:=(\xi_t)_{t\leq T}$ is a $\mathbb{R}$-valued, ${{\mathcal P}}$-measurable, continuous process such that $\mathsf{E}\big[\sup_{0\leq t\leq T}{\vert{\xi_t}\vert}^2\big]<\infty\}$;
\item[(v)] $\textbf{A}^2_T$ is the subspace of $\textbf{S}^2_T$ of non-decreasing processes which are null at $t=0$.
\item[(vi)] $\mathcal{C}^{1,2}([0,T] \times \mathbb{R}^d)$ (or simply $\mathcal{C}^{1,2}$) is the set of real-valued functions defined on $[0,T] \times \mathbb{R}^d$ which are once continuously differentiable in $t$ and twice continuously differentiable in $x$.
\end{itemize}
Let $X:=(X_s)_{s\leq T}$ be an $({\mathcal F}_t)_{t\le T}$-Markov process, valued in $\mathbb{R}^k$, $k\geq1$. For $(t,x)\in[0,T]\times\mathbb{R}^k$ fixed, we denote by $X^{t,x}$ the process $(X_s)_{s\in [t,T]}$ such that $\mathsf{P}(X^{t,x}_t=x)=1$, and by $\mu(t,x;s,dy)$ the law of $X^{t,x}_s$ (for $s\ge t$), i.e., $\mathsf{P}(X^{t,x}_s\in A)=\mu(t,x;s,A)$ for any $A\in {\mathcal B}(\mathbb{R}^k)$. We now introduce the following condition on the Markov process $X$.
\medskip
\noindent{\bf (A0)} [\emph{${L}^2$-domination condition}]. We say that the process $X$ satisfies the ${L}^2$-domination condition if the family of laws $\{\mu(t,x;s,dy),\ s\in [t,T],\,t\in[0,T],\,x\in\mathbb{R}^k\}$ verifies the following condition: There exists $x_0\in\mathbb{R}^k$ such that, for any $t\in[0,T]$ and $x\in \mathbb{R}^k$ and any $\delta>0$ (such that $\delta+t\leq T$) there exists an application $\phi_{t,x,x_0}^{\delta}: [t,T] \times \mathbb{R}^k \mapsto \mathbb{R}_+$ with the following properties:
\begin{itemize}
\item[(a)] $\mu(t,x;s,dy)ds=\phi_{t,x,x_0}^{\delta}(s,y)\mu(0,x_0;s,dy)ds$ for all $(s,y)\in [t+\delta,T]\times\mathbb{R}^k;$
\item[(b)] $\forall N\geq 1,\ \phi_{t,x,x_0}^{\delta}\in {L}^{2}([t+\delta,T]\times [-N,N]^k;\ \mu(0,x_0;s,dy)ds)$.
\end{itemize}
\medskip
\noindent \textbf{Example}. A Markov process fulfilling the
${L}^2$-domination condition is given by the solution of the
stochastic differential equation
\begin{equation}
\label{cmSDE}
X^{t,x}_s=x+\int_t^s b(r,X^{t,x}_r)dr+\int^s_t\sigma(r,X^{t,x}_r)dB_r,\,\,~s\in [t,T],
\end{equation}
with $(t,x)\in[0,T]\times\mathbb{R}^k$, under the conditions detailed below:
\begin{itemize}
\item[(E1)] We take $k=d$ (recall that $B$ is $d$-dimensional), and the functions $b: [0,T]\times \mathbb{R}^d \mapsto \mathbb{R}^d$ and $\sigma: [0,T]\times \mathbb{R}^d \mapsto \mathbb{R}^{d\times d}$ are jointly continuous in $(t,x)$. Moreover they are Lipschitz continuous in $x$, uniformly with respect to $t$, i.e.~there exists a non-negative constant $C_1$ such that for any $(t,x,x')\in [0,T]\times \mathbb{R}^{d+d}$ we have
\begin{align}\label{eq:Lip0}
|\sigma(t,x)-\sigma(t,x')|_{d\times d}+|b(t,x)-b(t,x')|_d\leq C_1|x-x'|_d.
\end{align}
The above property, together with the joint continuity, imply that
$b$ and $\sigma$ have sub-linear growth in $x$, i.e.~there
is $C_2>0$ such that
\begin{align}\label{eq:sublin}
|b(t,x)|_d+|\sigma(t,x)|_{d\times d}\leq C_2(1+|x|_d).
\end{align}
\item[(E2)]We assume further that $\sigma\sigma^\top$ is uniformly elliptic, i.e., that there exists a constant $\theta>0$ such that for any $(t,x) \in [0,T] \times \mathbb{R}^d$ (denoting by $\langle\cdot,\cdot\rangle_d$ the scalar product in $\mathbb{R}^d$) it holds
$$\theta^{-1}|\zeta|_d^2\leq \langle\sigma(t,x)\sigma(t,x)^\top\zeta,\zeta\rangle_d \leq \theta |\zeta|_d^2,\,\,\zeta \in \mathbb{R}^d.$$
\end{itemize}
Condition $(E1)$ guarantees that the solution of \eqref{cmSDE} exists and it is unique (see, e.g., Chapter 5 of \cite{KS} for more details). Moreover $(E2)$ implies that $\sigma$ is bounded and invertible, with bounded inverse $\sigma^{-1}$. Uniform ellipticity of $\sigma$ also implies (cf.\ \cite{aronson}) that for any $(t,x) \in [0,T] \times \mathbb{R}^d$ the law $\mu(t,x;s,dy)$ of $X^{t,x}_s$ has a density function $p(t,x;s,y)$ such that for every $s>t$ and $y\in \mathbb{R}^d$
\begin{equation}
\label{Aronson}
m(s-t)^{-\frac{d}{2}}\exp\Big\{-\frac{\Lambda|y-x|_d^2}{s-t}\Big\}\leq p(t,x;s,y)\leq M(s-t)^{-\frac{d}{2}}\exp\Big\{-\frac{\lambda|y-x|_d^2}{s-t}\Big\}.
\end{equation}
Here $m$, $M$, $\lambda$ and $\Lambda$ are positive constants such that $m\leq M$ and $\lambda\leq \Lambda$. It is then easily verified that the family $\{\mu(t,x;s,dy),\ s\in [t,T],\,t\in[0,T],\,x\in\mathbb{R}^k\}$ satisfies the $L^2$-domination condition (A0).
\medskip
For future reference we also recall that $(E1)$ above implies that
\begin{equation}
\label{estimx}
\mathsf{E}\big[\sup_{t\le s\leq T}|X^{t,x}_s|_d^{\gamma}\big]\leq C(1+|x|_d^{\gamma}),
\end{equation}
for any $\gamma \geq 1$ and with $C=C(T,\gamma, C_2)>0$, independent of $x$. Moreover, the infinitesimal generator of $X^{t,x}$, denoted by $\mathbb{L}_X$, reads
\begin{align}\label{eq:LX}
\big(\mathbb{L}_X\psi\big)(x)=\frac{1}{2}\sum_{i,j=1}^d((\sigma\sigma^\top)_{ij}\partial^2_{x_ix_j}\psi)(x)+
\sum_{i=1}^d(b_i\partial_{x_i}\psi)(x),
\end{align}
for $\psi \in {\mathcal C}^{2}(\mathbb{R}^d)$ and for any $x \in \mathbb{R}^d$.
At this point it is worth noticing that the results of this paper hold for a general Markov process $X$ provided that $X$ is a semi-martingale, it satisfies the ${L}^2$-domination condition and \eqref{estimx},
and the increments of the bounded variation part of the processes $(g_{ij}(t,X_t))_{t\in[0,T]}$ are non-positive (see Assumption \textbf{(A2)}-(b) below). However, in order to avoid technicalities and to improve readability of the paper, from now on we make the following standing assumption
\begin{assumption}\label{ass:1}
We assume that $k=d$ and that $X^{t,x}$ is the solution of \eqref{cmSDE} under conditions (E1) and (E2) above, hence satisfying the ${L}^2$-domination condition \textbf{(A0)}.
\end{assumption}
\subsection{A system of reflected BSDEs with interconnected obstacles}
\label{sec:coeff}
Here we formulate the problem object of our study, i.e.~a system of reflected BSDEs with interconnected obstacles. We begin by introducing $\Gamma:=\{1,2,...,m\}$ and functions $(f_i)_{i\in \Gamma}$, $(h_i)_{i\in \Gamma}$ and $(g_{ij})_{i,j\in \Gamma}$ which satisfy the requirements below.
\medskip
\noindent {\bf(A1)} For any $i\in \Gamma$, the function
\[
f_i:\,\,(t,x,(y_k)_{k\in \Gamma},(z_k)_{k\in \Gamma})\in [0,T]\times \mathbb{R}^{d+m+m\times d}\longmapsto f_i(t,x,(y_k)_{k\in \Gamma},(z_k)_{k\in \Gamma}) \in \mathbb{R}
\]
\begin{itemize}
\item[(a)] is Lipschitz continuous in the variables $(\vec{y},\vec{z}):=((y_k)_{k\in \Gamma},(z_k)_{k\in \Gamma})$, uniformly with respect to $(t,x)$; that is, there is $C>0$ such that
\begin{equation}
\label{growthcondition}
|f_i(t,x,\vec{y}_1,z_1)-f_i(t,x,\vec{y}_2,\vec{z_2})|\leq C(|\vec{y}_1-\vec{y}_2|_m+|\vec{z_1}-\vec{z_2}|_{m\times d}),
\end{equation}
for any $(t,x)\in [0,T]\times \mathbb{R}^d$, $(\vec{y}_1,\vec{y}_2)\in (\mathbb{R}^{m})^2$ and $(\vec{z}_1,\vec{z}_2)\in (\mathbb{R}^{m\times d})^2$;
\item[(b)] has sub-polynomial growth in $x$, uniformly with respect to $(t,\vec{y}, \vec{z})$; that is, there are $C>0$ and $q\ge 1$ such that
\begin{equation}
\label{f-bound}
|f_i(t,x,\vec{y},\vec{z})|\leq C(1+|x|_d^q),\qquad\text{for all $(t,x,\vec{y},\vec{z}) \in [0,T]\times \mathbb{R}^{d+m+m\times d}$}.
\end{equation}
\end{itemize}
\noindent{\bf(A2)} For $(i,j) \in \Gamma \times \Gamma$, the functions
\[
{g}_{ij}: (t,x)\in[0,T] \times \mathbb{R}^d \longmapsto g_{ij}(t,x)\in\mathbb{R}_+
\]
have the following properties:
\begin{itemize}
\item[(a)] let $i,j,\ell\in \Gamma$ with $\text{card}\{i,j,\ell\}=3$, then $g_{ij}(t,x)<g_{i\ell}(t,x)+ g_{\ell j}(t,x)$, for any $(t,x) \in [0,T] \times \mathbb{R}^d$. Moreover, $g_{ii}(t,x)=0$;
\item[(b)] for any $i,j\in \Gamma$, $g_{ij}$ belongs to ${\mathcal C}^{1,2}([0,T]\times \mathbb{R}^d)$ and
\[
\rho_{ij}(t,x):=(\partial_t\,g_{ij}+ \mathbb{L}_Xg_{ij})(t,x)\le 0,\qquad\text{for all $(t,x)\in[0,T]\times\mathbb{R}^d$}.
\]
\end{itemize}
\begin{remark}\
\begin{enumerate}
\item Notice that condition \textbf{(A2)}-(a) implies the so-called non-free loop property which is considered in several papers including \cite{hamadene-morlais, ishi-koike}, among others. Indeed, take a loop of $\Gamma$, i.e., a sequence $\{i_1,.....,i_{\ell}\}$ of $\Gamma$ such that ${\ell}\geq 3$, $\text{card}\{i_1,.....,i_{\ell}\}={\ell}-1$ and $i_{\ell}=i_1$. Then, under \textbf{(A2)}-(a) we have that for any $(t,x) \in [0,T] \times \mathbb{R}^d$
\begin{align*}
g_{i_1i_2}&(t,x)+g_{i_2i_3}(t,x)+\ldots+g_{i_{\ell-1}i_{\ell}}(t,x)\\
&>g_{i_1i_3}(t,x)+g_{i_3i_4}(t,x)+\ldots+g_{i_{\ell-1}i_{\ell}}(t,x)>\ldots >g_{i_1i_1}(t,x)=0.
\end{align*}
\item Conditions \textbf{(A2)} are satisfied if we take, for example, $g_{ij}$ independent of $x$ and of the form $g_{ij}(t,x)=\Phi(t)|i-j|$, with $\Phi$ continuously differentiable on $[0,T]$, non-increasing and positive.
\end{enumerate}
\end{remark}
\noindent {\bf(A3)} For any $i \in \Gamma$ the functions
\[
h_i:x\in \mathbb{R}^d \longmapsto h_i(x)\in \mathbb{R}
\]
are such that for every $x \in \mathbb{R}^d$
\begin{itemize}
\item[(a)] $|h_i(x)| \leq C(1 + |x|_d^p)$, for some non-negative constant $p$;
\item[(b)] $h_i(x)\geq \max_{j\neq i}(h_j(x)-g_{ij}(T,x))$.
\end{itemize}
Condition \textbf{(A3)}-(b) is usually referred to as a ``consistency condition''. This is needed in order for the process $Y$ in \eqref{main-system} below to be continuous on $[0,T]$ (provided that a solutions to \eqref{main-system} exists).
Assuming that conditions \textbf{(A0)}-\textbf{(A3)} hold, we now consider a system of reflected BSDEs with interconnected obstacles associated with
($(f_i)_{i\in \Gamma}$,$(h_i)_{i\in \Gamma}$,$(g_{ij})_{i,j\in \Gamma}$). More precisely we aim at finding a $m$-tuple of $(\mathcal F_t)_{t\le T}$-adapted processes $(Y ^i,Z^i,K^i)_{i\in \Gamma}$ which solves $\mathsf{P}$-a.s.~the following system: For any $i\in \Gamma$, any $(t,x)\in [0,T]\times\mathbb{R}^d$, and all $s\in[t,T]$ it holds
\begin{align}
\label{main-system}
\left\{
\begin{array}{ll}
\displaystyle Y^i\in \textbf{S}^2_T(\mathbb{R}),\,\,Z^i\in \textbf{H}^2_T(\mathbb{R}^d)\,\, \mbox{and}\,\, K^i\in \textbf{A}^2_T(\mathbb{R});\\[+4pt]
\displaystyle {Y}^{i}_s=h_i(X^{t,x}_T)\!+\!\!\int^T_s
\!\!f_i(r,X^{t,x}_r,({Y}_r^{k})_{k\in \Gamma},({Z}^{k}_r)_{k\in \Gamma})dr\!+\!{K}^{i}_T-\!{K}^{i}_s-\!\!\int^T_s\!\!{Z}^{i}_rdB_r; \\[+4pt]
\displaystyle {Y}^{i}_s \geq \max\limits_{j\neq i}\{{Y}^{j}_s-{g}_{ij}(s,X^{t,x}_s)\};\\[+4pt]
\displaystyle \int_t^T\Big( {Y}^{i}_s-\max\limits_{j\neq i}\{{Y}^{j}_s-{g}_{ij}(s,X^{t,x}_s)\}\Big)dK^i_s=0;
\end{array}
\right.
\end{align}
where we recall that $Z^idB:=\sum^d_{j=1}Z^{ij}dB^j$ with $Z^i:=(Z^{i1}, \ldots Z^{id})$.
The rest of the paper is devoted to proving existence of a solution to \eqref{main-system}.
\section{The main result}
\label{sec:mainresult}
In this section we perform an approximation of \eqref{main-system} via a sequence of penalized problems indexed by $n\in\mathbb N$. Each penalized problem admits a solution and we are able to show that, in the limit as $n\to\infty$, we obtain a solution for \eqref{main-system}.
Given $(t,x) \in [0,T]\times \mathbb{R}^d$ and $n \geq 1$ we introduce a system of BSDEs whose solution is a $m$-tuple of $(\mathcal F_t)_{t\le T}$-adapted processes $(Y^{i,n;t,x},Z^{i,n;t,x})_{i \in \Gamma}$ such that for any $i\in\Gamma$:
\begin{align}
\label{eq:penalisedBSDE}
\left\{
\begin{array}{l}
Y^{i,n;t,x}\in \textbf{S}^2_T(\mathbb{R})\:\text{and}\: Z^{i,n;t,x}\in \textbf{H}^2_T(\mathbb{R}^d); \\[+4pt]
\displaystyle Y^{i,n;t,x}_s = h_i(X^{t,x}_T) + \int_s^T \Big[f_i(r, X^{t,x}_r, (Y^{k,n;t,x}_r)_{k \in \Gamma}, (Z^{k,n;t,x}_r)_{k \in \Gamma}) \\
\qquad\qquad \displaystyle + n\,\sum_{j\neq i} \Big(Y^{i,n;t,x}_r - Y^{j,n;t,x}_r + g_{ij}(r,X^{t,x}_r)\Big)^{-}\Big] dr - \int_s^T Z^{i,n;t,x}_r dB_r,\\
\text{for every}\,\,s \in [t,T].
\end{array}
\right.
\end{align}
First we notice that \eqref{eq:penalisedBSDE} admits a unique solution $(Y^{i,n;t,x},Z^{i,n;t,x})_{i \in \Gamma}$ thanks to Pardoux-Peng's result \cite{pardoux-peng}. More precisely: for any $i\in \Gamma$, the random variable $h_i(X^{t,x}_T)$ is square integrable due to \textbf{(A3)} and \eqref{estimx}; moreover, the functions
$${f}^{(n)}_i(t,x,y,z):=f_i(t,x,y,z) + n \sum_{j\neq i} \big(y_i - y_j + g_{ij}(t,x)\big)^{-}$$
are uniformly Lipschitz in $(\vec{y},\vec{z})$ by \textbf{(A1)}.
Next the Markovian nature of our setting also implies that there
exist measurable deterministic functions $(u^{i,n})_{i\in \Gamma}$ and
$(v^{i,n})_{i\in \Gamma}$, with $u^{i,n}:[0,T]\times\mathbb{R}^d\to\mathbb{R}$ and
$v^{i,n}:[0,T]\times\mathbb{R}^d\to\mathbb{R}^d$, such that for any $(t,x) \in
[0,T]\times \mathbb{R}^d$ and $s \in [t,T]$,
\begin{equation}
\label{represent1}
Y^{i,n;t,x}_s=u^{i,n}(s,X^{t,x}_s)\quad \mbox{ and } \quad Z^{i,n;t,x}_s=v^{i,n}(s,X^{t,x}_s).
\end{equation}
One can refer to \cite{karoui-peng-quenez} (Theorem 4.1, p.\ 46) for more details.
Finally, the following representation holds: for any $i\in \Gamma$ and $(t,x) \in [0,T]\times \mathbb{R}^d$ one has
\begin{align}
\label{representation}
u^{i,n}(t,x)=&\mathsf{E}\bigg[h_i(X^{t,x}_T) + \int_t^T \Big\{f_i(r, X^{t,x}_r, (Y^{k,n;t,x}_r)_{k \in \Gamma}, (Z^{k,n;t,x}_r)_{k \in \Gamma}) \\
&\qquad\qquad\qquad+ n\sum_{j\neq i} \Big(Y^{i,n;t,x}_r - Y^{j,n;t,x}_r + g_{ij}(r,X^{t,x}_r)\Big)^{-}\Big\}dr\bigg].\nonumber
\end{align}
In order to simplify notation, from now and when no confusion may arise, we will drop the $(t,x)$-dependence of $(Y^{i,n;t,x},Z^{i,n;t,x})_{i\in \Gamma}$, and we will simply write $(Y^{i,n},Z^{i,n})_{i\in \Gamma}$. Moreover, we will simply denote $f_i(r, X^{t,x}_r, Y^{n}_r, Z^{n}_r)$ with the convention that $Y^{n}:=(Y^{k,n})_{k\in\Gamma}$ and $Z^{n}:=(Z^{k,n})_{k\in\Gamma}$ .
The next proposition provides a bound for the penalizing term in the driver of \eqref{eq:penalisedBSDE}, which is uniform with respect to $n$.
\begin{proposition}
\label{prop:penalisation}
Let $(t,x) \in [0,T] \times \mathbb{R}^d$ be given and fixed. Then, for $q \geq 1$ as in Assumption {\bf(A1)}-(b), there exists $C=C(q,T)>0$ such that, for any $i \in \Gamma$ and $n\geq 1$, one has
\begin{equation}
\label{eq:penalisation}
n \sum_{j\neq i} \Big(Y^{i,n}_s - Y^{j,n}_s + g_{ij}(s,X^{t,x}_s)\Big)^{-} \leq C\big(1 + |X^{t,x}_s|^{q}\big), \quad t\le s\le T.
\end{equation}
\end{proposition}
\begin{proof}
Fix $(t,x) \in [0,T] \times \mathbb{R}^d$, and for given $i,j \in \Gamma$ and $n \geq 1$, set
\begin{equation}
\label{eq:defxi1}
\xi^{ij,n}_{s}:=Y^{i,n}_{s} - Y^{j,n}_{s} + g_{ij}(s,X^{t,x}_{s}), \qquad s\in [t,T].
\end{equation}
By an application of It\^o-Tanaka's formula (cf.\ \cite{KS}, Chapter 3.7, Theorem 7.1), for every $s\in [t,T]$ we obtain
\begin{align}
\label{eq:stima1}
e^{-n(T-s)}\big(\xi^{ij,n}_{T})^- =&\, \big(\xi^{ij,n}_{s})^{-} -\! \int_s^T \!{{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} d\xi^{ij,n}_{u} - \int_s^T n e^{-n(u-s)} \big(\xi^{ij,n}_{u}\big)^- du \nonumber \\
& + \frac{1}{2}\int_s^T e^{-n(u-s)} dL^0_u(\xi^{ij,n}),
\end{align}
where $L^0(\xi^{ij,n})$ denotes the local-time at zero of the semimartingale $\xi^{ij,n}$. Noticing that the integral with respect to the local-time is nonnegative, we obtain from \eqref{eq:stima1} that for every $s\in [t,T]$
\begin{align}
\label{eq:stima2}
\big(\xi^{ij,n}_{s}\big)^{-} \leq&\, e^{-n(T-s)}\big(\xi^{ij,n}_{T})^- \\
&+\! \int_s^T \!\!{{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} d\xi^{ij,n}_{u} +\! \int_s^T \!\!n\, e^{-n(u-s)} \big(\xi^{ij,n}_{u}\big)^- du.\nonumber
\end{align}
We now want to find a convenient expression for $d\xi^{ij,n}_{u}$. In the definition of $\xi^{ij,n}$ (cf.\ \eqref{eq:defxi1}) we may express $Y^{i,n}$ and $Y^{j,n}$ in terms of their associated BSDEs \eqref{eq:penalisedBSDE}. This gives, for any $u \in [t,T]$
\begin{align}
\label{eq:defxi}
\xi^{ij,n}_{u} =& g_{ij}(u, X^{t,x}_u) + (h_i - h_j)(X^{t,x}_T) + \int_u^T (f_i - f_j)(r,X^{t,x}_r,Y^{n}_r,Z^{n}_r)dr \nonumber \\
& + n \sum_{k \neq i} \int_u^T \big(\xi^{ik,n}_{r}\big)^- dr - n \sum_{k \neq j} \int_u^T \big(\xi^{jk,n}_{r}\big)^- dr - \int_u^T (Z^{i,n}_r - Z^{j,n}_r) dB_r.
\end{align}
Then taking the differential with respect to the time variable $u$, and recalling $\rho_{ij}$ from {\bf(A2)}-(b), gives
\begin{align}
\label{eq:diffxi}
d\xi^{ij,n}_{u} =& \sum_{k =1}^d \frac{\partial g_{ij}}{\partial x_k}(u,X^{t,x}_u) \sigma_k(u,X^{t,x}_u)dB_u+ (Z^{i,n}_u - Z^{j,n}_u) dB_u \nonumber \\
&+\rho_{ij}(u,X^{t,x}_u)du - (f_i - f_j)(u,X^{t,x}_u,Y^{n}_u,Z^{n}_u)du \\[+4pt]
&- n \sum_{k \neq i} \big(\xi^{ik,n}_u\big)^- du + n \sum_{k \neq j} \big(\xi^{jk,n}_u\big)^- du . \nonumber
\end{align}
where we have also set $\sigma_k(u,X_u)dB_u:=\sum_{\ell}\sigma_{k\ell}(u,X_u)dB^\ell_u$ to simplify the notation. We multiply \eqref{eq:diffxi} by ${{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)}$ and integrate over $[s,T]$. Then adding
\[
\int_s^T n e^{-n(u-s)} \big(\xi^{ij,n}_{u}\big)^- du
\]
we obtain
\begin{align}
\label{eq:stima3}
\int_s^T& {{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} d\xi^{ij,n}_{u} + \int_s^T n e^{-n(u-s)} \big(\xi^{ij,n}_{u}\big)^- du \nonumber\\
=&\! \int_s^T \!\!\!{{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \Big[\rho_{ij}(u,X^{t,x}_u) - (f_i - f_j)(u,X^{t,x}_u,Y^{n}_u,Z^{n}_u)\Big] du \nonumber\\
& - n \sum_{k \neq i} \int_s^T\!\!\! {{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \big(\xi^{ik,n}_u\big)^- du + n \sum_{k \neq j} \int_s^T \!\!\! {{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \big(\xi^{jk,n}_u\big)^- du \\
& + \int_s^T n\,e^{-n(u-s)} \big(\xi^{ij,n}_u\big)^- du + M^{ij,n}_{s,T},\nonumber
\end{align}
where we have defined
\begin{equation}
\label{def:M}
M^{ij,n}_{s,T}:=\!\int_s^T \!\!{{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \Big[\sum_{k = 1}^d\frac{\partial g_{ij}}{\partial x_k}(u,X^{t,x}_u) \sigma_{k}(u,X^{t,x}_u)dB_u + (Z^{i,n}_u - Z^{j,n}_u)dB_u\Big].
\end{equation}
Notice in particular that $(M^{ij,n}_{t,s})_{s\in[t,T]}$ is indeed a martingale.
Next we provide upper bounds for some of the terms in \eqref{eq:stima3}. First we notice that for $u\in[t,T]$ it holds
\begin{align}
\label{eq:xi1}
& {{\bf}{1}}_{\{\xi^{ij,n}_{u} < 0\}}\Big(\big(\xi^{jk,n}_u\big)^- - \big(\xi^{ik,n}_u\big)^- \Big) \leq {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}}\Big(\xi^{jk,n}_u - \xi^{ik,n}_u \Big)^- \nonumber \\
& = {\bf}{1}_{\{Y^{j,n}_u > Y^{i,n}_u + g_{ij}(u,X^{t,x}_u)\}} \Big(Y^{j,n}_u + g_{jk}(u,X^{t,x}_u) - Y^{i,n}_u - g_{ik}(u,X^{t,x}_u)\Big)^- \\
& \leq {\bf}{1}_{\{Y^{j,n}_u > Y^{i,n}_u + g_{ij}(u,X^{t,x}_u)\}} \Big(g_{ij}(u,X^{t,x}_u) + g_{jk}(u,X^{t,x}_u) - g_{ik}(u,X^{t,x}_u)\Big)^- =0\nonumber
\end{align}
by Assumption \textbf{(A2)}-(a). Also we notice that
\begin{align}
\label{eq:xi2}
{\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} \big(\xi^{ji,n}_u\big)^- =& {\bf}{1}_{\{Y^{j,n}_u > Y^{i,n}_u + g_{ij}(u,X^{t,x}_u)\}} \Big(Y^{j,n}_u - Y^{i,n}_u + g_{ji}(u,X^{t,x}_u)\Big)^- \nonumber \\
\leq& {\bf}{1}_{\{Y^{j,n}_u > Y^{i,n}_u + g_{ij}(u,X^{t,x}_u)\}} \Big(g_{ij}(u,X^{t,x}_u) + g_{ji}(u,X^{t,x}_u)\Big)^- =0
\end{align}
because switching costs are non-negative.
Now, simple algebra and \eqref{eq:xi1}-\eqref{eq:xi2} give
\begin{align}
\label{eq:stima4}
& \sum_{k \neq j} \int_s^T \!\!\! {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \big(\xi^{jk,n}_u\big)^- du \nonumber\\
&\:\:- \sum_{k \neq i} \int_s^T\!\!\! {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \big(\xi^{ik,n}_u\big)^- du + \!\int_s^T \!\!\!e^{-n(u-s)} \big(\xi^{ij,n}_u\big)^- du\\
&= \sum_{k \neq i,j} \int_s^T\!\!\! {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)}\Big(\big(\xi^{jk,n}_u\big)^- - \big(\xi^{ik,n}_u\big)^- \Big)du \nonumber\\
&\:\:\:\:+\! \int_s^T\!\!\! {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \big(\xi^{ji,n}_u\big)^- du \leq 0.\nonumber
\end{align}
By feeding \eqref{eq:stima4} back into \eqref{eq:stima3} we obtain
\begin{align*}
\int_s^T& {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} d\xi^{ij,n}_{u} + \int_s^T \!\!n\, e^{-n(u-s)} \big(\xi^{ij,n}_{u}\big)^- du \\
\leq&\, M^{ij,n}_{s,T} +\! \int_s^T\!\!\! {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)} \Big[\rho_{ij}(u,X^{t,x}_u) - (f_i - f_j)(u,X^{t,x}_u,Y^{n}_u,Z^{n}_u)\Big] du. \nonumber
\end{align*}
The latter may be plugged in \eqref{eq:stima2} to yield
\begin{align}
\label{eq:stima5}
\big(\xi^{ij,n}_{s})^{-} \leq& e^{-n(T-s)}\Big(h_i(X^{t,x}_T) - h_j(X^{t,x}_T) + g_{ij}(T,X^{t,x}_T)\Big)^- + M^{ij,n}_{s,T} \nonumber\\
& + \int_s^T\!\! {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)}\Big[\rho_{ij}(u,X^{t,x}_u) - (f_i - f_j)(u,X^{t,x}_u,Y^{n}_u,Z^{n}_u)\Big] du,
\end{align}
for every $s \in [t,T]$.
By Assumption \textbf{(A3)}-(b) we have that $\Big(h_i(X^{t,x}_T) - h_j(X^{t,x}_T) + g_{ij}(T,X^{t,x}_T)\Big)^-=0$.
Moreover, our assumptions on the switching costs $g_{ij}$ (cf.\ Asssumption \textbf{(A2)}) and on the volatility $\sigma$ (cf.~Assumption \ref{ass:1}), imply that $\mathsf{E}\big[M^{ij,n}_{s,T}\big|{\mathcal F}_s\big]=0$ (see \eqref{def:M}) and $\rho_{ij}(u,X_u)\le0$. Then, taking conditional expectations with respect to $\mathcal{F}_s$ in \eqref{eq:stima5}, using the sub-polynomial growth of $f_i$ and $f_j$ (cf.\ Assumption \textbf{(A1)}-(b)) and \eqref{estimx}, we obtain that
\begin{align}
\label{eq:stima6}
\big(\xi^{ij,n}_{s})^{-} \leq& \mathsf{E}\bigg[\int_s^T {\bf}{1}_{\{\xi^{ij,n}_{u} < 0\}} e^{-n(u-s)}(f_j - f_i)(u,X^{t,x}_u,Y^{n}_u,Z^{n}_u)du\,\Big|\,\mathcal{F}_s\bigg] \nonumber \\
\leq& \int_s^T \!\!e^{-n(u-s)} c\,\big(1 + \mathsf{E}\big[\sup_{s \leq r \leq u} |X_r|^{q}|\mathcal{F}_s\big]\big) du \leq \frac{c}{n}\big(1 + |X_s|^{q}\big),
\end{align}
for every $s\in[t,T]$, with $q\geq 1$ and for a constant $c=c(T,q)>0$ changing from line to line and independent of $n$.
Recalling \eqref{eq:defxi1} we then conclude that for any $(i,j) \in \Gamma \times \Gamma$ and $n\geq 1$
\[
n \Big(Y^{i,n}_s - Y^{j,n}_s + g_{ij}(s,X^{t,x}_s)\Big)^{-} \leq c\,\big(1 + |X^{t,x}_s|^{q}\big), \quad t\le s\le T.
\]
Taking the summations over all $j\neq i$ and setting $C:=(m-1)c$, we finally obtain
\[
n \sum_{j \neq i}\Big(Y^{i,n}_s - Y^{j,n}_s + g_{ij}(s,X^{t,x}_s)\Big)^{-} \leq C\big(1 + |X^{t,x}_s|^{q}\big), \quad t\le s\le T.
\]
\end{proof}
From now on we denote
\begin{equation}
\label{eq:defK} K^{i,n}_{s}:= n \sum_{j \neq i} \int_t^s
\Big(Y^{i,n}_{r} - Y^{j,n}_{r} + g_{ij}(r,X^{t,x}_{r})\Big)^- dr,
\quad s\in [t,T].
\end{equation}
Thanks to Proposition \ref{prop:penalisation} we are able prove the next uniform estimate on the solution of the penalized problem.
\begin{proposition}
\label{prop:stimaL2}
Let $(t,x) \in[0,T]\times \mathbb{R}^d$ be arbitrary. For any $i \in \Gamma$ and $n\geq 1$ there exist constants $C>0$ and $\rho \geq 1$ independent of $n$ such that
\begin{equation}
\label{eq:stimaL2} \mathsf{E}\bigg[\sup_{t \leq s\leq
T}|Y^{i,n}_s|^2 + \int_t^T |Z^{i,n}_s|^2 ds + |K^{i,n}_T|^2 \bigg]
\leq C(1 + |x|^\rho).
\end{equation}
\end{proposition}
\begin{proof}
Applying It\^o's formula and recalling \eqref{eq:penalisedBSDE} we
obtain that for every $s \in [t,T]$
\begin{align}
\label{eq:stimaL2-1} |Y^{i,n}_s|^2 + \int_s^T\! |Z^{i,n}_r|^2 dr =&
|h_i(X^{t,x}_T)|^2 + 2 \int_s^T\!\!
Y^{i,n}_r f_i(r, X^{t,x}_r, Y^{n}_r, Z^{n}_r) dr\\
&- 2 \int_s^T\!\! Y^{i,n}_r Z^{i,n}_r dB_r + 2 \int_s^T\!\! Y^{i,n}_r dK^{i,n}_r.\nonumber
\end{align}
Taking expectations and using the sub-polynomial growth of $h_i$ and $f_i$ (cf.\ Assumptions \textbf{(A1)}-(b) and \textbf{(A3)}-(a)) we get
\begin{align}
\label{eq:stimaL2-2}
\mathsf{E}\bigg[|Y^{i,n}_s|^2 + \int_s^T |Z^{i,n}_r|^2 dr\bigg] \leq& c_1\Big(1 +
\mathsf{E}\big[|X^{t,x}_T|^{2p}\big]\Big) + 2 c_2 \mathsf{E}\bigg[ \int_s^T \!\!|Y^{i,n}_r|
\Big(1 + |X^{t,x}_r|^{q}\Big) dr\bigg]\nonumber\\
& + 2 \mathsf{E}\bigg[\int_s^T \!\!|Y^{i,n}_r|\, n \sum_{j \neq i}
\Big(Y^{i,n}_{r} - Y^{j,n}_{r} + g_{ij}(r,X^{t,x}_{r})\Big)^-
dr\bigg]
\end{align}
for suitable positive constants $c_1$ and $c_2$. We now use the classical inequality
$2|a b| \leq \varepsilon |a|^2 + \frac{1}{\varepsilon}|b|^2$, for any $a,b \in \mathbb{R}$ and $\varepsilon>0$, the bound \eqref{eq:penalisation} (notice that $q$ therein is the same as the one in \eqref{eq:stimaL2-2}) and \eqref{estimx} to obtain
\begin{align}
\label{eq:stimaL2-3} \mathsf{E}\bigg[|Y^{i,n}_s|^2 + \int_s^T
|Z^{i,n}_r|^2 dr\bigg]
\leq & C\Big( 1 + \mathsf{E}\big[|X^{t,x}_T|^{2p}\big] + \mathsf{E}\bigg[\int_s^T\!\! |Y^{i,n}_r|^2 dr+
\int_s^T\!\! |X^{t,x}_r|^{2q}dr\bigg]\Big)\nonumber\\
\leq& C\Big(1 + |x|^{\rho} + \mathsf{E}\bigg[\int_s^T |Y^{i,n}_r|^2
dr\bigg] \Big),
\end{align}
where $\rho=2(p\vee q)$ and $C=C(T,p,q,\varepsilon)>0$ varies from line to line and it is independent of $n$.
From \eqref{eq:stimaL2-3} and Gronwall's inequality we find $\forall
s\in [t,T]$
\begin{equation}
\label{eq:stimaL2-3bis} \mathsf{E}\big[|Y^{i,n}_s|^2 \big] \leq
C\big(1 + |x|^{\rho}\big),
\end{equation}
for all $n \geq 1$. Letting now $c>0$ be a constant varying from line to line but independent of $n$, using \eqref{eq:stimaL2-3}, \eqref{eq:stimaL2-3bis} and Proposition \ref{prop:penalisation}, we also get
\begin{equation}
\label{eq:stimaL2-3tris} \mathsf{E}\bigg[\int_t^T |Z^{i,n}_r|^2 ds +
|K^{i,n}_T|^2\bigg] \leq c\big(1 + |x|^{\rho}\big).
\end{equation}
The latter and \eqref{eq:stimaL2-3bis} then yield: for any $s\in
[t,T] \times \mathbb{R}^d$,
\begin{equation}
\label{eq:stimaL2-4} \mathsf{E}\bigg[|Y^{i,n}_s|^2 + \int_t^T
|Z^{i,n}_r|^2 dr + |K^{i,n}_T|^2\bigg] \leq c\big(1 +
|x|^{\rho}\big).
\end{equation}
In order to take the supremum of the process $Y^{i,n}$ inside the expectation we need a further bound for $\sup_{t\leq s \leq T}|Y^{i,n}_s|^2$. This can be obtained by using the expression \eqref{eq:penalisedBSDE} for $Y^{i,n}$ together with the sub-polynomial growth of $f_i$ and \eqref{eq:penalisation}, that is
\begin{align}
\label{eq:stimaL2-5}
\sup_{t\leq s \leq T}|Y^{i,n}_s|^2 \leq& 4\Big(|h_i(X^{t,x}_T)|^2 +\!\!
\int_t^T \!\!|f_i(r, X^{t,x}_r, Y^{n}_r, Z^{n}_r)|^2 ds \nonumber\\
&\hspace{+65pt} + |K^{i,n}_T|^2 + \sup_{t\leq s \leq T}\Big|\int_s^T \!\!Z^{i,n}_r dB_r
\Big|^2\Big)\nonumber\\
\le &C\Big(1+\sup_{t\le s\le T}|X^{t,x}_s|^\rho+\sup_{t\leq s \leq T}\Big|\int_s^T \!\!Z^{i,n}_r dB_r
\Big|^2\Big).
\end{align}
Taking the expected value, applying Burkholder-Davis-Gundy's inequality and \eqref{estimx} we finally obtain \eqref{eq:stimaL2}.
\end{proof}
Recall that for each $n\ge0$ we have $Y^{i,n;t,x}_s=u^{i,n}(s,X^{t,x}_s)$ (see \eqref{represent1} and \eqref{representation}). Next we show that the sequences $(u^{i,n})_{n\ge0}$ with $i\in\Gamma$ admit a converging subsequence.
\begin{proposition}
\label{prop:Cauchy-0}
There exists a subsequence $(n_j)_{j\geq 0}$ with $n_j\to\infty$ as $j\to\infty$, and
measurable functions $u^{i}:[0,T]\times\mathbb{R}^d\to\mathbb{R}$, $i\in\Gamma$, such that
\begin{align}\label{eq:lim-u}
\lim_{j\to\infty}u^{i,n_j}(t,x)=u^{i}(t,x)\quad\text{for all $i\in \Gamma$ and $(t,x) \in [0,T] \times \mathbb{R}^d$.}
\end{align}
Moreover there exist two constants $C>0$ and $\rho\ge 1$ (independent of $n_j$) such that for any
$i\in \Gamma$ and $j\ge 0$
\begin{align}
\label{eq:estim-u0}|u^{i,n_j}(t,x)|\leq C(1+|x|^\rho ),\quad \forall (t,x)\in [0,T] \times\mathbb{R}^d
\end{align}
and therefore
\begin{align}
\label{eq:estim-u}|u^i(t,x)|\leq C(1+|x|^\rho ),\quad \forall (t,x)\in [0,T] \times\mathbb{R}^d.
\end{align}
\end{proposition}
\begin{proof}
The proof is given in two steps.
\medskip
\emph{Step 1.} Let $x_0\in \mathbb{R}^d$ be given and fixed as in {\bf (A0)}. Consider the solution of \eqref{eq:penalisedBSDE} for $(t,x)=(0,x_0)$. By the sub-polynomial growth of $f_i$ (see \eqref{f-bound}), by \eqref{estimx} and \eqref{eq:stimaL2}, we can find $C=C(x_0)>0$ (independent of $n$ and $i\in\Gamma$) such that
\begin{align*}
\mathsf{E}\bigg[\int_0^T \Big|&\,f_i(r, X^{0,x_0}_r, Y^{n;0,x_0}_r, Z^{n;0,x_0}_r) \\
& + n\sum_{\ell\neq i} \Big(Y^{i,n;0,x_0}_r - Y^{\ell,n;0,x_0}_r + g_{i\ell}(r,X^{0,x_0}_r)\Big)^{-}\Big|^2 dr\bigg]\leq C.
\end{align*}
Using the representations \eqref{represent1} for $Y^{i,n;0,x_0}$ and $Z^{i,n;0,x_0}$, the above bound reads
\begin{align}
\label{L2bound-1}
\mathsf{E}\bigg[\int_0^T \Big|&\,f_i(r, X^{0,x_0}_r, (u^{k,n}(r,X_r^{0,x_0}))_{k \in \Gamma}, (v^{k,n}(r,X_r^{0,x_0}))_{k \in \Gamma}) \\
& + n\sum_{\ell\neq i} \Big(u^{i,n}(r,X_r^{0,x_0}) - u^{\ell,n}(r,X_r^{0,x_0}) + g_{i\ell}(r,X^{0,x_0}_r)\Big)^{-}\Big|^2 dr\bigg]\leq C. \nonumber
\end{align}
For simplicity we again set $u^{n}(\,\cdot\,):=(u^{k,n}(\,\cdot\,))_{k \in \Gamma}$ and $v^n(\,\cdot\,):=(v^{k,n}(\,\cdot\,))_{k \in \Gamma}$ inside the functions $f_i$, when no confusion may arise.
We can express the expectation in \eqref{L2bound-1} as an integral with respect to the law of $X^{0,x_0}_r$, $r\leq T$. This gives
\begin{align*}
\int_0^T\int_{\mathbb{R}^d}\Big|&\,f_i(r, y,u^{n}(r,y), v^{n}(r,y)) \\
&+ n\sum_{\ell\neq i} \Big(u^{i,n}(r,y) - u^{\ell,n}(r,y) + g_{i\ell}(r,y)\Big)^{-}\Big|^2\mu(0,x_0;r,dy)dr\leq C.
\end{align*}
If we now set
\[
F^{i}_n(r,y):=f_i(r, y,u^{n}(r,y), v^{n}(r,y)) +n\sum_{\ell\neq i} \Big(u^{i,n}(r,y) - u^{\ell,n}(r,y) + g_{i\ell}(r,y)\Big)^{-}
\]
we have that the map $F_n:=(F^{i}_n)_{i\in\Gamma}$, $F_n:[0,T]\times\mathbb{R}^d\to\mathbb{R}^m$ has all its components bounded in $L^2([0,T]\times \mathbb{R}^d, \mu(0,x_0;r,dy)dr)$ uniformly with respect to $n$. Therefore, the sequence $(F_n)_{n\ge 0}$ admits a subsequence $(F_{n_j})_{j\geq 0}$ such that $F^{i}_{n_j}\to F_i$
weakly in $L^2([0,T]\times \mathbb{R}^d, \mu(0,x_0;r,dy)dr)$ as $j\to\infty$, for each $i \in \Gamma$. Notice that the subsequence may depend on $x_0$.
\medskip
\emph{Step 2.} Here we want to prove that \eqref{eq:lim-u} holds along the subsequence $(n_j)_{j\ge0}$ found above. In particular, given $(t,x) \in [0,T] \times \mathbb{R}^d$ we will prove that the sequence $(u^{i,n_j}(t,x))_{j\geq 0}$ is of Cauchy type.
Let $\delta>0$ and $N>0$ be two constants (which will be taken small and large, respectively), and notice that by \eqref{representation} we have, for any non-negative $j,k$
\begin{align}
u^{i,n_j}(t,x)-u^{i,n_k}(t,x)=&\mathsf{E}\bigg[\int_t^{t+\delta} (F^{i}_{n_j}(r,X^{t,x}_r)-F^{i}_{n_k}(r,X^{t,x}_r))dr\bigg]\\
&+\mathsf{E}\bigg[\int_{t+\delta}^T(F^{i}_{n_j}(r,X^{t,x}_r)-F^{i}_{n_k}(r,X^{t,x}_r))1_{\{|X^{t,x}_r|\leq N\}}dr\bigg]\nonumber\\
&+\mathsf{E}\bigg[\int_{t+\delta}^T(F^{i}_{n_j}(r,X^{t,x}_r)-F^{i}_{n_k}(r,X^{t,x}_r))1_{\{|X^{t,x}_r|> N\}} dr\bigg]\nonumber\\
&=:\Theta_1^{jk}+\Theta_2^{jk}+\Theta_3^{jk}.\nonumber
\end{align}
In what follows we let $C=C(t,x)>0$ be a suitable constant (i.e.~sufficiently large for our purposes) independent of $\delta$ and $N$. Due to \eqref{f-bound} and \eqref{eq:penalisation} we easily get $|\Theta_1^{jk}|\leq C\cdot \delta$. Moreover, the bounds in \eqref{f-bound} and \eqref{eq:penalisation}, together with Cauchy-Schwarz and Markov inequalities yield $|\Theta_3^{jk}|\leq C/N$.
Now we use the law of $X^{t,x}$ to rewrite $\Theta_2^{jk}$ as
\begin{align*}
\Theta_2^{jk}=&\mathsf{E}\bigg[\int_{t+\delta}^T(F^{i}_{n_j}(r,X^{t,x}_r)-F^{i}_{n_k}(r,X^{t,x}_r))1_{\{|X^{t,x}_r|\leq N\}} dr\bigg]\\
=&\int_{t+\delta}^T\int_{\mathbb{R}^d}(F^{i}_{n_j}(r,y)-F^{i}_{n_k}(r,y))1_{\{|y|\leq N\}}\mu(t,x;r,dy) dr.\nonumber
\end{align*}
The $L^2$-domination condition \textbf{(A0)} implies
\begin{align}
\Theta_2^{jk}=\int_{t+\delta}^T\int_{\mathbb{R}^d}(F^{i}_{n_j}(r,y)-F^{i}_{n_k}(r,y))1_{\{|y|\leq N\}}\phi_{t,x,x_0}^{\delta}(r,y)\mu(0,x_0;r,dy)dr.
\end{align}
By assumption $\phi_{t,x,x_0}^{\delta}\in L^{2}([t+\delta,T]\times [-N,N]^d;\ \mu(0,x_0;r,dy)dr)$, hence weak convergence of the sequence $(F^{i}_{n_j})_{j\geq 0}$ implies $\limsup_{j,k\rightarrow \infty}|\Theta_2^{jk}|=0$.
Collecting the estimates for $\Theta_1^{jk}$, $\Theta_2^{jk}$ and $\Theta_3^{jk}$ we obtain
\[
\limsup_{j,k\rightarrow \infty}|u^{i,n_j}(t,x)-u^{i,n_k}(t,x)|\leq
C(\delta+ N^{-1})
\]
and, letting $\delta\to 0$ and $N\to\infty$, we complete the proof of \eqref{eq:lim-u}. Finally, estimates \eqref{eq:estim-u0} and \eqref{eq:estim-u} follow by using the representation formula \eqref{represent1} in \eqref{eq:stimaL2-3bis}, with $s=t$, and thanks to \eqref{eq:lim-u}.
\end{proof}
As a byproduct of the previous result we have the following.
\begin{corollary}
\label{cor:Cauchy} For any $i\in \Gamma$ one has
\begin{equation}
\label{eq:Cauchy} \lim_{j,k \rightarrow
\infty}\mathsf{E}\bigg[\int_t^T |Y^{i,n_j}_s - Y^{i,n_k}_s|^2 ds +
\int_t^T |Z^{i,n_j}_s - Z^{i,n_k}_s|^2 ds\bigg]=0.
\end{equation}
\end{corollary}
\begin{proof} Convergence of the first term in \eqref{eq:Cauchy} follows from the convergence result \eqref{eq:lim-u} and by using the dominated convergence theorem, which is enabled by \eqref{eq:estim-u0} and \eqref{estimx}. Convergence of the second term in
\eqref{eq:Cauchy} is obtained in a classical way. By It\^o's formula
and using the same estimates as in the proof of Proposition
\ref{prop:stimaL2} (and Lipschitz continuity of $f_i$) we get
\begin{align*}
\mathsf{E}&\left[\int_t^T|Z^{i,n_j}_s-Z^{i,n_k}_s|^2\right]\\
\le & 2c\,\varepsilon\,\mathsf{E}\left[\int_t^T\!\!\big(Y^{i,n_j}_s-Y^{i,n_k}_s\big)^2ds\right]+\frac{2c}{\varepsilon}\sum_{\alpha\in\Gamma}\mathsf{E}\left[\int_t^T\!\!\big(|Y^{\alpha,n_j}_s-Y^{\alpha,n_k}_s|^2+|Z^{\alpha,n_j}_s-Z^{\alpha,n_k}_s|^2\big)ds\right]\\
&+2c\,\mathsf{E}\left[\int_t^T|Y^{i,n_j}_s-Y^{i,n_k}_s|(1+|X^{t,x}_s|^q)ds\right],
\end{align*}
for a suitable constant $c>0$ and arbitrary $\varepsilon>0$. Taking the summation over $i\in\Gamma$ and picking $\varepsilon$ sufficiently large we may conclude that
\begin{align*}
\sum_{\alpha\in\Gamma}\mathsf{E}&\left[\int_t^T|Z^{\alpha,n_j}_s-Z^{\alpha,n_k}_s|^2\right]\\
\le& c_\varepsilon\sum_{\alpha\in\Gamma}\mathsf{E}\left[\int_t^T\!\!
|Y^{\alpha,n_j}_s-Y^{\alpha,n_k}_s|^2ds+\int_t^T\!\!|Y^{\alpha,n_j}_s-Y^{\alpha,n_k}_s|(1+|X^{t,x}_s|^q)ds\right]
\end{align*}
where $c_\varepsilon>0$ depends on $\varepsilon$ but is independent of $j,k$. Hence taking limits as $j,k\to\infty$ and using the above result we finally obtain \eqref{eq:Cauchy}.
\end{proof}
We can now prove the main result of this paper, which establishes the existence of a solution to system \eqref{main-system}. In what follows the subsequence $(n_j)_{j\ge 0}$ is the same as the one in Proposition \ref{prop:Cauchy-0}.
\begin{theorem}
\label{thm:main}
There exists a solution $(Y^i,Z^i,K^i)_{i \in \Gamma}$ to {\color{red}{\eqref{main-system}}}. Moreover, for any $i\in \Gamma$ and $t\in[0,T]$ it holds
\begin{equation}
\label{eq:convergence}
\lim_{j \rightarrow \infty}\mathsf{E}\bigg[\sup_{t\leq s \leq T}|Y^{i,n_j}_s - Y^{i}_s|^2 + \int_t^T |Z^{i,n_j}_s - Z^{i}_s|^2 ds + \sup_{t\leq s \leq T}|K^{i,n_j}_s - K^{i}_s|^2\bigg]=0.
\end{equation}
\end{theorem}
\begin{proof}
The proof is given in two steps. We first prove, in step 1, that
there exists $(Y^i,Z^i,K^i)_{i \in \Gamma}$ satisfying the first
equation in \eqref{main-system} and such that \eqref{eq:convergence}
holds. Then we prove, in step 2, that $(Y^i,Z^i,K^i)_{i \in \Gamma}$
fulfils the second and third conditions in \eqref{main-system} as
well.
\smallskip
\noindent \emph{Step 1.} Let $(t,x) \in [0,T] \times \mathbb{R}^d$ be fixed.
For any $i\in \Gamma$ let us set:
\begin{itemize}
\item[ i)] $Y^{i}_s=u^i(s,X^{t,x}_s), \,\,s\in [t,T]$, with $u^i$ as in \eqref{eq:lim-u};
\item[ii)] $(Z^i_s)_{s\in [t,T]}$ the limit in $\textbf{H}^2_T(\mathbb{R}^d)$ of
$(Z^{i,n_j}_s)_{s\in [t,T]}$ which exists thanks to
\eqref{eq:Cauchy}.
\end{itemize}
\noindent It is clear that
\[
Y^i_s=u^i(s,X^{t,x}_s)=\lim_{j\to\infty}u^{i,n_j}(s,X^{t,x}_s)=\lim_{j\to\infty}Y^{i,n_j}_s\quad\text{$\mathsf{P}$-a.s.~$\forall s\in[t,T]$}
\]
Let us now show that for any $i\in \Gamma$, the sequence
$(Y^{i,n_j})_{j\ge 0}$ is Cauchy in $\textbf{S}^2_T(\mathbb{R})$ so that it converges to $Y^i$ in $\textbf{S}^2_T(\mathbb{R})$.
By using It\^o's formula, Lipschitz property of $f_i$ and the bound
in Proposition \ref{prop:penalisation}, we can argue in a similar
way to the proof of Proposition \ref{prop:stimaL2} and obtain for
all $u\in[t,T]$
\begin{align}\label{eq:exist}
\sup_{t\le u\le T}&|Y^{i,n_j}_u-Y^{i,n_k}_u|^2+\int_t^T|Z^{i,n_j}_s-Z^{i,n_k}_s|^2\nonumber\\
\le & 2c\,\varepsilon\int_t^T\!\!\big(Y^{i,n_j}_s-Y^{i,n_k}_s\big)^2ds\nonumber\\
&+\frac{2c}{\varepsilon}\sum_{\alpha\in\Gamma}\int_t^T\!\!\big(|Y^{\alpha,n_j}_s-Y^{\alpha,n_k}_s|^2+|Z^{\alpha,n_j}_s-Z^{\alpha,n_k}_s|^2\big)ds\nonumber\\
&+2c\int_t^T|Y^{i,n_j}_s-Y^{i,n_k}_s|(1+|X^{t,x}_s|^q)ds\\
&+\sup_{t\le u\le T}\bigg|\int_u^T(Y^{i,n_j}_s-Y^{i,n_k}_s)(Z^{i,n_j}_s-Z^{i,n_k}_s)dB_s\bigg|,\nonumber
\end{align}
where $c>0$ is a suitable constant independent of $j,k$ and $\varepsilon>0$ is also arbitrary. Notice that by Burkholder-Davis-Gundy's inequality and $|ab|\le \varepsilon|a|^2+\varepsilon^{-1}|b|^2$ we have
\begin{align}\label{eq:exist1}
\mathsf{E}&\left[\sup_{t\le u\le T}\bigg|\int_u^T(Y^{i,n_j}_s-Y^{i,n_k}_s)(Z^{i,n_j}_s-Z^{i,n_k}_s)dB_s\bigg|\right]\nonumber\\
\le& C\mathsf{E}\left[\bigg(\int_t^T(Y^{i,n_j}_s-Y^{i,n_k}_s)^2(Z^{i,n_j}_s-Z^{i,n_k}_s)^2ds\bigg)^{\frac{1}{2}}\right]\\
\le& C\mathsf{E}\left[\sup_{t\le u\le T}|Y^{i,n_j}_u-Y^{i,n_k}_u|\bigg(\int_t^T(Z^{i,n_j}_s-Z^{i,n_k}_s)^2ds\bigg)^{\frac{1}{2}}\right]\nonumber\\
\le& C\varepsilon\mathsf{E}\left[\sup_{t\le u\le T}|Y^{i,n_j}_u-Y^{i,n_k}_u|^2\right]+\frac{C}{\varepsilon}\mathsf{E}\left[\int_t^T(Z^{i,n_j}_s-Z^{i,n_k}_s)^2ds\right],\nonumber
\end{align}
for a suitable $C>0$ independent of $j,k$ and any $\varepsilon>0$. Taking expectations in \eqref{eq:exist} and using \eqref{eq:exist1} (with $\varepsilon<1/C$), after rearranging terms we then obtain
\begin{align}\label{eq:exist2}
\mathsf{E}\bigg[\sup_{t\le u\le T}|Y^{i,n_j}_u-Y^{i,n_k}_u|^2\bigg]\le &\, c_{\varepsilon}\mathsf{E}\bigg[\sum_{\alpha\in\Gamma}\int_t^T\!\!\big(|Y^{\alpha,n_j}_s-Y^{\alpha,n_k}_s|^2+|Z^{\alpha,n_j}_s-Z^{\alpha,n_k}_s|^2\big)ds\bigg]\nonumber\\
&+2c\,\mathsf{E}\bigg[\int_t^T|Y^{i,n_j}_s-Y^{i,n_k}_s|(1+|X^{t,x}_s|^q)ds\bigg],
\end{align}
where $c_\varepsilon>0$ may depend on $\varepsilon>0$ but is independent of $j,k$.
Letting now $j,k\to\infty$ and using Corollary \ref{cor:Cauchy} we obtain that $Y^{i,n_j}$
forms a Cauchy sequence in $\textbf{S}^2_T(\mathbb{R})$ as claimed.
\vspace{+3pt}
Let us now define $K^i$, $i\in \Gamma$, as:
\begin{align}\label{eq:limK}
K^i_s:=Y^{i}_t-Y^{i}_s-\int_t^s\!\!f_i(u,X^{t,x}_u,(Y^{k}_u)_{k\in\Gamma},(Z^{k}_u)_{k\in\Gamma})du
+\int_t^s\!\!Z^{i}_udB_u,\,\,s\in [t,T].
\end{align}
Since $Y^{i,n_j}$ converges in $\textbf{S}^2_T(\mathbb{R})$, and upon recalling Lipschitz property of $f_i$ and \eqref{eq:Cauchy}, it is easy to verify that $K^i$ is the limit in $\textbf{S}^2_T(\mathbb{R})$ of the sequence $(K^{i,n_j})_{j\ge0}$ defined by (see \eqref{eq:defK} and
\eqref{eq:penalisedBSDE})
\[
K^{i,n_j}_s=Y^{i,n_j}_t-Y^{i,n_j}_s-\int_t^s\!\!f_i(u,X^{t,x}_u,(Y^{k,n_j}_u)_{k\in\Gamma},(Z^{k,n_j}_u)_{k\in\Gamma})du
+\int_t^s\!\!Z^{i,n_j}_udB_u,\,\,s\in [t,T].
\]
Hence \eqref{eq:convergence} holds and, by
\eqref{eq:limK}, $(Y^i,Z^i,K^i)_{i\in \Gamma}$ verify the first
equation of \eqref{main-system}.
\smallskip
\noindent \emph{Step 2.} It only remains to show that the second and
third conditions in \eqref{main-system} are satisfied by
$(Y^i,Z^i,K^i)_{i \in \Gamma}$. Proposition \ref{prop:penalisation}
implies that there exists $C>0$ for which
\begin{equation}
\label{faltoff1} \mathsf{E}\bigg[\int_t^T \sum_{j\neq i}
\big(Y^{i,n}_s - Y^{j,n}_s + g_{ij}(s,X^{t,x}_s)\big)^{-} ds \bigg]
\leq \frac{C}{n}.
\end{equation}
Using \eqref{eq:Cauchy} and letting $n \uparrow \infty$ (along the subsequence used in \eqref{eq:Cauchy}) we immediately obtain
\begin{equation}
\label{flatoff2} \mathsf{E}\bigg[\int_t^T \sum_{j\neq i}\big(Y^{i}_s
- Y^{j}_s + g_{ij}(s,X^{t,x}_s)\big)^{-} ds\bigg] =0.
\end{equation}
Hence, for all $i,j\in \Gamma$, $Y^{i}_s \geq Y^{j}_s +
g_{ij}(s,X^{t,x}_s)$, $\mathsf{P}$-a.s.\ for every~$s \in [t,T]$ (recall
that $s\mapsto Y^k_s$, $k\in\Gamma$ is indeed continuous as uniform
limit of continuous processes). In particular
\begin{equation}
\label{flatoff3} Y^{i}_s \geq \max_{j\neq i}\Big(Y^{j}_s -
g_{ij}(s,X^{t,x}_s)\Big), \quad \mathsf{P}-\text{a.s.} \quad \forall s \in
[t,T].
\end{equation}
Thanks to \eqref{eq:convergence}, by Tchebyshev's inequality we have that for any $i\in \Gamma$,
\begin{align}\label{eq:limP}
\lim_{j\to\infty}\mathsf{P}\left(\sup_{t\le s\le
T}\big(|Y^{i,n_j}_s-Y^{i}_s|+|K^{i,n_j}_s-K^{i}_s|\big)\ge
\varepsilon\right)=0
\end{align}
for any $\varepsilon>0$. Moreover, for a.e.~$\omega\in\Omega$ and
for each $j\ge 0$ the map $s\mapsto K^{i,n_j}_s(\omega)$ is
increasing and continuous, hence it is a (random) continuous measure
on $[t,T]$. The same holds for the limit process $K^i$. The uniform
convergence in \eqref{eq:limP} implies that (up to selecting a
subsequence) $K^{i,n_j}(\omega)\to K^{i}(\omega)$ as $j\to\infty$
\emph{in general} in the sense of measures (see \cite[Ch.~3]{S}).
Therefore, for $\mathsf{P}$-a.e.~$\omega\in\Omega$, it holds $d
K^{i,n_j}(\omega)\to d K^{i}(\omega)$ weakly as $j\to\infty$ (see
\cite[Thm.~1, Ch.~3]{S}) and
\begin{align}
\label{eq:flat1}
\lim_{j\to\infty}&\int_t^T \!\!\big[Y^{i,n_j}_s - \max_{k\neq i}\big(Y^{k,n_j}_s - g_{ik}(s,X^{t,x}_s)\big)\big]dK^{i,n_j}_s\nonumber\\
&\quad\quad\quad=\int_t^T \!\!\big[Y^{i}_s - \max_{k\neq
i}\big(Y^{k}_s - g_{ik}(s,X^{t,x}_s)\big)\big] dK^{i}_s,
\quad\mathsf{P}-a.s.
\end{align}
We now notice that the left-hand side of \eqref{eq:flat1} is
non-positive due to \eqref{eq:defK} and the fact that for any $\ell\neq i$ and all
$s\in[t,T]$
\[
\Big(Y^{i,n_j}_s - \max_{k\neq i}\big(Y^{k,n_j}_s - g_{ik}(s,X^{t,x}_s)\big)\Big)\Big(Y^{i,n_j}_s - Y^{\ell,n_j}_s + g_{i\ell}(s,X^{t,x}_s)\Big)^-\le0,\qquad \mathsf{P}-a.s.
\]
However, the right-hand side of \eqref{eq:flat1} is non-negative due to \eqref{flatoff3} and the fact that $K^i$ is increasing. Hence we get
\[
\int_t^T \!\!\big[Y^{i}_s - \max_{k\neq i}\big(Y^{k}_s -
g_{ik}(s,X^{t,x}_s)\big)\big] dK^{i}_s=0, \quad\mathsf{P}-a.s.,
\]
which completes the proof.
\end{proof}
We now provide a corollary of Proposition \ref{prop:Cauchy-0} and Theorem \ref{thm:main}.
\begin{corollary}\label{cor:uv}
There exist measurable deterministic functions $(u^i)_{i\in\Gamma}$ and $(v^i)_{i\in\Gamma}$ with $u^i:[0,T]\times\mathbb{R}^d\to\mathbb{R}$ and $v^i:[0,T]\times\mathbb{R}^d\to\mathbb{R}^d$ such that for any $(t,x)\in[0,T]\times\mathbb{R}^d$
\begin{align}
Y^{i;t,x}_s=u^i(s,X^{t,x}_s)\quad\text{and}\quad Z^{i;t,x}_s=v^i(s,X^{t,x}_s),\quad\mathsf{P}-a.s.~\text{for a.e.}~s\in[t,T].
\end{align}
\end{corollary}
\begin{proof}
It only remains to show the existence of $v^i$. Recall $v^{i,n}$ from \eqref{represent1} and set $v^i:=\limsup_{j\to\infty}v^{i,n_j}$, where the limit is taken along the subsequence introduced in Proposition \ref{prop:Cauchy-0}. Then, using that $Z^{i,n_j}_s\to Z^{i}_s$, $\mathsf{P}$-a.s.~for a.e.~$s\in[t,T]$, and choosing $(s,\omega)$ such that the convergence indeed holds we find
\[
v^i(s,X_s(\omega))=\limsup_{j\to\infty}v^{i,n_j}(s,X_s(\omega))=\limsup_{j\to\infty}Z^{i,n_{j}}_s(\omega)=Z^i_s(\omega).
\]
\end{proof}
\medskip
\textbf{Acknowledgement}: The authors acknowledge hospitality by
Center for Mathematical Economics (IMW) at Bielefeld University,
D\'epartement de Math\'ematiques at Le Mans Universit\'e and School
of Mathematics at University of Leeds. The first and third named
authors were partially supported by the Heilbronn Institute and the
School of Mathematics at Leeds during the research week
``\emph{Optimal stopping games for ambiguity-averse players}'' (4-9
September 2017). Financial support by the German Research Foundation
(DFG) through the Collaborative Research Centre `Taming uncertainty
and profiting from randomness and low regularity in analysis,
stochastics and their applications' is gratefully acknowledged by
the second named author.
|
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"redpajama_set_name": "RedPajamaArXiv"
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|
{"url":"https:\/\/kb.osu.edu\/handle\/1811\/18883?show=full","text":"dc.creator Bagnato, V. S. en_US dc.creator Marcassa, L. G. en_US dc.creator Telles, G. D. en_US dc.creator Muniz, S. R. en_US dc.creator Pereira, A. A. en_US dc.date.accessioned 2006-06-15T19:02:50Z dc.date.available 2006-06-15T19:02:50Z dc.date.issued 1998 en_US dc.identifier 1998-RC-07 en_US dc.identifier.uri http:\/\/hdl.handle.net\/1811\/18883 dc.description Author Institution: Instituto de Fisica de S\\^{a}o Carlos, Universidade de S\\~{a}o Paulo - Caixa Postal 369 en_US dc.description.abstract Using magneto optical trapping that simultaneously trap two different atomic species, we have performed experiments where trap loss due to the crossed specie effects are investigated. Our MOT operates in a room temperature vapor cell which can produce traps with spatial overlapping of two of any of the alkalis Na, K, Rb, Cs. To investigate the cross species cold collisions we use the traditional transient behaviour technique where either the loading or unloading of the MOT is observed for a simple or double specic trap. Measurements of the trap loss rate as a function of intensity show the existence of alternative mechanisms for the exocrgic collisions. A catalysis experiment was also performed two complement the understanding of the observed trap loss rates. Defining $\\beta$ as the trap loss rate for Na alone and $\\beta^{\\prime}$ the loss rate for Na due to collisions with Rb, the behavior of $\\beta$ and $\\beta^{\\prime}$ as a function of Na trap laser intensity is shown in fig.1. As the laser intensity decreases (so does the trap depth), $\\beta$ decreases while $\\beta^{\\prime}$ increases. This behavior is consistent with the fact that the main contribution to $\\beta^{\\prime}$ are collisions of the type $Na - Rb^{+}$. This behavior is confirmed by the catalysis data where the dependence of $\\beta^{\\prime}$ with intensity is very gentle. Other combinations of atoms indicate peculiar behavior inherent to each individual systems. Most the observation involving cross species exoergic collisions can not be explained using simple models. Studies involving mixed species cold collisions may find important applications towards the production of mixed Bose-Einstein Condensation. [FIGURE] en_US dc.format.extent 177008 bytes dc.format.mimetype image\/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title INVESTIGATION OF EXOERGIC COLLISIONS IN MIXED SPECIES TRAP: Na\/Rb, Na\/K, Rb\/K en_US dc.type article en_US\n\ufeff","date":"2021-08-04 07:50:03","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8449270725250244, \"perplexity\": 5255.78591863149}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-31\/segments\/1627046154796.71\/warc\/CC-MAIN-20210804045226-20210804075226-00158.warc.gz\"}"}
| null | null |
/* -*- C++ -*- */
// $Id: Supplier_Router.h 80826 2008-03-04 14:51:23Z wotte $
// The interface between a supplier and an Event Service ACE_Stream.
#ifndef _SUPPLIER_ROUTER_H
#define _SUPPLIER_ROUTER_H
#include "ace/UPIPE_Addr.h"
#if !defined (ACE_LACKS_PRAGMA_ONCE)
# pragma once
#endif /* ACE_LACKS_PRAGMA_ONCE */
#include "ace/UPIPE_Acceptor.h"
#include "ace/Map_Manager.h"
#include "ace/Svc_Handler.h"
#include "Peer_Router.h"
#if defined (ACE_HAS_THREADS)
// Forward declaration.
class Supplier_Handler;
// Type of search key for SUPPLIER_MAP.
typedef ACE_HANDLE SUPPLIER_KEY;
// Instantiated type for routing messages to suppliers.
typedef Peer_Router<Supplier_Handler, SUPPLIER_KEY> SUPPLIER_ROUTER;
class Supplier_Handler
: public Peer_Handler<SUPPLIER_ROUTER, SUPPLIER_KEY>
{
public:
Supplier_Handler (ACE_Thread_Manager *tm = 0);
virtual int open (void *);
};
class Supplier_Router : public SUPPLIER_ROUTER
{
public:
Supplier_Router (ACE_Thread_Manager *);
protected:
// ACE_Task hooks..
virtual int open (void *a = 0);
virtual int close (u_long flags = 0);
virtual int put (ACE_Message_Block *msg, ACE_Time_Value * = 0);
virtual int svc (void);
// Dynamic linking hooks inherited from Peer_Router.
virtual int info (ACE_TCHAR **info_string, size_t length) const;
};
#endif /* ACE_HAS_THREADS */
#endif /* _SUPPLIER_ROUTER_H */
|
{
"redpajama_set_name": "RedPajamaGithub"
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| 7,626
|
\section{Introduction}\label{sec:intro}
A recent development in Radiotherapy treatment planning is to use an optimization process to obtain a treatment plan. To do so one needs to formalize the clinical objectives in a so called object function whose optimum should correspond to an optimal treatment plan. Many current treatment planning systems rely on a quasi-Newton algorithm to perform the optimization. In order to guarantee that this process indeed converges to an optimum the object function should at least be twice continuously differentiable. This last property is the main subject of this study. As it turns out the object functions used in Radiotherapy do not always have this property so that convergence of the optimization process can not directly be guaranteed in those cases.
In section \ref{sec:optbg} we present some background for the use of optimization in Radiotherapy treatment planning. The object functions currently used are built from so called dose-volume and EUD constraints. By dose-volume constraints, the dose-volume histogram enters the object function. It is the finite differentiability of the former that causes the insufficient differentiability of the latter. In section \ref{sec:objdiff} this relationship is explained in more detail. A example illustrates how finite differentiability of the object function influences the optimization process. The practical consequences thereof are the subject of section \ref{sec:conc}. Where section \ref{sec:objdiff} is kept informal, section \ref{sec:proof} provides formal statements and proofs of the differentiability results. This last section does not contain new arguments but serves as a support for the exposition.
\section{Background for optimization in Radiotherapy}\label{sec:optbg}
\subsection{Optimization in Radiotherapy}\label{sec:optrt}
In Radiotherapy treatment planning a careful choice of values for treatment parameters is necessary to obtain a dose distribution in the patient that meets criteria for eradicating a malignant tumour. At the same time it should meet criteria so that healthy tissues and organs are not critically damaged. Due to the fact that dose is deposited in every irradiated part of the patient, these criteria are conflicting in almost every clinical situation. Experienced planners are usually very good at intuitively balancing such conflicting goals, still there is a need for more objective and quantifiable methods to devise and evaluate treatment plans. This has led to the construction of object functions, depending on treatment parameters, which formalize the objectives of dose on tumours and healthy tissues, see \cite{bra,tbn,wm}. By optimizing this object function with respect to treatment parameters, one hopes to get an optimal treatment plan. Another, very different, reason to use optimization in treatment planning is the emergence of IMRT where the sheer number of treatment parameters simply precludes manual planning\commentaar{refs}.
Even if we know that a function has a unique optimum, there is no simple recipe to find it. Therefore we have to resort to numerical, iterative methods to solve optimization problems, which, given a starting point, generate a sequence converging to an optimum. The latter can be guaranteed only if certain conditions on the object function are met. It seems, however, that if these conditions are considered at all, it is taken for granted that they are fulfilled.
Many treatment planning systems use a quasi-Newton method, like the LBFGS algorithm, to find an optimum of the object function, see for example \cite{blnz} or \cite{sb} as a general reference. One of the conditions guaranteeing that this algorithm yields a sequence locally converging to an optimum is that the object function is twice continuously differentiable. The object functions currently used in treatment planning systems are limited to so called \emph{EUD constraints} and \emph{dose-volume constraints}, or combinations of both, see for example \cite{tbn}, but also see section \ref{sec:objrt} remark \ref{rem:compro}. Our results show that the former do satisfy the differentiability condition. The latter however, do not, that is not for all values of the treatment parameters.
\subsection{Object functions in Radiotherapy}\label{sec:objrt}
The goal of Radiotherapy treatment planning is to devise a treatment plan achieving a high \emph{tumour control probability} (TCP) combined with a low \emph{normal tissue complication probability} (NTCP). A priori both TCP and NTCP are unrelated to treatment parameters. Even relating them to criteria on the dose distribution in the patient is a non-trivial task. To simplify matters one does not consider the 3-dimensional dose distribution, but a derived quantity namely the dose-volume histogram for each relevant region. The latter are tumours and healthy organs for which the dose-volume histogram tells which part of the volume receives at least a certain dose, see equation (\ref{eq:volfun}) below. Data from previously treated patients are used to relate TCP and NTCP to dose-volume histograms, see \cite{ehkms}. At this point further simplifications are necessary, because TCP and NTCP can only be related to certain points of dose-volume histograms or even only to averages. Generally for tumours this leads to dose-volume constraints, that is a minimum and a maximum dose for volumes $v_{\min}$ and $v_{\max}$ respectively, or in a more concise notation $(d_{\min},v_{\min})$ and $(d_{\max},v_{\max})$. In case of healthy tissues the situation is much more complicated, in practice one uses a mixture of dose-volume constraints and EUD constraints, see \cite{tbn}.
Criteria for the dose distribution or the derived dose-volume histograms are related to treatment parameters, thus the treatment planning objectives can be expressed with an object function of these treatment parameters. Now the object function used in treatment planning is a weighted sum of 'local' object functions, one for each dose-volume constraint or EUD constraint. Let us consider a tumour with a minimum and a maximum dose-volume constraint. Furthermore $V_{\sigma}$ is the dose-volume histogram of the tumour for a given dose distribution and treatment parameters $\sigma$, that is $V_{\sigma}(h)$ is the (relative) volume of the tumour, region $R$, receiving at least dose $h$
\begin{equation}\label{eq:volfun}
V_{\sigma}(h) = \textrm{vol} \{x \in R\;|\; f_{\sigma}(x) \geq h\}.
\end{equation}
In this expression $\textrm{vol}(A)$ is the volume of the region $A$ and $f_{\sigma}$ is the dose distribution. One constructs what we will call \emph{local object functions}. For example the local object function for a minimum does-volume constraint for a specific volume is a function $F_{\min}(\sigma)$ which increases the more the actual relative volume at $d_{\min}$ is below $v_{\min}$ and decreases or even becomes zero if the actual relative volume at $d_{\min}$ is above $v_{\min}$. Similarly a maximum dose object function $F_{\max}(\sigma)$ is constructed such that it increases when the relative volume at $d_{\max}$ is above $v_{\max}$ and decreases or becomes zero otherwise. Thus the object function for the tumour alone would be the weighted sum of two local object functions, usually of the following form
\begin{equation}\label{eq:objfun}
F(\sigma) = w_1\, F_{\min}(\sigma) + w_2\, F_{\max}(\sigma) = w_1\, G(v_{\min}-V_{\sigma}(d_{\min})) + w_2\, G(V_{\sigma}(d_{\max})-v_{\max}),
\end{equation}
where $w_1>0$ and $w_2>0$ are weights and
\begin{equation}\label{eq:locobjfun}
G(x) = \left\{\begin{array}{ll}
0, & x < 0\\
x^2, & x \geq 0,
\end{array}\right.
\end{equation}
Although the object function $F$ is a function of the treatment parameters $\sigma$ it does not depend on them directly, but only via the dose-volume histograms. The latter depend on the dose distribution which directly depends on the treatment parameters.
The EUD constraints are in a similar way expressed in the \emph{equivalent uniform dose}. A quantity used to measure the biological effect of radiation especially for healthy tissues, see \cite{nie}\commentaar{nog andere refs?}. We denote the equivalent uniform dose by $E_{\alpha}$, here it is a function of treatment parameters $\sigma$
\begin{equation}\label{eq:eud}
E_{\alpha}(\sigma) = \bigg[\frac{1}{\textrm{vol}(R)} \int_R f_{\sigma}(x)^{\alpha} \, dx\bigg]^{1/\alpha}.
\end{equation}
\textbf{Remarks}
\begin{enumerate}\topsep 0pt\itemsep 0pt
\item \label{rem:compro} The word \emph{constraint} in this context is somewhat misleading because a dose-volume nor an EUD constraint is used as a constraint in the sense of constrained optimization. If they were, many if not most optimization problems in treatment planning would have no solution. It is a common clinical situation that a tumour is located adjacent to a critical organ. For several reasons the region, called PTV, on which the treatment dose is prescribed is larger than the tumour. See \cite{icru} for a systematic description of XTV's, where X = G, C, I or P. Therefore the critical organ is possibly partly located in the PTV. Suppose for example that one third of the volume of the critical organ is located inside the PTV. Furthermore suppose that the minimal dose-volume constraint on the PTV is 60 Gy (Gy is unit of dose) and the maximum average dose constraint on the organ is 20 Gy. Since the dose distribution varies continuously in space and its gradient is bounded, this constrained optimization problem will have no solution. Due to the anatomy a minimum dose of 60 Gy on the PTV implies an average dose well above 20 Gy on the organ and vice versa. A weighted sum of local object functions is a way to find a clinically acceptable compromise for this situation. To avoid the term \emph{constraint} some authors use the term \emph{soft constraint}, but in my opinion this term is superfluous because we already have the notion of object function. Nevertheless we adhere to the word constraint, in order to stay in line with the existing literature.\commentaar{zij het tandenknarsend}
\item \label{rem:eud} The equivalent uniform dose, as defined in equation (\ref{eq:eud}), can be regarded as a generalized average, in fact for $\alpha=1$ it is the standard average of the dose distribution over the region $R$. Formally it is identical to the $L_p$-norm of the dose distribution $f_{\sigma}$ restricted to a region $R$ of the patient. But $p$ is called $\alpha$ and it has a biological interpretation. One associates a value of $\alpha$ to each tissue type, but there is no agreement yet in the literature about these values\commentaar{refs}.\commentaar{ Since the dose distribution is positive there is no need to restrict to $\alpha > 0$. Then there are two limiting cases, $\lim_{\alpha \to \infty} E_{\alpha}(\sigma)=\max_{x \in R}f_{\sigma}(x)$ and $\lim_{\alpha \to -\infty} E_{\alpha}(\sigma)=\min_{x \in R}f_{\sigma}(x)$. Both numbers exist because $f_{\sigma}$ is a smooth function and $R$ is compact.}\commentaar{moet dit er in? nee}
\item \label{rem:dvh} The dose-volume histogram refers to the graph of a function that is usually given the same name. This function, defined for each $\sigma$ in equation (\ref{eq:volfun}), can be regarded as a volume function in the sense of \cite{hvn}. Let us fix $\sigma$ for the moment. The dose-volume histogram of a region $R$ is a function $h \mapsto V_{\sigma}(h)$ whose value is the volume of $R$ receiving at least dose $h$. In other words it gives the volume of a region enclosed by two level sets of the dose distribution $f_{\sigma}$, namely those $x \in R$ where $f_{\sigma}(x) = h$ and the $x\in R$ where $f_{\sigma}(x) = \max_R(f_{\sigma})$ (this last level set will be a point).\commentaar{The results of \cite{hvn} imply that this function is smooth as a function of $h$ when $h$ is a regular value of $f_{\sigma}$ but only finitely differentiable when $h$ is a critical value of $f_{\sigma}$.}\commentaar{crypto!}
\item \label{rem:treatpar} Any quantity that influences the dose distribution may be considered as a \emph{treatment parameter}. Usually treatment parameters are \emph{beam intensities} and \emph{beam angles}, that is parameters of the treatment unit.
\item \label{rem:planpar} The object function $F(\sigma)$ in equation (\ref{eq:objfun}) also depends on $d_{\min}$, $d_{\max}$, $v_{\min}$, $v_{\max}$, $w_1$ and $w_2$. Such parameters might be called \emph{planning parameters}. Here we will keep them fixed.
\end{enumerate}
\section{Differentiability of object functions in Radiotherapy}\label{sec:objdiff}
The numerical optimization process based on a quasi-Newton method converges locally to an optimum if the object function is at least twice continuously differentiable. In the sequel we will simply say differentiable. In current Radiotherapy practice the object function is a function of treatment parameters, but only indirectly via dose-volume constraints and or EUD constraints. Here we will consider their differentiability properties.
Let us start with object functions only containing EUD constraints. Then $F(\sigma)$ is a weighted sum of local object functions involving only
\begin{displaymath}
G(d_{\min}-E_{\alpha}(\sigma))\;\; \text{and}\;\; G(E_{\alpha}(\sigma)-d_{\max})
\end{displaymath}
like in equations (\ref{eq:objfun}) and (\ref{eq:locobjfun}). In this case the object function $F$ is differentiable. Here we have to assume that the dose distribution is differentiable.
However, when the object function contains dose-volume constraints it is not always differentiable. This is due to the finite differentiability of the dose-volume histogram at critical values of the dose distribution, even when the latter is differentiable. Since it is not immediately obvious how the differentiability of the dose-volume histogram as a function of dose affects the differentiability of the object function as a function of treatment parameters, we present the following explanation. Essentially, the treatment parameters enter the argument of the volume function because in the presence of parameters, the critical values are parameter dependent.
When treatment parameters vary, the dose distribution $f_{\sigma}$ varies and in particular its critical values. Thus a critical value may pass through $h$. Suppose $x_{\sigma}=0$ is a critical point for $\sigma=\sigma_0$ (a critical point can always be translated to $0$). It can be shown, see section \ref{sec:proof}, that for values of $\sigma$ near $\sigma_0$ and $x$ near $0$, $f_{\sigma}$ can be transformed to a new function $\hat{f}_{\sigma}(x) = g(x) + f_{\sigma}(x_{\sigma})$. Where $g$ is a $\sigma$ independent local standard of $f_{\sigma}$ with critical point at zero and critical value zero and $f_{\sigma}(x_{\sigma})$ is the critical value of $\hat{f}_{\sigma}$ for $\sigma$ near $\sigma_0$.
Using the previous we will show in section \ref{sec:proof} that $V_{\sigma}(h) = V_{\textrm{glob}}(\sigma,h) + V_{\textrm{loc}}(\sigma,h)$ can be split into a global and a local part. The global part depends differentiably on $\sigma$. Let us consider the local part in more detail. On a neighbourhood $\mathcal{O} \subset \mathbb{R}^3$ of $0$ we have
\begin{align*}
V_{\textrm{loc}}(\sigma,h) &\propto \textrm{vol} \{x \in \mathcal{O} \;|\; \hat{f}_{\sigma}(x) \geq h\}\\
&= \textrm{vol} \{x \in \mathcal{O} \;|\; g(x) + f_{\sigma}(x_{\sigma}) \geq h\}\\
&= \textrm{vol} \{x \in \mathcal{O} \;|\; g(x) \geq h - f_{\sigma}(x_{\sigma})\}\\
&= V_g(h - f_{\sigma}(x_{\sigma})).
\end{align*}
The function $V_{\textrm{loc}}(\sigma,h)$ is proportional to the volume in the expression above and the proportionality factor is a differentiable function of $\sigma$. The last line in the formula above shows the volume function $V_g$ for a region enclosed by level sets of $g$. According to \cite{hvn} this function is not twice continuously differentiable at critical values of $g$. Thus $V_{\textrm{loc}}(\sigma,h)$ is not twice continuously differentiable at values of $\sigma$ for which
\begin{equation}\label{eq:hyper}
h = f_{\sigma}(x_{\sigma}).
\end{equation}
This, in turn, means that the object function $F$ is not twice continuously differentiable at values of $\sigma$ such that $h$ is a critical value of $f_{\sigma}$.
\textbf{Remarks}
\begin{enumerate}\topsep 0pt\itemsep 0pt
\item \label{rem:gnotsmo} The function $G$ in equation (\ref{eq:locobjfun}) is not differentiable at $0$. This alone would already imply that the object function in (\ref{eq:objfun}) is not differentiable. However we will assume that we are in a situation like in remark \ref{rem:compro} where we are looking for a compromise. This means we stay away from points where $G$ is not differentiable.
\item \label{rem:codim} Note that equation (\ref{eq:hyper}) defines a hyper surface $\Lambda$ in parameter space $\mathbb{R}^m$. It will be $m-1$-dimensional in most of its points, therefore we need to vary only one parameter to pass through $\Lambda$.
\item \label{rem:dosesmo} Throughout we assume that the dose distribution is a differentiable function. If it were less differentiable, the volume function would also be less differentiable for both critical and regular values.
\end{enumerate}
\subsection{Examples}\label{sec:xmpl}
The aim of this section is to provide two examples. One shows a differentiable object function, the other shows an object function that fails to be twice continuously differentiable at a certain value of the treatment parameters, here we take $\sigma=0$. In the first case the Newton iteration converges to an optimum, wheras in the second case the Newton iteration does not converge if one of the iterates becomes $0$. To avoid lengthy calculations obscuring the points we wish to illustrate we will make several simplifications.
\begin{figure}[htbp]
\setlength{\unitlength}{1mm}
\begin{picture}(100,25)(0,0)
\put(48, 0){\epsffile{distroos.pst}}
\put(70, 19){$d_+$}
\put(140,21){$d_+$}
\put(70, 10){$d_c$}
\put(140,10){$d_c$}
\put(70, 15){$d_{\max}$}
\put(140,15){$d_{\max}$}
\put(75, -3){$x$}
\put(140,-3){$x$}
\end{picture}
\caption{On the left the graph of $f_{\sigma_1}(x,0,0)$ and on the right the graph of $f_{\sigma_2}(x,0,0)$. $d_+$, $d_{\max}$ and $d_c$ are explained in the text.\label{fig:distroos}}
\end{figure}
Let $f_{\sigma}$ be a dose distribution on the region $R$. The object function consists of a single maximum dose-volume constraint $(d_{\max},v_{\max})$ so $F(\sigma) = G(V_{\sigma}(d_{\max}),v_{\max})$. The difference between the to examples will be the dose distribution. Suppose for $\sigma$ in a sufficiently large neighbourhood of $\sigma_1$ the dose distribution is a function with a single maximum and for a similar neighbourhood of $\sigma_2$ it is a function with two maxima for example:
\begin{displaymath}
f_{\sigma_1}(x,y,z) = \frac{1}{1+x^2+y^2+z^2}\;\;\text{and}\;\;
f_{\sigma_2}(x,y,z) = \frac{1}{1+x^2+y^2+z^2} + \frac{1}{2+(x-4)^2+y^2+z^2}.
\end{displaymath}
Here $\sigma_1$ and $\sigma_2$ are different parameter values. In figure \ref{fig:distroos} we draw the functions $f_{\sigma_1}(x,0,0)$ and $f_{\sigma_2}(x,0,0)$. For both dose distributions we call $d_+$ the value at the global maximum and $d_c$ is the value of the other local maximum of $f_{\sigma_2}$. Both $d_+$ and $d_c$ depend on the treatment parameters $\sigma$. Let us assume that always $d_c < d_{\max} < d_+$. Furthermore we assume that $V_{\sigma}(d_{\max}) < V_{\sigma}(d_c)$.
For $\sigma$ near $\sigma_1$ the dose distribution has only one critical value namely $d_+$ so for $d < d_+$, $V_{\sigma}(d)$ is a differentiable function of $\sigma$ which we call $U$. In the second case, where $\sigma$ is near $\sigma_2$, we will be particularly interested in $d=d_c$. For $d$ near $d_c$ but $d \neq d_c$, $V_{\sigma}(d)$ is again a differentiable function. At $d=d_c$, $V_{\sigma}(d) = U(\sigma) + V_g(d_c-d)$ can be split in a local and a global part, as in the previous section, where the global part is differentiable. We assume for simplicity that the global part is again $U$. The local part $V_g$ is essentially a function of the following form
\begin{displaymath}
V_g(h) = \left\{\begin{array}{ll}
\alpha h^{\frac{3}{2}},& h > 0\\
0 ,& h \leq 0,
\end{array}\right.
\end{displaymath}
where $g$ is a local standard form of the dose distribution, $\alpha$ is a constant. Since the volume function $V_g$ is not differentiable at $0$, the object function (that we will construct shortlye hereafter) is not differentiable at values of $\sigma$ for which $d=d_c$. The latter defines a hyper surface $\Lambda$ in parameter space of codimension one, which means that we need to vary only one parameter to cross $\Lambda$. Therefore we choose new parameters and the first one we take to be $d-d_c$. Only this parameter will be relevant for the example, we call it again $\sigma$, but now $\sigma \in \mathbb{R}$. Thus for the second dose distribution we finally have the following volume function:
\begin{equation}\label{eq:volfun2}\commentaar{sigma in twee betekenissen?!}
V_{\sigma}(d) = \left\{\begin{array}{ll}
U(\sigma) + \alpha (-\sigma)^{\frac{3}{2}},& \sigma < 0\\
U(\sigma) ,& \sigma \geq 0.
\end{array}\right.
\end{equation}
Now we construct the object function. First we scale such that $v_{\max}=1$. Then the object function for the first dose distribution is
\begin{displaymath}
F_1(\sigma) = (U(\sigma)-1)^2.
\end{displaymath}
The object function for the second dose distribution is given by almost the same formula, but $U$ is replaced by the volume function in equation (\ref{eq:volfun2}). The function $U$ must satisfy several conditions so that it is consistent with the dose distributions we chose. First $U$ must be differentiable and decreasing, furthermore $U(0) > 1$ since we assumed that $V_{\sigma}(d_{\max}) < V_{\sigma}(d_c)$. A possible choice is $U(\sigma) = \frac{15}{10+\sigma}$.
In the first optimization problem the object function $F_1$ is differentiable so there are no difficulties in applying the Newton algortihm. However, in the second optimization problem the object function $F_2$ is not twice continuously differentiable at $\sigma=0$. Recall that the 1-dimensional Newton iteration for $F$ is $\sigma_{n+1} = \phi(\sigma_n) = \sigma_n - \frac{F'(\sigma_n)}{F''(\sigma_n)}$. Fixed points $\sigma_0$ of $\phi$ correspond to extremal points of $F$ if $F''$ is continuous and nonzero at $\sigma_0$. Since the limit from the left of the second derivative of $F_2$ is unbounded, the Newton iteration has a fixed point at $\sigma=0$ which is not related to an minimum or maximum of $F_2$. But this 'spurious' fixed point is not stable for the Newton iteration $\phi$. Only if $\sigma_n=0$ the next iterates will also be zero. Therefore it does not lead to convergence to a non extremal point of $F$ in general. Although it may slow down the convergence of the Newton iterates. For a graphical explanation see figure \ref{fig:graphs}.
\begin{figure}[htbp]
\setlength{\unitlength}{1mm}
\begin{picture}(100,35)(0,0)
\put(0, 15){\epsffile{graphs.pst}}
\put(68, 16){$\sigma$}
\put(117,31){$\sigma$}
\put(95, 33){$\sigma_n$}
\put(107,33){$\sigma_{n+1}$}
\end{picture}
\caption{In the left figure, the graph of $F_2$ is indicated by a dashed line, $F'_2$ by a solid line and $F''_2$ by dotted line. The figure on the right shows a magnification of the graph of $F'_2$ near $\sigma=0$. The dashed lines indicate how the Newton iterate $\sigma_{n+1}$ is constructed from $\sigma_n$.\label{fig:graphs}}
\end{figure}
\section{Conclusion}\label{sec:conc}
Our results show that the Radiotherapy treatment planning optimization process based on a quasi-Newton method locally converges to an optimum of the object function for most values of the treatment parameters. Only when the object function contains a dose-volume constraint $(h,v)$, convergence can not be guaranteed. The reason is that the object function is not sufficiently differentiable for values of the treatment parameters such that $h$ is a critical value of the dose distribution $f_{\sigma}$. Although for all other values of $\sigma$, the object function is differentiable. The question remains whether this leads to practical consequences for optimization in treatment planning.
Let us take a closer look at the values of the treatment parameters for which the object function is not sufficiently differentiable. In parameter space these values of $\sigma$ are determined by the equation $h = f_{\sigma}(x_{\sigma})$, see section \ref{sec:objdiff}, expressing that $h$ is a critical value of the dose distribution. Geometrically this equation defines a hyper surface $\Lambda$ in parameter space. Roughly speaking $\Lambda$ is $m-1$-dimensional so it has measure zero. This means that given a starting point for optimization, the Newton iteration scheme generates a sequence of points that will lie on $\Lambda$ with probability zero. The points on $\Lambda$ are spurious fixed points of the Newton iteration like in the example of the previous section. The reason is that we can generalize this example. First all types of extreme points of the dose distribution, namely minima, saddles and maxima lead to the same type of non-differentiability of the volume function. Second, also in higher dimensions the second derivative of the object function becomes unbounded at points of insufficient differentiability. Third, the spurious fixed points of the Newton iteration scheme are unstable. Our final conclusion is that in practice the convergence of the optimization process is most likely only slowed down near $\Lambda$.
\section{Proof of differentiability statements}\label{sec:proof}
In order to make a statement about the differentiability of the object function, we need a few definitions. Let $\Sigma \subset \mathbb{R}^m$ be an open set of $m$ \emph{treatment parameters}. Let $D$ be a domain in $\mathbb{R}^3$ representing the patient and let $R \subset D$ represent a tumour region or a healthy organ. For each $\sigma \in \Sigma$, the \emph{dose distribution} is represented by a function $f_{\sigma} : D \to \mathbb{R}$ (here we do not consider $f_{\sigma}$ as a distribution in the sense of generalized functions). A function is called \emph{smooth} if it is infinitely many times continuously differentiable. We now assume that $f_{\sigma}$ is a smooth function of position $x \in D$ and moreover smoothly dependent on the parameters $\sigma \in \Sigma$. \emph{Critical points} of $f_{\sigma}$ are points in $D$ where the gradient of $f_{\sigma}$ vanishes, other points are called \emph{regular points}. The value $f_{\sigma}$ takes at a critical point is called a \emph{critical value}. If the level set $\{x \in R\;|\; f_{\sigma}(x) = h\}$ contains no critical points, $h$ is called a \emph{regular value}. We now state our results on the differentiability of object functions containing EUD constraints and dose-volume constraints respectively.
\begin{theorem}\label{the:diff1}
Let the object function $F : \Sigma \to \mathbb{R}$ be given by $F(\sigma) = G(E_{\alpha}(\sigma),h)$, where $G : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a smooth function and $E_{\alpha}(\sigma)$ is the equivalent uniform dose. Then $F$ is smooth with respect to $\sigma$.
\end{theorem}
\begin{theorem}\label{the:diff2} Let the object function $F : \Sigma \to \mathbb{R}$ be given by $F(\sigma) = G(V_{\sigma}(h),v)$, where $G : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a smooth function and $V_{\sigma}$ is a volume function for dose distribution $f_{\sigma}$. Then $F$ is smooth with respect to $\sigma$ provided that $h$ is not a critical value of $f_{\sigma}$. If $h$ is a critical value of $f_{\sigma_0}$, then $F$ is not twice continuously differentiable at $\sigma = \sigma_0$.
\end{theorem}
The proof of theorem \ref{the:diff1} is rather straightforward, but the proof of theorem \ref{the:diff2} is more involved. Since in the setting of theorem \ref{the:diff2} the differentiability of the object function $F$ is completely determined by the volume function, the proof is about the latter only. The theorem contains two statements which we here state and prove separately. The proofs heavily rely on the results in \cite{hvn}, but due to the presence of parameters we have to make several adjustments. Let us start with some definitions.
\textbf{Definitions}
\begin{enumerate}\topsep 0pt\itemsep 0pt
\item Let $f_{\sigma}: \mathbb{R}^n \to \mathbb{R}$ be a positive smooth function for all $\sigma \in \mathbb{R}^m$ and moreover $f_{\sigma}$ depends smoothly on $\sigma$. For each $\sigma \in \Sigma$, $f_{\sigma}$ vanishes at infinity in the following sense, for each $\varepsilon >0$ there is a compact $K \subset \mathbb{R}^n$ such that for all $x \in \mathbb{R}^n \setminus K$, $f_{\sigma}(x) < \varepsilon$. We assume that critical points of $f_{\sigma}$ are non-degenerate, that is if $x$ is a critical point of $f_{\sigma}$ then $\det(\textrm{Hess}\;f_{\sigma}(x)) \neq 0$.
\item Also $f: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}$ is a smooth function with non-degenerate critical points. Although we consider $f$ and $f_{\sigma}$ as different functions, their values are taken to be identical: $f(x,\sigma)=f_{\sigma}(x)$ for all $x \in \mathbb{R}^n$ and $\sigma \in \mathbb{R}^m$.
\item We denote the level sets of $f_{\sigma}$ by $N_{h,\sigma} = \{x \in \mathbb{R}^n \;|\; f_{\sigma}(x)=h\}$.
\item The level sets of $f$ are denoted by $\mathcal{N}_{h} = \{(x,\sigma) \in \mathbb{R}^n \times \mathbb{R}^m \;|\; f(x,\sigma)=h\}$.
\end{enumerate}
\textbf{Remark.} Note that critical points of $f$ are also critical points of $f_{\sigma}$, but the opposite is not necessarily true.
\textbf{Proof of theorem \ref{the:diff1}.}
The function $f_{\sigma}$ is a positive and smooth function, smoothly depending on parameters $\sigma$. Then $[f_{\sigma}]^{\alpha}$ for $\alpha > 0$ is also smooth in both variables an parameters. Now it immediately follows that $E_{\alpha}(\sigma)$ as defined in equation (\ref{eq:eud}) is again a smooth function of $\sigma$.\hfill $\square$
The following two propositions essentially cover theorem \ref{the:diff2}.
\begin{proposition}\label{pro:diff}
Let $f_{\sigma}$ and $f$ be as defined above. Then $V_{\sigma}(h) = \textrm{vol}_n \{x \in \mathbb{R}^n \;|\; f_{\sigma}(x) \geq h\}$ depends smoothly on $\sigma$ at $\sigma = \sigma_0$ if $h$ is a regular value of $f_{\sigma_0}$.
\end{proposition}
\begin{proposition}\label{pro:finidiff}
Let $f_{\sigma}$ and $V_{\sigma}(h)$ be as in the previous proposition. Then $V_{\sigma}(h)$ is finitely differentiable with respect to $\sigma$ at $\sigma = \sigma_0$ if $h$ is a critical value of of $f_{\sigma_0}$.
\end{proposition}
Let us begin with the nice situation where $h$ is a regular value of $f_{\sigma_0}$, so that $V_{\sigma}(h)$ smoothly depends on $\sigma$. In the proof of proposition \ref{pro:diff} we use a property of the level sets $N_{h,\sigma}$ of $f_{\sigma}$.
\begin{lemma}\label{lem:ndiffeo}
Let $f_{\sigma}$ and $f$ be as defined above and furthermore $h$ be a regular value of $f_{\sigma_0}$. Then an open neighbourhood $\Sigma$ of $\sigma_0$ exists, such that $N_{h,\sigma}$ is diffeomorphic to $N_{h,\sigma_0}$ for all $\sigma \in \Sigma$ and the diffeomorphism depends smoothly on $\sigma$.
\end{lemma}
\textbf{Proof of proposition \ref{pro:diff}.} The volume $V_{\sigma}(h)$ is given by an integral over a region bounded by $N_{h,\sigma}$. Since, according to lemma \ref{lem:ndiffeo}, the latter depends smoothly on $\sigma$ at $\sigma=\sigma_0$, the integral and thus $V_{\sigma}(h)$ depends smoothly on $\sigma$ at $\sigma=\sigma_0$.\hfill $\square$
It remains to prove lemma \ref{lem:ndiffeo}.
\textbf{Proof of lemma \ref{lem:ndiffeo}.} Let $\{\sigma=\sigma_0\}$ be a shorthand for $\{(x,\sigma) \in \mathbb{R}^n \times \mathbb{R}^m \;|\; \sigma=\sigma_0\}$. Note that $N_{h,\sigma_0} = \{\sigma=\sigma_0\} \cap \mathcal{N}_{h}$. Since $h$ is a regular value, $N_{h,\sigma_0}$ contains no critical points of $f_{\sigma_0}$. Thus the intersection of $\{\sigma=\sigma_0\}$ and $\mathcal{N}_{h}$ is transversal. Then a neighbourhood $\Sigma$ of $\sigma_0$ exists such that for all $\sigma_1 \in \Sigma$, the intersection of $\{\sigma=\sigma_1\}$ and $\mathcal{N}_{h}$ is transversal. Thus for all $\sigma_1 \in \Sigma$, $N_{h,\sigma_1}$ is a regular level set, consequently there are no critical points of $f$ in $\mathcal{N}_{h,\Sigma} = \{(x,\sigma) \in \mathbb{R}^n \times \mathbb{R}^m \;|\; x \in N_{h,\sigma}, \sigma \in \Sigma\}$. Suppose $\sigma_1 \in \Sigma$ and $\sigma_1 \neq \sigma_0$ then an index $i$ exists such that $\sigma_{1,i} \neq \sigma_{0,i}$. Now the gradient flow of the height function $\sigma_i$ on $\mathcal{N}_{h}$ is a regular flow on $\mathcal{N}_{h,\Sigma}$ thus defining a diffeomorphism $\Phi$ from $N_{h,\sigma_0}$ to $N_{h,\sigma_1}$. \hphantom{knudde}\hfill $\square$
Let us now turn to the case where $h$ is a critical value of $f_{\sigma_0}$. Then there is at least one critical point $x_{\sigma_0}$ in $N_{h,\sigma_0}$, but for the sake of simplicity we assume that $x_{\sigma_0}$ is unique. In the proof of proposition \ref{pro:finidiff} we will use a similar construction as in \cite{hvn} without providing all details.
\textbf{Proof of proposition \ref{pro:finidiff}.} Let us assume that $x_{\sigma_0}$ is the unique critical point of $f_{\sigma_0}$ on the critical level set $N_{h,\sigma_0}$. Since by assumption $x_{\sigma_0}$ is a non-degenerate critical point we can apply the implicit function theorem to conclude that an open neighbourhood $\Sigma$ of $\sigma_0$ exists such that the map $\Sigma \to \mathbb{R}^n : \sigma \mapsto x_{\sigma}$ and $x_{\sigma}$ is a critical point of $f_{\sigma}$ is smooth. Now we define a new function $g_{\sigma}(x) = f_{\sigma}(x_{\sigma}+x)$, which is a smooth function of $x$ and $\sigma \in \Sigma$. Moreover $g_{\sigma}$ has a critical point at $x=0$ with critical value $f_{\sigma}(x_{\sigma})$. Then we have $V_{\sigma}(h) = \textrm{vol}_n \{x \in \mathbb{R}^n \;|\; g_{\sigma}(x) \geq h\}$. Following the construction in \cite{hvn} we split the latter region in a part not containing the point $0$ and a remaining small part including $0$. The volume of the first part depends smoothly on $\sigma$ using the same arguments proving proposition \ref{pro:diff}. Since $V_{\sigma}(h)$ is the sum of both volumes, the differentiability of $V_{\sigma}(h)$ is determined by that of the volume of the second part.
Let us consider this second part contained in a small neighbourhood $\mathcal{O}$ of $0$. We may assume that $\mathcal{O}$ is small enough to put $g_{\sigma}$ into the standard form of \cite{hvn}. In order to keep notation simple we denote the standard form again by $g_{\sigma}$, but now we can write $g_{\sigma}(x) = g(x) + f_{\sigma}(x_{\sigma})$. The actual standard form is $g(x) = \sum_{i=1}^p x_i^2 - \sum_{i=1}^q x_{p+i}^2$, with $p+q=n$. Note that $g$ no longer depends on $\sigma$, it has a critical point at $x=0$ and the critical value is $0$. This form can be obtained by a smooth change of coordinates, even smoothly depending on $\sigma$. Here again we need the non-degeneracy of critical point of $f_{\sigma}$. In the following we use the results of \cite{hvn}. As a function of $k$, the volume function $V_{\textrm{loc}}(k) = \textrm{vol}_n \{x \in \mathcal{O} \subset \mathbb{R}^n \;|\; g(x) \geq k\}$ is only finitely differentiable with respect to $k$ at $k=0$.
Now the volume of the second part is proportional to $V_{\textrm{loc}}(h-f_{\sigma}(x_{\sigma})) = \textrm{vol} \{x \in \mathcal{O} \subset \mathbb{R}^n \;|\; g(x) \geq h-f_{\sigma}(x_{\sigma})\}$. The proportionality factor is of no concern because it originates from smooth coordinate changes and depends smoothly on $\sigma$. Thus we obtain that the differentiability of $V_{\sigma}(h)$ with respect to $\sigma$ is determined by that of $V_{\textrm{loc}}(h-f_{\sigma}(x_{\sigma}))$. The latter is only finitely differentiable with respect to $\sigma$ at those values of $\sigma$ where $h-f_{\sigma}(x_{\sigma})=0$. Or, put differently, where $h$ is a critical value of $f_{\sigma}$.\hfill $\square$
\textbf{Remarks}
\begin{enumerate}\topsep 0pt\itemsep 0pt
\item The statements and proofs in this section are valid for any dimension. Therefore we have not specialized them to the case $n=3$. In this respect finitely differentiable means less than two times continuously differentiable.
\item The critical points of a parameter family of functions will generally be degenerate for certain parameter values. However we excluded this possibility because the volume function will be even less differentiable at a critical value when the critical point on the level set of this value is degenerate.
\end{enumerate}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,523
|
\section{Introduction}
\label{sec:intro}
The D\O\ experiment was proposed in 1983 to study proton-antiproton collisions
at a center-of-mass energy of 1.8~TeV at the Fermilab Tevatron collider.
The focus of the
experiment was the study of high mass states and large $p_T$ phenomena. The
detector performed very well during Run I of the Tevatron, 1992--1996,
leading to the discovery of the top quark \cite{top_discovery} and measurement
of its mass \cite{top_mass1,top_mass2,top_mass3,top_mass4,top_mass5}, a
precision measurement of the mass of the $W$ boson
\cite{W_mass1,W_mass2,W_mass3,W_mass4,W_mass5,W_mass6,W_mass7},
detailed analysis
of gauge boson couplings \cite{gc3,gc4,gc5,gc6,gc7,gc8,gc9,gc10,gc11}, studies
of jet production \cite{jet1,jet2,jet3,jet4},
and greatly improved limits on
the production of new phenomena such as leptoquarks
\cite{lq1,lq2,lq3,lq4,lq5,lq6,lq7} and
supersymmetric particles \cite{susy1,susy2,susy3,susy4,susy5,susy6,susy7,susy8},
among many other accomplishments
\cite{d0_pub_list}.
During Run~I, the Tevatron operated using six bunches each of protons and
antiprotons, with 3500~ns between bunch crossings and a center-of-mass energy
of 1.8~TeV. The peak luminosity was typically
1--2$\times 10^{31}\ {\rm cm^{-2} s^{-1}}$ and approximately 120~pb$^{-1}$
of data were recorded by D\O. Following the completion of the new
Main Injector and associated Tevatron upgrades \cite{tev1,tev2},
the collider began running again in 2001. In Run~II, which began in March
2001, the Tevatron is operated with 36 bunches of protons
and antiprotons with a bunch spacing of 396~ns and at an increased
center-of-mass energy of 1.96~TeV. The instantaneous luminosity will increase
by more than a factor of ten to greater than $10^{32}\ {\rm cm^{-2} s^{-1}}$
and more than 4~fb$^{-1}$ of data are expected to be recorded.
To take advantage of these improvements in the Tevatron and to enhance the
physics reach of the experiment, we have significantly upgraded the D\O\
detector. The detector consists of three major subsystems: central tracking
detectors, uranium/liquid-argon calorimeters, and a muon spectrometer.
The original D\O\ detector is described in detail
in Ref.~\cite{d0_nim}. The central tracking system has been completely
replaced. The old system lacked a magnetic field and suffered from radiation
damage, and improved tracking
technologies are now available. The new system includes a silicon microstrip
tracker and a scintillating-fiber tracker located within a 2 T solenoidal
magnet. The silicon microstrip tracker is able to identify displaced vertices
for $b$-quark tagging. The magnetic field enables measurement of the
energy to momentum ratio ($E/p$)
for electron identification and calorimeter calibration,
opens new capabilities for tau lepton identification and hadron spectroscopy,
and allows precision muon
momentum measurement. Between the solenoidal magnet and the central
calorimeter and in front of the forward calorimeters, preshower detectors have
been added for improved electron identification. In the forward muon system,
proportional drift chambers have been replaced by mini drift tubes and trigger
scintillation counters that can withstand the
harsh radiation environment and additional shielding has been added. In the
central region, scintillation counters have been added for improved muon
triggering.
We have also added a forward proton detector for the study of diffractive
physics. A side view of the upgraded D\O\ detector is shown in
Figure~\ref{fig:detector}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure1.eps}}
\caption{Diagram of the upgraded D\O\ detector, as installed in the collision
hall and viewed from inside the Tevatron ring. The forward proton detector is
not shown. The detectors in the central region of the detector are shown in
Fig.~\ref{fig:tracker}.}
\label{fig:detector}
\end{figure}
The large reduction in the bunch spacing required the improvement of the
read-out electronics and the implementation of pipelining for the front-end
signals from the tracking, calorimeter, and muon systems. The calorimeter
preamplifiers and signal-shaping electronics have been replaced, as have
all of the electronics for the muon system. The trigger system has been
significantly upgraded, providing three full trigger levels
to cope with the higher collision rate and new hardware to identify displaced
secondary vertices for $b$-quark tagging.
Muon triggering has been enhanced by the addition of scintillation counters in
the central and forward regions.
A significant improvement to the detector's performance
resulted from the removal of the old
Main Ring beam pipe from the calorimeters. During Run~I, the Main Ring
was used to accelerate protons for antiproton production while the Tevatron
operated in collider mode. Losses from the Main Ring produced spurious
energy deposits in the calorimeters and muon system, and most triggers were
not accepted while Main Ring protons passed through the detector. Removal of
the Main Ring increased the livetime of the detector by approximately 10\%,
depending on the trigger.
In the following sections of this paper, we describe the design and performance
of the upgraded D\O\ detector. The new central tracking system and
solenoidal magnet are presented in Sections~\ref{sec:central_tracking} and
\ref{sec:solenoid}, respectively. The preshower detectors are described
in Section~\ref{sec:preshower}. The calorimeters are briefly described in
Section~\ref{sec:calorimeters} along with the new calorimeter electronics.
The muon system is discussed in Section~\ref{sec:muon}. The new forward
proton detector is presented in Section~\ref{sec:fpd}. The luminosity monitor
is described in Section~\ref{sec:lum-monitor}. The triggering
and data acquisition systems are described in Sections~\ref{sec:trigger} and
\ref{sec:daq}. Section~\ref{sec:controls} covers detector
controls and monitoring and Section~\ref{sec:software} contains an overview
of the software components of the experiment.
In the detector description and data analysis, we use a right-handed coordinate
system in which the $z$-axis is along the
proton direction and the $y$-axis is upward
(Figure~\ref{fig:detector}). The angles $\phi$ and $\theta$
are the azimuthal and polar angles, respectively. The $r$ coordinate denotes
the perpendicular
distance from the $z$ axis. The pseudorapidity, $\eta = -\ln[\tan(\theta/2)]$,
approximates the true rapidity, $y=1/2 \ln[(E+p_zc)/(E-p_zc)]$, for finite
angles in the limit that $(mc^2/E) \to 0$. We use the term ``forward'' to
describe the regions at large $|\eta|$.
\FloatBarrier
\section{Central tracking}
\label{sec:central_tracking}
Excellent tracking in the central region is necessary for studies of
top quark, electroweak, and $b$ physics and to search for new phenomena,
including the Higgs boson. The central tracking system consists of the silicon
microstrip tracker (SMT) and the central fiber tracker (CFT) surrounded by a
solenoidal magnet. It surrounds the D\O\ beryllium beam pipe, which has a wall
thickness of 0.508~mm and an outer diameter of 38.1~mm, and is 2.37~m long.
The two tracking detectors locate the primary interaction
vertex with a resolution of about 35~$\mu$m along the beamline. They can tag
$b$-quark jets with an impact parameter resolution of better than 15~$\mu$m
in $r-\phi$ for particles with transverse momentum $p_T > 10$~GeV/$c$ at
$|\eta| = 0$. The high resolution of the
vertex position allows good measurement of lepton $p_T$, jet transverse energy
($E_T$), and missing transverse energy \met. Calibration of the
electromagnetic calorimeter using $E/p$ for electrons is now possible.
Both the SMT and CFT provide tracking information to the trigger. The SMT
provides signals to the Level~2 and 3 trigger systems and
is used to trigger on displaced vertices from $b$-quark decay.
The CFT provides a fast and continuous readout of discriminator signals to the
Level~1 trigger system; upon a Level~1 trigger accept, track information based
on these signals is sent to Level~2. The Level~3 trigger receives
a slower readout of the CFT's digitized analog signals, in addition to the
discriminator information available at Level~1 and Level~2.
A schematic view of the central tracking system is shown in
Figure~\ref{fig:tracker}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure2.eps}}
\caption{Cross-sectional view of the new central tracking system in the $x-z$
plane. Also shown are the locations of the solenoid, the preshower detectors,
luminosity monitor, and the calorimeters.}
\label{fig:tracker}
\end{figure}
\subsection{Silicon microstrip tracker}
\label{sec:smt}
The SMT provides both tracking and vertexing over
nearly the full $\eta$ coverage of the calorimeter and muon
systems. Design of the detector, electronics, and cooling are, in large
part, dictated by the accelerator environment. The length of the interaction
region
($\sigma \approx 25$~cm) sets the length scale of the device. With
a long interaction region, it is a challenge to deploy detectors such that
the tracks are generally perpendicular to detector surfaces for all $\eta $.
This led us to a design of barrel modules interspersed with disks in the
center and assemblies of disks in the forward regions.
The barrel detectors primarily measure the $r-\phi$ coordinate and the disk
detectors measure $r-z$ as well as $r-\phi $. Thus vertices for particles at
high $\eta$ are reconstructed in three dimensions by
the disks, and vertices of particles at small values of $\eta$ are measured
in the barrels and central fiber tracker. This design poses difficult
mechanical challenges in arranging the detector components and minimizing
dead areas while providing sufficient space for cooling and cables.
An isometric view of the SMT is shown in Figure~\ref{fig:smt}.
The detector has six barrels in the central region. Each barrel has four
silicon readout layers. The silicon modules installed in the barrels are
called ``ladders.''
Layers 1 and 2 have twelve ladders each; layers 3 and 4 have twenty-four
ladders each, for a total of 432 ladders.
Each barrel is capped at high $\vert z\vert$ with a disk of twelve double-sided
wedge detectors, called an ``F-disk.'' Forward of the three disk/barrel
assemblies on each side is
a unit consisting of three F-disks. In the far forward regions,
two large-diameter disks, ``H-disks,'' provide tracking at high
$\vert \eta \vert$. Twenty-four full wedges, each consisting of two
back-to-back single-sided ``half'' wedges, are mounted on each H-disk.
There are 144 F-wedges and 96 full H-wedges in the tracker; each side of a
wedge (upstream and downstream) is read out independently.
There is a grand total of 912 readout modules, with 792,576 channels. The
centers of the H-disks are located at $\vert z\vert = 100.4,~121.0$~cm;
the F-disks are at $\vert z\vert = 12.5,~25.3,~38.2,~43.1,~48.1,$
and $53.1$~cm. The centers of the barrels are at $\vert z\vert =
6.2,~19.0,~31.8$~cm.
The SMT is read out by custom-made 128-channel SVXIIe readout chips.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure3.eps}}
\caption{The disk/barrel design of the silicon microstrip tracker.}
\label{fig:smt}
\end{figure}
\subsubsection{Sensors}
\paragraph{Sensor types}
The SMT uses a combination of single-sided (SS), double-sided (DS), and
double-sided double-metal (DSDM) technologies. Silicon sensors were obtained
from three manufacturers. All barrel sensors were produced by Micron
Semiconductor LTD~\cite{Micron}. The wedges for the F-disks were produced by
Micron Semiconductor LTD and Canberra Eurisys Mesures~\cite{EurisysMesures}.
The wedge sensors for the H-disks were manufactured by ELMA~\cite{elma}; these
sensors use intermediate strips for
charge interpolation. Single-sided and double-sided devices were
produced from high resistivity 4'' silicon wafers, with crystal orientation
$<$111$>$ and $<$100$>$. The 90$^\circ$ stereo sensors used in layers 1 and 3
of the four centermost barrels are DSDM sensors,
manufactured using $<$100$>$ 6'' wafers. Isolation on the n-side of all
double-sided sensors is provided by p-stop implants. All traces are biased
using polysilicon resistors. Table~\ref{tab:silicon} shows the sensor types
used in the SMT and their locations.
Disk sensors are trapezoids with readout strips arranged parallel to the
long edge of the devices. This provides an effective 30$^\circ$ stereo angle
for the double-sided F-disks. A wedge for the H-disks consists of a pair of
single-sided half-wedges mounted back-to-back, giving an effective stereo
angle of 15$^\circ$. This arrangement means that the strip length varies for
strips which originate past the base of the trapezoid. There are three types
of sensors in the central barrels. The second and fourth layers use
double-sided stereo sensors with the n-side implants at a 2$^\circ$ angle with
respect to the p-side axial strips. Two of these sensors are bonded together
to form a 12-cm-long ladder. The first and third layers of the outer barrels
use single-sided sensors with axial strips and, again, two sensors are bonded
together to make one 12-cm ladder. The inner four barrels use single
12-cm-long 90$^\circ$ stereo sensors. Ninety degree readout was achieved by
using a second metal layer on the n-side insulated from the first metal by
3~$\mu$m of PECVD (plasma enhanced chemical vapor deposition) silicon oxide.
Two readout strips on this side are multiplexed to a single readout channel.
Implants on the n-side are isolated by individual p-stop frames in addition
to a common p-stop enclosure.
\begin{table}
\begin{center}
\caption{Characteristics and deployment of various sensor types in the SMT.
{\it i} indicates the length of the inner H-disk sensor;
{\it o} is the length of the outer H-disk sensor.}
\label{tab:silicon}
\begin{tabular}{ l c c p{55pt} p{36pt} p{36pt} p{36pt}}
\hline
Module & Type & Layer & Pitch ($\mu$m) \par p/n &
Length \par (cm) & Inner \par radius (cm) & Outer \par radius (cm) \\
\hline
F-disks & DS & -- & 50/62.5 & 7.93 & 2.57 & 9.96 \\
\hline
H-disks & SS & -- & 40 \par 80 readout
& 7.63$^i$ \par 6.33$^o$ & 9.5 & 26 \\
\hline
\raisebox{-1.ex}[0pt]{Central}
& DSDM & 1, 3 & 50/153.5 & 12.0 & 2.715 & 7.582 \\
\cline{2-7}
\raisebox{1.ex}[0pt]{barrels (4)}
& DS & 2, 4 & 50/62.5 & 6.0 & 4.55 & 10.51 \\
\hline
\raisebox{-1.ex}[0pt]{Outer} & SS & 1, 3 & 50 & 6.0 & 2.715 & 7.582 \\
\cline{2-7}
\raisebox{1.ex}[0pt]{barrels (2)}
& DS & 2, 4 & 50/62.5 & 6.0 & 4.55 & 10.51 \\
\hline
\end{tabular}
\end{center}
\end{table}
\paragraph{Sensor testing}
All sensors were required to exhibit leakage current below 260~nA/cm$^2$,
and each channel was tested for leakage current, AC coupling
capacitance, and AC coupling leakage to 80~V. The AC coupling leakage test is
especially important for the double-sided sensors since these capacitors are
required to stand off the depletion voltage. Several classes of problems
were identified during the testing:
\begin{itemize}
\item Areas of low interstrip resistance --- Washing bad sensors in pure
water at the vendor sometimes cured this problem.
\item Isolation-implant shorts --- These occur in all sensor types but the
effects are particularly severe in the DSDM sensors where a single isolation
short can affect a region of approximately twenty channels. Strip-by-strip DC
scans were performed at the manufacturer to identify problems early in
processing.
\item Poorly controlled bias resistance --- Sensors with bias resistors as
high as 10~M$\Omega$ ($5\times$ nominal) were eventually accepted.
\item Microdischarge breakdown --- This breakdown was observed with modest
negative voltage applied to the p-side of double-sided sensors. The onset
voltage of breakdown was measured for each sensor and used to
set the split between p($-$)- and n($+$)-side bias voltage.
\end{itemize}
\paragraph{Radiation damage studies}
A set of radiation damage qualification studies was performed at the
Fermilab Booster and the Radiation Laboratory at the University of
Massachusetts Lowell. In general, the results of these studies conformed to
behavior expected from RD48~\cite{RD48} parameterizations. The exception was
the depletion voltage ($V_d$) behavior of the DSDM devices,
shown in Figure~\ref{fig:vdepl}. These sensors exhibit a rise in depletion
voltage almost twice as fast as that of the single-sided sensors. To
determine if the anomalous behavior is due to the bulk silicon properties,
photodiode test structures from the same wafer were irradiated in the booster.
These all showed normal behavior. It is likely that the rapid rise in
depletion voltage is related to the PECVD inter-metal isolation layer.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure4.eps}}
\caption{Depletion voltage as a function of fluence as measured
for SMT modules in the Fermilab Booster.}
\label{fig:vdepl}
\end{figure}
Microdischarge breakdown is caused by large fields near the junction
implant. This high-field region is expected to move with the junction from
the p-side to the n-side upon type inversion. In addition, fixed positive
charge in the oxide insulation layers will tend to increase the field on the
p-side and reduce it on the n-side. This is indeed what we observe. Breakdown
moves from the p-side to the n-side after type inversion and the threshold
increases to approximately 150~V after about 2~Mrad. Microdischarge in the
DSDM sensors will limit the lifetime of the SMT to an integrated luminosity
of 3.5--6 fb$^{-1}$.
\subsubsection{Mechanical design}
\paragraph{Ladders}
Sensor ladder assemblies were designed for low mass, precise alignment,
and good thermal performance. With the exception of the DSDM ladders, which
use a single 12-cm sensor, ladders were constructed using two 6-cm
sensors. All ladder assembly was done under the control of a coordinate
measuring machine (CMM). The two sensors were aligned using the fiducials on
the sensors. Alignment was achieved during assembly using locating notches in
precisely machined beryllium pieces on both the ``active'' (SVXIIe-readout-chip
carrying, see Section~\ref{sec:SVXII}) and ``passive'' sides of the ladder.
These notches correspond to posts on the support bulkheads and are aligned to
fiducials on the sensor. Carbon-boron fiber/Rohacell rails bridge the
beryllium pieces and maintain the flatness of the ladder assemblies.
A sketch of a double-sided 2$^\circ$ ladder with nine SVXIIe readout chips
is shown in Figure~\ref{fig:ldr_design}. Figure~\ref{fig:ldr_assy} shows
a similar ladder on an assembly fixture with the high density interconnect
(HDI, Section~\ref{sec:SVXII}) unfolded, spring-loaded
``pushers,'' and support rails. Each ladder and wedge was surveyed using an
automatic optical CMM after assembly.
\begin{figure}
\centerline{\includegraphics[width=4.in]{figure5.eps}}
\caption{Double-sided ladder design, n-side. The SVXIIe readout chips shown
as dashed lines are located on the p-side of the ladder.}
\label{fig:ldr_design}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=4.in]{figure6.eps}}
\caption{Double-sided ladder during assembly with the flex hybrid unfolded.
The spring-loaded ``pushers'' and support rails are also visible. }
\label{fig:ldr_assy}
\end{figure}
\paragraph{Wedges}
In contrast to the barrel assemblies, F- and H-disks are planar
modules, allowing for simple optical alignment during assembly. A major
design constraint was to provide minimum dead space between the disk/barrel
assemblies. F-wedges use double-sided sensors with separate 8- and 6-chip
HDIs for each side. The larger strip
pitch (62.5~$\mu$m vs 50~$\mu$m) on the n-side requires an adapter to match
the SVXIIe readout pitch. The 50~$\mu$m-thick copper-clad Kapton~\cite{kapton}
pitch adaptors are part of the cooling path to the p-side
and allow the six n-side SVXIIe chips to be located outboard of the cooling
channel. The double-sided hybrid assembly required a complex 12-layer vacuum
lamination with pitch adaptor, Tedlar~\cite{tedlar}, Ablefilm
adhesive~\cite{ablefilm}, Kapton~\cite{kapton}, beryllium, and HDI layers.
H-disk wedges are made from back-to-back single-sided sensors. In
this case, the precise front-to-back alignment was provided in the wedge
assembly process. A special dual-camera assembly fixture was developed to
simultaneously image fiducials on the top and bottom sensors. This
allowed front-to-back alignment of the two-sensor wedges of better than
15~$\mu$m. An aluminized silicon pitch adapter, fabricated on the same wafer
as the sensor, was used to match the 80~$\mu$m readout pitch to the SVXIIe
input.
\paragraph{Supports and assembly}
Barrel ladders are supported by beryllium bulkheads machined with posts and
pinholes for ladder support. Each barrel has a thick ``active'' bulkhead
containing the cooling channels and connections to the outer support
cylinder, and a thin ``passive'' bulkhead that serves to set the spacing
of the ends of the ladders without readout chips. Ladders were lowered onto
the posts using a special insertion fixture and fixed into place with pins.
Thermal contact to the cooled bulkhead edge is made with thermal grease. Each
ladder is grounded to the bulkheads with conducting epoxy. Edges of the
beryllium pieces were measured with the touch probe of the assembly CMM, and
these measurements were correlated to the beryllium and sensor fiducial
mark measurements taken with the optical CMM to provide the final ladder
position. Cables (HDI ``tails'') are routed between barrel sublayers inboard
of ladders, so no inter-module space is taken by the HDI tails. The tails
are coupled to ``card edge'' style Hirose connectors
\cite{hirose} on the low-mass, flexible, Kapton cables on the outer surface of
the support structure.
Disks are supported by beryllium rings. Wedges are located on alternate sides
of the ring with sufficient overlap to eliminate dead regions. Wedges were
manually aligned under a CMM and secured with screws. Finished F-disks were
then assembled into a disk/barrel central module or one of the three-disk
modules at the end of the central disk/barrel section.
Overall support of the SMT (exclusive of the H-disks) is provided by two
double-walled carbon fiber cylinders spaced by carbon fiber ribs
to eliminate differential contraction. North and south half-cylinders are
independent structures. This limits the size of the units, allowing
installation of the SMT in the limited space available in the
collision hall. The central upper section of each half-cylinder is removed
for placement of the disk/barrel modules. Each module is supported by
adjustable kinematic mounts. Cables and services are accessed through holes
in the cylinder whose outer surface is used for routing the low-mass cables
and water manifolds. Final alignment is provided by sapphire balls mounted
on the bulkheads, which are accessed with touch probes through additional
holes in the support cylinder. The disk/barrel half cylinders are supported
from the inner central fiber tracker barrel using mounts glued into place.
Figure~\ref{fig:barrel_disk} shows the disk/barrel module within its support
cylinder. H-disks are located on separate mounts suspended from the third
layer of the CFT.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure7.eps}}
\caption{Cross section of the SMT disk/barrel module showing
ladders mounted on the beryllium bulkhead, sample cable paths, three of twelve
F-disk wedges, carbon fiber support structure, and the low-mass cable stack.}
\label{fig:barrel_disk}
\end{figure}
\paragraph{Cooling}
The maximum operating temperature of the detector is limited to reduce bulk
damage to the silicon, which leads to type inversion and increasing
depletion voltage. Heat generated by
the SVXIIe chips must flow through the HDI, epoxy,
beryllium heat spreaders, and, for the top-side chips, through the silicon
and the bottom-side readout structure.
Coolant is a water/30\% ethylene glycol mixture supplied at $-10^{\circ}$~C
by two redundant 4.4 kW chillers. The coolant flows through channels
machined in the active beryllium bulkheads and ring supports and is maintained
below atmospheric pressure to minimize the effect of any possible leaks.
Pressure drop across detector elements is below 5~psi with a temperature
rise in the channel of 1--1.5$^{\circ}$~C. A manifold is used for each side of
the half-cylinder with flows through the modules controlled by restricting
apertures. The hottest point of the silicon at
the tip of the ladder is typically 10--15$^\circ$~C above the coolant
temperature, maintaining the silicon temperature below 5$^\circ$~C.
\subsubsection{Electronics}
\label{sec:SVXII}
\paragraph{SVXIIe chip and HDIs}
The SMT is read out using the 128-channel SVXIIe chip~\cite{SVXIIe}. The chip
includes preamp, analog delay, digitization, and data sparsification. Input
charge is integrated on the preamplifier for a train of beam crossings
(typically twelve) and reset during inter-bunch gaps. This charge is
delivered to a 32-cell analog pipeline. Upon a Level~1 trigger accept,
double-correlated sampling is performed on the appropriate
cells\footnote{Double correlated sampling is the process of subtracting the
analog baseline pedestal value from the value being measured.} and this
analog information is fed to a parallel set of Wilkinson ADCs. Digitization
utilizes both edges of the 53~MHz main accelerator clock, providing 8~bits of
analog information in 2.4~$\mu$s. Readout is half as fast. Typical noise
performance for a rise time setting of 100~ns is $490e + 50e$/pF. Fabrication
was done using the 1.2~$\mu$m UTMC radiation-hard process \cite{UTMC}
with yields of approximately 60\%.
To make the HDIs, chips were mounted on Kapton flex circuits laminated to
beryllium heat spreaders. Eight different types of HDI are necessary to
accommodate the various detector and readout geometries. A readout cable
``tail'' is part of each HDI. In the case of double-sided ladders, a
single HDI contains the readout for both the p- and n-sides with the
Kapton folded over to sandwich the ladder. The flex circuits are two-layer
50~$\mu$m Kapton with 125~$\mu$m line spacing and 50~$\mu$m plated-through
holes.
\paragraph{Readout}
Figure~\ref{fig:readout} shows the SMT readout chain. Trigger information is
received via the serial command link (SCL) by the sequencer crate
controller. The SVXIIe sequencer provides timing and control signals for the
SVXIIe chips on eight HDIs. These signals are regenerated by interface boards
that also control power and bias for the SVXIIe chips and provide interfaces
to the monitoring systems and individual HDI temperature and current
trips.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure8.eps}}
\caption{Diagram of the major components of the SMT readout
system. The data acquisition (DAQ) system is discussed in detail in
Section~\ref{sec:daq}. The horse shoe, cathedral, platform, MCH2, and MCH3 are
places where various pieces of electronics are located.}
\label{fig:readout}
\end{figure}
Data from the ladders and wedges are sent from the sequencers to VME readout
buffer (VRB) memories via optical link fibers. The VRBC (VRB controller)
receives trigger data from the SCL and uses that information to control the
operation of the VRBs. Two single-board computers are resident in each
readout crate. The primary purpose of the Power PC is slow monitoring and
calibration. The second computer (SBC)
collects data from the VRBs upon Level~2 accepts and then sends the data
to Level~3 for readout. Downloads and slow controls are provided by a
MIL-STD-1553B \cite{mil-1553} control system.
\paragraph{Cabling}
Clock, power, and signal quality and timing are critical to proper operation
of the SVXIIe chip. The SMT is read out using low-mass, flexible, Kapton
cables within the detector volume followed by high-mass 80- and 50-conductor
``pleated-foil'' cables on the outside. The cables carry both power (except
the 50-conductor high-mass cables)
and digital signals. A pair of coaxial cables carries the differential clock.
Low-mass cables are routed along the half cylinder and
coupled to the 80-conductor pleated-foil cables at a ring of adapter cards
mounted on the end of the central calorimeter cryostat.
\subsubsection{Production and testing}
Production of ladders and wedges and assembly into barrels and disks were
performed at the Silicon Detector Facility (SiDet) at Fermilab; the work took
slightly more than one year. Components
such as chips, HDIs, and cables were tested before shipment to Fermilab. A
number of sensor- and HDI-related problems were identified that necessitated
extensive quality control testing. Tests included:
\begin{itemize}
\item HDI burn-in --- Each HDI was operated at room temperature for 24 hours.
Current, pedestal, and gain were monitored.
\item Ladder/wedge debugging --- Pedestal and noise were measured for each
assembled module. Pinholes (shorted AC coupling capacitors) generated in the
wirebonding process were identified in 0.8\% of the channels and the
associated wirebonds pulled. The onset voltage for microdischarge was
determined on the n-side of double-sided sensors. Resistance between the
HDI ground and beryllium was measured. Grounds were remade if the resistance
was greater than 10~$\Omega$.
\item Ladder/wedge burn-in --- Assembled and tested ladders and wedges were
operated at $5^{\circ}$~C for 72 hours. Currents, pedestals, gain, data
integrity, and noise were monitored.
\item Laser test --- Modules were tested using a 1064~nm laser on a
computer-controlled $x-y$ table. This test allowed us to measure depletion
voltage and tabulate broken and noisy channels.
\item Assembly test --- Noise and readout were tested for each ladder (wedge)
as it was installed in a barrel (disk). Coherent noise was evident in badly
grounded ladders and grounds were remade.
\item Final cabling test --- Readout was verified after the low-mass cables
were attached. Upon leaving SiDet, 99.5\% of the sensors were functional.
\end{itemize}
This extensive testing was made possible by the development of a ``Stand
Alone Sequencer'' (SASEQ), a 3U VME module which could be read out simply
using a PC-VME interface and Microsoft EXCEL. This allowed for easy
replication of testing systems at Fermilab and universities.
A large scale ``10\%'' test was organized to test major detector
components with pre-production versions of final readout components. This
test was crucial for debugging readout hardware, testing termination
schemes, and adjusting sequencer timings. Features were identified
in the SVXIIe chip that caused pedestal jumps and readout errors.
Additional initialization states were added to the SVXIIe control sequence
to address these problems. An
assembled barrel and F- and H-disks were tested in the 10\% test to check
system integration and search for noise problems. Cosmic ray data were taken
with the barrel as a final check of system performance.
\subsubsection{Operation}
\paragraph{Operating experience}
\label{sec:smt-operation}
The SMT has been included in physics data taking since the start of Run II.
Full electronics debugging was completed in October 2001,
when 94\% of the sensors were functional. As of May 2005, 90\% of sensors
were functional (note that problems can be anywhere along the readout
chain). Most operational difficulties have been peripheral
to the silicon detector itself. These include latchup of operational
amplifiers on the
interface boards, low voltage power supply failures, and high leakage
currents in high voltage distribution boxes.
The most serious detector feature is ``grassy noise,'' shown in
Figure~\ref{fig:grassy-noise}, which is confined to the Micron-supplied
F-disk detectors (75\% of the F-disk sensors). This noise is
characterized by large charge spikes which cover 10--20 strips and occur in
about 20\% of the events for affected devices. Leakage currents
typically rise to greater than 100~$\mu$A within one hour of
turn-on at the beginning of a store.
Both the SMT and CFT observe pedestal shifts that depend on the phase of the
beam crossing with respect to the SVXIIe reset pulse. These shifts are
typically 1--2 counts (27 counts/MIP) in most ladders but can be as large as
10 counts in a few that presumably have poor ground connections. Digital
voltage
supply current in the SVXIIe chip rises steadily if the chip is not properly
initialized or not read out steadily and can cause individual HDIs to trip
off in 5--10 minutes. These currents are constantly displayed and a special
pulser is turned on if there is an extended interruption in data
taking.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure9.eps}}
\caption{Example of the grassy noise seen in the Micron-supplied F-disk
detectors. Ideally, the entire plot would look like the region above channel
800.}
\label{fig:grassy-noise}
\end{figure}
\paragraph{Alignment and calibration}
Signal/noise performance varies with detector type from 12:1 to 18:1.
Coherent noise is typically one-third of the random noise but varies due to
the crossing-dependent pedestal shifts described in
Section~\ref{sec:smt-operation}. Gains vary among detector
types with the n-sides 5--15\% lower than the p-sides due to the larger
load capacitance. Pulse height information from the SVXIIe is used to
calculate cluster centroids and can also be used for d$E$/d$x$ tagging of
low momentum tracks. Figure~\ref{fig:smt-dedx} shows d$E$/d$x$ distributions
after corrections for gain and incident angle are made.
Detector alignment was transferred from optical CMM measurements of detector
fiducials to ladder beryllium features to barrel sapphire balls to
half-cylinder targets to the D\O\ coordinate system.
Final alignment for the combined SMT-CFT tracking system
is better than 10~$\mu$m (Figure~\ref{fig:smt-align}).
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure10.eps}}
\caption{Distribution of energy loss for a kaon-enriched sample
of tracks showing $\pi$, $K$, and proton bands.}
\label{fig:smt-dedx}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure11.eps}}
\caption{Axial residual distribution using tracks with
$p_T > 3$~GeV/$c$ (a) upon initial installation and (b) after software
alignment of the central barrel detectors.
The residual is the distance between the SMT hit and the track. The track
fit was done using all SMT and CFT hits except the SMT hit in question.
The simulated resolution of the residual distribution for a perfectly aligned
detector is about 16~$\mu$m.}
\label{fig:smt-align}
\end{figure}
\subsection{Central fiber tracker}
\label{sec:CFT}
The CFT consists of scintillating
fibers mounted on eight concentric support cylinders and
occupies the radial space from 20 to 52~cm from the center of the
beampipe. To accomodate the forward SMT H-disks, the two innermost cylinders
are 1.66~m long; the outer six cylinders are 2.52~m long. The outer cylinder
provides coverage for $|\eta| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 1.7$. Each cylinder supports one doublet
layer of fibers oriented along the beam direction ($z$) and a second doublet
layer at a stereo angle in $\phi$ of $+3^\circ$ ($u$) or $-3^\circ$ ($v$).
Doublet layers with fibers oriented along the beam axis are referred to as
axial layers, while the doublet layers oriented at small angles are referred
to as stereo layers. From the smallest cylinder outward, the fiber doublet
orientation is $zu-zv-zu-zv-zu-zv-zu-zv$.
The scintillating fibers are coupled to clear fiber waveguides which carry the
scintillation light to visible light photon counters (VLPCs,
Section~\ref{sec:VLPC}) for read out. The small fiber diameter (835~$\mu$m)
gives the CFT an inherent doublet layer resolution of about 100~$\mu$m as
long as the location of the individual fibers is known to better than
50~$\mu$m. Scintillating fiber detectors are discussed in detail in
Ref.~\cite{scint_fiber_ann_rev}.
Discriminator signals from the axial doublet layers are used to form a fast
Level~1 hardware trigger based upon the number of track candidates above
specified $p_T$ thresholds (with a minimum threshold of 1.5 GeV/$c$). Level~1
track candidates are used by the Level~2 trigger, while the Level~3 trigger
uses the full CFT readout information.
\subsubsection{Fibers}
\label{sec:fibers}
The scintillating fibers, including the cladding, are 835~$\mu$m in diameter
and 1.66 or 2.52~m in length.
They are optically connected to clear fiber waveguides of identical diameter
which are 7.8 to 11.9~m long. The fibers were manufactured by
Kuraray \cite{Kuraray} and have a multi-clad structure consisting of a core
surrounded by two claddings. The scintillating fiber is structurally and
chemically similar to
the clear fiber, but contains fluorescent dyes. The CFT uses
about 200~km of scintillating fiber and 800~km of clear fiber.
Light production in the fibers is a multistep process. The base core material
is polystyrene (PS). The PS is doped with the organic fluorescent dye
paraterphenyl (pT) to about 1\% by weight. Excitations in the PS are rapidly
transferred to the pT via a non-radiative dipole-dipole interaction. pT has
a rapid fluorescence decay (a few nanoseconds) and a short emission wavelength
($\approx 340$~nm). The mean free path of the emitted light is only a few
hundred microns in the PS. To get the light out of the detector, a secondary
wave-shifter dye, 3-hydroxyflavone (3HF), is added at a low concentration
(1500~ppm). The 3HF is
spectrally matched to the pT but has minimal optical self-absorption. The 3HF
absorbs the 340~nm radiation from the pT and re-emits it at 530~nm which is
well-transmitted in PS.
Surrounding the PS core, whose refractive index is $n=1.59$, are two claddings,
each approximately 25~$\mu$m thick: an inner layer of polymethylmethacrylate
(PMMA) with $n=1.49$, and an outer layer of fluoro-acrylic with $n=1.42$.
The PMMA inner cladding serves as a mechanical
interface between the core and the outer cladding, which are mechanically
incompatible. The multiclad fiber is both mechanically and optically superior
to single-clad fiber and typical values of the attenuation length are
about 5~m for the scintillating fiber and about 8~m for the clear fiber.
We observe the light from only one end of each scintillating fiber.
The opposite end of each of the scintillating fibers was mirrored with a
sputtered aluminum coating that provides a reflectivity of about
90\%.
The scintillating fibers were assembled into ribbons consisting
of 256 fibers in two layers of 128 fibers each. Precisely spaced grooves were
machined into a long, 1/16"-thick piece of acetal. The spacing between
the grooves varies between 928 and 993~$\mu$m and depends on the
radius of the corresponding support cylinder. The grooved plastic was
inserted into a rigid, curved backing plate of the desired
radius, and the scintillating fibers were laid in and glued together to form the
doublet ribbons; the two layers of fiber are offset by one-half of the fiber
spacing. The technique is illustrated in
Figure~\ref{fig:CFT_ribbon_fabrication}.
It enables curved ribbons to match the curvature of each
support cylinder without machining precisely-spaced grooves into a curved
surface. Details on fiber lengths and spacings are provided in
Table~\ref{tab:CFT_params}.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure12.eps}}
\caption{Technique for curved scintillating fiber ribbon fabrication.}
\label{fig:CFT_ribbon_fabrication}
\end{figure}
\begin{table}
\begin{center}
\caption{Design parameters of the CFT; $u = +3^\circ$, $v = -3^\circ$. A
through H correspond to the eight axial layers of the CFT.}
\label{tab:CFT_params}
\begin{tabular}{lc c c c c}
\hline
Layer & Radius (cm) & Fibers/layer & Fiber separation ($\mu$m)
& Active length (m)\\
\hline
A & 20.04 & $1280 \times 2$ & 982.4 & 1.66 \\
A$u$ & 20.22 & $1280 \times 2$ & 990.3 & 1.66 \\
B & 24.93 & $1600 \times 2$ & 978.3 & 1.66 \\
B$v$ & 25.13 & $1600 \times 2$ & 985.1 & 1.66 \\
C & 29.87 & $1920 \times 2$ & 976.1 & 2.52 \\
C$u$ & 30.05 & $1920 \times 2$ & 980.9 & 2.52 \\
D & 34.77 & $2240 \times 2$ & 974.4 & 2.52 \\
D$v$ & 34.95 & $2240 \times 2$ & 979.3 & 2.52 \\
E & 39.66 & $2560 \times 2$ & 971.7 & 2.52 \\
E$u$ & 39.86 & $2560 \times 2$ & 976.3 & 2.52 \\
F & 44.56 & $2880 \times 2$ & 970.0 & 2.52 \\
F$v$ & 44.74 & $2880 \times 2$ & 974.3 & 2.52 \\
G & 49.49 & $3200 \times 2$ & 969.8 & 2.52 \\
G$u$ & 49.67 & $3200 \times 2$ & 973.3 & 2.52 \\
H & 51.97 & $3520 \times 2$ & 926.1 & 2.52 \\
H$v$ & 52.15 & $3520 \times 2$ & 927.8 & 2.52 \\
\hline
\end{tabular}
\end{center}
\end{table}
The readout ends of the fibers were carefully positioned and adhesively
bonded into v-groove connectors, which are located around the outer
perimeter of the detector, and then the mass-terminated ribbon and
connector were polished to facilitate high efficiency light transmission
across the connector joint. A polished curved connector is shown in
Figure~\ref{fig:CFT_v-groove-connector}. Each 256-fiber
waveguide bundle terminates in a matching curved connector.
The connectors for
each doublet fiber layer are different since the
connectors must have the proper
curvature for each layer. The light transmission through the v-groove
connectors, with optical grease between the fiber ends, is approximately 95\%.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure13.eps}}
\caption{A CFT fiber ribbon mass terminated via a v-groove connector.}
\label{fig:CFT_v-groove-connector}
\end{figure}
After the fiber ribbons were fabricated, a $^{57}$Co x-ray source was used to
verify the accuracy of the fiber placement, the responses of the fibers, and
the transmission efficiency of the connectors. The position of
each fiber within a ribbon was determined with an accuracy of better than
25~$\mu$m rms.
\subsubsection{Mechanical support structure}
The eight support cylinders are each double-walled with a 0.25"-thick core of
Rohacell~\cite{Rohacell}.
The walls are constructed from linear carbon fibers impregnated with about
40\% resin. To minimize sagging of the cylinders, the carbon fibers were
applied in layers in the following pattern:
$0^\circ/60^\circ/-60^\circ/{\rm core}/-60^\circ/60^\circ/0^\circ$
where the angles are those of the carbon fibers with respect
to the axis of the cylinder.
Each carbon fiber layer is about 0.0025" thick; the total
thickness of carbon per cylinder is 0.020". After fabrication, the outside
diameters of the cylinders were measured and compared to a perfect cylinder.
The radial deviation for all cylinders is approximately 100~$\mu$m rms.
The requirements of the Level~1 hardware trigger constrain the design of the
CFT, particularly in the placement of the scintillating fiber ribbons.
Offline reconstruction of the CFT data can incorporate corrections for
misalignment of fibers, but this is not possible in the hardware trigger;
the axial fibers had to be placed on the support cylinders such that any skew
from parallel to the beam axis is less than 200~$\mu$m from end to end.
Consequently, the individual fiber ribbons were precisely positioned and
bonded to the outer surface of each cylinder prior to nesting the cylinders.
The placement precision is approximately 35~$\mu$m rms for both axial and
stereo layers.
The connectors for the axial ribbons are all on one end of the cylinder,
while the connectors for the corresponding stereo ribbons are all on the
opposite end of the cylinder.
Successive cylinders are nested together by thin carbon-fiber annular
rings that connect the inner surface of the end ring of one cylinder to a
carbon-fiber ring mounted on the outer fiber surface of the cylinder
immediately inside. The fiber connectors and interlocking rings seal the
fiber volume.
For tracks traversing the detector at normal incidence, the thickness of each
cylinder can be described as follows: 0.28\% of a radiation length for the
scintillating fibers, 0.32\% for the carbon fiber support cylinder, 0.13\% for
the glue used to make ribbons out of fibers, and 0.17\% for the glue used to
attach the ribbons to the support cylinders.
\subsubsection{Waveguides}
\label{sec:waveguides}
The clear fiber waveguides range in length from 7.8 to 11.9~m, and
use the clear fibers described in Section~\ref{sec:fibers}. The waveguides
generally contain 256 clear fibers inside a flexible protective
plastic light shield. One end is mass terminated in a curved
acetal connector machined to mate to the corresponding connector at
the end of the scintillating fiber ribbon. The mass-terminated
curved connector end was polished using a diamond fly cutter.
About 40~cm from the other end of the waveguide, the 256 fibers are
separated in $\phi$ into two groups of 128 fibers.
Each group is covered with a flexible plastic protective light shield.
These fibers are individually routed into the appropriate locations
in two rectangular connectors
that are designed to mate to the VLPC cassettes (Section~\ref{sec:VLPC}).
These molded plastic rectangular connectors are composed of Noryl N190
\cite{noryl} with Celogen RA foam \cite{celogen} to
minimize distortions.
After the fibers were potted into the rectangular connectors, the
fiber-connector assembly was polished and the routing of the fibers was
verified.
The clear fiber waveguides are routed from the ends of the CFT through the
small gaps between the central and end calorimeter cryostats
to the VLPC cassettes located about 6~m below the central calorimeter cryostat.
These gaps also contain the forward preshower detectors
(Section~\ref{sec:fps}), the waveguides for the forward and central
(Section~\ref{sec:cps}) preshower detectors, and the readout cables for the
SMT. The narrowest region, between the forward preshower
detector and the solenoidal magnet, is about 1.5" wide in $z$, requiring the
waveguides to follow complex paths along the
central calorimeter cryostat. Moreover, the waveguide bundles must have the
correct lengths to provide the proper timing for signals at the
VLPCs. The waveguide routing is illustrated
in Figure~\ref{fig:CFT_fiber_routing}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure14.eps}}
\caption{Routing of the clear waveguide fibers on the south face of the central
cryostat.}
\label{fig:CFT_fiber_routing}
\end{figure}
\subsubsection{Visible light photon counter cassettes}
\label{sec:VLPC}
The light generated by the passage of charged particles through the
scintillating fibers of the CFT is converted
into electrical signals by the VLPCs housed in
the VLPC cassettes. VLPCs are impurity-band silicon avalanche photodetectors
that operate nominally at 9~K, and are capable of detecting
single photons \cite{VLPC1,VLPC2}. They provide fast response, excellent
quantum efficiency ($\geq$75\%), high gain (22,000 to 65,000), low gain
dispersion, and the capability of functioning in a high background
environment.
VLPCs are fabricated by growing a series of doped and undoped
silicon layers on silicon substrate wafers.
Individual wafers yield a maximum of 176 VLPC chips after dicing, and each chip
contains a two by four array of 1-mm-diameter pixels. Each
eight-pixel chip is soldered to an aluminum nitride substrate, and the outputs
from individual pixels are wirebonded to individual contact pads on the
substrate.
Non-uniformities in the production process result in variations
in characteristics such as gain, quantum efficiency, and thermal
noise rates among and across VLPC wafers~\cite{Bross}. Due to these variations,
the bias voltage at which the VLPCs operate at optimal signal-to-noise ratio
varies between 6 and 8 volts. To reduce the complexity of the bias voltage and
threshold implementations in the readout electronics,
the VLPC chips are carefully
sorted and assigned to specific cassettes to allow for optimal performance.
VLPC cassettes
mounted in cryostat slots provide the
mechanical support, optical alignment, and appropriate operating environment
for the VLPCs.
Figure~\ref{fig:cass_overview} shows an outside view of a cassette with a
readout board attached.
Each VLPC cassette houses 128 VLPC chips, and thus provides
1024 individual pixels of light-sensitive detector.
Details of the specially-designed VLPC cryostats are
available in Ref.~\cite{Rucinski}.
Cassettes are approximately 88~cm tall, 48~cm wide, and 4.4~cm thick.
Individual 0.965-mm-diameter fibers inside each cassette guide light from the
clear fiber waveguides to individual VLPC pixels, and flex circuits provide
paths for the electrical signals from the VLPCs to the preamplifiers on the
analog front-end boards (AFEs) that are mounted on the cassette body.
In addition to preamplifiers, the
AFEs also provide trigger discriminator signals, temperature control, and
bias-voltage control electronics.
Details on the AFEs are provided in
Section~\ref{sec:AFE}.
\begin{figure}
\centerline{\includegraphics[angle=180,width=6.in]{figure15.eps}}
\caption{A VLPC cassette supporting AFE readout boards as viewed
from the left side. The VLPC hybrids are located on the isotherms housed
inside the copper cup shown at the bottom of the figure.}
\label{fig:cass_overview}
\end{figure}
Each VLPC cassette consists of a cassette body housing eight modules.
A view of the internal structure of a partially assembled cassette is shown
in Figure~\ref{fig:cass_section}.
Each module is composed of a 128-fiber bundle and a cold-end assembly. The
cassettes are designed so that modules can be readily extracted to allow repair
of the cold-end assemblies. The 128-fiber bundle
terminates at the top end in a rectangular
molded optical connector (referred to as the warm-end optical connector), and
at the bottom (cold) end in sixteen groups of eight fibers. The ends of the
fibers are polished, and each group of eight fibers is glued into a
molded plug. The plugs and mating boxes are precision-molded
parts made of carbon-fiber-loaded polyphenylene sulfide (PPS) plastic.
A polyurethane feedthrough block is
cast around the 128 fibers and the flex circuits to form the barrier between
the warm and the cold ends of the cassette. The fibers accept light from the
clear fiber waveguides which are connected to the warm-end optical
connectors at
the top of the cassette and pipe the light to the VLPCs mounted in the cold-end
assemblies.
\begin{figure}
\centerline{\includegraphics[angle=-90,width=4.5in]{figure16.eps}}
\caption{VLPC cassette body with left side body panel and
side panels removed to show four installed modules.}
\label{fig:cass_section}
\end{figure}
The cold-end assembly hangs from the PPS plugs and consists of sixteen
8-channel VLPC hybrid assemblies supported on an isotherm. The inset in the
lower left corner of Figure~\ref{fig:cass_overview} shows a view of a
cold-end assembly supported on a 128-fiber bundle,
and Figure~\ref{fig:cass_ceexpl}
shows an exploded view of the components of
a cold-end assembly.
\begin{figure}
\centerline{
\includegraphics[angle=270,width=5.5in]{figure17.eps}}
\caption{Cold-end assembly for one 128-channel module of a VLPC cassette.}
\label{fig:cass_ceexpl}
\end{figure}
A VLPC hybrid assembly is composed of a PPS box adhesively bonded to the
aluminum nitride substrate upon which a VLPC chip is mounted.
The molded PPS box is precisely positioned (maintaining 25~$\mu$m tolerances)
so that the eight polished fibers in the PPS plug are aligned over the
individual VLPC pixels when the plug and box are mated, thus achieving good
light collection efficiency while minimizing optical crosstalk between
neighboring channels. Connectors are clamped across the contact pads on the
aluminum nitride substrates and mated to flex circuits which transmit
single-ended electrical signals between
the cold-end and the AFE. Each module contains two flex
circuits, and these flexible printed circuits are realized on 5-$\mu$m-thick
adhesiveless copper coatings on a 51-$\mu$m-thick
polyamide base material. The individual traces are 76~$\mu$m
wide with minimum 89~$\mu$m
spacings between circuit features. Each flex circuit is 41.1~cm
long.
The isotherm also supports a calibrated carbon resistor that
serves as a temperature sensor,
resistors used as heaters, and the required springs and
fasteners. The temperature sensor and heaters are employed to control the
temperature of the VLPCs to within 0.1~K.
The cassette body can be viewed as composed of cold-end and warm-end mechanical
structures. The cold-end structure (that portion of the cassette that is
inserted in the cryostat) comprises several sub-assemblies: the
feedthrough assembly, the G-10 side walls, the heat intercept assemblies, and
the cold-end copper cup. The cassette is mounted to the top plate of the
cryostat at the feedthrough assembly. This assembly provides the gas-tight
seal for the cold, $\leq2$~psig, stagnant helium gas
volume within the cryostat.
Side walls composed of G-10 support the top (or liquid nitrogen)
intercept which serves to cut off the flow of heat from the warm end.
Another set of G-10 side walls support the bottom (or liquid helium) intercept.
The bottom intercept supports the copper cup which surrounds the isotherms at
the ends of the modules.
The warm-end structure is made of parallel tin-plated
aluminum panels separated by spacer
bars that form a protective box for the optical fibers. Rails mounted on the
aluminum panels support the two AFEs.
The warm-end Cin::apse \cite{Cinapse} connector assembly
enables the connection between the flex circuits and the AFEs.
The AFEs mate to
backplane connectors mounted on a backplane support structure via card-edge
connectors. The backplane support structures are bolted to the VLPC cryostats,
and thus the combination of cassettes and
cryostat serve as the crates for the AFEs which are mounted
on the cassettes. This design, with connections on two orthogonal edges
of the AFE, allows the AFEs to be
removed for service without extracting the cassette from the cryostat.
The CFT requires 76,800 channels of VLPC readout,
and the central and forward preshower detectors (Section~\ref{sec:preshower})
are instrumented with
an additional 22,564 channels of VLPC readout. The ninety-nine cassettes
necessary to provide this readout are mounted in two custom-designed cryostats
located on the platform beneath the D\O\ detector.
Over 99.8\% of the individual VLPC channels in these cassettes met or
exceeded the desired performance specifications
during cryogenic qualification tests performed prior to installation
at D\O.
\subsubsection{CFT readout electronics}
\label{sec:AFE}
The CFT and the central and forward preshower detectors
(Section~\ref{sec:preshower}) share a VLPC-based readout as well as similar
Level~1 and Level~2 trigger electronics, and therefore all three use
the same front-end electronics to process the signals from the VLPCs.
The front-end electronics are custom printed circuit boards (the AFEs)
approximately 14" tall (9U) and 18" long which are mounted on the VLPC
cassettes inserted into cryostats
as described in Section~\ref{sec:VLPC}.
The AFE is a large and complex board that must perform a number of
functions with competing requirements. It has charge-sensitive
amplifiers to deal with the very small signals from the VLPCs. It is
part of the Level~3 readout, part of the Level~1 and Level~2 triggers, and part
of the slow control and monitoring system. It must also control the bias and
temperature of the VLPCs. This functionality must be embedded in the
AFE because it is the only piece of electronics that interfaces to the VLPCs.
It must serve three different sub-detectors with different dynamic range
requirements and comes in two types --- left-handed and right-handed,
depending on which side of the cassette body the board is mounted.
A number of features make it possible for the AFE to fulfill all
of the above requirements. First, only a single printed circuit board
was designed and laid out. The ``handedness'' of the board is determined by
the way a few key components are installed. So, for example, there
are mounting points and traces for the hard metric connectors on both
ends of the board. When the connector is mounted on one end of the
board, the board becomes left-handed and mounts on the left side of
the cassette. When the connector is mounted on the other end of the board,
the AFE is
right-handed. The great majority of the components are mounted exactly
the same way on both left- and right-handed AFEs.
Second, the most noise sensitive parts of the board,
including the front-end amplifier chips, are mounted on separate multi-chip
modules (MCMs), each with its own regulator for the power and
separate ground planes. The amplifier chips are wire bonded directly to
these much smaller (3.5" by 1.5") subassemblies. This allows
the very fine pitch required for wire bonding to be confined to only
the MCM substrate. Otherwise the AFE boards would not be manufacturable.
This also allows the MCM subassemblies to be tested separately from the
AFEs and to be removed as required for repair or replacement. There are
eight MCMs on each AFE, each serving sixty-four VLPC channels, to match the
construction of the cassette modules as described in Section~\ref{sec:VLPC}.
The MCMs are intended to be identical. To accommodate the different dynamic
ranges of the CFT, the central preshower detector, and the forward preshower
detector, capacitive charge division is used.
By properly sizing the AC coupling capacitors on the AFE, the
charge seen by the amplifiers is adjusted to accommodate the
required dynamic range
with the same electronics.
To further reduce the cost and simplify the system, the front-end
amplifier and digitizer is the same chip used for the SMT.
This also facilitates commonality further downstream in the
readout chain. The SVXIIe and its readout are described in
Section~\ref{sec:SVXII}. On the AFE, the SVXIIe provides for the integration
of the charge signals from the VLPCs, a pipeline for storing the signals
while the trigger is formed, and digitization of the signals and
sparsification of the digitized data for readout. There are eight SVXIIe chips
on every AFE, one on each MCM. The SVXIIe chips share a single 8-bit bus for
readout. However, because the signals from the AFE are also needed for
the trigger system and because the SVXIIe digitization speed is too slow to
generate a trigger decision, a second chip, the SIFT chip,
provides a trigger pickoff. A simplified schematic is shown in
Figure~\ref{fig:sift-schematic}.
Each SIFT chip has eighteen channels so there are four SIFT chips on every
MCM (eight channels are unused). The SIFT chips receive the signals from the
VLPCs before the SVXIIe does. For each channel, the SIFT has a preamplifier
that integrates the incoming charge and switched capacitors that are used to
split the amplified signal and send it along two paths: one to the SVXIIe,
for subsequent digitization and readout, the other to a
discriminator which fires if the charge exceeds a preset threshold
(typically 10 to 15~fC).
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure18.eps}}
\caption{Simplified schematic diagram of the SIFT chip.
Vclamp, Vthr, and VREF are externally-supplied DC voltage levels.}
\label{fig:sift-schematic}
\end{figure}
There are five clock signals that control the operation of the SIFT,
as can be seen from the schematic. One is the reset of the integrator,
PRST, another is the reset for the discriminator, DRST. The third, SH
(sample and hold) samples the discriminator output and the analog value
of the integrator at the end of the cycle. READ transfers the analog
charge to the SVXIIe. PCLMP resets the bias for the followers in the analog
chain. All of these signals are derived from the main accelerator clock
(53~MHz).
The digital signals from the discriminators serve as the
inputs to the Level~1 central track trigger (L1CTT, Section~\ref{sec:l1ctt}))
and a bit is sent to the trigger system for every axial fiber every 132~ns
(the time originally anticipated between beam crossings). This means that each
AFE board is transmitting data at about 4.1 Gbits/s.
The formatting and pipelining
of this data is accomplished by sixteen CPLDs (complex programmable logic
devices) in conjunction with four fast FIFO (first-in, first-out) memories.
The trigger data is sent over four LVDS (low-voltage differential signal)
links. Each LVDS link is driven by Texas Instruments SN65LVDS95 transmitter
chips~\cite{cft-transmitter-chip}. Each chip has 21 bits of parallel-load,
serial-out shift registers driven by a $7\times$ clock synthesizer
and drives four LVDS lines. Three line pairs are used to transmit 21 bits,
and the fourth carries the clock. With the data loaded into the transmitter
chips with the 53~MHz clock, the AFE can send $21\times7\times4= 588$\
bits every 132~ns. One bit on each link is used to carry the 53~MHz clock to
help synchronize the data frames at the receiver end. This leaves
$20\times7\times4= 560$\ bits for trigger information. The actual
discriminator information is carried by 512 bits. The other bits
are used to send control information to the L1CTT.
These include indication of which crossings had beam present, when a
Level~1 trigger accept was received from the trigger system, when the AFE is
in a test mode or in reset, and so on. The control bits are duplicated on each
of the four links and this redundancy is used to provide diagnostic
information about the integrity of the links from the AFE to the L1CTT.
There are twenty CPLDs on each AFE. Besides the sixteen used to handle
the discriminator data, one CPLD is used as a ``virtual'' SVXIIe chip to allow
stored discriminator data to be read out on the same bus as the SVXIIe chips.
Another CPLD is used to generate all the signals required to operate the SIFT
chips and other timing signals for the board. The last two CPLDs work in
conjunction with a microcontroller on the AFE to implement communications
with the monitoring and slow control system via the MIL-STD-1553B bus. The
microcontroller is a mixed signal controller PIC14000 from Microchip
\cite{Microchip}.
It has an integrated single-slope 16-bit ADC which is used to monitor
analog values on the board such as VLPC bias, VLPC current, and VLPC
temperature. The temperature of the VLPCs is measured via calibrated
carbon resistors mounted near the VLPC chips on each module. The value
of the resistance must be measured to better than one part in a thousand
and this drives the requirement for the ADC. The microcontroller
implements closed-loop control for individual heater resistors on each
VLPC module which allows the temperature at the VLPCs to be kept constant
to $9.00\pm0.05$~K despite larger cryostat temperature fluctuations. The
VLPC temperature monitoring and control is performed by the right-handed
AFE boards. The VLPC bias voltage (which varies between 6 and 8 volts
from module to module) must be controlled to approximately 50~mV to achieve
optimal detector performance.
The extensive use of programmable logic on the AFE (twenty CPLDs and a
microcontroller) greatly eased the tasks of designing, building, and
operating it. The design of the AFE could be completed before all details of
the interconnected detector systems were finalized. A single printed circuit
board is flexible enough to instrument three very different detectors.
Moreover, this intricate board operates with little external control:
once set points and operating parameters are
downloaded, firmware on the board controls the system using on-board DACs
and ADCs. We are also able to modify the firmware to optimize the
operation of the electronics as running conditions change.
The performance of the electronics is dictated by the physics requirements
of the detector. The small-diameter scintillating fibers used in the
CFT coupled with the long waveguides necessary
to carry the signals to the platform cause the signals generated by the
VLPCs to be small ($\approx$10 photoelectrons incident upon the VLPCs).
To assure acceptable efficiency for triggers and tracking, the
individual channel thresholds must therefore be set between 1.5 and
2~photoelectrons (pe). To maintain a low and stable threshold and
to distinguish individual photoelectron peaks
during calibration, the analog signal must be digitized with a noise of less
than 0.4 pe, which is equivalent to about 2~fC.
In fact, the front-end electronics achieve or exceed all requirements.
The mean pedestal width from fits to LED calibration spectra
(Section~\ref{sec:CFT-LED}) for all axial fibers is 0.24~pe or 1.6~fC,
and discriminator thresholds have
similar noise and offsets of less than 2~fC so that it is possible to set
discriminator thresholds below 10~fC for most channels. A sample spectrum
from calibration data
is shown in Figure~\ref{fig:LED-spectrum}. A summary of the important
parameters for the AFE is given in Table~\ref{tab:AFT-parameters}.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure19.eps}}
\caption{A typical LED spectrum for a single VLPC for an axial CFT fiber.
Every channel is fit automatically and the parameters of the fit are extracted
and used for monitoring. Typically, more than 97\% of the axial
channels are fit successfully. The solid histogram is the data; the smooth
curve is the fit.}
\label{fig:LED-spectrum}
\end{figure}
\begin{table}
\begin{center}
\caption{Summary of AFE board characteristics}
\label{tab:AFT-parameters}
\begin{tabular}{l p{7.cm} r}
\hline
\multicolumn{2}{l}{AFE overview} & \\
\hline
\phantom{X}
& Number of readout channels per AFE & 512 \\
& Number of MCMs per AFE & 8 \\
& Total number of individual components on AFE & 2300 \\
& \raggedright Threshold settings \\
(analog V generated by onboard DACs) & 4 per MCM \tabularnewline
& \raggedright Pedestal settings \\
(analog V generated by onboard DACs) & 1 per MCM \tabularnewline
& Output bandwidth (L1 trigger) & 4200 Mbit/s \\
& Output bandwidth (SVXIIe bus) & 159 Mbit/s \\
\multicolumn{2}{l}{AFE performance} & \\
\hline
& Noise (input referred) --- analog readout & 1.6 fC \\
& Noise (input referred) --- discriminator & 1.2 fC \\
& Threshold spread (rms) & 2 fC \\
& Pedestal spread (rms) & 3 fC \\
\multicolumn{2}{l}{AFE stability and reliability} & \\
\hline
& Mean time between failures (estimated) & 35000 hrs/board \\
& Calibration stability & Better than 1\% \\
\hline
\end{tabular}
\end{center}
\end{table}
The readout of the SVXIIe chips mounted on the AFE boards is coordinated
via signals from the sequencers and sequencer controllers.
The sparsified data is transferred
from the sequencers to VME transition modules (VTM)
via fiber optics cables. The VTMs transfer the data to buffers in
VRBs in VME VIPA \cite{VIPA} crates. A VRBC mounted in the same crate
controls the
assignment of buffers. A single-board computer mounted in that crate
serves as the interface to the higher level readout. The readout of the
CFT and the preshower detectors is buffered in four VME crates.
Details of the higher level readout system are provided in
Sections~\ref{sec:SVXII} and \ref{sec:daq}.
\subsubsection{Calibration}
\label{sec:CFT-LED}
The CFT and the preshower detectors (Section~\ref{sec:preshower}) are each
equipped~\cite{Baumbaugh} with fast Nichia \cite{Nichia} blue-emitting LED
pulsers that can be used to perform several functions. The pulsers certify
the proper connectivity and yes-no operation of the fiber channels and
attached readout chain, provide a means for monitoring the stability of the
VLPC readout over time, and calibrate the response of individual channels or,
more specifically, provide a relative channel-to-channel energy calibration
of the fiber-VLPC system. This monitoring and calibration is done
during detector commissioning or between collider stores. Although the
function of the LED system is the same for all three detectors, the method
of light injection differs for each.
The limited space at the ends of the CFT cylinders complicates
injection of light directly into the ends of the scintillating fibers.
However, a method has been devised to introduce light through the
cladding walls and into the active core of the fiber. The method takes
advantage of the fact that the 3HF dye in the CFT fibers has a small
absorption in the blue wavelength range ($\approx 450$~nm).
Blue light incident on several layers of the scintillating fiber
can be absorbed in the top layer, exciting the 3HF dye and
producing scintillation photons detectable by the VLPC.
To distribute the light over the cylindrical geometry of
the fiber tracker, flat optical panels consisting of a single
thin ribbon of 100 (typically) clear fibers, each
approximately 500~$\mu$m in diameter, are degraded in certain local
regions so that light is emitted through their cladding walls.
The fibers are bundled together at one end of the ribbon, potted, and
finished to allow coupling with an LED. The flat
optical panels are then mounted around every
fiber cylinder near each end of the cylinder and driven by pulsers.
The use of two bands provides not
only a margin of safety against failure of an individual optical
element but also allows comparison of the light from the two ends of the
tracker and monitoring of any changes in the quality
of the scintillating fibers over time.
Although similar flat optical panels could have been adapted to the preshower
detectors, the CPS and FPS use different designs. For the preshower detectors,
injection of LED light directly into the scintillator strips
complicates the design appreciably, so instead, light is injected into the
wavelength shifting (WLS) fibers.
Since the WLS fibers are grouped in sets of sixteen channels
and routed from the scintillator strip to connectors located at the perimeter
of the detectors, one LED
illuminates a group of sixteen WLS fibers. The mechanism, however, for
delivering light into these sixteen fibers differs for CPS and FPS.
In the CPS, two LEDs are mounted inside a machined Lexan light block,
directly before the outer perimeter of the detector, one on either side of
the connector holding the WLS fibers. Ten LEDs mounted on one side
of the connector are chained in series, allowing for an RG-178 cable to supply
power to each LED string. The series of grouping ten
LEDs continues across the perimeter of the CPS detector.
LEDs for the FPS system~\cite{patwa-thesis} are also mounted on the outer
perimeter of the detector.
However, due to the tight spatial constraints inside the FPS
scintillating wedges, a 0.835-mm-diameter clear fiber transports the
blue light to a group of sixteen WLS fibers, systematically arranged
within a hollow cylindrical cavity. The clear fiber is
inserted directly into the central hole of the cavity, allowing for
light to be delivered uniformly over a full solid angle to the sixteen
WLS fibers. The cavity holds the
same group of WLS fibers that are routed from their respective
scintillator strips to optical connectors coupling the clear fiber
waveguides. The LEDs themselves are mounted on two PC boards,
each providing the relevant electrical circuitry to drive the device.
PC boards located on the readout platform supply power to the LED system of
each detector. The entire system is
computer-controlled and can be operated directly from the control room,
allowing users to program the pulse height and width as well as
operate one LED in a detector at a time or the complete array.
\FloatBarrier
\section{Solenoidal magnet}
\label{sec:solenoid}
The superconducting solenoidal magnet was designed \cite{magnet_cdr,solenoid}
to optimize the momentum resolution,
$\delta p_T/p_T$, and tracking pattern recognition within the constraints
imposed by the Run~I detector.
The overall physical size of the magnet was determined by the available
space within the central calorimeter vacuum vessel: 2.73 m in length and
1.42 m in diameter. We selected a central field of 2 T after considering
the momentum resolution and tracking pattern recognition, the available space,
and the thickness of the cryostat which depends on the thicknesses of the
conductor and support cylinder. In addition, the magnet is required
{\it i}) to operate safely and stably at either polarity, {\it ii})
to have a uniform field over
as large a percentage of the volume as practical, {\it iii}) to be as thin as
possible to make the tracking volume as large as possible, {\it iv}) to
have an overall thickness of approximately $1X_0$ at $\eta = 0$ to optimize the
performance of the central preshower detector mounted on the outside of the
solenoid cryostat, and {\it v}) to quench safely without
a protection resistor (although one is installed to reduce the recool time
after an inadvertent fast dump). Services such as
cryogens, magnet current buses, and vacuum pumpout and relief must
reach the magnet from the control dewar through the narrow space (7.6~cm)
between the central and end
calorimeter vacuum vessels. The magnet system is controlled
remotely, including cool down, energization, de-energization for field
reversal, quench recovery, and warmup, without access to the magnet cryostat,
service chimney, or control dewar.
The major parameters of the solenoid design are listed in Table
\ref{tab:solenoid_params}. A perspective view of the solenoid inside the
central calorimeter with its chimney and control dewar is shown in Figure
\ref{fig:solenoid_perspective}.
\begin{table}
\begin{center}
\caption{Major parameters of the solenoid}
\label{tab:solenoid_params}
\begin{tabular}{ll}
\hline
Central field & 2.0 T \\
Operating current & 4749 A \\
Cryostat warm bore diameter & 1.067 m \\
Cryostat length & 2.729 m \\
Stored energy & 5.3 MJ \\
Inductance & 0.47 H \\
Cooling & Indirect, 2-phase forced flow helium \\
Cold mass & 1460 kg \\
Conductor & 18-strand Cu:NbTi, cabled \\
Conductor stabilizer & High purity aluminum \\
Thickness & 0.87 X$_0$ \\
Cooldown time & $\le 40$ hours \\
Magnet charging time & 15 minutes \\
Fast discharge time constant & 11 seconds \\
Slow discharge time constant & 310 seconds \\
Total operating heat load & 15 W plus 0.8 g/s liquefaction \\
Operating helium mass flow & 1.5 g/s \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure20.eps}}
\caption{Perspective view of the solenoid inside the central calorimeter.
One end calorimeter, several muon chambers, and parts of the toroids have been
omitted for clarity. Also shown are the service chimney and control dewar.}
\label{fig:solenoid_perspective}
\end{figure}
The solenoid, along with its cryostat, control dewar, and connecting service
chimney, was manufactured by Toshiba Corp.~\cite{Toshiba} in Yokohama, Japan.
The system was specified to operate safely and reliably over a twenty-year
lifetime with up to 150 cool-down cycles, 2500 energization cycles, and 400
fast dumps.
\subsection{Magnet construction}
The solenoid is wound with two layers of superconductor to achieve the required
linear current density for a 2 T central field. The support cylinder is located
on the outside of the windings to support the radial Lorentz forces on the
conductor, to provide axial rigidity to the finished coil, to provide a high
thermal conductivity path to the helium coolant piping, and to ensure reliable
quenchback for quench safety. To maximize the
field uniformity inside the bore of the magnet, the current density in the
windings is larger at the ends of the coil. This is accomplished by the use
of a narrower conductor at the ends of the coil.
Cross sectional views of the conductors are shown in
Figure~\ref{fig:solenoid_conductor}. Both conductors are made with a
superconducting Rutherford-type cable of multifilamentary
Cu:NbTi strands stabilized with pure aluminum. The basic strand has a
Cu:NbTi ratio of 1.34:1 and a diameter of 0.848~mm; eighteen strands are used
in each conductor.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure21.eps}}
\caption{Cross sections of the two conductors used in construction of
the solenoidal magnet.}
\label{fig:solenoid_conductor}
\end{figure}
Both conductors are used in both layers. The middle section of each
layer is wound with the wider conductor and the end sections with the narrower
conductor. The transition point between the two conductors in the inner layer
occurs at $z=\pm 0.941$~m, and at $z=\pm 0.644$~m in the outer
layer. At the four locations where the conductor width changes, the two types
of conductor are joined with a soldered lap joint one full turn in length.
There are no other joints in the coil.
\subsection{Magnet cryostat}
The magnet cryostat consists of four major components: the vacuum vessel, the
liquid-nitrogen-cooled radiation shield, the cold mass support system with
liquid-nitrogen-cooled intercepts, and the helium cooling tube on the outer
support cylinder of the superconducting coil.
The vacuum vessel consists of inner and outer coaxial aluminum shells with flat
annular
bulkheads welded to each end. The superconducting buses from the coil and the
cryogen pipes from the outer support cylinder and the radiation shields leave
the vacuum vessel through the service chimney nozzle welded in the bulkhead at
the south end of the cryostat. The cryostat is designed for full internal
vacuum and for an internal relieving pressure of 0.44 MPa (6.4 psig).
The magnet cold mass, the superconducting coil and outer support cylinder,
weighs 1.46 metric tons. The cold mass support system consists of axial members
which locate the coil axially and support it against axial thermal, decentering,
and seismic forces, and nearly tangential members that locate the coil radially
and provide radial support against thermal, gravitational, seismic, and
decentering forces. The support members connect the outer support cylinder of
the coil to the flat annular bulkheads of the vacuum vessel. All cold mass
supports have thermal intercepts that operate near 85~K. The magnet cryostat
is attached to the central calorimeter by support brackets that carry the
weight of the cryostat and the tracking detectors that are attached to it,
plus any residual magnetic decentering loads from the muon steel.
\subsection{Control dewar and refrigeration system}
The control dewar is the interface between the fixed cryogenic piping
and the detector, which must be moved from the assembly area into the collision
hall. It has bayonet connections for cryogenic lines
and contains the vapor-cooled current leads and a liquid-helium reservoir,
helium supply subcooler, helium flow control valve, and other valving and
instrumentation. The current leads carry the full operating current of
the magnet at the optimum cooling gas flow rate and are designed for safe
operation without cooling gas flow for at
least the full slow-discharge time constant of the magnet (approximately 310~s)
in the event that cooling flow is lost while the magnet is de-energized.
Cryogenic and electrical services are carried from the control dewar to the
magnet cryostat via the service chimney, which also
serves as the relief line for the solenoid vacuum space and provides a
path for pumpout of the magnet cryostat and control dewar vacuum spaces.
A cryogenic refrigeration system supplies liquid nitrogen and liquid helium to
the magnet and the VLPC cryostats for the CFT and central and
forward preshower detector readouts (Section~\ref{sec:CFT}).
The refrigeration requirements of the magnet and VLPC systems,
as well as operation of the detector in the collision hall and adjacent
assembly hall, were considered when the refrigeration system was designed.
A standard Fermilab satellite stand alone refrigerator (STAR) provides helium
refrigeration. The capacity of the STAR is sufficient for non-simultaneous
cooldown and simultaneous operation of both the solenoid and VLPC systems.
Liquid helium is stored in a 3000~L dewar which supplies the magnet control
dewar and the VLPC cryostats via separate transfer lines.
\subsection{Magnet energization and control}
A block diagram of the DC energization system is shown in Figure
\ref{fig:solenoid_energization}. The power supply is a special
Fermilab unit designed for superconducting loads. It is a
twelve-phase-thyristor water-cooled rectifier unit with precision feedback
current regulation.
The power supply taps are set at 15 V/5000 A for efficient
operation and reduced AC loading and DC ripple.
The power supply regulates the current to within 0.01\%
using an external precision Holec~\cite{holec}
5000~A direct current current transducer (also known as a DCCT or transductor)
installed downstream of the ripple filter and dump resistor.
\begin{figure}
\includegraphics[width=6.in]{figure22.eps}
\caption{Block diagram of the DC energization system for the solenoid.}
\label{fig:solenoid_energization}
\end{figure}
Magnet polarity is reversed using a 5000~A DC mechanical motorized polarity
reversing switch and a switch controller. Polarity reversal requires
complete discharge of the solenoid followed by recharging at the opposite
polarity. The controller confirms that the
polarity reversal occurs at zero current load and that the power supply is
turned off. Polarity reversal takes about forty minutes.
The Texas Instruments TI565T Programmable Logic Controller (PLC) (originally
installed to operate the liquid argon calorimeter refrigeration system, the
muon chamber gas systems, and various building utility and safety systems)
has been expanded to control and monitor the VLPC cryostats, solenoid, and
helium refrigeration systems. The PLC is a stand-alone system with an internal
battery-backed program memory that requires no host system to download the
control programs in the event of a power failure. It consists of two
independent TI565T processors operated in a ``hot backup'' configuration. One
PLC is online and actively controlling the system; the second is in standby,
running step-for-step with the online unit but with its input/output
communication disabled. Each PLC runs continuous internal diagnostics for
errors. When a fatal error is registered by the active PLC, it is taken offline
and the standby PLC is put online without intervention or disruption to the
system.
Dedicated control and monitoring of the magnet
energization and protection system is done by a new Texas Instruments TI555
PLC. A quench protection monitor (QPM) hardwired chassis and interlock logic
unit provide primary quench protection for the magnet. The QPM contains
filtering, signal averaging, and delay circuitry for the voltage taps and
thermometry which are used in the quench detection logic, vapor-cooled lead
fault detection, or power supply failure logic. It contains preset limits for
selected sensors which trigger fast or slow discharge of the magnet and it
preserves the time ordering of the detected fault signals which trigger magnet
discharge. The QPM defines a set of interlock conditions that must be
satisfied before the magnet can be energized or the reversing switch operated.
The cryogenic controls and PLCs are powered by a 10~kW uninterruptable power
supply that is backed up by an automatic diesel power generator. The total
power consumption is approximately 6~kW.
\subsection{Magnetic field}
The magnetic field of the full magnet system is modeled using the TOSCA
\cite{tosca} program. The calculated field map was compared with the measured
field in two locations: near the internal radius of the solenoid cryostat
($r\approx 54$~cm) at $z=4$~cm and in the gap (at $(x,y,z) = (0,372,105)$~cm)
at the top of the central muon toroid steel (Section~\ref{sec:toroids}).
Within the solenoid (operated at
4749~A), the measured field is $20.141\pm0.005$~kG;
the calculated field at this location is 20.158~kG. The calculated magnetic
field is scaled by 0.09\% to agree with the measurement. With full
operating current (1500~A) in the toroid coils, there is a 4.3\% difference
between the calculated and measured field at the central toroid gap,
requiring an adjustment in the calculated field for this magnet.
The $y-z$ view of the magnetic field with both the toroid and solenoid magnets
at full current is shown in Figure~\ref{fig:solenoid-map}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure23.eps}}
\caption{The $y-z$ view of the D\O\ magnetic field (in kG) with both the
toroidal and solenoidal magnets at full current (1500~A and 4749~A,
respectively). The field in the central toroid is approximately 1.8~T; that in
the end toroids is about 1.9~T. The field lines are projections
onto the $y-z$ plane; the left and right line ends differ by up to 2.5~m in
$x$.}
\label{fig:solenoid-map}
\end{figure}
\FloatBarrier
\section{Preshower detectors}
\label{sec:preshower}
The preshower detectors aid in electron identification and background rejection
during both triggering and off\-line reconstruction. They function as
calorimeters as well as tracking detectors, enhancing the spatial matching
between tracks and calorimeter showers~\cite{PS_cosmic_test}. The detectors
can be used offline to correct the electromagnetic energy measurement of the
central and end calorimeters for losses in the solenoid and upstream material,
such as cables and supports. Their fast energy and position measurements
allow preshower information to be included in the Level 1 trigger. The
central preshower detector (CPS) covers the region $|\eta| < 1.3$ and is
located between the solenoid and the central calorimeter with an inner radius of
28.25'' and an outer radius of 29.21''. The two
forward preshower detectors (FPS)~\cite{patwa-thesis}
cover $1.5 < |\eta| < 2.5$ and are attached to
the faces of the end calorimeters. The preshower detectors
can be seen in Figure~\ref{fig:tracker}.
\subsection{Common properties of the preshower detectors}
Both preshower detectors are made from triangular strips of
scintillator, as shown in Figure~\ref{fig:ps-cross-section}. Since the
triangles are interleaved, there is no dead space between strips and most tracks
traverse more than one strip, allowing for strip-to-strip interpolations and
improved position measurement. The strips are
made by extruding polystyrene plastic doped with 1\% p-terphenyl and
150~ppm diphenyl stilbene, with a light yield similar to that of commercial
Bicron BC-404 scintillator. Each scintillator strip is machine-wrapped in
aluminized mylar for optical isolation, and the ends are painted
white to enhance reflectivity. The
packing density is different for the CPS and the FPS modules, resulting in
different layer thicknesses and strip pitches. Because of the nesting process,
which requires epoxying the strips together to form a layer, the
measured pitch can differ by up to 20\% from the design dimensions shown in
Figure~\ref{fig:ps-cross-section}.
After extrusion and wrapping, the triangular strips have a tendency to bow. To
relieve stress in the plastic, making them easier to handle, the strips
were slumped to the required shapes by heating to about $180^\circ$~F for two
hours followed by gradual cooling at room temperature.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure24.eps}}
\caption{Cross section and layout geometry of the CPS and FPS scintillator
strips. The circles show the location of the embedded wavelength-shifting
fibers. Design dimensions are shown.}
\label{fig:ps-cross-section}
\end{figure}
Embedded at the center of each triangular strip is a wavelength-shifting
fiber that collects and carries the light to the end of the detector.
The non-readout ends of the WLS fibers are diamond-polished and silvered. At
the readout end, fibers are grouped into bunches of sixteen and potted into
connectors for transition to clear light-guide fibers. Light is transmitted
via the clear fibers to VLPC cassettes (Section~\ref{sec:VLPC}) for readout.
Both the WLS and clear fibers are 835-$\mu$m-diameter Kuraray multiclad fibers.
The preshower detectors share common elements with the central fiber tracker,
beginning with the waveguides and continuing through the entire readout
electronics system. The last elements which are unique to the preshower
detectors are the connections between the wavelength shifting fibers and the
waveguides. The waveguides, VLPC cassettes, and readout electronics are
described in Sections~\ref{sec:waveguides} --
\ref{sec:AFE}. The calibration systems for the preshower detectors are also
similar to that of the CFT and are described in Section~\ref{sec:CFT-LED}.
\subsection{Central preshower detector}
\label{sec:cps}
The CPS consists of three concentric cylindrical layers of triangular
scintillator strips and is located in the nominal 5~cm gap between the
solenoid and the central calorimeter. Between the solenoid and the CPS is a
lead radiator 7/32" thick (approximately 1 radiation length ($X_0$)) and 96"
long, covering $|\eta| < 1.31$. The lead is covered by stainless steel skins
1/32" thick and 103" long. The solenoid itself is $0.9X_0$ thick,
providing a total of about two radiation lengths of material for particles at
normal incidence, increasing to about four radiation lengths at the largest
angles.
The three layers of scintillator are arranged in an axial-$u$-$v$
geometry, with a $u$ stereo angle of $23.774^\circ$ and a $v$ stereo angle of
$24.016^\circ$. Each layer contains
1280 strips. The WLS fibers are split at $z=0$ and read out from each end
resulting in 2560 readout channels/layer.
The geometry of the CPS axial layer matches that of the CFT for Level~1
(Section~\ref{sec:l1})
triggering purposes. Each group of sixteen WLS fibers from the CPS axial layer
corresponds to one of the eighty CFT sectors in $\phi$. As with the CFT, the
stereo layers are not used in the Level~1 trigger. However, unlike the CFT,
the stereo layers of the CPS are used in the Level~2 trigger.
Each layer is made from eight octant modules. The modules consist of two
1/32"
stainless steel skins with the scintillator strips sandwiched in between. The
ends of the stereo modules align to the ends of alternating axial modules, as
shown in Figure~\ref{fig:cps-unwrapped}. The modules are attached directly to
the solenoid by bolts at each corner. Eight $\quart$" pins at each end of the
solenoid provide additional registration. Connector blocks are spot welded
between the stainless steel skins. The blocks provide structural integrity to
the end region of the modules and mounting locations for the WLS connector and
the cover for the light-guide connector.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure25.eps}}
\caption{The CPS unwrapped in a plane. Each rectangle (trapezoid) is one
octant module of the axial (stereo) layer. Note that the stereo octant edges
are precisely aligned with the axial octant edges.}
\label{fig:cps-unwrapped}
\end{figure}
\subsection{Forward preshower detector}
\label{sec:fps}
The two FPS detectors (north and south) are mounted on the spherical heads of
the end calorimeter cryostats, occupying the region between the luminosity
monitor (Section~\ref{sec:lum-monitor}) at the inner edge and the
intercryostat detectors (Section~\ref{sec:icd}) at the outer edge. Each
detector is made from two layers, at different $z$, of two planes of
scintillator strips. A $2X_0$-thick lead-stainless-steel absorber separates
the two layers, as shown in Figure~\ref{fig:fps-module}. The upstream layers
(those nearest the interaction region) are known as the minimum
ionizing particle, or MIP, layers while the downstream layers behind the
absorber are called the shower layers. Charged particles passing through
the detector will register minimum ionizing signals in the MIP layer, allowing
measurement of the location (in $\eta$, $\phi$, and $z$) of the track.
Electrons will readily shower in the absorber, leading to a cluster of energy,
typically on the order of three strips wide, in the shower layer that is
spatially matched with the
MIP-layer signal. Heavier charged particles are less likely to shower,
typically
producing a second MIP signal in the shower layer. Photons will not generally
interact in the MIP layer, but will produce a shower signal in the shower
layer. The shower layers cover the region $1.5 < |\eta| < 2.5$ and the
MIP layers and the absorber cover the region
$1.65 < |\eta| < 2.5$. The outer region of the FPS, $1.5 < |\eta| < 1.65$,
lies in the shadow of the solenoidal magnet coil, which provides up to $3X_0$
of material in front of the FPS. This amount of material readily induces
showers that can be identified in the shower layers of the detector.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure26.eps}}
\caption{Complete $\phi$-segment of a FPS module showing the overlapping
$u-v$ MIP and shower layers, separated by a lead and stainless steel absorber.}
\label{fig:fps-module}
\end{figure}
Each FPS detector has four measuring planes: MIP $u$ and $v$
and shower $u$ and $v$. Each measuring plane is constructed from two layers,
each containing eight 22.5$^\circ$ wedges (Figure~\ref{fig:fps-module-3d}) of
active material separated by eight
wedges of inactive material. The active material consists of two sublayers of
nested scintillator strips with a stereo angle of 22.5$^\circ$ with
respect to one another. The two layers are separated in $z$; the WLS
fibers are brought to the periphery of the detector in the space between the
layers where connection blocks facilitate the transition to clear waveguides
take the light to the VLPC cassettes. Each MIP plane has 206
scintillator strips
which are perpendicular to an edge at constant $\phi$. Each shower layer has
288 strips, also perpendicular to a constant $\phi$ edge. Four special
wedges in the vicinity of the solenoid cryogenics service pipes are notched to
allow these to enter and have 142 strips per wedge in both the MIP and shower
layers. The presence of these special wedges reduces the coverage to
$1.8 < |\eta| < 2.5$ in this area. Successive FPS layers are offset to prevent
projective cracks in $\phi$. An $r-\phi$ view of two layers of the north
FPS detector is shown in Figure~\ref{fig:fps-layer-2-4} and
the three distinct types of
wedges and their dimensions within the successive FPS $\phi$, $z$ layers are
described in Figure~\ref{fig:fps-module-overlap}.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure27.eps}}
\caption{Three dimensional view of an FPS module mounted within its supporting
frame.}
\label{fig:fps-module-3d}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=5.in]{figure28.eps}}
\caption{$r-\phi$ view of the north FPS detector. For clarity, only layers 2
(shower) and 4 (MIP) are shown; layers 1 (shower) and 3 (MIP) are rotated by
$22.5^\circ$ in $\phi$ with respect to these layers so their supports do not
overlap. The south FPS detector
is a mirror image of the north detector. The cutback near four o'clock
accommodates the solenoidal magnet cryogenic services on the south side.}
\label{fig:fps-layer-2-4}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure29.eps}}
\caption{Relative orientation of two modules in successive FPS layers.
The 12.7~mm overlap region is shown, along with the dimensions of the three
types of modules.}
\label{fig:fps-module-overlap}
\end{figure}
The absorber is divided into forty-eight wedge-shaped segments for
easier handling, each subtending $7.5^\circ$ in $\phi$ and weighing
approximately 5 lbs. Each segment consists of two lead absorber
elements epoxied to each side of a 1/8"-thick stainless-steel
plate. Each steel plate is approximately 0.5" longer
radially than the lead plates, to allow connection to inner and outer support
rings that are nested within the overall FPS layers. Similarly to the modules
occupying the MIP and shower layers, the absorber segments are
individually bent in three dimensions to conform to the spherical geometry of
the end calorimeters. The $\phi$ edges of each segment
are beveled to allow nesting of adjacent absorber segments with minimal
non-projective gaps between them. The total thickness of a
lead-stainless-steel-lead segment is 11 mm, or $2X_0$.
\FloatBarrier
\section{Calorimetry}
\label{sec:calorimeters}
The D\O\ calorimeter system consists of three sampling calorimeters
(primarily uranium/liquid-argon) and an intercryostat detector.
\subsection{Calorimeters}
The calorimeters were designed to provide energy measurements for electrons,
photons, and jets in the absence of a central magnetic field (as was the case
during Run I of the Tevatron), as well as assist in identification of
electrons, photons, jets, and muons and measure the transverse energy balance
in events. The calorimeters themselves are unchanged from Run I and are
described in detail in Ref.~\cite{d0_nim}. They are illustrated in
Figure~\ref{fig:cal-isometric}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure30.eps}}
\caption{Isometric view of the central and two end calorimeters.}
\label{fig:cal-isometric}
\end{figure}
As shown in Figure~\ref{fig:cal-side-eta}, the central calorimeter
(CC) covers $|\eta| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 1$ and the two end calorimeters, ECN (north) and ECS
(south), extend coverage to $|\eta| \approx 4$. Each calorimeter contains an
electromagnetic section closest to the interaction region followed by fine and
coarse hadronic sections.
The active medium for the calorimeters is liquid argon and each of the
three calorimeters (CC, ECN, and ECS) is located within its own cryostat that
maintains the detector temperature at approximately 90~K. Different absorber
plates are
used in different locations. The electromagnetic sections (EM) use thin plates
(3 or 4~mm in the CC and EC, respectively), made from nearly pure depleted
uranium. The fine hadronic
sections are made from 6-mm-thick uranium-niobium (2\%) alloy. The coarse
hadronic modules contain relatively thick (46.5 mm) plates of copper
(in the CC) or stainless steel (EC).
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure31.eps}}
\caption{Schematic view of a portion of the D\O\ calorimeters showing the
transverse and longitudinal segmentation pattern. The shading pattern
indicates groups of cells ganged together for signal readout. The rays
indicate pseudorapidity intervals from the center of the detector.}
\label{fig:cal-side-eta}
\end{figure}
A typical calorimeter cell is shown in Figure~\ref{fig:cal-cell}. The electric
field is established by grounding the metal absorber plates and connecting the
resistive surfaces of the signal boards to positive high voltage (typically
2.0 kV). The electron drift time across the 2.3~mm liquid argon gap is
approximately 450~ns. Signal boards for all but the
EM and small-angle hadronic modules in the EC are made from two 0.5~mm G-10
sheets. The surfaces of the sheets facing the liquid argon gap are coated with
carbon-loaded epoxy \cite{cal-epoxy} with a typical sheet resistivity of
40~M$\Omega/\Box$; these surfaces serve as the high voltage electrodes for
the gap. For one sheet, the other surface is bare G-10; the facing inner
surface of the second sheet,
originally copper-coated, is milled into the pattern necessary for segmented
readout. Several such pads at approximately the same $\eta$ and $\phi$ are
ganged together in depth to form a readout cell.
The two smallest-angle modules (EM and hadronic) in each EC have the added
problem that even small gaps between neighboring azimuthal sectors would give
undesirable dead regions. Thus, the signal boards for these monolithic
modules are made from multilayer printed circuit boards. The outer surfaces
are coated with the same resistive epoxy as for the other signal boards.
Etched pads on an interior surface give the desired segmentation. Signal
traces on another interior surface bring the signals to the outer periphery.
The pad and trace layers are connected by plated-through holes. The signals
from these multilayer boards in the EM and small-angle hadronic modules are
ganged together along the depth of the modules.
Calorimeter readout cells form pseudo-projective towers as shown in
Figure~\ref{fig:cal-side-eta}, with each tower subdivided in depth.
We use the term ``pseudo-projective'' because the centers of cells of
increasing shower depth lie on rays projecting from the center of the
interaction region, but the cell boundaries are aligned perpendicular to the
absorber plates.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure32.eps}}
\caption{Schematic view of the liquid argon gap and signal board unit cell for
the calorimeter.}
\label{fig:cal-cell}
\end{figure}
There are four separate depth layers for the EM modules in the
CC and EC. In the CC, the layers are approximately 1.4, 2.0, 6.8 and 9.8$X_0$
thick. In the EC, they are approximately 1.6, 2.6, 7.9 and 9.3$X_0$ thick.
The values given for the first layers include all material in the calorimeters
themselves from the outer warm walls to the first active liquid argon gap.
The detector components between the interaction region and the first active gap
in the CC at $\eta = 0$ provide about 4.0$X_0$ of material; those between the
interaction region and the first active gaps of the ECs at
$\eta = 2$ are 4.4$X_0$ thick.
In the CC, the fine hadronic modules have three longitudinal gangings of
approximately 1.3, 1.0, and 0.76$\lambda_A$. The single coarse hadronic module
has a thickness of about 3.2$\lambda_A$. The two EC inner hadronic modules
(Figure~\ref{fig:cal-isometric}) are cylindrical, with inner and outer radii of
3.92 and 86.4~cm. The fine hadronic portion consists of four readout cells,
each $1.1\lambda_A$ thick. The coarse hadronic portion has a single readout
cell $4.1\lambda_A$ thick. Each of the EC middle hadronic modules has four
fine hadronic readout cells of about $0.9\lambda_A$ each and a single coarse
hadronic section of $4.4\lambda_A$. The outer hadronic modules of the ECs are
made from stainless steel plates inclined at an angle of about $60^\circ$ with
respect to the beam axis (see Figure~\ref{fig:cal-side-eta}). The maximum
thickness is $6.0\lambda_A$. Each layer is offset from the others
to provide hermeticity.
The transverse sizes of the readout cells are comparable to the transverse
sizes of showers: 1--2~cm for EM showers and about 10~cm for hadronic showers.
Towers in both EM and hadronic modules are $\Delta\eta = 0.1$ and
$\Delta\phi = 2\pi/64 \approx 0.1$. The third layer of the EM modules, located
at the EM shower maximum, is segmented twice as finely in both $\eta$ and
$\phi$ to allow more precise location of EM shower centroids.
As can been seen in Figure~\ref{fig:ecem-segmentation}, cell sizes increase
in $\eta$ and $\phi$ at larger $\eta$ to avoid very small cells.
\begin{figure}
\centerline{\includegraphics[width=4.in]{figure33.eps}}
\caption{Layout of the EC electromagnetic readout cells for the four
longitudinal EM layers. EM1 is closest to the interaction region.
MH indicates the EC middle hadronic cells.}
\label{fig:ecem-segmentation}
\end{figure}
\subsubsection{Calorimeter electronics}
Figure~\ref{fig:chain} illustrates the main components
in the calorimeter readout chain.
There are 55,296 calorimeter electronics channels to be read out; 47,032
correspond to channels connected to physical readout modules in the cryostats.
The remaining electronics channels are not connected to the detector. (The
ADC cards are identical and contain enough channels to read out the most
populated regions of the detector.)
The readout is accomplished in three principal stages. In
the first stage, signals from the detector
are transported to charge preamplifiers located on the cryostats via
low impedance coaxial cable. In the second stage, signals from the
preamplifiers are transported on twisted-pair cables to the analog signal
shaping and storage circuits on baseline subtractor (BLS) boards.
The precision signals from the BLSs are transmitted on an analog bus and
driven by analog drivers over 130~m of twisted-pair cable to ADCs. These
signals then enter the data acquisition system for the Level~3
trigger decision (Section~\ref{sec:l3trigger}) and storage to tape. The
preamplifiers and BLSs are completely new for Run~II, and were necessary to
accomodate the significant reduction in the Tevatron's bunch spacing.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure34.eps}}
\caption{Readout chain of the calorimeter in Run~II indicating the
three major components: preamplifiers, baseline subtractor and storage
circuitry (BLS), and the ADCs.}
\label{fig:chain}
\end{figure}
\paragraph{Front-end electronics}
Signals from the calorimeter cells are transported
on 30~$\Omega$ coaxial cables (with a typical length of 10~m)
to a feedthrough port (the interface
between the cold region and the warm region) on the cryostat and then
on to the preamplifiers which are mounted as close as possible to the
feedthroughs at the top of the cryostat. The cables from the feedthrough port
to the preamplifier were replaced for Run~II to provide better impedance
matching to the
preamplifier input ($\approx 30~\Omega$) and to equalize the lengths to provide
better timing characteristics by minimizing the spread of signal arrival times.
Electron drift time across the 2.3~mm liquid-argon gap remains approximately
450~ns at 2.0~kV for Run~II, which provides a challenge for
signal charge integration with beam crossings occurring every 396~ns. The
calorimeter electronics were designed to maintain the good signal-to-noise
ratio obtained during Run~I in the expected high instantaneous luminosity
environment of the original Tevatron Run~II design, with a minimum bunch
crossing time of 132~ns.
The preamplifiers are individually packaged transimpedance
(i.e.\ charge to voltage) hybrid amplifiers on ceramic substrates. Forty-eight
individual preamplifier hybrids are mounted on a motherboard, with ninety-six
motherboards housed in a single preamplifier box. Twelve such boxes,
mounted on top of the calorimeter cryostats, make up the full calorimeter
preamplifier system. Since access to these preamplifier boxes
is very limited, the power supplies for the preamplifiers are
redundant with each preamplifier box having two supplies, a primary and a
backup that can be switched in to replace the primary supply. Low noise
commercial switching power supplies are used to provide the necessary
power.
The preamplifier shaping networks are modified for cells of different
capacitance to give approximately the same output waveform for all cells.
Given the large range of detector capacitances at the input to the
preamplifiers, there are fourteen species of preamplifier (plus one for
readout of the intercryostat detector (ICD), Section~\ref{sec:icd}) which
provide similarly shaped output signals to the BLS boards. This is important
to maintain good timing for the peak-sampling circuit.
The characteristics of the preamplifier species are
shown in Table~\ref{tab:cal_preamp}.
The ICD feedback capacitor is 22~pF to reduce the gain of the output
signals and thus preserve their dynamic range.
Dual parallel input JFETs help maintain low noise performance. Two
output driver stages provide the capability to drive a terminated
115~$\Omega$ twisted-pair cable. The power requirement of a single
preamplifier is 280~mW. The noise performance can be evaluated from
the width of the pedestal distribution, which is a function of the
detector input capacitance.
\begin{table}
\begin{center}
\caption{Characteristics of the different preamplifier species used for the
calorimeter and the ICD readout. Since the cell capacitances depend on the
size of the cell, with larger cells having higher capacitance, preamplifiers
A -- D are used in various locations in both the central and end calorimeters.
FH$n$ stands for fine hadronic, layer $n$;
CH$n$ is similar for the coarse hadronic layers;
and MG for the CC and EC massless gaps (Section~\ref{sec:icd}).}
\label{tab:cal_preamp}
\vskip 0.1 in
\begin{tabular}{ p{60pt} p{60pt} p{95pt} p{50pt} p{33pt} p{42pt} }
\hline
Preamplifier \par species & $<$Cell~cap.$>$ \par (nF) & Layer \par read out &
Feedback \par cap.\ (pF) & RC \par (ns) & Total \par preamp. \par channels \\
\hline
A & 0.26--0.56 & EM1,2, MG \par EC CH1,2 & 5 & 0 & 13376 \tabularnewline
B & 1.1--1.5 & MG \par EC FH1,2,4, CH1 & 5 & 26 & 2240 \tabularnewline
C & 1.8--2.6 & CC CH1 \par EC FH1--4, CH1--3 & 5 & 53 & 11008 \tabularnewline
D & 3.4--4.6 & CC FH1--3, CH1 \par EC FH1--4, CH1--3 & 5 & 109 & 8912
\tabularnewline
E & 0.36--0.44 & CC EM3 & 10 & 0 & 9920 \tabularnewline
F & 0.72--1.04 & EC EM3,4 & 10 & 14 & 7712 \tabularnewline
G & 1.3--1.7 & CC EM4 \par EC EM3,4 & 10 & 32 & 3232 \tabularnewline
Ha--Hg & 2--4 & EC EM3,4 & 10 & 47--110 & 896 \tabularnewline
I & -- & ICD & 22 & 0 & 384 \tabularnewline
\hline
\end{tabular}
\end{center}
\end{table}
The preamplifier motherboard is an eight-layer printed circuit board, with
ground or power planes of solid copper separating planes containing
signal traces to minimize noise pickup and cross-talk. The
motherboard houses the precision resistors (0.1\% 10~k$\Omega$ or
20~k$\Omega$, depending on the preamplifier feedback capacitor) for the
calibration voltage pulse. A single input line pulses six
preamplifiers at once using a low capacitance trace.
All of the electronics is located in an area where there can be a
residual magnetic field of a few hundred gauss, and so the switching
supplies are magnetically shielded. All other devices (including
cooling fans) have been verified to function in the residual field.
New heat exchange systems were installed in the existing preamplifier
boxes to deal with increased power dissipation.
\paragraph{Signal shaping and trigger}
The single-ended preamplifier signals are routed from the calorimeter
over approximately 25~m of twisted-pair cable
to the BLSs located below the cryostats, where access is significantly easier.
The BLSs use switched capacitor
arrays (SCAs) as analog storage devices to hold the signal for about
4~$\mu$s until the Level~1 trigger decision is made, and for an additional
2~ms (on average, with an allowed maximum of 25~ms) until the Level~2 trigger
decision is made. They also provide baseline
subtraction to remove any low frequency noise or pileup from the
signal. In addition, faster shaped analog sums of the signals are
picked off to provide prompt inputs to the calorimeter trigger system
for both the Level~1 and Level~2 calorimeter trigger decisions
(Sections~\ref{sec:l1cal} and \ref{sec:l2cal}).
To minimize the effects of pile-up (more than one event in the detector
due to multiple interactions in a single beam crossing or to interactions
in multiple beam crossings), only two-thirds of the charge
collected by the preamplifier circuit, corresponding to the first
$\approx 260$~ns of signal collection from the gap, is used in
the shaper circuit. The preamplifier output is an integral of the
detector signal characterized by a rise time of about 450~ns
and a recovery time of 15~$\mu$s. Shaped signals are sampled
every 132~ns.
The shaper circuit produces a unipolar signal with a peak
at about 320~ns and a return to zero after $\approx 1.2~\mu$s.
The shaped signal is sampled at $t_0 = 320$~ns, close to the
peak. To subtract the baseline, the signal three samples earlier
($t_0 - 396$~ns) is subtracted by the BLS circuitry.
The calorimeter Level~1 and Level~2 triggers are based on the energy
measured in trigger towers of size $0.2
\times 0.2$ in $\Delta\eta \times \Delta\phi$, which is obtained by
making appropriate sums (via resistor packs on the BLS boards) of the fast
pickoffs at the shaper inputs. The resistor packs have been tuned
based on a sampling weight optimization study which sought to
maximize electron resolution first, and jet resolution second, for the channels
included in the trigger (the coarse hadronic sections do not contribute to the
trigger).
\paragraph{Digitization}
A BLS board processes signals from the forty-eight channels from four
pseudo-projective calorimeter towers. Each tower corresponds to up to
twelve preamplifier signals.
There are signal shapers for each channel on the BLS motherboard, and
trigger pick-off and summation circuits tap the preamplifier signal
prior to the shaper circuitry. The shaped preamplifier signals are
fed to daughterboards, one per tower, each of which holds five SCA
chips. The SCAs contain an array of 48 capacitors to pipeline the
calorimeter signals. The first and last buffers are not used in the
readout to avoid edge effects in the chips. The SCA is not designed
for simultaneous read/write operations, therefore two SCA banks are
alternately employed to provide the capability to write and read
the integrated charges. This scheme provides the
4.2~$\mu$s buffering necessary prior to the arrival of the Level~1
(Section~\ref{sec:l1})
trigger decision. There are also two gain paths ($\times 1$ and
$\times 8$) to extend the ADC readout dynamic range, thus four of the SCAs are
used to store the signals for the twelve channels on a daughtercard
until the Level 1 trigger decision has been made. Once a positive
Level 1 decision is received, baseline subtractor circuitry on each
daughterboard decides channel-by-channel which gain path to use and
subtracts the stored baseline from
the peak signal. It then stores the result in the Level~2 SCA that
buffers the data until a Level~2 (Section~\ref{sec:l2}) trigger decision has
been made. Once a
Level~2 trigger accept is issued, the data are transferred from the Level 2
SCA to a sample-and-hold circuit on the daughterboard and an analog
switch on the BLS motherboard buses the data on the BLS backplane to
analog drivers which transfer the signal up to the ADC across 130~m of
twisted-pair cable. The gain
information is sent simultaneously on separate digital control cables.
The ADC successive approximation digitizers, reused from Run~I, have
a 12-bit dynamic range, but the low and high gain paths for each
readout channel effectively provide a 15-bit dynamic range. This matches the
measured accuracy of the SCA. The readout system effectively introduces no
deadtime up to a Level~1 trigger accept rate of 10~kHz, assuming
no more than one crossing per superbunch\footnote{A superbunch is a group of
twelve $p$ or \mbox{$\overline{p}$}\ bunches, where each bunch is separated by 396~ns; the
Tevatron contains three evenly-spaced superbunches of each particle.} is read
out.
A master control board synchronizes twelve independent controllers
(one for each readout quadrant of the calorimeter) in a shared VME crate.
These controllers
provide the timing and control signals that handle the SCA requirements and
interface to the Level~1 and Level~2 trigger systems. Each control
board houses three Altera FPGA (field programmable gate array)
chips~\cite{FLEX10K} (10K series, 208-pin packages).
Three FIFO pipelines buffer up to about forty events awaiting readout.
Timers on these event buffers ensure that
``stale'' data are appropriately flagged to the data acquisition
system. Other readout errors are also flagged in the readout. The
control boards permit the readout VME crate to be run in numerous
diagnostic and calibration modes.
\paragraph{Calibration system}
\label{sec:cal-calibration}
The calorimeter calibration system consists of twelve identical units
for the calorimeter and one slightly modified unit for the ICD
(Section~\ref{sec:icd}). Each unit is composed of one pulser board, its power
supply located in the BLS racks on the platform underneath the detector, and
six active fanout boards housed inside the preamplifier boxes mounted on the
cryostats. The pulser boards are controlled via a VME
I/O register to set the amplitude and the timing of the
calibration signal and enable the channels to be pulsed.
A DAC sets the current delivered by the pulser boards using 17 bits, with
an avarage current/DAC of 0.825~$\mu$A/DAC, and the timings are
controlled through a 8-bit delay line in steps of 2~ns.
Both the pulser board and the active fanout have been shown
to provide a pulser signal with a linearity at the per mil level, and
all the currents delivered are uniform within 0.2\% between all boards and
0.1\% within a board (Figure~\ref{fig:uni}).
All of the pulse shapes have been measured to
estimate systematic effects on the signal amplitude, the timing, and the
charge injected.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure35.eps}}
\caption{The mean current/DAC delivered by the pulser boards is
825~mA/DAC with a spread of (a) $\pm0.2$\% for all 15 pulser boards
produced and (b) $\pm0.1$\% within one board.}
\label{fig:uni}
\end{figure}
\paragraph{Gain determination}
For all calorimeter channels, the gain calibration factors are
determined from deviations of the slope ADC/DAC from its ideal
value of 0.25 ADC counts per DAC unit. The
dispersion of the coefficients for electromagnetic (hadronic) channels
is about 5\% (10\%). Systematic shifts of the slope values can be
observed for different preamplifier types.
Part of these differences is due to the injection of the calibration
signal at the preamplifier input, outside of the cryostat. A fraction
of the pulser signal travels down the signal cable and is reflected
from the calorimeter cell, depending on the capacitance of the cell,
and therefore has a different measured shape. The effects are largest
for hadronic cells with high capacitance, where differences in the
timing can produce large tails in the distribution of the calibration
coefficients. Corrections for these effects have been derived
for the calibration coefficients, and the delay settings of the pulsers
have been tuned to maximize the response in the electromagnetic layers.
Models of the electronics chain have been made to evaluate the
differences between the electronics response to a calibration signal
and a detector signal. To render these models realistic, all stable
parameters of the signal path from the detector to the preamplifier
input have been determined from reflection measurements. The reflected
response (in arbitrary voltage units) to a step function is shown in
Figure~\ref{fig:reflection} for three different channel
types. Quantitatively the values for the cable resistivity outside and
inside the cryostat, the inductance of the feedthrough and the
signal strips, as well as the capacitance of each cell, have been
determined and used in the simulation model. Correction factors to the
calibration coefficients indicate these effects are
below the percent level for electromagnetic channels, when sensed
close to the signal maximum.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure36.eps}}
\caption{Response to reflection measurement for three different channels.
The vertical voltage scale is in arbitrary units. Cd is the detector cell
capacitance.}
\label{fig:reflection}
\end{figure}
\subsubsection{Liquid argon monitoring}
The purity of the liquid argon is critical to the detector performance
as electronegative contaminants (particularly oxygen) can combine
with electrons traversing the gap and can severely impact the energy
measurement. The liquid argon was recovered from Run~I and, before
refilling the cryostats for Run~II, the purity was remeasured \cite{ar-purity}
with an external argon test cell equipped with $^{241}$Am and $^{106}$Ru
radioactive sources. The $^{241}$Am is an alpha source (5~MeV
$\alpha$-particles,
432~yr half-life) that was used for purity measurements during Run~I;
the $^{106}$Ru is a new beta source (maximum 3.5~MeV $\beta$-particles,
$\approx$1~yr half-life) with activity of 30--40~kBq.
Calibration of the test cell was
performed by injecting controlled amounts of $\rm{O}_2$ into pure argon.
The contamination of the liquid argon was measured to be less than
$0.30 \pm 0.12$~ppm for all three calorimeter cryostats. A 5\% signal loss is
expected for a contamination level of 1~ppm.
Radioactive sources are used to monitor the contamination levels {\it in situ}.
Each of the three cryostats is equipped with four
$^{241}$Am sources with an activity of 0.1~$\mu$Ci and four $^{106}\rm{Ru}$
sources. Three of
the beta sources in each cryostat now have very low levels of activity
($\mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 1$~Bq), fifteen years after the initial detector construction. One
stronger source has an activity of about 4~Bq. The charge
liberated in the liquid argon gap by the alpha sources is about 4~fC
(about 25,000 electrons) and about half as much for the beta source. The
trigger rate is about 500~Hz for the alpha source and 0.3~Hz
for the strong beta source. The argon purity of the three calorimeters has
been extremely stable over time.
A new readout board with a preamplifier, dual operational-amplifier, and
differential driver sends the amplified signals via a shielded
twisted-pair cable to a differential receiver board to be digitized by
a 12-bit ADC using a Xilinx FPGA (Spartan XCS40XL)~\cite{Xilinx}. Histograms
can be accumulated very rapidly (few kHz) on the receiver board which are
read out via CAN-bus to a PC running National Instruments LabView
\cite{labview}.
This design is a modification of the liquid argon monitoring system
\cite{atlas-lar} for ATLAS and sensitive to less than 1~ppm of oxygen
equivalent contamination.
The signal response is also a function of the liquid argon
temperature so this is monitored as well. The temperature of
the liquid argon is maintained at $90.7 \pm 0.1$~K.
\subsection{Intercryostat detector and massless gaps}
\label{sec:icd}
Since the calorimeter system is contained in three separate cryostats,
it provides incomplete coverage in the pseudorapidity region
$0.8<\mid\eta\mid<1.4$, as can be seen in Figure~\ref{fig:cal-side-eta}.
In addition, there is substantial unsampled material in this
region, degrading the energy resolution. To address this
problem, additional layers of sampling have been added.
Within the central and end cryostats are single-cell structures called
massless gaps --- calorimeter readout cells in front of the first layer of
uranium. The ICD provides scintillator sampling that is attached to the
exterior surfaces of the end cryostats. It covers the region
$1.1<\mid\eta\mid<1.4$; its position with respect to the rest of the detector
can be seen in Figure~\ref{fig:tracker}. The massless gaps are those
used in Run~I, while the ICD is a replacement for a similar detector in the
same location (the Run~I detector had approximately twice the coverage in
$\eta$; the space was needed for SMT and CFT cabling in Run~II).
The ICD, shown in Figure~\ref{fig:icd-tile-array}, is a series of 0.5"-thick
scintillating tiles (Bicron BC-400 \cite{bicron}) enclosed in light-tight
aluminum boxes. Each tile covers an area of
$\Delta\eta \times \Delta\phi \approx 0.3 \times 0.4$ and is
divided into twelve subtiles, as shown in Figure~\ref{fig:icd-tile-detail},
each covering $\Delta\eta \times \Delta\phi \approx 0.1 \times 0.1$.
Because of the cryogenic services for the solenoid, one half of a tile is
missing at the south end of the detector, giving a total of 378 channels.
The subtiles are optically isolated from one another by grooves filled with
epoxy mixed with highly reflective pigment. Each subtile is read out via
two 0.9-mm-diameter wavelength shifting fibers (Bicron BCF-91A) embedded in a
groove cut to a depth of 3.5~mm near the outside edge of each subtile. A third
fiber, used for calibration, is also located in the groove.
The wavelength-shifting fibers are mated to 1.0~mm clear optical fibers
(Bicron BCF-98) at the outer radius of the ICD tile enclosure.
The clear fibers terminate at a Hamamatsu R647 \cite{hamamatsu}
photomultiplier tube (PMT).
\begin{figure}
\centerline{\includegraphics[width=3.5in]{figure37.eps}}
\caption{The arrangement of the ICD tiles on the endcap cryostats.
The rectangles represent the iron block and fiber backplane assemblies in
which the ICD electronics and PMTs are installed. The beamline is
perpendicular to the page.}
\label{fig:icd-tile-array}
\end{figure}
\begin{figure}
\centerline{\includegraphics[angle=0,width=4.0in]{figure38.eps}}
\caption{Each ICD tile is subdivided into twelve straight-edged trapezoidal
subtiles. WLS fibers for readout are embedded into each subtile along the
curved trapezoids.}
\label{fig:icd-tile-detail}
\end{figure}
The ICD electronics are located in a low-magnetic-field region
away from the tiles and are contained in a drawer system.
An ICD drawer houses PMTs, PMT electronics, and preamplifiers for
six channels of readout.
Clear fibers from the ICD tiles are mated to clear fibers
contained in a light-tight enclosure called the ``fiber
backplane.'' Between the fiber backplane and the crate containing
the drawers is a block of iron, with holes drilled to accept the
PMTs. Fibers from the fiber backplane terminate in ``cookies'' that
hold the ends of the fibers in the correct location with respect to
the photocathodes of the PMTs. When a drawer is closed, six PMTs are
aligned with six fiber bundles (two per channel plus one for a
calibration signal in each bundle) from the backplane. The PMTs are
spring-loaded to ensure good contact with the fibers. The iron block, in
combination with tubular magnetic shields, protects the PMTs from
fringe magnetic fields. O-rings at the base of the PMTs and the drawers
themselves provide a dark environment for the PMTs.
The signal electronics are designed to be compatible with the calorimeter
BLS/ADC system and electronics calibration. The ICD uses a
modified version of the calorimeter preamplifiers,
designed to stretch the PMT signal into a signal similar to that
of the liquid argon readout. A modified form of the calorimeter electronic
pulser system is used for electronics commissioning. The responses of the
PMTs are monitored
using an LED calibration system, similar in design to that
used for the muon trigger scintillation counters
(Section~\ref{sec:muon-led-monitoring}). An LED pulser is housed in
each of the fiber backplanes, and the amount of light is software-controlled.
Individual channel responses were measured on a test stand using
cosmic rays. This test stand used the same combination of tiles,
fibers, and electronics that was mounted in the
experiment after testing.
Figure~\ref{fig:ICDmips} shows the distribution of cosmic ray peaks
from the test stand. ICD sampling weights are taken from the full
Monte Carlo simulation of the D\O\ detector and are tuned using
dijet and photon-jet $E_T$\ balance.
\begin{figure}
\centerline{\includegraphics[width=3.0in]{figure39.eps}}
\caption{Cosmic ray signals in all channels of the ICD.}
\label{fig:ICDmips}
\end{figure}
\FloatBarrier
\section{Muon system}
\label{sec:muon}
For muon triggering and measurement, the upgraded detector uses the original
central muon system proportional drift tubes (PDTs) and toroidal magnets
\cite{d0_nim}, central scintillation counters (some new and some installed
during Run~I), and a completely new forward muon system.
The central muon system provides coverage for $|\eta| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 1.0$. The new
forward muon system extends muon detection to $|\eta| \approx 2.0$, uses
mini drift tubes (MDTs) rather than PDTs, and includes trigger scintillation
counters and beam pipe shielding.
The small angle muon system \cite{d0_nim} of the original detector, including
its associated magnets, has been removed.
During Run I, a set of scintillation counters, the cosmic cap
\cite{cosmic_cap_nim}, was installed on the top and upper sides of the outer
layer of central muon PDTs. This coverage has been extended to the lower sides
and bottom of the detector, to form the cosmic bottom. These trigger
scintillation counters are fast enough to allow us to associate a muon in a
PDT with the appropriate bunch crossing and to reduce the cosmic ray
background.
Additional scintillation counters, the $A\phi$ counters, have been installed on
the PDTs mounted between the calorimeter and the toroidal magnet. The
$A\phi$ counters provide a fast detector for triggering and identifying muons
and for rejecting out-of-time background events.
The scintillation counters are used for triggering; the wire chambers are used
for precise coordinate measurements as well as for triggering. Both types of
detectors contribute to background rejection: the scintillator with timing
information and the wire chambers with track segments.
New detectors and the modifications made to the original system are discussed
in detail in the following sections; original components are described briefly.
Exploded views of the muon system are shown in Figures~\ref{fig:mudrift} and
\ref{fig:muscint}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure40.eps}}
\caption{Exploded view of the muon wire chambers.}
\label{fig:mudrift}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure41.eps}}
\caption{Exploded view of the muon scintillation detectors.}
\label{fig:muscint}
\end{figure}
\subsection{Toroidal magnets}
\label{sec:toroids}
The toroidal magnets are described in detail in Ref.~\cite{d0_nim} and visible
in Figures~\ref{fig:detector}, \ref{fig:solenoid_perspective}, and
\ref{fig:solenoid-map}. Having a stand-alone muon-system momentum measurement
{\it i}) enables a low-$p_T$ cutoff in the Level~1 muon trigger,
{\it ii}) allows for cleaner matching with central detector tracks,
{\it iii}) rejects $\pi/K$ decays, and
{\it iv}) improves the momentum resolution for high momentum muons.
The central toroid is a square annulus 109~cm thick whose inner surface is
about 318~cm from the Tevatron beamline; it covers the region
$|\eta| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 1$. To allow access to the
inner parts of the detector, it was constructed in three sections. The
center-bottom section is a 150-cm-wide beam, fixed to the detector platform,
which provides a base for the calorimeters and central tracking detectors. Two
C-shaped
sections, which can be moved perpendicularly to the center beam, complete the
central toroid. The magnet is wound using twenty coils of ten turns each.
The two end toroids are located at $454 \le |z| \le 610$~cm. In the center of
each end toroid is a 183~cm
square hole centered on the beamline; in $x$ and $y$ the magnets extend 426~cm
from the beamline. The end toroid windings are eight coils of
eight turns each.
During Run~I, the central and end toroid coils were operated in series at a
current of 2500~A; the internal fields in the central toroid were
approximately 1.9~T and those in the end toroids were approximately 2.0~T.
During Run II, the
magnets are again being operated in series, but at a current of 1500~A. The
magnetic field is about 6\% lower than that of Run I. Now, however, the
primary measurement of the muon momentum is performed using the new central
tracking system and the reduced current provides a substantial cost savings.
As in Run~I, the polarity of the magnets during data collection is regularly
reversed.
\subsection{Central muon detector}
The central muon system consists of a toroidal magnet
(Section~\ref{sec:toroids}), drift chambers, the
cosmic cap and bottom scintillation counters, and the $A\phi$ scintillation
counters.
\subsubsection{Central muon drift chambers}
The drift chambers are described in detail in Ref.~\cite{d0_nim}.
The three layers of drift chambers are located inside (A layer) and
outside (B and C layers) of the central toroidal magnet and cover
$|\eta| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 1$. Approximately 55\% of
the central region is covered by three layers of PDTs; close to 90\% is covered
by at least two layers. The drift chambers are large, typically $2.8 \times
5.6$~m$^2$, and made of rectangular extruded aluminum tubes. The PDTs outside
of the magnet have three decks of drift cells; the A layer has four decks with
the exception of the bottom A-layer PDTs which have three decks. The cells are
10.1~cm across; typical chambers are 24 cells wide and contain 72 or 96
cells. Along with an
anode wire at the center of each cell, vernier cathode pads are located above
and below the wires to provide information on the hit position along the wire.
The wires are ganged together in pairs within a deck and then read out by
electronics located at one end of each chamber.
For each PDT hit, the following information is recorded: the electron drift
time, the difference $\Delta T$ in the arrival time of the signal pulse at the
end of the hit cell's wire and at the end of its readout partner's
wire, and the
charge deposition on the inner and outer vernier pads. Both $\Delta T$ and the
charge deposition are used to determine the hit position along the wire. The
drift distance resolution is $\sigma \approx 1$~mm. The resolution of the
$\Delta T$ measurement varies depending on whether the muon passes through the
cell close to or far from the electronics. If the hit occurs far from the
electronics, the
resolution is approximately 10~cm. If it is close, the signal propagates two
wire lengths and the dispersion in the signal causes the resolution to degrade
to about 50~cm. Using charge division, the pad signal resolution is about
5~mm. However, only the A-layer pads are fully instrumented with electronics;
about 10\% of the B- and C-layer pads are instrumented. There are several
reasons for this: {\it i}) for tracks traversing all three layers, the pad
coordinate does not improve the pattern recognition or resolution significantly,
{\it ii}) for tracks that only reach the A-layer, the additional information
could help with track matching and background rejection, {\it iii}) the pad
signals are used to monitor the gain to track aging in the PDTs --- the
instrumented B- and C-layer pads serve this purpose, and {\it iv}) fully
instrumenting the B- and C-layer pads was too expensive.
To reduce the charge collection time,
we are using a faster gas mixture than we used during Run I. The new mixture
is 84\% argon, 8\% methane, and 8\% CF$_4$. The operating high voltage is
2.3~kV for the pads and 4.7~kV for the wires. The drift velocity
is approximately 10~cm/$\mu$s, for a maximum drift time of about 500~ns.
The contribution to the hit position uncertainty due to diffusion is about
0.4~mm, worse than the 0.3~mm achieved using a slower gas during Run I.
The poorer resolution is offset by the reduced occupancy and benefits to
triggering due to decreasing the number of crossings in the drift time interval
to two for 396~ns bunch spacing. The gas flow rate is 200 liters per minute,
providing an exchange rate of three volumes per day. The gas is recirculated
and filtered to remove contaminants.
Vapors from the copper-clad Glasteel (polyester and epoxy copolymer sheets with
chopped glass fibers) \cite{Glasteel} cathode pads are deposited
on the wires in a sheath whose thickness is proportional to the accumulated
integrated charge. This effect was observed after PDT chamber construction
was completed. As the coating
thickens, the gain of the chamber drops and the chamber becomes inefficient.
This aging during Run~II has been significantly reduced by the increased gas
flow, improved shielding to reduce background particles, and removal of the
Main Ring. The chamber wires could be cleaned if necessary,
but no aging has been observed up to an integrated luminosity of 0.8~fb$^{-1}$.
Since access to the
A-layer chambers and the four central B-layer chambers directly under the
detector is difficult, the cathode pads in these chambers have been replaced by
copper-clad G10 and no aging is anticipated.
\subsubsection{Cosmic cap and bottom counters}
The cosmic cap and bottom counters are installed on the top, sides and bottom
of the outer layer of the central muon PDTs. They provide a fast
timing signal to associate a muon in a PDT with the appropriate bunch
crossing and discriminate against the cosmic ray background.
The cosmic cap counters are described in detail in Ref.~\cite{cosmic_cap_nim}.
They are made from grooved 0.5" Bicron 404A scintillator with Bicron BCF 91A
and Kuraray Y11 wave-shifting fibers glued into the grooves using Bicron 600
optical epoxy. There are 240 counters, 25" wide, and 81.5"--113" long.
The counters are positioned with their width along $z$ and length
along $\phi$. The grooves are 1.75~mm deep and 4~mm wide; they run along the
length of the counter, from the end to just past the center. The grooves on
each half of the counter
are offset so they do not overlap at the center of the counter. They are
spaced 8~mm apart so that half of the counter surface is covered with fibers.
Each groove contains four fibers. The fibers are glued with five-minute epoxy
at the ends of the counters and polished using a diamond cutter. To
increase
the light yield, a 1/32" anodized aluminum sheet is attached to the ends with
aluminized mylar tape. The sides of the counters are milled. The fibers are
gathered at the center of the counters, divided into two bundles with two fibers
from each groove in one bundle and two in the other, and epoxied
into two acrylic plastic disks with holes in the center of each. The ends of
the disks are polished using the diamond cutter.
The scintillator is wrapped in a layer of Tyvek~\cite{Tyvek}
with a hole for the fibers and
disks; around the hole is an aluminum lip. A layer of 1/8" thick Styrofoam
is placed over the fibers on the counter top; aluminum sheets 0.020" thick
cover the bottom and top surfaces. An outer frame of Unistrut provides
support for the counter. A piece of black molded plastic fits over the outside
of the aluminum lip and covers the phototubes, cookies and fibers.
The fibers are read out using two 1.5" 10-stage EMI~\cite{EMI}
9902KA phototubes mounted on the counters. The
light yield varies depending on the distance from the phototube and the
proximity to the counter edge. It is typically 30 photoelectrons
per PMT for hits near the PMTs and 18 for hits near the distant corners.
The cosmic bottom counters complete the scintillator coverage of the central
toroidal magnet. There are 132 counters, of two different designs. The
forty-eight counters
located on the outside of the center bottom B layer of PDTs (where there is no
C layer) are nearly identical to the cosmic cap counters described above.
Some minor improvements were made in the placement of the edge fibers to
increase the light yield and the counter frames are made from 1/8" steel bent
into U-shaped channels. The counters are suspended from the B-layer PDTs.
The sixty-eight counters located on the undersides of the remaining
B and C layers of the PDTs are similar to the cosmic
cap counters except that the bottom counters have fewer fibers and they are
placed in vertical, rather
than horizontal, grooves. The grooves are approximately 6~mm deep and 6--10~cm
apart. This distribution of fibers results in the same light yield as the
horizontal arrangement used for the cap counters. These counters use
1"-diameter MELZ~\cite{melz} FEU-115M photomultiplier tubes.
They are 12-stage PMTs with a 2~ns risetime and good quantum efficiency and
uniformity. The PMTs are placed within 42-mm-diameter magnetic shields.
The B-layer counters are suspended from the
strong edges of the PDTs; the C-layer counters roll underneath the C-layer PDTs
with one set of wheels in a track to maintain the counter position.
An important difference between the cosmic cap and cosmic bottom counters is
that the bottom counters are positioned with their narrow dimension along
$\phi$ and their long dimension along $\eta$. This orientation has better
matching in $\phi$ with the central fiber tracker trigger. The widths of the
counters are approximately $4.5^\circ$ in $\phi$ and they
are as long or slightly longer than their respective PDTs are wide.
Table~\ref{tab:cosmic_counters} lists the location, number and size of the
cosmic cap and bottom counters.
\begin{table}
\begin{center}
\caption{Location, number, and size of cosmic cap and bottom scintillation
counters. All of the scintillator is 0.5" thick.}
\label{tab:cosmic_counters}
\begin{tabular}{lcccc}
\hline
Location & Number & Width (inches) & Length (inches) & PMT \\
\hline
Cosmic cap top & 80 & 25 & 113 & EMI \\
Cosmic cap upper sides & 80 & 25 & 108 & EMI \\
Cosmic cap lower sides & 80 & 25 & 81.5 & EMI \\
Central side B layer & 12 & 35 & 49 & EMI \\
Central side B layer & 4 & 35 & 67 & EMI \\
Central bottom B layer & 20 & 22.375 & 98.125 & EMI \\
Central bottom B layer & 20 & 15.75 & 98.125 & EMI \\
Central bottom B layer lap & 8 & 18.5 & 99.5 & EMI \\
End bottom B layer & 20 & 13.375 & 91.062 & MELZ \\
End bottom B layer & 12 & 19.25 & 91.062 & MELZ \\
Central bottom C layer & 20 & 22.062 & 68.062 & MELZ \\
Central bottom C layer & 16 & 29.3 & 68.062 & MELZ \\
\hline
\end{tabular}
\end{center}
\end{table}
\subsubsection{$A\phi$ scintillation counters}
The $A\phi$ scintillation counters cover the A-layer PDTs, those between
the calorimeter and the toroid. They provide a fast detector for triggering on
and identifying muons and for rejecting out-of-time backscatter from the
forward direction. In-time scintillation counter hits are matched with tracks
in the CFT in the Level~1 trigger (Section~\ref{sec:trigger}) for
high-$p_T$ single muon and low-$p_T$ dimuon triggers. The counters
also provide the time stamp for low-$p_T$ muons which do not penetrate the
toroid and thus do not reach the cosmic cap or bottom counters.
An end view of the $A\phi$ counter layout is shown in
Figure~\ref{fig:A_phi_layout}. The $\phi$ segmentation is approximately
$4.5^\circ$ which matches the central fiber tracker trigger
(Section~\ref{sec:l1ctt}) segmentation. The longitudinal segmentation is
33.25" which provides the necessary
time resolution and a match to the size of the PDTs; nine counters are
required along the detector in the $z$ direction. The nearly constant
segmentation in $\phi$ is accomplished through the use of three sizes of
counter: 14.46", 10.84", and 9.09" wide. The widest counters are located at
the corners of the detector, the narrowest at the center of each side. There is
a gap at the bottom of the detector where the calorimeter support is located.
The counters overlap an average of about 3\% in $\phi$ to reduce the
possibility of muons passing through cracks. Along the
length of the detector, the counters are butted end-to-end with a small gap
between each. There are 630 $A\phi$ counters.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure42.eps}}
\caption{End view of the layout of the $A\phi$ scintillation counters. The
inset box shows an enlarged view of four counters. Azimuthal coverage is shown
for seven of the counters. The bump on each counter represents the
photomultiplier tube attached to the counter case.}
\label{fig:A_phi_layout}
\end{figure}
The counters are made from Bicron 404A scintillator with G2 fiber embedded in
vertical grooves \cite{evdokimov} in a manner similar to the cosmic bottom
counters described above. The grooves are about 1.75" apart and run along the
length of each counter from the edge to the middle. Six fibers are spot glued
into each groove and taper out of the groove at the middle of the counter.
The fibers are routed to a MELZ FEU-115M photomultiplier tube which is secured
to the counter case.
The counter and fibers are wrapped in a layer of Tyvek, followed by a layer of
black TEDLAR~\cite{tedlar}. The counter case is an aluminum box with welded
corners. It
provides mechanical protection, support of the PMT and counter mounts, and
protection against light leaks. The counters are mounted on aluminum
cross-members attached to steel brackets which are fastened to the edges of the
A-layer PDTs.
The $A\phi$ counters operate in a magnetic field of 200--350~G due to the
residual magnetic field of the toroidal and solenoidal magnets. Magnetic
shields made of 1.2-mm-thick $\mu$-metal and 6-mm-thick soft iron with a
diameter of 48~mm provide
shielding from the fringe field (Section~\ref{sec:mu-trigger-counters}).
The effect of the magnetic field on the PMT gain is less than 10\% for any
field direction.
The performance of an $A\phi$ prototype counter was studied using cosmic rays.
The average muon signal corresponds to 50--60 photoelectrons.
Comparison of the amplitudes of signals from the ends of the counters and from
the center of the counter shows that the counters are uniform to $\pm7$\%. The
timing resolution of the counters is about 2~ns, due to photoelectron
statistics, amplitude variation along the length of the counter, variation in
the $z$-position of the vertex, variation in the $z$-position of the hit over
the length of the counter, and variation of the time of collision. Different
times-of-flight
for particles at different polar angles are compensated for by varying cable
lengths since the front-end electronics do not allow such timing adjustments.
\subsection{Forward muon system}
\label{sec:famus}
The layout of the forward muon system is shown in Figure~\ref{fig:detector}.
It covers $1.0 \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} |\eta| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 2.0$ and consists of four
major parts: the end toroidal magnets (Section~\ref{sec:toroids},
three layers of MDTs for muon track reconstruction, three layers
of scintillation counters for triggering on events with muons, and shielding
around the beam pipe.
\subsubsection{Mini drift tubes}
Mini drift tubes were chosen for their short electron drift time
(below 132~ns), good coordinate resolution (less
than 1~mm), radiation hardness, high segmentation, and low occupancy.
The MDTs are arranged in three layers (A, B, and C, with A closest to the
interaction region inside the toroidal magnet and C furthest away), each of
which is divided into eight octants, as illustrated in
Figure~\ref{fig:mudrift}. A layer consists of three (layers B and C)
or four (layer A) planes of tubes
mounted along magnetic field lines (the field shape
in the forward toroids is more ``square'' than ``circular''). The entire
MDT system contains 48,640 wires; the maximum
tube length is 5830~mm in layer C. Since the flux of particles
drops with increasing distance from the beam line, the occupancy of individual
tubes is the same within a factor of two over an entire layer.
An MDT tube consists of eight cells, each with a $9.4 \times 9.4$~mm$^2$
internal cross section and a 50~$\mu$m W-Au anode wire in the center, see
Figure~\ref{fig:mdt-xsec}. The
tubes are made from commercially available aluminum extrusion combs (0.6~mm
thick) with a stainless steel foil cover (0.15~mm thick) and are inserted
into PVC sleeves. They are closed by endcaps that provide accurate
positioning of the anode wires, wire tension, gas tightness, and electrical
and gas connections. The anode wires are supported by spacers; the
unsupported wire length never exceeds 1~m.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure43.eps}}
\caption{Cross-sectional view of a mini drift tube.}
\label{fig:mdt-xsec}
\end{figure}
The MDT system uses a CF$_4$-CH$_4$ (90\%-10\%) gas mixture. It is
non-flammable, fast, exhibits no radiation aging, and has
a wide operational plateau. The maximum drift time for tracks that are
perpendicular to the detector plane is 40~ns; for tracks inclined
at $45^\circ$, the maximum drift time is 60~ns. Figure~\ref{fig:mdt-t-to-d}
shows the time-to-distance relationship for inclinations of $0^\circ$ and
$45^\circ$ calculated using GARFIELD \cite{garfield} and for test beam data.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure44.eps}}
\caption{Time-to-distance relationship for a mini drift tube. The points
labeled 0~degrees and 45~degrees are calculated using GARFIELD. The crosses
indicate measurements done at $0^\circ$.}
\label{fig:mdt-t-to-d}
\end{figure}
Negative high voltage is applied to the cathode ($-3200$~V); the anode wire is
grounded at the amplifier. Each anode wire is connected to an amplifier and a
discriminator located as close as possible to the detector. Each
amplifier discriminator board (ADB) contains 32 channels and detects signals
with a 2.0~$\mu$A threshold. Output logical differential signals
from the ADB are sent to digitizing electronics which measure the signal
arrival time with respect to the beam crossing with an accuracy of 18.8~ns
(1.9~mm). This large time bin (equal to the Tevatron RF bucket size since the
main accelerator clock is 53~MHz) limits the
coordinate resolution of the MDTs and was selected to reduce the cost of almost
50,000 TDCs (time-to-digital converters). Hit information is sent to the
trigger and data acquisition systems; the data acquisition system also receives
drift times.
The efficiency of the MDTs is 100\% in the active area of the
cells for tracks that are perpendicular to the MDT plane. The overall plane
efficiency is less, due to the wall thickness and PVC sleeves, and is
approximately 95\%. The MDT efficiency is reduced near the
wire-support spacers as well. Although each spacer is only 5~mm wide along
the wire, the degradation in the electric field causes the efficiency to drop
to approximately 20\% for about 10~mm along the wire. An additional
inefficiency of about 5\% is caused by the tube
endcaps and gaps between octants for mounting, gas connections, and high
voltage and signal cables.
The momentum resolution of the forward muon spectrometer is limited by multiple
scattering in the toroid and the coordinate resolution of the tracking
detector. Although the MDT coordinate resolution measured in a test beam is
about 350~$\mu$m, the 18.8~ns time bin of the digitizing electronics leads to
a resolution of about 0.7~mm per hit. The standalone
momentum resolution of the forward muon system is approximately 20\% for muon
momentum below 40~GeV/$c$. The overall muon
momentum resolution is defined by the central tracking system for muons
with momentum up to approximately 100~GeV/$c$; the forward muon system
improves the resolution for higher momentum muons and is particularly
important for tracks with $1.6 \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} \eta \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 2.0$, i.e. those which
do not go through all layers of the CFT.
\subsubsection{Trigger scintillation counters}
\label{sec:mu-trigger-counters}
The muon trigger scintillation counters are mounted inside (layer A)
and outside (layers B and C) of the toroidal magnet
(Figure~\ref{fig:detector}). The C layer of
scintillation counters is shown in Figure~\ref{fig:c-layer-photo}. Each layer
is divided into octants containing about ninety-six counters. The $\phi$
segmentation is $4.5^\circ$
and matches the CFT trigger sectors. The $\eta$ segmentation is 0.12 (0.07)
for the first nine inner (last three) rows of counters. The largest counters,
outer counters in the C layer, are $60 \times 110$~cm$^2$. The B and C layers
have geometries similar to that of the A layer, but limited in places by the
collision hall ceiling and floor.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure45.eps}}
\caption{Photograph of the C layer of muon trigger scintillation counters
of the forward muon system.}
\label{fig:c-layer-photo}
\end{figure}
The counter design was optimized to provide good time resolution and amplitude
uniformity for background rejection, high muon detection efficiency, and
reasonable cost for the production of nearly five thousand counters. A
typical counter is shown in Figure~\ref{fig:mu-trig-counter}. The counters
are made of 0.5" Bicron 404A scintillator plate cut in a trapezoidal shape.
Kumarin 30 \cite{kumarin} WLS bars are
attached to two sides of the plate for light collection. The bars are 4.2~mm
thick and 0.5" wide and bent at $45^\circ$ to transmit light to a 1" phototube.
The phototubes are fast green-extended phototubes, FEU-115M, from MELZ.
They have a quantum efficiency of 15\% at 500~nm and a gain of about $10^6$.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure46.eps}}
\caption{Design of a scintillation counter for the forward muon trigger
system. Dimensions are in mm.}
\label{fig:mu-trig-counter}
\end{figure}
The sides of scintillator plates where the WLS bars are attached were left
unpolished after milling. The unpolished sides result in a larger number of
photoelectrons and better uniformity than polished sides do and cost less
\cite{evdokimov}.
The scintillator and WLS bars are wrapped in Tyvek for better light
collection and black paper for light tightness. The wrapped counters are placed
in 1-mm-thick aluminum containers with stainless steel transition pieces for
connection of the phototube assembly.
The fringe magnetic field due to the solenoidal and the toroidal magnets
reaches 300~G in the region where the A-layer phototubes are located
(Figure~\ref{fig:solenoid-map}). To
reduce the field in the area of the phototubes to approximately 1~G, they are
placed in 48-mm-diameter magnetic shields made of 1.2-mm-thick mu-metal and
3- or 6-mm-thick soft
iron. Since the field near the A layer is larger, the shields there use 6~mm
of iron. Those for the B and C layers use 3~mm of soft iron. The effect of
the magnetic field on the phototube signal is less than 10\% for fields up to
350~G for the 6-mm-thick shields for any field orientation with respect to the
phototube.
The performance of three counters of different sizes was studied in a
125~GeV/$c$ muon test beam. Figure~\ref{fig:mu-trig-res-eff} shows the
dependence of the counter efficiency and time resolution as a function of high
voltage for three counters: ``large'' $60 \times 106$ cm$^2$, ``typical''
$24 \times 34$ cm$^2$, and ``small'' $17 \times 24$ cm$^2$. A single
discrimination threshold of 10~mV was used for these measurements.
Time resolution
of better than 1~ns and detection efficiency above 99.9\% can be achieved at
appropriate high voltage settings for all counter sizes. The non-uniformity
of the counter response was measured by irradiating counters at different
points using a $^{90}$Sr source and checked via cosmic ray studies. For all
counter sizes, the rms non-uniformity is less than 10\%. Cosmic ray muons
were used to determine the average number of photoelectrons detected in the
counters. The largest counters have an average of 61
detected photoelectrons; the smallest counters give about three times as many
photoelectrons.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure47.eps}}
\caption{Time resolution and detection efficiency of muon trigger
scintillation counters. Large counters are $60 \times 106$~cm$^2$; typical
counters are $24 \times 34$~cm$^2$; small counters are $17 \times 24$~cm$^2$.}
\label{fig:mu-trig-res-eff}
\end{figure}
Phototube signals are sent to 48-channel VME-based scintillator front-end
(SFE) cards. Input 1:1 transformers at the SFE card isolate the SFE from DC
noise picked up between the phototubes and electronics modules which are
located about 10~m apart. After amplification, the signals are sent to a
10-bit ADC and to a discriminator. Discriminated and gated signals are passed
to the Level~1 trigger system (Section~\ref{sec:l1}) and to the SFE TDC with a
1.03~ns time bin.
After digitization, amplitude and time information is sent to the Level~2
trigger system (Section~\ref{sec:l2}) and to the data acquisition system
(Section~\ref{sec:daq}). The amplitude is measured
in one out of sixteen channels per event for counter response monitoring.
\subsubsection{Shielding}
Three sources contribute to non-muon background in the central and forward muon
systems: {\it i}) scattered proton and antiproton fragments that interact with
the end of the calorimeter or with the beampipe produce background in the
central and forward A layer; {\it ii}) proton and antiproton fragments
interacting with the Tevatron low-beta quadrupole magnets produce hits in
the B and C layers of the forward muon system; and {\it iii}) beam halo
interactions from the tunnel. Shielding installed in the accelerator tunnel
during Run~I \cite{run1-shielding} significantly reduced the background from
beam halo. New shielding has been installed for Run~II to reduce the
background due to proton and antiproton remnants. Reduction in backgrounds
along with the use of radiation-hard detectors helps ensure the long-term,
reliable operation of the muon system.
The shielding consists of layers of iron, polyethylene, and lead in a steel
structure surrounding the beam pipe and low-beta quadrupole magnets. Iron is
used as the hadronic and electromagnetic absorber due to its relatively short
interaction (16.8~cm) and radiation (1.76~cm) lengths and low cost.
Polyethylene is a good absorber of neutrons due to its high hydrogen
content. Lead is used to absorb gamma rays.
The position of the shielding is shown in Figure~\ref{fig:detector}. It
extends from the end calorimeter cryostat, through the end toroid magnet, to
the wall of the collision hall. It consists of three
rectangular cross section pieces that are 84", 85" and 60" long, starting at
the calorimeter and moving away from the detector center. The piece closest
to the toroid has a 20" square hole at the center
followed by 16" of iron, 6" of polyethylene, and 2" of lead. The two outer
pieces are identical except that the hole is 25" square followed by 20" of
iron. The most-forward section is split vertically down the center so that it
can be moved out of the way when the end toroid magnet is repositioned for
access to the central parts of the detector.
Figure~\ref{fig:c-layer-photo} shows the shielding in the ``open'' position.
Monte Carlo studies based on MARS code \cite{mars1,mars2}
show that the shielding provides a factor of 50--100 reduction in the energy
deposition in muon detector elements. Reduction of backgrounds reduces the
occupancy and detector aging, and provides almost background-free muon
triggering and reconstruction. MARS Monte Carlo predictions of the number of
hits in the muon detectors agree with the observed occupancies within 50\%.
\subsection{Muon scintillation counter monitoring}
\label{sec:muon-led-monitoring}
All muon scintillation counters are calibrated and
monitored using an LED-based pulser system~\cite{mu-led-monitoring}.
Given the large number of counters
involved, it is difficult to use cosmic ray and beam muons to check the
timing and do the PMT gain calibration quickly. The
LED system allows us to find dead PMTs, isolate the behavior of the PMTs
from the front-end electronics, adjust the relative timing between channels,
set initial PMT gains, monitor PMT gains, and track timing changes.
A pulser triggers an LED driver board that drives four LEDs embedded in an
acrylic block. The light is further mixed in two acrylic mixing
blocks. The upstream side of each block is frosted by sand-blasting; the sides
of the blocks are polished to maximize total internal reflection. The light is
then distributed to an array of clear optical fibers embedded in a
fourth acrylic block. These light distribution fibers are connected to
fibers which are butted up to each PMT. To match the peak wavelengths and
emission spectra of the wavelength-shifting fibers used in the counters,
different LEDs are used for the central and forward counters. In the central
region, blue-green NSPE510S LEDs from Nichia America Corp. are used; in the
forward region, blue NSPB320BS LEDs, also from Nichia America Corp., have been
installed. The clear fibers are Hewlett Packard HFBR-500 optical fibers.
The light intensity of the LEDs is monitored using a PIN diode (Hamamatsu S6775)
mounted at the downstream end of the first mixing block; this works well
because the mixing blocks
evenly distribute the light to the fiber array. As long as the PIN diode is
stable over time, variation in the gain of the PMTs can be measured
independently of variation in the light output of the LEDs.
Figure \ref{fig:amptime} shows results from the forward muon system monitoring
between July 2001 and January 2003. The ratio of the LED amplitudes
is stable within 4\% with a standard
deviation of about 7\%. The timing stability over this interval shows a peak
stability within 0.2~ns with a standard deviation of about 0.43~ns. The
typical time resolution of counters during data taking is 2~ns, so stability
is not a major factor in the overall time resolution of the system;
photoelectron statistics, beam spot size, and bunch timing are all more
important.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure48.eps}}
\caption{Results from the forward muon system monitoring by the LED calibration
system over one and one-half years; the smooth curves are Gaussian fits to the
data. a) shows the LED amplitude ratio, b) shows the absolute difference in the
LED timing.}
\label{fig:amptime}
\end{figure}
\FloatBarrier
\section{Forward proton detector}
\label{sec:fpd}
The forward proton detector (FPD)~\cite{FPD} measures protons and antiprotons
scattered at small angles (on the order of 1~mrad) that do not impinge upon
the main D\O\ detector. During Run I, such diffractive events were tagged
using a rapidity gap (the absence of particles in a region of the detector),
however a forward particle detector is necessary for access to the full
kinematics of the scattered particle.
\subsection{The detector}
The FPD consists of a series of momentum spectrometers that make use of
accelerator magnets in conjunction with position detectors along the beam line.
The position detectors operate a few millimeters away from the beam and have to
be moved out of the beamline during injection of protons or antiprotons
into the accelerator. Special
stainless steel containers, called Roman pots~\cite{ROMAN}, house the position
detectors, allowing them to function outside of the ultra-high vacuum of the
accelerator, but close to the beam. The scattered $p$ or $\bar{p}$ traverses
a thin steel window at the entrance and exit of each pot. The pots are remotely
controlled and can be moved close to the beam during stable conditions.
The Roman pots are housed in stainless steel chambers called castles. The FPD
consists of eighteen Roman pots arranged in six castles. The castles
are located at various distances from the D\O\ interaction point
and in locations that do not interfere with the accelerator.
The arrangement of the FPD is shown in Figure~\ref{fig:fpdlayout}.
Four castles are located
downstream of the low beta quadrupole magnets on each side of the collision
point: two on the $p$ side (P1 and P2) and two on the $\bar{p}$ side (A1 and
A2). Each of these quadrupole castles contains four Roman pots arranged to
cover most of the area around the beam. Two castles (D1 and D2) are located on
the outgoing $\bar{p}$ side after the dipole magnet. Each of these dipole
castles contains only one Roman pot.
There are nine spectrometers: the two dipole castles form one, and on each side
of the interaction region the two up, two down, two in, and two out pots are
paired to form the other eight (Figure~\ref{fig:fpdlayout}).
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure49.eps}}
\caption{FPD layout. Quadrupole castles are designated with a leading
P or A when placed on the
$p$ side or the $\bar{p}$ side, respectively; the number designates the station
location; while the final letter indicates pot potition (U for up, D down,
I in, O out). D1I and D2I are dipole castles.}
\protect\label{fig:fpdlayout}
\end{figure}
\subsubsection{The position detector}
The construction of the position detectors is illustrated in
Figure~\ref{fig:fpd-cartoon}. Each detector is made of 0.8-mm-thick
double-clad square scintillating fibers (Bicron BCF10 \cite{bicron}) bundled
in groups of four parallel fibers, forming a
scintillating structure measuring 0.8~mm $\times$ 3.2~mm. One end of the
detector element is aluminized (about a 3-$\mu$m-thick layer) to increase
the light yield and the other end of each scintillating fiber is spliced to a
double-clad clear fiber of square cross section (Bicron BCF98) with the
same dimensions. The use of square fibers gives an increase of
about 20\% in light output compared to round fibers.
The scattered $p$ or $\bar{p}$ goes through 3.2~mm of scintillating
material. The four clear fibers take the light of one element to
a single channel of the Hamamatsu H6568 16-channel multi-anode
photomultiplier (MAPMT), yielding approximately ten photoelectrons.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure50.eps}}
\caption{Detector fibers and the MAPMT.}
\label{fig:fpd-cartoon}
\end{figure}
As shown in Figure~\ref{fig:fpd-cartoon}, each detector consists of six planes
in three views ($u$, $x$ and $v$)
to minimize ghost hit problems and to reduce reconstruction ambiguities. Each
view is made of two planes ($u-u'$, $x-x'$, and $v-v'$), the primed layers
being offset by two-thirds of a fiber with respect to the unprimed layers.
The $u$ and $v$ planes are oriented at $\pm45^\circ$ with respect to the
horizontal bottom of the detector, while the $x$ plane is at $90^{\circ}$.
There are twenty channels in each layer of the $u$ and $v$ planes and sixteen
channels in each of the $x$ layers. There are 112 channels (each with four
fibers) per detector giving a total of 2016 channels in the eighteen Roman
pots. The readout of each detector requires seven MAPMTs and includes a
trigger scintillator read out by a
fast photomultiplier tube (Phillips XP2282 \cite{phillips}). The FPD uses CFT
electronics (Section~\ref{sec:AFE}) for the
scintillating fiber detector read out, luminosity monitor electronics
(Section~\ref{sec:lum-monitor})
for the trigger read out, and the muon LED system for calibration
(Section~\ref{sec:muon}). The trigger manager designed for the muon system
incorporates this information in making a Level~1 trigger decision on FPD
tracks (Section~\ref{sec:l1fpd}).
\subsubsection{The castle}
Figure~\ref{fig:castle} shows an FPD quadrupole castle. It has four arms
(dipole castles are similar, but have just one arm), each containing
a Roman pot housing a detector. The castles are made of 316L stainless steel
and, due to the ultra-high vacuum necessary in the interior of the castles,
all parts were cleaned with demineralized water and alkaline detergent in an
ultrasound bath and dried with a hot air jet before being TIG welded.
The quadrupole castles (dipole castles are not in ultra-high vacuum) are baked
at 150$^\circ$ C when the vacuum has been broken.
A set of hot cathode and convection Pirani-style sensors monitors
the vacuum in the chamber. Each castle has an associated ion pump
to provide the ultra-high vacuum.
\begin{figure}
\centerline{\includegraphics[width=4.5in]{figure51.eps}}
\caption{FPD quadrupole castle.}
\label{fig:castle}
\end{figure}
The castle sits on a stand that allows adjustment of its position in all
directions over a range of 15 mm with an accuracy of 0.1 mm.
The pot is connected to a driving system that makes it possible to move it
perpendicularly to the beam. A 200-$\mu$m-thick window separates the detector
(inside the pot) from the castle ultra-high vacuum. The system is operated by
a step motor and a set of reduction gears allows pot motion with a precision of
approximately $5~\mu$m. A system of cylindrical and conical bearings allows
adjustment of the pot alignment and a linear variable differential
transducer (LVDT) monitors the pot position.
\subsection{Acceptance}
The FPD acceptance is maximized by minimizing the distance
between the detectors and the beam axis. This distance is limited primarily by
interaction with the beam halo which increases as the pots are inserted closer
to the beam. FPD acceptance is determined as a function of $t$, the
four-momentum transfer squared of the scattered proton or antiproton, and
$\xi = 1 - x_p$ where $x_p$ is the fractional longitudinal momentum of the
scattered particle. For the dipole spectrometer, the acceptance is highest for
$|t| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 2$~GeV$^2$/$c^4$, $0.04 \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} \xi \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 0.08$ and extends to
$|t| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 4.3$~GeV$^2$/$c^4$,
$0.018 \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} \xi \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 0.085$ (coverage is incomplete). The acceptance in the
quadrupole spectrometers covers most of the region
$0.6 \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} |t| \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 4.5$~GeV$^2$/$c^4$, $\xi \mathrel{\rlap{\lower4pt\hbox{\hskip1pt$\sim$} 0.1$.
For elastic events, both particles must be detected by diagonally opposite
spectrometers with no activity detected in any other D\O\ subdetector. A
sample of elastic events collected during special runs was used to measure the
position resolution of the FPD by comparing the $x$ coordinate determined by
combining information from the $u$ and $v$ planes to the $x$ coordinate from
the $x$ plane. This process gives a resolution of 130~$\mu$m.
\FloatBarrier
\section{Luminosity monitor}
\label{sec:lum-monitor}
The primary purpose of the luminosity monitor (LM) is to determine
the Tevatron luminosity at the D\O\ interaction
region. This is accomplished by detecting inelastic \mbox{$p\overline{p}$}\
collisions with a dedicated detector. The LM also serves to measure beam halo
rates and to make a fast measurement of the $z$ coordinate of the interaction
vertex.
\subsection{The detector}
The LM detector consists of two arrays of twenty-four plastic
scintillation counters with PMT readout located at $z = \pm 140$~cm
(Figure~\ref{fig:lum-fig1}).
A schematic
drawing of an array is shown in Figure~\ref{fig:lum-fig2}.
The arrays are located in front of the end calorimeters and occupy the
radial region between the beam pipe and the forward preshower detector.
The counters are 15~cm long and cover the pseudorapidity range
$2.7 < |\eta | < 4.4$.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure52.eps}}
\caption{Schematic drawing showing the location of the LM detectors.}
\label{fig:lum-fig1}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure53.eps}}
\caption{Schematic drawing showing the geometry of the LM counters
and the locations of the PMTs (solid dots).}
\label{fig:lum-fig2}
\end{figure}
Scintillation light produced in the Bicron BC-408 scintillator is
detected by Hamamatsu \cite{hamamatsu} R5505Q fine mesh PMTs.
Due to space constraints and the characteristics of the PMTs,
they are mounted on the faces of the scintillators
with the axes of the PMTs parallel to the $z$ axis. They
have no magnetic shielding, and their gain is reduced by a
factor of about 30 when the solenoidal magnet is turned on due to the
approximately 1~T magnetic field in this region~\cite{LM-Ref1}.
The time-of-flight resolution for the counters is about 0.3~ns,
with the dominant contribution to the resolution being the variation in light
path length for particles striking different locations on the scintillator.
Radiation damage is a concern for detectors located this close to the
beams. Much of the radiation dose seen by these detectors comes from
the \mbox{$p\overline{p}$}\ collision products and is thus unavoidable.
The PMTs are exposed to a radiation flux of
about 25~krad/fb$^{-1}$, which is sufficient to cause darkening of
the borosilicate glass window typically used for PMTs.
The R5505Q PMTs have fused silica (quartz) windows which are
largely immune to radiation damage~\cite{hamamatsu}.
The radiation flux increases rapidly with decreasing radius, reaching a level
of approximately 300~krad/fb$^{-1}$ at the innermost scintillator edge.
Based on the radiation damage study in Ref.~\cite{LM-Ref3}, modest
($\approx 10$\%) light loss is expected for the innermost scintillator
edge after 3~fb$^{-1}$.
The scintillation counters are enclosed in light-tight enclosures, with
each enclosure holding twelve counters.
Preamplifiers inside the enclosures amplify the PMT signals
by a factor of five. The fused silica PMT windows are much
more permeable to helium gas than borosilicate glass \cite{LM-Ref2}.
To avoid damage from the widely fluctuating helium concentration
present in the collision hall, the enclosures are purged
with dry nitrogen.
For accurate timing of the PMT signals, low-loss
cables~\cite{LM-Ref4} are used to bring the signals from the detector to
the digitization and readout electronics.
The signals are equalized in time and split into two paths. On one
path, currently in use for luminosity measurements, analog sums are formed
from the PMT signals for
each of the two arrays, which are then timed using a digital TDC to
identify \mbox{$p\overline{p}$}\ collisions \cite{LM-Ref5}. On the other path, which is an
upgrade currently being installed, two types of
custom VME boards provide the required signal processing. Six LM-TDC
boards are used to digitize the time and charge for each PMT
and apply charge-slewing corrections to generate fully calibrated
time-of-flight measurements. A single LM-VTX board utilizes the
measurements made on the LM-TDC boards to calculate the average time
for each counter array and the $z$ coordinate of the interaction vertex.
\subsection{Luminosity determination}
The luminosity {$\mathcal{L}$} is determined from the average number of
inelastic collisions per beam crossing $\bar N_{LM}$ measured by the LM:
${\mathcal{L}} = {f \bar N_{LM} \over \sigma_{LM}}$\
where $f$ is the beam crossing frequency and $\sigma_{LM}$ is the effective
cross section for the LM that takes into account the
acceptance and efficiency of the LM detector \cite{LM-note}.
Since $\bar N_{LM}$ is typically greater than one, it is important to
account for multiple \mbox{$p\overline{p}$}\ collisions in a single beam crossing. This
is done by counting the fraction of beam crossings with no collisions and
using Poisson statistics to determine $\bar N_{LM}$.
To accurately measure the luminosity, it is necessary to distinguish
\mbox{$p\overline{p}$}\ interactions from the beam halo backgrounds. We separate these
processes by making precise time-of-flight measurements of particles
traveling at small angles with respect to the beams.
We first assume that particles hitting the LM detector
originate from a \mbox{$p\overline{p}$}\ interaction and estimate the $z$ coordinate
of the interaction vertex $z_{v}$ from the difference in time-of-flight:
$z_{v} = {c\over 2}(t_- - t_+)$ where $t_+$ and $t_-$ are the times-of-flight
measured for particles hitting the LM detectors placed at $\pm 140$~cm.
Beam-beam collisions are selected by requiring
$|z_v| < 100$~cm, which encompasses nearly all \mbox{$p\overline{p}$}\ collisions
produced by the Tevatron ($\sigma_{z} \approx 30$~cm).
Beam halo particles traveling in the $\pm \hat z$ direction will
have $z_v \approx \mp 140$~cm, and are eliminated
by the $|z_v| < 100$~cm requirement.
Level~1 triggers (Section~\ref{sec:l1}) are grouped together so that they
have common deadtime, i.e., common sources of enable, disable, and readout.
This allows the readout to be partitioned so that different triggers may
read out independent portions of the detector. The luminosity
associated with each trigger takes into account the instantaneous
luminosity, the deadtime, and losses in the data acquisition system.
The luminosity block is the fundamental unit of time for the
luminosity measurement. Each block is indexed by the luminosity block
number (LBN), which monotonically increases throughout Run~II. The LBN
is incremented upon run or store transitions, TFW or SCL initialization, by
request, or after 60~seconds have elapsed. The time period is short
enough so that the instantaneous luminosity is effectively constant
during each luminosity block, introducing negligible uncertainty into
the measurement of the luminosity due to the width of the time slice.
Raw data files are opened and closed on LBN boundaries. Luminosity
calculations are made independently for each LBN and
averaged over the luminosity block.
\subsection{Luminosity data acquisition system}
The luminosity data acquisition system (LDAQ) is a stand-alone
data acquisition system running on the online cluster. The LDAQ was designed
to collect sufficient data to measure, verify, and monitor the luminosity
delivered to and used by D\O. The LDAQ connects to the
following systems: L1~trigger framework (Section~\ref{sec:tfw}),
accelerator controls system,
D\O\ controls system (Section~\ref{sec:controls}), Level~3
(both ScriptRunner and DAQ, Sections~\ref{sec:l3trigger} and \ref{sec:daq}),
COOR (Section~\ref{sec:coor}), and the datalogger (Section~\ref{sec:daq}).
Data from different sources are correlated, loaded into a database, and
used for luminosity calculations.
\FloatBarrier
\section{Triggering}
\label{sec:trigger}
With the increased luminosity and higher interaction rate delivered by the
upgraded Tevatron, a significantly enhanced trigger is necessary to select the
interesting physics events to be recorded. Three distinct levels form this
new trigger system with each succeeding level examining fewer events but in
greater detail and with more complexity. The first stage (Level~1 or L1)
comprises a collection of hardware trigger elements that provide a trigger
accept rate of about 2~kHz. In the second stage (Level~2 or L2), hardware
engines and embedded microprocessors associated with specific
subdetectors provide information to a global processor
to construct a trigger decision based on individual objects as well as object
correlations. The L2 system reduces the trigger rate by a factor of about two
and has an accept rate of approximately 1~kHz.
Candidates passed by L1 and L2 are sent to a farm of Level~3 (L3)
microprocessors; sophisticated algorithms reduce the rate to about 50~Hz and
these events are recorded for offline reconstruction.
An overview of the D\O\ trigger and data acquisition system is shown in
Figure~\ref{fig:trigger-overview}. A block diagram of the L1 and L2 trigger
systems is shown in Figure~\ref{fig:l1-l2-layout}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure54.eps}}
\caption{Overview of the D\O\ trigger and data acquisition systems.}
\label{fig:trigger-overview}
\end{figure}
The trigger system is closely integrated with the
read out of data, as illustrated in Figure~\ref{fig:trigger-overview}.
Each event that satisfies the successive L1 and L2 triggers is
fully digitized, and all of the data blocks for the event are
transferred to a single commodity processor in the L3 farm.
The L1 and L2 buffers play an important role in minimizing the experiment's
deadtime by providing FIFO storage to hold event data awaiting a Level~2
decision or awaiting transfer to Level~3.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure55.eps}}
\caption{Block diagram of the D\O\ L1 and L2 trigger systems.
The arrows show the flow of trigger-related data.}
\label{fig:l1-l2-layout}
\end{figure}
The overall coordination and control of D\O\ triggering is handled by the
COOR package (Section~\ref{sec:coor})
running on the online host. COOR interacts directly with the trigger framework
(for L1 and L2 triggers) and with the DAQ supervising systems (for the L3
triggers).
The data acquisition system responsible for the data flow of the fully
digitized event into L3 is described in Section~\ref{sec:daq}.
\subsection{The Level 1 trigger}
\label{sec:l1}
L1 is implemented in specialized hardware and examines every event for
interesting features. The calorimeter trigger (L1Cal) looks for energy
deposition patterns exceeding programmed limits on transverse energy
deposits; the central track trigger (L1CTT) and the muon system trigger
(L1Muon) compare tracks, separately and together, to see if they exceed preset
thresholds in transverse momentum. The L1 forward proton detector trigger
(L1FPD) is used to select diffractively-produced events by triggering on
protons or antiprotons scattered at very small angles.
All events awaiting L1 trigger decisions are pipelined and thus make minimal
contributions to the deadtime. In order to participate in the trigger
decision, the L1 trigger decision must arrive at the trigger framework in
3.5~$\mu$s or less. The rate of L1 trigger accepts is limited by the maximum
readout rates of the participating subsystems and by a desire to minimize
the deadtime associated with the readout.
\subsubsection{Trigger framework}
\label{sec:tfw}
The trigger framework (TFW) gathers digital information from each of the
specific L1 trigger devices and chooses whether a particular event is to be
accepted
for further examination. In addition, it coordinates various vetoes that can
inhibit triggers, provides the prescaling of triggers too copious to pass on
without rate reduction, correlates the trigger and readout functions, manages
the communication tasks with the front-end electronics and the trigger
control computer (TCC), and provides a large number of scalers that allow
accounting of trigger rates and deadtimes.
The TFW for Run~II is built out of 9U 400~mm cards housed in
customized VME crates. All of the cards in the TFW use the same general
circuit board, the same front and rear panel layout, and the same
connectors and make extensive use of field programmable gate array (FPGA)
technology to implement different functions. A
block diagram of the principal functions of the TFW is shown
in Figure~\ref{fig:framework}.
\begin{figure}
\centerline{\includegraphics[width=5.in]{figure56.eps}}
\caption{Block diagram of the trigger framework.}
\label{fig:framework}
\end{figure}
The functions of the TFW are summarized below:
\begin{itemize}
\item It receives up to 256 ``AND-OR'' terms (bits) from various
parts of the experiment, which are used to form specific triggers.
\item
Up to 128 specific triggers can be programmed using the AND-OR terms.
The ``or'' of all of these triggers determines
whether or not a given crossing had a valid trigger. These are called
``physics'' triggers. Each of the 128 specific triggers has a separate
programmable ``beam condition'' trigger associated with it. For a trigger to
occur, both the ``physics'' trigger and ``beam condition'' conditions must be
satisfied.
\item
All the resources, including the triggers, are programmed from COOR
(Section~\ref{sec:coor}) via
text-based commands interpreted in the TCC which then configures the TFW.
\item
Any trigger can be prescaled under program control. The TFW
manages the prescale ratios.
\item
At the level of data acquisition, the
detector is divided into subsections or groups of subsections served
by a single serial command link (SCL) called geographic sectors.
Typical geographic sectors are detector front-ends but other systems such as
the TFW readout are also geographic sectors. The TFW supports up to 128
geographic sectors and any specific trigger can be programmed to
request digitization and readout of any subset of the 128 geographic sectors.
This permits a partitioning of the data
acquisition system into several separate systems, which is particularly
useful during setup, installation, and calibration. Each geographic sector
receives an L1Accept signal from the TFW over the SCL, followed by an
L2Accept or L2Reject from the L2 global processor.
\item
The TFW can be programmed so that only particular bunch crossings
contribute to a given trigger.
Beam crossing numbers and L1Accept numbers provide a unique identifier for
events.
\item
The TFW provides all of the scalers needed to count the triggers as well as
scalers to monitor the live time of combinations of these triggers
(known as exposure groups).
\item
The TFW offers an extensive menu of lower level and specialized commands for
diagnosis of the trigger hardware.
\item
TFW scalers and registers are accessed through the TCC to perform trigger
system programming, diagnostics, and monitoring.
\item
During Run~I, elements of the L1 trigger could only be ANDed to give a
specific trigger. For Run~II, the capability to allow both ANDs and ORs of
trigger terms has been added to the TFW, using resident firmware.
The resulting combinations are called pseudo-terms, since they are not
formed in hardware. The pseudo-terms have the ability
either to sharpen trigger turn-on curves, or to reduce combinatorial
backgrounds and thus reduce the rate of low-$p_T$ triggers.
\end{itemize}
\subsubsection{Level~1 calorimeter trigger}
\label{sec:l1cal}
The underlying architecture of the L1 calorimeter trigger for Run~II is
the same as it was for Run~I \cite{l1cal-run1}. There are, however, a number
of improvements in the way the data are received by the trigger and the way the
results are extracted.
\paragraph{Summary of operational principles}
The trigger inputs consist of electromagnetic (EM) and hadronic (H) trigger
tower energies made up from sums in depth and transverse coordinates
($\Delta\eta \times \Delta\phi = 0.2 \times 0.2$) of fast analog pickoffs from
the BLS circuits. There are 12 EM towers and 1280 H
towers: forty slices in $\eta$, covering the region $|\eta| < 4$, and
thirty-two slices in $\phi$, covering the full $2\pi$ of the azimuth.
The tower energies are converted to $E_T$ on input, and have pedestals
subtracted and energy scales adjusted if necessary. The variables used in
trigger calculations are actually the EM transverse energies and the total
transverse energies (EM+H) formed by adding the corresponding EM and H towers.
The triggers available for use in the experiment include:
\begin{itemize}
\item Global variables ~\\
There are two global variables: $\Sigma E_T$, the sum of all tower $E_T$s, with
four thresholds; and \met, again with four thresholds.
\item Local variables ~\\
Each EM tower and each EM+H tower is compared to four programmable $E_T$
threshold sets. A given threshold set could contain different values of $E_T$
for each tower although in practice they are all set to the same value. A bit
is set if a tower exceeds its reference value. (A feature vetoing the EM
tower if its corresponding EM+H tower exceeds a programmed reference value is
available but has never been used.) The number of EM towers
and EM+H towers exceeding their thresholds in a given reference set is counted
and trigger bits are generated if this count exceeds any of several programmed
count limits. We chose to use two such count limits for EM towers
and four for EM+H towers.
\item Large tiles ~\\
Since the trigger towers are small relative to the size of jets, we use
some of the partial $E_T$ sums needed for the \met\ calculation to trigger on
jets. Conveniently available are sums covering $4 \times 8$ trigger towers in
$\eta \times \phi$, corresponding to an area of 1.28 in
this space. For generating trigger bits, four reference sets are
available with two count thresholds for each set but only for EM+H and not
for EM-only large tiles.
\end{itemize}
\paragraph{Modifications for Run~II}
A number of changes were made to the calorimeter trigger in anticipation of
the higher crossing rates and luminosities expected during Run~II.
\begin{itemize}
\item{Signal receiver electronics} ~\\
In Run I, the conversion of the BLS energy signals to properly-scaled $E_T$
values was done on the calorimeter trigger front-end (CTFE) cards by a network
of hard-wired precision resistors, making gain adjustments difficult. For
Run~II, we bypassed this problem by building new front-end
receiver cards capable of gain adjustment up to a factor of two for
each channel under computer control. This change also allowed us to add
circuitry to decrease the trigger signal rise times to be compatible with the
132~ns accelerator bunch spacing originally anticipated for Run~II. Each card
processes the signals from two
trigger towers and two such cards were needed to upgrade each Run~I CTFE card.
\item{Readout electronics} ~\\
To speed up the readout of L1 calorimeter trigger data and
therefore improve the L1 trigger accept rate, the old readout system of
two eight-bit buses was replaced with ten sixteen-bit buses. With this
improvement, the L1Cal accept rate can exceed 10~kHz.
\item{Trigger coverage} ~\\
Trigger tower energy data are read out for the full calorimeter.
Because of signal to noise considerations, only the trigger towers for
$|\eta| < 3.2$ are used for triggering. In addition to the
calorimeter trigger towers, ICD towers are embedded in the data as well.
\end{itemize}
\subsubsection{Level~1 central track trigger}
\label{sec:l1ctt}
The L1CTT~\cite{l1ctt-1} reconstructs the trajectories of
charged particles using fast discriminator data provided by three
scintillator-based detectors: the central fiber tracker
(Section~\ref{sec:CFT}) and the central and forward preshower detectors
(Section~\ref{sec:preshower}).
Data processed by the L1CTT are used to make L1
trigger decisions, otherwise known as trigger terms. While the L1CTT is
optimized for making fast L1 trigger decisions, the electronics also store
more-detailed event data (e.g. sorted lists of tracks and preshower clusters)
for later L2/L3 readout, or for use as seeds by other D\O\ trigger systems.
\paragraph{Overall system design}
The input to the L1CTT system consists of the discriminator bits generated on
the AFE boards every 132~ns. These discriminator bits are sent from the AFE
boards over point-to-point low voltage differential signal (LVDS) links to
chains of digital front-end (DFE) boards. Each of the DFE boards in the
L1CTT system is built
using a common 6~U $\times$ 320~mm motherboard that supports one or
two daughterboards. Currently there are two different daughterboard layouts
with different sizes and numbers of FPGAs. A transition board allows a
DFE motherboard to drive LVDS, fiber
optic, and coaxial copper links. L1CTT-specific protocols define the data
format for communication between all DFE boards and consumers. This hardware
modularity, when coupled with the flexibility of FPGAs, enables
application-specific functionality to be located in firmware,
minimizing the number of unique DFE boards in the system.
The L1CTT system comprises three subsystems: CFT/CPS axial, CPS stereo, and
FPS. Of these, the CFT/CPS axial and FPS subsystems provide L1 trigger terms
to the trigger framework. All three subsystems participate
in L2/L3 readout by sending track and cluster lists to various preprocessor
and readout crates.
\paragraph{L1CTT subsystems}
\begin{itemize}
\item{CFT/CPS axial} ~\\
The CFT/CPS axial subsystem (Figure~\ref{fig:cft-cps-axial}) is designed to
provide triggers for charged particles with $p_T > 1.5$~GeV/$c$.
In addition to finding tracks, it
must also find CPS clusters, match tracks to clusters, and
report the overall occupancy of the CFT axial layers. Significant resources
are allocated for triggering on isolated tracks.
The CFT/CPS axial system also supplies
the L1Muon and L2STT systems with lists of seed
tracks, and sends track and cluster information to the L2CTT and L2PS
preprocessors (Figure~\ref{fig:l1-l2-layout}).
\begin{figure}
\centerline{\includegraphics[width=4.in]{figure57.eps}}
\caption{The CFT/CPS axial components of the L1CTT system. The numbers in the
top left corners of the boxes indicates the number of board of each type.}
\label{fig:cft-cps-axial}
\end{figure}
For mechanical reasons, the CFT and CPS axial fibers are grouped by cylinder
layer before routing to the AFE boards. However, the track finder algorithms
require the fiber information to be arranged in 4.5$^\circ$ sectors in the
transverse plane (Figure~\ref{fig:trig-sector}). Furthermore, each
track-finder daughterboard must receive information from each of its
neighboring sectors to find tracks that cross sector boundaries.
A set of twenty mixer boards \cite{ctt-mixer} handles this data reorganization
and duplication. After data duplication, the mixer output data rate is a
constant 475~Gbits/s. Total latency through the mixer system is
200~ns.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure58.eps}}
\caption{Transverse schematic view of a single 4.5$^\circ$ sector. A
hypothetical track is overlaid on the eight CFT axial doublet layers and CPS
axial layer. The track equations require a fiber hit on all eight CFT axial
layers. Note that the CPS is relatively farther from the CFT than shown.}
\label{fig:trig-sector}
\end{figure}
From the mixer system, the properly organized discriminator bits are sent to
the first tier of DFE boards. This first tier consists of
forty motherboards, with each motherboard having two daughterboards
(DFEAs). Each DFEA daughterboard unpacks the CFT data and compares the fiber
hits against approximately 20,000 predefined track equations. To
minimize latency, this operation is performed in parallel using combinatorial
logic in FPGAs. Device resources are balanced by grouping track equations into
four $p_T$ bins, with each $p_T$ bin corresponding to a single FPGA, as shown
in Table~\ref{tab:track-equation}.
\begin{table}
\begin{center}
\caption{Track equation distribution in each track-finder FPGA}
\label{tab:track-equation}
\begin{tabular}{lccc}
\hline
& $p_T$ range & & FPGA Resources \\
$p_T$ bin & (GeV/$c$) & Track equations & (system gates) \\
\hline
Maximum & $> 10$ & 3000 & 200k \\
High & 5 -- 10 & 3000 & 200k \\
Medium & 3 -- 5 & 4000 & 300k \\
Low & 1.5 -- 3 & 10000 & 500k \\
\hline
\end{tabular}
\end{center}
\end{table}
Each track-finder FPGA outputs the six highest-$p_T$ tracks it finds; these
tracks are passed to a fifth FPGA for additional processing which includes
sorting, matching tracks and CPS clusters, counting tracks, and calculating
sector occupancy and total $p_T$. Tracks from each sector are sent over
gigabit coaxial cables to L1Muon, where the tracks are matched
to hits in the muon detector. Counts of tracks and occupancy data are passed
downstream to the octant boards over LVDS links.
The next tier of DFE boards (CTOC) collects and sorts data within an octant
(ten sectors). The CTOC boards receive track counts and sector
occupancy data from the DFEAs and then sum up the number of
tracks, determine which sector had the most fibers hit, and check for isolated
tracks. This information is passed on to a single DFE board (CTTT)
where the individual trigger term bits are generated and sent to the trigger
framework within 2.5~$\mu$s of the beam crossing. The CTTT can provide up to
96 trigger terms of which 55 are currently defined. Two examples of CTTT
trigger terms are {\it i}) at least one track above a
$p_T$ threshold of 1.5, 3, 5, or 10~GeV/$c$ and {\it ii}) at least one
isolated track above 5 or 10~GeV/$c$ and at least one track with a
confirmation in the CPS for triggering on electrons.
The TFW issues an L1Accept control bit by
considering trigger term bits from many different
subsystems and makes a global L1 decision. Upon receiving an L1Accept, the
AFE boards digitize the fiber data from the appropriate analog buffers in the
analog pipeline with
8-bit resolution. When the L1Accept control bit embedded in the data
stream reaches the DFEA boards, they send a sorted list of tracks
and CPS clusters to the CTOC boards. At this processing stage, finer $p_T$
information is available than that used at L1. The lists are sorted by
the CTOC and CTQD boards and passed downstream to preprocessor crates for
more detailed analysis. These track lists, remapped onto the geometry of
the silicon tracker, are also used as seeds for the L2STT. Additionally
the CTOC and CTTT boards send copies of their input data to the L3 readout
system for monitoring and debugging.
\item{CPS stereo} ~\\
The CPS stereo subsystem (Figure~\ref{fig:cps-stereo}) provides
information on the clusters in the two stereo layers of the CPS. Unlike the
other L1CTT subsystems, CPS stereo does not generate L1 trigger terms. Rather,
the DFE stereo (DFES) boards store discriminator bits and begin processing
only after an L1 trigger decision has been made.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure59.eps}}
\caption{CPS stereo subsystem hardware.}
\label{fig:cps-stereo}
\end{figure}
Event buffers in the DFES FPGAs store discriminator bits sent from the AFE
boards. Upon receipt of the L1Accept control bit, the DFES boards extract
the discriminator bits and search for clusters of hits in the stereo layers.
Sorted lists of CPS clusters are sent on to the L2PS preprocessor for
additional analysis. The DFES boards also send a copy of either their
inputs (the discriminator bits) or their L2 outputs (the hit clusters)
to the L3 readout system.
\item{FPS} ~\\
The FPS subsystem (Figure~\ref{fig:ctt-fps}) produces its own set of L1
trigger terms, which are passed to the TFW. The overall
structure of the FPS subsystem is similar to the CFT/CPS axial subsystem in
that the FPS has three tiers of DFE boards: a finder (DFEF), a concentrator
(FPSS), and a trigger term generator (FPTT).
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure60.eps}}
\caption{FPS subsystem hardware.}
\label{fig:ctt-fps}
\end{figure}
The DFEF boards receive discriminator bits from the AFE boards and search for
clusters of hits in the FPS fiber layers. A list of clusters is stored in the
DFEF for later L2 readout while counts of clusters are passed downstream to
the FPSS boards. The FPSS boards sum these cluster counts and pass this
information to the FPTT, where the L1 trigger terms are produced.
When the TFW issues an L1Accept, the FPS subsystem switches
into readout mode. The AFE boards begin digitizing and reading out fiber data,
while the DFEF boards extract cluster lists from their buffers. These
lists of clusters are concatenated and sorted by the FPSS boards before being
sent to the L2PS preprocessor. The FPSS and FPTT boards send their L1 input
data to the L3 readout system.
\item{STT} ~\\
The STT subsystem, described in detail in Section~\ref{sec:l2stt},
is used to map the DFEA outputs onto the sixfold STT/SMT geometry.
After an L1Accept has been issued, lists of L1CTT seed tracks generated by
the DFEAs are concatenated and reformatted for the L2STT
by an L1CTT subsystem consisting of two types of boards:
six STOV (``overlay'' or ``overlap'') and six STSX (``sextant'') boards
(Figure~\ref{fig:cft-cps-axial}). Two types of board are necessary
because the maximum number of input/output links on each
board does not allow the remapping to be done in one set of boards.
The DFEAs have duplicate outputs: one set feeds the CTOCs,
the other the STOVs (DFEA outputs shared between L2STT sextants) and STSXs
(outputs unique to a sextant). The STOV outputs are fed into the STSXs; the
STSXs send the data (lists of L1CTT tracks) via six optical fibers to the
L2STT system.
\end{itemize}
\subsubsection{Level 1 muon trigger}
\label{sec:L1Muon}
L1Muon looks for patterns consistent with
muons using hits from muon wire chambers, muon scintillation counters,
and tracks from the L1CTT. Field
programmable gate arrays are used to perform combinatorial
logic on roughly 60,000 muon channels and up to 480 tracks from L1CTT
for every bunch crossing. Data from the detector front-ends are
transmitted on custom Gbit/s serial links over coaxial cable.
The serial link receivers and FPGAs are located on VME cards that
reside in four custom VME crates on the detector platform.
The muon system (and L1Muon) is divided into central, north, and south
regions. Each region is further divided into octants. Front-end data
from each octant are processed by two L1Muon trigger cards (Figure
\ref{L1Muon_overview}). The scintillator trigger cards (MTC05) match
central tracks to muon scintillator hits while the wire trigger cards
(MTC10) match scintillator-confirmed track stubs in wire chambers
between the two or three layers of the muon system. The octant decisions
from each MTC05/MTC10 pair in a region are summed in the muon trigger
crate managers (MTCMs) and sent to the muon trigger manager (MTM).
The MTM forms 256 global L1Muon triggers and sends up to 32 of these to
the TFW.
The download of the specific triggers is handled via EPICS software
(Section~\ref{sec:controls}).
The total latency of the L1Muon trigger is about 3.20~$\mu$s,
driven by the central wire chambers (PDTs) and tracks from L1CTT.
\begin{figure}
\centerline{\includegraphics[width=5.in]{figure61.eps}}
\caption{Level~1 muon trigger system overview. Each octant has
two trigger cards that process muon detector hit information and
L1CTT tracks for that octant. The octant triggers for a given
region are summed on the MTCM and sent to the MTM, which
combines the three regions and sends up to 32 triggers to the TFW.}
\label{L1Muon_overview}
\end{figure}
\paragraph{L1Muon hardware}
All detector inputs to L1Muon use Gbit/s serial links.
The transmitters and receivers are 1.5'' $\times$ 2.2''
daughter cards that are mounted on the muon front-end boards and on the L1Muon
trigger cards. Each serial link can transmit up to $16 \times 7 = 112$ bits
every 132~ns crossing. All MTC05, MTC10, and MTM trigger cards
use a common motherboard with sixteen serial links and different
flavor daughter cards that perform the MTC05, MTC10, and MTM logic.
The MTC05 cards match tracks from L1CTT to hits in the muon
scintillator system. Each octant trigger card receives tracks from
the L1CTT for the ten 4.5$^\circ$ sectors in that octant plus one
sector of overlap on either side. Each sector sends the six highest-$p_T$
tracks to L1Muon, and each track contains the CFT fiber position in the
outer layer, $p_T$ value, and sign of the track curvature in the central
magnetic field. The triggers formed by the MTC05 cards include loose (track
matched to A-layer scintillator) and tight (track matched to a
scintillator road using the A and B layers) for four $p_T$ thresholds
(roughly 1.5, 3, 5, and 10~GeV/$c$). Loose and tight scintillator-only
triggers are also formed.
The MTC10 cards form triggers based on wire hits. In the central
region, the hits from the wire chambers (PDTs) are sent directly to
the trigger cards. The hits for each layer are used to form track
stubs, or centroids, which are then used to confirm scintillator hits
in each layer. Triggers are formed by matching centroid-confirmed
scintillator hits between layers. In the forward region, the centroid
finding is done by separate centroid-finding cards (MCENs), which
subsequently send the centroids to the MTC10 cards. The MTC10 cards
then use the centroid-confirmed scintillator hits to form loose
(A-layer) and tight (A- and B-layer) triggers.
The data from the various front-end systems arrive asynchronously at
L1Muon and must be synchronized before triggers can be formed for a
given event. To accomplish this, all received data are written
directly into FIFOs which are initially empty. When all FIFOs are
not empty (i.e., they have all received data for the first bunch
crossing), the data are read from the FIFOs and sent to the MTC05,
MTC10, or MTM cards for trigger formation. In addition to
synchronizing the data for a given event, the trigger cards also
buffer the input data and trigger decisions pending global L1 and
L2 trigger decisions. The input data and trigger decisions are
stored in dual port memories and a pointer to the data is
written into a FIFO. When an L1 or L2 accept is received,
the pointer is used to read the data for a particular event. The L1Muon
trigger can also send all of the received input data from the detector
front-ends to aid debugging.
\subsubsection{Level 1 forward proton detector trigger}
\label{sec:l1fpd}
L1FPD selects events in which the outgoing beam particles pass through one
or a combination of the nine FPD spectrometers (Section~\ref{sec:fpd}).
A schematic view of the
L1FPD design is shown in Figure~\ref{fig:l1fpd-flow}.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure62.eps}}
\caption{A schematic view of the L1FPD trigger. The dashed boxes indicate
Run~I surplus boards; the bold boxes indicate Run~II standard; and the thin
boxes indicate custom FPD electronics. CuSL is a copper serial link.}
\label{fig:l1fpd-flow}
\end{figure}
The signals produced in the FPD scintillating fibers are read out via
16-channel MAPMTs, and are then amplified and shaped before being passed to
the AFE boards. In addition to providing analog fiber charge integration
information to the L3 data acquisition system, the AFE boards pass
discriminated signals via LVDS connections to
three DFE boards every 132~ns. At the DFE boards, pixel hit pattern
recognition (a pixel is defined to be a $u$, $x$, and $v$ fiber crossing)
is used to compare the fiber hit information with detector
hit patterns stored in FPGAs. Three
FPGAs containing between 600,000 and 1.5 million gates store
generated L1 hit patterns for the nine spectrometers. The DFE boards also
send their processed data to the L3 trigger system.
The relative positions of the fibers in the different planes of each
detector are used to define a finely segmented grid in which to search for
hit patterns with optimal spatial resolution. Hit trigger conditions are set
with different thresholds corresponding to different numbers of fiber layer
hit coincidences.
The FPD trigger manager (TM) searches for
coincidences between the discriminated hit signals of both position
detectors of any FPD spectrometer. We use a set of nine FPD spectrometer
(single diffractive) triggers with differing hit thresholds to
select events in which at least
one of the outgoing beam particles leaves the interaction region intact.
Coincidences between the spectrometers on different sides of the
interaction region are required to trigger on events in which both
outgoing beam particles leave the interaction region intact in
back-to-back configurations (elastic diffractive
triggers) and in the larger set of other two-spectrometer configurations
(double pomeron triggers).
To reduce the contribution from beam halo spray particles that mimic a trigger
signal, events with large hit multiplicities are rejected.
The FPD information can be combined with timing information from
the FPD trigger scintillator photomultiplier tubes (FPD LM), veto
counters, and the luminosity monitor (D\O\ LM).
The decision is then passed to the TFW.
\subsection{The Level~2 trigger}
\label{sec:l2}
The L2 trigger provides detector-specific preprocessing engines
and a global stage (L2Global) to test for correlations in physics signatures
across detector subsystems.
The L2 trigger system was designed to handle input rates of
up to 10~kHz with a maximum accept rate of 1~kHz.
L2 preprocessors collect data from the front-ends and L1
trigger system and analyze these data to form physics
objects. L2 can also combine data across detectors to form higher
quality physics objects and examine event-wide correlations in all L2 physics
objects. The
L2Global processor selects events based on the set of 128 selections
applied at L1 and additional script-controlled criteria.
Events passing L2 are tagged for full readout and further
analysis in the L3 trigger.
\subsubsection{Architecture and components of the L2 trigger}
L2 includes preprocessors for each detector subsystem and
a global processor for integration of the data. Preprocessor subsystems
include tracking, calorimeter, preshower, and muon systems. The subsystems
work in parallel and trigger decisions are made in the L2Global stage
based on physics objects reconstructed in the preprocessors.
Preprocessing is performed either with serial CPU-based cards or with
CPU cards plus highly parallelized DSP or programmable logic-based cards.
The preprocessor and global stages function as 2- or 3-stage
stochastic pipelines as illustrated in Figure~\ref{fig:l2_connections}.
Data arrive at the L2 system via three transmission protocols.
Calorimeter and tracker data and signals from the TFW are
transmitted by 1.3~Gbit/s serial G-Links~\cite{l2-glink} on optical fibers.
The muon system uses 160~Mbit/s Cypress Hotlink~\cite{l2-cypress-hotlink}
transmitters on coaxial cables or standard CAT/6 cables,
unshielded twisted pair (UTP) Hotlinks.
\begin{figure}
\centerline{\includegraphics[width=6.in]{figure63.eps}}
\caption{L2 data paths and connections. Data paths are labeled according
to transmission protocol, either 1.3~Gb/s fiber G-Links, 160~Mb/s copper
Cypress Hotlinks, or 1.4 Gb/s copper AMCC links. FIC indicates a fiber
converter card, MBT an Mbus transceiver card, and $\beta$ an L2$\beta$eta card.
The number of transmission cables is given for each pathway.}
\label{fig:l2_connections}
\end{figure}
\paragraph{L2 crates}
The L2 system occupies 9U VME crates built to the VME64/VIPA \cite{VIPA}
standard. In addition to a 64-bit VME bus, these crates are instrumented with
a custom high-bandwidth 128-bit bus (Magic Bus~\cite{l2-mbus} or Mbus)
for fast intra-crate data flow and communication. The Mbus
supports data rates of up to 320~Mbit/s.
The VME backplane is used to read out events accepted by L2 into L3 and for
control and monitoring operations.
A typical L2 preprocessor/global crate contains the following devices:
\begin{itemize}
\item VME controller and dual-port-memory (DPM) card used for
downloading runtime parameters and for reporting monitoring data to the TCC
\item a VME single-board computer (SBC) used to send data to L3
\item one or more Mbus transceiver (MBT) cards (Hotlink-to-Mbus
interface, Section~\ref{sec:mbt})
\item one or more sets of fiber converter (FIC) and VME transition (VTM)
cards (fiber-to-hotlink interface, Section~\ref{sec:mbt})
\item one or more L2$\beta$eta processor cards (VME and Mbus hosted
SBCs, Section~\ref{sec:beta}) for data processing.
\end{itemize}
The layout of the preprocessor/global crates is shown in
Figure~\ref{fig:l2-crate}. L2STT (Section \ref{sec:l2stt}) and
L2Muon (Section \ref{sec:l2muon}) use additional specialized cards for
processing their data before sending track
information to the L2CTT and L2Muon preprocessors respectively.
\begin{figure}
\centerline{\includegraphics[width=3.5in]{figure64.eps}}
\caption{Typical layout of cards in an L2 preprocessor crate.
Cards that could be installed in parallel to increase input channels
or processing power are designated by stacked boxes.}
\label{fig:l2-crate}
\end{figure}
\paragraph{Magic bus transceiver and data conversion cards}
\label{sec:mbt}
Data arriving at L2 must first be converted into the correct
physical format before processing can begin. The coaxial Hotlink
data from the muon system is received by analog cable input
converter (CIC) cards. These cards retransmit the data to
the first stage of L2Muon as differential UTP Hotlink signals. A relative
of the CIC card, the serial fanout (SFO), selectively fans out 1 to 12
UTP Hotlinks where signal duplication is necessary.
Similarly, FIC cards receive G-Link data
and buffer up to sixteen events while converting the inputs to differential
UTP Hotlink signals. The incoming fibers are received by a VME transition
module (VTM) located at the rear of the crate.
Data are passed to L2$\beta$eta processors via the Mbus.
An MBT card transfers data to and from the Mbus.
The MBT accepts up to seven Hotlink inputs, assembles the data into events,
and broadcasts them on the Mbus backplane to the L2$\beta$eta cards.
The high speed, 128-bit input path is equipped with a buffer
for up to sixteen events. There are two Hotlink
output paths that are targets of the 128-bit Mbus programmed
I/O on the L2$\beta$eta cards. These
are used by preprocessors to send their output to L2Global.
One MBT, designated the ``pilot,'' coordinates the event broadcast across all
MBTs and receives SCL information for the crate from the TFW, including the
L1Accept information, L2 decision information, and SCL initialize messages. The
pilot MBT in the L2 global crate transmits the L2 decision to the TFW over
sixteen PECL output channels. A demultiplexing card receives eight 16-bit words
and transmits the 128-bit decision to the TFW.
\paragraph{L2$\beta$eta processors}
\label{sec:beta}
L2$\beta$eta processor cards are used for analysis and control of
data flow in each preprocessor subsystem crate. The L2$\beta$eta processors
replaced the original L2Alpha boards\footnote{L2Alpha processors were similar
to those constructed by the CDF
experiment~\cite{l2-alpha-1,l2-alpha-2}. They were based on the PC164
motherboard manufactured by Digital Semiconductor (now Compaq). Each card
contained a 500~MHz 21164 64-bit Digital Alpha CPU and 21164 PCI
interface integrated with the necessary hardware interfaces.
The L2Alphas have been supplanted by the higher performance and
upgradable L2$\beta$eta cards.} early in Run~II and
are composed of commercially produced (Compact PCI
standard~\cite{l2-compact-pci}) SBCs mounted on 6U-to-9U VME adapter cards.
Each SBC provides dual 1~GHz Pentium processors.
The SBC connects to the adapter via a 64-bit, 33/66 MHz PCI bridge.
The 9U adapter, controlled by the SBC, implements
all D\O -specific protocols for Mbus and
TFW connections. Custom I/O functions on this card are
implemented in a single FPGA (Xilinx XCV405E) plus assorted logic converters
and drivers. The FPGA is used to implement
internal data FIFOs and address translation tables for
broadcasting data from the Mbus to CPU memory, reducing the complexity
of the adapter card. Mbus programmed I/O and various other trigger system
interfaces, via front panel connections and VME user-defined pins,
are also implemented.
The adapter also provides a 64-bit PCI-to-VME interface via
a Tundra Universe II chip~\cite{U2}.
The L2$\beta$etas run a GNU/Linux system and all programs
are written in C++. Real time trigger performance is achieved by
restricting system calls (including dynamic memory allocation)
and promoting the L2 executable to a non-interruptible
scheduling queue. The second CPU provides full access to the
SBC for runtime monitoring and control.
\subsubsection{L2 preprocessor subsystems}
\paragraph{L2Cal}
\label{sec:l2cal}
The calorimeter preprocessor system identifies jets and electrons/photons
and calculates event \met\ for
the global processor. The worker code may be combined to
run serially in a single processor or placed
in separate processor cards to increase throughput as data rates and
detector occupancies grow with luminosity.
Each processor uses the $E_T$ data from the 2560 calorimeter
trigger towers.
The input data arrives on ten input links which together transport
3~kB/event of tower transverse energy data, including both EM towers
and the EM+H tower sums.
The jet algorithm operates by clustering $n\times n$ (currently $n=5$)
groups of calorimeter trigger towers which are centered on seed towers.
The seed towers are $E_T$-ordered with $E_T \ge 2$~GeV.
Overlapping candidates may be reported as separate jets depending
on programmable criteria based on the number of shared towers (otherwise the
highest-$E_T$ jet of the overlapping candidates is kept).
The list of jets is passed to L2Global which applies jet requirements
as defined by the trigger menu.
The electron/photon algorithm begins by defining an $E_T$-ordered list of
EM towers with $E_T$ above 1~GeV. For each seed tower, the neighboring
tower with the largest $E_T$ is combined with the seed to make an EM cluster.
The EM energy fraction of the leading- and sub-leading-$E_T$ trigger towers
of the cluster and the amount of total $E_T$ in a $3 \times 3$ tower array
surrounding the seed tower of the cluster are used to reduce background.
The final list of electron candidates is sent to L2Global
to apply the trigger requirements.
The L2 calorimeter \met\ algorithm
calculates the vector sum $E_T$ from the individual trigger tower total-$E_T$
energies passed from L1. It is capable of calculating the \met\ for
different minimum tower $E_T$s and $\eta$ ranges.
\paragraph{L2Muon}
\label{sec:l2muon}
L2Muon uses calibration and more precise timing information to improve the
quality of the muon candidates~\cite{l2mu-1}. It receives the L1Muon output
and data from approximately 150 front-end modules
(from the PDTs, MDTs, and the scintillation counters). The muon
candidates contain the track $p_T$, $\eta$ and $\phi$
coordinates, and quality and timing information.
L2Muon implements one extra level of preprocessing in the
stochastic pipeline sequence. The first L2Muon stage incorporates eighty
200-MHz DSPs in a parallel processing scheme. Each DSP is
responsible for finding track segments in a small region of the
detector, so that the total execution time of the algorithms
is independent of the number of hits. The DSPs are geographically
organized in eleven central and five forward 9U VME boards (second level input
computers or SLICs), with each SLIC containing one administrator and four
worker DSP chips. The SLICs are programmable in C. The layout of an
L2Muon crate is shown in Figure~\ref{fig:slic_crate}. The DSP algorithms,
characterized by the detector region (central or forward) and the input or
sub-detector plane (L1, A or BC muon layers), make use of
detector symmetry to
run the same basic processing code. The MBT
sends the stubs found by the SLICs to the L2$\beta$eta processor.
The L2$\beta$eta board uses the track segments to construct
integrated muon candidates with an associated $p_T$
and quality.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure65.eps}}
\caption{Layout of cards in an L2Muon crate. Two stages of
processing are completed in a single crate using SLICs and
L2$\beta$eta processors. Central and forward muon regions are processed
in separate crates of similar configuration.}
\label{fig:slic_crate}
\end{figure}
\paragraph{L2PS}
Both the central and the forward preshower detectors are designed to provide
high electron detection efficiency, electron-photon separation
and high background (charged hadron) rejection at the trigger
level. This is accomplished by providing evidence
for early shower development and by giving a good spatial point
for comparison with calorimeter clusters or tracks.
At L2, the CPS and FPS are treated as separate detectors
and their data are processed independently.
Upon L1Accept, CPS axial clusters from each of the eighty
azimuthal trigger sectors are combined into
quadrants in azimuth and transmitted to the L2 preprocessor.
Stereo clusters are sent directly to the L2 preshower preprocessor.
Figure~\ref{fig:l2_connections} shows the L1 to L2 data transfer path into
the preshower subsystem.
The L2CPS preprocessor receives axial ($x$) clusters over four
G-Links, each serving one azimuthal quadrant. The CPS stereo
data are transmitted over four G-Links, two for each hemisphere
(positive/negative $\eta$), including both $u$ and $v$ layer clusters.
Axial clusters are tagged by L1 with the presence or
absence of a CFT track. Each of the G-Links can carry a maximum
of 48 axial CPS clusters or 96 stereo clusters.
In the L2$\beta$eta processor, the CPS cluster centroids are
compared to produce $\eta$ and $\phi$ coordinates for clusters
that match in three layers. The presence or
absence of CFT trigger tracks associated with CPS $x$ clusters is also
provided, and output clusters are flagged as electrons (when there is a track
associated with the cluster) or photons (no track). The $\eta$ and $\phi$
coordinates are binned to correspond to the calorimeter trigger tower geometry
of $\eta \times \phi = 0.2 \times 0.2$. A window of width 0.05 is drawn
around each calorimeter trigger tower, and any preshower hit in
this $\eta,\phi$ region is designated a calorimeter match.
The FPS provides similar functionality and is the only source of forward
tracking information available before the L3 trigger.
\paragraph{L2CTT}
The L2CTT preprocessor takes inputs from the L1CTT and the L2STT. This
preprocessor system has been designed to operate in two different modes:
{\it i}) with input tracks straight from L1CTT and {\it ii}) with input tracks
from L2STT which receives input from the L1CTT and SMT barrels.
In the first mode of operation, L2CTT reads in the track
lists from different $\phi$
regions of the L1 tracking trigger system and
concatenates these into a single $p_T$-sorted list. The $p_T$ measurements
are refined using additional hit and tracking information than is
available at L1. For each track, the azimuthal angle with respect to the
beamline, $\phi_0$, is determined. The value of the azimuthal
angle at the third layer of the EM calorimeter, $\phi_{em3}$, is also
calculated ($\phi_{em3}$ is different from $\phi_0$ due to the bending of tracks
in the solenoidal magnetic field). Finally, each track is evaluated according
to several isolation criteria to enhance the trigger capabilities for tau
leptons. The $p_T$-sorted list of L2 tracks
is reported to L2Global. In the second mode of operation, input data are
provided by the L2STT along with refined L2 track $p_T$s.
Only $\phi_0$, $\phi_{em3}$, and isolation are calculated for these data.
However two separate lists of L2 tracks are passed on to L2Global,
one sorted according to $p_T$ and another sorted according to impact parameter.
\paragraph{L2STT}
\label{sec:l2stt}
The L2STT performs online pattern
recognition in the data from the SMT. It reconstructs
charged particle tracks found in the CFT at L1
with increased precision by utilizing the much finer spatial resolution of
the SMT.
The L2STT improves the momentum measurement of charged particle tracks at the
trigger level. Requiring hits in the SMT helps reject spurious
L1 triggers from accidental track patterns in the CFT. The primary physics
justification of the L2STT is its ability to measure the impact parameter of
tracks precisely enough to tag the decays of long-lived particles,
specifically $B$ hadrons.
Figure \ref{fig:STT_idea} shows the basic principle of the L2STT. For each
event, the L1CTT sends a list of tracks to the L2STT.
A road is defined around each track, and the SMT hits within the road are
associated with the track. The L2STT uses only the hits in the axial strips
of the silicon ladders, which define points in the $r-\phi$ plane. The L2STT
uses the hits in the innermost and
outermost layers of the CFT and hits in at least three of the four layers
of the SMT to fit the track parameters. The results of the fits are sent to
L2Global.
The SMT barrel ladders are arranged in twelve sectors, each covering $30^\circ$
in azimuth. The ladders of adjacent sectors overlap slightly such that more
than 98\% of all tracks are contained in a single sector. The L2STT therefore
treats all $30^\circ$ sectors independently.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure66.eps}}
\caption{The definition of roads based on L1 tracks and SMT hit
selection in L2STT.}
\label{fig:STT_idea}
\end{figure}
The L2STT consists mainly of custom-designed digital electronics modules. All
custom modules are designed to plug into a motherboard, and a common
motherboard design is used throughout the system. Data input from the
SMT detector and the L1CTT is via optical fiber serial links into
receiver cards (VTMs) located in the
rear card cage of the VME64/VIPA crates that house the L2STT
electronics. The data are processed in large FPGAs and/or
DSPs on a logic daughterboard that sits on the motherboard. There are three
different types of such daughterboards in the system. Data are communicated
between modules in an L2STT crate using a serial link transmitter and
receiver cards. Each module is equipped with a daughterboard
that buffers data for readout through the data acquisition system once an
event has been accepted by the trigger system. The logic daughterboard is
connected to the buffer cards, the serial links, and to the VME backplane by
three PCI buses on the motherboard. The VME bus is used for initialization
and monitoring and to read data out of the buffer cards.
The three types of logic daughterboards are the fiber road card (FRC), silicon
trigger card (STC), and track fit card (TFC). The FRC receives data from
the L1CTT and the TFW that it fans out to the other
modules. The FRC also manages the storage of data in the buffer cards. The
STC receives the SMT data, clusters hits in adjacent strips, and associates
SMT clusters with roads. The TFC
performs the final hit selection and fits the tracks.
L2STT consists of six identical VIPA crates, each serving two $30^\circ$
sectors. Each crate is equipped with one FRC module, nine STC modules,
and two TFC modules. Each crate also has a CPU board to program the FPGAs and
DSPs, download constants, and monitor system performance. An SBC
is used to read data out of the buffer cards and
feed them into the data acquisition system.
Figure~\ref{fig:STT-flow} shows the flow of data through an L2STT crate.
Each of the twelve TFCs provides a list of tracks from its $30^\circ$ sector.
The track information includes track parameters and the $\chi^{2}$ of the fit
as well as additional information about the cluster selection and fit.
One TFC in each of the six crates also transmits a
list of the initial L1CTT tracks.
These data are transmitted to the L2CTT, where the
tracks are sorted by $p_T$ and impact parameter and passed to L2Global
to be used in the trigger decision.
The data are also sent to a buffer card for readout to L3.
\begin{figure}
\centerline{\includegraphics[width=3.5in]{figure67.eps}}
\caption{Data flow through an L2STT crate.}
\label{fig:STT-flow}
\end{figure}
The L2STT helps to select events with an enhanced heavy-flavor content by
measuring the impact parameter $b$ of reconstructed tracks with respect to
the beam. Figure~\ref{fig:mures} shows the impact
parameter resolution $\sigma_b$ obtained from simulated single muon events.
The impact parameter resolution has a $p_T$ dependence introduced by multiple
scattering. In the trigger, the effect of this
$p_T$ dependence can be reduced by using the impact parameter significance $S_b
\equiv b/\sigma_b$ instead of the impact parameter $b$. The uncertainty
$\sigma_b = \sqrt{(18\, \mu\mathrm{m})^2 +
[(50\, \mu{\mathrm{m\, GeV}/c})/p_T]^2}$
(for tracks with four SMT clusters in the fit) takes into account the effect
of multiple scattering.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure68.eps}}
\caption{Impact parameter resolution as a function of $p_T$ determined
using simulated single muon events.}
\label{fig:mures}
\end{figure}
\subsubsection{L2Global}
L2Global is the first level of the trigger to examine
correlations across all detector systems.
The L2Global worker is responsible for making trigger decisions
based on the objects identified by the L2 preprocessors. Trigger
decisions are made by creating global physics objects. These
objects can be based directly on the objects reported
by the preprocessors or can be created by combining objects
from different preprocessors. The L2Global worker imposes requirements
on the global physics objects according to configuration information
it receives from the TCC based on
the downloaded trigger menu.
\paragraph{Inputs to L2Global}
After an L1Accept is issued, the TFW sends a trigger decision mask
to L2Global. The SCL sends notification of an L1 accepted event to
every geographic sector. In the case of L2 preprocessors, the receipt of
the L1Accept SCL message means that the preprocessor
must send at least a header block to L2Global for this
event, and must prepare at least a header block for
eventual L3 readout in case the event passes L2. When
L2Global receives the L1Accept, L2Global performs a decision cycle
and prepares at least a header block for L3 readout if
the event passes (or if it is an Unbiased Sample as described
below).
For each event, the L2Global worker uses the L1 trigger decision mask and
preprocessor inputs to
decide which algorithms to run on the data from the preprocessors.
It then makes a trigger decision and returns this
decision to the TFW. The trigger list specifies which trigger
conditions L2 will impose for each run. The trigger list can
change as frequently as every run. The trigger list is downloaded to the
L2Global crate by the TCC, which receives its instructions from COOR.
A number of L1 qualifier bits may be sent along with a given L1Accept.
Of these L1 qualifiers, the ones of special interest to
L2Global are ``Unbiased Sample'' (UBS), ``Forced Write'',
and ``Collect Status.'' For a UBS event, L2Global sends the
event to L3 whether or not any of the L2 bits actually passed, marking
the condition in the L2Global event header. The actual L2 decision
is also recorded. Such events occur at a rate determined
by the trigger programming, for an independently
adjustable fraction of events passing each L1 bit. A
secondary effect of the UBS qualifier in
L2Global is that additional information is written to
L3 to assist in debugging the event. In particular,
the output from L2Global is expanded to
include more information to allow detailed checking of
the processing. Events marked with
the UBS qualifier are routed to a special data stream for system
monitoring.
The Forced Write qualifier provides a mechanism to test
new trigger definitions. Unlike the UBS qualifier,
Forced Write produces detailed output at every firing of
an L1 trigger bit that is marked with this qualifier. The Forced Write
qualifier has the same effect on L2 as the UBS qualifier, however
Forced Write events are routed to the standard output stream
for offline analysis.
After receipt of an event with a Collect Status
qualifier, L2Global (and all preprocessors) capture internal
scaler and other monitoring information (including distributions of
processing times and pipeline occupancies) for readout by the TCC.
The monitoring blocks are tagged with the L1 crossing number
of the event with the Collect Status qualifier, so the
TCC can assemble a consistent set of statistics in the L1 and L2
systems. Collect Status qualifiers are generated approximately
once every five seconds in a typical data run.
\paragraph{Trigger scripts}
The L2Global worker begins the processing of every event by checking
which of the L1 trigger bits fired. The trigger list specifies
which L2 script is associated with each L1 bit.
The L2 script is the trigger condition that must be satisfied
in order for the L2 trigger to fire for a given L1 bit.
The L2 script is
specified by a number of filters and a minimum number of objects
required to pass each filter.
An example of a script is an electromagnetic object filter
and a minimum of two objects. This script is satisfied if there are two
electromagnetic objects in the event that satisfied the
conditions of the filter. If any script is satisfied, the
event passes L2 and is sent to L3.
\paragraph{Tools and filters}
Tools and filters provide the main functions of the L2Global worker.
The filters make up the scripts described above and
in turn rely on tools. Tools are C++ classes that build a
specific type of L2Global object.
L2Global objects are based on preprocessor objects.
A tool starts with a list of preprocessor objects and applies
selection criteria to decide which preprocessor objects
should be made into global objects. A tool can also
correlate information from two preprocessors by combining
two preprocessor objects to form one global object. An example
of this is the combination of a track object and an EM object
that come from different preprocessors, but refer to the same
electron candidate. The tools are written
to be flexible and are configurable through the trigger list.
For the above example, one trigger list parameter specifies
whether or not EM objects from the preprocessor should be matched to tracks
from the tracking system.
The tools produce lists of global objects.
The filters then use these lists of global objects to make a trigger
decision by imposing trigger requirements on the objects.
For the case of an EM object, the electromagnetic fraction,
transverse momentum, and isolation can be required to have values above or
below specified thresholds to satisfy the filter.
The filter generates its own list of objects that
satisfy the trigger criteria. Tools and filters can also operate on
global objects from earlier filters to construct additional global objects of
greater complexity. At the conclusion of processing, the script
checks to see if there are at least the minimum number of objects
required to satisfy the script requirements.
\subsection{The Level 3 trigger}
\label{sec:l3trigger}
The L3 trigger provides additional rejection both to enrich
the physics samples and to maintain an acceptable throughput which can be
recorded to tape. A high level, fully programmable software trigger, L3
performs a limited reconstruction of events, reducing a nominal 1~kHz input
rate to 50~Hz for data recorded for offline analysis. Its decisions are
based on complete physics objects as well as on the relationships between such
objects (such as the rapidity or azimuthal angle separating physics objects or
their invariant mass). Candidate physics objects, or relations between them,
are generated by object-specific software algorithms (filter tools).
Tools perform the bulk of the work: unpacking raw data, locating hits, forming
clusters, applying calibration, and reconstructing electrons, muons, taus,
jets, vertices, and \met. Reference sets (refsets)
of programmable algorithm parameters are input to the tools via the
programmable trigger list. The refsets define the physics objects
precisely (jet refsets specify cone size, for example, and electron refsets,
the electromagnetic fraction, among other characteristics) for each invocation
of the filter tool. All tools cache their results to expedite possible multiple
calls within the same event, and if the event is accepted, add L3 object
parameters to the data block.
Individual calls to the tools are made by filters that define the specific
selection criteria employed by a tool or imposed on its results. These criteria
include the refset used by the tool, as well as thresholds and other
cuts applied by the filter on the results of a tool (for example, the
requirement of two jets within a given pseudorapidity range above a fixed $E_T$
threshold). Filter results are keyed for access by other filters, so in
addition a key to the results of a previous filter can be included in these
parameter sets. Part of the trigger programming, this information can be
changed with each trigger list download.
The trigger list programming includes blocks of filter scripts that specify
one or more filters and that define the L3 trigger condition for each L3
trigger or filter bit. Each L3 filter script is associated with a L2 bit;
multiple L3 scripts may be associated with each L2 bit. Failure to
pass an individual filter terminates execution of the script, calling no
further tools, and skipping to the script for the next filter bit. Only when
all filters in a script are satisfied, is the trigger satisfied and the event
sent to the host cluster to be recorded.
\subsubsection{ScriptRunner}
Each filter tool receives event data under the direction of ScriptRunner,
the interface of the L3 framework to the tools.
At the beginning of a run, ScriptRunner
initiates parsing of the tool refsets and the filter scripts. L3-relevant
pieces of the trigger list (and event data) are its input. Needed tools
initialize themselves with the necessary calibration constants
and refsets. ScriptRunner then processes any errors and parses the scripts to
build the execution tree. The output of ScriptRunner is filtered event data
dispatched to the data logger, monitoring information (individual and combined
filter rates and timing of both filter scripts and individual tools collected
in the event loop), and error messages. At the end of a run (and upon request)
ScriptRunner extracts and reports monitor information and rates.
In the order specified by the trigger list, the execution path proceeds
along the branches of each L2 trigger bit that has been set. Each filter
branch is traversed, with tools called in the given filter script order until
an event fails a filter or passes all filters; execution then returns to the
filter bit level and proceeds to a sister filter bit
branch, if one exists, or continues to the next trigger bit. Control is
returned to ScriptRunner as soon as an event fails a filter or after it passes
all filters.
\subsubsection{Available physics object tools}
\paragraph{Level 3 jets and electrons}
The L3 jet tool relies on the high-precision calorimeter readout and primary
vertex position available at L3 and the improved energy and position
resolution this information makes possible.
Implementing a simple cone algorithm and performing a suppression of hot
calorimeter cells, the L3 jet tool is able to sharpen the turn-on curve
dramatically. Applying offline cleanup cuts on collected data shows sharp
turn-ons to nearly 100\% efficiency for both jet and electron filters (see
Figure~\ref{fig:L3Jet}). Rejection factors of 20--50 have been realized for
the various jet triggers.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure69.eps}}
\caption{The L3 jet trigger turn-on. Efficiency is plotted as a function
of offline leading jet $p_T$. A selection cut of 15~GeV/$c$ has been
applied by the L3 filter. Statistical errors are included, although small
enough to be obscured by the symbol in most instances. }
\label{fig:L3Jet}
\end{figure}
A simple $\sqrt{(\Delta\eta)^2 + (\Delta\phi)^2} = 0.25$ jet cone, with
requirements on $E_T$, electromagnetic fraction
($>0.9$), and transverse shower shape, forms the basic electron tool.
An electron candidate match to a preshower detector signal can also be
required.
\paragraph{Level 3 muons}
Wire and scintillator hits are used in the reconstruction of muon
track segments inside and outside the toroid.
Track-finding links these segments to identify muon tracks in three dimensions.
Refsets determine the minimum number of hits in each detector layer required for
valid tracks. In addition to unpacking data and determining tracks in the
regions defined by L2, the L3 muon tool can call subordinate tools that
utilize inner tracking and calorimetry.
L3 improves on the ability of L2 to separate prompt and out-of-time hits
by fitting the available scintillator hits along a track to the velocity of the
particle. Remnant cosmic ray muons are recognized by the
out-of-time signals in addition to their association with a
penetrating track opposite their candidate track.
The availability of vertex constraints and the ability to match central tracks
makes possible improved momentum resolution in L3 compared with L2.
A track match tool extrapolates muon tracks to the central tracker,
defining a region to be searched. If more than
one central track is available, the track that minimizes the $\chi^2$
calculated using the angular distances in $\theta$ between the central
and extrapolated muon tracks is selected. Additionally, matching local muon
tracks to paths of minimum ionizing particles in the calorimeter distinguishes
isolated from non-isolated muons.
\paragraph{Missing transverse energy}
The L3 \met\ tool is based on calorimeter cells. Using corrected
calorimeter energy (assuming a nominal $(x,y,z) = (0,0,0)$ vertex),
the \met\ is
calculated through intermediate pseudorapidity ring sums (allowing quick
recalculation by geometrically corrected ring sums when the primary vertex tool
returns a vertex position). The tool calculates the \met, the azimuthal angle
of the \met, total scalar $E_T$, and a \met\ significance.
The filters that make up the L3 trigger are flexible and provide additional
rejection. For
example, adding a \met\ filter to loose electron or muon filters, in parallel
to single lepton scripts, provides redundancy for triggering on $W$ boson
events; adding \met\ to lepton + jets top quark filters provides additional
rejection factors (a factor of three for electron and eight for muon triggers).
\paragraph{Level 3 tracking}
CFT track-finding can be performed either in specified regions or across the
entire detector. At initialization, the calculated $x,y,\phi$ fiber positions
are stored in lookup tables. Adjacent hits are merged into clusters with their
$x,y,\phi$ position averaged. Tracking is first done using a fast circle fit
through the axial layers, identifying candidates as arcs through the origin and
defining an arc length $S$ in the $x-y$ plane.
A straight line fit in the $S-z$ plane then determines the track helix
parameters.
A link-and-tree algorithm joins clustered hits from different layers.
Starting from a link in the outer layer (and continuing recursively),
candidate tracks are built by adding links from adjacent layers, extending the
path length as long as the curvature is consistent with the preceding link.
The longest extended path found starting from the initial link is kept as a
track candidate. The radius defined by the circle through clusters (and the
origin) must exceed that corresponding to minimum trigger $p_T$.
SMT tracking modifies this method. Segments connect neighboring hits between
points within a specified $\Delta\phi$. Segment paths are linked when their
$z$-slope and $\Delta\phi$ are within (tunable) specifications. Unless
seeded by a location set by the candidates from an earlier tool, the algorithm
begins with the outermost SMT layer, looking for the longest paths toward the
center, fitting these to a helix. The path with the smallest $\chi^2$ is
selected.
The CFT-track-based $z$-vertex tool offers substantially higher
efficiency and purity than an SMT-hit-based algorithm.
Its use sharpens the turn-on curve for jet triggers, and
provides rejection against events with a primary vertex at large $|z|$.
Monte Carlo studies suggest that the tool achieves 0.5~mm $z$-vertex
position resolution. The CFT-track-based vertex is used for jet
identification and the \met\ calculation.
A global (SMT plus CFT) high-momentum-track finder starts from axial CFT tracks
(with matched stereo clusters) propagated into the SMT. If the CFT
axial/stereo match fails, the CFT-SMT match is done in $x,y$ only.
A stand-alone global track filter is a useful addition
to single muon and electron triggers. Starting from axial CFT tracks (with
matched stereo clusters) propagated into the SMT,
the L3 tracker makes an independent selection of individual charged high-$p_T$
tracks. This algorithm (with 60 $\mu$m distance-of-closest-approach (DCA)
resolution for central tracks) runs in less than 200 ms.
Studies of off-line track-matched electrons in a $Z\rightarrow e^+e^-$
data sample show a 60\% overall efficiency (95\% within
the CFT-axial acceptance).
The use of tracking information in the L3 filter reflects a strategy of
parallel single-electron filters which increase efficiency at low $E_T$ and
still provide redundancy at high $E_T$ (for cross checks). These parallel
triggers allow high-$E_T$
filters with loose cuts (where rejection is not critical) by introducing
tighter cuts at lower $E_T$. Similarly, L3 global track filters (run on
muon triggers) are complemented by a suite of L3 muon filters running the
stand-alone local muon filter for high-$p_T$ single muons.
Online monitoring keeps track of current beam spot information (the
mean position and spread in $x,y$ along with tilts in $x,z$ and $y,z$). Using
this information, L3 can calculate a fully 3-dimensional primary vertex for
each event.
By recalculating L3 track parameters using the 3-d vertex, L3 is capable of
triggering on the impact parameter of tracks. The DCA resolution with respect
to the primary vertex is 25~$\mu$m. With input provided by a tracking, jet and
vertex tool, event, jet and track $b$-quark-probabilities based on the signed
impact parameters of tracks can be calculated, and $b$-tagging implemented in
L3.
\paragraph{Relational filters}
Additional higher level selections, based on the relationship between physics
tool candidates, are implemented at the filter level. Examples include the
invariant mass filter, an acoplanarity tool that selects events
with the two leading-$p_T$ jet candidates separated by a polar angle between
$\phi_{min}$ and $\phi_{max}$, and an $H_T$ tool that applies a
selection cut to the scalar sum of the $E_T$s of all jet filter candidates.
\subsubsection{Special filter tools}
\label{sec:l3specialtools}
For monitoring purposes, events can be written out regardless of the L3 trigger
decision. Special ``mark and pass'' runs record every L2-accepted event and
mark them with the results of all filter
tools of every script under the L1/L2 bit(s) satisfied. Additionally, a
``pass one of $n$'' option for each script independently specifies a fraction
of events to be forced through marked but unfiltered. Every event selected in
this way passes into a special ``monitor'' stream, distinct from the physics
stream. Monitor stream data are collected continuously during regular
data collection.
When necessary to prescale events, L3 can mark a filter bit as failed
without running the filter script. L3Prescale
uses selection by random number generation (seeded uniquely by node) so that
collectively, successive short runs still see events in the correct prescale
fraction. This implementation by random numbers initialized differently in
each node is the same means by which ``pass 1 of $n$'' events are generated.
\subsubsection{Online monitoring}
Online monitoring of the L3 system is done in the control room during
regular data collection. Quick analysis of the event record from randomly
sampled online events produces distributions of
physics quantities (for failed as well as passed candidate events). Plots of
electron, jet, muon, tau, and global track multiplicity, $E_T$, $\phi$, and
$\eta$ are continuously reviewed. In addition, a comparator
package looks for discrepancies between online quantities and those computed
when the data are run through the trigger simulator offline.
\FloatBarrier
\section{Data acquisition system}
\label{sec:daq}
The data acquisition system (L3DAQ) transports detector
component data from the VME readout crates to the processing nodes of the
L3 trigger filtering farm. The online host receives event data from the L3
farm nodes for distribution to logging and monitoring tasks. Overall
coordination and control of triggering and data acquisition is handled by the
COOR program running on the online host system.
\subsection{L3DAQ}
The L3DAQ system's designed bandwidth is 250~MB/s,
corresponding to an average event size of about 200~kB at an L2
trigger accept rate of 1~kHz.
As shown in Figure~\ref{fig:l3daq-network}, the system is built around
a single Cisco 6509~\cite{CISCO} ethernet switch.
A schematic diagram of the communication and data flow in the system is shown
in Figure~\ref{fig:dataflow}. All nodes in the system are based on
commodity computers (SBCs) and run the Linux operating system.
TCP/IP sockets implemented via the ACE~\cite{ACE} C++ network and
utility library are used for all communication and data transfers.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure70.eps}}
\caption{The physical network configuration of the L3DAQ system. The moveable
counting house holds electronics for the detector and moves with the detector
between the assembly hall and the collision hall.}
\label{fig:l3daq-network}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure71.eps}}
\caption{Schematic illustration of the information and
data flow through the L3DAQ system.}
\label{fig:dataflow}
\end{figure}
Up to sixty-three VME crates are
read out for each event, each containing 1--20~kB of data distributed among VME
modules. An SBC (single board computer) in each VME crate reads out the VME
modules and sends the data to one or more farm nodes
specified by routing instructions received from the routing master (RM)
process. An event builder (EVB) process on each farm node builds a complete
event from the event fragments and makes it available to
L3 trigger filter processes.
The supervisor process provides the interface between the
main D\O\ run control program (COOR) and the L3DAQ system.
When a new run is configured, the supervisor passes run and general
trigger information to the RM and passes the COOR-provided L3 filter
configuration to the EVB process on relevant farm nodes,
where it is cached and passed on to the L3 filter processes.
The SBCs are single-board computers with
dual 100~Mbit/s Ethernet interfaces and a VME-to-PCI interface.
An expansion slot is occupied by a digital-I/O (DIO) module,
used to coordinate the readout of VME modules over the VME user (J3) backplane.
A custom kernel driver on the SBC handles
interrupt requests from the DIO module that are triggered by
readout requests from the crate-specific electronics.
On each readout request, the kernel module
performs the VME data transfers and stores the event
fragment in one
of several buffers in kernel memory.
A user-level process on the SBC receives route information from the RM
in the form of route tags that contain a
unique event identifier (L3 transfer number) and the indices of
the farm nodes to which that event should be sent.
If the L3 transfer number of the route tag matches that of the transfer number
embedded within the head event fragment in the kernel buffers, the event
fragment is sent to the specified farm nodes.
The EVB process on each farm node collates the event fragments
received from SBCs into complete events, keyed by L3 transfer
number. For each event, the EVB receives an expected-crate list
from the RM in order to determine when an event is complete.
Complete events are placed in shared memory buffers for
processing by the L3 filtering processes (Section~\ref{sec:l3trigger}).
The EVB process routinely informs the RM of the number
of free memory buffers that it has available.
The RM program executes on an SBC in a special VME crate
containing a hardware interface to the TFW.
The TFW provides trigger information and the L3 transfer number upon each L2
accept decision and allows the RM to asynchronously disable the firing
of L1 triggers. For each event, the RM program chooses a node for processing
based on the run configuration,
the trigger information, and the number of available buffers
in the set of nodes configured to process the type of event.
A node is chosen in a round-robin fashion from among the set of nodes
with the most free buffers.
If the number of available free buffers is too few, the RM instructs the
TFW to disable triggers so that the farm nodes have time to catch up.
\subsection{COOR and the online host}
\label{sec:coor}
Run control and detector configuration are handled by the central
coordination program COOR. COOR receives requests from users
to configure the detector or to start and stop runs and
sends the necessary commands to the rest of the system
to carry out those requests. COOR sends commands to
L1, L2, and L3 and to the manager processes for the SDAQ (see below)
and data logging
subsystems. COOR can also configure EPICS (Section~\ref{sec:controls})
devices via a connection to the COMICS (Section~\ref{sec:controls-config})
program. It also maintains a database of
name/value pairs accessible to the online system for recording the current
network addresses of various online services and for access by
trigger configurations; this allows for communicating time-dependent
values (such as the current beam spot position) to the trigger systems.
The online host system receives event data from the L3 farm nodes at a
combined rate of approximately 10~MB/s (50~Hz L3 accept rate of 200~kB
events) and distributes that data to logging and monitoring tasks.
The host system is capable of supporting multiple simultaneous
data runs.
Events that pass the L3 filters are tagged with a data stream
identification that is a function of the satisfied hardware and software
trigger components. Different streams are recorded independently;
events are assigned to only one stream, excepting some events which are
additionally assigned to a special monitoring stream.
The final repository for the raw event data is tape, maintained in a robotic
tape system located about 3~km from the detector. Data must be transmitted
to each tape drive at approximately 10~MB/s to keep the drive operating in
streaming mode, since the remote tape drive nodes have no intermediate disk
buffer. The online system is capable of simultaneous output to multiple tape
streams and of buffering in case of tape robot unavailability.
In addition to logging data, the online host system must supply between ten
and twenty data monitoring clients at anywhere from 1\% to 100\% of the full
data rate.
Figures~\ref{fig:host-hardware} and \ref{fig:host-software} illustrate the
physical and software architecture of the online host system.
Event data arrive from the L3 trigger nodes at collector processes.
The collector directs each event to the data logger appropriate for the
stream identifier determined for the event.
The collector also sends, on a
best-effort basis (there is no flow control backpressure to the L3
nodes), a copy of each event to a distributor process,
which is an event queueing system that provides event data in near real-time
to online analysis and monitoring clients (EXAMINE programs).
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure72.eps}}
\caption{Schematic of the physical architecture of the online host system.
We are using commodity computers running Linux connected to a Cisco 6509
switch.}
\label{fig:host-hardware}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure73.eps}}
\caption{Schematic of the software architecture of the online host system.}
\label{fig:host-software}
\end{figure}
A diagnostic secondary data path, SDAQ, makes possible the
processor-based readout of digitizing electronics. Information
from SDAQ is also routed via the collector processes, allowing all
downstream DAQ components to be shared. The SDAQ mechanism bypasses the
L3DAQ and L3 trigger and is valuable in the commissioning
and calibration of detector components. The detector-specific components of
an SDAQ application have access to a library of
SDAQ functions that handle queuing of data messages between components,
interrupt management with callbacks, run synchronization, and priority-based
scheduling. Several of the subdetectors use the SDAQ system:
the SMT for calibration and to monitor the performance of individual silicon
detector channels during a run,
and the CFT, CPS, FPS, and FPD to calibrate scintillator response.
The data logger writes data to files, grouped according to stream
classification tags. Each data logger is responsible for a set of streams.
The data logger also generates metadata information in file format for
storage in a database to enhance the offline access of data. The DLSAM
processes are the interfaces to the SAM/ENSTORE mass storage facility
(Section~\ref{sec:SAM} and the associated database descriptions). DLSAM
monitors the local data buffer disks and requests that files be stored in
the mass storage system (ENSTORE). This request is made through
the database interface (SAM) which negotiates the request for file storage
with the ENSTORE system.
All of the online computing systems are connected to
a single high-capacity network switch. The architecture
provides for parallel operation of multiple instances of the
bandwidth-critical components.
The high-level online software applications are predominately
constructed using the Python scripting language.
Network communication between the components is implemented with
the InterTask Communication (ITC) package, a multi-platform,
multi-threaded client/server messaging system developed at Fermilab and
based on the ACE~\cite{ACE} network and utility libary.
\FloatBarrier
\section{Controls and monitoring}
\label{sec:controls}
The D\O\ experiment has extended EPICS (Experimental Physics and
Industrial Control System) \cite{epics} to meet the control and monitoring
requirements of a large high energy physics detector. EPICS, an integrated
set of software building blocks for implementing a distributed control system,
has been adapted to satisfy the slow controls needs of the detector by {\it i})
extending the support for new device types and an additional field bus,
{\it ii}) the addition of a global event reporting system that augments the
existing EPICS alarm support, and {\it iii}) the addition of a centralized
database with supporting tools for defining the configuration of
the control system. Figure~\ref{fig:controls} shows the architecture and
components of the D\O\ controls and monitoring system.
EPICS uses a distributed client-server architecture consisting of
host-level nodes that run application programs (clients) and input/output
controller (IOC) nodes (servers) that interface directly with the detector
hardware. The two classes of nodes are connected by a local area network.
Clients access process variable (PV) objects on the servers using the
EPICS channel access protocol.
One of the unique properties of the D\O\ detector interface is the use of the
MIL-STD-1553B \cite{mil-1553} serial bus for control and monitoring operations
of the
electronics components located in the collision hall. Since the detector
is inaccessible for extended periods of time, a robust, high-reliability
communication field bus is essential. EPICS was extended by providing a
queuing driver for MIL-STD-1553B controllers and a set of device support
routines that provide the adaptive interface between the driver and the
standard EPICS PV support records. With these elements in place, all of
the features of EPICS are available for use with D\O's remote devices.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure74.eps}}
\caption{Organization of the control system components. The Hosts are
computers
running Linux or Tru64 Unix and the IOCs are embedded computers in VME crates.
Only one IOC is shown, and only a few of the online hosts.}
\label{fig:controls}
\end{figure}
\subsection{Global event reporting}
To process significant events from all experiment sources, a separate facility,
the significant event system (SES), collects and distributes all
changes of state. The SES has a central server that collects event messages
from sender clients and filters them, via a Boolean expression, for routing to
receiving clients. Sender clients, including the IOCs, connect to the server
and all state changes on those clients, including alarm transitions, are sent
to the server. The architecture of the SES and the flow of messages within
the system are illustrated in Figure~\ref{fig:ses}.
\begin{figure}
\centerline{\includegraphics[width=3.in]{figure75.eps}}
\caption{The major components of the SES are the server, logger, watcher, and
alarm display. COOR is a coordinating process that provides run control to the
experiment. Significant events originate in the detector hardware, run
control system, or online applications.}
\label{fig:ses}
\end{figure}
The alarm class of SES messages receives special handling in the server. The
SES server maintains the current alarm state of the entire experiment so that
receiving clients can obtain the current state when they first connect
to the server. In addition to specialized receiving clients that may connect
to the server, there are three standard clients: the SES logger, the SES alarm
display, and the SES alarm watcher. The logger has a pass-all filter so that
it receives all SES messages sent to the server and writes the messages
received to a disk file. The current state of the detector stored in the server
is relayed to users through the alarm display. For alarms that compromise
data quality, the alarm watcher automatically pauses the current run. In
addition to its monitoring and logging functions, the SES system provides the
means for distributing synchronizing messages to other components of the online
software system. Tools have been developed for mining data from the SES log
files. Hardware experts review the log files to understand which hardware
devices are unstable and collaborators performing data analysis can be sure
that the event they are examining is real and not caused by a fault in the
detector.
\subsection{Centralized device database}
The EPICS databases that configure the individual IOCs are flat ASCII files
containing record definitions, the database equivalent of a PV, that
are read by the
IOCs during startup. The EPICS system additionally provides a higher-level
construct, called a template, which is a parameterized collection of record
definitions. Generator files, which reference the templates, supply the
parameter values to produce instances of these templated devices. While these
collections of files are adequate for EPICS initialization, they are not easily
accessible to host-level processes, which require the same information.
To address this problem, we have centralized the relevant information
in a relational database (Oracle \cite{oracle}) and provided a family of
scripts to manage the transformation between the relational database and the
EPICS ASCII-format files. At the time this document was prepared, the
database contained approximately 6200 templated devices, corresponding to about
137000 process variables, and that number is constantly expanding.
In addition to the database management scripts, a WWW browser interface to the
relational database is available for the initial definition, modification, and
viewing of the relational database entries. With control system device
specifications centralized in the relational database, they are easily
accessible to other host-level processes.
\subsection{Detector configuration management}
\label{sec:controls-config}
One of the most complex tasks performed by the control system is the
configuration of the detector for specific run conditions. The set of distinct
configurations, both for physics data collection and for calibration runs, is
very large; the usual technique of uploading a specific detector
configuration and saving it as a file for subsequent downloading is impractical.
For ease of configuration management, the detector is represented as a tree
with nodes at successively deeper levels corresponding to smaller, more
specialized units of the detector. The terminal nodes of the
tree, called action nodes, each manage the configuration of a specific,
high-level device. They are instances of the high-level devices discussed
in the preceding database section. The intermediate nodes of the tree
primarily serve to
organize the traversal order of the subordinate nodes since the detector is,
in general, sensitive to the order in which devices are initialized.
A single server program, COMICS, manages the configuration of the
EPICS-accessible part of the detector. The tree nodes, both intermediate and
action, are all specialized instances of a base node class that defines the
majority of the methods that characterize node behavior. The detector tree
structure is defined by a set of configuration files that are Python program
segments that instantiate instances of these nodes.
\subsection{Operator interfaces and applications}
An application framework, in the Python scripting language,
assists in developing operator interfaces and provides a consistent ``look
and feel'' for all visual displays. This framework includes a collection of
specialized, graphical objects that construct updating displays of PV
values using a Python interface to the EPICS channel access application
program interface (APS). The experiment uses more than forty instances of
these monitoring displays in the control room to manage the detector
components.
\subsection{Archiving EPICS data}
While using EPICS records for control and monitoring tasks,
almost every detector group needs to maintain structured access
to archived PV values. There are two major archiving tools employed by D\O:
{\it i}) the channel archiver \cite{epics-archive}, for needs that require
sampling rates of 1~Hz or faster, but do not require frequent access
to historical data; and {\it ii}) the EPICS/Oracle Archiver, for long-term
studies that require slower sampling rates (once per minute or less
frequently), easy access to data at any moment, and minimal maintenance.
Many channel archivers are running concurrently, monitoring several thousand
PVs. About once a week, collected archives are sent to the central Fermilab
robotic tape storage via the SAM data management system
(Section~\ref{sec:SAM}). The channel archiver toolset has
interfaces, including web-based tools, that enable retrieval from an
archive of data in different formats and generation of time plots with
various options.
\subsection{ACNET gateway}
In the operation of the detector, it is vital to have a fast and reliable
messaging connection between D\O\ and accelerator operations to exchange
control and monitoring information. The D\O\ control system supplies cryogenic
and magnet data, the luminosity determined using the luminosity monitor
(Section~\ref{sec:lum-monitor}), as well as FPD pot positions and counter
rates. The accelerator control system (ACNET), in turn, sends information
about critical accelerator devices. A gateway between the D\O\ and ACNET
control systems, based on the XML-RPC protocol \cite{xml-rpc}, provides this
interconnection.
\FloatBarrier
\section{Computing and software}
\label{sec:software}
A large amount of software has been developed for data acquisition,
monitoring and controlling hardware, Monte Carlo event simulation, and
data and Monte Carlo event reconstruction. Early in the development
of software for Run II, we made the decision that all new software would be
written using the C++ programming language. Legacy Run I Fortran software and
programs from other sources (e.g. Monte Carlo event simulation programs) are
wrapped in C++ code. In this section, we give an overview of the computing and
software in use during Run II.
\subsection{Event data model}
\label{sec:edm}
The D\O\ event data model (EDM) is a library of C++ classes and
templates whose purpose is to support the implementation of
reconstruction and analysis software.
The central feature of the EDM is the event, a class that represents
the results of a single beam crossing. The event acts as a container
to manage all of the data associated with a single crossing: the raw
output of the detector, the results of trigger processing, and
the results of many different reconstruction tasks.
Each of the items in this collection contains both the
data describing the crossing in question as well as metadata that
describe the configuration of the program that constructed these
results. This allows us to run multiple instances of single algorithms
with different configuration information (for example, several
cone-based jet algorithms with different cone radii), and to
distinguish between the output of these different algorithms. The EDM
also provides a mechanism for access to the collected reconstruction
results, relieving the users from the burden of understanding the
somewhat complex internal organization required for the management of
the event data and the corresponding metadata.
\subsection{Data persistency}
\label{sec:d0om}
The conversion of the C++ objects used in the reconstruction
program to a persistent format is handled by the
D\O\ object model (D\O OM) \cite{d0om-chep}. This has several parts
(Figure~\ref{fig:d0om}). First, D\O OM maintains a dictionary
describing the layout of the C++ classes that are to be used
persistently, which is generated by running
a preprocessor over the C++ headers defining the classes.
This preprocessor is based on a modified version of the
CINT C/C++ interpreter (which is also used in the
ROOT system \cite{root}).
\begin{figure}
\centerline{\includegraphics[width=5.in]{figure76.eps}}
\caption{The structure of D\O OM. The shaded area shows the
components which are part of D\O OM proper.}
\label{fig:d0om}
\end{figure}
The actual translation between C++ objects and the persistent
format is handled by one of several I/O packages. User code
has no dependence on the I/O packages, so that new formats
can be added without changing any reconstruction code.
The I/O packages are typically built on top of external software
packages to do the actual I/O.
Finally, the external interface to the package is provided
by a set of stream classes. A set of D\O\ framework
packages that use these classes to read and write events
within the framework is also provided; in most cases,
a framework user need only to include these packages
to read and write events.
D\O OM includes numerous features to assist with schema evolution
and versioning of data. The dictionary information is maintained
along with the saved data, so the layout of the saved data
is always known. During reading, class members are matched
between the C++ and persistent forms based on name.
This allows D\O OM to handle the common cases of adding and
deleting data members with no explicit action on the part of the programmer.
For more complicated cases, it is possible to provide
conversion code that is automatically run when needed.
The D\O OM dictionary information may also be queried at run
time. The system can also make use of the persistent dictionary
information to build objects at run time for which no dictionary
information was compiled into the program. This is useful
for programs to dump or browse arbitrary D\O OM data files.
One I/O package is based on the DSPACK library, which originated with
the NA49 experiment \cite{dspack}.
DSPACK handles the conversion from C-like structures
to a serial data format; the D\O OM I/O package converts from
the C++ objects to the DSPACK structures. At D\O, DSPACK
data are usually encapsulated inside another, lightweight, format
called EVPACK, which provides data compression and random access
within files with keyed lookup. EVPACK-encapsulated DSPACK
records may also be sent over the network; this is used to
distribute data within the online host system.
All event data are stored in (EVPACK-encapsulated) DSPACK format.
In addition, this format is used for several static, structured
data files used by the reconstruction program, such as the
description of the detector geometry.
\subsection{Calibration databases}
\label{sec:databases}
Run-dependent information needed for reconstruction, such as
magnetic field polarities and calibration constants, is stored in a
database. To decouple the reconstruction program from any
particular proprietary database implementation, D\O\ makes use of a
three-tier architecture \cite{dbserver}, where the database and
the client code make up the top and bottom tiers
(Figure~\ref{fig:dbserver}). In between is a middle tier server which
accepts requests from the clients, makes the query to the database,
and returns the data to the client. Data can be cached by the server,
thus speeding up the response to the client and reducing the load on
the database.
\begin{figure}
\centerline{\includegraphics[width=8cm]{figure77.eps}}
\caption{The hierachy of the database, middle tier server, and client
applications. Each server can cache data, and the servers can be chained to
provide higher-speed access for clients located at remote processing sites.}
\label{fig:dbserver}
\end{figure}
This caching is especially effective for the large-scale reconstruction
of raw data. If runs that require the same calibration constants are
processed together, only one query to the database is made for
each calibration set; all other requests for this set are fulfilled
directly from the cache.
Servers can be chained together so that processing sites outside
Fermilab may run their own local servers which request their data from
the servers at the laboratory. This allows a remote processing site to
have its own cache to speed up queries without requiring the site to
access the database directly.
The particular software used for storing the calibration data is the
Oracle relational database. The middle tier servers are mostly
implemented in the Python programming language with some parts written
in C++ to improve performance. CORBA distributed object technology
\cite{CORBA} is used for the communication between the client and the
servers. The core part of the server is written to be independent of
the particular application it is serving data for; the customized
code which queries specific database tables and returns the results to
the client is automatically generated from the database schema for
each particular application. Each subsystem of the detector (SMT,
CFT, preshower detectors, calorimeter, and muon
detectors) has its own database schema and a separate server.
The client end of the system is an I/O package for D\O OM. The
necessary C++ class definitions are also automatically generated from
the database schema. Accessing the data is then a matter of navigating
a C++ object hierarchy: the client application need have no knowledge
of CORBA, Oracle, or the method by which it obtained the data. This
decoupling makes it relatively easy to replace the specific
implementation of the middle tier, if this becomes either necessary or
desirable, without requiring the whole of the client implementation to
change.
\subsection{Event simulation}
\label{sec:MC}
The generation of Monte Carlo (MC) events involves multiple stages and many
executables. To integrate all processes, all
programs use the EDM to carry data in memory and
D\O OM to store persistent event data. All code is
organized in independent packages running in a standard D\O\
framework and is written in C++ or embedded in C++ driving routines.
The first step in MC event generation is the simulation of a
physical process, a \mbox{$p\overline{p}$}\ collision producing a particular final
state. Nearly all existing event generator programs are written
in Fortran, but the StdHep code from the FNAL Computing Division can be used to
store the output in a standard common block format. This allowed us to write
a C++ wrapper that converts the StdHep Fortran format to C++ classes
satisfying the EDM requirements.
To trace the particles through the D\O\ detector, determine where their paths
intersect active areas, and simulate their energy deposition and secondary
interactions, we use the CERN program GEANT v3.21 \cite{geant}, which is also
written in Fortran.
A C++ wrapper is used to read files produced by the event generators and to
write the output of GEANT in D\O OM format. This executable is
called D\O GSTAR. All subsequent steps in the event simulation are
handled by programs written almost entirely in C++.
The D\O SIM program modifies the generated Monte Carlo data to account for
various detector-related effects.
After the particles from the simulated reaction have been traced through the
detector, the generated energy depositions must be converted to the form that
the real data takes when processed through the D\O\ electronics. Detector
inefficiencies and noise (from the detector and electronic readout) must be
taken into account, and more than one interaction may
occur during a beam crossing. In addition, some portions of the detector
(such as the calorimeter) remain sensitive to interactions over a period of
time that includes more than one beam crossing.
Simulation of the trigger electronics and the
effects of the trigger on data selection is performed by a separate
program, D\O TRIGSIM. D\O TRIGSIM contains simulation code only
for the L1 trigger. The L2 and L3 triggers are based on
filtering code, and exactly the same software runs in D\O TRIGSIM.
The output of D\O SIM and D\O TRIGSIM is in the same format as
the data recorded by the D\O\ data acquisition system, but contains additional
MC information to make it possible to correlate detector data with
the original generator output.
\subsection{Reconstruction}
\label{sec:reco}
The D\O\ offline reconstruction program D\O RECO is responsible for
reconstructing objects used for physics analysis. It
is a CPU-intensive program that processes events recorded
during data collection and simulated MC events. The executable is run on
the offline production farms and the results are placed into the central data
storage system (Section~\ref{sec:SAM}) for further
analysis. Information and results for each event are organized using the EDM.
The EDM manages information within the event in blocks called chunks.
The raw data chunk (RDC), created either by an L3 processor node
or the MC, contains the raw detector signals and is the primary
input to D\O RECO. The output from D\O RECO consists of many additional
chunks associated with each type of reconstructed object.
D\O RECO reconstructs events in several hierarchical steps. The
first involves detector-specific processing. Detector unpackers process
individual detector data blocks within the RDC, decoding the raw
information, associating electronics channels with physical detector elements,
and applying detector-specific calibration constants. For many of the
detectors, this information is then used to reconstruct cluster (for example,
from the calorimeter and preshower detectors) or hit (from the tracking
detectors) objects. These objects use geometry constants to associate detector
elements (energies and positions) with physical positions in space. The second
step in D\O RECO focuses on the output of the tracking detectors,
reconstructing global tracks from the hits in the SMT and CFT.
This process, involving several different tracking algorithms,
is the most CPU-intensive activity of D\O RECO.
The results are stored in corresponding track chunks, which are
used as input to the third step of D\O RECO, vertexing. First, primary
vertex candidates are found. These vertices indicate the locations of \mbox{$p\overline{p}$}\
interactions and are used in the calculation of various kinematical quantities
(e.g. $E_T$). Next, displaced secondary vertex candidates are
identified. Such vertices are associated with the decays of long-lived
particles. The results of the above algorithms are stored in vertex chunks,
and are then available for the final step of D\O RECO --- particle
identification.
Using a wide variety of algorithms, information from each of the preceding
reconstruction steps is combined and physics object candidates are
created. RECO first finds electron, photon, muon, neutrino (\met), and
jet candidates, after which it identifies candidates for heavy-quark and tau
decays.
\subsection{Data handling and storage}
\label{sec:SAM}
The sequential access via metadata \cite{SAM} (SAM) data handling system
gives users access to all the data created by the D\O\ experiment (both
detector data and simulation data), in a flexible and transparent manner.
The user does not need to know where the files are physically stored, nor
worry about exactly how they are delivered to her/his process.
SAM oversees
the functions of cataloging data (files and events, and associated metadata
regarding production conditions), transferring data in and out of mass storage
systems, transferring data among different computer systems (whether connected
via local or wide area network), allocating and monitoring computing resources
(batch slots, tape mounts, network bandwidth, disk cache space), and maintaining
file delivery status at the user process level. The bookkeeping
functions of the SAM system are provided by an Oracle \cite{oracle} database,
which is accessed via a client-server model utilizing CORBA technology.
Files are stored in SAM using interfaces that require appropriate metadata for
each file. The files are organized, according to the metadata provided, by
data tier, and by
production information (program version which produces the data, etc.). The
SAM system also provides file storage, file delivery, and file caching
protocols that permit the experiment to control and allocate the computing
resources. Tape resources can be guaranteed to high priority activities
(data acquisition and farm reconstruction), high usage files can be required
to remain in the disk cache, and different priorities and allocations for
resource usage can be granted to groups of users.
\ack
We would like to thank Ken Ford and Scott Baxter for their work on the
illustrations and to acknowledge in-kind contributions by
the Altera and Xilinx Corporations.
We thank the staffs at Fermilab and collaborating institutions,
and acknowledge support from the
DOE and NSF (USA);
CEA and CNRS/IN2P3 (France);
FASI, Rosatom and RFBR (Russia);
CAPES, CNPq, FAPERJ, FAPESP and FUNDUNESP (Brazil);
DAE and DST (India);
Colciencias (Colombia);
CONACyT (Mexico);
KRF (Korea);
CONICET and UBACyT (Argentina);
FOM (The Netherlands);
PPARC (United Kingdom);
MSMT (Czech Republic);
CRC Program, CFI, NSERC and WestGrid Project (Canada);
BMBF and DFG (Germany);
SFI (Ireland);
Research Corporation,
Alexander von Humboldt Foundation,
and the Marie Curie Program.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,822
|
Q: SQL query to select result based on data from column in multiple rows - phrase search based on tags So I have a little image gallery that I started to enhance using tags. I decided I go with simplest solution and I have a table just like:
describe photo_tags;
+---------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+---------+-------------+------+-----+---------+-------+
| photoid | bigint(20) | NO | PRI | NULL | |
| tag | varchar(32) | NO | PRI | NULL | |
+---------+-------------+------+-----+---------+-------+
It works, I have unique index for the photoid,tag pair to avoid duplicates and generally it does what's expected, spare for one annoying thing: I want to be able to search not just by a single tag, but a phrase.
The query (example below) is generated by PHP based on sanitized query string treated with str_word_count.
An example, here's a snippet from actual entries in DB
+---------+-----------------------+
| photoid | tag |
+---------+-----------------------+
| 8717 | red |
| 8717 | road |
| 8717 | sky |
| 8717 | tanker |
| 8717 | trees |
| 8717 | truck |
| 8717 | truck on truck action |
| 8717 | vehicle |
| 18858 | clouds |
| 18858 | green |
| 18858 | park |
| 18858 | sky |
| 18858 | trees |
| 18858 | truck |
| 18858 | vehicle |
| 18858 | walkway |
+---------+-----------------------+
Say I want to search the gallery based on tag "red truck":
This will not work, obviously
select photoid from photo_tags where tag="red truck" or (tag="red" and tag="truck");
This will sort of work:
select photoid from photo_tags where tag="red truck" or tag in('red','truck');
but it will basically select photoid that obviously have red or truck, not necessarily both of them.
Does anyone have idea how to improve the query so without modifying underlying table. Or maybe there's another way to achieve what I'm trying to do? I'm using MariaDB 10.3 and PHP 7.3 (basically what comes in Debian 10)
A: I think you want aggregation:
select photoid
from photo_tags
where tag in ('red', 'truck')
group by photoid
having count(*) = 2;
If you can have 'red truck' as well, then:
select photoid
from photo_tags
group by photoid
having sum(tag in ('red', 'truck')) = 2 or
sum(tag = 'red truck') > 0;
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,993
|
package com.tazine.structure.list.sequence;
/**
* Test
*
* @author frank
* @date 2018/01/19
*/
public class Client {
public static void main(String[] args) {
SequentialList<Player> list = new SequentialList<>(10);
Player p1 = new Player("kobe", "lakers", 24);
Player p2 = new Player("james", "cavs", 23);
Player p3 = new Player("iverson", "76ers", 3);
Player p4 = new Player("jordan", "bulls", 23);
list.add(p1);
list.add(p2);
list.add(p3);
list.insert(1, p4);
list.print();
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,180
|
First Day Hike draws New Year's Day crowd to Cacapon State Park
Jenni Vincent
jvincent@herald-mail.com
BERKELEY SPRINGS, W.Va. — New Year's Day morning was sunny and a little chilly, but not cold enough to stop a record number of people from participating in the First Day Hike at Cacapon State Park.
Park naturalist Valerie Chaney estimated about 100 participants gathered by 10 a.m. at the park's nature center to hike one of three trails available to them as part of the special event.
"Last year, we had 74 people, so this is exciting to see so many folks here this time, as well as multiple volunteers to help me," she said as more people filed into the nature center.
Participants were divided into three groups, including two offering easy-to-moderate hikes and a third group for individuals who don't mind the hiking being a little more strenuous on the Ziler trail, she said.
Each trail has its own distinct beauty and story, Chaney said.
"Ziler is an old mountain road, so that's more of a historic trail here. It was the only trail to go up and over the mountain ridge. So that is a neat one to hike and get those mountain views," she said.
"But not everyone wants to get that much elevation that quickly, so the rest will do about a 2-mile hike. Just going up and down the Ziler trail is about 5 miles."
Clark Dixon, a volunteer hike leader, said he enjoys starting off the new year outdoors.
"A lot of people do like doing this hike, so maybe it's like a resolution to get more active — and this is certainly a great way to get going," he said.
Angie McCusker, a Berkeley Springs resident, is a fan of the park's specialty hikes and was ready to kick off 2020 with some exercise.
"I exercise a lot anyhow, and this is a beautiful place to enjoy it. My favorite thing about Cacapon is the trails, so what a great way to start another year," she said.
Michele Mahoney was determined to hike despite an aching knee.
"I just moved here from Arlington, Va., and I'm hoping to do more hiking. I love this park and support anything this park does," she said. "We lived here to have a simpler life, and it's off to a great start."
West Virginia is one of the state park systems that participate in America's State Parks First Day Hike initiative, Sissie Summers of the parks and recreation section of the West Virginia Division of Natural Resources said in a telephone interview.
Attendance varies, but "the show always goes on," she said.
"The unofficial motto is, 'If you show, we go,' and that's what happens," she said.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 761
|
Q: Needing VueJS Getter to only return all values of the last JSON property as an array I am using a Django REST back-end to ship API data to a VueJS front-end.
I am trying to place daily counts in a ChartJS graph.
import { HorizontalBar } from '../BaseCharts'
export default {
extends: HorizontalBar,
data() {
return{
}
},
mounted () {
/* Need to write the API call for chart data here. */
this.$store.dispatch("DailyCountAction")
this.renderChart({
labels: [this.$store.state.DailyCount],
datasets: [
{
label: 'Existing Patients',
backgroundColor: [
'rgba(255, 99, 132, 0.2)',
'rgba(54, 162, 235, 0.2)',
'rgba(255, 206, 86, 0.2)',
'rgba(75, 192, 192, 0.2)',
'rgba(153, 102, 255, 0.2)',
'rgba(255, 159, 64, 0.2)'
],
borderColor: [
'rgba(255,99,132,1)',
'rgba(54, 162, 235, 1)',
'rgba(255, 206, 86, 1)',
'rgba(75, 192, 192, 1)',
'rgba(153, 102, 255, 1)',
'rgba(255, 159, 64, 1)'
],
borderWidth: 1,
data: [4,8,2,]
},
You can see in the "label" field I have a getter which returns the data:
{
"Check_In_Count": "7",
"Average_Time_Spent": "3",
"Average_Wait_Time": "2",
"Cancelations": "0",
"New_Patient_Count": "4",
"Existing_Patient_Count": "3",
"Current_Cycle_Date": "2062019"
},
{
"Check_In_Count": "4",
"Average_Time_Spent": "8",
"Average_Wait_Time": "6",
"Cancelations": "0",
"New_Patient_Count": "1",
"Existing_Patient_Count": "3",
"Current_Cycle_Date": "2072019"
},
{
"Check_In_Count": "7",
"Average_Time_Spent": "3",
"Average_Wait_Time": "9",
"Cancelations": "0",
"New_Patient_Count": "0",
"Existing_Patient_Count": "7",
"Current_Cycle_Date": "2082019"
},
{
"Check_In_Count": "8",
"Average_Time_Spent": "8",
"Average_Wait_Time": "1",
"Cancelations": "0",
"New_Patient_Count": "4",
"Existing_Patient_Count": "4",
"Current_Cycle_Date": "2092019"
},
I am needing to only return all the Current_Cycle_Date values into an array that goes into the "labels" field. I am very confused on how I could go about this.
In the end the label field should look like:
labels: ['2092019', '2082019', '2072019', '2062019']
I've seen examples of using the MAP in getters. Nothing has worked so far. Could this be accomplished using some logic in a Method?
Any recommendations would be greatly appreciated!
A: Maybe you could try returning it as an array:
const data = [{
"Check_In_Count": "7",
"Average_Time_Spent": "3",
"Average_Wait_Time": "2",
"Cancelations": "0",
"New_Patient_Count": "4",
"Existing_Patient_Count": "3",
"Current_Cycle_Date": "2062019"
},
{
"Check_In_Count": "4",
"Average_Time_Spent": "8",
"Average_Wait_Time": "6",
"Cancelations": "0",
"New_Patient_Count": "1",
"Existing_Patient_Count": "3",
"Current_Cycle_Date": "2072019"
},
{
"Check_In_Count": "7",
"Average_Time_Spent": "3",
"Average_Wait_Time": "9",
"Cancelations": "0",
"New_Patient_Count": "0",
"Existing_Patient_Count": "7",
"Current_Cycle_Date": "2082019"
},
{
"Check_In_Count": "8",
"Average_Time_Spent": "8",
"Average_Wait_Time": "1",
"Cancelations": "0",
"New_Patient_Count": "4",
"Existing_Patient_Count": "4",
"Current_Cycle_Date": "2092019"
}]
var ccd = data.map( i => i.Current_Cycle_Date )
console.log( ccd )
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 315
|
The 2020–21 Southern Conference men's basketball season began with practices in October 2020, followed by the start of the 2020–21 NCAA Division I men's basketball season in November. Conference play begins in January 2021 and will conclude in March 2021.
Preseason Awards
Preseason awards were announced by the league office on November 11, 2020.
Preseason men's basketball coaches poll
(First place votes in parenthesis)
UNC Greensboro (4) 76
Furman (5) 74
East Tennessee State (1) 63
Mercer 56
Wofford 47
Western Carolina 44
Chattanooga 38
Samford 23
VMI 15
The Citadel 14
Preseason men's basketball media poll
(First place votes in parenthesis)
Furman (16) 278
UNC Greensboro (9) 274
East Tennessee State (4) 211
Mercer (1) 190
Wofford 184
Western Carolina 175
Chattanooga 150
Samford 79
VMI 61
The Citadel 48
Conference matrix
All-Southern Conference awards
Southern Conference men's basketball weekly awards
References
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,482
|
{"url":"http:\/\/self.gutenberg.org\/articles\/eng\/Azuma%27s_inequality","text":"#jsDisabledContent { display:none; } My Account\u00a0|\u00a0 Register\u00a0|\u00a0 Help\n\n# Azuma's inequality\n\nArticle Id: WHEBN0000450515\nReproduction Date:\n\n Title: Azuma's inequality Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Azuma's inequality\n\nIn probability theory, the Azuma\u2013Hoeffding inequality (named after Kazuoki Azuma and Wassily Hoeffding) gives a concentration result for the values of martingales that have bounded differences.\n\nSuppose { Xk : k = 0, 1, 2, 3, ... } is a martingale (or super-martingale) and\n\n|X_k - X_{k-1}| < c_k, \\,\n\nalmost surely. Then for all positive integers N and all positive reals t,\n\nP(X_N - X_0 \\geq t) \\leq \\exp\\left ({-t^2 \\over 2 \\sum_{k=1}^N c_k^2} \\right).\n\nAnd symmetrically (when Xk is a sub-martingale):\n\nP(X_N - X_0 \\leq -t) \\leq \\exp\\left ({-t^2 \\over 2 \\sum_{k=1}^N c_k^2} \\right).\n\nIf X is a martingale, using both inequalities above and applying the union bound allows one to obtain a two-sided bound:\n\nP(|X_N - X_0| \\geq t) \\leq 2\\exp\\left ({-t^2 \\over 2 \\sum_{k=1}^N c_k^2} \\right).\n\nAzuma's inequality applied to the Doob martingale gives the method of bounded differences (MOBD) which is common in the analysis of randomized algorithms.\n\n## Contents\n\n\u2022 Simple example of Azuma's inequality for coin flips 1\n\u2022 Remark 2\n\u2022 References 4\n\n## Simple example of Azuma's inequality for coin flips\n\nLet Fi be a sequence of independent and identically distributed random coin flips (i.e., let Fi be equally likely to be -1 or 1 independent of the other values of Fi). Defining X_i = \\sum_{j=1}^i F_j yields a martingale with |Xk\u00a0\u2212\u00a0Xk\u22121|\u00a0\u2264\u00a01, allowing us to apply Azuma's inequality. Specifically, we get\n\n\\Pr[X_N > t] \\leq \\exp\\left(\\frac{-t^2}{2 N}\\right).\n\nFor example, if we set t proportional to N, then this tells us that although the maximum possible value of XN scales linearly with N, the probability that the sum scales linearly with N decreases exponentially fast with\u00a0N.\n\n## Remark\n\nA similar inequality was proved under weaker assumptions by Sergei Bernstein in 1937.\n\nHoeffding proved this result for independent variables rather than martingale differences, and also observed that slight modifications of his argument establish the result for martingale differences (see page 18 of his 1963 paper).\n\n## References\n\n\u2022 Alon, N.; Spencer, J. (1992). The Probabilistic Method. New York: Wiley.\n\u2022 Azuma, K. (1967). \"Weighted Sums of Certain Dependent Random Variables\" (PDF).\n\u2022 (vol. 4, item 22 in the collected works)\n\u2022 McDiarmid, C. (1989). \"On the method of bounded differences\". Surveys in Combinatorics. London Math. Soc. Lectures Notes 141. Cambridge: Cambridge Univ. Press. pp.\u00a0148\u2013188.\n\u2022 Hoeffding, W. (1963). \"Probability inequalities for sums of bounded random variables\". Journal of the American Statistical Association 58 (301): 13\u201330.\n\u2022 Godbole, A. P.; Hitczenko, P. (1998). \"Beyond the method of bounded differences\". DIMACS Series in Discrete Mathematics and Theoretical Computer Science 41: 43\u201358.\nThis article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.\n\nCrowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.","date":"2020-08-11 15:27:30","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.894871711730957, \"perplexity\": 3602.6642313002894}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-34\/segments\/1596439738816.7\/warc\/CC-MAIN-20200811150134-20200811180134-00493.warc.gz\"}"}
| null | null |
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<th class="colFirst" scope="col">Interface</th>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/BasketMarketParamRef.html" title="interface in org.drip.product.definition">BasketMarketParamRef</a></td>
<td class="colLast">
<div class="block">BasketMarketParamRef interface provides stubs for component's IR and credit curves that constitute the
basket.</div>
</td>
</tr>
<tr class="rowColor">
<td class="colFirst"><a href="../../../../org/drip/product/definition/BondProduct.html" title="interface in org.drip.product.definition">BondProduct</a></td>
<td class="colLast">
<div class="block">BondProduct interface implements the product static data behind bonds of all kinds.</div>
</td>
</tr>
<tr class="altColor">
<td class="colFirst"><a href="../../../../org/drip/product/definition/ComponentMarketParamRef.html" title="interface in org.drip.product.definition">ComponentMarketParamRef</a></td>
<td class="colLast">
<div class="block">ComponentMarketParamRef interface provides stubs for component name, IR curve, forward curve, credit
curve, TSY curve, and EDSF curve needed to value the component.</div>
</td>
</tr>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/BasketProduct.html" title="class in org.drip.product.definition">BasketProduct</a></td>
<td class="colLast">
<div class="block">BasketProduct abstract class extends BasketMarketParamRef.</div>
</td>
</tr>
<tr class="rowColor">
<td class="colFirst"><a href="../../../../org/drip/product/definition/Bond.html" title="class in org.drip.product.definition">Bond</a></td>
<td class="colLast">
<div class="block">Bond abstract class implements the pricing, the valuation, and the RV analytics functionality for the bond
product.</div>
</td>
</tr>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/CalibratableComponent.html" title="class in org.drip.product.definition">CalibratableComponent</a></td>
<td class="colLast">
<div class="block">CalibratableComponent abstract class provides implementation of Component's calibration interface.</div>
</td>
</tr>
<tr class="rowColor">
<td class="colFirst"><a href="../../../../org/drip/product/definition/Component.html" title="class in org.drip.product.definition">Component</a></td>
<td class="colLast">
<div class="block">Component abstract class extends ComponentMarketParamRef and provides the following methods:
- Get the component'sGet initial notional, notional, and coupon.</div>
</td>
</tr>
<tr class="altColor">
<td class="colFirst"><a href="../../../../org/drip/product/definition/CreditComponent.html" title="class in org.drip.product.definition">CreditComponent</a></td>
<td class="colLast">
<div class="block">CreditComponent is the base abstract class on top of which all credit components are implemented.</div>
</td>
</tr>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/CreditDefaultSwap.html" title="class in org.drip.product.definition">CreditDefaultSwap</a></td>
<td class="colLast">
<div class="block">CreditDefaultSwap is the base abstract class implements the pricing, the valuation, and the RV analytics
functionality for the CDS product.</div>
</td>
</tr>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/FXForward.html" title="class in org.drip.product.definition">FXForward</a></td>
<td class="colLast">
<div class="block">FXForward is the abstract class exposes the functionality behind the FXForward Contract.</div>
</td>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/FXSpot.html" title="class in org.drip.product.definition">FXSpot</a></td>
<td class="colLast">
<div class="block">FXSpot is the abstract class exposes the functionality behind the FXSpot Contract.</div>
</td>
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<td class="colFirst"><a href="../../../../org/drip/product/definition/RatesComponent.html" title="class in org.drip.product.definition">RatesComponent</a></td>
<td class="colLast">
<div class="block">RatesComponent is the abstract class that extends CalibratableComponent on top of which all rates
components are implemented.</div>
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{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,484
|
{"url":"https:\/\/andrescaicedo.wordpress.com\/2008\/04\/16\/","text":"## 116c- Homework 2\n\nApril 16, 2008\n\nDue Tuesday, April 22 at 2:30 pm.\n\n## 116c- Lecture 5\n\nApril 16, 2008\n\nWe defined addition, multiplication, and exponentiation of ordinals, and stated some basic properties of these operations. They extend into the transfinite the usual operations on natural numbers.\n\nI made a mistake when indicating how to define these operations \u201cintrinsically\u201d rather than as a consequence of the transfinite recursion theorem: In the definition of ordinal exponentiation $\\alpha^{\\cdot\\beta}$, we consider the set $\\{f\\in{}^\\beta\\alpha:f(\\xi)=0\\mbox{ for all but finitely many }\\xi<\\beta\\}$ and order it by setting $f, for $f\\ne g$ in this set, iff $f(\\xi) as ordinals, where $\\xi<\\beta$ is the largest ordinal such that $f(\\xi)\\ne g(\\xi)$.\u00a0In particular, there is such an ordinal $\\xi$. In\u00a0class, I mentioned that $\\xi$ was the smallest such ordinal, but this does not work.\n\nUsing Hartog\u2019s function and transfinite recursion we defined the long sequence of (well-ordered) cardinals, the alephs.\n\nRemark. It was asked in class whether one can make sense of well-orders longer than ${\\sf ORD}$ and if one can extend to them the operations we defined.\n\nOf course, one can define classes that are well-orders of order type longer than ${\\sf ORD}$ (for example, one can\u00a0define\u00a0the lexicographic ordering on $2\\times{\\sf ORD}$, which would correspond to the \u201clong ordinal\u201d ${\\sf ORD}+{\\sf ORD}$). In ${\\sf ZFC}$ this is cumbersome (since classes are formulas) but possible. There is an extension of ${\\sf GB}$ that allows these operations to be carried out in a more natural way, Morse-Kelley set theory ${\\sf MK}$, briefly discussed here.\n\nHowever, I do not know of any significant advantages of this approach. But a few general (and unfortunately vague) observations can be made:\n\n\u2022 Most likely, any way of extending well-orders beyond ${\\sf ORD}$ would also provide a way of extending $V$ to a longer \u201cuniverse of classes.\u201d The study of these end-extensions (in the context of large cardinals, where it is easier to formalize these ideas)\u00a0has resulted in an\u00a0interesting research area originated by Keisler and Silver with recent results by Villaveces and others.\n\u2022 I also expect that any systematic way of doing this\u00a0would translate with minor adjustments into a treatment of indiscernibles and elementary embeddings (which could potentially turn into\u00a0a motivation for the study of these important topics and would be interesting at least\u00a0from a pedagogical point of view).\n\u2022 As I said, however, I do not know of any systematic\u00a0attempt at doing something with these \u201clong ordinals.\u201d With one exception: the work of Reinhardt, with the caveat that\u00a0I couldn\u2019t make much sense of it in any productive way years ago.\u00a0But this is an excuse to recommend a couple of excellent papers by\u00a0Penelope Maddy, Believing the Axioms I and II, that originally appeared in The Journal of Symbolic Logic in 1988 and can be accessed through JSTOR. These papers discuss the intuitions\u00a0behind the axioms of set theory and end up discussing more recent developments (like large cardinal axioms and determinacy assumptions), and I believe you will\u00a0appreciate them. During her discussion of very large cardinals, Maddy mentions Reinhardt ideas, so this can also be a place to start if one is interested in the issue of \u201clong ordinals.\u201d","date":"2021-12-06 00:27:57","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 18, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8877779245376587, \"perplexity\": 546.2159627162747}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964363226.68\/warc\/CC-MAIN-20211205221915-20211206011915-00566.warc.gz\"}"}
| null | null |
Alamanda Motuga, né le , est un joueur de rugby à XV et rugby à sept international samoan, évoluant au poste de troisième ligne aile. Il joue avec la province néo-zélandaise des Counties Manukau en NPC depuis 2020, et avec la franchise des Moana Pasifika en Super Rugby depuis 2022.
Biographie
Après être passé par le lycée de Manurewa dans la banlieue d'Auckland en Nouvelle-Zélande, Alamanda Motuga commence à jouer avec le club de rugby local, le Manurewa RFC, dans le championnat amateur de la fédération des Counties Manaukau. Il joue également au rugby à sept avec l'équipe des Auckland Marist.
En février 2016, il est retenu avec l'équipe des Samoa de rugby à sept pour disputer les étapes de Las Vegas et Vancouver dans le cadre des World Rugby Sevens Series. Au mois de mai de la même année, il fait partie du groupe samoan qui remporte le Paris Sevens 2016, ce qui est la première victoire samoane dans un tournoi depuis 2012. Toujours en 2016, il participe au tournoi de qualification olympique de Monaco.
Avec la sélection samoane à sept, il dispute cinq éditions des Sevens Series, ainsi que les Jeux du Commonwealth de 2018 et la Coupe du monde 2018. Il joue un total de , et inscrit ().
Au mois de juillet 2019, il est sélectionné pour la première fois avec l'équipe des Samoa à XV pour disputer la Pacific Nations Cup. Il connait sa première sélection le à l'occasion d'un match contre l'équipe des Tonga à Apia.
Initialement non retenu pour disputer la Coupe du monde 2019, il rejoint les Manu Samoa en cours de compétition en remplacement d'Afa Amosa blessé lors du premier match de poule. Il ne dispute cependant aucune rencontre lors de la compétition.
En 2020, après que la saison de rugby à sept la saison ne soit interrompue à cause de la pandémie de Covid-19, il rejoint la province néo-zélandaise des Counties Manukau en NPC. Il effectue une bonne première saison avec sa nouvelle équipe, en inscrivant cinq essais en huit matchs, dont un triplé contre Manawatu.
En , il est retenu avec les Moana Pasifika, qui est alors une sélection représentant les îles du Pacifique, pour affronter les Māori All Blacks le . Il inscrit un essai lors de la rencontre, qui est perdue par son équipe sur le score de 28 à 21.
En 2022, il rejoint à nouveau les Moana Pasifika, qui sont entre-temps devenus une franchise et qui viennent de faire leur entrée en Super Rugby. Il joue son premier match le contre les Crusaders. Il joue un total de huitmatchs lors de la saison.
Palmarès
En équipe nationale
Rugby à XV
4 sélection depuis 2019.
5 points (1 essai).
Participation à la Coupe du monde 2019 (0 match).
Rugby à sept
Vainqueur du Paris Sevens 2016 avec l'équipe des Samoa à sept.
Notes et références
Liens externes
Joueur international samoan de rugby à XV
Joueur samoan de rugby à sept
Joueur de la Counties Manukau Rugby Union
Joueur des Moana Pasifika
Troisième ligne aile (rugby à XV)
Pilier (rugby à sept)
Talonneur (rugby à sept)
Naissance en septembre 1994
Lieu de naissance inconnu
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,578
|
Thomas Barquee and Satkirin collaborate together again in a soothing and upbeat Mantra collection with spoken English translations.
Yogi`s and yogini`s will love this music for yoga, meditating, relaxing, and joy filled city drives. Beautiful female vocals, mantra`s with English woven through for your entire family`s delight. Trumpet, keyboard, sitar, flute, guitar, tabla, Oud & more.
My friend, Dave Douglas, who happens to be one of the world's best trumpet players as well as a Kundalini Yogi, played on track #5 called WAHE GURU. I could not imagine trumpet accompany Mantra music. But I knew if anybody could make it work it would be Dave. He has studied Indian Classical Raga and traveled to Calcutta with Pundit Samir Chatterjee, the famous tabla master. So this is one of my favorite tracks as it exudes such joy and originality.
You may not know that I love Deva Premal's music when I need to relax, and I love to bike ride to Prem Joshua's DANCE OF SHAKTI. So I really put my heart into putting down some Sikh Shabds with all the bells and whistles of funky percussion of Tripp Dudley, sweet flute of Steve Gorn. So there is a lot of musical variety and merging of the Indian classical Raga type sounds and Western classical sounds. The track of "Ray Man eh Bidh Jog" is an old remake of the Shabd Yogi Bhajan asked Guru Shabd Singh Khalsa to compose for healing and for guiding the mind into understanding Yoga. There are so many memories of groups of us going to the bedside of someone needing healing that we would sing this "Re Man…." for.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 510
|
Kowalski Engineering, Inc. is a registered Engineering Corporation only in the State of Ohio.
*This list is current as of September 3, 2014. Contact us to confirm registration in these or additional states.
Copyright © 2019 Retaining Wall Videos.com. All Rights Reserved.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,090
|
/**
* Created by princesoni on 3/13/16.
*/
var server = require('../lib/server');
var should = require('should');
var token = 'xxxxxx';
/**
* Test cases for Server
*/
describe('Server', function () {
this.timeout(100000000);
/**
* Positive test cases
*/
describe('Positive cases for server', function () {
var serverObj = new server(token);
it('should get time of server ', function (done) {
serverObj.time(function (err, result) {
should.not.exist(err);
should.exist(result);
done();
});
});
it('should get timezones of server ', function (done) {
serverObj.timezones(function (err, result) {
should.not.exist(err);
should.exist(result);
done();
});
});
it('should get useragent of server ', function (done) {
serverObj.userAgent(function (err, result) {
should.not.exist(err);
should.exist(result);
done();
});
});
it('should get ip of server ', function (done) {
serverObj.ip(function (err, result) {
should.not.exist(err);
should.exist(result);
done();
});
});
});
/**
* Negative test cases
*/
describe('Negative cases', function () {
var serverObj = new server();
it('should give no access token found', function (done) {
serverObj.time(function (err, result) {
done();
})
});
});
});
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,114
|
Lee Mack
Follow Lee Mack
Lee Mack Quick Links News Pictures Press RSS
Lee Mack Quick Links
News Pictures Press RSS
Boy George, Miranda Hart And Geri Halliwell Set For Celebrity 'Gogglebox'
By Ed Biggs in Movies / TV / Theatre on 09 October 2015
Boy George Miranda Hart Geri Horner Nick Frost Jamie Dornan Lee Mack
The star-studded version of 'Gogglebox' is part of a special night of shows on Channel 4 designed to raise money for charity.
Channel 4 is to broadcast a special episode of runaway favourite 'Gogglebox', featuring celebrities such as Boy George, Miranda Hart and Geri Halliwell shouting things at the TV and each other.
As part of a night of one-off programming tonight (Friday October 9th) in order to raise money and awareness for the charitable organisation Stand Up To Cancer, a star-studded line-up will be sitting in front of their televisions for a night in and commenting on the week's shows.
Posted by Marilyn Peter Robinson on Wednesday, 7 October 2015
Continue reading: Boy George, Miranda Hart And Geri Halliwell Set For Celebrity 'Gogglebox'
Jack Whitehall Wins 'King Of Comedy' For Third Time
By Ed Biggs in Movies / TV / Theatre on 18 December 2014
Jack Whitehall Graham Norton Jonathan Ross Lee Mack Monty Python
The 'Bad Education' actor and popular stand-up comedian scooped the prize for the third year in a row.
Jack Whitehall has been named as "king of comedy" for the third consecutive year at the British Comedy Awards 2014.
The 26 year old comedian and actor won the publicly-voted prize at the ceremony in Wembley, north-west London, on Tuesday evening. He beat the other nominated contenders David Mitchell, Lee Mack, Graham Norton, Greg Davies and Jo Brand.
Jack Whitehall won 'King of Comedy' at the BBC Comedy Awards for a third time a row
Continue reading: Jack Whitehall Wins 'King Of Comedy' For Third Time
Lee Mack and Rob Brydon - Arqiva British Academy Television Awards held at the Theatre Royal, Drury Lane - Arrivals. - London, United Kingdom - Sunday 18th May 2014
Male Dominated BBC Panel Shows Facing A Change
By Jack de Aguilar in Movies / TV / Theatre on 09 February 2014
Victoria Wood Lee Mack
It's time for women to feature prominently in panel shows, the BBC say
BBC bosses have decided that males dominate their panel shows, and it's about time that changes. Starting soon, there will be "no excuse" for an all male panel on their various shows.
Lee Mack might not be too popular with his comments on the matter
BBC entertainment controller Mark Linsey said: "Comedy panel shows are always better for having a good mix of people and of course that must include women. I'm making it clear to production teams that there's just no excuse for delivering all male guest lists."
Continue reading: Male Dominated BBC Panel Shows Facing A Change
Lee Mack and Tara McKillop - The British Comedy Awards 2013 held at Fountain Studios - Arrivals - London, United Kingdom - Thursday 12th December 2013
British Comedy Awards 2013: Jack Whitehall, David Mitchell, Sarah Millican Up For Funny Crown
By Lauren James in Lifestyle / Showbiz on 27 November 2013
Alan Carr Graham Norton Sarah Millican Jack Whitehall David Mitchell Lee Mack
Which of the six nominees will you vote to become King or Queen of comedy?
The nominations are in for this year's British Comedy Awards King or Queen. Out of various awards for television performances, breakthrough comedians, and entertainment personalities, one comedian will rise to be crowned the year's King or Queen of Comedy at the ceremony hosted by Jonathan Ross.
Sarah Millican [L] & David Mitchell [R] Are Both Up For The Comedy Award.
Though there are 15 categories in total, only one - the King or Queen of Comedy - is open to public voting meaning British people will choose their favourite comedian of the year. Entertainers up for one of the country's most prestigious comedy awards include Jack Whitehall, David Mitchell, Sarah Millican, Graham Norton, Lee Mack and Alan Carr.
Continue reading: British Comedy Awards 2013: Jack Whitehall, David Mitchell, Sarah Millican Up For Funny Crown
Lee Mack And Catherine Tate Sign Up For UK 'Everybody Loves Raymond'
By Michael West in Movies / TV / Theatre on 09 May 2013
Lee Mack Catherine Tate Ray Romano
Can a UK version starring Lee Mack and Catherine Tate work?
British comedy stars Lee Mack and Catherine Tate have signed on to appear in a British remake of Everybody Loves Raymond, one of the most successful US sitcoms of all time. The Smiths will be set in Cheshire and follows a successful sportswriter who lives across the street from his overbearing parents and socially inept older brother. The BBC One pilot has been written by Mack and will be produced by Silver River, the team behind Pulling.
Image: Catherine Tate
Continue reading: Lee Mack And Catherine Tate Sign Up For UK 'Everybody Loves Raymond'
Twenty Twelve Snubbed At The British Comedy Awards, But Jack Whitehall Wins The Crown
By Lorna Greville in Movies / TV / Theatre on 13 December 2012
Jack Whitehall Vic Reeves Bob Mortimer Alan Carr Olivia Colman Jessica Hynes Rebecca Front Sarah Millican David Mitchell Graham Norton Jonathan Ross Lee Mack
The British Comedy Awards were held last night, presented once again by Jonathan Ross. It saw a room of the country's funniest people come under one roof to celebrate and ridicule one another, as is the British tradition.
Twenty Twelve and The Thick of It have been compared since the former hit screens in 2011. It satirises the shambolic organization of this year's the Olympic games, while The Thick of It satirises the inner workings of the British government. However, despite audiences loving Twenty Twelve and it receiving four nominations in the three categories that it's eligible for, it failed to take home anything. Both Olivia Colman and Jessica Hynes from the show were nominated for Best Actress, but they lost out to The Thick of It's Rebecca Front. It seems an odd snub given that everyone has been celebrating all things Olympic since it was revealed that London got the deal.
Another surprise was Jack Whitehall's voted win of King of Comedy for 2012. He was up against some serious comedy heavy weights that have done the circuit for years: Alan Carr, Lee Mack, David Mitchell, Graham Norton and last year's Queen of Comedy, Sarah Millican. Initial speculation had assumed David Mitchell would get the prize, for his many panel show appearances, Peep Show and That Mitchell and Webb Look. However, Whitehall has really broken through this year, with popular shows Bad Education and Fresh Meat, which are a lot better than they sound.
Continue reading: Twenty Twelve Snubbed At The British Comedy Awards, But Jack Whitehall Wins The Crown
Farce, Or Fair Enough? Olivia Colman To Battle Herself At British Comedy Awards
By Michael West in Movies / TV / Theatre on 03 December 2012
Olivia Colman Hugh Bonneville Rebecca Front Paddy Considine Steve Coogan Alan Carr David Mitchell Graham Norton Jack Whitehall Lee Mack Sarah Millican
Olivia Colman has been nominated twice in the same category for the upcoming British Comedy Awards. The 39-year-old, perhaps best known for playing Sophie on Peep Show, is up for Best Actress for two BBC shows, Rev and Twenty Twelve.
The decision has left some comedy fans bemused, though the general consensus is that Colman has put in two excellent performances and therefore has every right to be nominated twice. She played Hugh Bonneville's character's long-suffering assistant Sally Owen in the London Olympics comedy and plays Tom Hollander's wife Alex Smallbone on religious comedy Rev. Though having two chances to win the award, Colman still faces stiff competition to land the gong, with Twenty Twelve co-star Jessica Hynes also nominated. The Thick Of It's Rebecca Front, who plays the MP Nicola Murray, is also up for Best Actress. Colman - now considered one of the UK's top actresses - has already had a superb year, winning a slew of awards for playing an abused charity shop worker in Paddy Considine's gritty drama Tyrannosaur.
Elsewhere at the comedy awards, The Thick Of It is one of five shows to receive three nominations, along with The Graham Norton Show, Rev, Cardinal Burns and Harry Hill's TV Burp. Steve Coogan picked up a couple of nominations for Alan Partridge: Welcome To Places In My Life. This year's King or Queen of Comedy will be contested by Alan Carr, David Mitchell, Graham Norton, Jack Whitehall, Lee Mack and Sarah Millican. The award winners will be announced live on Channel 4 on Wednesday 12 December.
Continue reading: Farce, Or Fair Enough? Olivia Colman To Battle Herself At British Comedy Awards
4th August, 1968
@https://twitter.com/leemack
Lee Mack Tweets
People Index: 0 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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Can Your Business Benefit From A Targeted Media Campaign?
Almost every business can benefit from a targeted media campaign. Using all of the traditional pieces of a public relations approach, including publicity tactics, paid advertising, event marketing, press releases, and media pitches, can result in more attention, more followers, and eventually more sales.
How can businesses benefit from targeted media campaigns?
Grow your brand's awareness. These campaigns help spread the word about your business's products or services, the mission, and the key staff members.
Increase your website's traffic. You'll always want to include your URL in any piece of media. Chances are you'll be increasing your traffic to your website.
Generate conversation around your brand. Whether you're putting out social media posts, pitching story ideas to local media, or issuing a press release on a local event, people will start talking about your company or brand. Stay consistent, because once they start seeing your name appear in additional places, the more they will remember it.
You can build customer loyalty. Regularly having your name in the press, builds your brand/company's reputation. Through building that reputation, customer loyalty also comes into play.
They can help to establish your brand as an authority in the field. When the media is looking for an authoritative voice on a story in your industry, they will keep you in mind. Why? If your name pops up in search results, there is a likeliness they will reach out to you for a quote. You can become the go-to, trusted resource for information.
Targeted media campaigns can also keep your customers up to date on happenings within the company.
Strategies, like traditional press releases, allow you to use your targeted media campaign strategy to keep the public up-to-date on new products, services, or employment changes in your company.
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{
"redpajama_set_name": "RedPajamaC4"
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Best Time to go to Ko Samui - Climate, Weather, Where to go?
Best time to go to Ko Samui for a perfect weather and where to go?
When is the best time to go to Ko Samui?
The best months to go to Chaweng Beach are february and march.
The best months to go to Bang Kao (Laem Sor beach) are february and march.
The best months to go to Bophut are february and march.
The best months to go to Choeng Mon are february and march.
The best months to go to Koh Madsum are february and march.
The best months to go to Koh Phaluai are february and march.
On these 3 graphs, we present the evolution of temperatures of Ko Samui and month-by-month rainfall for the cities of Chaweng Beach, Bang Kao (Laem Sor beach), Bophut, Choeng Mon and Koh Madsum, as well as the month-by-month sea temperature for coastal cities.
Where to go in Ko Samui?
What can I do in Ko Samui?
Is this weather information for Ko Samui reliable?
Climate data for Ko Samui has been gathered every day since January 2009. The analysis of these meteorological data for Ko Samui allows us to determine the average for each month in Chaweng Beach, Bang Kao (Laem Sor beach), Bophut, Choeng Mon, Koh Madsum, Koh Phaluai, Koh Taen, Lamai beach, and 3 other cities.
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"redpajama_set_name": "RedPajamaC4"
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\section{Introduction}
\label{Sec:Intro}
In relativistic cosmology, the energy-matter content of the universe is often considered to be a perfect fluid in equilibrium. Perfect fluids manifest no dissipative effects and do not generate entropy, their dynamics being reversible. The early universe is commonly modelled by a radiation fluid, while for the later epoch a dust model is usually applied. However, the forces driving the fluid towards the equilibrium state will always be dissipative. Also, the transition between the epochs involves interactions between radiation and matter, which again implies the presence of dissipative processes~\cite{Udey1982,Zimdahl1996,Zimdahl1997}. Such processes are described in the framework of dissipative, irreversible thermo- and fluiddynamics.
\par
The potential importance of dissipative processes in cosmology was briefly discussed by R.~Tolman already in 1934~\cite{Tolman1934}. The relativistic formulation of fluid dynamics goes back to Eckart~\cite{Eckart1940} and Landau and Lifshitz~\cite{Landau1958}, who pointed out that all real substances, which should include cosmic fluids, do have transport properties such as viscosity and heat conductivity; for applications to cosmology, see e.\,g.~\cite{Kolb1990}. However, it is now well known that both variants ot the theory suffer from non-causality, including dissipative perturbations that propagate at infinite speeds, and instabilities~\cite{Hiscock1983, Hiscock1985, Hiscock1988a}; such features are clearly not acceptable in cosmology.
\par
These pathologies have been mediated in the more advanced Israel-Stewart~(IS) theory of irreversible thermodynamics~\cite{Israel1976, Israel1979}; for a brief review with applications to cosmology, see~\cite{Maartens1996}. Stability and causality are provided by including terms up to second order in the dissipative fluxes in the 4-vector of entropy production, while in the Eckart theory and the Landau-Lifshitz formulation, these terms are not included.
\par
A simplified version of the IS~transport equations is often employed instead of the full theory~\cite{Maartens1996}. In this truncated theory, several divergence terms are assumed to be negligible and are omitted, without violating causality and stability.
\par
Standard Friedmann-Robertson-Walker cosmologies have been studied using both truncated~\cite{Coley1995, Zimdahl1997a} and full~\cite{Maartens1995, Coley1996} versions of the IS~theory. In particular, reheating and inflation within causal thermodynamics have been considered respectively in~\cite{Zimdahl1997a} and~\cite{Maartens1995}. It was reported that some solutions obtained using the truncated theory are substantially different from those of the full theory, while the others are qualitatively similar.
\par
The geometry of spatially homogeneous and isotropic cosmological models restricts the dissipative processes to so-called scalar dissipation, i.\thinspace e. shear viscosity and heat conduction are not allowed in this case. The next step in investigating dissipative cosmologies therefore involves anisotropic models. Spatially homogeneous Bianchi cosmologies~\cite{Ellis2012, Groen2007, Christiansen2008} are of great theoretical interest and have been subject to thorough studies~(see e.\thinspace g.~\cite{Ellis2012} and references therein; for causal thermodynamics applied to simple anisotropic models, see~\cite{Belinskii1979, Hoogen1995}).
\par
In this paper we shall investigate the application of the IS theory to important Bianchi class~B cosmological models, namely types~IV and~V, and find out to which extent the truncated version of the IS equations represents the results of the full theory in anisotropic spacetimes. We shall use a simplified fluid model, which, however, retains the basic transport properties.
\par
The paper is organized as follows. In section~\ref{Sec:Model} we briefly describe the dissipative fluid model and the assumptions made. The system of equations governing the dynamics of the cosmologies is presented in section~\ref{Sec:Equations}. The future asymptotic states of the models without and with a positive cosmological constant are described in sections~\ref{Sec:AttractorsNoLambda} and~\ref{Sec:AttractorsLambda}, respectively. Conclusions are discussed in section~\ref{Sec:Conslusion}.
\section{The fluid model}
\label{Sec:Model}
We consider the matter content of the universe to be a dissipative fluid with energy-momentum tensor
\begin{equation}
\label{Eq:Model:Tensor}
T_{\alpha \beta} = (\rho+p+\pi)u_\alpha u_\beta+(p+\pi)g_{\alpha \beta}+q_{(\alpha}u_{\beta)}+\pi_{\alpha \beta},
\end{equation}
where~$\pi$ stands for the bulk viscous pressure, and~$\pi_{\alpha\beta}$ is the shear viscous stress tensor (with~$\pi_{\alpha\beta}u^\beta=\pi_{[\alpha\beta]}=\pi_\alpha^{~\alpha}=0$), the rest of the notations being standard. Pressure~$p$ and energy density~$\rho$ in~(\ref{Eq:Model:Tensor}) refer to the local equilibrium values of the corresponding functions.
\par
The energy-momentum tensor is decomposed with respect to a unit vector~$\mathbf{u}$. We apply the orthonormal frame approach~\cite{Elst1997}, where~$\mathbf{u}$ is chosen to be a unit vector, normal to spatial hypersurfaces. In the current paper we shall treat the case of so-called non-tilted fluids~\cite{King1973}, the 4-velocity of which, as defined by the particle flow, is aligned with~$\mathbf{u}$.
\par
For simplicity, we consider the so-called $\gamma$-fluids with barotropic pressure and temperature obeying
\begin{equation}
p=(\gamma-1)\rho, \quad T\propto \rho^{(\gamma-1)/\gamma}, \label{Eq:Model:Gamma-Law}
\end{equation}
the~$\gamma$-parameter being a positive constant. We assume~$1<\gamma \leq 4/3$, which is often used to model a dust-radiation mixture as a single fluid.
\par
We take into account that the linear transport equations of the IS~theory are derived with the assumption that the fluid is close to equilibrium. This implies that the dissipative fluxes are small and satisfy
\begin{equation}
\vert \pi \vert <<p, \qquad {(\pi_{ab}\pi^{ab})}^{1/2}<<p, \qquad {(q_aq^a)}^{1/2}<<\rho. \label{Eq:Model:EqulibriumConditions}
\end{equation}
Heat conductivity of the fluid is known to cause problems in the linear Israel-Stewart theory, since corresponding deviations from equilibrium~(\ref{Eq:Model:EqulibriumConditions}) may become infinitely large~\cite{Hiscock1988}. We therefore assume that the fluid is not heat-conducting, which in a non-tilted cosmological model leads to a vanishing energy flux vector $(q_a=0)$.
\par
In certain solutions the near-equilibrium conditions~(\ref{Eq:Model:EqulibriumConditions}) are broken by bulk or shear viscous stresses. For example, bulk viscous terms are large if the fluid undergoes accelerated expansion. Strictly speaking, a consistent theory of non-linear thermodynamics is required to provide an accurate description of out-of-equilibrium processes~\cite{Maartens1997,Chimento1997}. We nevertheless assume that in case of finite and relatively small deviations from equilibrium, the solutions obtained using linear thermodynamics are still physically reasonable.
The transport equations of IS~theory in their full form can be written as\footnote{Index 1 is reserved for heat conduction, which is neglected in the present paper}
\begin{align}
\tau_0\dot{\pi}+\pi &= -3\zeta H-\frac{1}{2}\tau_0\pi \left[3H+\frac{\dot{\tau}_0}{\tau_0}-\frac{\dot{\zeta}}{\zeta}-\frac{\dot{T}}{T} \right],\label{Eq:Model:IS-Full-0-Bulk} \\
\tau_2\dot{\pi}_{ab}+\pi_{ab} &= -2\eta \sigma_{ab}-\frac{1}{2}\tau_2\pi_{ab} \left[3H+\frac{\dot{\tau}_2}{\tau_2}-\frac{\dot{\eta}}{\eta}-\frac{\dot{T}}{T} \right], \label{Eq:Model:IS-Full-2-Shear}
\end{align}
where the (positive) bulk and shear viscosity coefficients $\zeta$ and $\eta$ are related to the corresponding relaxation times $\tau_0, \tau_2$ by
\begin{equation}
\tau_0=\zeta \beta_0, \qquad \tau_2=2\eta \beta_2, \label{Eq:Model:RelaxationTimes}
\end{equation}
$\beta_0, \beta_2$ being the non-negative thermodynamic coefficients for scalar and tensor dissipative contributions to the entropy density.
\par
The terms in square brackets on the right-hand sides of~(\ref{Eq:Model:IS-Full-0-Bulk})--(\ref{Eq:Model:IS-Full-2-Shear}) are often implicitly supposed to be negligible in comparison with the other terms in the equations. This leads to a widely used simplified version of equations~(\ref{Eq:Model:IS-Full-0-Bulk})--(\ref{Eq:Model:IS-Full-2-Shear}), which is commonly referred to as the truncated Israel-Stewart theory~\cite{Maartens1995}.
\par
For simplicity, we make the widely used assumption that the transport coefficients and corresponding relaxation times~(\ref{Eq:Model:RelaxationTimes}) are proportional to powers of the energy density of the fluid:
\begin{align}
\label{Eq:Model:ViscIndices}
\begin{split}
\zeta & \propto \rho^{m_0}, \qquad \frac{1}{\beta_0} \propto \rho^{r_0}; \\
\eta & \propto \rho^{m_2}, \qquad \frac{1}{\beta_2} \propto \rho^{r_2}.
\end{split}
\end{align}
There are some discussions in literature concerning what the exponents should actually be, see e.\thinspace g. \cite{Coley1995, Hoogen1995}. However, the only choice that results in values consistent with correct dimensional scaling properties is:
\begin{equation}
\label{Eq:Model:PhysicalIndices}
m_0=m_2=1/2, \qquad r_0=r_2=1.
\end{equation}
Finally, we note an important point. The hydrodynamic description must be valid for the considered matter model, i.\thinspace e. the mean interaction time~$t_c$ of fluid particles is required to be much shorter than the characteristic timescale for macroscopic processes. In cosmology, this means that
\begin{equation}
t_c<<H^{-1}. \label{Eq:Model:HydrodynamicDescription}
\end{equation}
Therefore, we assume the existence of some long-range interactions, which ensure that the fluid description~(\ref{Eq:Model:HydrodynamicDescription}) remains valid for the matter under accelerated expansion.
\section{Evolution equations and constraints}
\label{Sec:Equations}
We investigate the chosen cosmological models using the dynamical systems approach~\cite{Wainwright1997}. Scale-invariant dimensionless geometric and fluid variables are introduced using the common principles and notations of the field. Dimensionless bulk and shear viscosity are introduced by
\begin{align}
\frac{\pi}{3H^2} &= \Pi; \\
\frac{\pi_{ab}}{H^2} &= \Pi_{ab}= \left[
\begin{array}{ccc}
-2\Pi_+ & 0 & 0 \\
0 & \Pi_+ +\sqrt{3}\Pi_- & \sqrt{3}\Pi_{23} \\
0 & \sqrt{3}\Pi_{23} & \Pi_+-\sqrt{3}\Pi_-
\end{array} \right],
\end{align}
where we have taken into account that~$\Pi_{12}=\Pi_{13}=\Sigma_{12}=\Sigma_{13}=0$ for non-tilted cosmologies. With the choice~(\ref{Eq:Model:PhysicalIndices}) we can introduce positive constants~$a_0,~b_0$ and~$a_2,~b_2$ -- the bulk and shear viscosity parameters, respectively -- by
\begin{align}
\begin{split}
\frac{1}{H^2}\cdot\frac{1}{\beta_0} &= a_0 \Omega, \qquad \frac{1}{H}\cdot \frac{1}{\zeta \beta_0}=b_0\sqrt{\Omega};\\
\frac{1}{H^2}\cdot\frac{1}{\beta_2} &= a_2 \Omega, \qquad \frac{1}{H}\cdot \frac{1}{2\eta \beta_2}=b_2\sqrt{\Omega}.
\end{split}
\end{align}
The parameter~$a_0$ is restricted from above by $a_0<3\gamma(2-\gamma)$. This limitation comes from the expression for the speed of bulk viscous perturbations derived in~\cite{Maartens1996}.
\par
Introducing the quantities
\begin{equation}
\mathbf{X}_\Sigma = (\Sigma_+,\Sigma_-,\Sigma_{23}), \qquad
\mathbf{X}_\Pi = (\Pi_+,\Pi_-,\Pi_{23})
\end{equation}
as vectors in~$\mathbb{R}^3$, we can write the equations in a more compact manner. The most general case we consider is a Bianchi type~IV cosmology with a dissipative fluid and vacuum energy having the form of a cosmological constant. In this case, the equations of motion are:
\begin{align}
\Sigma^\prime_+ &= (q-2)\Sigma_+ -2N^2+\Pi_+, \label{Eq:Eqs:EvolSigmaPlus} \\
\Sigma^\prime_- &= (q-2-2\sqrt{3}\Sigma_{23})\Sigma_-+2AN+\Pi_-,\\
\Sigma_{23}^\prime &= (q-2)\Sigma_{23} +2\sqrt{3}\Sigma_-^2-2\sqrt{3}N^2+\Pi_{23},\\
N^\prime &= (q+2\Sigma_++2\sqrt{3}\Sigma_{23})N, \\
A^\prime &= (q+2\Sigma_+)A.
\end{align}
The evolution equations for the fluid can be written as:
\begin{align}
\Omega^\prime &= (2q+2-3\gamma)\Omega-3\Pi-2(\mathbf{X}_\Sigma\cdot \mathbf{X}_\Pi),\\
\Omega_\Lambda^\prime &= 2(q+1)\Omega_\Lambda,\\
\Pi^\prime &= \left[-\frac{3}{2}+\frac{1}{\gamma}(q+1)-b_0\Omega^{1/2}+\frac{2\gamma-1}{2\gamma}\frac{\Omega^\prime}{\Omega} \right]\Pi-a_0\Omega, \label{Eq:Eqs:EvolBulkVisk} \\
\mathbf{X}_\Pi^\prime &= \left[-\frac{3}{2}+\frac{1}{\gamma}(q+1)-b_2\Omega^{1/2}+\frac{2\gamma-1}{2\gamma}\frac{\Omega^\prime}{\Omega} \right]\mathbf{X}_\Pi-2a_2\mathbf{X}_\Sigma, \label{Eq:Eqs:EvolShearVisk}
\end{align}
with the deceleration parameter~$q$ being given by
\begin{equation}
q=2\vert\mathbf{X}_\Sigma\vert^2+\left(\frac{3}{2}\gamma-1\right)\Omega+\frac{3}{2}\Pi-\Omega_\Lambda.
\end{equation}
In the absence of tilt and heat conduction, only two algebraic constraint survives:
\begin{align}
1 &= \vert\mathbf{X}_\Sigma\vert^2+N^2+A^2+\Omega+\Omega_\Lambda, \label{Eq:Eqs:HamiltonianConstraint} \\
0 &= \Sigma_+A+\Sigma_-N. \label{Eq:Eqs:RConstraint}
\end{align}
Equations (\ref{Eq:Eqs:EvolSigmaPlus})--(\ref{Eq:Eqs:RConstraint}) form a complete system which will be used for analytical and numerical investigations. Replacing equations~(\ref{Eq:Eqs:EvolBulkVisk}) and~(\ref{Eq:Eqs:EvolShearVisk}) respectively with
\begin{align}
\Pi^\prime &= \left[2(q+1)-b_0\Omega^{1/2}\right]\Pi-a_0\Omega, \label{Eq:Eqs:EvolBulkVisk-trunc} \\
\mathbf{X}_\Pi^\prime &= \left[2(q+1)-b_2\Omega^{1/2}\right]\mathbf{X}_\Pi-2a_2\mathbf{X}_\Sigma\label{Eq:Eqs:EvolShearVisk-trunc}
\end{align}
yields the system with the truncated IS~transport equations.
\par
The equations for Bianchi type~V cosmologies are obtained by setting~$N=0$ in equations~(\ref{Eq:Eqs:EvolSigmaPlus})--(\ref{Eq:Eqs:RConstraint}). Note that it implies that for this model~$\Sigma_+=\Pi_+=0$. As for the case without vacuum energy, one simply assigns~$\Omega_\Lambda=0$.
\par
Independently of whether one is considering the system with the full or truncated transport equations, the state vector describing a Bianchi type~IV cosmology containing a dissipative $\gamma$-fluid and vacuum energy can be written as
\begin{equation}
\mathbf{X}=(\mathbf{X}_\Sigma,N,A,\Omega,\Omega_\Lambda,\Pi,\mathbf{X}_\Pi)
\end{equation}
modulo the constraints~(\ref{Eq:Eqs:HamiltonianConstraint})--(\ref{Eq:Eqs:RConstraint}). The evolution takes place on a 9-dimensional subspace of~$\mathbb{R}^{11}$. The physical state space is therefore of dimension~9. Considering the simpler Bianchi type~V model or the case without a cosmological constant reduces the dimension of the physical state space respectively by~2 and~1.
\par
The stability of equilibrium points for the system~(\ref{Eq:Eqs:EvolSigmaPlus})--(\ref{Eq:Eqs:RConstraint}) is investigated by standard analytical methods, see e.\thinspace g.~\cite{Hervik2005, Shogin2014a}, and numerical simulations are then used to make a conjecture about the global attractor of this dynamical system. We also use numerical runs to determine the relative magnitude of the dissipative fluxes. In the current paper we shall omit technical details, focusing on the main results.
\section{The futures of Bianchi type IV and V universes without vacuum energy}
\label{Sec:AttractorsNoLambda}
\subsection{The full transport equations}
The future stability of equilibrium points of the full system with the non-truncated transport equations depends critically on the bulk viscous stresses. Based on the precise relation between the bulk viscosity parameters~$a_0$ and~$b_0$, the system will asymptotically tend to one of the following future attractors:
\begin{enumerate}
\item
For
\begin{equation}
0<a_0<\frac{(3\gamma-2)^2}{6\gamma}
\end{equation}
the future asymptotic state is the Milne universe given by
\begin{align}
\begin{split}
\label{Eq:Results:Milne}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,1,0],\\
[\Omega,\Pi,q] &= [0,0,0].
\end{split}
\end{align}
The evolution of the universe is ultimately dominated by the three-space curvature, and the expansion is asymptotically uniform.
\item
For
\begin{equation}
\frac{(3\gamma-2)^2}{6\gamma}<a_0<\frac{(3\gamma-2)^2}{6\gamma}+\frac{3\gamma-2}{3} b_0
\end{equation}
we obtain the "transitionary" solution:
\begin{align}
\begin{split}
\label{Eq:Results:Transitionary}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,\bar{A},0],\\
[\Omega,\Pi,q] &= [\bar{\Omega},\bar{\Pi},0],
\end{split}
\end{align}
with
\begin{equation}
\bar{\Omega}={\left[\frac{a_0-\frac{(3\gamma-2)^2}{6\gamma}}{b_0(\gamma-\frac{2}{3})}\right]}^2, \qquad \bar{A}=\sqrt{1-\bar{\Omega}}, \qquad \bar{\Pi}=-\left( \gamma-\frac{2}{3} \right) \bar{\Omega}.
\end{equation}
The numerical simulations show that the geometric and viscous shear stresses decay extremely slowly in this case. The dynamics of the universe is anisotropic, but the system is slowly approaching isortopy at late times. The expansion is asymptotically uniform, the fluid, the shear stresses, and the three-space curvature all playing a significant role.
\item
For
\begin{equation}
a_0>\frac{(3\gamma-2)^2}{6\gamma}+\frac{3\gamma-2}{3} b_0
\end{equation}
the future attractor is
\begin{align}
\begin{split}
\label{Eq:Results:BVI}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Pi,q] &= [1,\bar{\Pi},\bar{q}],
\end{split}
\end{align}
with
\begin{align}
\bar{\Pi} &= \frac{1}{3}\left[\gamma b_0-\sqrt{\gamma^2 b_0^2+6\gamma a_0}\right],\\
\bar{q} &= -1+\frac{1}{2}\left[3\gamma+\gamma b_0-\sqrt{\gamma^2 b_0^2+6\gamma a_0}\right].
\end{align}
In this case, the universe is dominated by the fluid. The future asymptotic state corresponds to the bulk viscous inflationary scenario. The universe is asymptotically flat, its expansion being accelerated; the acceleration is caused by the (asymptotically constant) negative bulk viscous term.
\item
For the values on the border line between cases~1 and~2 above, where
\begin{equation}
a_0=\frac{(3\gamma-2)^2}{6\gamma},
\end{equation}
the future attractor is still the Milne universe given by~(\ref{Eq:Results:Milne}). However, as for the solutions tending to~(\ref{Eq:Results:Transitionary}), the dynamics of the universe is anisotropic. The system approaches isotropy very slowly at late times.
\item
Finally, for the values on the border line between cases~2 and~3, where
\begin{equation}
a_0=\frac{(3\gamma-2)^2}{6\gamma}+\frac{3\gamma-2}{3} b_0,
\end{equation}
the future attractor is the boundary state between the transitionary solution~(\ref{Eq:Results:Transitionary}) and bulk viscous inflation~(\ref{Eq:Results:BVI}):
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Pi,q] &= [1,-(\gamma-2/3),0].
\end{split}
\end{align}
The universe is asymptotically dominated by the fluid with negative bulk viscous pressure, the space three-curvature tends to zero, the expansion being uniform. The future attractor is approached extremely slowly, and the model remains significantly anisotropic during its evolution.
\end{enumerate}
Our numerical simulations show that deviations from equilibrium, caused by bulk viscosity, tend to a finite constant value for all the solutions. At the same time, deviations, caused by shear viscous stresses, decay in cases~3 and~5 above, are effectively finite for the transitionary solutions~(case 2), and increase without bound for solutions tending to the Milne universe~(cases~1 and~4). In the latter case the fluid state is evolving infinitely far from equilibrium, where the IS~theory is not valid anyhow.
\subsection{The truncated transport equations}
In constrast to the system with the full transport equations, in the truncated case the future stability of the equilibrium points is not dictated by the bulk viscosity parameters only. Shear viscosity can cause instabilities that yield solutions running into a singularity at finite times. The nature of this singularity is that the energy density~$\Omega$ is able to cross the vacuum boundary and become negative; the transport equations thus break down. This is a strict mathematical property of the truncated equations which is absent in the full IS~theory.
\par
In addition, the stable universes described by the system with the truncated transport equations have only three possible future attractors:
\begin{enumerate}
\item
For
\begin{equation}
0<a_0<\frac{1}{3}(b_0-2)(3\gamma-2)
\end{equation}
the future asymptotic state is
\begin{align}
\begin{split}
\label{Eq:Results:Transitionary-trunc}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,\bar{A},0],\\
[\Omega,\Pi,q] &= [\bar{\Omega},\bar{\Pi},0],
\end{split}
\end{align}
with
\begin{equation}
\bar{\Omega}={\left[\frac{a_0+2(\gamma-\frac{2}{3})}{b_0(\gamma-\frac{2}{3})}\right]}^2, \qquad \bar{A}=\sqrt{1-\bar{\Omega}}, \qquad \bar{\Pi}=-\left( \gamma-\frac{2}{3} \right) \bar{\Omega}.
\end{equation}
Note that this corresponds to the "transitionary" solution of the non-truncated case. As for the full equations, the universe is significantly anisotropic during the evolution, but slowly approaches isotropy at late times.
\par
Moreover, the following (significant) restriction should be imposed on the shear viscosity parameters to avoid instability of future equilibria and running into a singularity:
\begin{equation}
a_2>2-b_2\sqrt{\bar{\Omega}}.
\end{equation}
\item
For
\begin{equation}
a_0>\frac{1}{3}(b_0-2)(3\gamma-2)
\end{equation}
the universe evolves towards
\begin{align}
\begin{split}
\label{Eq:Results:BVI-trunc}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Pi,q] &= [1,\bar{\Pi},\bar{q}],
\end{split}
\end{align}
with
\begin{align}
\bar{\Pi} &= \frac{1}{6}\left[-(3\gamma-b_0)-\sqrt{(3\gamma-b_0)^2+12a_0}\right],\\
\bar{q} &= -1+\frac{1}{4}\left[3\gamma+b_0-\sqrt{(3\gamma-b_0)^2+12a_0}\right].
\end{align}
This solution corresponds to bulk viscous inflation. Note however, that despite certain similarities between solutions in the full and truncated cases, these solutions lead to different expressions for asymptotic values of the bulk viscous stress and the deceleration parameter, and thus to different late-time dynamics of the universe.
\par
Additional restrictions on the shear viscosity parameters are also required in this case. They can be written as
\begin{align}
a_2 & > \frac{3}{4}(\gamma+\bar{\Pi}-2)\left[b_2-3(\gamma+\bar{\Pi})\right],\\
b_2 & > -3+\frac{9}{2}(\gamma+\bar{\Pi}).
\end{align}
\item
For the border values between cases~1 and~2, given by the line
\begin{equation}
a_0=\frac{1}{3}(b_0-2)(3\gamma-2),
\end{equation}
the future attractor is
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Pi,q] &= [1,-(\gamma-2/3),0].
\end{split}
\end{align}
Similarly to the corresponding case of the full theory, this describes the boundary state between the "transitionary" solution~(\ref{Eq:Results:Transitionary-trunc}) and bulk viscous inflation~(\ref{Eq:Results:BVI-trunc}). The universe, as might be expected, isotropizes fairly slowly.
\par
To avoid the instability of the given attractor, the following additional requirement must be met:
\begin{equation}
a_2>2-b_2.
\end{equation}
\end{enumerate}
We see that the future equilibria of the truncated IS~theory are qualitatively similar to those of the full theory, although exact expressions, decay rates, and stability conditions are essentially different. In addition, the Milne universe, which is one of the possible future attractors in the full theory, is unstable in the truncated theory.
\par
The deviations from equilibrium are identical to those in the corresponding solutions in the full theory.
\section{The futures of Bianchi type IV and V universes with vacuum energy}
\label{Sec:AttractorsLambda}
\subsection{The full transport equations}
Similarly to the case with zero cosmological constant, the future of the universe is determined by bulk viscosity parameters alone. Depending on particular relation between~$a_0$ and~$b_0$, the solution tends to one of four possible future attractors.
\begin{enumerate}
\item
For
\begin{equation}
0<a_0\leq\frac{3\gamma}{2}
\end{equation}
the universe ends up in de Sitter state:
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [0,1,0,-1].
\end{split}
\end{align}
The expansion of the universe is accelerated, and the cosmological constant dominates the evolution; viscous terms do not play a significant role. Note that this is the future asymptotic state of Bianchi cosmologies with a {\it perfect} $\gamma$-fluid and a cosmological constant~\cite{Christiansen2008}. We also point at the fact that for the border value~$a_0=3\gamma/2$ the future attractor is the same, however the universe isotropizes at slower rates.
\item
At
\begin{equation}
\frac{3\gamma}{2}<a_0<\frac{3\gamma}{2}+\gamma b_0
\end{equation}
the future attractor is
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [\bar{\Omega},\bar{\Omega}_\Lambda,\bar{\Pi},-1],
\end{split}
\end{align}
with
\begin{equation}
\bar{\Omega}={\left[\frac{a_0-\frac{3\gamma}{2}}{\gamma b_0}\right]}^2, \qquad \bar{\Omega}_\Lambda=1-\bar{\Omega}, \qquad \bar{\Pi}=-\gamma \bar{\Omega}.
\end{equation}
This describes a "mixed" inflation with~$q\to-1$: both positive cosmological constant and fluid with negative bulk viscosity make a contribution to accelerating the expansion of universe.
\item
For
\begin{equation}
a_0>\frac{3\gamma}{2}+\gamma b_0
\end{equation}
the future attractor is given by
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [1,0,\bar{\Pi},\bar{q}],
\end{split}
\end{align}
with
\begin{align}
\bar{\Pi} &= \frac{1}{3}\left[\gamma b_0-\sqrt{\gamma^2 b_0^2+6\gamma a_0}\right],\\
\bar{q} &= -1+\frac{1}{2}\left[3\gamma+\gamma b_0-\sqrt{\gamma^2 b_0^2+6\gamma a_0}\right].
\end{align}
This corresponds to "pure" bulk viscous inflation, where the fluid is ultimately dominating over the cosmological constant. Note that for "large" viscosity parameters, the asymptotic futures of the models with and without a cosmological constant are exactly the same.
\item
Finally, for the border values between cases~2 and~3 above, given by
\begin{equation}
a_0=\frac{3\gamma}{2}+\gamma b_0,
\end{equation}
the future attractor is
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [\gamma^{-2},1-\gamma^{-2},-\gamma^{-1},-1],
\end{split}
\end{align}
which describes a special case of "mixed" inflation with~$q\to-1$.
\end{enumerate}
As in the case without a cosmological constant, the relative dissipative fluxe caused by bulk viscosity tends to a constant for all the solutions. However, the deviations from equilibrium caused by shear viscous stresses in presence of cosmological constant tend always to zero.
\subsection{The truncated transport equations}
Unlike the case without a cosmological constant, the presence of positive~$\Lambda$ does not allow the shear viscosity to affect the future of the universe. The stability of future attractors is determined by the bulk viscosity parameters. Namely, depending on the values of~$a_0$ and~$b_0$, the three future asymptotic states are possible:
\begin{enumerate}
\item
For
\begin{equation}
0<a_0<\gamma b_0
\end{equation}
the future asymptotic state corresponds to a "mixed" inflation:
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [\bar{\Omega},\bar{\Omega}_\Lambda,\bar{\Pi},-1],
\end{split}
\end{align}
with
\begin{equation}
\bar{\Omega}=\frac{a_0^2}{\gamma^2 b_0^2}, \qquad \bar{\Omega}_\Lambda=1-\bar{\Omega}, \qquad \bar{\Pi}=-\gamma \bar{\Omega}.
\end{equation}
\item
For
\begin{equation}
a_0>\gamma b_0
\end{equation}
the universe is undergoing "pure" bulk viscous inflation:
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [1,0,\bar{\Pi},\bar{q}],
\end{split}
\end{align}
with
\begin{align}
\bar{\Pi} &= \frac{1}{6}\left[-(3\gamma-b_0)-\sqrt{(3\gamma-b_0)^2+12a_0}\right],\\
\bar{q} &= -1+\frac{1}{4}\left[3\gamma+b_0-\sqrt{(3\gamma-b_0)^2+12a_0}\right].
\end{align}
\item
For the values on the border line between cases~1 and~2, given by
\begin{equation}
a_0=\gamma b_0,
\end{equation}
the universe approaches the state with
\begin{align}
\begin{split}
[\mathbf{X}_\Sigma,\mathbf{X}_\Pi,A,N] & = [0,0,0,0],\\
[\Omega,\Omega_\Lambda,\Pi,q] &= [1,0,-\gamma,-1],
\end{split}
\end{align}
which corresponds to the limiting case of bulk viscous inflation with~$q\to-1$.
\end{enumerate}
The consequences of using the truncated theory are somewhat similar to the case without a cosmological constant. The future equilibria of the truncated IS~theory, describing "pure" and "mixed" bulk viscous inflation, are qualitatively similar to the corresponding equilibria of the full theory. The quantitative difference is at the same time essential, and the stability conditions are substantially different. The de Sitter state, which is a possible future attractor in the full theory, is unstable in the truncated IS~theory.
\begin{figure}[ht!]
\begin{minipage}[h]{0.45\linewidth}
\includegraphics[width=1.0\linewidth]{FullNoLambda.eps}
\end{minipage}
\begin{minipage}[h]{0.45\linewidth}
\includegraphics[width=1.0\linewidth]{TruncNoLambda.eps}
\end{minipage}
\caption{The asymptotic futures of Bianchi type~IV and~V cosmological models without vacuum energy as functions of the bulk viscosity parameters, as predicted by the full (left) and truncated (right) IS~theories. Only non-singular solutions of the truncated theory are considered. Note that the state corresponding to the Milne universe is unstable in the truncated theory}
\label{Fig:NoLambda}
\end{figure}
\begin{figure}[ht!]
\begin{minipage}[h]{0.45\linewidth}
\includegraphics[width=1.0\linewidth]{FullLambda.eps}
\end{minipage}
\begin{minipage}[h]{0.45\linewidth}
\includegraphics[width=1.0\linewidth]{TruncLambda.eps}
\end{minipage}
\caption{The asymptotic futures of Bianchi type~IV and~V cosmologies with vacuum energy as functions of the bulk viscosity parameters, as predicted by the full (left) and truncated (right) IS~theories. Only non-singular solutions of the truncated theory are considered. Note that the de Sitter state becomes unstable in the truncated theory}
\label{Fig:Lambda}
\end{figure}
\section{Conclusion}
\label{Sec:Conslusion}
We have investigated the future dynamics, focusing on the equilibrum states and their stability conditions, for Bianchi type~IV and~V cosmologies with space filled with a dissipative~$\gamma$-fluid and, possibly, vacuum energy, which is modelled by a positive cosmological constant. We have considered the full transport equations of the IS~theory as well as the widely used truncated version of these equations. The future equilibrium states of the models as functions of the bulk viscosity parameters are presented in Figures~\ref{Fig:NoLambda} and~\ref{Fig:Lambda}. Multiple numerical runs allow us to make a conjecture that these states represent the global attractor for the system.
\par
We have found that two bifurcations are present in the full IS~theory, while the truncated theory has only one. As a result, the Milne universe, which is one of possible future asymptotic states for the models without a cosmological constant in the full theory, is unstable in the truncated theory. Correspondingly, for the models with a cosmological constant, the de Sitter universe becomes unstable in the truncated theory. The other future equilibria of both theories are qualitatively similar, but the quantitative expressions, the stability conditions, and, therefore, the dynamics of the models are substantially different.
\par
We have also found that the truncated transport equations possess mathematical properties which may have unphysical consequences. Shear viscosity can become a source of instability of the future equilibria, and cause universe models without a cosmological constant to run into a singularity. This phenomenon is not present in the full theory. In order to avoid singular solutions of the truncated theory, severe restrictions must be imposed on the shear viscosity parameters.
\par
In conclusion, great care must be taken if the truncated IS~theory is employed in a cosmological setting, in particular in the case of anisotropic cosmological models. It is true that the fluid description provided by the IS~theories is multiparametric, the future equilibrium states and their stability being dependent of the chosen equation of state and physical models of the transport coefficients. Still, this does not change our conclusion that the truncation of the IS~transport equations in general leads to very different asymptotic states and dynamics of the solutions.
\par
According to our numerical results, in all the solutions obtained, the fluid is strongly out of equilibrium, which implies that the underlying assumption of the IS~theory ceases to be valid. The deviations from the equilibrium, caused by bulk viscosity, are found to be always finite; the problem might therefore be solved by a modification of the~IS transport equations or of the equation of state, using the effective pressure instead of the local equilibrium value, at least in some cases. However, the relative dissipative fluxes, caused by shear viscosity, can either decay, be finite or increase without bound. In the latter case, the IS~theory is strictly inapplicable. This underlines the importance of a consistent theory of nonlinear thermodynamics and provides a possible direction for further investigations.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,310
|
The Waitsfield Village Historic District encompasses much of the main village center of Waitsfield, Vermont. Extending along Vermont Route 100 on either side of Bridge Street, it is a well-preserved example of a 19th-century village, with only a few sympathetic 20th-century additions. It was listed on the National Register of Historic Places in 1983.
Description and history
The town of Waitsfield was chartered in 1782 and settled by Benjamin Wait. Its early civic and commercial centers were at Waitsfield Common, and near Wait's house, which was located on a rise west of the present village. A road (now roughly Bridge Street), ran between the two, and a bridge spanned the Mad River on the site of the present Great Eddy Covered Bridge as early as 1797. In 1817, the present village center was essentially laid out by members of the Richardson family, who established a store at the crossroads of Bridge Street and what is now VT 100, and built mills to harness the river's water power. The village never achieved significant prominence in manufacturing, however, due to the lack of any nearby railroad connection, and development has been limited since the turn of the 20th century.
The historic district extends north, south, and east from the junction of Route 100 and Bridge Street. Its northern boundary is at Loop Road, and its southern boundary is at a geographic pinching point, where Route 100 passes between a steep rise on the west and the Mad River to the east. To the east, it extends across the Mad River (including the Great Eddy bridge, Vermont's oldest active covered bridge) to a point where the road crosses a dry branch of the river. Most of the buildings are wood-frame structures; there are six brick buildings. The most significant 20th-century building is the Joslin Library, prominently located at the central junction.
See also
National Register of Historic Places listings in Washington County, Vermont
References
Historic districts on the National Register of Historic Places in Vermont
National Register of Historic Places in Washington County, Vermont
Waitsfield, Vermont
Historic districts in Washington County, Vermont
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,341
|
Listed 1 sub titles with search on: UNESCO - World Heritage List for wider area of: "ATHENS Town ATTIKI" .
UNESCO - World Heritage List (1)
Acropolis, Athens
Illustrating the civilizations, myths and religions that flourished in Greece over a period of more than a thousand years, the Acropolis, which contains four of the greatest masterpieces of classical Greek art - the Parthenon, the Propylaea, the Erechtheum and the temple of Athena Nike - can be considered to symbolise the idea of world heritage.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 108
|
Trip reporting comes naturally to a photographer. You take photos of interesting things, take down details from the flight and then compile a report on how the journey went for everyone to see. Although a lot of time and effort is spent in making these trip reports, it is something I really enjoy doing.
Some information maybe obsolete as I started writing these in 2008.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,145
|
In some sections of our website you will have the option to share content (news, articles, promotions, etc.) with your friends and family. Here we will ask you for their names and email addresses just so we can share our content on your behalf.
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|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,453
|
Product information "Grounding Plate Screw GS"
Grounding plate for canopies, fabrics, fleeces, nettings, etc. Both plates are to be screwed together.
On this you have to make a hole in the material. 4 connection sockets.
Customer evaluation for "Grounding Plate Screw GS"
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 9,351
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\section*{Appendices}
Here we include additional technical details and results omitted from the main paper for brevity.
\cref{sec:supp-methoddetails} gives details on the main technical methods, \cref{sec:supp-exptdetails} gives details for experiments, and \cref{sec:supp-results} provides additional results to supplement those in the main paper.
\paragraph{Video Results} We encourage viewing the video results on the \href{https://nv-tlabs.github.io/STRIVE}{project page} to get a better sense of the kinds of scenarios produced by {{STRIVE}}\xspace.
\input{content/supp/text/02_details_method}
\input{content/supp/text/03_details_expts}
\input{content/supp/text/04_new_results}
\end{document}
\section{Analyzing and Using Generated Scenarios}
\subsection{Filtering and Collision Classification}
\label{sec:solopt}
\paragraph{Solution Optimization}
Adversarial optimization produces plausible scenarios, but it cannot guarantee they are \textit{solvable} and \textit{useful}: \eg a scenario in which the ego is squeezed by multiple cars produces an unavoidable collision and is therefore uninformative for evaluating or improving a planner. %
Therefore, we perform an additional optimization to identify an ego trajectory that avoids collision; if this optimization fails, the scenario is discarded for downstream tasks.
This solution optimization is initialized from the output of the adversarial optimization and essentially inverts the objectives described in \cref{sec:advopt}: non-ego latent $Z$ are tuned to maintain the adversarial trajectories while $\mathbf{z}^\text{plan}$ is optimized to avoid collisions and stay likely under the prior.
\paragraph{Clustering and Labeling}
To gain insight into the distribution of collision scenarios and inform their downstream use, we propose a simple approach to cluster and label them.
Specifically, scenarios are characterized by the explicit relationship between the planner and adversary at the time of collision: the relative direction and heading of the adversary are computed in the frame of the planner and concatenated to form a collision feature for each scenario. %
These features are clustered with $k$-means to form semantically similar groups of accidents that are labeled by visual inspection. %
Their distribution can then be visualized as in \cref{fig:analysis}.
\vspace{-2mm}
\subsection{Improving the Planner}
\label{sec:methodplanner}
With a large set of labeled collision scenarios, the planner can be improved in two main ways.
First, discrete improvements to functionality may be needed if many scenarios of the same type are generated.
For example, a planner that strictly follows lanes is subject to collisions from head on or behind as it fails to swerve, indicating necessary new functionality to leave the lane graph.
Second, scenarios provide data for tuning hyperparameters or learned parameters.
\paragraph{Rule-based Planner}
To demonstrate how {{STRIVE}}\xspace scenarios are used for these kinds of improvements, we introduce a simple, yet competent, rule-based planner that we use as a proxy for a real-world planner.
Our planner is ideal for evaluating STRIVE as it is easily interpretable, uses a small set of hyperparameters, and has known failure modes.
In short, it relies entirely on the lane graph to both predict future trajectories of non-ego vehicles and generate candidate trajectories for the ego vehicle.
Among these candidates, it chooses that which covers the most distance with a low ``probability of collision.''
Planner behavior is affected by hyperparameters such as maximum speed/acceleration and how collision probability is computed.
This planner has the additional limitation that it cannot change lanes, which scenarios generated by {{STRIVE}}\xspace exposes in \cref{sec:improveplanner}.
Full details of this planner are included in the supplementary. %
\section{Discussion}
\vspace{-1mm}
{{STRIVE}}\xspace enables automatic and scalable generation of plausible, accident-prone scenarios to improve a given planner.
However, remaining limitations offer potential future directions.
Our method assumes perfect perception and only attacks the planner, but using our traffic model to additionally attack detection and tracking is of great interest.
{{STRIVE}}\xspace generates scenarios from existing data and only considers collisions between vehicles, but other incidents involving pedestrians and cyclists are also important, and other kinds of adversaries like adding/removing assets and changing map topology will uncover additional AV weaknesses.
Our method is intended to make AVs safer by exposing them to challenging and rare scenarios similar to the real world.
However, our experiments expose the difficulty of properly balancing regular and challenging data when tuning a planner.
Care must be taken to integrate generated scenarios into AV testing and to design unified planners that robustly address highly variable driving conditions.
\section{Challenging Scenario Generation}
\label{sec:method}
{{STRIVE}}\xspace aims to generate high-risk traffic situations for a given \emph{planner}, which can subsequently be used to improve that planner (\cref{fig:overview}).
For our purpose, the planner encapsulates prediction, planning, and control, \ie we're interested in scenarios where the system misbehaves even with perfect perception.
The planner takes as input past trajectories of other agents in a scene and outputs the future trajectory of the vehicle it controls (termed the \emph{ego} vehicle).
It is assumed to be black-box: {{STRIVE}}\xspace has no knowledge of the planner's internals and cannot compute gradients through it.
Undesirable behavior includes collisions with other vehicles and non-drivable terrain, uncomfortable driving (\eg high accelerations), and breaking traffic laws.
We focus on generating \textbf{accident-prone scenarios} involving vehicle-vehicle collisions with the planner, though our formulation is general and in principle can handle alternative objectives.
\input{content/main/figures/arch}
Similar to prior work~\cite{wang2021advsim}, scenario generation is formulated as an optimization problem that perturbs agent trajectories in an initial scenario from real-world data.
The input is a planner $f$, map $\mathcal{M}$ containing semantic layers for drivable area and lanes, and a sequence from a pre-recorded real-world scene that serves as initialization for optimization. %
This initial scenario contains $N$ agents with trajectories represented in 2D BEV as $Y = \{\mathbf{Y}^i\}_{i=1}^{N}$, where $\mathbf{Y}^i = [ \mathbf{y}_1^i, \mathbf{y}_2^i, \dots, \mathbf{y}_{T}^i]$ is the sequence of states for agent $i$.
We let $\mathbf{Y}_t = [ \mathbf{y}_t^1, \mathbf{y}_t^2, \dots, \mathbf{y}_t^N]$ be the state of all agents at a single timestep.
An agent state $\y_t^i = [x_t, y_t, \theta_t, v_t, \Dot{\theta}_t]$ at time $t$ contains the 2D position $(x_t,y_t)$, heading $\theta_t$, speed $v_t$, and yaw rate $\Dot{\theta}_t$.
When rolled out within a scenario, at each timestep the planner outputs the next \textbf{ego} state $\mathbf{y}^\text{plan}_t = f (\mathbf{y}^\text{plan}_{<t}, \mathbf{Y}_{<t}, \mathcal{M})$ based on the \emph{past} motion of itself and other agents.
For simplicity, we will write the rolled out planner trajectory as $\mathbf{Y}^\text{plan} = f(Y, \mathcal{M})$ where $\mathbf{Y}^\text{plan} = [ \mathbf{y}^\text{plan}_1, \mathbf{y}^\text{plan}_2, \dots, \mathbf{y}^\text{plan}_{T}]$ for the remainder of this paper. %
Scenario generation perturbs trajectories for all non-ego agents to best meet an adversarial objective $\mathcal{L}_\text{adv}$ (\eg cause a collision with the planner):
\begin{equation}
\min_{Y} \mathcal{L}_\text{adv}(Y, \mathbf{Y}^\text{plan}), \quad \mathbf{Y}^\text{plan} = f(Y, \mathcal{M}).
\end{equation}
One may optimize a single or small set of ``adversaries'' in $Y$ explicitly, \eg through the kinematic bicycle model parameterization~\cite{wang2021advsim,kong2015bicycle,polack2017bicycle}.
While this enforces plausible single-agent dynamics, \textbf{interactions} must be constrained to avoid collisions between non-ego agents and, even then, the resulting traffic patterns may be unrealistic.
We propose to instead \emph{learn} to model traffic motion using a neural network and then use it at optimization time \textbf{(i) to parameterize} all trajectories in a scenario as vectors in the latent space, and \textbf{(ii) as a prior} over scenario plausibility.
Next, we describe this traffic model, followed by the ``adversarial'' optimization that produces collision scenarios.
\subsection{Modeling ``Realism'': Learned Traffic Model}
We wish to generate accident-prone scenarios that are assumed to develop over short time periods (\textless 10 sec)~\cite{najm2007pre}.
Therefore, traffic modeling is formulated as \emph{future forecasting}, which predicts future trajectories for all agents in a scene based on their past motion.
We learn $p_\theta(Y | X, \mathcal{M})$ to enable sampling a future scenario $Y$ conditioned on the fixed past $X = \{\mathbf{X}^i\}_{i=1}^{N}$ (defined similar to $Y$) and the map $\mathcal{M}$.
Two properties of the traffic model make it particularly amenable to downstream optimization: a low-dimensional latent space for efficient optimization, and a prior distribution over this latent space to determine the plausibility of a given scenario.
Inspired by recent work~\cite{casas2020implicit,suo2021trafficsim}, we design a conditional variational autoencoder (CVAE), shown in \cref{fig:arch}, that meets these criteria while learning accurate and scene-consistent joint future predictions.
We briefly introduce the architecture and training procedure here, and refer to the supplement for specific details.
\paragraph{Architecture}
To sample future motions at test time, the \emph{conditional prior} and \emph{decoder} are used; both are graph neural networks (GNN) operating on a fully-connected scene graph of all agents.
The \emph{prior} models $p_\theta(Z | X, \mathcal{M})$ where $Z = \{\mathbf{z}^i\}_{i=1}^{N}$ is a set of agent latent vectors.
Each node in the input scene graph contains a context feature $\mathbf{h}^i$ extracted from that agent's past trajectory, local rasterized map, bounding-box size, and semantic class.
After message passing, the \emph{prior} outputs parameters of a Gaussian $ p_\theta(\mathbf{z}^i | X, \mathcal{M}) = \mathcal{N} (\mu_\theta^i(X, \mathcal{M}), \sigma_\theta^i(X, \mathcal{M}))$ for each agent in the scene, forming a ``distributed'' latent representation that captures the variation in possible futures.
The deterministic \emph{decoder} $Y = d_\theta(Z, X, \mathcal{M})$ operates on the scene graph with both a sampled latent $\mathbf{z}^i$ and past context $\mathbf{h}^i$ at each node.
Decoding is performed autoregressively: at timestep $t$, one round of message passing resolves interactions before predicting accelerations $\Dot{v}_t, \Ddot{\theta}_t$ for each agent.
Accelerations immediately go through the kinematic bicycle model~\cite{kong2015bicycle,polack2017bicycle} to obtain the next state $\mathbf{y}_{t+1}^i$, which updates $\mathbf{h}^i$ before continuing rollout.
The determinism and graph structure of the decoder encourages scene-consistent future predictions even when agent $\mathbf{z}$'s are independently sampled.
Importantly for latent optimization, the decoder ensures plausible vehicle dynamics by using the kinematic bicycle model, even if the input $Z$ is unlikely.
\paragraph{Training}
Training is performed on pairs of $(X, Y_\text{gt})$ using a modified CVAE objective: %
\begin{equation}
\mathcal{L}_\text{cvae} = \mathcal{L}_\text{recon} + \w{KL} \mathcal{L}_\text{KL} + \w{coll}\mathcal{L}_\text{coll}.
\label{cvaeloss}
\end{equation}
To optimize this loss, a \emph{posterior} network $q_\phi(Z | Y_\text{gt}, X, \mathcal{M})$ is introduced similar to the prior, but operating jointly on past and future motion.
Future trajectory features are extracted separately, while past features are the same as used in the \emph{prior}.
The full training loss uses trajectory samples from both the posterior $Y_\text{post}$ and prior $Y_\text{prior}$:\\[-2mm]
\begin{align}
\mathcal{L}_\text{recon} &= \sum_{i=1}^{N} || \mathbf{Y}_\text{post}^i - \mathbf{Y}_\text{gt}^i ||^2 \\
\mathcal{L}_\text{KL} &= D_\text{KL} (q_\phi(Z | Y_\text{gt}, X, \mathcal{M}) || p_\theta(Z | X, \mathcal{M})) \\
\mathcal{L}_\text{coll} &= \mathcal{L}_\text{agent} + \mathcal{L}_\text{env}
\end{align}
where $\mathbf{Y}_\text{post}^i \in Y_\text{post}$, $\mathbf{Y}_\text{gt}^i \in Y_\text{gt}$, and $D_\text{KL}$ is the KL divergence.
Collision penalties $\mathcal{L}_\text{agent}$ and $\mathcal{L}_\text{env}$ use a differentiable approximation of collision detection as in \cite{suo2021trafficsim}, which represents vehicles by sets of discs to penalize $Y_\text{prior}$ for collisions between agents or with the non-drivable map area.
\subsection{Adversarial Optimization}
\label{sec:advopt}
To leverage the learned traffic model, the real-world scenario used to initialize optimization is split into the past $X$ and future $Y_\text{init}$.
Throughout optimization, past trajectories in $X$ (including that of the planner) are \emph{fixed} while the future is perturbed to cause a collision with the given planner $f$.
This perturbation is done in the learned latent space of the traffic model -- as described below, we optimize the set of latents for all $N$ \textit{non-ego} agents $Z = \{\mathbf{z}^i\}_{i=1}^{N}$ along with a latent representation of the planner $\mathbf{\z^\text{plan}}$.
Latent scenario parameterization encourages plausibility in two ways.
First, since the decoder is trained on real-world data, it will output realistic traffic patterns if $Z$ stays near the learned manifold.
Second, the learned prior network gives a distribution over latents, which is used to penalize unlikely $Z$.
This strong prior on behavioral plausibility enables jointly optimizing \emph{all agents} in the scene rather than choosing a small set of specific adversaries in advance.
At each step of optimization (\cref{fig:optimarch}), the perturbed scenario is decoded with $d_\theta(Z, \mathbf{\z^\text{plan}}, X, \mathcal{M})$ and non-ego trajectories $Y$ are passed to the (black-box) planner, which rolls out the ego motion before losses can be computed.
Adversarial optimization seeks two simultaneous objectives:
\input{content/main/figures/optim_arch}
\paragraph{1. Match Planner}\label{par:matchplanner}
Although optimization has no direct control over the planner's behavior -- an external function that is queried only when required -- it is still necessary to represent the planner within the traffic model (\ie include it in the scene graph with an associated latent $\mathbf{\z^\text{plan}}$) so that interactions with other agents are realistic.
In doing this, future predictions from the decoder include an estimate of the planner trajectory $\hat{\mathbf{Y}}^\text{plan}$ that, ideally, is close to the true planner trajectory $\mathbf{Y}^\text{plan} = f(Y, \mathcal{M})$.
Note that this gives a \emph{differentiable approximation} of the planner, enabling typical gradient-based optimization to be used for the second objective described below.
To encourage matching the real planner output with this ``internal'' approximation, we use
\begin{equation}
\min_\mathbf{\z^\text{plan}} || \hat{\mathbf{Y}}^\text{plan} - \mathbf{Y}^\text{plan} ||^2 - \alpha \log p_\theta(\mathbf{z}^\text{plan} | X, \mathcal{M})
\label{eqn:tgtmatch}
\end{equation}
where the right term lightly regularizes $\mathbf{\z^\text{plan}}$ to stay likely under the learned prior and $\alpha$ balances the two terms.
\paragraph{2. Collide with Planner}
The goal for non-ego agents is to cause the planner to collide with another vehicle:
\begin{equation}
\min_{Z} \mathcal{L}_\text{adv} + \mathcal{L}_\text{prior} + \mathcal{L}_\text{init} + \mathcal{L}_\text{coll}.
\end{equation}
\noindent The \textit{adversarial} term encourages a collision by minimizing the positional distance between controlled agents and the current traffic model approximation of the planner:
\begin{align}
\mathcal{L}_\text{adv} &= \sum_{i=1}^N \sum_{t=1}^T \delta_t^i \cdot || \y_t^i - \egoapprox_t ||^2 \label{eqn:lossadv}
\\
\delta_t^i &= \dfrac{\exp(-|| \y_t^i - \egoapprox_t ||)}{\sum_j \sum_t \exp(-|| \y_t^j - \egoapprox_t ||)}
\label{eqn:softmin}
\end{align}
where $\y_t$ here only includes the 2D position.
Intuitively, the $\delta_t^i$ coefficients defined by the \textit{softmin} in \cref{eqn:softmin} are finding a candidate agent and timestep to collide with the planner.
The agent with the largest $\delta_t^i$ is the most likely ``adversary'' based on distance, and \cref{eqn:lossadv} prioritizes causing a collision between this adversary and the planner while still allowing gradients to reach other agents.
This weighting helps $\mathcal{L}_\text{prior}$ to avoid all agents unrealistically colliding with the planner.
The \textit{prior} term encourages latents to stay likely under the learned prior network:
\begin{align}
\mathcal{L}_\text{prior} &= - \dfrac{1}{N} \sum_{i=1}^N \gamma^i \cdot \log p_\theta(\mathbf{z}^i | X, \mathcal{M}) \\
\gamma^i &= 1 - \sum_t \delta_t^i.
\label{eqn:priorweight}
\end{align}
The $\gamma^i$ coefficient will weight likely adversaries near zero, \ie agents close to colliding with the planner are allowed to deviate from the learned traffic manifold to exhibit rare and challenging behavior.
Because the traffic model training data does not contain collisions, we found it difficult for an agent to collide with the planner using a large prior loss, thus motivating the weighting in \cref{eqn:priorweight}.
Note that even when $\gamma^i$ is small, agents will maintain physical plausibility since the decoder uses the kinematic bicycle model.
$\mathcal{L}_\text{init}$ encourages staying close to the initialization in latent space, since it is already known to be realistic:
\begin{align}
\mathcal{L}_\text{init} = \dfrac{1}{N} \sum_{i=1}^N \gamma^i \cdot || \mathbf{z}^i - \mathbf{z}_\text{init}^i ||^2
\end{align}
where $\mathbf{z}_\text{init}^i \in Z_\text{init}$ are the latents that initialize optimization.
\noindent Finally, similar to CVAE training, $\mathcal{L}_\text{coll}$ discourages non-ego agents from colliding with each other and the non-drivable area.
In practice, all loss terms are balanced by manual inspection of a small set of generated scenarios.
\paragraph{Initialization and Optimization}
Given a real-world scene, $Z_\text{init}$ is obtained through the posterior network $q_\phi$, then further refined with an initialization optimization that fits to the input future trajectories of all agents (similar to \cref{eqn:tgtmatch}), including the initial planner rollout.
Optimization is implemented in PyTorch\cite{paszke2017automatic} using ADAM\cite{kingma2014adam} with a learning rate of $0.05$.
Runtime depends on the planner and number of agents; for our rule-based planner (see \cref{sec:methodplanner}), a 10-agent scenario takes 6-7 minutes.
\section{Introduction}
\label{sec:intro}
\input{content/main/figures/overview}
The safety of contemporary autonomous vehicles (AVs) is defined by their ability to safely handle complicated near-collision scenarios.
However, these kinds of scenarios are rare in real-world driving, posing a data-scarcity problem
that is detrimental to both the development and testing of data-driven models for perception, prediction, and planning.
Moreover, the better models become, the more rare these events will be, making the models even harder to train.
A natural solution is to synthesize difficult scenarios in simulation, rather than relying on real-world data, making it easier and safer to evaluate and train AV systems. This approach is especially appealing for \textit{planning}, where the appearance domain gap is not a concern.
For example, one can manually design scenarios where the AV may fail by inserting adversarial actors or modifying trajectories, either from scratch or by perturbing a small set of real scenarios.
Unfortunately, the manual nature of this approach quickly becomes prohibitively expensive when a large set of scenarios is necessary for training or comprehensive evaluation.
Recent work looks to \textit{automatically generate} challenging scenarios~\cite{wang2021advsim,abeysirigoonawardena2019generating,ding2021multimodal,ding2020learning,vemprala2020adversarial,o2018scalable,klischat2019generating}.
Generally, these approaches control a single or small group of ``adversaries'' in a scene, define an objective (\eg cause a collision with the AV), and then optimize the adversaries' behavior or trajectories to meet the objective.
While most methods demonstrate generation of only 1 or 2 scenarios~\cite{abeysirigoonawardena2019generating,chen2021adversarial,klischat2019generating,o2018scalable}, recent work~\cite{wang2021advsim} has improved scalability by starting from real-world traffic scenes and perturbing a limited set of pre-chosen adversaries.
However, these approaches \emph{lack expressive priors over plausible traffic motion}, which limits the realism and diversity of scenarios.
In particular, adversarial entities in a scenario are a small set of agents heuristically chosen ahead of time; surrounding traffic will not be reactive and therefore perturbations must be careful to avoid implausible situations (\eg collisions with auxiliary agents).
Furthermore, less attention has been given to determining if a scenario is ``unsolvable''~\cite{ghodsi2021generating}, \ie, if even an oracle AV is incapable of avoiding a collision.
In this degenerate case, the scenario is not \emph{useful} for evaluating/training a planner.
In this work, we introduce {{STRIVE}}\xspace~-- a method for generating challenging scenarios to \textbf{S}tress-\textbf{T}est d\textbf{RIVE} a given AV system.
{{STRIVE}}\xspace attacks the prediction, planning, and control subset of the AV stack, which we collectively refer to as the \emph{planner}.
As shown in \cref{fig:overview}, our approach perturbs an initial real-world scene through an optimization procedure to cause a collision between an arbitrary adversary and a given planner.
Our core idea is to measure the plausibility of a scenario during optimization by its likelihood under a learned generative model of traffic motion, %
which encourages scenarios to be challenging, yet realistic.
As a result, {{STRIVE}}\xspace does not choose specific adversaries ahead of time, rather it jointly optimizes all scene agents, enabling a diverse set of scenarios to arise.
Moreover, in order to accommodate for non-differentiable (or inaccessible) planners, which are widely used in practice, the proposed optimization uses a differentiable proxy representation of the planner within the learned motion model, thus allowing standard gradient-based optimization to be used.
We propose to identify and characterize generated scenarios that are \emph{useful} for improving a given planner. We first search for a ``solution'' to generated scenarios to determine if they are degenerate, and then cluster solvable scenarios based on collision properties. %
We test {{STRIVE}}\xspace on two AV planners, including a new rule-based planner, and show that it generates plausible and diverse collision scenarios in both cases.
We additionally use generated scenarios to improve the rule-based planner by identifying fundamental limitations of its design and tuning hyperparameters.
In short, our contributions are: (i) a method to automatically generate plausible challenging scenarios for a given planner, (ii) a solution optimization to ensure scenario utility, and (iii) an analysis method to cluster scenarios by collision type.
Supplementary videos and material for this work are available on the \href{https://nv-tlabs.github.io/STRIVE}{project webpage}.
\section{Experiments}
\label{sec:results}
\vspace{-1mm}
We next highlight the new capabilities that {{STRIVE}}\xspace enables.
\cref{sec:resultsadvgen} demonstrates the ability to generate challenging and useful scenarios on two different planners; these scenarios contain a diverse set of collisions, as shown through analysis in \cref{sec:resultsanalysis}.
Generated scenarios are used to improve our rule-based planner in \cref{sec:improveplanner}.
\paragraph{Dataset}
The nuScenes dataset~\cite{caesar2020nuscenes} is used both to train the traffic model and to initialize adversarial optimization.
It contains $20s$ traffic clips annotated at 2 Hz, which we split into $8s$ scenarios.
Only \textit{car} and \textit{truck} vehicles are used and the traffic model operates on the rasterized \textit{drivable area}, \textit{carpark area}, \textit{road divider}, and \textit{lane divider} map layers.
We use the splits and settings of the nuScenes prediction challenge which is $2s$ (4 steps) of past motion to predict $6s$ (12 steps) of future, meaning collision scenarios are $8s$ long, but only the future $6s$ trajectories are optimized.
\paragraph{Planners}
Scenario generation is evaluated on two different planners.
The \emph{Replay} planner simply plays back the ground truth ego trajectory from nuScenes data.
This is an open-loop setting where the planner's $6s$ future is fully rolled out without re-planning.
The \emph{Rule-based} planner, described in \cref{sec:methodplanner}, allows a more realistic closed-loop setting where the planner reacts to the surrounding agents during future rollout by re-planning at 5 Hz.
\paragraph{Metrics}
The \textbf{collision rate} is the fraction of optimized initial scenarios from nuScenes that succeed in causing a planner collision, which indicates the sample efficiency of scenario generation. \textbf{Solution rate} is the fraction of these colliding scenarios for which a solution was found, which measures how often scenarios are \emph{useful}.
\textbf{Acceleration} indicates how comfortable a driven trajectory is; challenging scenarios should generally increase acceleration for the planner, while the adversary's acceleration should be reasonably low to maintain plausibility.
If a scenario contains a collision, acceleration (and other trajectory metrics) is only calculated up to the time of collision.
\textbf{Collision velocity} is the relative speed between the planner and adversary at the time of collision; it points to the severity of a collision.
\input{content/main/figures/qualitative}
\input{content/main/tables/adv_gen_results}
\input{content/main/tables/adv_baselines}
\vspace{-1mm}
\subsection{Scenario Generation Evaluation}
\label{sec:resultsadvgen}
\vspace{-1mm}
First, we demonstrate that {{STRIVE}}\xspace generates challenging, yet solvable, scenarios causing planners to collide and drive uncomfortably.
Moreover, compared to an alternative generation approach that \emph{does not} leverage the learned traffic prior, {{STRIVE}}\xspace scenarios are more plausible.
Scenario generation is initialized from 1200 $8s$ sequences from nuScenes.
Before adversarial optimization, scenes are pre-filtered heuristically
by how likely they are to produce a useful collision, leaving \textless500 scenarios to optimize.
\paragraph{Planner-Specific Scenarios}
\cref{tab:advgen} shows that compared to rolling out a given planner on ``regular'' (unmodified) nuScenes scenarios, challenging scenarios from {{STRIVE}}\xspace produce more collisions and less comfortable driving.
For the \emph{Rule-based} planner, metrics on challenging scenarios are compared to the corresponding set of regular scenarios from which they originated (regular scenarios for \emph{Replay} are omitted since nuScenes data contains no collisions and, by definition, planner behavior does not change).
Collision and solution rates indicate that generated scenarios are accident-prone and \textit{useful} (solvable).
For the \emph{Rule-based} planner, adversarial optimization causes collisions in $27.4\%$ of scenarios compared to only $1.2\%$ in the regular scenarios. %
Generated scenarios also contain more severe collisions in terms of velocity, and elicit larger accelerations, \ie less comfortable driving.
The position and angle errors between approximate ($\hat{\mathbf{Y}}^\text{plan}$) and true ($\mathbf{Y}^\text{plan}$) planner trajectories at the end of adversarial optimization are shown on the right (see~\cref{par:matchplanner}).
The largest position error of $1.23m$ is reasonable relative to the $4.084m$ length of the planner vehicle. %
Qualitative results for the \emph{Rule-based} planner visualize 2D waypoint trajectories (\cref{fig:qualitative}); though not shown, {{STRIVE}}\xspace also generates speed and heading.
\paragraph{Baseline Comparison}
{{STRIVE}}\xspace is next compared to a baseline approach to demonstrate that
leveraging a learned traffic model is key to realistic and useful scenarios.
Previous works are not directly comparable as they focus on small-scale scenario generation (\eg \cite{abeysirigoonawardena2019generating,chen2021adversarial}) and/or attack the full AV stack rather than just the planner~\cite{wang2021advsim,o2018scalable}.
Therefore, in the spirit of AdvSim~\cite{wang2021advsim} we implement the \emph{Bicycle} baseline, which explicitly optimizes the kinematic bicycle model parameters (acceleration profile) of a single pre-chosen adversary in the scenario to cause a collision.
Rather than using the learned traffic model, it relies on the bicycle model, collision penalties, and acceleration regularization to maintain plausibility.
This precludes using the differentiable planner approximation from the traffic model, thus requiring gradient estimation (\eg finite differences) for the closed-loop setting, which we found is $\approx40\times$ slower and requires several hours to generate a single scenario.
Therefore, comparison is done only on the \emph{Replay} planner.
\cref{tab:baselines} shows that scenarios generated by \emph{Bicycle} exhibit more unrealistic adversarial driving, and are more difficult to find a solution for.
All metrics are reported only for scenarios where both methods successfully caused a collision.
In addition to higher accelerations, the \emph{Bicycle} adversary collides with the non-drivable area more often (\emph{Env Coll}), and exhibits less typical trajectories as measured by the distance to the nearest-neighbor ego trajectory in the nuScenes training split (\emph{NN Dist}).
After fitting the \emph{Bicycle}-generated scenarios with our traffic model, we see the adversary's behavior is also less realistic as measured by the negative log-likelihood (NLL) of its latent $\mathbf{z}$ under the learned prior.
These observations are supported qualitatively in \cref{fig:baselines}.
\input{content/main/figures/baselines}
\subsection{Analyzing Generated Scenarios}
\label{sec:resultsanalysis}
\vspace{-1mm}
Before improving a given planner, the analysis from \cref{sec:solopt} is used to identify useful scenarios by filtering out unsolvable scenarios and classifying collision types.
For classification, collision features are clustered with $k=10$ and clusters are visualized to manually assign the semantic labels shown in \cref{fig:analysis}.
The distribution of generated collision scenarios for both planners in \cref{sec:resultsadvgen} is shown in \cref{fig:analysis}(a) (see supplement for visualized examples from clusters).
{{STRIVE}}\xspace generates a diverse set of scenarios with solvable scenes found in all clusters.
``Head On'' is the most frequently generated scenario type, likely because \emph{Replay} is non-reactive and \emph{Rule-based} cannot change lanes.
``Behind'' exhibits the highest rate of unsolvable scenarios since being hit from behind is often the result of a negligent following vehicle, rather than undesirable planner behavior.
\emph{Replay} is much more susceptible to being cut off since it is open-loop, while the closed-loop \emph{Rule-based} can successfully react to avoid such collisions.
\input{content/main/figures/analysis}
\subsection{Improving Rule-Based Planner}
\label{sec:improveplanner}
\vspace{-1mm}
Now that we have a large set of labeled collision scenarios, in addition to the original nuScenes data containing ``regular'' scenarios (with few collisions), we can improve the \emph{Rule-based} planner to be better prepared for challenging situations.
Besides uncovering fundamental flaws that lead us to add new functionality, improvement is based on hyperparameter tuning via a grid search over possible settings.
For each set of hyperparameters, the planner is rolled out within all scenarios of a dataset, and the optimal tuning is chosen based on the minimum collision rate.
The planner is first tuned on regular scenarios before adversarial optimization is performed to create a set of challenging scenarios to guide further improvements.
Performance of this initial regular-tuned planner on held out regular (\emph{Reg}) and collision (\emph{Coll}) scenarios is shown in the top row of \cref{tab:tuneplanner}.
Before any improvements, the planner collides in $68.6\%$ of challenging and $4.6\%$ of regular scenarios.
Note that avoiding collisions altogether on regular scenarios is not possible: even if we choose optimal hyperparameters for each scenario separately, the collision rate is still $3.2\%$.
\input{content/main/tables/tune_planner}
\paragraph{Tuning on Challenging Scenes}
The first improvement, shown in the second row of \cref{tab:tuneplanner}, is to na{\"i}vely combine regular and challenging scenarios for tuning.
Combined tuning greatly reduces the collision rate on challenging scenarios, but negatively impacts performance on regular driving.
This points to a first \emph{fundamental issue}: the planner uses a single set of hyperparameters for all driving situations, %
causing it to drive too aggressively in regular scenarios when tuned on challenging ones.
\paragraph{Multi-Mode Operation}
To address this, we add a second set of parameters such that the planner has one for regular and one for accident-prone situations.
Using this second ``accident mode'' of operation requires a binary classification of the current scene during rollout.
For this, we augment the planner with a learned component that decides which parameter set to use based on a moving window of the past $2s$ of traffic; it is trained on scenarios generated by {{STRIVE}}\xspace.
The extra parameter set is tuned on collision scenarios only.
As shown in the third row of \cref{tab:tuneplanner}, this learned extra mode reduces the collision rate on challenging scenarios by $14.3\%$ compared to the vanilla planner without hindering performance on regular scenes.
We compare it to an \emph{oracle} version (bottom row) that automatically switches into accident mode $2s$ before a collision is supposed to happen on generated scenarios, showing the learned version is achieving near-optimal performance.
\paragraph{Lane Change Limitation}
The inability of the planner to switch lanes is another fundamental issue exposed by collision scenarios.
\cref{fig:analysis}(b) shows the distribution of tuning scenarios for the \emph{oracle} multi-mode version; red bars indicate ``impossible'' scenarios where all sets of evaluated hyperparameters collide.
A majority of ``Head On'' and ``Behind'' scenarios are impossible, pointing out the lane change limitation.
Adversarial optimization has indeed exploited the flaw and the proposed analysis made it visible.
\section{Related Work}
\label{sec:relwork}
\vspace{-1mm}
\paragraph{Traffic Motion Modeling} Scenario replay is insufficient for testing and developing AV planners as the motion of non-ego vehicles is strongly coupled to the actions chosen by the ego planner. Advances in deep learning have allowed us to replace traditional dynamic and kinematic models~\cite{wan2000unscented,kong2015bicycle,lefevre2014survey} or rule-based simulators~\cite{lopez2018microscopic,dosovitskiy2017carla} with neural counterparts that better capture traffic complexity \cite{bansal2018chauffeurnet,pal2020emergent}.
Efforts to predict future trajectories from a short state history and an HD map can generally be categorized according to the encoding technique, modeling of multi-modality, multi-agent interaction, and whether the trajectory is estimated in a single step or progressively.
The encoding of surrounding context of each agent is often done via a bird's-eye view (BEV) raster image~\cite{cui2019multimodal,chai2019multipath,fang2020tpnet}, though some work~\cite{liang2020learning,gao2020vectornet} replaces the rasterization-based map encoding with a lane-graph representation. SimNet~\cite{bergamini2021simnet} increased the diversity of generated simulations by initializing the state using a generative model conditioned on the semantic map.
To account for multi-modality, multiple futures have been estimated either directly \cite{cui2019multimodal} or through trajectory proposals \cite{chai2019multipath,fang2020tpnet, liu2021multimodal,learneval20}. %
Modeling multi-actor interactions explicitly using dense graphs has proven effective for vehicles~\cite{casas2020implicit, suo2021trafficsim}, lanes~\cite{liang2020learning}, and pedestrians~\cite{ivanovic2019trajectron, salzmann2020trajectron++, kosaraju2019social}. Finally, step-by-step prediction has performed favorably to one-shot prediction of the whole trajectory~\cite{djuric2020uncertainty}. We follow these works and design a traffic model that uses an inter-agent graph network \cite{johnson2020learning} to represent agent interaction and is variational, allowing us to sample multiple futures.
Our model builds on VAE-based approaches~\cite{casas2020implicit,suo2021trafficsim} that provide a learned prior over a controllable latent space~\cite{rempe2021humor}. Among other design differences, we incorporate a penalty for environment collisions and structure predictions through a bicycle model to ensure physical plausibility.
\paragraph{Challenging Scenario Generation}
Generating scenarios has the potential to exponentially increase scene coverage compared to relying exclusively on recorded drives.
Advances in photo-realistic simulators like CARLA~\cite{dosovitskiy2017carla} and NVIDIA's DRIVE Sim, along with the availability of large-scale datasets~\cite{caesar2020nuscenes,sun2020scalability,ettinger2021large,kesten2020lyftperception,houston2020lyftprediction}, have been instrumental to methods that generate plausible scene graphs to improve perception~\cite{kar2019meta,devaranjan2020metasim2,Resnick_2021_ICCV} and planning~\cite{bergamini2021simnet,casas2020implicit,suo2021trafficsim,Kim2021_DriveGAN}.
Our work focuses on generating challenging -- or ``adversarial''\footnote{we use ``challenging'' to denote generation procedures that do not explicitly attack a specific module in the perception or planning stack} -- scenarios, which are even more crucial since they are so rare in recorded data. While most works assume perfect perception and attack the planning module~\cite{chen2021adversarial,klischat2019generating,ding2021multimodal,ding2020learning,vemprala2020adversarial,ghodsi2021generating}, recent efforts exploit the full stack, including image or point-cloud perception~\cite{abeysirigoonawardena2019generating,o2018scalable,li2021fooling,wang2021advsim,tu2020physically}.
Our work focuses on attacking the planner only, though our scene parameterization as a learned traffic model could be incorporated into end-to-end methods. Unlike our approach, which uses gradient-based optimization enabled by the learned motion model, most adversarial generation works rely heavily on black-box optimization which may be slow and unreliable.
Our scenario generation approach is most similar to AdvSim~\cite{wang2021advsim}, however instead of optimizing acceleration profiles of a simplistic bicycle model we use a more expressive data-driven motion prior. This remedies the previous difficulty of controlling many adversarial agents simultaneously in a plausible manner. Moreover, we avoid constraining the attack trajectory to not collide with the playback AV by proposing a ``solution'' optimization stage to filter worthwhile scenarios.
Prior work \cite{chen2021adversarial} clusters lane-change scenarios based on trajectories of agents, while we cluster based on collision properties between the adversary and planner.
\paragraph{AV Planners}
Despite the recent academic interest in end-to-end learning-based planners and AVs~\cite{sadat2019jointly,zeng2019end,sadat2020perceive,casas2021mp3,bansal2018chauffeurnet}, rule-based planners remain the norm in practical AV systems~\cite{vitelli2021safetynet}.
Therefore, we evaluate our approach on a rule-based planner similar to the lane-graph-based planners used by successful teams in the 2007 DARPA Urban Challenge~\cite{stanforddarpa,Urmson-2007-9708} detailed in \cref{sec:methodplanner}.
\section{Method Details} \label{sec:supp-methoddetails}
In this section, we provide additional technical and implementation details about the methods introduced in Sec 3, 4, and 5 of the main paper.
Note that due to many formulated optimization problems being very similar, mathematical notation is often overloaded and may mean slightly different things depending on the section/optimization problem. This is always noted in the text.
\subsection{Learned Traffic Model}
\label{sec:supptrafficarch}
The learned traffic model is introduced in Sec 3.1 of the main paper.
Here we step through the main components of the architecture in more detail.
All neural network components use ReLU activations and layer normalization~\cite{ba2016layer} unless noted otherwise.
\paragraph{State Representation}
In practice, trajectories used as input and supervision for the traffic model are represented as $\mathbf{Y}^i = [ \mathbf{y}_1^i, \mathbf{y}_2^i, \dots, \mathbf{y}_{T}^i]$ for agent $i$ with $T$ timesteps where the state at time $t$ is $\y_t^i = [x_t, y_t, \theta_t^x, \theta_t^y, v_t, \Dot{\theta}_t] \in \mathbb{R}^6$ containing the 2D position $(x_t,y_t)$, heading unit vector $(\theta_t^x, \theta_t^y)$, speed $v_t$, and yaw rate $\Dot{\theta}_t$.
\paragraph{Feature Extraction}
Context features for each agent in the scene are first extracted based on: the past trajectory $\mathbf{X}^i$, the map $\mathcal{M}$,
a one-hot encoding of the semantic class $\mathbf{s}^i$ (either \emph{car} or \emph{truck} for the experiments in the main paper), and the agent's bounding box length/width $\mathbf{b}^i = (l, w)$.
The past trajectory feature for each agent $\mathbf{p}^i \in \mathbb{R}^{64}$ is encoded from $\mathbf{X}^i$, $\mathbf{s}^i$, and $\mathbf{b}^i$ using a 4-layer MLP with hidden size 128. To handle missing frames where an agent has not been annotated in nuScenes~\cite{caesar2020nuscenes}, the past encoding MLP is also given a flag for each input timestep indicating whether the agent state at that step is valid, \ie whether it is true nuScenes data or has been filled with dummy zeros.
The map feature $\mathbf{m}^i \in 64$ is extracted from a local map crop of size $256\times256$ around the agent ($17m$ behind, $60m$ in front, $38.5m$ to each side) at the last step of $\mathbf{X}^i$.
The map input contains one channel for each layer in the input (\textit{drivable area}, \textit{carpark area}, \textit{road divider}, and \textit{lane divider} in nuScenes); each layer is a binary image rasterized at 4 pixels/$m$.
The map extraction network is a CNN with 6 convolutional layers (stride 2, kernel sizes $[7, 5, 5, 3, 3, 3]$, and number of filters $[16, 32, 64, 64, 128, 128]$) followed by a single fully-connected layer.
Map extraction uses group normalization between layers~\cite{wu2018group}.
\paragraph{Prior Network}
The input to the prior is a fully-connected scene graph with a context feature placed at each node that is the concatenation of the previously detailed features $\mathbf{h}^i = [\mathbf{p}^i$, $\mathbf{m}^i, \mathbf{s}^i]$.
Before message passing, each $\mathbf{h}^i$ is further processed with a small 3-layer input MLP to be size 128.
The prior, posterior, and decoder are all graph neural networks (GNN) similar to a scene interaction module~\cite{casas2020implicit,suo2021trafficsim}.
They perform one round of message passing, which involves an \emph{edge network}, \emph{aggregation function}, and \emph{update network}.
Consider a single node $i$ in the scene graph.
First, interaction features are computed for every incoming edge.
For an edge from node $j \rightarrow i$, the edge feature is computed using the \emph{edge network} $\mathcal{E}$ as $\mathbf{e}^{ij} = \mathcal{E}(\mathbf{h}^i, \mathbf{h}^j, \mathcal{T}^{ij})$ where $\mathcal{T}^{ij}$ is the relative position and heading of agent $j$ in the local frame of agent $i$.
$\mathcal{E}$ is a 3-layer MLP with hidden and output size of 128.
After computing all edge features, they are aggregated into a single interaction feature $\mathbf{e}^i \in \mathbb{R}^{128}$ using maxpooling $\mathbf{e}^i = \max (\mathbf{e}^{i1}, \mathbf{e}^{i2}, \dots)$.
The update network then gives the output at each node $\mathbf{o}^i = \mathcal{U}(\mathbf{h}^i, \mathbf{e}^i)$; it is a 4-layer MLP with hidden size 128.
In the case of the prior network, the outputs at each node are the parameters of a Gaussian distribution.
In particular, $\mathbf{o}^i = [\mu^i, \sigma^i]$ where $\mu^i \in \mathbb{R}^{32}$ gives the mean and $\sigma^i \in \mathbb{R}^{32}$ parameterizes the diagonal covariance matrix so that $ p_\theta(\mathbf{z}^i | X, \mathcal{M}) = \mathcal{N} (\mu^i_\theta(X, \mathcal{M}), \sigma^i_\theta(X, \mathcal{M}))$ where $\theta$ are the learned parameters of the prior network.
\input{content/supp/figures/coll_loss}
\paragraph{Posterior (Encoder) Network}
The approximate posterior network (commonly referred to as the \emph{encoder} in CVAEs) is used (i) during training and (ii) to initialize adversarial optimization.
It is nearly the same as the prior, but takes in an additional future feature extracted from the future trajectory for each agent.
The future trajectory feature for each agent $\mathbf{f}^i \in \mathbb{R}^{64}$ is encoded from $\mathbf{Y}_\text{gt}^i$, $\mathbf{s}^i$, and $\mathbf{b}^i$ using a 4-layer MLP with hidden size 128.
Node $i$ of the scene graph input to the posterior contains a concatenation of $[\mathbf{f}^i, \mathbf{p}^i$, $\mathbf{m}^i, \mathbf{s}^i]$ where $\mathbf{p}^i$, $\mathbf{m}^i$ are the same past and map features given to the prior network.
Input processing and message passing is performed in the same way as the prior, and the final output is also a distribution over latents $ q_\phi(\mathbf{z}^i | Y_\text{gt}, X, \mathcal{M}) = \mathcal{N} (\mu^i_\phi(Y_\text{gt}, X, \mathcal{M}), \sigma^i_\phi(Y_\text{gt}, X, \mathcal{M}))$ with $\phi$ the learned parameters of the network.
\paragraph{Decoder Network}
As described in the main paper, the decoder progresses autoregressively, predicting one future step at a time.
When predicting step $t$, each node of the input scene graph contains a concatenation of $[\mathbf{z}^i, \mathbf{p}^i_{t-1}$, $\mathbf{m}^i_{t-1}, \mathbf{s}^i, \mathbf{b}^i]$ where $\mathbf{z}^i$ is sampled from either the prior or posterior output.
The past and map features are updated throughout autoregressive rollout, and therefore notated by a time subscript; initially these are the exact same as given to the prior, \ie $\mathbf{p}^i_0 = \mathbf{p}^i$ and $\mathbf{m}^i_0 = \mathbf{m}^i$.
At step $t$ of rollout, the concatenated input features are first processed by a 3-layer input MLP (hidden size 128) to get a single 64-dimensional feature at each node.
A single round of message passing proceeds in the same way as for the prior to get the output $\mathbf{o}^i_t = [\Dot{v}_t^i, \Ddot{\theta}_t^i] \in \mathbb{R}^2$ at each node, which contains current linear and angular acceleration.
The kinematic bicycle model~\cite{kong2015bicycle,polack2017bicycle} is then used to get the actual agent state $\y_t^i = \mathcal{K}(\mathbf{y}_{t-1}^i, \mathbf{o}^i_t, \mathbf{b}^i)$.
Before proceeding to the next step of rollout, the past and map context features must be updated according to the new state.
In particular, the past feature is updated using a gated-recurrent unit RNN $\mathbf{p}^i_{t} = \text{GRU}(\mathbf{p}^i_{t-1}, \y_t^i)$ with 3 layers (the GRU uses a hidden state of size 64 that is omitted here for brevity). The new map feature $\mathbf{m}^i_{t}$ is extracted using the same CNN as before, but with an updated map crop in the local frame of $\y_t^i$.
The feature at each node can then be updated to $[\mathbf{z}^i, \mathbf{p}^i_{t}$, $\mathbf{m}^i_{t}, \mathbf{s}^i, \mathbf{b}^i]$ before moving on to predict step $t+1$ in the exact same fashion.
Note the autoregressive nature of the decoder allows us to roll out further into the future than just the $6s$ training horizon if desired.
\paragraph{Training}
Training uses a modified CVAE objective:
\begin{align}
\mathcal{L}_\text{cvae} &= \mathcal{L}_\text{recon} + \w{KL} \mathcal{L}_\text{KL} + \w{coll}\mathcal{L}_\text{coll} \\
\mathcal{L}_\text{recon} &= \sum_{i=1}^{N} || \mathbf{Y}_\text{post}^i - \mathbf{Y}_\text{gt}^i ||^2 \label{eqn:suppreconloss}\\
\mathcal{L}_\text{KL} &= D_\text{KL} (q_\phi(Z | Y_\text{gt}, X, \mathcal{M}) || p_\theta(Z | X, \mathcal{M})) \\
\mathcal{L}_\text{coll} &= \mathcal{L}_\text{agent} + \mathcal{L}_\text{env}.
\end{align}
In practice, $\mathcal{L}_\text{recon}$ only supervises the position and heading, \ie $\mathbf{Y}_\text{post}^i$ and $\mathbf{Y}_\text{gt}^i$ in \cref{eqn:suppreconloss} contain only $[x_t, y_t, \theta_t^x, \theta_t^y]$. The reconstruction loss is applied on one sample from the posterior distribution, while collision losses use a sample from the prior.
$\mathcal{L}_\text{agent}$ is introduced in TrafficSim~\cite{suo2021trafficsim}.
It uses a differentiable approximation of vehicle-vehicle collision detection, which represents all $N$ vehicles by discs.
We estimate each agent vehicle $i$ by 5 discs with radius $r_i$, as shown in \cref{fig:collloss}(a), and compute the loss by summing over all pairs of agents $(i,j)$ over time as:
\begin{align}
\mathcal{L}_\text{agent} &= \dfrac{1}{N^2} \sum_{(i,j), i \neq j} \sum_{t=1}^{T} \mathcal{L}_\text{pair}(\y_t^i, \y_t^j) \label{eqn:vehcoll}\\
\mathcal{L}_\text{pair}(\y_t^i, \y_t^j) &= \begin{cases}
1 - \dfrac{d}{r_i + r_j}, & d \leq r_i + r_j \\
0, & \text{otherwise}
\end{cases}
\end{align}
where $d$ is the minimum distance over all pairs of discs representing agents $i$ and $j$.
$\mathcal{L}_\text{env}$ uses a similar idea to penalize collisions with the non-drivable area.
This penalty is only applied to the annotated \emph{ego} vehicle in the nuScenes~\cite{caesar2020nuscenes} sequences during training, since many non-ego vehicles appear off the annotated map.
At each step of rollout, collisions are detected between the ego vehicle and the non-drivable area (by checking for overlap between the rasterized non-drivable layer in $\mathcal{M}$ and the rasterized vehicle bounding box), and a collision point $\mathbf{c}$ is determined as the mean of all vehicle pixels that overlap with non-drivable area.
The loss is then calculated as:
\begin{align}
\mathcal{L}_\text{env} &= \dfrac{1}{T} \sum_{t=1}^{T} \mathcal{L}_\text{drivable}(\y_t, \mathcal{M}) \label{eqn:envcoll} \\
\mathcal{L}_\text{drivable}(\y_t, \mathcal{M}) &= \begin{cases}
1 - \dfrac{d}{d_\text{max}}, & \text{if partial collision} \\
0, & \text{otherwise}
\end{cases}
\end{align}
where $d$ is the distance between the vehicle position (center of bounding box) and collision point $\mathbf{c}$, and $d_\text{max}$ is half the ego bounding box diagonal as shown in \cref{fig:collloss}(b).
Note the loss is only applied if there is a partial collision, \ie only part of the bounding box overlaps with the non-drivable area -- this is because if the vehicle is completely embedded in the non-drivable area, the loss will not give a useful gradient.
The traffic model is implemented in PyTorch~\cite{paszke2017automatic} and trained using the ADAM optimizer~\cite{kingma2014adam} with learning rate $1e^{-5}$ for 110 epochs.
Losses are weighted with $\w{KL} = 4e^{-3}$ and $\w{coll}$ is split into $\w{agent} = 0.05$ for $\mathcal{L}_\text{agent}$ and $\w{env} = 0.1$ for $\mathcal{L}_\text{env}$.
The KL loss weight $\w{KL}$ is linearly annealed over time, starting from 0.0 at the start of training up to the full value at epoch 20.
\subsection{Initialization Optimization}
\label{sec:suppinitoptim}
As discussed in Sec 3.2 of the main paper, initialization of adversarial optimization is done by first performing inference with the learned approximate posterior $q_\phi(Z_\text{init} | Y_\text{init}, X, \mathcal{M})$ and then running an \emph{initialization optimization} to get the final $Z_\text{init}$ encapsulating both non-ego agents $Z = \{\mathbf{z}^i\}_{i=1}^{N}$ along with the latent planner $\mathbf{\z^\text{plan}}$.
The initialization optimization objective is nearly the same as Eqn 6 in the main paper.
It tries to match the initial future trajectories of non-ego agents (from the input nuScenes scenario) and the ego vehicle (from planner rollout within the initial scenario) that are contained in $Y_\text{init}$:
\begin{equation}
\min_{Z_\text{init}} \w{match} || Y - Y_\text{init} ||^2 - \w{prior} \log p_\theta(Z_\text{init} | X, \mathcal{M})
\end{equation}
where $Y = d_\theta(Z_\text{init}, X, \mathcal{M})$ is the decoded initial scenario but uses \emph{only} position and heading information (\ie no velocities).
For initialization optimization, we use $\w{match} = 10.0$, $\w{prior} = 0.01$, and run for 175 iterations.
\subsection{Adversarial Optimization}
\label{sec:suppadvoptim}
The adversarial optimization is introduced in Sec 3.2 of the main paper.
Here we provide details about each objective.
\paragraph{Match Planner}
In practice, the objective in Eqn 6 of the main paper only uses the position and heading information contained in $\mathbf{Y}^\text{plan}, \hat{\mathbf{Y}}^\text{plan}$ (\ie no velocities).
For all experiments, $\alpha = 1e^{-5}$.
\paragraph{Adversarial Loss}
The $\delta$ coefficients in the adversarial objective (Eqn 8 of the main paper) can be explicitly manipulated to discourage certain types of scenarios.
For all experiments, we dynamically set $\delta_t^i = 0$ if agent $i$ is ``behind'' the planner at time $t$ (\ie we ``mask out'' these agents).
``Behind'' is determined based on the current heading of the planner: if an agent is outside of the planner's $180 ^{\circ}$ field of view, it is considered behind.
This discourages degenerate scenarios with malicious collisions from behind.
We weight the adversarial loss in Eqn 8 of the main paper by $\w{adv} = 2.0$.
\paragraph{Prior Loss}
In practice, we do not use the $\gamma^i$ coefficients directly to weight the prior loss for each agent.
Instead, we use them to compute a dynamic weight for each agent by interpolating between minimum and maximum hyperparameters $\w{prior}^\text{min}, \w{prior}^\text{max}$.
Eqn 10 from the main paper becomes
\begin{align}
\mathcal{L}_\text{prior} &= - \dfrac{1}{N} \sum_{i=1}^N \log p_\theta(\mathbf{z}^i | X, \mathcal{M}) \cdot \w{prior} (\gamma^i)\\
\w{prior}(\gamma^i) &= \gamma^i \w{prior}^\text{max} + (1 - \gamma^i) \w{prior}^\text{min}
\end{align}
where we use $\w{prior}^\text{min} = 5e^{-3}$ and $\w{prior}^\text{max} = 1$.
This gives more fine-grained control over the prior loss weight rather than leaving it to $\gamma_i \in [0, 1]$.
In particular, if an agent is a likely adversary (\ie close to colliding with the planner), its weight will be near $\w{prior}^\text{min}$, whereas agents far away will be close to $\w{prior}^\text{max}$.
\paragraph{Initialization Loss}
Similar to the prior loss, the initialization loss (Eqn 12 in the main paper) actually uses $\gamma^i$ to interpolate between max and min hyperparameters and is written:
\begin{align}
\mathcal{L}_\text{init} &= \dfrac{1}{N} \sum_{i=1}^N || \mathbf{z}^i - \mathbf{z}_\text{init}^i ||^2 \cdot \w{init}(\gamma^i) \\
\w{init}(\gamma^i) &= \gamma^i \w{init}^\text{max} + (1 - \gamma^i) \w{init}^\text{min}
\end{align}
where we use $\w{prior}^\text{min} = 0.05$ and $\w{prior}^\text{max} = 0.5$.
\paragraph{Collision Losses}
The collision term for adversarial optimization is similar to that used to train the traffic model:
\[
\mathcal{L}_\text{coll} = \w{agent}\mathcal{L}_\text{agent} + \w{env}\mathcal{L}_\text{env} + \w{plan}\mathcal{L}_\text{plan}.
\]
$\mathcal{L}_\text{agent}$ is the same as defined in \cref{eqn:vehcoll} and is applied to all non-ego vehicles to avoid colliding with each other.
$\mathcal{L}_\text{env}$ is the same as defined in \cref{eqn:envcoll} and is applied to all non-ego vehicles (instead of the ego vehicle as in CVAE training) to encourage staying on the drivable area.
$\mathcal{L}_\text{plan}$ is similar to $\mathcal{L}_\text{agent}$, but instead of discouraging collisions between non-ego agents, it discourages collisions between the planner and non-ego agents with a large $\gamma^i$ (\ie unlikely adversaries). This term only affects agents that are close to the planner but have large $\gamma^i$ because they have been ``masked out'' as described previously.
Intuitively, we don't want these agents to ``accidentally'' collide with the planner from behind.
All collision losses are computed on trajectories that are upsampled by $\times 3$ to avoid missing collisions at the low nuScenes rate of 2 Hz. We use $\w{agent} = \w{env} = \w{plan} = 20$.
\paragraph{Optimization}
The adversarial optimization runtime reported in the main paper uses a machine with an NVIDIA Titan RTX GPU and 12x Intel i7-7800X@3.50GHz CPUs.
We optimize for 200 iterations using the ADAM~\cite{kingma2014adam} optimizer (we found L-BFGS is overly prone to local minima for this problem) with learning rate $0.05$.
\subsection{Solution Optimization}
\label{sec:suppsoloptim}
The solution optimization is introduced in Sec 4.1 of the main paper.
It attempts to find a trajectory for the planner that avoids the collision in the adversarial scenario.
Like the adversarial optimization, it optimizes $Z$ and $\mathbf{\z^\text{plan}}$ in the latent space of the traffic model and uses very similar objectives:
\paragraph{1. Match Adversarial Scenario}
All non-ego agents should maintain the same trajectories outputted from adversarial optimization.
Let $Y_\text{adv}$ be the set of non-ego trajectories from the collision scenario and $Y$ be the current scenario during solution optimization, then the objective is:
\begin{equation}
\min_Z || Y_\text{adv} - Y ||^2 - \alpha \log p_\theta(Z | X, \mathcal{M})
\end{equation}
with $\alpha = 1e^{-4}$.
Note the reason we need to actually optimize $Z$, instead of simply fixing it, is because the non-ego agent trajectories may change as $\mathbf{\z^\text{plan}}$ is optimized due to message passing in the traffic model decoder.
\paragraph{2. Avoid Collisions}
The goal for the planner is to avoid collisions with other agents and the environment while driving plausibly:
\begin{equation}
\min_{\mathbf{\z^\text{plan}}} \mathcal{L}_\text{prior} + \mathcal{L}_\text{coll}.
\end{equation}
The prior loss is
\begin{align}
\mathcal{L}_\text{prior} = - \w{prior}\log p_\theta(\mathbf{\z^\text{plan}} | X, \mathcal{M})
\end{align}
with $\w{prior} = 5e^{-3}$.
The collision loss is
\[
\mathcal{L}_\text{coll} = \w{agent}\mathcal{L}_\text{agent} + \w{env}\mathcal{L}_\text{env}
\]
where $\mathcal{L}_\text{agent}$ is the same as defined in \cref{eqn:vehcoll} but only discourages collisions between the planner and all non-ego agents.
$\mathcal{L}_\text{env}$ is the same as defined in \cref{eqn:envcoll} and is applied to the planner only.
We use $\w{agent} = \w{env} = 10$.
When computing collision losses, the planner is rolled out $8s$ into the future (instead of the $6s$ length of the scenario future) to ensure that it does not end up in an irrecoverable state at the end of the scenario.
The planner trajectory is upsampled $\times 3$ before collision checking.
\paragraph{Optimization}
Solution optimization uses the ADAM~\cite{kingma2014adam} optimizer for 200 iterations with a learning rate of $0.05$.
\subsection{Pre-Filtering Potential Scenarios}
\label{sec:suppfeasibility}
As discussed in Sec 5.1 of the main paper, before performing adversarial optimization, initial $8s$ scenarios from nuScenes are filtered to remove those that will be difficult or impossible to cause a collision.
For each potential initialization, 20 futures are sampled from the traffic model conditioned on the past trajectories.
A scenario is considered \emph{feasible} if any of the sampled futures meets the following heuristic conditions, which are designed to find an agent that could be made to collide with the planner:
\begin{itemize}[leftmargin=*] \itemsep0em
\item There is a non-ego agent that passes within $10m$ of the planner at some timestep $t$.
\item That agent is not behind the planner (where ``behind'' is determined in the same way as in \cref{sec:suppadvoptim}) at $t$.
\item That agent is not separated from the planner by non-drivable map area at $t$.
\item The ego vehicle must move $>1$ $m/s$ at some point, avoiding situations where the planner is a ``sitting duck'' with no hope of avoiding a collision.
\end{itemize}
If no samples meet these conditions, the initial scenario is not used for scenario generation.
After doing this feasibility check, we end up with a candidate non-ego agent that could reasonably collide with planner at time $t$.
Note, {{STRIVE}}\xspace does not use this information in any way -- the adversarial optimization can and does cause collisions with a different agent than the initial candidate.
However, this information is useful for the \emph{Bicycle} baseline which requires the adversary to be chosen before optimization.
\subsection{Bicycle Baseline Scenario Generation}
\label{sec:suppbike}
The \emph{Bicycle} baseline is introduced in Sec 5.1 of the main paper. %
This approach does not use the learned traffic model to parameterize scenarios, instead explicitly optimizing the acceleration profile of an adversary.
As it uses no strong priors on holistic traffic motion, only a single pre-chosen ``attacker'' is optimized (similar to prior work~\cite{wang2021advsim}).
For this, we use the candidate adversary determined by the feasibility check in \cref{sec:suppfeasibility} (in practice, using this feasibility check for \emph{Bicycle} would not be possible since it leverages samples from the traffic model, however we use it here to ensure a fair comparison to {{STRIVE}}\xspace).
Optimization is performed over the future acceleration profile of the adversary $\mathbf{A} = [\mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_T]$ with $\mathbf{a}_t = [\Dot{v}_t, \Ddot{\theta}_t]$.
Let $\mathbf{b}$ be the length and width of the adversary vehicle, then to get the adversary trajectory $\mathbf{Y} = [\mathbf{y}_1, \mathbf{y}_2, \dots, \mathbf{y}_T]$ when needed, the kinematic bicycle model is recursively applied $\y_t = \mathcal{K}(\mathbf{y}_{t-1}, \mathbf{a}_t, \mathbf{b})$ where $\mathbf{y}_0$ is the state at the last timestep of the fixed \emph{past}.
\paragraph{Initialization Optimization}
Before optimizing to cause a collision, the acceleration profile of the adversary is fit to its corresponding trajectory in the initial nuScenes scenario $\mathbf{Y}_\text{init}$ with
\begin{equation}
\min_\mathbf{A} || \mathbf{Y} - \mathbf{Y}_\text{init} ||^2.
\end{equation}
This loss is applied only to the position and heading part of the trajectories (\ie not velocities or accelerations).
\paragraph{Adversarial Optimization}
Next, the main adversarial optimization is performed, which encourages the adversary to collide with the planner using
\begin{equation}
\min_\mathbf{A} \w{adv}\mathcal{L}_\text{adv} + \w{accel}\mathcal{L}_\text{accel} + \mathcal{L}_\text{coll}.
\end{equation}
\noindent The adversarial term encourages a collision similar to {{STRIVE}}\xspace by minimizing positional distance:
\begin{align}
\mathcal{L}_\text{adv} &= \sum_{t=1}^T || \y_t - \mathbf{y}^\text{plan}_t || \cdot \delta_t \\
\delta_t &= \dfrac{\exp(-|| \y_t - \mathbf{y}^\text{plan}_t ||)}{\sum_t \exp(-|| \y_t - \mathbf{y}^\text{plan}_t ||)} \label{eqn:bikesoftmin}.
\end{align}
The \emph{softmin} in equation \cref{eqn:bikesoftmin} is only choosing a candidate timestep to cause a collision (as opposed to choosing both a candidate agent \emph{and} timestep as in {{STRIVE}}\xspace) since the colliding agent is fixed ahead of time.
Also note that $\mathbf{y}^\text{plan}_t$ is the actual planner trajectory, not a differentiable approximation from the traffic model as used in {{STRIVE}}\xspace.
This means that during optimization, it is necessary to compute gradients through the true planner if the setting is closed-loop (\eg the \emph{Rule-based} planner).
As discussed in Sec 5.1 of the main paper, we implemented an explicit gradient estimation using finite differences to be able to backpropagate through the planner.
However, this requires substantially more queries to the planner and is too slow to practically generate many scenarios.
Finite differences, while easy to implement, is inefficient compared to black-box approaches explored in prior work~\cite{wang2021advsim} -- however, it was out of our current scope to explore these options as well since {{STRIVE}}\xspace does not require them.
The acceleration term regularizes the optimized profile to prefer small accelerations:
\begin{align}
\mathcal{L}_\text{accel} &= \dfrac{1}{T} \sum_{t=1}^T || \mathbf{a}_t ||^2.
\end{align}
\noindent The collision loss is
\[
\mathcal{L}_\text{coll} = \w{agent}\mathcal{L}_\text{agent} + \w{env}\mathcal{L}_\text{env}
\]
where $\mathcal{L}_\text{agent}$ is the same as defined in \cref{eqn:vehcoll} but only discourages collisions between the adversary and other (fixed) non-ego agents.
$\mathcal{L}_\text{env}$ is the same as defined in \cref{eqn:envcoll} and is applied to the adversary only.
\paragraph{Optimization Details}
We use $\w{adv} = 1$, $\w{accel} = 1$, and $\w{agent} = \w{env} = 20$.
Initialization optimization uses L-BFGS (which was superior to ADAM for the simple objective) for 50 iterations.
Adversarial optimization uses ADAM for 300 iterations.
To fairly compare to {{STRIVE}}\xspace and evaluate certain metrics, %
after \emph{Bicycle} adversarial optimization we fit the output scenario within our learned traffic model using the procedure described in \cref{sec:suppinitoptim}.
We can then compute the likelihood of the adversary's $\mathbf{z}$ under the learned prior, and use the exact same solution optimization described in \cref{sec:suppsoloptim}.
\subsection{Rule-based Planner}
\label{sec:suppplanner}
Our rule-based planner is introduced in Sec 4.2 of the main paper and has the following structure:
\begin{enumerate}[leftmargin=*] \itemsep0em
\item Extract from the lane graph a finite set of splines that each vehicle might follow.
\item Generate predictions for the future motion of non-ego vehicles along each of the splines from (1).
\item Generate candidate trajectories for the ego vehicle and use the predictions from (2) to estimate the ``probability of collision'' $p_{\text{col}}(\tau)$ for each candidate $\tau$.
\item Among trajectories that are unlikely to collide $\{ \tau \mid p_{\text{col}}(\tau) < p_{\text{max}}\}$, choose the trajectory that covers the most distance. If no trajectories are unlikely to collide, choose the trajectory that is least likely to collide.
\item Repeat every $\Delta t$ seconds.
\end{enumerate}
Note the ``intent'' of the planner is deterministic, \ie it will always follow the same lane graph path (\eg choosing whether to turn left or right) when rollout starts from the same initialization.
As discussed in the main paper, nuScenes lane graphs do not contain information to switch lanes and since the rule-based planner strictly follows the lane graph, it cannot change lanes.
Planner behavior is affected by hyperparameters such as how $p_{\text{col}}(\tau)$ is computed, $p_{\text{max}}$, and the maximum speed and forward acceleration.
\section{Supplemental Experiments} \label{sec:supp-results}
In this section, we provide additional results and experiments omitted from the main paper for brevity.
\input{content/supp/tables/traffic_baselines}
\input{content/supp/tables/traffic_ablation}
\subsection{Traffic Motion Model}
We first evaluate the learned traffic model's ability to accurately predict future motion in a scene.
\paragraph{Data}
We evaluate on the test split of the nuScenes~\cite{caesar2020nuscenes} prediction challenge using $2s$ (4 steps) of past motion to predict $6s$ (12 steps) of future.
This data contains vehicles from the \emph{bus}, \textit{car}, \textit{truck}, \textit{construction}, and \textit{emergency} categories.
\paragraph{Metrics}
Evaluation is done with standard future prediction metrics including minimum average displacement error (\textbf{ADE}) and minimum final displacement error (\textbf{FDE}), which are measured over $K$ samples from the traffic model.
For a single agent being evaluated, these metrics are
\begin{align}
\text{ADE} &= \min_{k} \dfrac{1}{T} \sum_{t=1}^T || \hat{\mathbf{y}}_t^{(k)} - \mathbf{y}_t ||_2 \\
\text{FDE} &= \min_{k} || \hat{\mathbf{y}}_T^{(k)} - \mathbf{y}_T ||_2
\end{align}
where $\hat{\mathbf{y}}_t^{(k)}$ is the predicted \emph{position} of the agent in the $k$th sample at time $t$ and $\mathbf{y}_t$ is the ground truth.
In our experiments, we use $K=10$ samples.
For the ablation study, we also measure the \textbf{environment and vehicle collision rates}.
Environment collision rate is the fraction of predicted future trajectories where more than $5\%$ of the vehicle bounding box overlaps with the non-drivable area. This is measured over all $K$ samples.
The vehicle collision rate is measured over all agents in each scene (rather than only the single one specified at each data point in the prediction challenge test split) and all $K$ samples.
It is the same as used in TrafficSim~\cite{suo2021trafficsim}, which counts the number of agents in collision (\ie have a bounding box overlap more than IoU 0.02 with another agent).
\vspace{-2mm}
\subsubsection{Baseline Comparison}
\vspace{-1mm}
Prediction performance is compared to reported results for recent state-of-the art models AgentFormer~\cite{yuan2021agent}, Trajectron++~\cite{salzmann2020trajectron++}, DLow-AF~\cite{yuan2020dlow}, and LDS-AF~\cite{ma2021likelihood}.
Results are shown in \cref{tab:trafficbaselines}.
Our full model is trained only on $car$ and $truck$ vehicles to be used for scenario generation, so to evaluate on the prediction challenge test split, we modify the category of input vehicles to our model to be one of these (\eg \emph{bus} $\rightarrow$ \emph{truck}).
Our learned traffic model makes accurate predictions and is competitive with current SOTA methods as shown in \cref{tab:trafficbaselines}.
We also train an ablation of our model that does not use the kinematic bicycle model, instead the decoder directly predicts output position and headings.
This version is trained on all categories in the challenge dataset, and makes more accurate predictions according to ADE/FDE.
Note, however, that using the bicycle model is very important for adversarial and solution optimization to ensure output trajectories have reasonable dynamics even when the optimized $Z$ is off-manifold.
\input{content/supp/tables/advgen_ablation}
\subsubsection{Ablation Study}
\label{sec:supptrafficablation}
To evaluate key design differences from TrafficSim~\cite{suo2021trafficsim} and ILVM~\cite{casas2020implicit}, which our model is based on, we ablate various components of our traffic model design.
Results are shown in \cref{tab:trafficablation}, where all models are trained and evaluated only on the $car$ and $truck$ categories, since this is what we use in the main paper for scenario generation.
Same as \cref{tab:trafficbaselines}, \emph{No Bicycle} directly predicts the position and heading from the decoder rather than acceleration profiles that go through the kinematic bicycle model; again, this gives slightly improved performance but less realistic per-agent dynamics.
\emph{No $\mathcal{L}_\text{env}$} only uses vehicle collision penalties while training, similar to prior work~\cite{suo2021trafficsim}.
Removing the environment collision penalty results in a higher collision rate and lower predictive accuracy.
\emph{No Autoregress} uses a GNN decoder that predicts the entire future trajectory in one shot rather than as an autoregressive rollout.
This makes the future prediction task more difficult, substantially reducing accuracy.
\subsection{Adversarial Optimization Ablation Study}
Next, we study how various components of the adversarial optimization objective function (introduced in Sec 3.2 of the main paper and detailed in \cref{sec:suppadvoptim}) affect the generated scenarios.
We consider the following variations:
\begin{itemize}[leftmargin=*] \itemsep0em
\item No $\mathcal{L}_\text{prior}$ -- removes the prior loss which keeps agents likely under the learned prior.
\item No $\mathcal{L}_\text{coll}$ -- removes both environment and vehicle collision penalties.
\item No $\mathcal{L}_\text{init}$ -- removes the initialization loss which keeps agents near the initial (realistic) scenario.
\item No $\mathcal{L}_\text{init},\gamma$ -- removes the $\gamma$ weights from $\mathcal{L}_\text{prior}$, meaning likely adversaries will need to stay just as likely as all other agents in the scenario.
\item No $\mathcal{L}_\text{init},\gamma,\delta$ -- additionally removes the $\delta$ weighting scheme from $\mathcal{L}_\text{adv}$, meaning all agents will be simultaneously trying to collide with the planner at all timesteps, and it is left entirely up to $\mathcal{L}_\text{prior}$ to avoid unrealistic many-vehicle pileups.
\end{itemize}
Note that when ablating the $\delta$ and $\gamma$ weightings, we also remove $\mathcal{L}_\text{init}$ because adversaries should not be expected to stay very close to initialization when needing to collide.
Results are shown in \cref{tab:suppadvgenablation}.
We use the same metrics as for the baseline comparison in Sec 5.1 of the main paper.
In addition to computing metrics for the adversary (colliding agent), results for all other non-planner agents are also reported (except for environment collision since nuScenes contains many agents driving off the annotated drivable area).
These metrics are not perfect, and it can be hard to evaluate which optimization objective produces ``better'' scenarios (or even impossible since desired characteristics are dependent on downstream use), however they do give insight into the kinds of scenarios being produced.
The top section of \cref{tab:suppadvgenablation} computes metrics over scenarios where all methods caused a collision (same protocol as Tab 2 in the main paper).
However, since there are many variations, this is only 33 scenarios in total.
To give a more complete picture of each variation, the bottom section of the table computes metrics over all generated collision scenarios for each method. It also reports the collision rate to contextualize the solution rate and other metrics.
Because metrics are computed over a different set of scenarios for each method in the bottom section, numbers are not directly comparable, however trends often mirror those in the top part.
In the top of \cref{tab:suppadvgenablation} we see the full objective gives the highest solution rate, \ie it generates \emph{useful} scenarios at the highest frequency.
\emph{No $\mathcal{L}_\text{prior}$} tends to cause less likely trajectories for other agents in the scene since they are no longer constrained by the learned prior, and make environment collisions more common for the adversary as seen in the bottom section.
\emph{No $\mathcal{L}_\text{coll}$} similarly causes far more collisions in addition to adversaries with higher accelerations.
Despite this, trajectories maintain reasonable likelihoods since $\mathcal{L}_\text{prior}$ is still used and the traffic model does allows for some collisions as seen in \cref{sec:supptrafficablation}.
\emph{No $\mathcal{L}_\text{init}$} allows trajectories to stray far from the nuScenes initialization, which produces solvable scenarios at a lower rate and less plausible adversary motion in the bottom part of the table.
Note that \emph{NLL} for ``others'' is trivially very low since the only remaining regularization is $\mathcal{L}_\text{prior}$.
\emph{No $\mathcal{L}_\text{init},\gamma$} forces adversaries to stay more likely under the prior resulting in a low NLL, but lowered solution rate.
The reasonable results produced by this variation indicate the flexibility of our formulation to produce different kinds of scenarios.
By removing $\gamma$, we encourage scenarios where collisions happen within a more ``typical'' setting, rather than as a result of out-of-distribution, adversarial behavior.
This is shown qualitatively in \cref{fig:advgenablation}.
\emph{No $\mathcal{L}_\text{init},\gamma,\delta$} enables many adversaries to attack simultaneously producing many unsolvable scenarios with unrealistic trajectories.
\input{content/supp/figures/advgen_ablation}
\input{content/supp/figures/many_agent_scenes}
\subsection{Performance on Many-agent Scenes}
To evaluate the ability of {{STRIVE}}\xspace to effectively optimize large scenarios, we look specifically at scenes that contain many agents.
The mean number of agents in nuScenes is $11.4$, so it does not contain many massive scenes: only $2/496$ optimized scenarios in Tab 1 of the main paper for the \emph{Rule-based} planner contain $>$~$50$ agents.
For scenarios with $\geq$~$20$ agents, the collision rate is $21.2\%$, which is consistent with $27.4\%$ in Tab 1, indicating performance on large-scale scenes is similar to smaller ones.
This is because even when a scenario has many agents, only a handful near the planner will greatly affect the outcome (due to the optimization objective in Eq 9 of the main paper).
As shown in \cref{fig:manyagentresults}, adversarial optimization makes large changes to the latent vectors of agents within $10m$ of the planner much more frequently than for distant agents.
The change in latent $\mathbf{z}^i$ for each agent is measured as the Euclidean distance between the initialized latent and the final latent (after adversarial optimization), normalized by the maximum difference observed in the scene.
\subsection{Modeling Second-order Effects}
One interesting ability of adversarial optimization is to produce scenarios that cause collisions with the planner using so-called ``second-order effects."
In these scenarios, an adversary may not directly collide with the planner in an obviously malicious way -- instead, it performs some action that causes the planner or other vehicles to react, thereby causing a collision between the planner and some other agent.
For example, we have observed several instances of an adversary in front of the planner suddenly braking, causing the planner to brake and get hit by another agent from behind, rather than the initial adversary in front.
This prompted the regularizer $\mathcal{L}_\text{plan}$ introduced in \cref{sec:suppadvoptim}.
The second-order behavior is possible since {{STRIVE}}\xspace optimizes \textit{all} agents in a scene rather than choosing specific adversaries ahead of time.
Example videos of these second-order scenarios are shown in the supplementary webpage.
\subsection{Solution Optimization Evaluation}
To confirm the solution optimization is filtering out overly-difficult scenarios, we evaluate using the hyperparameter tuning procedure and generated challenging scenarios introduced in Sec 5.3 of the main paper.
In particular, we take the vanilla \emph{Rule-based} planner and perform two hyperparameter tuning sweeps: one on generated collision scenarios where the solution optimization found a solution, and one on collision scenarios where no solution could be found.
For each sweep, we measure the mean fraction of hyperparameter settings that succeed (\ie don't collide) per scenario in the tuning set.
When tuning on \textit{solution-failed} scenarios, \textbf{only $11.8\%$ of hyperparameter combinations succeed} in avoiding collisions for each scenario on average.
However for tuning on \textit{solution-found} scenarios, $29.8\%$ of hyperparameters succeed for each.
This indicates that finding a sufficient hyperparameter setting for scenarios where our solution optimization failed is more difficult than for those where a solution could be found.
\subsection{Additional Qualitative Results}
Additional qualitative results of {{STRIVE}}\xspace on the \emph{Rule-based} planner are shown in \cref{fig:advgenqual}.
Video examples are also included in the supplementary webpage.
\input{content/supp/figures/ped_examples}
\subsection{Pedestrian and Cyclist Adversaries}
Though our main focus in this work is generating scenarios involving vehicles, as a proof-of-concept we train the learned traffic model on all categories in the nuScenes dataset and generate scenarios for the \emph{Replay} planner where the adversary is a pedestrian or cyclist.
For this experiment, a pedestrian/cyclist adversary is specifically chosen before optimization using the procedure in \cref{sec:suppfeasibility}.
Results are shown in \cref{fig:pedexample} and on the webpage.
\input{content/supp/figures/advgen_qual_supp}
\subsection{Failure Cases and Limitations}
In addition to the limitations discussed in Sec 6 of the main paper, \cref{fig:failures} shows examples of other {{STRIVE}}\xspace limitations.
First, our proposed solution optimization is iterative and operates on the full temporal planner trajectory, therefore it has access to future information that sometimes allows performing evasive maneuvers even before an ``attack'' is apparent.
An example is in \cref{fig:failures}(a) where the optimized solution simply does not not pull into the roundabout where the collision occurs.
\cref{fig:failures}(b) shows that the adversary sometimes crosses non-drivable areas in order to collide with the planner.
Though this scenario is technically possible, it is extreme behavior that may not be desired.
However, these situations can be easily detected and discarded, and usually occur only when there is no other feasible adversary near the planner.
Finally, adversarial optimization can have difficulty exhibiting behavior that is very unlikely under the prior even when it is realistic, \eg a parked car pulling out as shown in \cref{fig:failures}(c), since these motions are rare in the traffic model training data.
\input{content/supp/figures/failures}
\section{Experimental Details} \label{sec:supp-exptdetails}
In this section, we provide details for the experiments performed in Sec 5 of the main paper.
\subsection{Data and Metrics}
\paragraph{Dataset}
The nuScenes~\cite{caesar2020nuscenes} dataset is used for all experiments; we use the scene splits and settings from the nuScenes prediction challenge.
All vehicle trajectories in the dataset are pre-processed to remove frames where the vehicle bounding box has a $>30\%$ overlap with either the non-drivable area or a car park area.
This avoids scenarios with many auxiliary agents that are off of the annotated map or not moving.
Trajectories are additionally annotated with velocity and yaw rate using finite differences on the provided positions/headings.
Finally, we flip the maps and trajectories for ``Singapore'' data about the $x$ axis so that vehicles across the whole dataset (both in Boston and Singapore) are consistently driving on the right-hand side of the road.
For training the learned traffic model, we use scenes from the training split of the nuScenes prediction challenge.
Note we use all available trajectory data in the training scenes for \emph{cars} and \emph{trucks}, including the ego trajectories and all agents not removed in our own pre-processing.
Scenario generation in all experiments is initialized from a set of 1200 $8s$ nuScenes scenarios (before pre-filtering in \cref{sec:suppfeasibility}) extracted from the train and val splits.
Since the original nuScenes data sequences are $20s$ long, some of these extracted $8s$ scenarios partially overlap.
\paragraph{Metrics}
Acceleration and collision velocity metrics are computed using finite differences.
The accelerations reported in Tab 1 and Tab 3 of the main paper are \emph{forward accelerations} (calculated using the change in speed) while those in Tab 2 are the full acceleration (encompassing both forward and lateral).
This is because Tab 1 and 3 focus on trajectories from the \emph{Rule-based} planner, which follows the lane graph, cannot change lanes, and has a deterministic route as discussed in \cref{sec:suppplanner}.
As a result, generated scenarios cannot make the planner swerve suddenly (which would require leaving the lane graph, changing lanes, or changing route), they can only cause a harsher slow down or speed up, which is captured by measuring forward acceleration.
For all experiments, acceleration (\emph{Accel}), environment collision rate (\emph{Env Coll}), and nearest-neighbor distance (\emph{NN Dist}) are only reported up to the time of collision, since any continuing motion is merely the result of not physically simulating the collision.
For the environment collision rate reported in Tab 2, a vehicle is considered in collision if there is more than 5\% overlap between its bounding box and the non-drivable area.
\subsection{Planner-Specific Scenario Generation}
\label{sec:suppadvgenexpt}
This experiment is presented in Sec 5.1 of the main paper.
During scenario generation, the \emph{Rule-based} planner uses default manually-set hyperparameters (\ie no large-scale tuning was done beforehand, as described in Sec 5.3 of the main paper).
These default hyperparameters were set by observing rollouts on a small subset of nuScenes.
In Tab 1, collision rate is reported over all scenarios for which adversarial optimization was performed (\ie the roughly 500 pre-filtered scenarios).
Solution rate is with respect to all scenarios where a collision was successfully caused, while planner trajectory and match planner metrics are computed over all useful scenarios (those where both a collision and solution were found).
\subsection{Baseline Comparison}
This experiment is presented in Sec 5.1 of the main paper, and compares {{STRIVE}}\xspace to the baseline scenario generation detailed in \cref{sec:suppbike}.
Metrics reported in Tab 2 are for a set of 139 scenarios where both methods were able to cause a collision from the same nuScenes initialization.
Please see \cref{sec:suppbike} for details on how solution rate and NLL are measured for \emph{Bicycle}.
\subsection{Scenario Analysis}
This experiment is presented in Sec 5.2 of the main paper.
The results shown in Fig 6 are collision labels assigned to the scenarios generated in Sec 5.1.
Note that $k$-means clustering was not performed directly on the scenarios generated in Sec 5.1, instead clustering was done beforehand with a large set of over 400 scenarios generated from various subsets of nuScenes and using many versions of our \emph{Rule-based} planner, giving a wide variety of collisions.
Clusters were assigned semantic labels by visual inspection of the collisions in each cluster.
Then, after generating new scenarios in Sec 5.1, collision types are assigned to each new scenario by simply associating their collision feature with the closest cluster (\ie clustering is \emph{not} done over again, scenarios are assigned to extant clusters).
Videos of representative scenarios assigned to each cluster are shown on the \textbf{supplementary webpage}.
We use $k=10$ for clustering. Finer-grained classification is possible, if necessary, using larger $k$ or additional features like collision velocity, however we found the described approach sufficient to analyze the \emph{Rule-based} planner's performance while being easily interpretable.
\subsection{Improving Rule-based Planner}
This experiment is presented in Sec 5.3 of the main paper.
Before any tuning is performed, the planner starts with the same set of ``default'' hyperparameters described in \cref{sec:suppadvgenexpt}.
When tuning the planner hyperparameters, the optimal set is chosen by lowest collision rate with ties broken by lowest acceleration.
The planner is first tuned on 800 $8s$ nuScenes scenarios (from the train/val splits), which gives initial optimal hyperparameters for ``regular'' driving scenarios.
Adversarial optimization is performed on the planner with both the \emph{default} and \emph{regular-tuned} hyperparameters to create a set of challenging scenarios to guide further improvements.
When tuning on challenging scenarios, ``Behind'' collisions are removed, which we found unrealistic as discussed in Sec 5.2 of the main paper.
The hyperparameters for the \emph{Rule-based} planner are described in \cref{sec:suppplanner}.
Tuning searches over $p_{\text{max}}$ in the range of $[0.05, 0.2]$, max speeds in the range $[12.5, 20.0]$ $m/s$, max accelerations in the range $[3.0, 4.5]$ $m/s^2$, and parameters related to computing $p_{\text{col}}(\tau)$. In total, tuning sweeps over 432 hyperparameter combinations.
Results in Tab 3 of the main paper are on scenarios from the held-out nuScenes test set.
Metrics over collision scenarios (\emph{Coll}) are reported for scenarios where some hyperparameter setting succeeded in avoiding a collision, \ie scenarios where all hyperparameter settings collide are considered impossible and discarded.
The best possible collision rate on regular scenarios is only $3.2\%$ (achieved by choosing a \emph{different} set of parameters for every scenario), making the $4.6\%$ of the regular-tuned planner version close to optimal.
The reason this collision rate cannot be 0 is the use of log replay (\ie rolling out the planner in pre-recorded scenarios), which results in some unavoidable collisions: (i) pre-recorded traffic is not reactive to the planner and (ii) the ego vehicle is sometimes initialized off the lane graph which it cannot robustly handle.
\paragraph{Learned Mode Classifier}
The multi-mode version of the planner uses a binary classifier that decides whether the ego vehicle is currently in a ``regular'' or ``accident-prone'' situation.
In the \emph{learned} version, this classifier is a neural network that has a very similar architecture to the learned traffic model described in \cref{sec:supptrafficarch}.
In particular, it takes in the past $2s$ trajectories for all agents in a scene, along with local map crops around each, and processes them in the same way as the traffic model.
These features are then placed into a scene graph and message passing is performed in the same way as done for the prior.
The output feature (size 64) at the ego node is given to 2-layer MLP that makes a binary classification for the scene.
This network is trained on regular nuScenes scenarios from the training split in addition to a diverse set of over $1000$ collision scenarios generated from train/val scenes using variations of both the \emph{Replay} and \emph{Rule-based} planners.
Training uses a typical binary cross entropy loss that is weighted to account for the data imbalance between regular and collision scenarios.
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{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,492
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Q: A question about Cauchy sequences Let $\langle a_n\rangle$ , $\langle b_n\rangle$ , $\langle c_n\rangle$ be Cauchy sequences of rational numbers, and $\langle c_n\rangle$ is equivalent to $\langle a_nb_n\rangle$. Prove or disprove that there are two Cauchy sequences $\langle a_n'\rangle$ , $\langle b_n'\rangle$ of rational numbers such that
(1) $\langle a_n\rangle$ is equivalent to $\langle a_n'\rangle$ ;
(2) $\langle b_n\rangle$ is equivalent to $\langle b_n'\rangle$ ;
(3) $\langle c_n\rangle=\langle a_n'b_n'\rangle$ .
If it is true, can we prove it intuitionistically?
A: Case 1: both $a_n,b_n$ converge to zero. Let $a'_n=r_n,b'_n=c_n/r_n$. Where $r_n$ is a sequence of rationals such that for all n $|r_n-\sqrt c_n|<1/n$
Case 2: one of the two sequences does not converge to zero
Let $lim_{n\rightarrow \infty} a_n=A$ . Assume WLOG that A is different from $0$. Let $a'_n$ be a sequence of nonzero rationals converging to $A$, and let $b'_n=c_n/a'_n$
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,441
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Q: Using DefaultRequestHeaders sends requests twice? I have a WebAPI that sends BASIC authorization information as following.
var client = new HttlpClient();
client.BaseAddress = new Uri(GlobalConstants.LdapUri);
var contentType = new MediaTypeWithQualityHeaderValue("application/json");
client.DefaultRequestHeaders.Accept.Add(contentType);
client.DefaultRequestHeaders.Authorization = new AuthenticationHeaderValue("Basic", Convert.ToBase64String(Encoding.ASCII.GetBytes(string.Format("{0}:{1}", userName, password))));
Task<HttpResponseMessage> results = client.GetAsync(GlobalConstants.FortressAPIUriDev);
var response = await results;
I've built this API using MVC Core 1.x and the receiving API is built using MVC5.
The problem is that this GetAsync sends two requests at the same time, and I have no clue how to resolve this. I've done some Googling myself to see if I can find a fix for this but so far no luck. Did anyone experience this problem and know how to resolve it?
Thank you very much in advance.
A: Long story short, found a solution as follows:
using (var client = new HttpClient())
{
var requestMessage = new HttpRequestMessage(HttpMethod.Get, GlobalConstants.LdapUri + GlobalConstants.FortressAPIUriDev);
requestMessage.Headers.Authorization = new AuthenticationHeaderValue("Basic", Convert.ToBase64String(Encoding.ASCII.GetBytes(string.Format("{0}:{1}", userName, password))));
var response = await client.SendAsync(requestMessage);
}
After replacing with this code, it is sending one request at a time.
Found a hint at :
Adding headers when using httpClient.GetAsync
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,215
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I'm always game to try out new #curlygirl hair products, so when I spotted Everyone's small hair styling line at a local health food store for a super inexpensive $5.99 each, I decided to pick up all three products to test out. Everyone is a budget-friendly offshoot of EO Products, specifically designed for younger consumers, large families, and others who might have less money to spend on quality, natural products. The products are made for "every budget" -- something you know really resonates with me -- and contain no GMO's, parabens, sulfates, synthethic fragrances, gluten, or synthetic polymers. The hair styling products are also phenoxyethanol-free.
I've been testing this hair trio out since the summer -- keep reading for my thoughts on each product!
The Style Gel is your basic no-frills hair gel. It has good hold, isn't sticky, and, just like the bottle says, doesn't flake or crumble throughout the day. The gel isn't specifically designed for curly hair, but it works well for curl definition nonetheless, and even leaves curls looking pretty great on the second (and sometimes third) day.
The scent on this one is my favorite of the three. It's a mild, sweet blend of citrus, vanilla, and coconut that smells nice as you use it, but isn't noticeable much beyond that.
The Hair Cream doesn't offer much in the way of hold, but it's fantastic for defining curls and taming frizz. It has a creamy consistency and gives hair a bit of moisture too, similar to a lightweight leave-in conditioner. For my hair, this is the best of the bunch for first day curls, but, because its hold is so light, by the second day (and beyond) my curls have fallen considerably.
Again, the scent is fairly mild -- this time a mix of lavender and coconut -- and I don't notice it in my hair other than when I'm first applying/distributing the product.
Of the three, the Volume Serum is the one I'd say you can skip. It has a thin gel consistency and claims to add volume, bounce and body, but I don't find it does much in my hair. On days that I air dry (which, let's be honest, is most of the time), the serum does basically nothing. When I blow dry instead, I do get more volume, of course, but I'm not convinced it's any more than blow drying without the serum, either with or without a different product.
The scent on this one is lemon and coconut and, again, it's very unobtrusive, although you can initially smell some alcohol -- the third ingredient in this one -- coming through too.
I'd pass on the Volume Serum -- I just haven't found a good use for it -- but I really like the Hair Cream and Hair Gel. Both give light to medium hold, offer good curl definition, and do well at controlling frizz and flyaways. The ingredients are also great and Everyone's inexpensive prices really can't be beat!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,970
|
<nav class="navbar navbar-fixed-top navbar-inverse">
<div class="navbar-header">
<a href="/#/dashboard" class="navbar-brand"><span class="brand-title">{{vm.title}}</span></a>
<a class="btn navbar-btn navbar-toggle" data-toggle="collapse" data-target=".navbar-collapse">
<span class="icon-bar"></span>
<span class="icon-bar"></span>
<span class="icon-bar"></span>
</a>
</div>
<div class="navbar-collapse collapse">
<div class="pull-right navbar-logo">
<ul class="nav navbar-nav pull-right">
<li>
<a href="/#/dashboard">
Created by Ken Vella
</a>
</li>
<li class="dropdown dropdown-big">
<a href="http://www.angularjs.org" target="_blank">
<img alt="angular logo" src="content/images/AngularJS-small.png" />
</a>
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</ul>
</div>
</div>
</nav>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,438
|
Q: How could you do this query in laravel? the goal is to get a list of posts with date of last comment
select
`posts`.*,
`follow_posts`.`follower_id` as `pivot_follower_id`,
`follow_posts`.`post_id` as `pivot_post_id`,
`follow_posts`.`created_at` as `pivot_created_at`,
`follow_posts`.`updated_at` as `pivot_updated_at`, (select max(updated_at) from `comments` where `commentable_id` = `posts`.`id`) as `comentario`
from
`posts`
inner join
`follow_posts` on `posts`.`id` = `follow_posts`.`post_id`
where
`follow_posts`.`follower_id` = '1'
order by
`comentario` desc
A: Assuming you have a Model called Post related to table posts:
use App\Post;
...
$posts = Post::innerJoin('follow_posts', 'posts.id', '=', 'follow_posts.post_id')
->where('follow_posts.follower_id', '=', '1')
->orderBy('comentario', 'desc')
->select([
'follow_posts.follower_id as pivot_follower_id',
'follow_posts.post_id as pivot_post_id',
'follow_posts.created_at as pivot_created_at',
'follow_posts.updated_at as pivot_updated_at',
])->selectRaw('
(select max(updated_at)
from comments
where commentable_id = posts.id) as comentario')
->get();
I think you have to specify all columns you want to select from table posts. Event if you don't need to do this, it's faster(you probably don't need all of them) and you know exactly what you are working on.
If you don't want to build this as mentioned above you can always use
$posts = DB::select('SELECT ...');
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,983
|
{"url":"https:\/\/www.calculatorsoup.com\/calculators\/discretemathematics\/combinationsreplacement.php","text":"Online\u00a0Calculators ...\n\n# Combination with Replacement Calculator\n\nCombinations Replacement CR(n,r)\n$C^R(n,r) = \\frac{(n + r - 1)!}{ r! (n - 1)! } = \\; ?$\n\n## Calculator Use\n\nFor a combination replacement sample of r elements taken from a set of n distinct objects, order does not matter and replacements are allowed.\n\nThe Combinations Replacement Calculator will find the number of possible combinations that can be obtained by taking a subset of items from a larger set. Replacement or duplicates are allowed meaning each time you choose an element for the subset you are choosing from the full larger set. This calculates how many different possible subsets can be made from the larger set. For this calculator, the order of the items chosen in the subset does not matter.\n\nCombinations with replacement, also called multichoose, for CR(n,r) = C(n+r-1,r) = (n+r-1)! \/ r! (n+r-1 - r)! = (n+r-1)! \/ r! (n - 1)!. For n >= 0, and r >= 0. If n = r = 0, then CR(n,r) = 1.\n\nFactorial\nThere are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r.\nCombination\nThe number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.\nCombination Replacement\nThe number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed.\nn\nthe set or population\nr\nsubset of n or sample set\n\n## Combination with Replacement Formula:\n\n$$C^R(n,r) = \\dfrac{(n + r - 1)!}{ r! (n - 1)! }$$","date":"2020-04-02 17:19:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.33850350975990295, \"perplexity\": 512.5535296590479}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-16\/segments\/1585370506988.10\/warc\/CC-MAIN-20200402143006-20200402173006-00396.warc.gz\"}"}
| null | null |
Q: TImedate stamp in MPLAB In my code I put a time date stamp with the version number.
const char prog_id[] = { __TIME__ " " __DATE__ "Foo project V1.3" } ;
Output:
11:09:52 May 10 2016 Foo project V1.3
This is the line of code when I was using the Freescale IDE. Due to a change in chipsets I have had to change to the MPLAB X IDE and I cannot find equivalents for the __TIME__ and __DATE__ macros.
Are there equivalents in the MPL X IDE or is there a way I can replicate these functions in C.
A: Your compiler has __DATE__ only.
Take a look at page 246 of your compiler manual
To emulate them you could, I'm not an mplab expert, use a pre-build step where a custom little program create a time_date.h file where it defines those defines with the correct value. Obviously this heather have to be included into your project.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,270
|
require 'dfp_api'
API_VERSION = :v201505
def create_creative_wrappers()
# Get DfpApi instance and load configuration from ~/dfp_api.yml.
dfp = DfpApi::Api.new
# To enable logging of SOAP requests, set the log_level value to 'DEBUG' in
# the configuration file or provide your own logger:
# dfp.logger = Logger.new('dfp_xml.log')
# Get the CreativeWrapperService.
creative_wrapper_service = dfp.service(:CreativeWrapperService, API_VERSION)
# Set the creative wrapper label ID.
label_id = 'INSERT_CREATIVE_WRAPPER_LABEL_ID_HERE'.to_i
# Create creative wrapper objects.
creative_wrapper = {
# A label can only be associated with one creative wrapper.
:label_id => label_id,
:ordering => 'INNER',
:header => {:html_snippet => '<b>My creative wrapper header</b>'},
:footer => {:html_snippet => '<b>My creative wrapper footer</b>'}
}
# Create the creative wrapper on the server.
return_creative_wrappers =
creative_wrapper_service.create_creative_wrappers([creative_wrapper])
if return_creative_wrappers
return_creative_wrappers.each do |creative_wrapper|
puts "Creative wrapper with ID: %d applying to label: %d was created." %
[creative_wrapper[:id], creative_wrapper[:label_id]]
end
else
raise 'No creative wrappers were created.'
end
end
if __FILE__ == $0
begin
create_creative_wrappers()
# HTTP errors.
rescue AdsCommon::Errors::HttpError => e
puts "HTTP Error: %s" % e
# API errors.
rescue DfpApi::Errors::ApiException => e
puts "Message: %s" % e.message
puts 'Errors:'
e.errors.each_with_index do |error, index|
puts "\tError [%d]:" % (index + 1)
error.each do |field, value|
puts "\t\t%s: %s" % [field, value]
end
end
end
end
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,604
|
import React from 'react';
import './LoadingPanel.css';
import Pint from '../pint.svg';
export default () => (
<div className="beer-tender-loading-panel">
<div className="beer-tender-loading-inner">
<img alt="beer pint" src={Pint} />
</div>
</div>
);
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,505
|
{"url":"https:\/\/questions.examside.com\/past-years\/gate\/question\/consider-a-machine-with-64-mb-physical-memory-and-a-32-bit-v-gate-cse-2001-marks-2-9nz5o2iwgnoscbrf.htm","text":"NEW\nNew Website Launch\nExperience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...\n1\n\n### GATE CSE 2001\n\nMCQ (Single Correct Answer)\nConsider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size is 4KB, what is the approximate size of the page table?\nA\n16 M\nB\n8 MB\nC\n2 MB\nD\n24 MB\n2\n\n### GATE CSE 2000\n\nMCQ (Single Correct Answer)\nSuppose the time to service a page fault is on the average $$10$$ milliseconds, while a memory access takes $$1$$ microsecond. Then a $$99.99$$% hit ratio results in average memory access time of.\nA\n$$1.9999$$ milliseconds\nB\n$$1$$ millisecond\nC\n$$9.999$$ microseconds\nD\n$$1.9999$$ microseconds\n3\n\n### GATE CSE 1999\n\nMCQ (More than One Correct Answer)\nWhich of the following is\/are advantage of virtual memory?\nA\nFaster access to memory on an average.\nB\nProcesses can be given protected address spaces.\nC\nLinker can assign addresses independent of where the program will be loaded in physical memory.\nD\nPrograms larger than the physical memory size can be run.\n4\n\n### GATE CSE 1998\n\nMCQ (Single Correct Answer)\nIf an instruction takes $${\\rm I}$$ microseconds and a page fault takes an additional $$j$$ microseconds, the effective instruction time if on the average a page fault occurs every $$k$$ instruction is:\nA\n$$i + \\left( {j\/k} \\right)$$\nB\n$$i + j{}^ * k$$\nC\n$$\\left( {i + j} \\right)\/k$$\nD\n$$\\left( {i + j} \\right){}^ * k$$\n\n### Joint Entrance Examination\n\nJEE Main JEE Advanced WB JEE\n\n### Graduate Aptitude Test in Engineering\n\nGATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN\n\nNEET\n\nClass 12","date":"2022-06-25 16:13:57","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.665631115436554, \"perplexity\": 13620.312020785093}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": false}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656103036077.8\/warc\/CC-MAIN-20220625160220-20220625190220-00199.warc.gz\"}"}
| null | null |
Hipurinska kiselina je organsko jedinjenje, koje sadrži 9 atoma ugljenika i ima molekulsku masu od 179,173 -{Da}-.
Osobine
Reference
Literatura
Spoljašnje veze
Карбоксилне киселине
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,459
|
\section{\label{sec1}Introduction}
In recent years spin carrying defects in solids have proved to be highly suitable for qubit \cite{quantumcomp, qubit} and nanoscale sensor applications \cite{sensor1}. So far the most investigated defect is the negatively charged nitrogen-vacancy defect (NV center) in diamond \cite{NVdiamond, NVdiamond1, nanosens} for which the afore-mentioned applications have been achieved. The exceptional properties of the NV center in diamond is related to its optically polarizable spin triplet ($S = 1$) ground state ($^3A_2$), a spin dependent radiative recombination from the excited $^3E$ triplet state and a parallel operating spin selective non-radiative decay via intermediate singlet states ($^1A_1$, $^1E$). This mechanism results in the strong spinpolarization of the $^3A_2$ ground state with a predominant population of the $m_s = 0$ state. In addition, the spinstate of the $^3A_2$ state can be read out optically even at room temperature due to the long spin-coherence times\cite{NVdiamond,
divacexp} via the optically detected magnetic resonance (ODMR) effect.
Despite the unique properties of NV center in diamond, the material properties of diamond are not optimal and difficult to integrate into existing semiconductor device technology. In addition, the visible emission of NV center in diamond is not favorable for quantum communication where the fiber optics provides the most efficient transmission at near infrared (NIR) wavelengths. Alternative qubits in technologically mature materials for various quantum technology application are much sought after. One of the most favorable candidates is silicon carbide with hosting divacancy defects that consist of adjacent carbon and silicon vacancies in the SiC lattice \cite{galipssb2011, quantumcomp, divacexp, Falk2013NatCom, arxivChristle}. This defect exhibits very similar properties to those of NV center in diamond, including the optical coherent control of this $S=1$ center \cite{quantumcomp}, and a relatively high contrast optical readout at resonant excitation \cite{arxivChristle}, but it emits in NIR region (around $1100$~nm of wavelength) not far from the telecom wavelengths. NIR emission is also desirable for \textit{in vivo} fluorescent biosensor applications where fabrication of nanocrystalline SiC hosting NIR color centers were already suggested to this end \cite{NIR, somogyi_JPC}.
Very recently, the equivalent of the NV center in diamond, the N$_\text{C}$V$_\text{Si}$ centers have been identified in different (3C,4H,6H) polytypes of SiC \cite{4Hexp, NVSiCPL, NVSiCour}. In the negative charge state they are spin S=1 centers with optical properties shifted to the NIR region (around $1200$~nm of wavelength) almost compatible with the transmission wavelength of optical fibers. However, some of their optical properties and excited state configurations have not yet been fully resolved either in experiments \cite{4Hexp,NVSiCour} or in theory \cite{quantumcomp, NVformation, NVgordon_PRB, NVSiCour} and thus required further investigations. Since qubits are individual quantum objects, thorough characterization of the individual NV defects in SiC is required to optimize the conditions of qubit operations. To this end, we carried out density functional theory (DFT) calculations of nitrogen-vacancy defects in 3C, 4H and 6H SiC. We provide a detailed results concerning their electronic structure, magneto-optical parameters, ionization energies and formation energies. We also discuss the possible formation processes of NV defects in SiC and briefly compare the formation and ionization energies of NV center and divacancy defects in SiC.
In the followings, we describe the computational methods in Section~\ref{sec:methods}. Our results are presented in detail in Section~\ref{sec:results} where we compare them to the experimental data if available. Here, we show new electron paramagnetic resonance (EPR) spectra for basal NV centers in 4H SiC. We discuss the photo-ionization and the formation of NV center in SiC in Section~\ref{sec:discussion}. Finally, we conclude our work in Section~\ref{sec:summary}.
\section{\label{sec2}Defect structure}
SiC exhibits various crystal structures, called polytypes, with 4H, 6H and 3C the most advanced from a material point of view. We have therefore calculated the properties of NV center in these technologically important polytypes. We found by \emph{ab initio} calculations that formation energy of N$_{\text{C}}$V$_{\text{Si}}$ defect is lower by about 2~eV than that of N$_{\text{Si}}$V$_{\text{C}}$ defect in each considered polytype in the neutral charged state implying that nitrogen preferentially substitutes carbon (N$_{\text{C}}$) in the SiC lattice, adjacent to a neighbor Si-vacancy (V$_{\text{Si}}$). We show the results on N$_{\text{C}}$V$_{\text{Si}}$ in detail that we also call NV center in the context.
Due to the special arrangement of the atoms in hexagonal polytypes, different varieties of NV center exist, depending on the lattice site of the N-atom and the adjacent Si-vacancy. In short, N$_{\text{C}}$V$_{\text{Si}}$ defect forms one configuration in 3C ($kk$) (cf. Fig. \ref{defstruct3c}), four configurations in 4H ($hh$, $kk$, $hk$, $kh$) (cf. Fig. \ref{defstruct4h}), and six configurations in 6H ($hh$, $k_1k_1$, $k_2k_2$, $hk_1$, $k_1k_2$, $k_2h$) (cf. Fig. \ref{defstruct6h}) polytype, where $h$ and $k_{\{1,2\}}$ labels the (quasi)cubic and hexagonal lattice sites in each polytype, respectively.
\begin{figure}
\includegraphics[width=0.5\textwidth]{defect_struct_3c.pdf}
\caption{Single configuration ($kk$) of N$_{\text{C}}$V$_{\text{Si}}$ defect in 3C SiC}
\label{defstruct3c}
\end{figure}
\begin{figure}
\includegraphics[width=0.5\textwidth]{defect_struct4H.pdf}
\caption{Possible configurations of N$_{\text{C}}$V$_{\text{Si}}$ defect in 4H SiC. The $hh$ and $kk$ configurations exhibiting C$_{\text{3v}}$ symmetry are called \textit{on-axis} configurations, since the axis of defects is parallel to the c-axis, while $kh$ and $hk$ configurations are \textit{off-axis} geometries with C$_{\text{1h}}$ symmetry.}
\label{defstruct4h}
\end{figure}
\begin{figure}
\includegraphics[width=0.5\textwidth]{defect_struct_6h.pdf}
\caption{Possible configurations of N$_{\text{C}}$V$_{\text{Si}}$ defect structure in 6H SiC. Three on-axis ($hh$, $k_1k_1$, $k_2k_2$) and three off-axis or basal configurations ($hk_1$, $k_1k_2$, $k_2h$) can be formed.}
\label{defstruct6h}
\end{figure}
\section{\label{sec:methods}Methodology}
\subsection{\label{subsec:comp}Computational approach}
Our calculations were carried out by means of HSE06 range-separated hybrid functional developed by Heyd, Scuseria and Ernzerhof \cite{HSE06}. In order to characterize ground and excited state zero-filed splitting (ZFS) arising from electron spin - electron spin dipole-diple interaction we calculated the \textit{D} and \textit{E} parameters employing the Perdew-Burke-Ernzerhof (PBE) \cite{PBE} functional as implemented by Iv\'ady \textit{et al.} \cite{ZFScalc}. In the excited state of on-axis defect configurations exhibits dynamic Jahn-Teller (JT) distortion due to an effective electron-phonon coupling, thus ZFS constants calculated under C$_\text{1h}$ symmetry were averaged to show a dynamic C$_\text{3v}$ symmetry. For basal configurations the natural symmetry is \emph{per se} C$_\text{1h}$ symmetry, nevertheless, similar electron-phonon coupling can occur in the excited state as that for the axial configurations. Therefore, we carried out a motional averaging procedure about the N-V axis for the basal NV configurations. In the calculation of zero-phonon lines (ZPL) we used the lowest total energy in the excited state corresponding to the JT geometry.
For atomistic modeling of defect structures, we applied 512-atom supercell for 3C, 576-atom supercell for 4H and 432-atom supercell for 6H polytype. For sampling the Brillouin-zone we employed $\Gamma$-point only for 3C and 4H polytypes, while $\Gamma$-centered $2\times2\times2$ Monkhorst-Pack \textit{k}-point mesh was used for 6H SiC. In addition, to reach sufficient accuracy for ZPL values $2\times2\times2$ \textit{k}-point mesh was employed in each case. Plane wave expansion of Kohn-Sham wavefunctions with a cutoff of 420~eV was applied. Relaxed geometries were achieved by minimizing the total energy with respect to the normal coordinates of the lattice using the force threshold of 0.01 eV/$\AA$. Core-electrons were treated by projector-augmented wave (PAW) \cite{PAW} potentials as implemented in VASP code \cite{VASP}. In the case of charged defects Freysoldt correction \cite{Freysoldt} in total energy was applied. The hyperfine couplings were calculated by HSE06 DFT calculations with taking into account the spin polarization of the core electrons \cite{hyperfine}.
\subsection{\label{subsec:form}Formation energies and charge transition levels}
Concentration of point defects in thermal equilibrium can be predicted via the formation energies. In addition, determining the adiabatic charge transition levels, i.e., ionization energies is also crucial to study the stability window of a given charge state of the defect that is applied as a qubit. We calculated the formation energies as \cite{ZhangPRL1991,aradimu}
\begin{equation}
\begin{aligned}
E_{\text{form}}^q = E_{\text{tot}}^q - n_{\text{SiC}}\mu_{\text{SiC}} - \frac{\mu_{\text{Si}} - \mu_{\text{C}} - \delta\mu}{2}(n_\text{Si} - n_\text{C})\\
- n_{\text{N}}\mu_{\text{N}} + q(E_{\text{f}}+E_\text{VBM}) + \Delta V(q),
\label{formeq}
\end{aligned}
\end{equation}
where $E_{\text{tot}}^q$ is the total energy of the defective system, $\mu_{\text{Si}}, \mu_{\text{C}}, \mu_{\text{N}}$ are the chemical potentials of Si atom in bulk Si, C atom in diamond and the N atom, respectively, $q$ is the charge state, $E_{\text{f}}$ is the Fermi-level, $E_\text{VBM}$ represents the valence band edge and $\Delta V(q)$ stands for the Freysoldt charge correction term \cite{Freysoldt}. If $\delta\mu$ is chosen to be the heat of formation of SiC, $ \mu_\text{SiC} - (\mu_\text{Si} +~\mu_\text{C})$, then this provides the formation energy of defects under stoichiometric conditions. In the actual VASP parameters and implementation we obtained $\mu_\text{SiC}$ = -17.47~eV, $\mu_\text{Si}$ = -6.43~eV, $\mu_\text{C}$ = -10.55~eV and $\delta\mu$ = -0.49~eV by HSE06. For determining $\mu_{\text{N}}$ the hexagonal $\beta$-Si$_3$N$_4$ was chosen as the most stable form of Si$_3$N$_4$ that can be considered as the solubility-limiting phase in SiC. We obtained $\mu_{\text{
N}}$ = -12.22~eV upon these condition. The adiabatic charge transition levels can be derived from Eq.~\ref{formeq} as follows,
\begin{equation}
E_{q+1/q} = E_{\text{tot}}^q - E_{\text{tot}}^{q+1} + \Delta V(q) - \Delta V(q+1).
\label{eq:CHLs}
\end{equation}
Binding energy ($E_\text{binding}$) of defects A and B forming the complex AB can be defined as
\begin{equation}
E_\text{binding}(E_\text{f}) = E_\text{form}^{\text{A}}(E_\text{f}) + E_\text{form}^{\text{B}}(E_\text{f}) - E_\text{form}^{\text{AB}}(E_\text{f}).
\label{eq:binde}
\end{equation}
According to this definition $E_\text{binding}>0$ implies that the formation of AB complex is favorable.
\subsection{\label{subsec:conc}Calculation of defect concentrations}
Defects may be introduced during growth of the crystal. High quality silicon carbide is typically grown via chemical vapor deposition process that may be considered as a quasi-equilibrium process. The concentration of defects can be then estimated in thermal equilibrium at the growth temperatures. The concentration of a defect D$^q$ with charge state $q$ is the sum of the concentration of individual configurations D$_i^q$, i.e. symmetry inequivalent forms of D$^q$. This can be calculated by multiplying the number of possible defect sites per cm$^3$ ($N_\text{sites}^{\text{D}}$) by the statistical weight $\omega_{D^q}$ coming from the spin multiplicity and the Boltzmann-factor
\begin{equation}
[\text{D}^{q}(E_\text{f})] = \sum_\text{i}[\text{D}_i^q(E_\text{f})] = \frac{1}{Z}N_\text{sites}^{\text{D}}\omega_{D^q}\mathbf{\sum_i}\text{exp}\bigg(-\frac{E_\text{form}^{\text{D}_i^{q}}(E_\text{f})}{k_\text{B}T}\bigg),
\label{eq:conceq}
\end{equation}
where
\begin{equation}
Z=1+\sum_{i,q}\text{exp}\bigg(-\frac{E_\text{form}^{\text{D}_i^{q}}(E_\text{f})}{k_\text{B}T}\bigg)
\label{eq:zeq}
\end{equation}
is the grand canonical partition function and the summation goes over all the individual configurations ($i$) and charge states ($q$) of defect D. In Eqs.~\ref{eq:conceq} and \ref{eq:zeq} $k_\text{B}$ is the Boltzmann constant and $T$ is the temperature. For calculating $E_\text{form}^{\text{D}_i^{q}}(E_\text{f})$ Fermi-energy has to be determined (Eq.~\ref{formeq}) which can be performed via solving the neutrality equation which reads as
\begin{equation}
\begin{aligned}
N_c(T)\text{exp}\bigg(-\frac{E_\text{CBM}-E_\text{f}}{k_\text{B}T}\bigg) + \sum_{\substack{\text{D}\\
q<0}} |q_\text{D}|[\text{D}^{q}(E_\text{f})] = \\
N_v(T)\text{exp}\bigg(-\frac{E_\text{f}-E_\text{VBM}}{k_\text{B}T}\bigg) + \sum_{\substack{\text{D}\\
q>0}} |q_\text{D}|[\text{D}^{q}(E_\text{f})],
\label{eq:neutraleq}
\end{aligned}
\end{equation}
where
\begin{equation}
N_c(T) = 2 \bigg(\frac{2m_\text{e}^*\pi k_\text{B}T}{h^2}\bigg)^{3/2}
\label{eq:neutraleqnc}
\end{equation}
and
\begin{equation}
N_v(T) = 2 \bigg(\frac{2m_\text{h}^*\pi k_\text{B}T}{h^2}\bigg)^{3/2}
\label{eq:neutraleqnv}
\end{equation}
are the effective densities of states of electrons in the conduction band edge and holes in the valence band edge, respectively, and $E_\text{CBM}, E_\text{VBM}$ labels the conduction and valence band edges, respectively. We applied the parameters of 4H SiC as we calculated the concentration of defects in this polytype Accordingly, for the effective masses in Eqs.~\ref{eq:neutraleqnc} and \ref{eq:neutraleqnv} we used $m_\text{h}^*$ = 1.26$m_\text{e}^0$ and $m_\text{e}^*$ = 0.39$m_\text{e}^0$, where $m_\text{e}^0$ is the electron rest mass. We note that the shallow substitutional nitrogen donors in SiC will be explicitly treated in Eq.~\ref{eq:conceq}.
In order to obtain Fermi energy the series of equations (Eqs.~\ref{eq:conceq} and \ref{eq:neutraleq}) has to be solved self-consistently. For the calculation thermal equilibrium and stoichiometric ratio of C and Si atoms were assumed. We considered infinite bulk material, thus, the effect of band bending near the surface and other kinetic effects were neglected. Bulk growth of SiC is usually carried out at temperatures between $1600 ^\circ$C and $2000 ^\circ$C \cite{temp, KordinaPSS1997}. Employing this technique different charge states of substitutional nitrogen (N$_\text{C}$), vacancies (V$_\text{C}$, V$_\text{Si}$), divacancies (V$_\text{C}$V$_\text{Si}$), carbon antisite vacancy pairs (CAV) and nitrogen-vacancy complexes ((N$_\text{C}$)$_k$V$_\text{Si}$, k=1,2,3,4) may be formed. For nitrogen-vacancy complexes we considered the case of $k$=1 (N$_\text{C}$V$_\text{Si}$) and the electrically inactive $k$=4 ((N$_\text{C}$)$_4$V$_\text{Si}$ defect) as the latter has an extremely low formation energy \cite{bockstedte1}. In this spirit, we calculated the concentrations of the these defects at temperatures of $1600 ^\circ$C, $1700 ^\circ$C, $1800 ^\circ$C, $1900 ^\circ$C and $2000 ^\circ$C considering all relevant charge states. To this end, we calculated the formation energies of all the defects in all configurations including V$_\text{Si}$, V$_\text{C}$ and CAV \cite{RaulsPSSB2011,singlephotonsourceNatMat,szaszCAV}.
\section{\label{sec:results}Results}
By using HSE06 functional we could reproduce the experimental band gap within 0.1~eV for 3C, 4H and 6H polytypes. This limits the accuracy of our method in the prediction of charge transition levels.
\subsection{\label{subsec32}Electronic structure}
Group theory analysis on the defect provides intriguing insights its electronic structure. Accordingly, we found that on-axis configurations of NV center with C$_{\text{3v}}$ symmetry introducing two $a_1$ levels and a degenerate $e$ level. Off-axis configurations exhibit C$_\text{1h}$ symmetry, and the degenerate $e$ level splits to an $a'$ and an $a''$ state while the $a_1$ states transform into $a'$ states. Our calculations revealed that one of the $a_1$ levels falls in the valence band, whereas the other is lying in the fundamental band gap and they both are fully-occupied. In the single negative charged state, the degenerate $e$ level is introduced in the band gap occupied by two electrons with parallel spins providing $S = 1$ triplet spin state. In summary, the one-electron structure of the ground state is $a_1(2)a_1(2)e(2)$ for on-axis and $a'(2)a'(2)a'(1)a''(1)$ for off-axis configurations in singly negative charge state (cf.\ Fig.~\ref{estruct}). Further application of grou
p theory implies that $^3A_2$, $^1E$ and $^1A_1$ multiplets can be formed by the $a_1(2)a_1(2)e(2)$ electron configurations for the on-axis defects. The $^3E$ bright triplet excited are realized by $a_1(2)a_1(1)e(3)$ electron configuration, and an $^1E$ multiplet occurs too for on-axis defects. The $^3E$ excited state is a dynamic Jahn-Teller system. We calculated this excited state by $\Delta$SCF method with allowing $C_{1h}$ symmetry distortion but the zero-field constant was calculated in the dynamic average of $C_{3v}$ symmetry. For the off-axis configurations, the electronic configurations and states are similar to those of on-axis configurations, just the degenerates states are split due to the $C_{1h}$ symmetry crystal field.
\begin{figure} [h!]
\centering
\includegraphics[width=0.4\textwidth]{estruct.pdf}
\caption{\small Scheme of ground state electronic structure of (a) on-axis and (b) off-axis NV center configurations exhibiting C$_\text{3v}$ and C$_\text{1h}$ symmetry, respectively.}
\vspace{- 20 pt}
\label{estruct}
\end{figure}
\subsection{Formation energies and charge state stability of NV centers in SiC}
\label{ssec:form}
We plot the formation energies of the NV defect in the considered polytypes of SiC in Fig.~\ref{CTLfig}. We find that the neutral and negative charged states are stable as function of the position of the Fermi-level in all the polytypes, whereas the double negative charged state exists in hexagonal polytypes. The negatively charged NV defect, i.e., NV center can occur in moderately or highly n-type 3C SiC, whereas NV center is stable in non-doped or moderately n-type doped hexagonal SiC. In highly n-type doped hexagonal SiC, the NV defect becomes double negatively charged. This result implies that different doping strategies should be applied to stabilize the single negative charge state of the NV defect in cubic and hexagonal polytypes.
\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{ctl_all.pdf}
\caption{Formation energies of the NV defects as a function of the position of the Fermi-level in (a) 3C, (b) 6H and (c) 4H polytypes. In 4H SiC we plot the formation energy of V$_{\text{C}}$V$_{\text{Si}}$ divacancy.}
\label{CTLfig}
\end{figure}
We find that the formation energies of N$_{\text{C}}$V$_{\text{Si}}$ defect configurations in 4H and 6H polytypes vary slightly, i.e., the values agree within $\sim0.1$~eV, whereas the maximum difference between the corresponding charge transition levels does not exceed $\sim0.2$~eV (see Table~\ref{CTLtab}). In Fig.~\ref{CTLfig}(c) formation energies of V$_{\text{C}}$V$_{\text{Si}}$ configurations in 4H polytype are also showed for discussion.
\begin{table}[t]
\begin{ruledtabular}
\caption{Charge transition levels of N$_{\text{C}}$V$_{\text{Si}}$ defects referenced to the valence band maximum ($E_{\text{VBM}}$).}
\label{CTLtab}
\begin{tabular}{@{}cccc@{}}
\multicolumn{1}{c}{Polytype}
& \multicolumn{1}{l} {conf.}
& \multicolumn{1}{c} {$E_{0/-}$ (eV)}
& \multicolumn{1}{c} {$E_{-/2-}$ (eV)}\\
\hline
3C & $kk$ & 1.48 & \\
\hline
\multirow{4}{*} {4H} & $hh$ & 1.54 & 2.65 \\
& $kk$ & 1.46 & 2.63 \\
& $hk$ & 1.49 & 2.42 \\
& $kh$ & 1.55 & 2.50 \\
\hline
\multirow{3}{*} {6H} & $hh$ & 1.54 & 2.59 \\
& $k_1k_1$ & 1.51 & 2.64 \\
& $k_2k_2$ & 1.53 & 2.62
\\
\end{tabular}
\end{ruledtabular}
\end{table}
\subsection{\label{ssec:DZPL}Magneto-optical properties of NV center in SiC}
The NV centers in all three polytypes have the same electronic structure with a $^3A_2$ triplet groundstate and a $^3E$ excited state. The paramagnetic groundstate make them suitable to EPR spectroscopy, which has been applied successfully for their assessment \cite{4Hexp, NVSiCPL, NVSiCour}. Our calculations show in accordance with the experimental results that each type of NV center is characterized by specific zero-field-splitting (ZFS) parameters, hyperfine interactions (HF) and optical properties as shown in Tables~\ref{tab:hypertable}-\ref{tab:ZFStable6H}.
\begin{figure}
\includegraphics[width=0.4\textwidth]{hyper_spindens_tot.pdf}
\caption{Isosurface of the spin density localized on the three C-atoms near the C-vacancy in the triplet ground state of NV center in SiC. The supercell structure is shown in perspective view where the lattice is depicted as a wire except for the atoms in the core of the defect that are represented by balls. The corresponding atom types are labeled.}
\label{fig:spindens}
\end{figure}
We first discuss the spin density distributions. In the ground state, the major spin density is localized on the three neighbor C atoms of the Si-vacancy (cf.\ Fig.~\ref{fig:spindens}) providing a strong HF interaction with the $^{13}$C nuclear spins. The HF interactions with the $^{14}$N neighbor and its adjacent three $^{29}$Si nuclei are small ($\approx 1$~MHz). Nevertheless due to the small EPR linewidth of $0.02$~mT they are already resolved and represent the fingerprint of NV centers which distinguish them from other defects in SiC. The Si-atoms near the nearest neighbor C-atoms have stronger HF constants of $\approx 10$~MHz also resolved in the EPR spectra. The small value of the $^{14}$N HF constants is due to the only indirect HF interaction; the calculational accuracy for these values is smaller than for the HF with the $^{13}$C and $^{29}$Si neighbors with direct HF interaction and estimated to 10\% according to our tests on related defects \cite{hyperfine}. Nevertheless,
the calculated values are in good agreement with the experimental findings, as shown in Table~\ref{tab:hypertable}.
\begin{table*}
\begin{ruledtabular}
\caption{Calculated hyperfine constants ($A_{ii}; i=x,y,z$) in the ground state of on-axis NV center configurations. Experimental data\cite{NVSiCour} available for $^{14}\text{N}$ isotope are also listed.}
\label{tab:hypertable}
\begin{tabular}{@{}cccccccc@{}}
\multirow{2}{*}{Polytype}
& \multirow{2}{*} {conf.}
& \multicolumn{2}{c} {$1\times^{14}\text{N}$}
& \multicolumn{2}{c} {$3\times^{29}\text{Si}$ + $6\times^{29}\text{Si}$}
& \multicolumn{1}{c} {$1\times^{13}\text{C}$ }\\
& & $A_\text{exp}^\text{iso}$(MHz) & $A_{xx}, A_{yy}, A_{zz}$ (MHz) & $A_{xx}, A_{yy}, A_{zz}$ (MHz) & $A_{xx}, A_{yy}, A_{zz}$ (MHz) & $A_{xx}, A_{yy}, A_{zz}$ (MHz) \\
\hline
3C & $kk$ & 1.26 & -1.67, -1.67, -1.72 & 11.93, 11.91, 12.31 & 9.64, 8.47, 10.48 & 48.23, 47.49, 119.38 \\
\hline
\multirow{2}{*} {4H} & $hh$ & 1.23 & -1.57, -1.57, -1.64 & 11.80, 11.72, 12.14 & 9.95, 8.77, 10.79 & 45.49, 44.86, 116.78 \\
& $kk$ & 1.12 & -1.71, -1.69, -1.71 & 12.77, 12.39, 13.26 & 10.56, 9.60, 11.32 & 42.23, 41.48, 112.79\\
\hline
\multirow{3}{*} {6H} & $hh$ & 1.32 & -1.61, -1.61, -1.80 & 11.95, 11.84, 12.30 & 9.81, 8.74, 10.69 & 42.40, 41.81, 113.84\\
& $k_1k_1$ & 1.21 & -1.76, -1.75, -1.76 & 12.94, 12.63, 13.48 & 10.93, 10.17, 11.74 & 37.00, 36.33, 108.14 \\
& $k_2k_2$ & 1.26 & -1.73, -1.73, -1.76 & 12.05, 12.00, 12.48 & 10.51, 9.51, 11.39 & 42.44, 41.78, 114.36 \\
\end{tabular}
\end{ruledtabular}
\end{table*}
We also determined the ZFS parameters ($D$, $E$) for NV centers by assuming electron spin - electron spin dipolar interaction. We calculated these parameters in the $^3A_2$ ground state for all the considered polytypes (Tables~\ref{tab:ZFStable3C4Ha}, \ref{tab:ZFStable6H}). In the point dipole spin - point dipole spin approximation, the $D$ varies as $D\sim\frac{1}{r^3}$ with $r$ being the distance between the dipoles. In the ground state, the spin density is localized on the C-atoms near the Si-vacancy, so the distance between these C-atoms gives the trend for the variation of the $D$-constants (see Tables~\ref{tab:ZFStable3C4Ha}, \ref{tab:ZFStable6H}).
Here, we also report new EPR spectra of the basal NV centers in 4H SiC which have been measured in the temperature range from $4$~K to $300$~K.
To allow comparison with the calculated values, we show in Table~\ref{tab:ZFStable3C4Ha} their parameters at T=$4$~K. The NV centers in the 4H polytype were created as reported in Refs.~\onlinecite{4Hexp, NVSiCPL,NVSiCour} by particle irradiation and thermal annealing. The observed EPR spectra are shown in Fig.~\ref{fig:EPRdata}. From these measurements, the $D$ and $E$ parameters for the basal NV configurations were extracted, and the low temperature $D$ parameters for all the four NV configurations in 4H SiC are shown in Table~\ref{tab:ZFStable3C4Ha}. We note that the previously published EPR data for axial configurations in 4H SiC are those measured at T=300K in Ref.~\onlinecite{NVSiCour}.
\begin{figure}
\includegraphics[width=\columnwidth]{EPR_data3.pdf}
\caption{(a) EPR spectrum of the NV center in 4H-SiC displaying the two axial and two basal related EPR spectra for $\textbf{B}\| c$ where $\textbf{B}$ is the applied external magnetic field and $c$ is the c-axis of 4H SiC; (b) simulated angular variation of the basal NV centers in 4H-SiC}
\label{fig:EPRdata}
\end{figure}
\begin{table}
\begin{ruledtabular}
\caption {Ground state zero-field-splitting constants ($D$, $E$) of NV center configurations (conf.) in 3C and 4H polytypes of SiC. The experimental data ($D_\text{e}$) in 3C was taken from Ref.~\onlinecite{NVSiCour}. The experimental data in 4H SiC recorded at cryogenic temperature are also provided as new results ($D_\text{e}$, $E_\text{e}$). In the ground state the two unpaired electrons are localized on the C atoms near Si-vacancy, hence, corresponding distances ($d_1$) are also given. For the off-axis configurations, two different C-C distances occur where the second distance is listed in the $d_2$ column. The larger the deviation is between $d_1$ and $d_2$ the larger is the $E$ parameter.}
\label{tab:ZFStable3C4Ha}
\begin{tabular}{@{}ccccccc@{}}
conf. & $D$(MHz)& $E$(MHz) & $d_1$(\AA)& $d_2$(\AA) & $D_{\text{e}}$(MHz) & $E_{\text{e}}$(MHz)\\
\hline
3C-$kk$ & 1409 & 0 & 3.34 & & 1303 & \\
\hline
4H-$kk$ & 1377 & 0 & 3.36 & & 1282 & 0\\
4H-$hh$ & 1427 & 0 & 3.33 & & 1331 & 0 \\
4H-$hk$ & 1331 & 110 & 3.38& 3.35 & 1193 & 104 \\
4H-$kh$ & 1404 & 44 & 3.34& 3.33 & 1328 & 15 \\
\end{tabular}
\end{ruledtabular}
\end{table}
\begin{table}
\begin{ruledtabular}
\caption {Calculated values of $D$ and $E$ parameters for NV center configurations (conf.) in 6H SiC. Experimental values ($D_\text{e}$) are only available for the axial configurations\cite{NVSiCour}. Distances ($d_1$) of C-C atoms near Si-vacancy are also listed. For the off-axis configurations, two different C-C distances occur where the second distance is listed in the $d_2$ column.}
\label{tab:ZFStable6H}
\begin{tabular}{@{}ccccccc@{}}
\multicolumn{1}{c} {conf.}
& \multicolumn{1}{c} {$D$(MHz)}
& \multicolumn{1}{c} {$E$(MHz)}
& \multicolumn{1}{c} {$d_1$(\AA)}
& \multicolumn{1}{c} {$d_2$(\AA)}
& \multicolumn{1}{c} {$D_\text{e}$(MHz)}\\
\hline
$hh$ & 1404 & 0 & 3.34 & & 1328 \\
$k_1k_1$ & 1348 & 0 & 3.36 & & 1278 \\
$k_2k_2$ & 1432 & 0 & 3.33 & & 1345 \\
$k_1k_2$ & 1404 & 145 & 3.35 & 3.33 & - \\
$k_2h$ & 1386 & 9 & 3.33 & 3.34 & - \\
$hk_1$ & 1352 & 14 & 3.34 & 3.35 & - \\
\end{tabular}
\end{ruledtabular}
\end{table}
We showed above that proximate nuclear spins may reside around the electron spin of NV centers. The transfer of electron spin polarization to neighboring nuclear spins is principally feasible for NV center in 4H SiC and thus allows the realization of quantum memories. One possible method to spin-polarize the proximate nuclear spins is the optical dynamic polarization via excited state level anticrossing (ESLAC)\cite{JacquesPRL2009,nucspinpol}. We calculated the ZFS parameters of the $^3E$ excited states from which the required magnetic fields for ESLAC \cite{Ivady2015} can be obtained (c.f.\ Table~\ref{tab:ZFStable3C4Hb}).
We emphasize that ZFS parameters in the excited state are subject to dynamic Jahn-Teller effect and can be temperature dependent, and our values are only valid at $T=0$~K. In the excited state, the spin density is partially localized on the N-atom, hence, the distance between one of the C-atoms and the N-atom is given.
\begin{table}
\begin{ruledtabular}
\caption {Calculated excited state $^3E$ zero-field-splitting constants (\textit{D}, \textit{E}) of NV center configurations (conf.) in 3C and 4H SiC. Distances between N and C atoms (N-C) around Si-vacancy are also provided.}
\label{tab:ZFStable3C4Hb}
\begin{tabular}{@{}ccccc@{}}
\multirow{1}{*} {Polytype}
& \multirow{1}{*} {conf.}
& $D$(MHz) & $E$(MHz) & N-C(\AA)\\
\hline
3C & $kk$ & 707.3 & 0 & 3.42 \\
\hline
\multirow{4}{*} {4H} & $kk$ & 483.0 & 0 & 3.46 \\
& $hh$ & 537.2 & 0 & 3.44 \\
& $hk$ & 471.9 & 47.8 & 3.49 \\
& $kh$ & 537.9 & 28.6 & 3.44\\
\end{tabular}
\end{ruledtabular}
\end{table}
The intracenter optical transition of the NV centers is a key parameter for all quantum applications. The zero-phonon-line (ZPL) energies have been recently identified experimentally for the four different NV centers in 4H-SiC\cite{NVSiCPL}. We calculated the ZPL energies of NV centers in SiC that is associated with the energy difference between the $^3E$ excited state and $^3A_2$ groundstate.
The results are shown in Table~\ref{tab:ZPLtable3C4H}, and once again we find a good agreement between calculated and experimental values. The discrepancy between the calculated and measured ZPL energies for each configuration is within 0.1 eV that is expected from HSE06 hybrid density functional method \cite{DeakPRB2010}. However, the calculated differences between the ZPL energies of the defect configurations are technically converged within few meV, in terms of the parameters of plane wave supercell DFT method. We found that 576-atom supercell with $2\times2\times2$ K-point sampling of Brillouin-zone is at least required, in order to obtain the correct order of ZPL energies of the four defect configurations in 4H SiC.
The accuracy of the calculated ZPL energies is estimated to $0.1$~eV. The accuracy is limited by the supercell methodology in technical terms and by the Born-Oppenheimer approximation in the $^3E$ excited state.
Optically induced groundstate spin polarization has been observed by EPR for all NV centers in the three polytypes \cite{4Hexp, NVSiCour}. The ZPL energies correspond to the low energy threshold of the optically induced groundstate spin polarization. This feature is a particularity of all NV centers in SiC and in diamond and allows the initialization of the groundstate spin configuration by an optical pulse. It is related to the existence of intermediate singlet states, which modify the recombination processes between the $^3E$ and $^3A_2$ states. The calculation of these highly-correlated multideterminant singlet states is out of the scope of this paper.
\begin{table}
\begin{ruledtabular}
\caption {Zero-phonon-line (ZPL) energies of individual NV center configurations in 3C and 4H SiC. Calculated and experimental ZPL data (see Ref.~\onlinecite{NVSiCPL}) are also presented.}
\label{tab:ZPLtable3C4H}
\begin{tabular}{@{}ccccc@{}}
\multirow{1}{*} {Polytype}
& \multirow{1}{*} {conf.}
& \multirow{1}{*} {ZPL(eV)}
& {Signal}
& \multirow{1}{*} {ZPL$_\text{exp}$(eV)}\\%\cline{3-6}\cline{7-9}
\hline
3C & $kk$ & 0.87 & - & - \\
\hline
\multirow{4}{*} {4H} & $hh$ & 0.966 & PLX1 & 0.998\\
& $kk$ & 1.018 & PLX2 & 0.999\\
& $hk$ & 1.039 & PLX3 & 1.014\\
& $kh$ & 1.056 & PLX4 & 1.051\\
\end{tabular}
\end{ruledtabular}
\end{table}
\section{\label{sec:discussion}Discussion}
In the following sections we discuss our results of the formation of NV centers in SiC (Sec.~\ref{subsec:discform}), the identification of individual NV centers in hexagonal SiC (Sec.~\ref{subsec:discident}), and the photo-ionization of NV centers in 4H SiC (Sec.~\ref{subsec:discphoto}). The results are compared to the properties of the closely related divacancy centers. Finally, we evaluate the stability of NV centers in the different SiC polytypes in Sec.~\ref{subsec:polytypes}.
\subsection{\label{subsec:discform}Formation of NV center in SiC}
We consider the formation of NV centers in 4H SiC in two scenarios: (i) formation of NV centers as native defects related to growth conditions, (ii) NV centers formed by thermal diffusion of radiation induced Si vacancies in nitrogen doped material. We neglect the kinetic effects on the surface in this study which might play a role in the first case.
Homo- and hetero-epitaxial high quality thin films of SiC are generally grown by a chemical vapor deposition (CVD) process. If one approximates this CVD process as a thermal equilibrium process then the concentration of the in-grown defects can be estimated from their formation energies (see Fig.~\ref{fig:ctlbind}).
\begin{figure*} [t]
\includegraphics[width=\textwidth]{bind_e_NV_VV11.pdf}
\caption{Formation energies of axial (a)-(c) NV centers and (d)-(f) divacancies as a function of the Fermi-level in 3C and 4H polytypes. Shaded areas represent the stability of the corresponding complexes.}
\label{fig:ctlbind}
\end{figure*}
We considered the formation of NV center by nitrogen doping of 4H SiC. We assumed the incorporation of basic intrinsic defects, C-vacancy, Si-vacancy, CAV complex, divacancy, as well as the substitutional nitrogen (N$_\text{C}$) and (N$_\text{C}$)$_4$V$_\text{Si}$ complexes beside NV center, alas, N$_\text{C}$V$_\text{Si}$ complexes. The simulations were carried out at different growth temperatures between 1600$^\circ$C and 2000$^\circ$C, typical to CVD growth of SiC. The results are plotted in Fig.~\ref{fig:defconc}.
\begin{figure*} [t]
\includegraphics[width=\textwidth]{def_conc.pdf}
\caption{Concentration of technologically important defects in 4H-SiC as a function of growth temperature. All the relevant charge states and configurations of each defect are considered. The concentration was calculated in thermal equilibrium assuming stochiometric SiC. Please, note the logarithmic scale on the concentration. }
\label{fig:defconc}
\end{figure*}
It follows, that the concentration of in-grown Si-vacancies, CAV complexes and divacancies is negligible. (N$_\text{C}$)$_4$V$_\text{Si}$ complexes have the lowest formation energy of $\sim$1~eV and exhibit the highest concentration (over $10^{19} 1/\text{cm}^3$) which explains the known doping limitation of nitrogen in SiC (see Ref.~\onlinecite{bockstedte1}). The (N$_\text{C}$)$_4$V$_\text{Si}$ defect is electrically inactive with exhibiting $S = 0$ spin state, thus it does not establish undesirable electron spin-bath for qubit application of N$_\text{C}$V$_\text{Si}$. The maximum concentration of neutral N$_\text{C}$ is around $10^{18} 1/\text{cm}^3$. The neutral N$_\text{C}$ has an $S=1/2$ spin. The C-vacancy has a higher concentration than that of NV complexes. Nevertheless, C-vacancies will be double negatively charged at these doping conditions with a diamagnetic groundstate.
It is important to note that the concentration of NV complex is about 7-9 orders of magnitude smaller than that of the N$_\text{C}$. In this condition, the majority of NV complex will be in the double negative charge state with $S=1/2$ spin and not in the desired single negative charge state. Whereas such low defect concentrations can be detected by PL spectroscopy, they are well below the detection limit of EPR spectroscopy. At lower temperature growth ($<1600^\circ$C) the total concentration of NV center will further decrease and be in the region where NV centers will occur as single NV centers. They can be detected by confocal PL microscopy but they have not yet been reported. However, the large concentration of neutral N$_\text{C}$ introduces a dense electron spin bath that is detrimental for single NV center ODMR measurements, because this can significantly reduce the electron spin coherence time.
The second approach, the one which has been successfully used in the past, is to start with lightly (10$^{16}$/cm$^3$) nitrogen doped SiC samples; then by ion implantation or particle irradiation Si-vacancies can be created; NV center formation is obtained by thermal annealing in a temperature range where Si vacancies become mobile. The first experimental results of NV centers in 4H-SiC were obtained on proton irradiated N-doped samples \cite{4Hexp, NVSiCour}. We note that irradiation or implantation creates vacancies and interstitials in both sublattices. Regarding the monovacancies, it is expected that due to the lower displacement energies of C atoms than that of Si atoms \cite{Choyke1977, SteedsMSF2001, SteedsPRB2002, LucasPRB2005}, more C-vacancies than Si-vacancies are created by irradiation or implantation. Thus divacancies can also be be formed when Si-vacancies become mobile during annealing at around $750 ^\circ\text{C}$ \cite{Sorman2000PRB, BockstedtePRB2003}. In order to study the formation of NV centers, we calculated the binding energies of the N$_\text{C}$ and Si-vacancy versus that of the C-vacancy and Si-vacancy (see Figs.~\ref{fig:ctlbind}(a)-(c) and the derived plots in Fig.~\ref{fig:binde}). Our results show that the binding energy for divacancies is always higher than for NV centers; consequently the formation of divacancies has a higher probability provided that C-vacancies are available for the mobile Si-vacancies. If the initial concentration of nitrogen is higher than that of C-vacancies then the relative concentration of divacancies and NV complexes may be tuned toward the preferential formation of NV centers. This scenario is the one used for the formation of large ensembles of NV centers. Indeed, the previous EPR and PL investigations always showed the presence of divacancies beside NV centers. \cite{4Hexp,NVSiCour}. Since divacancies have a non-zero spin in their various charge states and are optically active, they can influence the electron spin coherence time and the photo-stability of the NV center ZPL lines. The latter will be discussed bel
ow.
Clearly for the formation of individual NV centers different nanoscopic approaches have to be applied, like low dose implantation of $N_2$ ions into non-doped SiC.
\begin{figure}
\includegraphics[width=0.450\textwidth]{binde.pdf}
\caption{Binding energies as a function of Fermi-level for axial (a) NV center and (b) divacancy configurations in 3C and 4H SiC. Dashed lines represent the conduction band minimum of 3C polytype that lies lower than that of 4H SiC.}
\label{fig:binde}
\end{figure}
\subsection{\label{subsec:discident}Spectroscopy of individual NV centers in hexagonal SiC}
The magneto-optical parameters of large ensemble of NV centers in 4H SiC are known from EPR and PL studies \cite{4Hexp, NVSiCPL, NVSiCour}, however, future quantum technology applications will be based on the spectroscopy of individual NV centers. Therefore, it is of high importance to identify the individual NV centers in 4H SiC.
We study first the magnetic properties. As the spin density associated with hyperfine couplings, so the spin density matrix associated with ZFS, is mostly localized on the carbon dangling bonds of the Si-vacancy in the ground state of the NV center (see Fig.~\ref{fig:spindens}). The experimentally determined\cite{4Hexp, NVSiCour} $g$-tensor anisotropy implies that the second-order spin-orbit coupling may be not negligible that may contribute to the ZFS. Furthermore, the first order spin-orbit coupling might contribute to the ZFS in the off-axis configurations with C$_{1h}$ symmetry. Nevertheless, the vast majority of ZFS involves the electron spin - electron spin dipolar interaction, that we can fairly well calculate within Kohn-Sham DFT.
First, we discuss the on-axis configurations. The distance between carbon dangling bonds is shorter at $hh$ site than that at $kk$ site (see Table~\ref{tab:ZFStable3C4Ha}), and correspondingly, the $D$-constant is larger at $hh$ site than that at $kk$ site. Indeed, we find the calculated $D$-constant at $hh$ configuration to be the highest one. This is very similar to the case of the divacancy in 4H SiC \cite{Falk2014}. Regarding the off-axis configurations, the $E$ constant is a good parameter to distinguish the N$_\text{C}$V$_\text{Si}$ $kh$ and $hk$ configurations. The $kh$ configuration has always a smaller $E$ constant than that of $hk$ configuration (see Table~\ref{tab:ZFStable3C4Ha}). The $hh$ and $kk$ sites can be also distinguished by their strong $^{13}$C and $^{29}$Si hyperfine interactions. Basically, the spin density is more localized in the $hh$ configuration than that in $kk$ configurations. Accordingly, the corresponding $^{13}$C hyperfine constants are higher whereas the $^{29}$Si hyperfine constants are smaller in $hh$ configuration than those in $kk$ configuration. In summary, the groundstate parameters of NV centers in 4H SiC allow a simple distinction between the four configurations. The $\approx 100$~MHz systematic discrepancy between the calculated and experimentally determined $D$-constants might be partially attributed to the neglect of second-order spin-orbit interaction.
Another important fingerprint of the NV centers is their photoluminescence spectrum. At low temperatures, the ZPL energies in the PL spectrum of NV centers have been identified in 4H SiC \cite{NVSiCPL}. The calculated absolute values of the ZPL energies are close to the experimental ones (see Table~\ref{tab:ZPLtable3C4H}). However, the calculated site dependence of the ZPL energies is not so well reproduced by the calculations. The ZPL energy difference between the PLX1 and PLX2 NV centers is only $1$~meV. It is extremely challenging to reach such an accuracy in the calculation. We find that the $hh$ and $kk$ NV center configurations have the lowest ZPL energies, as was previously suggested \cite{4Hexp,NVSiCour}. Thus, PLX1 and PLX2 should be associated with the axial configurations. Our calculations imply that the $hh$ configuration has the lowest ZPL energy, nevertheless, the identification of PLX1 does not stand on a solid ground based on solely the calculated ZPL. In analogy with
the divacancy, for which ODMR measurements under resonant excitation have been performed, $hh$ configuration should have the lowest ZPL energy and the largest $D$ constant \cite{Falk2014}. By assuming the same trends for the NV center, we deduce that the $hh$ configuration should be associated with the PLX1 spectrum and $kk$ with the PLX2 spectrumr. The PLX3 and PLX4 ZPL should be associated with the off-axis NV center configurations. Our calculations imply that N$_\text{C}$V$_\text{Si}$ $hk$ and $kh$ configurations can be associated with PLX3 and PLX4 PL centers, respectively.
We also show the calculated ZFS parameters for the ground state of NV centers in 6H SiC (see Table~\ref{tab:ZFStable6H}). Interestingly, the $k_2k_2$ configuration has the largest $D$-constant which is followed by that of $hh$ and $k_1k_1$ configurations in descending order. This can be directly compared to the experimental assignments \cite{NVSiCour}. In the 6H polytype the $k_1k_2$ off-axial configuration has the largest $E$ value which is followed by that of $hk_1$ and $k_2h$ configurations in descending order. The PL spectrum of the NV centers in 6H SiC has not yet been reported.
\subsection{\label{subsec:discphoto}Photo-ionization of NV centers in 4H SiC}
The photo and spectral stability of solid state qubits is of high importance. The photostability of NV centers might be compromised by simultaneous photo-ionization. The photon ionization energies are given in Table~\ref{CTLtab}: the NV center in a negative charge state NV($-$) can be photo-ionized to neutral charge state NV($0$) by promoting an electron to the conduction band edge with an energy of about 1.7-1.8~eV. This energy is sufficient to re-ionize NV($0$) to NV($-$) by promoting an electron from the valence band to the in-gap defect level. In the case of single defect spectroscopy, confocal microscopy is applied with high excitation power that can result in two-photon absorption processes. According to our results, two-photon absorption process will also give rise to photoionisation. Unfortunately, the excitation energy of NV($0$) is not known and cannot be calculated by Kohn-Sham DFT. Therefore, the threshold energy to reionize NV($0$) to NV($-$) by two-photon absorption
is not known. If the ZPL energy of NV($0$) is higher than that of NV($-$) then NV($-$) should be excited into the phonon sideband, in order to re-ionize NV($0$) to NV($-$) by two-photon absorption.
Spectral diffusion in the emission of single NV($-$) center might occur upon photo-excitation when nearby defects are simultaneously excited resulting in fluctuating charges in the proximity of the NV($-$) center. These fluctuating charges can shift the ZPL energy of NV($-$) by small amounts which is detrimental for quantum applications. We showed above that divacancy defect will form together with NV centers under implantation or irradiation induced formation. Our calculations show, if divacancies are excited by about 1.1~eV two-photon absorption can also occur which leads to the ionization of the neutral divacancy [see Fig.~\ref{CTLfig}(c)]. The same photon energy is sufficient to excite NV($-$) in the phonon side band. Our calculations imply that excitation energy lower than 1.1eV should be applied, in order to avoid the photo-ionization of the divacancies.
We may conclude by emphasizing the two conditions for optimal readout process of NV qubits in 4H SiC, (i) efficient re-ionization of NV($-$) and (ii) conserving spectral stability by avoiding ionization of the divacancy by two-photon absorption.
\subsection{\label{subsec:polytypes}Comparison of SiC polytypes to host NV qubits}
We previously discussed the photo and spectral stability of NV qubits in 4H SiC in detail. Here, we further discuss this issue for NV qubits in 3C and 6H SiC.
In 3C SiC, the calculated acceptor level of NV defect lies at about 1.5~eV with respect to valence band edge, similarly to the NV defects in the hexagonal polytypes. Since the band gap of 3C SiC is smaller, 2.4~eV at low temperatures, the ZPL energy for the intra center transition and the ionization energy of NV($-$) almost coincide.
Thus excitation of NV center in 3C-SiC results in an excited state resonant with the conduction band edge. In this case the photostability of NV($-$) in 3C SiC may be difficult to maintain, particularly, at elevated temperatures. We note that the ensembles of NV centers could be efficiently optically spinpolarized in N-doped 3C SiC \cite{NVSiCour}. Nevertheless, this process might involve recapture of free electrons created by photoexcitation, that may not efficiently work at single defect level. On the other hand, if a spin-selective photo-ionization occurs for this defect then photo-ionization based detection of magnetic resonance (PDMR) could be the appropriate methodology \cite{PDMR} to read out the spin of single NV qubit.
The band gap of 6H SiC is about 0.25~eV lower than that of 4H SiC. The acceptor levels with respect to the valence band edges and excitation energies of the NV defects are very similar in the two polytypes. As a consequence, the two-photon absorption of NV($-$) defect in 6H SiC may be more effective than that in 4H SiC because of the larger density states in the conduction bands of 6H SiC. Thus, the relative rates of ionization and re-ionization of NV($-$) centers in 6H SiC may shift toward the ionization and compromise the stability of NV centers in 6H SiC. If NV($-$) is excited with an energy above the ZPL energy of divacancy then two-photon ionization of neutral divacancy can take place because of the relatively low-lying conduction band edge of 6H SiC. We conclude that our calculations imply that 4H SiC with the largest band gap is the optimal host for NV center qubit applications with optical read out.
\section{\label{sec:summary}Summary}
In summary, we carried out DFT calculations of N$_{\text{C}}$V$_{\text{Si}}$ defects in 3C, 4H and 6H SiC. We focused on the negatively charged N$_{\text{C}}$V$_{\text{Si}}$ defect, i.e., the NV center with potential qubit applications. We
discussed the formation of NV center in SiC, and found the divacancy inevitably forms when NV center is produced by implantation or irradiation. We calculated the ground state and excited state magneto-optical properties of the NV centers and compared them with the experimental magneto-optical data. We identified the individual NV qubits in hexagonal polytypes. We also discussed the photo-ionization and spectral stability of NV center in SiC. Our results show the importance of selective photoexcitation avoiding simultaneous excitation of divacancies in the two-photon absorption process.
\section*{\label{sec:ackn} Acknowledgments}
Andr\'as Cs\'or\'e acknowledges the Ministry of Human Resources for financial support in the framework of National Talent Program (NTP-NFT\"O-16). Adam Gali acknowledges Hungarian NKFIH grant No.~NVKP\_16-1-2016-0152958.
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Q: Trying to create script to port GPO [SOLVED]: When creating GPO programmatically, the gPCMachineExtensionNames property in the LDAP entry for the policy isn't set. We put some code in the script to export the key to a txt file and then reimported it at the client sites and everything went fine after that.
I created a script to create a GPO that we use during onboarding of new clients. Unfortunately, the way GPO is handled, some of the settings are saved in XML files. The XML file I'm battling with right now is for service settings. Unfortunately, this file isn't portable because of the UID in the NTService line. I believe this is equal to a UID for the user account that this service is set to run under (NetworkService, in this case).
I just want to be able to scoot the script and a directory with the XML files over to the client DC and execute the script (which will change out that UID so the XML files are valid). The script will create the GPO, insert the registry settings into the GPO, find the GPO GUID, copy the XML files from the directory to the GPO directory based on the GPO GUID it found.
\DomainController\SYSVOL\domain.local\Policies{XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX}\Machine\Preferences\Services
So far all of that works except the finding and replacing the UID bit. Unfortunately, that bit prevents it from applying the service settings to the computers at the client site.
<?xml version="1.0" encoding="utf-8"?>
<NTServices clsid="{2CFB484A-4E96-4b5d-A0B6-093D2F91E6AE}"><NTService clsid="{AB6F0B67-341F-4e51-92F9-005FBFBA1A43}" name="WinRM" image="2" changed="2018-06-13 20:23:16" uid="{XXXXXXXX-XXXX-XXXX-XXXX-XXXXXXXXXXXX}"><Properties startupType="AUTOMATIC" serviceName="WinRM" serviceAction="START" timeout="30"/>`</NTService>
</NTServices>
I've been at it for a couple days now so any help would be greatly appreciated.
A:
Unfortunately, this file isn't portable because of the UID in the NTService line. I believe this is equal to a UID for the user account that this service is set to run under (NetworkService, in this case).
According to the XML spec for group policy preferences, the uid value is not related to the user you're configuring the service to run as. From the link, it is:
A unique GUID generated when the preferences element is created and used to uniquely identify the preference item for tracing and reporting.
As a demonstration, try creating a new GPO via GPMC that contains two different service pref entries, both running as NETWORK SERVICE. Then examine the generated XML. The uid values will be completely different. Record the current values for reference later.
Modify the GPO and switch the accounts to LOCAL SERVICE. Then notice that both values have changed and are still different. Now modify the GPO again and set them back to NETWORK SERVICE. The values will have changed again and are not the same as the original values you recorded.
Unfortunately, I don't think the uid value is the root of your problem unless you've accidentally duplicated the values anywhere. That might screw things up if they're meant to be unique.
As a side note, I could've sworn there were more elegant ways to programmatically create a GPO that might be less problematic...possibly via libraries that come with GPMC.
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{"url":"https:\/\/socratic.org\/questions\/5a0ee47eb72cff5cddcbc838","text":"Question #bc838\n\nNov 17, 2017\n\nWe can't, because that is not the limit. If you intended to write ${\\lim}_{x \\to 10} 3 - \\frac{4 x}{5} = - 5$, then see explanation below.\n\nExplanation:\n\nOne of the first steps in checking the limit of a function at an x-value is seeing if the function itself is defined for that value, followed by whether it will be defined for nearby x-values.\n\nIn this case our function is:\n\n$\\frac{3 - 4 x}{5}$\n\nBecause this is a simple linear function, it is self evident that the function will be continuous throughout. Thus, the limit shall exist everywhere, and will be equal to the value $f \\left({x}_{o}\\right)$ for any ${x}_{0}$ in the domain.\n\nKnowing this, we plug in $x = 10$\n\n$\\frac{3 - 4 \\left(10\\right)}{5} = \\frac{3 - 40}{5} = - \\frac{37}{5} = - 7 \\frac{2}{5} = - 7.4$\n\nThis is not equal to the $- 5$ you sought, ergo, $- 5$ is not the appropriate limit.\n\nIt is possible that you mis-wrote the function, but for the function as you have written it, you cannot prove $- 5$ is the limit because it is not the limit.\n\nIf you instead meant to write:\n\n$3 - \\left(4 \\frac{x}{5}\\right)$\n\nYou can perform the same process as above. Again, since the function is linear and continuous throughout, the function will be equal to its own limit at any point.\n\n$f \\left(10\\right) = 3 - \\frac{4 \\left(10\\right)}{5} = 3 - \\frac{40}{5} = 3 - 8 = - 5$","date":"2019-05-22 09:33:27","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 11, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9092572927474976, \"perplexity\": 211.8881996089178}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-22\/segments\/1558232256778.29\/warc\/CC-MAIN-20190522083227-20190522105227-00414.warc.gz\"}"}
| null | null |
ERICKSON, WINDRAM IN COMPETITION CANDIDATES OFFER CONTRAST OF EXPERIENCE, 'NEW BLOOD'
Ramsey Campbell of The Sentinel StaffTHE ORLANDO SENTINEL
Experience versus "new blood" continues to be the theme of the race for the Republican nomination for the Lake County Commission District 4 race, which is the only local contest on Tuesday's primary ballot.
Incumbent Tom Windram, an 18-year veteran of the commission, is facing challenger Bob Erickson, a former pollution control staff member.
Windram was the leading vote-getter in the Sept. 2 primary with 3,889 votes to Erickson's 2,889. A third candidate, Tom Wetherington, was out of the running with 2,061 votes.
For both primary races, Windram has concentrated on his commission experience and his background in working on county problems. Erickson has focused on the need for a fresh face on the commission, which he said is in the doldrums.
Erickson has stressed the need for the county to be more energetic in drawing up long-range plans and in adopting a formal land-use plan. He said his campaign has prompted the county commission to take action by hiring additional planners.
"I think I've already done some good," he said.
During the past three weeks Erickson, 41, said he has had a chance to improve his name recognition around the county.
"We're facing some dynamic changes," he said. "We need to get off center."
Calling himself an environmentalist, Erickson said his background in pollution control regulations would be an asset.
But Windram, 56, said voters need only compare himself and his opponent. Windram originally won his commission seat as an environmentalist.
During his tenure, the county adopted a controversial ban on phosphates in detergents, which has since been recognized nationally. The county also began prohibiting municipalities and industry from dumping effluents into area lakes.
Windram also was involved in creating the current countywide rural fire system and has been viewed as a diplomat in dealing with thorny issues before the commission.
Windram said it would be easy for the county to draw up a land-use plan, but difficult to implement fairly. He said too many people expect simple answers to complex problems.
Windram said voters need to decide CONTRAST, 15
which candidate would be better at handling issues as diverse as the environment, road construction and management of a $50 million county budget. "I think if people will look at what we've done, not just promises, I'll come out all right. I feel good about it," he said.
Both candidates said the key to the election will be who has the most supporters in what is likely to be a low turnout Tuesday.
The winner of Tuesday's race will meet Mount Dora Councilwoman Margaret "Peggy" Curtis, a Democrat, in the November 4 general election.
The county commission seat pays $23,452 a year. Commissioners serve four- year terms.
|
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\section*{Acknowledgement}
We thank Xiangbing Huang, Xudong Sun, Sam Cheng, Jack Chen, and Darko Marinov
for discussions.
The research was mainly conducted when Zhang was a visiting student at
UIUC, supported by China Scholarship Council.
Zhang, He, Li, and Dong were supported in part of National Key R\&D Program of
China No. 2017YFB1001802; NSFC No. 61872373 and 61872375.
Xu was supported in part of NSF 1816615.
\section*{Appendix B: Replication Package}
We release our research artifacts of this paper at:
\begin{center}
\fbox{\url{https://github.com/xlab-uiuc/open-cevo}}
\end{center}
The artifacts include:
\begin{itemize}
\item the script code for collecting commits and JIRA issues;
\item the annotated dataset of each categories (interface, usage, and document);
\item documents for running the code and navigating the data.
\end{itemize}
We believe our study can be reproduced by different teams based on the taxonomy
described in
Figure~\ref{sec:overview} and Table~\ref{tab:taxonomy}, together with the
methodology described in Section~\ref{sec:meth}.
\vspace{5pt}
\noindent
{\bf Author experience.} All the authors have been working on software configuration
research for several years, ranging from two to nine years.
The authors have a good understanding of the four systems under study (HDFS, HBase,
Spark, and Cassandra)---they have used those
studied systems as subjects of evaluation in their prior research.
We expect that a similar level of experiences and expertises are needed for a team to
reproduce the analysis (mainly categorization), including the understanding
of the designs of the systems under study and the implementations
in Java/Scala programming languages to understand
the evolution. We believe a fair understanding of software configuration
design and implementation is required, too.
\vspace{5pt}
\noindent
{\bf Heuristics for commit identification and analysis.} The heuristics are
described in Section~\ref{sec:diff:analysis}. The script code that implements the heuristics
can be found at:
\begin{center}
\fbox{\url{https://github.com/xlab-uiuc/open-cevo/tree/main/code}}
\end{center}
We documented how to run the code and the basic regular expressions used at the \texttt{README.md} file.
\vspace{5pt}
\noindent
{\bf Identification/categorization of the rationales for parameterization.}
We identify and categorize the rationales for parameterization based on the descriptions
(from the reporter) and the discussions (among the developers) documented in the JIRA issues.
We developed our categories bottom-up instead of imposing a taxonomy {\it ex ante}.
Specifically, we list all the rationales/reasons and then summarizing them into categories.
We do not claim completeness of the categories.
We present an example from Spark, in which
the commit (\href{https://github.com/apache/spark/pull/21705/commits/8ee8e00f1}{link}) changed a constant value \texttt{100}
to a configuration parameter, \texttt{spark.sql.codegen.cache.maxEntries};
the commit is associated with the JIRA issue,~\href{https://issues.apache.org/jira/browse/SPARK-24727}{SPARK-24727}.
Quoting the JIRA issue description,
\begin{quote}
``{\it The cache 100 in CodeGenerator is too small for realtime streaming calculation, although is ok for offline calculation.
Because realtime streaming calculation is mostly more complex in one driver, and performance sensitive.
I suggest spark support config for user with default 100, such as spark.codegen.cache=1000.}''
\end{quote}
Based on the description, we conclude that the rationale behind the parameterization is that the constant value
cannot meet the performance requirement of real-time application.
So, it is categorized as ``performance'' in Table~\ref{tab:purpose_constant2config}.
\vspace{5pt}
\noindent
{\bf Identification on how developers find constants to parameterize.}
It is straightforward to identify the constants that were parameterized in a selected commit.
The following commit (\href{https://github.com/apache/spark/commit/565e7a8d}{link}) illustrates this point; the commit comes
from ~\href{https://issues.apache.org/jira/browse/SPARK-20950}{SPARK-20950}.
We can see that the constant (\texttt{1024*1024}) was parameterized---the value
was replaced by a program variable
\texttt{diskWriteBufferSize} read from a configuration parameter.
\lstinputlisting[language=diff]{./figures/parameterization.java}
For each commit that parameterized constants, we identify the intent of the parameterization,
i.e., ``how developers find constants to parameterize.''
Similar to the identification/categorization of the rationales for parameterization,
we identify and categorize the reasons for how developers find constants to parameterize
based on the descriptions (from the reporter) and the discussions (among the developers)
documented in the JIRA issues. The process is identical---we list all the rationales/reasons and then summarizing them into categories.
We do not claim completeness of the categories.
We present an example from Spark, in which the commit (\href{https://github.com/apache/spark/commit/126310ca}{link})
parameterized the thread number of broadcast-exchange thread pool from a constant \texttt{128}.
Based on the associated JIRA issue, \href{https://issues.apache.org/jira/browse/SPARK-26601}{SPARK-26601}:
\begin{quote}
``{\it Currently, thread number of
broadcast-exchange thread pool is fixed and keepAliveSeconds is also fixed as 60s. But some times,
if the thread object do not GC quickly it may caused server(driver) OOM. In such case, we need to make this thread pool configurable.}''
\end{quote}
Therefore, we conclude that the parameterization is to avoid OOMs, a type of ``failures'' (Section~\ref{sec:parameterization}).
\vspace{5pt}
\noindent
{\bf Identification/categorization of the reasons for changing default values.}
Similar to the identification/categorization of the rationales for parameterization,
we identify and categorize the reasons for changing default values
based on the descriptions (from the reporter) and the discussions (among the developers)
documented in the JIRA issues. The process is identical---we list all the
rationales/reasons and then summarizing them into categories.
We do not claim completeness of the categories.
We present an example from HDFS. The commit (\href{https://github.com/apache/hadoop/commit/afb42aeab}{link})
changed the default value of \texttt{dfs.namenode.edits} \texttt{.asynclogging} from \texttt{off} to \texttt{on}.
The commit is associated with the JIRA issue,~\href{https://issues.apache.org/jira/browse/HDFS-12603}{HDFS-12603},
which in turn was linked to~\href{https://issues.apache.org/jira/browse/HDFS-7964}{HDFS-7964}.
Quoting the discussion:
\begin{quote}
``{\it It was off by default due to concerns about correctness.
We have been running it in production for quite a while with no issues so far.}''
\end{quote}
We conclude that the commit enabled a previously-disabled feature flag (Section~\ref{sec:default_value}).
\vspace{5pt}
\noindent
{\bf Definitions of four levels of messages and feedback quality.}
The definitions are developed based on how the messages are changed, i.e.,
how developers improved the original messages.
Specifically, we summarized what additional information is added to the original
message, which can be categorized into the four levels.
The following is a commit (\href{https://github.com/apache/hbase/commit/4b84ab32}{link})
that changed the message which improved the message quality from L3 to L4
defined in Table~\ref{tab:msg_quality}.
\lstinputlisting[language=diff]{./figures/feedbackMSG.java}
\vspace{5pt}
\noindent
{\bf Handling commits that change multiple categories.}
One commit could revise multiple categories in our taxonomy (Figure~\ref{sec:overview} and Table~\ref{tab:taxonomy}).
In total, there are 26 commits that revise multiple categories.
For those commits, we study them in each category independently.
For example, the following commit (\href{https://github.com/apache/spark/commit/10248758}{link})
changes the default value of \texttt{spark.master.rest.enabled};
meanwhile, it also adds the checking code for \texttt{SPARK\_AUTH\_SECRET\_CONF}.
\lstinputlisting[language=diff]{./figures/countCase2.java}
Therefore, we study this commit in both Mod$\rm_{\text{DefaultValue}}$
(Section~\ref{sec:default_value}) and Check (Section~\ref{sec:usage:check}).
\section{Taxonomy}
\label{sec:background}
\begin{figure}
\centering
\includegraphics[width=0.495\textwidth]{overview_cevo}
\caption{Three parts of our taxonomy of software configuration design and implementation,
and their components.}
\label{sec:overview}
\end{figure}
Figure~\ref{sec:overview} shows the three parts of our taxonomy of
cloud-system configuration engineering: {\it interface}, {\it usage}, and {\it documentation}.
We focus on aspects
that affect how system users interact
with configurations, and not on developer-focused aspects
like variability and testability.
We organize our study along the categories\Space{ of changes in each of
three part, as} shown in Table~\ref{tab:taxonomy}.
\vspace{5pt}
\mypara{Interface.}The configuration interface
that a system exposes to
users consists primarily of {\it configuration parameters} (parameters, for short).
As shown in Fig.~\ref{sec:overview}, a parameter is identified
by a {\it name} and it typically has a {\it default value}.
Users can customize system configuration
by changing parameter values
in configuration files or by using
command line interfaces (CLIs).
Each parameter places {\it constraints} (correctness rules)
on its values,
e.g., type, range, dependency on other parameters.
Values that violate the constraints lead to misconfigurations.
\Space{As shown }In Table~\ref{tab:taxonomy}, changes that contribute to configuration interface evolution include adding\Space{ new} parameters, removing\Space{ existing}
parameters, and modifying\Space{ existing} parameters.
\vspace{5pt}
\mypara{Usage.}Fig.~\ref{sec:overview} presents the configuration usage model.
To use
a parameter, the software program first
reads its value from a configuration file or CLI,
{\it parses} the value and stores it in a program variable.
The variable is then used when the program executes.
In principle, the program {\it checks} the value against the parameter's constraints
before {\it using} it.
If checks fail, the program needs to {\it handle}
the error and provide user with {\it feedback messages}.
In Table~\ref{tab:taxonomy}, configuration usage evolution consists of
changes to all parts of the usage model.
\vspace{5pt}
\mypara{Documentation.}These are natural language descriptions
related to configurations. We consider changes to user manuals and\Space{
source} code comments---the former are written for system users
while the latter are written for developers.
\begin{table}[]
\footnotesize
\caption{Our taxonomy of configuration engineering evolution.}
\vspace{-2.5pt}
\setlength{\tabcolsep}{2pt}
\begin{tabular}{ll}
\toprule
\rowcolor{VeryLightGray}\multicolumn{2}{c}{\header{INTERFACE (Section~\ref{sec:interface:evolution}})} \\
{\bf AddParam} & {\bf Add new configuration parameters} \\
\quad Add$\rm_{\text{NewCode}}$ & \quad Add new parameters when introducing new modules \\
\quad Add$\rm_{\text{CodeChange}}$ & \quad Add new parameters due to changes in existing code \\
\quad Add$\rm_{\text{Parameterization}}$ & \quad Convert constant values to configuration parameters \\
{\bf RemoveParam} & {\bf Remove existing configuration parameters} \\
\quad Rmv$\rm_{\text{RmvModule}}$ & \quad Remove parameters when removing existing modules \\
\quad Rmv$\rm_{\text{Replace}}$ & \quad Replace parameters with constants or automation \\
{\bf ModifyParam} & {\bf Modify existing configuration parameters} \\
\quad Mod$\rm_{\text{Naming}}$ & \quad Change the name of a configuration parameter \\
\quad Mod$\rm_{\text{DefaultValue}}$ & \quad Change the default value of a configuration parameter \\
\quad Mod$\rm_{\text{Constraint}}$ & \quad Change the constraints of a configuration parameter \\
\midrule
\rowcolor{VeryLightGray}\multicolumn{2}{c}{\header{USAGE (Section~\ref{sec:behavior:evolution}})} \\
{\bf Parse} & \textbf{Change configuration parsing code} \\
{\bf Check} & \textbf{Change configuration checking code} \\
{\bf Handle} & \textbf{Change configuration error handling code} \\
\quad Handle$\rm_{\text{Action}}$ & \quad Handling actions (correction and exceptions) \\
\quad Handle$\rm_{\text{Message}}$ & \quad Feedback messages (log and exception messages) \\
{\bf Use} & \textbf{Change how configuration values are used} \\
\quad Use$\rm_{\text{Change}}$ & \quad Change existing code that uses parameters \\
\quad Use$\rm_{\text{Add}}$ & \quad Add code to reuse a configuration parameter \\
\midrule
\rowcolor{VeryLightGray}\multicolumn{2}{c}{\header{DOCUMENTATION (Section~\ref{sec:document}})} \\
{\bf User manual} & \textbf{Change configuration-related user manual content} \\
{\bf Code comment} & \textbf{Change configuration-related source code comments} \\
\bottomrule
\end{tabular}
\label{tab:taxonomy}
\vspace{15pt}
\end{table}
\section{Configuration Usage Evolution}
\label{sec:behavior:evolution}
\begin{table}
\footnotesize
\setlength{\tabcolsep}{4pt}
\caption{Statistics on configuration usage evolution.}
\begin{tabular}{lccccc}
\toprule
& \headerHDFS & \headerHBase & \headerSpark & \headerCassandra & \header{total} \\
\midrule
{\bf Parse} & \checked{5} & \checked{14} & \checked{59} & \checked{7} & \checked{85} \\
\midrule
{\bf Check} & \checked{7} & \checked{20} & \checked{29} & \checked{11} & \checked{67} \\
\midrule
{\bf Handle} & \checked{12} & \checked{18} & \checked{20} & \checked{2} & \checked{52} \\
\quad Handle$_\text{Action}$ & \checked{8} & \checked{6} & \checked{4} & \checked{1} & \checked{19} \\
\quad Handle$_\text{Message}$ & \checked{4} & \checked{12} & \checked{16} & \checked{1} & \checked{33} \\
\midrule
{\bf Use} & \checked{34} & \checked{35} & \checked{74} & \checked{12} & \checked{155} \\
\quad Use$_\text{Change}$ & \checked{7} & \checked{10} & \checked{25} & \checked{3} & \checked{45} \\
\quad Use$_\text{Add}$ & \checked{27} & \checked{25} & \checked{49} & \checked{9} & \checked{110} \\
\bottomrule
\end{tabular}
\label{tab:behavior}
\end{table}
\begin{table*}
\begin{center}
\footnotesize
\centering
\caption{Examples of consequences that can be prevented by adding configuration checking code.}
\begin{tabular}{lll}
\toprule
\header{Consequence} & \header{Example Parameter} & \header{Description} \\
\toprule
Runtime Error & {\scriptsize hbase.bucketcache.bucket.sizes} & If value is not aligned with 256, instantiating a bucket cache throws IOException~(\href{https://issues.apache.org/jira/browse/HBASE-16993}{HBase-16993}) \\
\midrule
Early Termination & {\scriptsize commitlog\_segment\_size\_in\_mb} & If value $\ge$ 2048, Cassandra throws an exception when creating commit logs~(\href{https://issues.apache.org/jira/browse/CASSANDRA-13565}{Cassandra-13565}) \\
\midrule
Service unavailability & {\scriptsize spark.dynamicAllocation.enabled} & Running barrier stage with dynamic resource allocation may cause deadlocks~(\href{https://issues.apache.org/jira/browse/SPARK-24954}{Spark-24954}) \\
\midrule
Unexpected Results & {\scriptsize spark.sql.shuffle.partitions} & If the value is 0, the result of a table join will be an empty table~(\href{https://issues.apache.org/jira/browse/SPARK-24783}{Spark-24783}) \\
\bottomrule
\end{tabular}
\label{tab:conseq_check}
\end{center}
\end{table*}
We present results on configuration usage evolution (recall the configuration usage model described in Fig.~\ref{sec:overview} and Section~\ref{sec:background}).
\Comment{To use a configuration parameter, the software program
{\it parses} its value and stored it in a program variable.
The value is then {\it checked} before {\it using} it during
execution. If checks fail, the program {\it handles}
the configuration error and provides user {\it feedback} by either (1)~throwing an exception,
or (2)~logging and correcting the error then resuming execution.}
Across the four cloud systems,
\checked{26.2\%--36.8\%} of commits changed parameter usage (Table~\ref{tab:behavior}).
We describe changes to checking, error handing, and
use of parameters.
We omit changes to parsing APIs (e.g.,~\href{https://github.com/apache/spark/commit/dad2d826}{Spark-23207}).
\subsection{Evolution of Parameter Checking Code}
\label{sec:usage:check}
Proactively checking parameter values is key to preventing\Space{ runtime failures
and performance anomalies caused by} misconfigurations~\cite{xu:16}.
However, we find that many parameters had no \emph{checking code} when they were introduced.
Checking code was added {\it reactively}\Space{ in the aftermath}:
1)~\checked{74.6\% (50/67)} of commits that changed checking code occurred after users reported
runtime failures, service unavailability,
incorrect/unexpected results, startup failures, etc.
(Table~\ref{tab:conseq_check} shows examples).
2)~\checked{In 14.9\% (10/67)} commits, developers {\it proactively}
added or improved the checking code; 2 of them\Space{ added checks
for six parameters by} applied reactively-added
checking code to other parameters with similar types (e.g.,~\href{https://issues.apache.org/jira/browse/CASSANDRA-13622}{Cassandra-13622}).
3)~We did not find sufficient information of the other \checked{7 commits}.
\begin{figure}
\begin{minipage}{0.97\linewidth}
\centerline{\lstinputlisting[language=diff]{./figures/simpleCheck.java}}
\vspace{3.5pt}
\centerline{\small (a) Add a \CodeIn{NOT-NULL} check (\href{https://github.com/apache/hbase/commit/293cb87d52d6ccc98f0d03387f9dc07dc4522042}{HBase-18161})}
\vspace{3.5pt}
\end{minipage}
\begin{minipage}{0.97\linewidth}
\centerline{\lstinputlisting[language=diff]{./figures/semanticCheck.java}}
\vspace{3.5pt}
\centerline{\small (b) Add a semantic check (\href{https://github.com/apache/hbase/commit/bc93b6610b349d38502290af27da0ae0b5fd4936}{HBase-16993})}
\vspace{5.5pt}
\end{minipage}
\begin{minipage}{0.97\linewidth}
\centerline{\lstinputlisting[language=diff]{./figures/controlDependency.java}}
\vspace{3.5pt}
\centerline{\small (c) Add a check for parameters dependency (\href{https://github.com/apache/spark/commit/10248758438b9ff57f5669a324a716c8c6c8f17b}{Spark-25088})}
\vspace{5.5pt}
\end{minipage}
\begin{minipage}{0.97\linewidth}
\centerline{\lstinputlisting[language=diff]{./figures/runtimeCheck.java}}
\vspace{3.5pt}
\centerline{\small (d) Add a check for execution context (\href{https://github.com/apache/spark/commit/92b48842b944a3e430472294cdc3c481bad6b804}{Spark-24954})}
\end{minipage}
\caption{Examples of configuration checking code.}
\label{fig:code:CheckingCode}
\end{figure}
\subsubsection{Adding new checking code}
\label{sec:add:checking:code}
\checked{79} new checks were added in {\checked{83.6\% (56/67)}} of checking-code related commits.
\checked{87.3\% (69/79)} of these new checks were for specific parameters, while
the others were applied to groups of configuration parameters (e.g., read-only parameters).
Surprisingly, for specific parameter checks (\checked{69 checks in 46 commits}), 58.0\% (40/69) were\Space{ simple and} basic checks:
\CodeIn{NOT-NULL} \checked{(20/69)}, value range \checked{(15/69)} and deprecation checks \checked{(5/69)}.
An example\Space{ of such checks} is\Space{ shown} in Fig.~\ref{fig:code:CheckingCode}(a).
Majority of new checking code were added reactively,
corroborating that simple checks can
prevent many severe failures~\cite{xu:16,yuan:14}. More of such checks could be
automatically added and invoked at system startup.
The other \checked{29} checks were more complex: \checked{9} value semantic checks
(e.g., file/URI properties and data alignment, Fig.~\ref{fig:code:CheckingCode}(b)),
2)~\checked{13} checks for parameter dependencies (e.g., Fig.~\ref{fig:code:CheckingCode}(c)),
and 3)~\checked{7} checks for execution context (e.g., Fig.~\ref{fig:code:CheckingCode}(d)).
\Comment{
\yuanliang{the distribution is actually average}
\haochen{We find checks for time-related (10/69) and path-related (11/69) parameters occupy the highest proportion}
}
\subsubsection{Improving existing checking code}
\checked{11} commits improved existing checking code:
\checked{eight}\Space{ of them} made checks more strict, e.g.,
a \CodeIn{NOT-NULL} check was improved to ``{\it only allow\Space{s} table replication
for sync replication}''~(\href{https://issues.apache.org/jira/browse/HBASE-19935}{HBase-19935}), and
\checked{three} moved checking code to be invoked earlier instead
of during subsequent execution, e.g., ``{\it \Space{when using dynamic
executor allocation, if we set spark.executor.cores smaller than
spark.task.cpus, an exception will be thrown. But, if dynamic executor
allocation not enabled, Spark{} will hang ...}\Space{ when submit new job as
TaskSchedulerImpl will not schedule a task in an executor which
available cores is small than spark.task.cpus. So, }when starting task
scheduler, spark.task.cpus should be checked}''~(\href{https://issues.apache.org/jira/browse/SPARK-27192}{Spark-27192}).
\Space{We believe that checks should be performed earlier rather than later.}
\begin{shadedbox}
{\header{Discussion:}} Checks for parameter values
are often added as afterthoughts. Proactively generating checking code
can help prevent failures due to misconfigurations.
\end{shadedbox}
Two possible directions are automatically
learning checking code (we find that newly-added checking code is
often simple) and automatically applying checking code for one
parameter to other parameters both in the same software (which
developers are already doing manually) and across software
projects.
A direction is to co-learning checking code and usage code.
Techniques for extracting complex constraints
and specifications can reduce
manual effort for reasoning about and implementing checking code.
A few recent works show promise for inferring parameter constraints through analysis
of source code and documentation~\cite{xu:13,chen:2020,xiang:2020,Li:2020}.
Techniques for extracting feature constraints could be extended and applied
to runtime configurations~\cite{NadiTSE2015,nadi:14,kang:90,SheICSE11}.
\Comment{\haochen{(\it HHC: I inspected if there are data that can support this impl., but I fond this is a bit ``semantic'', depending on context. And some checks just can not move to startup stage.)}}
\begin{table*}[]
\begin{center}
\footnotesize
\caption{\small Four levels of message feedback quality in commits that changed exception or logging messages.}
\begin{tabular}{lll}
\toprule
\header{Level} & \header{Description} & \header{Example} \\
\midrule
\multirow{2}{*}{L4} & Contain parameter names and & ``Barrier execution mode does not support dynamic resource allocation... You can disable dynamic \\
& provide guidance for fixing & resource allocation by setting\Space{ Spark{} conf}...\ttvar{spark.dynamicAllocation.enabled} to false.''~(\href{https://github.com/apache/spark/commit/92b4884}{Spark-24954}) \\
\midrule
L3 & Contain parameter names & ``Failed to create SSL context using \ttvar{server_encryption_options}.''~(\href{https://github.com/apache/cassandra/commit/16ef9ac}{Cassandra-14991}) \\
\midrule
L2 & Do not contain parameter names & ``This is commonly a result of insufficient YARN configuration.''~(\href{https://github.com/apache/hbase/commit/2773510f}{HBase-18679}) \\
\midrule
L1 & No mention of configuration & ``Could not modify concurrent moves thread count.''~(\href{https://issues.apache.org/jira/browse/HDFS-14258}{HDFS-14258}) \\
\bottomrule
\end{tabular}
\label{tab:msg_quality}
\end{center}
\end{table*}
\Comment{@Owolabi, I feel that misconfiguration correction is a direction to go.}
\subsection{Evolution of Error-Handling Code}
\label{sec:feedback:evolution}
We discuss changes to misconfiguration-related exception-handling code
and to messages that provide user feedback.
\subsubsection{Changes to configuration error handling}
\checked{19} commits dealt with error handling: \checked{10}
added new handling code to \CodeIn{try-catch} blocks
or \CodeIn{throw} new exceptions; 9 commits changed handling code.
Among the 9 commits,
(1)~\checked{four} changed misconfiguration-correction code: three of these
added logic to handle a misconfiguration, e.g.
``{\it if secret file specified in httpfs.authentication.signature.secret.file does not exist,
random secret is generated}''~(\href{https://issues.apache.org/jira/browse/HDFS-13654}{HDFS-13654}) and one\Space{, another commit do the opposite thing that developers} changed buggy
misconfiguration-correction code to simply log errors~(\href{https://issues.apache.org/jira/browse/HDFS-14193}{HDFS-14193})
(showing that auto-correcting misconfigurations is not always easy),
(2)~\checked{three} changed the exception type as it was
``{\it dangerous to throw statements whose exception class does not
accurately describe why they are thrown...since it makes correctly handling
them challenging}''~(\href{https://issues.apache.org/jira/browse/HDFS-14486}{HDFS-14486}), and
(3)~\checked{two} replaced exception throwing with logging the errors and resuming the execution.
\Space{Besides the commits that changed error handling,}
We also studied the newly added handling code in the \checked{79} commits
that added new checking code in Section~\ref{sec:add:checking:code}\Comment{---what is the handling
code when the newly-added checks fail?}.
In \checked{73.4\% (58/79)} of the cases,
the handling code threw runtime exceptions\Space{ with message} or logged error messages.
The expectation is that users should
handle the errors.
In the remaining \checked{26.6\% (21/79)} cases, developers attempted to correct
the misconfigurations, e.g.,
``{\it it's developers' responsibility
to make sure the configuration don't break code.}'' (\href{https://github.com/apache/spark/pull/21601}{Spark-24610}).
Developers corrected misconfigurations by changing to the
closest value in the valid range (\checked{11/21}),
reverting to the default value (\checked{3/21}), and
using the value of another parameter with similar semantics (\checked{7/21}).
\begin{shadedbox}
{\header{Discussion:}} \Space{We found that }Developers want to make code
more robust in the presence of misconfigurations, but their manual
efforts are often ad hoc. There is need for new techniques for
generating misconfiguration correction
code and improving existing handling code.
\end{shadedbox}
Techniques for fixing compile-time configuration errors,
such as range fixes~\cite{xiong:12,xiong:15}, may be applicable
for generating correction strategies for some types of runtime parameters.
A key challenge is to attribute runtime errors (e.g., exceptions) to
misconfigurations and to rerun the related execution with the corrected configurations.
\subsubsection{Changes to feedback messages}
Feedback (error log or exception) messages
are important for users to
diagnose and repair misconfigurations\Space{ or to handle problems by tuning parameter values}.
We investigated commits that modified feedback messages and categorize the level of feedback
that they provided in Table~\ref{tab:msg_quality}, where
L4 messages provide the highest-quality feedback and L1 messages provide the lowest-quality feedback.
Among \checked{33} commits that modified messages, \checked{18}
enhanced feedback quality by adding configuration-specific information.
After enhancement, \checked{8} messages became \CodeIn{L3}, and \checked{7} became L4.
Changes in the other 15 commits improved
1)~correctness (\checked{9/15})---half changed imprecise parameter
boundary values, e.g., from ``no less'' to ``greater'' (\href{https://issues.apache.org/jira/browse/SPARK-26564}{Spark-26564}),
2)~readability (\checked{3/15}), such as fixing typographic errors\Space{ and rewriting sentences},
3)~the log level (\checked{2/15})\Space{ of the commits, 4/16 of the messages)}, and
4)~security (\checked{1/15}), i.e., removing potentially sensitive value.
\Comment{
How many of the 32 commits are about exception messages and how
many are about logging during error correction? Also, why do we have 32 commits in this section but 34 in Table~\ref{tab:behavior}?
\needCheck{25} of the \needCheck{30} commits\Space{, 26/31 of the messages} involved exception messages\Space{ when problem happen. (e.g.
``{\it spark.network.timeout must be greater than the value of spark.executor.heartbeatInterval}'' ~\href{https://issues.apache.org/jira/browse/SPARK-26564}{- Spark-26564})};\Space{ the other}
\needCheck{5}\Space{ of the 30 commits}\Space{, 5/31 of the messages are for} involved error-correction log messages.\Space{, e.g.(``{\it No HREGION\_COLUMNFAMILY\_FLUSH\_SIZE\_LOWER\_BOUND
set in table descriptor, using region.getMemStoreFlushSize of families instead.}'' ~\href{https://issues.apache.org/jira/browse/HBASE-19647}{- HBase-19647})}
}
\begin{shadedbox}
{\bf \header{Discussion:}} Future work could study the
feedback level in \emph{all} messages related to misconfiguration handling code.
If most messages are not L4, then future work should
automatically detect deficient messages and
automatically enhance them to L4.
\end{shadedbox}
Moreover, configuration-related logging is not as mature as
logging for\Space{ regular} debugging~\cite{yuan:12, Yuan_icse12, asplos-logenhancer,
yuan:10, Ding:15, Fu:15}. Improving configuration-related logging requires
logging related parameters, erroneous
values, and, where feasible, possible fixes.\Space{ Such future work is
needed:} Poor-quality feedback from tools
hinders developers~\cite{JohnsonEtAlICSE2013} and
techniques exist for dealing with
message errors in other domains~\cite{zhendong-su-finding-and-analyzing-compiler-warning-defects, BarikEtAlCompilerErrorsICSE14}.
\Comment{
\yuanliang{
Here, I want to discuss about the Configurator tool in SPL field.
(not sure to put it here or in related work)
Actually 5.1 and 5.2 are about misconfiguration checking and
handling. One reviewer mentions that the configurator tool can
already solve the issues we found.
Kconfig and CDL (Component Definition Language) are equipped with GUI-based configurators,
Each of the two configurators takes a different approach to ensure that the user retains a valid configuration.
The Kconfig configurator prevents the user from modifications that violate constraints;
the CDL configurator allows such modifications, but it detects violations and helps in resolving them.
However, from our discussion,
configurator tool can't solve the issue is our study for the following reasons:
1. Both configurator tool works for build-time configuration, so the invoke
time and place is very clear. However, for runtime configuration, when and where
to add the checks are not easy to figure out. Some basic check can bee added in the parsing
API (e.g. type check). However, for some complex check that have runtime constraint, it is hard
to add them uniformly at start-up phase.
2. The constraints for runtime configuration is more complex. Majority of build time configuration are
feature selection while runtime configuration has larger value space. So the correction for
runtime configuration is difficult. This is also declared in the paper~\cite{Berger} ``{\it The resolution
is incomplete as The inference rules are incomplete. For example, the engine has rules for
handling cardinality constraints on interfaces of 0 or 1, but not for arbitrary bounds. }''
3. The feedback message in configurator tool just need to tell users how to build the system. However
the feedback message for runtime configuration need to help users debugging and diagnosis runtime
failure and fix the problem, which need more informative guidance. That's why we use 4 levels to qualify
a feedback massage for runtime configuration.
}
}
\subsection{Evolution of Parameter Value Usage}
\Space{The way parameters and their values are used evolve over
time. }Software developers change how existing
parameters are used (``Use$_\text{Change}$'' in Table~\ref{tab:behavior})
and reuse existing parameters for different purposes
(``Use$_\text{Add}$'' in Table~\ref{tab:behavior}).
\Comment{This section also falls in short -- there are two aspects to look
into: (1) do developers change existing uses (i.e., a parameter used
in one way but changed into another)?
(2) do developers reuse an existing parameter (i.e., add a new use)
which is currently covered.}
\subsubsection{Changing how existing parameters are used}
\Comment{Actually ``use'' doesn't have a clear scope here.
So I will write what I have. Please help me to justify my mistakes.}
\Comment{I leave it to @Owolabi}
\checked{45 commits} changed parameter usage for
the following reasons:
\vspace{5pt}
\mypara{Fine-grained control.} In \checked{12/45} commits,\Space{14/41 params of the cases. The} developers previously used one parameter for multiple purposes,
due to poor design---``e.g., {\it CompactionChecker and PeriodicMemStoreFlusher
execution period are bound together\Space{, both use \CodeIn{hbase.server.thread.wakefrequency}}}'' (\href{https://issues.apache.org/jira/browse/HBASE-22596}{HBase-22596})---or
for reuse---e.g., ``{\it \Space{spark.sql.execution.}arrow.enabled was added...\Space{ when we added} with PySpark\Space{ arrow optimization}...
Later, SparkR...\Space{ arrow optimization} was added...\Space{ and it's} controlled by the same parameter. Suppose users want to share some JVM between PySpark and SparkR...
They\Space{ are currently forced to} use the optimization for all or none.}'' (\href{https://issues.apache.org/jira/browse/SPARK-27834}{Spark-27834}).
Developers resolved both categories by creating separate parameters\Space{ (sometime will leave the original one) and} for fine-grained control.
\vspace{5pt}
\mypara{Domain/scope.} \checked{8/45} commits changed the
usage domain or scope of a parameter\Space{ (\needCheck{9/41 params})}.
For example, HDFS{} developers changed a\Space{ the \ttvar{dfs.namenode.decommission.interval}} parameter, which was previously only used in the decommission phase to also be used in the maintenance phase, so ``{\it lots of code can be shared\Space{ between the\Space{ existing}
decommission functionality and to-be-added maintenance state support for datanodes}}'' (\href{https://issues.apache.org/jira/browse/HDFS-9388}{HDFS-9388}).
\Comment{So . This param is used in DecommissionManager and
now this becomes DatanodeAdminManager. Both decommission and maintenance
will use this manager.}
\vspace{5pt}
\mypara{Parameter overriding}\checked{9/45} commits changed parameter override priority\Space{(\needCheck{7/41 params})}, e.g., ``{\it We need to support both table-level parameters. Users might also use session-level parameter\Space{ \CodeIn{\seqsplit{spark.sql.parquet.compression.codec}}}...\Space{
The priority rule will be like If other compression codec configuration was found
through hive or parquet,} the precedence would be...\Space{ \CodeIn{compression}, \CodeIn{\seqsplit{parquet.compression}},
\CodeIn{\seqsplit{spark.sql.parquet.compression.codec}}}}'' (\href{https://issues.apache.org/jira/browse/SPARK-21786}{Spark-21786}).
\vspace{5pt}
\mypara{Semantics}\checked{6/45} commits
changed what a parameter is used for\Space{(\needCheck{5/41 params})}, e.g., in
\href{https://issues.apache.org/jira/browse/SPARK-21871}{Spark-21871}, developers started using
\CodeIn{\seqsplit{spark.sql.codegen.hugeMethodLimit}} as the maximum\Space{
bytecode} compiled function size instead of
\CodeIn{spark.sql.codegen.maxLinesPerFunction}.
\vspace{5pt}
\mypara{Parameter replacement}\checked{6/45} commits
swapped one parameter for another
because the previous one was outdated or
wrong \Space{(\needCheck{4/41 params})},
e.g., in \href{https://issues.apache.org/jira/browse/SPARK-24367}{Spark-24367},
a use of \CodeIn{\seqsplit{parquet.enable.summary-metadata}} was replaced with
a use of \CodeIn{\seqsplit{parquet.summary.metadata.level}} because the former was deprecated.
\vspace{5pt}
\mypara{Buggy parameter values}\checked{4/45}
commits changed parameter values that were buggy\Space{ (\needCheck{2/41 params})},
e.g., the value of a parameter\Space{ \CodeIn{\seqsplit{spark.ui.filters}}}
changed because, ``{\it user specified filters are not applied in YARN
mode\Space{, as it is overridden by the yarn AmIp filter. So}...we need...\Space{ to append} user provided filters\Space{ with yarn filter}}'' (\href{https://issues.apache.org/jira/browse/SPARK-26255}{Spark-26255}).
\Comment{
using \ttvar{spark.sql.codegen.maxLinesPerFunction} to restrict the maximum lines of a single Java
function generated by whole-stage codegen. Developers now use \ttvar{spark.sql.codegen.hugeMethodLimit}
to restrict The maximum bytecode size of a single compiled Java function generated by whole-stage codegen.
However. Both of those tow cases change the parameter's name after changing its functional usage.
}
\subsubsection{Reusing existing parameters}
\label{sec:reuse:parameters}
To avoid growing the configuration space unnecessarily, developers
sometimes reuse existing parameters that are similar to their new use
case, instead of introducing a new parameter. \checked{110}
commits reused \checked{151} existing parameters for
different purposes. However, parameter reuse comes at a cost.
We find two main problems. First, reusing a parameter and code
that it controls can result in subtle inconsistencies that can lead to bugs
or user confusion.
\checked{19.2\% (29/151)} parameter reuses had such inconsistencies.
Second,\Space{ even when there are no
inconsistencies,} developers often\Space{ resort to} clone existing code to enable
reuse.\Space{ In the rest of this section, } We focus on
inconsistencies. Problems
of code cloning are the subjects of other
research.
\Space{To find inconsistencies in parameter reuse we manually compared the old
and new uses in the commit where we saw the reuse. First we
check the instant version \yuanliang{this is a bad word, I mean the
version of this commit} of the commit that use existing
parameter. And try to find whether there are inconsistency between
the new use and old use. Finally, we find \needCheck{22} inconsistent code
behavior among \needCheck{125} parameter new usage. (17.6\%).}
We manually checked for inconsistencies by comparing the newly-added code in a target commit
with code that used the parameter in \blue{existing code base}.
We found \checked{29} inconsistencies in HDFS{} \checked{(9/29)}, HBase{}
\checked{(9/29)} and Spark{} \checked{(11/29)}.\Space{ We did not find inconsistent
parameter reuse in Cassandra.}
Inconsistencies manifest in different ways.
We categorized them based on the sources of inconsistencies during reuse:
1) feedback message \checked{(9/29)}, e.g.,~\href{https://issues.apache.org/jira/browse/SPARK-18061}{Spark-18061};
2) checking code \checked{(4/29)} e.g.,~\href{https://issues.apache.org/jira/browse/HBASE-20590}{HBase-20590};
3) new uses of deprecated parameters \checked{(6/29)}, e.g.,~\href{https://issues.apache.org/jira/browse/HDFS-12895}{HDFS-12895};
4) default values \checked{(3/29)}, e.g.,~\href{https://issues.apache.org/jira/browse/HBASE-21809}{HBase-21809};
and 5) use statements \checked{(7/29)}, e.g.,~\href{https://issues.apache.org/jira/browse/HBASE-20586}{HBase-20586}.
Fig.~\ref{fig:code:InconsistentUse} shows examples of inconsistencies in reuse of checking code
and use statements, where added lines start with $+$.
In Fig.~\ref{fig:code:InconsistentUse}(a), the new parameter usage did not check for parameter value emptiness as the old usage did.
In Fig.~\ref{fig:code:InconsistentUse}(b), the new usage of \CodeIn{hbase.security.authentication} checked case-insensitive equality; the old usage was case-sensitive.
\begin{shadedbox}
{\header{Discussion:}} Inconsistencies in
parameter usage can confuse users (the same values are
used in different ways) or lead to bugs. Ideas for detecting bugs as deviations from
similar program behavior~\cite{engler:01,tan:08} could
be starting points for addressing this problem.
\end{shadedbox}
\begin{figure}
\begin{minipage}{0.97\linewidth}
\centering
\centerline{\lstinputlisting[language=diff]{./figures/InconsistentCheckCode.java}}
\vspace{3.5pt}
\centerline{\small (a) Inconsistent checking (\href{https://github.com/apache/hbase/commit/7da0015a3b58a28ccbae0b03ba7de9ce62b751e1}{HBase-20590})}
\vspace{3.5pt}
\end{minipage}
\begin{minipage}{0.97\linewidth}
\centering
\centerline{\lstinputlisting[language=diff]{./figures/InconsistentParse.java}}
\vspace{3.5pt}
\centerline{\small (b) Inconsistent parameter usage (\href{https://github.com/apache/hbase/commit/cd61bcc01eb29eb3509f72cf72326605afefabc8}{HBase-20586})}
\vspace{3.5pt}
\end{minipage}
\caption{Examples of configuration inconsistent reuse.}
\vspace{-10pt}
\label{fig:code:InconsistentUse}
\end{figure}
\subsection{Summary}
\Space{Based on our findings on configuration usage evolution,}
We advocate that improving software qualities---resilience, diagnosability,
and consistency---should be
first-class principles in software configuration engineering.
We find that even in mature,
production-quality cloud systems,
checking, error handling, feedback, and parameter usage are often not designed
or implemented in a principled manner.
More research effort should be put on enhancing these essential qualities
of configurable software to defend against misconfigurations,
in addition to detection
and diagnosis tools that are external to the cloud system~\cite{attariyan:10,
attariyan:12,attariyan:08,zhang:13,zhang:14,
wang:04,wang:03,santo:16,santo:17,dong:15,
tang:15,huang:15,Baset:2017,sun:osdi:20}.
\Comment{@Owolabi, I think the implications of S5 is very clear!
@Tianyin, I have sprinkled these implications and more in the rest
of the section, so readers can see them as soon as they read each
section. We can move them all here if needed.
\begin{itemize}
\item Checks are often added as afterthoughts -- how to proactively adding
checking code to prevent misconfigurations from
runtime exceptions/failures is still a big problem with very few work~\cite{xu:16}.
\item Inconsistencies of different usage of the same configuration parameters
are another problem -- it confuses users as the same values are used
in different ways. Ideas like such as~\cite{engler:01,tan:08} should be helpful
(there are definitely more papers to cite).
Figures 3-7 are all about inconsistencies and I think we should kill some
of them and find the others a place to live.
\item How to correct misconfigurations is an interesting problem,
but very few work (I don't even know any);
\item Configuration related logging needs to be systematically improved ---
compared with normal logging for
debugging~\cite{yuan:12,Yuan_icse12,asplos-logenhancer,asplos-sherlog,Ding:15,Fu:15},
configuration related logging has to go extra miles --- need to
print out the related configuration parameters and the erroneous values
to enable actions.
\end{itemize}
}
\section{Conclusions}
\label{sec:conclusions}
We presented present an evolutionary study of
configuration design and implementation in cloud systems. To the best of our knowledge, ours is the first evolutionary study on code-level
runtime configuration design and implementation{} in these systems.
We analyze rationales and practices for revising configuration design and implementation{}
decisions, especially in response to consequences of
misconfigurations. Our study yields several new insights into
the configuration engineering process,
and research opportunities for
reducing misconfigurations.
Our hope is to inspire researchers and developers to treat
configuration engineering
as a first-class software engineering endeavor.
\section*{Appendix A: Runtime versus SPL Configuration}
\label{sec:appendix}
A very frequent request is to compare runtime configuration (the type of
configuration studied in this paper) with software product lines (SPL) configuration
(often referred to as ``feature flags'' or ``feature toggles'')
and to position the work in the area of SPL
and variability modeling. We explicitly discuss a few fundamental differences:
First, runtime configurations are changed by software users (operators/sysadmins
in our context); SPL configurations are managed by developers.
Since users are unfamiliar with code, the configuration specifications
become the interfaces (Section~\ref{sec:interface:evolution}).
Moreover, as users are prone to misconfigurations, checking and providing feedback are critical (Section~\ref{sec:behavior:evolution}).
Second, runtime configurations are implemented differently than SPL configurations.
Runtime configurations are mostly in the form of configuration parameters
that load values from files or command lines; SPL
configurations are typically in the form of preprocessors
that determine modules to be included in the released binary.
Third, runtime configurations of cloud software are changed frequently
(hundreds to thousands of times a day~\cite{tang:15,mehta2020rex,maurer:15});
SPL configurations are typically changed with product release cycles.
This higher velocity of runtime configuration changes increases
misconfiguration occurrences and makes checking, error handling, and logging critical.
Fourth, runtime configurations depend on the deployment environment,
including machine resources (e.g., CPU, memory, and storage),
operating systems (e.g., files, IP addresses, and ports),
and workloads (data size and requests per seconds). In contrast, SPL configurations
are often determined before software release or system deployment.
Lastly, runtime configurations have more complex data types (e.g., string and numeric)
with different error patterns; SPL configurations are mostly boolean or enumerative types.
Certainly, ideas in SPL and variability modeling can be extended and applied
to runtime configuration.
We have discussed them in context of our analysis throughout the paper.
\section{Configuration$\:$Document$\:$Evolution}
\label{sec:document}
We very briefly discuss configuration document evolution:
\checked{114} commits made \checked{149} changes to user manuals or code comments.
100 of these commits changed user
manuals and the rest\Space{
\checked{12.3\% (14/114)}} changed code comments.
We discuss why configuration documents were changed and the changed content.
\vspace{5pt}
\mypara{\bf Reasons for changing configuration documents.}
The \checked{149} changes\Space{ (in \checked{114} commits)} to configuration documents resolved five types of
problems\Space{ (and their proportions): previous information was}:
1)~\checked{63} were inadequate\Space{ \checked{(63/149)}} for users to understand parameters or
to set values correctly,\Space{. A developer making one such change in Spark{} said} e.g.,
``{\it\Space{ Couple of }users wondered why spark.sql.shuffle.partitions...unchanged when they changed the config\Space{ value after running the query. It's}...worth to explain it in guide doc}'' (\href{https://issues.apache.org/jira/browse/SPARK-25245}{Spark-25245});
2)~\checked{29} were outdated\Space{ \checked{(29/149)}}} after
configuration design and implementation{} changed (Section~\ref{sec:interface:evolution}
and Section~\ref{sec:behavior:evolution});
3)~\checked{21} were incorrect\Space{ \checked{(21/149)}}\Space{Developers found that the previous information was not correct. An example from HDFS{}:}, e.g., ``{\it LazyPersistFileScrubber will be disabled if\Space{ scrubber
interval is}... configured to zero. But the document was incorrect\Space{---setting it to a negative value to disable this behavior}}'' (\href{https://issues.apache.org/jira/browse/HDFS-12987}{HDFS-12987});
4)~\checked{17} had readability issues\Space{ \checked{(17/149)}}, e.g.,
``{\it Client rpc timeouts are not easy to understand from\Space{ the} documentation}'' (\href{https://issues.apache.org/jira/browse/HBASE-21727}{HBase-21727}); and
5)~\checked{19} improved content\Space{ \checked{(19/149)}}\Space{. The developer try to extend the document by adding configuration contents.},
e.g., ``{\it Add thrift scheduling\Space{ pool}... config to scheduling docs}'' (\href{https://issues.apache.org/jira/browse/SPARK-20220}{Spark-20220}).
\begin{shadedbox}
{\bf \header{Discussion:}}
Document-as-code techniques can be applied to eliminate inconsistencies between
configuration documents and configuration design/implementation.
\end{shadedbox}
\vspace{5pt}
\mypara{Content added to enhance documents.}
Inadequate information was the most common problem resolved by
configuration document changes.
We put the \checked{63} changes that enhanced inadequate documents in
six categories \Comment{(and their proportion)}based on the content added:\Comment{
were added by developers, as an indication of the types of information
that should be provided in the first place.}
1)~\checked{16} changed constraints on parameter values\Space{ \checked{(16/63)}},
e.g., ``{\it This should be positive and less than 2048}'' (\href{https://github.com/apache/cassandra/commit/a586f6c88dab173663b765261d084ed8410efe81}{Cassandra-13622});
2)~\checked{10} explained dependence on other parameters\Space{ \checked{(10/63)}},
e.g., ``{\it This property works with dfs.namenode.invalidate.work.pct.per.iteration}''
(\href{https://github.com/apache/hadoop/commit/b0560e0624756e2b3ce7b6bc741eee3c18d2a873}{HDFS-12079});
3)~\checked{6} changed parameter value types and units\Space{ \checked{(6/63)}};
4)~\checked{6} changed parameter scope\Space{ \checked{(6/63)}}, e.g.,
``{\it Timeout\Space{ for scan operations}... is controlled differently. Use hbase.client.scanner.timeout.period
property to set this timeout}'' (\href{https://github.com/apache/hbase/commit/51c58e083ca89a33de79c8531a16f7072c488d6d}{HBase-21727}),
5)~\checked{22} provided use cases and guidance\Space{ \checked{(22/63)}}, e.g.,
\Space{ \texttt{\small dfs.image.compress} in}
``{\it enabling this will be very helpful if dfs image is large}'' (\href{https://issues.apache.org/jira/browse/HDFS-13884}{HDFS-13884}); and
6)~\checked{3} warned about deprecation,
e.g., ``{\it this config will be removed in Spark 3.0}''~\href{https://issues.apache.org/jira/browse/SPARK-25384}{(Spark-25384)}.
\Space{We recommend these categories of information to be systematically documented and maintained.}
\Comment{
\begin{table}
\scriptsize
\caption{Statistics of configuration document evolution.}
\begin{tabular}{lccccc}
\toprule
& \header{HDFS} & \header{HBase} & \header{Spark} & \header{Cassandra} & \header{total} \\
\midrule
{\bf User manual} & \checked{27} & \checked{18} & \checked{52} & \checked{3} & \checked{100} \\
{\bf Code comments} & \checked{0} & \checked{3} & \checked{9} & \checked{2} & \checked{14} \\
\bottomrule
\end{tabular}
\label{tab:Documentation}
\end{table}
}
\begin{shadedbox}
{\bf \header{Discussion:}}
Ethnographic studies could help understand the gaps between documented
configuration information and configuration obstacles faced by users.
\end{shadedbox}
\vspace{3pt}
\mypara{Summary}
Correctness and effectiveness of technical documentation is a long-lasting
problem in software engineering.
Configuration documentation is no exception.
Specialized techniques for maintaining and improving configuration documentation are needed.
For example, checking for inconsistencies between documents
and source code~\cite{tan:07,tcomment,yzhou:18,hzhong:13}
could help detect defects in configuration-related code
or documents.
Also, techniques for auto-generating documents, especially
using structured data,
can be applied to generating per-parameter
comments and manual entries~\cite{autocomment,cpc,Giriprasad:10}.
\Comment{
\red{Yuanliang: We need to cite papers on feature models that deal with document--please find some.}
\red{
I find and select some content from~\cite{meinicke:20}:
In the product line community, notations such as feature diagrams~\cite{Apel:13:book,Czarnecki:00} are widely adopted to
document features and especially to describe constraints on possible configurations. (most prominently
documenting multiple features to be optional, mutually exclusive, depending on another,
or in a hierarchical relationship where child features depend on parent features. Those constraints are different from runtime configuration).
Beyond tools focusing on product lines like FeatureIDE~\cite{FeatureIDE}.
An example is Linux kernel's variability model~\cite{She:10}, for which the kernel developers have built their own
domain-specific language to describe and document options. (However, the natural language doc still needs human efforts. e.g., ---help---)
Academic research has invested effort into tools that can work with such documented
feature models, for example, detecting inconsistencies among constraints~\cite{Batory:05,Benavides:10},
analyzing the evolution of model changes~\cite{Lotufo:10,Thum:09}, resolving conflicts in configurations~\cite{xiong:15} and
guiding humans through the configuration process~\cite{Hubaux:13,Schmid:11}. (Those research are not about how to document the feature well but how to use the doc,
in this section we actually try to discuss what is the defects of current document, so I think they are different)
\cite{meinicke:20} recommend set and enforce clear documentation standards for feature flags and configuration options.}
}
\section{Configuration Interface Evolution}
\label{sec:interface:evolution}
Changes to the configuration
{\it interface} were the most common\Space{ evolution types},
compared with behavior or documentation changes (Table~\ref{tab:categ_num}).
\Space{As shown in Table~\ref{tab:categ_num}, interface evolution
contributes to \red{XX\%--YY\%} of all the changes across the
studied software projects.} We focus on analyzing changes to \emph{configurability}---the level of user-facing configuration flexibility\Space{
that the system}---(Section~\ref{sec:configurability})
and default values (Section~\ref{sec:default_value}).
We omit other kinds of configuration
interface changes which are often routine and cannot directly lead to misconfigurations.
\subsection{Evolution of User-Facing Configuration}
\label{sec:configurability}
Table~\ref{tab:interface} shows our categorization of changes to configurability.\Space{Most configurability-related} Most changes
add or remove parameters;\Space{ for comparison,}
per project, removal\Space{ existing parameters} is 5.1$\times$ to
21.2$\times$ less frequent than addition (with an average of 8.4$\times$).
\Space{Upon further analysis, }We find that adding or removing parameters occur naturally\Space{ consequence of} during software
evolution---\Space{new }parameters are added with new code, and\Space{ parameters}
removed with code deletion.
\Space{In the rest of this section. So,}We do not focus on co-addition or
co-removal of parameters with code.
Rather, we focus on changes\Space{ developers make} that revise
previous configuration engineering decisions by
1)~parameterizing constants and 2)~eliminating parameters or converting them to constants.
\Comment{Owolabi: why should we not study rationales for adding and deleting more deeply in the future?}
Our data corroborates a prior finding~\cite{xu:15:2} that configuration interface complexity
increases rapidly over time, as more parameters are added than are removed.
Complexity measures the size of the configuration space (number
of parameters multiplied by the number of all their possible values).
\Space{Techniques and practices}Approaches for dealing with the rapid growth rate\Space{
interface complexity} are\Space{ urgently} desired.
\Space{Specifically, }Variability modeling~\cite{Lotufo:10,She:10,Damir2019FSE,Berger2013TSE,Liebig:10} which is extensively
researched for compile-time configurations,
can potentially be extended to understand and manage runtime configuration complexity.
\subsubsection{Parameterization}
\label{sec:parameterization}
\Space{We observe that software }Developers often\Space{ parameterize constants---they} convert constants into
parameters\Space{---} after discovering that one constant cannot satisfy all use cases. We find \checked{142} commits
that parameterize \checked{169} constants (\checked{169} parameterizations). We report on
1)~rationales for the parameterizations,
2)~how developers identify constants to parameterize,
3)~use cases that made constants insufficient,
and 4)~how developers\Space{ currently use to} balance increase in configuration complexity
(caused by adding new parameters) with the need for flexibility (which necessitates parameterization).
Our results have\Space{ significant} ramifications for
configuration interface design:
we provide understanding\Space{ necessary} for managing the
configurability versus simplicity tradeoff.
The rationales for parameterization also\Space{ has implications that} motivate\Space{ the need for}
configuration parameter auto-tuning.
\vspace{5pt}
\mypara{\bf Rationales for parameterization.}These
include: performance
tuning, reliability, environment setup, manageability, debugging,
compatibility, testability, and security.
Table~\ref{tab:purpose_constant2config} shows, for each rationale, the number of commits\Space{ (\header{\#Commits})} and parameters\Space{ (\header{\#Params})}\Space{ that
we encountered}, an example parameter, and a description.
We discuss the top two rationales, due to space limits.
Performance tuning was the top rationale for parameterizing constants,
involving \checked{39.6\%
(67/169)} of parameters in \checked{56}\Space{ Add$\rm_{const2config}$} commits\Space{ in
Table~\ref{tab:purpose_constant2config}}.\Space{. Parameterizing for performance tuning makes sense:}
Different workloads need different values, so it is hard to
find one-size-fits-all constants.\Space{ So, developers parameterize
constants\Space{ that are then exposed} to allow users
customize\Space{ those parameters} for their workloads.}
\Space{Among performance-tuning-related parameterization,}
Resource-related (e.g., buffer size and thread number),
feature selection (turning on/off features with performance impact, e.g.,
monitoring),
and timing logic (mostly timeouts and intervals) were the main resulting parameter types, with
20.9\% (14/67), 37.3\% (25/67), and 14.9\% (10/67) new parameters, respectively.
Others 26.9\% (18/67) set algorithm-specific parameters (e.g., weights and sample sizes).
\begin{table}[]
\small
\centering
\setlength{\tabcolsep}{4pt}
\caption{Statistics on configuration interface changes.}
\scalebox{0.9}{
\begin{tabular}{lccccc}
\toprule
& \headerHDFS & \headerHBase & \headerSpark & \headerCassandra & \header{Total} \\
\midrule
{\bf AddParam} & \checked{106} & \checked{122} & \checked{277} & \checked{42} & \checked{547} \\
\quad Add$_\text{NewCode}$ & \checked{54} & \checked{55} & \checked{143} & \checked{23} & \checked{275} \\
\quad Add$_\text{CodeChange}$ & \checked{16} & \checked{34} & \checked{72} & \checked{8} & \checked{130} \\
\quad Add$_\text{Parameterization}$ & \checked{36} & \checked{33} & \checked{62} & \checked{11} & \checked{142} \\
\midrule
{\bf RemoveParam} & \checked{5} & \checked{24} & \checked{30} & \checked{6} & \checked{65} \\
\quad Rmv$_\text{RmvModule}$ & \checked{3} & \checked{16} & \checked{25} & \checked{4} & \checked{48} \\
\quad Rmv$_\text{Replace}$ & \checked{2} & \checked{8} & \checked{5} & \checked{2} & \checked{17} \\
\midrule
{\bf ModifyParam} & \checked{28} & \checked{25} & \checked{60} & \checked{6} & \checked{119} \\
\quad Mod$_\text{Naming}$ & \checked{5} & \checked{8} & \checked{30} & \checked{1} & \checked{44} \\
\quad Mod$_\text{DefaultValue}$ & \checked{19} & \checked{14} & \checked{20} & \checked{3} & \checked{56} \\
\quad Mod$_\text{Constraint}$ & \checked{4} & \checked{3} & \checked{10} & \checked{2} & \checked{19} \\
\bottomrule
\end{tabular}
}
\label{tab:interface}
\end{table}
\begin{table*}[t]
\footnotesize
\setlength{\tabcolsep}{2pt}
\caption{Statistics and examples of developers' rationales for parameterization (excluding two commits that lacks information). }
\scalebox{0.95}{
\begin{tabular}{lccll}
\toprule
\header{Rationale} & \header{\#Commit} & \header{\#Param} & \header{Example New Parameter} & \header{Limitation of previous constant} \\
\toprule
Performance & 56 & 67 & {\scriptsize spark.sql.codegen.cache.maxEntries} & The cache size does not work for online stream processing~(\href{https://issues.apache.org/jira/browse/SPARK-24727}{Spark-24727})\\
\midrule
Reliability & 28 & 37 & {\scriptsize spark.sql.broadcastExchange.maxThreadThreshold} & Out of memory if thread-object garbage collection is too slow~(\href{https://issues.apache.org/jira/browse/SPARK-26601}{Spark-26601}) \\
\midrule
Manageability & 20 & 20 & {\scriptsize dfs.federation.router.default.nameservice.enable} & Enable the default name service to store files\Space{ at specified locations}~(\href{https://issues.apache.org/jira/browse/HDFS-13857}{HDFS-13857}) \\
\midrule
Debugging & 8 & 9 & {\scriptsize spark.kubernetes.deleteExecutors} & Disable auto-deletion of pods for debugging and diagnosis~(\href{https://issues.apache.org/jira/browse/SPARK-25515}{Spark-25515}) \\
\midrule
Environment & 8 & 13 & {\scriptsize dfs.cblock.iscsi.advertised.ip} & \Space{An iSCSI}Allows server and target addresses to be different~(\href{https://issues.apache.org/jira/browse/HDFS-13018}{HDFS-13018}) \\
\midrule
Compatibility & 13 & 13 & {\scriptsize spark.network.remoteReadNioBufferConversion} & Add the parameter to fall back to an old code path~(\href{https://issues.apache.org/jira/browse/SPARK-24307}{Spark-24307}) \\
\midrule
Testability & 3 & 4 & {\scriptsize spark.security.credentials.renewalRatio} & May not need to be set in production but can make testing easier~(\href{https://github.com/apache/spark/pull/20657}{Spark-23361}) \\
\midrule
Security & 4 & 4 & {\scriptsize spark.sql.redaction.string.regex} & The output of query explanation can contain sensitive information~(\href{https://issues.apache.org/jira/browse/SPARK-22791}{Spark-22791}) \\
\bottomrule
\end{tabular}
}
\label{tab:purpose_constant2config}
\end{table*}
Reliability, with\Space{ \checked{21.9\%} (} \checked{37} of \checked{169} of the parameterizations, was the second most
common rationale.
\Space{\checked{45.9\% (17/37)}}Of these, \checked{17} were caused by hardcoded timeout values
that led to constant request failures in the reported deployments,
so developers made them configurable.
Note that new timing parameters were created for
both reliability and performance tuning.
For example, in HDFS{}, a new timing\Space{-related} parameter\Space{,
\CodeIn{\seqsplit{regionserver.hfilecleaner.thread.timeout.msec}}}
was created to improve performance. The previous constant was causing a ``{\it \Space{
While deleting a file never complete (strange but possible),
then }delete file task\Space{ needs} to wait for\Space{ a max
of 60 seconds which might be}... too long}''~(\href{https://issues.apache.org/jira/browse/HBASE-20401}{HBase-20401}).
But, another HDFS{} timing\Space{-related} parameter\Space{, \CodeIn{\seqsplit{cblock.rpc.timeout.seconds}},} was created to improve reliability. The previous constant\Space{value}
{was too small,}
\Space{of 300s}causing ``{\it \Space{Volume creation }timeouts while creating 3TB
volume\Space{ because of too many containers}}''~(\href{https://issues.apache.org/jira/browse/HDFS-12210}{HDFS-12210}).
\begin{shadedbox}
{\bf \header{Discussion:}} Configuration auto-tuning techniques
that consider reliability and functionality are needed, in addition to
performance-only optimization~\cite{aken:17,
wang:18, zheng:07, yu:18, nair:17, siegmund:15,
Nair:2018, Hsu:2018, duan:09, zhu:17,
Jamshidi:17, Jamshidi:18, xi:04, ye:03, Herodotou:2011, osogami:06,
Krishna:19, hoffmann:11, Herodotou:11}.
Specifically, timing\Space{ related} parameters have important implications to both reliability
and performance; however, not much work has been done on auto-tuning
timing parameters (e.g., timeouts and intervals).
\end{shadedbox}
\Comment{@Owolabi: the point I wanted to make is that there are multiple
objectives for setting configuration values --- it seems to me that
all the existing auto-tuning techniques only focus on performance.}
\vspace{8pt}
\mypara{\bf How developers find constants to parameterize.}
\checked{54.4\% (92/169)} of parameterizations\Space{ (\red{56.3\% (63/112)} of commits),}
were {\it postmortem} to
severe consequences\Space{ in the field}, e.g., system failures, performance
degradation, resource overuse, and incorrect results.
Among previous constants for these,
\checked{40.2\% (37/92)}\Space{ of the constants} led to performance
degradation\Space{ such as \red{prolonged execution time}, \red{excessive
end-to-end latency}, and \red{low throughput}};
\checked{35.9\% (33/92)}\Space{ of the constants} caused
severe failures\Space{ such as job-level and system-level failures};
\checked{19.6\% (18/92)}\Space{ of the constants}
led to incorrect or unexpected results (e.g., data
loss and wrong output);
and \checked{4.3\% (4/92)}\Space{ of the constants} resulted in
resource overuse.
\Comment{
\vspace{5pt}
{\color{gray}
\mypara{\bf \header{Implications}}Over 50\% of parameterizations were
found deficient---not satisfying all use cases---after
negative consequences ensued.\Comment{may not be "after failure". I guess is "after bad consequence" }
Future work should develop \emph{proactive} techniques for
detecting\Space{ and warning developers about} deficient constants
and automate parameterization; the latter could\Space{ also} assist\Space{ with}
performance testing of cloud and datacenter software. \Comment{Any other deep stuff?}
}
}
\begin{shadedbox}
{\bf \header{Discussion:}}
Despite the efforts in parameterization,
developers still overlook deficient constants that may lead to severe consequences (e.g., failures and performance issues).
\textit{Proactive} techniques for detecting deficient constants and for automating
parameterization are needed; the latter could\Space{ also} assist\Space{ with}
performance testing of cloud systems.
\end{shadedbox}
\vspace{5pt}
\mypara{\bf Describing use cases that prompt parameterization.}\Space{As discussed above, The main reason for introducing
parameters is that a constant does not satisfy
all use cases and workloads. Thus, to improve understanding of the
resulting configuration design, }Use cases where constants
were deficient should be described fully\Space{ (if not defined)} to help users set
correct values.
But,\Space{ we find that} developers described the {\it concrete}
use cases that prompt parameterization for only
\checked{37.9\% (64/169) parameters}. Others discussed use cases
either {\it vaguely} \checked{(45.0\% or 76/169 parameters)} or provided \emph{no information}
\checked{(17.1\% or 29/169 parameters)}\Space{, thus failing to provide guidance
for improving the understanding of
resulting parameters}. Table~\ref{tab:des_level} shows examples.
\Comment{
\tianyin{@Owolabi: One potential direction is to automatically identify the
scenarios and apply the corresponding configuration values.}
}
\begin{shadedbox}
{\bf \header{Discussion:}} Future work should
identify and document use cases and workloads,
including which parameters can be tuned, and suggest beneficial configuration
values that are designed for concrete use cases.
\end{shadedbox}
\begin{table}
\footnotesize
\centering
\setlength{\tabcolsep}{0pt}
\caption{Use-case description of parameterization changes.}
\scalebox{0.85}{
\begin{tabular}{ll}
\toprule
\header{Level} & $\ \ \ $\header{Example} \\
\toprule
Concrete & $\ $``{\it \scriptsize Volume creation times out while creating \textbf{3TB} volume}''~(\href{https://issues.apache.org/jira/browse/HDFS-12210}{HDFS-12210}). \\
\midrule
Vague & ``{\it \scriptsize If \textbf{many} regions on a RegionServer, the default will be not enough}''~(\href{https://issues.apache.org/jira/browse/HBASE-21764}{HBase-21764}). \\
\midrule
No Info & ``{\it \scriptsize It \textbf{would be better} if the user has the option instead of a constant}''~(\href{https://issues.apache.org/jira/browse/SPARK-25233}{Spark-25233}). \\
\bottomrule
\end{tabular}
}
\label{tab:des_level}
\end{table}
\vspace{5pt}
\mypara{\bf Balancing flexibility and simplicity}Configuration interface design must balance flexibility
(i.e., configurability) with simplicity~\cite{xu:15:2}. New
parameters increase flexibility (by handling additional use cases),
but increase interface complexity (thus reducing usability).
\checked{16.2\% (23/142)} of parameterization commits\Space{issue discussion, PR discussion} contained developer discussions on the
flexibility-simplicity tradeoff. Most\Space{ of them} discussed\Space{
the} estimated prevalence of use cases for parameterization---it
is not worth increasing interface complexity
for rare\Space{ly-occurring} use cases---and typically involve\Space{ one or
two} advanced users, e.g., ``{\it
admittedly, this...\Space{ [\texttt{.blockWhenSyncLagsNanos}]}is\Space{ only} an
expert-level setting,\Space{ but} useful in some cases}''~(\href{https://issues.apache.org/jira/browse/CASSANDRA-14580}{Cassandra-14580}).
\Comment{
\begin{itemize}
\item
``{\it From an HDFS user hat on, I think this is a good improvement to have.
I would expect HDFS to refuse to format a cluster with data.
(avoid a disaster for some poor user.)
But from a sysadmin/developer hat on, I do like that fact that I can format a cluster with data.
I do that when I test and develop.
So in my mind, the question boils down to easier dev/ops cycles vs. user safety.}''
Finally, the developer add a Boolean option \ttvar{dfs.reformat.disabled} to solve this problem.
\item SPARK-27870,
one reporter try to flush batch timely because the current default buffer size is 65535,
which is too large for ML cases as they try to make real-time prediction
(if the buffer size is too big, the flush will lag behind).
One developer against that ``{\it This change might improve perf for your case
where a pandas\_udf takes massive amount of time on small data,
but my concern is it could have a negative effect on other cases
where IO is the main bottleneck.}''
After the discussion, they make the buffer size configurable as an alternate solution.
However, the reporter himself still have some doubts:
``{\it I also consider about tuning ``buffer size'', but one of my concern is,
user may mix ML prediction with other simple udf (such as data preprocessing).
If we tune the ``buffer size'' globally to be small, then it will hurt the first simple udf performance.}''
\end{itemize}
}
We also found\Space{ subjective developer discussions and} developers' debates on whether\Space{
a constant should be} to parameterize
(\href{https://issues.apache.org/jira/browse/CASSANDRA-12526}{Cassandra-12526},
\href{https://issues.apache.org/jira/browse/HDFS-12496}{HDFS-12496},
\href{https://github.com/apache/spark/pull/23090}{Spark-26118}).
A {\it middle-ground} solution\Space{ in these discussions} is to
parameterize without documenting or exposing the parameter, e.g.,
``{\it although...\Space{ maybe}not widely
used, I could see allowing control\Space{ of this}...via an undocumented
parameter}''~(\href{https://github.com/apache/spark/pull/21433}{Spark-23820}).
With this practice, not all but the most advanced users know of such parameters.
We found that
\checked{58.0\% (98/169)} of the
newly added parameters were not documented in the parameterization commit, indicating
that these parameters were first added as middle-ground solutions.
\begin{shadedbox}
{\bf \header{Discussion:}} Further studies are needed on 1) if and why
undocumented parameters are eventually documented, and 2) how often and why (expert) users
modify un-exposed parameters, in order to understand the intent and utility of such parameters.
\end{shadedbox}
Specifically, visibility conditions from variability modeling~\cite{Damir2019FSE,Berger2013TSE,Berger:10}
can be extended to manage the tradeoff of flexibility versus simplicity,
which can benefit navigation support and user guidance~\cite{barrett:04,takayama:06,haber:07,jin:14}.
Currently, visibility conditions are mainly designed for Boolean feature flags based
on dependency specifications (e.g., in {CDL and KConfig);
complexity metrics and variability analysis for other parameter types (e.g., numeric and strings)
are needed.
\subsubsection{Removing Parameters}
\label{sec:removing:parameters}
\Space{Parameter removal
is infrequent relative to addition (Table~\ref{tab:interface}).} Understanding
parameter removal can yield insights on reducing configuration
interface complexity~\cite{xu:15:2,meinicke:20}.
We examined all \checked{17} configuration-related commits{} that removed a parameter (not co-removal with code).
All \checked{17} removed parameters were
converted to constants or code logic was added that obviated them.
\checked{14}\Space{ of \checked{17}} removed parameters were converted to constants: 11 to their default values
and 3 to safe values.
Developers
mentioned that \checked{8} of the \checked{17} parameters had no clear use case
(e.g.,~\href{https://issues.apache.org/jira/browse/HBASE-8518}{HBase-8518},
\href{https://issues.apache.org/jira/browse/HBASE-18786}{HBase-18786}),
or required users to understand implementation details (e.g.,~\href{https://issues.apache.org/jira/browse/CASSANDRA-14108}{Cassandra-14108}).
\checked{Three} parameters\Space{ even} confused users or might lead to severe
errors~(e.g.,~\href{https://issues.apache.org/jira/browse/SPARK-26362}{Spark-26362}).
\checked{Three} of \checked{17} removed parameters were obviated by new automation logic.
For example, in \href{https://issues.apache.org/jira/browse/HBASE-21228}{HBase-21228},
\CodeIn{\seqsplit{hbase.regionserver.handler.count}} which
specified the number of concurrently updating threads to be garbage
collected in a Java \CodeIn{ConcurrentHashMap}, was removed after
developers switched to
\CodeIn{\seqsplit{ThreadLocal<SyncFuture>}} which automatically garbage collects terminated threads.
This example shows how implementation choices could affect configuration complexity.
\begin{shadedbox}
{\bf \header{Discussion:}} Future studies can evaluate the utility and impact
of each parameter (e.g., by analyzing if and how often deployed
values are equal or similar to the default values).
Configurations with low utility can be replaced with constants (e.g., default values).
\end{shadedbox}
\Comment{@Owolabi: is this an implication: basically adding a configuration
parameter should be a metric to consider in the implementation.}
\Comment{@Owolabi: perhaps we can make a point that
those two principles can be generalized to further eliminate and
simplify existing configuration design. Owolabi: that would require
a way to first know the use cases under which values of all
parameters are useful; I don't think that's feasible}
\subsection{Evolution of Default Values}
\label{sec:default_value}
Parameter default values are important to the usability of\Space{ highly}
configurable systems; they provide users with good
starting points for setting parameters without needing to understand the
entire configuration space. Thus,\Space{ to minimize user configuration
efforts,} developers usually choose default values that satisfy
common use cases. Ideally, a default value applies under most common
workloads, without causing failures\Space{ and does not lead to software failure (including
performance degradation) in any workload}
(\href{https://issues.apache.org/jira/browse/HBASE-16417}{HBase-16417},
\href{https://issues.apache.org/jira/browse/HBASE-20390}{HBase-20390},
\href{https://issues.apache.org/jira/browse/HDFS-11998}{HDFS-11998}).
The ``Mod$\rm_{DefaultValue}$'' row in Table~\ref{tab:interface} shows\Space{ that
there were} \checked{56} commits that changed \checked{81} default\Space{ parameter}
values\Space{ in all four subjects}\Space{ during the two-year range of
configuration-related commits{} that we studied, 2--19 default values are changed
across the studied modules}. We discuss\Space{ here the reasons} why default
values changed and how new default values were chosen.
\Comment{Tianyin:This is a very deep question but the answer is not
answering the question at all. Let me elaborate the question. --- as
we discussed over Skype, the simplest way is to focus on why the old
default starts to break things and need the new values --- something
has to be changed, either the workload, or the hardware, or the
scale, or the environment.}
\vspace{3pt}
\mypara{Reasons for changing default values.}
We observe proactive and reactive default value changes.
\checked{38.3\% (31/81)} default-value changes were proactive, including
1) enabling a previously disabled feature flag (\checked{32.3\% (10/31)}),
e.g., ``{\it running the feature in production for a while with no issues,
so enabled the feature by default}'' (\href{https://issues.apache.org/jira/browse/HDFS-7964}{HDFS-7964}),
2) performance reasons (\checked{35.4\%, 11/31}), e.g.,
``{\it sets properties at values yielding optimal performance}'' (\href{https://issues.apache.org/jira/browse/HBASE-16417}{HBase-16417}), and
3) supporting new use cases (\checked{32.3\%, 10/31}), e.g., {\it ``it may be a common use case to ...list
queries on these values}'' (\href{https://issues.apache.org/jira/browse/Cassandra-14498}{Cassandra-14498}).
The remaining \checked{61.7\% (50/81)} of default value changes were reactive to
user-reported issues, including
1) system failures and performance anomalies due to not supporting
new workloads, deployment scale, hardware, etc (\checked{50.0\%, 25/50}),
2) inconsistencies with user manual (\checked{38.0\%, 19/50}),
and 3) working around software bugs (\checked{12.0\%, 6/50}), e.g.,
{\it ``we set the parameter to false by default for Spark 2.3 and re-enable
it after addressing the lock congestion issue''} (\href{https://issues.apache.org/jira/browse/SPARK-23310}{Spark-23310}).
\Comment{After discussing each of these 6 cases,
we reached an converged opinion to put each of them in one category}
\vspace{3pt}
\mypara{Choosing new values.}It is straightforward to change
new default values for
Boolean and enumerative parameters, given their small value ranges.
So, we describe how new default
values of \checked{32} {\it numeric} parameters were chosen (excluding
those that fix default value inconsistency (e.g., \href{https://issues.apache.org/jira/browse/HBASE-18662}{HBase-18662}).
Only \checked{28.1\% (9/32)} numeric parameters had systematic
performance testing and benchmarking mentioned in the JIRA/GitHub
issues\Space{ accompanying the default-value change}.
Later commits reset these new default values,
despite the initial testing and benchmarking. For example, HBase{}
developers\Space{ said they had} performed ``{\it write-only workload
evaluation...read performance in read-write workloads. We
investigate several settings...\Space{ of hardware (SSD, HDD), key
distribution (Zipf, uniform), with multiple settings\Space{ of the
system, and compare measures like write throughput, read latency,
write volume, total gc time, etc.}}}''~(\href{https://issues.apache.org/jira/browse/HBASE-16417}{HBase-16417}). Yet,
we found three later commits that changed the default value of the
same parameter to different numbers.
For \checked{31.3\% (10/32)} of numeric parameters, new default values were
chosen by adjusting the previous default values to resolve production
failures. In many of these cases, usually without high
confidence in the new default values, developers simply chose
values that resolve the\Space{ reported} problem(s). Examples: ``{\it It
probably makes sense to set it to something\Space{ even} lower\Space{
still}}''~(\href{https://issues.apache.org/jira/browse/SPARK-24297}{Spark-24297}), or ``{\it I'm thinking something like
3000 or 5000 would be\Space{ a} safer\Space{ bet.}}''~(\href{https://issues.apache.org/jira/browse/HBASE-18023}{HBase-18023}).
We found no\Space{ concrete} information on\Space{ how new default values
were selected for} the remaining \checked{40.6\% (13/32)} numeric
parameters.
We observe that backward compatibility and safety are common considerations
in selecting new default values. New default values that radically
change system behavior are often considered inappropriate
(e.g., \href{https://issues.apache.org/jira/browse/HBASE-18662}{HBase-18662}).
\Comment{@Yuanliang, we need a concrete example to show what
performance tests/benchmarks are used -- this needs more
investigation beyond the data you have. Without the understanding
of the benchmarks, it is really hard to dig any insights out.
\begin{itemize}
\item HDFS-11814: using ErasureCodeBenchmarkThroughput to test IO (read/write) performance.
and the Time to Complete(metric) decrease when cell size in the range of 32KB to 1MB.
So change it from 32KB to 1 MB. (No change later in the study period)
\item HBASE-21000: used PE sequentialWrite to write 1M rows x 10 threads and
monitored when compaction activity started and stopped to test compaction max throughput bound.
But as developer said, this is a simple and quick test. (No change later in the study period)
\item HBASE-16417 ``{\it It presents the result of write-only workload evaluation as well
as read performance in read-write workloads. We investigate several settings of hardware (SSD, HDD),
key distribution (Zipf, uniform), with multiple settings of the system, and compare measures like write throughput,
read latency, write volume, total gc time, etc.}`` However, the parameter \ttvar{hbase.memstore.inmemoryflush.threshold.factor}
changed three times later in HBASE-19282, HBASE-20390 and HBASE-20542. All these cases have conducted IO performance
test on benchmarks (didn't give the name). The problem is that in HBASE-19282, developers find that 10\% is better than 2\%
under write-only workload.(More ops/second) So they change the default value from 2\% to 10\%. However, later in
HBASE-20390, developers find that the new default value is a bit better on writes but seems worse for mixed load.
So they use 1.4\% which show performance improvement in all test workloads.
Finally they turn it to 0 in HBASE-20542(equal to disable this feature bu default) because of heap under-utilization problem.
But users can still tune this parameter.
The same thing happened to \ttvar{hbase.hregion.compacting.pipeline.segments.limit} in HBASE-16417 and HBASE-20390.
(changed from 1 to 4 to 2). \yuanliang{Basically HBASE-20390 is doing the right thing: Running different workload.
(100\% write, 50\%write50\%read, 100\%read) }
\end{itemize}
\yuanliang{Here are the things I can think of: First, the parameters in those cases mainly influence the I/O operation,
so developers reuse some I/O tests and benchmarks to find the new value.
This can be implemented to other parameters, first we need to figure out what aspects actually
influenced and decided by the parameter and then we can conduct corresponding performance test.
(I believe there are already lots of tests and benchmarks for different aspects of performance.)
Second, performance test is obviously a more convincing approach to choose the new value
comparing to ``resolving issues'' or based on developers experience and bet.
But developers should still keep in mind that default value should work for most cases
(e.g. workload, environment). Otherwise there may be performance regression.
}
}
\begin{shadedbox}
{\bf \header{Discussion:}} \Space{In the studied cloud systems,
}Default value changes are often reactive to the reported issues,
without systematic assessment.
Systematic testing and evaluation of
new (and\Space{ even} existing)
default values are needed.
\end{shadedbox}
Dynamic workloads and heterogenous deployments necessitate continuous and incremental changes to default values.
Future work could maintain a set of
default values (instead of one) for\Space{ various} typical
workloads, hardware, and scale.
\Comment{
\haochen{
\myitpara{\bf \header{Implications}} Developers set default values in a conservative manner:
they always try to keep the default values free from any potential bugs/performance issue
(i.e., disabling newly developed features, workaround defects).
However, developers are usually sloppy when choosing new numeric parameters' default values,
which sometimes even cause additional unexpected results.
Therefore, better ways of systematically testing and benchmarking new (and maybe even existing) default values are needed.
Also, even though developers has made effort to proactively improving the default values,
severe consequences are still happening, so there is still a large space to be filled in this topic.
{(\cite{Damir2019FSE, Berger2013TSE} have emphasized the importance of defining a default value for features.
``{\it It can be beneficial to define default feature values if the configuration space of a feature model is very large.}''
However, they didn't discuss how to define one.
Also, defining default value is different for those two kinds of configuration.
One example is that the vast majority of features are of type Boolean~\cite{Damir2019FSE}, so choosing a
default value is easier because of the limited value space.
Another example is that compile-time features don't need to do ``proactive performance exploration'') and worry about
runtime workload. }
}
}
\Comment{
\tianyin{@Owolabi, This is an interesting
observation -- what is the implications?} \yuanliang{In my opinion,
there should be some systematic approaches to \textbf{validate} that
the new default value is better than the old one. I don't study this
one but seems that developers now use production result and user
feedback as validation. But it can actually bring some damage to
the system. Also, we can't change default value just because it can
fix a single bug. (Probably break this wall to fix that wall) This
is not what default value should do. It actually should fit for
common workload. Performance test is good approach, but it just for
performance, we also need some test for reliability, security, etc.}
}
\Comment{
\subsection{Other Configuration Interface Changes}
\label{sec:other:interface}
For completeness, we briefly describe the other kinds of configuration
interface changes (Mod$\rm_{Contraints}$ and Mod$\rm_{Naming}$ in
Table~\ref{tab:interface}). These changes are rather routine and, in
most cases, cannot directly lead to misconfigurations.
\vspace{3pt}
\mypara{Changes to Parameter Constraints}\checked{19} commits
changed the constraint on \checked{35} parameters in order to
support new value formats, add more options to enumeration-type
parameters, increase the range of allowable values, or change the
parameter type. These changes occurred during changes to code modules,
to make it easier for users to change the configurations, or because
the original constraint did not support a valid use case.
\vspace{3pt}
\mypara{Changes to Parameter Names}\checked{44} commits changed the
names of \checked{71} parameters for consistency with naming
conventions or rename refactoring in the code, or to fix typographic
errors.
}
\Comment{
There are \needCheck{14} commits that change \needCheck{30} params' constraint.
\needCheck{12/30} params change the acceptable value. Including
(1) value format (\needCheck{5/12}) e.g., ``{\it support wildcard \* for \CodeIn{\seqsplit{spark.yarn.dist.forceDownloadSchemes}}}''~\href{https://issues.apache.org/jira/browse/SPARK-24646}{- Spark-24646}
(2) choosable value (\needCheck{5/12}, this is for parameter has clear space)
e.g., ``{\it support ZSTD Compressor for commitlog\_compression.}''~\href{https://issues.apache.org/jira/browse/CASSANDRA-14482}{CASSANDRA-14482}
(3) value range (\needCheck{2/12}) e.g. ``{\it Change \ttvar{dfs.datanode.failed.volumes.tolerated}' support minimum number}''~\href{https://issues.apache.org/jira/browse/HDFS-12716}{- HDFS-12716}
Other \needCheck{18/20} params change the type. e.g., ``{\it Modify \ttvar{spark.kubernetes.allocation.batch.delay} to take time instead of int.}''~\href{https://issues.apache.org/jira/browse/SPARK-22845}{- Spark-22845}
The reason why developer change constraint to those params. \needCheck{7/14} of the commits
change because of the feature/module change. e.g., ``{\it Removing the on heap Bucket cache feature.
The config \ttvar{hbase.bucketcache.ioengine} no longer support the 'heap' value.}''~\href{https://issues.apache.org/jira/browse/HBASE-19187}{- HBase-19187}.
\needCheck{5/14} of the commits change for user's easier configuring. e.g., ``{\it Currently, if we want to configure
\ttvar{spark.sql.files.maxPartitionBytes} to 256 megabytes,
we must set it as 268435456, which is very unfriendly to users.}''~\href{https://issues.apache.org/jira/browse/SPARK-27256}{- Spark-27256}
\needCheck{2/14} of the commits change because the original constraint doesn't fit for some scenario.
e.g., ``{\it \ttvar{dfs.datanode.failed.volumes.tolerated} should support negative value n (means totalNum - n)
because datanode volumes can be changed dynamically.}''~\href{https://issues.apache.org/jira/browse/HDFS-12716}{- HDFS-12716}.
\subsection{Param rename}
There are \needCheck{44} commits that change \needCheck{71} parameters' name.
There are 5 reasons that developer rename the parameter.
(1) \needCheck{10/30} of the commits are because of the consistency naming convention.
e.g., ``{\it Might rename this to "enabled" rather than "activate" to align with other previous config keys.}''~\href{https://issues.apache.org/jira/browse/HDFS-12214}{- HDFS-12214}.
(2) \needCheck{8/30} of the commits are because the feature/module which hold this param has changed. e.g., \ttvar{hbase.rpc.server.nativetransport}
rename to \ttvar{hbase.netty.nativetransport} because the EventLoopGroup class will be shared for NettyRpcServer and NettyRpcClient. ~\href{https://issues.apache.org/jira/browse/HBASE-18307}{- HBase-18307}
(3)\needCheck{4/30} of the commits are because the name is not precise or even wrong.
e.g., ``{\it The parameter name should be improved because it's not about spark files.}''~cite{SPARK-22233}
(4) \needCheck{5/30} of the commit are due to typo.
(5) \needCheck{3/30} don't give concrete reason.
We then try to find what developer usually made mistakes when they naming a parameter.
\yuanliang{Here is my own thought. I didn't look at some citation}
There are 3 major parts of parameter name. Usually a name looks like this: \textbf{domain.somePrefix.function/type}.
``Domain'' can either be the software name or component name. (e.g., ``.server'') ``somePrefix'' can show some
characteristics of the param. (e.g., ``.legacy'' means this param is for compatibility) ``function/type" is try to show the
functionality of this param. Usually users will briefly know what the param is and the syntactic based on the last part. (e.g., ``sync\_batch\_window\_in\_ms'')
We first exclude those renamed parameters that caused by feature changed and typo because they are not
due to the original bad name design. We have \needCheck{27} params left.
\needCheck{14/27} param names was changed with ``function/type''. (e.g. change ``.docker.image'' to ``.container.image''.
``{\it Since with CRI, we are not limited to running only docker images.}''~\href{https://issues.apache.org/jira/browse/SPARK-22807}{- Spark-22807}.
\needCheck{9/27} param names was changed with the ``domain''. (e.g., ``{\it Kafka related Spark parameters has to start
with ``spark.kafka.' and not with ``spark.sql.'}''~\href{https://issues.apache.org/jira/browse/SPARK-27687}{- Spark-27687})
\needCheck{4/27} param names was changed with ``somePrefix''. (e.g., ``{Makes our legacy backward compatibility configs more consistent}''~\href{https://issues.apache.org/jira/browse/SPARK-23549}{- Spark-23549})
}
\subsection{Summary}
There is an unmet need for {\it practical} configuration
automation techniques and tools for
choosing and testing parameter values---why do cloud system developers still change parameter values statically
rather than using parameter automation?
There is also need
for automatic ways of identifying workloads or use cases for which
default values (and even constants) are ill-suited.
Such automatic workload identification
approaches can help developers to better 1)~decide which constant
values need to be parameterized, 2)~understand when their
current default values will lead to system failures,
and 3)~come up with better tests and benchmarks for default values.
\Comment{
\tianyin{@Owolabi, @Yuanliang, the implication should connect with auto
tuning given that there are a lot of work there~\cite{aken:17,alipo:17,wzheng:11,wang:18,zheng:07,
yu:18,nair:17,siegmund:15,Nair:2018,Nair:2018:2,Nair:18:3,Hsu:2018,duan:09,
zhu:17,Jamshidi:17,Jamshidi:18,xi:04,ye:03,Herodotou:2011,osogami:06,
Gencer:2015,Krishna:19,Bu:09,Liao:2013,Liu:17,hoffmann:11,
Herodotou:11,Wang:2016,Cheng:2014,Bei:2016,TChen:16}.
We should take a look at what exiting approaches are doing and what they are
missing based on our data.
For example, what types of configuration parameters they are tuning?
Are those parameters that made into configurations tuned by existing tools?}
\tianyin{A simple future work is that we need to automatically know what
are the use cases or workloads that clearly cannot be supported well
by the default values. If we are able to know the precise intent of
each configuration parameter -- we can do much better -- we know
what parameters to tune and what are not (as default values are
perhaps already the best).}
\tianyin{The ultimate question: why these systems are not adopting auto-tuning
techniques but still use static values?}
}
\section{Introduction}
\label{sec:intro}
\input{findings_table}
Software configuration design and implementation have
significant impact on the functionality, reliability, and performance
of large-scale cloud systems\Space{, especially
those operate at the cloud and datacenter scale}.
The idea behind configuration is to expose\Space{ a number of}
{\it configuration parameters} which enable\Space{
system users to perform} {\it deployment}-time system customization.
Using different parameter values, system users (e.g., operators, sysadmins,
and DevOps engineers) can port a software system
to different environments,
accommodate different workloads,
or satisfy new user requirements.
In cloud systems,
configuration parameters are changed constantly.
For example, at Facebook, configuration changes are
committed thousands of times a day,
significantly outpacing source-code changes~\cite{tang:15}.
With the high velocity of configuration changes,
misconfigurations (in the form of erroneous parameter values) inevitably become
a major cause of
system failures,\Space{ and anomalies that lead to} severe service outages, and downtime.
For example, misconfigurations were
the second largest cause of service-level disruptions
in one of Google's main production services~\cite{barroso2018}.
Misconfigurations also contribute to 16\% of production incidents at Facebook~\cite{tang:15},
including
the worst-ever outage of Facebook and Instagram that occurred in March 2019~\cite{Shieber:2019}.
Similar statistics and incidents were reported in other\Space{ types of}
systems~\cite{amvro:16,yin:11,kendrick:12,rabkin:13,gunawi:16,maurer:15,
oppenheimer:03,nagaraja:04}.
\Comment{Many of those incidents are newsworthy, such as
the recent worst-ever outage of
Facebook and Instagram in March 2019~\cite{Shieber:2019}.}
Software configurations\Space{ issues} also impose significant
total cost of ownership on software vendors, who
need to diagnose user-reported
failures or performance issues caused by misconfigurations.
Vendors may even have to compensate users,
if the failures lead to outages and downtime.
Software vendors
also need to support and help users with configuration-related
questions, e.g., how to find the right parameter(s) and set the right
value(s)~\cite{xu:15:2}.
Note that system users
are often not\Space{ software} developers; they may not understand
implementation details or they may not be able to debug code~\cite{xu:13,xu:chi:17,xu:16:2}.
Unfortunately,
configuration design and implementation{} have been largely overlooked as first-class software engineering endeavors in cloud systems,
except for few recent studies (Section~\ref{sec:relatedwork}).
\Space{So far, }The\Space{ main research and development} focus has been on
detecting misconfigurations and diagnosing their consequences\Space{ (e.g.,
failures and performance anomalies)}~\cite{xu:13,xu:16,attariyan:10,
attariyan:12,attariyan:08,zhang:13,zhang:14,
wang:04,wang:03,santo:16,santo:17,dong:15,
huang:15,Baset:2017,sayagh:17,sun:osdi:20}.
These efforts tremendously improve {\it system}-level
defenses against misconfigurations, but they do not address the fundamental need for better
{\it software} configuration design and implementation.
Yet, better configuration design can effectively reduce
user difficulties, reduce configuration complexity while maintaining flexibility,
and proactively reduce misconfigurations~\cite{norman:83a,norman:83b,maxion:05,xu:15:2}.
Also,\Space{ rigorous, systematic} better configuration implementation can help detect and correct
misconfigurations earlier to prevent failure damage~\cite{xu:16,xu:13}.
\Comment{
A few recent studies have demonstrated that
better configuration engineering
could significantly improve the reliability and usability
of software systems.
As shown in~\cite{xu:16}, many catastrophic failures
could be prevented if the software is engineered with the
practice of rigorous configuration
checking at the initialization phase.
It is also reported that software configuration design
is often overly complex and error-prone~\cite{xu:15:2};
a user-centric design can effectively reduce the complexity
while maintaining the desired flexibility.
}
The understanding of what constitutes software configuration engineering in cloud systems is preliminary in the literature,
compared with other aspects of
engineering these software systems
(e.g., software architecture, modeling, API design, and testing) which are well studied\Space{~\cite{}}.
Meanwhile, we observed that developers struggle to design and implement configurations. For example, we found that developers raise many configuration-related concerns and questions---``{\it is the configuration helpful?}''~(\href{https://github.com/apache/spark/pull/22823}{Spark-25676}),
``{\it can we reuse an existing parameter\Space{ purpose}?}''~(\href{https://issues.apache.org/jira/browse/HDFS-13735}{HDFS-13735}),
``{\it what is a reasonable default value?}''~(\href{https://issues.apache.org/jira/browse/HBASE-19148}{HBase-19148}).
Furthermore, we found that
developers frequently revise
configuration design/implementation decisions,
usually after observing severe consequences (e.g., failures and performance issues)
induced by the initial decisions (Sections~\ref{sec:parameterization} and~\ref{sec:removing:parameters}).
\Comment{
\subsection{Contributions}
\label{sec:contribution}
}
This paper presents a\Space{ comprehensive} source-code level study of
the evolution of configuration design and implementation{} in cloud systems, towards filling the knowledge gap
and better understanding the needs that configuration engineering must meet.
Specifically, we study \checked{1178} configuration-related commits{} spanning 2.5 years (2017.6--2019.12) in four large-scale, widely-used, and actively-maintained open-source cloud systems (HDFS, HBase, Spark, and Cassandra).
\Space{The evolution is embodied in version-control {\it commits} related to
software configuration in the revision history of the four selected projects;
furthermore, in these four project, }Each commit that we study is associated with a JIRA/GitHub issue or a Pull Request link which provides more context about the change and the discussions among developers.
(Section~\ref{sec:meth} describes our methodology for selecting
configuration-related commits{}\Space{ from the entire revision history}).
Our goal is to\Space{ provide} understand\Space{ing of} current
configuration engineering practices,
identify developer pain points,
and highlight future research opportunities\Space{ for\Space{ achieving} more
rigorous\Space{, principled} configuration design and implementation{}}.
\Space{In particular, }We focus on analyzing commits
that revise or refine
initial configuration design or implementation decisions,
instead of commits that add or remove parameters as code evolves.
These revisions or refinements were driven by\Space{ real-world}
consequences of misconfigurations\Space{ such as failures and performance issues}.
Our analysis helps to 1)~understand the rationale for the changes,
2)~learn design lessons and engineering principles\Space{ qualitatively and quantitatively},
and 3)~motivate future automated solutions that can prevent such consequences.
To systematically analyze configuration-related commits,
we propose a taxonomy of configuration design and implementation{} changes in cloud systems along three dimensions:
\myitpara{1)~interface:}why and how developers change the configuration interface
(parameters, default values and constraints)\Space{ over time}.
\myitpara{2)~usage:}how developers change and improve parameter value
checking, error-handling, and uses.
\myitpara{3)~documentation:}how developers improve configuration\Space{ information in different forms
of} documentation.
Note that this paper focuses on cloud systems, instead
of desktop software or mobile apps,
because misconfiguring cloud systems results in more far-reaching impact\Space{ due to the scale}.
Moreover, we focus on {\it runtime} configurations~\cite{sayagh18}
whose values can be changed {\it post-deployment} without\Space{ the need to} re-compiling the software.
Runtime configurations fundamentally differ from
{\it compile-time} configurations such as \CodeIn{\#ifdef}-based
feature flags~\cite{meinicke:20}.
But runtime and compile-time configurations have similar problems.
So, there are opportunities to extend techniques that solve problems for one to the other.
This paper makes the following contributions:
\begin{itemize}[topsep=.2ex,itemsep=.2ex,leftmargin=1.75em]
\item[\Contrib{}]\mypara{Study and Insights.}We study code changes to
understand the evolution of configuration design and implementation in cloud systems. We find
insights that motivate future research on
reducing misconfigurations in these systems.
\item[\Contrib{}]\mypara{Taxonomy.}We develop a taxonomy of
cloud system configuration design and implementation evolution.
\item[\Contrib{}]\mypara{Data.}We release our dataset and scripts at
``\url{https://github.com/xlab-uiuc/open-cevo}''
to help followup research (see Appendix B for the replication package).
\end{itemize}
Table~\ref{tab:findings} summarizes our findings and their implications.
\section{Related Work}
\label{sec:relatedwork}
\Space{We presented one of the first studies that systematically analyze
different dimensions in the evolution
of software configuration design and implementation.}
A prior study~\cite{sai:14} found that software evolution necessitates
resetting parameter values and\Space{ the
usage of configuration parameters; thus, when
upgrading to a new software version,
users may need to re-configure the software with different parameter values.
They} built \CodeIn{ConfigSuggester} to identify parameters whose values need to be
changed after a software updates. We study how the configuration interface and parameter
usage change across (a portion of) version control history to draw insights for
better configuration design and implementation{}.
Sayagh et al.~\cite{sayagh18} studied
software configuration engineering in practice using interviews,
user surveys, and a literature review.
Our work is complementary: we perform a \emph{code}-level study of configuration\Space{
engineering with respect to software} evolution, which\Space{.
Our analysis of code-level configuration evolution
from developers' experiences and practices} yields new insights.
There have been many studies
on misconfigurations in a wide variety of software
systems~\cite{amvro:16,tang:15,barroso2018,yin:11,kendrick:12,
rabkin:13,gunawi:16,maurer:15,
oppenheimer:03,nagaraja:04}.
Our work does not focus on detecting misconfigurations
or diagnosing failures caused by misconfigurations.
We focus on current configuration engineering practices,
with the goal to understand how to improve configuration design and implementation.
\Space{
on reducing misconfigurations (as well as other configuration
problems) in the first place, with better
design and implementation.}
Recently, a few studies investigated automated techniques or
engineering practices to enhance configuration
checking code~\cite{xu:16},
diagnosability~\cite{zhang:15},
interface~\cite{xu:15}, security~\cite{Meng:18,xiang:19,xu:chi:17},
configuration data analysis~\cite{xu:18},
configuration libraries~\cite{SayaghSCAM,raab:17},
and correlations or coupling in
configuration and code~\cite{Horton:19,Wen:20,mehta2020rex}.
Our work corroborates and complements the aforementioned work from the perspective
of software evolution. Specifically, our work studies the practices of
{\it software developers} and reveals
how software configuration design and implementation are revised and evolved.
Despite the differences (Appendix A), \Space{software} runtime configurations share commonalities
with compile-time configurations or SPL configurations, such as \CodeIn{\#ifdef}-based feature
flags~\cite{meinicke:20}. It is possible that techniques and methodologies designed for compile-time
configurations, especially feature and variability modeling~\cite{Lotufo:10,Passos:18,Berger2013TSE,Damir2019FSE,Berger:10,Liebig:10},
could be adapted for use with runtime configurations.
Such adaptation needs to address unique challenges of runtime configuration parameters,
such as dependencies on deployment environments, as well as their complex data types
and misconfiguration patterns.
Configuration design and implementation have significant implications
on software testing and debugging~\cite{yilmaz:06,jin2:14,fouche:09,qu:08,reisner:10,song:12}.
For example, introducing new parameters
enlarges the configuration space and thus makes it more costly to comprehensively
test software.
In this paper, we focus on understanding how to improve configuration design and implementation{} so that fewer misconfigurations occur,
and not on software bugs that can occur under different parameter value combinations.\Space{
Understanding the impact of configuration design/implementation on software code quality is our future work.}
\Comment{
We limit our study scope to deployment-time parameters that can be
changed via configuration files or CLIs.
We do not study build-time configurations studied in the variability
and software product line research.
Our work is also complementary to configuration-aware combinatorial testing.
}
\section{Study Setup}
\label{sec:meth}
To understand how configuration design/implementation evolve,
we identified and analyzed {\it configuration-related commits{}} that
modify configuration design and implementation{}.
Following~\cite{sayagh18}, we refer to the design, implementation, and maintenance of software configuration
as \emph{configuration engineering}.
We start from commits instead of bug databases (e.g., JIRA and GitHub issues)
because configuration design and implementation evolution is not limited to bug fixing.\Space{
Thus, bug databases likely contain less information about configuration evolution than commits.
In fact,}
All cloud systems that we study\Space{ enforce the practice of} record
related issue or Pull Request ID(s) in commit messages\Space{, if the commit is part of addressing an
issue} (Section~\ref{sec:meth:software}).
We found detailed context about changes in the configuration-related commits through developer discussions.
Moreover, commits allow us to analyze the ``{\it diffs}''---the actual changes.
\subsection{Target Software and Version Histories}
\label{sec:meth:software}
\Space{In this section, we describe the projects that we study and the portions of their version histories that we use.}
We study configuration design and implementation in four open-source cloud systems, shown in
Table~\ref{tab:software}\Space{ (``Subject{}'' column)}: HDFS{},
HBase{}, Spark{}, and Cassandra{}.\Space{ We selected} These projects\Space{
because they} 1)~have many configuration parameters and
configuration-related commits{}\Space{ Configuration activities are
frequent when developing, deploying and using these software.},
2)~are mature, actively-developed and widely-used, with
well-organized GitHub repositories and bug databases,\Space{.
The information is up-to-date and sufficient for us
to study the evolution of configuration,
3)~use git as their
version-control system and are hosted on GitHub,} (3)~link
to issue IDs in commit messages\Space{ when applicable},
and 4)~are commonly used subjects in cloud and
datacenter systems research.
In these subjects, we studied configuration-related commits{} from June 2017 to
December 2019, a 2.5--year time span.
In Table~\ref{tab:software}, ``\header\#Params{}'' is the total number of documented parameters in the
most recent version, ``\header\#AllC{}'' is the total number of
commits in the 2.5--year span, and ``\header\#StudiedC{}'' is the
number of configuration-related commits{} that we studied. We excluded configuration-related commits{}
that only added or modified test cases; we expected such commits
to yield less insights on design/implementation evolution.
In total, we
studied \checked{1178} configuration-related commits{}. \Space{(\checked{1164} contained
issue IDs)}
\begin{table}
\begin{center}
\footnotesize
\caption{Software and their commits that we study.}
\begin{tabular}{llrrr}
\toprule
\headerSubject{} & \header\#Description{} & \header\#Params{} & \header\#AllC{} & \header\#StudiedC{} \\
\midrule
HDFS{} & File system & \checked{560} & \checked{1618} & \checked{221} \\
HBase{} & Database & \checked{218} & \checked{3516} & \checked{268} \\
Spark{} & Data processing & \checked{442} & \checked{6194} & \checked{602} \\
Cassandra{} & Database & \checked{220} & \checked{1868} & \checked{87} \\
\bottomrule
\end{tabular}
\label{tab:software}
\vspace{-10pt}
\end{center}
\end{table}
\Comment{
\begin{table}
\begin{center}
\footnotesize
\caption{Commits collected by different analysis.}
\vspace{-5pt}
\begin{tabular}{ll}
\toprule
Analysis & Num \\
\midrule
Commit Messages & 384 \\
Code Diff & 794 \\
total & 1178 \\
\bottomrule
\end{tabular}
\label{tab:commit_num}
\vspace{-15pt}
\end{center}
\end{table}
}
\subsection{Data Collection and Analysis}
\label{sec:meth:data_collection}
To find {\it configuration-related commits{}} within our chosen time span,
we\Space{ manually analyzed the commit message and the diff\Space{ (both obtained using the \CodeIn{git show} command)}.
We} wrote scripts to automate the analysis of commit messages
and diffs, filter out irrelevant commits,
and select likely configuration-related commits{}.
Then, we manually inspected each resulting commit and its associated issue.
Overall, we collected 384 commits by analyzing commit messages and
794 commits by analyzing the commit diffs,
yielding a total of 1178 configuration-related commits.
\subsubsection{Analysis of Commit Messages}
\label{sec:message:analysis}
Keyword search on commit messages is commonly used to find
related commits\Space{ in a project's version history}, e.g.,~\cite{sai:14,Luo:2014,Bernardo:18,Rigger:19,Hattori:08,dutta2018testing}.
We manually performed a formative study with hundreds of commit messages and found that three strings commonly occur in configuration-related commits{}: ``config'', ``parameter'' and ``option''.
These strings were previously used in keyword searches\Space{ in previous research}~\cite{sai:14,sayagh18}, and matched 525 times in all four subjects.
We manually inspected these 525 commits and removed commits that did not change configurations, yielding 384 configuration-related commits.
\Comment{Owolabi: (1)~these numbers do not match Table~\ref{tab:software} (2) It would be good to give one or two broad classes of false positives in the keyword search.}
\Comment{
Why number do not match? This number is just for commit log, we also have commit diff analyze.
Two broad reason:
(1) Most are because ``option'' or ``parameter'' in message which is not configuration option. It is actually easy to tell whether
this option is configuration option from the title and commit log. e.g., ``{\it Add -E option in hdfs ``count'' command to...}'' \href{https://issues.apache.org/jira/browse/HDFS-11647}{- HDFS-11647}
So basically such FP won't waste much time. However, there are still some commit that just use option/parameter without using config.
So we still keep those two keywords.
(2) Some Commit log mentions ``config'' but the commit is not changing configuration.
e.g., ``{\it Right now spark lets go of executors when they are idle for the 60s (or configurable time) ...}'' \href{https://issues.apache.org/jira/browse/SPARK-21656}{- Spark-21656}
}
\subsubsection{Analysis of Commit Diffs}
\label{sec:diff:analysis}
\Space{We observed that searching commit messages by keywords is
insufficient---}Many commit messages do not match during keyword search, even
though the diffs show configuration-related changes.
So, we further analyzed diffs\Space{ (the content that developers changed in each commit)} to
find more configuration-related commits{}\Space{ in a more systematic way},
and found additional 794 configuration-related commits{}\Space{, representing between 1.1$\times$ and 5.4$\times$ more configuration-related commits{} per project than the keyword-based search (\S\ref{sec:meth:keyword_search})}.
Our diff analysis determines whether diffs
modify how parameters are defined, loaded, used, or described.
Accurate automated diff analysis requires applying precise taint tracking---treating parameter
values as initial taints that are propagated along control- and data-flow
paths~\cite{xu:13,xu:16,rabkin:11,rabkin2:11,attariyan:10,attariyan:12,zhang:13,sai:14,dong:15,Lillack:14}---to
each commit and its predecessor
and comparing the taint results in \emph{both} commits.
Such\Space{ precise and} pairwise analysis does not scale well to the 13196 commits in all four projects (Table~\ref{tab:software}).
To scale diff analysis, we used a simple text-based\Space{ approach}
\Space{ which implements the principle of analyzing if a given commit
changes how configuration is defined, loaded, and used in the
program. that}search\Space{es Our analysis requires the
understanding of the} of configuration metadata\Space{ in target
software projects}, including the 1)~configuration interface
(including how configurations are defined and loaded), 2)~default
configuration file\Space{ structure}, and 3)~message that contains
configuration information.
\Space{configuration }Metadata are expected to be stable in the mature cloud systems
that we study; commits that modify them may yield good insights on configuration evolution.
\Comment{In our experience, these configuration metadata rarely change
in mature software projects and has been consistent
in the four software projects used in this study.}
\vspace{3pt}
\mypara{Finding commits that change parameter definitions.}We start
from commits that change default configuration files or parameter
descriptions in those files. These two locations are key
user-facing parts of configuration design\Space{As configuration is an
interface for system users}~\cite{xu:13,xu:15}. Thus,\Space{ any}
modification of parameters (introduction, deprecation, changes to default values, etc.)
likely requires changes to either. This
heuristic was effective: it found 272 additional
configuration-related commits{}
with\Space{ a low} an average false positive rate of\Space{ (} 3.2\%\Space{ on average)}.
\vspace{3pt}
\mypara{Finding commits that change parameter loading or setting.}Here, we
leverage knowledge of configuration APIs\Space{ in the subject programs}. As
reported in prior studies~\cite{xu:16,xu:13,Lillack:14,behrang:15,rabkin:11,rabkin2:11} and validated in
our study, mature software projects have unified, well-defined APIs
for retrieving and assigning parameter values. For instance, HDFS{} has\Space{ all
parameter values are retrieved or assigned using}\Space{ a few} getter or setter methods (e.g.,
\texttt{getInt}, \texttt{getBoolean}, declared in a Java class;
each of which has a corresponding setter method (e.g.,
\texttt{setInt}, \texttt{setBoolean}). The other evaluation subjects\Space{
also} follow this pattern.\footnote{This is common in Java and Scala
projects: the configuration interface typically wraps around
core library APIs such as \texttt{java.util.Properties}
to provide configuration getter and setter
methods.} So, identifying commits that changed code containing getter or setter method usage
requires a few lines of\Space{ scripting} code using regular expressions. This heuristic\Space{ was
effective: it} found 457 additional
configuration-related commits{}
with a 19.9\% average false positive rate.
\vspace{3pt}
\mypara{Finding commits that change parameter value data flow.}If a commit changes code
with variables that store parameter values, then that commit is likely related to the data flow of
parameter values.
We implemented a simple text-based taint tracking to track such
variables\Space{. The text-based taint tracking works} as follows.
Once a configuration value is stored in a variable, we add the
variable name to a global taint set.\Space{ The variable name is simply the
left value of an assignment statement, obtained using regular
expressions.} We perform the tracking for every commit in the
time span that we studied. We do not remove variables from our taint set\Space{ for
completeness}. We output candidate commits where a modified statement
contains a variable name in the taint set. Taint tracking found 31 additional
configuration-related commits{}
with an average false positive rate of 26.2\%.
\begin{table}[]
\footnotesize
\caption{Configuration-related commits{} by category. Some commits contain changes in multiple
categories.}
\begin{tabular}{lccccc}
\toprule
& \header{Interface} & \header{Behavior} & \header{Document} & \header{Commit} \\
\midrule
HDFS & \checked{139 (62.9\%)} & \checked{58 (26.2\%)} & \checked{27 (12.2\%)} & \checked{221} \\
HBase & \checked{171 (63.8\%)} & \checked{87 (32.5\%)} & \checked{21 (7.8\%)} & \checked{268} \\
Spark & \checked{367 (61.0\%)} & \checked{182 (30.2\%)} & \checked{61 (10.1\%)} & \checked{602} \\
Cassandra & \checked{54 (62.1\%)} & \checked{32 (36.8\%)} & \checked{5 (5.7\%)} & \checked{87} \\
\midrule
Total & \checked{731 (62.1\%)} & \checked{359 (30.5\%)} & \checked{114 (9.7\%)} & \checked{1178} \\
\bottomrule
\end{tabular}
\label{tab:categ_num}
\end{table}
\vspace{3pt}
\mypara{Identifying other configuration-related commits{}}We applied the same keyword
search on commit messages
(Section~\ref{sec:message:analysis}) to messages that occur in diffs, to capture commits that change
related exception or log messages\Space{. Developers
may change these messages} without modifying any other code.
We found 34 additional
configuration-related commits{}
with an average false positive rate of 29.2\%.
\subsubsection{Inspection and Categorization}
\label{sec:inspection_cat}
At least two authors independently studied each configuration-related commit and its corresponding issue.
They independently categorized each commit based on the taxonomy\Space{
presented} in Section~\ref{sec:background},
and then met to compare their categorization.
When they diverged, a third author was consulted for additional
discussion until consensus was reached.
Further, in twice-weekly project meetings, the inspectors met with a fourth
author to review their categorization of 15\% of commits inspected during the week.
These meetings helped check that understanding of the taxonomy is consistent.
Our experience shows that consistently checking a taxonomy like Figure~\ref{sec:overview}
with concrete examples significantly improves inter-rater reliability
and categorization efficiency.
\Space{We would like to}Note that we categorized each commit based on how it revised
the original configuration design/implementation.
If a commit adds a new parameter and also a manual entry to document this
new parameter,
we treat this commit as Add$_\text{Param}$ (Table~\ref{tab:taxonomy})---the
commit revises the configuration interface instead of documentation.
Some commits modify multiple (sub-)parts in our taxonomy.
\Comment{
\blue{
\mypara{Analysis of commits}
We manually analyze and categorize each commit based on the commit log and diff code.
The key principle is to understand how each commit changes the configuration in
different dimensions described in Table~\ref{tab:taxonomy}. Here we illustrate two examples
to show the process:
The variable diskWriteBufferSize is load from configuration by using a configuration getter method.
Then we track how it is used in the commit diff and find out that the value pass to writeBuffer which
is a hard-coded value (1024*1024). Therefore, we categorize it as Parameterization.
Another example is CASSANDRA-14991.
This commit adds error-handling code for using \texttt{server\_encryption\_options} and \texttt{client\_encryption\_options}.
We categorize this commit into "Evolution of Error-handling Code (Section 5.2)."
}
\vspace{3pt}
\mypara{Analysis of JIRA/GitHub Issues}All \checked{1178} configuration-related commits{}
that we collected\Space{ in this (\S\ref{sec:message:analysis} and
\S\ref{sec:diff:analysis})} have associated JIRA/GitHub issues
(\checked{1164} commits) or are linked to Pull Requests (\checked{14} commits).
We\Space{ manually} inspected the description and developers discussion
in these issues and PRs, to
obtain more\Space{ detailed background and} context about each commit\Space{s},
which helps us validate the categorization and finding insights.
}
\subsubsection{Data Collection Results}
Table~\ref{tab:categ_num} shows the studied configuration-related commits{}
along the three parts of our taxonomy\Space{ in Table~\ref{tab:taxonomy}
(interface, behavior, and documentation)}\Space{ across the four
studied software projects}. There is a significant number of
commits in each part.
The rest of this paper summarizes
our\Space{ per-commit} analysis and provides insights on how configuration design and implementation{} evolve
along these three parts.
\section{Threats to Validity}
We studied cloud systems.
Some of our findings may not generalize to other kinds of software\Space{ (e.g. mobile applications, embedded software)}.
We chose these projects because they are widely used, highly configurable
with lots of parameters, mature, and well maintained.
They also have issue-tracking systems that help us understand the context of configuration-related commits.
Though we selected candidate commits from version control history, we may have missed some configuration-related commits{} due to two limitations.
First, our regular expressions assume standard coding conventions and will not match if developers do not follow these conventions.
Second, our simple, text-based tainting may miss some changes to the data flow of variables that store parameter values.
However, as we mentioned in Section~\ref{sec:meth}, precise pairwise tainting does not scale to all the commits in the range that we studied---we traded off precision for scalability.
All commits selected were manually inspected and categorized through a rigorous
quality-assurance process (Section~\ref{sec:inspection_cat}).
\Space{ the simple approach recognizes configuration-related diff,
the} False positives came mainly from commits that touched lines containing configuration-related variables but
did not change the configuration.
\Comment{CASSANDRA-13578, simplify mx4j configuration,2017-06-07}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,993
|
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An incomplete LU decomposition is used for the solution of the transport-type equations and a globally minimized residual method resolves the pressure correction equation. Turbulence is resolved through the $$k-\\varepsilon$$ model. Dealing with turbomachinery applications, results are presented for two-dimensional compressor and turbine cascades under design and off-design conditions.\n\n##### MSC:\n 76M20 Finite difference methods applied to problems in fluid mechanics 76F10 Shear flows and turbulence 76N10 Existence, uniqueness, and regularity theory for compressible fluids and gas dynamics 76H05 Transonic flows\nFull Text:\n##### References:\n [1] Issa, AIAA J. 15 pp 182\u2013 (1977) \u00b7 Zbl\u00a00354.76052 \u00b7 doi:10.2514\/3.7313 [2] Karki, AIAA J. 27 pp 1167\u2013 (1989) \u00b7 doi:10.2514\/3.10242 [3] McGuirk, AIAA J. 28 pp 1751\u2013 (1990) \u00b7 doi:10.2514\/3.10470 [4] Zhou, Computational Fluid Dynamics 2 pp 749\u2013 (1992) [5] Demirdzic, Int. j. numer. methods fluids 16 pp 1029\u2013 (1993) \u00b7 Zbl\u00a00774.76066 \u00b7 doi:10.1002\/fld.1650161202 [6] Lien, J. Fluids Eng. 115 pp 717\u2013 (1993) \u00b7 doi:10.1115\/1.2910204 [7] Giannakoglou, Int. j. numer. methods fluids 21 pp 1067\u2013 (1995) \u00b7 Zbl\u00a00863.76051 \u00b7 doi:10.1002\/fld.1650211105 [8] Numerical Computation of Internal and External Flows, Wiley, Chichester, (1990). [9] and , \u2019The prediction of laminarization with a two-equation model of turbulence\u2019, Int. J. Heat Mass Transfer, 15, (1972). [10] Majumdar, Numer. Heat Tranfer 13 pp 125\u2013 (1988) [11] Leonard, Comput. Methods Appl. Mech. Eng. 19 (1979) \u00b7 Zbl\u00a00423.76070 \u00b7 doi:10.1016\/0045-7825(79)90034-3 [12] Hayase, J. Comput. Phys. 98 (1992) \u00b7 Zbl\u00a00743.76054 \u00b7 doi:10.1016\/0021-9991(92)90177-Z [13] Schneider, Numer. Heat Transfer 4 pp 1\u2013 (1981) \u00b7 doi:10.1080\/01495728108961775 [14] and , \u2019Design and testing of a controlled diffusion airfoil cascade for industrial axial flow compressor applications\u2019, ASME Paper 90-GT-140, (1990). [15] , and , \u2019Losses prediction in axial flow compressor cascades using an explicit k-$$\\epsilon$$ Navier-Stokes solver\u2019, Proc. 85th PEP\/AGARD Symp. on Losses Mechanisms and Unsteady Flows in Turbomachines, Derby, May (1995). [16] , and , \u20192-D transonic bump flow calculations using an explicit fractional-step method\u2019, Proc. Efficient Turbulence Models for Aeronautics (ETMA) Workshop, Manchester, November (1994) Vieweg, Braunschweig, (1997). [17] and , \u2019Experimental analysis data on the transonic flow past a plane turbine cascade\u2019, ASME Paper 90-GT-313, (1990).\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.","date":"2021-06-15 14:00:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4360013008117676, \"perplexity\": 5845.823558416569}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-25\/segments\/1623487621273.31\/warc\/CC-MAIN-20210615114909-20210615144909-00050.warc.gz\"}"}
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Q: Control to drag and drop buttons in WPF I just want to know which in which control (wrapPanel, listView etc..) its possible to make items draggable? When I say items i mean buttons - want to have x buttons in control and change theirs positions, so if Ill drag button 1 and drop in index 10 i want rest of buttons to change their positions (from 2 to 10, id/position -1).
If that possible in both cotnrols, in which one is easier?
A: This article has a dragging canvas example.Dragging Elements in a Canvas
For the positions you can save the positions of each element in a property and then when you drop an element you get his position, calculate its boundaries. You would then loop through all the elements and modify their position depending on what you need
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
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La cattedrale di Västerås () è la cattedrale luterana di Västerås, in Svezia, e sede della diocesi di Västerås. La cattedrale fu costruita nel XIII secolo e consacrata il 16 agosto 1271, ma successive modifiche ne hanno cambiato l'aspetto originale. La guglia è stata costruita dall'architetto svedese Nicodemus Tessin il Giovane.
La cattedrale contiene tre pale d'altare fiamminghe, tra cui una nella cappella dei battesimi, eseguita ad Anversa agli inizi del XVI secolo, ed una nella cappella degli apostoli, opera di Jan Borman e databile intorno al 1500.
Note
Voci correlate
Chiesa di Svezia
Cattedrali in Svezia
Altri progetti
Collegamenti esterni
Västerås
Architetture del gotico baltico della Svezia
Västerås
Architetture della contea di Västmanland
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"redpajama_set_name": "RedPajamaWikipedia"
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Whether you have just joined the Company and are enrolling in a Company-sponsored medical plan for the first time or you are considering changing your existing medical coverage, you should evaluate the medical plan options based on your needs and experience. Be sure to consider your future as well as your present medical needs when selecting your plan.
When you first begin working at the Company, you will be eligible to enroll in a medical plan (described in the Health Care Participation section) and elect coverage for your Eligible Dependents. Your medical coverage elections for you and your Eligible Dependents begins on the first day of the month following receipt of your elections, provided you complete and return an enrollment form within 31 days of your date of hire or transfer into a Management or Administrative & Technical employee position. For example, if you were hired on February 1 and you return a completed form on February 15, you would be eligible for health care benefits for you and any Eligible Dependents you choose to cover on March 1.
New hires who do not enroll within 31 days will be required to wait until the next Open Enrollment period to enroll, with coverage effective January 1 of the following year.
Depending on where you live, you are eligible to enroll in one of the following two medical plan options: the Health Account Plan (HAP) administered by Anthem Blue Cross (the Anthem HAP) or the Health Account Plan (HAP) administered by Kaiser Permanente Insurance Company (the KPIC HAP). As an alternative, you may elect to decline (opt out of) medical coverage.
You are not eligible for medical coverage if you are an intern, contract worker, agency worker, or hiring hall employee.
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{
"redpajama_set_name": "RedPajamaC4"
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Paspalum pauciciliatum är en gräsart som först beskrevs av Parodi, och fick sitt nu gällande namn av Wilhelm Guillermo Gustav o Franz Francis Herter. Paspalum pauciciliatum ingår i släktet tvillinghirser, och familjen gräs. Inga underarter finns listade i Catalogue of Life.
Källor
Externa länkar
Tvillinghirser
pauciciliatum
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{
"redpajama_set_name": "RedPajamaWikipedia"
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Sage 50 US Announcements, News, and Alerts Year-end question? Sage Live Tutor available weekdays January 2 – 31, 2019 for Sage 50--U.S.
Year-end question? Sage Live Tutor available weekdays January 2 – 31, 2019 for Sage 50--U.S.
Got a year-end question? Instead of calling in and waiting on hold, get live help through Sage Live Tutor—an online "meeting room" with customer questions answered live by a Sage expert.
Can I create a corrected W2 in Sage 50?
Please note that the Sage Live Tutor times are weekdays from 9am to 5pm ET (6am to 2pm PT). The last day that this service is offered is tomorrow (Jan. 31).
http://www.sagecity.com/resources, (scroll to Sage 50 US and click on the Live Chat link in the fifth column from the right).
Select "Correct Completed W-2s" and proceed through the wizard. You'll get to a screen that lists all of your employee's W-2 information, just like when you did the original W-2s. Make whatever changes are needed. It will only print corrected W-2s for the ones that had changes.
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"redpajama_set_name": "RedPajamaC4"
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Jack Wilshere out for further several weeks Arsenal midfielder Jack Wilshere has been ruled out for further few weeks as the reports revealed that he requires a surgery on his left leg.
Arsenal midfielder Jack Wilshere has been ruled out for further few months as the reports revealed that he requires a surgery on his left leg.
The English midfielder has been suffering with a continuous injury problems. He missed most part of last season with an ankle injury which he sustained against Manchester United and after returning from the ankle injury he suffered a new fracture in his left leg in the pre-season and is yet to recover.
The reports revealed that the he was recovering as per the schedule and would have returned on time but the recent comments from Arsenal officials revealed that he has to get a surgery.
Arsenal confirmed in a statement that due to the slow healing of his injury the medical team has decided to do an operation. Furthermore the Gunners confirmed that Wilshere will be out for further three months.
The 23-year-old would be again struggling to get a regular chance in the England senior team for the coming European competition next summer.
Wilshere has suffered so many injuries in his career so far. In 2011, he was injured for 14 months while then last season he spent six months on the sidelines and now again he will be completing six months on the sidelines.
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\section{Introduction}
\label{sec:introduction}
The past century has witnessed a steady rise in atmospheric Green House Gas (GHG) levels with nearly 584 Gt $CO_2$ from fossil fuels, land use change and industrial activity contributing to 0.9\degree C of global temperature increase since 1960 \cite{mora2018bitcoin}. $CO_2$ levels have increased from 250ppm in 1960 to 400ppm in 2020 and current average trends show a rise in natural disasters caused by high temperatures and droughts \cite{zandalinas2021global}. Day and night temperatures have increased worldwide, and the average global temperatures are expected to go up by 3-5\degree C by 2100 according to the Intergovernmental Panel on Climate Change (IPCC) \cite{IPCC}. Evidence suggests a change in the lengths of seasons across the globe due to global warming. Summers in the mid-high latitudes have lengthened while the winters have shortened, as well as shorter spring and autumn periods \cite{wang2021changing}. It has been predicted that even if the GHG levels do not increase beyond current levels, summers will last for nearly half a year while winters will be less than two months long by 2100.
In 2015, leaders from 197 countries settled upon the Paris Agreement with the aim to keep global warming caused by human beings under 2\degree C \cite{rogelj2016paris}. This is already a difficult task given the increase in population, energy consumption and the lack of environment friendly policies by governments worldwide. USA, China, Japan, Germany, and India which have been the main ecological footprint hotspots since 2019, also correspond to the top GHG emission nations across the world \cite{sarkodie2021environmental}. Since 2009, various cryptocurrencies have emerged starting with Bitcoin which was the first well-known application of Satoshi Nakamoto's blockchain technology introduced in 2008 \cite{nakamoto2008bitcoin}. It soon became the biggest cryptocurrency in the world with a market capitalization of USD \$ 614.9 billion as of July 2021 among the 5,655 known cryptocurrencies \cite{coin2021crypto}: Ethereum, Tether, Binance, Cardano, and Dogecoin to name a few; and together they account for to a total market capitalization of USD \$1.39 trillion. Millions of transactions are made every single day to exchange these currencies and their stock markets and operations run 24/7 \cite{stoll2019carbon}. The electrical energy consumption of cryptocurrencies is over-proportionate compared to their technical performance \cite{sedlmeir2020energy} and despite their promising applications, cryptocurrencies have also been contributors responsible for global warming due to their high carbon footprint \cite{de2021bitcoin}. It has been predicted that Bitcoin alone can raise the global temperatures by 2\degree C within the next three decades \cite{mora2018bitcoin}.
\begin{figure*}[!t]
\vspace{-0.1in}
\centering
\includegraphics[width = \textwidth]{images/content_flow.pdf}
\caption{The flow of this review.}
\label{flow}
\vspace{-0.1in}
\end{figure*}
Due to the distributed nature of cryptocurrency networks, obtaining close estimates of electrical energy consumption and carbon footprints is a difficult task. The main source of uncertainty is the mining equipment used \cite{houy2019rational} and the source of energy \cite{koomey2019estimating}. The minimum and maximum power demand estimates for the Bitcoin network according to various studies conducted between 2014 and 2018 have been compiled in \cite{kufeouglu2019bitcoin}. Estimates in the range of 2.5 GW to 7.67 GW \cite{de2018bitcoin}, 1.3 GW to 14.8 GW and 15.47 TWh to 50.24 TWh \cite{kufeoglu2019energy}, and 22 TWh to 105 TWh \cite{zade2019bitcoin} were made for Bitcoin in 2018. The power consumption was later estimated to be 4.3 GW in March 2020, nearly a 68\% share of the top 20 cryptocurrencies drawing a total of 6.5 GW \cite{gallersdorfer2020energy}. This was done without considering auxiliary losses caused by the cooling and mining equipment, and with that premise, the true power draw is expected to be higher. With the consideration that only 20 cryptocurrencies were used in this study, the actual cryptocurrency network power consumption of the 5,654 cryptocurrencies \cite{coin2021crypto} would be much higher than their estimate. Among the latest data on consumption, the University of Cambridge Bitcoin Energy Consumption Index (CBECI) \cite{uoc2021bitcoin} shows theoretical maximum and minimum power consumptions of 26.09 TWh to 174.82 TWh respectively, with an estimate of 69.63 TWh. A study from early 2021 \cite{sedlmeir2020energy} showed a range of 60 TWh to 125 TWh per year for Bitcoin, 15 TWh for Ethereum and 100 TWh for Bitcoin Cash. A sensitivity-based method used by Alex de Vries in early 2021 \cite{de2021bitcoin} factored into the Bitcoin market cost, electricity cost and the percentage of miners' income spent on electricity. The results showed the Bitcoin network energy consumption to be up to 184 TWh. His famous blog, Digiconomist founded in 2014 \cite{digi} estimates the energy consumption of Bitcoin and Ethereum to be 135.12 TWh and 55.01 TWh respectively as of July 2021.
As a consequence of high electrical energy consumption, cryptocurrencies have also been found to have high carbon footprints. The carbon footprint of Bitcoin alone was estimated to be 63 Mt$CO_2$ in 2018 \cite{kohler2019life} and 55 Mt$CO_2$ in 2019 \cite{stoll2019carbon}. Another study in 2018 \cite{sriman2021blockchain} stated a footprint of 38.73 Mt$CO_2$ which was equivalent to Denmark, over 700,000 Visa transaction and nearly 49,000 hours of YouTube viewing. Alex de Vries showed the consumption to be up to 90.2 Mt$CO_2$ \cite{de2021bitcoin} early in 2021 with an estimate of 64.18 Mt$CO_2$ \cite{digi2021bitcoin}. Along similar lines, Digiconomist also calculated a 26.13 Mt$CO_2$ footprint for Ethereum in July 2021. The 3rd Global Cryptoasset Benchmarking Study (GCBS) conducted by the University of Cambridge in 2020 \cite{blandin20203rd} found an average of 39\% of renewable energy share in Proof of Work (PoW) mining while a contesting result was found in a 2018 study \cite{bendiksen2018bitcoin} with a 78\% share of renewable energy. But considering the high carbon footprints for these cryptocurrencies, we can infer that there is still a considerable load on non-renewable sources of energy such as fossil fuels.
From the discussion so far, we found that the energy consumption and carbon footprint of cryptocurrencies are very high. We show later in this work that these metrics are close to if not more than those of several countries and much of this high energy consumption stems from mechanisms used by many of the cryptocurrency implementations. Figure-\ref{flow} presents the organization of this review paper. We summarize the main research contributions of this review as follows:
\begin{itemize}
\item{We present a global perspective on energy consumption and carbon footprints by the two most popular cryptocurrencies namely, Bitcoin and Ethereum. We also present a comparison of energy consumption and carbon emissions of Bitcoin, Ethereum, and the card payment system Visa.}
\item{We identify four underlying factors responsible for high energy consumption and carbon emissions of Bitcoin and Ethereum.}
\item{We discuss possible solutions to address the factors that result in high energy consumption and carbon emissions for cryptocurrencies such as Bitcoin and Ethereum. Additionally, we discuss two case studies on work-in-progress solutions.}
\end{itemize}
\begin{figure*}[!t]
\centering
\includegraphics[width = \textwidth]{images/block_chain.pdf}
\caption{A block diagram depicting the structure of a blockchain.}
\label{blockchain}
\vspace{-0.1in}
\end{figure*}
\begin{table*}[t]
\centering
\caption{Ranking Bitcoin and Ethereum among countries based on annual electrical energy consumption as of July 2021 \cite{digi2021bitcoin,world2021pop,digi2021ethereum,eia2019energy,wiki2019energy} (Note: N.A. stands for Not Available).}
\resizebox{1.5\columnwidth}{!}{
\begin{tabular}{|l|l|r|r|r|}
\hline
\multicolumn{1}{|c|}{\textbf{Rank}} & \multicolumn{1}{c|}{\textbf{Country}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Population \\ (Millions) \cite{world2021pop}\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Energy (TWh) \\ \cite{digi2021bitcoin,digi2021ethereum,eia2019energy,wiki2019energy}\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Share \\ (\%)\end{tabular}}} \\ \hline
0 & World & 7,878.2 & 23,398.00 & 100.00 \\ \hline
1 & China & 1,444.9 & 7,500.00 & 32.05 \\ \hline
2 & U.S.A & 332.9 & 3,989.60 & 17.05 \\ \hline
3 & India & 1,366.4 & 1,547.00 & 6.61 \\ \hline
20 & Taiwan & 23.8 & 237.55 & 1.01 \\ \hline
21 & Vietnam & 98.2 & 216.99 & 0.92 \\ \hline
22 & South Africa & 60.1 & 210.30 & 0.89 \\ \hline \rowcolor{LightCyan}
23 & Bitcoin + Ethereum & N.A. & 190.13 & 0.81 \\ \hline
24 & Thailand & 69.9 & 185.85 & 0.79 \\ \hline
25 & Poland & 37.80 & 153.00 & 0.65 \\ \hline
26 & Egypt & 104.3 & 150.57 & 0.64 \\ \hline
27 & Malaysia & 3.1 & 147.21 & 0.62 \\ \hline \rowcolor{LightCyan}
28 & Bitcoin & N.A. & 135.12 & 0.57 \\ \hline
29 & Sweden & 10.2 & 131.79 & 0.56 \\ \hline
49 & Switzerland & 8.7 & 56.35 & 0.24 \\ \hline \rowcolor{LightCyan}
50 & Ethereum & N.A. & 55.01 & 0.24 \\ \hline
51 & Romania & 19.1 & 55.00 & 0.23 \\ \hline
\end{tabular}}
\label{energy-comp}
\end{table*}
\begin{table*}[t]
\centering
\caption{Ranking of Bitcoin and Ethereum among countries based on annual carbon footprint as of July 2021 \cite{digi2021bitcoin,digi2021ethereum,iea2018carbon,world2021pop}.}
\resizebox{1.5\columnwidth}{!}{
\begin{tabular}{|l|l|r|r|r|}
\hline
\multicolumn{1}{|c|}{\textbf{Rank}} & \multicolumn{1}{c|}{\textbf{Country}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Population \\ (Millions) \cite{world2021pop} \end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Emission (Mt$CO_2$) \\
\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Share \\ (\%)\end{tabular}}} \\ \hline
0 & World & 7,878.2 & 37,077.40 & 100.00 \\ \hline
1 & China & 1,444.9 & 10,060.00 & 27.13 \\ \hline
2 & U.S.A & 332.9 & 5410.00 & 14.59 \\ \hline
3 & India & 1,336.4 & 2,300.00 & 6.2 \\ \hline
38 & Nigeria & 211.3 & 104.30 & 0.28 \\ \hline
39 & Czech Republic & 10.7 & 100.80 & 0.27 \\ \hline
40 & Belgium & 11.6 & 91.20 & 0.24 \\ \hline \rowcolor{LightCyan}
41 & Bitcoin + Ethereum & N.A. & 90.31 & 0.24 \\ \hline
42 & Kuwait & 4.3 & 87.80 & 0.23 \\ \hline
43 & Qatar & 2.9 & 87.00 & 0.23 \\ \hline
49 & Oman & 5.2 & 68.80 & 0.18 \\ \hline \rowcolor{LightCyan}
50 & Bitcoin & N.A. & 64.18 & 0.17 \\ \hline
51 & Greece & 10.3 & 61.60 & 0.16 \\ \hline
76 & Tunisia & 11.94 & 26.20 & 0.07 \\ \hline \rowcolor{LightCyan}
77 & Ethereum & N.A. & 26.13 & 0.07 \\ \hline
78 & SAR & 17.9 & 25.80 & 0.06 \\ \hline
\end{tabular}}
\label{$CO_2$-comp}
\end{table*}
\section{Background}
\label{sec:background}
This section presents a brief overview of blockchain technology and cryptocurrencies. It provides a global perspective of energy consumptions and carbon emissions of the two biggest cryptocurrencies namely, Bitcoin and Ethereum, and compares them to the centralized banking system, Visa.
\subsection{Blockchain, Bitcoin and Cryptocurrencies}
Blockchain is a disruptive technology of distributed ledgers which was created by Satoshi Nakamoto in 2008 \cite{nakamoto2008bitcoin}. A blockchain is a database that chronologically stores information in "blocks". These blocks have a storage capacity of information that consists of the stored information, a time-stamp, the hash value of the previous block, and a unique identification number called the nonce. Once a block has been filled, it is added or "chained" onto the previously filled block thereby creating a "blockchain" as Figure-\ref{blockchain} shows. In addition, any changes to a block are detected by the hash value for that block making it easy to identify fraud \cite{nofer2017blockchain}. Blockchain offers many benefits. First, it stores data chronologically and securely, with a copy of the ledger stored on every node in the cryptocurrency network. Second, the functionality of the network is maintained even if a few participating nodes are removed or malfunction. Third, peer-peer trust is maintained through the consensus mechanism, which removes the need for intermediaries that may not be trustworthy. Blockchain finds applications in various areas such as logistics and supply chain \cite{zeadally2019blockchain, hassija2020survey}, e-commerce \cite{ometov2020overview}, education \cite{grather2018blockchain}, healthcare \cite{ismail2019lightweight}, governance \cite{olnes2017blockchain} and others \cite{pilkington2016blockchain}. It can also be used in telecommunication technology \cite{praveen2020blockchain}, stock exchange \cite{bansal2019smart}, industrial IoT \cite{alladi2019blockchain}, smart city development \cite{hassija2020traffic, hassija2020parking}, energy management \cite{miglani2020blockchain}, Unmanned Aerial Vehicles (UAV) \cite{alladi2020applications}, and smart grids \cite{alladi2019blockchain2}. But the most successful application has been in the banking sector \cite{hileman2017global} with the rise of over 5,000 cryptocurrencie as of July 2021 \cite{coin2021crypto}.
Bitcoin, as described by Satoshi Nakamura, is a peer-peer electronic cash system in which the double spending prevention process is decentralized across various nodes through a consensus protocol. All Bitcoin transactions are time-stamped, and any double spending attempts are rejected. "Bitcoin Miners" play a major role in maintaining consensus over the ledger's state through the PoW (discussed in depth in Section-\ref{sec:problems}) in which they compete with others on the cryptocurrency network to solve resource intensive cryptographic problems to earn the right to add their proposed block onto the chain. The difficulty of the puzzle changes over time to maintain the time to mine a block at nearly 10 minutes \cite{antonopoulos2014mastering}. The miners invest in higher computational power in order to not be left behind in the race of pushing their blocks onto the ledger. Successful attempts are awarded a certain quantity of Bitcoin (BTC) as a reward for each block solved. The reward is halved after every 210,000 blocks, in order to maintain a steady synthetic inflation until the 21 million possible BTC is in circulation \cite{berg2020proof,mora2018bitcoin}. The reward per block has been 6.25 BTC since the most recent halving that occurred on May 11, 2020 \cite{bitcoin2021clock}. With nearly 140,000 blocks left to mine, the next halving is expected to occur on March 26, 2024.
Another popular blockchain network, Ethereum, introduced the concept of a programmable network. Ethereum supports the cryptocurrency Ether (ETH) which has the second highest market capitalization \cite{coin2021crypto}. With the development of the Ethereum Virtual Machine (EVM), the concept of smart contracts (i.e. the automatic execution of contracts when certain conditions are met) was proposed. However, as is the case in Bitcoin, Ethereum is also based on the PoW consensus algorithm and therefore it is associated with the same issues of electrical energy consumption and carbon footprints. Ethereum has proposed Ethereum 2.0 in order to address most of the issues with BTC and ETH which we discuss in more detail in Section-\ref{sec:case}.
\subsection{A Global Perspective: Energy Consumption and $CO_2$ Emissions}
Table-\ref{energy-comp} shows the comparison of electrical energy consumption of Bitcoin and Ethereum obtained from Digiconomist \cite{digi2021bitcoin,digi2021ethereum}. We obtained the country-wise consumption and population data from the U.S. Energy Information Administration database \cite{eia2019energy} and Worldometer \cite{world2021pop} respectively. We calculated the percentage Share of energy consumption as follows:
\begin{equation}
Share = \frac{Energy_i}{Energy_{w}}\times100
\end{equation}
where $Energy_i$ is the energy consumption of the country at rank $i$ and $Energy_w$ corresponds to the total energy consumption of the world as the table shows. Accordingly, the estimates of the total electrical energy consumption share of Bitcoin and Ethereum are 0.58\% and 0.23\%. They rank 28th and 50th with 135.12 TWh and 55.01 TWh of consumption respectively. The University of Cambridge has also arrived at a close estimate of 0.6\% for Bitcoin \cite{uoc2021bitcoin} which supports these calculations. The consumption by Bitcoin is comparable to Sweden (131.79 TWh, 0.56\%), while that by Ethereum is nearly the same as Romania (55 TWh, 0.23\%). Considering the high rated power share of 79.85\% for these two cryptocurrencies among all in circulation as of March 2020 \cite{gallersdorfer2020energy}, the data for the two cryptocurrencies as a single entity has also been considered to obtain a holistic representation. It is worth noting that they together rank 23rd in the world and consume a total of 190.13 TWh of energy annually with a share of 0.81\%, which is equivalent to Thailand (185.85 TWh, 0.79\%).
Table-\ref{$CO_2$-comp} presents a similar ranking, but this time based on the annual $CO_2$ emissions. Data on the emissions of various countries was obtained from the International Energy Agency database \cite{iea2018carbon}. The percentage Share has been calculated in the same manner as for energy consumption. It can be observed from the table that Bitcoin ranks 50th in emissions among the 143 countries in this database, with 64.18 Mt$CO_2$ of emissions and a share of 0.17\%. These values are close to those of Oman (68.8 Mt$CO_2$, 0.18\%) and Greece (61.6 Mt$CO_2$, 0.16\%). The statistics for Ethereum are also significant, with a rank of 77, emissions of 26.13 Mt$CO_2$ and a 0.07\% global share, which is comparable to Tunisia (26.2 Mt$CO_2$, 0.07\%). When the two cryptocurrencies are considered together, they rank 41 in the world with emissions of 90.31 Mt$CO_2$ and a global share of 0.24\% which is nearly the same as Belgium (91.2 Mt$CO_2$, 0.24\%).
\begin{table}[t]
\centering
\caption{Energy consumption and carbon footprints of Bitcoin, Ethereum and Visa (total) as of July 2021 \cite{digi2021bitcoin,digi2021ethereum,impakter2018Visa}.}
\resizebox{\columnwidth}{!}{
\begin{tabular}{|l|r|r|r|c|}
\hline
\multicolumn{1}{|c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Transaction \\ method\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Market cap \\ (\$ Billion)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Transactions/day \\ (Million)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}} Emission \\ (Mt$CO_2$)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Energy consumption \\ (TWh)\end{tabular}}} \\ \hline
Bitcoin \cite{digi2021bitcoin} & 617.05 & 0.4 & 64.18 & 135.12 \\ \hline
Ethereum \cite{digi2021ethereum} & 247.8 & 1.23 & 26.13 & 55.01 \\ \hline
Visa \cite{impakter2018Visa} & 520.62 & 500 & 62,400 & 197.57 \\ \hline
\end{tabular}}
\label{comp-total}
\end{table}
\begin{table}[t]
\centering
\caption{Comparison of energy consumption and carbon footprints per transaction for Bitcoin, Ethereum and Visa as of July 2021 \cite{digi2021bitcoin,digi2021ethereum}.}
\resizebox{0.8\columnwidth}{!}{
\begin{tabular}{|l|r|c|}
\hline
\multicolumn{1}{|c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Transaction \\ method\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Emission \\ (Kg$CO_2$)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Energy consumption \\ (kWh)\end{tabular}}} \\ \hline
Bitcoin \cite{digi2021bitcoin} & 844.13 & 1777.11 \\ \hline
Ethereum \cite{digi2021ethereum} & 59.55 & 125.36 \\ \hline
Visa \cite{digi2021ethereum} & 0.00045 & 0.0015 \\ \hline
\end{tabular}}
\label{comp-trans}
\end{table}
\subsection{Comparison with Visa}
Table-\ref{comp-total} and Table-\ref{comp-trans} present the data available on the energy consumption and $CO_2$ emissions of Bitcoin \cite{digi2021bitcoin}, Ethereum \cite{digi2021ethereum} and Visa \cite{impakter2018Visa,digi2021ethereum}.
Table-\ref{comp-total} shows the annual energy consumption and emission values for the three transactions methods,considering all sources of consumption in Visa. While at first glance it may seem that the total $CO_2$ emission and energy consumption are comparatively high for Visa, it is worth pointing out that the number of daily transactions occurring in the Bitcoin and Ethereum networks is 0.4 million and 1.25 million, i.e. 0.08\% and 0.25\% respectively of the 500 million daily Visa transactions. This implies the over-proportionate consumption in cryptocurrencies which are relatively nascent transaction methods. In addition, the total metrics for Visa have been calculated considering all requirements to run the cooperation offices such as office and server electricity, and commute. Table-\ref{comp-trans} shows the per-transaction estimates for the three transaction methods considering only the computational costs. From the table we observe that the energy consumption and $CO_2$ emission per transaction are very high for Bitcoin and Ethereum. Figure-\ref{btc-eth-Visa} presents a visual comparison between these metrics per transaction. Energy consumption and $CO_2$ emissions for Visa have been plotted by raising their values by a factor of $10^5$. Accordingly, Table-\ref{BE} shows the Break Even ($BE$) values that correspond to the number of Visa transactions that can occur to have total energy consumption and $CO_2$ emission equal to a single transaction of these cryptocurrencies. We calculate $BE$ as follows:
\begin{equation}
\label{eq:BE}
BE_{Visa/i}^M = \frac{M_i}{M_{Visa}}
\end{equation}
\begin{table}[b]
\centering
\caption{Break Even (BE) count for the number of Visa transactions per Bitcoin and Ethereum transaction as of July 2021, obtained from Equation-\ref{eq:BE}.}
\resizebox{0.95\columnwidth}{!}{
\begin{tabular}{|l|l|l|}
\hline
\textbf{Category} & \textbf{$BE^{Energy consumption}$} & \textbf{$BE^{CO_2 emission}$} \\ \hline
Visa/Bitcoin & 1,195,657 & 1,870,875 \\ \hline
Visa/Ethereum & 83,574 & 132,334 \\ \hline
\end{tabular}}
\label{BE}
\end{table}
where $BE_{Visa/i}^M$ is the $BE$ value for Visa with cryptocurrency $i$, which is either Bitcoin or Ethereum. $M$ corresponds to the metric in consideration, energy consumption or $CO_2$ emissions. As Table-\ref{BE} shows, it takes 1,195,657 Visa transactions to use the same amount of electrical energy as one transaction of Bitcoin. Similarly, it takes 83,574 Visa transactions to generate the same carbon footprint as a single transaction of Bitcoin. Similarly the $BE$ counts of Visa to Ethereum are 83,574 for energy consumption and 132,334 for carbon footprint.
\begin{figure}[!t]
\vspace{-0.1in}
\centering
\includegraphics[width = \columnwidth]{images/bit_eth_visa.pdf}
\caption{Electrical energy consumption and $CO_2$ emissions per transaction for Bitcoin, Ethereum and Visa \cite{digi2021bitcoin,digi2021ethereum}.}
\label{btc-eth-Visa}
\vspace{-0.1in}
\end{figure}
\begin{figure}[b]
\vspace{-0.1in}
\centering
\includegraphics[width = \columnwidth]{images/mining_flow_chart.pdf}
\caption{The mining process in Bitcoin.}
\label{mining-flow}
\vspace{-0.1in}
\end{figure}
\section{Problems}
\label{sec:problems}
Based on a review of past studies, we have identified four major responsible for the high energy consumption and $CO_2$ emissions in cryptocurrencies, namely: the Proof of Work consensus mechanism, redundancy in operation and traffic, mining devices, and the energy sources. This section discusses these issues so that future development in cryptocurrencies can take them into consideration.
\subsection{Consensus Mechanism: Proof of Work}
PoW was the first consensus mechanism proposed for blockchain networks \cite{nakamoto2008bitcoin}. Paul Haunter, a contributor of Ethereum, acknowledged the high energy requirements of PoW \cite{ieee2019ethereum} being the reason for the development of Ethereum 2.0 which we discuss in more detail in Section-\ref{sec:case}. While redundancy in the operation and traffic of cryptocurrency networks is also a contributor to energy consumption (as we discuss in the next subsection), the transactions themselves do not consume as much energy as the PoW process does. It has been proven that PoW mining has high computational needs and thus imposes major limitations on the continuous use and scalability of cryptocurrencies \cite{mishra2017energy,hassijaframework}. Recent research estimates that PoW mining in Bitcoin consumes nearly 18GW of power for 100 million transactions a week \cite{mishra2017energy} making the practical use Bitcoin questionable. Based on current trends, a study from 2021 has predicted that, because of the rapid growth of cryptocurrencies, PoW mining processes in China alone will consume nearly 300 TWh of electrical energy and generate 130 Mt$CO_2$ by 2024 \cite{jiang2021policy}. To understand why it is an important energy issue, we need to first understand its operation.
Figure-\ref{mining-flow} shows the mining process in Bitcoin using the PoW. Each new block proposed every T minutes is given a hash that is computed using the 256-bit hash of the previous block, the Nonce and the Merkle root using the equation:
\begin{equation}
SHA256(H_{prev} + M_{B} + Nonce) \leq Target
\end{equation}
where SHA256 is the hash function, $H_{prev}$ refers to the 256-bit hash of previous block, Nonce is a one time use positive number and $M_{B}$ is the Merkle root. Once the hash has been calculated, it is compared with the target hash value. This target value is set to increase the difficulty of mining so as to maintain a constant time for the block to be added to the chain. This time is set to 10 minutes for Bitcoin. If the hash is higher than the target, the Merkel root is changed, the nonce is re-calculated and another hash is generated. This process is repeated until the miner reaches a hash value below the target value set.
It is computationally expensive to find the nonce and therefore provides the proof of the amount of computational power put in by the miner, thereby giving this consensus mechanism the name PoW. Since the solution searching process cannot be sped up by parallelization and alternative algorithms \cite{wang2019survey}, a miner's share of reward can be equated to the share of computational power owned in the cryptocurrency network \cite{de2021bitcoin}. As mining becomes harder over time, the PoW becomes an arms race of computational power and resources because miners with more powerful devices compute more hashes per second.
\subsection{Redundancy in Traffic and Operation}
While PoW blockchains have energy problems which stem mainly from the consensus mechanism, energy consumption due to redundant operations and network traffic becomes more relevant in non-PoW blockchains. It arises from the system storing the complete ledger on all nodes in the network \cite{jia2018elasticchain}. In addition, each node performs operations associated with the transactions independently, based on the available transaction information. Additionally, redundant network traffic is another contributor to this problem \cite{zhang2021traffic}. Redundancy reduces the efficacy of the system \cite{zhang2021traffic} while also increasing the total electrical energy consumption \cite{sedlmeir2020energy}.
As stated in \cite{sedlmeir2020energy}, redundancy in the network arises from the number of nodes and the workload on each node. In \cite{zhang2021traffic}, simulation results obtained for network traffic redundancy showed its impact with network size, number of peers, and routing length. A linear relation was found between the total number of peers and the traffic redundancy in the Bitcoin network, with over 98\% of network traffic being redundant showing inefficiency in the current Bitcoin broadcasting algorithm. Every 1000 nodes were shown to increase the effective traffic by 0.3 GB, while the total traffic increased by 24 GB showing a redundancy of 23.7 GB. In addition, the study found a positive correlation between the routing path length and traffic redundancy in the network, demonstrating that denser networks consisting of shorter routing lengths have less redundant traffic.
\subsection{Mining Devices}
In \cite{economist}, the authors argued that if all mining facilities utilized the highly efficient ASIC-based mining devices as done in the KnCMiner Facility in Sweden, the overall Bitcoin mining process would consume nearly 1.46 TWh worldwide which is much lower than the current estimates of 184 TWh \cite{de2021bitcoin}, 135.12 TWh \cite{digi2021bitcoin} and 69.63 TWh \cite{uoc2021bitcoin} for 2021.This discrepancy demonstrates that inefficient mining devices are being used worldwide. Thus, a major contributor of energy consumption is the use of inefficient mining devices \cite{houy2019rational,kufeouglu2019bitcoin}.
\begin{table}[!t]
\vspace{0.05in}
\centering
\caption{Performance metrics of different mining devices (Sources: \cite{stoll2019carbon,kufeouglu2019bitcoin,economist}).}
\resizebox{\columnwidth}{!}{
\begin{tabular}{|l|l|l|l|} \hline
\textbf{Hardware type} & \textbf{Mining rate (GH/s)} & \textbf{Efficiency (J/GH)} & \textbf{mEC (TWh)} \\ \hline
CPU & 0.01 & 9000 & 11,000 \\ \hline
GPU & 0.2 – 2 & 1500 – 400 & 3,000 \\ \hline
FPGA & 0.1 – 25 & 100 – 45 & 250 \\ \hline
ASIC & 44,000 & 0.05 & 1.46 \\ \hline
\end{tabular}}
\label{devices}
\end{table}
\begin{figure}[t]
\vspace{-0.1in}
\centering
\includegraphics[width = \columnwidth]{images/log_mining.pdf}
\caption{{Logarithmic plot of electrical energy versus hashes calculated for Application-Specific Integrated Circuit (ASIC), Field-Programmable Gate Array (FPGA), Graphics Processing Unit (GPU) and Central Processing Unit (CPU) devices \cite{houy2019rational}.}}
\label{log-device}
\vspace{-0.1in}
\end{figure}
\begin{figure*}[!t]
\vspace{-0.1in}
\centering
\includegraphics[width = 2\columnwidth]{images/new_eecs.pdf}
\caption{Energy consumption per transaction for various cryptocurrencies and their consensus mechanisms.}
\label{fig:cryptocomp}
\vspace{-0.1in}
\end{figure*}
Due to the increasing difficulty of mining, several devices have been used since the introduction of Bitcoin starting from CPU in 2009, to GPU in 2010, FPGA in 2011 and ASIC since 2013 \cite{stoll2019carbon}. Table-\ref{devices} presents these devices along with their hash rates, efficiencies \cite{houy2019rational}, and minimum total energy consumption \cite{kufeouglu2019bitcoin,economist}. The total energy consumption corresponds to the amount of energy used when only that type of device is used for Bitcoin mining worldwide. From the table, we note that CPU provides the least computational power calculated in giga-hashes (GH) at 0.01 GH/s at 9000 J/GH, while consuming the most energy per GH, while ASIC-based devices used in the study provide the highest computational power with 44,000 GH/s at an efficiency of 0.05 J/GH. Figure-\ref{log-device} is a logarithmic-plot of log-energy consumption against the number of GH computed. The plot depicts the high energy consumption of CPU devices followed by GPU, FPGA and ASIC being the most efficient among them. It is important to note that ASIC-based devices provide 40,000 to 200,000 times the computational power of GPUs which can be seen from in Table-\ref{devices}. They create a problem of centralization of computational power \cite{mariem2020all}. ASIC-resistant algorithms remove the benefit of using ASIC-based devices, because reaching a solution for these algorithms with ASIC deices is either impossible or comparable to GPUs. Such algorithms, for instance, X16Rv2 in Ravencoin and Ethash in Ethereum, force miners to use general purpose and cheaper devices such as GPUs, which cause an over-proportionate amount of energy consumption \cite{gallersdorfer2020energy}.
In 2019, the authors of \cite{kufeouglu2019bitcoin} investigated the power demand of Bitcoin mining by considering the performance of 269 mining hardware devices (111 CPU, 111 GPU, 4 FPGA and 43 ASIC) in a 160 GB Bitcoin network. They used data published by the manufacturer in whitepapers corresponding to the device, and also the user-benchmark \cite{userbenchmark} and pass-mark \cite{passmark} websites for manufacturer reliability. The study also considered mining pools \cite{miningpools} to make estimates using the regional electricity costs. Two metrics were defined namely the Minimum Energy Consumption (mEC) and Maximum Energy Consumption (MEC), corresponding to the energy consumption of the most efficient and the least efficient devices respectively relative to context. Calculations of the mEC showed that in comparison to the global energy demand of 23,000 TWh, continued use for only CPU devices alone would consume a minimum of 11,000 TWh of electrical energy. GPU and FPGA devices were shown to consume a minimum of nearly 3,000 TWh, 250 TWh respectively while ASIC devices consumed the least among all devices. The minimum and maximum power demands for all devices were shown to be 2 GW and 6 GW respectively.
\subsection{Sources of Energy}
The annual carbon footprint of Aluminium mining has been estimated at 90 Mt$CO_2$ \cite{de2021bitcoin} and that of Oman is 68.8 Mt$CO_2$ as Table-\ref{$CO_2$-comp} shows. From our earlier discussion on global comparisons of carbon footprints of cryptocurrencies in Section-\ref{sec:background}, considering the latest emission estimates of upto 90 Mt$CO_2$ \cite{de2021bitcoin} and 64.18 Mt$CO_2$ \cite{digi2021bitcoin} for Bitcoin and 26.13 Mt$CO_2$ for Ethereum \cite{digi2021ethereum} respectively, it is alarming to see nation-level and industry-level carbon emissions from relatively nascent transaction systems.
The earliest research on the impact of energy consumption on ecological footprints \cite{chen2007ecological} found the negative impacts of fossil fuels on the environment. Subsequently, research indicated a negative impact on the ecological footprint due to the excessive use of fossil fuels by the pulp production industry in the Canadian Prairies \cite{kissinger2007wood}. It is important to note that the results of these studies can also extend to non-renewable energy based cryptocurrency mining. While the sources of energy themselves do not cause the over-proportionate electrical energy consumption in PoW blockchains, the use of non-renewable energy sources leads to high carbon footprints \cite{liu2021selection}.
Due to the cryptocurrency networks being distributed, it is difficult to obtain an accurate share of renewable and non-renewable sources of energy during mining \cite{koomey2019estimating}. Additionally, there is also uncertainty in estimations based on the mining devices used \cite{houy2019rational}. While some studies \cite{bendiksen2018bitcoin,li2018life} argue that the main source of energy for cryptocurrencies is renewable with a share of nearly 80\%, the 3rd GCBS \cite{blandin20203rd} shows a 61\% reliance on non-renewable sources of energy. Statistics of cryptocurrency mining in China show a 58\% and 42\% split of hydro-energy and coal-heavy power generation respectively according to a recent study \cite{stoll2019carbon}. The research has estimated an adjustment emission factor of 550 g/kWh for China by considering a weighted average of the hydro-rich and coal-heavy provinces of Sichuan and Inner Mongolia. Considering the mining pool share based on hashrate (number of hashes computed per second) of 46\% in China \cite{uoc2021bitcoin} and the prediction of 130 Mt$CO_2$ of the Bitcoin network by 2021 in China alone \cite{jiang2021policy}, along with the continued use of fossil fuels, there is a real threat to the environment all of which make the expected rise of 2\degree C contributed by Bitcoin within the next few decades a real possibility \cite{mora2018bitcoin}. It is therefore necessary to find green solutions to minimize the $CO_2$ emissions of cryptocurrencies.
\section{Solutions}
\label{sec:solutions}
As we have discussed in the previous section, current day cryptocurrencies impose problems of high energy consumption and $CO_2$ emissions because of various reasons such as the PoW, redundancy, device efficiency and sources of energy. This section explores and recommends solutions to address these issues by providing examples from past works in individual areas of alternative consensus mechanisms and redundancy reduction. It also discusses some of the most popular and effective mining devices as of July 2021, and conducts an in depth analysis of renewable energy sources in top mining areas as alternatives to fossil fuels to reduce the carbon footprints of cryptocurrencies.
\begin{table*}[t]
\centering
\caption{Proposed ASIC-based mining devices for cryptocurrency mining as of March 2021 \cite{tech2021mining}.}
\resizebox{\textwidth}{!}{
\begin{tabular}{|l|r|r|r|r|c|}
\hline
\multicolumn{1}{|c|}{\textbf{Device}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Cost (\$)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Hashrate (TH/s)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Power (W)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}}Efficiency (J/TH)\end{tabular}}} & \multicolumn{1}{c|}{\textbf{\begin{tabular}[c]{@{}c@{}} MAC (GWh)\end{tabular}}} \\ \hline
Whatsminer M32-70 & 6,200 & 70 & 3,360 & 48 & 29.43 \\ \hline
Antminer S7 & Variable & 4.73 & 1,293 & 273.36 & 11.32 \\ \hline
AvalonMiner 1246 & Variable & 90 & 3,420 & 38 & 29.95 \\ \hline
WhatsMiner M32-62T & 1,075 & 62 & 3,348 & 54 & 29.32 \\ \hline
AvalonMiner A1166 Pro & 2,199 & 81 & 3,400 & 41.97 & 29.78 \\ \hline
\end{tabular}}
\label{asic-device}
\end{table*}
\subsection{Alternative Consensus Mechanisms}
Out of the four issues we have discussed above, the consensus mechanism is the biggest contributor to the energy consumption in current cryptocurrencies that use PoW. Thus, one viable option would be to explore other consensus mechanisms which are more energy-efficient than PoW. Figure-\ref{fig:cryptocomp} provides the Electrical Energy Consumption per transaction (EEC/trans) for various cryptocurrencies compiled from the studies \cite{digi2021bitcoin,digi2021ethereum,tgr2021data,platt2021energy}.
One of the most promising substitutes for the PoW is the Proof of Stake (PoS) consensus mechanism which was first used in Peercoin \cite{kingppcoin} as an energy saving alternative of PoW. It has also been proposed in Ethereum 2.0 which is discussed in Section-\ref{sec:case}. In PoS, the proof is derived from stakes, i.e. contributions of miners to the blockchain, instead of computational power. This removes the computational race involved in the PoW thereby reducing energy consumption and $CO_2$ emissions during mining \cite{nguyen2019proof}. In a study from 2014 \cite{bentov2014proof}, the authors extended the PoW using PoS and proposed the Proof of Activity (PoA) which provided reduced network communication and storage requirements without compromising on security. Proof of Burn (PoB) is another low energy consuming consensus mechanism \cite{karantias2020proof}. Miners reach a consensus by "burning" coins and permanently remove them from circulation. This process is initiated by miners on virtual mining rigs instead of physical mining devices. A miner's mining power increases when the number of coins burned, and not based on computational power. PoB has been proven to be sustainable and highly decentralized, and is implemented in cryptocurrencies such as SlimCoin.
Hedera is an exo-friendly cryptocurrency with a highly efficient consensus mechanism called Hashgraph \cite{baird2016swirlds}, based on the gossip protocol. Participants in the blockchain relay novel information (called gossip), and the collaborative gossip history is stored as a hashgraph, which each member in the network uses to comes to a consensus based on their knowledge of what another node might know. The authors f \cite{popov2021fpc} proposed probabilistic mechanism called the Fast Probabilistic Consensus (FPC). It is used in the cryptocurrency IOTA. It is a highly efficient and secure binary voting protocol wherein a set of nodes can come to consensus on the value of an individual bit instead of consensus through computation. A trust-based mechanism called the XRP Consensus \cite{chase2018analysis} has also been proposed, in which the participants reach an agreement without complete consensus among all members of the network. Hashgraph, FPC and XRP do not require high computational power and therefore consume substantially lower energy than PoW, which can also be seen in Figure-\ref{fig:cryptocomp}.
The authors of \cite{luu2015scp} proposed the Stellar Consensus Protocol (SCP) based on the Byzantine Agreement \cite{dolev1983authenticated}. It removes time-limitations for the processing of blocks by enabling flexibility in the PoW-difficulty parameters and processes several blocks in parallel. Increased computational power therefore increases the throughput of the system, thereby increasing the scalability and sustainability because there are more blocks processed in the same amount of time, making the energy consumption proportionate to the outcome obtained. SCP is used in the Pi Network \cite{pinet} which we discussed in more detail in Section-\ref{sec:case}.
\begin{figure*}[!t]
\vspace{-0.1in}
\centering
\includegraphics[width = 2\columnwidth]{images/asic.pdf}
\caption{Energy efficiency (energy consumption per TH) of various ASIC devices.}
\label{asic}
\vspace{-0.1in}
\end{figure*}
Several storage-based consensus mechanisms have been proposed. The authors of \cite{miller2014permacoin} proposed a consensus mechanism based on distributed storage called the Proof of Retrievability (PoR). However, because the proposed scheme lacks a leader node election method, a similar PoR-based approach was proposed in \cite{moran2019simple} called the Proof of SpaceTime (PoST). PoST proves that useful data was stored for a certain amount of time, and it is thus a storage power-based consensus mechanism. PoST consumes less energy because the difficulty of the proof can be changed by extending the time-period of data stored instead of computational capacity. Another storage share based consensus mechanism called the Proof of Space (PoSpace) was adopted in SpaceCoin \cite{park2015spacecoin} and CHIA \cite{cohen2019chia}. PoSpace requires little computation power and can be run on any free computer with free disk space and an Internet connection.
The authors of \cite{ball2017proofs} recommend the adoption of useful Proofs of Work (uPoW) based on the Orthogonal Vectors (OV) problem. They explain usefulness as the allocation of computational tasks to the miners such that the solutions for the tasks can be reconstructed verifiably and quickly from the miners' response. uPoW converts the amount of wasteful work in PoW into useful work without compromising on hardness. Research on Resource Efficient Mining (REM) \cite{zhang2017rem} for Bitcoin proposed the REM framework using trusted hardware (Intel SGX) and developed the first complete implementation of SGX-blockchain with a computational overhead of 5-15\%. This mechanisms is similar to the uPoW \cite{wang2019survey}. Clients supply their workloads as tasks to the SGX protected enclave. The truthfulness guaranteed feature of the attestation service in SGX verifies and measures the software running in the enclave. The enclave randomly decides which computational task leads to a valid proof for the block.
\subsection{Redundancy Reduction Techniques}
Among the methods proposed in the literature for reducing storage redundancy in blockchain networks, a promising one relies on "sharding", i.e. breaking the network into sub-parts called "shards" based on the consensus mechanism and updating the transactions within the bounds of each shard \cite{sedlmeir2020energy}. In \cite{zamani2018rapidchain}, the authors conducted research on scaling blockchain via sharding, and proposed a stable sharding technique with a low failure rate. The concept of sharding has also been proposed for Ethereum 2.0 which we discuss in Section-\ref{sec:case}. While the division of blockchain networks into shards is difficult because of the decentralization of computational power in the PoW, it can be done based on the proportions of stakes and storage in the case of PoS and PoSpace respectively \cite{nguyen2019proof,wang2019survey}. In \cite{jia2018elasticchain}, the authors proposed another method (called ElasticChain) to reduce redundancy. In ElasticChain, the nodes of the chain store a part of the complete ledger based on a duplicate ratio regulation algorithm. The research shows stability, security and fault tolerance at the same level as the current blockchain design, while improving its storage scalability. The authors of \cite{xue2020semantic} proposed a different approach with Semantic Differential Transaction (SDT) to reduce redundancy in the integration of Building Information Modeling (BIM) and blockchain. SDT captures local changes in an information model as BIM Change Contracts (BCC) at 0.02\% the size of Industry Foundation Classes (IFC) (the standard of ensuring interoperability across BIM platforms, safeguarding them in a blockchain and restoring them when needed). SDT thus reduces redundancy in BIM-blockchain systems. A study on network traffic redundancy \cite{zhang2021traffic} recommends reducing the average routing path lengths between two nodes in order to reduce traffic redundancy in the Bitcoin network.
Another category of methods proposed to reduce operational redundancy in blockchains lies in the use of Zero Knowledge Proofs (ZKP) such as SNARKS \cite{yang2020zero,pinto2020introduction}. ZKP does not require complex encryption. It increases privacy of users by avoiding the disclosure of personal information as is the case in public blockchains such as Bitcoin. Additionally, it provides security while increasing the scalability and throughput of the cryptocurrency network, thereby making it more energy-efficient. The methodology proposed by the authors of \cite{ben2019scalable} uses ZKP to reduce the time needed to prove and verify large sequential computations in comparison to other current ZKP implementations \cite{ishai2015zero}.
\begin{figure*}[t]
\vspace{-0.1in}
\centering
\includegraphics[width = 1.5\columnwidth]{images/pie_chart_mine_share.pdf}
\caption{Mining share based on hashrates \cite{uoc2021bitcoin}.}
\label{mining-share}
\vspace{-0.1in}
\end{figure*}
\subsection{Choice of Mining Device}
While efficient devices will help reduce energy costs regardless of the consensus mechanism, if the PoW continues to be in use, it becomes imperative to use highly efficient devices such as ASIC \cite{houy2019rational}. As Table-\ref{devices} and Figure-\ref{log-device} show, ASIC-based devices consume the least amount of energy per hash, and provide the highest computational power with a hash rate of 40,000 GH/s at 0.05 J/GH. Studies have shown that the use of ASIC devices as done in the KnCMiner facility in Boden, Sweden can reduce the worldwide annual energy consumption by mining to 1.46 TWh \cite{economist,kufeouglu2019bitcoin}.
In \cite{tech2021mining}, the author discusses the top five ASIC-based mining devices as of March 2021 which include Whatsminer M32-70, Antminer S7, AvalonMiner 1246, WhatsMiner M32-62T, and AvalonMiner A1166 Pro. Table-\ref{asic-device} presents the cost, hashrate, power consumption, efficiency and Maximum Annual Consumption (MAC) for each of these devices. Figure-\ref{asic} shows that among the most popular available options, Antminer S7 is the least efficient device, with the energy consumption per terra-hash (TH) of 273.36 J/TH. The other four devices have comparable efficiencies, with AvlonMiner 1246 being the most energy-efficient at 38 J/TH. In addition to the efficiencies, we have also calculated the MAC (in GWh) of these devices as follows:
\begin{equation}
MAC = \frac{P\times24\times365}{10^6}
\end{equation}
wherein, $P$ (in W) is the power consumption of the device which is multiplied with the total number of hours in a year as $24\times365$ to obtain the annual energy consumption equivalent. The table shows that Antminer S7 consumed the least amount of energy, while providing the least efficiency among the five options. The other four ASIC-based mining devices have comparable MAC ranging from 29.32 GWh to 29.95 GWh, thereby further demonstratingthat the best choice is AvlonMiner 1247 based on its efficiency.
\begin{table*}[!t]
\centering
\caption{Mining shares, renewable energy capacities installed and evaluation metrics: Estimated Energy Consumptions (EEC), Maximum Energy Generation (MEG), Renewable Capacity Ratio (RCR) for major Bitcoin mining regions.}
\resizebox{\textwidth}{!}{
\begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|r|}
\hline
\multicolumn{1}{|c|}{\multirow{2}{*}{\textbf{Region}}} & \multicolumn{2}{c|}{\textbf{Bitcoin mining \cite{uoc2021bitcoin,digi2021bitcoin}}} & \multicolumn{7}{c|}{\textbf{Renewable energy capacity installed (MW \cite{irena2020renewable}}} & \multicolumn{1}{c|}{\multirow{2}{*}{\textbf{\begin{tabular}[c]{@{}c@{}} MEG \\ (TWh)\end{tabular}}}} & \multicolumn{1}{c|}{\multirow{2}{*}{\textbf{RCR}}} & \multicolumn{1}{c|}{\multirow{2}{*}{\textbf{Relative RCR}}} \\ \cline{2-10}
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{\textbf{Share (\%)}} & \multicolumn{1}{c|}{\textbf{EEC (TWh)}} & \multicolumn{1}{c|}{\textbf{Total capacity}} & \multicolumn{1}{c|}{\textbf{Hydropower}} & \multicolumn{1}{c|}{\textbf{Wind}} & \multicolumn{1}{c|}{\textbf{Solar}} & \multicolumn{1}{c|}{\textbf{Bioenergy}} & \multicolumn{1}{c|}{\textbf{Geothermal}} & \multicolumn{1}{c|}{\textbf{Marine}} & \multicolumn{1}{c|}{} & \multicolumn{1}{c|}{} & \multicolumn{1}{c|}{} \\ \hline
World & 100 & 135.12 & 2,799,094 & 1,331,889 & 733,267 & 713,970 & 126,557 & 14,050 & 527 & 24520.06 & - & - \\ \hline
China & 46 & 62.15 & 894,879 & 370,160 & 281,993 & 254,335 & 18,687 & 0 & 5 & 7839.14 & 126.12 & 0.4134 \\ \hline
U.S.A & 16.8 & 22.70 & 292,065 & 103,058.00 & 117,744 & 75,572 & 12,372 & 2,587 & 0 & 2558.48 & 112.70 & 0.3694 \\ \hline
Kazakhstan & 8.2 & 11.07 & 4,997 & 2,785 & 486 & 1,719 & 8 & - & - & 43.77 & 3.95 & 0.0129 \\ \hline
Russia & 6.8 & 9.18 & 54,274 & 51,811 & 945 & 1,428 & 1,370 & 74 & 2 & 475.44 & 51.74 & 0.1696 \\ \hline
Iran & 4.6 & 6.21 & 12,922 & 13,233 & 303 & 414 & 12 & - & - & 113.19 & 18.21 & 0.0597 \\ \hline
Malaysia & 3.4 & 4.59 & 8,699 & 6,275 & & 1,493 & 931 & - & - & 76.20 & 16.58 & 0.0543 \\ \hline
Canada & 3 & 4.05 & 101,188 & 81,058 & 13,577 & 3,325 & 3,383 & - & 20 & 886.40 & 218.67 & 0.7168 \\ \hline
Germany & 2.8 & 3.78 & 131,739 & 10,720 & 62,184 & 53,783 & 10,364 & 40 & - & 1154.03 & 305.02 & 1.0000 \\ \hline
Ireland & 2.3 & 3.10 & 4,685 & 529 & 4,300 & 40 & 107 & - & - & 41.04 & 13.20 & 0.0432 \\ \hline
Other & 6 & 8.24 & 1,293,646 & 692,260 & 251,735 & 321,861 & 79,323 & 11,349 & 500 & 11332.33 & - & - \\ \hline
\end{tabular}}
\label{renewable}
\end{table*}
\subsection{Renewable Sources of Energy}
Considering the high electrical energy consumption in current PoW blockchains, we need to address the impact of their emissions and the deterioration of the ecological footprint. Research shows a reduction in $CO_2$ emission by using renewable sources of energy \cite{dong2020renewable}. Sustainable Development Goals (SDG) for economic growth and trade provided by a study on renewable and non-renewable energy and their impact \cite{destek2020renewable} recommends the transition from fossil fuels to renewable energy sources, implementation of environmental friendly production processes, enforcement of green trade, education, and creating awareness. While these recommendations have been provided for sustainable economic growth and trade in general, they are also applicable to cryptocurrencies. In \cite{liu2021selection}, the authors show that legal criteria, and the continuity and cost of electrical energy supply are the most important factors considered to decide the location of cryptocurrency mining operations. The study concluded that wind and solar energy are the best energy alternatives for blockchain networks. The use of these renewable energy sources will make the high energy consumption in PoW cryptocurrencies more environmental friendly. Subsequently it is highly recommended that countries with high cryptocurrency mining activity should invest in the use of renewable energy.
Figure-\ref{mining-share} shows the distribution of mining shares based on hashrates as of July 2021. We obtain the data from the Cambridge Bitcoin Energy Consumption Index \cite{uoc2021bitcoin}. The major Bitcoin mining countries of the world are China (46\%), U.S.A (16.8\%), Kazakhstan (8.2\%), Russia (6.8\%), Iran (4.6\%), Malaysia (3.4\%), Canada (3\%), Germany (2.8\%) and Ireland (2.3\%). Considering the Digiconomist \cite{digi2021bitcoin} estimate of 135.12 TWh for Bitcoin, and energy shares of these to be equal to the hashrate shares, we can calculate their Estimated Energy Consumption ($EEC$) as follows:
\begin{equation}
EEC = \frac{135.12\times Share (\%)}{100}
\end{equation}
From Table-\ref{renewable}, we note that China alone consumed 62.15 TWh of electrical energy, which is comparable to the electrical energy consumptions of Switzerland (56.35 TWh). It is therefore important for these major mining regions to focus on measures to minimize the environmental degradation and global warming caused by the PoW mining processes.
\begin{figure*}[!t]
\vspace{-0.1in}
\centering
\includegraphics[width = 2\columnwidth]{images/ratio_renewable.pdf}
\caption{Renewable Capacity Ratio for major Bitcoin mining regions.}
\label{ratio}
\vspace{-0.1in}
\end{figure*}
Table-\ref{renewable} presents the data, provided by the International Renewable Energy Agency (IREA) \cite{irena2020renewable}, on various infrastructures based on renewable energy sources as of 2020. The table also presents the Maximum energy Generaction ($MEG$) from renewable sources based on installed renewable capacities for each country. The $MEG$ (in TWh) is calculated using the following equation:
\begin{equation}
MEG = \frac{Total Capacity\times24\times365}{10^6}
\end{equation}
It is worth noting that the $MEG$ is calculated as the highest possible energy generation per annum using the installed capacity. Additionally the Renewable Capacity Ratio ($RCR$) for each country is calculated as the ratio of the MEG to the EEC following the equation below:
\begin{equation}
RCR = \frac{MEG}{EEC}
\end{equation}
The $RCR$ provides a proportion of renewable energy available per TWh of energy consumption in Bitcoin mining. High $RCR$ values indicate a higher capacity to allocate renewable energy toward the mining process. From Figure-\ref{ratio}, we deduce that countries such as Germany, Canada, China, and U.S.A have high renewable energy capacities relative to their mining energy consumption in comparison with those such as Kazakhstan, Ireland, Malaysia, Iran, and Russia which do not, making it imperative for these countries to further invest in renewable energy.
\section{Case Studies}
\label{sec:case}
Sections-\ref{sec:problems} and \ref{sec:solutions} have discussed implementation factors that cause high energy consumption and carbon footprints of cryptocurrencies, and proposed possible solutions respectively. In this section, we explore a few cryptocurrency networks that aim to solve some of the practical limitations of cryptocurrencies such as Bitcoin and Ethereum. As we have discussed earlier, several alternative consensus mechanisms such as the PoS, and redundancy reduction techniques such as sharding, have been proposed to reduce the energy consumption of cryptocurrencies. In this section, we discuss how some recently developed cryptocurrency networks such as Ethereum 2.0 and the Pi Network have adapted these solutions to solve the cryptocurrency energy consumption and carbon footprint problems in the real world. These case studies will provide more insight into the ongoing active research and development, and will shed light on future research directions in this area.
\subsection{Ethereum 2.0}
We briefly described Ethereum in Section-\ref{sec:background}. Several alternate cryptocurrencies have been introduced over time, but none of them have gained as much traction as Bitcoin, with the exception of the PoW cryptocurrency, Ethereum. However, since it also suffers from energy and scalability issues, Ethereum has come up with a major upgrade, called Ethereum 2.0 \cite{eth2.0}. This version aims to resolve issues related to sustainability, scalability, and security. The security aspects are beyond the scope of this paper, hence we focus our discussion on sustainability and scalability:
i. \textbf{Sustainability}: Ethereum 2.0 attempts the energy problem by shifting from the PoW consensus mechanism to the PoS. PoS consumes significantly lower amount of energy because it involves much fewer mathematical calculations and hence has lesser computational requirements. It also provides security against attacks like 51\% attack, and prevents over-centralization of miners as ownership of coins is considered as opposed to share of computational power for reward payouts. This change in consensus algorithm is expected to consume less than 99\% of the current consumption of the PoW algorithm. \cite{ieee2019ethereum}.
ii. \textbf{Scalability}: The current version of Ethereum is not very scalable due to the increase in network congestion and data redundancy with the addition of nodes and transactions. This increases the energy consumption of the cryptocurrency network in addition to slowing down the speed of the transaction process. With Ethereum 2.0, Ethereum plans to introduce the "Beacon Chain" which implements the concept of sharding. Sharding is a concept where the load on a network is distributed amongst nodes or groups of nodes to reduce network congestion and increase throughput. The release will also include the introduction of 64 new chains, with each chain consisting of a fraction of the nodes validating the transactions. Hence more transactions can be processed in parallel, with the requirement to share the transaction details with only a fraction of the nodes. This reduces redundancy, congestion and energy consumption.
\subsection{Pi Network}
In \cite{pinet}, the authors present an introduction to the Pi Network which addresses the two issues that the Bitcoin network suffers from namely, high energy consumption and centralization of miners.
i. \textbf{Energy efficiency}: The Pi Network uses a modified version of the Stellar Consensus Protocol (SCP) \cite{scprotocol} instead of the highly energy intensive PoW consensus mechanism. While such networks need multiple exchanges among the nodes to reach consensus and can lead to network congestion, they have significantly lower energy requirements.
ii. \textbf{Decentralization}: While the original goal of Bitcoin was to provide a decentralized transaction method, the increase in price and better payoffs has made the network extremely centralized to the extent that around 87\% of the BTCs are owned by 1\% of the nodes. The Pi network allows any user with a mobile phone to mine coins without any need for expensive ASIC devices. Hence it makes mining inexpensive and more widely accessible.
\section{Conclusion}
\label{sec:conc}
This review has shown the alarmingly high electrical energy consumption and carbon footprints of PoW cryptocurrencies such as Bitcoin and Ethereum. When compared the energy consumption of countries around the world, we found that Bitcoin and Ethereum consumed nearly as much energy as countries such as Sweden and Romania respectively. We also found that their $CO_2$ emissions were close to those of Greece and Tunisia respectively. Our analysis of centralized transaction methods has revealed that Visa is much more energy-efficient and has a lower carbon footprint per transaction compared to the cryptocurrencies discussed in this review. The review identified four underlying issues causing these problems, namely, the PoW consensus mechanism, network redundancy, mining devices and sources of energy. We found that, among other possible solutions such as PoSpace, PoST, PoA, uPoW and REM, PoS proves to be the most promising alternative to PoW. We discussed redundancy reduction methods and popular ASIC devices for efficient mining. We compiled a list of popular mining devices available on the market that would be useful to various stakeholders working in the cryptocurrency area. We calculated the maximum possible energy consumption using $MAC$. Additionally, we presented renewable energy capacities for major Bitcoin mining areas, and the defined $RCR$ has showed that it would be easier for major mining countries such as China, U.S.A, Germany and Canada to allocate renewable energy compared to countries such as Russia, Iran, Malaysia, Ireland and Kazakhstan. Finally, we presented two case studies on Ethereum 2.0 and the Pi-Network which plan to use consensus algorithms such as $PoS$ and $SCP$, and concepts such as sharding to distribute the load and reduce redundancy in the cryptocurrency network to reduce the overall energy consumption and carbon footprints. While these networks are still under development, they demonstrate that considerable efforts are being made in this direction to address the real world energy consumption and $CO_2$ issues associated with cryptocurrencies to make them more sustainable and widely acceptable.
\section{Acknowledgment}
This work was also supported by the SERB ASEAN prject CRD/2020/000369 received by Dr. Vinay Chamola. Sherali Zeadally was supported by a 2021-2022 Fulbright U.S. scholar grant award administered by the U.S. Department of State Bureau of Educational and Cultural Affairs, and through its cooperating agency the Institute of International Education ("IIE"). Further, we thank the anonymous reviewers for their valuable comments which helped us improve the quality and presentation of this work.
\bibliographystyle{elsarticle-num}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 8,500
|
{"url":"http:\/\/rivista.math.unipr.it\/vols\/2021-12-2\/04-chaira-etal.html","text":"Riv. Mat. Univ. Parma, Vol. 12, No. 2, 2021\n\nKarim Chaira [a], Soumia Chaira [b] and Samih Lazaiz [c]\n\nBest proximity point theorems in Banach Algebras\nPages: 245-266\nAccepted in revised form: 8 April 2021\nMathematics Subject Classification: 47H08, 47H09, 47H10, 54H25.\nKeywords: Weakly sequentially continuous, Weakly condensing, Measure of weak noncompactness, Best proximity point, Relatively nonexpansive.\n[a]: L3A Laboratory, Department of Mathematics and Computer Sciences, Faculty of Sciences Ben M'sik, University of Hassan II Casablanca, Morocco.\n[b]: Laboratory of Mathematics and Applications, Faculty of Sciences and Technologies Mohammedia, University Hassan II Casablanca, Morocco.\n[c]: LaSMA Laboratory, Department of Mathematics, Faculty of Sciences Dhar El Mahraz, University Sidi Mohamed Ben Abdellah, Fes, Morocco.\n\nAbstract: The aim of this paper is to obtain best proximity point theorems for weakly sequentially continuous mappings in Banach algebras. An example has also been given to support the usability of our results.\n\nReferences\n[1]\nA. Abkar and M. Gabeleh, Best proximity points for asymptotic cyclic contraction mappings, Nonlinear Anal. 74 (2011), 7261-7268. MR2833709\n[2]\nR. P. Agarwal, N. Hussain and M.-A. Taoudi, Fixed point theorems in ordered Banach spaces and applications to nonlinear integral equations, Abstr. Appl. Anal. 2012, Art. ID 245872, 15 pp. MR2947757\n[3]\nC. Angosto and B. Cascales, Measures of weak noncompactness in Banach spaces, Topology Appl. 156 (2009), 1412-1421. MR2502017\n[4]\nJ. Banas and M. Lecko, Fixed points of the product of operators in Banach algebra, PanAmer. Math. J. 12 (2002), 101-109. MR1895774\n[5]\nJ. Bana\u015b and L. Olszowy, On a class of measures of noncompactness in Banach algebras and their application to nonlinear integral equations, Z. Anal. Anwend. 28 (2009), 475-498. MR2550700\n[6]\nJ. Bana\u015b and J. Rivero, On measures of weak noncompactness, Ann. Mat. Pura Appl. 151 (1988), 213-224. MR0964510\n[7]\nJ. Bana\u015b and K. Sadarangani, Solutions of some functional-integral equations in Banach algebra, Math. Comput. Modelling 38 (2003), 245-250. MR2004993\n[8]\nJ. Bana\u015b and M.-A. Taoudi, Fixed points and solutions of operator equations for the weak topology in Banach algebras, Taiwanese J. Math. 18 (2014), 871-893. MR3213392\n[9]\nS. Sadiq Basha, N. Shahzad and R. Jeyaraj, Best proximity points: approximation and optimization, Optim. Lett. 7 (2013), 145-155. MR3017101\n[10]\nA. Ben Amar, S. Chouayekh and A. Jeribi, New fixed point theorems in Banach algebras under weak topology features and applications to nonlinear integral equations, J. Funct. Anal. 259 (2010), 2215-2237. MR2674112\n[11]\nA. Ben Amar, S. Derbel, D. O\u2019Regan and T. Xiang, Fixed point theory for countably weakly condensing maps and multimaps in non-separable Banach spaces, J. Fixed Point Theory Appl. 21 (2019), Paper No. 8, 25 pp. MR3884834\n[12]\nA. Ben Amar, A. Jeribi and R. Moalla, Leray-Schauder alternatives in Banach algebra involving three operators with applications, Fixed Point Theory 15 (2014), 359-372. MR3235657\n[13]\nD. W. Boyd and J. S. W. Wong, On nonlinear contractions, Proc. Amer. Math. Soc. 20 (1969), 458-464. MR0239559\n[14]\nM. S. Brodski\u012d and D. P. Mil'man, On the center of a convex set, (Russian), Doklady Akad. Nauk SSSR (N.S.) 59 (1948), 837-840. MR0024073\n[15]\nT. A. Burton, A fixed-point theorem of Krasnoselskii, Appl. Math. Lett. 11 (1998), 85-88. MR1490385\n[16]\nK. Chaira and S. Lazaiz, Best proximity point theorems for cyclic relatively $$\\rho$$-nonexpansive mappings in modular spaces, Abstr. Appl. Anal. 2018, Art. ID 8084712, 8 pp. MR3864593\n[17]\nF. S. De Blasi, On a property of the unit sphere in a Banach space, Bull. Math. Soc. Sci. Math. R. S. Roumanie (N.S.) 21(69) (1977), 259-262. MR0482402\n[18]\nB. C. Dhage, A fixed point theorem in Banach algebras involving three operators with applications, Kyungpook Math. J. 44 (2004), 145-155. MR2040753\n[19]\nB. C. Dhage, On a fixed point theorem in Banach algebras with applications, Appl. Math. Lett. 18 (2005), 273-280. MR2121036\n[20]\nI. Dobrakov, On representation of linear operators on $$\\mathcal{C}_0 ({T,X})$$, Czechoslovak Math. J. 21(96) (1971), 13-30. MR0276804\n[21]\nR. E. Edwards, Functional Analysis, Theory and Applications, Holt, Rinehart and Winston, New York, 1965. MR0221256\n[22]\nA. A. Eldred, W. A. Kirk and P. Veeramani, Proximal normal structure and relatively nonexpansive mappings, Studia Math. 171 (2005), 283-293. MR2188054\n[23]\nA. A. Eldred and P. Veeramani, Existence and convergence of best proximity points, J. Math. Anal. Appl. 323 (2006), 1001-1006. MR2260159\n[24]\nG. Emmanuele, Measure of weak noncompactness and fixed point theorems, Bull. Math. Soc. Sci. Math. R. S. Roumanie (N.S.) 25(73) (1981), 353-358. MR0654030\n[25]\nM. A. Krasnosel'skii, Two remarks on the method of successive approximations, (Russian), Uspekhi Mat. Nauk 10 (1955), 123-127. MR0068119\n[26]\nA. Kryczka and S. Prus, Measure of weak noncompactness under complex interpolation, Studia Math. 147 (2001), 89-102. MR1853479\n[27]\nA. Kryczka, S. Prus and M. Szczepanik, Measure of weak noncompactness and real interpolation of operators, Bull. Austral. Math. Soc. 62 (2000), 389-401. MR1799942\n[28]\nB. N. Sadovskii, A fixed-point principle, Funct. Anal. Its Appl. 1 (1967), 151-153. DOI\n[29]\nV. Sankar Raj and P. Veeramani, Best proximity pair theorems for relatively nonexpansive mappings, Appl. Gen. Topol. 10 (2009), 21-28. MR2602599\n\nHome Riv.Mat.Univ.Parma","date":"2022-06-26 05:07:22","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5020337104797363, \"perplexity\": 3183.5587770462434}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-27\/segments\/1656103037089.4\/warc\/CC-MAIN-20220626040948-20220626070948-00442.warc.gz\"}"}
| null | null |
{"url":"http:\/\/mathhelpforum.com\/algebra\/30450-please-help-me-print.html","text":"\u2022 March 9th 2008, 07:46 AM\nGhaidaa\nIf the F(x)=2x+1 , g(x)=\u221aX Founding the following\n\na) F(x)- g(x)\n\nb) F(x) + g(x)\n\u2022 March 9th 2008, 08:22 AM\ntopsquark\nQuote:\n\nOriginally Posted by Ghaidaa\nIf the F(x)=2x+1 , g(x)=\u221aX Founding the following\n\na) F(x)- g(x)\n\nb) F(x) + g(x)\n\nWhat is the problem here?\n\n$F(x) - g(x) = (2x + 1) - \\sqrt{x}$\n\n$F(x) + g(x) = (2x + 1) + \\sqrt{x}$\n\n-Dan\n\u2022 March 9th 2008, 08:53 AM\nGhaidaa\nI want to know the final product\n\nReached a solution but not the resolve convincingly\n\u2022 March 9th 2008, 10:52 AM\ntopsquark\nQuote:\n\nOriginally Posted by Ghaidaa\nI want to know the final product\n\nReached a solution but not the resolve convincingly\n\nWhat do you mean? As far as simplifying the expressions, it can't be done. The forms I gave you are as simplified as they get. They are the \"final product.\"\n\n-Dan\n\u2022 March 10th 2008, 06:25 AM\njvignacio\nthe values for x are not given so you are not able to simplify more.\n\u2022 March 10th 2008, 09:18 AM\nGhaidaa\nThank you\n\nGhaidaa\n:)","date":"2016-05-04 12:03:30","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 2, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8553365468978882, \"perplexity\": 7189.234859009579}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-18\/segments\/1461860123023.37\/warc\/CC-MAIN-20160428161523-00015-ip-10-239-7-51.ec2.internal.warc.gz\"}"}
| null | null |
using System.Web.Http;
using Microsoft.Azure.Mobile.Server.Config;
namespace AuctionHouseService.Controllers
{
// Use the MobileAppController attribute for each ApiController you want to use
// from your mobile clients
[MobileAppController]
public class ValuesController : ApiController
{
// GET api/values
public string Get()
{
return "Hello World!";
}
// POST api/values
public string Post()
{
return "Hello World!";
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,914
|
Éric Dupond-Moretti (Maubeuge, 20 d'abril de 1961) és un advocat penalista i polític francès, famós pel seu nombre rècord d'absolucions. El 6 de juliol de 2020 es va convertir en ministre de Justícia de França al govern del primer ministre Jean Castex.
Origen
Dupond és l'únic fill de Jean-Pierre Dupond, un treballador del metall d'Avesnois i Elena Moretti, una dona d'Itàlia. Els seus avis paterns, Achille i Louise, també eren obrers. Orfe de pare als quatre anys, la seva mare el va criar sol. Com molts famosos advocats penalistes sense pare (Robert Badinter, Georges Kiejman, Hervé Temime), la seva infància li va atorgar un sentiment d'injustícia. Va cursar l'educació secundària al liceu catòlic Notre-Dame de Valenciennes on va obtenir el batxillerat.
Trajectòria
Després de prestar jurament com a advocat l'11 de desembre de 1984 a Douai, es va inscriure al Col·legi de l'Advocacia de Lilla. Treballant per a un despatx d'advocats de Lilla, va començar la seva carrera a les Corts laborals i, després com a advocat, nomenat per un tribunal sota la tutoria de l'advocat de Lilla Jean Descamps i l'advocat de Tolosa de Llenguadoc, Alain Furbury. Pels seus resultats (més de 145 absolucions el 2019), va rebre el sobrenom d'«Acquittator» (acquittor + matador) als tribunals.
El febrer de 2006 va obtenir l'absolució de Jean Castela, acusat de ser col·laborador necessari de l'assassinat del prefecte Claude Érignac, a Còrsega, quan en primera instància va ser condemnat a trenta anys de presó.
El 2013 va rebutjar la condecoració de la Legió d'Honor.
Cinema
Va interpretar-se a si mateix a la pel·lícula de 2013 de Claire Denis Bastards. Anys després, al 2017, Va interpretar el paper d'un jutge a la pel·lícula de Claude Lelouch, Chacun sa vie.
Política
Va presidir el comitè de suport a Martine Aubry a les eleccions municipals de 2008 a Lilla. També va signar una carta a favor seu a Libération abans de les primàries presidencials del Partit Socialista de 2011.
El maig de 2015 va declarar el seu suport a la prohibició del Front Nacional.
El 6 de juliol de 2020 va ser nomenat ministre de Justícia francès (Garde des Sceaux) pel president Emmanuel Macron al govern del primer ministre Jean Castex.
Vida personal
El 1991 es va casar amb Hélène, una exadvocada que havia conegut durant un judici, i van tenir dos fills.
Caçador apassionat, posseeix una masia flamenca amb gossos cobradors i ocells entrenats per a la falconeria.
Posteriorment es va divorciar i des d'abril de 2016 manté una relació amb la cantant Isabelle Boulay.
Publicacions
Bête noire, éditions Michel Lafon, 2012 (amb Stéphane Durand-Souffland)
Le Calvaire et le Pardon, éditions Michel Lafon, 2013 (amb Loïc Sécher)
Directs du droit, éditions Michel Lafon, 2017 (amb Stéphane Durand-Souffland)
Le Droit d'être libre, éditions de l'Aube, 2018 (amb Denis Lafay)
Ma Liberté, éditions de l'Aube, 2019 (amb Denis Lafay)
Referències
/
Advocats francesos
Persones de Maubeuge
Ministres francesos
Polítics dels Alts de França
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,384
|
var read = require('./read');
var update = require('./updateFB');
var fips = require('./fips');
module.exports = function() {
read(__dirname+'/data/water.csv', function(err, rows){
rows.splice(0, 1);
rows.forEach(add);
});
};
function add(row) {
var data = {
name : 'Water',
authority : 'AHB',
type : 'simple',
description : 'water cost from SWAP',
price : parseFloat(row[2].replace('$','')),
units : 'us$',
locality : [row[1].toLowerCase()]
};
//console.log(data);
update(data);
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,392
|
using System;
namespace Fiddler.Kerberos.NET
{
public class KerberosResponseInspector : KerberosInspector, IResponseInspector2
{
public HTTPResponseHeaders headers { get; set; }
public byte[] body { get; set; }
public bool bDirty { get; }
public bool bReadOnly { get; set; }
protected override bool IsResponse => true;
public override void Inspect(Session session)
{
var authz = session.ResponseHeaders["WWW-Authenticate"];
if (!string.IsNullOrWhiteSpace(authz))
{
TryParseHeader(authz);
}
TryDetectKdcProxy(session, session.responseBodyBytes);
}
private void TryParseHeader(string authz)
{
var split = authz.Split(new[] { ' ' }, 2);
if (split.Length == 1)
{
View.Warning = $"WWW-Authenticate: {split[0]}";
return;
}
try
{
var message = Convert.FromBase64String(split[1]);
View.ProcessMessage(message, $"WWW-Authenticate: {split[0]}");
}
catch (FormatException)
{
View.Warning = "Message is not formatted correctly";
}
catch (Exception ex)
{
View.Warning = ex.ToString();
}
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,652
|
My mission has hit a bump in the road. I have clearly made a pigs ear of my Riemman and Graphite exercises – I have been totally spammed by emails from one of my EC2 instances. Thousands of emails, so many that I think Google has disabled my alerting email account. That hasn't helped as I now get spammed by the bounce backs.
This has lead me to the conclusion that I have gone too deep into the weeds. A little too much too soon.
As a result I have made a decision to tear down my six instances (three for Riemann, three for Graphite – this will make more sense to those who have gone through The Art of Monitoring) and instead restart The Docker Book. It is motivating me as I will not be starting from scratch.
As per my previous blogs, I intend to get to the point where I have Docker images for Riemann and Graphite, test them, then roll out across multiple instances.
It's not been a lot of fun copying and pasting config from one iTerm window to another. Of course, there have been many lessons learnt from building these by hand.
Adding to a feeling of being burnt out, I have decided to change approach with regards to my reading material. I had got into the habit of reading mission-related materials on every commute apart from Monday mornings. I commute Monday to Thursday, that's seven session a week.
So this week I gave up and read for fun. Next week I am going to start The Goal and read it alongside David Bowie, A Life and try and set a more sustainable pace.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,517
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Q: Stop Apache mod_rewrite affecting Redirect rules I have the following .htaccess file:
Redirect 301 /test/example-1 /test/example-2
RewriteEngine On
RewriteBase /test/
RewriteRule ^(.*)$ index.php?_url=$1 [QSA,L]
I want the redirect to happen without mod_rewrite interfering with it, but at the moment when you hit /test/example-1 it redirects to /test/example-2?_url=example-1.
Is there any way to stop the query parameter from the rewrite rule getting append to the redirect URL? Thanks.
A: Have your rewrite rule as:
RewriteEngine On
RewriteBase /test/
RewriteCond %{REQUEST_URI} !/(example-1|index\.php) [NC]
RewriteRule ^(.*)$ index.php?_url=$1 [QSA,L]
But in general you shouldn't mix mod_alias rules with mod_rewrite rules. Your code can be rewritten as:
RewriteEngine On
RewriteBase /test/
RewriteCond %{THE_REQUEST} /example-1[\s?/] [NC]
RewriteRule ^ example-2 [R=301,L]
RewriteRule ^(.*)$ index.php?_url=$1 [QSA,L]
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 543
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Q: how to detect android studio file modification events I need to visualize how a project develops over time.
I have used this: https://docs.oracle.com/javase/tutorial/essential/io/notification.html#overview
as basis for detecting when any project file has been modified locally.
The current setup is Android Studio on windows 7. And my problem is that the java.nio.file.StandardWatchEventKinds from the example only gets fired when the programmer manually does a "ctrl-s".
How do I detect Android Studios file modifications realtime?.
My fallback approach is to to set a timer an traverse all files. But i'd rather if I could just catch the appropriate file.
EDIT: solved with an IntelliJ plugin, took me 1/2 day.
first the tutorial from: https://confluence.jetbrains.com/display/IDEADEV/Getting+Started+with+Plugin+Development#GettingStartedwithPluginDevelopment-anchor3
and using the menuaction to start/stop a BulkFileListener:
public class Watcher implements BulkFileListener {
private final MessageBusConnection connection;
public Watcher() {
connection = ApplicationManager.getApplication().getMessageBus().connect();
}
public void initComponent() {
connection.subscribe(VirtualFileManager.VFS_CHANGES, this);
}
public void disposeComponent() {
connection.disconnect();
}
@Override
public void before(List<? extends VFileEvent> events) {
//logging done
}
@Override
public void after(List<? extends VFileEvent> events) {
//logging done
}
}
placed the . jar in ...\JetBrains\IntelliJ IDEA Community Edition 14.1.3\plugins
and imported as instructed in the tutorial.
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,661
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In this problem, you'll create a program that guesses a secret number!
The program works as follows: you (the user) thinks of an integer between 0 (inclusive) and 100 (not inclusive).
The computer makes guesses, and you give it input - is its guess too high or too low?
Using bisection search, the computer will guess the user's secret number!
Note: your program should use input to obtain the user's input!
Be sure to handle the case when the user's input is not one of h, l, or c.
When the user enters something invalid, you should print out a message to the user explaining you did not understand their input.
Then, you should re-ask the question, and prompt again for input.
```python
low = 0
high = 100
answer = ''
print('Please think of a number between 0 and 100!')
while answer != 'c':
guess = (low + high)//2
print('Is your secret number', guess, end='?')
print('')
answer = input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly.")
if answer not in ('lhc'):
print('Sorry, I did not understand your input.')
elif answer == 'l':
low = guess
elif answer == 'h':
high = guess
print('Game over. Your secret number was:', guess)
```
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{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,297
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\section{Introduction}
In this paper, we rigorously study the two-dimensional problem of reflection and transmission of elastic waves at an interface between a homogeneous solid and a band-gap metamaterial modeled as a Cauchy medium and a relaxed micromorphic medium, respectively.
We propose a systematic revision of the reflection/transmission problem at a Cauchy/Cauchy interface in order to set up our notation and to carry out clearly all analytical results concerning the critical incident angles, giving rise to Stoneley interface weaves. This systematic presentation allows us to readily generalize the adopted techniques to the case of reflection and transmission at an interface separating a Cauchy medium from a relaxed micromorphic one. We observe in this latter case the existence of high-frequency critical angles of incidence, which are able to determine the onset of microstructure-related Stoneley interface waves. We clearly show that both low and high-frequency interface waves are directly related to the relative stiffnesses of the two media and can discriminate between total reflection and total transmission phenomena.
\textcolor{black}{Recent years have seen the abrupt development of acoustic metamaterials whose mechanical properties allow unorthodox material behaviors such as band-gaps (\cite{wang2014harnessing,xiao2011longitudinal,liu2000locally}), cloaking (\cite{misseroni2016cymatics,chen2007acoustic,valentine2009optical,bueckmann2015mechanical}), focusing (\cite{guenneau2007acoustic,cummer2016controlling,tallarico2017tilted}), wave-guiding (\cite{kaina2014slow,tallarico2017edge}) etc. The bulk behavior of these metamaterials has gathered the attention of the scientific community via the refinement of mathematical techniques, such as Bloch-Floquet analysis or homogenization techniques (\cite{lions1978asymptotic,craster2010high}). More recently, filtering properties of bulk periodic media, i.e. the onset of so-called band-gaps and non-linear dispersion, has been given attention in the framework of enriched continuum models (\cite{neff2014unifying,dagostino2018effective,madeo2014band,madeo2015wave,madeo2016complete,madeo2016first,madeo2016reflection,madeo2017modeling,madeo2017relaxed,madeo2017review,madeo2017role,barbagallo2017transparent,neff2015relaxed,neff2017real}). Although bulk media are investigated in detail, the study of the reflective/refractive properties at the boundary of such metamaterials is far from being well understood. Good knowledge of the reflective and transmittive properties of such interfaces could be a key point for the conception of metamaterials systems, which would completely transform the idea we currently have about reflection and transmission of elastic waves at the interface between two solids. It is for this reason that many authors convey their research towards so-called ``metasurfaces'' (\cite{xie2014wavefront, liang2018wavefront, li2014acoustic}), i.e. relatively thin layers of metamaterials whose microstructure is able to interact with the incident wave-front in such a way that the resulting reflection/transmission patterns exhibit exotic properties, such as total reflection or total transmission, conversion of a bulk incident wave in interface waves, etc.}
Notwithstanding the paramount importance these metasurfaces may have for technological advancements in the field of noise absorption or stealth, they show limitations in the sense that they work for relatively small frequency ranges, for which the wavelength of the incident wave is comparable to the thickness of the metasurface itself. This restricts the range of applicability of such devices, above all for what concerns low frequencies which would result in very thick metasurfaces.
In this paper, we choose a different approach for modeling the reflective and diffractive properties of an interface which separates a bulk homogeneous material from a bulk metamaterial. This interface does not itself contain any internal structure, but its refractive properties can be modulated by acting on the relative stiffnesses of the two materials on each side. The homogeneous material is modeled via a classical linear-elastic Cauchy model, while the metamaterial is described by the linear relaxed micromorphic model, an enriched continuum model which already proved its effectiveness in the description of the bulk behavior of certain metamaterials (\cite{madeo2016first,madeo2017modeling,madeo2017relaxed}).
We are able to clearly show that when the homogeneous material is ``stiffer'' than the considered metamaterial, zones of very high (sometimes total) transmission can be found both at low and high frequencies. More precisely, we find that high-frequency total transmission is discriminated by a critical angle, beyond which total transmission gradually shifts towards total reflection. Engineering systems of this type could be fruitfully exploited for the conception of wave filters and polarizers, for non-destructive evaluation or for selective cloaking.
On the other hand, we show that when the homogeneous system is ``softer'' than the metamaterial, broadband total reflection can be achieved for almost all frequencies and angles of incidence. This could be of paramount importance for the conception of wave screens able to isolate from noise and/or vibrations. We are also able to show that such total reflection phenomena are related to the onset of classical Stoneley interface waves\footnote{Interface waves propagating at the interface between an elastic solid and air are called Rayleigh waves, after Lord Rayleigh, who was the first to show their existence (\cite{rayleigh1885onwaves}). Interface waves propagating at the surface between two solids are called Stoneley waves.} at low frequencies (\cite{stoneley1924elastic}) and of new microstructure-related interface waves at higher frequencies.
We explicitly remark that no precise microstructure is targeted in this paper, since the presented results could be generalized to any specific metamaterial without changing the overall results. This is due to the fact that the properties we unveil here only depend on the ``relative stiffnesses'' of the considered media and not on the absolute stiffness of the metamaterial itself.
The considerations drawn in this paper thus open the way for the conception of new metastructures for wave front manipulation with all the possible applications aforementioned applications.
\subsection{Notational agreement}
Let $\mathbb{R}^{3\times 3}$ be the set of all real $3\times 3$ second order tensors (matrices) which we denote by capital letters. A simple and a double contraction between tensors of any suitable order is denoted by $\cdot$ and $:$ respectively, while the scalar product of such tensors by $\left\langle \cdot, \cdot \right\rangle$.\footnote{For example, $(A\cdot v)_i=A_{ij}v_j, (A\cdot B)_{ik}=A_{ij}B_{jk}, A:B=A_{ij}B_{ji},(C\cdot B)_{ijk}=C_{ijp}B_{pk},(C:B)_{i}=C_{ijp}B_{pj},\left\langle v,w\right\rangle = v\cdot w = v_i w_i, \left\langle A, B \right\rangle = A_{ij}B_{ij}$, etc.} The Einstein sum convention is implied throughout this text unless otherwise specified. The standard Euclidean scalar product on $\mathbb{R}^{3 \times 3}$ is given by $\left\langle X, Y \right\rangle_{\mathbb{R}^{3\times 3}} = \tr (X\cdot Y^T)$ and consequently the Frobenius tensor norm is $\norm{X}^2 = \left\langle X, X\right\rangle_{\mathbb{R}^{3\times 3}}$. From now on, for the sake of notational brevity, we omit the scalar product indices $\mathbb{R}^3, \mathbb{R}^{3\times 3}$ when no confusion arises. The identity tensor on $\mathbb{R}^{3\times 3}$ will be denoted by $\mathds{1}$; then, $\tr(X)=\left\langle X, \mathds{1} \right\rangle$.
Consider a body which occupies a bounded open set $B_L \subset \mathbb{R}^3$ and assume that the boundary $\d B_L$ is a smooth surface of class $C^2$. An elastic material fills the domain $B_L$ and we denote by $\Sigma$ any material surface embedded in $B_L$. The outward unit normal to $\d B_L$ will be denoted by $\nu$ as will the outward unit normal to a surface $\Sigma$ embedded in $B_L$ (see Fig. \ref{fig:BL}).
\begin{figure}[H]
\centering
\includegraphics[scale=0.4]{BL}
\caption{\small Schematic representation of the body $B_L$, the surface $\Sigma$ and the sub-bodies $B_L^-$ and $B_L^+$.}
\label{fig:BL}
\end{figure}
Given a field $a$ defined on the surface $\Sigma$, we set
\begin{equation}
[[a]] = a^+ - a^-,
\end{equation}
which defines a measure of the jump of $a$ through the material surface $\Sigma$, where
\begin{equation}
[\cdot ]^{-} := \lim_{\substack{x \in B_L^{-}\setminus \Sigma \\ x \to \Sigma}} [\cdot ], \quad [\cdot ]^{+} := \lim_{\substack{x \in B_L^{+}\setminus \Sigma \\ x \to \Sigma}} [\cdot ],
\end{equation}
with $B_L^{-}, B_L^{+}$ being the two subdomains which result from splitting $B_L$ by the surface $\Sigma$ (see again Fig. \ref{fig:BL}).
The Lebesgue spaces of square integrable functions, vectors or tensors fields on $B_L$ with values on $\mathbb{R}, \mathbb{R}^3, \mathbb{R}^{3\times 3}$ respectively, are denoted by $L^2(B_L)$. Moreover we introduce the standard Sobolev spaces\footnote{The operators $\nabla$, $\curl$ and $\div$ are the classical gradient, curl and divergence operators. In symbols, for a field $u$ of any order, $(\nabla u)_i=u_{,i}$, for a vector field $v$, $(\curl v)_i = {\varepsilon} _{ijk}v_{k,j}$ and for a field $w$ of order greater than $1$, $\div w = w_{i,i}$. }
\begin{align}
&H^1(B_L)=\left\lbrace u \in L^2(B_L)| \,\nabla u \in L^2(B_L), \norm{u}^2_{H^1(B_L)}:=\norm{u}^2_{L^2(B_L)} + \norm{\nabla u}^2_{L^2(B_L)}\right\rbrace, \nonumber \\
&H(\curl ;B_L)=\left\lbrace v \in L^2(B_L)|\, \curl v \in L^2(B_L), \norm{v}^2_{H(\curl ;B_L)}:=\norm{v}^2_{L^2(B_L)} + \norm{\curl v}^2_{L^2(B_L)}\right\rbrace, \nonumber \\
&H(\div ;B_L)=\left\lbrace v \in L^2(B_L)|\, \div v \in L^2(B_L), \norm{v}^2_{H(\div ;B_L)}:=\norm{v}^2_{L^2(B_L)} + \norm{\div v}^2_{L^2(B_L)}\right\rbrace, \label{SobolevSpaces}
\end{align}
of functions $u$ and vector fields $v$ respectively.
For vector fields $v$ with components in $H^1(B_L)$ and for tensor fields $P$ with rows in $H(\curl ; B_L)$ (resp. $H(\div ; B_L)$), i.e.
\begin{equation}
v=(v_1,v_2,v_3)^T, \text{ } v_i \in H^1(B_L), \quad P=(P_1^T,P_2^T,P_3^T)^T, \text{ } P_i\in H(\curl ; B_L) \text{ resp. }P_i \in H(\div ; B_L), \quad i=1,2,3,
\end{equation}
we define
\begin{align}
\nabla v &=((\nabla v_1)^T,(\nabla v_2)^T,(\nabla v_3)^T)^T,\nonumber \\
\Curl P &= ((\curl P_1)^T,(\curl P_2)^T,(\curl P_3)^T)^T, \label{eq:nablacurldiv}
\\ \Div P &=(\div P_1, \div P_2, \div P_3)^T. \nonumber
\end{align}
The corresponding Sobolev spaces are denoted by $H^1(B_L)$, $H(\Div ; B_L)$, $H(\Curl ; B_L)$.
\section{Governing equations and energy flux}\label{sec:governingeqs}
Let $t_0>0$ be a fixed time and consider a bounded domain $B_L \subset \mathbb{R}^3$. The action functional of the system at hand is defined as
\begin{equation}\label{eq:actionfunct}
\mathcal{A}=\int_0^{t_0} \int_{B_L} (J-W) dX dt,
\end{equation}
where $J$ is the kinetic and $W$ the potential energy of the system.
\subsection{Governing equations for the classical linear elastic Cauchy model}
Let $u$ be the classical macroscopic displacement field and $\rho$ the macroscopic mass density. Then, the kinetic energy for a classical, linear-elastic, isotropic Cauchy medium takes the form\footnote{Here and in the sequel, we denote by the subscript $, t$ the partial derivative with respect to time.}
\begin{equation}\label{eq:Cauchykinetic}
J=\frac{1}{2}\rho \norm{u_{,t}}^2.
\end{equation}
The linear elastic strain energy density for the isotropic case is given by
\begin{equation}\label{eq:Cauchyenergy}
W = \mu \norm{\sym \nabla u}^2 + \frac{\lambda}{2}(\tr(\sym \nabla u))^2,
\end{equation}
where $\mu$ and $\lambda$ are the classical Lam\'e parameters. Applying the least action principle allows us to derive the \emph{equations of motion in strong form} which, expressed equivalently in compact and index form, are respectively given by
\begin{equation}\label{eq:Cauchystrong}
\rho \, u_{,tt} = \Div \sigma, \qquad \rho \, u_{i,tt} = \sigma_{ij,j},
\end{equation}
where
\begin{equation}\label{eq:Cauchystress}
\sigma = 2 \mu \sym \nabla u + \lambda \tr(\sym \nabla u)\mathds{1}, \qquad \sigma_{ij} = \mu(u_{i,j}+u_{j,i})+\lambda u_{k,k}\delta_{ij},
\end{equation}
is the classical symmetric Cauchy stress tensor for isotropic materials.
\subsection{Governing equations for the relaxed micromorphic model}
The kinetic energy in the isotropic relaxed micromorphic model takes the following form \cite{madeo2016complete,madeo2016reflection}
\begin{equation}\label{eq:relaxedkinetic}
J = \frac{1}{2}\rho \norm{u_{,t}}^2+\frac{1}{2}\eta\norm{P_{,t}}^2,
\end{equation}
where $u$ is the macroscopic displacement field and $P\in \mathbb{R}^{3\times 3}$ is the non-symmetric micro-distortion tensor which accounts for independent micro-motions at lower scales, $\rho$ is the apparent macroscopic density and $\eta$ is the micro-inertia density.
The strain energy for an isotropic relaxed micromorphic medium is given by \cite{madeo2016complete,madeo2016reflection}
\begin{align}
W = &\,\mu_e \norm{\sym(\nabla u - P)}^2 +\frac{\lambda_e}{2}(\tr(\nabla u - P))^2 + \mu_c \norm{\skew(\nabla u - P)}^2 \nonumber \\
&+ \mu_{\text{micro}} \norm{\sym P}^2+\frac{\lambda_{\text{micro}}}{2}(\tr P)^2 + \frac{\mu_e L_c^2}{2}\norm{\Curl P}^2. \label{eq:relaxedenergy}
\end{align}
Minimizing the action functional (i.e. imposing $\delta \mathcal{A} = 0$, integrating by parts a suitable number of times and taking arbitrary variations $\delta u$ and $\delta P$ for the kinematic fields), allows us to obtain the \emph{strong form of the bulk equations of motion} (\cite{ghiba2014relaxed,madeo2015wave,madeo2014band,neff2014unifying})
\begin{equation}\label{eq:relaxedeqns}
\begin{array}{clccl}
\rho \, u_{tt} = & \Div \widetilde{\sigma}, & &\quad \rho \, u_{i,tt} = & \widetilde{\sigma}_{ij,j},\\
\eta \,P_{tt} = & \widetilde{\sigma}-s-\Curl m, & & \quad \eta \, P_{ij,tt} = &\widetilde{\sigma}_{ij} - s_{ij} - m_{ik,p} {\varepsilon} _{jkp},
\end{array}
\end{equation}
where
\begin{align}
\widetilde{\sigma}&=2 \mu_e \sym(\nabla u - P) + \lambda_e \tr (\nabla u - P)\,\mathds{1} + 2\mu_c \skew(\nabla u - P), \nonumber\\
s&=2\mu_{\text{micro}} \sym P + \lambda_{\text{micro}} \left(\tr P\right) \, \mathds{1}, \label{eq:relaxedRHS} \\
m&= \mu_e L_c^2 \Curl P, \nonumber
\end{align}
$ {\varepsilon} _{jkp}$ is the Levi-Civita tensor and the equivalent index form of the introduced quantities reads
\begin{align}
\widetilde{\sigma}_{ij} &= \mu_e (u_{i,j} - P_{ij} + u_{j,i} - P_{ji}) + \lambda_e (u_{k,k} - P_{kk})\delta_{ij} + \mu_c(u_{i,j} - P_{ij} -u_{j,i} + P_{ji}),\nonumber\\
s_{ij} &= \mu_{\text{micro}}(P_{ij}+P_{ji}) + \lambda_{\text{micro}} P_{kk}\delta_{ij}, \\
m_{ik} &= \mu_e L_c^2 P_{ia,b} {\varepsilon} _{kba}.\nonumber
\end{align}
Here, $m$ is the moment stress tensor, $\mu_c\geq 0$ is called the Cosserat couple modulus and $L_c \geq 0$ is a characteristic length scale.
\subsection{Conservation of total energy and energy flux}
The mechanical system we are considering is conservative and, therefore, the energy must be conserved in the sense that the following differential form of a continuity equation must hold:
\begin{equation}
E_{,t}+\div H=0,\label{EnnergyConservation}
\end{equation}
where $E=J+W$ is the total energy of the system and $H$ is the energy flux vector, whose explicit expression is computed in the following subsections for a classical Cauchy medium and a relaxed micromorphic one.
The conservation of energy \eqref{EnnergyConservation} plays an important role when considering a surface of discontinuity between two continuous media, i.e. it prescribes continuity of the normal component of the energy flux $H$ across the considered interface. The tangent part of the energy flux vector $H$ is not subjected to the same restriction, so that we can observe the onset of Stoneley waves along the interface. Such interface waves do not have to satisfy the conservation of energy at the interface. On the other hand, when considering waves with non-vanishing normal component, it is necessary to check that they contribute to the normal part of the energy flux in such a way that it is conserved at the interface. The definition of the normal part of the flux for different waves will allow us to introduce the concepts of reflection and transmission coefficients as a measure of the partition of the energy of the incident wave between reflected and transmitted waves. Nevertheless, one must keep in mind that such reflection and transmission coefficients do not contain any information about the possible presence of Stoneley waves.
\subsubsection{Classical Cauchy medium}
In the case of a Cauchy medium we have, differentiating expressions \eqref{eq:Cauchykinetic} and \eqref{eq:Cauchyenergy} with respect to time and using definition \eqref{eq:Cauchystress}
\begin{equation*}
E_{,t}=J_{,t}+W_{,t} = \rho \left\langle u_{,t},u_{,tt}\right\rangle+\left\langle(2\mu \sym \nabla u + \lambda \tr (\sym \nabla u)\mathds{1}),\sym \nabla (u_{,t})\right\rangle = \rho \left\langle u_{,t},u_{,tt}\right\rangle + \left\langle \sigma,\sym \nabla (u_{,t})\right\rangle.
\end{equation*}
We now replace $\rho \, u_{,tt}$ from the equations of motion \eqref{eq:Cauchystrong} and we use the fact that, given the symmetry of the Cauchy stress tensor $\sigma$, $\left\langle \sigma, \sym \nabla (u_{,t})\right\rangle = \left\langle \sigma, \nabla (u_{,t})\right\rangle = \Div (\sigma \cdot u_{,t})-\Div \sigma \cdot u_{,t}$ to get
\begin{equation}\label{eq:EdotCauchy}
E_{,t}=\Div \sigma \cdot u_{,t}+\Div(\sigma\cdot u_{,t})-\Div \sigma \cdot u_{,t} = \Div (\sigma \cdot u_{,t}).
\end{equation}
Thus, by comparing \eqref{eq:EdotCauchy} to the conservation of energy \eqref{EnnergyConservation}, we can deduce that the energy flux in a Cauchy continuum is given by (see e.g. \cite{achenbach1973wave})
\begin{equation}\label{eq:Cauchyflux}
H = -\sigma \cdot u_{,t}, \qquad H_k = - \sigma_{ik} u_{k,t}.
\end{equation}
\subsubsection{Relaxed micromorphic model}
In the case of a relaxed micromorphic medium we have, differentiating equations \eqref{eq:relaxedkinetic} and \eqref{eq:relaxedenergy} with respect to time\footnote{Let $\psi$ be a vector field and $A$ a second order tensor field. Then
\begin{equation}\label{eq:vectorid1}
\langle \nabla \psi, A\rangle = \Div (\psi \cdot A) - \langle \psi, \Div A \rangle.
\end{equation}
Taking $\psi = u_{,t}$ and $A = \widetilde{\sigma}$ we have
\begin{equation}\label{eq:vectorid2}
\left\langle \nabla u_{,t}, \widetilde{\sigma} \right\rangle = \Div(u_{,t}\cdot \widetilde{\sigma}) - \left\langle u_{,t},\Div \widetilde{\sigma} \right\rangle.
\end{equation}
Furthermore, we have the following identity
\begin{equation}\label{eq:vectorid3}
\left\langle m, \Curl P_{,t} \right\rangle = \Div\left( (m^T\cdot P_{,t}): {\varepsilon} \right)+\left\langle \Curl m , P_{,t}\right\rangle,
\end{equation}
which follows from the identity $\div( v \times w)=w \cdot\curl v - v \cdot \curl w$, where $v, w$ are suitable vector fields, $\times$ is the usual vector product and $:$ is the double contraction between tensors.}
\begin{align}
E_{,t}=&\,\rho \left\langle u_{,t},u_{,tt}\right\rangle + \eta \left\langle P_{,t},P_{,tt}\right\rangle + \left\langle2\mu_e \sym (\nabla u - P),\sym(\nabla u_{,t}-P_{,t})\right\rangle + \left\langle \lambda_e \tr (\nabla u - P) \mathds{1},\nabla u_{,t}-P_{,t}\right\rangle \nonumber \\
&+\left\langle 2\mu_c \skew(\nabla u - P),\skew (\nabla u_{,t}-P_{,t})\right\rangle + \left\langle 2\mu_{\text{micro}} \sym P, \sym P_{,t}\right\rangle + \left\langle \lambda_{\text{micro}} (\tr P)\mathds{1},P_{,t}\right\rangle \nonumber\\
&+ \left\langle \mu_e L_c^2 \Curl P, \Curl P_{,t}\right\rangle,\label{eq:EdotRMM1}
\end{align}
or equivalently,\footnote{We explicitly recall that given a symmetric tensor $S$, a skew-symmetric tensor $A$ and a generic tensor $C$, we have $\left\langle S, C\right\rangle = \left\langle S, \sym C\right\rangle$ and $\left\langle A, C\right\rangle = \left\langle A, \skew C\right\rangle$.} suitably collecting terms and using definitions \eqref{eq:relaxedRHS} for $\widetilde{\sigma}, s$ and $m$:
\begin{align}
E_{,t}=&\,\rho \left\langle u_{,t},u_{,tt}\right\rangle + \eta \left\langle P_{,t},P_{,tt}\right\rangle + \left\langle2\mu_e \sym (\nabla u - P)+\lambda_e \tr (\nabla u - P),\sym(\nabla u_{,t}-P_{,t})\right\rangle \nonumber\\
&+\left\langle 2\mu_c \skew(\nabla u - P),\skew (\nabla u_{,t}-P_{,t})\right\rangle + \left\langle 2\mu_{\text{micro}} \sym P+\lambda_{\text{micro}} (\tr P)\mathds{1}, \sym P_{,t}\right\rangle + \left\langle \mu_e L_c^2 \Curl P, \Curl P_{,t}\right\rangle \nonumber\\
=&\,\rho \left\langle u_{,t},u_{,tt}\right\rangle + \eta \left\langle P_{,t},P_{,tt}\right\rangle + \left\langle \widetilde{\sigma},\nabla u_{,t}\right\rangle - \left\langle \widetilde{\sigma}-s,P_{,t}\right\rangle + \left\langle m,\Curl P_{,t}\right\rangle.\label{eq:EdotRMM2}
\end{align}
We can now replace $\rho \, u_{,tt}$ and $\eta \, P_{,tt}$ by the governing equations \eqref{eq:relaxedeqns} and use \eqref{eq:vectorid2} and \eqref{eq:vectorid3} to finally get
\begin{align}
\hspace{-0.4cm}
E_{,t}&=\left\langle u_{,t},\Div \widetilde{\sigma}\right\rangle + \left\langle P_{,t}, \widetilde{\sigma} - s - \Curl m\right\rangle + \Div\left(u_{,t} \cdot \widetilde{\sigma}\right) -\left\langle u_{,t},\Div \widetilde{\sigma} \right\rangle- \left\langle \widetilde{\sigma}-s,P_{,t}\right\rangle + \left\langle m, \Curl P_{,t}\right\rangle \nonumber
\\
&=\Div (\widetilde{\sigma}^T\cdot u_{,t})-\left\langle \Curl m , P_{,t}\right\rangle + \Div \left(\left(m^T\cdot P_{,t}\right): {\varepsilon} \right) + \left\langle\Curl m, P_{,t}\right\rangle = \Div\left(\widetilde{\sigma}^T\cdot u_{,t} + \left(m^T\cdot P_{,t}\right): {\varepsilon} \right). \label{eq:EdotRMM3}
\end{align}
Thus, by comparison of \eqref{eq:EdotRMM3} with \eqref{EnnergyConservation}, the energy flux for a relaxed micromorphic medium is given by
\begin{equation}\label{eq:relaxedflux}
H =-\widetilde{\sigma}^T\cdot u_{,t} - \left(m^T\cdot P_{,t}\right): {\varepsilon} , \qquad H_{k} = -u_{i,t} \widetilde{\sigma}_{ik} - m_{ih}P_{ij,t} {\varepsilon} _{jhk},
\end{equation}
where $:$ is the double contraction between tensors.
\section{Boundary conditions}\label{sec:BoundCond}
\subsection{Boundary conditions on an interface between two classical Cauchy media}\label{sec:BoundCondCC}
As it is well known (see e.g. \cite{achenbach1973wave,graff1975wave,madeo2016reflection}), the boundary conditions which can be imposed at an interface between two Cauchy media are continuity of displacement or continuity of force.\footnote{It is also well known that if no surface forces are present at the considered interface, imposing continuity of displacement implies continuity of forces and vice versa. Both such continuity conditions must then be satisfied at the interface.} For the displacement, this means
\begin{equation}\label{displcont}
[[u]] =0 \Rightarrow u^{-} = u^{+},
\end{equation}
where $u^{-}$ is the macroscopic displacement on the ``minus'' side (the $x_1<0$ half-plane) and $u^{+}$ is the macroscopic displacement on the ``plus'' side (the $x_1>0$ half-plane).
As for the jump of force we have
\begin{equation}\label{jumpforcevec}
[[f]] = 0 \Rightarrow f^{-} = f^{+},
\end{equation}
where $f^{-}$ and $f^{+}$ are the surface force vectors on the ``minus'' and on the ``plus'' side, respectively. We recall that in a Cauchy medium, $f = \sigma \cdot \nu$, $\nu$ being the outward unit normal to the surface and $\sigma$ being the Cauchy stress tensor given by \eqref{eq:Cauchystress}.
\subsection{Boundary conditions on an interface between a classical Cauchy medium and a relaxed micromorphic medium}\label{sec:BoundCondCRM}
In this article we will be imposing two kinds of boundary conditions between a Cauchy and a relaxed micromorphic medium (see \cite{madeo2016reflection} for a detailed discussion). The first kind is that of a free microstructure, i.e. we allow the microstructure to vibrate freely. The second kind is that of a fixed microstructure, i.e. we do not allow any movement of the microstructure at the interface. For a derivation of these types of boundary conditions see \cite{madeo2016reflection}.
As mentioned, we impose continuity of displacement, i.e.
\begin{equation}
u^{-} = u^{+},
\end{equation}
where now the ``plus'' side is occupied by the relaxed micromorphic medium.\footnote{In the following, it will be natural to collect variables of the relaxed medium in two vectors $v_1$ and $v_2$, so that the components of the displacement $u^{+}$ will be denoted by $u^{+}=(v^1_1,v^1_2,v^1_1)^T$, see equations \eqref{eq:v1}, \eqref{eq:v2}.} Imposing continuity of displacement also implies continuity of force,\footnote{This fact can be deduced from the minimization of the action functional. Indeed, imposing the variation to be zero, we find the boundary condition $[[\langle F,\delta u\rangle]]=0\Rightarrow f\, u^{+} - t\, u^{-} = 0$, which implies $f=t$ since we impose continuity of displacement.} i.e.
\[
f = t,
\]
where $f = \sigma \cdot \nu$ is the surface force calculated on the Cauchy side and the force for the relaxed micromorphic model is given by
\begin{equation}\label{eq:forcemicromorphic}
t_i = \widetilde{\sigma}_{ij}\nu_j,
\end{equation}
where $\nu$ is the outward unit normal vector to the interface and $\widetilde{\sigma}$ is given in \eqref{eq:relaxedRHS}.
The least action principle provides us with another jump duality condition for the case of the connection between a Cauchy and a relaxed micromorphic medium. This extra jump condition specifies what we call ``free microstructure'' or ``fixed microstructure'' (see \cite{madeo2016reflection}).
In order to define the two types of boundary conditions we are interested in, we need the concept of double force $\tau$ which is the dual quantity of the micro-distortion tensor $P$ and is defined as \cite{madeo2016reflection}
\begin{equation}\label{eq:doubleforcedef}
\tau=-m\cdot {\varepsilon} \cdot \nu,\quad \tau_{ij}=-m_{ik} {\varepsilon} _{kjh}\nu_{h},
\end{equation}
where the involved quantities have been defined in \eqref{eq:relaxedRHS}.
\subsubsection{Free microstructure}
In this case, the macroscopic displacement is continuous while the microstructure of the relaxed micromorphic medium is free to move at the interface (\cite{madeo2016first,madeo2016reflection,madeo2017relaxed}). Leaving the interface free to move means that $P$ is arbitrary, which, on the other hand, implies the double force $\tau$ must vanish. We have then
\begin{equation}\label{eq:BCfreemicro}
[[u_i]]=0, \quad t_i -f_i = 0,\quad \tau_{ij} = 0, \quad i=1,2,3, \text{ } j=2,3.
\end{equation}
Figure \ref{fig:freemicro} gives a schematic representation of this boundary condition.
\\
\begin{figure}[H]
\begin{centering}
\begin{picture}(248,125)
\put(0,10){\line(0,1){124}}
\put(0,10){\line(1,0){248}}
\put(0,134){\line(1,0){248}}
\linethickness{0.7 mm}
\put(124,10){\line(0,1){124}}
\thinlines
\put(248,10){\line(0,1){124}}
\multiput(131,16)(11,0){11}{
\multiput(0,0)(0,11){11}{\circle{7.2}}}
\put(35,140){\footnotesize{Cauchy medium $\ominus$}}
\put(126,140){\footnotesize{relaxed micromorphic medium $\oplus$}}
\end{picture}
\par\end{centering}
\caption{\small Schematic representation of a macro internal clamp with free microstructure at a Cauchy/relaxed-micromorphic interface \cite{madeo2016first,madeo2016reflection}.}
\label{fig:freemicro}
\end{figure}
\subsubsection{Fixed microstructure}
This is the case in which we impose that the microstructure on the relaxed micromorphic side does not vibrate at the interface. The boundary conditions in this case are (see \cite{madeo2016reflection})\footnote{Let us remark again that the first condition (continuity of displacement) implies the second one when no surface forces are applied at the interface. On the other hand, imposing the tangent part of $P$ to be equal to zero implies that the double force $\tau$ is left arbitrary.}\footnote{We remark here that only the tangent part of the double force in \eqref{eq:BCfreemicro} or of the micro-distortion tensor in \eqref{eq:BCfixedmicro} must be assigned (\cite{neff2015relaxed,neff2014unifying}). This is peculiar of the relaxed micromorphic model and is related to the fact that only $\Curl P$ appears in the energy. In a standard Mindlin-Eringen model, where the whole $\nabla P$ appears in the energy, the whole double force $\tau$ (or alternatively the whole micro-distortion tensor $P$) must be assigned at the interface. Finally, in an internal variable model (no derivatives of $P$ in the energy), no conditions on $P$ or $\tau$ must be assigned at the considered interface.}
\begin{equation}\label{eq:BCfixedmicro}
[[u_i]]=0,\quad t_i-f_i=0,\quad P_{ij} = 0,\quad i=1,2,3, \text{ } j=2,3.
\end{equation}
Observe that in equations \eqref{eq:BCfreemicro} and \eqref{eq:BCfixedmicro} the components tangent to the interface do not have to be assigned when considering a relaxed micromorphic medium. This is explained in detail in \cite{madeo2016reflection,neff2014unifying,neff2015relaxed}, where the peculiarities of the relaxed micromorphic model are presented. In Figure \ref{fig:fixedmicro} we give a schematic representation of the realization of this boundary condition between a homogeneous material and a metamaterial.\\
\begin{figure}[H]
\begin{centering}
\begin{picture}(248,125)
\multiput(124,10)(0,6){20}{\line(-1,1){10}}
\put(0,10){\line(0,1){124}}
\put(0,10){\line(1,0){248}}
\put(0,134){\line(1,0){248}}
\linethickness{0.7 mm}
\put(124,10){\line(0,1){124}}
\thinlines
\put(248,10){\line(0,1){124}}
\multiput(131,16)(11,0){11}{
\multiput(0,0)(0,11){11}{\circle{7.2}}}
\put(35,140){\footnotesize{Cauchy medium $\ominus$}}
\put(126,140){\footnotesize{relaxed micromorphic medium $\oplus$}}
\end{picture}
\par\end{centering}
\caption{\small Schematic representation of a macro internal clamp with fixed microstructure at a Cauchy/relaxed-micromorphic interface \cite{madeo2016first,madeo2016reflection}.}
\label{fig:fixedmicro}
\end{figure}
We explicitly remark (and we will show this fact in all detail in the remainder of this paper) that the boundary condition of Fig. \ref{fig:freemicro} is the only one which allows to obtain an equivalent Cauchy/Cauchy system when considering low frequencies. Indeed, in this case, since the tensor $P$ is left free, it can adjust itself in order to recover a Cauchy medium in the homogenized limit. On the other hand, the boundary condition of Fig. \ref{fig:fixedmicro} imposes an artificial value on $P$ along the interface, so that the effect of the microstructure is forced to be present in the considered system. It is for this reason that a Cauchy/Cauchy interface cannot be recovered as a homogenized limit of the system shown in Fig. \ref{fig:fixedmicro}.
\section{Wave propagation, reflection and transmission at an interface between two Cauchy media}
We now want to study wave propagation, reflection and transmission at the interface between two Cauchy media in two space dimensions. To that end, we agree on the following conventions.
When we discuss reflection and transmission, we assume that the surface of discontinuity (the interface between the two media) from which the wave reflects and refracts is the $x_2$ axis ($x_1=0$). Furthermore, we assume that the incident wave hits the interface at the origin $O (0,0)$.
\begin{figure}[H]
\centering
\includegraphics[scale=0.4]{CauchyCauchy}
\caption{\small Schematic representation of a Cauchy/Cauchy interface with an incident, a reflected and a transmitted wave for the case of an incident out-of-plane SH wave. The indices $i,r,t$ stand for ``incident'', ``reflected'' and ``transmitted'' respectively.}
\label{fig:CauchyCauchy}
\end{figure}
Incident waves propagate from $-\infty$ in the $x_1<0$ half-plane towards the interface, reflected waves propagate from the interface towards $-\infty$ in the $x_1<0$ half-plane and refracted (or, equivalently, transmitted) waves propagate from the interface towards $+\infty$ in the $x_1>0$ half-plane. Remark that, since we consider $2$D wave propagation, the incident wave can hit the interface at an arbitrary angle. In Figure \ref{fig:CauchyCauchy} we graphically present the schematics of the reflection and transmission of elastic waves at a Cauchy/Cauchy interface for the simple case of an incident out-of-plane SH wave.
As is classical, we will consider both in-plane (in the $(x_1 x_2)-$ plane) and out-of-plane (along the $x_3$ direction) motions. Nevertheless, all the considered components of the displacement, namely $u_1, u_2$ and $u_3$ will only depend on the $x_1, x_2$ space components (plane wave hypothesis). We hence write
\begin{equation}\label{eq:u2D}
u = (u_1(x_1,x_2,t),u_2(x_1,x_2,t),u_3(x_1,x_2,t))^T.
\end{equation}
As will be evident, depending on the type of wave (longitudinal, SV shear or SH shear)\footnote{Following classical nomenclature (\cite{achenbach1973wave}) we call longitudinal (or L) waves those waves whose displacement vector is in the same direction of the wave vector. Moreover SV waves are shear waves whose displacement vectors is orthogonal to the wave vector and lies in the $(x_1x_2)-$plane. SH waves are shear waves whose displacement vector is orthogonal to the wave vector and lies in a plane orthogonal to the $(x_1x_2)-$plane.} some components of these fields will be equal to zero.
We now make a small digression on wave propagation in classical Cauchy media. These results are of course well known (see e.g. \cite{achenbach1973wave,graff1975wave}), however, we present them here in detail following our notation, so that we can carry all the results and considerations over to the relaxed micromorphic model as a natural extension.
\subsection{Wave propagation in Cauchy media}\label{sec:Cauchywaveprop}
We start by writing the governing equations \eqref{eq:Cauchystrong} for the special case where the displacement only depends on the in-plane space variables $x_1$ and $x_2$. Plugging $u$ as in \eqref{eq:u2D} into \eqref{eq:Cauchystrong} gives the following system of PDEs:
\begin{align}
&\left.
\begin{tabular}{cc}
$\rho \, u_{1,tt}$ &$= (2\mu + \lambda)u_{1,11} + (\mu + \lambda) u_{2,12} + \mu \, u_{1,22}$ \\
$\rho \, u_{2,tt}$ &$= (2\mu + \lambda)u_{2,22} + (\mu + \lambda) u_{1,12}+ \mu \, u_{2,11}$
\end{tabular}
\right\}\label{eq:comp12}\\
&\hspace{2.8mm} \rho \, u_{3,tt} \hspace{3.6mm} = \mu (u_{3,11} + u_{3,22}) \label{eq:comp3} .
\end{align}
We remark that the first two equations \eqref{eq:comp12} are coupled, while the third \eqref{eq:comp3} is uncoupled from the first two.
We now formulate the plane wave ansatz, according to which the displacement vector $u$ takes the same value at all points lying on the same orthogonal line to the $(x_1x_2)-$plane (no dependence on $x_3$). Moreover, we also consider that the displacement field is periodic in space. In symbols, the plane-wave ansatz reads
\begin{align}
u &= \widehat{\psi}\, e^{i(\left\langle x,k \right\rangle- \omega t)} \hspace{1.5mm} =\widehat{\psi}\, e^{i(x_1k_1 + x_2k_2- \omega t)},\label{eq:planeansatzCauchy}\\
u_3 &= \widehat{\psi}_3\, e^{i\left(\langle x,k\rangle - \omega t\right)} = \widehat{\psi}_3 \,e^{i\left(x_1k_1 + x_2 k_2- \omega t\right)}, \label{eq:planewaveCauchy3}
\end{align}
where $\widehat{\psi}= (\widehat{\psi}_1,\widehat{\psi}_2)^T$ is the vector of amplitudes, $k = (k_1,k_2)^T$ is the wave-vector, which fixes the direction of propagation of the considered wave and $x=(x_1,x_2)^T$ is the position vector. Moreover, $\widehat{\psi}_3$ is a scalar amplitude for the third component of the displacement. We explicitly remark that in equation \eqref{eq:planeansatzCauchy} we consider (with a slight abuse of notation) $u=(u_1,u_2)^T$ to be the vector involving only the in-plane displacement components $u_1$ and $u_2$ which are coupled via equations \eqref{eq:comp12}. We start by considering the first system of coupled equations and we plug the plane-wave ansatz \eqref{eq:planeansatzCauchy} into \eqref{eq:comp12}, so obtaining
\begin{align}
(\omega^2 - c_L^2 k_1^2 -c_S^2 k_2^2)\widehat{\psi}_1 - c_V^2 k_1 k_2 \widehat{\psi}_2 &=0, \nonumber \\
-c_V^2 k_1 k_2 \widehat{\psi}_1 + (\omega^2 -c_L^2 k_2^2 - c_S^2 k_1^2)\widehat{\psi}_2 &=0, \label{eq:systemCauchy}
\end{align}
where we made the abbreviations
\begin{equation}\label{eq:speedsCauchy}
c_L^2 = \frac{2 \mu + \lambda}{\rho}, \quad c_S^2 = \frac{\mu}{\rho}, \quad c_V^2 = c_L^2 - c_S^2 = \frac{\mu + \lambda}{\rho},
\end{equation}
where $\mu, \lambda$ are the Lam\'e parameters of the material and $\rho$ is its apparent density. This system of algebraic equations can be written compactly as $A \cdot \widehat{\psi} =0 $, where $A$ is the matrix of coefficients
\begin{equation}\label{eq:coeffsmatrixCauchy}
A = \left(\begin{array}{cc}
\omega^2 - c_L^2 k_1^2 - c_S^2 k_2^2 & -c_V^2 k_1 k_2 \\
-c_V^2 k_1 k_2 & \omega^2 -c_L^2 k_2^2 - c_S^2 k_1^2
\end{array}\right).
\end{equation}
For $A \cdot \widehat{\psi} = 0$ to have a solution other than the trivial one, we impose $\det A = 0$. We have (see Appendix \ref{appendixCauchy} for a detailed calculation of this expression)
\begin{equation}\label{eq:detACauchy}
\det A =\frac{\left((2\mu + \lambda) \left(k_1^2 + k_2^2\right) - \rho \omega^2\right)\left( \mu\left(k_1^2 + k_2^2\right) - \rho \omega^2\right)}{\rho^2}.
\end{equation}
We now solve the algebraic equation $\det A = 0$ with respect to the first component $k_1$ of the wave-vector $k$ (as will be evident in section \ref{sec:ReflTransCC}, the second component of the wave-vector $k_2$ is always known when imposing boundary conditions)
\begin{align}
\left((2\mu + \lambda) \left(k_1^2 + k_2^2\right) - \rho \omega^2\right)\left( \mu \left(k_1^2 + k_2^2\right) - \rho \omega^2\right) = 0, \nonumber\\
(2\mu + \lambda) \left(k_1^2 + k_2^2\right) - \rho \omega^2 = 0 \quad \text{or} \quad \mu \left(k_1^2 + k_2^2\right) -\rho \omega^2 = 0,\nonumber\\
k_1^2 = \frac{\rho \,\omega^2}{2\mu + \lambda}-k_2^2 \quad \text{or} \quad k_1^2 = \frac{\rho}{\mu}\omega^2 - k_2^2, \label{eq:k1Cauchysquared}\\
k_1 =\pm \sqrt{\frac{\omega^2}{c_L^2}-k_2^2} \quad \text{or} \quad k_1 =\pm \sqrt{\frac{\omega^2}{c_S^2}-k_2^2}, \label{eq:k1Cauchy}
\end{align}
As we will show in the remainder of this subsection, the first or second solution in \eqref{eq:k1Cauchy} is associated to what we call a longitudinal or SV shear wave, respectively. The choice of sign for these solutions is related to the direction of propagation of the considered wave (positive for incident and transmitted waves, negative for reflected waves).\footnote{As a matter of fact, the sign $+$ or $-$ in expressions \eqref{eq:k1Cauchy} is uniquely determined by imposing that the solution preserves the conservation of energy at the considered interface. For Cauchy media, as well as for isotropic relaxed micromorphic media, it turns out that one must choose positive $k_1$ for transmitted and incident waves and negative $k_1$ for reflected ones (according to our convention). On the other hand, there exist some cases, as for example the case of anisotropic relaxed micromorphic media, for which these results are not straightforward and negative $k_1$ may appear also for transmitted waves, giving rise to so-called negative refraction phenomena. These unorthodox reflective properties are not the object of the present paper and will be discussed in subsequent works, where applications of the anisotropic relaxed micromorphic model will be presented.}
We will show later on in detail that, once boundary conditions are imposed at a given interface between two Cauchy media, the value of the component $k_2$ of the wave-vector $k$ can be considered to be known. We will see that $k_2$ is always real and positive, which means that, according to \eqref{eq:k1Cauchy}, the first component $k_1$ of the wave-vector can be either real or purely imaginary, depending on the values of the frequency and of the material parameters. Two scenarios are then possible:
\begin{enumerate}
\item both $k_1$ and $k_2$ are real: This means that, according to the wave ansatz \eqref{eq:planeansatzCauchy}, we have a harmonic wave which propagates in the direction given by the wave-vector $k$ lying in the $(x_1x_2)-$ plane. The wave-vector of the considered wave has two non-vanishing components in the $(x_1x_2)-$plane. A simplified illustration of this case is given in Fig. \ref{fig:propagative}.
\begin{figure}[H]
\centering
\includegraphics[scale=0.7]{propagative}
\caption{\small Schematic representation of an incident wave which propagates after hitting the interface. For graphical simplicity, only normal incidence and normal transmission are depicted.}
\label{fig:propagative}
\end{figure}
\item $k_2$ is real and $k_1$ purely imaginary: In this case, according to equation \eqref{eq:planeansatzCauchy}, the wave continues to propagate in the $x_2$ direction (along the interface), but decays with a negative exponential in the $x_1$ direction (away from the interface). Such a wave propagating only along the interface is known as a Stoneley interface wave (\cite{stoneley1924elastic,auld1973acoustic2}). An illustration of this phenomenon is given in Fig. \ref{fig:evanescent}.
\begin{figure}[H]
\centering
\includegraphics[scale=0.7]{evanescentfinal2}
\caption{\small Schematic representation of an incident wave which is transformed into a interface wave along the interface and decays exponentially away from it (Stoneley wave).}
\label{fig:evanescent}
\end{figure}
\end{enumerate}
In conclusion, we can say that when considering wave propagation in $2-$dimensional Cauchy media, it is possible that, given the material parameters, some frequencies exist for which all the energy of the incident wave can be redirected in Stoneley waves traveling along the interface, thus inhibiting transmission across the interface. Such waves are also possible at the interface between a Cauchy and a relaxed micromorphic medium, in a generalized setting.
Assuming the second component $k_2$ of the wave-vector to be known, we now calculate the solution $\widehat{\psi}$ to the algebraic equations $A \cdot \widehat{\psi} = 0$;
these solutions make up the kernel (or nullspace) of the matrix $A$ and are essentially eigenvectors to the eigenvalue $0$. Hence, as is common nomenclature (\cite{madeo2016reflection}), we call them eigenvectors.
Using the first solution of \eqref{eq:k1Cauchy}, we see that it also implies
\begin{equation}
k_1^2 = \frac{\omega^2}{c_L^2}-k_2^2.
\end{equation}
Using such relations between $k_1$ and $k_2$ in the first equation of \eqref{eq:systemCauchy}, we obtain
\begin{align}
\left(\omega^2 - c_L^2\left(\frac{\omega^2}{c_L^2}-k_2^2\right)-c_S^2k_2^2\right)\widehat{\psi}_1 - c_V^2 \sqrt{\frac{\omega^2}{c_L^2}-k_2^2}~k_2 \widehat{\psi}_2 &= 0, \nonumber \\
(\omega^2-\omega^2+c_L^2k_2^2-c_S^2k_2^2)\widehat{\psi}_1-c_V^2 \sqrt{\frac{\omega^2}{c_L^2}-k_2^2}~k_2 \widehat{\psi}_2 &= 0,\nonumber\\
(\underbrace{c_L^2-c_S^2}_{=c_V^2})k_2^2\widehat{\psi}_1-c_V^2 \sqrt{\frac{\omega^2}{c_L^2}-k_2^2}~k_2 \widehat{\psi}_2 &= 0, \label{eq:alphsa2long0}
\end{align}
this implies
\begin{equation}\label{eq:alpha2long}
\widehat{\psi}_2 =\frac{k_2}{\sqrt{\frac{\omega^2}{c_L^2}-k_2^2}}~\widehat{\psi}_1 \Leftrightarrow
\widehat{\psi}_2 =\frac{c_Lk_2}{\sqrt{\omega^2-c_L^2k_2^2}}~\widehat{\psi}_1.
\end{equation}
So, the eigenvector of the matrix $A$ in this case is given by\footnote{We explicitly remark that the definition of $\widehat{\psi}^L$ given by the first equality allows us to compute the vector $\widehat{\psi}^L$ once $k_2$ is known, i.e. when imposing boundary conditions. On the other hand, the equivalent definition $\widehat{\psi}^L=\left(1, k_2/k_1\right)^T$ allows us to remark that $k_1 \widehat{\psi}^L=\left(k_1,k_2\right)^T$ is still a solution of the equation $A\cdot \widehat{\psi}=0$. In this case, the vector of amplitudes is collinear with the wave-vector $k$. This allows us to talk about \textbf{longitudinal waves}, since the displacement vector $(u_1,u_2)^T$ given by expression \eqref{eq:planeansatzCauchy} (and hence the motion) is parallel to the direction of propagation of the traveling wave. We also notice that, given the eigenvector $\widehat{\psi}^L$, all vectors $a \widehat{\psi}^{L}$, with $a \in \C$, are solutions to the equation $A\cdot \widehat{\psi}=0$.}
\begin{equation}\label{eq:nullspacelong}
\widehat{\psi}^{L}
:=\left(\begin{array}{c}
1\\
\frac{c_Lk_2}{\sqrt{\omega^2-c_L^2k_2^2}}
\end{array}\right)=\left( \begin{array}{c}
1\\
\frac{k_2}{k_1}
\end{array}\right).
\end{equation}
Analogous considerations can be carried out when considering the second solution of \eqref{eq:k1Cauchy}, which also implies
\begin{equation}
k_1^2=\frac{\omega^2}{c_S^2}-k_2^2.
\end{equation}
Using this relation between $k_1$ and $k_2$ in the second equality of \eqref{eq:systemCauchy} gives
\begin{align}
\left(\omega^2 - c_L^2\left(\frac{\omega^2}{c_S^2}-k_2^2\right)-c_S^2k_2\right)\widehat{\psi}_1 - c_V^2 \sqrt{\frac{\omega^2}{c_S^2}-k_2^2}~k_2 \widehat{\psi}_2 &= 0, \nonumber\\
\left(\omega^2-\frac{c_L^2}{c_S^2}\omega^2+c_L^2k_2^2-c_S^2k_2^2\right)\widehat{\psi}_1-c_V^2\sqrt{\frac{\omega^2}{c_S^2}-k_2^2}~k_2\widehat{\psi}_2 &=0,\nonumber \\
\left(\omega^2\frac{c_S^2-c_L^2}{c_S^2}+(c_L^2-c_S^2)k_2^2\right)\widehat{\psi}_1-c_V^2\sqrt{\frac{\omega^2}{c_S^2}-k_2^2}~k_2\widehat{\psi}_2&=0,\nonumber\\
\left(-\omega^2\frac{c_V^2}{c_S^2}+c_V^2k_2^2\right)\widehat{\psi}_1-c_V^2\sqrt{\frac{\omega^2}{c_S^2}-k_2^2}~k_2\widehat{\psi}_2&=0,\label{eq:alpha2sv0}\\
\end{align}
this implies
\begin{equation}\label{eq:alpha2sv}
k_2\sqrt{\omega^2-k_2^2c_S^2}~\widehat{\psi}_2 = \frac{k_2^2c_S^2-\omega^2}{c_S}~\widehat{\psi}_1 \Leftrightarrow \widehat{\psi}_2=\frac{k_2^2c_S^2-\omega^2}{k_2c_S\sqrt{\omega^2-k_2^2c_S^2}}~\widehat{\psi}_1.
\end{equation}
Therefore, the eigenvector of the matrix $A$ in this case is given by\footnote{Analogously to the case of longitudinal waves, we can remark that the vector $\left(k_2,-k_1\right)^T$ is still a solution of the equation $A\cdot \widehat{\psi}=0$. This means that, in this case, the vector of amplitudes lies in the $(x_1x_2)-$plane and is orthogonal to the direction of propagation given by the wave-vector $k$. This allows us to introduce the concept of \textbf{transverse in-plane waves}, or ``shear vertical'' SV waves, since the displacement $(u_1,u_2)^T$ given by \eqref{eq:planeansatzCauchy} (and hence the motion) is orthogonal to the direction of propagation of the traveling wave, but lying in the $(x_1x_2)-$plane. We also remark that any vector $a\, \widehat{\psi}^{SV}$, with $a \in \C$, is solution to the equation $A\cdot \widehat{\psi}=0$. The first equality defining $\widehat{\psi}$ in \eqref{eq:nullspaceshear} allows to directly compute $\widehat{\psi}$ when $k_2$ is known, i.e. when imposing boundary conditions.}
\begin{equation}\label{eq:nullspaceshear}
\widehat{\psi}^{SV}
:=\left(\begin{array}{c}
1\\
\frac{k_2^2c_S^2-\omega^2}{k_2c_S\sqrt{\omega^2-k_2^2c_S^2}}
\end{array}\right)=\left(
\begin{array}{c}
1\\
-\frac{k_1}{k_2}
\end{array} \right).
\end{equation}
Finally, replacing the plane-wave ansatz \eqref{eq:planewaveCauchy3} for the third component $u_3$ of the displacement in equation \eqref{eq:comp3} gives\footnote{The component $u_3$ of the displacement being orthogonal to the $(x_1x_2)-$plane and thus to the direction of propagation of the wave, allows us to talk about \textbf{transverse, out-of-plane waves} or, equivalently, ``shear horizontal'' SH waves. Such waves have the same speed $c_S$ as the SV waves.}
\begin{align}
- \omega^2 \rho \, \widehat{\psi}_3 e^{i\left(\langle x,k\rangle - \omega t\right)} &= \mu(-k^2_1 -k_2^2) \widehat{\psi}_3 e^{i\left(\langle x,k\rangle - \omega t\right)}, \nonumber \\
\rho \,\omega^2 &= \mu (k_1^2+k_2^2), \nonumber \\
k_1 &= \pm \sqrt{\frac{\omega^2}{c_S^2}-k_2^2}. \label{eq:k1Cauchy3}
\end{align}
This relation, compared to the second of equations \eqref{eq:k1Cauchy}, tells us that the same relation between $k_1$ and $k_2$ exists when considering SV or SH waves.
\subsubsection*{Plane-wave ansatz: solution for the displacement field in a Cauchy medium}
We have seen in section \ref{sec:Cauchywaveprop} how to write the displacement field making use of the concepts of longitudinal, SV and SH waves.
According to equations \eqref{eq:planeansatzCauchy} and \eqref{eq:planewaveCauchy3} and considering the $2$D eigenvectors \eqref{eq:nullspacelong} and \eqref{eq:nullspaceshear}, we can finally write the solution for the displacement field $u=(u_1,u_2,u_3)^T$ as
\begin{equation}\label{eq:sollongsv}
u = u^{L}+u^{SV} = a^L \psi^L e^{i\left(x_1 k_1^L + x_2 k_2^L - \omega t\right)} + a^{SV} \psi^{SV} e^{i\left(x_1 k_1^{SV} + x_2 k_2^{SV} - \omega t\right)},
\end{equation}
when we consider a longitudinal or an SV wave, or
\begin{equation} \label{eq:solsh}
u = u^{SH} = a^{SH} \psi^{SH} e^{i\left(x_1 k_1^{SH} + x_2 k_2^{SH} - \omega t\right)},
\end{equation}
when we consider an SH wave. In these formulas, starting from equations \eqref{eq:nullspacelong} and \eqref{eq:nullspaceshear}, we defined the unit vectors
\begin{equation}\label{eq:psiLSVSH}
\psi^L = \frac{1}{\left| \widehat{\psi}^L\right|}\left( \begin{array}{c}
\widehat{\psi}^L_1\\
\widehat{\psi}^L_2\\
0
\end{array}\right), \quad
\psi^{SV} = \frac{1}{\left| \widehat{\psi}^{SV}\right|}\left( \begin{array}{c}
\widehat{\psi}^{SV}_1\\
\widehat{\psi}^{SV}_2\\
0
\end{array}\right), \quad \text{ and } \quad
\psi^{SH} = \left( \begin{array}{c}
0\\
0\\
1
\end{array}\right).
\end{equation}
Finally, $a^L, a^{SV}, a^{SH} \in \C$ are arbitrary constants.
We also explicitly remark that in equations \eqref{eq:sollongsv} and \eqref{eq:solsh}, $k_1^L$ and $k_2^L$ are related via the first equation of \eqref{eq:k1Cauchy}, $k_1^{SV}$ and $k_2^{SV}$ via the second equation of \eqref{eq:k1Cauchy} and $k_1^{SH}$ and $k_2^{SH}$ via \eqref{eq:k1Cauchy3}.
As we already mentioned, $k_2$ will be known when imposing boundary conditions, so the only unknowns in the solution \eqref{eq:sollongsv} (resp \eqref{eq:solsh}) remain the scalar amplitudes $a^L, a^{SV}$ (resp. $a^{SH}$). We will show in the following subsection how, using the form \eqref{eq:sollongsv} (resp. \eqref{eq:solsh}) for the solution in the problem of reflection and transmission at an interface between Cauchy media, the unknown amplitudes can be computed by imposing boundary conditions.
\subsection{Interface between two Cauchy media}\label{sec:ReflTransCC}
We now turn to the problem of reflection and transmission of elastic waves at an interface between two Cauchy media (cf. Fig. \ref{fig:CauchyCauchy}). We assume that an incident longitudinal wave\footnote{The exact same considerations can be carried over when the incident wave is SV transverse.} propagates on the ``minus'' side, hits the interface (which is the $x_2$ axis) at the origin of our co-ordinate system and is subsequently reflected into a longitudinal wave and an SV wave and is transmitted into the second medium on the ``plus'' side into a longitudinal wave and an SV wave as well, according to equations \eqref{eq:sollongsv}. If we send an incident SH wave, equation \eqref{eq:solsh} tells us that it will reflect and transmit only into SH waves.
\subsubsection{Incident longitudinal/SV-transverse wave}\label{sec:CCLSV}
According to our previous considerations, and given the linearity of the considered problem, the solution of the dynamical problem \eqref{eq:Cauchystrong} on the ``minus'' side can be written as\footnote{With a clear extension of the previously introduced notation, we denote by $a^{L,i}, a^{L,r}, a^{L,t},a^{SV,r},a^{SV,t}$ and $\psi^{L,i},
\psi^{L,r},\psi^{L,t},\psi^{SV,r},\psi^{SV,t}$ the amplitudes and eigenvectors of longitudinal incident, reflected, transmitted, in-plane transverse incident, reflected and transmitted waves respectively. Analogously, $a^{SV,i}$ and $\psi^{SV,i}$ will denote the amplitude and eigenvector related to in-plane transverse incident waves.}
\small
\begin{equation}\label{eq:solplus}
u^{-}(x_1,x_2,t) = a^{L,i} \psi^{L,i} e^{i\left(\left\langle x, k^{L,i}\right\rangle - \omega t\right)} + a^{L,r} \psi^{L,r} e^{i\left(\left\langle x, k^{L,r}\right\rangle - \omega t\right)} + a^{SV,r} \psi^{SV,r} e^{i\left(\left\langle x,k^{SV,r}\right\rangle - \omega t\right)} =: u^{L,i} + u^{L,r} + u^{SV,r}.
\end{equation}
\normalsize
As for the ``plus'' side, the solution is
\begin{equation}\label{eq:solminus}
u^{+}(x_1,x_2,t) = a^{L,t} \psi^{L,t} e^{i\left(\left\langle x, k^{L,t} \right\rangle - \omega t\right)} + a^{SV,t} \psi^{SV,t}e^{i\left(\left\langle x,k^{SV,t}\right\rangle - \omega t\right)} =: u^{L,t} + u^{SV,t}.
\end{equation}
The vectors $\psi^{L,i}, \psi^{SV,r}$ and $\psi^{SV,t}$ are as in \eqref{eq:psiLSVSH}.
Now the new task is given an incident wave, i.e. knowing $a^{L,i}$ and $k^{L,i}$, to calculate all the respective parameters of the ``new'' waves.
We see that the jump condition \eqref{displcont} can be further developed considering that $u^{-} = u^{L,i} + u^{L,r} + u^{SV,r}$ and $u^{+} = u^{L,t} + u^{SV,t}$. We equate the first components of \eqref{displcont}
\begin{equation*}
u_1^{L,i} + u_1^{L,r} +u_1^{SV,r} = u_1^{L,t} +u_1^{SV,t},
\end{equation*}
which, evaluating the involved expression at $x_1=0$ (the interface), implies
\small
\begin{equation*}
a^{L,i} \psi_1^{L,i} e^{i\left( x_2 k^{L,i}_2 - \omega t\right)} + a^{L,r}\psi^{L,r}_1 e^{i\left(x_2 k^{L,r}_2 - \omega t\right)} + a^{SV,r} \psi_1^{SV,r} e^{i\left(x_2 k^{SV,r}_2 - \omega t\right)} = a^{L,t} \psi_1^{L,t}e^{i\left( x_2 k_2^{L,t} - \omega t\right)} + a^{SV,t} \psi_1^{SV,t} e^{i\left(x_2 k_2^{SV,t} - \omega t\right)},
\end{equation*}
\normalsize
or, simplifying everywhere the time exponential
\begin{equation}
a^{L,i} \psi_1^{L,i} e^{i x_2 k^{L,i}_2 } + a^{L,r}\psi^{L,r}_1 e^{i x_2 k^{L,r}_2} + a^{SV,r} \psi_1^{SV,r} e^{i x_2 k^{SV,r}_2 }
= a^{L,t} \psi_1^{L,t}e^{i x_2 k_2^{L,t}} + a^{SV,t} \psi_1^{SV,t} e^{i x_2 k_2^{SV,t}}. \label{expfactor}
\end{equation}
This must hold for all $x_2 \in \mathbb{R}$. The exponentials in this expression form a family of linearly independent functions and, therefore, we can safely assume that the coefficients $a^{L,i},a^{L,r},a^{SV,r},a^{L,t},a^{SV,t}$ are never all zero simultaneously. This means, than in order for \eqref{expfactor} to hold, we must require that the exponents of the exponentials are all equal to one another. Canceling out the imaginary unit $i$ and $x_2$, we deduce the fundamental relation\footnote{It is now clear why we said previously that the second components of all wave-vectors are known, since they are all equal to the second component of the wave-vector of the prescribed incident wave, which is known by definition.}
\begin{equation}
\label{eq:SnellCauchy}
\boxed{
k_2^{L,i}=k_2^{L,r}=k_2^{SV,r}=k_2^{L,t}=k_2^{SV,t}
}\, ,
\end{equation}
which is the well-known \textbf{Snell's law} for in-plane waves (see \cite{achenbach1973wave,graff1975wave,zhu2015study,auld1973acoustic2}).
Using \eqref{eq:SnellCauchy} we see that the exponentials in \eqref{expfactor} can be canceled out leaving only
\begin{equation}\label{amplitudes1}
a^{L,i} \psi^{L,i}_1 + a^{L,r} \psi^{L,r}_1 + a^{SV,r} \psi^{SV,r}_1 = a^{L,t} \psi^{L,t}_1 +a^{SV,t} \psi^{SV,t}_1.
\end{equation}
Analogously, equating the second components of the displacements in the jump conditions \eqref{displcont}, using \eqref{eq:SnellCauchy} and the fact that this must hold for all $x_2 \in \mathbb{R}$, gives
\begin{equation}\label{amplitudes2}
a^{L,i} \psi^{L,i}_2 + a^{L,r} \psi^{L,r}_2 + a^{SV,r} \psi^{SV,r}_2 = a^{L,t} \psi^{L,t}_2 +a^{SV,t} \psi^{SV,t}_2.
\end{equation}
As for the jump of force, we remark that the total force on both sides is $
F^{-} = f^{L,i} + f^{L,r} + f^{SV,r}$, $F^{+} = f^{L,t} + f^{SV,t}$,
with the $f$'s being evaluated at $x_1=0$. The forces are vectors of the form
\begin{equation}\label{forcevec}
f = \left(f_1,f_2,0\right)^T,
\end{equation}
with
\begin{equation}\label{forcecomp}
f_i = \sigma_{ij} \nu_j,
\end{equation}
where $\nu = (\nu_1,\nu_2,\nu_3)^T = (1,0,0)^T$ is the normal vector to the interface, i.e. to the $x_2$ axis.\footnote{We immediately see that the only components of the stress tensor which have a contribution in the calculation of the force jump are $\sigma_{11}$ and $\sigma_{21}$.}
The force jump condition \eqref{jumpforcevec} can now be written component-wise as
\begin{align}
&\sigma^{L,i}_{11} + \sigma^{L,r}_{11} + \sigma^{SV,r}_{11} = \sigma^{L,t}_{11} + \sigma^{SV,t}_{11}, \label{forcejumpone} \\
&\sigma^{L,i}_{21} + \sigma^{L,r}_{21} + \sigma^{SV,r}_{21} = \sigma^{L,t}_{21} + \sigma^{SV,t}_{21}. \label{forcejumptwo}
\end{align}
Calculating the stresses according to eq. \eqref{eq:Cauchystress}, where we use the solutions \eqref{eq:solplus} and \eqref{eq:solminus} for the displacement and again using \eqref{eq:SnellCauchy} gives
\small
\begin{align}
a^{L,i} &\left( (2\mu+\lambda)\psi_1^{L,i} k_1^{L,i} + \lambda\psi_2^{L,i}k_2^{L,i}\right) + a^{L,r} \left( (2\mu+\lambda)\psi_1^{L,r} k_1^{L,r} + \lambda\psi_2^{L,r}k_2^{L,r}\right) + a^{SV,r} \left( (2\mu+\lambda)\psi_1^{SV,r} k_1^{SV,r} + \lambda\psi_2^{SV,r}k_2^{SV,r}\right)\nonumber \\
&= a^{L,t} \left( (2\mu^{+}+\lambda^{+})\psi_1^{L,t} k_1^{L,t} + \lambda^{+}\psi_2^{L,t}k_2^{L,t}\right) + a^{SV,t} \left( (2\mu^{+}+\lambda^{+})\psi_1^{SV,t} k_1^{SV,t} + \lambda^{+}\psi_2^{SV,t}k_2^{SV,t}\right), \label{eq:jumpforce1}
\end{align}
\normalsize
and
\begin{align}
a^{L,i} \mu \left(\psi_1^{L,i} k_2^{L,i} + \psi_2^{L,i} k_1^{L,i}\right) &+ a^{L,r} \mu \left(\psi_1^{L,r} k_2^{L,r} + \psi_2^{L,r} k_1^{L,r}\right) + a^{SV,r} \mu \left(\psi_1^{SV,r} k_2^{SV,r} + \psi_2^{SV,r} k_1^{SV,r}\right) \nonumber \\
&=a^{L,t} \mu^{+} \left(\psi_1^{L,t} k_2^{L,t} + \psi_2^{L,t} k_1^{L,t}\right) + a^{SV,t} \mu^{+} \left(\psi_1^{SV,t} k_2^{SV,t} + \psi_2^{SV,t} k_1^{SV,t}\right). \label{eq:jumpforce2}
\end{align}
Thus, equations \eqref{amplitudes1}, \eqref{amplitudes2}, \eqref{eq:jumpforce1}, \eqref{eq:jumpforce2} form an algebraic system for the unknown amplitudes $a^{L,r},$ $a^{SV,r},$ $a^{L,t},$ $a^{SV,t}$ from which we can fully calculate the solution.
\subsubsection{Incident SH-transverse wave}\label{sec:CCSH}
In this case, the solution on the ``minus'' side of the interface is
\begin{equation}\label{eq:solSHminus}
u^{-}(x_1,x_2,t)= a^{SH,i} \psi^{SH,i} e^{i\left(\left\langle x,k^{SH,i}\right\rangle-\omega t\right)} + a^{SH,r} \psi^{SH,r} e^{i\left(\left\langle x,k^{SH,r}\right\rangle-\omega t\right)},
\end{equation}
and on the ``plus'' side
\begin{equation}\label{eq:solSHplus}
u^{+}(x_1,x_2,t)= a^{SH,t} \psi^{SH,t} e^{i\left(\left\langle x,k^{SH,t}\right\rangle-\omega t\right)},
\end{equation}
where the vectors $\psi^{SH,i},\psi^{SH,r}, \psi^{SH,t}$ are all equal to $(0,0,1)^T$, according to the third equation of \eqref{eq:psiLSVSH}. Following the same reasoning as in section \ref{sec:CCLSV}, the continuity of displacement condition now only involves the $u_3$ component, which is the only non-zero one, and reads (evaluating again at $x_1=0$)
\begin{equation}\label{eq:expfactorSH}
a^{SH,i} e^{i\left(x_2 k_2^{SH,i}\right)} + a^{SH,r} e^{i\left(x_2 k_2^{SH,r}\right)} = a^{SH,t} e^{i\left(x_2 k_2^{SH,t}\right)},
\end{equation}
which, since the exponentials build a family of linearly independent functions and if we exclude the case where all amplitudes $a^{SH,i}, a^{SH,r} , a^{SH,t}$ are identically equal to zero, becomes \textbf{Snell's law} for out-of-plane motions
\begin{equation}\label{eq:SnellCauchySH}
\boxed{
k_2^{SH,i} = k_2^{SH,r} = k_2^{SH,t}
}\,.
\end{equation}
Using that, we see that the exponentials in \eqref{eq:expfactorSH} cancel out leaving only
\begin{equation}\label{eq:displSH}
a^{SH,i} + a^{SH,r} = a^{SH,t}.
\end{equation}
As for the jump in force in the case of SH waves, the total force on both sides is $F^{-} = f^{SH,i} + f^{SH,r}$, $F^{+} = f^{SH,t}$, with the $f$'s being evaluated at $x_1=0$. Now the forces are vectors of the form
\[
f = (0,0,f_3)^T,
\]
where, once again, $f_i = \sigma_{ij}\nu_j$, where $\nu = (1,0,0)^T$ is the vector normal to the interface.\footnote{We now see that the component of the stress which has an influence in this boundary condition is $\sigma_{31}$.} Condition \eqref{jumpforcevec} can now be written as
\begin{equation}
\sigma_{31}^{SH,i} + \sigma_{31}^{SH,r} = \sigma_{31}^{SH,t}
\end{equation}
The stresses are calculated again by \eqref{eq:Cauchystress} and using the solutions \eqref{eq:solSHplus} and \eqref{eq:solSHminus} for the displacement and using \eqref{eq:SnellCauchySH} gives
\begin{equation}\label{eq:stressesSH}
\mu \left(a^{SH,i} k_1^{SH,i}+a^{SH,r} k_1^{SH,r} \right) = \mu^{+} a^{SH,t} k_1^{SH,t}.
\end{equation}
Equations \eqref{eq:displSH} and \eqref{eq:stressesSH} build a system for the unknown amplitudes $a^{SH,r},a^{SH,t}$, which we can solve and fully determine the solution to the reflection-transmission problem.
\subsubsection{A condition for the onset of Stoneley waves at a Cauchy/Cauchy interface}
In this subsection we show how we can find explicit conditions for the onset of Stoneley waves at Cauchy/Cauchy interfaces. This can be done by simply requesting that the quantities under the square roots in equation \eqref{eq:k1Cauchy} become negative.
Assume that the incident wave is longitudinal. This means that its speed is given by $c_L^-=\sqrt{(2\mu^-+\lambda^-)/\rho^-}$ and that the wave vector $k$ can now be written as $k=(k_1,k_2)=|k|(\sin \theta_i, -\cos \theta_i)$, where $|k| = \omega/c_L^-$ and $\theta_i$ is the angle of incidence. This incident longitudinal wave gives rise to a longitudinal and a transverse wave both on the ``$-$'' and on the ``$+$'' side. Setting the quantity under the square root in the first equation in \eqref{eq:k1Cauchy} to be negative and using the fact that $k_2 = -|k| \cos \theta_i$, gives a condition for the appearance of Stoneley waves in the case of an incident longitudinal wave:
\begin{align}
\frac{\omega^2}{(c_L^{+})^2}-k_2^+ <0 \Leftrightarrow \frac{\omega^2}{(c_L^{+})^2}-k_2^2 <0 &\Leftrightarrow \frac{\omega^2}{(c_L^{+})^2} <|k|^2 \cos^2 \theta_i \nonumber \\
&\Leftrightarrow \frac{\omega^2}{(c_L^{+})^2} < \frac{\omega^2}{(c_L^-)^2} \cos^2 \theta_i \nonumber \\
&\Leftrightarrow \cos^2 \theta_i > \left( \frac{c_L^-}{c_L^{+}}\right)^2 \nonumber\\
&\Leftrightarrow \cos^2 \theta_i > \frac{\rho^{+}(2 \mu^{-} + \lambda^{-})}{\rho^{-}(2\mu^{+} + \lambda^{+})}. \label{eq:StoneleyLong}
\end{align}
To establish the previous relation we also used the fact that $k_2^+ = k_2$, as established by \textbf{Snell's law} in \eqref{eq:SnellCauchy}.
Similar arguments can be carried out when considering all other possibilities for incident, transmitted and reflected waves, as detailed in Tables \ref{table:StoneleyT} and \ref{table:StoneleyR}.
\begin{table}[H]
\centering
\begin{tabular}{c|c|c|c}
Incident Wave & Transmitted L & Transmitted SV &Transmitted SH \\
\hline
L & $\cos^2 \theta_i > \frac{\rho^+(2\mu^- + \lambda^-)}{\rho^-(2 \mu^+ + \lambda^+)}$ & $\cos^2 \theta_i > \frac{\rho^+(2\mu^- + \lambda^-)}{\rho^- \mu^+}$ & $-$ \\
\hline
SV & $\cos^2 \theta_i > \frac{\rho^+\mu^-}{\rho^-(2 \mu^+ + \lambda^+)}$ & $\cos^2 \theta_i > \frac{\rho^+\mu^-}{\rho^- \mu^+}$ & $-$ \\
\hline
SH & $-$& $-$ &$\cos^2 \theta_i > \frac{\rho^+\mu^-}{\rho^- \mu^+}$
\end{tabular}
\caption{\small Conditions for appearance of transmitted Stoneley waves for all types of waves at a Cauchy/Cauchy interface.}
\label{table:StoneleyT}
\end{table}
\begin{table}[ht!]
\centering
\begin{tabular}{c|c|c}
Incident Wave & Reflected L & Reflected SV \\
\hline
L &$-$ & $-$\\
\hline
SV &$\cos^2 \theta_i > \frac{\mu^-}{2\mu^- + \lambda^-}$ & $-$ \\
\hline
SH & $-$& $-$
\end{tabular}
\caption{\small Conditions for appearance of reflected Stoneley waves for all types of waves at a Cauchy/Cauchy interface.}
\label{table:StoneleyR}
\end{table}
\normalsize
The conditions in Tables \ref{table:StoneleyT} and \ref{table:StoneleyR} establish that the square of the cosine of the angle of incidence must be greater than a given quantity (this happens for smaller angles) for Stoneley waves to appear. This means that it is most likely to observe Stoneley waves when the angle of incidence is smaller than the normal incidence angle, i.e. for incident waves which are inclined with respect to the surface upon which they hit. Moreover, if there exists a strong contrast in stiffness between the two sides and the ``$-$'' side is by far stiffer than the ``$+$'' side, then Stoneley waves could be observed for angles closer to normal incidence. On the other hand, if the ``$-$'' side is only slightly stiffer than the ``$+$'' side, then Stoneley waves will be observed only for smaller angles (far from normal incidence). We refer to Appendix \ref{appendixCauchy3} for an explicit calculation of all these conditions.
In order to fix ideas, assuming that both materials on the left and on the right have the same density, i.e. $\rho^{-} = \rho^{+}$ and examining Table \ref{table:StoneleyT}, we can deduce that Stoneley waves appear only when the expressions on the right of each inequality are less than one, i.e:\footnote{In order to respect positive definiteness of the strain energy density, we must require that $2 \mu^+ + \lambda^+>0$ and $\mu^+>0$.}
\begin{itemize}
\item For an incident L wave, the transmitted L mode becomes Stoneley only if the Lam\'e parameters of the material are chosen in such a way that $2\mu^{+} + \lambda^{+} > 2\mu^{-} + \lambda^{-}$.
\item For an incident L wave, the transmitted SV mode becomes Stoneley only if the Lam\'e parameters of the material are chosen in such a way that $\mu^{+} > 2\mu^{-} + \lambda^{-}$.
\item For an incident SV wave, the transmitted L mode becomes Stoneley only if the Lam\'e parameters of the material are chosen in such a way that $2\mu^{+} + \lambda^{+} > \mu^{-} $.
\item For an incident SV wave, the transmitted SV mode becomes Stoneley only if the Lam\'e parameters of the material are chosen in such a way that $\mu^{+} > \mu^{-} $.
\item For an incident SH wave, the transmitted SH mode (which is the only transmitted mode) becomes Stoneley only if the Lam\'e parameters of the material are chosen in such a way that $\mu^{+} > \mu^{-} $.
\item Reflected waves can become Stoneley waves only when the incident wave is SV. The only mode which can be converted into a Stoneley wave is the L mode when $\mu^{-}>-\lambda^{-}$.
\end{itemize}
All these cases are shown graphically in Figures \ref{fig:Stoneley1}, \ref{fig:Stoneley2}, \ref{fig:Stoneley3}. Figures \ref{fig:StoneleyTransmittedL1} and \ref{fig:StoneleyTransmittedL2} demonstrate the case of an incident L wave. From Table \ref{table:StoneleyT} we deduce that either only the L transmitted mode will be Stoneley or both L and SV (when $\rho^- = \rho^+$). The same is true for the case of an incident SV wave as shown in Figures \ref{fig:StoneleyTransmittedSV1} and \ref{fig:StoneleyTransmittedSV2}. For incident SH waves, we only have one transmitted mode which can also be Stoneley as shown in Figure \ref{fig:StoneleyTransmittedSH}. Figure \ref{fig:StoneleyReflected} shows the only possible manifestation of a reflected Stoneley wave, which is the L mode for the case of an incident SV wave.
\vspace{-0.5cm}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.28]{RayleighTransmittedL1.png}
\caption{}
\label{fig:StoneleyTransmittedL1}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.28]{RayleighTransmittedL2.png}
\caption{}
\label{fig:StoneleyTransmittedL2}
\end{subfigure}
\caption{\small Possible manifestations of Stoneley waves at a Cauchy/Cauchy interface for L and SV transmitted modes in the case of incident L wave. The vectors in black represent propagative modes, while the vectors in red represent modes which propagate along the surface only (Stoneley modes).}
\label{fig:Stoneley1}
\end{figure}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.28]{RayleighTransmittedSV1.png}
\caption{}
\label{fig:StoneleyTransmittedSV1}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.28]{RayleighTransmittedSV2.png}
\caption{}
\label{fig:StoneleyTransmittedSV2}
\end{subfigure}
\caption{\small Possible manifestations of Stoneley waves at a Cauchy/Cauchy interface for L and SV transmitted modes in the case of incident SV wave. The vectors in black represent propagative modes, while the vectors in red represent modes which propagate along the surface only (Stoneley modes).}
\label{fig:Stoneley2}
\end{figure}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.28]{RayleighTransmittedSH.png}
\caption{}
\label{fig:StoneleyTransmittedSH}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.28]{RayleighReflected}
\caption{}
\label{fig:StoneleyReflected}
\end{subfigure}
\caption{\small Possible manifestations of Stoneley waves at a Cauchy/Cauchy interface for the only SH transmitted mode in the case of an incident SH wave (a) and for the reflected L mode in the case of an incident SV wave (b). The vectors in black represent propagative modes, while the vectors in red represent modes which propagate along the surface only (Stoneley modes).}
\label{fig:Stoneley3}
\end{figure}
\subsubsection{Determination of the reflection and transmission coefficients}
We denote by $H_1$ the first (normal) component of the flux vector and we introduce the following quantities
\begin{equation}\label{JCauchy}
J^i=\frac{1}{T}\int_0^{T}H^i_1(x,t) dt,\quad J^r=\frac{1}{T}\int_0^{T}H^r_1(x,t) dt,\quad J^t=\frac{1}{T}\int_0^{T}H^t_1 (x,t)dt,
\end{equation}
where $H^i_1=H^{L,i}_1$ (or, $H^i_1=H^{SV,i}_1$ if we consider an incident SV wave), $H^r_1 = H^{L,r}_1 + H^{SV,r}_1$ and \\$H^t_1 = H^{L,t}_1 + H^{SV,t}_1$, $T$ being the period of the wave.\footnote{Or, in the case of an incident SH wave $H^i_1=H^{SH,i}_1, H^r_1 = H^{SH,r}_1$ and $H^t_1 = H^{SH,t}_1$.} Then, the \textbf{reflection} and \textbf{transmission coefficients} are defined as
\begin{equation}\label{refltranscoeff}
\mathcal{R} = \frac{J^r}{J^t}, \quad \mathcal{T}=\frac{J^t}{J^i}.
\end{equation}
These coefficients tell us what part of the average normal flux of the incident wave is reflected and what part is transmitted; also, since the system is conservative, we must have $\mathcal{R}+\mathcal{T}=1$.
The integrals involved in these expressions are the average normal fluxes of the respective waves (incident, reflected or transmitted). We use Lemma \ref{Lemma1} (provided with proof in Appendix \ref{appendixCauchy}) in the computations of these coefficients.
In order to have a physical meaning, the final solution for the displacement $u$ must be real, so that we consider only the real parts of the displacements and stresses for the computation of the flux. For the first component of the flux vector for a longitudinal or SV wave, we have, according to equation \eqref{eq:Cauchyflux} and using the plane-wave ansatz
\begin{align}
\frac{1}{T}\int_0^{T} H_1 dt &= \frac{1}{T}\int_0^{T} \Re\left(-u_{1,t}\right)\Re\left(\sigma_{11}\right) + \Re\left(-u_{2,t}\right)\Re\left(\sigma_{12}\right) dt \nonumber\\
&= \frac{1}{T} \int_0^{T} \Re\left(i\omega a \psi_1 e^{i\left(\langle x,k\rangle -\omega t\right)}\right)\Re\left( \left[(2 \mu + \lambda) \psi_1 k_1 + \lambda \psi_2 k_2\right]i a e^{i\left(\langle x,k\rangle -\omega t\right)}\right) \nonumber\\
&\hspace{0.965cm} +\Re\left( i\omega a \psi_2 e^{i\left(\langle x,k\rangle -\omega t\right)}\right)\Re\left(\mu(\psi_1 k_2 + \psi_2 k_2)i a e^{i\left(\langle x,k\rangle -\omega t\right)}\right)dt \nonumber\\
&\hspace{-0.39cm}\underset{\mathrm{ Lemma \ref{Lemma1}}}{=}\frac{1}{2} \Re\left(\left[(2\mu + \lambda) |\psi_1|^2 k_1 + \lambda \psi_1^{*}\psi_2 k_2 + \mu \left(\psi_1\psi_2^{*}k_2 + |\psi_2|^2k_1\right)\right]|a|^2\omega\right) \label{eq:fluxLSV}.
\end{align}
As for the case of an SH wave, we have
\begin{align}
\frac{1}{T}\int_0^{T} H_1 dt &= \frac{1}{T}\int_0^{T} \Re\left(-u_{3,t}\right)\Re\left(\sigma_{13}\right)dt=\frac{1}{T}\int_0^{T} \Re\left(i\omega ae^{i\left(\langle x,k\rangle -\omega t\right)} \right)\Re\left(ik_1\mu a e^{i\left(\langle x,k\rangle -\omega t\right)}\right)dt \nonumber\\
&\underset{\mathrm{ Lemma \ref{Lemma1}}}{=} \frac{1}{2}\Re\left(\mu k_1 |a|^2 \omega\right) \label{eq:fluxSH}.
\end{align}
Such expressions for the fluxes, together with the linear decompositions given for $H_1^r$ and $H_1^t$, allow us to explicitly compute the reflection and transmission coefficients.
\subsection{The particular case of propagative waves}
We have seen that, when considering two Cauchy media with an interface, two cases are possible, namely waves which propagate in the two considered half-planes and Stoneley waves, which only propagate along the interface but decay away from it. Stoneley waves do not propagate in the considered media and are related to imaginary values of the first component of the wave number. When considering fully propagative waves ($k_1$ and $k_2$ both real) the results provided before can be interpreted on a more immediate physical basis, which we detail in the present section.
The previous ansatz and calculations were performed without any hypothesis on the nature of the components of $k$: they were assumed to be either real or imaginary.
However, when we consider a fully propagating wave we will demonstrate that we can recover some classical formulas and results which are usually found in the literature by considering the vector of direction of propagation of the wave, instead of the wave-vector $k$.
For a fully propagative wave, the plane-wave ansazt can be written as
\begin{equation}\label{eq:planeansatzCauchy2}
u=\widehat{\psi}\,e^{i(|k| \left\langle x ,\xi\right\rangle - \omega t)} = \widehat{\psi}\,e^{i(|k|( x_1\xi_1 + x_2\xi_2) - \omega t)},
\end{equation}
where $|k|$ is now the wave-number, which is defined as the modulus of the wave-vector $k$ and $\xi=(\xi_1,\xi_2)^T:=\frac{k}{|k|}$ is the so-called vector of propagation. This real vector $\xi$ has unit length ($\xi_1^2+\xi_2^2=1$).
By inserting $k_1 = |k| \xi_1$ and $k_2 = |k| \xi_2$ in \eqref{eq:k1Cauchysquared} we find
\begin{equation}\label{eq:k1Cauchy2squared}
|k|^2 = \frac{\omega^2}{c_L^2}\text{ or }|k|^2 = \frac{\omega^2}{c_S^2},
\end{equation}
or,
\begin{equation} \label{eq:k1Cauchy2}
|k| = \pm \frac{\omega}{c_L} \text{ or } |k| =\pm \frac{\omega}{c_S},
\end{equation}
where, again, the signs in \eqref{eq:k1Cauchy2} must be chosen depending on what kind of wave we consider (positive for incident and transmitted waves, negative for reflected waves). Expressions \eqref{eq:k1Cauchy2} give the well-known linear dependence between the frequency $\omega$ and the wave-number $|k|$ through the speeds $c_L$ and $c_S$ for longitudinal and shear waves respectively. Such behavior is known as a ``non-dispersive'' behavior, which means that in a Cauchy medium longitudinal and shear waves propagate at a constant speed ($c_L$ for longitudinal and $c_S$ for shear waves).
Choosing the first solution in \eqref{eq:k1Cauchy2}, so that $\omega = |k| c_L$ and inserting $k=|k| \xi$ into \eqref{eq:nullspacelong} we can can find the nullspace in the case of a propagative longitudinal wave
\begin{equation}\label{eq:nullspaceCauchylongReal}
\widehat{\psi} = \left(\begin{array}{c}
1 \\
\frac{c_L |k| \xi_2}{\sqrt{|k|^2 c_L^2 - c_L^2 |k| \xi_2^2}}
\end{array}\right)=
\left(
\begin{array}{c}
1\\
\frac{\xi_2}{\xi_1}
\end{array}\right)
\end{equation}
Equivalently, choosing the second solution in \eqref{eq:k1Cauchy2squared} so that $\omega = |k| c_S$ and again inserting $k=|k| \xi$ into \eqref{eq:nullspaceshear} we find for the second component of the eigenvector
\begin{equation}
\frac{k_2^2c_S^2-\omega^2}{k_2c_S\sqrt{\omega^2-k_2^2c_S^2}}=\frac{|k|^2 \xi_2^2 c_S^2-|k|^2 c_S^2}{|k| \xi_2 c_S \sqrt{|k|^2 c_S^2 -|k|^2 \xi_2^2 c_S^2}}=-\frac{\xi_1^2}{\xi_2 \xi_1},
\end{equation}
so the eigenvector for a propagative shear wave is\footnote{We neglected the sign of the $|k|$ in the above calculations. Fixing the direction of propagation will automatically impose the sign of both $|k|$ and $\xi_1$, which we then plug into equations \eqref{eq:nullspaceCauchylongReal} or \eqref{eq:nullspaceCauchyShearReal}.}
\begin{equation}\label{eq:nullspaceCauchyShearReal}
\widehat{\psi}=\left(
\begin{array}{c}
1\\
-\frac{\xi_1}{\xi_2}
\end{array}\right).
\end{equation}
The forms \eqref{eq:nullspaceCauchylongReal} and \eqref{eq:nullspaceCauchyShearReal} for the eigenvectors of L and SV propagative waves are suggestive because they allow to immediately visualize the vector of propagation $\xi$ and, thus, the eigenvectors $\psi$ themselves in terms of the angles formed by the considered propagative wave and the interface (see Figure \ref{fig:SnellCauchy} and Table \ref{table:Table1}).
\begin{figure}[H]
\centering
\includegraphics[scale=0.35]{SnellCauchy.png}
\caption{\small Illustration of reflection and transmission patterns for propagative waves and Snell's law at a Cauchy/Cauchy interface. The second components of all wave-vectors are equal to each other according to \eqref{eq:SnellCauchy}, forcing the reflected and transmitted wave-vectors to form angles with the interface as shown here according to \eqref{eq:SnellCauchyReal}.}
\label{fig:SnellCauchy}
\end{figure}
In the propagative case, the solutions \eqref{eq:solplus} and \eqref{eq:solminus} particularize into
\begin{align}
u^{-}(x_1,x_2,t) &= a^{L,i} \psi^{L,i} e^{i\left(\left\langle x, |k| \xi^{L,i}\right\rangle - \omega t\right)} + a^{L,r} \psi^{L,r} e^{i\left(\left\langle x, |k| \xi^{L,r}\right\rangle - \omega t\right)} + a^{SV,r} \psi^{SV,r} e^{i\left(\left\langle x,|k| \xi^{SV,r}\right\rangle - \omega t\right)},\label{eq:solplusprop}\\
u^{+}(x_1,x_2,t) &= a^{L,t} \psi^{L,t} e^{i\left(\left\langle x, |k| \xi^{L,t} \right\rangle - \omega t\right)} + a^{SV,t} \psi^{SV,t}e^{i\left(\left\langle x,|k| \xi^{SV,t}\right\rangle - \omega t\right)}. \label{eq:solminusprop}
\end{align}
\begin{table}[H]
\centering
\begin{tabular}{c|c|c|c}
Wave & $\xi$ & $\psi$ & Speed of propagation \\
\hline
$L,i$ & $\left(\sin \theta^{L}_i, -\cos\theta^{L}_i,0\right)^T$ & $\left(\sin \theta^{L}_i, -\cos\theta^{L}_i,0\right)^T$ & $c_L$\\
\hline
$SV,i$ & $\left(\sin \theta^{SV}_i, -\cos\theta^{SV}_i,0\right)^T$ & $\left(\cos \theta^{SV}_i, \sin\theta^{SV}_i,0\right)^T$ & $c_S$ \\
\hline
$L,r$ & $\left(-\sin \theta^{L}_r, -\cos \theta^{L}_r,0\right)^T$ & $\left(-\sin \theta^{L}_r, -\cos \theta^{L}_r,0\right)^T$ & $c_L$\\
\hline
$SV,r$ & $\left(-\sin \theta^{SV}_r, -\cos \theta^{SV}_r,0\right)^T$ & $\left(\cos \theta^{SV}_r, -\sin \theta^{SV}_r,0\right)^T $ & $c_S$\\
\hline
$L,t$ & $\left(\sin \theta^{L}_t, -\cos \theta^{L}_t,0\right)^T $ & $\left(\sin \theta^{L}_t, -\cos \theta^{L}_t,0\right)^T $ & $c_L^+$\\
\hline
$SV,t$ & $\left(\sin \theta^{SV}_t, -\cos \theta^{SV}_t,0\right)^T $ & $\left(\cos \theta^{SV}_t, \sin \theta^{SV}_t,0\right)^T $ & $c_S^+$\\
\end{tabular}
\caption{\small Summary of the vectors of direction of propagation and of vibration for all different waves produced at a Cauchy/Cauchy interface.}
\label{table:Table1}
\end{table}
Using in \eqref{eq:solplusprop}, \eqref{eq:solminusprop} and the forms given in Table \ref{table:Table1} for the propagation vectors $\xi$ and the eigenvectors $\psi$ as well as expressions \eqref{eq:k1Cauchy2} and \eqref{eq:nullspaceCauchylongReal}, we can remark that the only unknowns are the amplitudes $a$ and the angles $\theta$. The angles $\theta$ of the different waves can be computed in terms of the angle of the incident wave by using boundary conditions. Indeed, condition \eqref{eq:SnellCauchy} can be rewritten in the propagative case as
\begin{equation}\label{eq:SnellCauchyProp}
|k^{L,i}| \xi_2^{L,i} = |k^{L,r}| \xi_2^{L,r} = |k^{SV,r}| \xi_2^{SV,r} = |k^{L,t}| \xi_2^{L,t} =|k^{SV,t}| \xi_2^{SV,t},
\end{equation}
which, using equations \eqref{eq:k1Cauchy2} and \eqref{eq:nullspaceCauchylongReal}, as well as the $\xi_2$ given in Table \ref{table:Table1} and simplifying the frequency, gives the well-known \textbf{Snell's law} (Fig. \ref{fig:SnellCauchy})\footnote{Equation \eqref{eq:SnellCauchyReal} clarifies why the angles of a different reflected and transmitted waves are chosen to be as in Fig. \ref{fig:SnellCauchy}, instead of choosing the opposite ones.}
\begin{equation}\label{eq:SnellCauchyReal}
\boxed{\frac{\cos \theta^{L}_i}{c_i} = \frac{\cos \theta^{L}_r}{c_L} = \frac{\cos \theta^{SV}_r}{c_S} = \frac{\cos \theta^{L}_t}{c_L^{+}} = \frac{\cos \theta^{SV}_t}{c_S^{+}}}\, ,
\end{equation}
where we have to choose the speed of the incident wave $c_i=c_L$ if it is longitudinal or $c_i=c_S$ if it is shear.
As already remarked, once the angles of the different propagative waves are computed, the only unknowns in the solutions \eqref{eqsolplusprop} and \eqref{eq:solminusprop} are the scalar amplitudes $a$, which can be computed as done in subsections \ref{sec:CCLSV} and \ref{sec:CCSH}, by imposing boundary conditions. The treatise made in this section does not add new features to the previous considerations made in section \ref{sec:ReflTransCC}, but allows us to visualize the traveling waves according to the classical \textbf{Snell's law} and to recover classical results concerning the dispersion curves. Clearly, such reasoning cannot be repeated for Stoneley waves, for which the more general digression made in section \ref{sec:ReflTransCC} must be addressed.
\section{Basics on dispersion curves analysis for bulk wave propagation in relaxed micromorphic media}\label{sec:dispersion}
Since it is useful for the comparison with the literature, we recall here the classical analysis of dispersion curves for the considered relaxed micromorphic medium (see \cite{madeo2016reflection,madeo2016complete,dagostino2017panorama}).
To that end, we make the hypothesis of propagative waves (see eq. \eqref{relaxedplanewave1}) and we show the plots of the frequency $\omega$ against the wave-number $|k|$ which are known as dispersion curves.
We will show that some frequency ranges, known as band-gaps, exist, such that for a given frequency in this range, no real value of the wave-number $|k|$ can be found. This basically means that the hypothesis of propagative wave is not satisfied in this range of frequencies and the solution must be more generally written as
\begin{equation}e^{i\left( x_1 k_1 + x_2 k_2 - \omega t\right)},\end{equation} where $k_1$ and $k_2$ are both imaginary, giving rise to evanescent waves (i.e. waves decaying exponentially in both the $x_1$ and the $x_2$ direction).
We explicitly remark here, that this treatise on dispersion curves in the relaxed micromorphic model has already been performed in \cite{madeo2016reflection,madeo2016complete,dagostino2017panorama}, but we recall it here for the $2$D case in order to have a direct idea on the band-gap region of the considered medium. The reader who is uniquely interested in the reflective properties of interfaces and not in the bulk properties of the relaxed micromorphic model can entirely skip this section.
From now on, we set the following parameter abbreviations for characteristic speeds and frequencies:
\small
\begin{align}
&c_m = \sqrt{\frac{\mu_e L_c^2}{\eta}},\quad c_s = \sqrt{\frac{\mu_e + \mu_c}{\rho}},\quad c_p=\sqrt{\frac{\lambda_e +2\mu_e}{\rho}},\quad c_l=\sqrt{\frac{\lambda_e + \mu_e - \mu_c}{\rho}},\nonumber \\
&\omega_s = \sqrt{\frac{2(\mu_e + \mu_{\text{micro}})}{\eta}}, \quad \omega_r =\sqrt{\frac{2\mu_c}{\eta}},\quad \omega_t = \sqrt{\frac{\mu_{\text{micro}}}{\eta}},\quad \omega_l = \sqrt{\frac{\lambda_{\text{micro}} + 2 \mu_{\text{micro}}}{\eta}},\label{eq:charseppesfreqRMM}\\
&\omega_p = \sqrt{\frac{(3\lambda_e + 2\mu_e) + (3\lambda_{\text{micro}} + 2\mu_{\text{micro}})}{\eta}}. \nonumber
\end{align}
\normalsize
As done for the Cauchy case, we suppose that the involved kinematic fields only depend on $x_1$ and $x_2$ (no dependence on the out-of-plane variable $x_3$), i.e.
\begin{equation}\label{eq:uP2D}
u = (u_1(x_1,x_2,t),u_2(x_1,x_2,t),u_3(x_1,x_2,t))^T,\quad P=(P_1^T(x_1,x_2,t),P_2^T(x_1,x_2,t),P_3^T(x_1,x_2,t))^T,
\end{equation}
where we recall that, according to our notation, $P^T_i, \text{ } i=1,2,3$ are the rows of the micro-distortion tensor $P$.
We plug $u$ and $P$ from \eqref{eq:uP2D} into \eqref{eq:relaxedeqns}. The resulting system of equations is presented in Appendix \ref{appendixRelaxed1} in component-wise notation. We proceed to make the following change of variables which are motivated by the Cartan-Lie decomposition of the tensor $P$:
\begin{align}
&P^S = \frac{1}{3}(P_{11} + P_{22} + P_{33}), \quad P^{D}_1 = P_{11} - P^S, \quad P^D_2 = P_{22} - P^S, \quad P_{(1\gamma)} = \frac{1}{2}(P_{1\gamma} + P_{\gamma 1}), \nonumber \\
&P_{[1\gamma]} = \frac{1}{2}(P_{1\gamma} - P_{\gamma 1}), \quad P_{(23)} = \frac{1}{2}(P_{23} + P_{32}), \quad P_{[23]} = \frac{1}{2}(P_{12} - P_{21}), \label{eq:CartanLie}
\end{align}
with $\gamma = 2,3$.
We can then collect the variables which are coupled (see the equations presented in Appendix \ref{appendixRelaxed2}) as
\begin{align}
v^1 &= \left(u_1, u_2, P_1^D, P_2^D, P^S, P_{(12)}, P_{[12]}\right)^T, \label{eq:v1}\\
v^2 &= \left(u_3, P_{(13)}, P_{[13]}, P_{(23)}, P_{[23]}\right)^T. \label{eq:v2}
\end{align}
We make the following plane-wave ansatz:
\begin{align}
v^1 &= \widehat{\phi}\,e^{i(\left\langle x,k \right\rangle - \omega t)}=\widehat{\phi}\,e^{i(x_1 k_1 + x_2 k_2 - \omega t)},\label{eq:planewavemicromorphic1}\\
v^2 &= \widehat{\chi}\,e^{i(\left\langle x,\widetilde{k} \right\rangle - \omega t)}=\widehat{\chi}\,e^{i(x_1\widetilde{k}_1 +x_2 \widetilde{k}_2 - \omega t)},\label{eq:planewavemicromorphic2}
\end{align}
and end up with two mutually uncoupled systems of the form (see Appendix \ref{appendixRelaxed3} for the explicit form of the matrices $A_1$ and $A_2$)
\begin{equation} \label{eq:systemmicro}
A_1 \cdot \widehat{\phi} = 0,\quad
A_2 \cdot \widehat{\chi} = 0,
\end{equation}
where $A_1 \in \C^{7\times 7}$, $A_2\in \C^{5\times 5}$, $\widehat{\phi} \in \C^7$ and $\widehat{\chi}\in \C^5$. Closer examination of the first system reveals that the components of the kinematic fields involved in these equations are the first and second only, while in the second system only components involving the out-of-plane direction $x_3$ are always present in every equation. This means that, in analogy to the case of Cauchy media, we have the same kind of uncoupling between movement in the $(x_1x_2)-$ plane (in-plane) and in the $(x_2x_3)-$ plane (out-of-plane). There is, however, no immediate distinction of longitudinal and shear waves.
\subsection{In-plane variables}\label{sec:planewaverelaxedA1}
We assume a propagating wave in which case the plane-wave ansatz is
\begin{equation} \label{relaxedplanewave1}
v^1 =\widehat{\phi}\, e^{i\left(|k|\left(x_1\xi_1 + x_2\xi_2 \right)-\omega t\right)},
\end{equation}
where $\xi=(\xi_1,\xi_2)^T$ is a real unit vector and $\widehat{\phi}$ is the vector defined of amplitudes.
The polynomial $\det A_1$ is of degree $14$ in $\omega$ and of degree $10$ in $|k|$. Solving the equation $\det A_1=0$ with respect to $\omega$ gives fourteen solutions of the form
\begin{equation}\label{eq:omRMM1}
\omega(|k|)\, =\, \pm \omega_1(|k|),\,\,\pm \omega_2(|k|), \,\,\pm \omega_3(|k|), \,\,\pm \omega_4(|k|), \,\, \pm \omega_5(|k|),\,\,\pm \omega_6(|k|),\,\, \pm \omega_7(|k|);
\end{equation}
while solving $\det A_1 = 0$ with respect to to $|k|$, gives ten solutions of the form
\begin{equation}\label{eq:kRMM1}
|k|(\omega)\,=\,\pm |k^{(1)}|(\omega),\,\,\pm |k^{(2)}|k(\omega),\,\,\pm |k^{(3)}|(\omega),\,\,\pm |k^{(4)}|(\omega),\,\,\pm |k^{(5)}|(\omega);
\end{equation}
in both cases, we keep only the positive values because the wave is traveling in the $x_1>0$ direction.
Plotting the functions $\omega_i(|k|)$, $i=1,\ldots,7$ in the $(\omega,|k|)-$ plane gives us the dispersion curves for plane waves propagating in a relaxed micromorphic medium in two space dimensions (see Figure \ref{fig:dispA1}). The results concerning the dispersive behavior of relaxed micromorphic media are presented in \cite{madeo2016reflection,madeo2016complete} and are recalled in Appendix \ref{appendixRelaxed41} for the sake of completeness.
\begin{figure}[H]
\centering
\includegraphics[scale=0.65]{dispersionA1}
\caption{\small The in-plane dispersion curves of a plane wave propagating in an isotropic relaxed micromorphic continuum in two space dimensions. The gray region denotes the band-gap, in which propagation cannot occur.}
\label{fig:dispA1}
\end{figure}
\subsection{Out-of-plane-variables}
Analogously to the case of in-plane variables, we assume a propagating wave in which case the plane-wave ansatz is
\begin{equation} \label{relaxedplanewave2}
v^2 =\widehat{\chi}\, e^{i\left(|\widetilde{k}|\left(x_1\xi_1 + x_2\xi_2 \right)-\widetilde{\omega} t\right)},
\end{equation}
where $\xi=(\xi_1,\xi_2)^T$ is a real unit vector and $\widehat{\chi}$ is the vector of amplitudes.
The polynomial $\det A_2$ is of degree $10$ in $\widetilde{\omega}$ and of degree $8$ in $|\widetilde{k}|$. Solving the equation $\det A_2=0$ with respect to $\widetilde{\omega}$ gives ten solutions of the form
\begin{equation}\label{eq:omRMM2}
\widetilde{\omega}(|\widetilde{k}|)\, =\, \pm \widetilde{\omega}_1(|\widetilde{k}|),\,\,\pm \widetilde{\omega}_2(|\widetilde{k}|),\,\,\pm \widetilde{\omega}_3(|\widetilde{k}|),\,\,\pm \widetilde{\omega}_4(|\widetilde{k}|),\,\, \pm \widetilde{\omega}_5(k);
\end{equation}
while solving $\det A_2 = 0$ with respect to to $\widetilde{k}$, gives eight solutions
\begin{equation}\label{eq:kRMM2}
|\widetilde{k}|(\widetilde{\omega})\,=\,\pm|\widetilde{k}^{(1)}|(\widetilde{\omega}),\,\,\pm|\widetilde{k}^{(2)}|(\widetilde{\omega}),\,\,\pm|\widetilde{k}^{(3)}|(\widetilde{\omega}),\,\,\pm|\widetilde{k}^{(4)}|(\widetilde{\omega}).
\end{equation}
Once again, we only consider the positive values of the $\widetilde{\omega}$'s and $|\widetilde{k}|$'s since the waves are traveling in the $x_1>0$ direction. The dispersion curves for the out-of-plane variables are presented in Figure \ref{fig:dispA2}.
\begin{figure}[h!]
\centering
\includegraphics[scale=0.65]{dispersionA2}
\caption{\small The out-of-plane dispersion curves of a wave propagating in a relaxed micromorphic continuum in two space dimensions. The gray region depicts the band-gap for out-of-plane quantities.}
\label{fig:dispA2}
\end{figure}
Extra details are given in \cite{madeo2016reflection} and recalled in Appendix \ref{appendixRelaxed42}.
\section{Reflective properties of a Cauchy/relaxed micromorphic interface}\label{sec:reflCauchyRMM}
In this section, closely following what was done for the Cauchy case in section \ref{sec:Cauchywaveprop}, we look for non-trivial solutions of the equations \eqref{eq:systemmicro} imposing $\det A_1= 0$ and $\det A_2 = 0$. Once again, we fix the second component $k_2$ (resp. $\widetilde{k}_2$) of the wave-vector and solve these equations with respect to $k_1$ (resp. $\widetilde{k}_1$).\footnote{We will show later on that also in the case of an interface between a Cauchy and a relaxed micromorphic medium, the component $k_2$ of the wave-vector can be considered to be known when imposing the jump conditions holiding at the interface} The expressions for the solutions of these equations are quite complex and we do not present them explicitly here. As discussed before, we find five and four solutions for the two systems respectively\footnote{Since, as already stated, in this case we cannot a priori distinguish which modes are longitudinal and which are shear, we slightly shift our notation and use numbers in parentheses instead of describing the nature of the mode.}
\begin{equation}\label{eq:pmk}
\pm k_1^{(1)}(k_2,\omega),\,\,\pm k_1^{(2)}(k_2,\omega),\,\,\pm k_1^{(3)}(k_2,\omega),\,\,\pm k_1^{(4)}(k_2,\omega),\,\,\pm k_1^{(5)}(k_2,\omega),
\end{equation}
for the in-plane problem and
\begin{equation}\label{eq:pmktil}
\pm \widetilde{k}_1^{(1)}(\widetilde{k}_2,\omega),\,\,\pm \widetilde{k}_1^{(2)}(\widetilde{k}_2,\omega),\,\,\pm \widetilde{k}_1^{(3)}(\widetilde{k}_2,\omega),\,\,\pm \widetilde{k}_1^{(4)}(\widetilde{k}_2,\omega),
\end{equation}
for the out-of-plane problem.
Such solutions for $k_1$ (resp. $\widetilde{k}_1$) depend on the second component $k_2$ (resp. $\widetilde{k}_2$) of the wave-vector and on the frequency, but of course also on the values of the material parameters of the relaxed micromorphic model. We plug these solutions of the characteristic polynomials into the matrix $A_1$ (resp. $A_2$) and calculate for each different $k$ (resp. $\widetilde{k}$) the five (resp. four) nullspaces of the matrix. We find
\begin{equation}\label{eq:phiRMM}
\widehat{\phi}^{(1)},\widehat{\phi}^{(2)},\widehat{\phi}^{(3)},\widehat{\phi}^{(4)},\widehat{\phi}^{(5)},
\end{equation}
\begin{equation} \label{eq:chiRMM}
\widehat{\chi}^{(1)},\widehat{\chi}^{(2)},\widehat{\chi}^{(3)},\widehat{\chi}^{(4)},
\end{equation}
as solutions to the equations $A_1 \cdot \widehat{\phi} = 0$ and $A_2 \cdot \widehat{\chi} =0$, respectively. We normalize these vectors, thus introducing the normal vectors
\begin{equation}\label{eq:phichinormal}
\phi^{(i)} = \frac{1}{|\widehat{\phi}^{(i)}|}\widehat{\phi}^{(i)}, \quad \chi^{(j)} = \frac{1}{|\widehat{\chi}^{(j)}|}\widehat{\chi}^{(j)},
\end{equation}
$i=1,\ldots 5$, $j=1,\ldots 4$. Finally, we can write the solution to equations \eqref{eq:relaxedeqns} as
\small
\begin{equation}\label{eq:solmicro1}
v^1 = \alpha_1 \phi^{(1)} e^{i\left(\left\langle x,k^{(1)} \right\rangle - \omega t\right)} + \alpha_2 \phi^{(2)} e^{i\left(\left\langle x,k^{(2)} \right\rangle - \omega t\right)}+\alpha_3 \phi^{(3)} e^{i\left(\left\langle x,k^{(3)} \right\rangle - \omega t\right)}+\alpha_4 \phi^{(4)} e^{i\left(\left\langle x,k^{(4)} \right\rangle - \omega t\right)}+\alpha_5 \phi^{(5)} e^{i\left(\left\langle x,k^{(5)} \right\rangle - \omega t\right)},
\end{equation}
\begin{equation}\label{eq:solmicro2}
v^2 = \beta_1 \chi^{(1)} e^{i\left(\left\langle x,\widetilde{k}^{(1)} \right\rangle - \omega t\right)} + \beta_2 \chi^{(2)} e^{i\left(\left\langle x, \widetilde{k}^{(2)} \right\rangle - \omega t\right)}+\beta_3 \chi^{(3)} e^{i\left(\left\langle x,\widetilde{k}^{(3)} \right\rangle - \omega t\right)}+\beta_4 \chi^{(4)} e^{i\left(\left\langle x,\widetilde{k}^{(4)} \right\rangle - \omega t\right)},
\end{equation}\normalsize
where $\alpha_i, \beta_j \in \C$ for $i=1,\ldots 5$, $j=1,\ldots, 4$ are the unknown amplitudes of the different modes of propagation.\footnote{We recall again that, once the eigenvalue problem is solved (i.e. once the wave-vector $k$ (resp. $\widetilde{k}$ is known) and the eigenvectors $\phi$ (resp. $\chi$) are computed, the only unknowns of the problem remain the five (resp. four) amplitudes $\alpha$ (resp $\beta$), which can be computed by imposing boundary conditions.}
We explicitly remark that expressions \eqref{eq:pmk} and \eqref{eq:pmktil} for the first component $k_1$ and $\widetilde{k}_1$ of the wave-vectors, can give rise, similarly to the Cauchy case, to different scenarios when varying the value of the frequency $\omega$ and the material parameters. As a matter of fact, we briefly remarked before that $k_2$ can be considered to be known when imposing jump conditions. Indeed, following analogous steps to those performed to obtain equation \eqref{eq:SnellCauchy} for the interface between two Cauchy media, we can impose the continuity of displacements between a Cauchy and a relaxed micromorphic medium. Considering the first component of the vector equation for the continuity of displacement, in which the plane-wave ansatz has been used, one can find, when imposing a longitudinal incident wave on the Cauchy side\footnote{When imposing a longitudinal incident wave on the Cauchy side, $k_2^{L,i}$ is considered to be known. The same reasoning holds when imposing an incident SV wave; in this case, $k_2^{L,i}$ must be replaced by $k_2^{SV,i}$ in eq. \eqref{eq:SnellmicromorphicLSV}.}
\begin{empheq}[box=\fbox]{gather}
k_2^{L,i} = k_2^{L,r} = k_2^{SV,r} = k_2^{(1),t} = k_2^{(2),t} = k_2^{(3),t} = k_2^{(4),t} = k_2^{(5),t}. \label{eq:SnellmicromorphicLSV}
\\\notag\text{\textbf{Generalized in-plane Snell's Law}}
\end{empheq}
On the other hand, when imposing an out-of-plane shear incident wave, the continuity of displacement at the interface gives
\begin{empheq}[box=\fbox]{gather}
k_2^{SH,i} = k_2^{SH,r} = \widetilde{k}_2^{(1),t} = \widetilde{k}_2^{(2),t} = \widetilde{k}_2^{(3),t} = \widetilde{k}_2^{(4),t}. \label{eq:SnellmicromorphicSH}
\\\notag\text{\textbf{Generalized out-of-plane Snell's Law}}
\end{empheq}
Equations \eqref{eq:SnellmicromorphicLSV} and \eqref{eq:SnellmicromorphicSH} tell us that, when fixing the incident wave in the Cauchy medium to be longitudinal ($k_2^{L,i}$ known), in-plane shear ($k_2^{SV,i}$ known) or out-of-plane shear ($k_2^{SH,i}$ known), the second components of all the reflected and transmitted wave-vectors are known. They are the \textbf{generalized Snell's law} for the case of a Cauchy/relaxed micromorphic interface. As before, this traces two possible scenarios, given that the value of $k_2$ for the incident wave is always supposed to be real and positive (propagative wave)
\begin{enumerate}
\item both $k_1$ and $k_2$ (resp. $\widetilde{k}_1, \widetilde{k}_2$) are real (when computing $k_1$ or $\widetilde{k}_1$ via \eqref{eq:pmk} or \eqref{eq:pmktil} respectively) so that one has propagative waves.
\item $k_2$ (resp. $\widetilde{k}_2$) is real and $k_1$ (resp. $\widetilde{k}_1$), when computed via \eqref{eq:pmk} (resp. \eqref{eq:pmktil}) is imaginary, so that one has Stoneley waves propagating only along the interface and decaying away from it.
\end{enumerate}
\begin{figure}[H]
\centering
\includegraphics[scale=0.7]{RayleighMetamaterial.png}
\caption{\small Simplified representation of the onset of a interface wave (in red) propagating along the interface between a homogeneous solid and a metamaterial. Depending on the relative stiffnesses of the two media, each of the existing low and high-frequency modes can either become Stoneley or remain propagative.}
\label{fig:StoneleyMetamaterial}
\end{figure}
Depending on the values of the frequency, of the material parameters and of the angle of incidence, each of the five in-plane waves, or of the four out-of-plane waves, can be either propagative or Stoneley. Thus, Stoneley waves can appear at the considered homogeneous solid/metamaterial interface (see Fig. \ref{fig:StoneleyMetamaterial} for a simplified illustration), both for low and for high-frequency modes.
\subsection{Determination of the reflection and transmission coefficients in the case of a relaxed micromorphic medium}
As for the flux, the normal outward pointing vector to the surface (the $x_2$ axis) is $\nu = (-1,0,0)$. This means that in the expression \eqref{eq:relaxedflux} for the flux, we need only take into account the first component. According to our definition \eqref{eq:relaxedflux}, we have
\begin{equation}\label{eq:relaxedflux1}
H_1=-u_{i,t}\widetilde{\sigma}_{i1}-m_{ih}P_{ij,t} {\varepsilon} _{jh1}, \quad i,j,h\in\{1,2,3\}.
\end{equation}
This equation for the flux must now be written with respect to the new variables $v^1$ and $v^2$. It is a tedious but easy calculation to see that the following holds
\begin{equation}\label{eq:relaxedfluxmatrixinplane}
\widetilde{H}:=H_1=v^1_{,t}\cdot (H^{11}\cdot v^1_{,1} +H^{12}\cdot v^1_{,2}+H^{13}\cdot v^1),
\end{equation}
for the in-plane problem and
\begin{equation}\label{eq:relaxedfluxmatrixoutofplane}
\widetilde{H}:=H_1=v^2_{,t}\cdot (H^{21}\cdot v^2_{,1} +H^{22}\cdot v^2_{,2}+H^{23}\cdot v^2),
\end{equation}
for the out-of plane problem, where $H^{11}, H^{12}, H^{13},$ and $H^{21}, H^{22}, H^{23}$ are matrices of suitable dimensions (found in Appendix \ref{appendixRelaxed6}).
Having calculated the ``transmitted'' flux, we can now look at the reflection and transmission coefficients for the case of a Cauchy/relaxed micromorphic interface.
To that end, we again define
\begin{equation}
J^i = \frac{1}{T}\int_0^{T} H^i(x,t) dt, \quad J^r = \frac{1}{T}\int_0^{T} H^r(x,t) dt, \quad J^t = \frac{1}{T}\int_0^{T} H^t(x,t) dt,
\end{equation}
where $H^i=H^{L,i}$, $H^r=H^{L,r}+H^{SV,r}$ and $H^t = \widetilde{H}$. Then the reflection and transmission coefficients are
\begin{align}\label{eq:reflcoeff2}
\mathcal{R}=\frac{J^r}{J^i}, \quad \mathcal{T}=\frac{J^t}{J^i}.
\end{align}
In order to easily compute these coefficients, we again employ Lemma \ref{Lemma1}.
Finally, once again we have that $\mathcal{R} + \mathcal{T}=1$.
In the case of a Cauchy/relaxed micromorphic interface, the dependency of the fluxes on the frequency $\omega$ is maintained. This is due to the fact that the amplitudes needed to calculate the flux depend on $\omega$ (dispersive response), something which is not the case in the Cauchy/Cauchy interface, as was evident in the previous section.
\section{Results}\label{sec:Results}
In this section we present our results concerning the reflective properties of an interface between a Cauchy medium and a relaxed micromorphic medium. We will show that, at low frequencies, the considered interface can be regarded as an interface between a Cauchy medium and a second Cauchy medium, equivalent to the relaxed micromorphic oneand with macroscopic stiffnesses $\lambda_{\text{macro}}$ and $\mu_{\text{macro}}$, when suitable boundary conditions are imposed.
Moreover, we will be able to show that critical angles for the incident wave can be identified in the low-frequency regime, beyond which we can observe the onset of Stoneley waves. These angles are computed from the relations established in Table \ref{table:StoneleyT}.
In order to present explicit numerical results for the reflective properties of the interface between a Cauchy and a relaxed micromorphic medium, we chose the values for the parameters of the relaxed micromorphic medium as shown in Table \ref{table:parameters}. We explicitly remark that other values of such parameters could be chosen, which would be more or less close to real metamaterials parameters (\cite{madeo2016first,madeo2017relaxed,dagostino2018effective}). Nevertheless, the basic results which we want to show in the present paper are not qualitatively affected by this choice since they only depend on the relative stiffness of the two media which are considered on the two sides and not on the absolute values of such stiffnesses.
\begin{table}[ht]
\centering
\begin{tabular}{c c c c c c c c}
$\rho$ [$\mathrm{kg/m^3}$]& $\eta$ [$\mathrm{kg/m}$] & $\mu_c$ [$\mathrm{Pa}$]& $\mu_e$ [$\mathrm{Pa}$] & $\mu_{\text{micro}}$ [$\mathrm{Pa}$] & $\lambda_{\text{micro}}$ [$\mathrm{Pa}$] & $\lambda_e$ [$\mathrm{Pa}$] & $L_c$ [$\mathrm{m}$]\\
\hline
$2000$& $10^{-2}$ & $2\times 10^9$ & $2\times 10^8$ & $10^8$ & $10^8$ & $4\times 10^8$ & $10^{-2}$
\end{tabular}
\caption{\small Numerical values of the constitutive parameters chosen for the relaxed micromorphic medium.}
\label{table:parameters}
\end{table}
We can now use the following homogenization formulas, presented in \cite{barbagallo2017transparent,dagostino2018effective}, to compute the equivalent macroscopic coefficients of the Cauchy medium which is approximating the relaxed micromorphic medium at low frequencies
\begin{equation}\label{eq:macroparameters}
\mu_{\text{macro}}=\frac{\mu_e \,\mu_{\text{micro}}}{\mu_e +\mu_{\text{micro}}}, \qquad
2\mu_{\text{macro}} + 3\lambda_{\text{macro}} = \frac{(2\mu_e + 3\lambda_e)(2\mu_{\text{micro}} + 3\lambda_{\text{micro}})}{2(\mu_e+\mu_{\text{micro}}) + 3(\lambda_e + \lambda_{\text{micro}})}.
\end{equation}
Note that the Cosserat couple modulus $\mu_c$ does not appear in the homogenization formulas \eqref{eq:macroparameters}.
Using formulas \eqref{eq:macroparameters}, we compute the stiffnesses $\lambda_{\text{macro}}$ and $\mu_{\text{macro}}$ of the Cauchy medium which is equivalent to the relaxed micromorphic medium of Table \ref{table:parameters} in the low-frequency regime, as in the following Table:
\begin{table}[H]
\centering
\begin{tabular}{c c c}
$\rho$ [$\mathrm{kg/m^3}$]& $\lambda_{\text{macro}}$ [$\mathrm{Pa}$] & $\mu_{\text{macro}}$ [$\mathrm{Pa}$] \\
\hline
$2000$& $8.25397 \times 10^7$& $6.66667 \times 10^7$
\end{tabular}
\caption{\small Macro parameters of the equivalent Cauchy medium corresponding to the relaxed medium of Table \ref{table:parameters} at low frequencies.}
\label{table:macroparameters}
\end{table}
At this point, we will consider the two cases in which the Cauchy medium on the ``$-$'' side (where the incident wave is traveling) is stiffer or softer than the equivalent Cauchy medium on the ``$+$'' side. We will show how, as expected, this difference in stiffness affects the onset of Stoneley waves at low frequencies and, as a consequence, the transmission patterns across the considered interface.
We will also show that the relaxed micromorphic model is able to predict the appearance of Stoneley waves at higher frequencies, which are substantially microstructure-related.
We will finally show that the relaxed micromorphic model also allows for the description of wide frequency bounds, for which extraordinary reflection is observed. Such frequency bounds go beyond the band-gap region and are related to the presence of the interface, as well as the relative mechanical properties of the considered media. In some cases, high-frequency critical angles discriminating between total transmission and total reflection can also be identified.
\subsection{Cauchy medium which is ``stiffer'' than the relaxed micromorphic one}\label{sec:ResultsStiffer}
In this section we present the reflective properties of a Cauchy/relaxed micromorphic interface for which we consider that the Cauchy medium on the left side is ``stiffer'' than the corresponding macroscopic parameters of the relaxed micromorphic medium in Table \ref{table:macroparameters} on the right side. To that end, we chose the material parameters of the left Cauchy medium to be those presented in Table \ref{table:Cauchyparameters} and we explicitly remark that these values are greater than those of Table \ref{table:macroparameters}, which are relative to the equivalent Cauchy medium corresponding to the considered relaxed micromorphic one.
\begin{table}[ht]
\centering
\begin{tabular}{c c c}
$\rho$ [$\mathrm{kg/m^3}$] &$\lambda$ [$\mathrm{Pa}$] & $\mu$ [$\mathrm{Pa}$] \\
\hline
$2000$ & $4\times 10^8$ & $2\times 10^8$
\end{tabular}
\caption{\small Lam\'e parameters and mass density of the Cauchy medium on the left side of the considered Cauchy/relaxed micromorphic interface.}
\label{table:Cauchyparameters}
\end{table}
For the chosen values of the constitutive parameters, the critical angles of the incident wave giving rise to Stoneley waves can be calculated using Tables \ref{table:StoneleyT} and \ref{table:StoneleyR}. As already mentioned, this approximation for the Cauchy/relaxed micromorphic interface is valid in the low-frequency regime (see \cite{neff2017real}), where the relaxed micromorphic medium is well approximated by its Cauchy counterpart with macroscopic stiffnesses $\lambda_{\text{macro}}$ and $\mu_{\text{macro}}$. We explicitly identify in the following figures \ref{fig:Tfree} and \ref{fig:Tfixed} the ``low-frequency regime'', where the aforementioned approximation is valid, as well as the critical angles governing the onset of Stoneley waves.
The explicit computed values of these critical angles are given in Table \ref{table:criticalangles1}.
\begin{table}[H]
\centering
\begin{tabular}{c c c c c}
Incident wave & $\theta_{\text{crit}}^{L,r}$ & $\theta_{\text{crit}}^{L,t}$& $\theta_{\text{crit}}^{SV,t}$& $\theta_{\text{crit}}^{SH,t}$ \\
\hline
L& $-$ & $-$ & $-$ & $-$ \\
\hline
SV & $\frac{33 \pi}{100}$ & $\frac{17\pi}{200}$ &$-$ & $-$ \\
\hline
SH & $-$ & $-$ & $-$ & $-$ \\
\end{tabular}
\caption{\small Critical angles governing the onset of Stoneley waves at the Cauchy/equivalent Cauchy interface between the two Cauchy media given in Tables \ref{table:Cauchyparameters} and \ref{table:macroparameters}, respectively. These values are computed according to the formulas given in Tables \ref{table:StoneleyT} and \ref{table:StoneleyR}. The superscripts $r$ and $t$ stand for ``reflected'' and ``transmitted''.}
\label{table:criticalangles1}
\end{table}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FreeLfinal.png}
\caption{}
\label{fig:FreeL}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FreeSVfinal.png}
\caption{}
\label{fig:FreeSV}
\end{subfigure}\\
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45 ]{FreeSHfinal.png}
\caption{}
\label{fig:FreeSH}
\end{subfigure}
\caption{\small Transmission coefficients as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for L (a), SV (b) and SH (c) incident waves for the case of macro-clamp with free microstructure. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The band-gap region is highlighted by two dashed horizontal lines, where, as expected, we observe no transmission. The low-frequency regime is highlighted by the bottom horizontal dashed line, while the critical angles for the onset of Stoneley waves are denoted by vertical dashed lines. The dark blue zone shows that no transmission takes place, while the gradual change from dark blue to red shows the increase of transmission, red being total transmission.}
\label{fig:Tfree}
\end{figure}
Figure \ref{fig:Tfree} shows the transmission coefficient for the considered Cauchy/relaxed micromorphic interface, as a function of the angle of incidence and of the frequency, when the microstructure is free to move at the interface ($P$ is left arbitrary at the interface). The coloring of this plot is such that the dark blue regions mean zero transmission, while the gradual change towards red is the increase in transmission (red is total transmission). Before commenting on the details of the behavior of the transmission coefficient, we recall that the case of free microstructure boundary condition is the only one which allows us to precisely obtain a Cauchy/equivalent Cauchy interface in the low-frequency regime, something which is not possible when imposing the fixed microstructure boundary condition ($P_{ij}=0, \text{ } i=2,3, \text{ } j=1,2,3$) at the interface. As a matter of fact, it is firmly established that a relaxed micromorphic continuum is equivalent to a Cauchy continuum with stiffnesses $\lambda_{\text{macro}}$ and $\mu_{\text{macro}}$ when considering the low-frequency regime (see \cite{neff2017real}), but this is proven only for the bulk medium. When considering an interface between a Cauchy and a relaxed micromorphic medium, the latter will behave exactly as an equivalent Cauchy medium at low frequencies only if the micro-distortion tensor $P$ is left free at the interface. Indeed, this tensor will arrange its values at the interface in order to let the low-frequency reflective properties of the Cauchy/relaxed micromorphic interface be equivalent to those of a Cauchy/equivalent Cauchy interface. On the other hand, if we impose the fixed microstructure boundary conditions, the tangential components of the tensor $P$ are forced to vanish at the interface, so that the effect of the microstructure is artificially introduced in the response of the material even for those low frequencies for which the bulk material would tend to behave as an equivalent Cauchy medium.
\vspace{-0.5cm}
Having drawn such preliminary conclusions, we can now comment Figures \ref{fig:Tfree} and \ref{fig:Tfixed} in detail. For the set of numerical values of the parameters given in Table \ref{table:Cauchyparameters} and \ref{table:macroparameters}, we established that Stoneley waves can appear in the low-frequency regime only when imposing the incident wave to be SV. In particular, the onset of Stoneley waves in the low-frequency regime can be observed in this case only for longitudinal reflected and transmitted waves when the angles of incidence are beyond $\theta_{\text{crit}}^{L,r}$ and $\theta_{\text{crit}}^{L,t}$, respectively. This fact can be retrieved in Figure \ref{fig:FreeSV}, in which an increase of the transmission coefficient can be observed in the low-frequency regime corresponding to $\theta_{\text{crit}}^{L,r}$ (Stoneley reflected waves are created, producing a decrease of the reflected normal flux and, due to energy conservation, a consequent increase of the transmitted normal flux). On the other hand, we can notice in the same figure a decrease of the transmitted energy in the low-frequency regime beyond the critical angle $\theta_{\text{crit}}^{L,t}$. This is sensible, given that beyond the value of $\theta_{\text{crit}}^{L,t}$, transmitted Stoneley waves are created, which do not contribute to propagative transmitted waves in the relaxed micromorphic continuum. We can also explicitly remark that such a decrease of transmitted energy beyond $\theta_{\text{crit}}^{L,t}$ in the low-frequency regime is much more pronounced than in the corresponding Figures \ref{fig:FreeL} and \ref{fig:FreeSH}. This means that the creation of transmitted Stoneley waves contributes to a decrease of the transmitted energy in the low-frequency regime, but a decreasing trend for the transmission coefficient is observed also for the other cases, when considering angles which are far from normal incidence. This goes along the common feeling, according to which the more inclined the incident wave is with respect to the interface, the less transmission one can expect.
The same behavior, even if qualitatively and quantitatively different, can be found in Figure \ref{fig:FixedSV}, in which an increase of transmission can be observed after $\theta_{\text{crit}}^{L,r}$ and a decrease after $\theta_{\text{crit}}^{L,t}$, also for the case of fixed microstructure boundary conditions.
\vspace{-0.3cm}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FixedLfinal.png}
\caption{}
\label{fig:FixedL}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FixedSVfinal.png}
\caption{}
\label{fig:FixedSV}
\end{subfigure}\\
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FixedSHfinal.png}
\caption{}
\label{fig:FixedSH}
\end{subfigure}
\caption{\small Transmission coefficients as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for L (a), SV (b) and SH (c) incident waves for the case of macro-clamp with fixed microstructure. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The band-gap region is highlighted by two dashed horizontal lines, where, as expected, we observe no transmission. The low-frequency regime is highlighted by the bottom horizontal dashed line, while the critical angles for the onset of Stoneley waves are denoted by vertical dashed lines. The dark blue zone shows that no transmission takes place, while the gradual change from dark blue to red shows the increase of transmission, red being total transmission.}
\label{fig:Tfixed}
\end{figure}
Direct comparison of Figures \ref{fig:Tfree} and \ref{fig:Tfixed} allows us to identify the effect that the chosen type of boundary conditions has on the transmission properties of the interface. We already remarked that, at low frequencies, common trends can be identified which are related to critical angles determining the onset of Stoneley waves at the Cauchy/equivalent Cauchy interface. Nevertheless, some differences can also be remarked which are entirely related to the choice of boundary conditions.
Surprisingly, the effect of boundary conditions intervenes already for low frequencies, meaning that the fact of imposing the value of $P$ at the interface introduces a tangible effect of the interface microstructured properties on the overall behavior of the considered system. In particular, we can notice that the fact of forcing $P=0$ at the interface globally reduces the low-frequency transmission for angles which are much closer to normal incidence, than for the case of free microstructure. This means that the fact of considering a microstructure which is not free to vibrate at the interface, allows for microstructure-related reflections, even if the frequency is relatively low.
Such additional reduction of transmission takes place for incident waves which are very inclined with respect to the surface ($\theta_i\leq \pi/4$).
Up to now, we only discussed the transmittive properties of the considered Cauchy/relaxed micromorphic interface on the low-frequency regime. Some of the features that we discussed on Stoneley waves
can be retrieved by observing Figures \ref{fig:k1LF}, \ref{fig:k1SVF} and \ref{fig:k1SHF} in which the plots of the imaginary part of the first component of the wave vector $k_1$ are given for each mode of the relaxed micromorphic medium, for L, SV and SH incident waves respectively.
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L2F.png}
\caption{}
\label{fig:k1L2F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L4F.png}
\caption{}
\label{fig:k1L4F}
\end{subfigure}\\
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L1F.png}
\caption{}
\label{fig:k1L1F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L3F.png}
\caption{}
\label{fig:k1L3F}
\end{subfigure}
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L5F.png}
\caption{}
\label{fig:k1L5F}
\end{subfigure}
\caption{\small Values of $\Im(k_1)$ as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for the five modes of the relaxed micromorphic medium and for the case of an incident L wave. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The first two modes (a) and (b) correspond to the L and SV modes for the equivalent Cauchy continuum at low frequencies. The red color in these plots means that the mode is Stoneley and does not propagate, while blue means that the mode is propagative.}
\label{fig:k1LF}
\end{figure}
The blue region denotes $\Im(k_1)=0$ (which implies that $k_1$ is real), while $\Im(k_1)$ is not vanishing in the red regions. In other words, we can say that for each mode, the red color means that there are Stoneley waves associated to that mode. The first two modes in Figures \ref{fig:k1LF} and \ref{fig:k1SVF} correspond to L and SV Cauchy-like modes, while the first mode in Figure \ref{fig:k1SHF} is the SH Cauchy-like mode in the low-frequency regime. Since we are considering a relaxed micromorphic medium, three additional modes with respect to the Cauchy case are present both for the in-plane (Figures \ref{fig:k1LF} and \ref{fig:k1SVF}) and for the out-of-plane problem (Fig. \ref{fig:k1SHF}). For the Cauchy-like modes we can observe that at low frequencies they are always propagative, except in the case of an incident SV wave, for which Stoneley longitudinal waves appear beyond $\theta_{\text{crit}}^{L,t}$ (also Stoneley reflected waves can be observed in this case, but we do not present the plots of $\Im(k_1)$ for reflected waves to avoid overburdening).
We can note by inspecting Figures \ref{fig:k1LF}, \ref{fig:k1SVF} and (Fig. \ref{fig:k1SHF} that the presence of Stoneley waves at high frequencies is much more widespread than at low frequencies for all $5$ (resp. $4$) modes.
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S2F.png}
\caption{}
\label{fig:k1S2F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S4F.png}
\caption{}
\label{fig:k1S4F}
\end{subfigure}\\
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S1F.png}
\caption{}
\label{fig:k1S1F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S3F.png}
\caption{}
\label{fig:k1S3F}
\end{subfigure}
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.4]{k1S5F.png}
\caption{}
\label{fig:k1S5F}
\end{subfigure}
\caption{\small Values of $\Im(k_1)$ as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for the five modes of the relaxed micromorphic medium and for the case of an incident SV wave. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The first two modes (a) and (b) correspond to the L and SV modes for the equivalent Cauchy continuum at low frequencies. The red color in these plots means that the mode is Stoneley and does not propagate, while blue means that the mode is propagative.}
\label{fig:k1SVF}
\end{figure}
\begin{figure}[H]
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH3F.png}
\caption{}
\label{fig:k1SH3F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH1F.png}
\caption{}
\label{fig:k1SH1F}
\end{subfigure}\\
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH2F.png}
\caption{}
\label{fig:k1SH2F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH4F.png}
\caption{}
\label{fig:k1SH4F}
\end{subfigure}
\caption{\small Values of $\Im(k_1)$ as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for the four modes of the relaxed micromorphic medium and for the case of an incident SH wave. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The first mode (a) corresponds to the SH mode for the equivalent Cauchy continuum at low frequencies. The red color in these plots means that the mode is Stoneley and does not propagate, while blue means that the mode is propagative.}
\label{fig:k1SHF}
\end{figure}
We can observe by direct observation of Figures \ref{fig:k1LF}, \ref{fig:k1SVF} and \ref{fig:k1SHF} that high-frequency critical angles exist for each mode corresponding to which a transition from Stoneley to propagative waves takes place. The value of such critical angles depends on the frequency for the medium-frequency regime and become constant for higher frequencies.
The influence of the existence of such high-frequency critical angles can be directly observed on the patterns of the transmission coefficient in Figures \ref{fig:Tfree} and \ref{fig:Tfixed}, in which high frequency transmission is observed for angles closer to normal incidence and no transmission is reported for smaller angles due to the simultaneous presence of Stoneley waves for all modes. We can call such zones in which transmission is equal to one ``extraordinary transmission regions'' (see e.g. \cite{misseroni2016cymatics}). Such extraordinary transmission can be used as a basis for the conception of innovative systems such as selective cloaking and non-destructive evaluation.
We can finally remark that the influence of the choice of boundary conditions on the high-frequency behavior of the transmission coefficient is still present, but do not determine drastic changes on the transmission patterns (see Figures \ref{fig:Tfree} and \ref{fig:Tfixed}).
\subsection{Cauchy medium which is ``softer'' than the relaxed micromorphic one}
In this section we present the reflective properties of a Cauchy/relaxed micromorphic interface for which we consider that the Cauchy medium on the left is ``softer'' than the relaxed micromorphic medium on the right in the same sense as in the previous section. To that end, we choose the material parameters of the left Cauchy medium to be those presented in the following Table and we explicitly remark that these values are smaller than those of Table \ref{table:macroparameters}.
\begin{table}[ht]
\centering
\begin{tabular}{c c c}
$\rho$ [$\mathrm{kg/m^3}$] & $\lambda$ [$\mathrm{Pa}$] & $\mu$ [$\mathrm{Pa}$] \\
\hline
$2000$ & $2\times 10^7$ & $0.7\times 10^7$
\end{tabular}
\caption{\small Lam\'e parameters of the ``softer'' Cauchy medium on the left side of the considered Cauchy/relaxed micromorphic interface.}
\label{table:CauchyparametersSofter}
\end{table}
With these new parameters we can compute again, following Tables \ref{table:StoneleyT} and \ref{table:StoneleyR}, the critical angles for the appearance of Stoneley waves at low frequencies. We present these values in Table \ref{table:criticalangles2}:
\begin{table}[H]
\centering
\begin{tabular}{c c c c c}
Incident wave & $\theta_{\text{crit}}^{L,r}$ & $\theta_{\text{crit}}^{L,t}$& $\theta_{\text{crit}}^{SV,t}$& $\theta_{\text{crit}}^{SH,t}$ \\
\hline
L& $-$ & $\frac{37 \pi}{100}$ & $\frac{49 \pi}{200}$ & $-$ \\
\hline
SV & $\frac{7 \pi}{20}$ & $\frac{11\pi}{25}$ &$\frac{39 \pi}{100}$ & $-$ \\
\hline
SH & $-$ & $-$ & $-$ & $\frac{39 \pi}{100}$ \\
\end{tabular}
\caption{\small Critical angles governing the onset of Stoneley waves at a Cauchy/equivalent Cauchy interface between the two Cauchy media given in Tables \ref{table:CauchyparametersSofter} and \ref{table:macroparameters}, respectively. These values are computed according to the formulas given in Tables \ref{table:StoneleyT} and \ref{table:StoneleyR}. The superscripts $r$ and $t$ stand for ``reflected'' and ``transmitted'', respectively.}
\label{table:criticalangles2}
\end{table}
Figures \ref{fig:Tfreeg} and \ref{fig:Tfixedg} show the transmission coefficient for the softer Cauchy/relaxed micromorphic interface as a function of the angle of incidence and of frequency for both boundary conditions. The coloring of this plot is again such that the dark blue regions mean zero transmission, while the gradual change towards red is the increase in transmission (red being total transmission).
Table \ref{table:criticalangles2} shows that when the incident wave travels in a soft medium and hits the interface separating this medium from a stiffer one, many critical angles exist which determine the onset of Stoneley waves for all types of incident wave at low frequencies. Since many more Stoneley waves are created with respect to the previous case of section \ref{sec:ResultsStiffer}, we would expect less transmission in the low-frequency regime than before. This is indeed the case if we inspect Figures \ref{fig:Tfreeg} and \ref{fig:Tfixedg}: the presence of low-frequency Stoneley waves induces a wides zero-transmission zone in the low-frequency regime. We can also detect a certain role of boundary conditions in widening these zero-transmission zones when considering the fixed microstructure boundary condition (see Figure \ref{fig:Tfixedg}).
Figures \ref{fig:k1LgF}, \ref{fig:k1SVgF} and \ref{fig:k1SHgF} once again show the imaginary part of the first component of the wave-vector $k_1$ for each mode of the relaxed micromorphic medium on the right. We see that Stoneley waves are observed almost everywhere both at low and high frequencies, with the exception of angles which are very close to normal incidence. Once again, the blue region denotes $\Im(k_1)=0$ (which implies that $k_1$ is real), while $\Im(k_1)$ is not vanishing in the red regions, which means that for each mode, the red color denotes that there are Stoneley waves associated to that mode. The first two modes in Figures \ref{fig:k1LgF} and \ref{fig:k1SVgF} correspond to L and SV Cauchy-like modes, while the first mode in Figure \ref{fig:k1SHgF} is the SH Cauchy-like mode in the low-frequency regime. Since we are considering a relaxed micromorphic medium, three additional modes with respect to the Cauchy case are present both for the in-plane (Figures \ref{fig:k1LgF} and \ref{fig:k1SVgF}) and for the out-of-plane problem. For this choice of parameters which make the left-side medium ``softer'' than the corresponding Cauchy medium on the right, we see that the Cauchy-like modes for all incident waves become Stoneley after a critical angle (clearly denoted on the plots with a vertical dashed line), something which is in accordance with Table \ref{table:criticalangles2}. Also Stoneley reflected waves can be observed in this case, but we do not present the plots of $\Im(k_1)$ for reflected waves to avoid overburdening.
We can conclude that it is possible to create an almost perfect total screen, which completely reflects incident waves for almost all angles of incidence and wave frequencies. This extraordinary possibility can be obtained by simply tailoring the properties of the left Cauchy medium which has to be chosen to be suitably softer than the right equivalent Cauchy medium. Regions of extraordinary transmission for very wade ranges of incident angles and wave frequencies can be engineered, opening the door to exciting applications.
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FreeL2Fig}
\caption{}
\label{fig:FreeLg}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FreeS2Fig}
\caption{}
\label{fig:FreeSVg}
\end{subfigure}\\
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FreeSH2Fig}
\caption{}
\label{fig:FreeSHg}
\end{subfigure}
\caption{\small Transmission coefficients as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for L (a), SV (b) and SH (c) incident waves for the case of macro-clamp with free microstructure and for a ``softer'' Cauchy medium on the left. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The band-gap region is highlighted by two dashed horizontal lines, where, as expected, we observe no transmission. The low-frequency regime is highlighted by the bottom horizontal dashed line, while the critical angles for the onset of Stoneley waves are denoted by vertical dashed lines. The dark blue zone shows that no transmission takes place, while the gradual change from dark blue to red shows the increase of transmission, red being total transmission.}
\label{fig:Tfreeg}
\end{figure}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FixedL2Fig}
\caption{}
\label{fig:FixedLg}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FixedS2Fig}
\caption{}
\label{fig:FixedSVg}
\end{subfigure}\\
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.45]{FixedSH2Fig}
\caption{}
\label{fig:FixedSHg}
\end{subfigure}
\caption{\small Transmission coefficients as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for L (a), SV (b) and SH (c) incident waves for the case of macro-clamp with fixed microstructure and for a ``softer'' Cauchy medium on the left. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The band-gap region is highlighted by two dashed horizontal lines, where, as expected, we observe no transmission. The low-frequency regime is highlighted by the bottom horizontal dashed line, while the critical angles for the onset of Stoneley waves are denoted by vertical dashed lines. The dark blue zone shows that no transmission takes place, while the gradual change from dark blue to red shows the increase of transmission, red being total transmission.}
\label{fig:Tfixedg}
\end{figure}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L2gF.png}
\caption{}
\label{fig:k1L2gF}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L4gF.png}
\caption{}
\label{fig:k1L4gF}
\end{subfigure}\\
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L1gF.png}
\caption{}
\label{fig:k1L1g1F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L3gF.png}
\caption{}
\label{fig:k1L3gF}
\end{subfigure}
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1L5gF.png}
\caption{}
\label{fig:k1L5gF}
\end{subfigure}
\caption{\small Values of $\Im(k_1)$ as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for the five modes of the relaxed micromorphic medium for the case of an incident L wave and a ``softer'' Cauchy medium on the left. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence isparallel to the interface. The first two modes (a) and (b) correspond to the L and SV modes for the equivalent Cauchy continuum at low frequencies. The red color in these plots means that the mode is Stoneley and does not propagate, while blue means that the mode is propagative.}
\label{fig:k1LgF}
\end{figure}
\begin{figure}[H]
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S2gF.png}
\caption{}
\label{fig:k1SV2gF}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S4gF.png}
\caption{}
\label{fig:k1SV4gF}
\end{subfigure}\\
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S1gF.png}
\caption{}
\label{fig:k1SV1g1F}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S3gF.png}
\caption{}
\label{fig:k1SV3gF}
\end{subfigure}
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1S5gF.png}
\caption{}
\label{fig:k1SV5gF}
\end{subfigure}
\caption{\small Values of $\Im(k_1)$ as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for the five modes of the relaxed micromorphic medium for the case of an incident SV wave and a ``softer'' Cauchy medium on the left. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The first two modes (a) and (b) correspond to the L and SV modes for the equivalent Cauchy continuum at low frequencies. The red color in these plots means that the mode is Stoneley and does not propagate, while blue means that the mode is propagative.}
\label{fig:k1SVgF}
\end{figure}
\begin{figure}[H]
\centering
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH3gF.png}
\caption{}
\label{fig:k1SH3gF}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH1gF.png}
\caption{}
\label{fig:k1SH1gF}
\end{subfigure}\\
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH2gF.png}
\caption{}
\label{fig:k1SH2gF}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
\centering
\includegraphics[scale=0.35]{k1SH4gF.png}
\caption{}
\label{fig:k1SH4gF}
\end{subfigure}
\caption{\small Values of $\Im(k_1)$ as a function of the angle of incidence $\theta_i$ and of the wave-frequency $\omega$ for the four modes of the relaxed micromorphic medium for the case of an incident SH wave and a ``softer'' Cauchy medium on the left. The origin coincides with normal incidence ($\theta_i=\pi/2$), while the angle of incidence decreases towards the right until it reaches the value $\theta_i=0$, which corresponds to the limit case where the incidence is parallel to the interface. The first mode (a) corresponds to the SH mode for the equivalent Cauchy continuum at low frequencies. The red color in these plots means that the mode is Stoneley and does not propagate, while blue means that the mode is propagative.}
\label{fig:k1SHgF}
\end{figure}
\section{Conclusions}
In this paper we present the detailed study of the reflective and refractive properties of a two-dimensional interface separating a classical Cauchy medium from a relaxed micromorphic medium. Both media are assumed to be semi-infinite.
We show in great detail that critical angles of incidence exist, beyond which classical Stoneley waves appear at low frequencies. It is shown that these critical angles directly depend on the relative mechanical properties of the two media. Moreover, we unveil the existence of critical angles which give rise to Stoneley waves at higher frequencies. These Stoneley waves are clearly related to the presence of an underlying microstructure in the metamaterial.
We show that, due to the onset of low and high-frequency Stoneley waves, wide frequency bounds where total reflection and/or total transmission occur can be engineered. This total reflection/transmission phenomenon is appealing for applications, in which total screens for elastic waves, such as cloaks or wave-filters, are desirable. It is clear that the ability of widening the frequency and incident angle intervals for which total reflection/transmission occur, would be of paramount importance for conceiving new devices which are more and more performant for wavefront manipulation.
We also clearly show that the simple fact of suitably tailoring the relative stiffnesses of the two media allows for the possibility of conceiving almost perfect total screens which do not transmit elastic waves for any kind of incident wave (longitudinal, in-plane and out-of-plane shear) and for almost all (low and high) frequencies and angles of incidence. Acting on such relative stiffnesses allows to achieve also the opposite situation, where total transmission occurs for large frequency bounds before a microstructure-related critical angle. This could be exploited for the conception of selective cloaks which make objects transparent to waves dependently on the angle of incidence.
Further work will be devoted to extending the present results to the case of anisotropic media and to considering refection/transmission properties at interfaces between finite media.
\newpage
{\footnotesize{}\bibliographystyle{plain}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 8,060
|
## Designing with CSS Grid Layout
Copyright © 2017 SitePoint Pty. Ltd.
* **Cover Design:** Natalia Balska
## Notice of Rights
All rights reserved. No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without the prior written permission of the publisher, except in the case of brief quotations embodied in critical articles or reviews.
## Notice of Liability
The author and publisher have made every effort to ensure the accuracy of the information herein. However, the information contained in this book is sold without warranty, either express or implied. Neither the authors and SitePoint Pty. Ltd., nor its dealers or distributors will be held liable for any damages to be caused either directly or indirectly by the instructions contained in this book, or by the software or hardware products described herein.
## Trademark Notice
Rather than indicating every occurrence of a trademarked name as such, this book uses the names only in an editorial fashion and to the benefit of the trademark owner with no intention of infringement of the trademark.
Published by SitePoint Pty. Ltd.
48 Cambridge Street Collingwood
VIC Australia 3066
Web: www.sitepoint.com
Email: books@sitepoint.com
## About SitePoint
SitePoint specializes in publishing fun, practical, and easy-to-understand content for web professionals. Visit <http://www.sitepoint.com/> to access our blogs, books, newsletters, articles, and community forums. You'll find a stack of information on JavaScript, PHP, design, and more.
# Preface
Layout in CSS has always been a tricky task: hacking solutions using positioning, floats, and the one-dimensional flexbox has never been very satisfactory. Fortunately, there is a new tool to add to our arsenal: CSS Grid Layout. It is an incredibly powerful layout system that allows us to design pages using a two-dimensional grid - offering the kind of fine-grained layout control that print designers take for granted!
Grid Layout's been in development for a while, but has recently been made a W3C candidate recommendation and has been added to most of the major browsers, so is ready for prime time.
This short selection of tutorials, hand-picked from SitePoint's HTML & CSS channel, will get you up and running with Grid Layout and using it in your own sites in no time.
## Conventions Used
You'll notice that we've used certain layout styles throughout this book to signify different types of information. Look out for the following items.
### Code Samples
Code in this book is displayed using a fixed-width font, like so:
<h1>A Perfect Summer's Day</h1>
<p>It was a lovely day for a walk.</p>
### Tips, Notes, and Warnings
#### Hey, You!
Tips provide helpful little pointers.
#### Ahem, Excuse Me ...
Notes are useful asides that are related—but not critical—to the topic at hand. Think of them as extra tidbits of information.
#### Make Sure You Always ...
... pay attention to these important points.
#### Watch Out!
Warnings highlight any gotchas that are likely to trip you up along the way.
# Chapter 1: An Introduction to the CSS Grid Layout Module
### by Ahmad Ajmi
As web applications become more and more complex, we need a more natural way to do advanced layouts easily without hacky solutions that use floats and other less burdensome techniques. An exciting new solution for creating layouts comes with the CSS Grid Layout Module.
In this introductory tutorial, I'll introduce you to this relatively new CSS feature and I'll show you using some examples how the CSS Grid Layout Module works.
## What is the CSS Grid Layout Module?
The core idea behind the Grid Layout is to divide a web page into columns and rows, along with the ability to position and size the building block elements based on the rows and columns we have created in terms of size, position, and layer.
The grid also gives us a flexible way to change the position of elements with only CSS without any change to the HTML. This can be used with media queries to alter the layout at different breakpoints.
## A Grid Layout Example
Let's start with an example to see the power of Grid Layout, and then I'll explain some new concepts in more detail.
Imagine you want to create a Twitter app with four full height columns layout (Tweets, Replies, Search, and Messages), something abstracted and similar to the screenshot below.
Here is our HTML:
<div class="app-layout">
<div class="tweets">Tweets</div>
<div class="replies">Replies</div>
<div class="search">Search</div>
<div class="messages">Messages</div>
</div>
Then we will apply some CSS to the `.app-layout` container element:
.app-layout {
.app-layout {
display: grid; /* 1 */
grid-template-columns: 1fr 1fr 1fr 1fr; /* 2 */
grid-template-rows: 100vh; /* 3 */
}
}
**View a demo here**
Here is the explanation of what we've done in the previous CSS:
1. Set the display property to `grid`.
2. Divide the container element into four columns, each column is `1fr` (one fraction) of the free space within the grid container.
3. Create one row and set the height to be `100vh` (full viewport height).
As you can see, the Grid Layout Module adds a new value to the `display` property which is `grid`. The `grid` value is responsible for setting the `.app-layout` element to be a grid container, which also establishes a new grid formatting context for its contents. This property is required to start using Grid Layout.
The `grid-template-columns` property specifies the width of each grid column within the Grid, and in our case it divides the `.app-layout` container to four columns; each one is `1fr` (25%) of the available space.
The `grid-template-rows` specifies the height of each grid row, and in our example we only created one row at `100vh`.
A layout with two columns and two rows would look like this:
And we would use the following CSS:
.app-layout {
display: grid;
grid-template-columns: 1fr 1fr;
grid-template-rows: 50vh 50vh;
}
**View a demo here**
We can also achieve the above example only on small screens by wrapping the code inside a media query. This opens up a great opportunity for us to customize the layout differently in different viewports. For example, we can create the previous layout only on viewports under `1024px` as follows:
@media screen and (max-width: 1024px) {
.app-layout {
display: grid;
grid-template-columns: 1fr 1fr;
grid-template-rows: 50vh 50vh;
}
}
**View a demo here**
## Grid Layout Module Concepts
Now that you've seen a simple example, there are some new concepts that I'd like to cover to give you a better understanding of Grid Layout. Although there are a lot of new concepts, I will only take a look at a few of them.
**Grid Item**
Grid items are the child elements of the grid container. In the above example, the `.tweets`, and `.replies` elements would qualify as grid items.
**Grid Lines**
A Grid Line is a line that exists on either side of a column or a row. There are two sets of grid lines: One set defining columns (the vertical axis), and another set defining rows (the horizontal axis).
From the above screenshot, which represents the first example, I've created four columns at `1fr` each, which will give us five vertical lines. I also created one row, which gives us two horizontal lines.
Let's see how we can position a grid item inside the grid container.
## Position Items by Using a Line Number
You can refer to an exact line number in a grid by using the properties `gridcolumn-start` and `gridcolumn-end`. We then give these properties the start and end line numbers.
Looking at the previous example, this is how the browser positions the elements by default for us:
.tweets {
gridcolumn-start: 1;
gridcolumn-end: 2;
gridrow: 1;
}
.replies {
gridcolumn-start: 2;
gridcolumn-end: 3;
gridrow: 1;
}
.search {
gridcolumn-start: 3;
gridcolumn-end: 4;
gridrow: 1;
}
.messages {
gridcolumn-start: 4;
gridcolumn-end: 5;
gridrow: 1;
}
Looking at the code for the `.tweet` column, this is what each of the three lines in the CSS does:
1. Position the child element starting from the first vertical line on the left.
2. End the element's position at the second vertical line.
3. Position the element inside the whole row.
You can change this by changing the order of elements with different positions, so the order of the elements will be: `.search`, `.replies`, `.messages`, and `.tweets`.
And we can do it as follows:
.tweets {
gridcolumn-start: 4;
gridcolumn-end: 5;
gridrow: 1;
}
.replies {
gridcolumn-start: 2;
gridcolumn-end: 3;
gridrow: 1;
}
.search {
gridcolumn-start: 1;
gridcolumn-end: 2;
gridrow: 1;
}
.messages {
gridcolumn-start: 3;
gridcolumn-end: 4;
gridrow: 1;
}
We can also use the `gridcolumn` shorthand property to set the start and end lines in one line:
.tweets {
grid-column: 4 / 5;
grid-row: 1;
}
.replies {
grid-column: 2 / 3;
grid-row: 1;
}
.search {
grid-column: 1 / 2;
grid-row: 1;
}
.messages {
grid-column: 3 / 4;
grid-row: 1;
}
**View a demo here**
This has changed the layout structure with only CSS while the markup is still as it was without any changes. This is a huge advantage of using the Grid Layout Module. We can rearrange the layout of elements independent of their source order, so we can achieve any desired layout for different screen sizes and orientations.
## Position Items by Using Named Areas
A grid area is the logical space used to lay out one or more grid items. We can name a grid area explicitly using the `grid-template-areas` property, then we can place a grid item into a specific area using the `gridarea` property.
To make this concept more clear, let's redo the four-column example with the `search` column placed first:
.app-layout {
display: grid;
grid-template-columns: 1fr 1fr 1fr 1fr;
grid-template-rows: 100vh;
grid-template-areas: "search replies messages tweets";
}
In the last line, we divide the grid container into four named grid areas, each name for a column. The next step is to position each grid item into a named area:
.search {
gridarea: search;
}
.replies {
gridarea: replies;
}
.messages {
gridarea: messages;
}
.tweets {
gridarea: tweets;
}
**View a demo here**
## Slack Example
What about using the Grid Layout Module to implement a more complex example, for example, creating the building blocks of the Slack layout. Since we are talking about layouts, we will abstract and simplify the Slack design to the building blocks represented in the grid. Something like this:
From this layout we will create three vertical columns, and three horizontal rows, and we can visualize it using the grid lines as follows:
Here is the HTML:
<div class="app-layout">
<div class="teams">Teams</div>
<div class="channels">Channels</div>
<div class="header">Header</div>
<div class="messages">
<ul class="message-list">
<li></li>
<li></li>
</ul>
</div>
<div class="input">
<input type="text" placeholder="CSS Grid Layout Module">
</div>
</div>
And the CSS:
.app-layout {
display: grid;
height: 100vh;
grid-template-columns: 100px 250px 1fr;
grid-template-rows: auto 1fr auto;
}
Here I'm using the `grid-template-columns` property to create three columns at 100px, 250px, and the third column takes up the remaining available space. The last line creates three rows: The first and third rows with auto height while the middle row takes up the remaining available space.
The remainder of the CSS looks like this:
.teams {
grid-column: 1;
grid-row: 1 / 4;
}
.channels {
grid-column: 2;
grid-row: 1 / 4;
}
.header {
grid-column: 3;
grid-row: 1;
}
.messages {
grid-column: 3;
grid-row: 2;
}
.input {
grid-column: 3;
grid-row: 3;
}
**View a demo here**
We can also create the slack layout using named areas, which you can see in this demo.
## Grid Layout Module vs Flexbox
Since many of you have started using Flexbox, you might wonder: When would it be appropriate to use Flexbox and when would it be more appropriate to use Grid Layout?
I found a good explanation from Tab Atkins:
> Flexbox is appropriate for many layouts, and a lot of "page component" elements, as most of them are fundamentally linear. Grid is appropriate for overall page layout, and for complicated page components which aren't linear in their design.
>
> The two can be composed arbitrarily, so once they're both widely supported, I believe most pages will be composed of an outer grid for the overall layout, a mix of nested flexboxes and grid for the components of the page, and finally block/inline/table layout at the "leaves" of the page, where the text and content live
Also, Rachel Andrew says:
> Grid Layout for the main page structure of rows and columns.
>
> Flexbox for navigation, UI elements, anything you could linearize.
## CSS Grid Layout Module Resources
I have not covered all the Grid Layout concepts and syntax, so I recommend you check out the following resources to go deeper:
* CSS Grid Layout Module spec
* CSS Grid Layout Examples
* Grid by Example
* The future of layout with CSS: Grid Layouts
* Follow Rachel Andrew for updates and resources. She is doing a lot of great work in relation to Grid Layout.
## Conclusion
As you've seen, the CSS Grid Layout Module is powerful because of its code brevity and the fact that you have the power to change the layout order without touching the markup. These features can help us permanently change the way we create layouts for the web.
# Chapter 2: Seven Ways You Can Place Elements Using CSS Grid Layout
### by Nitish Kumar
In this chapter, the focus will be on specific ways in which you can lay out elements on the web using CSS Grid. Now, let's go over each one of them.
## #1 Specifying Everything in Individual Properties
This is the version we have been using to place the elements in our previous articles. This method is verbose but easy to understand. Basically, the left/right and top/bottom bounds of an element are specified using `gridcolumn-start`/`gridcolumn-end` and `gridrow-start`/`gridrow-end` properties. If an element is only going to span one row or column, you can omit the `-end` properties, this way you will have to write a little less CSS.
In the demo below, element A has been placed in the second row and second column using the following CSS:
.a {
gridcolumn-start: 2;
gridcolumn-end: 3;
gridrow-start: 2;
gridrow-end: 3;
}
The same effect could be achieved by using:
.a {
gridcolumn-start: 2;
gridrow-start: 2;
}
See the demo Specifying Everything in individual Properties.
## #2 Using `gridrow` and `gridcolumn`
Even though the CSS in our first example was readable and easy to understand, we had to use four different properties to place a single element. Instead of using four properties, we can just use two — `gridcolumn` and `gridrow`. Both these properties will take two values separated by a slash where the first value will determine the start line and the second value will determine the end line of our element.
Here is the syntax you need to use with these properties:
.selector {
grid-row: row-start / row-end;
grid-column: col-start / col-end;
}
To place item C in the bottom right corner of our grid, we can use the following CSS:
.c {
grid-row: 2 / 4;
grid-column: 2 / 4;
}
See the demo Using gridrow and gridcolumn.
## #3 Using `gridarea`
Technically, the item we are laying out covers a specific area of the webpage. The boundary of that item is determined by the values we provide for the grid lines. All of these values can be supplied at once using the `gridarea` property.
This is what your CSS would look like when using this property:
.selector {
grid-area: row-start / col-start / row-end / col-end;
}
If you have trouble remembering the correct order of these values, just keep in mind that first you have to specify the position of the top-left ( `row-start` \- `col-start` ) corner and then the bottom-right ( `row-end` \- `col-end` ) corner of your element.
Just like the previous example, to place item C in the bottom right corner of our grid, we can use the following CSS:
.c {
grid-area: 2 / 2 / 4 / 4;
}
See the demo Using gridarea.
## #4 Using the `span` Keyword
Instead of specifying the end line while laying out elements, you can also use the `span` keyword to set the number of columns or rows a particular element will span.
This is the proper syntax when using the `span` keyword:
.selector {
grid-row: row-start / span row-span-value;
grid-column: col-start / span col-span-value;
}
If your element spans across only one row or column you can omit both the `span` keyword and its value.
This time let's place item C in the top left corner of our grid. We can use the following CSS to do so.
.c {
grid-row: 1 / span 2;
grid-column: 1 / span 2;
}
See the demo Using span with grid lines.
## #5 Using Named Lines
Until now, we have been using raw numbers to specify grid lines and it is easy to use when we are working with simple layouts. However, when you have to place several elements, it can get a bit confusing. Most of the times, an element on your page will fall under a specific category. For example, the header may go from column line `c1` to column line `c2` and from row line `r1` to row line `r2`. It would be a lot easier to properly name all the lines and then place your elements using these names instead of numbers.
Let's create a very basic layout to make the concept clearer. First, we will have to modify the CSS applied to our grid container:
.container {
display: grid;
grid-gap: 10px;
grid-template-columns: [head-col-start] 180px [content-col-start] 180px [content-col-mid] 180px [head-col-end];
grid-template-rows: [head-row-start] auto [head-row-end] auto [content-row-end] auto [footer-row-end];
}
What I have done above is assign names to all the lines based on the type of content that they will enclose. The idea here is to use names that provide us with some insight into the placement of different elements. In this particular example, our header element spans across all the columns. Therefore, assigning the name "head-col-start" and "head-col-end" to the first and last column line respectively will make it clear that these lines represent the left and right end of our header. All other lines can be named in a similar fashion. After all the lines have been named, we can use the following CSS to place all our elements.
.header {
grid-column: head-col-start / head-col-end;
grid-row: head-row-start / head-row-end;
}
.sidebar {
grid-column: head-col-start / content-col-start;
grid-row: head-row-end / content-row-end;
}
.content {
grid-column: content-col-start / head-col-end;
grid-row: head-row-end / content-row-end;
}
.footer {
grid-column: head-col-start / head-col-end;
grid-row: content-row-end / footer-row-end;
}
Although we had to write more CSS than usual, now just looking at the CSS will give us an idea of where an element is located.
See the demo Using Named Lines.
## #6 Using Named Lines with a Common Name and the `span` Keyword
In the previous method, all lines had different names marking the starting, midpoint or the end of an element. For example, "content-col-start" and "content-col-mid" marked the starting and mid point of the content section of our webpage. If the content section covered a few more rows, we would have to come up with additional line names like "content-col-mid-one", "content-col-mid-two" and so on.
In situations like these, we can just use a common name like "content" for all grid lines of our content section and then use the `span` keyword to specify how many of those lines an element spans. We can also just mention a number along with the line name to set the number of rows or columns an element would span.
Using this method, the CSS would look like this:
.selector {
gridrow: row-name row-start-number/ row-name row-end-number;
gridcolumn: col-name col-start-number / span col-name col-to-span;
}
Like the last method, this one also requires you to modify the CSS of your grid container.
.container {
display: grid;
grid-gap: 10px;
grid-template-columns: [one-eighty] 180px [one-eighty] 180px [one-eighty] 180px;
grid-template-rows: [head-row] auto [content-row] auto [content-row] auto [content-row] auto [footer-row] auto;
}
Each of the named column lines has the same name representing their width in pixels and each named row line represents the rows covered by a specific section of the webpage. In this demo, I have introduced an advertisement section just under the sidebar. Here is the CSS:
.header {
grid-column: one-eighty 1 / one-eighty 4;
grid-row: head-row / content-row 1;
}
.sidebar {
grid-column: one-eighty 1 / one-eighty 2;
grid-row: content-row 1 / content-row 2;
}
.advert {
grid-column: one-eighty 1 / one-eighty 2;
grid-row: content-row 2 / content-row 3;
}
.content {
grid-column: one-eighty 2 / one-eighty 4;
grid-row: content-row 1 / span content-row 2;
}
.footer {
grid-column: one-eighty 1 / span one-eighty 3;
grid-row: content-row 3 / footer-row;
}
See the demo Using Named Lines with Span.
## #7 Using Named Grid Areas
Instead of using lines, we can also place elements by assigning names to different areas. Again, we will have to make some changes to the CSS of our grid container.
The CSS for our container should now look like this:
.wrapper {
display: grid;
grid-gap: 10px;
grid-template-columns: 180px 180px 180px;
grid-template-areas: "header header header"
"content content advert"
"content content ......"
"footer footer footer";
}
A single dot (`.`) or a sequence of dots will create an empty cell with nothing in it. All the strings need to have the same number of columns. That's why we had to add the dots instead of leaving it completely blank. For now, the named grid area can only be rectangular. However, this may change in future versions of the spec. Let's take a look at the CSS of all our elements.
.header {
gridarea: header;
}
.content {
gridarea: content;
}
.advert {
gridarea: advert;
}
.footer {
gridarea: footer;
}
Once you have defined all the grid areas, assigning them to various elements is pretty straightforward. Keep in mind that you can't use any special characters while assigning names to areas. Doing so will make the declaration invalid.
See the demo Using Named Grid Areas.
# Chapter 3: How to Order and Align Items in Grid Layout
### by Nitish Kumar
In this tutorial, you will learn how to control the order in which items are placed using the Grid Layout module. After that, we will discuss how to control the alignment of different items in the grid.
## How the Order Property Works in Grid Layout
The `order` property can be used to specify the order in which different items should be placed inside a grid. By default, the items are placed in the order in which they appear in the DOM. For example, if item A is above item B in the actual source document, it will also be placed in the grid before item B. Depending on your project, this may or may not be the desired behavior.
The `order` property can be very useful, especially when there are lots of items or the items are being added dynamically. In such cases, if you want an item to be placed always at the end of the grid, you can do so easily using the `order` property.
Items with the **lowest order value are placed first** in the grid. Items with **higher values are placed later**. Items which have **the same order value will be placed in the order in which they appear in the source document**.
Let's take a look at an example.
This is our markup:
<div class="container">
<div class="item a">A</div>
<div class="item b">B</div>
<div class="item c">C</div>
<div class="item d">D</div>
<div class="item e">E</div>
<div class="item f">F</div>
<div class="item g">G</div>
<div class="item h">H</div>
<div class="item i">I</div>
<div class="item j">J</div>
</div>
Here is the CSS for placing the grid items:
.c {
gridrow-start: 1;
gridrow-end: 2;
}
.e {
gridrow-start: 1;
gridrow-end: 3;
}
.b, .j {
order: 2;
}
.a, .i {
order: 3;
}
If you recall the steps from the autoplacement algorithm tutorial, you know that the algorithm will place the items with an explicitly specified row position before placing items with no definite position. So, even though item D, without a definite row or column position, comes before item E in the actual document, it will still be placed after item E (which has a definite row position: `gridrow-start: 1` and `gridrow-end: 3`).
Among the items with no definite position, items with the lowest `order` value will be placed first. That's why items D, F, G and H are placed before items A and B. B and J have the same `order` value, therefore they are placed in the same order in which they appear in the document. Please note that both B and J are still placed before placing A and I because they have a lower `order` value.
See the demo The order Property in Grid Layout.
You need to keep **accessibility** in mind before reordering grid items using the `order` property. This property does not affect ordering of items in non-visual media. It also does not affect the default traversal order when navigating the document using sequential navigation modes like tabs. Therefore, don't use this property unless the visual order of elements needs to be out-of-sync with the speech and navigation order.
**Performing logical reordering of the grid items using the order property will make the style sheets non-conforming**.
## Aligning Content Along the Row Axis in Grid Layout
You can control the alignment of different grid items along the row axis using `justify-self` and `justify-items` .
The `justify-self` property aligns the content of a single grid item while `justify-items` aligns all the items in the grid.
The `justify-self` property can have four possible values:
* `end` aligns content to the right of the grid area
* `start` aligns content to the left of the grid area
* `center` aligns content to the center of the grid area
* `stretch` fills the whole width of the grid area
The default value of `justify-self` is `stretch`.
The `justify-items` property also aligns items with respect to the row axis and takes the same four values as the `justify-self` property. The default value is also `stretch`.
See the demo Aligning Content Along Row Axis.
## Aligning Content Along the Column Axis in Grid Layout
You can also align content along the column axis using `align-self` for single grid items and `align-items` for all the items in the grid.
Just like the previous two properties, both `align-self` and `align-items` can have four possible values: `start`, `end`, `center` and `stretch`.
The `start` and `end` values align the content to the top and bottom of the grid area respectively. The `center` value aligns the content to the center of the grid area and `justify` fills the whole height of the grid area.
The extra space between the first and second row in the demo occurs because the items in the first row are aligned to the top and the items in the second row are aligned to the bottom.
Below is `align-self` to align content along the column axis in action.
See the demo Aligning Content Along Column Axis.
## Aligning the Whole Grid
Sometimes, the outer edges of the grid may not align with the outer edges of the container. This can happen when the grid rows or columns have a fixed size. In such cases, you may want to align the grid in a specific direction.
Just like the previous properties, alignment along the row axis can be achieved using the `justify-content` property and alignment along the column axis can be achieved using the `align-content` property.
Both these properties are applied to the container grid and besides the usual `start`, `end`, `center` and `stretch` values, they also accept `space-around`, `space-between` and `space-evenly` as valid values.
* `space-around` places an equal amount of space between each grid track and half of that space between the outer edges of the grid and its container
* `space-between` places an equal amount of space between each grid track and no space between the outer edges of the grid and its container
* `space-evenly` places an equal amount of space between each grid track as well as the outer edges of the grid and its container.
The default value of both `justify-content` and `align-content` is `start`.
I have used the CSS below to align the grid items in the following demo:
.container {
justify-content: space-around;
align-content: space-evenly;
}
See the demo Aligning the Whole Grid.
Aligning items this way can also have some unintended side effects like adding extra width or height to grid items that span more than one column or row respectively. The amount of width and height that gets added to a grid item in this case depends on the number of gutters that the item crosses in each respective direction.
## Conclusion
In this tutorial, we have covered all the basics of ordering and aligning items in Grid, which can help you to gain precise control over your Grid-based layouts.
# Chapter 4: A Step by Step Guide to the AutoPlacement Algorithm in CSS Grid
### by Nitish Kumar
In this tutorial, I'll be going over all the steps the **autoplacement algorithm of the CSS Grid Layout module** follows when positioning elements. These steps are controlled by the `gridauto-flow` property.
In Seven Ways You Can Place Elements Using CSS Grid Layout, I explained all the different ways Grid lets you arrange elements on the web. However, in my previous articles I explicitly specified the position of just one element in the grid. As for the rest of the items, they got placed properly based on an algorithm.
Here, I am going to show you how this algorithm works. This way, the next time an element ends up in an unexpected location, you are not left scratching your head wondering what just happened.
## Basic Concepts for a Better Grasp of the Autoplacement Algorithm
Let's go over some fundamental concepts before diving into the workings of the algorithm.
* **Anonymous grid items** \- If you place some text directly inside a grid container without wrapping it in any tag, it will form its own anonymous grid item. You can't style an anonymous grid item because there is no element for you to style, but it still inherits style rules from its parent container. On the other hand, note that white space inside the grid container will not create its own anonymous grid item
* **Value of grid spans** \- Unlike grid positions, the algorithm has no special rules to resolve the value of grid spans. If not explicitly specified, their value is set to 1 (the item only occupies its own cell)
* **Implicit grid** \- The grid built on the basis of the value of properties like `grid-template-rows`, `grid-template-columns` and `grid-template-areas` is called **explicit grid**. Now, if you specify the position of a grid item in such a way that it lies outside the bounds of the explicit grid, the browser will generate additional grid lines to accommodate the item. These lines, along with the explicit grid, form the implicit grid. You can read more about it in Where Things Are at in the CSS Grid Layout Working Draft. The autoplacement algorithm can also result in the creation of additional rows or columns in the implicit grid.
Finally, I'd like to make the following preliminary point. The default value of the `gridauto-flow` property, which is the property controlling the algorithm, is `row`. This is also the value I am going to assume in the following explanation of the autoplacement algorithm. On the other hand, if you explicitly set the above property to `column`, remember to replace instances of the term _row_ with the term _column_ in my explanation of the algorithm. For example, the step "Placement of Elements With a Set Row Position but No Set Column Position" will become "Placement of Elements With a Set Column Position but No Set Row Position".
Now, enable the experimental features flag in your favorite modern browser to follow along and let's go over the details of all the steps the algorithm follows to build the layout.
## Step #1: Generation of Anonymous Grid Items
The first thing that happens when the algorithm is trying to place all the items inside a grid is the creation of anonymous grid items. As I mentioned earlier, you cannot style these elements because there is no item to apply the style to.
The markup below generates an anonymous grid item from the inter-element text:
<div class="container">
<span class="nonan">1</span>
Anonymous Item
<div class="nonan floating">2</div>
<div class="nonan">3</div>
<div class="nonan floating">4</div>
<div class="nonan">5</div>
</div>
Besides the generation of an anonymous item, one more thing to notice is that the grid placement algorithm ignores the CSS floats applied to _div 2_ and _div 4_ .
See the demo Anonymous Grid Items.
## Step #2: Placement of Elements with an Explicitly Specified Position
For this and the next few steps, I will be using a grid of nine different items to show how they are going to be placed.
Here's the relevant markup:
<div class="container">
<div class="item a">A</div>
<div class="item b">B</div>
<div class="item c">C</div>
<div class="item d">D</div>
<div class="item e">E</div>
<div class="item f">F</div>
<div class="item f">G</div>
<div class="item f">H</div>
<div class="item f">I</div>
</div>
The first items to be placed in the grid are those with an **explicitly set position** , which in the example below are items A and B. Just ignore all other items in the grid for now. Here is the CSS that explicitly sets the position of A and B:
.a {
grid-area: 1 / 2 / 2 / 3;
}
.b {
grid-area: 2 / 1 / 4 / 3;
}
The algorithm positions items A and B according to the values of their respective `gridarea` property. In particular, the algorithm:
* Sets the position of the top left corner of both A and B using the first and second value of the `gridarea` property
* Sets the position of the bottom right corner of both A and B using the third and fourth value of the`gridarea` property.
See the demo Placing Elements with Explicit Positions.
## Step #3: Placement of Elements With a Set Row Position but No Set Column Position
Next, the algorithm places the elements whose **row position** has been **set explicitly** by using the `gridrow-start` and `gridrow-end` properties.
For this example, I will be setting only the gridrow values of item C and item D to give them a definite row position. Here is the CSS for placing both of them:
.c {
gridrow-start: 1;
gridrow-end: 3;
}
.d {
gridrow-start: 1;
gridrow-end: 2;
}
To determine the column position, which is not explicitly set, the algorithm behaves in one of two ways, according to the **packing mode** :
* sparse packing (default)
* dense packing
### Sparse Packing in Step #3
This is the default behavior. The column-start line of our item will be set to the smallest possible line index which ensures that there won't be any overlap between the item's own grid area and the cells already occupied by other items. The column-start line also needs to be past any other item already placed in this row _by this step_. Please note that I wrote _by_ this step and not _until_ this step.
To clarify this point further, item D in the demo below did not move to the left of item A, even though it could fit in there without any overlap. This is due to the fact that the algorithm does not allow any item with an **explicitly set row position** but **not set column position** to be placed **before any other similarly positioned item in that specific row** (item C in this case). In fact, if you remove the gridrow rules applied to item C, then item D will move to the left of item A.
In short, item D, which has a definite row position but no explicitly set column position, can end up being placed before item A, but only if C does not interfere (interference takes place in this case because C, like D, has a definite row position but no set column position and is in the same row as D).
See the demo Placing Elements with Definite Grid Positions.
### Dense Packing in Step #3
If you want to fill that empty space before item A with item D, you will have to set the value of the `gridauto-flow` property to `row dense`.
.container {
gridauto-flow: row dense;
}
In this case too, the column-start line is placed at the smallest index which does not cause any overlap with other grid items. The only difference is that this time, if there is some empty space in a row where our element can fit without any overlap, it will be placed in that position, without considering the previous item in the same row with the same position rules (in this case item C).
See the demo Elements with Definite Grid Positions - Dense.
## Step #4: Determining the Number of Columns in the Implicit Grid
Next, the algorithm tries to determine the number of columns in the implicit grid. This is done by following a series of steps:
* The algorithm starts with the number of columns in the explicit grid.
* Then it goes through all the grid items with a definite column position and adds columns to the beginning and end of the implicit grid to accommodate all these items.
* Finally, it goes through all the grid items without a definite column position. If the biggest column span among these items is greater than the width of the implicit grid, it adds columns at the end of the grid to accommodate that column span.
## Step #5: Placement of Remaining Items
At this point, the algorithm has placed all the items whose position has been explicitly specified as well as items whose row positions are known. Now, it will position all other remaining items in the grid.
Before we go into more detail, you should know about the term **autoplacement cursor**. It defines the current insertion point in the grid, specified as a pair of row and column grid lines. To start off, the autoplacement cursor is placed at the start-most row and column in the implicit grid.
Once more, the packing mode, determined by the value of the `gridauto-flow` property, controls how these items are positioned.
### Sparse Packing in Step #5
By default, the remaining items are placed **sparsely**. Here is the process the algorithm follows to place them.
If the item has no definite position in either axis:
* The algorithm increments the cursor's column position until: a) either there is no overlap between already placed items and the current item b) or the cursor's column position plus the item's column span overflow the number of columns in the implicit grid.
* If such a non-overlapping position is found, the algorithm sets the item's `row-start` and `column-start` values to the cursor's current position. Otherwise, it increases the rowposition by 1, sets `column-start` to the start-most line in the implicit grid and repeats the previous step.
If the item has a definite column position:
* The column position of the auto placement cursor is set to the item's `column-start` line. If the value of this new position is less than the cursor's previous column position, the row position is increased by one.
* Next, the row position is increased by one until the algorithm reaches a value where the grid item does not overlap with any of the occupied grid cells. If needed, extra rows can be added to the implicit grid. Now the item's row start line is set to the cursor's row position and the item's row end line is set according to its span.
Let's clarify the steps listed above with what's happening in the following demo.
#### Placement of Items E and F without Definite Position in Either Axis
When the algorithm deals with item **E** , which has **no explicitly set column or row position** , the autoplacement cursor is set to row 1 and column 1. Item E occupies just one cell and it can fit in the top left corner without any overlap. Therefore, the algorithm simply places **item E at position row 1 / column 1**.
The next item without an explicitly set column or row position is **F**. At this point, the column position for the autoplacement cursor is increased to 2. However, the position row 1 / column 2 is already occupied by item A. This means that the algorithm will have to keep increasing the column position. This goes on until the autoplacement cursor reaches column 4. Since there are no more columns, the row position for the autoplacement cursor is increased by 1 and the column position is set to 1. Now the row position is 2 and the column position is 1. The algorithm again starts increasing the column position by 1 until it reaches column 4. The space at **row 2 and column 4 is currently empty and can be occupied by item F**. The algorithm places item F there and moves on to the next item.
#### Placement of Items G and H with a Definite Column Position
The remaining items with a **definite column position** in our demo are G and H. Let's examine **G**. The column position of the autoplacement cursor is set to be equal to the value of the `gridcolumn-start` property for item G, which is 3. Since this new value is less than the previous column value (which was 4), the row position is increased by one. The current row and column positions are now 3 and 3 respectively. The space at **row 3 and column 3** is currently empty and item G can be placed there without any overlap. Therefore, the algorithm places item G there. The same steps are then repeated for the placement of item H.
See the demo Positioning Remaining Items by SitePoint (@SitePoint).
### Dense Packing in Step #5
Things are handled a bit differently when the `gridauto-flow` property is set to `row dense`. When placing a grid item with no definite position, the cursor's current position is set to the start-most row and column line in the implicit grid **before** the item's position is determined.
Unlike the previous example, item I is placed to the left of item H because the cursor position gets **reset to the start-most row and column line** in the implicit grid **instead of starting from the last item placed**. At this point, while looking for a suitable position to place item I without any overlap, the cursor finds the spot to the left of item H and places it there.
See the demo Positioning Remaining Items - Dense.
## Conclusion
In this tutorial, I went over all the steps followed by the autoplacement algorithm of the CSS Grid Layout module, controlled by the `gridauto-flow` property.
Have a go at figuring out the final position of different items in a few other layouts of your own to get a better understanding of the algorithm.
# Chapter 5: How I Built a Pure CSS Crossword Puzzle
### by Adrian Roworth
Recently I created a pure CSS crossword puzzle implemented using CSS grid that does not need JavaScript in order to work. It gained heavy interest pretty quickly on CodePen. As of this writing, it has more than 350 hearts and 24,000+ page views!
The great CSS Grid Garden tutorial inspired me to build something with Grid Layout features. I wondered if these features could be put to good use in building a crossword puzzle — then I thought, let's try to create the whole thing without using JavaScript.
## Building the Board/Grid
So, first thing's first, let's create the board itself!
I ended up with the following basic structure, with HTML comments included to show what the different sections will accomplish:
<div class="crossword-board-container">
<div class="crossword-board">
<!-- input elements go here. Uses CSS Grid as its layout -->
<div class="crossword-board crossword-board--highlight crossword-board--highlight--across">
<!-- highlights for valid 'across' answers go here. Uses CSS Grid as its layout -->
</div>
<div class="crossword-board crossword-board--highlight crossword-board--highlight-down">
<!-- highlights for valid 'down' answers go here. Uses CSS Grid as its layout -->
</div>
<div class="crossword-board crossword-board--labels">
<!-- row and column number labels go here. Uses CSS Grid as its layout -->
</div>
<div class="crossword-clues">
<dl class="crossword-clues__list crossword-clues__list--across">
<!-- clues for all the 'across' words go here -->
</dl>
<dl class="crossword-clues__list crossword-clues__list--down">
<!-- clues for all the 'down' words go here -->
</dl>
</div>
</div>
</div>
That puts our basic skeleton in place so we can add more elements and start styling things.
## Using Form Elements for the Squares
The crossword puzzle I'm creating is a 13x13 grid with 44 blank spaces so I need to create 125 `input` elements each with its own ID in the format `item{row number}-{column number}`, _i.e._ `item4-12`. Here's what the grid will look like:
Each of the inputs will get have a `minlength` and `maxlength` of "1" to emulate the behaviour of a crossword puzzle (i.e. one letter per square). Each input will also have the `required` attribute so that HTML5 form validation will be used. I take advantage of all of these HTML5 attributes using CSS.
## Using the General Sibling Selector
The input elements are visually laid out in groups (exactly how a crossword puzzle is). Each group of input elements represents a word in the crossword. If each of the elements in that group is valid (which can be verified using the `:valid` pseudo selector), then we can use CSS to style an element that appears later in the DOM (using an advanced CSS selector called the general sibling selector) that will indicate that the word is correct.
Due to how sibling selectors work, and how CSS works in general, this element has to appear later in the DOM. CSS can only style elements that are after the currently selected element. It cannot look backwards in the DOM (or up the DOM tree) and style something before the current element (at the moment at least anyway).
This means I can use the `:valid` pseudo-class to style valid elements:
.input:valid {
border: 2px solid green;
}
.input:invalid {
border: 2px solid red;
}
See the demo Valid Pseudo Selector Example.
To style an element later on in the DOM that is a sibling of another element, I can use the ~ (tilde/general sibling) selector, _e.g._ `A ~ B`. This selector will select all elements that match B, that are a sibling of A and appear after A in the DOM. For example:
#input1:valid ~ #input2:valid ~ #input3:valid ~ #input4:valid ~ #input5:valid ~ .valid-message {
display: block;
}
With this code, if all these input elements are valid, the `valid-message` element will be displayed.
See the demo Using Sibling Selector to Display a Message.
The general sibling selector is extremely useful here. To make the crossword work, I needed to make sure that everything was laid out in a way that allowed me to take advantage of the general sibling selector.
The finished crossword example is using the above technique, starting at line 285. I've separated it out in the code block below:
#item1-1:valid ~ #item1-2:valid _#item1-3:valid_
#item1-4:valid _#item1-5:valid_ #item1-6:valid ~
.crossword-board--highlight .crossword-board__item-highlight--across-1 {
opacity: 1;
}
This part of the CSS ensures that if all these input elements are valid, then the opacity of the `.crossword-board__item-highlight--across-1` element will be changed. `.crossword-board--highlight` is a sibling of all the input elements, and `.crossword-board__item-highlight--across-1` is a child of `.crossword-board--highlight` so it's selectable with CSS!
## Indicating Correct Answers
Each crossword answer (i.e. group of input elements) has a corresponding "correct answer indicator" (`.crossword-board__item-highlight--across-{{clue number}}`) grid item. These grid items are placed behind the input elements on the z-axis, and are hidden using `opacity: 0`. When a correct word is entered, then the correct answer indicator grid item is displayed by changing the `opacity` to `1`, as the pseudo-class selector snippet above demonstrates.
This technique is repeated for each "word" group of input elements. So this means manually creating each CSS rule for each of the input elements in the word group and then selecting the corresponding correct answer indicator grid item. As you can imagine, this makes the CSS get big fast!
So the logical approach would be to create all the CSS rules that show/hide the correct answer indicator grid items for all the horizontal (across) clue answers. Then you would do the same for the vertical clue answers.
## Challenges of the Grid System
If, like me, you are trying to use as little CSS as possible, you will quickly realise that you cannot have overlapping grid areas within the same grid system without having to explicitely declare it. They can only sit next to each other (1 across, and 1 down share a square at the top right of the board and this is not possible when using one CSS grid to layout all the correct answer indicator items).
The solution is to wrap each horizontal (across) correct answer indicator grid item in its own grid system, and each vertical (down) correct answer indicator grid item in another. This way I can still use CSS to select them (using the general sibling selector), and they will not interfere with each other and ruin the layout of the grids.
CSS Grid Layout items act similarly to inline-block elements. This basically means that if you specify two grid items to occupy the same space, then the second item will flow around the first item and appear after it in the grid.
See the demo Grid Layout Module Example.
In the above example, the first grid item is seven columns wide and spans from the first column to the seventh column. The second grid item is meant to start at the 4th column and span to the 9th column. CSS grid doesn't like this so it wraps it to the next row. Even if you specify `gridrow: 1/1` in the second item, that will take priority and then move the first grid item to the second row.
As explained above, I avoided this by having multiple grids for horizontal and vertical items. This situation can be avoided by specifying the row and column span for each element, but I used the above method to reduce the amount of CSS, and also to have a more maintainable HTML structure.
## Checking for Valid Letter Input
Each input element has a `pattern` attribute with a regular expression as its value. The regular expression matches an uppercase or lowercase letter for that square:
<input id="item1-1" class="crossword-board__item"
type="text" minlength="1" maxlength="1"
pattern="^[sS]{1}$" required value="">
This was not ideal because the answers are in the HTML. I wanted to hide the answers in the CSS, but I could not find a way of doing this successfully. I attempted the following technique:
.input#item1-1[value="s"],
.input#item1-1[value="S"] {
/* do something... */
}
But this won't work. The attribute selector will select the element based on what is actually inside the HTML, and doesn't consider live changes. So I had to resort to the `:valid` pseudo-class approach detailed above, and consequently (and annoyingly) exposing the answers in the HTML itself.
## Highlighting the Clues on Hover
All horizontal (across) clues are wrapped in a `div`, as are the vertical (down) clues. These wrapping `div` elements are siblings of the `input` elements in the crossword grid. This is demonstrated in the HTML structure listed above in a previous code block. This makes it easy to select the correct clue(s) depending on which input element is being focused/hovered.
For this, each `input` element needs `:active`, `:focus`, and `:hover` styles to highlight the relevant clue by applying a background color when the user interacts with an `input` element.
#item1-1:active ~ .crossword-clues .crossword-clues__list-item--across-1,
#item1-1:focus _.crossword-clues .crossword-clues__list-item--across-1,
#item1-1:hover_ .crossword-clues .crossword-clues__list-item--across-1 {
background: #ffff74;
}
## Numbering the Clues
The numbers for the clues are positioned using a CSS Grid pattern. Here's an example of the HTML, abbreviated:
<div class="crossword-board crossword-board--labels">
<span id="label-1" class="crossword-board__item-label crossword-board__item-label--1">
<span class="crossword-board__item-label-text">1</span></span>
<span id="label-2" class="crossword-board__item-label crossword-board__item-label--2">
<span class="crossword-board__item-label-text">2</span></span>
<!-- rest of the items here..... -->
</div>
Then the CSS looks something like this:
.crossword-board__item-label--1 {
gridcolumn: 1/1;
}
.crossword-board__item-label--2 {
gridcolumn: 4/4;
}
/* etc... more items here... */
Each number is placed at the start position of its related group of input elements (or word). The number is then made to be the width and height of 1 grid square so that it takes up as little space as possible within the grid. It could take up even less room by implementing CSS Grid differently here, but I opted to do it this way.
## The "Check for Valid Squares" Checkbox
At the top of the crossword, you'll notice a checkbox labelled "Check for valid squares". This allows the user to check if certain letters are correct, even if a given word is wrong.
Creating this is rather beautiful as it's one CSS rule that makes all the valid squares get highlighted. It's using the checkbox hack to select all valid input elements that are after the checked checkbox in the DOM.
Here is the code:
#checkvaliditems:checked ~ .crossword-board-container .crossword-board__item:valid {
background: #9aff67;
}
## Conclusion
That covers all the main techniques used in the demo. As an exercise, this shows you how far CSS has come in recent years. There are plenty of features we can get creative with. I for one can't wait to try and push other new features to the limits!
If you want to mess around with the CSS from this article, I've put all the code examples into a CodePen collection. The full working CSS crossword Puzzle can be found here.
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Here is the list of solved easy to difficult trigonometric limits problems with step by step solutions in different methods for evaluating trigonometric limits in calculus.\n\nEvaluate $\\displaystyle \\large \\lim_{x\\,\\to\\,0}{\\normalsize \\dfrac{1-\\cos{mx}}{1-\\cos{nx}}}$\n\nEvaluate $\\displaystyle \\large \\lim_{x\\,\\to\\,0}{\\normalsize \\dfrac{\\log_{\\displaystyle e}{\\big(\\cos{(\\sin{x})}\\big)}}{x^2}}$\n\nEvaluate $\\displaystyle \\large \\lim_{x\\,\\to\\,\\Large \\frac{\\pi}{4}}{\\normalsize \\dfrac{\\sin{x}-\\cos{x}}{x-\\dfrac{\\pi}{4}}}$\n\nEvaluate $\\displaystyle \\large \\lim_{x\\,\\to\\,0}{\\normalsize \\dfrac{\\sin^3{x}}{\\sin{x}-\\tan{x}}}$\n\nEvaluate $\\displaystyle \\large \\lim_{x\\,\\to\\,0}{\\normalsize \\dfrac{\\log_{e}{(\\cos{x})}}{\\sqrt[\\Large 4]{1+x^2}-1}}$\n\nEvaluate $\\displaystyle \\large \\lim_{x \\,\\to\\, 0}{\\normalsize \\dfrac{e^x-e^{x\\cos{x}}}{x+\\sin{x}}}$\n\nEvaluate $\\displaystyle \\large \\lim_{x\\,\\to\\,0}{\\normalsize 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\\dfrac{\\cos{3x}-\\cos{4x}}{x\\sin{2x}}}$\n\n$\\displaystyle \\large \\lim_{\\theta \\,\\to\\, 0}{\\normalsize \\dfrac{\\sin{5\\theta}-\\sin{3\\theta}}{\\theta}}$\n\nLatest Math Problems\n\nA best free mathematics education website for students, teachers and researchers.\n\nMaths Topics\n\nLearn each topic of the mathematics easily with understandable proofs and visual animation graphics.\n\nMaths Problems\n\nLearn how to solve the maths problems in different methods with understandable steps.\n\nLearn solutions\n\nSubscribe us\n\nYou can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.","date":"2023-01-31 07:11:58","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, 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| null | null |
Cliff Barrows (1923-2016): A man of song. And the trombone.
Clifford "Cliff" Barrows, long-time song leader for evangelist Billy Graham, died yesterday at the age of 93. He was part of a trio – along with Graham and singer George Beverly Shea – who defined large-scale Christian evangelism in the second half of the twentieth century. Graham, Shea and Barrows preached, sang and led singing before millions of people since they first worked together in 1946. The photo above shows Cliff Barrows leading singing at the 1946 Youth for Christ meeting in Seattle, Washington.
The newspapers today are full of tributes to Cliff Barrows and a good summary of his life and career is found in his obituary in the Charlotte (North Carolina) Observer. This was a Godly man who changed lives in many ways and he is more than deserving of all of the warm remembrances that are being written about him today.
But several years ago, I learned about a side of Cliff Barrows that most people had either not ever known about or had long forgotten: he played the trombone.
I first learned that Cliff Barrows played trombone while touring the Billy Graham Center Museum at my undergraduate alma mater, Wheaton College (Illinois). As I came around a corner, I saw photographs of two men that were holding trombones: Homer Rodeheaver (I had never heard of him before) and Cliff Barrows (I didn't know he had played trombone). I learned quickly that Rodeheaver was the trombone-playing song leader for evangelist Billy Sunday in the first third of the 20th century. And this realization – that the two most influential Christian evangelists of the 20th century were both named "Billy" and both had song leaders that played the trombone – sent me running to learn more.
I turned my attention to Rodeheaver, a man who was a household name for decades but today has been largely forgotten. Here was a man who played the trombone for over 100 million people; his tremendous influence as a trombonist is incalculable. "Surely," I thought, "there must be a story in all of this." And indeed there was. It first led to my writing an article, "Homer Rodeheaver: Reverend Trombone" that was published in the Historic Brass Society Journal earlier this year. And, happily, it has now led to my co-authoring a book about Rodeheaver for University of Illinois Press with my friend, Kevin Mungons. We are, at this moment, deep into the process of writing the book and when it appears in a few years, it will be accompanied by a two-CD set of recordings of Rodeheaver singing, speaking and playing trombone. More on this in time! But while doing research about Rodeheaver at the Billy Graham Center Archives and at Grace College in Winona Lake, Indiana (where Billy Sunday and Rodeheaver had their homes and archives for both Sunday and Rodeheaver are found) and the Winona History Center, photos of Cliff Barrows kept popping up. I needed to know more.
The Barrows' specialize in piano and trombone arrangements, and their duets and solos have made them friends of everyone who has attended their performances. It was ventured by one who attended the great Billy Sunday campaigns that Mr. Barrows is the equal of Homer Rodeheaver, song leader for the late evangelist, so skillfully does he lead the large crowds in congregational singing of hymns and choruses. Billie Barrows, who, by the way, has been Cliff's wife for just 13 days, has thrilled young and old with her renditions of favorite songs at the piano.
Unfortunately, there are no known recordings of Cliff Barrows playing the trombone. But there is a brief moment where he is seen on film with the trombone in his hands. The screenshot above is from a video of Cliff Barrows playing the trombone at the 1949 Christ for Greater Los Angeles Billy Graham Crusade. You'll find the footage of Barrows at around 5:00 in the video (click on the link in the text above to view the complete film).
I would play with the choir and bring the downbeat with my horn and when I would hold a long note, I'd hold it out with them and the horn was just a part of me. I felt so natural with it hanging on my arm.
Of all of the photographs I have seen of Cliff Barrows, it is the one above, taken at the 1954 Billy Graham Crusade at the Olympic Stadium in Helsinki, Finland, that I like the best. Look at the tens of thousands of people sitting in the stadium. The infield is empty. And on the platform is Cliff Barrows, playing his trombone accompanied by an upright piano (see the enlargement, below). Two people playing a hymn tune. They are minuscule and nearly lost in the enormity of the crowd. But when a trombone was in his hand, Cliff Barrows knew how to make it sing.
You're welcome, Brother Yeo. God bless you brother. Thank you for letting me visit with you.
And with that, two trombone players named Cliff Barrows and Douglas Yeo hung up the phone. Today, as I reflect on the life and ministry of Cliff Barrows, I am so grateful my life intersected with his for a brief moment, where our shared love of Jesus Christ, music and the trombone came together. It was Homer Rodeheaver who led me to Cliff Barrows, the same Homer Rodeheaver who was such an encouragement when Cliff Barrows was just beginning his ministry with Billy Graham. And like Rodeheaver, shown below with Billy Sunday (in a white suit standing behind Rodeheaver) at Winona Lake, Indiana in 1931, Cliff Barrows used the trombone as a tool for leading singing and for bringing the Good News of Jesus Christ to millions of people. It's something I'll be doing this Sunday when, with my wife at the piano, I pick up my trombone and play the great song by George Beverly Shea and Rhea Miller, I'd Rather Have Jesus Than Silver or Gold as the offertory in our church's Sunday morning service. At that moment, I certainly will be thanking God for the life, ministry and influence of Cliff Barrows, a man of song. And the trombone.
Last week I had the great pleasure of traveling to University of Illinois to take part in several immensely rewarding activities.
Over the years I have been a guest artist at dozens of schools, colleges and universities around the world. The opportunity to engage with students – whether in a lecture, performance, masterclass or, as was the case at University of Illinois, something completely different – is exceptionally rewarding and I always enjoy becoming part of the local musical culture when I am visiting.
The invitation to travel to Champaign-Urbana came from Scott Schwartz, Archivist for Music and Fine Arts and Director of the Sousa Archives and Center for American Music on the University of Illinois campus. Scott and I had met many years ago at the Great American Brass Band Festival in Danville, Kentucky, where I had presented a paper about the use of serpent and ophicleide in brass bands and I performed a solo on ophicleide accompanied by the Athena Brass Band.
Scott asked if I would be interested in coming to Illinois to give a lecture/demonstration about early American trombone makers, their innovations and marketing strategies. The Sousa Archives had set up a very nice exhibit of six late-nineteenth and early-twentieth trombones as well as mouthpieces, catalogs, advertisements and other ephemera. In addition, we had selected six other instruments for me to play and demonstrate. Oh, and not to be lost in the moment is that the Chicago Cubs had just won baseball's World Series and it seemed appropriate to make my Cubs hat part of the display.
I always enjoy getting my hands on, talking about and playing old instruments, such as the alto valve trombone pictured above. The time at the Sousa Archives was very rewarding and was made more so because of the engaged audience and their great questions. the trombone exhibit will run for many more months along with an exhibit about the great American composer and bandmaster, Henry Fillmore. If you're in the area, don't miss it.
From the Sousa Archives I went to the University of Illinois School of Music where I gave a trombone masterclass. I worked with three talented students and also enjoyed getting together with my friend, Jim Pugh, who teaches jazz trombone and composition at University of Illinois. That was fun.
I have known Jim for decades and have the utmost respect for him as a player and a person. Several years ago I reviewed his superb CD, X Over Trombone, and I consider him to be one of the most creative players – and composers – on the scene today. Despite our long friendship, we had never played together, so we started the masterclass with a performance of Charles Small's duet Conversation.
The third piece of my University of Illinois trip was a performance with the Marching Illini Band under the direction of Barry Houser. As an event with another connection to my trombone lecture and masterclass, I led a group of 75 trombone players – both members of the Marching Illini Band and students from local high schools – in a performance of Meredith Wilson's 76 Trombones to start the halftime show of the Illinois/Michigan State football game. 75 + me = 76 Trombones. That doesn't happen every day. Click the video image below to see the whole halftime show; it begins with 76 Trombones, and continues with a tribute to the Chicago Cubs and much more.
Now, when you put 76 trombones on a football field accompanied by a marching band, that is one impressive sight and sound. My hat is off to the Marching Illini for inviting local high school trombone players to join with the 40 trombonists of the Illini Band to get us up to 76 trombone players. This is one fine band, and I was caught up in many of their great traditions. School spirit was alive and well; it was a great day of interaction for all of us and, hey, Illinois won the game. It must have been the trombones.
I want to send a big THANK YOU to Scott Schwartz for making all of this happen, to Jim Pugh for his setting up the trombone masterclass and for playing Conversation with me, and to Barry Houser and all of the members of the Marching Illini Band for a great few days where we all came together in Illini blue and orange and celebrated the trombone. This was a memorable and very satisfying trip. Go Illini! I – L – L – – I – N -I !
When you pick up a hymnal or book of songs, you usually find the composer and author of the lyrics listed at the top of the page, near the title. When both music and lyrics are written by the same person, often just the person's initials appear. Today, November 7, marks what would have been the 100th birthday of one of the most popular and beloved writers of gospel songs, Beatrice Bush Bixler, or B.B.B.
Bea was a gifted and prolific song writer. After attending Houghton College for two years, she toured with evangelist Burton B. Bosworth in the late 1930s and early 1940s; Bosworth was a trombone player in addition to a speaker and it was there that Bea first accompanied a trombonist. It was not to be her last time doing so. She continued and finished her studies at Nyack Missionary Training Institute, later renamed Nyack College, where my wife's father met her when he was in school there after World War II. She went on to publish dozens of gospel songs including Life is a Symphony, I Am Not Worthy, It May Be Today and The Breaking of the Bread, many of these were featured in the song books by Singspiration, Favorites. These songs were particularly popular in the 1940s through 1970s in church meetings and revivals. Singspiration published two books of her music, now out of print, that contain many of her most beloved songs.
Bea Bixler had a style that was all her own. I've often described her voice as that of a rough-hewn Kate Smith. Her piano playing was, at times, typical of the style used among evangelical Christian pianists of her time – fingers running up and down the keyboard, huge waves of sound that swept over you like a mighty wave. But Bea also had a beautiful, poetic piano touch that could melt your heart. She recorded two LPs of her singing, also long out of print.
Bea died in 2013 but her influence is still felt today by those who knew her and heard her music. The wife of a Christian and Missionary Alliance pastor, she loved Jesus deeply, and she was vivacious and energetic. I treasure the times our family had with her, enjoying the stories of how she came to write her songs, collaborating with her in music, and listening to her robust laugh.
Click here to hear a track from my CD, Cornerstone, with me on bass trombone and Bea Bixler at the piano, playing her song I Am Not Worthy. She improvised her accompaniment at the recording session; it makes me smile as I remember this remarkable woman. Beatrice Bush Bixler, B.B.B. A dear friend and saint of the Lord who would have been 100 years old today.
I know I'm not the only person who thinks that Conversation by Charles Small is the finest duet ever written for tenor and bass trombone. I've played this piece dozens of times with dozens of tenor trombonists over the year, beginning with the first time I played it in a recital at Peabody Conservatory of Music in Baltimore with my then Baltimore Symphony trombone colleague, Jim Olin in 1983. I recorded Conversation in 1996 with my Boston Symphony colleague, Ronald Barron and that recording was released on my CD, Proclamation, and Ron's CD, In the Family. This past spring, I played it at Arizona State University on a doctoral recital given by my one of my students, Tim Hutchens. In fact, I will be playing it again this Friday, November 4, with jazz trombonist Jim Pugh at a masterclass I'm giving at University of Illinois School of Music at 2:30 PM in the University of Illinois School of Music Auditorium (the class is free and open to the public). Charlie wrote the duet for himself and bass trombonist David Taylor and has become a regular feature on student and professional trombone recitals for many years.
I have always played Conversation off set of parts that were done from Charlie Small's original manuscript by a copyist (shown at the top of this article). But that version was not widely available, and in 1993, Conversation was published by Kagarice Brass Editions. The duet saw several printings by KBE but each had particular problems – things that were missing or not clear. Kagarice Brass Editions is now part of Ensemble Publications, and I asked the owner of Ensemble Publications, my friend Chuck DePaolo, if he would be interested in making a new printing of Conversation with Charlie Small's approval. Chuck thought it would be great to to this so a few days ago, I called up Charlie Small – he lives here in Arizona, not far from where my wife and I live – and yesterday, we spent a several hours going over Conversation with a fine toothed comb.
It should be said that spending time with Charlie Small is a rare and truly wonderful thing. A member of the Tommy Dorsey Orchestra in the 1940s when he was a teenager, Charlie has been a major influence on the trombone scene for many decades. Conversation is only one of his many compositions and it was great not only to sit down and talk about his piece, but hear Charlie tell some great stories as well. We went through Conversation measure by measure. After several hours of working together, Charlie felt confident that we had identified all of the things that needed to be changed/fixed in this new edition. It's now been sent off to the publisher and very shortly, a version of Conversation that reflects the composer's intentions will soon be available. At last!
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E-text prepared by Bethanne M. Simms, Diane Monico, and the Online
Distributed Proofreading Team (http://www.pgdp.net)
Everyman's Library
Edited by Ernest Rhys
Romance
AUCASSIN & NICOLETTE
And Other Mediaeval Romances and Legends.
With an Introduction by
EUGENE MASON
* * * * *
THE PUBLISHERS OF _EVERYMAN'S LIBRARY_ WILL BE PLEASED TO SEND FREELY
TO ALL APPLICANTS A LIST OF THE PUBLISHED AND PROJECTED VOLUMES TO BE
COMPRISED UNDER THE FOLLOWING THIRTEEN HEADINGS:
TRAVEL . SCIENCE . FICTION
THEOLOGY & PHILOSOPHY
HISTORY . CLASSICAL
FOR YOUNG PEOPLE
ESSAYS . ORATORY
POETRY & DRAMA
BIOGRAPHY
REFERENCE
ROMANCE
[Illustration]
IN FOUR STYLES OF BINDING: CLOTH,
FLAT BACK, TOP; LEATHER,
ROUND CORNERS, GILT TOP; LIBRARY
BINDING IN CLOTH, & QUARTER PIGSKIN
LONDON: J. M. DENT & SONS, LTD.
NEW YORK: E. P. DUTTON & CO.
* * * * *
[Illustration:
A ROMANCE,
AND IT
ME TOOK
TO READ
& DRIVE
THE NIGHT
AWAY
CHAUCER]
AUCASSIN & NICOLETTE
and other Mediaeval Romances and Legends
translated from the French by
EUGENE MASON
London & Toronto
Published By J. M. Dent
& Sons Ltd & in New York
by E. P. Dutton & Co
First Issue Of This Edition 1910
Reprinted 1912, 1915
INTRODUCTION
The little tales brought together in this volume are drawn from the
literature of the Middle Ages, and in many cases were written in France
of the thirteenth century. I hope that they may be found interesting in
themselves, but to appreciate them fully they should be considered in
their relations to a definite historical background. Their conceptions
of society, of religion, of politics, of humour--that precious gift
which always dies so young--are not common to all of us to-day. They
are of the thirteenth century, and we of the twentieth. We may not be
better than our forefathers, but a great chasm of seven hundred years
yawns between us and them. To enjoy their work without reserve it is
necessary for a time to breathe the same air that was breathed--roughly
speaking--by the subjects of St. Louis of France.
It is possible to love the period known as the Middle Ages, or it is
possible to detest it. But you cannot ignore it, nor find it
flavourless on the palate, because that period possesses character,
"character, that personal quality, that idiosyncrasy which, no doubt,
you are the richer for possessing, be it morally bad or good--for it is
surely better to have a bad character than none, and if you are a
church, better to be like the Badia than the City Temple." Indeed, it
is evident that the personal equation must largely determine what any
writer's conception of the Middle Ages is. A great modern poet, for
instance, loved the Middle Ages because economic conditions pressed
less hardly on the poor; because London was small and white and clean;
because chivalry afforded opportunity for that decorative treatment of
knightly episodes which makes his poetry so attractive. Yet across the
Channel, much at the same time, an equally distinguished poet treated
of the same period in a book of poems which it is instructive to
consider side by side with the work of William Morris, and the
Frenchman's verse is lurid with fire and bigotry, and the tale of
man's inhumanity to man. And the strange point is that both writers
could give chapter and verse for the very different type of story they
selected. Again, the religious temperament is apt to look back fondly
to the Middle Ages as the "Age of Faith." To such minds mediaevalism is
a period of easy acquiescence in spiritual authority, a state of health
before the world grew sick with our modern disease of doubt. Certainly
these centuries produced saints whose arresting examples and haunting
words must always be the glory of Christianity, and it is equally
certain that the offices and doctrines of the Church entered far more
intimately into the lives of the common folk than they do to-day. But
side by side with faith there was a "spirit of rebellion and revolt
against the moral and religious ideas of the time." It may be found in
many strangely different shapes: in the life of Abelard; in the
extraordinary spread of witchcraft; and--in its supreme literary
expression, perhaps--in a famous passage of "Aucassin and Nicolette."
And, to take a third illustration of the same difficulty, were the
Middle Ages years of sheer lyric beauty, or rather years of
inexpressible ugliness and filth? "If you love the very words 'Middle
Age'; if they conjure up to your mind glowing old folios of black
letter with gilt and florid initials; crimson and green and blue pages
in which slim ladies with spiked head-dresses walk amid sparse flowers
and trees like bouquets, or where men-at-arms attack walled cities no
bigger than themselves, or long-legged youths with tight waists and
frizzed hair kiss girls under apple-trees; or a king is on a dais with
gold lilies for his background, minstrels on their knees before him,
lovers in the gallery"--well, if you accept all this dainty
circumstance, you get sheer lyric beauty, and nothing else. Only there
is another side, a side not very pleasant to dwell upon, and it may
perhaps be hinted at by saying that such a necessary of the toilet as a
pocket-handkerchief was not generally known in this Age of Beauty.
Perhaps it would be truer to hold that the Middle Ages comprised all
these things--the knight-errant and the tormentor; the altar and the
witch's Sabbath; a dream of loveliness having its roots in slime and
squalor. These centuries were both "enormous and delicate." They were
great enough to include opposites, and to square the circle. You may
love them, or you may hate them; but they cannot be given the go-by.
The philosophy of the Middle Ages--that is to say, the idea which
governed their political and theological conceptions--was both simple
and profound. The Emperor or King was considered to be the guardian of
the temporal order of things, just as the Pope was held to be the
supreme authority in matters of eternal and spiritual concern. It was
an idea fascinating in its simplicity, but life is a tangled and
complex matter, and in practice, planets, which in theory moved
strictly in their own orbits, were continually striking across each
other's path. Even St. Louis, the King, who carried saintliness to the
extreme limit permitted to man, was involved in frequent political
quarrels with the supreme head of his Church, and by one of the little
ironies of fate came within measurable distance of excommunication. The
King--again in theory--was the owner of all his realm. This was
necessary to abolish Heptarchies. But for the support of the Crown he
parcelled out his realm amongst great lords, and thus established
Heptarchies again. The great barons, in their turn, divided their
estates amongst knights, bound to assist them in their quarrels, and to
furnish a certain number of soldiers to their service. Amongst these
knights sprang up one of the supreme institutions of the Middle
Ages--the institution of chivalry. "It took its birth in the interior
of the feudal mansions, without any set purpose beyond that of
declaring, first, the admission of the young man to the rank and
occupation of the warrior; secondly, the tie which bound him to his
feudal superior--his lord, who conferred upon him the arms of
knighthood. But when once the feudal society had acquired some degree
of stability and confidence, the usages, the feelings, the
circumstances of every kind which attended the young man's admission
among the vassal warriors, came under two influences, which soon gave
them a fresh direction, and impressed them with a novel character.
Religion and imagination, poetry and the Church, laid hold on chivalry,
and used it as a powerful means of attaining the objects they had in
view, of meeting the moral wants which it was their business to provide
for." Throughout a long apprenticeship, in a castle which contained
practically but one woman, the wife of his lord and she removed how
infinitely from him in distance and in station, the young squire was
trained to feel towards all women something of the dreamy devotion with
which art and religion taught him to regard Our Lady herself. And the
apprenticeship culminated in the ceremony of knighthood, with all the
mystical significance of the symbolism preserved for us in the little
story of Sir Hugh of Tabarie and the Sultan Saladin, carefully
calculated to impress the recipient in the highest degree. Devotion to
God, to his king, and to his lady--these were the ideals of knighthood,
not always, unfortunately, its realities. But ideals are difficult of
realization in so faulty a world as ours. The Black Prince was the very
pattern of chivalry in his youth, yet Froissart remarks in his account
of the battle of Poitiers that "the Prince of Wales, who was as
courageous and cruel as a lion, took great pleasure this day in
fighting and chasing his enemies." The conduct of that perfect gentle
knight, Sir Graelent, towards the lady he discovered bathing in the
fountain, was far from chivalrous, according to modern notions, and yet
I can assure the reader that I have walked delicately as Agag, and gone
to the verge of weakness, in recounting the incident. Finally, here is
a passage from a letter written by a knight of the fourteenth century
to the Tyrant of Mantua, relating to a French girl, Jeannette, which is
sufficiently explicit. "Let her be detained at my suit, for if you
should have a thousand golden florins spent for her, I will pay them
without delay, for if I should have to follow her to Avignon I will
obtain this woman. Now, my lord, should I be asking a trifle contrary
to law, yet ought you not to cross me in this, for some day I shall do
more for you than a thousand united women could effect; and if there be
need of me in a matter of greater import, you shall have for the asking
a thousand spears at my back." Ah, well, ideals that are realized cease
to be ideals.
Just as this worship of woman was the great social note of the Middle
Ages, so the devotion to the Blessed Virgin was the distinguishing
religious feature of those times. In honour of Our Lady were erected
the magnificent Gothic cathedrals--those masterpieces of moral
elevation--which stud the fair land of France like painted capitals
upon a written page. In these buildings the genius of the Middle Ages
found its supreme expression. Above the crowded market-place and narrow
mediaeval street rose those incomparable churches, "like Gothic queens
at prayer, alone, silent and adorned." In her honour, too, they were
made beautiful with glass and statuary, so that never before nor since
were churches filled with such an entrancing congregation, never had
buildings such wonderful eyes. And at a time when masons built to her
honour and theologians defined her position, the story-tellers were not
slack in her praise. The three legends relating to the Virgin, which I
have included in this book of translations, are but specimens of an
immense literature devoted to her service. "Our Lady's Tumbler" is, to
the modern taste, one of the most appealing of all these legends, but
there are others nearly, if not quite, so beautiful. Once upon a time
there was a monk who was so ignorant that he was exposed to the rebuke
of his brethren. But in his devotion to Our Lady he took for his
meditation five psalms, each commencing with a letter of her name. And
when it pleased God that his end should come, there happened a very
beauteous miracle, for from his mouth came forth five fresh roses,
sweet, crimson and leafy, in honour of the five letters of the name of
Maria. Again, how exquisite is the story of the nun who by frailty of
heart fled from her cloister to give herself over to sin. After many
long years she returned to the nunnery, having lost her innocence, but
not her faith, for during all her wanderings she had never omitted her
habit of prayer to Our Lady. But, to her surprise, always she was
addressed by her sisters as if she had never gone from amongst them.
For the Blessed Virgin, having clothed herself with the vesture and
seeming of the truant who loved her, even in sin, took also upon her
the duties of a sacristan from which she had fled, so that no single
person had noticed the absence from her cloister of the faithless nun.
Yet, after all, the Middle Ages delighted to honour Our Lady as the
tender Mother rather than as the Queen of Heaven. In numberless
miniatures, and on the portals of the cathedrals raised to her glory,
she stands presenting her Child to the adoration of men. It is as the
instrument of the Incarnation that her ultimate dignity consists.
Indeed, the religion of the Middle Ages can only be appreciated by
regarding it in the light of the doctrine of the Incarnation. Christ is
God. The Mass--the popular service instituted by Himself--is an
extension of His Incarnation. The Blessed Virgin is to be held in
peculiar veneration as the Mother of God. The two threads can easily be
seen twined together in that story of how Our Lady tourneyed whilst the
knight was at Mass. But belief in the Incarnation is the keystone of
mediaeval theology, and the only explanation of the lives of those
saints who poured out their years like water in the service of God and
man.
The authors of the stories brought together in this book from various
sources are, in some cases, identified, but in others are unknown. They
may, perhaps, be regarded as representative of the three classes who
are responsible for this kind of fiction--the monk, the trouvere and
the professional minstrel. The monk, for his part, wrote in French
seldom enough. He was a scholar, and when he had something to say,
preferred to deliver himself in Latin, the language common to all
educated men. But, for once, in the thirteenth century, a monk of
Soissons, named Gautier de Coinci, translated into French verse a great
collection of the miracles of Our Lady. From this garner I have
selected the legend "Of a Jew who took as Surety the Image of Our
Lady." Gautier de Coinci may not have been a supreme poet--that saving
grace comes seldom enough--but his industry was certainly abnormal. His
labour of love must have been the occupation of a lifetime, and it is
pleasant to recall the old monk, in silent scriptorium and shady
cloister, turning the Latin legends into fluent and pious verse.
The trouvere was drawn from the same class as the troubadour, and the
circumstances of their lives were in essentials much the same. He lived
very probably in some nobleman's castle, where he composed his stories
as a sort of amateur, and recited the verses to an audience more or
less select. His pride forbade him to appear personally before the
populace, but it permitted him to provide wandering minstrels with
copies of these poems, and so entertain the common folk by deputy. In
the lord's castle it was, of course, another matter. On summer
afternoons he would recite before the baron's household, where they
were seated on the steps of the garden terrace, each in his order and
degree. You can feel the hush and heat of the Provencal evening, whilst
the sombre cypresses spire into the sky, and the olives whisper, and,
far below, the broad stretches of the Rhone are suffused with the
lovely light and colour of southern France. Or, in winter, after
supper, when the tables were cleared, the trouvere would recite in
hall. At the feet of the ladies sat their knights on silken cushions,
fettered with silver chains, each to his friend. It was an audience
rich and idle, familiar with the fantastic lives of the troubadours,
and with the wanton judgments of the Courts of Love. For such a company
no flower of sentiment could be too highly scented, and no tale come
amiss, save only that it spoke of love.
If the trouveres were "the aristocracy of this literature," the
minstrel was its "democracy." Sometimes he rose almost to the status of
the trouvere, composing his own stories, and reciting them even in
kings' houses. Generally, however, the minstrel was but a strolling
player, speaking other men's thoughts, and wandering over the length
and breadth of the land. Occasionally he went alone with his viol. At
other times he was accompanied by bears, or a little troupe of singing
boys or dancing girls. The minstrel might have the good fortune to give
his entertainment before some knight or count. At any rate, the common
folk heard him gladly, before the church or on the village green. If he
was lucky, the homeless minstrel got free lodgings for the night at
some hospitable monastery, but occasionally he was turned from the
door, with hard words, because of St. Bernard's saying that "the tricks
of the jongleurs can never please God." Once upon a time such a
minstrel as this knocked at a monastery door, and asked for
hospitality. He was received without indecent joy, and the
guest-master, forgetting that a grace conferred unwillingly is no
favour at all, provided the guest with black bread, salted vegetables,
cold water, and a hard and dirty pallet. The abbot obviously felt no
passion for strolling vagabonds, and had appointed a guest-master after
his own heart. On the morrow, when the minstrel was leaving the
monastery, he met the abbot returning from a short journey. To revenge
himself, at any rate, on one of the two, the minstrel accosted him
effusively. "My lord," said he, "I thank you and all the community from
the bottom of my heart, for Brother such-an-one has welcomed me like
Christ Himself last night. He lighted a fire in my chamber, and served
me with choice wines, excellent fish, and more dishes than I am able to
recall. And this morning when he bade me farewell he gave me shoes,
and these leathern laces, and a knife." When the abbot heard this he
was filled with anger, and, parting shortly from the minstrel, he
hastened to the monastery, and promptly relieved the guest-master of
his office, before the latter could offer a word of explanation. Thus
was the minstrel revenged on this grudging heart.
If, however, any reader would like to see closer the actual life of a
minstrel of the thirteenth century, I would suggest that he obtain the
excellent little book on Rutebeuf, one of the most famous of them all,
published in the Grands Ecrivains Francais series. There he may read of
the poet's bare cupboard, and the unfurnished lodging, where he lived
with his ugly and dowerless old wife, who brought him but fifty years
with her soup. He coughs with cold and gapes with hunger. He has no
mattress, but only straw, and a bed of straw is not a bed. He fears to
face his wife without money for food and rent. If he cannot dig,
emphatically, to beg he is not ashamed. All his goods are in pawn, and
his time is wasted in the tavern, playing dice, which are his curse and
his downfall. Well, Rutebeuf is not the first nor the last to be ruined
by dice. How the Devil must smile! Do you remember the legend of the
making of these little figures? A merchant who sold himself to the
Devil was bidden by him to make a six-sided piece of bone, and to mark
each side with a number. One point was to insult the only true God. Two
points were to insult God and the Blessed Virgin. Three points to
insult the Holy Trinity. Four points to insult the four Evangelists.
Five points to insult the Sacred Wounds; and six points to insult the
Days of Creation. From that hour the little figures spread rapidly
through the world, to man's confusion. Such is the picture Rutebeuf
paints of his life--a life curiously anticipative of that of many a
Bohemian poet since. It is not a very attractive picture, and though
for artistic and other reasons the shadows may be unduly darkened, yet
in the main it is doubtless substantially correct.
The stories written by such men as these are racy of their soil, and
give the very form and pressure of their times. I have tried to make my
little selection representative, and have included in this book not
only romances of love and chivalry, but legends of devotion and
moralities. Greatly daring, I have translated a specimen of their
humour even--not too characteristic, I hope, of the robust merriment
of the feudal period. These stories will be found illustrative of some
of the ideas with which the mind of the Middle Ages was concerned. The
devotion to the Blessed Sacrament and to Our Lady; the languid and
overwrought sentiment of love; the mystical ceremonies of knighthood;
all these things are illuminated by the tales which follow this
Introduction. Bound up with them are customs and ideas which to the
modern mind are, perhaps, less happy. It seems odd, for instance, that
the feudal knight should see nothing repugnant in accepting money and
clothing from the lady who had given him already the supreme favour of
her love. It is possible to entertain a high ideal of friendship
without being prepared to cut the throats of your children for the sake
of your friend. Yet this is what Amile did for Amis in the great epic
of friendship of the Middle Ages. In its stark adherence to a
superhuman standard, it puts one in mind of the animal-like patience of
Griselda--which story (not included here) may perhaps be regarded as
the modest ideal of the mediaeval husband. It is strange, too, to find
in stories so concerned with the knightly exercises of the tourney and
the joust, no hint of the singular disfavour in which these games (or,
perhaps, pursuits) were held by the Church. Popes prohibited them; St.
Louis forbade them. Those slain therein were refused burial in
consecrated ground. The Church testified, "Of those who fell in
tournament there is no question but that they go down to hell, unless
they are aided by the great benefit of absolution." At Cologne sixty
knights and squires were killed, and the cries were heard all about of
demons carrying off their souls to perdition. Apparently all this
tremendous machinery failed utterly in its purpose. The most pious
knights strove in tournaments equally with the most reckless,
and--according to Miss Knox, to whose admirable _Court of a Saint_ I am
indebted--a son of St. Louis himself was thrown at a tourney, and was
afterwards weak in intellect as a consequence.
Nor is it only with the lives of the rich that the mediaeval minstrel
was concerned. He dealt, too, with the lives and aspirations of that
yet more numerous class, the poor. Such a story as "The Three Thieves"
is indeed a picture of the home of the hind. We see the mean mud and
timber hovel, into which the thieves broke so easily, with its
cauldron upon the fire of fagots, its big bedstead, and the little
lean-to byre. The peasant's tools stood around the wall, whilst outside
was the garden, in which a wise ordinance of St. Louis required that
pot-herbs should be planted. And if the tale of "The Three Thieves"
shows us the home of the peasant, his soul is stripped for us to the
quick, in--of all places in the world of literature--"Aucassin and
Nicolette." Amongst the full-blown flowers of sentiment in that
incomparable love-story is placed an episode which, in its violence and
harsh realism, has been likened to a spot of blood and mud on a silver
ground. Possibly it was inserted merely to show the hero's good heart,
or is simply an instance of that artistic use of contrast so noticeable
throughout the book. Any way, there are few things in feudal literature
more striking than the meeting of the "dansellon" with the tattered,
hideous ploughman, the one weeping in delicate grief, the other
telling, dry-eyed, the sordid story of the abject poor. It is very far
from being the happiest incident in the romance, but it is certainly
one of the most memorable. One wonders how it was taken by an audience
that concerned itself so little with the interests of the serf, and
whose literature never mentioned that class, except in scorn. Was the
author possibly of the ploughman's kindred, like Chaucer's parish
priest in _The Canterbury Tales_? Had the stinging whips of captivity
taught him sympathy with unpoetical sorrows; or is this an early hint
of the coming storm! "They are clothed in velvet, and warm in their
furs and their ermines, while we are covered with rags. They have wine
and spices and fair bread; and we, oat-cake and straw, and water to
drink. They have leisure and fine houses; we have pain and labour, the
rain and the wind in the fields." We cannot tell; but comparing this
dainty make-belief with that tragic misery, we feel the significance of
the peasant's cry, "Woe to those who shall sorrow at the tears of such
as these."
I hope I have not dwelt unduly on these stories considered as pictures
of the customs and philosophy of their times. Perhaps, after all, these
matters are of interest to the archaeologist and the ecclesiologist
rather than to the general reader. Not being a scholar myself, I have
no pretension to write for scholars. My object is more modest. I have
tried to bring together a little garland for the pleasure of the
amateurs of beautiful tales. To me these mediaeval stories are
beautiful, and I have striven to decant them from one language into
another with as little loss as may be. To this end I have refined a
phrase, or, perhaps, softened an incident here and there. I do not
pretend that they are perfect works of art. "All poets are unequal,
except the bad, and they are uniformly bad." Sometimes a story drags,
or there are wearisome repetitions. The psychology occasionally strikes
a modern reader as remarkably summary. When Amis, for example, became a
leper, we are gravely told that his wife held him in bitter hatred, and
many a time strove to strangle him. Here is an author who, obviously,
is astonished at nothing. But in reading these narratives you will
remember how they have delighted, and been used by, writers in some
cases greater than their own authors. Is it possible, for instance, to
peruse "The Lay of the Little Bird" without recalling Shelley's
"Sensitive Plant"? The tale of "The Divided Horsecloth" is told, in
another version, both by Montaigne and Browning. The principal incident
of "King Florus and the Fair Jehane" is used by Shakespeare in
"Cymbeline." "Our Lady's Tumbler" and "A Jew who took as Surety the
Image of Our Lady" have been re-written by Monsieur Anatole France with
such perfection of art and artistry as to be the admiration and despair
of all who come after him.
It should not be forgotten that the majority of these stories were
intended to be recited, and not read. Repetition, therefore, is the
more easily excused. This also accounts for the dramatic use of
dialogue, so noticeable in "The Palfrey" and in "Aucassin and
Nicolette." But it is evident that this Introduction, already
over-long, will not permit me to go _seriatim_ through these tales,
"Item, a grey eye or so. Item, two lips, indifferent red." Let me
therefore content myself with appreciating the most lovely of them all,
"Aucassin and Nicolette."
A single copy of "Aucassin and Nicolette," transcribed in the
thirteenth century, and preserved as by miracle, has retained for us
not only a charming tale, but also an unique specimen of the minstrel's
craft. Without it we could not have gathered that so elaborate a
blending of prose and verse was possible to a strolling player of
mediaeval France. The cante-fable was designed for recitation, with
illustrative gesture, to the accompaniment of viol and pipes. In this,
and not only in this, it seems to suggest an Eastern origin, and
to-day, in any Moorish coffee-house, the tales of the _Arabian Nights_
are delivered in a manner very similar to that witnessed in Provence
seven hundred years ago. The peculiar quality of pleasure afforded by
"Aucassin and Nicolette" is not to be found in the story itself. That,
indeed, is very simple, and, perhaps, a trifle hackneyed. Aucassin, the
only son of the Count of Beaucaire, is passionately in love with
Nicolette, a beautiful girl of unknown parentage, bought of the
Saracens, whom his father will not permit him to marry. The story turns
on the adventures of these fond lovers, until at the end their common
fidelity is rewarded. Portions have faded sadly, like old tapestry, and
the laughter sounds especially hollow, for of all precious things fun
dies soonest. But in "Aucassin" the part is emphatically greater than
the whole, and its charm must rather be sought in its graceful turns of
speech--jewels, five words long--and in the pictorial quality which
makes it more a series of pictures than a narrative. Who can forget the
still night of May on which Nicolette escapes from her prison, and
hurries through the garden, kilting her skirt against the dew; or the
ruined tower in whose kindly shadow she remains hidden, whilst the
watch march along the moonlit street, their swords beneath their
mantles; or that bower of branches, built by her own white hands,
through the trellis-work of which her lover looks upon the stars! In
such felicitous picture-making the dainty little classic is equalled by
no work of its period.
May I express the pious wish that every reader may find it all as
delightful to read as I have found it to transcribe?
EUGENE MASON.
NOTE.--The originals of these narratives are to be found in
Romania; in the _Fabliaux et Contes des Poetes Francois_,
edited by Barbazan et Meon; in two volumes of the _Nouvelles
Francoises en prose_, edited by Moland and D'Hericault; and
in _Les Miracles de la Sainte Vierge_, by Gautier de Coinci.
CONTENTS
PAGE
'TIS OF AUCASSIN AND OF NICOLETTE 1
THE STORY OF KING CONSTANT, THE EMPEROR 39
OUR LADY'S TUMBLER 53
THE LAY OF THE LITTLE BIRD 67
THE DIVIDED HORSECLOTH 75
SIR HUGH OF TABARIE 85
THE STORY OF KING FLORUS AND OF THE FAIR JEHANE 91
OF THE COVETOUS MAN AND OF THE ENVIOUS MAN 129
OF A JEW WHO TOOK AS SURETY THE IMAGE OF OUR LADY 133
THE LAY OF GRAELENT 145
THE THREE THIEVES 161
THE FRIENDSHIP OF AMIS AND AMILE 173
OF THE KNIGHT WHO PRAYED WHILST OUR LADY TOURNEYED
IN HIS STEAD 195
THE PRIEST AND THE MULBERRIES 199
THE STORY OF ASENATH 203
THE PALFREY 213
MEDIAEVAL ROMANCE
'TIS OF AUCASSIN AND OF NICOLETTE
Who will deign to hear the song
Solace of a captive's wrong,
Telling how two children met,
Aucassin and Nicolette;
How by grievous pains distraught,
Noble deeds the varlet wrought
For his love, and her bright face!
Sweet my rhyme, and full of grace,
Fair my tale, and debonair.
He who lists--though full of care,
Sore astonied, much amazed,
All cast down, by men mispraised,
Sick in body, sick in soul,
Hearing shall be glad and whole,
So sweet the tale.
Now they say and tell and relate:
How the Count Bougars of Valence made war on Count Garin of Beaucaire,
war so great, so wonderful, and so mortal, that never dawned the day
but that he was at the gates and walls and barriers of the town, with a
hundred knights and ten thousand men-at-arms, on foot and on horse. So
he burned the Count's land, and spoiled his heritage, and dealt death
to his men. The Count Garin of Beaucaire was full of years, and frail;
he had long outworn his day. He had no heir, neither son nor daughter,
save one only varlet, and he was such as I will tell you. Aucassin was
the name of the lad. Fair he was, and pleasant to look upon, tall and
shapely of body in every whit of him. His hair was golden, and curled
in little rings about his head; he had grey and dancing eyes, a clear,
oval face, a nose high and comely, and he was so gracious in all good
graces that nought in him was found to blame, but good alone. But Love,
that high prince, so utterly had cast him down, that he cared not to
become knight, neither to bear arms, nor to tilt at tourneys, nor yet
to do aught that it became his name to do.
His father and his mother spake him thus--
"Son, don now thy mail, mount thy horse, keep thy land, and render aid
to thy men. Should they see thee amongst them the better will the
men-at-arms defend their bodies and their substance, thy fief and
mine."
"Father," said Aucassin, "why speakest thou in such fashion to me? May
God give me nothing of my desire if I become knight, or mount to horse,
or thrust into the press to strike other or be smitten down, save only
that thou give me Nicolette, my sweet friend, whom I love so well."
"Son," answered the father, "this may not be. Put Nicolette from mind.
For Nicolette is but a captive maid, come hither from a far country,
and the Viscount of this town bought her with money from the Saracens,
and set her in this place. He hath nourished and baptized her, and held
her at the font. On a near day he will give her to some young bachelor,
who will gain her bread in all honour. With this what hast thou to do?
Ask for a wife, and I will find thee the daughter of a king, or a
count. Were he the richest man in France his daughter shalt thou have,
if so thou wilt."
"Faith, my father," said Aucassin, "what honour of all this world would
not Nicolette, my very sweet friend, most richly become! Were she
Empress of Byzantium or of Allemaigne, or Queen of France or England,
low enough would be her degree, so noble is she, so courteous and
debonair, and gracious in all good graces."
Now is sung:
Aucassin was of Beaucaire,
Of the mighty castle there,
But his heart was ever set
On his fair friend, Nicolette.
Small he heeds his father's blame,
Or the harsh words of his dame.
"Fool, to weep the livelong day,
Nicolette trips light and gay.
Scouring she from far Carthage,
Bought of Paynims for a wage.
Since a wife beseems thee good
Take a wife of wholesome blood."
"Mother, naught for this I care,
Nicolette is debonair;
Slim the body, fair the face,
Make my heart a lighted place;
Love has set her as my peer,
Too sweet, my dear."
Now they say and tell and relate:
When the Count Garin of Beaucaire found that in nowise could he
withdraw Aucassin his son from the love of Nicolette, he sought out the
Viscount of the town, who was his man, and spake him thus--
"Sir Count, send Nicolette your god-child straightly from this place.
Cursed be the land wherefrom she was carried to this realm; for because
of her I lose Aucassin, who will not become knight, nor do aught that
it becometh knight to do. Know well that were she once within my power
I would hurry her to the fire; and look well to yourself, for you stand
in utmost peril and fear."
"Sire," answered the Viscount, "this lies heavy upon me, that ever
Aucassin goes and he comes seeking speech with my ward. I have bought
her with my money, and nourished and baptized her, and held her at the
font. Moreover, I am fain to give her to some young bachelor, who will
gain her bread in all honour. With this Aucassin your son had nought to
do. But since this is your will and your pleasure, I will send her to
so far a country that nevermore shall he see her with his eyes."
"Walk warily," replied the Count Garin, "for great evil easily may fall
to you of this."
So they went their ways.
Now the Viscount was a very rich man, and had a rich palace standing
within a garden. In a certain chamber of an upper floor he set
Nicolette in ward, with an old woman to bear her company, and to watch;
and he put there bread and meat and wine and all things for their need.
Then he placed a seal upon the door, so that none might enter in, nor
issue forth, save only that there was a window looking on the garden,
strict and close, whereby they breathed a little fresh air.
Now is sung:
Nicolette is prisoned fast,
In a vaulted chamber cast,
Shaped and carven wondrous well,
Painted as by miracle.
At the marble casement stayed
On her elbow leaned the maid;
Golden showed her golden hair,
Softly curved her eyebrows rare,
Fair her face, and brightly flushed,
Sweeter maiden never blushed.
In the garden from her room
She might watch the roses bloom,
Hear the birds make tender moan;
Then she knew herself alone.
"'Lack, great pity 'tis to place
Maid in such an evil case.
Aucassin, my liege, my squire,
Friend, and dear, and heart's desire,
Since thou dost not hate me quite
Men have done me foul despite,
Sealed me in this vaulted room,
Thrust me to this bitter doom.
But by God, Our Lady's Son,
Soon will I from here begone,
So it be won."
Now they say and tell and relate:
Nicolette was prisoned in the chamber, as you have heard and known. The
cry and the haro went through all the land that Nicolette was stolen
away. Some said that she had fled the country, and some that the Count
Garin of Beaucaire had done her to death. Whatever man may have
rejoiced, Aucassin had no joy therein, so he sought out the Viscount of
the town and spake him thus--
"Sir Viscount, what have you done with Nicolette, my very sweet friend,
the thing that most I love in all the world? Have you borne her off, or
hidden her from my sight? Be sure that should I die hereof, my blood
will be required of you, as is most just, for I am slain of your two
hands, since you steal from me the thing that most I love in all the
world."
"Fair sire," answered the Viscount, "put this from mind. Nicolette is a
captive maid whom I brought here from a far country. For her price I
trafficked with the Saracens, and I have bred and baptized her, and
held her at the font. I have nourished her duly, and on a day will give
her to some young bachelor who will gain her bread in honourable
fashion. With this you have nought to do; but only to wed the daughter
of some count or king. Beyond this, what profit would you have, had
you become her lover, and taken her to your bed? Little enough would be
your gain therefrom, for your soul would lie tormented in Hell all the
days of all time, so that to Paradise never should you win."
"In Paradise what have I to do? I care not to enter, but only to have
Nicolette, my very sweet friend, whom I love so dearly well. For into
Paradise go none but such people as I will tell you of. There go those
aged priests, and those old <DW36>s, and the maimed, who all day long
and all night cough before the altars, and in the crypts beneath the
churches; those who go in worn old mantles and old tattered habits; who
are naked, and barefoot, and full of sores; who are dying of hunger and
of thirst, of cold and of wretchedness. Such as these enter in
Paradise, and with them have I nought to do. But in Hell will I go. For
to Hell go the fair clerks and the fair knights who are slain in the
tourney and the great wars, and the stout archer and the loyal man.
With them will I go. And there go the fair and courteous ladies, who
have friends, two or three, together with their wedded lords. And there
pass the gold and the silver, the ermine and all rich furs, harpers and
minstrels, and the happy of the world. With these will I go, so only
that I have Nicolette, my very sweet friend, by my side."
"Truly," cried the Viscount, "you talk idly, for never shall you see
her more; yea, and if perchance you spoke together, and your father
heard thereof, he would burn both me and her in one fire, and yourself
might well have every fear."
"This lies heavy upon me," answered Aucassin.
Thus he parted from the Viscount making great sorrow.
Now is sung:
Aucassin departed thus
Sad at heart and dolorous;
Gone is she his fairest friend,
None may comfort give or mend,
None by counsel make good end.
To the palace turned he home,
Climbed the stair, and sought his room.
In the chamber all alone
Bitterly he made his moan,
Presently began to weep
For the love he might not keep.
"Nicolette, so gent, so sweet,
Fair the faring of thy feet,
Fair thy laughter, sweet thy speech,
Fair our playing each with each,
Fair thy clasping, fair thy kiss,
Yet it endeth all in this.
Since from me my love is ta'en
I misdoubt that I am slain;
Sister, sweet friend."
Now they say and tell and relate:
Whilst Aucassin was in the chamber lamenting Nicolette, his friend, the
Count Bougars of Valence, wishful to end the war, pressed on his
quarrel, and setting his pikemen and horsemen in array, drew near the
castle to take it by storm. Then the cry arose, and the tumult; and the
knights and the men-at-arms took their weapons, and hastened to the
gates and the walls to defend the castle, and the burgesses climbed to
the battlements, flinging quarrels and sharpened darts upon the foe.
Whilst the siege was so loud and perilous the Count Garin of Beaucaire
sought the chamber where Aucassin lay mourning, assotted upon
Nicolette, his very sweet friend, whom he loved so well.
"Ha, son," cried he, "craven art thou and shamed, that seest thy best
and fairest castle so hardly beset. Know well that if thou lose it
thou art a naked man. Son, arm thyself lightly, mount to horse, keep
thy land, aid thy men, hurtle into the press. Thou needest not to
strike another, neither to be smitten down, but if they see thee
amongst them, the better will they defend their goods and their bodies,
thy land and mine. And thou art so stout and strong that very easily
thou canst do this thing, as is but right."
"Father," answered Aucassin, "what sayest thou now? May God give me
nought that I require of Him if I become knight, or mount to horse, or
thrust into the press to strike knight or be smitten down, save only
thou givest me Nicolette, my sweet friend, whom I love so well."
"Son," replied the father, "this can never be. Rather will I suffer to
lose my heritage, and go bare of all, than that thou shouldest have
her, either as woman or as dame."
So he turned without farewell. But when Aucassin saw him part he stayed
him, saying--
"Father, come now, I will make a true bargain with thee."
"What bargain, fair son?"
"I will arm me, and thrust into the press on such bargain as this, that
if God bring me again safe and sound, thou wilt let me look on
Nicolette, my sweet friend, so long that I may have with her two words
or three, and kiss her one only time."
"I pledge my word to this," said the father.
Of this covenant had Aucassin much joy.
Now is sung:
Aucassin the more was fain
Of the kiss he sought to gain,
Rather than his coffers hold
A hundred thousand marks of gold.
At the call his squire drew near,
Armed him fast in battle gear;
Shirt and hauberk donned the lad,
Laced the helmet on his head,
Girt his golden-hilted sword,
Came the war-horse at his word,
Gripped the buckler and the lance,
At the stirrups cast a glance;
Then most brave from plume to heel
Pricked the charger with the steel,
Called to mind his absent dear,
Passed the gateway without fear
Straight to the fight.
Now they say and tell and relate:
Aucassin was armed and horsed as you have heard. God! how bravely
showed the shield about his neck, the helmet on his head, and the
fringes of the baldric upon his left thigh. The lad was tall and
strong, slender and comely to look upon, and the steed he bestrode was
great and speedy, and fiercely had he charged clear of the gate. Now
think not that he sought spoil of oxen and cattle, nor to smite others
and himself escape. Nay, but of all this he took no heed. Another was
with him, and he thought so dearly upon Nicolette, his fair friend,
that the reins fell from his hand, and he struck never a blow. Then the
charger, yet smarting from the spur, bore him into the battle, amidst
the thickest of the foe, so that hands were laid upon him from every
side, and he was made prisoner. Thus they spoiled him of shield and
lance, and forthwith led him from the field a captive, questioning
amongst themselves by what death he should be slain. When Aucassin
marked their words,
"Ha, God," cried he, "sweet Creature, these are my mortal foes who lead
me captive, and who soon will strike off my head; and when my head is
smitten, never again may I have fair speech with Nicolette, my sweet
friend, whom I hold so dear. Yet have I a good sword, and my horse is
yet unblown. Now if I defend me not for her sake, may God keep her
never, should she love me still."
The varlet was hardy and stout, and the charger he bestrode was right
fierce. He plucked forth his sword, and smote suddenly on the right
hand and on the left, cutting sheer through nasal and headpiece,
gauntlet and arm, making such ruin around him as the wild boar deals
when brought to bay by hounds in the wood; until he had struck down ten
knights, and hurt seven more, and won clear of the _melee_, and rode
back at utmost speed, sword in his hand.
The Count Bougars of Valence heard tell that his men were about to hang
Aucassin, his foe, in shameful wise, so he hastened to the sight, and
Aucassin passed him not by. His sword was yet in hand, and he struck
the Count so fiercely upon the helm, that the headpiece was cleft and
shattered upon the head. So bewildered was he by the stroke that he
tumbled to the ground, and Aucassin stretched forth his hand, and took
him, and led him captive by the nasal of the helmet, and delivered him
to his father.
"Father," said Aucassin, "behold the foe who wrought such war and
mischief upon you! Twenty years hath this war endured, and none was
there to bring it to an end."
"Fair son," replied his father, "better are such deeds as these than
foolish dreams."
"Father," returned Aucassin, "preach me no preachings; but carry out
our bargain."
"Ha, what bargain, fair son?"
"How now, father, hast thou returned from the market? By my head, I
will remember, whosoever may forget; so close is it to my heart. Didst
thou not bargain with me when I armed me and fared into the press, that
if God brought me again safe and sound, thou wouldst grant me sight of
Nicolette, my sweet friend, so long that I might have with her two
words or three, and kiss her once? Such was the bargain, so be thou
honest dealer."
"I," cried the father, "God aid me never should I keep such terms. Were
she here I would set her in the flames, and thou thyself might well
have every fear."
"Is this the very end?" said Aucassin.
"So help me God," said his father; "yea."
"Certes," said Aucassin, "grey hairs go ill with a lying tongue."
"Count of Valence," said Aucassin, "thou art my prisoner?"
"Sire," answered the Count, "it is verily and truly so."
"Give me thy hand," said Aucassin.
"Sire, as you wish."
So each took the other's hand.
"Plight me thy faith," said Aucassin, "that so long as thou drawest
breath, never shall pass a day but thou shalt deal with my father in
shameful fashion, either in goods or person, if so thou canst!"
"Sire, for God's love make me not a jest, but name me a price for my
ransom. Whether you ask gold or silver, steed or palfrey, pelt or fur,
hawk or hound, it shall be paid."
"What!" said Aucassin; "art thou not my prisoner?"
"Truly, sire," said the Count Bougars.
"God aid me never," quoth Aucassin, "but I send thy head flying, save
thou plight me such faith as I said."
"In God's name," cried he, "I plight such affiance as seems most meet
to thee."
He pledged his troth, so Aucassin set him upon a horse, and brought him
into a place of surety, himself riding by his side.
Now is sung:
When Count Garin knew his son
Aucassin still loved but one,
That his heart was ever set
Fondly on fond Nicolette;
Straight a prison he hath found,
Paved with marble, walled around,
Where in vault beneath the earth
Aucassin made little mirth,
But with wailing filled his cell
In such wise as now I tell.
"Nicolette, white lily-flow'r,
Sweetest lady found in bow'r;
Sweet as grape that brimmeth up
Sweetness in the spiced cup.
On a day this chanced to you;
Out of Limousin there drew
One, a pilgrim, sore adread,
Lay in pain upon his bed,
Tossed, and took with fear his breath,
Very dolent, near to death.
Then you entered, pure and white,
Softly to the sick man's sight,
Raised the train that swept adown,
Raised the ermine-bordered gown,
Raised the smock, and bared to him
Daintily each lovely limb.
Then a wondrous thing befell,
Straight he rose up sound and well,
Left his bed, took cross in hand,
Sought again his own dear land.
Lily-flow'r, so white, so sweet,
Fair the faring of thy feet,
Fair thy laughter, fair thy speech,
Fair our playing each with each.
Sweet thy kisses, soft thy touch,
All must love thee over much.
'Tis for thee that I am thrown
In this vaulted cell alone;
'Tis for thee that I attend
Death, that comes to make an end,
For thee, sweet friend."
Now they say and tell and relate:
Aucassin was set in prison as you have heard tell, and Nicolette for
her part was shut in the chamber. It was in the time of summer heat, in
the month of May, when the days are warm, long and clear, and the
nights coy and serene. Nicolette lay one night sleepless on her bed,
and watched the moon shine brightly through the casement, and listened
to the nightingale plain in the garden. Then she bethought her of
Aucassin, her friend, whom she loved so well. She called also to mind
the Count Garin of Beaucaire, her mortal foe, and feared greatly to
remain lest her hiding-place should be told to him, and she be put to
death in some shameful fashion. She made certain that the old woman who
held her in ward was sound asleep. So she rose, and wrapped herself in
a very fair silk mantle, the best she had, and taking the sheets from
her bed and the towels of her bath, knotted them together to make so
long a rope as she was able, tied it about a pillar of the window, and
slipped down into the garden. Then she took her skirt in both hands,
the one before, and the other behind, and kilted her lightly against
the dew which lay thickly upon the grass, and so passed through the
garden. Her hair was golden, with little love-locks; her eyes blue and
laughing; her face most dainty to see, with lips more vermeil than ever
was rose or cherry in the time of summer heat; her teeth white and
small; her breasts so firm that they showed beneath her vesture like
two rounded nuts; so frail was she about the girdle that your two hands
could have spanned her, and the daisies that she brake with her feet in
passing, showed altogether black against her instep and her flesh, so
white was the fair young maiden.
She came to the postern, and unbarring the gate, issued forth upon the
streets of Beaucaire, taking heed to keep within the shadows, for the
moon shone very bright, and thus she fared until she chanced upon the
tower where her lover was prisoned. The tower was buttressed with
pieces of wood in many places, and Nicolette hid herself amongst the
pillars, wrapped close in her mantle. She set her face to a crevice of
the tower, which was old and ruinous, and there she heard Aucassin
weeping within, making great sorrow for the sweet friend whom he held
so dear; and when she had hearkened awhile she began to speak.
Now is sung:
Nicolette, so bright of face,
Leaned within this buttressed place,
Heard her lover weep within,
Marked the woe of Aucassin.
Then in words her thought she told,
"Aucassin, fond heart and bold,
What avails thine heart should ache
For a Paynim maiden's sake.
Ne'er may she become thy mate,
Since we prove thy father's hate,
Since thy kinsfolk hate me too;
What for me is left to do?
Nothing, but to seek the strand,
Pass o'er sea to some far land."
Shore she then one golden tress,
Thrust it in her love's duress;
Aucassin hath seen the gold
Shining bright in that dark hold,
Took the lock at her behest,
Kissed and placed it in his breast,
Then once more his eyes were wet
For Nicolette.
Now they say and tell and relate:
When Aucassin heard Nicolette say that she would fare into another
country, he was filled with anger.
"Fair sweet friend," said he, "this be far from thee, for then wouldst
thou have slain me. And the first man who saw thee, if so he might,
would take thee forthwith and carry thee to his bed, and make thee his
leman. Be sure that if thou wert found in any man's bed, save it be
mine, I should not need a dagger to pierce my heart and slay me.
Certes, no; wait would I not for a knife; but on the first wall or the
nearest stone would I cast myself, and beat out my brains altogether.
Better to die so foul a death as this, than know thee to be in any
man's bed, save mine."
"Aucassin," said she, "I doubt that thou lovest me less than thy words;
and that my love is fonder than thine."
"Alack," cried Aucassin, "fair sweet friend, how can it be that thy
love should be so great? Woman cannot love man, as man loves woman; for
woman's love is in the glance of her eye, and the blossom of her
breast, and the tip of the toe of her foot; but the love of man is set
deep in the hold of his heart, from whence it cannot be torn away."
Whilst Aucassin and Nicolette were thus at odds together, the town
watch entered the street, bearing naked swords beneath their mantles,
for Count Garin had charged them strictly, once she were taken, to put
her to death. The warder from his post upon the tower marked their
approach, and as they drew near heard them speaking of Nicolette,
menacing her with death.
"God," said he, "it is great pity that so fair a damsel should be
slain, and a rich alms should I give if I could warn her privily, and
so she escape the snare; for of her death Aucassin, my liege, were
dead already, and truly this were a piteous case."
Now is sung:
Brave the warder, full of guile,
Straight he sought some cunning wile;
Sought and found a song betime,
Raised this sweet and pleasant rhyme.
"Lady of the loyal mind,
Slender, gracious, very kind,
Gleaming head and golden hair,
Laughing lips and eyes of vair!
Easy, Lady, 'tis to tell
Two have speech who love full well.
Yet in peril are they met,
Set the snare, and spread the net.
Lo, the hunters draw this way,
Cloaked, with privy knives, to slay.
Ere the huntsmen spie the chace
Let the quarry haste apace
And keep her well."
Now they say and tell and relate.
"Ah," said Nicolette, "may the soul of thy father and of thy mother
find sweetest rest, since in so fair and courteous a manner hast thou
warned me. So God please, I will indeed keep myself close, and may He
keep me too."
She drew the folds of her cloak about her, and crouched in the darkness
of the pillars till the watch had passed beyond; then she bade farewell
to Aucassin, and bent her steps to the castle wall. The wall was very
ruinous, and mended with timber, so she climbed the fence, and went her
way till she found herself between wall and moat. Gazing below, she saw
that the fosse was very deep and perilous, and the maid had great fear.
"Ah, God," cried she, "sweet Creature, should I fall, my neck must be
broken; and if I stay, to-morrow shall I be taken, and men will burn
my body in a fire. Yet were it better to die, now, in this place, than
to be made a show to-morrow in the market."
She crossed her brow, and let herself slide down into the moat, and
when she reached the bottom, her fair feet and pretty hands, which had
never learned that they could be hurt, were so bruised and wounded that
the blood came from them in places a many; yet knew she neither ill nor
dolour because of the mightiness of her fear. But if with pain she had
entered in, still more it cost her to issue forth. She called to mind
that it were death to tarry, and by chance found there a stake of
sharpened wood, which those within the keep had flung forth in their
defence of the tower. With this she cut herself a foothold, one step
above the other, till with extreme labour she climbed forth from the
moat. Now the forest lay but the distance of two bolts from a crossbow,
and ran some thirty leagues in length and breadth; moreover, within
were many wild beasts and serpents. She feared these greatly, lest they
should do her a mischief; but presently she remembered that should men
lay hands upon her, they would lead her back to the city to burn her at
the fire.
Now is sung:
Nicolette the fair, the fond,
Climbed the fosse and won beyond;
There she kneeled her, and implored
Very help of Christ the Lord.
"Father, King of majesty,
Where to turn I know not, I.
So, within the woodland gloom
Wolf and boar and lion roam,
Fearful things, with rav'ning maw,
Rending tusk and tooth and claw.
Yet, if all adread I stay,
Men will come at break of day,
Treat me to their heart's desire,
Burn my body in the fire.
But by God's dear majesty
Such a death I will not die;
Since I die, ah, better then
Trust the boar than trust to men.
Since all's evil, men and beast,
Choose I the least."
Now they say and tell and relate:
Nicolette made great sorrow in such manner as you have heard. She
commended herself to God's keeping, and fared on until she entered the
forest. She kept upon the fringes of the woodland, for dread of the
wild beasts and reptiles; and hiding herself within some thick bush,
sleep overtook her, and she slept fast until six hours of the morn,
when shepherds and herdsmen come from the city to lead their flocks to
pasture between the wood and the river. The shepherds sat by a clear,
sweet spring, which bubbled forth on the outskirts of the greenwood,
and spreading a cloak upon the grass, set bread thereon. Whilst they
ate together, Nicolette awoke at the song of the birds and the
laughter, and hastened to the well.
"Fair children," said she, "God have you in His keeping."
"God bless you also," answered one who was more fluent of tongue than
his companions.
"Fair child," said she, "do you know Aucassin, the son of Count Garin
of this realm?"
"Yes, we know him well."
"So God keep you, pretty boy," said she, "as you tell him that within
this wood there is a fair quarry for his hunting; and if he may take
her he would not part with one of her members for a hundred golden
marks, nor for five hundred, nay, nor for aught that man can give."
Then looking upon her steadfastly, their hearts were troubled, the maid
was so beautiful.
"Will I tell him?" cried he who was readier of word than his
companions. "Woe to him who speaks of it ever, or tells Aucassin what
you say. You speak not truth but faery, for in all this forest there is
no beast--neither stag, nor lion, nor boar--one of whose legs would be
worth two pence, or three at the very best, and you talk of five
hundred marks of gold. Woe betide him who believes your story, or shall
spread it abroad. You are a fay, and no fit company for such as us, so
pass upon your road."
"Ah, fair child," answered she, "yet you will do as I pray. For this
beast is the only medicine that may heal Aucassin of his hurt. And I
have here five sous in my purse, take them, and give him my message.
For within three days must he hunt this chace, and if within three days
he find not the quarry, never may he cure him of his wound."
"By my faith," said he, "we will take the money, and if he comes this
way we will give him your message, but certainly we will not go and
look for him."
"As God pleases," answered she.
So she bade farewell to the shepherds, and went her way.
Now is sung:
Nicolette as you heard tell
Bade the shepherd lads farewell,
Through deep woodlands warily
Fared she 'neath the leafy tree;
Till the grass-grown way she trod
Brought her to a forest road,
Whence, like fingers on a hand,
Forked sev'n paths throughout the land.
There she called to heart her love,
There bethought her she would prove
Whether true her lover's vows.
Plucked she then young sapling boughs,
Grasses, leaves that branches yield,
Oak shoots, lilies of the field;
Built a lodge with frond and flow'r,
Fairest mason, fairest bow'r!
Swore then by the truth of God
Should her lover come that road,
Nor for love of her who made
Dream a little in its shade,
'Spite his oath no true love, he,
Nor fond heart, she.
Now they say and tell and relate:
Nicolette builded the lodge, as you have heard; very pretty it was and
very dainty, and well furnished, both outside and in, with a tapestry
of flowers and of leaves. Then she withdrew herself a little way from
the bower, and hid within a thicket to spy what Aucassin would do. And
the cry and the haro went through all the realm that Nicolette was
lost. Some had it that she had stolen away, and others that Count Garin
had done her to death. Whoever had joy thereof, Aucassin had little
pleasure. His father, Count Garin, brought him out of his prison, and
sent letters to the lords and ladies of those parts bidding them to a
very rich feast, so that Aucassin, his son, might cease to dote. When
the feast was at its merriest, Aucassin leaned against the musicians'
gallery, sad and all discomforted. No laugh had he for any jest, since
she, whom most he loved, was not amongst the ladies set in hall. A
certain knight marked his grief, and coming presently to him, said--
"Aucassin, of such fever as yours I, too, have been sick. I can give
you good counsel, if you are willing to listen."
"Sir knight," said Aucassin, "great thanks; good counsel, above all
things, I would hear."
"Get to horse," said he; "take your pleasure in the woodland, amongst
flowers and bracken and the songs of the birds. Perchance, who knows?
you may hear some word of which you will be glad."
"Sir knight," answered Aucassin, "great thanks; this I will do."
He left the hall privily, and went down-stairs to the stable where was
his horse. He caused the charger to be saddled and bridled, then put
foot in stirrup, mounted, and left the castle, riding till he entered
the forest, and so by adventure came upon the well whereby the shepherd
lads were sitting, and it was then about three hours after noon. They
had spread a cloak upon the grass, and were eating their bread, with
great mirth and jollity.
Now is sung:
Round about the well were set
Martin, Robin, Esmeret;
Jolly shepherds, gaily met,
Frulin, Jack and Aubriet.
Laughed the one, "God keep in ward
Aucassin, our brave young lord.
Keep besides the damsel fair,
Blue of eye and gold of hair,
Gave us wherewithal to buy
Cate and sheath knife presently,
Horn and quarter staff and fruit,
Shepherd's pipe and country flute;
God make him well."
Now they say and tell and relate:
When Aucassin marked the song of the herdboys he called to heart
Nicolette, his very sweet friend, whom he held so dear. He thought she
must have passed that way, so he struck his horse with the spurs and
came quickly to the shepherds.
"Fair children, God keep you."
"God bless you," replied he who was readier of tongue than his fellows.
"Fair children," said he, "tell over again the song that you told but
now."
"We will not tell it," answered he who was more fluent of speech than
the others; "sorrow be his who sings it to you, fair sir."
"Fair children," returned Aucassin, "do you not know me?"
"Oh yes, we know well that you are Aucassin, our young lord; but we are
not your men; we belong to the Count."
"Fair children, sing me the song once more, I pray you!"
"By the Wounded Heart, what fine words! Why should I sing for you, if I
have no wish to do so? Why, the richest man in all the land--saving the
presence of Count Garin--would not dare to drive my sheep and oxen and
cows from out his wheatfield or his pasture, for fear of losing his
eyes. Wherefore, then, should I sing for you, if I have no wish to do
so?"
"God keep you, fair children; yet you will do this thing for me. Take
these ten sous that I have here in my purse."
"Sire, we will take the money; but I will not sing for you, since I
have sworn not to do so; but I will tell it in plain prose, if such be
your pleasure."
"As God pleases," answered Aucassin; "better the tale in prose than no
story at all."
"Sire, we were in this glade between six and nine of the morn, and were
breaking our bread by the well, just as we are doing now, when a girl
came by, the loveliest thing in all the world, so fair that we doubted
her a fay, and she brimmed our wood with light. She gave us money, and
made a bargain with us that if you came here we would tell you that you
must hunt in this forest, for in it is such a quarry that if you may
take her you would not part with one of her members for five hundred
silver marks, nor for aught that man can give. For in the quest is so
sweet a salve that if you take her you shall be cured of your wound;
and within three days must the chace be taken, for if she be not found
by then, never will you see her more. Now go to your hunting if you
will, and if you will not, let it go, for truly have I carried out my
bargain with her."
"Fair children," cried Aucassin, "enough have you spoken, and may God
set me on her track."
Now is sung:
Aucassin's fond heart was moved
When this hidden word he proved
Sent him by the maid he loved.
Straight his charger he bestrode,
Bade farewell, and swiftly rode
Deep within the forest dim,
Saying o'er and o'er to him;
"Nicolette, so sweet, so good,
'Tis for you I search this wood;
Antlered stag nor boar I chase,
Hot I follow on your trace.
Slender shape and deep, blue eyes,
Dainty laughter, low replies,
Fledge the arrow in my heart.
Ah, to find you, ne'er to part!
Pray God give so fair an end,
Sister, sweet friend."
Now they say and tell and relate:
Aucassin rode through the wood in search of Nicolette, and the charger
went right speedily. Do not think that the spines and thorns were
pitiful to him. Truly it was not so; for his raiment was so torn that
the least tattered of his garments could scarcely hold to his body, and
the blood ran from his arms and legs and flanks in forty places, or at
least in thirty, so that you could have followed after him by the blood
which he left upon the grass. But he thought so fondly of Nicolette,
his sweet friend, that he felt neither ill nor dolour. Thus all day
long he searched the forest in this fashion, but might learn no news of
her, and when it drew towards dusk he commenced to weep because he had
heard nothing. He rode at adventure down an old grass-grown road, and
looking before him saw a young man standing, such as I will tell you.
Tall he was, and marvellously ugly and hideous. His head was big and
blacker than smoked meat; the palm of your hand could easily have gone
between his two eyes; he had very large cheeks and a monstrous flat
nose with great nostrils; lips redder than uncooked flesh; teeth yellow
and foul; he was shod with shoes and gaiters of bull's hide, bound
about the leg with ropes to well above the knee; upon his back was a
rough cloak; and he stood leaning on a huge club. Aucassin urged his
steed towards him, but was all afeared when he saw him as he was.
"Fair brother, God keep you."
"God bless you too," said he.
"As God keeps you, what do you here?"
"What is that to you?" said he.
"Truly, naught," answered Aucassin. "I asked with no wish to do you
wrong."
"And you, for what cause do you weep?" asked the other, "and make such
heavy sorrow? Certainly, were I so rich a man as you are, not the whole
world should make me shed a tear."
"Do you know me, then?" said Aucassin.
"Yes, well I know you to be Aucassin, the son of the Count, and if you
will tell me why you weep, well, then I will tell you what I do here."
"Certes," said Aucassin, "I will tell you with all my heart. I came
this morning to hunt in the forest, and with me a white greyhound, the
swiftest in the whole world. I have lost him, and that is why I weep."
"Hear him," cried he, "by the Sacred Heart, and you make all this
lamentation for a filthy dog! Sorrow be his who shall esteem you more.
Why, there is not a man of substance in these parts who would not give
you ten or fifteen or twenty hounds--if so your father wished--and be
right glad to make you the gift. But for my part I have full reason to
weep and cry aloud."
"And what is your grief, brother?"
"Sire, I will tell you. I was hired by a rich farmer to drive his
plough, with a yoke of four oxen. Now three days ago, by great
mischance, I lost the best of my bullocks, Roget, the very best ox in
the plough. I have been looking for him ever since, and have neither
eaten nor drunk for three days, since I dare not go back to the town,
because men would put me into prison, as I have no money to pay for my
loss. Of all the riches of the world I have nought but the rags upon my
back. My poor old mother, too, who had nothing but one worn-out
mattress, why, they have taken that from under her, and left her lying
on the naked straw. That hurts me more than my own trouble. For money
comes and money goes; if I have lost to-day, why, I may win to-morrow;
and I will pay for my ox when pay I can. Not for this will I wring my
hands. And you--you weep aloud for a filthy cur. Sorrow take him who
shall esteem you more."
"Certes, thou art a true comforter, fair brother, and blessed may you
be. What is the worth of your bullock?"
"Sire, the villein demands twenty sous for his ox. I cannot beat the
price down by a single farthing."
"Hold out your hand," said Aucassin; "take these twenty sous which I
have in my purse, and pay for your ox."
"Sire," answered the hind, "many thanks, and God grant you find that
for which you seek."
So they parted from each other, and Aucassin rode upon his way. The
night was beautiful and still, and so he fared along the forest path
until he came to the seven cross-roads where Nicolette had builded her
bower. Very pretty it was, and very dainty, and well furnished both
outside and in, ceiling and floor, with arras and carpet of freshly
plucked flowers; no sweeter habitation could man desire to see. When
Aucassin came upon it he reined back his horse sharply, and the
moonbeams fell within the lodge.
"Dear God," cried Aucassin, "here was Nicolette, my sweet friend, and
this has she builded with her fair white hands. For the sweetness of
the house and for love of her, now will I dismount, and here will I
refresh me this night."
He withdrew his foot from the stirrup, and the charger was tall and
high. He dreamed so deeply on Nicolette, his very sweet friend, that he
fell heavily upon a great stone, and his shoulder came from its socket.
He knew himself to be grievously wounded, but he forced him to do all
that he was able, and fastened his horse with the other hand to a
thorn. Then he turned on his side, and crawled as best he might into
the lodge. Looking through a crevice of the bower he saw the stars
shining in the sky, and one brighter than all the others, so he began
to repeat--
Now is sung:
Little Star I gaze upon
Sweetly drawing to the moon,
In such golden haunt is set
Love, and bright-haired Nicolette.
God hath taken from our war
Beauty, like a shining star.
Ah, to reach her, though I fell
From her Heaven to my Hell.
Who were worthy such a thing,
Were he emperor or king?
Still you shine, oh, perfect Star,
Beyond, afar.
Now they say and tell and relate:
When Nicolette heard Aucassin speak these words she hastened to him
from where she was hidden near by. She entered in the bower, and
clasping her arms about his neck, kissed and embraced him straitly.
"Fair sweet friend, very glad am I to find you."
"And you, fair sweet friend, glad am I to meet."
So they kissed, and held each other fast, and their joy was lovely to
see.
"Ah, sweet friend," cried Aucassin, "it was but now that I was in
grievous pain with my shoulder, but since I hold you close I feel
neither sorrow nor wound."
Nicolette searched his hurt, and perceived that the shoulder was out of
joint. She handled it so deftly with her white hands, and used such
skilful surgery, that by the grace of God (who loveth all true lovers)
the shoulder came back to its place. Then she plucked flowers, and
fresh grass and green leafage, and bound them tightly about the setting
with the hem torn from her shift, and he was altogether healed.
"Aucassin," said she, "fair sweet friend, let us take thought together
as to what must be done. If your father beats the wood to-morrow, and
men take me, whatever may chance to you, certainly I shall be slain."
"Certes, fair sweet friend, the sorer grief would be mine. But so I may
help, never shall you come to his hands."
So he mounted to horse, and setting his love before him, held her fast
in his arms, kissing her as he rode, and thus they came forth to the
open fields.
Now is sung:
Aucassin, that loving squire,
Dainty fair to heart's desire,
Rode from out the forest dim
Clasping her he loved to him.
'Laced upon the saddle bow
There he kissed her, chin and brow,
There embraced her, mouth and eyes.
But she spake him, sweetly wise;
"Love, a term to dalliance,
Since for us no home in France
Seek we Rome or far Byzance?"
"Sweet my love, all's one to me,
Dale or woodland, earth or sea;
Nothing care I where we ride
So I hold you at my side."
So, enlaced, the lovers went,
Skirting town and battlement,
Rocky scaur, and quiet lawn;
Till one morning, with the dawn,
Broke the cliffs down to the shore,
Loud they heard the surges roar,
Stood by the sea.
Now they say and tell and relate:
Aucassin dismounted upon the sand, he and Nicolette together, as you
have heard tell. He took his horse by the bridle, and his damsel by the
hand, and walked along the beach. Soon they perceived a ship,
belonging to merchants of those parts, sailing close by, so Aucassin
made signs to the sailors, and presently they came to him. For a
certain price they agreed to take them upon the ship, but when they had
reached the open sea a great and marvellous storm broke upon the
vessel, and drove them from land to land until they drew to a far-off
country, and cast anchor in the port of the castle of Torelore. Then
they asked to what realm they had fared, and men told them that it was
the fief of the King of Torelore. Then inquired Aucassin what manner of
man was this king, and whether there was any war, and men answered--
"Yes, a mighty war."
So Aucassin bade farewell to the merchants, and they commended him to
God. He belted his sword about him, climbed to horse, taking his love
before him on the saddle bow, and went his way till he came to the
castle. He asked where the King might be found, and was told that he
was in child-bed.
"Where, then, is his wife?"
And they answered that she was with the host, and had carried with her
all the armed men of those parts. When Aucassin heard these things he
marvelled very greatly. He came to the palace door and there
dismounted, bidding Nicolette to hold the bridle. Then, making his
sword ready, he climbed the palace stair, and searched until he came to
the chamber where the King lay.
Now is sung:
Hot from searching, Aucassin
Found the room and entered in;
There before the couch he stayed
Where the King, alone, was laid,
Marked the King, and marked the bed,
Marked this lying-in, then said,
"Fool, why doest thou this thing?"
"I'm a mother," quoth the King:
"When my month is gone at length,
And I come to health and strength,
Then shall I hear Mass once more
As my fathers did before,
Arm me lightly, take my lance,
Set my foe a right fair dance,
Where horses prance."
Now they say and tell and relate:
When Aucassin heard the King speak thus he took the linen from the bed,
and flung it about the chamber. He saw a staff in the corner, so he
seized it, returned to the bed, and beat the King so rudely therewith,
that he was near to die.
"Ha, fair sire," cried the King, "what do you require of me? Are you
mad that you treat me thus in my own house?"
"By the Sacred Heart," said Aucassin, "bad son of a shameless mother, I
will strike with the sword if you do not swear to me that man shall
never lie in child-bed in your realm again."
He plighted troth, and when he was thus pledged, "Sire," required
Aucassin, "bring me now where your wife is with the host."
"Sire, willingly," said the King.
He got to horse, and Aucassin mounted his, leaving Nicolette at peace
in the Queen's chamber. The King and Aucassin rode at adventure until
they came to where the Queen was set, and they found that the battle
was joined with roasted crab-apples and eggs and fresh cheeses. So
Aucassin gazed upon the sight and marvelled greatly.
Now is sung:
Aucassin hath drawn his rein,
From the saddle stared amain,
Marked the set and stricken field,
Cheered the hearts that would not yield.
They had carried to the fight
Mushrooms, apples baked aright,
And for arrows, if you please,
Pelted each with good fresh cheese.
He who muddied most the ford
Bore the prize in that award.
Aucassin, the brave, the true,
Watched these deeds of derring do,
Laughed loudly too.
Now they say and tell and relate:
When Aucassin saw this strange sight he went to the King and asked of
him--
"Sire, are these your foes?"
"Yea, sire," answered the King.
"And would you that I should avenge you on them?"
"Yea," answered he, "right willingly."
So Aucassin took sword in hand, and throwing himself in the _melee_,
struck fiercely on the right and on the left, and slew many. When the
King saw the death that Aucassin dealt he snatched at his bridle and
cried--
"Hold, fair sire, deal not with them so cruelly."
"What," said Aucassin, "was it not your wish that I should avenge you
on your enemies?"
"Sire," replied the King, "too ready is such payment as yours. It is
not our custom, nor theirs, to fight a quarrel to the death."
Thereon the foemen fled the field.
The King and Aucassin returned in triumph to the castle of Torelore,
and the men of the country persuaded the King that he should cast
Aucassin forth from the realm, and give Nicolette to his son, for she
seemed a fair woman of high lineage. When Nicolette heard thereof she
had little comfort, so began to say--
Now is sung:
Simple folk, and simple King,
Deeming maid so slight a thing.
When my lover finds me sweet,
Sweetly shapen, brow to feet,
Then know I such dalliance,
No delight of harp, or dance,
Sweetest tune, or fairest mirth,
All the play of all the earth
Seems aught of worth.
Now they say and tell and relate:
Aucassin abode in the castle of Torelore in ease and great delight,
having with him Nicolette his sweet friend, whom he loved so well.
Whilst his days passed in so easy and delightful a manner a great
company of Saracens came in galleys oversea and beset the castle, and
presently took it by storm. They gathered together the spoil, and bore
off the townsfolk, both men and women, into captivity. Amongst these
were seized Nicolette and Aucassin, and having bound Aucassin, both
hands and feet, they flung him into one vessel, and bestowed Nicolette
upon another. Thereafter a great tempest arose at sea, and drove these
galleys apart. The ship whereon Aucassin lay bound, drifted idly, here
and there, on wind and tide, till by chance she went ashore near by the
castle of Beaucaire, and the men of that part hurrying to the wreck,
found Aucassin, and knew him again. When the men of Beaucaire saw their
lord they had much joy, for Aucassin had lived at the castle of
Torelore in all ease for three full years, and his father and his
mother were dead. They brought him to the castle of Beaucaire, and
knelt before him; so held he his realm in peace.
Now is sung:
Aucassin hath gained Beaucaire,
Men have done him homage there;
Holds he now in peace his fief,
Castellan and count and chief.
Yet with heaviness and grief
Goeth he in that fair place,
Lacking love and one sweet face;
Grieving more for one bright head
Than he mourneth for his dead.
"Dearest love, and lady kind,
Treasure I may never find,
God hath never made that strand
Far o'er sea or long by land,
Where I would not seek such prize
And merchandize."
Now they say and tell and relate:
Now leave we Aucassin and let us tell of Nicolette. The ship which
carried Nicolette belonged to the King of Carthage, and he was her
father, and she had twelve brothers, all princes or kings in the land.
When they saw the beauty of the girl, they made much of her, and bore
her in great reverence, and questioned her straitly as to her degree,
for certainly she seemed to them a very gracious lady and of high
lineage. But she could not tell them aught thereof, for she was but a
little child when men sold her into captivity. So the oarsmen rowed
until the galley cast anchor beneath the city of Carthage, and when
Nicolette gazed on the battlements and the country round about, she
called to mind that there had she been cherished, and from thence borne
away when but an unripe maid; yet she was not snatched away so young
but that she could clearly remember that she was the daughter of the
King of Carthage, and once was nourished in the city.
Now is sung:
Nicolette, that maid demure,
Set her foot on alien shore;
Marked the city fenced with walls,
Gazed on palaces and halls.
Then she sighed, "Ah, little worth
All the pomp of all the earth,
Since the daughter of a king,
Come of Sultan's blood, they bring
Stripped to market, as a slave.
Aucassin, true heart and brave,
Sweet thy love upon me steals,
Urges, clamours, pleads, appeals;
Would to God that peril past
In my arms I held you fast;
Would to God that in this place
We were stayed in one embrace,
Fell your kisses on my face,
My dear, my fere."
Now they say and tell and relate:
When the King of Carthage heard Nicolette speak in this wise he put his
arms about her neck.
"Fair sweet friend," said he, "tell me truly who you are, and be not
esmayed of me."
"Sire," answered she, "truly am I daughter to the King of Carthage, and
was stolen away when but a little child, full fifteen years ago."
When they heard her say this thing they were assured that her words
were true, so they rejoiced greatly, and brought her to the palace in
such pomp as became the daughter of a king. They sought to give her
some king of those parts as husband and baron, but she had no care to
marry. She stayed in the palace three or four days, and considered in
her mind by what means she might flee and seek Aucassin. So she
obtained a viol, and learned to play thereon; and when on a certain day
they would have given her in marriage to a rich king among the Paynim,
she rose at night and stole away secretly, wandering until she came to
the seaport, where she lodged with some poor woman in a house near the
shore. There, by means of a herb, she stained her head and face, so
that her fairness was all dark and discoloured; and having made herself
coat and mantle, shirt and hose, she equipped her in the guise of a
minstrel. Then, taking her viol, she sought out a sailor, and persuaded
him sweetly to grant her a passage in his ship. They hoisted sail, and
voyaged over the rough seas until they came to the land of Provence;
and Nicolette set foot on shore, carrying her viol, and fared playing
through the country, until she came to the castle of Beaucaire, in the
very place where Aucassin was.
Now is sung:
'Neath the keep of strong Beaucaire
On a day of summer fair,
At his pleasure, Aucassin
Sat with baron, friend and kin.
Then upon the scent of flow'rs,
Song of birds, and golden hours,
Full of beauty, love, regret,
Stole the dream of Nicolette,
Came the tenderness of years;
So he drew apart in tears.
Then there entered to his eyes
Nicolette, in minstrel guise,
Touched the viol with the bow,
Sang as I will let you know.
"Lords and ladies, list to me,
High and low, of what degree;
Now I sing, for your delight,
Aucassin, that loyal knight,
And his fond friend, Nicolette.
Such the love betwixt them set
When his kinsfolk sought her head
Fast he followed where she fled.
From their refuge in the keep
Paynims bore them o'er the deep.
Nought of him I know to end.
But for Nicolette, his friend,
Dear she is, desirable,
For her father loves her well;
Famous Carthage owns him king,
Where she has sweet cherishing.
Now, as lord he seeks for her,
Sultan, Caliph, proud Emir.
But the maid of these will none,
For she loves a dansellon,
Aucassin, who plighted troth.
Sworn has she some pretty oath
Ne'er shall she be wife or bride,
Never lie at baron's side
Be he denied."
Now they say and tell and relate:
When Aucassin heard Nicolette sing in this fashion he was glad at
heart, so he drew her aside, and asked--
"Fair sweet friend," said Aucassin, "know you naught of this Nicolette,
whose ballad you have sung?"
"Sire, truly, yes; well I know her for the most loyal of creatures, and
as the most winning and modest of maidens born. She is daughter to the
King of Carthage, who took her when Aucassin also was taken, and
brought her to the city of Carthage, till he knew for certain that she
was his child, whereat he rejoiced greatly. Any day he would give her
for husband one of the highest kings in all Spain; but rather would she
be hanged or burned than take him, however rich he be."
"Ah, fair sweet friend," cried the Count Aucassin, "if you would return
to that country and persuade her to have speech with me here, I would
give you of my riches more than you would dare to ask of me or to take.
Know that for love of her I choose not to have a wife, however proud
her race, but I stand and wait; for never will there be wife of mine
if it be not her, and if I knew where to find her I should not need to
grope blindly for her thus."
"Sire," answered she, "if you will do these things I will go and seek
her for your sake, and for hers too; because to me she is very dear."
He pledged his word, and caused her to be given twenty pounds. So she
bade him farewell, and he was weeping for the sweetness of Nicolette.
And when she saw his tears--
"Sire," said she, "take it not so much to heart; in so short a space
will I bring her to this town, and you shall see her with your eyes."
When Aucassin knew this he rejoiced greatly. So she parted from him,
and fared in the town to the house of the Viscountess, for the
Viscount, her god-father, was dead. There she lodged, and opened her
mind fully to the lady on all the business; and the Viscountess
recalled the past, and knew well that it was Nicolette whom she had
cherished. So she caused the bath to be heated, and made her take her
ease for fully eight days. Then Nicolette sought a herb that was called
celandine, and washed herself therewith, and became so fair as she had
never been before. She arrayed her in a rich silken gown from the
lady's goodly store; and seated herself in the chamber on a rich stuff
of broidered sendal; then she whispered the dame, and begged her to
fetch Aucassin, her friend. This she did. When she reached the palace,
lo, Aucassin in tears, making great sorrow for the long tarrying of
Nicolette, his friend; and the lady called to him, and said--
"Aucassin, behave not so wildly; but come with me, and I will show you
that thing you love best in all the world; for Nicolette, your sweet
friend, is here from a far country to seek her love."
So Aucassin was glad at heart.
Now is sung:
When he learned that in Beaucaire
Lodged his lady, sweet and fair,
Aucassin arose, and came
To her hostel, with the dame:
Entered in, and passed straightway
To the chamber where she lay.
When she saw him, Nicolette
Had such joy as never yet;
Sprang she lightly to her feet
Swiftly came with welcome meet.
When he saw her, Aucassin
Oped both arms, and drew her in,
Clasped her close in fond embrace,
Kissed her eyes and kissed her face.
In such greeting sped the night,
Till, at dawning of the light,
Aucassin, with pomp most rare,
Crowned her Countess of Beaucaire.
Such delight these lovers met,
Aucassin and Nicolette.
Length of days and joy did win,
Nicolette and Aucassin,
Endeth song and tale I tell
With marriage bell.
THE STORY OF KING CONSTANT, THE EMPEROR
Now telleth the tale that once upon a time there lived an Emperor of
Byzantium, the which town is now called Constantinople, but in ancient
days it was called Byzantium. In days long since there reigned in this
city an Emperor; a Paynim he was, and was held to be a great clerk in
the laws of his religion. He was learned in a science called astronomy,
and knew the courses of the stars, the planets and the moon; moreover,
in the stars he read many marvels; he had knowledge of many things
which the Paynims study deeply, and had faith in divinations, and in
the answers of the Evil One--that is to say, the Adversary. He knew,
besides, much of enchantments and sorceries, as many a Paynim doth to
this very day.
Now it chanced that the Emperor Muselin fared forth one night, he and a
certain lord of his together, and went their ways about this city of
Constantinople, and the moon shone very clear. They heard a Christian
woman, travailing of child, cry aloud as they passed before her house;
but the husband of this dame was set in the terrace upon his roof, and
now he prayed God to deliver her from her peril, and again he prayed
that she might not be delivered. When the Emperor had listened to his
words for a long time, he said to the knight--
"Have you heard this caitif who prays now that his wife may not be
delivered of her child, and again that she may be delivered? Surely he
is viler than any thief, for every man should show pity to woman, and
the greater pity to her in pain with child. But may Mahound and
Termagaunt aid me never if I hang him not by the neck, so he give me
not fair reason for this deed. Let us now go to him."
So they went, and the Emperor spake him thus, "Caitif, tell me truly
why thou prayest thy God in this fashion, now that He should deliver
thy wife in her labour, and again that she should not be delivered;
this must I know!"
"Sire," answered he, "I will tell you readily. Truly I am a clerk, and
know much of a science that men call astrology. I have learned, too,
the courses of the stars and the planets, and thus I knew well that
were my wife delivered in that hour when I prayed God to close her
womb, then the child must be for ever lost, and certainly would he be
hanged, or drowned, or set within the fire. But when I saw the hour was
good, and the case fair, then I prayed God that she might be delivered;
and I cried to Him, so that of His mercy He heard my prayer, and now
the boy is born to a goodly heritage; blessed be God and praised be His
Name."
"Now tell me," said the King, "to what fair heritage is this child
born?"
"Sire," said he, "with all my heart. Know, sire, of a truth that the
child born in this place shall have to wife the daughter of the Emperor
of this town, she who was born but eight days since, and shall become
Emperor and lord of this city, and of the whole world."
"Caitif," cried the Emperor, "never can it come to pass as thou
sayest."
"Sire," answered he, "so shall it be seen, and thus behoveth it to be."
"Certes," said the Emperor, "great faith hath he who receives it."
Then they went from the house, but the Emperor commanded his knight
that he should bear away the child in so privy a manner, if he were
able, that none should see the deed. The knight came again to the
house, and found two women in the chamber, diligently tending the
mother in her bed, but the child was wrapt in linen clothes, and was
laid upon a stool. Thereupon the knight entered the room, and set hands
upon the child, and placed him on a certain table used for chess, and
carried him to the Emperor, in so secret a fashion that neither nurse
nor mother saw aught thereof. Then the Emperor struck the child with a
knife, wounding him from the stomach to the navel, protesting to the
knight that never should son of such a miscreant have his daughter to
wife, nor come to sit upon his throne. He would even have plucked the
heart from out the breast, but the knight dissuaded him, saying--
"Ah, sire, for the love of God, what is this thing that you would do!
Such a deed becomes you naught, and if men heard thereof, great
reproach would be yours. Enough have you done, for he is more than dead
already. But if it be your pleasure to take further trouble in the
matter, give him to me, and I will cast him in the sea."
"Yea," cried the Emperor, "throw him in the water, for I hate him too
much."
The knight took the child, wrapped him in a piece of broidered silk,
and went with him towards the water. But on his way, pity came into his
heart, and he thought within himself that never should new-born babe be
drowned by him; so he set him, swathed in the silken cloth, on a warm
muck-heap, before the gate of a certain abbey of monks, who at that
hour were chanting matins. When the monks kept silence from their
singing, they heard the crying of the child, and carried him to the
Lord Abbot, who commanded that so fair a boy should be cherished of
them. So they unswathed him from the piece of stuff, and saw the grisly
wound upon his body. As soon, therefore, as it was day the Abbot sent
for physicians, and inquired of them at what cost they would cure the
child of his hurt; and they asked of him one hundred pieces of gold.
But he answered that such a sum was beyond his means, and that the
saving of the child would prove too costly. Then he made a bargain with
the surgeons to heal the child of his wound for eighty golden pieces;
and afterwards he brought him to the font, and caused him to be named
COUSTANT, because of his costing the abbey so great a sum to be made
whole.
Whilst the doctors were about this business, the Abbot sought out a
healthy nurse, in whose breast the infant lay till he was healed of his
hurt, for his flesh was soft and tender, and the knife wound grew
together quickly, but ever after on his body showed the gash. The child
grew in stature, and to great beauty. When he was seven years old the
Abbot put him to school, where he proved so fair a scholar that he
passed all his class-mates in aptness and knowledge. When he was twelve
years of age the boy had come to marvellous beauty; no fairer could you
find in all the land; and when the Abbot saw how comely was the lad and
how gracious, he caused him to ride in his train when he went abroad.
Now it chanced that the Abbot wished to complain to the Emperor of a
certain wrong that his servants had done to the abbey. So the Abbot
made ready a rich present, for the abbey and monastery were his
vassals, although this Emperor was but a Saracen. When the Abbot had
proffered his goodly gift, the Emperor appointed a time, three days
thence, to inquire into the matter, when he would lie at a castle of
his, some three miles out from the city of Byzantium. On the day fixed
by the Emperor, the Abbot got to horse, with his chaplain, his squire,
and his train; and amongst them rode Constant, so goodly in every whit
that all men praised his exceeding beauty, and said amongst themselves
that certainly he came of high peerage, and would rise to rank and
wealth. Thus rode the Abbot towards the castle where the Emperor lay,
and when they met, he greeted him and did him homage, and the Emperor
bade him to enter within the castle, where he would speak with him of
his wrong. The Abbot bowed before him and answered--
"Sire, as God wills."
The Abbot called Constant to him, for the lad carried the prelate's hat
of felt, whilst he talked with the Emperor, and the Emperor gazed on
the varlet, and saw him so comely and winning, that never before had he
seen so fair a person. Then he asked who the boy was; and the Abbot
answered that he knew little, save that he was his man, and that the
abbey had nourished him from his birth--"and truly were this business
of ours finished, I could relate fine marvels concerning him."
"Is this so?" said the Emperor; "come now with me to the castle, and
there you shall tell me the truth."
The Emperor returned to the castle, and the Abbot was ever at his side,
as one who had a heavy business, and he made the best bargain that he
might, for the Emperor was his lord and suzerain. But the matter did
not put from the Emperor's mind the great beauty of the lad, and he
commanded the Abbot to bring the varlet before him. So the boy was sent
for, and came with speed. When Constant stood in the presence, the
Emperor praised his beauty, and said to the Abbot that it was a great
pity that so fair a child should be a Christian. The Abbot replied that
it was rather a great happiness, for one day he would render to God an
unspotted soul. When the Emperor heard this thing he laughed at his
folly, saying the laws of Christ were of nothing worth, and that hell
was the portion of such as put faith in them. Sorely grieved was the
Abbot when he heard the Paynim jest in this fashion, but he dared not
to answer as he wished, and spake soft words to him right humbly.
"Sire, so it pleases the Almighty, such souls are not lost, for, with
all sinners, they go to the mercy of the Merciful."
The Emperor inquired when the boy came to his hands, and the Abbot
replied that fifteen years before he was found by night on the
muck-heap before the abbey door.
"Our monks heard the wail of a tiny child as they came from chanting
matins, so they searched for him, and carried him to me. I looked on
the child, and he was very fair, so that I bade them to take him to the
font and to cherish him duly. He was swathed in a rich stuff of scarlet
silk, and when he was unwrapped I saw on his stomach a grievous wound;
so I sent for doctors and surgeons, and bargained with them to cure him
of his hurt for eighty pieces of gold. Afterwards we baptized him, and
gave him the name of COUSTANT, because of his costing so great a sum to
be made whole. Yet, though he be healed of his wound, never will his
body lose the mark of that grisly gash."
When the Emperor heard this story he knew well that it was the child
whom he had sought to slay in so felon a fashion; so he prayed the
Abbot to give the lad to his charge. Then replied the Abbot that he
would put the matter before his Chapter, but that for his own part the
boy should be given to the King very willingly. Never a word, for good
or evil, spake the King; so the Abbot took leave, and returned to the
monastery, and calling a Chapter of his monks, told them that the
Emperor demanded Constant from their hands.
"But I answered that I must speak to you to know your pleasure therein.
Now answer if I have done aright."
"What, sire, done rightly!" cried the gravest and wisest of all the
monks; "evilly and foolishly have you done in not giving him just what
he asked at once. If you will hear our counsel, send Constant to him
now as he requires, lest he be angry with us, for quickly can he do us
much mischief."
Since it seemed to all the Chapter good that Constant should be sent to
the Emperor, the Abbot bade the prior to go upon this errand, and he
obeyed, saying, "As God pleases."
He got to horse, and Constant with him, and riding to the Emperor,
greeted him in the name of the Abbot and the abbey; then taking
Constant by the hand, gave him to the Emperor formally, in such names
and in their stead. The Paynim received him as one angered that a
nameless man and vagabond must have a king's daughter to wife, and well
he thought in his heart to serve him some evil turn.
When the Emperor held Constant in his power, he pondered deeply how he
might slay him, and no man speak a word. It chanced at this time that
the Emperor had business which called him to the frontier of his realm,
a very long way off, a full twelve days' journey. He set forth,
carrying Constant in his train, yet brooding how to do him to death;
and presently he caused letters to be written in this wise to the
castellan of Byzantium.
"I, the Emperor of Byzantium, and lord of Greece, make him,
the governor of my city, to know that as soon as he shall
read this letter he shall slay, or cause to be slain, the
bearer of this letter, forthwith, upon the delivery thereof.
As your proper body to you is dear, so fail not this
command."
Such was the letter Constant carried, and little he knew that it was
his death he held in hand. He took the warrant, which was closely
sealed, and set out upon his way, riding in such manner that in less
than fifteen days he reached Byzantium, the town we now call
Constantinople. When the varlet rode through the gate it was the
dinner-hour, so (by the will of God) he thought he would not carry his
letter to table, but would wait till men had dined. He came with his
horse to the palace garden, and the weather was very hot, for it was
near to Midsummer day. The pleasaunce was deep and beautiful, and the
lad unbitted his horse, loosened the saddle, and let him graze; then he
threw himself down beneath the shelter of a tree, and in that sweet and
peaceful place presently fell sound asleep.
Now it happened that when the fair daughter of the Emperor had dined,
she entered the garden, and with her four of her maidens, and soon they
began to run one after the other, in such play as is the wont of
damsels when alone. Playing thus, the fair daughter of the Emperor
found herself beneath the tree where Constant lay sleeping, and he was
flushed as any rose. When the Princess saw him, she would not willingly
withdraw her eyes, saying to her own heart that never in her life had
she beheld so comely a person. Then she called to her that one of her
companions who was her closest friend, and made excuses to send the
others forth from the garden. The fair maiden took her playfellow by
the hand, and brought her towards the slumbering youth, saying--
"Sweet friend, here is rich and hidden treasure. Certes, never in all
my days have I seen so gracious a person. He is the bearer of letters,
and right willingly would I learn his news."
The two damsels came near the sleeping lad, and softly withdrew the
letter. When the Princess read the warrant she began to weep very
bitterly, and said to her companion, "Certainly this is a heavy
matter."
"Ah, madame," said her fellow, "tell me all the case."
"Truly," answered the Princess, "could I but trust you fully, such
heaviness should soon be turned to joy."
"Lady," replied she, "surely you may trust me; never will I make known
that which you desire to be hid."
So that maiden, the daughter of the Emperor, caused her fellow to
pledge faith by all that she held most dear, and then she revealed what
the letter held; and the girl answered her--
"Lady, what would you do herein?"
"I will tell you readily," said the Princess. "I will put within his
girdle another letter from my father in place of this, bidding the
castellan to give me as wife to this comely youth, and to call all the
people of this realm to the wedding banquet; for be sure that the youth
is loyal and true, and a man of peerage."
When the maiden heard this she said within herself that such a turn
were good to play.
"But, Lady, how may you get the seal of your father to the letter?"
"Very easily," answered the Princess; "ere my father left for the
marches he gave me eight sheets of parchment, sealed at the foot with
his seal, but with nothing written thereon, and there will I set all
that I have told you."
"Lady," said she, "right wisely have you spoken; but lose no time, and
hasten lest he awake."
"I will go now," said the Princess.
The fair maiden, the daughter of the Emperor, went straight to her
wedding chest, and drew therefrom one of the sealed parchments left her
by her father, so that she might borrow moneys in his name should
occasion arise. For, always was this king and his people at war with
felon and mighty princes whose frontiers were upon his borders. Thereon
she wrote her letter in such manner as this--
"I, King Muselin, Emperor of Greece and of Byzantium the
great city, to my Castellan of Byzantium greeting. I command
you to give the bearer of this letter to my fair daughter in
marriage, according to our holy law; for I have heard, and
am well persuaded, that he is of noble descent and right
worthy the daughter of a king. And, moreover, at such time
grant holiday and proclaim high festival to all burgesses of
the city, and throughout my realm."
In such fashion wrote and witnessed the letter of that fair maiden the
daughter of the Emperor. So when her letter was finished she hastened
to the garden, she and her playmate together, and finding Constant yet
asleep, placed privily the letter beneath his girdle. Then the two
girls began to sing and to make such stir as must needs arouse him. The
lad awoke from his slumber, and was all amazed at the beauty of the
lady and her companion. They drew near, and the Princess gave him
gracious greeting, whereupon Constant got to his feet and returned her
salutation right courteously. She inquired of him as to his name and
his business, and he answered that he was the bearer of letters from
the Emperor to the governor of the city. The girl replied that she
would bring him at once to the presence of the castellan; so she took
him by the hand and led him within the palace; and all within the hall
rose at the girl's approach, and did reverence to their Lady.
The demoiselle sought after the castellan, who was in his chamber, and
there she brought the varlet, who held forth his letter, and added
thereto the Emperor's greeting. The seneschal made much of the lad,
kissing his hand; but the maid for her part kissed both letter and
seal, as one moved with delight, for it was long since she had learned
her father's news. Afterwards she said to the governor that it were
well to read the dispatch in counsel together, and this she said
innocently as one who knew nothing of what was therein. To this the
castellan agreed, so he and the maiden passed to the council chamber
alone. Thereupon the girl unfolded the letter, and made it known to the
governor, and she seemed altogether amazed and distraught as she read.
But the castellan took her to task.
"Lady, certainly the will of my lord your father must be done;
otherwise will his blame come upon us with a heavy hand."
But the girl made answer to this--
"How, then, should I be married, and my lord and father far away? A
strange thing this would be; and certainly will I not be wed."
"Ah, lady," cried the castellan, "what words are these? Your father's
letter biddeth you to marry, so give not nay for yea."
"Sire," said the demoiselle, to whom time went heavy till all was
done--"speak you to the lords and dignitaries of this realm, and take
counsel together. So they deem that thus it must be, who am I to
gainsay them?"
The castellan approved such modest and becoming words, so he took
counsel with the barons, and showed them his letter, and all agreed
that the letter must be obeyed, and the commandment of the Emperor
done. Thus was wedded according to Paynim ritual Constant, that comely
lad, to the fair daughter of the Emperor. The marriage feast lasted
fifteen days, and all Byzantium kept holiday and high festival; no
business was thought of in the city, save that of eating and drinking
and making merry. This was all the work men did.
The Emperor tarried a long time in the borders of his land, but when
his task was ended he returned towards Byzantium. Whilst he was about
two days' journey from the city, there met him a messenger with letters
of moment. The King inquired of him as to the news of the capital, and
the messenger made answer that there men thought of nought else but
drinking and eating and taking their ease, and had so done for a whole
fortnight.
"Why is this?" asked the Emperor.
"Why, sire, do you not remember?"
"Truly, no," said the Emperor; "so tell me the reason."
"Sire," replied the varlet, "you sent to your castellan a certain
comely lad, and he bore with him letters from you commanding that he
should be wed to your daughter, the fair Princess, since after your
death he would be Emperor in your stead, for he was a man of lineage,
and well worthy so high a bride. But your daughter refused to marry
such an one, till the castellan had spoken with the lords; so he showed
the council your letter, and they all advised him to carry out your
will. When your daughter knew that they were all of one mind, she dared
no longer to withstand you, and consented to your purpose. In just such
manner as this was your daughter wedded, and a merrier city than yours
could no man wish to see."
When the Emperor heard this thing from the messenger, he marvelled
beyond measure, and turned it over in his thoughts; so presently he
inquired of the varlet how long it was since Constant had wedded his
daughter, and whether he had bedded with her.
"Yea, sire," answered the varlet, "and since it is more than three
weeks that they were married, perchance one day will she be mother as
well as wife."
"Truly it were a happy hazard," said the Emperor, "and since the thing
has fallen thus, let me endure it with a smiling face, for nothing else
is left to do."
The Emperor went on his way until he reached Byzantium, and all the
city gave him loyal greeting. Amongst those who came to meet him was
the fair Princess with her husband, Constant, so gracious in person
that no man was ever goodlier. The Emperor, who was a wise prince, made
much of both of them, and laid his two hands on their two heads, and
held them so for long, for such is the fashion of blessing amongst the
Paynim. That night the Emperor considered this strange adventure, and
how it must have chanced, and so deeply did he think upon it that well
he knew that the game had been played him by his daughter. He did not
reproach her, but bade them bring the letter he sent to the governor,
and when it was shown him he read the writing therein, and saw that it
was sealed with his very seal. So, seeing the way in which the thing
had come to pass, he said within himself that he had striven against
those things which were written in the stars.
After this the Emperor made Constant, his newly wedded son, a belted
knight, and gave and delivered to him his whole realm in heritage after
his death. Constant bore himself wisely and well, as became a good
knight, bold and chivalrous, and defended the land right well against
all its foes. In no long while his lord the Emperor died, and was laid
in the grave, according to Paynim ritual, with great pomp and ceremony.
The Emperor Constant reigned in his stead, and greatly he loved and
honoured the Abbot who had cherished him, and he made him Chancellor of
his kingdom. Then, by the advice of the Abbot, and according to the
will of God, the All Powerful, the Emperor Constant brought his wife to
the font, and caused all men of that realm to be converted to the law
of Jesus Christ. He begot on his wife an heir, whom he christened
Constantine, and who became true Christian and a very perfect knight.
In his day was the city first called Constantinople, because of
Constant his father, who cost the abbey so great a sum, but before then
was the city known as Byzantium.
So endeth in this place the story of King Constant the Emperor.
OUR LADY'S TUMBLER
Amongst the lives of the ancient Fathers, wherein may be found much
profitable matter, this story is told for a true ensample. I do not say
that you may not often have heard a fairer story, but at least this is
not to be despised, and is well worth the telling. Now therefore will I
say and narrate what chanced to this minstrel.
He erred up and down, to and fro, so often and in so many places, that
he took the whole world in despite, and sought rest in a certain Holy
Order. Horses and raiment and money, yea, all that he had, he
straightway put from him, and seeking shelter from the world, was
firmly set never to put foot within it more. For this cause he took
refuge in this Holy Order, amongst the monks of Clairvaux. Now, though
this dancer was comely of face and shapely of person, yet when he had
once entered the monastery he found that he was master of no craft
practised therein. In the world he had gained his bread by tumbling and
dancing and feats of address. To leap, to spring, such matters he knew
well, but of greater things he knew nothing, for he had never spelled
from book--nor Paternoster, nor canticle, nor creed, nor Hail Mary, nor
aught concerning his soul's salvation.
When the minstrel had joined himself to the Order he marked how the
tonsured monks spoke amongst themselves by signs, no words coming from
their lips, so he thought within himself that they were dumb. But when
he learned that truly it was by way of penance that speech was
forbidden to their mouths, and that for holy obedience were they
silent, then considered he that silence became him also; and he
refrained his tongue from words, so discreetly and for so long a space,
that day in, day out, he spake never, save by commandment; so that the
cloister often rang with the brothers' mirth. The tumbler moved amongst
his fellows like a man ashamed, for he had neither part nor lot in all
the business of the monastery, and for this he was right sad and
sorrowful. He saw the monks and the penitents about him, each serving
God, in this place and that, according to his office and degree. He
marked the priests at their ritual before the altars; the deacons at
the gospels; the sub-deacons at the epistles; and the ministers about
the vigils. This one repeats the introit; this other the lesson;
cantors chant from the psalter; penitents spell out the Miserere--for
thus are all things sweetly ordered--yea, and the most ignorant amongst
them yet can pray his Paternoster. Wherever he went, here or there, in
office or cloister, in every quiet corner and nook, there he found
five, or three, or two, or at least one. He gazes earnestly, if so he
is able, upon each. Such an one laments; this other is in tears; yet
another grieves and sighs. He marvels at their sorrow. Then he said,
"Holy Mary, what bitter grief have all these men that they smite the
breast so grievously! Too sad of heart, meseems, are they who make such
bitter dole together. Ah, St. Mary, alas, what words are these I say!
These men are calling on the mercy of God, but I--what do I here! Here
there is none so mean or vile but who serves God in his office and
degree, save only me, for I work not, neither can I preach. Caitif and
shamed was I when I thrust myself herein, seeing that I can do nothing
well, either in labour or in prayer. I see my brothers upon their
errands, one behind the other; but I do naught but fill my belly with
the meat that they provide. If they perceive this thing, certainly
shall I be in an evil case, for they will cast me out amongst the dogs,
and none will take pity on the glutton and the idle man. Truly am I a
caitif, set in a high place for a sign." Then he wept for very woe, and
would that he was quiet in the grave. "Mary, Mother," quoth he, "pray
now your Heavenly Father that He keep me in His pleasure, and give me
such good counsel that I may truly serve both Him and you; yea, and may
deserve that meat which now is bitter in my mouth."
Driven mad with thoughts such as these, he wandered about the abbey
until he found himself within the crypt, and took sanctuary by the
altar, crouching close as he was able. Above the altar was carved the
statue of Madame St. Mary. Truly his steps had not erred when he sought
that refuge; nay, but rather, God who knows His own had led him thither
by the hand. When he heard the bells ring for Mass he sprang to his
feet all dismayed. "Ha!" said he; "now am I betrayed. Each adds his
mite to the great offering, save only me. Like a tethered ox, naught I
do but chew the cud, and waste good victuals on a useless man. Shall I
speak my thought? Shall I work my will? By the Mother of God, thus am I
set to do. None is here to blame. I will do that which I can, and
honour with my craft the Mother of God in her monastery. Since others
honour her with chant, then I will serve with tumbling."
He takes off his cowl, and removes his garments, placing them near the
altar, but so that his body be not naked he dons a tunic, very thin and
fine, of scarce more substance than a shirt. So, light and comely of
body, with gown girt closely about his loins, he comes before the Image
right humbly. Then raising his eyes, "Lady," said he, "to your fair
charge I give my body and my soul. Sweet Queen, sweet Lady, scorn not
the thing I know, for with the help of God I will essay to serve you in
good faith, even as I may. I cannot read your Hours nor chant your
praise, but at the least I can set before you what art I have. Now will
I be as the lamb that plays and skips before his mother. Oh, Lady, who
art nowise bitter to those who serve you with a good intent, that which
thy servant is, that he is for you."
Then commenced he his merry play, leaping low and small, tall and high,
over and under. Then once more he knelt upon his knees before the
statue, and meekly bowed his head. "Ha!" said he, "most gracious Queen,
of your pity and your charity scorn not this my service." Again he
leaped and played, and for holiday and festival, made the somersault of
Metz. Again he bowed before the Image, did reverence, and paid it all
the honour that he might. Afterwards he did the French vault, then the
vault of Champagne, then the Spanish vault, then the vaults they love
in Brittany, then the vault of Lorraine, and all these feats he did as
best he was able. Afterwards he did the Roman vault, and then, with
hands before his brow, danced daintily before the altar, gazing with a
humble heart at the statue of God's Mother. "Lady," said he, "I set
before you a fair play. This travail I do for you alone; so help me
God, for you, Lady, and your Son. Think not I tumble for my own
delight; but I serve you, and look for no other guerdon on my carpet.
My brothers serve you, yea, and so do I. Lady, scorn not your villein,
for he toils for your good pleasure; and, Lady, you are my delight and
the sweetness of the world." Then he walked on his two hands, with his
feet in the air, and his head near the ground. He twirled with his
feet, and wept with his eyes. "Lady," said he, "I worship you with
heart, with body, feet and hands, for this I can neither add to nor
take away. Now am I your very minstrel. Others may chant your praises
in the church, but here in the crypt will I tumble for your delight.
Lady, lead me truly in your way, and for the love of God hold me not in
utter despite." Then he smote upon his breast, he sighed and wept most
tenderly, since he knew no better prayer than tears. Then he turned him
about, and leaped once again. "Lady," said he, "as God is my Saviour,
never have I turned this somersault before. Never has tumbler done such
a feat, and, certes, it is not bad. Lady, what delight is his who may
harbour with you in your glorious manor. For God's love, Lady, grant me
such fair hostelry, since I am yours, and am nothing of my own." Once
again he did the vault of Metz; again he danced and tumbled. Then when
the chants rose louder from the choir, he, too, forced the note, and
put forward all his skill. So long as the priest was about that Mass,
so long his flesh endured to dance, and leap and spring, till at the
last, nigh fainting, he could stand no longer upon his feet, but fell
for weariness on the ground. From head to heel sweat stood upon him,
drop by drop, as blood falls from meat turning upon the hearth. "Lady,"
said he, "I can no more, but truly will I seek you again." Fire
consumed him utterly. He took his habit once more, and when he was
wrapped close therein, he rose to his feet, and bending low before the
statue, went his way. "Farewell," said he, "gentlest Friend. For God's
love take it not to heart, for so I may I will soon return. Not one
Hour shall pass but that I will serve you with right good will, so I
may come, and so my service is pleasing in your sight." Thus he went
from the crypt, yet gazing on his Lady. "Lady," said he, "my heart is
sore that I cannot read your Hours. How would I love them for love of
you, most gentle Lady! Into your care I commend my soul and my body."
In this fashion passed many days, for at every Hour he sought the crypt
to do service, and pay homage before the Image. His service was so much
to his mind that never once was he too weary to set out his most
cunning feats to distract the Mother of God, nor did he ever wish for
other play than this. Now, doubtless, the monks knew well enough that
day by day he sought the crypt, but not a man on earth--save God
alone--was aware of aught that passed there; neither would he, for all
the wealth of the world, have let his goings in be seen, save by the
Lord his God alone. For truly he believed that were his secret once
espied he would be hunted from the cloister, and flung once more into
the foul, sinful world, and for his part he was more fain to fall on
death than to suffer any taint of sin. But God considering his
simplicity, his sorrow for all he had wrought amiss, and the love which
moved him to this deed, would that this toil should be known; and the
Lord willed that the work of His friend should be made plain to men,
for the glory of the Mother whom he worshipped, and so that all men
should know and hear, and receive that God refuses none who seeks His
face in love, however low his degree, save only he love God and strive
to do His will.
Now think you that the Lord would have accepted this service, had it
not been done for love of Him? Verily and truly, no, however much this
juggler tumbled; but God called him friend, because he loved Him much.
Toil and labour, keep fast and vigil, sigh and weep, watch and pray,
ply the sharp scourge, be diligent at Matins and at Mass, owe no man
anything, give alms of all you have--and yet, if you love not God with
all your heart, all these good deeds are so much loss--mark well my
words--and profit you naught for the saving of your soul. Without
charity and love, works avail a man nothing. God asks not gold, neither
for silver, but only for love unfeigned in His people's hearts, and
since the tumbler loved Him beyond measure, for this reason God was
willing to accept his service.
Thus things went well with this good man for a great space. For more
years than I know the count of, he lived greatly at his ease, but the
time came when the good man was sorely vexed, for a certain monk
thought upon him, and blamed him in his heart that he was never set in
choir for Matins. The monk marvelled much at his absence, and said
within himself that he would never rest till it was clear what manner
of man this was, and how he spent the Hours, and for what service the
convent gave him bread. So he spied and pried and followed, till he
marked him plainly, sweating at his craft in just such fashion as you
have heard. "By my faith," said he, "this is a merry jest, and a fairer
festival than we observe altogether. Whilst others are at prayers, and
about the business of the House, this tumbler dances daintily, as
though one had given him a hundred silver marks. He prides himself on
being so nimble of foot, and thus he repays us what he owes. Truly it
is this for that; we chant for him, and he tumbles for us. We throw him
largesse: he doles us alms. We weep his sins, and he dries our eyes.
Would that the monastery could see him, as I do, with their very eyes;
willingly therefore would I fast till Vespers. Not one could refrain
from mirth at the sight of this simple fool doing himself to death with
his tumbling, for on himself he has no pity. Since his folly is free
from malice, may God grant it to him as penance. Certainly I will not
impute it to him as sin, for in all simplicity and good faith, I firmly
believe, he does this thing, so that he may deserve his bread." So the
monk saw with his very eyes how the tumbler did service at all the
Hours, without pause or rest, and he laughed with pure mirth and
delight, for in his heart was joy and pity.
The monk went straight to the Abbot and told him the thing from
beginning to end, just as you have heard. The Abbot got him on his
feet, and said to the monk, "By holy obedience I bid you hold your
peace, and tell not this tale abroad against your brother. I lay on you
my strict command to speak of this matter to none, save me. Come now,
we will go forthwith to see what this can be, and let us pray the
Heavenly King, and His very sweet, dear Mother, so precious and so
bright, that in her gentleness she will plead with her Son, her Father,
and her Lord, that I may look on this work--if thus it pleases Him--so
that the good man be not wrongly blamed, and that God may be the more
beloved, yet so that thus is His good pleasure." Then they secretly
sought the crypt, and found a privy place near the altar, where they
could see, and yet not be seen. From there the Abbot and his monk
marked the business of the penitent. They saw the vaults he varied so
cunningly, his nimble leaping and his dancing, his salutations of Our
Lady, and his springing and his bounding, till he was nigh to faint. So
weak was he that he sank on the ground, all outworn, and the sweat fell
from his body upon the pavement of the crypt. But presently, in this
his need, came she, his refuge, to his aid. Well she knew that
guileless heart.
Whilst the Abbot looked, forthwith there came down from the vault a
Dame so glorious, that certainly no man had seen one so precious, nor
so richly crowned. She was more beautiful than the daughters of men,
and her vesture was heavy with gold and gleaming stones. In her train
came the hosts of Heaven, angel and archangel also; and these pressed
close about the minstrel, and solaced and refreshed him. When their
shining ranks drew near, peace fell upon his heart; for they contended
to do him service, and were the servants of the servitor of that Dame
who is the rarest Jewel of God. Then the sweet and courteous Queen
herself took a white napkin in her hand, and with it, gently fanned her
minstrel before the altar. Courteous and debonair, the Lady refreshed
his neck, his body and his brow. Meekly she served him as a handmaid in
his need. But these things were hidden from the good man, for he
neither saw nor knew that about him stood so fair a company.
The holy angels honour him greatly, but they can no longer stay, for
their Lady turns to go. She blesses her minstrel with the sign of God,
and the holy angels throng about her, still gazing back with delight
upon their companion, for they await the hour when God shall release
him from the burden of the world, and they possess his soul.
This marvel the Abbot and his monk saw at least four times, and thus at
each Hour came the Mother of God with aid and succour for her man.
Never doth she fail her servants in their need. Great joy had the Abbot
that this thing was made plain to him. But the monk was filled with
shame, since God had shown His pleasure in the service of His poor
fool. His confusion burnt him like fire. "Dominus," said he to the
Abbot, "grant me grace. Certainly this is a holy man, and since I have
judged him amiss, it is very right that my body should smart. Give me
now fast or vigil or the scourge, for without question he is a saint.
We are witnesses to the whole matter, nor is it possible that we can be
deceived." But the Abbot replied, "You speak truly, for God has made us
to know that He has bound him with the cords of love. So I lay my
commandment upon you, in virtue of obedience, and under pain of your
person, that you tell no word to any man of that you have seen, save to
God alone and me." "Lord," said he, "thus I will do." On these words
they turned them, and hastened from the crypt; and the good man, having
brought his tumbling to an end, presently clothed himself in his habit,
and joyously went his way to the monastery.
Thus time went and returned, till it chanced that in a little while the
Abbot sent for him who was so filled with virtue. When he heard that he
was bidden of the Abbot, his heart was sore with grief, for he could
think of nothing profitable to say. "Alas!" said he, "I am undone; not
a day of my days but I shall know misery and sorrow and shame, for well
I trow that my service is not pleasing to God. Alas! plainly doth He
show that it displeases Him, since He causes the truth to be made
clear. Could I believe that such work and play as mine could give
delight to the mighty God! He had no pleasure therein, and all my toil
was thrown away. Ah me, what shall I do? what shall I say? Fair, gentle
God, what portion will be mine? Either shall I die in shame, or else
shall I be banished from this place, and set up as a mark to the world
and all the evil thereof. Sweet Lady, St. Mary, since I am all
bewildered, and since there is none to give me counsel, Lady, come thou
to my aid. Fair, gentle God, help me in my need. Stay not, neither
tarry, but come quickly with Your Mother. For God's love, come not
without her, but hasten both to me in my peril, for truly I know not
what to plead. Before one word can pass my lips, surely will they bid
me 'Begone.' Wretched that I am, what reply is he to make who has no
advocate? Yet, why this dole, since go I must?" He came before the
Abbot, with the tears yet wet upon his cheeks, and he was still weeping
when he knelt upon the ground. "Lord," prayed he, "for the love of God
deal not harshly with me. Would you send me from your door? Tell me
what you would have me do, and thus it shall be done." Then replied the
Abbot, "Answer me truly. Winter and summer have you lived here for a
great space; now, tell me, what service have you given, and how have
you deserved your bread?" "Alas!" said the tumbler, "well I knew that
quickly I should be put upon the street when once this business was
heard of you, and that you would keep me no more. Lord," said he, "I
take my leave. Miserable I am, and miserable shall I ever be. Never yet
have I made a penny for all my juggling." But the Abbot answered, "Not
so said I; but I ask and require of you--nay, more, by virtue of holy
obedience I command you--to seek within your conscience and tell me
truly by what craft you have furthered the business of our monastery."
"Lord," cried he, "now have you slain me, for this commandment is a
sword." Then he laid bare before the Abbot the story of his days, from
the first thing to the last, whatsoever pain it cost him; not a word
did he leave out, but he told it all without a pause, just as I have
told you the tale. He told it with clasped hands, and with tears, and
at the close he kissed the Abbot's feet, and sighed.
The holy Abbot leaned above him, and, all in tears, raised him up,
kissing both his eyes. "Brother," said he, "hold now your peace, for I
make with you this true covenant, that you shall ever be of our
monastery. God grant, rather, that we may be of yours, for all the
worship you have brought to ours. I and you will call each other
friend. Fair, sweet brother, pray you for me, and I for my part will
pray for you. And now I pray you, my sweet friend, and lay this bidding
upon you, without pretence, that you continue to do your service, even
as you were wont heretofore--yea, and with greater craft yet, if so you
may." "Lord," said he, "truly is this so?" "Yea," said the Abbot, "and
verily." So he charged him, under peril of discipline, to put all
doubts from his mind; for which reason the good man rejoiced so greatly
that, as telleth the rhyme, he was all bemused, so that the blood left
his cheeks, and his knees failed beneath him. When his courage came
back, his very heart thrilled with joy; but so perilous was that
quickening that therefrom he shortly died. But theretofore with a good
heart he went about his service without rest, and Matins and Vespers,
night and day, he missed no Hour till he became too sick to perform his
office. So sore was his sickness upon him that he might not rise from
his bed. Marvellous was the shame he proved when no more was he able to
pay his rent. This was the grief that lay the heaviest upon him, for of
his sickness he spake never a word, but he feared greatly lest he
should fall from grace since he travailed no longer at his craft. He
reckoned himself an idle man, and prayed God to take him to Himself
before the sluggard might come to blame. For it was bitter to him to
consider that all about him knew his case, so bitter that the burden
was heavier than his heart could bear, yet there without remedy he must
lie. The holy Abbot does him all honour; he and his monks chant the
Hours about his bed, and in these praises of God he felt such delight
that not for them would he have taken the province of Poitou, so great
was his happiness therein. Fair and contrite was his confession, but
still he was not at peace; yet why say more of this, for the hour had
struck, and he must rise and go.
The Abbot was in that cell with all his monks; there, too, was company
of many a priest and many a canon. These all humbly watched the dying
man, and saw with open eyes this wonder happen. Clear to their very
sight, about that lowly bed, stood the Mother of God, with angel and
archangel, to wait the passing of his soul. Over against them were set,
like wild beasts, devils and the Adversary, so they might snatch his
spirit. I speak not to you in parable. But little profit had they for
all their coming, their waiting, and their straining on the leash.
Never might they have part in such a soul as his. When the soul took
leave of his body, it fell not in their hands at all, for the Mother of
God gathered it to her bosom, and the holy angels thronging round,
quired for joy, as the bright train swept to Heaven with its burthen,
according to the will of God. To these things the whole of the
monastery was witness, besides such others as were there. So knew they
and perceived that God sought no more to hide the love He bore to His
poor servant, but rather would that his virtues should be plain to each
man in that place; and very wonderful and joyful seemed this deed to
them. Then with meet reverence they bore the body on its bier within
the abbey church, and with high pomp commended their brother to the
care of God; nor was there monk who did not chant or read his portion
that day within the choir of the mighty church.
Thus with great honour they laid him to his rest, and kept his holy
body amongst them as a relic. At that time spake the Abbot plainly to
their ears, telling them the story of this tumbler and of all his life,
just as you have heard, and of all that he himself beheld within the
crypt. No brother but kept awake during that sermon. "Certes," said
they, "easy is it to give credence to such a tale; nor should any doubt
your words, seeing that the truth bears testimony to itself, and
witness comes with need; yea, without any doubt have we full assurance
that his discipline is done." Great joy amongst themselves have all
within that place.
Thus endeth the story of the minstrel. Fair was his tumbling, fair was
his service, for thereby gained he such high honour as is above all
earthly gain. So the holy Fathers narrate that in such fashion these
things chanced to this minstrel. Now, therefore, let us pray to God--He
Who is above all other--that He may grant us so to do such faithful
service that we may win the guerdon of His love.
Here endeth the Tumbler of Our Lady.
THE LAY OF THE LITTLE BIRD
Once upon a time, more than a hundred years ago, there lived a rich
villein whose name I cannot now tell, who owned meadows and woods and
waters, and all things which go to the making of a rich man. His manor
was so fair and so delightsome that all the world did not contain its
peer. My true story would seem to you but idle fable if I set its
beauty before you, for verily I believe that never yet was built so
strong a keep and so gracious a tower. A river flowed around this fair
domain, and enclosed an orchard planted with all manner of fruitful
trees. This sweet fief was builded by a certain knight, whose heir sold
it to a villein; for thus pass baronies from hand to hand, and town and
manor change their master, always falling from bad to worse. The
orchard was fair beyond content. Herbs grew there of every fashion,
more than I am able to name. But at least I can tell you that so sweet
was the savour of roses and other flowers and simples, that sick
persons, borne within that garden in a litter, walked forth sound and
well for having passed the night in so lovely a place. Indeed, so
smooth and level was the sward, so tall the trees, so various the
fruit, that the cunning gardener must surely have been a magician, as
appears by certain infallible proofs.
Now in the middle of this great orchard sprang a fountain of clear,
pure water. It boiled forth out of the ground, but was always colder
than any marble. Tall trees stood about the well, and their leafy
branches made a cool shadow there, even during the longest day of
summer heat. Not a ray of the sun fell within that spot, though it were
the month of May, so thick and close was the leafage. Of all these
trees the fairest and the most pleasant was a pine. To this pine came a
singing bird twice every day for ease of heart. Early in the morning he
came, when monks chant their matins, and again in the evening, a little
after vespers. He was smaller than a sparrow, but larger than a wren,
and he sang so sweetly that neither lark nor nightingale nor blackbird,
nay, nor siren even, was so grateful to the ear. He sang lays and
ballads, and the newest refrain of the minstrel and the spinner at her
wheel. Sweeter was his tune than harp or viol, and gayer than the
country dance. No man had heard so marvellous a thing; for such was the
virtue in his song that the saddest and the most dolent forgot to
grieve whilst he listened to the tune, love flowered sweetly in his
heart, and for a space he was rich and happy as any emperor or king,
though but a burgess of the city or a villein of the field. Yea, if
that ditty had lasted a hundred years, yet would he have stayed the
century through to listen to so lovely a song, for it gave to every man
whilst he hearkened, love, and riches, and his heart's desire.
But all the beauty of the pleasaunce drew its being from the song of
the bird; for from his chant flowed love which gives its shadow to the
tree, its healing to the simple, and its colour to the flower. Without
that song the fountain would have ceased to spring, and the green
garden become a little dry dust, for in its sweetness lay all their
virtue.
The villein, who was lord of this domain, walked every day within his
garden to hearken to the bird. On a certain morning he came to the well
to bathe his face in the cold spring, and the bird, hidden close
within the pine branches, poured out his full heart in a delightful
lay, from which rich profit might be drawn.
"Listen," chanted the bird in his own tongue, "listen to my voice oh,
knight, and clerk, and layman, ye who concern yourselves with love, and
suffer with its dolours: listen, also, ye maidens, fair and coy and
gracious, who seek first the gifts and beauty of the world. I speak
truth and do not lie. Closer should you cleave to God than to any
earthly lover, right willingly should you seek His altar, more firmly
should you hold to His commandment than to any mortal's pleasure. So
you serve God and Love in such fashion, no harm can come to any, for
God and Love are one. God loves sense and chivalry; and Love holds them
not in despite. God hates pride and false seeming; and Love loveth
loyalty. God praiseth honour and courtesy; and fair Love disdaineth
them not. God lendeth His ear to prayer; neither doth Love refuse it
her heart. God granteth largesse to the generous; but the grudging man,
and the envious, the felon and the wrathful, doth He abhor. But
courtesy and honour, good sense and loyalty, are the leal vassals of
Love, and so you hold truly to them, God and the beauty of the world
shall be added to you besides."
Thus told the bird in his song.
But when he saw the villein beneath the pine hearkening to his words,
straight he changed his note, for well he knew him to be covetous and
disloyal, and so he sang in quite another fashion.
"Oh, river, cease to flow; crumble, thou manor, keep and tower; let the
grass wither with the rose, and the tall tree stand bare, for the
gentle dames and knights come no more who once delighted in my song,
and to whom this fountain was dear. In place of the brave and generous
knights, set upon honour, stands this envious churl, greedy of naught
but money. Those came to hear my song for solace, and for love of love;
he but to eat and drink the more, and for ease of his gluttony."
And when the bird had thus spoken he took his flight.
Now the villein, who had listened to this song, thought within himself
that might he snare so marvellous a bird, very easily could he sell him
at a great price; or if he might not sell him, at least he could set
him fast in a cage and hearken his lay at pleasure both early and late.
So he climbed within the tree and sought and searched and pried until
he marked the branch from whence the bird was wont to sing. There he
set a cunning snare, and waited to see what time should make clear. At
the hour of vespers the bird returned to the orchard, and lighting upon
the branch was fast taken in the net. Then the villein came forth, and
mounting quickly, joyously seized him in his hand.
"Small profit will you have of your labour," said the bird, "for I can
pay but a poor ransom."
"At least I shall be paid in songs," answered the villein. "You were
wont to sing for your own pleasure, now you will carol for mine."
"Think not so," replied the bird. "He who is used to the freedom of
wood and meadow and river cannot live prisoned in a cage. What solace
may I find there, or joy? Open your hand, fair sweet friend, for be
assured no captive has a heart for songs."
"By my faith, then, you shall be served at table."
"Never will you have dined worse, for there is nothing of me. I pray
you to let me go, for it were a sin to slay me."
"By my faith, you talk and talk; the more you plead, the less will I
grant."
"Certes," answered the bird, "you are in your right, for such is the
law. Many a time have I heard tell that the uncharitable granteth no
alms. But there is a proverb that teaches that often man gives in his
own interest what cannot be taken from him by force. Now, if you
release me from this net I will make you free of three secrets which
are little known to men of your lineage, and from which you may draw
much profit."
"Tell me these secrets," said the villein, "and I will open my hand."
"Such faith have I in you," answered the bird, "that I will speak only
when you free me from the snare."
The villein opened his hand, and the bird flew to a place of surety.
His feathers were all ruffled, for he had been grossly handled by a
glove not of silk but of wool, so he preened and plumed himself
carefully with his beak. But the villein grew impatient, and urged him
to pay his ransom. Now the bird was full of guile, so presently he made
answer to the churl.
"Hear now the first of my three weighty secrets--Do not believe all
that you may hear."
The villein frowned with anger, and answered that he knew it well.
"Fair friend, forget it never," replied the bird.
"Much I fear that I did foolishly in letting you from the snare. This
secret was plain to me before; but now tell me the two others."
"They are fair and wise," said the bird. "Listen well to my second
weighty secret--Do not regret what you have never lost."
"You mock me," cried the villein, "and do wrong to the faith you
plighted with me. You pledged your word to tell me three secrets known
but little to men of such lineage as mine, and you give me musty
proverbs told over by all the world. Certes, what manner of man is he
who weeps over what he has never had!"
"Shall I tell it once again," replied the bird, "for great fear have I
lest it should travel from your mind."
"By my head," answered the villein, "I am a fairer scholar than you
think. These two proverbs have naught to teach me; but hold truly to
our covenant and bargain, and let the third secret contain a graver
matter."
"Listen well to my third secret," said the bird, "for he who receives
it shall never be poor."
"Ah, tell me this secret quickly," cried the churl, "for it draws near
the hour of meat, and truly, beyond all things, do I desire to grow
rich."
Now when the bird heard him--
"This be thy punishment, oh, thou false churl--What you hold in your
hand, never throw between your feet."
Then was the villein all wrathful; but when words came to him to speak,
he said--
"And are these your three mighty secrets! Why, these are but children's
riddles, which I have known ever since I was born. You have but lied to
me, and of all your teaching had I full knowledge long before."
"By my faith," responded the bird, "had you known my third secret never
would you have let me from your hand."
"You say well," said the villein, "but at least knew I the two other
proverbs."
"Ah," said the bird, with malice, "but this proverb was worth a hundred
of the others."
"In what manner?" inquired the villein.
"What, know you not what has chanced to you? Had you slain me when I
was in your power that day would have been the happiest of your life.
For in my body is a jewel, so precious and so rare, that it weighs at
least three ounces. Yea, the virtue of this stone is such that he who
owns it has but to wish, and lo, his desire is fulfilled."
When the villein heard this thing he beat upon his breast, he tore his
raiment, and disfigured his face with his nails, crying out that he was
wretched and undone. The bird from his refuge in the tree rejoiced
greatly to observe the churl's miserable plight, and said nothing till
his enemy's clothes were torn to rags, and his hands sore wounded in
many places. Then he spake--
"Miserable churl, when you held me fast in your rude hand, easy was it
to know that I was no larger than a sparrow or a finch, and weighed
less than half an ounce. How, then, could a precious stone, three
ounces in weight, be hid in my body? Now will I prove to you that of my
three secrets you understood not a single one. You asked me what man
was fool enough to weep over that which he had never lost, and even now
I watch your tears fall for a jewel which was never yours, nor will be
ever. You had faith in all that I was pleased to tell you, trusting all
you heard; and in your folly you flung the bird you held in hand
between your very feet. Fair friend, con over my three secrets, and
learn wisdom even from the counsel of a bird."
When he had spoken thus he took his flight, and from that hour the
orchard knew him no more. With the ceasing of his song the leaves
withered from the pine, the garden became a little dry dust, and the
fountain forgot to flow. Thus the rich villein lost his pleasaunce,
which once was fair beyond content. And remember well, fair lords and
dames, that truly speaks the proverb, "He who covet another's good, oft
loses his own," as we may learn from the "Lay of the Little Bird."
THE DIVIDED HORSECLOTH
Each owes it to his fellows to tell as best he may, or, better still,
to write with fair enticing words, such deeds and adventures as are
good and profitable for us to know. For as men come and go about their
business in the world, many things are told them which it is seemly to
keep in remembrance. Therefore, it becomes those who say and relate,
diligently and with fair intent to keep such matters in thought and
study, even as did our fathers before us. Theirs is the school to which
we all should pass, and he who would prove an apt scholar, and live
beyond his day, must not be idle at his task. But the world dims our
fine gold: the minstrel is slothful, and singers forget to sing,
because of the pain and travail which go to the finding of their songs.
So without waiting for any to-morrow, I will bring before you a certain
adventure which chanced, even as it was told to me.
Some seven years ago it befell that a rich burgess of Abbeville
departed from the town, together with his wife, his only son, and all
his wealth, his goods and plenishing. This he did like a prudent man,
since he found himself at enmity with men who were stronger and of more
substance than he. So, fearing lest a worse thing should bechance him,
from Abbeville he went up to Paris. There he sought a shop and
dwelling, and paying his service, made himself vassal and burgess of
the King. The merchant was diligent and courteous, his wife smiling and
gracious, and their son was not given over to folly, but went soberly,
even as his parents taught him. Much were they praised of their
neighbours, and those who lived in the same street often set foot in
their dwelling. For very greatly are those loved and esteemed by their
fellows who are courteous in speech and address. He who has fair words
in his mouth receives again sweet words in his ear, and foul words and
foul deeds bring naught but bitterness and railing. Thus was it with
this prudent merchant. For more than seven years he went about his
business, buying and selling, concerning himself with matters of which
he had full knowledge, putting by of his earnings a little every day,
like a wise and worthy citizen. So this wealthy merchant lived a happy
blameless life, till, by the will of God, his wife was taken from him,
who had been his companion for some thirty years. Now these parents had
but one only child, a son, even as I have told you before. Very
grievously did he mourn the death of her who had cherished him so
softly, and lamented his mother with many tears, till he came nigh to
swoon. Then, to put a little comfort in his heart, his father said to
him--
"Fair son, thy mother is dead, and we will pray to God that He grant
her mercy in that day. But dry now thine eyes and thy face, for tears
can profit thee nothing. By that road we all must go, neither can any
man pass Death upon the way, nor return to bring us any word. Fair son,
for thee there is goodly comfort. Thou art a young bachelor, and it is
time to take thee a wife. I am full of years, and so I may find thee a
fair marriage in an honourable house I will endow thee with my
substance. I will now seek a bride for thee of birth and breeding--one
of family and descent, one come of ancient race, with relations and
friends a gracious company, a wife from honest folk and from an honest
home. There, where it is good and profitable to be, I will set thee
gladly, nor of wealth and moneys shalt thou find a lack."
Now in that place were three brethren, knights of high lineage, cousins
to mighty lords of peerage, bearing rich and honourable blazons on
their shields. But these knights had no heritage, since they had pawned
all that they owned of woods and houses and lands, the better to take
their pleasure at the tourney. Passing heavy and tormented were these
brethren because in no wise might they redeem their pledge. The eldest
of these brothers had a daughter, but the mother of the maid was dead.
Now this damsel owned in Paris a certain fair house, over against the
mansion of the wealthy merchant. The house was not of her father's
heritage, but came to her from her mother, who had put the maid in ward
to guardians, so that the house was free from pledge. She received in
rent therefrom the sum of twenty Paris pounds every year, and her dues
were paid her right willingly. So the merchant, esteeming her a lady of
family and estate, demanded her hand in marriage of her father and of
all her friends. The knight inquired in his turn of the means and
substance of the merchant, who answered very frankly--
"In merchandise and in moneys I have near upon fifteen hundred pounds.
Should I tell you that I had more, I should lie, and speak not the
truth. I have besides one hundred Paris pounds, which I have gained in
honest dealings. Of all this I will give my son the half."
"Fair sir," made answer the knight, "in no wise can this be agreed to.
Had you become a Templar, or a White or a Black monk you would have
granted the whole of your wealth either to the Temple or your Abbey. By
my faith, we cannot consent to so grudging an offer, certes, sir
merchant, no."
"Tell me then what you would have me do."
"Very willingly, fair, dear sir. We would that you grant to your son
the sum and total of your substance, so that he be seised of all your
wealth, and this in such fashion that neither you, nor any in your
name, may claim return of any part thereof. If you consent to this the
marriage can be made, but otherwise he shall never wed our child and
niece."
The merchant turned this over for a while, now looking upon his son,
now deep in thought. But very badly he was served of all his thought
and pondering. For at the last he made reply to him and said--
"Lord, it shall even be done according to your will. This is our
covenant and bargain, that so your daughter is given to my son I will
grant him all that I have of worth. I take this company as witness that
here I strip myself of everything I own, so that naught is mine, but
all is his, of what I once was seised and possessed."
Thus before the witnesses he divested himself utterly of all his
wealth, and became naked as a peeled wand in the eyes of the world, for
this merchant now had neither purse nor penny, nor wherewithal to break
his fast, save it were given him by his son. So when the words were
spoken and the merchant altogether spoiled, then the knight took his
daughter by the hand and handfasted her with the bachelor, and she
became his wife.
For two years after this marriage the husband and the dame lived a
quiet and peaceful life. Then a fair son was born to the bachelor, and
the lady cherished and guarded him fondly. With them dwelt the merchant
in the same lodging, but very soon he perceived that he had given
himself a mortal blow in despoiling himself of his substance to live on
the charity of others. But perforce he remained of their household for
more than twelve years, until the lad had grown up tall, and began to
take notice, and to remember that which often he heard of the making of
his father's marriage. And well he promised himself that it should
never go from mind.
The merchant was full of years. He leaned upon his staff, and went bent
with age, as one who searches for his lost youth. His son was weary of
his presence, and would gladly have paid for the spinning of his
shroud. The dame, who was proud and disdainful, held him in utter
despite, for greatly he was against her heart. Never was she silent,
but always was she saying to her lord--
"Husband, for love of me, send your father upon his business. I lose
all appetite just for the sight of him about the house."
"Wife," answered he, "this shall be done according to your wish."
So because of his wife's anger and importunity, he sought out his
father straightway, and said--
"Father, father, get you gone from here. I tell you that you must do
the best you can, for we may no longer concern ourselves with you and
your lodging. For twelve years and more we have given you food and
raiment in our house. Now all is done, so rise and depart forthwith,
and fend for yourself, as fend you must."
When the father heard these words he wept bitterly, and often he cursed
the day and the hour in which he found he had lived too long.
"Ah, fair, sweet son, what is this thou sayest to me! For the love of
God turn me not from thy door. I lie so close that thou canst not want
my room. I require of thee neither seat in the chimney corner, nor soft
bed of feathers, no, nor carpet on the floor; but only the attic, where
I may bide on a little straw. Throw me not from thy house because I
eat of thy bread, but feed me without grudging for the short while I
have to live. In the eyes of God this charity will cover all thy sins
better than if thou went in haircloth next the flesh."
"Fair father," replied the bachelor, "preach me no preachings, but get
you forth at once, for reason that my wife would have you gone."
"Fair son, where then shall I go, who am esteemed of nothing worth?"
"Get you gone to the town, for amongst ten thousand others very easily
you may light on good fortune. Very unlucky you will be if there you
cannot find a way to live. Seek your fortune bravely. Perchance some of
your friends and acquaintance will receive you into their houses."
"Son, how then shall men take me to their lodging, when you turn me
from the house which I have given you? Why should the stranger welcome
that guest whom the son chases from his door? Why should I be received
gladly by him to whom I have given naught, when I am evilly entreated
of the rich man for whose sake I go naked?"
"Father," said he, "right or wrong, I take the blame upon my own head;
but go you must because it is according to my will."
Then the father grieved so bitterly that for a little his very heart
would have broken. Weak as he was, he raised himself to his feet and
went forth from the house, weeping.
"Son," said he, "I commend thee to God; but since thou wilt that I go,
for the love of Him give me at least a portion of packing cloth to
shelter me against the wind. I am asking no great matter; nothing but a
little cloth to wrap about me, because I am but lightly clad, and fear
to die for reason of the cold."
Then he who shrank from any grace of charity made reply--
"Father, I have no cloth, so neither can I bestow, nor have it taken
from me."
"Fair, sweet son, my heart trembles within me, so greatly do I dread
the cold. Give me, then, the cloth you spread upon your horse, so that
I come to no evil."
So he, seeing that he might not rid himself of his father save by the
granting of a gift, and being desirous above all that he should part,
bade his son to fetch this horsecloth. When the lad heard his father's
call he sprang to him, saying--
"Father, what is your pleasure?"
"Fair son," said he, "get you to the stable, and if you find it open
give my father the covering that is upon my horse. Give him the best
cloth in the stable, so that he may make himself a mantle or a habit,
or any other sort of cloak that pleases him."
Then the lad, who was thoughtful beyond his years, made answer--
"Grandsire, come now with me."
So the merchant went with him to the stable, exceedingly heavy and
wrathful. The lad chose the best horsecloth he might find in the
stable, the newest, the largest, and the most fair; this he folded in
two, and drawing forth his knife, divided the cloth in two portions.
Then he bestowed on his grandfather one half of the sundered
horsecloth.
"Fair child," said the old man, "what have you done? Why have you cut
the cloth that your father has given me? Very cruelly have you treated
me, for you were bidden to give me the horsecloth whole. I shall
return and complain to my son thereof."
"Go where you will," replied the boy, "for certainly you shall have
nothing more from me."
The merchant went forth from the stable.
"Son," said he, "chastise now thy child, since he counts thy word as
nothing but an idle tale, and fears not to disobey thy commandment.
Dost thou not see that he keeps one half of the horsecloth?"
"Plague take thee!" cried the father; "give him all the cloth."
"Certes," replied the boy, "that will I never do, for how then shall
you be paid? Rather will I keep the half until I am grown a man, and
then give it to you. For just as you have chased him from your house,
so I will put you from my door. Even as he has bestowed on you all his
wealth, so, in my turn, will I require of you all your substance.
Naught from me shall you carry away, save that only which you have
granted to him. If you leave him to die in his misery, I wait my day,
and surely will leave you to perish in yours."
The father listened to these words, and at the end sighed heavily. He
repented him of the evil that he purposed, and from the parable that
his child had spoken took heed and warning. Turning himself about
towards the merchant, he said--
"Father, return to my house. Sin and the Enemy thought to have caught
me in the snare, but, please God, I have escaped from the fowler. You
are master and lord, and I render all that I have received into your
hands. If my wife cannot live with you in quiet, then you shall be
served and cherished elsewhere. Chimney corner, and carpet, pillow and
bed of feathers, at your ease you shall have pleasure in them all. I
take St. Martin to witness that never will I drink stoup of wine,
never carve morsel from dish, but that yours shall be the richer
portion. Henceforth you shall live softly in the ceiled chamber, near
by a blazing fire, clad warmly in your furred robe, even as I. And all
this is not of charity, but of your right, for, fair sweet father, if I
am rich it is because of your substance."
Thus the brave witness and the open remonstrance of a child freed his
father from the bad thoughts that he harboured. And deeply should this
adventure be considered of those who are about to marry their children.
Let them not strip themselves so bare as to have nothing left. For he
who gives all, and depends upon the charity of others, prepares a rod
for his own back.
SIR HUGH OF TABARIE
In the years when Saladin was King, there lived a Prince in Galilee,
who was named Sir Hugh of Tabarie. On a day he was with other Christian
men who gave battle to the Turks, and, since it pleased God to cast His
chivalry behind Him, Sir Hugh was taken prisoner, and many another
stout knight with him. When dusk closed down on the field, the Prince
was led before Saladin, who, calling him straightway to mind, rejoiced
greatly and cried--
"Ah, Sir Hugh, now are you taken."
"Sire," answered the brave knight, "the greater grief is mine."
"By my faith, Hugh, every reason have you for grief, since you must
either pay your ransom or die."
"Sire, I am more fain to pay ransom than to die, if by any means I may
find the price you require of me."
"Is that truly so?" said the King.
"Sire," said Sir Hugh, "in the fewest words, what is the sum you demand
of me?"
"I ask of you," replied the King, "one hundred thousand besants."
"Sire, such a sum is too great a ransom for a man of my lands to pay."
"Hugh," said the King, "you are so good a knight, and so hardy, that
there is none who hears of your prison and this ransom, but will gladly
send of his riches for your ease."
"Sire," said he, "since thus it must be, I promise to pay the sum you
require, but what time do you grant me to find so mighty a ransom?"
"Hugh," said the King, "I accord you the grace of one year. If within
the year you count me out the tale of these besants, I will take it
gladly; but if you fail to gain it, then must you return to your
prison, and I will hold you more willingly still."
"Sire, I pledge my word and my faith. Now deliver me such a safe
conduct that I may return in surety to my own land."
"Hugh, before you part I have a privy word to speak to you."
"Sire, with all my heart, and where?"
"In this tent, close by."
When they were entered into the pavilion, the Emperor Saladin sought to
know of Sir Hugh in what fashion a man was made knight of the Christian
chivalry, and required of him that he should show it to his eyes.
"Sire, whom then should I dub knight?"
"Myself," answered the King.
"Sire, God forbid that I should be so false as to confer so high a gift
and so fair a lordship even upon the body of so mighty a prince as
you."
"But wherefore?" said the King.
"For reason, sire, that your body is but an empty vessel."
"Empty of what, Sir Hugh?"
"Sire, of Christianity and of baptism."
"Hugh," said he, "think not hardly of me because of this. You are in my
hand, and if you do the thing that I require of you, what man is there
to blame you greatly when you return to your own realm. I seek this
grace of you, rather than of another, because you are the stoutest and
most perfect knight that ever I may meet."
"Sire," said he, "I will show you what you seek to know, for were it
but the will of God that you were a christened man, our chivalry would
bear in you its fairest flower."
"Hugh," said he, "that may not be."
Thereupon Sir Hugh made ready all things necessary for the making of a
knight; and having trimmed the hair and beard of the King in seemly
fashion, he caused him to enter within a bath, and inquired--
"Sire, do you understand the meaning of this water?"
"Hugh, of this I know nothing."
"Sire, as the little child comes forth from the waters of baptism clean
of sin, so should you issue from this bath washed pure of all stain and
villainy."
"By the law of the Prophet, Sir Hugh, it is a fair beginning."
Then Sir Hugh brought the Sultan before an untouched bed, and having
laid him therein, he said--
"Sire, this bed is the promise of that long rest in Paradise which you
must gain by the toils of chivalry."
So when the King had lain softly therein for a little space, Sir Hugh
caused him to stand upon his feet, and having clothed him in a fair
white vesture of linen and of silk, said--
"Sire, this spotless stole you first put on is but the symbol of a body
held and guarded clean."
Afterwards he set upon the King a gown of scarlet silk, and said--
"Sire, this vermeil robe keeps ever in your mind the blood a knight
must shed in the service of his God and the defence of Holy Church."
Then taking the King's feet in his hands, he drew thereon shoes of
brown leather, saying--
"Sire, these brown shoes with which you are shod, signify the colour of
that earth from which you came, and to which you must return; for
whatever degree God permits you to attain, remember, O mortal man, that
you are but dust."
Then Sir Hugh raised the Sultan to his feet, and girt him with a white
baldrick, saying--
"Sire, this white cincture I belt about your loins is the type of that
chastity with which you must be girded withal. For he who would be
worthy of such dignity as this must ever keep his body pure as any
maid."
After this was brought to Sir Hugh a pair of golden spurs, and these he
did upon the shoes with which the Sultan was shod, saying--
"Sire, so swiftly as the destrier plunges in the fray at the prick of
these spurs, so swiftly, so joyously, should you fight as a soldier of
God for the defence of Holy Church."
Then at the last Hugh took a sword, and holding it before the King,
said--
"Sire, know you the three lessons of this glaive?"
"What lessons are these?"
"Courage, justice and loyalty. The cross at the hilt of his sword gives
courage to the bearer, for when the brave knight girds his sword upon
him he neither can, nor should, fear the strong Adversary himself.
Again, sire, the two sharp edges of the blade teach loyalty and
justice, for the office of chivalry is this, to sustain the weak
against the strong, the poor before the rich, uprightly and loyally."
The King listened to all these words very heedfully, and at the end
inquired if there was nothing more that went to the making of a
knight.
"Sire, there is one thing else, but that I dare not do."
"What thing is this?"
"It is the accolade."
"Grant me now this accolade, and tell me the meaning thereof."
"Sire, the accolade is a blow upon the neck given with a sword, and the
significance thereof is that the newly made knight may always bear in
mind the lord who did him that great courtesy. But such a stroke will I
not deal to you, for it is not seemly, since I am here your prisoner."
That night Saladin, the mighty Sultan, feasted in his chamber, with the
fifty greatest lords of his realm, emirs, governors and admirals, and
Sir Hugh of Tabarie sat on a cushion at his feet. At the close of the
banquet Sir Hugh rose up before the King and said--
"Sire, grant me grace. I may not forget that you bade me to seek out
all fair and honourable lords, since there is none who would not gladly
come to my help in this matter of my ransom. But, fair Sir King, in all
the world shall I never find a lord so wise, so hardy, and so courteous
as yourself. Since you have taught me this lesson, it is but just and
right that I should pray you to be the first to grant me aid herein."
Then Saladin laughed loudly out of a merry heart, and said--
"Pray God that the end be as sweet as the beginning. Truly, Sir Hugh, I
will not have it on my conscience that you miss your ransom because of
any meanness of mine, and therefore, without guile, for my part I will
give you fifty thousand good besants."
Then the great Sultan rose from his throne, and taking Prince Hugh with
him, came to each of the lords in turn--emir, governor and
admiral--and prayed of him aid in the business of this ransom. So all
the lords gave largely out of a good heart, in such measure that Sir
Hugh presently acquitted himself of his ransom, and returned to his own
realm from amongst the Paynim.
THE STORY OF KING FLORUS AND OF THE FAIR JEHANE
Here begins the story of a certain King who was named King Florus of
Ausay. This King Florus was a very stout knight, and a gentleman of
proud descent. He was wedded to the daughter of the Prince of Brabant,
a gentlewoman of high lineage. Very fair was the maid when she became
his dame, slender of shape and dainty of fashion, and the story telleth
that she was but fifteen summers old when King Florus became her lord,
and he was but of seventeen years. A right happy life they passed
together, as becometh bride and groom who wed fondly in their youth;
yet because he might have no child of her King Florus was often dolent,
and she for her part was vexed full grievously. This lady was very
gracious of person, and very devout towards God and Holy Church. She
gave alms willingly, and was so charitable that she nourished and
clothed the needy, kissing their hands and feet. Moreover, so constant
and private in service was she to the lepers of the lazar house, both
men and women, that the Holy Ghost dwelt within her. Her lord, King
Florus, so long as his realm had peace, rode forth as knight-errant to
all the tournaments in Allemaigne and France and many other lands of
which the noise reached him; thereon he spent much treasure, and gained
great honour thereby.
But now my tale ceases to speak of him, and telleth of a knight who
dwelt in the marches of Flanders and of Hainault. This knight was wise
in counsel, and brave of heart, very sure and trusty. He had to wife a
right fair lady, of whom he had one daughter, young and fresh, named
Jehane, a maid of some twelve years. Many sweet words were spoken of
this maiden, for in all the country round was none so fair. Her mother
prayed often to her lord that he should grant the girl in marriage, but
so given were all his thoughts to the running of tourneys that he
considered nothing of the trothing of his child, though his wife
admonished him ever on his return from the jousts.
This knight had for squire a man named Robert, the bravest squire in
any Christian realm. His prowess and his praise were such that oft he
aided his lord to bear away the prize from the tournaments whereat he
ran. So great was his praise that his lady spake him thus--
"Robert, more careth my lord for these joustings than for any words I
speak, which thing is grievous to me, for I would that he gave care and
pains to wed this daughter of mine. I pray you, therefore, for love of
me, that if you may, you tell him that very ill he does, and is greatly
to be blamed, not to marry his own fair child, for there is no knight
of these parts, however rich his state, who would not gladly welcome
such a bride."
"Lady," said Robert, "you have well spoken. Very readily will I speak
thereof, and since my lord asks often of my counsel, every hope have I
that he will take heed to my words."
"Robert," said the lady, "you will find me no niggard, so you do this
task."
"Lady," said Robert, "your prayer is guerdon enough for me. Be assured
I will do all that I may."
"I am content," returned the lady.
Now within a little space the knight made ready to fare to a
tournament very far from his land. When he came to the field, he (with
a certain knight in whose company he rode) was joined to one party, and
his banner was carried to the lodging of his lord. The tilting began,
and such deeds did the knight, by the cunning service of his squire,
that he bore off the honour and the prize of that tourney from the one
side and the other. On the second day the knight prepared to return to
his own country; so Robert took him often to task and blamed him
greatly that he had not bestowed his fair daughter in marriage. Having
heard this many times, at the end his lord replied--
"Robert, thou and thy lady give me no peace in the matter of the
marriage of my daughter; but at present I see and know of none in my
parts to whom I am content to give her."
"Ah, sir," cried Robert, "there is no knight in your realm who would
not receive her right joyously."
"Robert, fair friend, they are worth nothing, not one of them; neither
will I bestow her there with my good will. I know of no man in the
world who is worthy of her, save one man only, and he, forsooth, is no
knight."
"Sir, tell me his name," answered Robert, "and I will find means to
speak to him so privily that the marriage shall be made."
"Certes, Robert," returned the knight, "meseems thou art very desirous
that my daughter shall be wedded."
"Sir," quoth Robert, "you speak truly, for it is full time."
"Robert," said the knight, "since thou art so hot to carol at her
wedding, she shall soon enough be married if thou accord thereto."
"Certes, sir," said Robert, "right willingly will I consent thereto."
"To that you pledge your word?" demanded the knight.
"Truly, sir, yes," answered Robert.
"Robert, thou hast served me very faithfully, and ever have I found
thee skilled and true. Such as I am, that thou hast made of me; for by
thine aid at the tourneys have I gained five hundred pounds of rent.
'Twas but a short time since that I had but five hundred; whereas now I
have one thousand pounds from rent of land. This, therefore, I owe to
thee, and I acquit me of my debt by giving thee my fair daughter, so
thou art willing to take her at my hand."
"Ah, sir," cried Robert, "for the pity of God, say not thus. I am too
low a man to snatch at so high a maiden, nor dare I pretend to one so
rich and gracious as my demoiselle, since there is no knight in all the
realm, whate'er his breeding, who would not count it honour to be her
lord."
"Robert, know of a surety that never shall knight of this country call
her his; but I will bestow her on thee, if thou refusest her not, and
for her dowry shall she bring thee four hundred pounds from rent of my
lands."
"Ah, sir," said Robert, "you are pleased to make a mock of me."
"Robert," said the knight, "be assured this is no jest."
"Ah, sir, neither my lady nor her mighty kin will endure to consent
thereto."
"Robert," said the knight, "this matter concerns none of them. Hold, I
give thee my glove, and I invest thee with four hundred pounds of my
land, and this is my warrant for the delivery thereof."
"Sir," said Robert, "I will not refuse so goodly a gift, since it is
given with so true a heart."
"Robert," replied the knight, "the grant is sealed."
So the knight granted him his glove, and invested him with rights in
that fair maiden and her land.
Thus they passed upon their ways until it fortuned that this knight
returned to his own house. When he was entered therein, his wife--that
comely dame--received him right sweetly, and said--
"Husband, for the love of God, give thought at this time to the
marriage of our maid."
"Dame," said her lord, "thou hast spoken so often of this matter that I
have trothed her already."
"Sir," inquired the lady, "to whom?"
"Certes, dame, I have pledged her to a man who will ever be loyal and
true. I have given her to Robert, my squire."
"To Robert! Alas the day," quoth the lady. "Robert is but a naked man,
nor is there a knight, however noble, in all this realm who would not
have taken her gladly. Certainly Robert shall have none of her."
"Dame, have her he shall, for I have delivered to him as my daughter's
portion four hundred pounds in rent of land, and all his rights therein
I warrant and will maintain."
When the lady heard this thing she was sore troubled, and said to her
lord that of a surety should Robert never possess her maid.
"Dame," said her husband, "have her he shall, with good will or with
bad will, for I have made a covenant with him, and will carry out my
bargain."
When the lady heard these words of her lord she sought her chamber, and
wept and lamented very grievously. After her tears were shed then she
sent to seek her brothers and other kinsmen of her house, and showed
them of that thing her lord would do, and they said--
"Lady, what have we to do herein? We have no care to go counter to your
lord, for he is a stout knight, weighty of counsel and heavy of hand.
Moreover, can he not do as he will with his daughter, and his land
besides? Know you well that for this cause will none of us hang shield
about his neck."
"Alas," said the lady, "never may my heart find happiness again, if
thus I lose my child. At the least, fair lords, I pray and require you
to show him that should he make this marriage he acts not rightly, nor
after his own honour."
"Lady," said they, "this we will do full willingly."
So they sought out the knight and acquitted themselves of their task,
and he answered them in courteous wise--
"Fair lords, I will tell you what I can do for your love. So it be your
pleasure, I will defer this marriage on such understanding as I now
declare. You are great lords, and are rich in gold and lands. Moreover,
you are near of kin to this fair maid of mine, whom very tenderly I
love. If on your part you will endue her with four hundred pounds of
rent on your lands, I, on mine, will disavow this bond of marriage, and
will wed the girl according to your wise counsel."
"In the name of God," answered they with one accord, "would you spoil
us of all the wealth in our wallets?"
"Since, then," replied the knight, "you may not do this thing, suffer
me to do as I will with my own."
"Sir, with right good mind," answered they.
Then the knight sent for his chaplain, and before him affianced Robert
and his fair daughter together, appointing a certain day for the
marriage. But on the third day Robert prayed his lord that he would
dub him knight, since it was not seemly that he should take a wife so
fair and of such high station till he was of her degree. His lord
agreed thereto with a glad heart, and on the morrow granted him his
desire; therefore after the third day he married the fair maid with
great joy and festival.
At the hour Messire Robert was made knight he spake thus to his lord--
"Sir, once when I was in grievous peril of death, I vowed to seek St.
James's shrine on the morrow of that day I gained my spurs. I pray you
be not wroth with me if to-morrow morn it becomes my honour to wend
thither directly after this marriage, for in no wise will I fail to
observe my vow."
"Certes, Messire Robert, if you do this despite to my daughter, and go
lonely upon your road, very rightly will you be held to blame."
"Sir," said he, "so it pleases God, I shall soon return, but go I must
on peril of my soul."
When a certain knight of the lord's household heard these words,
greatly he reproached Messire Robert for parting from his bride at such
an hour, but Robert answered him that he durst not break his oath.
"Truly," said the knight, who was named Raoul, "truly if you wend thus
to St. James's shrine, leaving so fair a bride but a wedded maid, very
surely will I win her love ere you return. Certain proofs, moreover,
will I give that I have had my way with her; and to this will I pledge
my lands against the lands our lord has granted you, for mine are fully
worth the rents of yours."
"My wife," answered Messire Robert, "does not come of a race to deal me
so shrewd a wrong, and since I give no credence to your words,
willingly will I make the wager, if so it pleases you."
"Yes," said Raoul, "and to this you pledge your faith?"
"Yea," said Messire Robert, "willingly. And you?"
"I, too, pledge my faith. Now let us seek our lord forthwith, and set
before him our bargain."
"That is my desire also," said Messire Robert.
Then they went straight to their lord and laid before him this wager,
and plighted troth to observe their covenant. So in the morning Messire
Robert was married to the fair maiden, and when the bridal Mass was
ended, incontinent he parted from the hall, without tasting the wedding
meats, and set forth on his way, a pilgrim to Compostella.
Now ceaseth the tale to speak of him, and telleth of Raoul, who was hot
in thought as to how he might gain the wager and have to do with the
fair lady. So relateth the tale that the lady behaved very discreetly
whilst her husband was on pilgrimage, for she spent much time upon her
knees in church, praying God to bring her lord again. For his part
Messire Raoul was in a heat in what manner he might win the wager, for
more and more it seemed to him that he should lose his land. He sought
speech with an old dame who attended on the lady, promising that so she
brought him in such a place and hour that he might speak privily to
Madame Jehane, and have his will, then he would deal so largely with
her, that never in her life should she be poor.
"Certes, sir," said the crone, "you are so lovely a knight, so sweet in
speech and so courteous, that verily it is my lady's duty to set her
love upon you, and it will be my pleasure to toil in your service."
So the knight took forty sous from his pouch, and gave them to her that
she might buy a kirtle. The old woman received them greedily, and
hiding the money in a secret place promised to speak to her lady. The
knight bade farewell, and went his way, but the crone tarried in that
place, and when her lady entered from the church said straitly--
"Lady, for God's love, tell me truly, when my lord went to Compostella
did he leave you a maid?"
"Why ask you such a question, Dame Hersent?"
"Because, lady, I believe you to be a virgin wife!"
"Certes, Dame Hersent, and that I am, nor do I know woman who would be
aught else in my case."
"Lady," returned Dame Hersent, "ah, the pity of it! If you but knew the
joy that women have in company of the man they love, you would say that
there is no fonder happiness to be found on earth. Greatly I marvel,
therefore, that you love not, _par amours_, seeing that every lady
loveth with her friend. Were the thing but pleasing to you, fair
falleth the chance, for well I know a knight, comely of person, sweet
and wise of speech, who asks naught better than to set on you his love.
Very rich is he, and lovelier far than the shamed recreant who has left
you in this plight. If you are not too fearful to grant him grace, you
can have of him all that you please to ask, and such joy moreover as no
lady can hope for more."
Whilst the crone was speaking, the lady, who was but a woman, felt her
senses stir within. Curiously she inquired who this knight should be.
"Who is he, lady? God above! one has no fear to cry his name! Who
should it be but that lovely lord, so courteous, so bold, Messire
Raoul, of your father's house, the sweetest heart of all the world."
"Dame Hersent," said the lady, "you will do well to let these words be,
for I have no wish to do myself such wrong, neither come I of such
stock as goes after shame."
"Dame," replied the old woman, "I know it well; but never can you have
the joy of maid with man."
Thus ended their discourse; but presently Sir Raoul came again to the
crone, and she made plain to him how she had spoken to her lady, and in
what fashion she was answered.
"Dame Hersent," said the knight, "so should a virtuous lady reply; but
I pray you speak again with her of this matter, for the archer does not
wing the bird with a first arrow; and, stay, take these twenty sous,
and buy a lining to your coat."
So that ancient dame took the gift, and wearied the lady with enticing
words, but nothing came of all her proffers.
Slowly or quickly thus passed the days, till came the tidings that Sir
Robert was on his way from Compostella, and was already near to Paris.
Very speedily this news was noised abroad, and Sir Raoul, fearing
greatly to lose his lands, again sought speech with the crone. Then
said the old woman that in no wise could she snare the bird, but that
for the great love she bore him this thing she would do--so he would
recompense her service--namely, that she would put matters in such a
case that none should be in the house save himself and the lady, and
then he could act according to his pleasure, whether she would or
whether she would not. So Raoul answered that he desired no other
thing.
"This I will do," said the old woman. "Messire shall come again in
eight days, and on that day shall my lady bathe within her bower. I
will see that all her household are forth from the castle, so may you
come privily to her chamber, and have your desire of her, whether she
cry yea or whether she cry nay."
"You have fairly spoken," answered he.
Hard upon this came letters from Messire Robert that he would be at
the castle on Sunday. On the Thursday, therefore, the crone caused the
bath to be heated in the bower, and the lady disarrayed herself to
enter therein. Then the old woman sent messages to Sir Raoul that he
should come speedily, and moreover she caused all the household to go
forth from that place. Sir Raoul came to the bower, and entering,
saluted the lady, but she deigned no reply to his greeting, and said--
"Sir Raoul, of a truth I thank you for this courtesy, yet you might
have asked if such a visit would be according to my wish. Accursed may
you be for a most ungentle knight."
But Sir Raoul made reply--
"Madame, for God's sake have pity upon me, for I die for love of you.
Lady, as you hope for grace, so grant grace to me."
"Sir Raoul," cried she, "never for pity will I grant you this day, or
any day, the grace of my love. Know well that if you do not leave me
alone in peace certainly will I tell your lord, my father, the honour
that you require of me, for I am no such woman as you think."
"Nay, lady, is it so indeed?"
"Yes, and very surely," replied she.
Then Sir Raoul sprang forward, and clasping her in his arms (for he was
very mighty) bore her towards her bed. As they strove he saw beneath
her right breast a black spot upon the groin, and thought within
himself that here was certain proof that he had had to do with her. But
as he carried her towards the bed his spurs caught within the serge
valence about the foot thereof, so that they fell together, the lord
below and the lady above; whereupon she rose lightly to her feet, and
seizing a billet of wood from the hearth, smote him upon the head so
shrewdly that the blood dropped upon the rushes from the wound. When
Sir Raoul knew his wound to be both deep and large no more he desired
to play, so he arose from the floor and departed straightway from that
chamber to his own lodging, a long mile thence, and sought a surgeon
for his hurt. For her part the faithful lady called upon Dame Hersent,
and returning to her bath, complained to her of this strange adventure
with the knight.
Very great and rich was the feast that the father of the fair lady
ordained against the home-coming of Sir Robert. Many a lord was bidden
to his hall, and amongst these my lord, Sir Raoul, his knight; but he
sent messages that he might not come, for reason of his sickness. On
the Sunday came Sir Robert, and was sweetly welcomed of all; but the
father of the fair lady sought out Sir Raoul, nor would hold him
excused from the feast because of his grievous wound. Therefore he
tired his face and his wound the best that he was able, and went to
hall, where all day long the lords and ladies sat at meat and drink,
and rose for morris and to dance.
When closed the night Sir Robert sought his chamber, and very
graciously the lady received him, as it becometh every wife to receive
her husband. On the morrow again the guests were gathered about the
board, but after dinner uprose Sir Raoul demanding that Messire Robert
should pay his wager, since he had had to do with his wife, by sign and
token of a certain black spot beneath her right breast.
"Of that I know nothing," answered Sir Robert, "for I have not looked
so boldly upon her."
"I require you by the faith that you have pledged me to take heed, and
to do me justice herein."
"That will I, truly," answered Sir Robert.
When came the night once more, then Sir Robert observed his wife
curiously, and marked the black spot upon her white body, whereat the
greater grief was his. In the morning he sought out Sir Raoul, and
owned before his lord that he had lost the bet. Sick at heart was he
throughout the day. When darkness came he went to the stable, and
saddling his palfrey, issued forth from the courtyard, taking with him
what he might carry of his wealth. So he set forth on the road to
Paris, and coming to the city sojourned therein for some three days.
There the tale ceaseth to speak of him, and telleth of his wife.
Very dolent and right heavy was the fair lady that thus her lord had
fled his house. Very long and right greatly she considered the reason
of his flight. She wept and lamented her widowhood, even till such time
as her father entered her chamber, and said that it were much better
that she had never wed, since she had brought him to shame, him, and
all her house, and told her how and why. When she heard this thing she
was sick of heart, and swore that never had she done such deed; but her
words profited her nothing, for though a woman gave her body to be
burned, yet would none believe her clean of sin, once such blame is set
upon her.
Very early in the night the lady rose from the bed, and taking what
wealth she had in her coffer, saddled a palfrey and took the road. She
had sheared her dainty tresses to the shoulder, and in all points was
clad as a boy. In this manner came she to Paris, seeking for her
husband, for to her heart she declared that never would she give over
her search until they were met together once more. So she rode at
adventure, a squire searching for her lord. Now on a morning she
departed from Paris, and riding on the way to Orleans came to Tombe
Isoire, and there met with Sir Robert, her husband. Her heart was very
full as she drew close and saluted him, and he rendered her greeting
for greeting, saying--
"Fair friend, God give you heart's desire."
"Sir," said she, "from whence come you?"
"Certes, fair friend, I am of Hainault."
"Sir, and whither go you?"
"Forsooth, fair friend, little I know where my path may lead me, nor
have I home where I may dwell. Where Fortune hales me, thither I must
go, and the Dame looks not kindly on me, for I have lost the thing that
most I loved in all the world, and she hath lost me. Moreover with her
went house and lands that were fair and deep. But tell me, what is your
name, and whither doth God bring you?"
"Certes, sir," answered Jehane, "I purpose to seek Marseilles, near by
the sea, where as I hope there is noise of war. There, if I may, will I
enter the service of some hardy captain and learn the trade of arms, so
it be God's pleasure. For such is my plight that in nowise can I stay
in my own country. To my eyes, sir, you seem a knight whom I would
serve very gladly, if such was your will, nor of my fellowship could
you take any harm."
"Fair friend," answered Messire Robert, "truly am I a belted knight,
and in what place the battle is set, there would I gladly ride. But
tell me now, what is your name?"
"Sir, my name is John."
"It is right welcome," said the knight.
"And you, sir, what is your name?"
"John, my name is Robert."
"Sir Robert, join me to your company as squire, and I will serve you to
the utmost of my power."
"John, so would I do gladly, but I have so little money in my pouch,
that ere three days are gone I must sell my very steed; therefore I
may take no squire."
"Sir," said John, "be not troubled thereat, for God will provide, if so
it seems good to Him. But where are you set to dine?"
"John, my dinner is a simple business, for I have nothing in my purse
save three sous of Paris."
"Sir, be not troubled thereat, for on my part I have with me nearly ten
pounds of Tournay money, and these are as your own, since your wallet
is not heavy to your wish."
"Fair friend, thanks, and thanks again."
The two comrades rode at a brisk pace to Montlhery, where John found
meat for his lord, and they ate together. When they had eaten they
sought their chamber, the knight lying in a fair bed, and John sleeping
in another, at his feet. Refreshed with sleep, John rose and did the
harness upon their horses, so they mounted and passed upon their way.
Journeying thus at last they lighted at Marseilles upon the Sea, but to
their grief they might not hear the rumour of any war. There for the
time my story ceases to speak of the two of them, and returns to
Messire Raoul, that false knight, who, by leasing, had wrongly gained
the land of Sir Robert.
For more than seven years did Messire Raoul hold the lands of Sir
Robert against law and right. Then a sore sickness took hold upon him,
and afflicted him so grievously that very near he came to death. Much
he feared the wrong he had wrought to that fair lady, the daughter of
his lord, and to her husband besides, for by reason of his malice were
they utterly undone. So great was his sin that he dared not show the
matter to the priest, but tossed upon his bed in utter unrest. On a
certain day when his sickness lay too heavy upon him he bade his
chaplain draw near his bed, for this priest was a wise confessor, loyal
and true, and very close to the sick man's heart. Then he spake--
"Father--my father in God, if not according to the flesh--the time is
come when I must die. For God's love give me now your counsel, as you
are a ghostly man, for on my soul there lies a sin so ugly and so black
that scarcely may I hope to be anealed."
The priest prayed him to speak more plainly, so that he might aid him
to the utmost of his power, wherefore Sir Raoul brought himself to tell
the story that you have heard. At the end he begged the chaplain for
the love of God to show him what he must do to obtain the grace of
pardon for a sin so dark.
"Sir," said the priest, "be not altogether cast down, for so you are
willing to do such penance as I lay upon you, I will take your sin on
me and on my own soul, and you shall be clean."
"Now tell me of this penance," said the knight.
"Sir, within a year of your recovery from this sickness must you take
the cross and pass beyond the sea, and in all places where men ask the
reason of your pilgrimage, there you must tell the story of this bitter
wrong. Moreover, this day must you give hostages to God that thus you
will do."
"All this will I do gladly."
"Sir, what rich pledge can you offer, therefore?"
"The best," replied the knight. "You, yourself, shall be hostage and
surety for me; and on my honour as a knight well will I redeem my
pledge."
"Sir," said the priest, "in the hand of God am I set as your pledge."
The sick man turned from death to life, and soon was altogether healed.
A full year passed away, and yet he had not taken the cross. Right
often the holy man reminded him of his bond, but he treated the
covenant as a jest. Then the chaplain told him straitly that except he
discharged him as his surety before God, he would tell the whole matter
to the father of the fair lady whom he had utterly destroyed. When the
knight heard this he said to the chaplain that within six months would
he seek the sea for the springtide crossing, and thereto he plighted
faith. But now the story ceases to speak of Messire Raoul, and returns
to King Florus of Ausay, of whom it has told nought for a great while.
A right happy life led King Florus and his wife together, as becomes
bride and groom who wed fondly in their youth, but very dolent and sore
of heart were they that they might get no child. The lady caused Masses
to be sung, and was urgent in prayer for her desire, but since it was
not according to the will of God, no gain she got thereby. On a day
there came to the castle of King Florus a holy hermit who dwelt deep
within the great forest of Ausay, in a very desolate place. The queen
received him very gladly, and because he was a wise man and a holy,
would be shriven by him of her sins. So she bared him her secret wound,
and told him of her grief that she might have no child by her lord.
"Ah, madame," said the holy man, "it becometh you patiently to suffer
the pleasure of our Lord. When it is His will, then shall the barren
become a joyful mother of children."
"Certes, sir," said the lady, "would that it were now, for less dear am
I to my lord therefor. Moreover the high barons of this realm cast the
thing against me, and give counsel to my lord that he should put the
barren woman away and take another bride."
"Truly, madame," said the holy man, "grievously would he sin against
God and Holy Church by such a deed."
"Ah, sir, pray you to God for me that I may bear a child to my lord,
for much I doubt that he will put me away."
"Madame," said the holy man, "prayers of mine are little worth, save by
the will of God, yet such as they are you shall have them willingly."
Hardly had this holy man departed from the lady, when the barons of the
realm drew together before the King, and counselled him that he should
put away his wife, since by her he might have no child, and take
another bride. Moreover, if he would not abide by their counsel, then
would they withdraw their fealty, for in no case should the kingdom
remain without an heir. King Florus feared his barons greatly, and gave
credence to their word, so he promised to send his wife to her kindred,
and prayed the lords to seek him another queen, which thing was
accorded between them. When the lady knew thereof she was stricken to
the heart, but nothing might she do, for well she understood that her
lord was purposed to forsake her. Therefore she sent to seek that
hermit who was her confessor, and when he was come she set before him
this business of the barons, and how they would bring another wife to
her husband. "So I pray you, fair father, to aid me with counsel as to
what I must do."
"Lady," said the holy man, "if it be thus, you must suffer it as best
you may, for against king and barons you can make no head."
"Sir," said the gentle lady, "you speak truly; so, if it pleases God, I
will dwell as an anchoress near to you, for then shall I serve God all
the days of my life, and yet draw some stay and comfort from your
presence."
"Lady," said the prudent man, "that were too hazardous a thing, for
you are too tender in years, and fair and fresh. But I will tell you
what to do. Near by my hermitage is a convent of White Nuns, very quiet
and devout. If you go thither, right gladly will they receive you, as
well by reason of your blameless life as of your high degree."
"Sir," said she, "wisely have you spoken, and this I will do, since so
you counsel me."
On the morrow King Florus spake to his wife, and said--
"Since you may have no child by me, needs must we say farewell. I tell
you truly that the parting presses hardly upon me, for never again
shall woman lie so nearly to my heart as you have lain."
Then might he speak no more because of tears, and the lady wept with
him.
"Husband," said she, "for God's love have pity upon me, for where may I
hide myself, and what may I find to do?"
"Wife, so it pleases God, your good days are not yet past, for
honourably and in rich estate shall you return to your own friends in
your own land."
"Lord," said the dame, "I require none of this gear. So it please you,
I will bestow me in a certain convent of nuns, if it will receive me,
and there I will serve God all my life; for since I lose your love I am
she whose heart shall never harbour love again."
So King Florus and the lady wept together very bitterly.
On the third day the Queen set forth to her convent; and the fresh
Queen came to the palace in great pomp, and held revel and festival
with her friends. For four years did King Florus possess this lady, yet
never might he get an heir. So now the story ceases to speak of King
Florus, and turns again to Messire Robert and to John, who were come
to Marseilles.
Very sad was Sir Robert when he came to the city that he might hear of
no arming in all the land; so he said to John--
"What shall we do? You have lent me much money, for the which I owe you
more than thanks. I will give it you again, though I have to sell my
very palfrey, to discharge me of the debt."
"Sir," said John, "if it please you give heed to me, and I will show
you what we have to do. There remain yet to us one hundred Tournay
sous. If you grant me leave, I will turn our two good horses into
better money. With this I will make French bread, for I am the lightest
baker of whom you have heard, and I doubt but little that we shall gain
our money and our livelihood besides."
"John," said Sir Robert, "I am content that you should do according to
your will."
The next day John sold his two horses for ten pounds Tournay. With
these he bought corn, and carried it to the mill. Afterwards he bought
baskets and set to work at his oven to bake good French bread. So white
and so fresh were these loaves of his baking that he sold more than the
best baker of the town, and prospered so greatly that within two years
he had put by well one hundred pounds for their need.
Then said John to his lord--
"Would it not be good to hire a fair large house, with cellarage for
wine, that we might offer hostelry and lodging to wealthy folk from
home?"
"John," answered Sir Robert, "your will is mine, for every reason have
I for content with you."
Then John hired a house, both fair and great, and there gave lodging to
honest folk, gaining money very plenteously. He clad his lord in costly
raiment, so that Sir Robert bestrode his own palfrey, and sat at meat
and drink with the most honourable of the town. Moreover John caused
his board to be furnished with all manner of wines and store, so that
his companions marvelled greatly at the abundance thereof. With all
this so bravely did John prosper that within four years he had put by
more than three hundred pounds, besides the furnishing of inn and
bakery, which very well was worth another fifty pounds. But here the
story ceases to speak of John and Sir Robert, and turneth again to tell
of Messire Raoul.
Now telleth the tale that the chaplain pressed Sir Raoul right
earnestly that he should pass beyond the sea, and thus discharge his
surety from the bond, for much he feared that the knight would yet find
reason to remain. So instant was the priest in pleading, that Sir Raoul
saw well that go he must. He made him ready for his journey, spending
money without stint, and at the end set forth upon the road, him and
his three squires. He drew presently to Marseilles-on-Sea, and there
sought lodging at the French Hostelry owned by Sir Robert and by John.
When John set eyes upon him he knew him well, because he had seen him
many times, and for reason of the scar of the wound that he had given
him. The knight sojourned in the town for fifteen days, till he might
find passage in some vessel going oversea. Whilst he was dwelling at
the inn John took him apart and asked him of the purpose of his
journey, whereat Messire Raoul told him openly all the occasion
thereof, just as the tale hath related already. John listened to his
story, but answered naught for good or evil. Presently Sir Raoul caused
his harness and his gear to be bestowed on the nave, and mounted in the
ship, but for eight days it might not depart from forth the harbour. On
the ninth day the vessel sailed from port on its way to the Holy
Sepulchre. Thus Sir Raoul did his pilgrimage, and there made honest
confession of his sins. In sign of penitence his confessor charged him
strictly to restore to the knight and his lady the fief he held in
scorn of law and right; and Sir Raoul promised straitly that when he
came again to his own land he would carry out the wishes of his heart.
So parting from Jerusalem he voyaged to Acre, and took passage in the
first homing ship, as a man who desires above all things to look upon
the face of his own country. He adventured on the sea, and fared so
speedily, by night as by day, that in less than three months he cast
anchor at the port of Aigues Mortes. Parting from the harbour he stayed
not till he was come to Marseilles, where he rested eight days at the
inn owned by Sir Robert and John, which inn men called the French
Hostelry; but Sir Robert did not recall him to mind, for he thought but
little of Sir Raoul. At the end of eight days he set forth from
Marseilles with his three squires, and at length returned to his own
home, where his household received him gladly, for he was a great lord,
very rich in land and in store. His chaplain inquired of him if any had
asked the reason for his journey.
"Yes," said he, "in three places, to wit, Marseilles, Acre, and
Jerusalem. Moreover that priest who shrived me counselled me to give
back his lands to my lord, Sir Robert, so I may find him, or if I may
not hear of him, to his wife or his heirs."
"Certes," said the chaplain, "he gave you godly counsel."
So Messire Raoul dwelt in his own house for a great while in peace and
ease; and there the tale ceaseth to speak of him, and returns to
Messire Robert and to John.
Sir Robert and John dwelt as citizens in Marseilles for the space of
six years. At the end of six years had they put by in a sure place the
sum of six hundred pounds. John and his business prospered exceedingly,
for so gentle was he and diligent, that he was beloved of all his
neighbours. Men spake almost too well of him, and he maintained his
lord in such estate and worship that it was marvellous to see. When the
end of the seventh year drew near, John sought occasion to speak
soberly to Sir Robert his lord, and said--
"Sir, we have dwelt a great while in this city, and have been so
fortunate in our dealings that we have gained nearly six hundred pounds
in money and in silver vessels."
"Certes," said Sir Robert, "all this, John, is not mine, but yours, for
you have earned it."
"Sir," said John, "saving your grace, it is not mine, but yours, for
you are my own true lord, and never, please God, will I take another."
"John, I thank you heartily," said Robert. "I hold you not as servant,
but as comrade and as friend."
"Sir," said John, "all my days have I given you loyal service, and so
will I ever do."
"By my faith," said Sir Robert, "what you require of me, that is my
pleasure. But as to returning to my own country, I know not what to
say. So much have I lost there that never can it be made up to me."
"Sir," answered John, "fret not over your loss, for, so God pleases,
you shall hear good news when you come into your own land. And be not
fearful of anything, for in whatever place we shall be, please God, I
shall gather enough for me and for you."
"Certes, John," said Sir Robert, "I will do that which pleases you, and
lodge wheresoe'er you will."
"Sir," said John, "now will I sell our goods and make ready for the
journey, for we shall part within fifteen days."
So John sold all the fair furnishing of his houses, and bought thereout
three horses, a palfrey for his lord, another for himself, and a pack
horse for the road. Then they bade farewell to their neighbours and to
the most worshipful citizens of the town, who grieved sorely at their
going.
Sir Robert and John travelled so hardily that in less than three weeks
they drew to their own country, and Sir Robert caused it to be told to
his lord, whose daughter he had wedded, that he was near at hand. The
lord was merry at heart, for much he hoped that his daughter might be
with her husband; and so she was, but hid in the trappings of a squire.
The lord greeted Robert warmly, but when he could learn no tidings of
his daughter, his mirth was turned into sorrow; nevertheless he made a
rich banquet for Sir Robert, and bade his knights and his neighbours to
the feast. Amongst these came Sir Raoul who held Sir Robert's land in
his despite. Great was the merriment on that day and the morrow, and
during all this joy Sir Robert told to John the story of his wager, and
of the manner in which Sir Raoul spoiled him of his land.
"Sir," said John, "challenge him to combat as a false traitor, and I
will fight the battle in your stead."
"John," said Sir Robert, "this you shall not do."
Thus they left the matter till the morrow, when John came to Sir
Robert, and said that he was purposed to speak to the father of his
wife. So they sought the lord, and John spake him thus--
"Sir, you are, after God, the lord of my master Sir Robert, who in the
years that are gone married your child. As you know, a wager was made
between him and Sir Raoul, who said that ere Sir Robert came home from
St. James's shrine he would gain the lady to his wish. Sir Raoul spake
falsely, and is a most disloyal and traitor knight, for never had he
part or share in your daughter's love. All which I am ready to prove
upon his body."
Then Robert strode forth and said--
"John, fair friend, this business is mine alone, nor because of it
shall you hang shield about your neck."
So Sir Robert held forth his gage to his lord, and Sir Raoul tendered
gage of battle in return, though but fearfully; for needs must he
defend himself, or be proclaimed recreant and traitor. Thus were the
pledges given, and the day for the ordeal by battle pronounced to be
fifteen days thence without appeal.
Now hearken well to this strange story of John, and what he did. John,
who more sweetly was named Madame Jehane, had in the house of her
father a certain cousin, who was a fair demoiselle of some twenty-five
years. To this cousin Jehane went and discovered the whole matter,
telling her all the story, from the first thing to the last. She prayed
her, moreover, to keep the business hidden, until such time and hour as
she should make herself known to her father. The cousin--to whom Jehane
was very well known--promised readily to conceal the matter, saying
that never should the secret be made plain by her fault. Then was the
chamber of her cousin made fresh and ready for Madame Jehane. Therein
for the two weeks before the battle Jehane bathed and perfumed her, and
took her ease as best she might, for well had she reason to look her
fairest. Also she caused women to shape closely to her figure four
goodly gowns; one was of scarlet, one of vair, one of peacock blue,
and one of trailing silk. Thus with rest and peace she came once more
to the fulness of her beauty, and was so dainty, fresh and fair, that
no lady showed her peer in all the world.
As for Sir Robert, very greatly was he discomforted during all these
fifteen days at the loss of John his squire, for he knew nothing of his
fate. Nevertheless on the appointed day he got himself into his
harness, and prepared him for the battle stoutly and with a good heart.
On the appointed day the two knights entered within the lists together.
Drawing apart for a little space, they rushed furiously the one on the
other, and gave such mighty strokes with the blades of their great
swords that their horses were borne to the ground beneath them. Sir
Raoul was wounded lightly in the left side, so Sir Robert getting first
upon his feet came swiftly to him, and smote him with all his force
upon the helm. So mighty was the blow that the sword sheared clear
through the helmet to the coif of steel, but the coif was so strong
that the head was not wounded; nevertheless of that stroke he reeled so
that had he not caught at his saddle, certainly he had fallen to the
earth. Then Sir Raoul, who was a very stout champion, struck Sir Robert
so fiercely upon the headpiece that he was all bemused, and the sword
glancing downwards upon the shoulder hacked off the mail of the
hauberk, but did him no hurt. Thereat Sir Robert smote him again with
all the strength that he was able, and the blow lighting upon the
buckler carried away a quarter of the shield. When Sir Raoul knew the
hardiness of his foe much he feared for the issue of the combat, and
well he wished himself once more beyond the sea, and Sir Robert settled
safely on his land. However, he put forward all his prowess, and
pressed Sir Robert so grimly that with one great stroke he clove to the
boss upon the very middle of Sir Robert's shield. For his part Sir
Robert struck fairly at Sir Raoul's helm, but he thrust his shield
before him, and that mighty blow passing clean through the buckler came
full upon the charger's neck, so that horse and rider tumbled to the
ground. Messire Raoul climbed stoutly to his feet, as a valiant man who
had often ridden with the spears, but Sir Robert lighted from his
steed, for he would not deign to fight at vantage with a foe on foot.
Now strove the two knights together, hand to hand, in such fashion that
shield and helm and hauberk were hewn in pieces, and the blood ran from
their bodies by reason of their trenchant glaives. Had they been able
to deal such blows as in the first passage of their arms, very quickly
both one and the other had been slain, for of their shields scarce
enough held together to cover their gauntlets. The fear of death or
shame was now before their eyes, and the nearness of their persons
summoned them to bring this judgment to an end. Sir Robert gripped his
sword in both hands, and with all the greatness of his strength smote
Sir Raoul upon the helm. Half the shattered headpiece fell upon his
shoulders, and the sword cutting through the coif made a grisly wound.
So bewildered was Sir Raoul at the stroke that he was beaten to the
knee; but he rose lightly again, though, since he knew that his head
was naked, very fearful was he of death. He ran therefore at Sir
Robert, smiting with all his power at the remnants of his shield.
Through shield and helmet went the glaive to the depth of full three
fingers, but the wearied sword coming full upon the coif of steel brake
in pieces, for the armourer's work was very strong. When Sir Raoul
looked upon the shards of his sword, and remembered that his head was
naked, much he doubted of his end. Nevertheless he stooped to the
ground, and seizing a great stone in both his hands flung it at Sir
Robert with all his might. Sir Robert stepped aside quickly, avoiding
the cast, and ran in upon his adversary, who turned his back and took
to flight about the lists. So Sir Robert cried that save his foe
admitted himself recreant and shamed he would slay him with the sword.
"Gentle knight," answered Sir Raoul, "I yield thee what remaineth of my
sword, and throw myself entirely on thy grace. Show mercy on me, gentle
knight, and pray thy lord and mine that he have pity upon me, and spare
my life. Take back thy land that I have held against both law and
right, and therewith take my own; for all I said against that fair and
spotless lady was just foul lies."
When my lord, Sir Robert, heard these words he thought within himself
that Sir Raoul might do no more. Therefore he prayed his lord so
urgently to pardon Sir Raoul for this felony, that his prayer was
accorded on such terms that Sir Raoul should abide over sea for all his
days.
In such fashion Sir Robert won back his land, and added that of Sir
Raoul besides. But in this thing he found little comfort, for grief of
heart over the fair and faithful lady from whom he had parted.
Moreover, in no wise could he forget John, his squire, who was lost to
him also. His lord, too, shared in his sorrow, for reason that he might
never gain tidings of his one fair child.
But Madame Jehane, who had spent two weeks in her cousin's chamber in
all ease and comfort, when she heard that her husband had gained the
battle, was greatly content. As we know, she had caused her women to
shape closely to her person four goodly gowns, and of these she arrayed
herself in the most rich, which was of cloth of silk, banded with fine
Arabian gold. So shapely was she of body, so bright of face, and so
gracious of address that nothing more lovely could be found in all the
world, so that her very cousin, even, marvelled at her exceeding
beauty. For the bathing, the tiring, and ease of mind and body of the
past fifteen days had given her back her early freshness, as was
wonderful to see. Very sweet, very ravishing showed Madame Jehane in
her silken robe banded with gold. So when she was ready she called to
her cousin, and said--
"How seem I to thee?"
"Why, dame, the prettiest person in all the world."
"Now, fair cousin, I will tell thee what thou shalt do. Go thou
straight to my father, and tell him to be heavy no more, but rather
merry and glad, because thou bringest him good news of his daughter.
Tell him that she is sound and well, and that so he come with thee, he
shall see her with his eyes. Then lead him here, and he will greet me
again, I deem, right willingly."
The maiden answered that gladly would she give the message, so she
sought out the father of Madame Jehane, and said as she was bidden.
When the lord heard thereof he wondered at this strange thing, and
going after the damsel found his daughter in her chamber. When he saw
her face he cast his arms about her neck, shedding tears of joy and
pity, yea, such was his happiness that scarcely could he find a word.
When he might speak he asked where she had been so long a while.
"Fair father," said the lady, "you shall hear it in good time. But, for
the love of God, cause my mother to come to me speedily, for I die
till I see her once again."
The lord sent incontinent for his wife, and when she was come into the
chamber where her daughter lay, and saw and knew her face, straight she
fell down in a swoon for joy, and might not speak for a great space.
But when her senses were come to her again no man could conceive the
joy and festival she made above her child.
Whilst mother and daughter held each other fast, the father of the fair
lady went in quest of Sir Robert, and meeting him said thus--
"Fair sweet son, very joyful news have I to share with you."
"Certes," said Sir Robert, "of joy have I great need, but God alone can
help my evil case, for sad at heart am I for the loss of my sweet wife,
and sad, besides, for the loss of him who did me more good than any
other in the world, for John, my faithful squire."
"Sir Robert," said the lord, "spoil not your life for John; squires can
be met with at every turning. But as to your wife, I have a certain
thing to tell, for I come from her but now, and know well that she is
the most peerless lady in all the world."
When Messire Robert heard this he fell a-trembling with joy, and said
to his lord--
"Ah, sir, for God's love bring me to see that this is true!"
"Right willingly," said the lord, "come now with me."
The lord went before and Robert followed after, till they were come to
the chamber where mother and daughter yet clasped each other close,
weeping with joy the one upon the other. When they knew their husbands
near they drew apart, and as soon as Sir Robert saw his wife he ran to
her with open arms, and embraced her. So they kissed each the other
with many little kisses, and wept for joy and pity. Yea, they held each
to the other in this fashion whilst a man might run ten acres of land,
nor ceased enlacing. Then the lord commanded that the tables should be
spread for supper; so they ate with mirth and merriment.
After supper, when the songs and the dances were done, they went to
their beds, neither was Sir Robert parted from the Lady Jehane, for
they were right happy to be met together again, and talked of many
things. At the last Sir Robert asked of her where she had been so great
a time, and she said--
"Husband, it is over long a story to tell, but you shall hear it all at
a more convenient season. Tell me, rather, what you have done, and
where you have been all this while."
"Wife," said Sir Robert, "I will tell you gladly."
So he told her all the tale she knew by rote, and of John his squire,
who gained him bread, and said that so distressed was he at the loss of
his companion that never would he give over the search till he had
found him, yea, that he would saddle with the morn and part.
"Husband," said the lady, "that would be madness. Are you set again to
leave me, and what shall I do thereof?"
"Certes, lady, I can do none other; for never man did such things for
his friend as he has done for me."
"Husband," said the wife, "what he did for you was but his duty; he did
no more than what he should have done."
"Wife," said Messire Robert, "by your speech you should have known
him."
"Truly," answered the lady, "truly, I should know him well, for never
aught of what he did was hid from me."
"Lady," said Sir Robert, "I marvel at such words."
"Sir," said she, "there is no need for wonder. If I tell you, yea and
verily, that such a thing is true, will you honestly believe my word?"
"Wife," said he, "on my honour."
"Believe, then, what I am about to tell you, for know assuredly that I
am that very John whom you would seek and this is how it happed. When I
was told the matter of the wager, and of the treason of Messire Raoul;
when, too, I knew that you were fled because of your grief at my
faithlessness, and by reason of the land that for ever you had lost,
then was I more cast down than any woman since woman first was made. So
I clipped my hair close to my head, and taking all the money in my
chest, about ten pounds Tournay, I arrayed me in the guise of a squire,
and followed after you to Paris, coming up with you at Tombe Isoire.
From there we companied together, even to Marseilles, where I served
you as my own liege lord for near seven years, nor do I grudge you
varlet's service. And know for truth that I am innocent and clean of
that deed the foul knight fastened upon me, as clearly now appears, for
he has been put to shame in open field, and has publicly confessed his
treason."
Having spoken thus, Madame Jehane embraced Sir Robert, her lord, and
kissed him very sweetly on the mouth. When Messire Robert was persuaded
that she, indeed, was John, his faithful squire, his joy was greater
far than thought or words may express, and much he marvelled that so
high a lady could prove so lowly and so serviceable. For which thing he
loved her the more dearly all the days of his life.
Thus came together these two parted lovers; thus, on their own domain,
which was both broad and fair, they lived a happy life, as becometh
lovers in their youth. Often Sir Robert rode to tournaments in the
train of his lord, and much honour he gained and such wealth, moreover,
that his land became twice as great as that he had. After the death of
the father and mother of Lady Jehane he became the heir to all their
substance. So stout a knight was he, that by his prowess he was made a
double banneret, and was worth four thousand pounds in land. Yet always
must he be a childless man, to his exceeding grief, though for more
than ten years he was with his wife after the combat with Sir Raoul.
After the term of ten years, by the will of God--which is mightier than
the strength of man--the pains of death gat hold upon him. He met death
like a brave knight, assoiled by the rites of Holy Church, and was laid
in his grave with great honour. His wife, the fair lady, mourned so
grievously upon him, that all about her felt pity for her sorrow. Yet,
during the days, the sharpness of her grief was assuaged, and she came
to take a little comfort, though as yet it was but a little.
The Lady Jehane bore herself during her widowhood as a devout and
kindly lady, devoted to God and Holy Church. Very humble was she and
right charitable, dearly cherishing the poor and needy. So good was she
that no tongue might say aught of her but praise; and so fair that all
who looked upon her owned that she was the mirror of all ladies in the
world for beauty and for virtue. But now for a little space the tale
ceases to speak of her, and returns to tell of King Florus, for it has
been dumb of him o'erlong.
King Florus of Ausay lay at his own castle sorely grieved and vexed at
the departure of his first wife, for she whom the barons had seated in
her chair, though fresh and gracious, might not bring that peace of
heart which was that lady's gift. Four years they lived together, yet
never might have an heir. At the end thereof the pains of death seized
the lady, so she was buried amidst the weeping of her friends, and with
such fair state and service as were fitting to the dignity of a queen.
King Florus remained a widower for above two years. He was yet a young
man, for he was no more than forty-five years of age, and his barons
prayed him that he would seek another wife.
"Certes," answered King Florus, "I desire not greatly to do this thing,
for I have had two wives, yet might not get an heir by either. Moreover
the first wife that I had was so virtuous and so fair, and so dearly
did I love her in my heart for her exceeding goodlihead, that never is
she absent from my thoughts. I tell you truly that never again will I
wed till I may meet a woman sweet and good as she. God rest her soul,
for as I hear she passed away in that White convent where she was
withdrawn."
"Ah, sire," said a knight who was in his private counsel, "many a
comely dame goes about the realm whom you have never seen. One at least
I know who for kindness and for beauty has not her like in all the
world. If you but saw her fairness, if you but knew her worth, you
would own that fortunate indeed were he--yea, though a king--who might
own such rich treasure. She is a gentlewoman, discreet, and rich in
money and in lands, and, if you will, I can tell you many a tale of her
discretion and of her worth."
The King replied that gladly would he hear; so the knight related how
the lady set out to follow after her lord, how she came up with him and
brought him to Marseilles, and the many kindnesses and the great
services she rendered him, just as the tale hath told before. Thereat
King Florus marvelled much, and said privily to the knight that very
gladly would he become the husband of such a wife.
"Sire," answered the knight, who was near neighbour to Madame Jehane,
"I will seek the lady, if such is your good pleasure, and will speak
her so fairly, if I may, that in marriage you twain may be one."
"Yea," said King Florus, "get you speedily to horse, and I pray you to
be diligent in your embassy."
The knight passed straightway upon his errand, and without any tarrying
came to the land where dwelt that lovely lady whom the tale calls
Madame Jehane. He found her in a certain castle of hers, and she
welcomed him gladly as a neighbour and a friend. When they might have
some private speech together, the knight conveyed to her the
commandment of King Florus, that she should ride to him and be wedded
as his wife. When the lady heard his word she smiled more sweetly than
ever siren sang, and answered softly to the knight--
"Your king knows less of women, nor is he so courteous, as fame has
bruited, to command that I should hasten to him that he may take me as
his wife. Certes, I am not a handmaid to ride to him for wages. But
tell your king rather to come to me if he finds my love so desirable
and sweet, and woo me to receive him as husband and as spouse. For
truly the lord should pray and require the lady, and not the lady the
lord."
"Lady," answered the knight, "all that you have told me will I tell
him again; but I doubt that he will come for pride."
"Sir knight," said the lady, "he will do the thing that pleases him;
but in this matter he shows neither courtesy nor reason."
"Lady," said the knight, "in God's name, so let it be. With leave I
take farewell to seek my lord the King, and will tell him as I am
bidden. So if there is any over-word give it me before I part."
"Yea," said the lady. "Take to him my greeting, and add my fairest
thanks for the honour to which he calls me."
The knight parted from the lady forthwith, and on the fourth day
returned to King Florus of Ausay, whom he found in his chamber, deep in
business with his privy council. The knight saluted the King, who gave
him his salutation again, and seating him by his side, asked how it
chanced in this matter of the lady. Then the knight gave the message
with which she charged him; how she would not come, for she was no
kitchen-maid to haste at his bidding for her wages; but that rather
should a lord pray and require of a lady; how that she sent him her
fairest greeting, and her sweetest thanks for the honour he craved of
her.
When King Florus heard these words, he pondered in his seat, nor did
any man speak for a great space.
"Sire," said a knight, who was of his inmost mind, "what do you
consider so deeply? Certes, all these words most richly become a
discreet and virtuous lady, and--so help me God--she is both wise and
brave. In good faith you will do well to fix upon a day when you can
seek her, and send her greetings and letters that on such a day you
will arrive to do her honour, and to crave her as your bride."
"Certes," said King Florus, "I will send her letters that I will lie at
her castle for Easter, and that she make all ready to receive her
husband and her King."
Then King Florus bade the knight who was his messenger to prepare
himself within three days to carry these tidings to his lady. On the
third day the knight set forth, and, riding hard, brought messages to
the lady that the King would spend Easter at her castle. So she
answered that since it was God's will it was woman's too, and that she
would take counsel with her friends, and would array herself to receive
him as the honour of a lady and his greatness required. At these words
the knight returned to his lord, King Florus, and gave him the answer
of the fair lady as you have heard. So King Florus of Ausay made him
ready for his journey, and with a great company set forth to the
country of this fair dame. When he was come there he took and married
her with great pomp and festival. Then he brought her to his own realm,
where she was welcomed of all most gladly. And King Florus joyed
exceedingly over his wife because of her great beauty, and because of
the right judgment and high courage that were in her.
Within the year that the King had taken her to wife the fair Jehane was
delivered of a daughter, and afterwards she rejoiced as the mother of a
son. The boy was named Florence, and the girl Flora. The boy Florence
was very goodly to see, and after he was made knight was esteemed the
hardiest warrior of his day, insomuch that he was chosen to be Emperor
of Constantinople. A mighty prince was he, and wrought great mischief
and evil to the Paynims. As to the Princess Flora, she became the Queen
of her father's realm, and the son of the King of Hungary took her as
wife, so was she lady of two kingdoms.
Such honour as this God gave to the fair lady because of her true and
loyal heart. For many years King Florus lived happily with his virtuous
wife, and when it was the will of God that his days should end, he took
back to his Maker a stainless soul. The lady endured to live but six
months after him, and departed from this world as became so good and
loyal a dame with a quiet mind.
Here finishes the tale of King Florus and the fair Jehane.
OF THE COVETOUS MAN AND OF THE ENVIOUS MAN
Once upon a time, more than one hundred years ago, there lived two
companions, who spent their days together very evilly. The one of these
comrades was so brimmed with envy, that you might find no heart so rank
with the gall of bitterness. The other was so filled with covetousness,
that nothing sufficed of all that could be given to him. Now
covetousness is so foul a vice, that often she bringeth many men to
shame. Covetousness lendeth out her money upon usury, and deceiveth
with her balances, so that he who lendeth may have the greater gain.
But envy is the worser sin, since she grudges joy to others, and is
desirous of all the wealth of all the world.
On a day the envious man and the covetous man were about their business
together, and they came upon St. Martin walking in the fields. But the
saint had been but a little space in their company when he perceived
very clearly the evil desires that were rooted in the hidden places of
their hearts. Thus they fared till they lighted on two beaten paths,
one going this way, and the other that, and a chapel stood between the
ways. There St. Martin stayed his steps, and beckoned to these
evil-minded men.
"Lords," said he, "I take this path to the right that I may enter
within the church. I am St. Martin, who bestowed his cloak on the
beggar, and that you may always keep in mind this meeting I will give,
in turn, to each of you a gift. He who makes known to me his prayer
shall have his desire granted forthwith. But to him who refrains from
words, straightway shall be given twice as much as is bestowed upon his
fellow."
So when St. Martin was gone, the covetous man considered within himself
that if he left his companion to require a gift, he would receive twice
as much as him, and sweetly enjoy a double gain.
"Make your prayer, fair fellow, to the holy saint," said he, "for very
surely you will receive of him all that you may ask. Ask largely of
him, for he will largely give. If you go prudently about the matter you
will be wealthy all your life."
But he whose heart was brimmed with venom and envy dared not to ask
according to his desire, for reason that he feared to die of grief and
malice that his comrade's portion should be larger than his. Thus for a
great while they kept their tongues from speech, turning the business
over, this way and that.
"Wait no longer, lest a mischief befall you," cried at last the
covetous man. "Yea you or nay you, I must have the double of your
share, for all your cunning and caution. Ask, or I will beat you more
grievously than ever yet was beaten donkey at Pont."
"Sire," answered the envious man, "pray I will, since it is better to
receive a gift than stripes. If I require of the saint, money or houses
or lands, very surely will you receive of his bounty twice that he
giveth to me. But, so I am able, of all these shall you get nothing.
Holy St. Martin, I pray and require of your clemency that I may lose
one of my eyes, so that my fellow may lose both of his; thus shall he
be pained and grieved in double measure."
Very careful was the saint to observe his covenant, and of four eyes
these comrades lost three, since the envious man became one-eyed, and
the covetous man a poor blind beggar. Thus these fair friends were
ruined by their gain. But sorrow may he have who lets his heart be
troubled by their wretched plight, for these men were not of sterling
gold, but of false alloy.
OF A JEW WHO TOOK AS SURETY THE IMAGE OF OUR LADY
So many marvels are written of the sweet miracles of Our Lady, that
amongst them I scarce know which to choose. Yet, alas! I have not that
long leisure to set them forth before you every one. Then must it be
with me as with him who walks abroad through summer meadows deep in
flowers. Before, behind, on either side, he sees the countless blossoms
of the field. Blue, vermeil, gold, they dance upon the green. Then,
since he may not gather all, he plucks a rose, a lily, here and there,
as he may find them to his hand. So from amongst the number of Our
Lady's lovely deeds I pluck a leaf, one here, one there, and wreathe
this artless garland, lest I have naught to cast before her feet.
* * * * *
In days long past--as the scribe hath truly written--there lived in the
strong city of Byzantium a certain citizen, who held Our Lady very
dear. Rich he was, and of great worship, because of his wealth and of
the praise of men. To keep his station in the eyes of his fellows, he
spent his substance so largely, and thereto so wantonly, that in a
little while he had wasted all his goods, and naught remained to do but
that he must sell his very lands. Yet so rich of heart was this burgess
that not poverty even might make him knot his purse. He still kept open
house, and gave goodly cheer to all, ever borrowing more and more,
spending and vending, wasting and hasting to scatter everything he had.
For of poverty he had neither heed nor fear, so long as he might find
a man to lend. But at the last he was utterly undone. All his friends
passed him by when they saw how deeply he was sunk in debt, and that to
no lender did he e'er repay what he owed. For he who borrows, never
paying back again, neither seeking from others that which is his due,
very quickly loses his credit, yea, though he be the King of France.
The good citizen was sore vexed, and knew not what to do or say, when
he found that his creditors pressed him hardly, and that he was wholly
disappointed of those friends in whom he put his trust. Sore was his
sorrow, deep his distress, and bitter his shame, his wrath and sadness,
when by no means whatever might he grant his customary bounty, nor of
his charity give alms and benefits to the poor. So long as he was a man
of worship, with store of gold and silver, great were his doles to
those of low estate. But such was the malice wherewith Dame Fortune
pursued him, and such the shame and loss she set upon him, that he had
nothing left to give to others, or to keep for himself. And since Dame
Fortune looked upon him with a frowning countenance, there was none to
welcome him with a smiling face.
Now this unhappy burgess knew not what to do, for some of his
acquaintance gazed the other way, whilst men, to whom he had done
naught but good, jested upon him openly in the street. Doubtless such
is the way of the world to those honest folk who are cast beneath at
the turn of Fortune's wheel. Therefore those to whom he had shown the
greatest kindness requited him with the utmost despite, counting him
viler than a dog; and those, who in his day of prosperity loved and
affected his company, were the very men who now mocked and despised
him. Well say the Scriptures, Put not your trust in man. For in these
days faith is so rare and so forgetful, that the son fails the father
in his peril, and the mother may not count upon her maid. Mad is he who
strips himself for others, for so soon as he comes before them naked,
then they cry, "Beggar, begone!"
When this citizen, who for so long a while had known such great honour,
saw himself so scorned, and found that in all the town he had neither
kinsman nor friend, he knew not what to say or do, nor whom to take for
counsel in his need. So, by the will of God, he turned in his despair
to a certain Jew, the richest in all the city. Him he sought out
straightway, with a face aflame, and said--
"Master Jew, here is my case. All my daughters, all my sons, all my
friends, and, very surely, all of those to whom I have done most good
in this world, have failed me utterly and every one. I am stripped of
all my substance. Foolish have I been, and unlucky, since I wasted all
I had on those very clerks and laymen who desert me now. I am a
merchant of great knowledge, and so you will lend me of your treasure,
I count to gain so largely, that never shall I have to pray another for
a loan; for of your wealth will I make such usage that all will think
the more of me thereby."
"Because you have dealt so generously with others," answered the Jew,
"in this very hour will I lend you freely of my moneys if you can give
me pledge or surety for them."
But the Christian made answer to him--
"Fair, sweet friend, all my kinsfolk and acquaintance have cast me
utterly behind them, neither care they for me any more, notwithstanding
that they thrive by reason of my gifts and toil. I can offer no kinsman
as surety, nor have I a friend in the world. But though I can give
neither pledge nor surety, strive how I may, yet I swear to you now on
my faith and conscience, that, without fail, I will repay you your loan
and your substance on the very day that the debt becomes due."
"If things are thus, I can lend you nothing," answered the Jew; "for
grievously I doubt that you may not carry out your bargain."
"Fair, sweet friend," he made reply, "since then I neither have, nor
think to have, a pledge to offer, take now in pledge, I pray you, my
Maker, in whom is all my faith, this Jesus Christ, the King of Heaven,
the King of kings, the God of gods. If you have not your money returned
on the very day that you shall name, I swear to you by God, fair
brother Jew, and by His Mother, so tender and so dear, that I will
become your villein and your serf, in such wise and fashion as any
other slave of yours; so that with a ring about my neck you may sell me
in the market-place, just as any brute beast."
Now in his heart the Jew greatly desired and longed to make this
Christian his bondsman. Therefore, laughing, he replied--
"I believe but little that Jesus Christ, the son of Mary, whom our
forefathers crucified on a cross of wood, was truly God. But inasmuch
as He was doubtless a holy man, and a prophet of mighty name, if you
will put Him in pledge in such manner that you will serve me all your
life should you fail me in this our bargain, why, I will take your
pledge without demur."
"Fairly have you spoken," said he, "by my soul. Let us go straightway
to the church of Our Lady, the most glorious Mother of God."
A great company of Christians and of Jews went with them to the church,
and many a clerk and layman was witness to their device and covenant.
Without any delay, the wretched merchant kneeled him down before the
Statue, whilst the hot tears rushed to his eyes, and over-ran and
wetted all his face, because of the poverty which drove him to this
deed. The unhappy man knew not what to do in his plight, but he cast
his burden upon the Lord, and, weeping, prayed God's precious Mother
that she would deign to set wretchedness and bondage far from him. But
very fearful was he, and sore adread in his heart.
When he had prayed his prayer to Our Lady, he sprang lightly to his
feet, and said--
"Friend Jew, by my soul see here my Surety. In giving you this Child
and this Image, I give you Jesus Christ, Himself, as pledge. He created
me, and He fashioned me. 'Tis He Who is my bond for your moneys. A
richer pledge you may not think to have, so help me God, now and for
evermore."
He placed the hand of the Child in that of the Jew, and forthwith
delivered the pledge and plighted faith. Then, yet upon his knees, most
pitifully, with eyes all wet, he cried aloud in the hearing of
Christian and of Jew--
"Fair Lord God, most merciful, most mighty and most sure, at the end of
this business, I beseech Thee with clasped hands, fair, kindest Father,
that by the pleadings of Thy sweet Mother, if it should happen that on
the appointed day for any cause I may not give again the wealth I owe
the Jew, then of Thy courtesy pay Thou my debt, and without an hour's
delay redeem Thy pledge and faith. For if but one single day I fail to
keep faith, then his serf must I be all the days of my life, save only
that I break my oath sworn on this Image."
He rose lightly to his feet, though with a tearful face, and the Jew
straightway counted out to him a great sum of money, to deal with in
the future as he had dealt with his own. But he had lost the desire to
play, for he remembered too plainly that of such mirth comes
bitterness. The scalded man hates boiling water, and well he knew, and
clearly he perceived, that he who is in rags goes shivering in the
wind.
The honest merchant--whom God kept in charge--went forth with a light
heart, that leaped and fluttered in his breast, because of the wealth
he had in seisin. He bargained for a bark that lay in harbour, and
since he had much skill in such business, he stored the ship with
divers kinds of merchandise. Then putting his trust in God, and
commending body and goods to His keeping, he hoisted sail, and set
forth upon the water. He voyaged to divers lands, and trafficked with
the merchants thereof to such purpose, that before the year had gone by
he was no more in dread of beggary. God increased his store, so that he
prospered in every market. But the gains and riches of the merchant in
nowise made him grudging of his substance. Freely was given to him,
freely he gave to others, for the love of God Who for every man ripens
His harvest.
In a short while the merchant became very rich. One market opened
another market, and money made more money. So greatly did his substance
multiply that at the end, the story tells us, he might not keep the
count of his wealth. So to set field by field, and house by house, he
travelled in many strange lands. One day darkened, and the next day
dawned, but he never gave thought to that certain day when he must
return to the Jew the loan of which he had made so fruitful a use. He
called it not to mind until there was but one single day between him
and the appointed time, and as it chanced he bethought him thereof
when he was at sea. He well-nigh swooned when the day came to his
heart and memory.
"Ah, gentle Lady of the King of Glory! sweet Maid and debonair!" cried
he; "unhappy wretch, what can I do?"
So sore was his grief that with clenched hands he beat upon his breast,
and presently with locked teeth fell fainting to the deck, where he lay
senseless for a great space. The sailors ran to his succour, and,
pressing about him, cried out and lamented his evil case, for certainly
they deemed that he was dead. Passing heavy were they at this sad
mischance, for not one word could they draw from his lips, nor for all
their pains might they find in him either pulse or breath. When he was
returned a little from his swoon, he addressed himself to prayer,
weeping and sighing for a great while, because for grief he found no
words to say.
"Alas!" cried he, "alas, my luckless lot! What a besotted merchant have
I been! How foully has misfortune stolen upon me! How has the Adversary
beguiled me, and snared my thoughts, that I might not better mark the
appointed day! Surely on the tables of my heart should it have been
written that for pledge I gave Jesus Christ, and His Mother, sweet and
dear. Alas! very right is it that I should go heavy, and that my heart
should be sick and sad, since never by day nor by night have I taken
thought how to return that mighty debt which so affrights me now.
Affrighted, alas! much cause have I to fear. Were a bird now to quit
the ship, yet should he not wing to Byzantium in thirty days--no, nor
in forty. Foul fall the day, for I am quite undone. Alas! for the shame
I have brought upon my kin. Very great riches are very little worth,
since thus am I snared and taken."
In this manner the good merchant made his complaint, and with many
sighs bewailed his wretched plight. But when he had eased his heart
with words, the Holy Spirit wrought upon him, so that his courage came
to him again, and he said--
"What is here for tears? Rather should I take comfort in that He, Who
hath power over all, is holden as my pledge. Let me place the matter in
His mighty hand, nor concern myself overmuch with what is His business
more than mine. I owe the money, but He will pay my debt; and thus by
His balm shall I be healed. On the morrow must I repay the money that I
owe, but there is yet a full night before the money need be counted to
the Jew. I will not concern myself greatly with this matter, but
commend myself humbly to His will. No other thing is there to do, for
none can deliver me from my trouble, save Him alone. He is my Surety,
and very surely will He discharge me from this debt, for without Him
there is no redemption."
Then straightway the merchant took a strong, clamped coffer, and sealed
within it the debt which he must now restore the Jew. Without waiting
for the morrow, he cast it into the sea with his own hand, and with
tears commended it to that great Lord and God Who holds every man in
His good keeping, and to Whom earth and sea are ministers and servants
alike. So He Who is of such high and puissant majesty, that naught He
wills to do is burdensome or heavy to Him, was pleased to steer that
coffer with its precious load of besants through the waters, so that it
made more than a thousand leagues in that one night. Thus with the dawn
it drew right to Byzantium, and on the appointed day the casket and the
treasure came to the shore.
Now by the will of God it chanced that the rich Jew, who lent the
Christian of his moneys, lived in a fair dwelling near by the sea. A
certain servant of his rose early from his bed to walk on the shore in
the cool summer dawn, and spied the casket, which had but just drawn to
land. So, without taking off his raiment, he sprang into the sea that
he might lay hold upon it; but he was not able, for the coffer tossed
grievously whenever he would make it his own. Very covetous was the
varlet of this coffer, yet might he never set his hand upon it. For the
casket moved warily, as though it would say, "Go your road, since in
nowise am I yours."
So presently the servant sought his lord, and returned with him to the
shore. And forthwith the coffer drew to the very feet of the Jew, and
seemed to him to say--
"Fair Sir Jew, receive your own. By me God redeems the merchant from
his debt, and henceforth he is free, quite free of you."
Then the Jew entered swiftly within his door, bearing the casket with
him, and when he had counted over its great riches, he hid the treasure
in a privy place at the foot of his bed, so that none might know of the
matter. Moreover, he found within a certain letter news that, very
soon, this merchant, who so far had voyaged in so many lands, would
seek Byzantium in ships laden with tissues and broideries and all
manner of stuffs and merchandise. So the friends and acquaintance of
the merchant rejoiced greatly at his prosperity, and the whole city
welcomed him with mirth and festival. All men made much of his
home-coming, and clerk and layman joined alike in the feast.
When the Jew heard the noise of the joyous greeting vouchsafed to the
citizen, he rose up quickly, and sought him out without delay. They
spoke at great length together, and many words passed between the
twain. At the last the Jew made mention of his money, as if he sought
to know when payment should be made. For presently in his merry talk,
yet laughing, he took the Christian by the hand, and wagging his head
from side to side, said--
"Oh, faithful Christian! oh, faithful Christian!"
Thereat the burgess began to smile, and made reply that he would learn
the meaning of those words.
"By the Law, it means that I have lent you monies in heaped-up measure
from my wealth, to be repaid me on a day now gone. Since you have
failed in bond and faith, now holds the bargain, that should you break
your covenant, though but for one single day, then all the years of
your life must you labour as my serf. If now you throw me back your
bond, then I must reckon your Holy Faith and your plighted word as
worth just two grains of dust upon a balance."
Then he, whose only hope was in God, made answer to the Jew, and said--
"I owe you nothing, since all that was your due has been paid to the
uttermost doit."
Very cunning was this Jew; therefore he replied--
"Many an honest man was witness to the loan, but what witness can you
bring to the payment of the debt? There is little new in such a plea as
yours."
"Right easily can I find proof of quittance, and to spare. All this
would make me fear, indeed, were not such a mighty Surety at my side.
But so you will come with me to the church, where my pledge was taken,
very surely will I show you proof of the redemption of my bond."
So they, and a great company with them, went to the minster, which was
filled altogether with the press.
Then the citizen, hoping all things of his God, and rooted deeply in
his trust, bowed himself down with clasped hands right humbly to the
floor before the Image of Our Lady. From his very heart, with all his
soul, he prayed and required of her that she would obtain of her sweet
Son to hearken to his prayer, and his words were broken by his sighs.
Afterwards he cried with a clear voice in the hearing of them all, and
said--
"Lord Jesus, so truly as Thou art the very Son of God, witness for me
to this Hebrew of the truth as it is known to Thee. Very God of Very
God, exalt now Thine honour, and for the glory of Thy Name make clear
whether I have discharged me of this debt or not."
Then the Image made answer in these very words--
"It is a true testimony that to the appointed day this Jew has been
paid in full whatsoever you have had of him. In proof whereof the
casket yet remains hid in a privy place beneath his bed, from whence he
took the debt I paid him in your place."
When the Jew heard this marvel he was filled with confusion, and was
greatly astonied, so that he knew not what to say, nor what to do. So
by the grace and lovingkindness of the Holy Spirit that very day he was
baptized, and became a christened man, nor did he ever after waver in
that faith.
So every year it was the gracious custom of all good citizens to keep
this wonder in remembrance with dances and midnight revelry, with feast
and high solemnity. And this holy day was observed in Byzantium, the
mighty city, which Constantine, the noble Emperor, afterwards called
Constantinople.
THE LAY OF GRAELENT
Now will I tell you the adventure of Graelent, even as it was told to
me, for the lay is sweet to hear, and the tune thereof lovely to bear
in mind.
Graelent was born in Brittany of a gentle and noble house, very comely
of person and very frank of heart. The King who held Brittany in that
day made mortal war upon his neighbours, and commanded his vassals to
take arms in his quarrel. Amongst these came Graelent, whom the King
welcomed gladly, and since he was a wise and hardy knight, greatly was
he honoured and cherished by the Court. So Graelent strove valiantly at
tourney and at joust, and pained himself mightily to do the enemy all
the mischief that he was able. The Queen heard tell the prowess of her
knight, and loved him in her heart for reason of his feats of arms and
of the good men spake of him. So she called her chamberlain apart, and
said--
"Tell me truly, hast thou not often heard speak of that fair knight,
Sir Graelent, whose praise is in all men's mouths?"
"Lady," answered the chamberlain, "I know him for a courteous
gentleman, well spoken of by all."
"I would he were my friend," replied the lady, "for I am in much unrest
because of him. Go thou and bid him come to me, so he would be worthy
of my love."
"Passing gracious and rich is your gift, lady, and doubtless he will
receive it with marvellous joy. Why, from here to Troy there is no
priest even, however holy, who in looking on your face would not lose
Heaven in your eyes."
Thereupon the chamberlain took leave of the Queen, and seeking Graelent
within his lodging, saluted him courteously, and gave him the message,
praying him to come without delay to the palace.
"Go before, fair friend," answered the knight, "for I will follow you
at once."
So when the chamberlain was gone, Graelent caused his grey horse to be
saddled, and mounting thereon, rode to the castle, attended by his
squire. He descended without the hall, and passing before the King,
entered within the Queen's chamber. When the lady saw him she embraced
him closely, and cherished and honoured him sweetly. Then she made the
knight to be seated on a fair carpet, and to his face praised him for
his exceeding comeliness. But he answered her very simply and
courteously, saying nothing but what was seemly to be said. Then the
Queen kept silence for a great while, considering whether she should
require him to love her for the love of love; but at the last, made
bold by passion, she asked if his heart was set on any maid or dame.
"Lady," said he, "I love no woman, for love is a serious business, not
a jest. Out of five hundred who speak glibly of love, not one can spell
the first letter of his name. With such it is idleness, or fulness of
bread, or fancy, masking in the guise of love. Love requires of his
servants chastity in thought, in word and in deed. If one of two lovers
is loyal, and the other jealous and false, how may their friendship
last, for love is slain! But sweetly and discreetly love passes from
person to person, from heart to heart, or it is nothing worth. For what
the lover would, that would the beloved; what she would ask of him,
that should he go before to grant. Without accord such as this, love is
but a bond and a constraint. For above all things love means sweetness,
and truth, and measure; yea, loyalty to the loved one and to your word.
And because of this I dare not meddle with so high a matter."
The Queen heard Graelent gladly, finding him so tripping of tongue, and
since his words were wise and courteous, at the end she discovered to
him her heart.
"Friend, Sir Graelent, though I am a wife, yet have I never loved my
lord. But I love you very dearly, and what I have asked of you, will
you not go before to grant?"
"Lady," said he, "give me pity and forgiveness, but this may not be. I
am the vassal of the King, and on my knees have pledged him loyalty and
faith, and sworn to defend his life and honour. Never shall he have
shame because of me."
With these words Sir Graelent took his leave of the Queen, and went his
way.
Seeing him go in this fashion, the Queen commenced to sigh. She was
grieved in her very heart, and knew not what to do. But whatever
chanced she would not renounce her passion, so often she required his
love by means of soft messages and costly gifts, but he refused them
all. Then the Queen turned from love to hate, and the greatness of her
passion became the measure of her wrath, for very evilly she spoke of
Graelent to the King. So long as the war endured, Graelent remained in
that realm. He spent all that he had upon his company, for the King
grudged wages to his men. The Queen persuaded the King to this,
counselling him that by withholding the pay of the sergeants, Graelent
might in no wise flee the country, nor take service with another lord.
So at the end Graelent was wonderfully downcast, nor was it strange
that he was sad, for there remained nothing which he might pledge, but
one poor steed, and when this was gone, no horse had he to carry him
from the realm.
It was now the month of May, when the hours are long and warm. The
burgess with whom Graelent lodged had risen early in the morning, and
with his wife had gone to eat with neighbours in the town. No one was
in the house except Graelent, no squire, nor archer, nor servant, save
only the daughter of his host, a very courteous maid. When the hour for
dinner was come she prayed the knight that they might sit at board
together. But he had no heart for mirth, and seeking out his squire,
bade him bridle and saddle his horse, for he had no care to eat.
"I have no saddle," replied the squire.
"Friend," said the demoiselle, "I will lend you bridle and saddle as
well."
So when the harness was done upon him, Graelent mounted his horse, and
went his way through the town, clad in a cloak of sorry fur, which he
had worn overlong already. The townsfolk in the street turned and
stared upon him, making a jest of his poverty, but of their jibes he
took no heed, for such act but after their kind, and seldom show
kindliness or courtesy.
Now without the town there spread a great forest, thick with trees, and
through the forest ran a river. Towards this forest Graelent rode, deep
in heavy thought, and very dolent. Having ridden for a little space
beneath the trees, he spied within a leafy thicket a fair white hart,
whiter even than snow on winter branches. The hart fled before him, and
Graelent followed so closely in her track that man and deer presently
came together to a grassy lawn, in the midst of which sprang a
fountain of clear, sweet water. Now in this fountain a demoiselle
disported herself for her delight. Her raiment was set on a bush near
by, and her two maidens stood on the bank, busied in their lady's
service. Graelent forgot the chase at so sweet a sight, since never in
his life had he seen so lovely a dame. For the lady was slender in
shape and white, very gracious and dainty of colour, with laughing eyes
and an open brow--certainly the most beautiful thing in all the world.
Graelent dared not draw nigh the fountain for fear of troubling the
dame, so he came softly to the bush to set hands upon her raiment. The
two maidens marked his approach, and at their fright the lady turned,
and calling him by name, cried with great anger--
"Graelent, put my raiment down, for it will profit you little even if
you carry it away, and leave me naked in this wood. But if you are
indeed too greedy of gain to remember your knighthood, at least return
me my shift, and content yourself with my mantle, since it will bring
you money, as it is very good."
"I am not a merchant's son," answered Graelent merrily, "nor am I a
huckster to sell mantles in a booth. If your cloak were worth the spoil
of three castles I would not now carry it from the bush. Come forth
from your bathing, fair friend, and clothe yourself in your vesture,
for you have to say a certain word to me."
"I will not trust myself to your hand, for you might seize upon me,"
answered the lady; "and I tell you frankly that I put no faith in your
word, nor have had any dealings with your school."
Then Graelent answered still more merrily--
"Lady, needs must I suffer your wrath. But at least I will guard your
raiment till you come forth from the well; and, fairest, very dainty
is your body in my eyes."
When the lady knew that Graelent would not depart, nor render again her
raiment, then she demanded surety that he would do her no hurt. This
thing was accorded between them, so she came forth from the fountain,
and did her vesture upon her. Then Graelent took her gently by the left
hand, and prayed and required of her that she would grant him love for
love. But the lady answered--
"I marvel greatly that you should dare to speak to me in this fashion,
for I have little reason to think you discreet. You are bold, sir
knight, and overbold, to seek to ally yourself with a woman of my
lineage."
Sir Graelent was not abashed by the dame's proud spirit, but wooed and
prayed her gently and sweetly, promising that if she granted him her
love he would serve her in all loyalty, and never depart therefrom all
the days of his life. The demoiselle hearkened to the words of
Graelent, and saw plainly that he was a valiant knight, courteous and
wise. She thought within herself that should she send him from her,
never might she find again so sure a friend. Since then she knew him
worthy of her love, she kissed him softly, and spoke to him in this
manner--
"Graelent, I will love you none the less truly, though we have not met
until this day. But one thing is needful that our love may endure.
Never must you speak a word by which this hidden thing may become
known. I will furnish you with deniers in your purse, with cloth of
silk, with silver and with gold. Night and day will I stay with you,
and great shall be the love between us twain. You shall see me riding
at your side, you may talk and laugh with me at your pleasure, but I
must never be seen of your comrades, nor must they know aught
concerning your bride. Graelent, you are loyal, brave and courteous,
and comely enough to the view. For you I spread my snare at the
fountain; for you shall I suffer heavy pains, as well I knew before I
set forth on this adventure. Now must I trust to your discretion, for
if you speak vainly and boastfully of this thing, then am I undone.
Remain now for a year in this country, which shall be for you a home
that your lady loves well. But noon is past, and it is time for you to
go. Farewell, and a messenger shortly shall tell you that which I would
have you do."
Graelent took leave of the lady, and she sweetly clasped and kissed him
farewell. He returned to his lodging, dismounted from his steed, and
entering within a chamber, leaned from the casement, considering this
strange adventure. Looking towards the forest, he saw a varlet issue
therefrom riding upon a palfrey. He drew rein before Graelent's door,
and taking his feet from the stirrup, saluted the knight. So Graelent
inquired from whence he rode, and of his name and business.
"Sir," answered he, "I am the messenger of your lady. She sends you
this destrier by my hand, and would have me enter your service, to pay
your servitors their wages and to take charge of your lodging."
When Graelent heard this message he thought it both good and fair. He
kissed the varlet upon the cheek, and accepting his gift, caused the
destrier--which was the noblest, the swiftest and the most speedy under
the sun--to be led to the stable. Then the varlet carried his baggage
to his master's chamber, and took therefrom a large cushion and a rich
coverlet which he spread upon the couch. After this he drew thereout a
purse containing much gold and silver, and stout cloth fitting for the
knight's apparel. Then he sent for the host, and paying him what was
owing, called upon him to witness that he was recompensed most largely
for the lodging. He bade him also to seek out such knights as should
pass through the town to refresh and solace themselves in the company
of his lord. The host was a worthy man. He made ready a plenteous
dinner, and inquired through the town for such poor knights as were in
misease by reason of prison or of war. These he brought to the hostelry
of Sir Graelent, and comforted them with instruments of music, and with
all manner of mirth. Amongst them sat Graelent at meat, gay and
debonair, and richly apparelled. Moreover, to these poor knights and
the harpers Graelent gave goodly gifts, so that there was not a citizen
in all the town who did not hold him in great worship, and regard him
as his lord.
From this moment Graelent lived greatly at his ease, for not a cloud
was in his sky. His lady came at will and pleasure; all day long they
laughed and played together, and at night she lay softly at his side.
What truer happiness might he know than this? Often, besides, he rode
to such tournaments of the land as he was able, and all men esteemed
him for a stout and worthy knight. Very pleasant were his days and his
love, and if such things might last for ever he had nothing else to ask
of life.
When a full year had passed by, the season drew to the feast of
Pentecost. Now it was the custom of the King to summon at that tide his
barons and all who held their fiefs of him to his Court for a rich
banquet. Amongst these lords was bidden Sir Graelent. After men had
eaten and drunk the whole day, and all were merry, the King commanded
the Queen to put off her royal robes, and to stand forth upon the
dais. Then he boasted before the company--
"Lord barons, how seems it to you? Beneath the sky is there a lovelier
queen than mine, be she maid, dame or demoiselle?"
So all the lords made haste to praise the Queen, and to cry and affirm
that in all the world was neither maid nor wife so dainty, fresh and
fair. Not a single voice but bragged of her beauty, save only that of
Graelent. He smiled at their folly, for his heart remembered his
friend, and he held in pity all those who so greatly rejoiced in the
Queen. So he sat with covered head, and with face bent smiling to the
board. The Queen marked his discourtesy, and drew thereto the notice of
the King.
"Sire, do you observe this dishonour? Not one of these mighty lords but
has praised the beauty of your wife, save Graelent only, who makes a
mock of her. Always has he held me in envy and despite."
The King commanded Graelent to his throne, and in the hearing of all
bade the knight to tell, on his faith as vassal to his liege, for what
reason he had hid his face and laughed.
"Sire," answered Graelent to the King, "sire, hearken to my words. In
all the world no man of your lineage does so shameful a deed as this.
You make your wife a show upon a stage. You force your lords to praise
her just with lies, saying that the sun does not shine upon her peer.
One man will tell the truth to your face, and say that very easily can
be found a fairer dame than she."
Right heavy was the King when he heard these words. He conjured
Graelent to tell him straightly if he knew a daintier dame.
"Yes, sire, and thirty times more gracious than the Queen."
The Queen was marvellously wrathful to hear this thing, and prayed her
husband of his grace to compel the knight to bring that woman to the
Court of whose beauty he made so proud a boast.
"Set us side by side, and let the choice be made between us. Should she
prove the fairer, let him go in peace; but if not, let justice be done
on him for his calumny and malice."
So the King bade his guards to lay hands on Graelent, swearing that
between them never should be love nor peace, nor should the knight
issue forth from prison, until he had brought before him her whose
beauty he had praised so much.
Graelent was held a captive. He repented him of his hasty words, and
begged the King to grant him respite. He feared to have lost his
friend, and sweated grievously with rage and mortification. But though
many of the King's house pitied him in his evil case, the long days
brought him no relief, until a full year went by, and once again the
King made a great banquet to his barons and his lieges. Then was
Graelent brought to hall, and put to liberty, on such terms that he
would return bringing with him her whose loveliness he had praised
before the King. Should she prove so desirable and dear as his boast,
then all would be well, for he had nought to fear. But if he returned
without his lady, then he must go to judgment, and his only hope would
be in the mercy of the King.
Graelent mounted his good horse and parted from the Court, sad and
wrathful. He sought his lodging, and inquired for his servant, but
might not find him. He called upon his friend, but the lady did not
heed his voice. Then Graelent gave way to despair, and preferred death
to life. He shut himself within his chamber, crying upon his dear one
for grace and mercy, but from her he got neither speech nor comfort.
So, seeing that his love had withdrawn herself from him by reason of
his grievous fault, he took no rest by night or day, and held his life
in utter despite. For a full year he lived in this piteous case, so
that it was marvellous to those about him that he might endure his
life.
On the day appointed, the sureties brought Graelent where the King was
set in hall with his lords. Then the King inquired of Graelent where
was now his friend.
"Sire," answered the knight, "she is not here, for in no wise might I
find her. Now do with me according to your will."
"Sir Graelent," said the King, "very foully have you spoken. You have
slandered the Queen, and given all my lords the lie. When you go from
my hands never will you do more mischief with your tongue."
Then the King spoke with a high voice to his barons.
"Lords, I pray and command you to give judgment in this matter. You
heard the blame that Graelent set upon me before all my Court. You know
the deep dishonour that he fastened on the Queen. How may such a
disloyal vassal deal honestly with his lord, for as the proverb tells,
'Hope not for friendship from the man who beats your dog!'"
The lords of the King's household went out from before him, and
gathered themselves together to consider their judgment. They kept
silence for a great space, for it was grievous to them to deal harshly
with so valiant a knight. Whilst they thus refrained from words a
certain page hastened unto them, and prayed them not to press the
matter, for (said he) "even now two young maidens, the freshest maids
in all the realm, seek the Court. Perchance they bring succour to the
good knight, and, so it be the will of God, may deliver him from
peril." So the lords waited right gladly, and presently they saw two
damsels come riding to the palace. Very young were these maidens, very
slender and gracious, and daintily cloaked in two fair mantles. So when
the pages had hastened to hold their stirrup and bridle, the maidens
dismounted from their palfreys, and entering within the hall came
straight before the King.
"Sire," said one of the two damsels, "hearken now to me. My lady
commands us to pray you to put back this cause for a while, nor to
deliver judgment therein, since she comes to plead with you for the
deliverance of this knight."
When the Queen heard this message she was filled with shame, and made
speed to get her from the hall. Hardly had she gone than there entered
two other damsels, whiter and more sweetly flushed even than their
fellows. These bade the King to wait for a little, since their mistress
was now at hand. So all men stared upon them, and praised their great
beauty, saying that if the maid was so fair, what then must be the
loveliness of the dame. When, therefore, the demoiselle came in her
turn, the King's household stood upon their feet to give her greeting.
Never did woman show so queenly to men's sight as did this lady riding
to the hall. Passing sweet she was to see, passing simple and gracious
of manner, with softer eyes and a daintier face than girl of mother
born. The whole Court marvelled at her beauty, for no spot or blemish
might be found in her body. She was richly dressed in a kirtle of
vermeil silk, broidered with gold, and her mantle was worth the spoil
of a king's castle. Her palfrey was of good race, and speedy; the
harness and trappings upon him were worth a thousand livres in minted
coin. All men pressed about her, praising her face and person, her
simplicity and queenlihead. She came at a slow pace before the King,
and dismounting from the palfrey, spoke very courteously in this
fashion--
"Sire," said she, "hearken to me, and you, lord barons, give heed to my
pleading. You know the words Graelent spake to the King, in the ears of
men, when the Queen made herself a show before the lords, saying that
often had he seen a fairer lady. Very hasty and foolish was his tongue,
since he provoked the King to anger. But at least he told the truth
when he said that there is no dame so comely but that very easily may
be found one more sweet than she. Look now boldly upon my face, and
judge you rightly in this quarrel between the Queen and me. So shall
Sir Graelent be acquitted of this blame."
Then gazing upon her, all the King's household, lord and lackey, prince
and page, cried with one voice that her favour was greater than that of
the Queen. The King himself gave judgment with his barons that this
thing was so; therefore was Sir Graelent acquitted of his blame, and
declared a free man.
When judgment was given the lady took her leave of the King, and
attended by her four damsels departed straightway from the hall upon
her palfrey. Sir Graelent caused his white horse to be saddled, and
mounting, followed hotly after her through the town. Day after day he
rode in her track, pleading for pity and pardon, but she gave him
neither good words nor bad in answer. So far they fared that at last
they came to the forest, and taking their way through a deep wood rode
to the bank of a fair, clear stream. The lady set her palfrey to the
river, but when she saw that Graelent also would enter therein she
cried to him--
"Stay, Graelent, the stream is deep, and it is death for you to
follow."
Graelent took no heed to her words, but forced his horse to enter the
river, so that speedily the waters closed above his head. Then the lady
seized his bridle, and with extreme toil brought horse and rider back
again to land.
"Graelent," said she, "you may not pass this river, however mightily
you pain yourself, therefore must you remain alone on this shore."
Again the lady set her palfrey to the river, but Graelent could not
suffer to see her go upon her way without him. Again he forced his
horse to enter the water; but the current was very swift and the stream
was very deep, so that presently Graelent was torn from his saddle, and
being borne away by the stream came very nigh to drown. When the four
maidens saw his piteous plight they cried aloud to their lady, and
said--
"Lady, for the love of God, take pity on your poor friend. See how he
drowns in this evil case. Alas, cursed be the day you spake soft words
in his ear, and gave him the grace of your love. Lady, look how the
current hurries him to his death. How may your heart suffer him to
drown whom you have held so close! Aid him, nor have the sin on your
soul that you endured to let the man who loved you die without your
help."
When the lady heard the complaint of her maidens, no longer could she
hide the pity she felt in her heart. In all haste she turned her
palfrey to the river, and entering the stream clutched her lover by the
belt. Thus they won together to the bank. There she stripped the
drowned man of his raiment, and wrapping him fast in her own dry mantle
cherished him so meetly that presently he came again to life. So she
brought him safely into her own land, and none has met Sir Graelent
since that day.
But the Breton folk still hold firmly that Graelent yet liveth with his
friend. His destrier, when he escaped him from the perilous river,
grieved greatly for his master's loss. He sought again the mighty
forest, yet never was at rest by night or day. No peace might he find,
but ever pawed he with his hoofs upon the ground, and neighed so loudly
that the noise went through all the country round about. Many a man
coveted so noble a steed, and sought to put bit and bridle in his
mouth, yet never might one set hands upon him, for he would not suffer
another master. So each year in its season, the forest was filled with
the cry and the trouble of this noble horse which might not find its
lord.
This adventure of the good steed and of the stout knight, who went to
the land of Faery with his love, was noised abroad throughout all
Brittany, and the Bretons made a lay thereof which was sung in the ears
of many people, and was called a Lay of the Death of Sir Graelent.
THE THREE THIEVES
This story tells that once upon a time there were three thieves faring
together, who had robbed many people, both church folk and lay. One of
these thieves was named Travers, but though he was in the company of
two robbers, yet he was not altogether such as they. They, indeed, were
thieves by descent as well as by choice, for their father was hanged
for his misdeeds. The one was called Haimet, and the other Barat, but
which was the more cunning workman at his trade it would be hard to
tell.
The three companions were passing one day through a high and leafy
wood, when Haimet spied a magpie's nest hidden within an oak. He went
beneath the tree, and his sharp eyes quickly perceived that the bird
was sitting upon her eggs. This thing he showed to Travers, and
afterwards to his brother.
"Friends," said he, "would not he be a good thief who might take these
eggs, and so softly descend the tree that the magpie knew nought
thereof?"
"There is no man in the world who can do such a feat," answered Barat.
"Certes, there is such a man," said Haimet, "and you shall see him at
his task, if you will only look at me."
Haimet set hands upon the oak, and climbed lightly up the great tree,
as one who had no fear to fall. He came to the nest, and parting the
straw softly from beneath, drew forth the eggs coyly and delicately.
Then he descended to the ground with a merry heart, and addressing
himself to his comrades, showed the eggs that he had stolen.
"Friends," said he, "here are the eggs, ready for boiling upon a fire!"
"Truly," said Barat, "no man's fingers are nimbler than yours, and if
you can only return the eggs to the nest, why I will own freely that
you are the most cunning thief of us all."
"Certes," answered Haimet, "they shall be set again beneath the bird,
and not a shell of them all shall be broken."
So he came again to the oak, and mounted swiftly into the tree, hand
over hand. Now he had gone but a little way when Barat hastened to the
tree, and climbed therein even more lightly and surely than his
brother. He followed him secretly from branch to branch, for Haimet was
intent upon his task, and gave no thought to those he had left below.
Then, whilst Haimet returned the eggs to the rifled nest, he stole the
very breeches from his legs, and forthwith descended to the ground.
When Travers saw this he was sick at heart, because he knew well he
might never do such feats as these. Presently Haimet came down to his
companions, and said--
"Friends, how seems it to you? Fingers like mine should pick up a good
living."
"I know not how it looks to me," answered Barat. "Your fingers are
quick enough, but your brains must be very dull, since they cannot
procure you even hosen for your legs."
"Yes, truly, I have hosen, and those altogether new, for it was but the
other day I laid hands upon the cloth, and they reach to my very
ankles."
"Are they so long as that?" said Barat; "shew them to us, and hide them
not away."
Then Haimet lifted his tunic and stared upon his legs, for he was
without breeches.
"Lord!" said he, "how can this have chanced? Where, then, are my
hosen?"
"I do not think that you have any, fair fellow," said Travers. "There
is no such thief as Barat, from here to Nevers, or so it seems to me.
Cunning indeed is the thief who can steal from a thief. But for my part
I am not meant for your trade, for I cannot spell even its A B C. A
hundred times should I be taken in my simplicity, where you would
escape by guile. I will return to my own village where I was married to
my wife. Mad must I have been to forsake it to become a thief. I am
neither fool nor idler, and know well how to toil in the fields, to
winnow and to reap. With the help of God I am yet strong enough to gain
my bread, so I go my way, and commend you to God His keeping."
So Travers parted from the company of the two thieves, and travelled by
hill and dale till he came at last to his own country. His comely wife,
Dame Maria, bore him no grudge for his absence, but welcomed his return
with much joy, as was her husband's due. He settled down amongst his
friends and acquaintance, and earned his living honestly and well. He
prospered greatly, so that he had enough and to spare, both of this and
of that. Now, towards Christmas, Travers killed a pig which he had
fattened all the year. He hung the bacon from a rafter of his house,
but better had he done, and much trouble would he have escaped, had he
sold it in the village, as you will see who read this story.
On a day when Travers was cutting fagots within a coppice, Haimet and
Barat, seeking what they might find, lighted on his house, and found
Dame Maria spinning at her wheel. Then said these rogues whose business
it was to cozen the simple--
"Dame, where is your husband?"
"Gentles," answered she, unknowing of these cheats, "he is in the wood,
gathering fagots for the fire."
"May God prosper his work," said they devoutly.
So they seated themselves, and looked about the house, high and low, at
larder and hearth-stone, in every nook and corner. Presently Barat,
raising his head, saw the side of bacon hanging from the rafters. He
drew the attention of Haimet to the meat, saying--
"Travers pains himself greatly to hide this bacon in his room. He fears
lest we should live a little at his cost, or taste his savoury meat.
Yet taste we will, if so we may."
Then they took their leave, and going a short distance, hid themselves
behind a hedge, where each set to work upon the sharpening of a stake.
When Travers returned to his home--
"Husband," said his wife, Dame Maria, "two men have sought you who
frightened me greatly, for I was alone in the house, and they would not
tell me their business. They were mean and shifty to look upon, and
there is not a thing in all the room that they have not taken stock
of--not the bacon, nor anything else--knife, reaping-hook, nor axe, for
their eyes were in every place at once."
"Well I know who they are and what they want of me," said Travers, "for
they have seen me often. We have lost our bacon, I promise you, since
Barat and Haimet have come to seek it for themselves. It is to no
purpose that we have cured it in the smoke, of that I am very sure. In
an evil hour I killed my pig, and certainly it were better to have sold
it last Saturday when I was able."
"Husband," answered the wife, "if you take the bacon down from the
ceiling, perchance these thieves may not find it when they come."
Therefore, because of the importunity of his wife, Travers mounted on a
stool and cut the cord, so that the bacon fell upon the floor. But not
knowing where to bestow the meat, they let it remain even where it had
fallen, having first covered it with the vessel in which they kneaded
their bread. Then, sad at heart, they went to bed to take what rest
they might.
When the night was come, those who were so desirous of the bacon came
to the house, and with their stakes made a hole in the wall near to the
threshold, a hole so large that you might have trundled a mill-stone
therein. Thereby they entered softly, and groped warily about the
house. Now Barat went from stool to table till he came beneath the
rafter from whence the bacon hung. He knew by touch that the cord was
severed, and he whispered in his brother's ear that he had not found
the meat, "But," said the thief, "Travers is a fool if he thinks to
conceal it for long."
Then they listened in the darkness of the room to the breathing of
those upon the bed.
Travers did not dare to sleep, and finding that his wife was becoming
drowsy, roused her, saying--
"Wife, this is no time for sleep. I shall go about the house to see
that all is fast."
"Do not leave me," answered his wife.
But Travers, who was a prudent man, rose from his bed to make sure of
all his goods. He came to the kneading trough, and raising it a little
from the ground, felt the bacon safely beneath. Then taking a great axe
in his hand he went out to visit his cow in her byre.
Barat came swiftly to the bed, like the bold and cunning thief he was.
"Marion," said he, "fair sister, I have a certain thing to ask you, but
dare not do so, for fear you think me mad."
"That I will never deem you, husband, by St. Paul; but I will counsel
you to the best of my power."
"I slept so soundly that I cannot remember where we bestowed the bacon
yester night, so bemused am I with dreams."
"God help you, husband, to find more seasonable jests; is it not hid
beneath the bin upon the floor?"
"In God's name, sister, you speak truly, and I will go to feel if it is
yet there."
Being desirous to keep his word, Barat lifted the trough and drew forth
the bacon. Then he rejoined Haimet, who was near by, and the two
thieves hastened towards the coppice, making much of each other because
of the success of their trick.
Now Travers returned to his bed, first carefully fastening his doors.
"Certes," said his wife, "dazed you must have been to ask me what had
become of our bacon."
"God help me," cried Travers, "when did I ask you this question?"
"Why, but now, husband."
"Sister, our bacon has walked off. Never shall we see it more, unless I
may steal it from these thieves. But they are the most cunning robbers
in all the land."
Travers went out forthwith in quest of the rogues who had carried off
his bacon. He took a short cut through a field of wheat, and following
the path very swiftly, presently found himself between the tricksters
and the wood. Haimet was very near to cover, but Barat went more
heavily, seeing that his load was right heavy. So Travers, being
anxious to take his own again, quickened his steps, and coming to him
said--
"Give it to me, for you are weary, seeing you have carried it so long a
road. Sit down now, and take a little rest."
Barat, thinking that he had met with Haimet, gladly placed the bacon on
the shoulders of Travers, and went his way. But Travers turned him back
to his own house, and hastened towards his home by the nearest path.
Now Barat, deeming that Haimet followed after, ran towards the wood
until he overtook his brother. When he knew him again he had great
fear, because he thought him behind. But when Haimet saw him stagger,
he cried out, "Let me bear the bacon for a while. I think it little
likely that I shall fall beneath its weight, as you are near to do.
Certainly you are overdone."
"God give me health," answered Barat, "for Travers has made a fool of
us. It is he who carries his bacon on his own shoulders. But the game
is not finished yet, and I have yet a throw to make."
Travers proceeded on his way in quietness and peace, as one who had
nought to fear from any man. But Barat, wet with haste, overtook him in
the end. He had taken off his shirt and wrapped it about his head like
a coif, and as much as he was able bore himself in the semblance of a
woman.
"Alas," cried he, "very nearly am I dead by reason of the loss and
mischief dealt me by these wicked men. God, what has become of my
husband, who has suffered so many things at their hands?"
Thinking that his wife was speaking to him, Travers held forth the
bacon.
"Sister," said he, "God is yet above the Devil. You see we have again
our own."
Then he, who never thought to lay hands upon the meat, seized upon it
greedily.
"Do not wait for me, husband, but get to bed as quickly as you can, for
now you may sleep without any fear."
So Travers returned to his own house, and Barat hastened to his
brother, bearing the bacon with him.
When Travers found his wife in tears--
"Certes, Mary," said he, "all this has come upon us by reason of our
sins. I thought to charge your shoulders with our bacon in the garden,
but now I know well that these rogues have bestowed it upon theirs.
Heavens, I wonder where he learned to play the part of a woman so
bravely in manner and in speech! Hard is the lesson I am set to learn
in school, because of a flitch of bacon. But, please God, I will find
them this night, yea, though I walk till I have no sole to my shoe, and
supplant them yet."
Travers took the path leading to the wood, and entering in the coppice,
saw the red blaze of a fire which these two thieves had litten. He
heard their voices lifted in dispute, so he concealed himself behind an
oak, and listened to their words. At the end Barat and Haimet agreed
that it were better to eat the bacon forthwith, lest a new cast of the
dice should go against them. Whilst they went to seek dry cones and
brushwood for the fire, Travers crept privily to the oak beneath which
it was burning. But the wood was damp and green, so that more smoke and
smother came from that fire than flame. Then Travers climbed into the
tree, and by the aid of bough and branch came at last to the place
where he would be. The two thieves returned presently with cones and
brambles. These they threw upon the fire in handfuls, saying that very
soon it would grill their bacon, and Travers hearkened to their speech.
He had stripped himself to his shirt, and hung from a limb of the oak
by his arm. Now, in a while, Haimet lifted his eyes to the tree, and
saw above him the hanged man, tall, grotesque and horrible to see,
naked in his very shirt.
"Barat," whispered he, "our father is spying upon us. Behold him
hanging from this branch in a very hideous fashion. Surely it is he
come back to us, is it not?"
"God help me," cried Barat, "it seems to me that he is about to fall."
Then because of their fear the two thieves fled from that place,
without leisure to eat, or to bear away, the bacon they had stolen.
When Travers marked their flight he tarried no longer in the oak, but
taking his bacon, returned straightway to his house, with none to give
him nay. His wife praised him to his face, saying--
"Husband, you are welcome home, for you have proven your worth. Never
did there live a braver man than you."
"Sister," said he, "take wood from the cellar, and make a fire.
Certainly we must eat our bacon, if we would call it our own."
Dame Maria lighted a fire with fagots upon the hearth; she put water in
the cauldron, and hung it on the hook above the fire. Travers for his
part carefully cut the bacon for which he had suffered so great
trouble, and put it in the pot till all was full. When this was done--
"Fair sister," said he, "watch by the fire, if you can keep awake. I
have not slept this night, and will rest a little on the bed. But I
will not take off my clothes, because I still am troubled of these
thieves."
"Husband," answered she, "ill luck go with them. Sleep soundly and in
peace, for there is none to do you wrong."
So Dame Maria kept vigil whilst Travers slept, for very greatly had he
need of rest.
During this time Barat lamented in the wood, for well he knew, when he
found the bacon gone, that Travers had played this trick upon them.
"Certes," said he, "we have lost the meat because of our fearful
hearts, and it belongs to Travers by right of courage. A good breakfast
he will make, for he deems that none can take it from him. He will look
upon us as dirt, if we leave it in his hands. Let us go to his house
and mark where he has bestowed it."
The two thieves hastened to the door of Travers' house. Barat set his
eye to a crevice therein, and saw a sight which gave him little joy,
for the pot was boiling upon the fire.
"Haimet," said he, "the bacon is cooking, and much I grieve that there
is none for us."
"Let it boil in peace till it is fit for eating," answered Haimet. "I
shall not give Travers quittance in this matter till he has paid me
wages for my toil."
Haimet sought a long stake which he cut from a hazel tree, and
sharpened it with his knife. Then he climbed upon the roof of the
house, and uncovered a little space above the spot where the cauldron
boiled upon the fire. Through this opening he could see the wife of
Travers sound asleep, for she was weary of her vigil, and nodded over
the hearth. Haimet lowered the rod, which he had sharpened like a dart,
and struck it in the pot so adroitly that he drew forth a portion of
the bacon from out the cauldron. This he raised cunningly to the roof,
and had great joy of his fishing. Then awoke Travers from his sleep,
and saw this thing, and marked the thief, who was both malicious and
strong.
"Gossip, upon my roof," said he, "it is not reasonable of you to strip
the covering from over my head. In this manner we shall never come to
an end. Climb down; let us give and take. Let each of us have his share
of the bacon."
So Haimet descended from the roof, and the bacon was taken from the
cauldron. Dame Maria divided the meat into three portions, for the
thieves had no care to let Travers part the lots. The two brothers took
two portions, and Travers one; but his was not the best, for all that
he had nourished the pig.
For this reason was the proverb made, oh, gentles, that "Bad is the
company of thieves."
THE FRIENDSHIP OF AMIS AND AMILE
In the days of Pepin, King of the Franks, a boy was born in the Castle
of Bericain to a father of Allemaigne, of noble descent and of great
holiness. His father and mother, who had no other child, vowed to God
and to St. Peter and St. Paul that if God vouchsafed him breath he
should be carried to Rome for his baptism. At the same hour a vision
was seen of the Count of Alverne--whose wife was near her day--in which
he saw the Apostle of Rome, who baptized many children in his palace,
and confirmed them with the anointing of holy oil. When the Count awoke
from his sleep he inquired of the wise men of those parts what this
thing might mean. Then a certain wise old man, having heard his words,
by the counsel of God made answer, and said--
"Rejoice greatly, Count, for a son shall now be born to thee great in
courage and in virtue, and thou shalt carry him to Rome, so that he may
be baptized by the Apostle."
So the Count rejoiced in his heart, and he and his people praised the
counsel of that ancient man.
The child was born, and cherished dearly, and when he was of the age of
two years his father prepared to carry him to Rome, according to his
purpose. On his way he came to the city of Lucca, and there fell in
with a certain nobleman of Allemaigne who was on pilgrimage to Rome,
that there he might baptize his son. Each greeted the other, and
inquired of his name and business; and when they knew they were in the
like case, and bound on the same errand, they took each other as
companion with a kind heart, and voyaged together to Rome. The two
children, also, loved so dearly, that one would not eat save the other
ate with him; so that they fed from the same dish, and lay in the one
bed. In such manner as this the fathers carried the boys before the
Apostle at Rome, and said to him--
"Holy Father, whom we believe and know to be seated in the chair of St.
Peter the Apostle, we, the Count of Alverne, and the Chatelain of
Castle Bericain, humbly pray your Holiness that you would deign to
baptize the sons they have carried here from a distant land, and to
accept this humble offering from their hands."
Then the Pope made answer--
"It is very meet to come with such a gift before me, but of such I have
no need. Give it, therefore, to the poor, who cry for alms. Right
willingly will I baptize the children, and may the Father, the Son and
the Holy Ghost ever fold them close in the love of the Holy Trinity."
So at that one time the Apostle baptized the two children in St.
Saviour's Church, and he gave to the son of the Count of Alverne the
name of Amile, and to the son of the Chatelain of Castle Bericain gave
he the name of Amis. Many a knight of Rome held them at the font, and
answered in their name as god-parents, according to the will of God.
Then, when the Sacrament of Baptism was at an end, the Apostle
commanded to be brought two wooden cups, fair with gold and set with
costly stones, of one workmanship, size and fashion, and these he
handed to the children, saying--
"Take this gift in witness that I have baptized you in St. Saviour's
Church."
So the knights received the cups with great joy, and rendered him
grace for his gift, and parting from thence repaired each to his own
home in all comfort and solace.
To the child of the Knight of Bericain God also gave a gift, the gift
of such wise understanding that men might almost believe that he was
another Solomon.
When Amis was of the age of thirty years a fever seized upon his
father, and he began to admonish his son in words such as these--
"Fair, dear son, my end is near at hand, but thou shalt tarry for a
season, and be thine own lord. Firstly, fair son, observe the
commandments of God, and be of the chivalry of Jesus Christ. Keep faith
with thy overlords, and turn not thy back on thy companions and thy
friends. Defend the widow and the orphan; be pitiful to the captive and
to all in need; think every day upon that day which shall be thy last.
Forsake not the society and friendship of the son of the Count of
Alverne, for the Apostle of Rome baptized you together on one day, and
graced you with one gift. Are you not alike in all things--in beauty,
in comeliness, and in strength, so that whosoever sees you, thinks you
to be sons of one mother?"
Having spoken these words, he was houselled of the priest, and died in
our Lord; and his son gave him fitting burial, and paid him all such
service as is meetly required for the dead.
After the death of his father divers evil persons wrought Amis much
mischief, because of the envy they felt towards him; but nevertheless
he bore them no ill will, and patiently suffered all the wrong and
malice that they did. Let me tell you, then, without more words, that
such was his case that he and his servants were cast forth from the
heritage of his fathers, and driven from the gate of his own keep. But
when he had called to mind the words of his father, he said to those
who journeyed with him in the way--
"The wicked have spoiled me wrongfully of my inheritance, yet have I
good hope that the Lord is on my side. Come now, let us seek the Court
of Count Amile, my comrade and my friend. Peradventure he will give us
of his goods and lands; but if not, then will we gather to Hildegarde,
the Queen, wife of King Charles of France, the stay and support of the
disinherited."
So those of his company made answer that they would follow where he
led, and would serve him as his men. They rode, therefore, to the court
of the Count, but might not find him, for reason that he had passed to
Bericain to comfort Amis, his companion, because of the death of his
father. When Amile might not find Amis, he departed from the castle,
greatly vexed, and resolved within himself that he would not solace
himself in his own fief until he had met with Amis, his friend.
Therefore he rode on this quest through France and Allemaigne, seeking
news of him from all his kindred, but finding none.
Now Amis, together with his company, for his part sought diligently for
Amile his friend, until it chanced that on a day a certain lord gave
him harbourage, and at his bidding Amis told him of this adventure.
Then said the nobleman--
"Dwell ye with me, sir knights, and I will give my daughter to your
lord, because of the wisdom men report of him, and you, for your own
part, shall be made rich in silver, in gold and in lands."
They rejoiced greatly at his word, and the wedding feast was celebrated
with marvellous joy. But when they had tarried in that place for one
year and six months, Amis called together his ten companions and spake
to them.
"We are recreant, inasmuch as we have forgotten all this while to seek
for Amile."
So he left two men-at-arms, together with his precious cup, and set
forth towards Paris.
Now for the space of nearly two years Amile had sought for Amis without
pause or rest. Drawing near to Paris he lighted upon a pilgrim and
asked of him if perchance he knew aught of Amis, whom evil men had
hunted from his lands. The palmer said "Nay," wherefore Amile divested
himself of his cloak, and gave it to the pilgrim, saying--
"Pray thou to our Lord and His saints for me that they give me grace to
meet Amis, my friend."
So he saluted the pilgrim, and went his way to Paris, seeking in every
place for news of Amis his friend, and finding none. But the pilgrim,
passing swiftly upon his road, came upon Amis about the hour of
vespers, and they saluted each the other. Then Amis inquired of the
palmer whether he had seen or heard, in any land or realm, aught of
Amile, the son of the Count of Alverne.
"What manner of man art thou," answered the palmer all astonied, "that
thou makest mock of a pilgrim? Thou seemest to me that very Amile who
but this morn sought of me if I had seen Amis, his friend. I know not
for what reason thou hast changed thine apparel, thy company, thy
horses and thy arms, nor why thou askest of me the same question thou
didst require at nine hours of the morn when thou gavest me this
cloak."
"Be not angry with me," said Amis, "for I am not the man you deem; but
I am Amis who searches for his friend Amile."
So he gave him money from his pouch, and prayed him that he would
require of our Lord that He might grant him grace to find Amile.
"Hasten quickly to Paris," said the pilgrim, "and there shalt thou
find him whom so fondly thou seekest."
So Amis hastened instantly to the city.
It chanced upon the morrow that Amile departed from Paris, and took his
ease within a daisied meadow near by the pleasant waters of the Seine.
Whilst he ate there with his knights there came that way Amis with his
men-at-arms. So Amile and his company armed themselves forthwith, and
rode forth before them at adventure. Then Amis said to his companions--
"Behold these French knights who seek to do us a mischief. Stand
stoutly together, and so shall we defend our lives. If we but escape
this peril soon shall we be within the walls of Paris, and sweetly
shall we be entreated at the palace of the King."
Then drew the two companies together with loosened rein, with lance in
rest, and with brandished sword, in such fashion that it seemed as if
none might escape alive from the fury of that onset. But God, the all
powerful, Who knoweth all, and bringeth to a good end the travail of
the just, suffered not that spears should meet in that encounter. So
when they were near at hand Amis cried aloud--
"Who are you, knights, that are so eager to slay Amis the Banished and
his companions?"
When Amile heard these words he knew well the voice of Amis, his
comrade, so he answered him--
"Oh thou, Amis, most dear, sweet as rest to my labour, know me for
Amile, son of the Count of Alverne, who have not given over my quest
for thee these two whole years."
Then forthwith they lighted from their steeds, and clasped and kissed
each the other, giving grace to God Who granteth the treasure to the
seeker. Moreover, upon the guard of Amile's sword, wherein was set a
holy relic, they swore faith, and friendship, and fellowship to death,
the one with the other. So set they forth from that place, riding
together to the Court of Charles, the King of France. There they moved
amongst the lords, young, discreet and wise, fair to see, shapen
wondrously alike in form and face, beloved of all men and held of all
in honour. There, too, the King received them with much courtesy,
making of Amis his treasurer, and to Amile gave he the office of
seneschal.
In this fashion they tarried long with the King, but at the end of
three years Amis said to Amile--
"Fair, sweet companion, I desire greatly to see my wife, whom I have
left so long. Stay thou at Court, and for my part I will return so soon
as I may. But have thou no dealings with the daughter of the King, and,
more than all, beware and keep thee from the malice of Arderay the
felon knight."
"I will observe thy bidding," answered Amile, "but make no long
tarrying from my side."
On these words Amis departed from the Court; but Amile for his part saw
with his eyes that the daughter of the King was fair, and knew the
princess, in love, as soon as he was able. Thus the commandment and the
warning of Amis, his companion, passed quickly from his mind; yet think
not too hardly of the young man, forasmuch that he was not more holy
than David, nor wiser than Solomon, David's son.
Whilst Amile was busied with these matters there came to him Arderay,
the traitor, full of envy, and said--
"Thou dost not know, comrade, thou dost not know that Amis has stolen
gold from the King's treasury, and therefore hath he taken flight.
Since things are thus I require that you swear to me fealty of
friendship and of brotherhood, and I will swear to you the like oath on
the holy Gospels."
Having pledged such troth as this, Amile feared not to betray his
secret to the felon knight. Now when Amile bore bason and ewer to the
King, that he might wash his hands, then said that false Arderay to his
lord--
"Take no water from the hands of this recreant, Sir King, for he is
worthier of death than of life, since he has plucked from the Queen's
daughter the flower of her maidenhood."
When Amile heard this thing he was so fearful that he fell upon the
floor, and answered not a word, so that the courteous King raised him
to his feet, and said--
"Have no fear, Amile, but stand up and acquit thee of this blame."
Then Amile stood upon his feet and said--
"Sir King, give no ready credence to the lies of this traitor Arderay,
for well I know that you are an upright judge, turning neither for love
nor hate out of the narrow way. Grant me, therefore, time for counsel
with my friends, so that I may purge myself of this charge before you,
and in single combat with Arderay, the traitor, prove him to be a liar
before all your Court."
The King gave to both champions till three hours after noon that each
might take counsel with his friends, and bade that at such time they
should stand before him to fulfil their devoir. At the appointed hour
they came before the King. With Arderay for friend and witness came
Herbert the Count; but Amile found none to stand at his side, save only
Hildegarde, the Queen. So sweetly did the lady plead his cause that she
prevailed upon her lord to grant Amile such further respite for counsel
that he might seek Amis, his friend; yet nevertheless only on such
covenant that if Amile returned not on the appointed day the lady
should be banished ever from the royal bed.
Whilst Amile was on his way to take counsel with his friend, he chanced
on Amis, his comrade, who repaired to the Court of the King. So he
alighted from his steed, and kneeling at the feet of his companion,
said--
"Oh thou, my one hope of surety, I have not obeyed the charge you laid
upon me, and am truly blamed by reason of my dealings with the daughter
of the King. Therefore must I endure ordeal of battle with the false
Arderay."
"Let us leave here our companions," returned Amis, sighing, "and enter
in this wood to make the matter clear."
Then Amis, having heard, reproached Amile, and said--
"Let us now exchange our garments and our horses, and thou, for thy
part, get thee gone to my house, whilst I ride to do judgment by combat
for thee upon this traitor."
But Amile answered him--
"How then may I go about thine house, seeing that I know not thy wife
nor thy household, nor ever have looked upon their face?"
And Amis replied--
"Very easily mayest thou do this thing, so thou dost but walk
prudently; but take thou good heed to have no dealings with my wife."
Thereupon the two companions departed one from the other, with tears;
Amis riding to the Court of the King in the guise of Amile, and Amile
to the house of his comrade in the guise of Amis. Now the wife of Amis,
seeing him draw near, hastened to embrace him whom she thought was her
lord, and would have kissed him. But Amile said--
"Is this a time for play? I have matter for tears rather than for
claspings, for since I parted from thee have I suffered many bitter
griefs, yea, and yet must suffer."
And that night as they made ready to lie together in one bed, Amile set
his naked sword between the twain, and said to his brother's wife--
"Beware lest thy body draw near in any wise to mine, for then will I
slay thee with this sword."
In such fashion passed the night, and every night, until Amis repaired
secretly to the castle to know certainly whether Amile kept faith and
word in this matter of his wife.
The day appointed for the combat now was come, and the Queen awaited
Amile, sick of heart; for Arderay, that traitor, cried aloud, that
certainly ought she never to come near the King's bed, since she had
suffered and consented to Amile's dealings with her maid. Whilst
Arderay boasted thus, Amis entered within the Court of the King at the
hour of noon, clad in the apparel of his comrade, and said--
"Right debonair and Lord Justicier of this realm, here stand I to seek
ordeal of battle with this false Arderay, because of the blame he has
laid upon me, the Queen, and the Princess, her child."
Then answered the King right courteously--
"Be stout of heart, oh Count, for if you prove Arderay to be false I
will give thee my daughter Belisant to wife."
On the morning of the morrow Arderay and Amis rode into the lists,
armed from plume to heel, in the presence of the King and of much
people. But the Queen with a great company of maidens and widows and
dames went from church to church, giving gifts of money and of torches,
and praying God for the safety of the champion of her daughter. Now
Amis considered in his heart that should he slay Arderay he would be
guilty of his blood before the eyes of God, and if he were overthrown
then would it be a shame to him for all his days. So he spake in such
manner as this to Arderay.
"Foul counsel hast thou followed, Sir Count, so ardently to seek my
death, and to thrust this life of thine into grievous peril of hurt. So
thou wilt withdraw the reproach thou hast fastened upon me, and avoid
this mortal strife, thou canst have of me friendship and loyal
service."
But Arderay was right wroth at these words, and replied--
"No care have I for friendship or service of thine; rather will I swear
to the truth as that truth is, and smite thy head from thy shoulders."
Then Arderay swore that his foe had done wrong to the daughter of the
King, and Amis made oath that he lied. Thereupon, incontinent they
drove together, and with mighty strokes strove one against the other
from the hour of tierce till it was nones. And at nones Arderay fell
within the lists; and Amis struck off his head.
The King lamented that Arderay was dead, but rejoiced that his daughter
was proved clean from stain. He gave the Princess to Amis for dame, and
with her, as dowry, a mighty sum in gold and silver, and a city near by
the sea where they might dwell. So Amis rejoiced greatly in his bride;
and returned as quickly as he might to the castle where he had hidden
Amile, his companion. When Amile saw him hastening homewards with many
horsemen, he was sore adread that Amis was overthrown, and made ready
to escape. But Amis sent messages to him that he should return in all
surety, since he had avenged him upon Arderay, and thus, by proxy, was
he married to the daughter of the King. So Amile repaired from that
place, and dwelt with his dame in that city which was her heritage.
Now Amis abode with his wife, but by the permission of God he became a
leper, and his sickness was so heavy upon him that he could not leave
his bed, for whom God loveth him He chasteneth. His wife--who was named
Obias--for this cause hated him sorely, and sought his death many a
time in shameful fashion. When Amis perceived her malice he called to
him two of his men-at-arms, Azonem and Horatus, and said to them--
"Deliver me from the hands of this wicked woman, and take with you my
cup secretly, and bear us to the tower of Bericain."
When they drew near to the castle men came out before them asking of
the sickness and of the man whom they carried there. Then they answered
that this was Amis, their lord, who was a leper, for which cause they
prayed them to show him some pity. But mercilessly they beat the
sergeants, and tumbled Amis forth from the litter in which he was
borne, crying--
"Flee swiftly from hence, if ye care aught for your lives."
Then Amis wept grievously, and said--
"Oh Thou, God most pitiful and compassionate, grant me to die, or give
me help in this my extremity."
Again he said to the men-at-arms--
"Carry me now to the church of the Father of Rome; perchance God of His
loving kindness will there give alms to the beggar."
When they were come to Rome, Constantine the Apostle, full of pity and
of sanctity, together with many a knight of those who had held Amis at
the font, came before him and supplied the wants of Amis and his
servants. But after three years a great famine came upon the city--a
famine so grievous that the father put his very offspring from the
door. Then Azonem and Horatus spake to Amis--
"Fair sir, bear witness how loyally we have served you from the death
of your father, even to this day, and that never have we done against
your bidding. But now we dare no longer to bide with you, since we have
no heart to die of hunger. For this cause we pray you to acquit us of
our service, so that we may avoid this mortal pestilence."
Then answered Amis in his tears--
"Oh, my dear children, not servants but sons, my only comfort, I pray
you for the love of God that you forsake me not here, but that you bear
me to the city of my comrade, Count Amile."
And these, willing to obey his commandment, carried him to that place
where Amile lay. Now when they came before the court of Amile's house
they began to sound their clappers, as the leper is wont to do; so when
Amile heard the sound thereof he bade a servitor of his to carry to the
sick man bread and meat, and the cup which was given to him at Rome
brimmed with rich wine. When the man-at-arms had done the bidding of
his lord, he came to him again, and said--
"Sir, by the faith which is your due, if I held not your cup within my
hand, I should believe it to be the cup that the sick man beareth even
now, for they are alike in workmanship and height."
And Amile said to him--
"Go quickly, and bring him hither to me."
When the leper was come before his comrade, Amile inquired of him who
he was, and how he came to own such a cup.
"I am of Castle Bericain," said he, "and the cup was given me by the
Apostle of Rome who baptized me."
When Amile heard these words he knew within himself that this was Amis,
his comrade, who had delivered him from death, and given him the
daughter of the King of France as dame. So at once he fell upon his
neck, and began to weep and lament his evil case, kissing and
embracing him. When his wife heard this thing she ran forth with fallen
hair, weeping and making great sorrow, for she bore in mind that this
was he who had done judgment on Arderay. Forthwith they set him in a
very fair bed, and said to him--
"Tarry with us, fair sir, until the will of God is done on you, for all
that we have is as thine own."
So he dwelt with them, he and his two men-at-arms likewise.
Now on a night when Amis and Amile lay together in a chamber, without
other company, God sent Raphael, His angel, to Amis, who spake him
thus--
"Amis, sleepest thou?"
And he, deeming that Amile had called him, answered--
"I sleep not, fair dear companion."
And the angel said to him--
"Thou hast well spoken, for thou art the companion of the citizens of
Heaven, and like Job and Tobit hast suffered all things meekly and with
patience. I am Raphael, an angel of our Lord, who am come to show thee
medicine for thy healing, for God hath heard thy prayers. Thou must bid
Amile, thy comrade, to slay his two children with the sword, and wash
thee in their blood, that thus thy body may become clean."
Then Amis replied--
"This be far from me, that my comrade be blood-guilty for my health."
But the angel said--
"It is meet that he should do this thing."
On these words the angel departed from him.
Now Amile also, in his sleep, had heard these words, and he awoke, and
said--
"Comrade, who is this who hath spoken to thee?"
And Amis answered that no man had spoken. "But I prayed our Lord, as is
my wont."
But Amile said--
"It is not thus, but some one hath spoken with thee."
Then he rose from the bed, and went to the door of the chamber, and
finding it fast, said--
"Tell me, fair brother, who hath said to thee these hidden words."
Then Amis began to weep bitterly, and denied not that it was Raphael,
the angel of our Lord, who had said to him, "Amis, our Lord sends word
to thee that thou biddest Amile to slay his two children with the
sword, and to wash thee in their blood, that thou mayest be clean of
thy leprosy."
And Amile was sorely distressed on hearing these words, and said--
"Amis, gladly have I given thee sergeant and damsel and all the riches
that I had, and in fraud thou feignest that the angel hath bidden me to
slay my two little ones with the sword."
Then Amis broke out into weeping, and said--
"I know that I have told thee of a grievous matter, but not of mine own
free will; I pray thee therefore that thou cast me not forth from thy
house."
And Amile answered him that the covenant he had made with him he would
not depart from till the hour of death. "But I adjure thee by the faith
between me and thee, and by our fellowship, and by the baptism given to
us twain at Rome, that thou tell me truly whether it was man or angel
who spoke to thee of this thing."
And Amis made reply--
"So truly as the angel hath held converse with me this night, so may
God make me clean of my infirmity."
Then Amile began to weep privily, and to consider within his heart. "If
this man was willing to die in my stead before the King, why then
should I not slay mine own for him! He hath kept faith with me even
unto death: shall I not therefore keep faith with him! Abraham was
saved by faith, and by faith have the saints proved mightier than
kings. Yea, God saith in the Gospel, 'Whatsoever ye would that men
should do unto you, even so do unto them.'"
Then Amile delayed no more, but went to his wife's chamber, and bade
her to attend the Divine Office; so the Countess sought the church, as
was her wont to do, and the Count took his sword and went to the bed
where lay the children, and they were asleep. And bending above them he
wept bitterly, and said--
"Hath any man heard of such father who was willing to slay his child?
Alas, alas, my children, no longer shall I be your father, but your
cruel murderer."
The children awoke because of their father's tears which fell upon
them, and looking upon his face began to laugh. Since therefore they
were about the age of three years he said to them--
"Your laughter will turn to tears, for now your innocent blood shall be
shed."
He spoke thus, and cut off their heads; and making straight their limbs
upon the bed, he set their heads to their bodies, and covered all with
the coverlet, as if they slept. So he washed his companion with the
blood of that slaying, and said--
"Lord God, Jesus Christ, Who hast bidden men to keep faith on earth,
and didst cleanse the leper with Thy word, deign Thou to make clean my
comrade, for love of whom I have shed the blood of my children."
Straightway was Amis made whole of his leprosy, and they gave grace to
our Lord with great joy, saying--
"Blessed be God, the Father of our Lord Jesus Christ, who saveth those
who put their trust in Him."
And Amile clad his comrade from his own rich apparel; and passing to
the church to render thanks in that place, the bells rang without
ringers, as was the will of God. When the people of the city heard
thereof they hastened to behold this marvel. Now the wife of the Count,
when she saw the twain walking together, began to question which was
her husband, and said, "Well I know the vesture which they wear, but
which is Amile, that I know not," and the Count said--
"I am Amile, and this, my companion, is Amis, who is healed."
Then the Countess marvelled greatly, and said--
"Easy is it to see that he is healed, but much desire I to know the
manner of that healing."
"Render thanks to our Lord," returned the Count, "nor seek curiously of
the fashion of that cleansing."
The hour of tierce was now come, and neither of the parents had yet
entered in the chamber where the children lay, but the father went
heavily for reason of their death. The Countess asked therefore for her
sons that they might share in the joy, but the Count replied--
"Nay, dame, but let the children sleep."
Then entering by himself within the chamber to bewail his children, he
found them playing in the bed and about their necks, in the place of
that mortal wound, showed as it were a crimson thread. So he clasped
them in his arms, and bore them to their mother, saying--
"Dame, rejoice greatly, for thy sons whom I had slain with the sword,
at the bidding of the angel, are alive, and by their blood is Amis
cleansed and healed."
When the Countess heard this thing she said--
"Count, why was I not with thee to gather the blood of my children,
that I too might have washed Amis, thy comrade and my lord?"
And the Count answered her--
"Dame, let be these words; rather let us dedicate ourselves to our
Lord, who hath wrought such marvels in our house."
So from that day, even unto their deaths, they lived together in
perfect chastity; and for the space of ten days the people of that city
held high festival. But on that very day that Amis was made clean, the
devil seized upon his wife, and breaking her neck, carried off her
soul.
After these things Amis rode to the castle of Bericain, and laid siege
thereto, and sat before it for so long a time that those within the
castle yielded themselves into his hand. He received them graciously,
forgetting his anger against them, and forgiving them the wrongs that
they had done, so that from thenceforth he dwelt peaceably amongst
them, and with him, in his own house, lived the elder son of Count
Amile. There he served our Lord with all his heart.
Now Adrian, being at this time Pope of Rome, sent letters to Charles,
King of France, praying him to come to his aid against Didier, King of
the Lombards, who wrought much mischief to him and the Church. Now
Charles lay in the town of Thionville, and to that place came Peter,
the envoy of the Apostle, with messages from the Pope praying him to
hasten to the succour of Holy Church. For this cause Charles sent
letters to the said Didier requiring him to render to the Holy Father
the cities and all other things which he had wrongfully seized, and
promising that if he would do this thing the said Charles would send
him in return the sum of forty thousand pieces of gold, in gold and
silver. But he would not do right, neither for prayers nor for gifts.
Then the stout King Charles summoned to his aid all his men--bishops,
abbots, dukes, princes, marquises, and other stout knights. Divers of
these he sent to Cluses to guard the pass, and of this number was
Albin, Bishop of Angers, a man of great holiness.
King Charles himself, with a large company of spears, drew towards
Cluses by the way of Mont Cenis, and he sent Bernard, his uncle, with
other knights, thither by way of Mont Saint-Bernard. The vanguard of
the host said that Didier, with all his strength, lay at Cluses, which
town he had made strong with iron chains and works of stone. Whilst
Charles approached to Cluses he sent messengers to Didier, requiring
him to render to the Holy Father the cities which he had taken, but he
would not heed his prayer. Again Charles sent him other letters
demanding three children of the Justices of Lombardy as hostages, until
such time as he had yielded up the cities of the Church; in which case
for his part he would return to France with all his spears, without
battle and without malice. But neither for this nor for that would he
stint.
When God the All-powerful had beheld the hard heart and the malice of
this Didier, and found that the French desired greatly to return, He
put so fearful a trembling in the hearts of the Lombards that they took
to flight, though there was none that pursued, leaving behind them
their tents and all their harness. So Charles and his host followed
after them, and Frenchman, German, Englishman and divers other people
entered hot after them into Lombardy.
Amis and Amile were of the host, and very near to the person of the
King. Always they strove to follow our Lord in good works, and were
constant in fast, in vigil, in giving of alms, in succouring the widow
and the orphan, in assuaging often the wrath of the King, in patient
suffering of evil men, and in piteous dealings within the Roman realm.
But though Charles had a great army drawn together in Lombardy, King
Didier feared not to come before him with his little host--for there
where Didier had a priest, Charles had a bishop; where one had a monk,
the other had an abbot; if this had a knight, that had a prince; if
Didier had a man-at-arms, then Charles had a duke or a count. What
shall I tell you; for a single knight on the one side Charles could
number thirty pennons. And the two hosts fell each upon the other with
a tumult of battle cries, and with banners in array; and the stones and
arrows flew from here and there, and knights were smitten down on every
side.
For the space of three days the Lombards strove so valiantly that they
slew a very great company of Charles's men. But on the third day
Charles set in order the hardiest and bravest of his host and said to
them--
"Go now, and win this battle, or return no more."
So King Didier together with the host of the Lombards fled to the place
called Mortara, which was then known as Belle-Foret, because the
country was so fair, there to refresh themselves and their horses. On
the morning of the next day King Charles with his army drew near the
town, and found the Lombards arrayed for the battle. So fierce was the
combat that a great multitude of men were slain, both of one party and
the other, and for reason of this slaying was the place named Mortara.
There, too, on that field died Amis and Amile, for as it had pleased
God to make their lives lovely and pleasant together, so in their
deaths they were not divided. There also many another hardy knight was
slain with the sword. But Didier, together with his Justiciary, and all
the multitude of the Lombards, fled to Pavia; and King Charles followed
closely after him and lay before the city, and invested it on every
side; and lying there he sent to France to seek the Queen and his
children. But St. Albin, the Bishop of Angers, and many another bishop
and abbot counselled the King and Queen that they should bury those who
fell in that battle, and build in that place a church. This counsel
greatly pleased the King, so that on the field were built two churches,
one by bidding of Charles in honour of St. Eusebius of Verceil, and the
other by bidding of the Queen in honour of St. Peter.
Moreover the King caused to be brought the two coffins of stone wherein
were buried Amis and Amile, and Amile was carried to the church of St.
Peter, and Amis to the church of St. Eusebius. But on the morrow the
body of Amile in his coffin of stone was found in the church of St.
Eusebius near by the coffin of his comrade, Amis. So have you heard the
story of this marvellous fellowship which could not be dissevered, even
by death. This miracle did God for His servants--that God Who gave such
power to His disciples that in His strength they might move even
mountains. Because of this wonder the King and Queen tarried there for
thirty days, giving fit burial to the bodies of the slain, and
honouring those ministers with many rich gifts.
But all this while the host of Charles toiled mightily for the taking
of the city before which it lay. Our Lord also tormented those within
the walls so grievously that they might not bear their harness by
reason of weakness and of death. At the end of ten months Charles took
Didier the King, and all those who were with him, and possessed himself
of the city and of all that realm. So Didier the King and his wife were
led as captives into France.
But St. Albin, who in his day gave life to the dead and light to the
blind, ordained clerks, and priests and deacons in the aforesaid church
of St. Eusebius, and bade them always to hold in tireless keeping the
bodies of those two comrades, Amis and Amile, who suffered death under
Didier, King of Lombardy, the 12th day of October, and are now with our
Lord Jesus Christ, Who liveth and reigneth with the Father and the Holy
Ghost, world without end. Amen.
OF THE KNIGHT WHO PRAYED WHILST OUR LADY TOURNEYED IN HIS STEAD
Sweet Jesus, what brave warfare doth he make, and how nobly doth he
joust, whose feet devoutly seek the church where the Divine Office is
rendered, and who assists at the holy mysteries of Him, the spotless
Son of the Mother Maid. For this cause will I tell you a certain story,
even as it was told to me, for a fair ensample.
There was once a knight, esteemed of all as a wise and courteous lord,
stout and of great valour, who dearly loved and honoured the Virgin
Mary. The fame of this knight was bruited about all chivalry; so to
make proof alike of lisping squire and burly man-at-arms, he set forth
to a tourney, together with a strong company. Now by the will of God it
chanced that when the day of the tournament was come he fared speedily
towards the field, because he would be first at the breaking of the
spears. Near by the road was builded a little church, and the bells
thereof rang loud and clear to call men to the singing of the holy
Mass. So without doubt or hesitation this knight dismounted at the
door, and entered within the church to hearken to the service of God.
At an altar therein a priest chanted meetly and with reverence a Mass
of the holy Virgin Mary. Then another Mass was begun, the good knight
yet kneeling devoutly on his knees, and praying our Lady with an
earnest heart. When the second Mass came to its appointed end,
straightway a third Mass was commenced, forthwith and in the selfsame
place.
"Sir, by the holy Body of God," said the squire to his lord, "the hour
to tourney hurries by. Why tarry you here? Depart from hence, I pray
you. Let us keep to our own trade, lest men deem you hermit or
hypocrite, or monk without the cowl!"
"Friend," answered the knight, "most worshipfully doth he tourney who
hearkens to the service of God upon his knees. When the Masses are
altogether at an end, we will go upon our way. Till then, please God,
part from here will I not. But so that all are said, then will I joust
to the very utmost of my might, according to the will of God."
With these words the knight refrained from further speech, and turning
himself again towards the altar took refuge in the holy liturgy, till
the last prayer came to a close with the last chant. Then they got to
horse, as was their bounden duty, and rode with speed towards that
place where the lists were set for the great play. So, presently, the
knights who were returning from the tournament, discomfited and
overborne, met him who had carried off all the prizes of the game. They
saluted the knight who was on his way from the Divine Offerings, and,
joining themselves to his company, praised him to his face, affirming
that never before had knight done such feats of arms as he had wrought
that day, to his undying fame. Moreover many amongst them drew near and
yielded themselves his captives, saying--
"We are your prisoners, for truly we may not deny that you have
overthrown us in the field."
Then, taking thought, the knight was amazed no more, for quickly he
perceived that She had been upon his business in the press, about whose
business he had been within the chapel.
So he called these knights and his fellowship around him, and said
right courteously--
"I pray you, one and all, to hearken to my words, for I have that to
tell you which never has been heard of ears."
Then he told over to them, word for word, how that he had not jousted
in the tournament, neither had broken lance nor hung shield about his
neck, by reason of those Masses he had heard, but verily he believed
that the Maiden, whom humbly he had besought within the chapel, had
worn his harness in the lists. "Altogether lovely in my eyes is this
tournament wherein She has done my devoir; but very foully shall I
requite such gracious service if I seek another Lady, or in my folly
return to the vanities of the world. Therefore I pledge my word to God
in truth, that henceforth I will never fight, save in that tourney
where He sits, the one true Judge, Who knows the loyal knight, and
recompenses him according to his deeds."
Then he bade them farewell right piteously, and many of his company
wept tenderly as they took their leave. But he, parting from them, went
his way to an abbey, to become the servant of the Handmaid of the Lord,
and to follow in that path which leadeth to a holy end.
So, clearly we may perceive from this ensample, that the gracious God,
in Whom we put our faith, loves, cherishes, and delights to honour that
man who gladly tarries before His holy altar at the offering of the
Mass, and who willingly serves His Mother, so gentle and so dear. Of
much profit is this custom, and he who is quiet in the land and wise,
will always continue to walk in the way his feet were set in youth,
yea, even to that time when he is old and grey-headed.
THE PRIEST AND THE MULBERRIES
A certain priest having need to go to market, caused his mare to be
saddled and brought to his door. The mare had carried her master for
two years, and was high and well nourished, for during these years
never had she known thirst nor hunger, but of hay and of oats ever had
she enough and to spare. The priest climbed to the saddle and set out
upon his journey, and well I remember that it was the month of
September, for in that season mulberries grow upon the bushes in great
plenty and abundance. The priest rode upon his way repeating his hours,
his matins and his vigils. As he drew near the gate of the town the
path ran through a certain deep hollow, and raising his eyes from his
book the priest marked a bush thick with mulberries, bigger, blacker
and more ripe than any he had ever seen. Desire entered his heart, for
very covetous was he of this fair fruit, and gradually checking the
pace of his mare, he presently caused her to stand beside the bush. Yet
one thing still was wanting to his delight. The mulberries near the
ground were set about with spines and thorns, whilst the sweetest of
all hung so high upon the tree that in no wise could he reach them from
his seat. This thing the priest saw, so in a while he climbed up, and
stood with his two feet upon the saddle, whence by leaning over a
little he could pluck the fruit. Then he chose the fairest, the ripest,
and the sweetest of all these mulberries, eating them as swiftly and
greedily as he might, whilst the mare beneath him moved never a whit.
Now, when this priest had eaten as many mulberries as he was able, he
glanced downwards, and saw that the mare was standing still and coy,
with her head turned towards the bank of that deep road. Thereat the
priest rejoiced very greatly, for his two feet were yet upon the
saddle, and the mare was very tall.
"God!" said he, "if any one now should cry 'Gee up!'" He thought and
spoke the words at the same moment, whereat the mare was suddenly
frighted, and springing forward on the instant tumbled the luckless
priest into the bush where the thorns and briars grew sharpest and
thickest. There he lay in that uneasy bed, nor might move from one side
to the other, backwards or forwards, for all the money in the mint.
The mare galloped straight to her own stable, but when the priest's
household saw her return in this fashion they were greatly
discomforted. The servants cursed her for an evil and a luckless jade,
whilst the cook maid swooned like any dame, for well she believed that
her master was dead. When they were returned a little to themselves
they ran to and fro, here and there, about the country searching for
the priest, and presently on their way to the market town they drew
near to that bush where their master yet lay in much misease. On
hearing their words bewailing his piteous case, the priest raised a
lamentable voice, and cried--
"Diva, Diva, do not pass me by. This bush is an uneasy bed, and here I
lie very hurt and troubled and utterly cast down. Do you not see how my
blood is staining these thorns and briars a vermeil red?"
The servants hurried to the bush, and stared upon the priest.
"Sir," said they, "who has flung you herein?"
"Alas," answered he, "'tis sin that has undone me. This morning when I
rode this way reading in my Book of Hours, I desired over greatly to
eat of the mulberries growing hereon, and so I fell into the sin of
gluttony. Therefore this bush gat hold upon me. But help me forth from
this place, for I wish now for no other thing but to have a surgeon for
my hurts, and to rest in my own house."
Now by this little story we may learn that the prudent man does not cry
aloud all he may think in his heart, since by so doing many an one has
suffered loss and shame, as we may see by this fable of the Priest and
the Mulberries.
THE STORY OF ASENATH
In the first of the seven years of great plenty Pharaoh sent forth
Joseph to lay up corn, and gather food within the cities. So Joseph
went out over all the land of Egypt, and came in the country of
Heliopolis, where lived Poti-pherah, the priest, and chief counsellor
of the great King. His daughter, Asenath, was the fairest of all the
virgins of the earth; and seemed rather to be a daughter of Israel than
an Egyptian. But Asenath was scornful and proud, and a despiser of men.
No man of all the sons of men had seen her with his eyes, for she
lodged within a strong tower, tall and wide, near by the habitation of
Poti-pherah, the priest. Now high upon this tower were ten chambers.
The first chamber was fair and great, and was builded of marble blocks
of divers colours; the walls were of precious stones set in a chasing
of gold, and the ceiling thereof was golden. There stood the gods of
the Egyptians in metal of silver and gold, and Asenath bowed before
them and offered sacrifice, every day of all the days. The second
chamber was the habitation of Asenath, and was adorned cunningly with
ornaments of gold and silver, with costly gems, and with arras and
stuffs most precious. In the third chamber was brought together the
wealth of all the world, and in that place also were set the aumbries
of Asenath. Seven virgins, her fellows, lodged in the seven other
chambers. They were very fair, and no man had spoken with them, nor any
male child.
The chamber of Asenath was pierced with three windows; the first,
which was very wide, looked towards the east, the second looked towards
the south, and the third was set towards the north. Here was spread a
couch of gold, covered with a purple coverlet, embroidered with golden
thread, and hemmed with jacinths. There slept Asenath, with no
bed-fellow, neither had man sat ever upon her bed. About this house was
a goodly garden, closed round with a very strong wall, and entered by
four iron gates. Each door had for warders eighteen men, very mighty
and young, well armed and full of valour. At the right side of the
garden sprang a fountain of living water, and near by the fountain a
cistern which gave of this water to all the trees of the garden, and
these trees bore much fruit. And Asenath was queenly as Sarah, gracious
as Rebecca, and fair as Rachel.
_How Joseph rebuked Asenath because she worshipped idols._
Joseph sent a message to Poti-pherah that he would come to his house.
So Poti-pherah rejoiced greatly, saying to his daughter, "Joseph, the
friend of God, enters herein. I would give thee to him as his wife."
But Asenath was sore vexed when she heard these words, and said--
"No captive shall ever be my husband, but only the son of a king."
Whilst they spake thus together, a messenger came before them and
cried, "Joseph is here"; so Asenath fled to her chamber high within the
tower. Now Joseph was seated in Pharaoh's own chariot of beaten gold,
and it was drawn by four horses, white as snow, with bridles and
harness of gold. Joseph was clad in a vesture of fine linen, white and
glistering, and his mantle was of purple, spun with gold. He wore a
golden circlet upon his head, and in this crown were set twelve stones,
most precious, each stone having for ornament a golden star. Moreover
he held in his hand the royal sceptre, and an olive branch charged with
fruit. Poti-pherah and his wife hastened to meet him, and bowed before
him to the ground. They led him within the garden, and caused the doors
to be shut. But when Asenath regarded Joseph from on high the tower,
she repented her of the words she spoke concerning him, and said--
"Behold the sun and the chariot of the sun! Certainly this Joseph is
the child of God; for what father could beget so fair an offspring, and
what womb of woman could carry such light."
Joseph entered in the house of Poti-pherah, and whilst they washed his
feet he asked what woman had looked forth from the window of the tower.
"Let her go forth from the house," he commanded.
This he said because he feared lest she should desire him, and should
send him messages and divers gifts, even as other women of her nation,
whom he had refused with holy indignation. But Poti-pherah replied--
"Sire, this is my daughter, who is a virgin, and hateth men; neither
hath she seen any man save me, her father, and thyself this very day.
If thou wilt, she shall come before thee and salute thee."
Then Joseph thought within himself, "Since she hateth man, she will not
cast her eyes upon me." So he answered to her father--
"Since your daughter is a virgin, I will cherish her even as my
sister."
Then her mother went out to seek Asenath, and brought her before
Joseph.
"Salute thy brother," said Poti-pherah, "who hateth the strange woman,
even as thou hatest man."
"God keep thee," replied Asenath, "for thou art blessed of God most
high."
And Joseph answered, "May the God of life bless thee evermore."
Then commanded Poti-pherah that she should kiss Joseph; but as she drew
near Joseph set his hand against her breast and said--
"It is not meet that a man who worships the living God, and eateth the
bread of life and drinketh from the chalice without corruption, should
embrace the strange woman, who bows down before deaf and dumb idols;
who serves them with the kisses of her mouth; is anointed with their
reprobate oil, and eats an accursed bread, and drinks unsanctified wine
from their table."
_Of the penitence of Asenath, and of the consolation of an
angel; how he came from Heaven to the chamber of Asenath,
and spake with her and sweetly comforted her._
When Asenath heard Joseph speak these words she was sore vexed, even
unto tears; wherefore Joseph took pity upon her and blessed her, laying
his hand upon her head. Asenath rejoiced greatly at the benediction.
She sought her bed, sick with fear and joy, and renounced the gods
before whom she bowed, and humbled herself to the ground. So Joseph ate
and drank, and when he rose to go Poti-pherah prayed him to tarry till
the morrow; but he might not, and parted, having promised to return
within eight days.
Then Asenath put on sad raiment, such as she wore at the death of her
brother, and went clothed in a garment of heaviness. She closed the
doors of her chamber upon her and wept. Moreover she flung forth all
her idols by the window set towards the north; all the royal meat she
gave to the dogs; she put dust upon her head, lay upon the ground, and
lamented bitterly for seven days.
But the eighth morning, at the hour when the cock crows and the dogs
howl at the breaking of the day, Asenath looked forth from the window
giving to the east, and saw a star shining clear, and the heavens open,
and there appeared a great light. She fell to earth with her face in
the dust, and a man descended from the heavens and stood by her head,
calling on her by her name. But Asenath answered nothing, because of
the greatness of her fear. Then the man called her a second time,
saying, "Asenath! Asenath!" and she replied--
"Lord, here am I. Tell me whom thou art."
And he said--
"I am Prince of the House of God and Captain of His Host. Rise, stand
upon thy feet, for I have to speak with thee."
Then Asenath raised her head, and saw a man by her side who in all
points was, as it were, Joseph. He was clad in a white stole, and bore
the royal sceptre in his hand, and a crown was upon his brow. His face
was as the lightning, his eyes as rays of the sun, and the hair of his
head like a flame of fire. At the sight of him Asenath was sore afraid,
and hid her face upon the ground. But the Angel raised her to her feet,
and comforted her, saying--
"Put off this black raiment with which thou art clothed, and this
girdle of sadness. Remove the sackcloth from thy body, and the dust
from thine head; cleanse thy face and thy hands with living water, and
adorn thee with fair apparel, for I have somewhat to say to thee."
So she adorned herself with speed, and when she came to him again he
said--
"Asenath, take off this ornament from thine head, for thou art virgin.
Rejoice, and be of good cheer, for thy name is written in the Book of
Life, and shall never be taken away. Thou art born again this very day
and quickened anew. For thou shalt receive the Bread of Blessing, and
drink of the Wine without corruption; and be anointed with the Holy
Chrism. Yea, I have given thee for wife to Joseph, and thou no more
shall be called Asenath, but a name shall be given thee of fair refuge,
for thy Penitance hath come before the High King, of whom she is the
daughter, and thou shalt ever live before Him in mirth and gladness."
Then inquired she of the Angel his name, but he answered--
"My name is written by the finger of God in the Book of the most high
King, but all that is written therein may not be told, neither is it
proper for the hearing of mortal man."
_Of the table and of the honey that Asenath set before the
Angel, and how the Angel blessed Asenath._
But Asenath caught the angel by his mantle, and said--
"If I have found favour in thine eyes, sit for a little space upon this
bed, where never man has sat, and I will spread the table before my
lord."
And the Angel replied, "Do quickly."
So Asenath set a fair linen cloth upon the table, and put thereon new
bread of a sweet savour. Then said the Angel--
"Give me also a little honey in the honeycomb."
So Asenath was grievously troubled because she had no honey to set
before her guest. But the Angel comforted her, saying--
"Look within thine aumbrey, and thou shalt find withal to furnish thy
table."
Then she hastened thereto, and found a store of virgin honey, white as
snow, of sweetest savour. So she spake to the Angel--
"Sire, I had no honey, but thou spakest the word, and it is there, and
the perfume thereof is as the breath of thy mouth."
The Angel smiled at the understanding of Asenath, and placed his hand
upon her head, and said--
"Blessed be thou, O Asenath, because thou hast forsaken thy idols, and
believed in our living Lord. Yea, blessed are they whom Penitence
bringeth before Him, for they shall eat of this honey gathered by the
bees of Paradise from the dew of the roses of Heaven; and those who eat
thereof shall never see death, but shall live for evermore."
Then the Angel stretched forth his hand and took of the honeycomb and
break it; and he ate a little, and gave the rest to the mouth of
Asenath, saying--
"This day hast thou eaten of the Bread of Life, and art anointed with
the Holy Chrism. Beauty is given thee for ashes; for virtue shall never
go from thee, neither shall thy youth wither, nor thy fairness fail;
but thou shalt be as the strong city builded as a refuge for the
children of our Lord, Who is King for ever more."
Then the Angel touched the honeycomb, and it became unbroken as before.
Again he stretched forth his hand, and with his finger signed the cross
thereon, and there where his finger touched came forth blood. So he
spake to Asenath, and said--
"Behold this honey!"
Whilst she gazed thereon, she saw bees come forth from that honey, some
white as snow, others vermeil as jacinths, and they gathered about her,
and set virgin honey in the palm of her hand; and she ate thereof, and
the Angel with her.
"Bees," said the Angel, "return now to your own place."
So they passed through that window which gave upon the east, and took
their way to Paradise.
"Faithful as these bees are the words which I have spoken."
Then the Angel put forth his hand three times, and touched the honey,
and fire came forth and consumed the honey without singeing the table,
and the perfume which came from the honey and the fire was very sweet.
_Of the blessing of the seven maidens, and of the marriage of Asenath,
as set forth in the story._
Asenath said to the Angel--
"Lord, I have with me seven virgins, born in one night, and nourished
with me from my childhood until now. I will seek them, and thou shalt
bless them, even as thou hast blessed me."
So she brought them before him, and he blessed them, saying--
"May the most high God bless you, and make you to be seven strong
columns of the City of Refuge."
Afterwards he bade Asenath to carry forth the table, and whilst she
went about her task, the Angel vanished from her eyes. But looking
towards the east she saw, as it were, a chariot drawn by four horses
ascending towards Heaven. So Asenath prayed to God right humbly that He
would pardon the boldness with which she had spoken to the Captain of
His Host.
Whilst she prayed thus a messenger came to Poti-pherah saying that
Joseph, the friend of God, sought his house, and was even then at his
door. Asenath hastened to meet him, and awaited his coming before the
offices of the house. When Joseph entered the garden she bowed herself
before him, and washed the dust from his feet, telling him the words
which the Angel had spoken concerning her. The next day Joseph prayed
Pharaoh that he might have Asenath to wife, and Pharaoh gave him the
woman. He set also garlands of gold upon their heads, the fairest that
cunning smiths could fashion, and caused them to embrace in the sight
of men. So for seven days was kept high feast and festival, nor might
any man labour for those days. He also gave them new names, calling
Joseph, the Son of God, and Asenath, Daughter of the Most High King.
Before the time of the seven lean years Asenath bore two sons. And
Joseph called the name of the firstborn Manasseh, which is to say
Forgetfulness; "For," said he, "God hath made me to forget all my toil,
and all my father's house." And the name of the second was called
Ephraim, which is to say Fruitfulness; "For," said he, "God hath caused
me to be fruitful in the land of my affliction."
THE PALFREY
That men may bear in mind the fair deeds that woman has done, and to
tell of her sweetness and frankness, this tale is here written. For
very right it is that men should hold in remembrance the excellent
virtues that can so easily be perceived in her. But grievous is it, and
very heavy to me, that all the world does not laud and praise women to
the height which is their due. Ah, God, if but they kept their hearts
whole and unspotted, true and strong, the world would not contain so
rich a treasure. The greater pity and sorrow, then, that they take not
more heed to their ways, and that so little stay and stability are to
be found in them. Too often the heart of a woman seems but a
weathercock upon a steeple, whirled about in every wind that blows; so
variable is woman's heart, and more changeable than any wind. But the
story that I have taken upon me to narrate shall not remain untold
because of the fickle-hearted, nor for reason of those who grudge
praise to the frank and pure; therefore, give ear to this Lay of the
Marvellous Palfrey.
Once upon a time a certain knight, courteous and chivalrous, rich of
heart, but poor in substance, had his dwelling in the county of
Champagne. So stout of heart was this lord, so wise in counsel, and so
compact of honour and all high qualities, that had his fortune been
equal to his deserts he would have had no peer amongst his fellows. He
was the very pattern of the fair and perfect knight, and his praise
was ever in the mouth of men. In whatever land he came he was valued at
his proper worth, since strangers esteemed him for the good that was
told of him, and rumour but increased his renown. When he had laced the
helmet on his head, and ridden within the lists, he did not court the
glances of the dames, nor seek to joust with those who were of less
fame than he, but there where the press was thickest he strove mightily
in the heart of the stour. In the very depths of winter he rode upon
his horse, attired in seemly fashion (since in dress may be perceived
the inclinations of the heart) and this although his substance was but
small. For the lands of this knight brought him of wealth but two
hundred pounds of rent, and for this reason he rode to tourneys in hope
of gain as well as in quest of honour.
This knight had set all his earthly hope and thoughts on gaining the
love of a certain noble lady. The father of the damsel was a puissant
Prince, lacking nought in the matter of wealth, and lord of a great
house furnished richly as his coffers. His fief and domain were fully
worth one thousand pounds a year, and many an one asked of him his fair
daughter in marriage, because her exceeding beauty was parcel of the
loveliness of the world. The Prince was old and frail; he had no other
child than the maiden, and his wife had long been dead. His castle was
builded in a deep wood, and all about it stretched the great forest,
for in the days of my tale Champagne was a wilder country then than
now.
The gentle knight who had set his heart on the love of the fair lady
was named Messire William, and he lived within the forest in an ancient
manor some two miles from the palace of the Prince. In their love they
were as one, and ever they fondly dreamed one upon the other; but the
Prince liked the matter but little, and had no mind that they should
meet. So when the knight would gaze upon the face of his mistress, he
went secretly by a path that he had worn through the profound forest,
and which was known of none save him. By this path he rode privily on
his palfrey, without gossip or noise, to visit the maiden, many a time.
Yet never might these lovers see each other close, however great was
their desire, for the wall of the courtyard was very high, and the
damsel was not so hardy as to issue forth by the postern. So for their
solace they spoke together through a little gap in the wall, but ever
between them was the deep and perilous fosse, set thickly about with
hedges of thorn and spine, so that never closer might they meet. The
castle of the Prince was builded upon a high place, and was strongly
held with battlement and tower; moreover bridge and portcullis kept his
door. The ancient knight, worn by years and warfare, seldom left his
lodging, for he might no longer get him to horse. He lived within his
own house, and ever would have his daughter seated at his side, to
cheer his lonely age with youth. Often this thing was grievous to her,
for she failed to come to that fair spot where her heart had taken
root. But the brave knight in nowise forgot the road that he had worn,
and asked for nothing more than to see her somewhat closer with his
eyes.
Now the tale tells that in spite of his poverty the knight owned one
thing that was marvellously rich. The palfrey on which he rode had not
his like in all the world. It was grey and of a wonderful fair colour,
so that no flower was so bright in semblance, nor did any man know of
so beautiful a steed. Be assured that not in any kingdom could be
discovered so speedy a horse, nor one that carried his rider so softly
and so surely. The knight loved his palfrey very dearly, and I tell you
truly that in nowise would he part with him for any manner of wealth,
though the rich folk of that country, and even from afar, had coveted
him for long. Upon this fair palfrey Messire William went often to his
lady, along the beaten path through the solitary forest, known but to
these two alone. Right careful was he to keep this matter from the
father of the demoiselle; and thus, though these two lovers had such
desire one of the other, they might not clasp their arms about the
neck, nor kiss, nor embrace, nor for their solace, even, hold each
other by the hand. Nought could they do but speak, and hearken softly
to such sweet words, for well they knew that should the old Prince know
thereof, very swiftly would he marry his daughter to some rich lord.
Now the knight considered these things within himself, and day by day
called to remembrance the wretched life that was his, for he might not
put the matter from his mind. So at the end he summoned all his
courage, and for weal or woe resolved that he would go to the aged
Prince and require of him his daughter for his wife, let that betide
what may. For very clear it was to him that such a life he could not
lead for every day of the week. On a certain day he made himself ready,
and repaired to the castle where the demoiselle dwelt with her father.
He was welcomed very gladly by the Prince and his company, for he was
esteemed a courteous and gentle knight, and bragged of by all men as a
valiant gentleman, who was lacking in no good qualities.
"Sire," said the knight, "I ask you of your grace to listen to my
words. I enter in your house to crave of you such a gift as may God put
it in your heart to bestow."
The old man looked upon him fixedly, and afterwards inquired--
"What is it you would have? Tell me now, for by my faith I will aid
you if I may, yet in all things saving my honour."
"Yea, sire, very easily you may do this thing, if so you please. May
God but grant that such is your pleasure."
"I will grant you the gift if it seems to me well, and I will refuse
you the boon if it seems to me ill. Nothing will you get from me,
neither gift, nor promise, nor blame, that it is not fitting for me to
bestow."
"Sire," answered he, "I will tell you the gift I crave at your hand.
You know who I am, and right well you knew my father. Well, too, you
know my manor and my household, and all those things wherein I take my
pleasure and delight. In guerdon of my love, I pray--so it may please
you, sire--your daughter as my wife. God grant that my prayer may not
disturb your heart, and that my petition may not be refused to my
shame. For I will not hide from you that although I am not of her
fellowship, yet have I spoken from afar with my demoiselle, and
perceived those fair virtues which all men praise. Greatly is my lady
loved and esteemed in these parts, for truly there is not her like in
all the world. I have been too rash, since I have dared to require so
gracious a gift, but so you deign to give to my asking, joyous and
merry shall I go for all my days. Now have I told you my petition; so
answer me according to my hope and your good pleasure."
The old man had no need for counsel in this matter, so without delay he
made answer to the knight--
"I have heard with patience what you had to tell. Certes, and without
doubt, my daughter is fair, and fresh, and pure, and a maiden of high
descent. For myself, I am a rich vavasour, and come of noble ancestry,
having fief and land worth fully one thousand pounds each year in
rent. Think you I am so besotted as to give my daughter to a knight who
lives by play! I have no child but one, who is close and dear to my
heart, and after I am gone all my wealth will be hers. She shall wed no
naked man, but in her own degree; for I know not any prince of this
realm, from here even to Lorraine, however wise and brave, of whom she
would not be more than worthy. It is not yet a month agone since such a
lord as this prayed her at my hand. His lands were worth five hundred
pounds in rent, and right willingly would he have yielded them to me,
had I but hearkened to his suit. But my daughter can well afford to
wait, for I am so rich that she may not easily lose her price, nor miss
the sacrament of marriage. Too high is this fruit for your seeking, for
there is none in this realm, nor from here to Allemaigne, however high
his race, who shall have her, save he be count or king."
The knight was all abashed at these proud words. He did not wait for
further shame, but took his leave, and went as speedily as he might.
But he knew not what to do, for Love, his guide, afflicted him very
grievously, and bitterly he complained him thereof. When the maiden
heard of this refusal, and was told the despiteful words her father had
spoken, she was grieved in her very heart, for her love was no girl's
light fancy, but was wholly given to the knight, far more than any one
can tell. So when the knight--yet heavy and wrathful--came to the
accustomed trysting place to speak a little to the maiden, each said to
the other what was in the mind. There he opened out to her the news of
his access to her father, and of the disaccord between the twain.
"Sweet my demoiselle," said the knight, "what is there to do? It seems
better to me to quit my home, and to dwell henceforth amongst
strangers in a far land, for my last hope is gone. I may never be
yours, neither know I how these things will end. Cursed be the great
wealth with which your father is so puffed up. Better had it been that
you were not so rich a dame, for he would have looked upon my poverty
with kinder eyes if his substance were not so great."
"Certes," answered she, "very gladly would I be no heiress, but only
simple maid, if all things were according to my will. Sire, if my
father took heed only to your good qualities, by my faith he would not
pain himself to prevent your coming to me. If he but weighed your
little riches in the balance against your great prowess, right soon
would he conclude the bargain. But his heart cannot be moved: he does
not wish what I would have, nor lament because I may wring my hands. If
he accorded with my desire, right speedily would this matter be ended.
But age and youth walk not easily together, for in the heart is the
difference between the old and young. Yet so you do according to my
device, you shall not fail to gain what you would have."
"Yea, demoiselle, by my faith, I will not fail herein; so tell me now
your will."
"I have determined on a thing to which I have given thought many a time
and oft. Very surely you remember that you have an uncle who is right
rich in houses and in goods. He is not less rich than my father; he has
neither child, wife nor brother, nor any kindred of his blood nearer
than you. Well is it known that all his wealth is yours when he is
dead, and this in treasure and in rent is worth sixty marks of virgin
gold. Now go to him without delay, for he is old and frail; tell him
that between my father and yourself is such a business that it may not
come to a good end unless he help therein. But that if he would promise
you three hundred pounds of his land, and come to require grace of my
father, very soon can the affair be ended. For my father loves him
dearly, and each counts the other an honourable man. Your uncle holds
my father as prudent and wise: they are two ancient gentlemen, of ripe
years, and have faith and affiance the one in the other. Now if for
love of you your uncle would fairly seek my father and speak him thus,
'I will deliver to my nephew three hundred pounds of my lands, so that
you give him your child,' why, the marriage will be made. I verily
believe that my father would grant your uncle his request, if only he
would ask me of him. And when we are wedded together, then you can
render again to your uncle all the land that he has granted you. And so
sweetly do I desire your love, that right pleasing I shall find the
bargain."
"Fairest," cried the knight, "verily and truly there is nothing I crave
in comparison with your love; so forthwith I will find my uncle, and
tell him this thing."
The knight bade farewell, and went his way, yet thoughtful and
bewildered and sad, by reason of the shame which had been put upon him.
He rode at adventure through the thick forest upon his grey palfrey.
But as he rode fear left him, and peace entered in his heart, because
of the honest and wise counsel given him by the fair maiden. He came
without hindrance to Medet, where his uncle had his dwelling, but when
he was entered into the house he bewailed his lot, and showed himself
all discomforted. So his uncle took him apart into a privy chamber, and
there he opened out his heart, and made plain to him all this business.
"Uncle," said he, "if you will do so much as to speak to her sire, and
tell him that you have granted me three hundred pounds of your land, I
will make this covenant with you, and plight you my faith, my hand in
yours, that when I have wedded her who is now denied me, that I will
render again and give you quittance for your land. Now I pray that you
will do what is required of you."
"Nephew," answered the uncle, "this I will do willingly, since it
pleases me right well. By my head, married you shall be, and to the
pearl of all the country, for good hope have I to bring this matter to
an end."
"Uncle," said the knight, "put your hand to my task, and so press on
with the business that time may go swiftly to the wedding. For my part
I will arm me richly, and ride to the tournament at Galardon, where, by
the aid of God, I trust to gain such ransom as will be helpful to me.
And I pray you to use such diligence that I may be married on my
return."
"Fair nephew, right gladly," answered he, "for greatly it delights me
that so gracious and tender a lady shall be your bride."
So without further tarrying Messire William went his way, merry of
heart because of his uncle's promise that without let he should have as
wife that maid whom so dearly he desired. For of other happiness he
took no heed. Thus blithe and gay of visage he rendered him to the
tournament, as one who had no care in all the world.
On the morrow, very early in the morning, the uncle got to horse, and
before the hour of prime came to the rich mansion of that old Prince,
and of her whose beauty had no peer. He was welcomed with high
observance, for the ancient lord loved him very dearly, seeing that
they were both of the same years, and were rich and puissant princes,
near neighbours in that land. Therefore he rejoiced greatly that one so
high in station did honour to his house, and spread before him a fair
banquet, with many sweet words, for the old Prince was frank and
courteous of heart, and knew to praise meetly where honour was due.
When the tables were cleared, the two spake together of old faces and
old stories, shields, and swords and spears, and of many a doughty
deed, in the most loving fashion. But the uncle of the good knight
would not forget his secret thought, and presently discovered it to the
Prince in saying--
"What go I now to tell you? I love you very truly, as you may easily
perceive. I am come to require a favour at your hand. May God put it
into your heart to lend your ear to my prayer in such a fashion that
the matter may be brought to a right fair end."
"By my head," answered the old Prince, "you are so near to my heart
that you are not likely to be refused aught that you may ask of me.
Tell me, that I may grace you with the gift."
"Sire, thanks and thanks again, for I would do the same by you,"
returned the uncle of the knight, who no longer cared to hide his privy
mind. "I am come to pray of you, fair sire, the hand of your virtuous
maid in marriage. When we once were wed I would endow her with my
wealth to the utmost of my power. You know well that I have no heir of
my body, which troubles me sorely; and I will keep good faith with you
herein, for I am he who loves you dearly. When your daughter is
bestowed upon me, it would not be my care to separate father and child,
nor to withdraw my wealth from yours, but all our substance should be
as one, and we would enjoy together in common that which God has given
us."
When he whose heart was crafty heard these words, he rejoiced greatly,
and made reply--
"Sire, I will give her to you right gladly, for you are a loyal and an
honourable man. I am more content that you have required her of me than
if the strongest castle of these parts had been rendered to my hand.
To none other in the world would I grant my maid so willingly as to
you, for you are prudent and hardy, and many a time have I proved ere
now that I may have confidence in your faith."
Then was promised and betrothed the damsel to a husband of whom she had
little envy, for she was persuaded that another had asked her as his
wife. When the maiden knew the truth thereof she was altogether amazed
and sorrowful, and often she swore by St. Mary that never should she be
wedded of him. Right heavy was she, and full of tears, and grievously
she lamented her fate.
"Alas, unhappy wretch, for now I am dead. What foul treason has this
old traitor done, for which he justly should be slain! How shamefully
he has deceived that brave and courteous knight, whose honour is
untouched by spot. By his wealth this aged, ancient man has bought me
at a price. May God requite it to his bosom, for he purposes to commit
a great folly, since the day we are wed he takes his mortal foe to
wife. How may I endure that day! Alas, may God grant that I shall never
see that hour, for too great is the anguish that I suffer because of
this treason. If I were not fast in prison, right swiftly would I get
quit of this trouble, but nought is there for me to do, since in no
wise can I flee from this manor. So stay I must, and suffer as my
father wills, but truly my pain is more than I can bear. Ah, God, what
will become of me, and when shall he return who so foully is betrayed.
If he but knew the trick his uncle has set on him, and how, too, I am
taken in the snare, well I know that he would die of grief. Ah, if he
but knew! Sure I am that he would ride with speed, and that soon these
great woes would be as they had never been. Too sorely is my heart
charged with sorrow, and better I love to die than to live. Alas, that
this old man ever should cast his thought upon me, but none may deliver
me now, for my father loves him because of his wealth. Fie on age! Fie
upon riches! Never may bachelor wed with loving maid save he have money
in his pouch. Cursed be the wealth which keeps me from him wherein I
have my part, for truly my feet are caught in a golden net."
In this wise the maiden bewailed her lot, by reason of her great
misease. For so sweetly was her heart knit in the love of her fair
bachelor, that in nowise might she withdraw her thoughts from him.
Therefore she held in the more despite him to whom her father had given
her. Old he was, very aged, with a wrinkled face, and red and peering
eyes. From Chalons to Beauvais there was no more ancient knight than
he, nor from there to Sens a lord more rich, for that matter. But all
the world held him as pitiless and felon; whilst so beautiful and brave
was the lady, that men knew no fairer heiress, nor so courteous and
simple a maiden, no, not within the Crown of France. How diverse were
these twain. On one side was light, and on the other darkness; but
there was no spot in the brightness, and no ray within the dark. But
the less grief had been hers had she not set her love on so perilous a
choice.
Now he to whom the damsel was betrothed, because of his exceeding
content, made haste to appoint some near day for the wedding. For he
knew little that she was as one distraught by reason of the great love
she bore his nephew, as you have heard tell. So her father made all
things ready, very richly, and when the third day was come he sent
letters to the greybeards, and to those he deemed the wisest of that
land, bidding them to the marriage of his daughter, who had bestowed
her heart elsewhere. Since he was well known to all the country round,
a great company of his friends came together to the number of thirty,
to do honour to his house, since not one of them but owed him service
for his lands. Then it was accorded between them that the demoiselle
should be wedded early on the morrow, and her maidens were bidden to
prepare their lady for the wedding on the appointed day and hour. But
very wrathful and troubled in heart were the maidens by reason of this
thing.
The Prince inquired of the damsels if his daughter was fitly arrayed
against her marriage, and had content therein, or was in need of aught
that it became her state to have.
"Nothing she needs, fair sire," made answer one of her maidens, "so far
as we can see; at least so that we have palfreys and saddles enow to
carry us to the church, for of kinsfolk and of cousins are a many near
this house."
"Do not concern yourself with the palfreys," replied the Prince, "for I
trow we shall have to spare. There is not a lord bidden to the wedding
whom I have not asked to lend us from his stables."
Then, making no further tarrying, he returned to his own lodging, with
peace and confidence in his heart.
Messire William, that brave and prudent knight, had little thought that
this marriage was drawing so near its term. But Love held him so fast
that he made haste to return, for ever the remembrance of her face was
before his eyes. Since love flowered so sweetly within his heart, he
parted from the tournament in much content, for he deemed that he rode
to receive the gift he desired beyond all the world. Such he hoped was
the will of God, and such the end of the adventure. Therefore he
awaited in his manor, with what patience he might, the fair and
pleasant tidings his uncle must presently send him, to hasten to the
spousal of his bride. Since he had borne off all the prizes of the
tourney, he bade a minstrel to his hall, and sang joyously to the
playing of the viol. Yet, though all was revelry and merriment, often
he looked towards the door to see one enter therein with news. Much he
marvelled when the hour would bring these welcome words, and often he
forgot to mark the newest refrains of the minstrel, because his
thoughts were otherwhere. At the time hope was growing sick a varlet
came into the courtyard. When Messire William saw him the heart in his
breast leaped and fluttered for joy.
"Sire," said the varlet, "God save you. My lord, your friend, whom well
you know, has sent me to you in his need. You have a fair palfrey, than
which none goes more softly in the world. My lord prays and requires of
you that for love of him you will lend him this palfrey, and send it by
my hand forthwith."
"Friend," answered the knight, "for what business?"
"Sire, to carry his lady daughter to the church, who is so dainty-sweet
and fair."
"For what purpose rides she to church?"
"Fair sire, there to marry your uncle to whom she is betrothed. Early
to-morrow morn my lady will be brought to the ancient chapel deep
within the forest. Hasten, sire, for already I tarry too long. Lend
your palfrey to your uncle and my lord. Well we know that it is the
noblest horse within the realm, as many a time has been proved."
When Messire William heard these words--
"God," said he, "then I am betrayed by him in whom I put my trust; to
whom I prayed so much to help me to my hope. May the Lord God assoil
him never for his treasonable deed. Yet scarcely can I believe that he
has done this wrong. It is easier to hold that you are telling me
lies."
"Well, you will find it truth to-morrow at the ringing of prime; for
already is gathered together a company of the ancient lords of these
parts."
"Alas," said he, "how, then, am I betrayed and tricked and deceived."
For a very little Messire William would have fallen swooning to the
earth, had he not feared the blame of his household. But he was so
filled with rage and grief that he knew not what to do, nor what to
say. He did not cease lamenting his evil case till the varlet prayed
him to control his wrath.
"Sire, cause the saddle to be set forthwith on your good palfrey, so
that my lady may be carried softly to the church."
Then Messire William considered within himself to know whether he
should send his grey palfrey to him whom he had cause to hate more than
any man.
"Yea, without delay," said he, "since she who is the soul of honour has
nothing to do with my trouble. My palfrey shall bear her gladly, in
recompense of the favours she has granted me, for naught but kindness
have I received of her. Never shall I have of her courtesies again, and
all my joy and happiness are past. Now must I lend my palfrey to the
man who has betrayed me to my death, since he has robbed me of that
which I desired more than all the world. No man is bound to return love
for treason. Very rash is he to require my palfrey of me, when he
scrupled not to take the sweetness, the beauty and the courtesy with
which my demoiselle is endowed. Alas, now have I served her in vain,
and my long hope is altogether gone. No joy in my life is left, save to
send her that thing which it breaks my heart to give. Nevertheless,
come what may, my palfrey shall go to the most tender of maidens. Well
I know that when she sets her eyes upon him she will bethink her of
me; of me and of my love, for I love and must love her all the days of
my life, yea, though she has given her heart to those who have wounded
mine. But sure am I that this thing is not seemly to her, for Cain, who
was brother to Abel, wrought no fouler treason."
In this manner the knight bewailed his heavy sorrow. Then he caused a
saddle to be set upon the palfrey, and calling the servitor delivered
the horse to his keeping. So the varlet forthwith went upon his way.
Messire William, yet heavy and wrathful, shut himself fast within his
chamber to brood upon his grief. He charged his household that if there
was a man so bold as to seek to hearten him in his sorrow he would
cause him to be hanged. For his part he had no care for mirth, and
would live withdrawn from men, since he might never lose the pain and
sorrow that weighed upon his heart.
But whilst the knight was in this case, the servant in custody of the
palfrey returned with all the speed he might to the castle of the old
Prince, where all was merriment and noise.
The night was still and serene, and the house was filled with a great
company of ancient lords. When they had eaten their full, the Prince
commanded the watch that, without fail, all men should be roused and
apparelled before the breaking of the day. He bade, too, that the
palfrey and the horses should be saddled and made ready at the same
hour, without confusion or disarray. Then they went to repose
themselves and sleep. But one amongst them had no hope to sleep,
because of the great unrest she suffered by reason of her love. All the
night she could not close her eyes. Others might rest: she alone
remained awake, for her heart knew no repose.
Now shortly after midnight the moon rose very bright, and shone
clearly in the heavens. When the warder saw this thing, being yet giddy
with the wine that he had drunken, he deemed that the dawn had broken.
"Pest take it," said he, "the lords should be about already."
He sounded his horn and summoned and cried--
"Arouse you, lords, for day is here."
Then those, yet drowsy with sleep, and heavy with last night's wine,
got them from their beds all bewildered. The squires, too, made haste
to set saddles upon the horses, believing that daybreak had come,
though before the dawn would rise very easily might the horses go five
miles, ambling right pleasantly. So when the company which should bring
this demoiselle to the chapel deep within the forest were got to horse,
her father commended his maid to the most trusty of his friends. Then
the saddle was put upon the grey palfrey; but when it was brought
before the damsel her tears ran faster than they had fallen before. Her
guardian recked nothing of her weeping, for he knew little of maidens,
and considered that she wept because of leaving her father and her
father's house. So her tears and sadness were accounted as nought, and
she mounted upon her steed, making great sorrow. They took their way
through the forest, but the road was so narrow that two could not ride
together side by side. Therefore the guardian put the maiden before,
and he followed after, because of the straitness of the path. The road
was long, and the company were tired and weary for want of sleep. They
rode the more heavily, because they were no longer young, and had the
greater need for rest. They nodded above the necks of their chargers,
and up hill and down dale for the most part went sleeping. The surest
of this company was in charge of the maiden, but this night he had
taken so little sleep in his bed that he proved an untrusty warder,
for he forgot everything, save his desire to sleep. The maiden rode,
bridle in hand, thinking of nought except her love and her sorrow.
Whilst she followed the narrow path, the barons who went before had
already come forth upon the high road. They dozed in their saddles, and
the thoughts of those few who were awake were otherwhere, and gave no
heed to the demoiselle. The maiden was as much alone as though she
fared to London. The grey palfrey knew well this ancient narrow way,
for many a time he had trodden it before. The palfrey and the maiden
drew near a hillock within the forest, where the trees stood so close
and thick that no moonlight fell within the shadow of the branches. The
valley lay deeply below, and from the high road came the noise of the
horses' iron shoes. Of all that company many slept, and those who were
awake talked together, but none gave a thought to the maiden. The grey
palfrey knew nothing of the high road, so turning to the right he
entered within a little path which led directly to the house of Messire
William, But the knight, in whose charge the damsel was placed had
fallen into so heavy a slumber that his horse stood at his pleasure on
the way. Therefore she was guarded of none--save of God--and dropping
the rein upon the palfrey's neck, she let him have his will. The
knights who preceded her rode a great while before they found that she
was not behind them, and he who came after kept but a poor watch and
ward. Nevertheless she had not escaped by her choice, for she recked
nought of the path that she followed, nor of the home to which she
would come. The palfrey followed the track without hesitation, for many
a time he had journeyed therein, both winter and summer. The weeping
maiden looked this way and that, but could see neither knight nor
baron, and the forest was very perilous, and all was dark and obscure.
Much she marvelled what had become of all her company, and it was no
wonder that she felt great fear. None regarded her safety, save God and
the grey palfrey, so she commended herself to her Maker, whilst the
horse ambled along the road. Nevertheless she had dropped the rein from
her fingers, and kept her lips from uttering one single cry, lest she
should be heard of her companions. For she chose rather to die in the
woodlands than to endure such a marriage as this. The maiden was hid in
thought, and the palfrey, in haste to reach his journey's end, and
knowing well the path, ambled so swiftly, that soon he came to the
borders of the forest. A river ran there both dark and deep, but the
horse went directly to the ford, and passed through as quickly as he
was able. He had won but little beyond when the maiden heard the sound
of a horn, blown from that place where she was carried by the grey
palfrey. The warder on his tower blew shrilly on his horn, and the
demoiselle felt herself utterly undone, since she knew not where she
had come, nor how to ask her way. But the palfrey stayed his steps on a
bridge which led over the moat running round the manor. When the watch
heard the noise of the palfrey thereon, he ceased his winding, and
coming from the tower demanded who it was who rode so hardily on the
bridge at such an hour. Then the demoiselle made reply--
"Certes, it is the most unlucky maid of mother born. For the love of
God give me leave to enter in your house to await the day, for I know
not where to go."
"Demoiselle," answered he, "I dare not let you or any other in this
place, save at the bidding of my lord, and he is the most dolorous
knight in all the world, for very foully has he been betrayed."
Whilst the watch spoke of the matter he set his eye to a chink in the
postern. He had neither torch nor lantern, but the moon shone very
clear, and he spied the grey palfrey, which he knew right well. Much he
marvelled whence he came, and long he gazed upon the fair lady who held
the rein, and was so sweetly clad in her rich new garnishing. Forthwith
he sought his lord, who tossed upon his bed with little delight.
"Sire," said he, "be not wrath with me. A piteous woman, tender of
years and semblance, has come forth from the woodland, attired right
richly. It seems to me that she is cloaked in a scarlet mantle, edged
with costly fur. This sad and outworn lady is mounted on your own grey
palfrey. Very enticing is her speech; very slim and gracious is her
person. I know not, sire, if I am deceived, but I believe there is no
maiden in all the country who is so dainty, sweet and fair. Well I deem
that it is some fay whom God sends you, to bear away the trouble which
is spoiling your life. Take now the gold in place of the silver you
have lost."
Messire William hearkened to these words. He sprang forth from his bed
without further speech, and with nothing but a surcoat on his back
hastened to the door. He caused it to be opened forthwith, and the
demoiselle cried to him pitifully in a loud voice--
"Woe is me, gentle lord, because of the sorrow I have endured this
night. Sire, for the love of God turn me not away, but suffer me to
enter in your house. I beg for shelter but a little while. But much I
fear by reason of a company of knights who are pained greatly, since
they have let me from their hands. Sir Knight, be surety for the maid
whom Fortune has guided to your door, for much am I sorrowful and
perplexed."
When Messire William heard her voice he was like to swoon with joy. He
knew again the palfrey which was so long his own. He gazed upon the
lady, and knew her in his heart. I tell you truly that never could man
be more happy than was he. He lifted her from the palfrey and brought
her within his home. There he took her by the right hand, kissing her
more than twenty times; and for her part the lady let him have his way,
because she had looked upon his face. When the two sought each other's
eyes, very great was the joy that fell between the twain, and all their
sorrow was as if it had never been. So when the damsel had put aside
her mantle, they seated themselves merrily on silken cushions, fringed
with gold. They crossed their brows again and yet again, lest they
should wake and find this thing a dream. Then the maiden told her
bachelor this strange adventure, and said--
"Blessed be the hour in which God brought me to this place, and
delivered me from him who sought to add my marriage chest to his own
coffers."
When morning was come Messire William arrayed himself richly, and led
the demoiselle within the chapel of his own house. Then, without delay,
he called his chaplain to him, and was forthwith wedded to the fair
lady by a rite that it was not lawful to call in question. So when the
Mass was sung, blithe was the mirth of that household, squire and
maiden and man-at-arms.
Now when that company which so lightly had lost the maiden came
together at the ancient chapel, they were very weary by having ridden
all the night, and were sore vexed and utterly cast down. The old
Prince demanded his daughter of him who had proved so untrusty a
guardian. Knowing not what to say, he made answer straightly--
"Sire, because of the strictness of the way I put the maid before, and
I followed after. The forest was deep and dark, and I know not where
she turned from the path. Moreover I nodded in my saddle till I was
waked by my companions, for I deemed that she was yet in my company,
but she was altogether gone. I cannot tell what has become of the
damsel, for very basely have we kept our trust."
The old Prince sought his daughter in every place, and inquired of her
from every person, but he might not find her whereabouts, nor hear of
any who had seen the maid. Yet all men marvelled at her loss, for none
was able to bring him any news. The ancient bridegroom, that the
demoiselle should have wed, grieved yet more at the loss of his bride,
but to no purpose did he seek her, for the hind had left no slot. Now
as the two lords were riding with their company in such fear as this,
they saw upon the road a certain squire making towards them in all
haste. When he was come to them he said--
"Sire, Messire William sends by me assurance of the great friendship he
bears you. He bids me say that early this morning, at the dawn of day,
he married your daughter, to his great happiness and content. Sire, he
bids you welcome to his house. He also charged me to say to his uncle,
who betrayed him so shamefully, that he pardons him the more easily for
his treason, since your daughter has given him herself as a gift."
The old Prince hearkened to this wonder, but said no word in reply. He
called together all his barons, and when they were assembled in hall,
he took counsel as to whether he should go to the house of Sir William,
and bring with him the lord to whom his daughter was betrothed. Yet
since the marriage was done, nothing could make the bride again a maid.
So, making the best of a bad bargain, he got to horse forthwith, and
all his barons with him. When the company came to the manor they were
welcomed with all fair observance, for right pleasing was this to
Messire William, since he had all things to his own desire. Whether he
would, or whether he would not, nought remained to the old Prince but
to embrace his son-in-law; whilst as to that greybeard of a bridegroom,
he consoled himself with what crumbs of comfort he could discover.
Thus, since it was the will of God that these lovers should be wed, it
pleased the Lord God also that the marriage should prove lasting.
Messire William, that courteous and chivalrous knight, lost not his
hardihood in marriage, but ever sought advancement, so that he was
esteemed of the counts and princes of his land. In the third year of
his marriage the old Prince (as the tale tells us) died, because his
time was come. So all that he died possessed of in wealth and lands and
manors, together with the rich garnishing thereof, became the heritage
of the knight. After this, Death laid hands upon his uncle, who, too,
was very rich. And Sir William, who was not simple, nor grudging of
heart, nor little of soul, nor blusterous with his neighbours,
inherited all the goods that were his.
So the story which I have told you endeth in this fashion, in
accordance with the truth, and to your pleasure.
THE END
[Illustration:
THE
TEMPLE PRESS
LETCHWORTH
ENGLAND]
[Illustration:
EVERYMAN,
I WILL GO WITH
THEE
& BE THY GVIDE
IN THY MOST NEED
TO GO BY THY SIDE]
* * * * *
Transcriber's note:
Page 23: Added closing quotes: Sister, sweet friend."
***
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 9,601
|
Produced by David Edwards, Mike Zeug, Lisa Reigel, and the
Online Distributed Proofreading Team at http://www.pgdp.net
(This file was produced from images generously made
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Transcriber's Notes: No corrections of typographical or other errors
have been made to this text. Words in italics in the original are
surrounded by _underscores_. Words in bold in the original are
surrounded by =equal signs=. On pages 6 and 7 of the original, a note
was typed vertically in the margin. These notes have been treated as
footnotes and an anchor has been added in the text. The letter that
occurs at the end of the text was not bound into the original book. It
was an insert included with the book and has been reproduced here.
[Illustration: CAPT. JOHN BROWN]
The Raid of John Brown at Harper's
Ferry As I Saw It.
BY
REV. SAMUEL VANDERLIP LEECH, D. D.
_Author of "Ingersoll and The Bible," "The Three Inebriates," "From West
Virginia to Pompeii," "Seven Elements in Successful Preaching," Etc._
PUBLISHED BY THE AUTHOR.
THE DESOTO
WASHINGTON, D. C.
1909
Copyright by S. V. Leech, 1909.
THE RAID OF JOHN BROWN AT HARPER'S FERRY AS I SAW IT.
_By REV. SAMUEL VANDERLIP LEECH, D. D._
The town of Harper's Ferry is located in Jefferson County, West
Virginia. Lucerne, in Switzerland does not excel it in romantic grandeur
of situation. On its northern front the Potomac sweeps along to pass the
national capital, and the tomb of Washington, in its silent flow towards
the sea. On its eastern side the Shenandoah hurries to empty its waters
into the Potomac, that in perpetual wedlock they may greet the stormy
Atlantic. Across the Potomac the Maryland Heights stand out as the tall
sentinels of Nature. Beyond the Shenandoah are the Blue Ridge mountains,
fringing the westward boundary of Loudon County, Virginia. Between these
rivers, and nestling inside of their very confluence, reposes Harper's
Ferry. Back of its hills lies the famous Shenandoah Valley, celebrated
for its natural scenery, its historic battles and "Sheridan's Ride." At
Harper's Ferry the United States authorities early located an Arsenal
and an Armory.
Before the Civil War, the Baltimore Conference of the Methodist
Episcopal Church was constituted of five extensive districts in
Virginia, stretching from Alexandria to Lewisburg and two great
districts north of the Potomac, including the cities of Washington and
Baltimore. The first three years of my ministerial life I spent on
Shepherdstown, West Loudon and Hillsboro Circuits, being then all in
Virginia. The State of West Virginia, now embracing Harper's Ferry, had
not been organized by Congress as a war measure out of the territory of
the mother State. Our Methodist Episcopal Church was theoretically an
anti-slavery organization; but our Virginia and Maryland members held
thousands of inherited and many purchased slaves. These were generally
well-cared for and contented. Being close to the free soil of
Pennsylvania they could have gotten there in a night had they wished to
escape bondage, and then they could have easily reached Canada by that
Northern aid, called the "Underground Railroad."
On the Sunday night when John Brown and his men invaded Virginia, I
slept within a half mile of Harper's Ferry. That day I inaugurated
revival services at my westward appointment called "Ebenezer," in Loudon
County two miles from Harper's Ferry. I was twenty-two years of age.
Three months before this raid Captain John Brown with two of his sons,
Owen and Oliver, and Jeremiah G. Anderson, calling themselves "Isaac
Smith and Sons" rented a small farm on the Maryland side of the Potomac
four miles from Harper's Ferry. It was known as the "Booth-Kennedy
Place." They also carried on across the mountains at Chambersburg,
Pennsylvania, a small hardware store managed by John H. Kagi. It was a
depot for the munitions of war to be hauled to their Maryland farm.
Another of Brown's men, John E. Cook, sold maps in the vicinity. He was
a relative of Governor Willard of Indiana who secured the services of
Hon. Daniel W. Voorhees, Attorney General of Indiana, to defend Cook at
his after trial in Virginia. It was a time of profound national peace.
Brown and his men represented themselves as geologists, miners and
speculators. They had a mule and wagon with which to haul their boxes
from Chambersburg. A wealthy merchant of Boston, Mr. George Luther
Stearns, Chairman of the Massachusetts Aid Society had financed Brown's
Kansas border warfare work, as well as his approaching Harper's Ferry
raid. Other Northern friends assisted. Brown had completed his
preparations and collected his twenty-one helpers early in October,
1859. He had hidden in an old log cabin on the place 200 Sharpe's
rifles, 13,000 rifle cartridges, 950 long iron pikes, 200 revolving
pistols, 100,000 pistol caps, 40,000 percussion caps, 250 pounds of
powder, 12 reams of cartridge paper and other warlike materials. He
organized his twenty-two men, himself included, into a "=Military
Provisional Government=" to superintend the possible uprising of the
slaves of Virginia. Thirteen of these men had engaged in border warfare
in Kansas, in a successful effort to prevent Kansas from becoming a
slave state. He, sixteen other white men and five <DW64>s, constituted
his entire Virginia army. The white men were Captain John Brown,
Adjutant General John H. Kagi, Captains Owen Brown, Oliver Brown, Watson
Brown, Aaron D. Stephens, John E. Cook, Dauphin Adolphus Thompson,
George P. Tidd, William Thompson and Edwin Coppoc. The Lieutenants were
Jeremiah G. Anderson, Albert Hazlitt and William Henry Leeman. The
privates numbered eight. Three of them were white men and five were
<DW64>s. The whites were Francis J. Merriam, Barclay Coppoc and Steward
Taylor. The <DW64>s were Dangerfield Newby, Osborne P. Anderson, John A.
Copeland, Sherrard Lewis Leary and Shields Green.
On Sunday morning, October 16th, 1859, Brown assembled his men and
informed them that on that night their invasion into Virginia would take
place. They took the oath of allegiance to the "Provisional Government."
Adjutant General Kagi presented to each officer his commission.
The contents of the Armory, Arsenal and Hall's Rifle Works were daily
open to public inspection. Captain John Brown well knew that Daniel
Whelan was the only watchman, during the night time, at the Armory
grounds. He believed that if he could secure the arms and ammunition in
these buildings, carry them into the fastnesses of the adjacent
mountains, and then unfurl the flag of freedom for all slaves who would
flock to his standard, the result would be a general uprising of the
<DW64> population throughout the border states. A more idiotic and
senseless theory never entered an American mind. In the superlative
degree it was unreasonable and ridiculous. I personally know of the
general loyalty of the slaves to their masters in that locality, at that
period in our national history. Federal generals were astonished at the
devotion of the <DW64>s to their masters everywhere in the South after
the war had begun. This was especially true along the border states. But
John Brown--honest, enthusiastic and intensely fanatical on the slavery
question--issued his commands. On this Sunday he assigned to each his
earliest work. Captain Owen Brown, Barclay Coppoc and Francis J. Merriam
were to remain at the farm to guard the arms and ammunition. Hence only
nineteen left the Kennedy farm. They were to walk down the river road on
the Maryland side to the Maryland end of the Baltimore and Ohio railroad
bridge. The Virginia end was close to the depot, hotel, Armory and the
Arsenal. Captain John Brown was to ride in the wagon with the necessary
guns, pistols and tools. Captains Cook and Tidd were to go in advance
and cut the telegraph wires on the Maryland side. Captain Stephens and
Adjutant General Kagi were to capture Mr. Williams, the guard of the
bridge. Captain Watson Brown and Taylor were to hold up the passenger
train due from the west at 1:40 A. M. It would be bound for Washington
and Baltimore. Captain Oliver Brown and Thompson were to hold the
bridges spanning the two rivers. Captain Dauphin Adolphus Thompson and
Lieutenant Anderson were to hold the first building in the Armory[6:1]
grounds popularly known afterwards as "=John Brown's Fort=." It was the
engine house where Brown held his most distinguished prisoners. From the
portholes of it that they made after his entrance, his men did their
final fighting. Captain Coppoc and Lieutenant Hazlitt were to hold the
Arsenal outside and opposite the Armory gates. Adjutant General Kagi and
Copeland were to seize and retain Hall's Rifle Works. They were half of
a mile up the western shore of the Shenandoah. Captain Stephens, and
such men as he might select, were to go out to the home of Colonel Lewis
W. Washington, the grand nephew of General George Washington, and bring
him and some of his adult male slaves, to the engine house. They were
also to secure the swords presented to General George Washington by
Frederick the Great and by General Lafayette. For this object Stephens
selected as his helpers Captains Tidd and Cook and privates Leary, Green
and Anderson. Brown made the raid at 11:30 that night. Mr. Williams the
bridge guard was captured by Stephens and Kagi. The watchman at the
Armory[7:1], Daniel Whelan, refused Brown and his men admission to the
grounds. They broke the locks with tools, captured Whelan, and took
possession of the Armory and also of the Arsenal outside. The following
prisoners were brought in early on Monday and placed in the engine
house: Jesse W. Graham who was master workman, Colonel Lewis W.
Washington, Terance Byrne, John M. Allstadt, John Donohue, who was clerk
of the railroad company; Benjamin F. Mills, the master armorer; Armstead
M. Ball, the master machinist; Archibald M. Kitzmiller, assistant
superintendent; Isaac Russell, a Justice of the Peace; George D. Shope,
of Frederick and J. Bird, Arsenal armorer. The white prisoners were to
be held as hostages and the blacks were to be armed and placed in
Brown's army. Cook and Tidd evidently mistrusted their surroundings.
During the night they made their way back to the farm and hastily
escaped into Pennsylvania. Captain Watson Brown and Taylor held up the
train bound for Baltimore, detaining it for three hours. The
porter of the depot, Shepherd Hayward, went out on the bridge to hunt
for Williams. He was brutally shot by one of Brown's bridge guards.
Hayward managed to crawl to the baggage room where he died at noon on
Monday. Dr. John Starry dressed his wounds and ministered to his every
want. The physician was under the impression that a band of train
robbers had captured the depot. He told this to Mr. Kitzmiller before
Kitzmiller's imprisonment. Captain E. P. Dangerfield, clerk to the
paymaster, entered the grounds and was hustled into the engine house
quite early in the morning. Numerous arriving workmen were imprisoned in
an adjoining building. Colonel Washington said that fully sixty men were
imprisoned by eight o'clock on Monday morning. The citizens were hearing
of the situation. Newby and Green, <DW64>s, were stationed at the
junction of High and Shenandoah streets. Newby shot at and killed
Captain George W. Turner, a graduate of West Point. Green shot and
killed Mr. Thomas Boerley, a grocer. Dr. Claggett attended Boerley, who
also soon died. After the mulatto had shot Turner, a man named Bogert
entered the residence of Mrs. Stephenson by a rear door. Having no
bullet he put a large nail into his gun, went up stairs and shot Newby,
the nail cutting his throat from ear to ear. He was also shot in the
stomach by some one else. I saw him die, in great agony, with an
infuriated crowd around him. About ten o'clock in the morning, armed
citizens crossed the Potomac and Shenandoah rivers to prevent the escape
by the bridges, or by water, of any of the raiders. Some walked down the
Maryland river road and wounded Captain Oliver Brown on the bridge. He
reached the engine house but soon died beside his father. Citizens
seized the uninjured prisoner, Captain Thompson, and put him under guard
at the Galt hotel. Captain Stephens tried to reach the hotel to propose,
as he stated, terms of surrender. George Chambers wounded him, and then
assisted him into the Galt hotel, where his wounds were dressed. About
eleven o'clock in the morning the Jefferson Guards from Charlestown
commanded by Captain J. W. Rowen arrived. A half hour passed and the
Hamtramck guards under Captain V. M. Butler came to the Ferry. They were
followed by the Shepherdstown Mounted Troop commanded by Captain Jacob
Reinhart. Then a military company from Martinsburg twenty miles distant
reached the place, under the command of Captain Alburtis. Colonels W. R.
Baylor and John T. Gibson took the general direction of the military
affairs. Some soldiers crossed the Shenandoah along with armed citizens
to intercept the four raiders Kagi, Leary, Leeman and Copeland, when
they should be driven out of Hall's Rifle Works. These raiders also had
in these works one of Colonel Washington's slaves pressed into their
service. All of them ran out into the river to swim across to the Loudon
County shore. All were shot to death in the river with the exception of
Copeland. He threw up his hands and surrendered. During the excitement
Hazlitt and the <DW64> Anderson left the Arsenal and, undetected, escaped
into Pennsylvania. Early in the morning Captain Owen Brown, Barclay
Coppoc and Merriam had deserted the Kennedy farm and gone north. Thus
seven of the twenty-two men fled to the North. Cook and Hazlitt were
captured. They were returned to Virginia, tried and executed.
By 2 o'clock P. M., the town and hills swarmed with militia and
citizens. Brown had barricaded the engine house doors with the engine
and reel. Inside were Captains John Brown and his son Watson; also
Captain Oliver Brown, who was soon dead; Shields Green, Captain Edwin
Coppoc, Lieutenant Jeremiah G. Anderson, Captain Dauphin Adolphus
Thompson and ten white prisoners. The numerous prisoners, mostly
workmen, in the adjoining structure had all escaped from the grounds,
Brown having no port-holes on that side of his fort. The militia were
afraid to fire into the port-holes for fear of killing some of the
prominent prisoners. About 4 o'clock the Mayor, Mr. Fontaine Beckham,
aged sixty years, who was also station agent of the railroad company,
went out on the platform unarmed. He was shot dead by the <DW64> Shields
Green. Captain Watson Brown in the engine house received his death wound
soon afterwards. Mayor Beckham was very much beloved by the people. A
number of citizens hurried into the hotel and brutally seized Captain
Thompson, threw him over the wall into the Potomac and riddled him with
bullets. Mrs. Foulke of the hotel, and her porter, went to the
platform and brought in the dead body of the Mayor.
As night was settling on the excited city a military company from
Winchester, Virginia, commanded by Captain B. B. Washington, arrived by
a Shenandoah Valley train. Shortly thereafter a Baltimore and Ohio
railroad train brought several companies of soldiers from Frederick,
Maryland. They were commanded by Colonel Shriver. Soon several
independent companies from Baltimore, accompanied by the Second Light
Brigade, arrived under the general command of General Charles C.
Edgerton. Colonel Robert E. Lee of the United States army, overtook
these troops at Sandy Hook, a mile and a half below the Ferry on the
Maryland side. He had come from Washington with several companies of
marines. He was accompanied by Lieutenant J. E. B. Stuart, afterwards a
famous Confederate Cavalry General; also by Major Russell and by
Lieutenant Israel Green, who died several months ago in the West. All
were regular army officers. Colonel Lee regarded it as unwise to attack
the engine house that night, fearing that Colonel Lewis W. Washington or
other prisoners might be killed. Early in the morning he sent Lieutenant
J. E. B. Stuart, who had once held Brown as a prisoner in Kansas, to
demand an immediate and unconditional surrender. Brown refused to trust
himself and men to the United States officers. About this time Colonel
Robert E. Lee got within range of Captain Coppoc's rifle. Prisoners said
that Mr. Graham knocked the muzzle aside. Lee's life was saved. Had he
been then killed who knows that the battles of Antietam, Gettysburg, and
the final conflicts north of the Appomattox would have ever been fought?
On the Confederate side no abler general or more magnificent man, ever
sat on a saddle than Robert E. Lee. He was the son of "Light Horse Harry
Lee," a brave Major General of the Revolutionary War. He was the father
of William Henry Fitzhugh Lee, who became a Major General of the
Confederate forces of Virginia, at a later date. General Robert E. Lee
made a brilliant record in the Mexican war as Chief Engineer of the
United States army. After surrendering his decimated army to General
Ulysses S. Grant, at Appomattox, he accepted the political situation
with dignity. He became President of the Washington University at
Lexington, Virginia. The South lavished on him every possible honor.
During the late summer the Virginia legislature placed in the National
Hall of Fame, at the United States Capitol, two fine statues of two
representative men of their state. One was the statue of General George
Washington; the other that of General Robert E. Lee.
By the advice of Colonel Lewis W. Washington all of Brown's prisoners
mounted the fire engine and the reel carriage and lifted up their hands
when the attack began. Three marines undertook to batter down the doors
with heavy sledge hammers. They were not successful. Then twelve
marines struck the doors with the end of a strong ladder. They opened.
Lieutenant Green entered first of all amidst a shower of bullets.
Discovering Brown reloading his rifle he sprang on him with his sword
and cut his head and stomach. The raider Captain Anderson rose to shoot
Green. A marine named Luke Quinn ran his bayonet through him. Another
raider shot Luke Quinn who soon died. Two other marines were wounded. I
saw Captains Anderson and Watson Brown as they lay dying on the grass
after their capture. The dead body of Captain Oliver Brown lay beside
them. Captain Watson Brown had been dying for sixteen hours. Captain
John Brown, bleeding profusely, and Captain Stephens from the hotel,
were carried into the paymaster's office. Brown's long grey beard was
stained with wet blood. He was bare headed. His shirt and trousers were
grey in color. His trousers were tucked into the top of his boots.
Captain Coppoc and the <DW64> Green were also taken prisoners. They were
not wounded.
As Brown lay on the floor of the paymaster's office he was very cool and
courageous. Governor Henry A. Wise, United States Senator J. M. Mason of
Virginia and Honorable Clement L. Vallandingham of Ohio plied him with
many questions. To all he gave intelligent and fearless replies. He
refused to involve his Northern financiers and advisers. He took the
entire responsibility on himself. He told Governor Wise that he, Brown,
was simply "An instrument in the hands of Providence." He said to some
newspaper correspondents and others: "I wish to say that you had
better--all you people of the South--prepare for a settlement of this
question. You may dispose of me very easily. I am nearly disposed of
now. But this question is yet to be settled--this <DW64> question I mean.
The end is not yet." Before thirteen months had passed one of the
greatest Americans of any century, Abraham Lincoln, had been elected
President of the United States; the Republican party was for the first
time dominating national affairs and, soon thereafter, the Civil War
was begun which culminated in the physical freedom of every slave in
this Republic.
On Wednesday Captains John Brown, Stephens and Coppoc, along with
Copeland and Green, were removed to the county jail at Charlestown, ten
miles south of Harper's Ferry. Being acquainted with the jailor, Captain
John Avis, I was permitted to visit Brown on one occasion. Captain Aaron
D. Stephens was lying on a cot in the same room. I was told that Brown
had ordered out of his room a Presbyterian minister named Lowrey when he
had proposed to offer prayer. He had also said to my first colleague,
Rev. James H. March, "You do not know the meaning of the word
Christianity. Of course I regard you as a gentleman, but only as a
=heathen= gentleman." I was advised to say nothing to him about prayer. He
had told other visitors that he wanted no minister to pray with him who
would not be willing to die to free a slave. I was not conscious that I
was ready for martyrdom from Brown's standpoint. I have never been
anxious to die to save the life of any body. My life is as valuable to
me and my family as any other man's is to him and his family. But young
as I was I hated American slavery. I was a "boy minister" of a great
anti-slavery denomination of Christians. For more than a century the
Methodist Episcopal Church has carried in its Disciplines its printed
testimony against slavery. It is to-day the largest fully organized
anti-slavery society on earth. I would have gladly offered prayer in
Brown's room at Charlestown if an honorable opportunity had been
afforded.
At his preliminary examination before five justices, Colonel Davenport
presiding, Brown said: "Virginians! I did not ask for quarter at the
time I was taken. I did not ask to have my life spared. Your governor
assured me of a fair trial. If you seek my blood you can have it at any
time without this mockery of a trial. I have no counsel. I have not been
able to advise with any one. I know nothing of the feelings of my fellow
prisoners and am utterly unable to attend to my own defense. If a fair
trial is to be allowed there are mitigating circumstances to be urged.
But, if we are forced with a mere form, a trial for execution, you
might spare yourselves that trouble. I am ready for my fate."
Two very able Virginia attorneys were assigned as a matter of State form
as counsel for Brown. They were Honorable Charles J. Faulkner of
Martinsburg, afterwards United States Envoy Extraordinary to France, and
Judge Green, Ex-Mayor of Charlestown. The county grand jury indicted
Brown on three separate charges: first, conspiracy with slaves for
purposes of insurrection; second, treason against the commonwealth of
Virginia; third, murder in the first degree. Mr. Faulkner withdrew from
the case and Mr. Lawson Botts took his place. Mr. Samuel Chilton a
learned lawyer of Washington, D. C., and Judge Henry Griswold of Ohio,
another distinguished attorney, volunteered their services as counsel
for John Brown and were accepted. Some of Brown's friends sent an
excellent young lawyer named George H. Hoyt from Boston, as additional
counsel. These attorneys made an able defense, whatever may have been
their private opinion as to Brown's guilt or innocence. The prosecuting
attorney for the State of Virginia was Andrew Hunter, an exceptionally
brilliant orator and able lawyer. He was a courtly and commanding
speaker. He was gifted with a rich and powerful voice. After the
indictment of Brown by the court of justices, the prosecuting attorney
of Jefferson county, Mr. Charles B. Harding left the prosecution almost
exclusively to Mr. Andrew Hunter, who represented the State. So too,
after the arrival of Brown's chosen outside counsel, Judge Green and Mr.
Lawson Botts withdrew, in good taste, from his defense.
At the regular trial Brown's counsel requested a postponement on account
of the prisoner's health. But Dr. Mason, his physician, attested the
physical ability of his patient to undergo the strain. The State was
spending almost a thousand dollars a day for military guards and other
items. When Brown's counsel presented telegrams from his relatives
asking for delay until they could forward proofs of his insanity, Brown
said, "I will say, if the court will allow me, that I look on this as a
miserable artifice and trick of those who ought to take a different
course in regard to me if they take any at all. I view it with contempt
more than otherwise. I am perfectly unconscious of insanity and I
reject, so far as I am capable, any attempts to interfere in my behalf
on that score."
On the last day of the trial, October 31st, after six hours of argument
by Hunter, Chilton and Griswold, the jury delivered the following
verdict: "Guilty of treason, and of conspiring and advising with slaves
and others to rebel; and of murder in the first degree." On Wednesday,
November the 2nd, he was brought into court to receive his sentence. The
County Clerk, Robert H. Brown, asked: "Have you anything to say why
sentence should not be passed on you?" Brown, leaning on a cane, slowly
arose from his chair and with plaintive emphasis addressed Judge Parker
as follows:
"I have, may it please the court, a few words to say. In the first place
I deny everything but what I have all along admitted, the design on my
part to free the slaves. I certainly intended to have made a clean thing
of that matter as I did last winter when I went into Missouri and took
slaves without the snapping of a gun on either side, moved them through
the country and finally left them in Canada. I designed to have done the
same thing again on a larger scale. That was all I intended. I never did
intend murder or treason, or the destruction of property, or to excite
or incite slaves to rebellion or to make insurrection. I have another
objection and that is that it is unjust that I should suffer such a
penalty. Had I interfered in the manner which I admit, and which I admit
has been fairly proved, for I admire the truthfulness and candor of the
greater portion of the witnesses who have testified in this case,--had I
so interfered in behalf of the rich, the powerful, the intelligent, the
so-called great; or in behalf of any of their friends, either father,
mother, sister, brother, wife or children, or any of that class, and
suffered and sacrificed what I have in this interference, it would have
been all right and every man in this court would have deemed it an act
worthy of reward rather than punishment. This court acknowledges as I
suppose the validity of the law of God. I see a book kissed here which I
suppose is the Bible, or at least the New Testament. That teaches me
that all things, whatsoever I would that men should do to me I should do
even unto them. It teaches me further to 'Remember them that are in
bonds as bound with them.' I endeavored to act up to that instruction. I
say that I am yet too young to understand that God is any respecter of
persons. I believe that to have interfered as I have done, as I have
always admitted freely I have done, in behalf of His despised poor was
not wrong but right. Now if it is deemed necessary that I should forfeit
my life for the furtherance of the ends of justice, and mingle my blood
further with the blood of my children and with the blood of millions in
this slave country whose rights are disregarded by wicked, cruel and
unjust enactments, I submit. So let it be done.
"Let me say one word further. I feel entirely satisfied with the
treatment I have received on my trial. Considering all the circumstances
it has been more generous than I expected. But I feel no consciousness
of guilt. I never had any design against the life of any person, nor any
disposition to commit treason or excite slaves to rebellion or make any
general insurrection. I never encouraged any man to do so but always
discouraged any idea of the kind.
"Let me say a word in regard to the statements made by some of those
connected with me. I hear it has been stated by some of them that I
induced them to join me. But the contrary is true. I do not say this to
injure them, but as regards their weakness. There is not one of them but
joined me of his own accord and the greater part of them at their own
expense. A number of them I never saw, and never had a word of
conversation with, till the day they came to me and that was for the
purpose I have stated. Now I am done."
Brown's statement was not exactly sustained by the facts. Why had he
collected the Sharpe's rifles, the pikes, the kegs of powder, many
thousands of caps and much warlike material at the Kennedy farm? Why did
he and other armed men, break into the United States Armory and
Arsenal, make portholes in the engine house, shoot and kill citizens and
surround their own imprisoned persons with prominent men as hostages?
But everybody in the court house believed the old man when he said that
he did everything with a solitary motive, the liberation of the slaves.
Judge Parker could, under his oath, do nothing else than to sentence him
to be hung. He fixed the date for Friday, the second of December.
Brown's counsel appealed to the Supreme Court of Virginia. Its five
judges unanimously sustained the action of the Jefferson county court.
Brown was hung on the bright and beautiful morning of December 2nd at
11:15 o'clock. At his request Andrew Hunter wrote his will. He then
visited his fellow prisoners who were all executed at a later date. He
rode to his death between Sheriff Campbell and Captain Avis in a
furniture wagon drawn by two white horses. He did not ride seated on his
coffin as some of his chief eulogists have affirmed. The wagon was
escorted to the scaffold by State military companies. No citizens were
allowed near to the jail. Hence he did not kiss any <DW64> baby as he
emerged from his prison, as Mr. Whittier has described in a poem on the
event and as artists have memorialized in paintings. The utter absurdity
of such an incident occurring under such surroundings any Virginian will
see. Avis, Campbell and Hunter publicly denied it. But the story will
doubtless have immortality. In one of the companies of soldiers walked
the actor John Wilkes Booth, the infamous assassin of Abraham Lincoln.
At the head of the Lexington cadets walked Professor Thomas Jefferson
Jackson, who became an able Confederate General and is best known to the
world as "Stonewall Jackson." As the party neared the gallows Brown
gazed on the glorious panorama of mountain and landscape scenery. Then
he said: "This is a beautiful country." He wore a black slouch hat with
the front tipped up. Reaching the scaffold the numerous State troops
formed into a hollow square. Brown mounted the platform without
trepidation. Standing on the drop he said to the sheriff and his
assistants: "Gentlemen! I thank you for your kindness to me. I am ready
at any time. Do not keep me waiting." The drop fell and in ten minutes
Dr. Mason pronounced him dead. That evening Mrs. Brown and her friends
received the casket at Harper's Ferry and accompanied it to the old home
at North Elba, N. Y. His funeral, as reported by the metropolitan
papers, took place there six days after his execution. An immense
concourse was in attendance. The conspicuous and brilliant orator,
Wendell Phillips, delivered the address. He closed it with these words:
"In this cottage he girded himself and went forth to battle. Fuller
success than his heart ever dreamed of God had granted him. He sleeps in
the blessings of the crushed and the poor. Men believe more firmly in
virtue now that such a man has lived." Personally I remained in
Virginia.
On the day that Brown was hung =Martyr Services=, as they were called,
were held in many Northern localities. At Concord, Dr. Edmund Sears read
a poem in which are these stanzas:
"Not any spot, six feet by two
Will hold a man like thee:
John Brown will tramp the shaking earth
From Blue Ridge to the sea
Till the strong angel comes at length
And opes each dungeon door:
And God's Great Charter holds and waves
O'er all the humble poor.
And then the humble poor may come
In that far distant day,
And from the felon's nameless grave
Will brush the leaves away:
And gray old men will point the spot
Beneath the pine tree's shade,
As children ask with streaming eyes
Where old John Brown was laid."
Before he was executed many threatening communications were received by
the Virginia State and Jefferson County officers. Large numbers of E. C.
Stedman's poem, entitled "John Brown of Ossawattamie," were scattered
about Charlestown. One stanza reads as follows:
"But Virginians! Don't do it, for I tell you that the flagon,
Filled with blood of Old Brown's offspring, was first poured by
Southern hands;
And each drop from Old Brown's life veins, like the red gore of the
dragon,
May spring up, a vengeful Fury, hissing through your slave-worn
lands;
And Old Brown,
Ossowattamie Brown,
May trouble you more than ever,
When you've nailed his coffin down."
Whether they be from the North or the South, fair-minded men, who are
thoroughly conversant with the history of this raid, can hardly cherish
any doubt concerning the turpitude of the invasion, the fairness of
Brown's trial and the justice of his conviction and execution. He fell
under the direction of a misguided conscience. The noble endowment that
philosophers call conscience that gives its verdicts as to the moral
merit or demerit of actions and affections, was strangely warped in
Brown's intense and brave character. The possession of this faculty of
conscience is the massive foundation of all human responsibility.
Illustrations of the moral enormities that a perverted conscience can
perpetrate are manifold along the pages of sacred and secular history.
When Jesus looked down the aisles of the future, He said to His
disciples that the men who would finally transfigure them into martyrs
would murder them in the belief that they were rendering acceptable
service to God.
Paul declared that he regarded himself as meeting the divine approval
when he was persecuting and murdering the primitive Christians.
When the officers of the Spanish Inquisition saw the agonies of the
victims who refused to renounce their religious creeds they joyfully
exclaimed, "Let God be glorified."
Charles the Ninth of France said he was conscientious in ordering the
Saint Bartholomew massacre that resulted in the murder in French cities
of tens of thousands of Christian Hugenots.
The Bloody Queen, Mary Tudor, said she had a pure conscience when she
sent to the scaffold the learned and gentle young Ex-Queen Lady Jane
Grey. Thousands of criminals have sheltered their crimes in the temple
of Conscience.
The trend of Brown's constant defence was that he obeyed his conscience.
His lawless conduct, the death of many of his party and the murder of
Virginia citizens gave him very little apparent intellectual unrest. He
sowed to the wind and reaped the logical harvest, if it is the
appropriate word, the whirlwind.
Brown's high Calvinism bordered on fatalism. Oliver Cromwell never
believed more radically in the foreordination of all human actions than
did he. When questioned concerning the failure of this invasion he
replied: "All of our actions, even all of the follies that led to this
disaster, were decreed to happen ages before the world was made." When
Judge Russell visited him he said: "I know that the very errors by which
my scheme was marred were decreed before the world was made. I had no
more to do with the course I pursued than a shot leaving a cannon has to
do with the spot where it shall fall."
It is when patriotic men read the story of "John Brown's Raid" by the
torches of President Lincoln's early election, the Civil War and the
Emancipation of all American slaves, that they seem to become blind to
the terrible criminal features of the invasion and look only at the
national results and the magnificient courage, benevolent motives and
supreme self-sacrifice of this martyr. Multitudes of visionary men
regard him as a divinely appointed John the Baptist raised up to usher
in the day of physical freedom for every slave on American soil and
their posterity to the end of time. They claim that in this instance
"The End has justified the Means." His raid made the North solid against
the slave system and the South as solid against anti-slavery theories
and agitators. Before the Brown raid the vote for John C. Fremont, the
Republican candidate for President, was 1341000. James Buchanan had
496000 majority. The year after the raid Abraham Lincoln received
1886000 votes for President and had 491000 majority over Stephen A.
Douglas, when the South voted for another Democrat. Fremont had 114
votes in the Electoral College. Lincoln had 180. Under his presidency
the emancipation of every slave on the national soil took place. The
nations of Europe learned for the first time the important lesson that
the United States was able to maintain its national unity. This raid
beyond question hastened in the Civil War. I have seen Federal regiments
marching on to battle enthusiastically singing:
"John Brown's body lies a mould'ring in the grave,
But his soul is marching on."
A few weeks after Brown's execution Victor Hugo said, "What the South
slew last December was not John Brown but slavery." His statement
developed into a colossal historical truth. The great statesman, orator
and senator, John J. Ingalls of Kansas, closed an oration with these
remarkable words:
"Carlyle says that when any great change in human society is to be
wrought God raises up men to whom that change is made to appear as the
one thing needful and absolutely indispensable. Scholars, orators,
poets, philanthropists, play their parts, but the crisis comes at last
through some one who is stigmatized as a fanatic by his contemporaries,
and whom the supporters of the systems he assails crucify between
theives or gibbet as a felon. The man who is not afraid to die for an
idea is the most potential and convincing advocate.
"Already the great intellectual leaders of the movement for the
abolition of slavery are dead. The student of the future will exhume
their orations, arguments and state papers, as a part of the
subterranean history of the epoch. The antiquarian will dig up their
remains from the alluvial drift of the period, and construe their
relations to the great events in which they were actors. But the three
men of this era who will loom forever against the remotest horizon of
time, as the pyramids against the voiceless desert, or mountain peaks
over the subordinate plains, are Abraham Lincoln, Ulysses S. Grant and
old John Brown of Ossowattamie."
Senator Ingalls well knew that Brown had no such intellectual
massiveness, or splendid culture, as had Webster, Clay, Jefferson,
Sumner, and many other eminent Americans. He referred to the majesty of
personal achievements. From this standpoint men like Garabaldi, Morse,
Harriman, Edison, Roosevelt and Cook, the Arctic explorer have been
great. Brown's life was a perpetual sacrifice for the annihilation of
American slavery. Very defective as a military leader he was always
ready to do, dare and die to assist in this work. Even today tens of
thousands of educated men regard him as a monomaniac concerning the
abolition of slavery. For many years, in the state of Kansas, he had
permitted his own life, and the life of each of his sons, to be in
continual peril that they might assist in placing Kansas in the
constellation of free States. Men like Gerrit Smith and John L. Stearns
financed his schemes from their wealth. Men like Henry Ward Beecher,
Ralph Waldo Emerson, George B. Cheever, William Lloyd Garrison, Wendell
Phillips and Theodore Parker, delivered eulogies on Brown after he had
been hung. They most eloquently denounced slavery from pulpits and
platforms; but they lived in the limelight of oratorical popularity and
flourished amidst luxurious ease. To Brown's immortal credit be it said
that he gave domestic security, his humble fortune, his perillous work,
the lives of his cherished sons and his own blood and life for the
anti-slavery opinions that were anchored in his soul. His prison letters
to many friends are full of intrepidity, submission to the divine
providence and heroic anticipations of immortal blessedness. Ten minutes
before he left his jail cell for the gallows he handed to a prison
official a sheet of paper on which he had written these words: "I, John
Brown, am quite certain that the crimes of this guilty land will never
be purged away but with blood, I had, as I now think, vainly flattered
myself that without very much bloodshed it might be done."
His surpassing bravery and self-sacrificing candor profoundly impressed
eminent Virginians. Governor Henry A. Wise said: "He is a bundle of the
best nerves I ever saw, cut and thrust; and bleeding and in bonds. He is
a man of clear head, of courage, fortitude and simple ingenuousness. He
is cool, collected, indomitable; and it is but just to him to say that
he was humane to his prisoners. He is a fanatic, but firm, and truthful
and intelligent." Colonel Lewis W. Washington and Captain John E. P.
Dangerfield bore testimony to his courage.
Brown's wonderful moral heroism became resplendent after Judge Richard
Parker had sentenced him to death. Many of his letters to his friends,
collected and published by Mr. F. B. Sanford, would have done honor to
the pen of Paul. He was exultant from the standpoint of a happy
spiritual experience and triumphant as he gazed beyond this mortal life.
In one of his last letters he wrote these words: "I sleep as peacefully
as an infant, or if I am wakeful glorious thoughts come to me
entertaining my mind. I do not believe I shall deny my Lord and Master,
Jesus Christ, in this prison or on the scaffold. But I should do so if I
denied my principles against slavery." Surely he must have been sincere
as he faced eternity.
As early as 1820 John Quincy Adams said of the overthrow of American
slavery, "The object is vast in its compass, awful in its prospects and
sublime and beautiful in its issues. A life devoted to it would be nobly
spent or sacrificed." John Brown, along illegal and criminal lines,
placed before the world such a life and death. He saw clearly what
American statesmen of his period saw but dimly. Beyond all question he
died as emphatically for the overthrow of slavery as Paul died for the
honor of Christianity. Three of his favorite books were the life stories
of men of great achievements:--"The Life of Oliver Cromwell," "The Life
of Marco Bozarris," and "The Life of William Wallace."
Some years ago, in an oration delivered at Harper's Ferry, the
distinguished freedman and orator, the late Frederick Douglass, said:
"If John Brown did not end the war that ended slavery he did at least
begin the war that ended slavery. If we look over the dates, places and
men for which this honor is claimed we shall find that not Carolina, but
Virginia; not Fort Sumter, but Harper's Ferry and the United States
Arsenal; not Major Anderson, but John Brown, began the war that ended
American slavery and made this a free republic. Until this blow was
struck the prospect was dim, shadowy and uncertain. The irrepressible
conflict was one of words, votes and compromises. When John Brown
stretched forth his arm the sky was cleared, the time for compromise was
gone, the armed hosts stood face to face over the chasm of a broken
Union and the clash of arms was at hand."
And let it be remembered that when Brown had told Douglass the details
of his proposed invasion at Harper's Ferry, Douglass begged him to
abandon his plans and assured him that they would end, as they did, in
untold disaster.
The chief authors who have written concerning John Brown and his
invasion were not in Virginia during the forty-four days intervening
between the raid and his execution. They were destitute of any personal
knowledge of the facts. They were bitter enemies of the South and most
intense admirers of the intrepid man executed at Charlestown. Their
narratives are replete with errors and contain much romance. They are,
generally, saturated with misrepresentation of the Virginia people and
are burdened with eulogistic apologies for Brown's conduct in Virginia.
Because I was on the ground and saw things as they occurred; because I
have kept in touch with Brown literature; and because I am in love with
the Truth I believe that my story is worthy of public confidence.
I have known Virginians, personally, for over fifty years. My long
career, as a minister of Christ, was begun among them. They have not
deserved the traduction Brown's eulogists have heaped on them. His
unfortunate execution was the logical result of his criminal and bloody
raid. The Virginia people have been noble in chivalry, bounteous in
hospitality, sublime in kindness of heart and life and models of high
social and moral purity.
Spartacus led the way for the destruction of Roman slavery. John Brown
performed a similar service for the American slaves. He mingled in his
strange character fanaticism and courage--eccentricity and a prophetical
insight into future events--a warped conscience and a sublime martyr
heroism. But whether in safety or peril, at home or in prison, in battle
or on the scaffold, this mysterious man intensely cherished the
conviction that Joanna Baillie imbedded into poetry:
"The strength of man sinks in the hour of trial,
But there doth live a power that for the battle
Girdeth the weak."
FOOTNOTES:
[6:1] For Armory read Arsenal.
[7:1] For Armory read Arsenal.
+------------------------------------------------------------------+
| |
| |
| THE DE SOTA |
| |
| Washington, D. C., November 18, 1909. |
| |
| My Dear Sir: |
| |
| There has just been issued a small volume copyrighted, |
| entitled., "The Raid Of John Brown As I Saw It" from the pen |
| of "Rev. Samuel Vanderlip Leech, D. D., of Washington, D. C.," |
| who has been a Methodist Episcopal Minister for 52 years. For |
| this book The Maurice Engraving Company furnished the latest |
| portrait of Captain John Brown. The edition is limited to four |
| hundred copies. They are not sold at any store. The object of |
| the publication is to place on the shelves of Libraries, |
| Colleges, Universities and Historical societies, from the |
| southern standpoint, an accurate narrative of the raid, and |
| the events associated with it. I was 22 years of age, was |
| preaching close to Harper's Ferry, saw the fighting and |
| capture and visited Brown in his prison. I was a witness of |
| the events of the forty four days intervening between the raid |
| and his execution. |
| |
| His partisan biographers were not in Virginia at that time. |
| Their books contain historical errors and much romance. Their |
| abuse of the Virginians is unfair. I am a Republican, and have |
| steadily voted for Republican Presidents. But I think the time |
| has come when a truthful version of this famous raid should |
| find a place in national literature. I think that you will |
| agree with me. On receipt of a money order for 45 cents I will |
| mail to you a post-paid copy of this small volume. |
| |
| With Respect, |
| |
| S. V. LEECH. |
| |
+------------------------------------------------------------------+
End of the Project Gutenberg EBook of The Raid of John Brown at Harper's
Ferry as I Saw It, by Rev. Samuel Vanderlip Leech
***
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 9,601
|
Q: Is it necessary to use in the array NavLink, Interface? I'm new to react and typescript, I have a DupNavLink component that creates an array of NavLink. There is also an Interface faceNavLink which I use for the typing of the object in the map.
Do I need a faceNavLink interface? And if so, how should the structure of my interface be?
Here is my example on codesandbox: https://codesandbox.io/s/x6l89n75o
A: I don't think you need to define all those values in your faceNavLink interface.
You can just extend the interface NavLinkProps provided by react-router-dom and extend it with your changes.
In your Interface.tsx you can do something like:
import { NavLinkProps } from 'react-router-dom';
export default interface faceNavLink extends NavLinkProps {
readonly id: string;
}
In this case you'll have all the properties from NavLink plus your custom id: string.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,607
|
From the Town of Leesburg, VA – At 11:00 a.m., on Saturday, October 9, 2010, Town of Leesburg officials will hold a ribbon cutting ceremony for Legacy Orthodontics located at 705-C East Market Street in the Prosperity Center shopping center. Mayor Umstattd will be officiating.
Legacy Orthodontics is the practice of Dr. Markus Niepraschk. A Virginia native, Dr. Niepraschk graduated first in his class from Virginia Commonwealth University School of Dentistry and completed an additional two-year advanced educational program in the exclusive study of orthodontics at Marquette University.
Family Fun Today at 4 p.m.: Loudoun Symphony Presents "Movie Magic"
The Loudoun Symphony invites you and your family to join them today as they celebrate the opening of their 2010-11 Season.
This year, the Loudoun Symphony and the Loudoun String Workshop will be presenting its concerts in Virginia Academy's state-of-the-art 800-seat auditorium. The Virginia Academy is Loudoun County's largest private school, offering classes from preschool to 8th grade. The school enrolls over 700 students and is located in the familiar, steepled Community Church, which is located on Route 7 in Ashburn.
South Riding Community is hosting it's 11th Annual Oktoberfest this Saturday, October 2 from 12:00-7:00 on the Town Green.
SR transforms it's Town Green into a local biergarten filled with incredible bratwursts cooking and cold brews to follow. Come dressed in your best Lederhosen and let loose!
There will also be the 1st "unofficial" Volksmarch starting at 11:30. It will be a quick 5.7 mile walk. Starting point is at the church parking lot across from the Town Green. Rain or shine come ready to celebrate!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8,951
|
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