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Q: Directly proportional to which? This might be a simple question, but the exact wording of this statement is important. If I know that the relationship between two variables is this:
$$\gamma \propto \frac{1}{\sqrt{k}}$$
Is the appropriate way to write that in a sentence: 'y is inversely proportional to k' or should it be 'y is inversely proportional to the square root of k'.
If somebody could explain why whichever is correct is correct, that would be great.
A: I would take "$y$ is inversely proportional to $k$" to mean $y=c/k$ for some constant $c$, so for your situation I'd definitely say "$y$ is inversely proportional to the square root of $k$."
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A great collection of books to add to your kindle.
Writing Rules – Do Authors Have to Adhere to Them?
Yet another great post from Tanya.
Reference to 'Writing Rules' regularly appears in author and writer related discussions, posts and articles. Nevertheless, the so called 'rules' are not usually listed, leaving readers with no option but to surmise what they are.
Have to say I've always liked WordPress and if I can work it, anyone can!
When I first started working on the Internet, back in the mid-90s, we had to code everything by hand. And I mean, everything. You couldn't even write a simple "Hello World" without a dozen lines of code.
Fast-forward to today's WordPress-powered blogs and websites and the difference couldn't be starker. Even though my coding experience has proven invaluable time and again, it's now possible to develop in a single day a website that would have taken me at least a week back then. Months, if I wanted to use some of the fancier elements that are now available at the click of a button!
And it still amazes me that it's all free! So, this is my way of saying "thank-you" to Automattic and the WordPress team for making my life so much easier–and productive. Many thanks to Hosting Tribunalfor the excellent infographic!
Now, on a serious note.
graciously allowed me to reprint (aka copy/paste) her post.
I need to tell a story – it's going to be a long one, so settle in if you're interested in hearing it.
Food for thought, lots of good points.
I refrain from calling what we do self-publishing. I am an independent author. My publisher is Amazon. Instead of having services provided to me by a traditional publisher, I outsource them to providers that fit within my budget and style.
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I recently had the wonderful experience of being part of one of the largest projects I have ever worked on. It was something that began as a simple conversation at a Whole Foods with a local Lululemon store, and ended with a full series of images of some of the most inspiring and active men in the Houston community. Not only that, but we took those images and installed them across two galleries in the Houston area to help raise funds and awareness for an amazing program being sponsored by the KIPP Academy.
I won't go into much detail now because the majority of it is being covered at length in the upcoming issue of Inside RV due out in just a few weeks. (Be sure to hit Subscribe so you don't miss out on all of the planning, shooting, and details on the galleries and so much more!) But, I did want to give a quick peek into the Behind The Scenes of shooting the images and video associated with the project in the above video, and give y'all an idea of what the project entailed with some of my favorite images. So, be sure to check out the full gallery of images available on the site, and don't forget to click that Subscribe button to catch the full story in Inside RV Issue 2!
Modern Day Man was one of my main projects for 2017. Partnering with local Lululemon stores, we created a setting where we could tell the stories of 29 amazing men from the Houston community. These men are doing so much to help others in the community and act as role models for the next generation not necessarily by following in the footsteps of the men before them, but by forging their own path and combining the strength of men past and present with the vulnerability needed to be a great man in today's society. The project culminated in a one-night gallery showing and a follow up gallery installation at Silver Street Studios to raise money for the Community Connect program operated by the KIPP Foundation.
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Although you can find ways to do so, this Marketing aspect is easier said than done. Think about the costs of building a ecommerce website, finding a network of people to purchase from guest and you posting on blogs to expand your reach to the ideal people. It takes time and cash, if you do not have the ideal tools to begin doing all this, and you are bound to burn out yourself.
Imagine if you had a straightforward tool for fast turning Your normal site or social media station into an online store, while at the exact same time sharing your products in search of content like yours with a huge network of users?
This tool could be Sellfy, and upon first Glance, it has somewhat of a Kickstarter feel for its interface (without the crowdfunding,) where you can generate your own storefront on the Sellfy platform, peddle your products to the tens in the community, while also having the opportunity to post Sell buttons on your own website.
Although this service does not exactly cut out the Money or time you'll spend on making your website, it can produce the process a little simpler. That's pretty enticing to us, so join us as we take a deeper look into what Sellfy has to offer.
This is a simple instrument the Sellfy documentation Only consists of approximately 30 or so things. It's easy to hunt, and you do not need to worry about getting overwhelmed.
The only way to contact the support team right Is via an email form. You can get pushed with a plan that is paid, although so, no phone service.
With this inspection, we figured it'd Be Smart to test Out selling a digital product that we're familiar with. Thus, we gave the Sellfy comes with a look by uploading and submitting Ecommerce Platform's Guide to Creating a Successful Online Shop.
Here's another motive Sellfy is Far Better than simply Placing a PayPal button on your site. Rather than managing a MailChimp accounts that is completely different, Sellfy gives you custom purchase emails which send out automatically.
Alright, a few of those other features come into play You are contemplating a PayPal button on your site. Basically, discount codes can be generated by you for every one of your products right in the Sellfy dashboard.
Offering free products is a choice, and you can Even develop social discount codes to give out if your page is shared by folks.
I see the benefit of going with Sellfy as the Fact that you're able to get a little storefront up and running in seconds. You are able to upload 2GB per file, and you don't need to worry about designing your own website and use bandwidth. Simply speaking, you can forget about hosting.
Each of the storefronts are optimized for the mobile World, and it lets you take Buy Now widgets and buttons and execute them.
And being a proponent of automating many As you can processes, I have always loved Zapier. Sellfy streamlines the process that is advertising through automation and also includes a Zapier integration now.
PayPal and Stripe are the choices you have when selling on Sellfy. These are the two most frequent payment processors, but they cut out some people who aren't allowed to use these options in other nations. The integration functions seamlessly, but we'd love to view offerings.
Aside from that, Sellfy allows for Payouts, international taxes, multiple currencies and monies, EU VAT accounts and customer invoices. They supply webhooks that enable users to add customized functionalities.
There also possess a free day trial. I Personally prefer not to have to cover transaction fees, but newbies can get started for cheap together with the fundamental plan. I wonder if folks may be better off with a solution like Easy Digital Downloads.
Sellfy seems to work best for sellers with large Digital libraries who need the tools for ignoring, PDF stamping and analytics. Additionally, you increase the amount Via the Sellfy network.
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Meet the winners of the 2016 Track League!
Jack Escritt was the overall winner with Matthew Fostun 2nd and Andrew Magnier 3rd.
Joseph White won Group B and the rider with the most points in Group C was Simon Hopkins.
It was very close in Group D, the women only group with Tiffany Fletcher finally getting her hands on the trophy, closely followed by Laura Clode and Emily Capewell.
The most improved rider was Ian Upcott.
Congratulations to you all and thank you to all the riders for making it such a friendly league.
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Saint Alferio Pappacarbone (né vers 930 à Salerne et mort le à l'abbaye de Cava) est un religieux catholique italien du Moyen Âge, vénéré comme un saint par l'Église catholique et dans les mémoires comme le fondateur de l'abbaye de la Sainte Trinité de Cava.
Biographie
Né d'une famille noble lombarde, Alferio a été envoyé comme ambassadeur par son prince (Guaimar III de Salerne) auprès de l'empereur Henri II solliciter une aide militaire contre les Byzantins qui menaçaient les frontières de la Principauté de Salerne.
Ayant atteint les Alpes, il tomba très malade et demanda l'hospitalité au monastère de Chiusa di San Michele ; il promit que s'il se remettait, il renoncerait à sa carrière diplomatique et deviendrait un religieux bénédictin. Une fois guéri, il respecta sa promesse et adopta l'habit de Saint Benoît de Nursie, dans la grande abbaye de Cluny en France, vers l'an 991 où il a été ordonné prêtre.
Cependant après un peu moins d'une année, Guaimar III, prince de Salerne, le rappela à lui pour réformer les nombreux monastères de cet état. Alferio se mit à l'œuvre, mais après un certain temps, se sentant attiré par une vie de solitude, il quitta secrètement Salerno et se réfugia dans une grotte au pied du mont Finestra, aujourd'hui dans la ville de Cava de' Tirreni. Ici, avec deux compagnons, il s'adonna complètement à la prière, à la pénitence et au travail manuel.
Il mourut le (Jeudi Saint), à l'âge de 120 ans (sic), après avoir fêté les services liturgiques, stimulé par une vision supposée du Rédempteur qui aurait prédit sa mort imminente.
Il a été enterré dans la grotte même, qui depuis lors est devenue le cœur de l'Abbaye de Cava.
Bibliographie
Ratzinger Joseph., Santi. Gli autentici apologeti della Chiesa, Lindau Edizioni, Torino 2007
Voir aussi
Articles connexes
Abbaye de Cava
Cava de' Tirreni, Italie
Liens externes
San Alferio
Naissance à Salerne
Ermite du Xe siècle
Ermite du XIe siècle
Personnalité italienne du Xe siècle
Personnalité italienne du XIe siècle
Moine du Xe siècle
Moine du XIe siècle
Abbé du XIe siècle
Religieux catholique italien
Bénédictin italien
Saint bénédictin
Saint catholique italien
Saint catholique du XIe siècle
Date de naissance non renseignée (Xe siècle)
Décès à Cava de' Tirreni
Décès en 1050
Diplomate du haut Moyen Âge
Personnalité du haut Moyen Âge par nom
Diplomate du Xe siècle
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\section{Introduction}
There are many generalizations of the classical Schwarz Lemma on holomorphic maps between unit balls via the work of Ahlfors, Chen-Cheng-Look, Lu, Mok-Yau, Royden, Yau, etc (see \cite{Kobayashi-H} and \cite{Roy, Yau-sch} and references therein). The one obtained by Royden \cite{Roy} states:
\begin{theorem}\label{thm-sch-roy}
Let $f: M^m\to N^n$ be a holomorphic map. Assume that the holomorphic sectional curvature of $N$, $H(Y)\le -\kappa |Y|^4, \, \forall Y\in T'N$ and the Ricci curvature of $M$, $\Ric^M(X, \overline{X})\ge -K |X|^2, \, \forall X\in T'M$ with $\kappa, K>0$. Let $d=\dim(f(M))$. Then
\begin{equation}\label{eq:sch-roy1}
\|\partial f\|^2(x) \le \frac{2d}{d+1}\frac{K}{\kappa}.
\end{equation}
\end{theorem}
In \cite{Ni-1807} the author proved a new version which only involves the holomorphic sectional curvature of domain and target manifolds. Recall that for the tangent map $\partial f: T_x'M \to T'_{f(x)}N$ we define its maximum norm square to be
\begin{equation}\label{eq:1}
\|\partial f\|^2_0(x)\doteqdot \sup_{v\ne 0}\frac{|\partial f(v)|^2}{|v|^2}.
\end{equation}
\begin{theorem}\label{thm:sch1} Let $(M, g)$ be a complete K\"ahler manifold such that the holomorphic sectional curvature $H^M(X)/|X|^4 \ge -K$ for some $K\ge0$. Let $(N^n, h)$ be a K\"ahler manifold such that $H^N(Y)<-\kappa |Y|^4$ for some $\kappa>0$. Let $f:M\to N$ be a holomorphic map. Then
\begin{equation}\label{eq:sch-ni}
\|\partial f\|^2_0(x) \le \frac{K}{\kappa}, \forall x\in M,
\end{equation}
provided that the bisectional curvature of $M$ is bounded from below if $M$ is not compact. In particular, if $K=0$, any holomorphic map $f: M\to N$ must be a constant map.
\end{theorem}
The assumption on the bisectional curvature lower bound can be replaced with the existence of an exhaustion function $\rho(x)$ which satisfies that
\begin{equation}\label{eq:2}
\limsup_{\rho\to \infty} \left(\frac{|\partial \rho|+[\sqrt{-1}\partial \bar{\partial} \rho]_{+}}{\rho}\right)=0.
\end{equation}
The proof uses a viscosity consideration from PDE theory. It is also reminiscent of Pogorelov's Lemma \cite{Pogo} (cf. Lemma 4.1.1 of \cite{Gu}) for Monge-Amp\`ere equation, since the maximum eigenvalue of $\nabla^2 u$ is the $\|\cdot\|_0$ for the normal map $\nabla u$ for any smooth $u$.
A consequence of Theorem \ref{thm:sch1} asserts that {\it the equivalence of the negative amplitude of the holomorphic sectional curvature implies the equivalence of the metrics}. Namely if $M^m$ admits two K\"ahler metrics $g_1$ and $g_2$ satisfying that
$$
-L_1|X|_{g_1}^4\le H_{g_1}(X)\le -U_1|X|_{g_1}^4, \quad -L_2|X|_{g_2}^4 \le H_{g_2}(X)\le -U_2|X|_{g_2}^4
$$
then for any $v\in T_x'M$ we have the estimates:
$$
|v|^2_{g_2}\le \frac{L_1}{U_2}|v|^2_{g_1};\quad |v|^2_{g_1}\le \frac{L_2}{U_1}|v|^2_{g_2}.
$$
This result can be viewed as a stability statement of the classical result asserting that a complete K\"ahler manifold with the negative constant holomorphic sectional curvature must be a quotient of the complex hyperbolic space form. Motivated by Rauch's work which induces much work towards the $1/4$-pinching theorem and, the above stability of K\"ahler metrics it is natural to ask {\it whether or not a K\"ahler manifold $M$ with its homomorphic sectional curvature being close to $-1$ is biholomorphic to a quotient of the complex hyperbolic space.} Besides the Liouville type theorem for holomorphic maps into manifolds with negative holomorphic sectional curvature, we shall show in Section 5 further implications of this estimate towards the structure of the fundamental groups of manifolds with nonnegative holomorphic sectional curvature.
Before we state another recent result of the author we first recall some basic notions from Grassmann algebra \cite{Fede, Whit}. Let $\mathbb{C}^m$ be a complex Hermitian space (later we will identify the holomorphic tangent spaces $T_x'M$ and $T'_{f(x)}N$ with $\mathbb{C}^m$ and $\mathbb{C}^n$). Let $\wedge^\ell \mathbb{C}^m$ be the spaces of $\ell$-multi-vectors $\{v_1\wedge\cdots \wedge v_\ell\}$ with $v_i\in \mathbb{C}^m$. For ${\bf{a}}=v_1\wedge\cdots \wedge v_\ell, {\bf{b}}=w_1\wedge \cdots w_\ell$, the inner product can be defined as $\langle {\bf{a}}, \overline{{\bf{b}}}\rangle =\det(\langle v_i, \bar{w}_j\rangle)$. This endows $\wedge^\ell \mathbb{C}^n$ an Hermitian structure, hence a norm $|\cdot|$. There are also other norms, such as the {\it mass} and the {\it comass}, which shall be denoted as $|\cdot|_0$ as in \cite{Whit}, and could be useful for some problems. We refer \cite{Whit} Sections 13, 14 for detailed discussions. Assume that $f: (M^m, g)\to (N^n, h)$ is a holomorphic map between two K\"ahler manifolds. Let $\partial f: T'M\to T'N$ be the tangent map. Let $\Lambda^\ell \partial f:\wedge^\ell T_x'M \to \wedge^\ell T_{f(x)}'N$ be the associated map defined as
$\Lambda^\ell \partial f( v_1\wedge \cdots \wedge v_\ell)=\partial f(v_1)\wedge\cdots\wedge \partial f(v_\ell)$. Define $\|\cdot\|_0$ as $$\|\Lambda^\ell \partial f\|_0(x) \doteqdot\sup_{{\bf a}=v_1\wedge\cdots\wedge v_\ell\ne 0, {\bf a}\in \wedge^\ell T_x'M} \frac{|\Lambda^\ell \partial f({\bf a})|}{|{\bf a}|}.$$
The notion $\|\cdot\|_0$ is adapted to be consistent with the comass notion in \cite{Whit}.
By the singular value decomposition, we may choose normal coordinates centered at $x_0$ and $f(x_0)$ such that at $x_0$, $df\left(\frac{\partial\, }{\partial z^{\alpha}}\right)=\lambda_\alpha \delta_{i\alpha} \frac{\partial\, }{\partial w^i}$. If we order $\{\lambda_\alpha\}$ such that $|\lambda_1|\ge |\lambda_2|\ge \cdots \ge |\lambda_m|$, $\|\Lambda^\ell \partial f\|_0(x_0)=|\lambda_1\cdots\lambda_\ell|$. It is also easy to see that $\|\partial f \|^2 \doteqdot g^{\alpha\bar{\beta}}h_{i\bar{j}}\frac{\partial f^i}{\partial z^\alpha} \overline{ \frac{\partial f^j}{\partial z^\beta}}=\sum_{\alpha=1}^m |\lambda_\alpha|^2$. The following was proved in Corollary 3.4 of \cite{Ni-1807}.
\begin{theorem}\label{thm-schni2} Let $f:M^m\to N^n$ ($m\le n$) be a holomorphic map with $M$ being a complete manifold. Assume that $\Ric^M $ is bounded from below and the scalar curvature $S^M(x)\ge -K$. Assume further that $\Ric^N_m(x)\le -\kappa<0$. Then we have the estimate
$$
\|\Lambda^m \partial f\|^2_0(x)\le \left(\frac{K}{m\kappa}\right)^m.
$$
\end{theorem}
Here recall that in \cite{Ni-1807} $\Ric(x, \Sigma)$ is defined as the Ricci curvature of the curvature tensor restricted to the $k$-dimensional subspace $\Sigma\subset T_x'M$. Precisely for any $v\in \Sigma$, $\Ric(x, \Sigma)(v,\bar{v})\doteqdot \sum_{i=1}^k R(E_i,\overline{E}_i, v,\bar{v})$ with $\{E_i\}$ being a unitary basis of $\Sigma$. We say that $\Ric_k(x)<0$ if $\Ric(x, \Sigma)<0$ for every k-dimensional subspace $\Sigma$. Clearly $\Ric_k(x)<0$ implies that $S_k(x)<0$, and it coincides with $H$ when $k=1$, with the Ricci curvature $\Ric$ when $k=\dim(N)$. Here $S_k(x, \Sigma)$ is defined to be the scalar curvature of the curvature operator restricted to $\Sigma\subset T'_x N$. One can refer to \cite{Ni-1807, Ni, Ni-Zheng2} for the definitions and related results on the geometric significance of $\Ric_\ell$ and $S_\ell$.
Note that Theorem \ref{thm-schni2} has at least two limits in studying the holomporphic maps. The first it applies only to the case that $\dim(N)$, the dimension of the target manifold is at least as big as the dimension of the domain. The second limit is that it can only be applied to detect whether or not the map is full-dimensional, namely $\dim(f(M))=\dim(M)$ or not.
The first goal of this paper is to prove a family of estimates for holomorphic maps between K\"ahler manifolds containing the above three results as special cases. The result below removes the above mentioned constraints of Theorem \ref{thm-schni2}.
\begin{theorem}\label{thm:main1} Let $f:M^m\to N^n$ be a holomorphic map with $M$ being a complete manifold. When $M$ is noncompact assume either the bisectional curvature is bounded from below or (\ref{eq:2}) holds for some exhaustion function $\rho$. Let $\ell\le \dim(M)$ be a positive integer.
(i) Assume that the holomorphic sectional curvature of $N$, $H^N(Y)\le -\kappa |Y|^4$ and $M$ has, $\Ric_\ell^M\ge -K $, for some $K\ge 0, \kappa >0$. Then
$$
\sigma_\ell(x)\le \frac{2\ell'}{\ell'+1}\frac{K}{\kappa}
$$
where $\sigma_\ell(x)=\sum_{\alpha=1}^\ell |\lambda_\alpha|^2(x)$, and $\ell'=\min\{\ell, \dim(f(M))\}$. In particular, if $K=0$, the map $f$ must be a constant.
(ii) Assume that $S^M_\ell(x)\ge -K$ and that $\Ric^N_\ell(x)\le -\kappa$ for some $K\ge 0, \kappa >0$. Then
$$
\|\Lambda^\ell \partial f\|^2_0(x)\le \left(\frac{K}{\ell\kappa}\right)^\ell.
$$
In particular, if $K=0$, the map $f$ has rank smaller than $\ell$.
\end{theorem}
Note that part (i) above recovers Theorem \ref{thm-sch-roy} for $\ell=\dim(M)$, and recovers Theorem \ref{thm:sch1} for $\ell=1$. Hence it provides a family of estimates interpolating between Theorem \ref{thm-sch-roy} and \ref{thm:sch1}. Similarly part (ii) recovers Theorem \ref{thm-schni2} when $\ell=\dim(M)$, and recovers Theorem \ref{thm:sch1} for $\ell=1$, noting that in the case $\ell=\dim(M)$, the assumption on the lower bound of bisectional curvature can be weakened to a lower bound of the Ricci curvature (from the proof this is obvious). Hence part (ii) provides a family of estimates interpolating between Theorem \ref{thm:sch1} and \ref{thm-schni2}. Part (ii) also implies that any K\"ahler manifold with $\Ric_\ell\le -\kappa<0$ must be $\ell$-hyperbolic, a result proved in \cite{Ni-1807}. Moreover it can also be applied to $M$ with $\dim(M)>\ell$ or even $\dim(M)>\dim(N)$ concluding more detailed degeneracy information of the map, re-enforcing the relationship between the $\ell$ dimensional ``holomorphic" area of $N$ and the $\Ric^N_\ell$.
The proof of the result (in Section 4) is built upon extensions of $\partial\bar{\partial}$-Bochner formulae of \cite{Ni-1807}, which are proved in Section 3 after some preliminaries in Section 2. In Section 5 we show that the estimates can be used to rule out the existence of certain holomorphic mappings under some curvature conditions (cf. Theorem \ref{thm:51}). In particular Theorem \ref{thm:sch1} (cf. Corollary 5.4 of \cite{Ni-1807}) implies that {\it if a compact K\"ahler manifold $(M, g)$ has $H\ge 0$, then there is no onto homomorphism from its fundamental group to the fundamental group of any oriented Riemann surface (complex curve) of genus greater than one.} The more flexible Theorem \ref{thm:main1} extends this statement to include all K\"ahler manifolds with $\Ric_\ell \ge0$ (for some $\ell\in \{1, \cdots, m\}$). Note that a similar statement was proved for Riemannian manifold with positive isotropic curvature in \cite{FW}. In \cite{Tsu, Ni} it was proved that if the holomorphic sectional curvature $H>0$ or more generally $\Ric_\ell>0$ then $\pi_1(M)=\{0\}$. The result here provides some information for the nonnegative case. Note that the examples in \cite{Hitchin} indicate that the class of K\"ahler manifolds with $H>0$ (most of them are not Fano) seems to be much larger than that with $\Ric>0$. There has been very little known for manifold $M$ with $H\ge 0$ (or $\Ric_\ell\ge 0$ for $\ell<\dim(M)$) comparing with the situation for compact manifolds with $\Ric\ge 0$. In fact when $M$ is a compact K\"ahler manifold with nonnegative bisectional curvature, Mok's classification result \cite{Mok} implies that the fundamental group $\pi_1(M)$ must be a Bieberbach one. In Corollary 5.1 of \cite{Ni-Tam} a paper by Tam and the author, this was extended (as a result of F. Zheng) to the case when $M$ is a non-compact complete K\"ahler manifold, but under the nonnegativity of sectional curvature. For compact Riemannian manifolds with nonnegative Ricci curvature Cheeger-Gromoll \cite{CG} proved that $\pi_1(M)$ must be a finite extension of a Bieberbach group. {\it Could this be proven for a compact K\"ahler manifold with $\Ric_\ell \ge 0$ with $\ell<\dim(M)$}? Note that such a statement can not be possibly true for K\"ahler manifold with $B^\perp\ge 0$ (hence nor with $\Ric^\perp\ge 0$). In a recent preprint \cite{Mu}, the question has been answered positively for $H\ge 0$, assuming additionally that $M$ is a projective variety. Given that there are many non-algebraic K\"ahler manifolds with $H\ge 0$, our result for general K\"ahler manifolds is not contained in \cite{Mu}.
In \cite{ABCKT}, two invariants were defined for a K\"ahler manifold $M$. One is the so-called Albanese dimension $a(M)\doteqdot \dim_{\mathbb{C}}(Alb(M))$ (we use the complex dimension instead), the dimension of the image of the Albanese map $Alb: M\to \mathbb{C}^{\dim(H^{1,0}(M))}/H_1(M, \mathbb{Z})$. The other invariant is the genus of $M$, $g(M)$ which is defined as the maximal $\dim(U)$ with $U$ being an isotropic subspace of $H^1(M, \mathbb{C})$. The above consequence of Theorem \ref{thm:51} can be rephrased as that for $M$ with $H^M(X)\ge 0$, or more generally $\Ric_\ell\ge 0$, we must have $g(M)\le 1$. The same conclusion is obtained in Section 6 for K\"ahler manifold $M$ with the Picard number $\rho(M)=1$ and $S_2>0$, or $h^{1,1}(M)=1$. A corollary of Theorem \ref{thm:51} concludes that if $S^M_\ell>0$, then $a(M)\le \ell-1$. ( This is also a consequence of the vanishing theorem proved in \cite{Ni-Zheng2}.) These results endow the curvature $\Ric_\ell$ and $S_\ell$ some algebraic geometric/topological implications.
In Section 5 we also illustrate that the $C^2$-estimate for the complex Monge-Amp\`ere equation is a special case of our computation in Section 3. In Section 6 we derive some estimates on the minimal ``energy" needed for a non-constant holomorphic map between certain K\"ahler manifolds extending earlier results in \cite{Ni-1807}.
\section{Preliminaries}
We collect some needed algebraic results. For holomorphic map $f: (M^m, g)\to (N^n, h)$, let $\partial f(\frac{\partial\ }{\partial z^{\alpha}})=\sum_{i=1}^n f^i_{\alpha} \frac{\partial\ }{\partial w^i}$ with respect to local coordinates $(z^1, \cdots, z^m)$ and $(w^1, \cdots, w^n)$. The Hermitian form $A_{\alpha\bar{\beta}}dz^\alpha \wedge dz^{\bar{\beta}}$ with
$A_{\alpha\bar{\beta}}=f^i_{\alpha} \overline{f^j_\beta} h_{i\bar{j}}$ is the pull-back of K\"ahler form $\omega_h$ via $f$. By the singular value decomposition for $x_0\in M$ and $f(x_0)\in N$ we may choose normal coordinates centered at $x_0$ and $f(x_0)$ such that $\partial f(\frac{\partial\ }{\partial z^\alpha})=\lambda_\alpha \delta^i_{\alpha} \frac{\partial\ }{\partial w^i}$. Then $|\lambda_\alpha|$ are the singular values of $\partial f: (T'_{x_0}M, g) \to (T_{f(x_0)}'N, h)$. It is easy to see that $|\lambda_1|^2 \ge \cdots \ge |\lambda_m|^2$ are the eigenvalues of $A$ (with respect to $g$).
\begin{proposition} \label{prop:21} For any $1\le \ell\le m$ the following holds
$$
\sigma_\ell\doteqdot \sum_{\alpha=1}^\ell |\lambda_\alpha|^2 \ge \sum_{1\le \alpha, \beta\le \ell} g^{\alpha \bar{\beta}}A_{\alpha\bar{\beta}}\doteqdot U_\ell.
$$
\end{proposition}
\begin{proof} Arguing invariantly we choose unitary basis of $T'_{x_0}M$ with respect to $g$. Then the left hand side is the partial sum of the eigenvalues of $A$ in descending order, and the right hand side is the trace of the first $\ell\times \ell$ block of $(A_{\alpha\bar{\beta}})$. Hence the result is well-known (cf. \cite{Horn-Johnson}, Corollary 4.3.34).
\end{proof}
For a linear map $L: \mathbb{C}^m\to \mathbb{C}^n$ between two Hermitian linear spaces, $\Lambda^\ell L: \wedge^\ell \mathbb{C}^m \to \wedge^\ell \mathbb{C}^n$ is define as the linear extension of the action on simple vectors: $\Lambda^\ell L({\bf{a}})\doteqdot L(v_1)\wedge\cdots\wedge L(v_\ell)$ with ${\bf{a}}=v_1\wedge\cdots \wedge v_\ell$. The metric on $\wedge^\ell \mathbb{C}^m$ is defined as $\langle {\bf{a}}, \overline{{\bf{b}}}\rangle =\det(\langle v_i, \bar{w}_j\rangle)$. If $\{e_\alpha\}$ is a unitary frame of $\mathbb{C}^m$, the $\{e_{\lambda}\}$, with $\lambda=(\alpha_1, \cdots, \alpha_\ell)$, $\alpha_1\le \cdots \le \alpha_\ell$, being the multi-index, and $e_{\lambda}=e_{\alpha_1}\wedge \cdots\wedge e_{\alpha_\ell}$, is a unitary frame for $\wedge^\ell \mathbb{C}^m$. The Binet-Cauchy formula implies that this is consistent with the Hermitian product $\langle {\bf{a}}, \overline{{\bf{b}}}\rangle$ defined in the previous section. The norm $\|\Lambda^\ell L\|_0$ is the operator norm with respect to the Hermitian structures of $\wedge^\ell \mathbb{C}^m $ and $\wedge^\ell \mathbb{C}^m$ defined above, which equals to the Jacobian of a Lipschitz map $f$, when $\ell=m$ or $n$, applying to $L=\partial f$ (cf. Section 3.1 of \cite{Fede}).
For the local Hermitian matrices $A=(A_{\alpha\bar{\beta}})$ and $G=(g_{\alpha\bar{\beta}})$ we denote $A_\ell$ and $G_{\ell}$ be the upper-left $\ell\times \ell$ blocks of them.
\begin{proposition}\label{prop:22} For any $1\le \ell\le m$ the following holds:
\begin{eqnarray}
\|\Lambda^\ell \partial f\|_0^2=\Pi_{\alpha=1}^\ell |\lambda_\alpha|^2&\ge& \frac{\det(A_\ell)}{\det(G_\ell)}\doteqdot W_\ell; \label{eq:21}
\end{eqnarray}
\end{proposition}
\begin{proof} For the inequality in (\ref{eq:21}), as in the above proposition we may choose a unitary frame of $T'_{x_0}M$ such that $G=\operatorname{id}$. Then the claimed result is also a well-known statement about the partial products of the descending eigenvalues. The result can be seen by applying 4.1.6 of \cite{MM} to $(A+\epsilon G)^{-1}$ and let $\epsilon \to 0$ (see also Problem 4.3.P15 of \cite{Horn-Johnson}).
For the equality (\ref{eq:21}), first observe that
$$
\|\Lambda^\ell \partial f\|^2_0(x) \ge \frac{| \partial f\left(v_1\right)\wedge \cdots \wedge \partial f\left(v_\ell\right)|^2}{|v_1\wedge\cdots \wedge v_\ell|^2}=\Pi_{\alpha=1}^\ell |\lambda_\alpha|^2
$$
if $\{v_\alpha \}$ are the eigenvectors of $A$ with eigenvalues $\{|\lambda_\alpha|^2\}$. On the other hand for general orthonormal vectors $\{v_\alpha\}$, the above paragraph implies $ \frac{| \partial f\left(v_1\right)\wedge \cdots \wedge \partial f\left(v_\ell\right)|^2}{|v_1\wedge\cdots \wedge v_\ell|^2}\le \Pi_{\alpha=1}^\ell |\lambda_\alpha|^2
$. Combining them we have the equality in (\ref{eq:21}). \end{proof}
\section{$\partial\bar{\partial}$-Bochner formulae}
Here we generalize the $\partial\bar{\partial}$-Bochner formula derived in \cite{Ni-1807} on $\|\partial f\|^2$ and $\|\Lambda^m \partial f\|_0^2$ to $\sigma_\ell$ and $\|\Lambda^\ell \partial f\|_0^2$. Since both $\sigma_\ell(x)$ and $\|\Lambda^\ell \partial f\|_0^2(x)$ are only continous in general we first derive formula on their barriers supplied by Proposition \ref{prop:21}, \ref{prop:22}.
\begin{proposition}\label{prop:31} Under the normal coordinates near $x_0$ and $f(x_0)$ such that $\partial f(\frac{\partial\ }{\partial z^\alpha})=\lambda_\alpha \delta^i_{\alpha} \frac{\partial\ }{\partial w^i}$ with $|\lambda_1|\ge \cdots\ge |\lambda_\alpha|\ge \cdots \ge |\lambda_m|$ being the singular values of $\partial f: (T'_{x_0}M, g) \to (T_{f(x_0)}'N, h)$, let $U_\ell(x)$ and $W_\ell(x)$ be the functions defined in the last section in a small neighborhood of $x_0$. Then at $x_0$, for $v\in T'_{x_0}M$, and nonzero $U_\ell$ and $W_\ell$,
\begin{eqnarray}
\langle \sqrt{-1}\partial \bar{\partial} \log U_\ell, \frac{1}{\sqrt{-1}}v\wedge \bar{v}\rangle &=&\frac{U_\ell \sum_{1\le i\le n, 1\le \alpha \le \ell} |f^i_{\alpha v}|^2-|\sum_{\alpha =1}^\ell \overline{\lambda_\alpha}f^\alpha_{\alpha v}|^2}{U_\ell^2}\label{eq:31}\\
&\quad&+\sum_{\alpha=1}^\ell \frac{|\lambda_\alpha|^2}{U_\ell}(-R^N(\alpha, \bar{\alpha}, \partial f(v), \overline{\partial f(v)})+R^M(\alpha, \bar{\alpha}, v, \bar{v}));\nonumber \\
\langle \sqrt{-1}\partial \bar{\partial} \log W_\ell, \frac{1}{\sqrt{-1}}v\wedge \bar{v}\rangle &=& \sum_{\alpha=1}^\ell \sum_{ \ell+1\le i \le n} \frac{|f^i_{\alpha v}|^2}{|\lambda_\alpha|^2}\label{eq:32} \\
&\quad& + \sum_{\alpha=1}^\ell (- R^N(\alpha, \bar{\alpha}, \partial f(v), \overline{\partial f(v)})+R^M(\alpha,\bar{\alpha}, v, \bar{v})). \nonumber
\end{eqnarray}
\end{proposition}
\begin{proof} The calculation is similar to that of \cite{Ni-1807}. Here we include the details of the first. Choose holomorphic normal coordinate $(z_1, z_2, \cdots, z_m)$ near a point $p$ on the domain manifold $M$, correspondingly $(w_1, w_2, \cdots, w_n)$ near $f(p)$ in the target. Let $\omega_g=\sqrt{-1}g_{a\bar{\beta}}dz^\alpha\wedge d\bar{z}^{\beta}$ and $\omega_h=\sqrt{-1}h_{i\bar{j}}dw^i\wedge d\bar{w}^{j}$ be the K\"ahler forms of $M$ and $N$ respectively. Correspondingly, the Christoffel symbols are given
$$
^M\Gamma_{\alpha \gamma}^\beta =\frac{\partial g_{\alpha \bar{\delta}}}{\partial z^{\gamma}}g^{\bar{\delta}\beta}=\Gamma_{\gamma \alpha }^\beta; \quad \quad ^N\Gamma_{i k}^j =
\frac{\partial h_{i \bar{l}}}{\partial w^{k}}h^{\bar{l}k}=\Gamma_{k i }^j.
$$
We always uses Einstein's convention when there is an repeated index. The symmetry in the Christoffel symbols is due to K\"ahlerity. If the appearance of the indices can distinguish the manifolds we omit the superscripts $^M$ and $^N$. Correspondingly the curvatures are given by
$$
^MR^\beta_{\alpha \bar{\delta} \gamma}=-\frac{\partial}{\partial \bar{z}^{\delta}} \Gamma_{\alpha \gamma}^\beta; \quad \quad \quad \,^NR^j_{i \bar{l} k}=-\frac{\partial}{\partial \bar{w}^{l}} \Gamma_{i k}^j.
$$
At the points $x_0$ and $f(x_0)$, where the normal coordinates are centered we have that
$$
R_{\bar{\beta}\alpha \bar{\delta} \gamma}=-\frac{\partial^2 g_{\bar{\beta}\alpha}}{\partial z^{\gamma}\partial \bar{z}^{\delta}}; \quad \quad R_{\bar{j}i \bar{l} k}=-\frac{\partial^2 h_{\bar{j}i}}{\partial w^{k}\partial \bar{w}^{l}}.
$$
Direct calculation shows that at the point $x_0$ (here repeated indices $\alpha, \beta $ are summed from $1$ to $\ell$, while $i, j, k, l$ are summed from $1$ to $n$)
\begin{eqnarray*}
(\log U_\ell)_\gamma &=&\frac{g^{\alpha \bar{\beta}}_{\quad, \gamma} A_{\alpha\bar{\beta}}+g^{\alpha \bar{\beta}}f^i_{\alpha \gamma}h_{i\bar{j}}\overline{f^j_\beta}+g^{\alpha \bar{\beta}}f^i_{\alpha }\overline{f^j_\beta}f^k_\gamma h_{i\bar{j}, k} }{U_\ell}=\frac{f^i_{\alpha\gamma}\overline{f^i_\alpha}}{U_\ell};\\
(\log U_\ell)_{\bar{\gamma}} &=&\frac{g^{\alpha \bar{\beta}}_{\quad, \bar{\gamma}} A_{\alpha\bar{\beta}}+g^{\alpha \bar{\beta}}f^i_{\alpha}h_{i\bar{j}}\overline{f^j_{\beta\gamma}}+g^{\alpha \bar{\beta}}f^i_{\alpha }\overline{f^j_\beta}\overline{f^k_\gamma} h_{i\bar{j}, \bar{k}} }{U_\ell}=\frac{\overline{f^i_{\alpha\gamma}}f^i_\alpha}{U_\ell};\\
\left(\log U_\ell\right)_{\gamma \bar{\gamma}}&=& \frac{R^M_{\alpha\bar{\beta}\gamma\bar{\gamma}}f^i_\alpha \overline{f^i_\beta}+|f^i_{\alpha \gamma}|^2-R^N_{i\bar{j}k\bar{l}}f^i_\alpha \overline{f^j_\beta}f^k_\gamma \overline{f^l_\gamma}}{U_\ell}-\frac{|\sum_{1\le \alpha\le \ell; 1\le i\le n} f^i_{\alpha \gamma}\overline{f^i_\alpha}|^2}{U_\ell^2}.
\end{eqnarray*}
The claimed equation then follows.
\end{proof}
\begin{corollary} Let $f: M\to N$ be a holomorphic map between two K\"ahler manifolds.
(i) If the bisectional curvature of $N$ is non-positive and the bisectional curvature of $M$ is nonnegative, then $\log \sigma_\ell(x)$ is a plurisubharmonic function.
(ii) Assume that $\Ric^N_\ell\le 0$ and $\Ric^M_\ell\ge0$. If $\|\Lambda^\ell \partial f\|^2_0$ not identically zero, then for every $x$, there exists a $\Sigma \subset T_x'M$ with $\dim(\Sigma)\ge \ell$ such that $\log \|\Lambda^\ell \partial f\|^2_0(x)$ is plurisubharmonic on $\Sigma$.
\end{corollary}
\section{Proof of Theorem \ref{thm:main1}}
Since in general $\sigma_\ell$ and $\|\Lambda^\ell \partial f\|_0$ are not smooth we adopt the viscosity consideration as in Section 5 of \cite{Ni-1807} to prove the result. We also need to modify the algebraic argument in the Appendix of \cite{Ni-1807} for some point-wise estimates needed. Another difference of the argument is that we shall apply the maximum principle to a degenerate operator. First we need a Royden type lemma.
\begin{lemma}\label{lem:41} If the holomorphic sectional curvature $R^N$ has a upper bound $-\kappa$, with respect
to the normal coordinates as in Proposition \ref{prop:21} at $x_0$ (and $f(x_0)$),
$$
\sum_{1\le \alpha,\beta, \gamma, \delta\le \ell} g^{\alpha \bar{\beta}}g^{\gamma \bar{\delta}}R^N_{i\bar{j}k\bar{l}}f^i_{\alpha} \overline{f^{j}_\beta} f^k_\gamma \overline{f^l_\delta} \le -\frac{\ell'+1}{2\ell'} \kappa U^2_\ell, \mbox{ when }\kappa>0; \quad \le -\kappa U^2_\ell \mbox{ when } \kappa\le 0.
$$
Here $\ell'=\min\{ \ell, \dim(f(M))\}$.
\end{lemma}
\begin{proof} We follow the argument in Appendix of \cite{Ni-1807}, which is due to F. Zheng. The left hand side can be written as $ \sum_{1\le \alpha, \beta\le \ell'} R^N_{\alpha\bar{\alpha}\beta\bar{\beta}}|\lambda_\alpha|^2|\lambda_\beta|^2$. In the space $$\Sigma\doteqdot \operatorname{span} \{ \partial f\left( \frac{\partial\ }{\partial z^1}\right), \cdots, \partial f\left( \frac{\partial\ }{\partial z^{\ell'}}\right)\}, $$ consider the vector $Y=\sum_{1\le i \le \ell'} w^i\lambda_i \frac{\partial \ }{\partial w^i}$ with $(w^1, \cdots, w^{\ell'})\in \mathbb{S}^{2\ell'-1}\subset \Sigma$. Then direct calculations show that
\begin{eqnarray*}
\sum_{1\le \alpha, \beta\le \ell'} R^N_{\alpha\bar{\alpha}\beta\bar{\beta}}|\lambda_\alpha|^2|\lambda_\beta|^2&=&\frac{\ell' (\ell'+1)}{2}\cdot \frac{1}{Vol (\mathbb{S}^{2\ell'-1})}\int_{\mathbb{S}^{2\ell'-1}} R^N(Y, \overline{Y}, Y,\overline{Y})\\
&\le & -\kappa \frac{\ell' (\ell'+1)}{2}\cdot \frac{1}{Vol (\mathbb{S}^{2\ell'-1})}\int_{\mathbb{S}^{2\ell'-1}} |Y|^4\\
&=& \frac{-\kappa }{2}\left(U_\ell^2+\sum_{1\le \alpha \le \ell'} |\lambda_\alpha|^4\right).
\end{eqnarray*}
The result follows from elementary inequalities $\sum_{1\le \alpha \le \ell'} |\lambda_\alpha|^4\le U_\ell^2 \le\ell'\, \sum_{1\le \alpha \le \ell'} |\lambda_\alpha|^4$.
\end{proof}
To prove part (i), let $\eta(t):[0, +\infty)\to [0, 1]$ be a function supported in $[0, 1]$ with $\eta'=0$ on $[0, \frac{1}{2}]$, $\eta' \le 0$, $\frac{|\eta'|^2}{\eta}+(-\eta'')\le C_1$. The construction of such $\eta$ is elementary.
Let $\varphi_R(x)=\eta(\frac{r(x)}{R})$. When the meaning is clear we omit subscript $R$ in $\varphi_R$. Clearly $\sigma_\ell \cdot \varphi$ attains a maximum somewhere at $x_0$ in $B_p(R)$. With respect to the normal coordinates near $x_0$ and $f(x_0)$, $(U_\ell \varphi)(x_0)=(\sigma \varphi)(x_0)$, and $(U_\ell\varphi)(x)\le (\sigma_\ell \varphi)(x)\le (\sigma\varphi)(x_0)\le (U_\ell \varphi)(x_0)$ for $x$ in the small normal neighborhood. The maximum principle then implies that at $x_0$
$$
\nabla (U_\ell \varphi)=0; \sum_{1\le \alpha \le \ell} \frac{1}{2}(\nabla_{\alpha}\nabla_{\bar{\alpha}} +\nabla_{\bar{\alpha}}\nabla_{\alpha}) \log (U_\ell \varphi) \le 0.
$$
Now applying the $\partial \bar{\partial}$ formula (\ref{eq:31}), the above Lemma and the complex Hessian comparison theorem of Li-Wang \cite{LW}, together with the argument in \cite{Ni-1807}, imply the result. It is clear from the proof that if $\ell=m=\dim(M)$, only the Laplacian comparison theorem is needed. Hence one only needs to assume that the Ricci curvature of $M$ is bounded from below.
The proof of part (ii) is similar. For the sake of the completeness we include the argument under the assumption (\ref{eq:2}). In this case we let $\varphi=\eta\left(\frac{\rho}{R}\right)$. Now $\varphi$ has support in $D(2R)\doteqdot \{\rho\le 2R\}$. Hence $W_\ell \cdot \varphi$ attains its maximum somewhere, say at $x_0 \in D(2R)$. Now at $x_0$ we have \begin{eqnarray*}
0&\ge& \sum_{\gamma=1}^\ell \frac{\partial^2}{\partial z^\gamma \partial z^{\bar{\gamma}}}\, \left(\log (W_\ell \, \varphi)\right) \ge \sum_{\alpha,\gamma=1}^\ell R^M_{\alpha\bar{\alpha}\gamma \bar{\gamma}}-R^N_{\alpha\bar{\alpha }\gamma\bar{\gamma}}|\lambda_\gamma|^2 + \sum_{\gamma=1}^\ell \frac{\partial^2 \log \varphi}{\partial z^\gamma \partial z^{\bar{\gamma}}} \\
&\ge& -K +\ell \cdot \kappa \cdot W_\ell^{1/\ell}+\frac{ \eta''}{R^2\varphi} |\nabla \rho|^2+\frac{\ell \eta'}{R \varphi}\left([\sqrt{-1}\partial \bar{\partial} \rho]_{+}\right)-\frac{|\eta'|^2}{\varphi^2 R^2}\cdot |\nabla \rho|^2\\
&\ge& -K +\ell \cdot \kappa \cdot W_\ell^{1/\ell} -\frac{C_1}{\varphi R^2}|\nabla \rho|^2 -\frac{C_1}{\varphi R} \cdot C(m)\left( [\sqrt{-1}\partial \bar{\partial} \rho]_{+}\right).
\end{eqnarray*}
Multiplying $\varphi$ on both sides of the above we have that
$$
\sup_{D(R)}\|\Lambda^\ell \partial f\|_0^2(x)\le \left(\frac{K+ +\frac{C_1}{\varphi R^2}|\nabla \rho|^2 +\frac{C_1}{\varphi R} \cdot C(m)\left( [\sqrt{-1}\partial \bar{\partial} \rho]_{+}\right)}{\ell \kappa}\right)^\ell.
$$
The result follows by observing that $\frac{|\nabla \rho|^2}{R^2}\le \frac{4|\nabla \rho|^2}{\rho^2} \to 0$ and $\frac{ [\sqrt{-1}\partial \bar{\partial} \rho]_{+}}{R}\le 2 \frac{ [\sqrt{-1}\partial \bar{\partial} \rho]_{+}}{\rho}\to 0$ as $R\to \infty$.
\section{Applications}
First we show that the Pogorelov type estimate of \cite{Ni-1807} can be adapted to derive the $C^2$-estimate for the Monge-Amp\`ere equation related to the existence of K\"ahler-Einstein metrics and prescribing the Ricci curvature problem. Recall that the geometric problems reduce to a complex Monge-Amp\`ere equation
$$
\frac{\det(g_{\alpha\bar{\beta}}+\varphi_{\alpha\bar{\beta}})}{\det(g_{\alpha\bar{\beta}})}=e^{t\varphi +f}
$$
with $t\in [-1, 1]$, $f$ being a fixed function with prescribed complex Hessian. $g'_{\alpha\bar{\beta}}=g_{\alpha\bar{\beta}}+\varphi_{\alpha\bar{\beta}}$ is another K\"ahler metric with $[\omega_{g'}]=[\omega_g]$. We apply our previous setting to the map $\operatorname{id}:(M, g)\to (M, g')$. The computation in \cite{Aub, Yau} (See also the exposition in \cite{Siu}) is on $\mathcal{L} \|\partial f\|^2$. By the computation from Section 3 and 4, at the point where $\|\partial \operatorname{id}\|^2_0$ is attained we have that
$$
0\ge \frac{\partial^2}{\partial z^\gamma \partial \bar{z}^\gamma} \log (1+\varphi_{1\bar{1}})\ge R_{1\bar{1}\gamma\bar{\gamma}}-R'_{1\bar{1}\gamma \bar{\gamma}}\left(1+\varphi_{\gamma\bar{\gamma}}\right).
$$
Here $R'$ is the curvature of $g'$ and $|\lambda_\gamma|^2=1+\varphi_{\gamma\bar{\gamma}}$. Since we do not have information on $R'$ in general, but only $\Ric^{g'}(\frac{\partial}{\partial z^1}, \frac{\partial}{\partial \bar{z}^1})=\Ric^g_{1\bar{1}}-t\varphi_{1\bar{1}}-f_{1\bar{1}}$, we multiply $\frac{1}{1+\varphi_{\gamma\bar{\gamma}}}$ on the both sides of the above inequality and then sum $\gamma$ from $1$ to $m$ arriving at
\begin{eqnarray*}
0&\ge& \sum_{\gamma=1}^m\frac{1}{1+\varphi_{\gamma\bar{\gamma}}}R^g_{1\bar{1}\gamma\bar{\gamma}}-
\frac{\Ric^g_{1\bar{1}}}{1+\varphi_{1\bar{1}}}
+t\frac{\varphi_{1\bar{1}}}{1+\varphi_{1\bar{1}}}+\frac{f_{1\bar{1}}}{1+\varphi_{1\bar{1}}}\\
&\ge& -C(M, g, f)\sum_{\gamma=1}^m\frac{1}{1+\varphi_{\gamma\bar{\gamma}}}-1.
\end{eqnarray*}
Now we apply/repeat the same consideration/calculation to $Q\doteqdot \log \sigma_1 -(C(M, g, f)+1)\varphi$. Then at the point $x_0$, where $Q$ attains its maximum, we have that
$$
0\ge -C(M, g, f)\sum_{\gamma=1}^m\frac{1}{1+\varphi_{\gamma\bar{\gamma}}}-(C(M, g, f)+2) +(C(M, g, f)+1)\sum_{\gamma=1}^m\frac{1}{1+\varphi_{\gamma\bar{\gamma}}},
$$
which then implies that
$$
\sum_{\gamma=1}^m\frac{1}{1+\varphi_{\gamma\bar{\gamma}}}\le C(M, g, f)+2.
$$
This implies that at the maximum point of $\sigma_1 e^{-(C(M, g, f)+1)\varphi}$,
\begin{eqnarray*}
\sigma_1 e^{-(C(M, g, f)+1)\varphi}&=&\sigma_1\frac{\omega^m_g}{\omega^m_{g'}}e^{t\varphi +f}e^{-(C(M, g, f)+1)\varphi}\\
&\le& \left(\frac{1}{m-1} \sum_{\gamma=2}^m \frac{1}{1+\varphi_{\gamma\bar{\gamma}}}\right)^{m-1} e^{t\varphi +f}e^{-(C(M, g, f)+1)\varphi}\\
&\le& \left(\frac{C(M, g, f)+2}{m-1} \right)^{m-1} e^{t\varphi +f}e^{-(C(M, g, f)+1)\varphi}.
\end{eqnarray*}
If we write $K=\left(\frac{C(M, g, f)+2}{m-1} \right)^{m-1}$, $\kappa=C(M, g, f)+2$, the above implies
\begin{equation}
1+\varphi_{\gamma\bar{\gamma}}(x)\le \sigma_1(x)\le Ke^{\kappa (\varphi(x)-\varphi(x_0))} e^{t\varphi(x_0) +f(x_0)}, \quad \forall \gamma\in \{1, \cdots, m\}.
\end{equation}
As mentioned in the introduction, Theorem \ref{thm:main1} removes the constrains that $\dim(M)\le \dim(N)$ in the previous results proved in \cite{Ni-1807}. As in \cite{NZ} we denote by $B^\perp$ the orthogonal bisectional curvature. We say $B^\perp\le \kappa$ if for any $X, Y\in T'N$ with $\langle X, \overline{Y}\rangle=0$, $R(X, \bar{X}, Y, \bar{Y})\le \kappa |X|^2|Y|^2$. The following is a corollary of the proof Theorem \ref{thm:main1}.
\begin{theorem}\label{thm:51} Let $f: (M, g)\to (N, h)$ be a holomorphic map.
(i) Assume that $M$ is compact. Under the assumptions either $\Ric^M_\ell >0$, and the holomorphic sectional curvature $H^N\le 0$, or $\Ric^M_\ell\ge 0$ and $H^N < 0$, $f$ must be constant. The same result also holds if $(B^M)^\perp>0$ and $(B^N)^\perp\le 0$ or $(B^M)^\perp\ge0$ and $(B^N)^\perp< 0$.
(ii) If $M$ is compact with $S^M_\ell\ge 0$ and $\Ric^N_\ell <0$, or $S^M_\ell >0$, $\Ric^N_\ell \le 0$ then $\dim(f(M))< \ell$. The same result holds if $\Ric^M_\ell\ge 0$ and $S^N_\ell <0$, or $\Ric^M_\ell> 0$ and $S^N_\ell \le 0$.
\end{theorem}
\begin{proof} Since $M$ is compact $\sigma_\ell$ attains a maximum somewhere, say at $x_0$. If $f$ is not constant, $\sigma_\ell(x_0)>0$. Applying (\ref{eq:31}), using the normal coordinates around $x_0$ and $f(x_0)$ specified as in the last two sections we have that
$$
0\ge \sum_{\gamma=1}^\ell \frac{\partial^2\ }{\partial z^\gamma, \partial \bar{z}^\gamma}\left(\log U_\ell\right)\ge \sum_{1\le \alpha, \gamma \le \ell} \frac{-R^N_{\alpha\bar{\alpha}\gamma\bar{\gamma}}|\lambda_\alpha|^2|\lambda_\gamma|^2}{U_\ell}+ \sum_{\alpha =1}^\ell \frac{\Ric^M(x_0, \Sigma)(\alpha, \bar{\alpha})|\lambda_\alpha|^2}{U_\ell}.
$$
Here $\Sigma=\operatorname{span}\{ \frac{\partial\ }{\partial z^1}, \cdots, \frac{\partial\ }{\partial z^\ell} \}$. By Lemma \ref{lem:41}, if $H^N<0$, the first term is positive, the second one is nonnegative since $\Ric^M_\ell\ge 0$. Hence a contradiction. From the proof, the same holds if $H^N\le0$ and $\Ric^M_\ell>0$. For the case concerning $B^\perp$ the proof is similar.
For (ii), if $\operatorname{rank}(f)\ge \ell$, $\|\Lambda^\ell\partial f\|_0$ has a nonzero maximum somewhere, say at $x_0$. Then applying (\ref{eq:32}), using the normal coordinates around $x_0$ and $f(x_0)$ specified as in the last two sections we have that
$$
0\ge \sum_{\gamma=1}^\ell \frac{\partial^2\ }{\partial z^\gamma, \partial \bar{z}^\gamma}\left(\log W_\ell\right)\ge \sum_{1\le \gamma \le \ell} (-\Ric^N_\ell(x_0, \Sigma) |\lambda_\gamma|^2)+\operatorname{Scal}^M(x_0, \Sigma).
$$
This leads to a contradiction under the assumptions either $S^M_\ell\ge 0$ and $\Ric^N_\ell <0$, or $S^M_\ell >0$, $\Ric^N_\ell \le 0$. For the second part, we introduce the operator:
$$
\mathcal{L}_\ell=\sum_{\gamma=1}^\ell\frac{1}{2|\lambda_{\gamma}|^2}\left(\nabla_{\gamma}\nabla_{\bar{\gamma}}
+\nabla_{\bar{\gamma}}\nabla_{\gamma}\right).
$$
Since at $x_0$ $\W_\ell\ne 0$, the above operator is well defined in a small neighborhood of $x_0$. As before applying $\mathcal{L}$ at $x_0$ implies that
$$
0\ge \mathcal{L}_\ell \left(\log W_\ell\right)\ge (-\operatorname{Scal}^N(x_0, \partial f(\Sigma))+\sum_{1\le \gamma \le \ell} \frac{\Ric^M(x_0, \Sigma)(\gamma, \bar{\gamma})}{|\lambda_\gamma|^2}.
$$
The above also induces a contradiction under either $\Ric^M_\ell\ge 0$ and $S^N_\ell <0$, or $\Ric^M_\ell> 0$ and $S^N_\ell \le 0$.
\end{proof}
This can be combined with the following result of Siu-Beauville (cf. Theorem 1.5 of \cite{ABCKT}) to infer information regarding the fundamental group of the manifolds with $\Ric_\ell\ge0$.
\begin{theorem}[Siu-Beauville] Let $M$ be a compact K\"ahler manifold. There exists a compact Riemann surface $C_g$ of genus greater than one and a surjective holomorphic map $f: M \to C'$ with $g(C')\ge g(C)$ with connected fibers if and only if there exists a surjective homomorphism $h: \pi_1(M)\to \pi_1(C_g)$.
\end{theorem}
\begin{corollary} (i) Let $(M, g)$ be a compact K\"ahler manifold with $\Ric_\ell\ge 0$ for some $1\le \ell\le m$. Then
there exists no surjective homomorphism $h: \pi_1(M)\to \pi_1(C_g)$. Furthermore, there is no subspace $V\subset H^1(M, \mathbb{C})$ with $\wedge^2 V=0$ in $H^2(M, \mathbb{C})$ and $\dim(V)\ge 2$. Namely $g(M)\le 1$. Similarly, if $\Ric_\ell\ge 0$, $\pi_1(M)$ can not be of the type of an amalgamated product $\Gamma_1*_{\Delta}\Gamma_2$ with the index of $\Delta$ in $\Gamma_1$ greater than one and index of $\Delta$ in $\Gamma_2$ greater than two.
(ii) Let $(M, g)$ be a compact K\"ahler manifold with $S^M_\ell> 0$ for some $1\le \ell\le m$. Then $a(M)\le\ell-1$.
(iii) If $S^M_n\ge 0$, then any harmonic map $f: M\to N$ with $N$ being a locally Hermitian symmetric space, can not have $\operatorname{rank}(f)=\dim(N)$.
\end{corollary}
\begin{proof} The first part of (i) follows from part (i) of Theorem \ref{thm:51}. Namely apply it to $N=C_g$ and combine it with the above Siu-Beauville's result. The second part follows by combining Theorem \ref{thm:51} with Theorem 1.4 of \cite{ABCKT} due to Catanese (cf. Theorem 1.10 of \cite{Cat}). For the second part involving the amalgamated product, apply Theorem 6.27 of \cite{ABCKT}, namely a result of Gromov-Schoen below instead, to conclude that there exists an equivariant holomorphic map from $\widetilde{M} $ into the Poincar\'e disk. This induce a contradiction with part (i) of Theorem \ref{thm:51} since the maximum principle argument still applies (see also \cite{NR}). The statement of (ii) is an easy consequence of part (ii) of Theorem \ref{thm:51}.
For part (iii), by Siu's result on the holomorphicity of the harmonic maps between K\"ahler manifolds, namely Theorem 6.13 of \cite{ABCKT}, any such a harmonic map must be holomorphic. Then part (ii) of Theorem \ref{thm:51} induces a contradiction noting that the canonical metric on $N$ is K\"ahler-Einstein with negative Einstein constant.
\end{proof}
\begin{theorem}[Gromov-Schoen]
Let $M$ be a compact K\"ahler manifold with fundamental group $\Gamma=\Gamma_1*_{\Delta}\Gamma_2$ with the index of $\Delta$ in $\Gamma_1$ greater than one and index of $\Delta$ in $\Gamma_2$ greater than two. Then there exists a representation $\rho: \pi_1(M)\to \operatorname{Aut}(\mathbb{D})$, where $\mathbb{D}=\{z\, |\, |z|=1\}$, with discrete cocompact image, and a holomorphic equivariant map from the universal cover $\widetilde{M}\to \mathbb{D}$, which also descends to a surjective map $M\to \rho(\Gamma)/\mathbb{D}$.
\end{theorem}
In fact the vanishing theorem of \cite{Ni-Zheng2} implies that for K\"ahler manifolds with $S_\ell>0$, there does not exist a $k$-wedge subspace in $H^{1, 0}$ (in the sense of \cite{Cat}) for any $k\ge \ell$. Moreover, such manifolds have to be Albanese primitive for $k\ge \ell$.
For noncompact manifolds, Theorem \ref{eq:sch-ni} and Theorem \ref{thm:main1} can also be applied, together with Theorem 4.14 and 4.28 of \cite{ABCKT}, to infer some restriction on K\"ahler manifolds with nonnegative holomorphic sectional curvature or with $\Ric_\ell\ge 0$.
\begin{corollary} Assume that $M$ is a complete K\"ahler manifold with bounded geometry with $\Ric^M_\ell\ge 0$. Then
(i) $H^1(M, \mathbb{C})=\{0\}$ implies that $\mathcal{H}^1_{L^2}(M)=\{0\}$;
(ii) And $\dim (\mathcal{H}^1_{L^2, ex}(M))\le 1$.
\end{corollary}
Here $\mathcal{H}_{L^2}(M)$ is the space of the harmonic $L^2$-forms and $\mathcal{H}^1_{L^2, ex}(M)$ is the space of the $L^2$ harmonic exact forms. The statements are trivial when $M$ is compact.
\section{Mappings from positively curved manifolds}
In \cite{NZ}, the orthogonal $\Ric^\perp$ was studied. Recall that $\Ric^\perp (X, \overline{X})=\Ric(X, \overline{X})-H(X)/|X|^2$. We say $\Ric^\perp\ge K$ if $\Ric^\perp (X, \overline{X})\ge K|X|^2$. It is easy to see that $B^\perp\ge \kappa$ implies that $\Ric^\perp \ge (m-1)\kappa$. Similar upper estimate also holds if $B^\perp$ is bounded from above. It was also shown in \cite{NZ} via explicit examples that $B^\perp$ is independent of the holomorphic sectional curvature $H$, as well as the Ricci curvature. Similarly $\Ric^\perp$ is independent of $\Ric$, as well as $H$. It was proved in \cite{NZ} that for manifold whose $\Ric^\perp$ has a positive lower bound, the manifold is compact with an effective diameter uppper bound. (See \cite{Tsu} for the corresponding result for holomorphic sectional curvature.) It is not hard to see that for K\"ahler manifolds with $\Ric_\ell\ge K>0$, they must be compact with an upper diameter estimate.
Applying $\partial\bar{\partial}$-Bochner formulae we have the following estimates in the spirit of \cite{Ni-1807}.
\begin{theorem}\label{thm:hoop}
(i) Assume that $\Ric^M_\ell (X, \overline{X})\ge K|X|^2$, and $H^N(Y)\le \kappa |Y|^4$, with $K, \kappa>0$. Then for any nonconstant $f: M\to N$
$$
\max_{x\in M} \sigma_\ell(x)\ge \frac{K}{\kappa}.
$$
(ii) Assume that $(B^M)^{\perp}\ge K$, and $(B^N)^\perp\le \kappa$, with $K, \kappa>0$. Then for any nonconstant $f: M\to N$, $\dim(f(M))=m$. Moreover for any $\ell<\dim(M)$
$$
\max_{x\in M} \sigma_\ell(x)\ge \ell \frac{K}{\kappa}.
$$
(iii) Assume that $\Ric^M_\ell\ge K$, and that $\Ric^N_\ell \le \kappa $, with $K,\kappa>0$. Then for any holomorphic map $f:M\to N$ with $\dim(f(M))\ge \ell$
$$
\max_{x}\|\Lambda^\ell \partial f\|_0^2(x) \ge \left(\frac{K}{\kappa}\right)^\ell.
$$
(iv) Assume that $(\Ric^M)^\perp \ge K$, and that $(B^N)^\perp \le \kappa $, with $K,\kappa>0$. Then for any holomorphic map $f:M\to N$ with $\dim(f(M))\ge m-1$, $\dim(f(M))=m$. Moreover
$$
\max_{x}\|\Lambda^m \partial f\|_0^2(x) \ge \left(\frac{K}{(m-1)\kappa}\right)^{m}.
$$
In the case $\dim(M)=\dim(N)$, only $(\Ric^N)^{\perp}\le (m-1)\kappa$ is needed. In general $(B^N)^\perp \le \kappa $ can be weakened to $(\Ric^N_m)^{\perp}\le (m-1)\kappa$. Here $(\Ric^N_\ell)^{\perp}$ is the orthogonal Ricci curvature of the curvature tensor $R^N$ restricted to $m$-dimensional subspaces.
\end{theorem}
\begin{proof} First observe that under any assumption of the above theorem $M$ is compact. From Lemma \ref{lem:41} and (\ref{eq:31}), part (i) follows. For part (ii), at the point $x_0$ where $\sigma_\ell(x)$ attains its maximum, applying (\ref{eq:31}) to $v=\frac{\partial\ }{\partial z^m}$, we have that
$$
0\ge -\kappa |\lambda_m|^2+K
$$
which implies that $|\lambda_m|^2\ge \frac{K}{\kappa}$. Then claimed estimate follows from $\sigma_\ell\ge \ell |\lambda_m|^2$.
For part (iii), we apply (\ref{eq:32}) at the point $x_0$, where $\|\Lambda^\ell \partial f\|_0^2(x)$ attains its maximum. In particular we apply it to $v=\frac{\partial\ }{\partial z^\ell}$ and let $\Sigma=\operatorname{span}\{ \frac{\partial\ }{\partial z^1}, \cdots, \frac{\partial\ }{\partial z^\ell}\}$. Hence at $x_0$
$$
0\ge -\Ric^N(x_0, f(\Sigma)) |\lambda_\ell|^2+\Ric^M (x_0, \Sigma).
$$
Hence we derive that $|\lambda_\ell|^2\ge \frac{K}{\kappa}$. The claimed result then follows.
The part (iv) can be proved similarly.
\end{proof}
The part (ii) of the theorem is not as strong as it appears, since $B^\perp>0$ implies that $h^{1,1}(M)=1$. On the other hand we have the following observation.
\begin{proposition}
Let $M$ be a K\"ahler manifold with $h^{1,1}(M)=1$. Then any holomorphic map $f: M\to N$, with $\dim(f(M))<\dim(M)$ must be a constant map. Hence $g(M)\le 1$, if $\dim(M)\ge 2$. In particular, if the Picard number $\rho(M)=1$ and $S_2^M>0$, any holomorphic map $f: M\to N$, with $\dim(f(M))<\dim(M)$ must be a constant map.
\end{proposition}
\begin{proof}
In fact $f^*\omega_h$, with $\omega_h$ being the K\"ahler form of $N$, is a $d$-closed positive $(1,1)$-form. By the assumption $[f^*\omega_h]$ proportional to $[\omega_g]$. Hence it must be either zero or a positive multiple of $[\omega_g]$. Since the second case implies that $\dim(f(M))=m$, only the first case can occur, which implies that $f$ is a constant map.
Note that this implies that for any K\"ahler manifold $M$ with $\dim(M)\ge 2$ and $h^{1,1}(M)=1$, the genus $g(M)\le 1$, in view of the result of Catanese (cf. Theorem 1.10 of \cite{Cat}) since otherwise there exists a nonconstant holomorphic map $f: M\to C_g$ with $C_g$ being a Riemann surface of genus $g(M)$. Since the first Chern class map $c_1: H^{1}(M, \mathcal{O}^*)\to \mathcal{H}^{1,1}(M)\cap H^2(M, \mathbb{Z})$ is onto, and $S^M_2>0$ implies that $H^2(M, \mathbb{C})=\mathcal{H}^{1,1}(M)$, the assumption then implies $h^{1,1}(M)=1$. The last result then follows from the first.
\end{proof}
Taking $\kappa\to 0$, the part (ii) of Theorem \ref{thm:hoop} also implies that any holomorphic map from a compact manifold with $B^\perp>0$ into one with $B^\perp\le 0$ must be a constant map (cf. Theorem \ref{thm:51}). Given that $B^\perp$ is independent of $H$ and $\Ric$, this does not follow from Yau-Royden's estimate Theorem \ref{thm-sch-roy}, nor from Theorem \ref{thm:sch1}. The part (iv) provides an additional information on compact K\"ahler manifolds with $\Ric^\perp>0$.
\section*{Acknowledgments} {We would like to thank James McKernan (particularly bringing my attention to the work \cite{Laz}) and Fangyang Zheng for conversations regarding holomorphic maps from $\mathbb{P}^m$. We are also grateful to Yanyan Niu for informing \cite{Mu}.}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,808
|
A shield I personally have, it is extremly unique as the only way to obtain it is via the Warlock Dreadsteed quest (http://www.wowhead.com/?quest=7631) as the only way to summon the boss that drops this sheild is via that quest. I love the looks of the shield, and it has a decently high armor, block, and pretty decent stats. The Stamina is a bit on the low side but I slapped a +7 Stam enchant on it and bam! I was good to go.
It also greatly resembles the Horde icon. Good for RPing maybe? :) I think so.
Seems to be no longer available since 4.0 and the removal of the Dreadsteed quest.
Edit: Ok, so its still available but good luck finding a warlock who actually has the mats!
If anyone wants to run this on Horde side, you can add me, jellobiafra#11155. More than happy to help out!
Any Warlock who can summon it on EU server to help me get the Shield for Xmog maybe? I can give some pet to sell for help or something.
This shield was by dropped by Lord Hel'nurath in Dire Maul, who was involved in the warlock epic mount quest chain. The quest chain was removed upon the release of the Cataclysm expansion, but Lord Hel'nurath can still be summoned provided you can find a warlock who still has the required summoning materials.
|
{
"redpajama_set_name": "RedPajamaC4"
}
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\section{Exact analytical dispersion relation for isolated system in the thermodynamic limit}
In the derivation of our main results, we have used the bare retarded Green's function of the isolated system in the thermodynamic limit. In this section we derive the dispersion relation for the clean long-range isolated system. First, we take our system Hamiltonian in Eq.(7) and re-label the sites to $r\rightarrow r-\frac{N}{2}-1$, assuming, for simplicity, that $N$ is even,
\begin{align}
\label{Hs2}
\hat{\mathcal{H}}_S = -\sum_{m=1}^N \left(\sum_{r=-N/2}^{N/2-1-m} \frac{1}{m^\alpha} \left(\hat{c}_r^\dagger \hat{c}_{r+m} + \hat{c}_{r+m}^\dagger \hat{c}_{r}\right)\right).
\end{align}
Next, we take $N\rightarrow \infty$ in the above equation. This gives,
\begin{align}
\label{Hs3}
\hat{\mathcal{H}}_S = -\sum_{r=-\infty}^{\infty} \sum_{m=1}^\infty \frac{1}{m^\alpha} \left(\hat{c}_r^\dagger \hat{c}_{r+m} + \hat{c}_{r+m}^\dagger \hat{c}_{r}\right),
\end{align}
where we have neglected some boundary terms. Now the system has translational invariance. We can diagonalize the above Hamiltonian by going to momentum space via a Fourier transform,
\begin{align}
& \hat{\tilde{c}}(k) = \sum_{r=-\infty}^\infty \hat{c}_r e^{i rk},~~k \in [-\pi, \pi] \nonumber \\
& \hat{c}_r = \frac{1}{2\pi} \int_{-\pi}^{\pi} dk e^{-i rk} \hat{\tilde{c}}(k).
\end{align}
This gives
\begin{align}
\hat{\mathcal{H}}_S = \int_{-\pi}^{\pi} dk~\varepsilon(k,\alpha) \hat{\tilde{c}}^\dagger(k) \hat{\tilde{c}}(k),
\end{align}
with the dispersion relation
\begin{align}
\varepsilon(k,\alpha) = -2\sum_{m=1}^\infty \frac{\cos(m k)}{m^\alpha} .
\end{align}
The retarded Green's function of the isolated system in the thermodynamic limit in the momentum-frequency space is then given by
\begin{align}
g^+(k,\omega) = \lim_{\epsilon\rightarrow 0}\frac{1}{\omega - \varepsilon(k,\alpha)-i\epsilon}.
\end{align}
Returning to the site basis $(p,q=\{1,N\})$, we get the bare retarded Green's function as
\begin{align}
\mathbf{g}_{pq}^+(\omega) = \lim_{\epsilon\rightarrow 0} \frac{1}{2\pi} \int_{-\pi}^{\pi} dk \frac{e^{-ik |p-q|}}{\omega - \varepsilon(k,\alpha)-i\epsilon}.
\end{align}
\section{Properties of the dispersion relation}
\begin{figure}
\includegraphics[width=\textwidth]{numerical_dispersion.pdf}
\caption{(Color online) In the first row, numerically we have plotted $\varepsilon(k,\alpha)+2\zeta(\alpha)$ in small $k$ regime for $\alpha=1.5, 2.5$ and $3.5$ respectively. For $\alpha<3$, $\varepsilon(k,\alpha)+2\zeta(\alpha)\sim k^{\alpha-1}$ and for $\alpha>3$, $\varepsilon(k,\alpha)+2\zeta(\alpha)\sim k^2$. This also matches with our analytical result eq.~\ref{small_k_dispersion}. Similarly, in the second row, we have plotted $\vert \varepsilon(k\pm \pi)-2\eta(\alpha)\vert$ in small $k$ regime for three different values of $\alpha$. Here we can see $\vert \varepsilon(k\pm \pi)-2\eta(\alpha)\vert$ always goes as $k^2$. This also matches with our analytical result eq.~\ref{k_pi_dispersion}.}
\label{fig:dispersion}
\end{figure}
In this section, we state the relevant properties of the dispersion relation which are crucial for the proof of the scaling in conductance. The infinite series giving the dispersion relation is absolutely convergent for all $k$ when $\alpha>1$. The following relation holds connecting $\varepsilon(k,\alpha)$ with $\varepsilon(k \pm \pi,\alpha)= 2\sum_{m=1}^\infty (-1)^{m-1} \frac{\cos(m k)}{m^\alpha}$,
\begin{align}
\label{dispersion_property}
\varepsilon(k \pm \pi,\alpha) = - \varepsilon(k,\alpha) + 2^{1-\alpha} \varepsilon (2k, \alpha)
\end{align}
The minimum of the dispersion relation is given at $k=0$,
\begin{align}
\varepsilon_{min}(\alpha)=\varepsilon(0,\alpha)=-2 \, \zeta(\alpha),
\end{align}
where $\zeta(\alpha)=\sum_{m=1}^\infty \frac{1}{m^\alpha}$ is the Reimann-zeta function. Similarly, the maximum of the dispersion relation is at $k=\pm \pi$,
\begin{align}
\varepsilon_{max}(\alpha)=\varepsilon(\pm \pi,\alpha)=2 \, \eta(\alpha),
\end{align}
where $\eta(\alpha)$ is the Dirichlet eta function, $\eta(\alpha)=\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m^\alpha}=(1-2^{1-\alpha})\zeta(\alpha)$, also known as the alternating zeta function. Importantly, the dispersion relation is non-analytic at $k=0$ for all values of $\alpha$. This is because,
\begin{align}
\lim_{k\rightarrow 0} \frac{\partial^p \varepsilon(k,\alpha)}{\partial k^p} \rightarrow \infty,~~\forall~p>\alpha-1.
\end{align}
Thus a Taylor expansion around $k=0$ does not exist. However, we still need to find the small $k$ behavior of $\varepsilon(k,\alpha)$. To do this we define
\begin{align}
f(s k,\alpha)=\zeta(\alpha)+ \frac{\varepsilon(s k,\alpha)}{2}=\sum_{m=1}^\infty \frac{1-\cos(s m k)}{m^\alpha}.
\end{align}
Here, $s$ is an integer. Next, we divide the summation in the right-hand-side into two parts,
\begin{align}
& f(s k,\alpha)=\sum_{m=1}^{1/s |k|} \frac{1-\cos(s m k)}{m^\alpha} + |k|^\alpha B, \nonumber \\
& B=\sum_{y\geq 1/s} \frac{1-\cos(s y)}{y^\alpha} \simeq s^{\alpha -1} \int_1^\infty dy^{\prime} \frac{1-\cos(y^{\prime})}{y^{\prime \alpha}} \nonumber \\
&\simeq s^{\alpha -1} B^{\prime}, ~~B^{\prime}=\int_1^\infty dy^{\prime} \frac{1-\cos(y^{\prime})}{y^{\prime \alpha}} \nonumber
\end{align}
where we have used $y=m|k|$ and $y^{\prime}=s y$. The expression in the definition of $B^{\prime}$ converges. Thus $B^{\prime}$ is a real number which depends on $\alpha$. Now we expand the cosine to obtain
\begin{align}
f(sk,\alpha)=s^{\alpha -1}|k|^\alpha B^{\prime} - \sum_{p=1}^\infty \frac{(-1)^p s^{2p} k^{2p}}{(2p)!} \sum_{m=1}^{1/s|k|} \frac{1}{m^{\alpha-2p}}.
\end{align}
Till now, the expression is exact. After this we make some approximations and assumptions. We replace the summation over $m$ by an integration and further assume that $\alpha$ is not an integer. This then gives,
\begin{align}
\label{general}
& f(sk,\alpha)\simeq s^{\alpha -1}|k|^\alpha B^{\prime} - \sum_{p=1}^\infty \frac{(-1)^p s^{2p} k^{2p}}{(2p)!} \frac{(1/s|k|)^{2p-\alpha+1} -1}{2p-\alpha+1} \nonumber \\
& = s^{\alpha -1}|k|^\alpha B^{\prime} + s^{\alpha-1}|k|^{\alpha-1} a_1 - \sum_{p=1}^\infty \frac{(-1)^{p+1} s^{2p}k^{2p}}{(2p)!~(2p-\alpha+1)} \nonumber ,\\
& a_1 = \sum_{p=1}^\infty \frac{(-1)^{p+1} }{(2p)!~(2p-\alpha+1)},~~\alpha>1, \alpha \notin \mathbb{Z},
\end{align}
where $\mathbb{Z}$ is the set of all integers.
It can be checked by ratio test that the infinite series in the definition of $a_1$ converges. So, $a_1$ is a real number which depends on $\alpha$. Now considering $s=1$, we have an approximate series expansion for $\varepsilon(k,\alpha)$ around $k=0$,
\begin{align}
\label{kequation}
\varepsilon(k,\alpha) & \simeq -2\Big[\zeta(\alpha)-|k|^{\alpha-1} a_1-|k|^\alpha B^{\prime} \nonumber \\
&+\sum_{p=1}^\infty \frac{(-1)^{p+1} k^{2p}}{(2p)!~(2p-\alpha+1)} \Big]
\end{align}
It is interesting to note that $a_1$, which is the coefficient of $|k|^{\alpha-1}$, has contribution from all terms coming from the expansion of the cosine. This is consistent with the fact that Taylor series expansion around $k=0$ is invalid. The presence of absolute values and the terms raised to non-integer powers, both of which make $k=0$ non-analytic, clearly distinguishing the above series expansion from a Taylor series expansion. Armed with the series expansion, we obtain the small $k$ behavior of the dispersion relation by keeping the lowest order terms with non-integer and integer powers,
\begin{align}
\label{small_k_dispersion}
& \varepsilon(k,\alpha)\simeq-2\left[ \zeta(\alpha)- a_1 |k|^{\alpha-1}- a_2 k^2 \right],~~|k| \ll 1, \\
& a_2 = \frac{1}{2(\alpha-3)}. \nonumber
\end{align}
Similarly putting $s=2$ in Eq.(\ref{general}) one can compute,
\begin{align}
\label{2kequation}
\varepsilon(2k,\alpha) & \simeq -2\Big[\zeta(\alpha)-2^{\alpha-1}|k|^{\alpha-1} a_1- 2^{\alpha -1}|k|^\alpha B^{\prime} \nonumber \\
&+\sum_{p=1}^\infty \frac{(-1)^{p+1} 2^{2p}k^{2p}}{(2p)!~(2p-\alpha+1)} \Big]
\end{align}
Further, using Eq.(\ref{kequation}) and Eq.(\ref{2kequation}) in Eq.(\ref{dispersion_property}) and considering the leading order term $p=1$ we can also obtain an equivalent expansion around $k=\pm \pi$,
\begin{align}
\label{k_pi_dispersion}
& \varepsilon(k\pm \pi,\alpha)\simeq 2\,\eta(\alpha)- 2 a_2 (1-2^{3-\alpha}) k^2,\nonumber \\
&|k| \ll 1.
\end{align}
which is analytic as the expansion contains only integer powers. The above two expansions are used to obtain the scaling of current with system size. We have also checked this two expansions eq.~\ref{small_k_dispersion} and eq.~\ref{k_pi_dispersion} numerically in fig.\ref{fig:dispersion}. Though the above expansions are obtained for non-integer values of $\alpha$, the results can be analytically continued to include integer values of $\alpha>1$ by making the fractional part arbitrarily small.
{
\section{The NEGF formalism}
In the main text we have used the expression of conductance at zero temperature as obtained from the non-equilibrium Green's function (NEGF) approach. Here we give details of this approach. This pedagogical section follows standard texts and references \cite{Jauho_book,di_Ventra_book,Dhar_2006}.
We want to describe an open system set-up of the form $\hat{\mathcal{H}}=\hat{\mathcal{H}}_S+\hat{\mathcal{H}}_{SB}+\hat{\mathcal{H}}_{B_1}+\hat{\mathcal{H}}_{B_N}$,
\begin{align}
\hat{\mathcal{H}}_S = \sum_{\ell m=1}^N \mathbf{H}_{\ell m} \hat{c}_\ell^\dagger \hat{c}_m, ~~\hat{\mathcal{H}}_{B_1}=\sum_{r=1}^{N_B} \Omega_{r1} \hat{B}_{r1}^\dagger\hat{B}_{r1} ,~~\hat{\mathcal{H}}_{B_N}=\sum_{r=1}^{N_B} \Omega_{rN} \hat{B}_{rN}^\dagger\hat{B}_{rN},~~\hat{\mathcal{H}}_{SB}=\sum_{\ell=1,N}\sum_{r=1}^{N_B} (\kappa_{r\ell} \hat{c}_\ell^\dagger\hat{B}_{r\ell} +~\kappa_{r\ell}^* \hat{B}_{r\ell}^\dagger\hat{c}_\ell),
\end{align}
where $\hat{c}_r$ is the fermionic annihilation operator at the $r$th site of the system, $\hat{B}_{r1}$ ($\hat{B}_{rN}$) is the fermionic annhilation operator of the $r$th mode of the left (right) bath and $N_B$ is the number of modes in the baths, which will shortly be taken to infinity. We will assume that $\mathbf{H}$ is a real symmetric matrix. The Hamiltonian of the entire set-up can be written in the form
$
\hat{\mathcal{H}}= \sum_{p,q=1}^{N+2N_B} \mathbf{H}_{p,q}^{\rm tot} \hat{d}_p^\dagger \hat{d}_q,
$
where $\hat{d}_p$ is the fermionic annihilation operator of either a system or a bath site. The retarded single-particle Green's function of the entire set-up in frequency space is given by the $(N+2N_B) \times (N+2N_B)$ matrix,
\begin{align}
\mathbf{G}^{{\rm tot} +}(\omega) = \big[(\omega-i\epsilon) \mathbb{I}-\mathbf{H}^{\rm tot}\big]^{-1} \Rightarrow \left[(\omega-i\epsilon) \mathbb{I}-\mathbf{H}^{\rm tot}\right]\mathbf{G}^{{\rm tot} +}(\omega)=\mathbb{I},
\end{align}
where $\mathbb{I}$ is the identity matrix of the corresponding dimension and $\epsilon$ is a small positive number that takes care of the causality condition of the retarded Green's function. Breaking $\mathbf{G}^{{\rm tot} +}(\omega)$ into various blocks, the above expression can be re-written in the following form
\begin{align}
\left(
\begin{array}{ccc}
(\omega-i\epsilon)\mathbb{I}-\mathbf{H} & -\mathbf{\kappa}_1 &-\mathbf{\kappa}_N \\
\mathbf{\kappa}_1^\dagger & (\omega-i\epsilon)\mathbb{I}-\mathbf{\Omega}_1 & 0 \\
\mathbf{\kappa}_N^\dagger & 0 & (\omega-i\epsilon)\mathbb{I}-\mathbf{\Omega}_N \\
\end{array}
\right)
\left(
\begin{array}{ccc}
\mathbf{G}^+(\omega) & \mathbf{G}^{SL+}(\omega) & \mathbf{G}^{SR+}(\omega) \\
\mathbf{G}^{LS+}(\omega) & \mathbf{G}^{L+}(\omega) & \mathbf{G}^{LR+}(\omega) \\
\mathbf{G}^{RL+}(\omega) & \mathbf{G}^{RS+}(\omega) & \mathbf{G}^{R+}(\omega) \\
\end{array}
\right)=\mathbb{I},
\end{align}
where $\mathbf{\Omega_1}$ ($\mathbf{\Omega_N}$) is a $N_B$ dimensional diagonal matrix whose elements are the mode frequencies of the left (right) bath, and $\mathbf{\kappa}_1$ ($\mathbf{\kappa}_N$) is a matrix whose elements are hopping between the system sites and the various modes of the left (right) bath. Since only the first (last) site is attached to the left (right) bath, the only the first (last) row of $\mathbf{\kappa}_1$ ($\mathbf{\kappa}_N$) is non-zero. In the above equation, the $N \times N$ matrix $\mathbf{G}^+(\omega)$ is the retarded non-equilibrium Green's function (NEGF) of the system. By solving the above equation for $\mathbf{G}^+(\omega)$, one obtains
\begin{align}
\mathbf{G}^+(\omega)=\big[(\omega -i\epsilon)\mathbb{I}-\mathbf{H}-\Sigma^{(1)}(\omega)-\Sigma^{(N)}(\omega)\big]^{-1}, \textrm{ with } \Sigma^{(1)}(\omega)=\mathbf{\kappa}_1^\dagger \mathbf{g}^{L+}(\omega)\mathbf{\kappa}_1,~~\Sigma^{(N)}(\omega)=\mathbf{\kappa}_N^\dagger \mathbf{g}^{R+}(\omega)\mathbf{\kappa}_N,
\end{align}
where $\mathbf{g}^{L+}(\omega)=\left[(\omega-i\epsilon)\mathbb{I}-\mathbf{\Omega}_L\right]^{-1}$ ($\mathbf{g}^{R+}(\omega)=\left[(\omega-i\epsilon)\mathbb{I}-\mathbf{\Omega}_R\right]^{-1}$) is the bare retarded Green's function of the left (right) bath is absence of coupling with the system. Here the $N \times N$ matrix $\Sigma^{(1)}(\omega)$ ($\Sigma^{(N)}(\omega)$) is the self-energy of the left (right) bath. Since only the first (last) site of the system is coupled to the left (right) bath, the form of $\kappa_1$ ($\kappa_N$) enforces that only the top left (bottom right) corner element of $\Sigma^{(1)}(\omega)$ ($\Sigma^{(N)}(\omega)$) is non-zero. Now upon taking the number of bath modes to infinity ($N_B \rightarrow \infty$) such that the bath spectral functions become continuous, we can obtain the following expressions for the only non-zero elements of the self-energy matrices as
\begin{align}
\label{self_energy_elements}
\Sigma^{(\ell)}_{\ell \ell}(\omega)= -i\frac{\mathfrak{J}_\ell(\omega)}{2}-\mathcal{P}\int \frac{d\omega^\prime}{2\pi}\frac{\mathfrak{J}_\ell(\omega^\prime)}{\omega-\omega^\prime},~~\mathfrak{J}_\ell(\omega)=\sum_{r=1}^\infty |\kappa_{r \ell}|^2 \delta(\omega- \Omega_\alpha),~~\ell=\{1,N\},
\end{align}
$\delta(\omega)$ being the Dirac delta function.
We are interested in the non-equilibrium steady state (NESS) of the system, starting from an arbitrary initial state of the system and thermal states of the baths. The correlation functions involving system operators can be expressed in terms of the NEGF as \cite{Jauho_book,di_Ventra_book,Dhar_2006}
\begin{align}
\langle \hat{c}_p^\dagger \hat{c}_q \rangle_{\rm NESS}=& \int \frac{d\omega}{2\pi} \Big[ \mathbf{G}_{p 1}^*(\omega) \mathbf{G}_{q 1}(\omega) \mathfrak{J}_1(\omega) \mathfrak{n}_1(\omega)+ \mathbf{G}_{p N}^*(\omega) \mathbf{G}_{q N}(\omega) \mathfrak{J}_{N}(\omega) \mathfrak{n}_{N}(\omega) \Big],
\end{align}
where $\mathfrak{n}_1(\omega)=[e^{\beta_1(\omega-\mu_1)}+1]^{-1}$ ($\mathfrak{n}_N(\omega)=[e^{\beta_N(\omega-\mu_N)}+1]^{-1}$) is the Fermi distribution corresponding to the initial temperatures and chemical potentials of the left (right) bath. In NESS, the particle current from the left bath is the same as the particle current into the right bath and its general expression can be written as \cite{Jauho_book,di_Ventra_book,Dhar_2006}
\begin{align}
\label{current}
I = \int \frac{d\omega}{2\pi} \mathcal{T}(\omega) \left(\mathfrak{n}_1(\omega) - \mathfrak{n}_N(\omega)\right),~~\mathcal{T}(\omega)={\rm Tr}\left(\mathbf{\Gamma}^{(1)}(\omega) \mathbf{G}^{+*}(\omega)\mathbf{\Gamma}^{(N)}(\omega)\mathbf{G}^{+}(\omega)\right)
\end{align}
where $\mathbf{\Gamma}^{(\ell)}(\omega)= {\rm Im}\left(\Sigma^{(\ell)}(\omega)\right)$, $\ell=\{1,N\}$. The above expression has the form of a Landauer-Buttiker formula for current, with $\mathcal{T}(\omega)$ being the transmission function. Since the only non-zero elements of the self-energy matrices are as given in Eq.(\ref{self_energy_elements}), the transmission function simplifies to
\begin{align}
\mathcal{T}(\omega) = \mathfrak{J}_1(\omega)\mathfrak{J}_N(\omega) |\mathbf{G}^{+}_{1N}(\omega)|^2.
\end{align}
Going to zero temperature limit, $\beta_1,\beta_N \rightarrow \infty$, the expression for current reduces to
\begin{align}
I = \int_{\mu_1}^{\mu_N} \frac{d\omega}{2\pi} \mathcal{T}(\omega)
\end{align}
Writing $\mu_1=\mu$ and $\mu_N = \mu - \Delta \mu$, the conductance at zero temperature is given by
\begin{align}
G(\mu) = \lim_{\Delta \mu \rightarrow 0} \frac{I}{\Delta \mu}= \frac{\mathcal{T}(\mu)}{2\pi} = \frac{\mathfrak{J}_1(\mu)\mathfrak{J}_N(\mu) |\mathbf{G}^{+}_{1N}(\mu)|^2}{2\pi}.
\end{align}
This is the expression used in the main text to calculate the system-size scaling of conductance.
}
\section{Analytical Scaling of $\mathbf{G}^{+}_{1N}(\mu)$ with system size}
Our main conjecture is that for large $N$ the system size scaling of $\mathbf{G}^{+}_{1N}(\mu)$ will be same as that of the bare retarded Green's function $\mathbf{g}^{+}_{1N}(\mu)$, i.e, $\mathbf{G}^{+}_{1N}(\mu) \propto \mathbf{g}_{1N}(\mu)$, with proportionality constant being independent of $N$. If we further assume that $\mathbf{g}^{+}_{1N}(\mu)$ is evaluated in the thermodynamic limit, we get
\begin{align}
\label{conjecture2}
\mathbf{G}^{+}_{1N}(\mu) \sim \lim_{\epsilon\rightarrow 0} \frac{1}{2\pi} \int_{-\pi}^{\pi} dk \frac{e^{-ik N}}{\mu - \varepsilon(k,\alpha)-i\epsilon}=\mathbf{\mathcal{G}}^{+}_{1N}(\mu).
\end{align}
The major contribution to the integral on the right-hand-side comes from the singularities of the integrand.
{\it Case 1: $-2\,\zeta(\alpha)<\mu<2 \, \eta(\alpha)$} : Clearly, if $\mu$ lies within the bandwidth of the system, $-2\zeta(\alpha)<\mu<2 \eta(\alpha)$, then the integrand will have poles on the real line. Poles on the real line can, at best, generate terms which oscillate with $N$, and not any scaling behavior with $N$. So, as far as system size scaling is concerned, we can infer,
\begin{align}
\mathbf{G}^{+}_{1N}(\mu) \sim N^0~~\forall~~ -2\zeta(\alpha)<\mu<2\eta(\alpha).
\end{align}
This is what leads to the ballistic behavior of current.
\begin{figure}
\includegraphics[scale=0.5]{Contour.pdf}
\caption{The contour chosen to carry out the integration in Eq.(\ref{contour_int}). }
\label{fig:contour}
\end{figure}
{\it Case 2: $\mu < -2\zeta(\alpha)$:} Next, we consider the case when $\mu$ lies below the lower band edge i.e., $\mu \leq-2\,\zeta(\alpha)$ which is our main regime of interest. In this case, the maximum contribution to the integral comes from small values of $k$. Therefore we use Eq.(\ref{small_k_dispersion}) to obtain
\begin{align}
& \frac{1}{\mu - \varepsilon(k,\alpha)-i\epsilon}\simeq -\frac{1}{a_0(\omega)+2 a_1 |k|^{\alpha-1}+2 a_2 k^2 + i \epsilon }, \\
& a_0(\mu) = -2\zeta(\alpha)-\mu \geq 0. \nonumber
\end{align}
Then we have,
\begin{align}
\mathbf{G}^{+}_{1N}(\mu) \sim -\frac{1}{2 \pi} \, \int_{-\infty}^\infty dk \frac{e^{-ikN}}{a_0(\mu)+2 a_1 |k|^{\alpha-1}+2 a_2 k^2 + i \epsilon} = -\frac{1}{2 \pi} \big(A_{+} + A_{-}\big) ,
\end{align}
where we have extended both the upper and the lower limit of the integral to infinity since we have already assumed that large values of $k$ gives a negligible contribution. Here
\begin{align}
A_{\pm} = \int_{0}^\infty dk \frac{e^{\mp ikN}}{a_0(\mu)+2 a_1 |k|^{\alpha-1}+2 a_2 k^2 + i \epsilon}
\label{contour_int}
\end{align}
The integration in Eq.(\ref{contour_int}) can be carried out using contour integration techniques, by choosing a proper contour as shown in Fig.(\ref{fig:contour}). This is a valid contour for computing $A_{-}$, whereas for $A_{+}$ a valid contour is the one enclosing the lower half of the complex plane. Let us first focus on computing $A_{-}$. Depending on the value of $\alpha$, the integrand may or may not have branch point singularities in the right upper half plane. In Fig.(\ref{fig:contour}), we assume there is one such singularity. For $\alpha>3$, it can be argued that this will be case, since the $k^{\alpha-1}$ term in the denominator will be sub-leading. By carrying out the integration along the curves CD, DE and EF in the contour, it can be checked that the contribution from them scales with system size as $e^{-a N}$, $(a>0)$. So, the contribution from any branch point singularity in the right upper half plane is exponentially decaying with system size. The contribution from BC and FG is zero, as is standard. The leading contribution then comes from the line GA of the contour. The integral along the line GA, after some simplification is
\begin{align}
&A_{-}|_{\rm GA} = \frac{i}{N} \int_0^\infty dy \, \frac{e^{-y}} { x_R + i x_I}, & A_{+}|_{\rm GA} = - \frac{i}{N} \int_0^\infty dy \, \frac{e^{-y}} { x_R + i \tilde{x}_I},
\end{align}
where
\begin{eqnarray}
x_R &=& a_0 \!+\!2\, a_1 (\frac{y}{N})^{\alpha-1} \cos\Big[{\frac{\pi (\alpha\!-\!1)}{2}}\Big]\!-\!2 a_2 \left(\frac{y}{N}\right)^2, \nonumber \\
x_I &=& \epsilon + 2 \, a_1 (\frac{y}{N})^{\alpha-1} \sin\Big[{\frac{\pi (\alpha\!-\!1)}{2}}\Big],\nonumber \\
\tilde{x}_I &=& \epsilon - 2 \, a_1 (\frac{y}{N})^{\alpha-1} \sin\Big[{\frac{\pi (\alpha\!-\!1)}{2}}\Big],\nonumber \\
\end{eqnarray}
One can finally write
\begin{align}
\mathbf{G}^{+}_{1N}(\mu) \sim - \frac{2 \, a_1 \sin\Big[{\frac{\pi (\alpha\!-\!1)}{2}\Big]}}{N^{\alpha} \, \pi } \, \int_0^\infty dy \, \frac{e^{-y} \, y^{ \alpha-1}}{(x_R + i x_I)\, (x_R + i \tilde{x}_I)}
\end{align}
Given that $a_0$ is finite, one can ignore $N$ dependent terms in $x_R, x_I, \tilde{x}_I$ in the thermodynamics limit ($N \to \infty$) leading to $1/N^{\alpha}$ dependence for the NEGF and thus conductance $G(\mu) $ scaling as $\sim N^{-2\alpha}$.
\begin{figure}
\includegraphics[width=\textwidth]{scalings.pdf}
\caption{(Color online). Here numerically we have shown system size scaling of three different retarded Green's functions (i) central system's bare retarded Green's function $\mathbf{g}^{+}_{1N}(\mu)$, (ii) actual retarded Green's function $\mathbf{G}^{+}_{1N}(\mu)$ that appears in the conductance formula (iii) further approximated bare retarded Green's function $\mathbf{\mathcal{G}}^{+}_{1N}(\mu)$. The first row is for $\alpha=1.8$ and the second row is for $\alpha=2.8$. For the calculation of $\mathbf{G}^{+}_{1N}(\mu)$, the bath spectral functions are chosen to be $\mathfrak{J}_1(\omega)=\mathfrak{J}_N(\omega)=\Gamma \sqrt{1-\left(\frac{\omega}{\Lambda}\right)^2}$, with $\Lambda=10$ and $\Gamma=1.6$. We can see that all the retarded Green's functions can capture the $1/N^{2\alpha}$ scaling outside the band edge for both $\alpha=1.8$ and $2.8$. At the edges $1/N^2$ scaling can not be captured by $\mathbf{\mathcal{G}}^{+}_{1N}(\mu)$ but it is captured by systems's bare retarded Green's function $\mathbf{g}^{+}_{1N}(\mu)$. It also ensures that all these sub-diffusive scalings are actually the property of the central system.}
\label{fig:scaling}
\end{figure}
{\it Case 3: $\mu > 2\,\eta(\alpha)$:}
A similar analysis like above can be done also in the case when $\mu$ lies above the maximum band energy $2 \,\eta(\alpha)$ corresponding to $k=\pm \pi$. Interestingly, as $\varepsilon(k\pm \pi)$ is analytic around $k = \pm \pi$, (Eq.~(\ref{k_pi_dispersion})), it is easy to show that this leads to an exponential contribution in the system size, i.e., $\mathbf{G}^{+}_{1N}(\mu) \sim e^{-b N}, b>0$. The leading order contribution in $N$ once again arises from the non-analyticity behavior of the dispersion relation at $k=0$ and following similar contour integration steps as above one obtains exactly the same scaling
\begin{align}
& \mathbf{G}^{+}_{1N}(\mu) \sim N^{-\alpha},~~\forall~\mu<-2\zeta(\alpha), \mu>2\eta(\alpha)
\end{align}
{\it Case 4: $\mu = 2\,\eta(\alpha)$, and $\mu=-2\zeta(\alpha)$:} At any finite $N$, these values of $\mu$ do not correspond to any eigenvalue of $\mathbf{H}$, but the minimum and the maximum eigenvalues of $\mathbf{H}$ tend to these values with increase in $N$. We find that this case is difficult to obtain from scaling of $\mathcal{G}_{1N}(\omega)$ defined in Eq.(\ref{conjecture2}). In other words, we cannot use the expression for $\mathbf{g}_{1N}(\mu)$ in the thermodynamic limit. However, direct numerical evaluation gives $\mathbf{G}^{+}_{1N}(\mu) \propto \mathbf{g}_{1N}(\mu)$, confirming the original conjecture, as we show in the next section.
\section{Numerical Scaling of $\mathbf{g}^{+}_{1N}(\mu),\mathbf{G}^{+}_{1N}(\mu), \mathbf{\mathcal{G}}^{+}_{1N}(\mu)$ with system size}
In the previous section, we have analytically calculated the approximated bare retarded Green's function $\mathbf{\mathcal{G}}^{+}_{1N}(\mu)$ for different cases like inside the band and outside the band. The analytical results give clear understanding of sub-diffusive behaviour $(1/N^{2\alpha})$ outside the band and ballistic behaviour $N^0$ inside the band. But, this approximated retarded Green's function $\mathbf{\mathcal{G}}^{+}_{1N}(\mu)$ can not capture the sub-diffusive scaling $1/N^2$ at the two band edges. Thus, numerically we have plotted system size scaling of all three different retarded Green's functions (i) central system's bare retarded Green's function $\mathbf{g}^{+}_{1N}(\mu)$, (ii) actual retarded Green's function $\mathbf{G}^{+}_{1N}(\mu)$ that appears in the conductance formula (iii) approximated bare retarded Green's function $\mathbf{\mathcal{G}}^{+}_{1N}(\mu)$ in Fig.~\ref{fig:scaling} for $\alpha=1.8$ and $\alpha=2.8$. Here, we can numerically also see that all the retarded Green's function can capture the sub-diffusive scaling ($1/N^{2\alpha}$) outside the band. The scaling at band-edge can not be captured by $\mathbf{\mathcal{G}}^{+}_{1N}(\mu)$ but can be captured by system's bare retarded Green's function $\mathbf{g}^{+}_{1N}(\mu)$.
\section{Effect of system-bath coupling}
In the main text, we have said that the system-size scaling of conductance is independent of the strength of system-bath coupling.
In this section, we explicitly check this numerically. In Fig.~\ref{fig:coupling}, we have shown the system size scaling of $\mathbf{G}^{+}_{1N}(\mu)$ for two widely different strengths of system-bath coupling. We clearly see that all the scaling properties are unaffected by the system-bath coupling strength.
\begin{figure}
\includegraphics[width=\textwidth]{scaling2.pdf}
\caption{(Color online). We have plotted the system size scaling of exact retarded Green's function $\mathbf{G}^{+}_{1N}(\mu)$ for different system-bath couplings for $\alpha=2$. Here, we can see all the scaling properties are unaffected by the strength of system-bath coupling. For the plots, the bath spectral functions are chosen to be $\mathfrak{J}_1(\omega)=\mathfrak{J}_N(\omega)=\Gamma \sqrt{1-\left(\frac{\omega}{\Lambda}\right)^2}$, with $\Lambda=8$.}
\label{fig:coupling}
\end{figure}
{
\section{Inability of local and global Lindblad approaches to describe the sub-diffusive phases}
Even though the system size scaling of conductance is independent of the strength of system-bath coupling, standard quantum master equations like Lindblad equations in local and global forms, which are often used to describe weak-system-bath coupling situations, are unable to capture the sub-diffusive behavior. In this section, we explicitly discuss this.
\subsubsection{The local Lindblad approach}
The commonly used local Lindblad approach corresponds to the following quantum master equation \cite{Breuer_book,Landi_2021,Archak6}
\begin{align}
\label{LLE}
\frac{\partial \hat{\rho}}{\partial t}=i\left[\hat{\rho},\hat{\mathcal{H}}_S + \hat{\mathcal{H}}_{LS} \right] + \sum_{\ell=1,N}\left[\mathfrak{J}_\ell(\varepsilon_\ell)\Big(1-\mathfrak{n}_\ell(\varepsilon_\ell)\Big) \left(\hat{c}_\ell \hat{\rho} \hat{c}_\ell^\dagger - \frac{1}{2}\{\hat{c}_\ell^\dagger\hat{c}_\ell, \hat{\rho}\}\right)+\mathfrak{J}_\ell(\varepsilon_\ell)\mathfrak{n}_\ell(\varepsilon_\ell) \left(\hat{c}_\ell^\dagger \hat{\rho} \hat{c}_\ell - \frac{1}{2}\{\hat{c}_\ell\hat{c}_\ell^\dagger, \hat{\rho}\}\right) \right],
\end{align}
where the so-called Lamb-Shift Hamiltonian $\hat{\mathcal{H}}_{LS}$ is given by
\begin{align}
\hat{\mathcal{H}}_{LS}=\sum_{\ell=1,N} \mathfrak{J}_\ell^H(\varepsilon_\ell) \hat{c}_\ell^\dagger\hat{c}_\ell, \textrm{ where } \mathfrak{J}_\ell^H(\omega) = \frac{1}{\pi}\int d\omega^\prime \frac{\mathfrak{J}_\ell(\omega^\prime)}{\omega-\omega^\prime},
\end{align}
is the Hilbert transform of $\mathfrak{J}_\ell(\omega)$. Here $\varepsilon_\ell$ is the on-site energy of the site attached to the bath, which, for our set-up is $\varepsilon_1=\varepsilon_N=0$. The first dissipative term in Eq.(\ref{LLE}) is called often called the loss Lindblad term. It describes the process that results in loss of particle and energy due to coupling with bath. The second dissipative term is often called the gain Linblad term. It describes the process that results in gain of particle and energy. At zero temperature, and for chemical potentials $<0$, from Eq.(\ref{LLE}), we see that only the loss Lindblad term survives. As a consequence, the system looses all its particles and its steady state is empty. Exactly similarly, for chemical potentials $>0$, from Eq.(\ref{LLE}) we see that only the gain Lindblad term survives. As a consequence, the steady state is completely filled. There can be no transport in either of these cases and hence current is zero. Thus, not only is the local Lindblad equation unable to describe the sub-diffusive phases, but also it cannot describe the ballistic transport at zero temperature when both chemical potentials are either positive or negative. These observations can be also checked by direct calculation.
Such limitations of local Lindblad approach are known \cite{Walls1970,Wichterich_2007,Rivas_2010,barranco_2014,Levy2014,Archak6,
Trushechkin_2016,
Eastham_2016,Hofer_2017,Gonzalez_2017,Mitchison_2018,
Cattaneo_2019,Hartmann_2020_1,konopik_2020local,
Scali_2021}. Microscopic derivations suggest that it can only describe situations either at infinite temperature or at infinite voltage bias or when the connections within the system are small enough that the sites do not hybridize well with one another. The physics we are describing in this paper are far from all these regimes. So this physics is beyond the regime that can be described by a local Lindblad equation.
\subsubsection{Global Lindblad approach}
To circumvent some of the drawbacks of the local Lindblad approach, a different approach often advocated is the global Lindblad or the eigenbasis Lindblad approach \cite{Breuer_book,Landi_2021,Archak6}. In deriving this quantum master equation, we first need to diagonalize the system Hamiltonian. For non-interacting systems (quadratic Hamiltonians) that we are discussing, this can be done by diagonalizing the single-particle Hamiltonian,
\begin{align}
\Phi^T \mathbf{H} \Phi = \mathbf{D},~~\mathbf{D}=diag\{\omega_\alpha\}.
\end{align}
Here $\{\omega_\alpha\}$ are the single particle eigenvalues, and the columns of $\Phi$ are the single particle eigenvectors. The system Hamiltonian can be written in the form \cite{Archak6},
\begin{align}
\hat{\mathcal{H}}_S = \sum_{\ell,m=1}^N \mathbf{H}_{\ell m} \hat{c}^\dagger_\ell \hat{c}_m = \sum_{\alpha=1}^N \omega_\alpha \hat{A}_\alpha^\dagger \hat{A}_\alpha,~~\hat{A}_\alpha = \sum_{\ell=1}^N \Phi_{\ell \alpha } \hat{c}_\ell.
\end{align}
The eigenbasis Lindblad equation is given by
\begin{align}
\label{ELE}
\frac{\partial \hat{\rho}}{\partial t}=i\left[\hat{\rho},\hat{\mathcal{H}}_S + \hat{\mathcal{H}}_{LS} \right] + \sum_{\alpha=1}^N\sum_{\ell=1,N}|\Phi_\ell \alpha|^2\Big[ & \mathfrak{J}_\ell(\omega_\alpha)\Big(1-\mathfrak{n}_\ell(\omega_\alpha)\Big) \left(\hat{A}_\alpha \hat{\rho} \hat{A}_\alpha^\dagger - \frac{1}{2}\{\hat{A}_\alpha^\dagger\hat{A}_\alpha, \hat{\rho}\}\right)\nonumber \\
&+\mathfrak{J}_\ell(\omega_\alpha)\mathfrak{n}_\ell(\omega_\alpha) \left(\hat{A}_\alpha^\dagger \hat{\rho} \hat{A}_\alpha - \frac{1}{2}\{\hat{A}_\alpha\hat{A}_\alpha^\dagger, \hat{\rho}\}\right) \Big],
\end{align}
where the Lamb-shift Hamiltonian is given by $\hat{\mathcal{H}}_{LS}=\sum_{\alpha=1}^N\sum_{\ell=1,N} |\Phi_\ell \alpha|^2 \mathfrak{J}_\ell^H(\omega_\alpha) \hat{A}_\alpha^\dagger\hat{A}_\alpha$. As in the local Lindblad case, the first dissipative term in above equation is a loss Lindblad term, and describes loss of particles from the system eigenmodes due to coupling with the baths. Likewise, the second dissipative term in above equation is a gain Lindblad term and describes the gain of particles into the system eigenmodes due to coupling with the baths. At zero temperature and for $\mu_1,\mu_N \leq -2\zeta(\alpha)$, the Fermi distributions appearing the above equation dictate that only the loss term survives and the gain term is zero. Thus, the system in steady state is completely empty. Likewise, for $\mu_1,\mu_N \geq 2\eta(\alpha)$, the only the gain term survives and the loss term is zero. Thus, in this case, the state state is completely full. Either of these cases cannot have any transport so currents from the baths will be zero. For $-2\zeta(\alpha)<\mu_1,\mu_N<2\eta(\alpha)$, however, the currents from the baths will capture the ballistic behavior. These statements can also be verified via direct calculation. Therefore, we see that the global Lindblad approach also fails to capture the sub-diffusive phases and the critical points.
There are a variety of more refined quantum master equation approaches \cite{ule,Kleinherbers_2020,Davidovic_2020,mozgunov2020,mccauley2020,
kirvsanskas2018}, including the Redfield equation. Whether such quantum master equations can capture the sub-diffusive behavior remains to be seen and requires further investigation.
}
\begin{figure}
\includegraphics[width=0.8\textwidth]{particle_density.pdf}
\caption{{\bf (a)} The particle density $\gamma$ in the system is plotted as a function of chemical potential at zero temperature. The vertical black line corresponds to $\mu=-2\zeta(\alpha)$. For $\mu\leq-2\zeta(\alpha)$, the particle density decays with system size. {\bf (b)} The hole density in the system, given by $1-\gamma$, is plotted as function of chemical potential at zero temperature. The vertical line corresponds to $\mu=2\eta(\alpha)$. For $\mu\geq 2\eta(\alpha)$, the hole density decays with system size. For the plots, the bath spectral functions are chosen to be $\mathfrak{J}_1(\omega)=\mathfrak{J}_N(\omega)=\Gamma \sqrt{1-\left(\frac{\omega}{\Lambda}\right)^2}$, with $\Lambda=8$, $\Gamma=10$. \label{fig:particle_density} }
\end{figure}
\section{Particle density in the system}
In the main text, we have remarked that there is a sub-extensive number of particle in the system for $\mu\leq-2\zeta(\alpha)$, while there is a sub-extensive number of holes for $\mu\geq 2\eta(\alpha)$. For $-2\zeta(\alpha)<\mu<2\eta(\alpha)$, there is an extensive number of both particles and holes. Here we explicitly check this. The particle density in the system is defined as
\begin{align}
\gamma=\frac{1}{N}\sum_{\ell=1}^N \langle \hat{c}_\ell^\dagger \hat{c}_\ell \rangle.
\end{align}
The occupation at $\ell$th site in NESS is given in terms of the NEGF as
\begin{align}
\langle \hat{c}_\ell^\dagger \hat{c}_\ell \rangle = \int_{-\Lambda}^\mu \frac{d\omega}{2\pi} \Big[\left|\mathbf{G}_{1\ell}^+(\omega)\right|^2 \mathfrak{J}_1(\omega) +
\left|\mathbf{G}_{N\ell}^+(\omega)\right|^2 \mathfrak{J}_N(\omega) \Big],
\end{align}
where $-\Lambda$ is the minimum energy of the band of the bath. We numerically calculate $\gamma$ and check its behavior with $\mu$ and $N$, as shown in Fig.~\ref{fig:particle_density}.
When there is a sub-extensive number of particles in the system, $\gamma$ decays with $N$, which happens for $\mu\leq-2\zeta(\alpha)$. When there is an extensive number of particles in the system, $\gamma$ is independent of $N$, which happens for $\mu>-2\zeta(\alpha)$ . When there is a sub-extensive number of holes in the system, $1-\gamma$ decays with $N$, which happens for $\mu\geq 2\eta(\alpha)$, while if there is an extensive number of holes in the system, $1-\gamma$ is independent of system-size, which happens for $\mu<2\eta(\alpha)$.
{
\section{Relation to experiments}
In a number of experiments in various platforms like trapped ions \cite{expt_trapped_ions6,
Experiment_transport}, polar molecules \cite{expt_polar_molecules1,
expt_polar_molecules3}, dipolar gas \cite{expt_dipolar_gas1}, nuclear spins \cite{expt_nuclear_spins} the spin Hamiltonians of the following form has been realized,
\begin{align}
\label{spin_hamiltonian}
\hat{\mathcal{H}}_S = -\sum_{m=1}^N \left(\sum_{r=1}^{N-m} \frac{1}{m^\alpha} \left(\hat{\sigma}_r^+ \hat{\sigma}_{r+m}^- + \hat{\sigma}_{r+m}^+ \hat{\sigma}_{r}^- + \Delta \hat{\sigma}_{r+m}^z \hat{\sigma}_{r}^z \right)\right),
\end{align}
where $\hat{\sigma}_r^{\pm}=\left(\hat{\sigma}_r^x \pm i\hat{\sigma}_r^y \right)/2$, and $\hat{\sigma}_r^{x,y,z}$ are the Pauli spin operators at site $r$. This Hamiltonian conserves the total magnetization, $\hat{M}_z= \sum_{r=1}^N \hat{\sigma}_{r}^z$, i.e, $[\hat{M}_z,\hat{\mathcal{H}}]=0$. Let us perform Jordan-Wigner transformation to convert this Hamiltonian into a fermionic one. The Jordan-Wigner transformation is given by
\begin{align}
\hat{\sigma}_r^{+} = \hat{c}_r^\dagger e^{-i \pi \sum_{p=1}^{r-1} \hat{n}_{p}},~~ \hat{\sigma}_r^{-} = e^{i \pi \sum_{p=1}^{r-1} \hat{n}_{p}}\hat{c}_r,~~\hat{n}_r = \hat{c}_r^\dagger\hat{c}_r = \frac{\hat{\sigma}_r^z +1}{2}.
\end{align}
Using these, we receive,
\begin{align}
\hat{\sigma}_r^+ \hat{\sigma}_{r+m}^- = \hat{c}_r^\dagger e^{i\pi \sum_{p=r}^{r+m-1} \hat{n}_p} \hat{c}_{r+m} = \hat{c}_r^\dagger \prod_{p=r}^{r+m-1} \left(1-2\hat{n}_p\right) \hat{c}_{r+m}, \textrm{ and } \hat{\sigma}_r^z = 2\hat{n}_r -1,
\end{align}
where we have also used the result $e^{i\pi \hat{n}_p} = 1-2\hat{n}_p$, which can be proven by expanding the exponential. Substituting these into Eq.(\ref{spin_hamiltonian}) we see that the resulting fermionic Hamiltonian is of the form
\begin{align}
\hat{\mathcal{H}}_S=-\sum_{m=1}^N \left(\sum_{r=1}^{N-m} \frac{1}{m^\alpha} \left(\hat{c}_r^\dagger \hat{c}_{r+m} + \hat{c}_{r+m}^\dagger \hat{c}_{r}\right)\right) + \hat{H}_{\rm int},
\end{align}
where $\hat{H}_{\rm int}$ contains the many-body interacting terms, i.e, the higher-than-quadratic terms. The total magnetization operator $\hat{M}_z = \sum_{r=1}^N (2\hat{n}_r -1)$. Therefore, the conservation of net magnetization guarantees that in the fermionic picture the Hamiltonian is particle number conserving. Thus, upon Jordan-Wigner transformation, the spin Hamiltonians realized in several controlled experimental platforms can be mapped into number conserving fermionic Hamiltonians with power-law-decaying hopping and many-body interactions. Since, as argued in the main text, the sub-diffusive phases and critical points are expected to be robust against arbitrary number conserving many-body interaction terms, the physics described here is relevant to these experimental set-ups at low temperatures.
}
\end{document}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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\section{Introduction}
The vacuum of Quantum Chromodynamics (QCD) is a prominent nonperturbative
system, whose strongly interacting nature persists even above the transition
to the quark-gluon plasma. To get insight into its mechanism, spectral
properties of the QCD Dirac operator are very useful. A nonzero density of
eigenvalues at zero gives rise to chiral symmetry breaking via the
Banks-Casher formula \cite{Banks:1980}. Moreover, exact zero modes are
related to the topological charge via index theorems.
In recent years, localization properties of the Dirac eigenmodes have
attracted attention as they can be used to draw analogies to condensed matter
phenomena: concepts like the mobility edge and Anderson localization can be
studied in QCD lattice simulations. In this spirit the chiral transition at
finite temperature\footnote{At zero temperature the situation is not clear due
to the continuum limit, see \cite{Forcrand:2007a} and references therein.}
has been conjectured to be an Anderson (metal-insulator) transition.
This goes hand in hand with different random matrix theory (RMT) descriptions
of the Dirac spectra. In the low temperature phase, the existence of the
chiral condensate connects QCD to chiral perturbation theory and random matrix
theory (becoming exact in the epsilon-regime), which explains the statistics
of the low lying part of the Dirac spectrum \cite{Verbaarschot:2000}.
The spectral gap in the high temperature phase, on the other hand, seems to
call for the ``soft edge'' description of RMT, which, however, could not be
supported by lattice data \cite{Farchioni:2000,Damgaard:2000,Narayanan:2006sd,Narayanan:2006ek}. Instead, a
transition to independent eigenmodes obeying Poisson statistics has been
suggested \cite{Garcia-Garcia:2007}. A refined analysis by one of us has
shown, that the bulk of the spectrum is still delocalized and subject to RMT,
while the lowest lying eigenmodes display a transition to localization and
Poissonian behavior of the eigenvalues \cite{Kovacs:2010,Kovacs:2010a}. This
effect has been confirmed to be universal in the sense that it does not depend
on the resolution of the lattice. Observables, like e.g.\ the number of
localized modes, rather scale with the physical volume. On the other hand,
Ref.\ \cite{Gavai:2008} found that localization is essentially a finite volume
artifact. However, they used a different, less restrictive definition of
localization and thus their results are not in conflict with the rest of the
above cited literature.
In this work we give another crucial ingredient of Anderson localization in
QCD, namely we identify the ``defects'' causing it. We show that at high
temperature local Polyakov loop fluctuations trap low-lying modes. The
average Polyakov loop as an order parameter of the deconfinement (or center)
phase transition approaches 1 with increasing temperature.
Locally, however, the Polyakov loop takes on other values, in
particular close to other center elements, $-1$ in the case of gauge group $SU(2)$. The phase transition can actually
be viewed as a percolation of the physical sector embracing such islands
where the Polyakov loop is close to other center elements
\cite{Yaffe:1982,Gattringer:2010}.
The locations of these Polyakov loop fluctuations (which are similar to Weiss
domains in ferromagnetism) are correlated with maxima in the profile of
low-lying Dirac modes. We will demonstrate this for eigenmodes of both the
overlap and staggered operator. Similarities between the spectrum of the
overlap and staggered Dirac operator have already been found in the Schwinger
model \cite{Durr:2003xs} and also in QCD \cite{Durr:2004as}. Our present
study is, however, the first one when similarity of staggered and overlap
Dirac eigen{\it modes} is seen in lattice simulations and we consider this an
important side result of our study.
The observed localization can be understood via Polyakov loops compensating the
twist caused by the antiperiodic boundary conditions in the temporal
direction. The corresponding Matsubara frequency is effectively lowered
(locally) which results in lower eigenvalues. Actually, our finding has been
inspired by a similar localization effect in the spectrum of the
gauge-covariant Laplace operator \cite{Bruckmann:2005c}. The Laplacian is the
square of the Dirac operator in the free case, so the twist picture
applies. Otherwise this operator does not share important chiral features like
topological zero modes and condensates.
Likewise, our finding is consistent with the existence of a chiral condensate
in the Polyakov loop sector close to other center elements
\cite{Chandrasekharan:1995gt,Stephanov:1996he,Meisinger:1995ih,Chandrasekharan:1995nf,Gattringer:2002dv,Gattringer:2002tg,Bornyakov:2008bg,Kovacs:2008sc}
(also needed for center symmetry breaking \cite{Bilgici:2008}), which implies
low Dirac eigenvalues in islands of such Polyakov loops.
The connection of these islands to topological excitations like magnetic
monopoles is attractive, but in its naive form contradicts the observed
topological susceptibility quantitatively, see below.
For the construction of random matrix models valid at high temperature we
investigate the distribution of local Polyakov loops and find them to be
uncorrelated to a good approximation. Hence, Polyakov loops in fact provide
the Poissonian ingredient for the Dirac spectrum. This is built into a novel
Anderson-like random matrix model through supplementing it by random matrix
entries that represent nearest neighbor hoppings in three-dimensional space. We motivate this
model and show, that with a few parameters it reproduces the main features of
the Dirac modes: chirality, spectral gap, RMT-Poisson transition and
localization (to the analogue of local Polyakov loops).
Our findings are based on quenched lattice simulations with the SU(2) gauge group, we strongly believe that the described phenomena are present in more realistic gauge theories, too.\\
The paper is organized as follows. In the next section we describe the Dirac
spectra at high temperature including the RMT-Poisson (chaotic to integrable)
transition and the similarity of the staggered and overlap
modes. Sect.~\ref{sect polyakov} is devoted to the connection between local
Polyakov loops and low-lying modes. In Sect.~\ref{sect interpretation}
thereafter we investigate two possible interpretations of this finding,
effective Matsubara frequencies and topological objects. In Section
\ref{sect rmt} we introduce and explore our
random matrix model and finally Section \ref{sect conc} contains our conclusions.
\begin{figure}[b]
\includegraphics[width=1.05\linewidth]{eigenvaluedensity_staggeredandoverlap.pdf}
\caption{Logarithm of the spectral density along the imaginary axis for
the staggered (red, at smaller $\lambda$) and overlap (blue) Dirac operator
from the lowest $256$ eigenvalues.}
\label{fig spec density}
\end{figure}
\section{Dirac spectra at high temperature}
\label{sect dirac}
We analyze quenched SU(2) lattice configurations generated with Wilson action
on a $24^3\cdot 4$ lattice at $\beta=2.6$ which amounts to a temperature of
$2.6 T_c$. The average Polyakov loop of $\,0.37\,$ signals deconfinement (by
Polyakov loop we refer to the trace of the products of all temporal links
$L(\vec{x})=1/2\cdot {\rm Tr}\prod_{x_0=1}^{N_t}U_0(x_0,\vec{x})$ and we selected the physical sector of positive Polyakov loops by hand).
We measured the $256$ lowest eigenvalues with positive imaginary parts
of the overlap \cite{Neuberger:1997fp,Neuberger:1998wv} (with pa\-ra\-me\-ter $s=0.4$ cf.\ \cite{Kovacs:2010}) and staggered
Dirac operator on 1136 (overlap)/
3149 (staggered) configurations. For a set of 1102 configurations, we also
measured the 12 lowest eigenmodes of both staggered and overlap
operator. In all cases the quark mass was set to zero.
The eigenvalues are ordered according to their imaginary parts and the corresponding
eigenvalue densities\footnote{Due to chirality, the nonzero eigenvalues come
in pairs of opposite imaginary part and we restrict ourselves to the half
with positive imaginary part. In other words, all plots can be extended
symmetrically around $\lambda=0$.} are plotted in Fig.~\ref{fig spec
density}. They display a gap-like behavior with the eigenvalue density
starting to differ from zero considerably at $a\lambda\simeq 0.15$ and $a\lambda\simeq 0.5$,
respectively. In addition, the overlap operator possesses exact zero modes, which we use to
determine the topological charge of the configuration.
To describe the RMT vs.\ Poissonian behavior of the Dirac spectra we will scan
windows in the range of available eigenvalues and measure the level spacing
distributions $P(s)$ on unfolded eigenvalues \cite{Verbaarschot:2000}, a
typical quantity describing the eigenvalue statistics. The Gaussian RMT ensembles
provide predictions for it, which also apply to systems with chiral symmetry,
depending only on the universality class. For gauge group
$SU(2)$ the latter are the Gaussian Orthogonal Ensemble (GOE) for the overlap operator
(like for the continuum Dirac operator) and the Gaussian Symplectic Ensemble (GSE) for
the staggered operator. The main difference of those $P(s)$ formulae lies in
the different repulsion strength of nearby eigenvalues, which results in
a linear and quartic behavior of $P(s)$ near $s=0$,
respectively. Independent eigenvalues, on the
other hand, lack such a repulsion and the unfolded level spacing is a Poissonian distribution, i.e.\ $P(s)=\exp(-s)$.
\begin{figure*}[!t]
\includegraphics[height=1.1\threepicwidth]{spacings_overlap_spectrum_0,45_to_0,52.pdf}\hspace*{-0.9cm}
\includegraphics[height=1.1\threepicwidth]{spacings_overlap_spectrum_0,53_to_0,545.pdf}\hspace*{-0.9cm}
\includegraphics[height=1.1\threepicwidth]{spacings_overlap_spectrum_0,59_to_0,605.pdf}
\caption{Spacing distributions of the overlap spectrum in spectral windows
indicated by the insets showing the spectral density. The pure RMT
(GOE) prediction and the Poissonian distribution are plotted for comparison.}
\label{fig transition overlap}
\end{figure*}
Fig.~\ref{fig transition overlap} shows the level spacing distribution of overlap eigenvalues (see \cite{Kovacs:2010a} for staggered spectra). It
clearly reveals that the level spacing agrees with the associated RMT predictions in the bulk and moves
towards Poissonian when the spectral window is shifted towards lower eigenvalues. To observe such a
level {\it spacing} at least a few
independently (Poissonian) distributed eigenvalues are needed on each
configuration. Since independent modes occur only at the very low end of the
spectrum where the spectral density is low, large enough volumes are required
for that. In \cite{Kovacs:2010} also the independence of these data of the
lattice spacing has been demonstrated for the staggered case.
The properties of the independent and localized modes at the lower ends of
the spectra are the main subject of the rest of the paper. We therefore check
first, whether the modes of the overlap and the staggered operator see
similar physical effects, meaning that their profiles are correlated and
localized to similar locations.
\begin{figure}[b]
\includegraphics[width=0.9\linewidth]{scatter_overlap_vs_staggered_excluded2_dark.pdf}
\caption{Scatter plot of staggered (horizontal) vs.\ overlap (vertical)
amplitudes for the lowest modes, $|\psi^{{\rm st}}_1(x)|$ vs. $|\psi^{{\rm ov}}_1(x)|$,
on a $Q=0$ configuration over the whole lattice. Data points with small amplitudes -- as indicated by the circle -- have been excluded to avoid overcrowding the plot.}
\label{fig scatter overlap staggered}
\end{figure}
First of all we remark that for every overlap zero mode we find a staggered
eigenvalue with unusually small value. In the topological sector $Q=0$, the
average smallest eigenvalue is 0.175(1), whereas it is 0.109(3) for $|Q|=1$
(and 0.098(8)/0.141(6) for $|Q| = 2$, where we have two small
eigenvalues). This can also be seen in Fig.~\ref{fig polloop modes one}
bottom.
Next, the scatter plot of Fig.~\ref{fig scatter overlap staggered} gives a strong indication
for the correlation of the local mode amplitudes in a typical example configuration, which
is further visualized by the two-dimensional profiles in Fig.~\ref{fig profile modes}.
\begin{figure}[b]
\includegraphics[width=0.9\linewidth]{profile_overlap_Z20_T4.pdf}\\
\includegraphics[width=0.9\linewidth]{profile_staggered_Z20_T4.pdf}
\caption{Profile of the overlap (top) and staggered (bottom) lowest mode
of the configuration of Fig.~\protect\ref{fig scatter overlap staggered},
in a lattice plane where the overlap mode takes on its maximum.
(The absolute maximum of the staggered mode is separated from the overlap one by $\sqrt{2}$ lattice spacings in the remaining directions.)}
\label{fig profile modes}
\end{figure}
In order to quantify the similarity and localization of two modes we propose the following ``interlocalization''
\begin{equation}
I := N\sum_{x} \left|\psi_m^{{\rm ov}}(x)\right|^2\left|\psi_n^{{\rm st}}(x)\right|^2\,,
\label{eqn interlocalisation}
\end{equation}
where $N$ is the number of lattice sites, and $|\psi_m^{{\rm ov,st}}(x)|$ is
the absolute value ($L^2$-norm) of the $m$th overlap/staggered eigenmode summed up
over gauge -- and in the case of overlap also spinor -- indices at lattice site $x$.
This is a positive quantity that receives large contributions when both modes are considerably large at some locations. Moreover, for two identical modes it becomes their inverse
participation ratio (IPR). The latter is a well-known measure for the localization, taking
on a value of $N$ for modes localized on a single point (on the lattice) and
$1$ for constant modes. In fact, we find $I\simeq 1$ also for two normalized
modes with independent Gaussian distributed amplitudes at each site. Only
modes that are similar {\it and} localized generate large values of $I$.
\enlargethispage{\baselineskip}
We utilize $I$ for matching the overlap and staggered modes\footnote{In \cite{Hoellwieser:2009} the positions of the highest peaks were used to reveal similarities between overlap and staggered modes.}. We start by
taking the lowest overlap mode and pair it with the staggered mode that has
the largest interlocalization with it. Going up in the overlap spectrum we
continue this matching procedure in the same way, but use only those staggered
modes that have not yet been paired up with a lower overlap mode. In
Fig.~\ref{fig matched modes interlocalisation} we plot the interlocalization
values for the lowest modes matched in this way as a function of the
corresponding overlap eigenvalue $\lambda_m$. From this plot it is clear that
the zero modes (of the overlap operator, near-zero modes of the staggered
operator) are matching close to perfectly. The value of 300 is actually in the same order of magnitude as the IPR of the
individual overlap zero modes\footnote{The associated low-lying staggered
modes have a slightly larger IPR, around 400, presumably because in contrast
to the overlap operator the staggered operator is ultralocal.}.
Then $I$ drops quickly and after a few modes it reaches the
reference value $1$ discussed above.
\begin{figure}[t]
\includegraphics[width=0.9\linewidth,clip=true,trim=170 \value{scutbottom} \value{scutright} \value{scuttop}]{interlocalization.pdf}
\caption{The interlocalization $I$, Eq.~(\ref{eqn interlocalisation}), for
matched modes (see text) as a function of the averaged overlap eigenvalue
$\langle a\lambda_m\rangle$ (in lattice units), on a logarithmic scale and ensemble averaged. The
maximal possible value for $I$ on our lattice is $24^3\cdot 4\simeq 5.5\cdot 10^4$,
whereas delocalized modes yield $I\simeq 1$, indicated by the dashed gray
line. Horizontal error bars visualize the spreads of the eigenvalues.}
\label{fig matched modes interlocalisation}
\end{figure}
\section{Polyakov loops as defects/traps}
\label{sect polyakov}
As an appetizer of our main finding we show in Fig.~\ref{fig profile polloops}
the Polyakov loops (of one example configuration) in the lattice plane, where
the lowest overlap and staggered eigenmodes take on their maximum, as shown
in Fig.~\ref{fig profile modes}. The Polyakov loop is dominated by UV
fluctuations at the scale of the lattice spacing as almost every lattice
observable. Therefore it is virtually impossible to see any structures in it.
We applied 6 sweeps of APE smearing \cite{Albanese:1987,Falcioni:1985} with
$\alpha=0.55$ to the configuration, which leads to a smoother Polyakov loop
landscape. Indeed, an island of Polyakov loop with opposite sign emerges at
the location of the maximum of the lowest Dirac mode on the original unsmeared
configuration. A similar profile becomes visible after simply averaging the
(traced) Polyakov loops with their neighbors.
\begin{figure}[b]
\includegraphics[width=0.9\linewidth]{profile_polloop_unsm_Z20_T4.pdf}\\
\includegraphics[width=0.9\linewidth]{profile_polloop_sm_Z20_T4.pdf}
\caption{Profile of the unsmeared (top) and smeared (bottom) Polyakov loop for the same configuration and in
the same plane as in Fig.~\protect\ref{fig profile modes}. By the naked eye nothing seems
particular in this plane for the unsmeared case, whereas the smeared
Polyakov loop actually takes its minimum (-0.68, compared to an average
smeared Polyakov loop of 0.80) at the hotspot visible in the fermion modes
in Fig.~\protect\ref{fig profile modes}. (A correlation of the unsmeared Polyakov loop to fermion modes is exposed by virtue of statistical measurements, see text.)}
\label{fig profile polloops}
\end{figure}
Let us stress, that the lower panel of Fig.~\ref{fig profile polloops} is the
only occasion that we present a smeared result. We now return to correlation
functions of unsmeared Polyakov loops for the rest of this paper.
\begin{figure}[!b]
\includegraphics[width=\linewidth,clip=true,trim=170 \value{scutbottom} \value{scutright} \value{scuttop}]{polyoverlap_overlap_figure.pdf}
\includegraphics[width=\linewidth,clip=true,trim=170 \value{scutbottom} \value{scutright} \value{scuttop}]{polyoverlap_staggered_figure.pdf}
\caption{The ratio of ``Polyakov loops as averaged by low-lying modes'' $L_m$,
Eq.~(\protect\ref{eqn loops seenby modes}), to the average Polyakov loop
for overlap (top) and staggered modes (bottom)
as a function of the corresponding averaged eigenvalue
$\langle a \lambda_m\rangle$. Horizontal error bars visualize the spreads of the eigenvalues.}
\label{fig polloop modes one}
\end{figure}
To check the correlation of Polyakov loop islands and low Dirac modes in
a quantitative way, we define ``Polyakov loops as averaged by a
particular mode'', i.e. Polyakov loops weighted with the density of a
normalized Dirac mode, cf.\ \cite{Bruckmann:2005c},
\begin{equation}
L_m := \sum_x |\psi_m(x)|^2 L(x)\,.
\label{eqn loops seenby modes}
\end{equation}
This quantity is restricted to the interval $[-1,1]$, just like the
Polyakov loop $L$. It is clear that a wave-like mode $\psi_m$ with
approximately constant amplitude yields an $L_m$ close to the average Polyakov
loop $\sum_x L(x)/V$, whereas a strongly localized mode picks the Polyakov
loop in that region. If the latter happens to be an island of ``wrong''
Polyakov loops, $L_m$ tends to zero or even becomes negative.
Fig.~\ref{fig polloop modes one} shows the ratio of $L_m$ averaged over
different configurations and the average Polyakov loop, both for the low-lying
overlap and low-lying staggered modes. The Polyakov loops averaged by the
low-lying modes are indeed smaller than on average. Higher modes, on the
contrary, tend to see the average Polyakov loop.
\begin{figure}[!]
\includegraphics[height=1.1\twopicwidth]{polyloopdist_zero.pdf}
\caption{The ``Polyakov loop distribution as seen by the lowest overlap mode'', $p_1(L)$ from
Eq.~(\protect\ref{eqn new probability}), in
the $Q=0$
sector
compared to the Polyakov loop
distribution and the Haar measure (valid in the low temperature phase).
}
\label{fig polloop modes two}
\end{figure}
The connection between local Polyakov loops and low-lying Dirac modes
is confirmed from a slightly different perspective by the following
``Polyakov loop distribution as seen by a mode''. For that we weight the probability of Polyakov loops $L$ by the amplitudes
of the low-lying modes at those positions $x$ where $L(x)=L$,
\begin{equation}
p_m(L)=\sum_x\delta_{\epsilon}(L-L(x))|\psi_m(x)|^2\,,
\label{eqn new probability}
\end{equation}
in continuous notion with some smeared delta-function $\delta_{\epsilon}$ (in practice we rescale histograms). The quantity $L_m$ is recovered by the $L$-expectation value with this probability,
\begin{equation}
L_m=\int dL\, p_m(L)\, L\,,\qquad \big(\int dL\, p_m(L)=1\big)\,.
\end{equation}
Thus the probability $p_m$ visualizes how the global quantity $L_m$ is
generated by modifying the distributions of the local Polyakov loops.
In Fig.~\ref{fig polloop modes two} we show this probability for the lowest modes of the overlap operator, $p_1(L)$, on $Q=0$ configurations. One can clearly see that low fermion modes enhance low Polyakov loops down to $L\simeq -1$.
\section{Interpretations}
\label{sect interpretation}
\subsection{Effective Matsubara frequencies}
In a background of constant temporal and vanishing spatial gauge fields
the influence of the Polyakov loop on the Dirac spectra is very clear. We
first diagonalize the Polyakov loop introducing its phase
$\varphi$,
\begin{eqnarray}
\prod_{x_0=1}^{N_t}U_0(x_0) & = & \exp\big(i\varphi\left(\begin{array}{cc} 1 & \\ & -1\end{array}\right)\big)\,,
\label{eqn polloop phase first}\\
L & = & \cos\varphi\qquad \varphi \in [0,\pi]\,.
\label{eqn polloop phase second}
\end{eqnarray}
We gauge it into the last time slice, where it effectively changes the
temporal boundary condition.
The lowest modes of the free Dirac operator in the presence of this Polyakov loop
are constant in space and
plane waves in time, $\exp(i p x_0 T)$. The
quantum numbers
$p$ are governed by a
combination of the antiperiodic temporal boundary condition for fermions and
the Polyakov loop phase, namely $p=\pi\pm\varphi+2\pi\mathbb{Z}$, where the
different signs emerge from the different color components, see
Eq.~(\ref{eqn polloop phase first}).
The eigenvalues of the free Dirac operator are these
numbers
multiplied by the
temperature. The lowest ones,
\begin{equation}
\lambda_{\rm M}^{\mbox{cont.}}=(\pi-\varphi)T\,,
\label{eqn eff MF}
\end{equation}
we name {\it effective Matsubara frequencies}. These hold in the limit $N_t \to\infty$, whereas on the lattice one has
\begin{equation}
\lambda_{\rm M} = \frac1a \sin\left(\frac{\pi-\varphi}{N_t}\right)\,.
\label{eqn eff MF latt}
\end{equation}
At high temperatures the Polyakov loop (at fixed lattice spacing) becomes
trivial, $L\to 1$, hence $\varphi\to 0$ and the Matsubara frequency is
$\lambda_{\rm M}=\frac1a \sin\left(\pi/N_t\right)$. A more realistic estimate
is obtained by using the average Polyakov loop at our temperature $\langle L
\rangle =0.37$, from which we obtain the effective Matsubara frequency in
lattice units ($T=1/N_t a$) as
\begin{equation}
\lambda_{{\rm M}} a = \sin\left(\frac{\pi-\arccos 0.37}{4}\right)=0.47\,,
\end{equation}
which is the same order of magnitude as the lower end of the bulk of eigenvalues we measured (consistent with the findings of \cite{Gavai:2008}).
The main point of these considerations is that ``wrong'' Polyakov loops, $L=-1$ with $\varphi=\pi$, would lead to a vanishing effective Matsubara frequency, $\lambda_{\rm M}=0$, and thus to the lowest Dirac eigenvalues.
Of course, in realistic configurations one has to take into account that the Polyakov loop varies in space and that nontrivial spatial links are present.
Both will change the Dirac eigenvalues away from the free ones. Nonetheless, the tendency that ``wrong'' Polyakov loops give rise to smaller eigenvalues
persists and explains our finding about their pinning nature.
\subsection{Topological objects}
Topological excitations of the gauge field and their zero modes are an
attractive hypothesis to explain low-lying Dirac modes in Yang-Mills theory
(and QCD). The chiral condensate at zero temperature is thought
of as due to instantons of realistic size and density and the first conjecture about the metal-insulator transition at finite temperature was based on instanton ensembles \cite{Diakonov:1984vw,GarciaGarcia:2005vj}.
The natural topological excitations at finite temperature are magnetic
monopoles. They appear as self\-dual or antiselfdual solutions of the
Yang-Mills equations of motion at finite temperature. As they are also
electrically charged, they are called dyons. Dyons can be constituents of
calorons (finite temperature instantons) \cite{Kraan:1998,Lee:1998}, but may
exist in isolation as well \cite{Gross:1981}.
In gauge group $SU(2)$, there are two dyons (and two antidyons) with opposite
magnetic charge. One sort of these dyons is characterized by properties that
exactly match our findings: the Polyakov loop at their core is $-1$ and they
possess zero modes with the physical antiperiodic boundary conditions
\cite{Callias:1978,Nye:2000} (just like the constant configurations discussed
above). The other sort has Polyakov loop $+1$\footnote{Monopoles also
appear as defects \protect\cite{Hooft:1981} of the Polyakov gauge
\protect\cite{Weiss:1981}, where they have $L=-1$ or $L=+1$ by
definition.} and zero modes with periodic boundary conditions.
In dilute ensembles of dyons most of the zero modes should remain low-lying
modes. Then the first sort of dyons could explain the
localized modes we analyzed, while the second sort could
be responsible for low-lying periodic modes. At a positive average Polyakov
loop one expects fluctuations to $-1$ much less frequent than those to $+1$,
cf.\ Fig.~\ref{fig polloop properties}. Hence the first sort of dyons
could give independent low-lying modes, while the second sort
could yield a condensate at periodic boundary conditions. This different
appearance can be made quantitative by the different fractions of unit
topological charge the two dyon sorts have, which are such that the $-1$
dyons are indeed heavier as they have a larger (classical) action, see
\cite{Bornyakov:2008im} for lattice evidence of this picture and
\cite{Bruckmann:2008d} for a simple model.
From the relation of magnetic monopoles to both low-lying modes and
topological charge a crucial test of this picture is to compare these two
quantities at given temperature and volume. Above the finite temperature
transition fluctuations of the topological charge decrease sharply. This can
be clearly seen by looking at how the topological susceptibility decreases at
higher temperatures. As a result, at high temperature,
topological objects presumably form a dilute gas of non-interacting
objects. From the index of the overlap Dirac operator we have full information
about the fluctuations of the {\it total topological charge}. Assuming that in
the dilute gas topological objects are uncorrelated, this can be used to
compute the density of topological objects that in turn can be compared to the
density of localized Poissonian Dirac eigenmodes.
In Table \ref{tab:charge_distribution} we show the probability of different
topological sectors extracted from the index of the overlap Dirac
operator. From the low occurrence of the topological charge sector $\pm 1$ and
in particular of the sector $\pm 2$ it is indeed clear that topological
objects form a very dilute gas. In these volumes it very rarely happens that
there is more than one topological object on any given configuration. Since we
need only an order of magnitude estimate of the density of topological objects
we will ignore the probability of two or more objects occurring on any single
configuration. In this approximation we can shortcut the calculation of the
density of topological objects, and the average number of topological
objects per configuration is just given by the probability of the $|Q|=1$
sector. On our volumes it is between $0.04$ and $0.1$ which is clearly far too
small to account for the few localized Poissonian modes we found on average
per configuration.
This comparison rules out models based on uncorrelated gluonic objects that carry both (O(1)) topological charge and (antiperiodic) zero modes. However, it does not exclude combinations of topological objects in which the topological charge cancels, like
instanton-antiinstanton molecules
originally suggested to be present with light dynamical quarks \cite{Ilgenfritz:1994nt} or molecules of dyons called 'bions' carrying one near zero mode \cite{Unsal:2007jx}.
\begin{table}
\begin{tabular}{|c|c|c||c|c|c|}
\hline
$|Q|$ & 0 & 1 & 0 & 1 & 2 \\
\hline
${\cal P}_Q$ & 0.958(6) & 0.042(6) & 0.89(1) & 0.105(9) & 0.009(3) \\
\hline
\end{tabular}
\caption{The probability of different charge sectors in the $16^3\times 4$
(first two columns) and $24^3\times 4$ ensemble (last three columns). The
probabilities of charge sectors with the same magnitude but opposite sign
have been added.
\label{tab:charge_distribution}}
\end{table}
\begin{figure}[!t]
\includegraphics[width=0.9\linewidth]{polyloopcorrelations.pdf}\\
\includegraphics[width=0.9\linewidth]{polyloopsspacingdistribution.pdf}
\caption{Independence of local Polyakov loops. Top: the auto-correlation
$A:=\left(\langle L(x)L(y)\rangle - \langle L(x)\rangle\langle L(y)\rangle \right)/$ $\left(\langle L^2(x)\rangle - \langle L(x)\rangle^2 \right)$
as a function of the lattice distance $|x-y|/a$
(containing the same information as the free energy).
Bottom: the level spacing distribution (of the Polyakov loop trace $L$) compared to the Poisson distribution.}
\label{fig polloop properties}
\end{figure}
\section{Random matrix model using the staggered Dirac operator}
\label{sect rmt}
\begin{figure*}
\includegraphics[height=1.1\threepicwidth]{spacings_rmtmodel_spectrum_0,05_to_0,085.pdf}\hspace*{-0.9cm}
\includegraphics[height=1.1\threepicwidth]{spacings_rmtmodel_spectrum_0,09_to_0,1.pdf}\hspace*{-0.9cm}
\includegraphics[height=1.1\threepicwidth]{spacings_rmtmodel_spectrum_0,11_to_0,12.pdf}
\includegraphics[height=1.1\threepicwidth]{spacings_rmtmodel_spectrum_0,12_to_0,13.pdf}\hspace*{-0.9cm}
\includegraphics[height=1.1\threepicwidth]{spacings_rmtmodel_spectrum_0,14_to_0,15.pdf}\hspace*{-0.9cm}
\includegraphics[height=1.1\threepicwidth]{spacings_rmtmodel_spectrum_0,155_to_0,165.pdf}
\caption{Spacing distributions in several parts of our RMT spectrum plotted along with
the GSE prediction and the Poissonian distribution. The parameters were fixed as discussed in the text.
The data was obtained by an ensemble average over $2000$ random matrices.}
\label{sparsermtspacfigure}
\end{figure*}
Transitions between correlated and uncorrelated eigenvalues like the one observed in
the lattice data can be described by RMT models in various ways. One of the
simplest possible ans\"atze is to start with a diagonal matrix, whose entries are
uncorrelated random numbers, and add a matrix taken from one of the Gaussian
ensembles. This model
shows the desired transition in between parts of its spectrum with different
eigenvalue density,
however, the interpolating spacing distributions
are different
from the ones in the spectrum of the staggered operator \cite{Schierenberg:x}.
A possible explanation is the
sparseness of this operator and the spatial information that is contained in
the next-neighbor interactions. In contrast, the full matrices from
the Gaussian ensembles blindly connect all diagonal elements with equal
strength.
\subsection{Motivation}
We construct a better suited random matrix model, based on sparse matrices, in the following. This model can be nicely motivated by our previous
findings. Concerning the Polyakov loop, which will be connected to a random potential, three properties are relevant:
(i) the distribution of local Polyakov loops extends to negative values and thus small effective Matsubara frequencies, cf. Fig.~\ref{fig polloop modes two},
(ii) Polyakov loops become independent quickly with their distance and therefore
(iii) the level spacings of the Polyakov loop trace $L$ -- neglecting the spatial information -- are distributed according to a Poissonian distribution.
The last two properties are depicted in Fig.~\ref{fig polloop properties}.
Hence the Polyakov loops could provide the Poissonian ingredient for the statistics of the Dirac eigenvalues. To become more concrete, we
remind the reader of the definition of the staggered Dirac operator,
\begin{equation}
D_{xx'} = \frac1{2a} \sum_{\mu=1}^4 \eta_\mu(x)\left[\delta_{x+\hat\mu, x'}U_\mu(x) - \delta_{x-\hat\mu, x'}U_\mu^\dagger(x')\right]\,,
\end{equation}
with $\eta_\mu(x) = \left(-1\right)^{\sum_{\nu<\mu}x_\nu}$ and $U\in SU(2)$. We split this operator in the temporal and spatial part
\begin{equation}
D = D^{TE} + D^{SP}\,.
\end{equation}
$D^{TE}$ contains the hopping terms in the temporal direction, whereas spatial hoppings are included in $D^{SP}$.
\begin{figure}
\includegraphics[width = 0.9\linewidth]{localizations_rmtmodel_to_averageEV_err_is_var.pdf}
\includegraphics[width = 0.9\linewidth]{polyoverlap_rmtmodel_to_averageEV_err_is_var.pdf}
\caption{Top: Inverse participation ratios of the 25 lowest-lying eigenmodes for the RMT model, plotted versus the average eigenvalue of
that mode. The data was obtained by an ensemble average over $2000$ random matrices.
Bottom: diagonal entries ``as averaged by the lowest modes'' of the RMT model, $\vartheta_m$ from Eq.~(\protect\ref{eqn theta averaged}), divided by the average diagonal entry, to be compared with Fig.~\protect\ref{fig polloop modes one}.
Horizontal error bars visualize the spreads of the eigenvalues.}
\label{sparsermtlocfigure}
\end{figure}
The temporal part has a block-diagonal structure, with one block for each
spatial site. We want to approximate each block and thus the whole Dirac operator
by restricting it to the subspace of the smallest
eigenvalue quadruplet of the temporal operator at each spatial lattice
site. The quadruplet consists of two eigenvalue pairs with opposite sign. The
plus-minus degeneracy is necessary to conserve chiral symmetry, whereas the
exact two-fold degeneracy\footnote{In RMT language, this is Kramers degeneracy
of the GSE.} has to be kept in order to have the right RMT universality
class we find in the SU(2) staggered spectra.
In this basis, the temporal part of the Dirac operator can be brought in the form
\begin{equation}
D^{TE(n=0)}_{\vec x \vec x} = \begin{pmatrix} -\theta_{\vec x} & 0 & 0 & 0 \\ 0 & -\theta_{\vec x} & 0 & 0 \\
0 & 0 & \theta_{\vec x} & 0 \\ 0 & 0 & 0 & \theta_{\vec x}\end{pmatrix}\,,
\end{equation}
with
\begin{equation}
\theta_{\vec x} = \frac1a \sin\left(\frac{\pi-\varphi_{\vec x}}{N_t}\right)\,,
\end{equation}
which resemble the effective Matsubara frequencies on the lattice, as given in Eq.~(\ref{eqn eff MF latt}), however as a function of the
local rather than averaged Polyakov loop phase $\varphi_{\vec x}$. Therefore, $D^{TE(n=0)}$ is a diagonal matrix with very weakly coupled entries, as argued above.
Physically speaking, the temporal part represents a (chiral) random potential.
The spatial part becomes in the restricted basis
\begin{equation}\label{sp0dirac}
D^{SP(n=0)}_{\vec x, \vec x + \hat{\imath} } = \begin{pmatrix} U_{\vec x, \vec x + \hat{\imath} } & V_{\vec x, \vec x + \hat{\imath} } \\
V_{\vec x, \vec x + \hat{\imath} } & U_{\vec x, \vec x + \hat{\imath} }\end{pmatrix}\,,
\end{equation}
where the 2-by-2 matrices $U$ and $V$ can be shown to represent real quaternions.
These have a concrete meaning:
$U$ connects eigenvalue pairs of equal sign, and is therefore responsible for the GSE-like level repulsion between nearest neighbors.
$V$ on the other hand generates the gap around zero in the spectrum as it connects eigenvalues of different signs.
\subsection{Explicit construction of the model}
We propose a random matrix model based on matrices
\begin{equation}
M = M^{TE} + M^{SP},
\end{equation}
that consist of two parts. $M^{TE}$ has the same diagonal structure like $D^{TE(n=0)}$, i.e.
\begin{equation}
M^{TE}_{\vec x \vec x} = \begin{pmatrix} -\vartheta_{\vec x} & 0 & 0 & 0 \\ 0 & -\vartheta_{\vec x} & 0 & 0 \\
0 & 0 & \vartheta_{\vec x} & 0 \\ 0 & 0 & 0 & \vartheta_{\vec x}\end{pmatrix}\,,
\end{equation}
where the diagonal entries
\begin{equation}
\vartheta_{\vec x} = t(\pi-\phi_{\vec x})
\end{equation}
are random numbers constructed with an overall scale $t$ and a random angle $\phi_{\vec x}\in[0,\pi]$. These quantities are equivalent to the effective Matsubara frequencies $\theta_{\vec x}$ in the continuum, Eq.~(\ref{eqn eff MF}), the temperature $T$ and the angle $\varphi_{\vec x}$ of the local Polyakov loop, respectively.
For $\phi_{\vec x}$ we have taken the empirical distribution of the angle $\arccos L(\vec{x})$ of local Polyakov loops, by converting the fine histogram in Fig.~\ref{fig polloop modes two} accordingly. This yields an asymmetric distribution of $\phi_{\vec x}$ between $0$ and $\pi$ with a maximum below $\pi/2$ (i.e.\ at positive Polyakov loop). Its most important feature, however, seems to be the tail towards the ``trapping'' $\phi_{\vec x}\simeq \pi$ (negative Polyakov loop locally), since we have observed that most of the features of the model persist when using the Haar measure $\sin^2\phi$ as distribution [not shown].
For the spatial part, one first of all needs to fix the periodicity of the underlying space, i.e. an integer $N_s$ such that $\vec x$ is identified with
$\vec x + N_s \hat{\imath}$, with unit vectors $\hat{\imath}$ in each of the spatial directions. $M^{SP}$, like $D^{SP(n=0)}$, has nonvanishing entries only at positions
that connect next neighbors in that space. Its blocks
\begin{equation}
M^{SP}_{\vec x, \vec x + \hat{\imath} } = \begin{pmatrix} u_{\vec x, \vec x + \hat{\imath} } & v_{\vec x, \vec x + \hat{\imath} } \\
v_{\vec x, \vec x + \hat{\imath} } & u_{\vec x, \vec x + \hat{\imath} }\end{pmatrix}\,,
\end{equation}
consist of random real quaternions $u$ and $v$, that are RMT counterparts of $U$ and $V$ from Eq.~(\ref{sp0dirac}). We have taken them to be Gaussian distributed around zero with their mean deviations $\sigma_{u},\sigma_{v}$ as parameters of the model.
By rescaling all the random matrices $M$ it is clear, that only the ratios of the scales $t$, $\sigma_{u}$ and $\sigma_{v}$ are relevant parameters. To fix them we have measured the ratio of the average determinants\footnote{For real quaternions, the determinant is just the sum over all squared components.} of the empirical quaternionic hopping terms finding $\langle \det V\rangle/ \langle \det U\rangle \approx 1.6^2$ and took this ratio over for the ratio of $\sigma_{u}^2/\sigma_{v}^2$. The remaining ratio $\sigma_u/t=0.2$ ($\sigma_v/t=0.32$)
was put in by hand to obtain desired properties of the RMT model, namely a gap at zero and a
transition between a Poissonian and a GSE spacing distribution.
The eigenvalue density and spacing distribution of this model are plotted in Fig.~\ref{sparsermtspacfigure} for a spatial extent $N_s = 12$ and $t=1/4$.
We observe a similar transition like in the spectrum of the staggered Dirac operator. Another feature that this model shares with the staggered operator is the
increasing localization of eigenmodes as the corresponding eigenvalues decrease,
quantified by the IPR's in Fig.~\ref{sparsermtlocfigure} top.
In order to measure the correlation of the lowest eigenmodes of this RMT model to the diagonal entries, we again define the latter ``as averaged by a particular mode''
\begin{equation}
\vartheta_m:=\sum_{\vec x} |\psi_m({\vec x})|^2\vartheta_{\vec x}\,.
\label{eqn theta averaged}
\end{equation}
One can see in Fig.~\ref{sparsermtlocfigure} bottom, that the low modes are indeed localized to ``islands'' of low $\vartheta$, which are equivalent to low Polyakov loops. Hence this important effect is shared by our random matrix model, too.
\section{Conclusions}
\label{sect conc}
In the present paper we found a possible explanation for the emergence of
localized Poissonian modes at the low end of the high temperature QCD Dirac
spectrum. We showed that the localized modes are strongly correlated with
large fluctuations of the Polyakov loop. This lowers the effective
Matsubara frequencies for modes concentrated there. We argued that the lowest part of the Dirac spectrum
consists of this type of eigenmodes. We verified this picture for eigenmodes
of both the staggered and the overlap Dirac operator. As a side result we also
demonstrated that the spatial structure of the lowest overlap and staggered
modes is highly correlated. This shows that different discretizations of the
Dirac operator are sensitive to the same type of gauge field
fluctuations.
We also looked at the topological charge fluctuations as
given by the index of the overlap. Assuming that at high temperature
topological objects form a dilute gas and are uncorrelated,
we could safely rule them out in creating the localized low modes.
Finally, we proposed a dimensionally reduced random matrix
model. It is based on sparse matrices encoding the three dimensional nature of the problem through
nearest neighbor couplings from a lattice Dirac operator. To the best of our knowledge this is an example of a new kind of random matrix models for QCD, where so far only full matrices have been used. Beside chirality and the spectral gap our model reproduces the transition from
localized Poissonian to delocalized random matrix type modes observed in the
lattice Dirac spectrum as well as the correlation of the localized modes to ``islands'' of low on site ``potential''.
It is instructive to compare the Dirac operator to the
Hamiltonian of Anderson type models, see \cite{Lee:1985zzc,Evers:2008zz} for reviews.
In the latter case usually diagonal (on
site) disorder is responsible for creating the transition to localized
eigenmodes. In the case of the Dirac operator, the on site terms do not seem
to be relevant. In fact, in the staggered operator they are exactly
zero. However, in our dimensionally reduced three dimensional effective model
the non-zero fluctuating on site terms are dynamically generated by
fluctuations of the local Matsubara frequency resulting from fluctuations of
the Polyakov loop. In this way our dimensionally reduced effective random
matrix model is analogous to the Anderson model. It would be interesting to
study further how the presence and details of the transition depend on the
parameters and matrix size defining the random matrix ensemble.
\section{Acknowledgements}
We thank helpful discussions to Jacques Bloch, Antonio M.\ Garcia-Garcia, Ferenc Pittler, Mithat \"Unsal and Tilo Wettig.
FB and SS are supported by DFG (BR 2872/4-2) and TGK by EU Grant
(FP7/2007-2013)/ERC n$^o$208740.
|
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Excellence Every Day Without Excuses
CHOOSE MT. PISGAH MIDDLE SCHOOL
Electives & Library
Main Office & Support Team
Bachelors degree in Elementary Education, with a minor in Sociology Master of Arts in Teaching Special Education Master's degree in Educational Leadership & Administration
Mr. ROBERT DAVIS, II
Robert Davis II, is a Memphis native, from the Whitehaven area. He attended Hillcrest High School, and went on to attend The University of Memphis, where he obtained a Bachelors degree in Elementary Education, with a minor in Sociology. During his tenure as an undergrad, he also became a member of Kappa Alpha Psi Fraternity, Inc., Kappa Beta Chapter and is currently active with the Memphis Alumni Chapter. For graduate school, he attended Christian Brothers University, where he secured a Master of Arts in Teaching Special Education, and an additional Master's degree in Educational Leadership & Administration. He's now attending the University of Mississippi working towards a Doctorate in Education, which includes a focus in Educational Leadership and Policy Studies. Robert was recently named one of Memphis' "Top 40 under 40" Urban Elite Professionals in 2018. He spends most of his time participating in the Leadership Memphis program to become a more effective change agent for the City of Memphis and serving as a mentor within a non-profit organization called "DNA" (Developing Noble Men). Professionally, Robert previously served as Principal at Ida B. Wells Academy, an Assistant Principal, Professional Learning Coach, Special Education Teacher, and ISS Coordinator, all within Memphis City & Shelby County Schools
Bachelor of Arts Degree in English Literature Master of Arts in Education Doctorate in Educational Leadership and Policy
Dr. RHONDA M. ANTHONY
Dr. Rhonda Anthony has served as an educator in the legacy Memphis City Schools/Shelby County Schools system for eleven years. She began her career as a 9th grade English teacher and later served as a 7th and 8th Grade ELA teacher. She has also had many teacher-leader roles, including Common Core Lead, Grade Level Chairperson, Master Teacher, and ILT Content Lead. Prior to assuming the role of Assistant Principal at Mt. Pisgah, Dr. Anthony served as a PLC Coach, RTI Lead, Title 1 Coordinator, and ILT Admin Lead for ELA.
Dr. Anthony obtained a Bachelor of Arts Degree in English Literature with a minor in History from the University of Oklahoma. She went on to obtain a Master of Arts in Education from Union University, and recently defended her dissertation to obtain a Doctorate in Educational Leadership and Policy from the University of Memphis. Dr. Anthony has a passion for equity in education and culturally responsive pedagogy in urban schools.
PLC COACH
Bachelor of Arts degree in Political Science with a minor in History Master in the Art of Teaching degree in Special Education Education Specialist degree in Curriculum, Instruction, and Supervision
Mrs. TARVIS L. MULL
Tarvis Williams-Mull has been an educator for eleven years. Mrs. Mull's service as an educator spans across three local school districts, both as a classroom teacher and in various teacher support roles. Mrs. Mull has earned a Bachelor of Arts degree in Political Science with a minor in History, a Master in the Art of Teaching degree in Special Education, and an Education Specialist degree in Curriculum, Instruction, and Supervision.
Mrs. Mull began her teaching career as a high school, special education algebra I and English 9 inclusion teacher. Before leaving the classroom, Mrs. Mull also served as a middle school content inclusion teacher, a 7th and 8th-grade science teacher, and an AVID elective teacher.
Before joining the Mt. Pisgah faculty in 2016, Mrs. Mull served as the Director of Special Education for the five direct-run Achievement Schools. In this role, Mrs. Mull was responsible for developing and monitoring an annual budget, developing and evaluating special programs, planning and facilitating staff development, as well as serving on various teams and committees.
1444 Pisgah Road
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Monaco officials say Crown Prince Albert has taken over regency of the Mediterranean city-state, in light of the failing health of his father, Prince Rainier.
The announcement Thursday comes two days after a health bulletin called the 81-year-old ruler's prognosis "extremely reserved."
Prince Rainier, Europe's longest-reigning monarch, was hospitalized three weeks ago for heart, lung and kidney problems.
He ascended to the throne in 1949. His Grimaldi family has ruled the tiny principality on the French Riviera since the 13th century.
The prince is widely credited with turning the city-state, of less than two square kilometers in area, into a fashionable and wealthy tourist resort. His wife, Hollywood film star Grace Kelly, died in a car crash in 1982.
The 47-year-old Albert is the eldest of the couple's three children.
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Skitter TV
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{"url":"https:\/\/www.themathdoctors.org\/at-least-one-six-right-and-wrong-ways\/","text":"# Rolling a 6 on Three Dice\n\n#### (An archive question of the week)\n\nOne of the discussions we looked at last time involved rolling three dice and getting at least one six. I didn\u2019t go into detail on the calculation there; but I found another place where we discussed it at length. We\u2019ll look at that here.\n\n## A wrong way and a right way\n\nThe question came in 2013:\n\nAt Least One \"6\" in Three Dice Rolls, Summed SimplyI know this is wrong, but not why.\n\nIf you have 3 fair six-sided dice, and you wish to find the probability of\u00a0rolling at least one 6, why can't you add the individual probabilities of getting a 6?\n\nIn other words, if 1\/6 is the probability of getting a 6 in one throw, why is 1\/6 + 1\/6 + 1\/6 not the probability of getting a 6 in three throws?\n\nI know it is wrong to do so, but why, exactly? Where is the logical error in reasoning when adding the probabilities?\n\nIn case you wonder whether or not I know how to actually solve such a simple problem, the actual solution is\n\nP(at least one 6 out of three throws) = 1 - P(no six)\n= 1 - (5\/6)^3\n\nAnd I know that calculating the probability for 6 tosses by adding them would give a probability of 1, which is impossible.\n\nBut again, where is the logical error in adding them?\n\nFirst, we can tell that adding probabilities can\u2019t be right, because no matter what the probability of an event happening on one trial, if you add up enough of them, the total will exceed 1, and no probability can be greater than 1.\n\nSecond, the quickest way to find an \u201cat least one\u201d probability is to first find the probability of \u201cnone\u201d. These are complementary events: If the outcome is not \u201cat least one\u201d, then it is \u201cnone\u201d, and if it is not \u201cnone\u201d, then it is \u201cat least one\u201d. So the probability we want is 1 minus the probability of none, which is easily found by multiplication if the individual trials are independent.\n\n## Why the wrong is wrong\n\nI replied, starting with what you can\u2019t do:\n\nYou can only add the probabilities of mutually exclusive events (unless you make adjustments, as I'll show). This is because if you add events that \"overlap,\" you are counting some possible outcomes more than once.\n\nI\u2019ll have more to say about this below. But I can use a Venn diagram to illustrate the issue. Rather than put the number of items in each region of the diagram, as we usually do, I will put an example of a roll; for example, (1, 2, 3) will mean we roll first a 1, then a 2, then a 3, so that we rolled no sixes.\n\nBecause of the overlaps, adding the number (or probability) of rolls with the first, the second, and the third being six will greatly overcount the rolls. For example, the roll (6,6,3) would be counted twice, for the first roll and for the second roll.\n\n## More right ways\n\nBut the Venn diagram suggests several other ways to do it:\n\nThere are, however, alternative ways that do involve adding. One way to break up the event \"at least one 6\" into mutually exclusive events is:\n\nP(exactly one 6) + P(exactly two 6's) + P(exactly three 6's)\n\nThis gives\n\nC(3,1)*(1\/6)^1*(5\/6)^2\n+ C(3,2)*(1\/6)^2*(5\/6)^1\n+ C(3,3)*(1\/6)^3*(5\/6)^0\n\n= 3*25\/216 + 3*5\/216 + 1*1\/216\n= 91\/216\n\nWhat does this mean? The probability that we roll exactly one six is 3 (the number of ways to choose which one will be a six \u2014 the number of one-set regions in the diagram), times the probability that that roll will be a six (1\/6), times the probability that the other two rolls will not be six (5\/6*5\/6). Similarly for the other two probabilities.\n\nThis takes some work, but is very straightforward.\n\nThis agrees with your calculation:\n\n1 - (5\/6)^3 = 216\/216 - 125\/216\n= 91\/216\n\nThis is what Eway had done, which is the method we usually teach. The probability that none of the three rolls is a six is 5\/6 (the probability that an individual roll is not a six), raised to the 3rd power. The probability that this does not happen (that is, that at least one roll is a six) is 1 minus that.\n\nSo now we have two ways to calculate the same answer.\n\nYou asked about calculating this probability this way:\n\nP(first is 6) + P(second is 6) + P(third is 6)\n\nThis would count rolls with more than one 6 more than once; e.g., the roll 3, 6, 6 is counted as having the second roll be 6 AND having the third roll be 6.\n\nIn order to correct for this, you would have to use the inclusion-exclusion principle, subtracting out the outcomes counted twice, but then adding back in the outcome that was added three times and then subtracted out three times:\n\nP(1st is 6) + P(2nd is 6) + P(3rd is 6)\n- P(1st and 2nd are 6) - P(2nd and 3rd are 6) - P(1st and 3rd are 6) + P(all three are 6)\n\n= 1\/6 + 1\/6 + 1\/6 - 1\/36 - 1\/36 - 1\/36 + 1\/216\n= 36\/216 + 36\/216 + 36\/216 - 6\/216 - 6\/216 - 6\/216 + 1\/216\n= 91\/216\n\nSo there are three valid ways to do it. Which do you prefer?\n\nLooking back at our Venn diagram, we are adding the three circles together, which counts each of the two-circle regions twice, and counts the middle region three times. Then we subtract each of the intersections of two circles to compensate for the overlaps; but this subtracts the middle region three times, totally removing it from our count (as it had been counted three times initially). So we have to add it back in.\n\nThis is an extension of the formula for the union of two sets:\n\n$$P(A\\cup B) = P(A)+P(B)-P(A\\cap B)$$\n\nFor three sets, it is\n\n$$P(A\\cup B\\cup C) = P(A)+P(B)+P(C)-P(A\\cap B)-P(B\\cap C)-P(C\\cap A)+P(A\\cup B\\cap C)$$\n\nFor more on inclusion-exclusion, see\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.","date":"2022-05-20 09:46:30","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\": 1, \"mathjax_asciimath\": 1, \"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\": 0.7196267247200012, \"perplexity\": 437.74455576559063}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 5, \"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-2022-21\/segments\/1652662531779.10\/warc\/CC-MAIN-20220520093441-20220520123441-00171.warc.gz\"}"}
| null | null |
Q: Bulk rename files in Unix with current date as suffix I am trying to bulk rename all the files in the current folder with date suffix:
rename 's/(.*)/$1_$(date +%F)/' *
But that command is renaming info.txt to info.txt_1000 4 24 27 30 46 113 128 1000date +%F). I want the result to be info.txt_2016-10-13
A: You want $1 to be passed literally to rename, yet have $(date +%F) to be expanded by the shell. The latter won't happen when you use single quotes, only with double quotes. The solution is to use double quotes and escape $1 so the shell doesn't expand it.
rename "s/(.*)/\$1_$(date +%F)/" *
A: Portable Posix shell solution
Since you said "in Unix" and the rename command isn't portable (it's actually a part of the perl package), here is a solution that should work in more environments:
for file in *; do mv "$file" "${file}_$( date +%F )"; done
This creates a loop and then moves each individual file to the new name. Like the question, it uses date +%F via shell process substitution to insert the "full date" (YYYY-mm-dd). Process substitution must use double quotes (") and not single quotes (') because single quotes shut off shell interpretations (this goes for variables as well), as noted in John's answer.
Argument list too long
Your shell will complain if the directory has too many files in it, with an error like "Argument list too long." This is because * expands to the contents of the directory, all of which become arguments to the command you're running.
To solve that, create a script and then feed it through xargs as follows:
#!/bin/sh
if [ -z "$1" ]; then set *; fi # default to all files in curent directory
for file in "$@"; do mv "$file" "${file}_$( date +%F )"; done
Using ls to generate a file list isn't always wise for scripts (it can do weird things in certain contexts). Use find instead:
find . -type f -not -name '*_20??-??-??' -print0 |xargs -0 sh /path/to/script.sh
Note, that command is recursive (change that by adding -maxdepth 1 after the dot). find is extremely capable. In this example, it finds all files (-type f) that do not match the shell glob *_20??-??-?? (* matches any number of any characters, ? matches exactly one of any character, so this matches abc_2016-10-14 but not abc_2016-10-14-def). This uses find … -print0 and xargs -0 to ensure spacing is properly preserved (instead of spaces, these use the null character as the delimiter).
You can also do this with perl rename:
find . -type f -not -name '*_20??-??-??' -print0 |xargs -0 \
rename "s/(.*)/\$1_$( date +%F )/"
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Since Dec. 22, the University and other research institutions have scrambled to adapt to the unprecedented political stalemate over the federal government shutdown, which has dried up funds for federally supported research projects and now threatens long-term scientific advancement across the nation.
When the government shut down before the holidays, major agencies like the National Science Foundation, the National Endowment for the Humanities, the National Oceanic and Atmospheric Administration and the National Aeronautics and Space Administration all closed without approved appropriations for 2019. Research institutions rely on these agencies not only for funding but also for technological, data-based and research support.
"The federal shutdown is causing uncertainty and complications about research funding at Brown … In some cases the shutdown is slowing or halting research itself," Vice President for Research Jill Pipher wrote in a statement to The Herald.
"We've been in a state of pretty perpetual anxiety since the shutdown started," said postdoctoral researcher Christopher Horvat, who works as a polar oceanographer at the Institute at Brown for Environment and Society, where he studies sea ice in the Arctic and Antarctic Oceans in relation to climate change. Horvat is not only a post-doc at the University but also a government contractor — he holds a second postdoctoral fellowship with the NOAA Climate and Global Change Program. Though Horvat's paycheck has not yet been cut, his financial situation could change if the shutdown continues for more than a few more weeks, he said.
Many research endeavours, especially in relation to the work of government contractors, have been put on hold. Conference travel cannot be completed during the shutdown since it is a "large capital investment that takes time to plan" and cannot be scheduled when future sources of funding are uncertain, Hovart said. Meeting with other experts is crucial for scientists who use these trips to forge collaborations with their peers and discuss new findings from institutions scattered across the globe.
"Some researchers are having to pay for travel on their own, hoping for later reimbursement, and some have had to cancel trips because they couldn't afford to pay the expenses on their own. In addition, federal program officers are unable to attend scientific meetings and conferences that inform their work," Pipher wrote.
In order to serve on grant review panels for the NSF, many University researchers must travel to the locations where the panels are held. All scheduled panels have been "canceled and will likely be rescheduled to a later date," according to the NSF. As the schedule continues to be delayed, the pace of scientific research at the University and across the nation could slow. "This will cause problems for many months after the shutdown ends," Pipher wrote. If panels cannot convene, then grants previously submitted to the NSF cannot be reviewed and researchers who otherwise would have been funded will see a delay in their project timelines. In addition, no new grants will be reviewed or approved during the shutdown.
Professor Emeritus of Geological Sciences Peter Schultz, who directs the Northeast Planetary Data Center and previously worked at NASA, expressed concerns about back pay for contractors working at NASA and other agencies. Though President Trump has signed an act guaranteeing back pay for most federal employees affected by the shutdown, this legislation does not usually apply to government contractors. As the shutdown continues, this could equate to the loss of a significant percentage of their yearly salary, Schultz said. The "human toll" is detrimental, and "NASA may lose some really good people" as a result, he added.
Beyond the immediate impact, the University and other research institutions may feel a number of long-term ripple effects.
Time-sensitive work can be compromised by a lack of resources during a shutdown, Horvat said, especially as this shutdown continues to stretch into the longest in American history. For example, reports for the United Nations Intergovernmental Panel on Climate Change, which involves groups working for years to develop accurate predictive climate models, could theoretically be permanently handicapped by the pause in data collection. "Those labs in America may have their computers shut down. If they are not able to do that simulation, they won't be included in the next climate report," Horvat said. Without funding, valuable data can be lost and years of work compromised.
At NASA, "missions are still going to operate," Schultz said, but data analysis from ongoing missions or preparations for upcoming projects may be postponed. Committees populated by contractors or civil servants who are not able to work during the shutdown will be behind schedule when they return to work, he added.
"Brown is monitoring the situation closely and will continue to report any potential implications to the University community," Pipher wrote in her statement. It has yet to be determined when the shutdown will end, but for these scientists, like many federal employees, the effects are already being felt.
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Q: Page-turning number of a graph Motivation. As I was travelling in the UK, I used a physical copy of the "A-Z Road Atlas BRITAIN" for getting around. I was impressed that whenever I wanted to go from the map segment shown on page 23, say (showing a part of the West Midlands), to the segment east of the one I was looking at, I only had to turn unexpectedly few pages. (For every page, the page showing the next part of the country South, East, West, or North, was indicated.) Which motivated the following concept.
Formalization. We regard any non-negative integer $n$ as an ordinal, so $n$ is the set of all its predecessors, i.e. $0 = \emptyset$, and $n = \{0,\ldots,n-1\}$ for any positive integer $n$.
Let $n$ be a positive integer, and let $G = (V,E)$ be a finite, simple, undirected graph with $|V| = n$. We define the "page-turning number" of $G$ by $$\pi(G) = \min\big\{\max\{|\varphi(v) - \varphi(w)|: \{v,w\}\in E\} \; : \; \varphi: V \to n \text{ is bijective}\}.$$ The intuition behind this is the following: we
regard $\varphi\in S_n$ as a "page-number assignment" to the vertices of the graph and the $\max$ part denotes the largest number of pages we have to turn to get from one vertex to any adjacent one. We want to minimize on this $\max$ part.
Note that for the complete graph $K_n$ we have $\pi(K_n) = n-1$. It seems that the page-turning number might be loosely related to coloring, but it is certainly not the same. I would be grateful for hints to an official name for $\pi(G)$.
Question. For any positive integer $n$, let the $n\times n$-grid graph be given by $G_n = (n\times n, E)$ where $$E = \big\{\{(a,b), (c,d)\}: a,b,c,d \in n \text{ and } |a-c| + |b-d| = 1\big\}.$$ In terms of $n$, what is the value of $\pi(G_n)$?
A: The page-turning number of a graph $G$ is also known as the bandwidth of $G$ (https://en.wikipedia.org/wiki/Graph_bandwidth).
The Wikipedia page also contains values of the bandwidth for some special graphs, including the $m\times n$ grid graph, in which case it is equal to $\min\{m,n\}$. This was proved by Jarmila Chvátalová: https://doi.org/10.1016/0012-365X(75)90039-4.
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Home//Blog List//Government Announces €130m for Programmes to Target Fuel Poverty
Government Announces €130m for Programmes to Target Fuel Poverty
by brianflax | Oct 5, 2015 |
Minister for Energy, Alex White, announced new plans to help reduce fuel poverty in Ireland with €130m in new funding as part of the Government's Capital Investment Plan. This funding is in addition to the €310m already allocated for energy efficiency and renewable energy programmes. The new funding will specifically target those experiencing or at risk of fuel poverty across the state.
Speaking at Energy Action's Fuel Poverty Conference, Minister White addressed a number of key points that will be included in the new Affordable Energy Strategy, to be published before the end of the year:
The Affordable Energy Strategy, as part of the Capital Investment Plan, will include retrofitting initiatives for those in the private rental sector
The plan will target those with health conditions that could be affected by unhealthy/cold homes
Health care professionals will be trained on programmes and schemes available to those they provide care – such as the Warmer Homes Scheme
Minister White also expressed the need for wholesale utilities to pass rate decreases on to consumers as quickly as possible. The Affordable Energy Strategy will go into further detail when it is published before the end of 2015.
Key Facts and Numbers – Energy Efficiency
Government-funded programmes have assisted in upgrading over 300,000 households across the State
The Warmer Homes Scheme has upgraded over 120,000 energy poor homes to date, and continues to do so
Better Energy Programmes account for more than 2,300 construction jobs per annum
Ireland spends over €6B per annum on fossil fuel imports. Energy efficient buildings will help to reduce this amount and associated greenhouse gas emissions
While addressing fuel poverty and energy efficient buildings is extremely important, the Capital Investment Plan includes many additional programmes and measures aimed at mitigating climate change, including using renewable heating in industry, utilising renewable energy sources throughout the commercial sector and flood mitigation for at-risk areas.
Funding will also be provided to benefit the following industries:
Other areas such as justice, sports, culture and the National Broadband Plan
You can learn more about the funding for the Better Energy Homes Scheme and Warmer Homes Scheme directly on the KORE website.
published:05 Oct 2015
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{"url":"https:\/\/chemistry.stackexchange.com\/questions\/143960\/need-reactants-for-a-rocket","text":"# Need reactants for a rocket [closed]\n\nI am doing a project in chemistry at the moment to build a rocket out of materials that can be easily bought and that react strongly to create thrust. I was wondering if anyone knew a good chemical formula for my rocket fuel? I know lots of good fuels and reactants used in amateur rocketry however commercial fuels have been banned from the project. Any help would be appreciated.\n\n\u2022 I remember school days where sodium chlorate NaClO3 has been widely used as total herbicide called Travex . We used as the solid propelent the filtration paper, soaked by concentrated Travex solution and dried. It had to be kept dry, as sodium chlorite is hygroscopic. Not sure if it is still available. Generally, in context of terorism, public access to resources for violent chemical reactions are rather limited. For similar purposes could be used KNO3, but solution must be hot because of solubility. Other option could be classical black gun powder. \u2013\u00a0Poutnik Dec 18 '20 at 8:16\n\u2022 See this excellent site about all things rocketry-related for suggestions. There are also other answers on this site so poke around. \u2013\u00a0Todd Minehardt Dec 18 '20 at 17:59\n\u2022 en.wikipedia.org\/wiki\/V-2_rocket used oxygen and climate neutral ethanol. Whatever you use, it must burn fast enough to be a good propellant, but slow enough not to burst your rocket. A cardboard rocket for new years day will need different propellant from a high performance steel casing. \u2013\u00a0Gyro Gearloose Dec 18 '20 at 23:10\n\nPut some potassium nitrate $$\\ce{KNO3}$$ plus powdered charcoal and sulfur in a mortar. Wet it carefully and grind the whole until you get a black homogeneous paste. Let it dry overnight in the mortar. The obtained black gun powder may be detached from the mortar with a wooden spoon. It is a safe rocket fuel. The proportions of the powders are defined by stoichiometry. They must allow the chemical reaction : $$\\ce{2 KNO3 + 3 C + S -> K2S + 3 CO2 + N2}$$\n\nI have done it many times with my students. It's a good exercice of stoichiometry. Never had an accident ! And remember : Never ! Never grind dry powders ! Grind as wet powders as possible ! In case of doubt add more water ! Pastes are even a better choice for grinding purposes !\n\nThe chemistry is a safety issue, but the mechanics of rocket guidance is important too. Take it easy with grinding: I was grinding a little mass (maybe 10 grams) of propellant when it ignited (did not explode!), and I jerked back and banged my head.\n\nI made small rockets by winding paper around a pencil, then filling the tube with $$KClO_3$$ + sulfur + sugar. Ignited in a copper tube about 2 feet long, they would fly straight out and go about 200 feet, except one that made a right angle turn right out of the tube and sailed right by an observer, hitting a garage door, where the rest of the propellant exploded, leaving a clean spot on the rusty door about 6 inches in diameter. It probably burned thru the paper tube unsymmetrically, deviating the thrust from a straight flight. The propellant burned well, but had an unexpected impact sensitivity.\n\nWith the high energy reactions, scale-up is problematic: you know that at some point, your mix may change from burning fast to exploding very fast. Do things cautiously, small scale at first and do not scale up too fast. Have several layers of safety measures, because you know that you will not stop building bigger rockets until you have a catastrophic failure. Plan to have the catastrophe some considerable distance from you and others.\n\ntry 65g of KNO3 per 35g of sugar. Dissolve the mix in the smallest possible amount of hot water on a pan, and boil off the water, once you start seeing large bubbles and it turns into a paste start mixing. Make sure it stops sizzling when you push it down with a spatula. Sounds dangerous but really it's reliable, just do it outside. Coalesce it into a large blob with the spatula while it's still hot. It will harden within a few minutes so you better put it into the motor immediately.\n\nPush a pencil in the middle of the fuel to create a hollow cyllinder, that will make it burn much faster.\n\nAlso, it's much easier to set off with an ember than flame","date":"2021-04-16 12:23:29","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\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 3, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.4782482981681824, \"perplexity\": 2117.685616449098}, \"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\/1618038056325.1\/warc\/CC-MAIN-20210416100222-20210416130222-00190.warc.gz\"}"}
| null | null |
Q: Update each button item inside a list instead of all buttons at once in Swift / SwiftUI I'm currently trying to change a button from one view to another depending on what the user presses. They would get a list/scrollview of pending requests and if the user clicks on accept, it should change that button to an accepted button and if the user clicks on reject, it should change that button to a rejected button. I'm running into a problem where if a user clicks on accept, all the pending request buttons gets changed to an accepted button and similarly for the reject case.
import SwiftUI
import Firebase
import SDWebImageSwiftUI
import Foundation
import Combine
struct BottomSheet: View {
// @ObservedObject var userData : UserViewModel
var edges = UIApplication.shared.windows.first?.safeAreaInsets
@StateObject var userData = UserViewModel()
@State var declinedRequest: Bool = false
@State var acceptedRequest: Bool = false
var body: some View {
VStack{
Spacer()
VStack(spacing: 12){
Divider()
// here are the buttons inside the scrollview
ScrollView{
ForEach(userData.pendingFriendUsers){ person in
HStack{
if person.pic != ""{
WebImage(url: URL(string: person.pic)!)
.resizable()
.aspectRatio(contentMode: .fill)
.frame(width: 60, height: 60)
.clipShape(Circle())
.padding(.leading, 30)
.padding(.trailing, 10)
}else{
Circle()
.stroke(Color.black.opacity(0.8), lineWidth: 2)
.frame(width: 60, height: 60)
.padding(.leading, 20)
.padding(.trailing, 10)
}
VStack(alignment: .leading){
Text("\(person.name)")
.font(.custom("Helvetica Neue", size: 16))
.foregroundColor(Color.white).bold()
Text("@\(person.username)")
.font(.custom("Helvetica Neue", size: 16))
.foregroundColor(Color.white)
.opacity(0.8)
}
Spacer()
if person.isFriends == 2 {
RoundedRectangle(cornerRadius: 15, style: .continuous)
.fill(Color("Dark-Grey"))
.frame(width: 150, height: 40)
.padding(.trailing, 25)
.overlay(
Text("Request removed")
.font(.custom("Helvetica Neue", size: 14))
.foregroundColor(Color.white)
.padding(.trailing, 25)
)
}else if person.isFriends == 1 {
RoundedRectangle(cornerRadius: 15, style: .continuous)
.fill(Color.white)
.frame(width: 150, height: 40)
.padding(.trailing, 25)
.overlay(
Text("Request accepted")
.font(.custom("Helvetica Neue", size: 14))
.foregroundColor(Color.black)
.padding(.trailing, 25)
)
}else {
Button(action: {
print("declined friend request for \(person.uid)")
withAnimation(){declinedRequest = true}
userData.declineFriendRequest(otherUserUID: person.uid)
}){
RoundedRectangle(cornerRadius: 12, style: .continuous)
.fill(Color("Dark-Grey"))
.frame(width: 55, height: 26.5)
.padding(.trailing, 13)
.overlay(
Image("x")
.renderingMode(.template)
.resizable()
.foregroundColor(Color.white)
.opacity(0.8)
.frame(width: 12, height: 12)
.padding(.trailing, 12)
)
}
Button(action: {
print("accepted friend request for \(person.uid)")
withAnimation(){acceptedRequest = true}
userData.acceptFriendRequest(otherUserUID: person.uid)
}){
RoundedRectangle(cornerRadius: 15, style: .continuous)
.fill(Color.white)
.frame(width: 50, height: 25)
.padding(.trailing, 35)
.overlay(
Image("check")
.resizable()
.aspectRatio(contentMode: .fill)
.frame(width: 20, height: 20)
.padding(.trailing, 33)
)
}
}
}.padding(.top, 12.5)
}
}
Spacer()
.contentShape(Rectangle())
}
.frame(width: UIScreen.main.bounds.width, height: UIScreen.main.bounds.height / 1.15)
.padding(.top)
.background(Color("gray")
.clipShape(CustomCorner(corners: [.topLeft,.topRight])))
.offset(y: offset)
// bottom sheet remove swipe gesture....
.gesture(DragGesture().onChanged(onChanged(value:)).onEnded(onEnded(value:)))
.offset(y: showSheet ? 0 : UIScreen.main.bounds.height)
}
.ignoresSafeArea()
.background(
Color.black.opacity(showSheet ? 0.3 : 0).ignoresSafeArea()
.onTapGesture(perform: {
withAnimation{showSheet.toggle()}
})
)
}
}
When declinedRequest or acceptedRequest get modified, its not mapped to the button so everything in the for view gets changed instead of the individual button. I've added my basic code below but some things I tried are making a published variable in my user data model class but it doesn't get updated in the for each as it moves on by then and I also tried making a published variable here. It looks like I have something to do with indices and mapping but I'm unsure what the best approach is. Thanks for your help
Edit:
Here's relevant sections from my user view model:
import SwiftUI
import Firebase
import Combine
import Foundation
struct pendingFriendUser: Identifiable {
var id: Int
var uid: String
var name: String
var username: String
var pic: String
var isFriends: Int
init(id: Int, uid: String, name: String, username: String, pic: String, isFriends: Int){
self.id = id
self.uid = uid
self.name = name
self.username = username
self.pic = pic
self.isFriends = isFriends
}
}
class UserViewModel : ObservableObject{
@Published var userInfo = UserModel(username: "", pic: "", name: "", age: 1, uid: "", phoneNumber: "")
let ref = Firestore.firestore()
let uid = Auth.auth().currentUser!.uid
@Published var pendingFriendUsers: [pendingFriendUser]
//show add friends sheet
@AppStorage("showSheet") var showSheet = false
//check friendship variabe
@Published var isFriend = 0
init() {
self.searchedUsers = []
self.pendingFriendUsers = []
fetchUser(uid: uid) { (user) in
self.userInfo = user
}
}
func getPendingRequests(){
//check if friends has any false memberships
var pendingFriendRequests: [String: Bool] = [:]
self.ref.collection("Users").document(uid).getDocument(){
(document, err) in
if let err = err {
print("Error getting documents \(err)")
} else {
if document!.data()!["friends"] != nil {
pendingFriendRequests = document!.data()!["friends"] as! [String : Bool]
}
//filter based on false pending friend requests
self.pendingFriendUsers.removeAll()
var friendUserID = 0
for key in pendingFriendRequests.keys {
if pendingFriendRequests[key] == false {
self.ref.collection("Users").document(key).getDocument(){
(friendDocument, err) in
if let err = err {
print("Error getting documents \(err)")
} else {
let pendingFriendUsername = (friendDocument?.data()?["username"]) as! String
let pendingFriendUID = (friendDocument?.data()?["uid"]) as! String
let pendingFriendName = (friendDocument?.data()?["name"]) as! String
let pendingFriendPic = (friendDocument?.data()?["imageurl"]) as! String
self.pendingFriendUsers.append(pendingFriendUser(id: friendUserID, uid: pendingFriendUID , name: pendingFriendName , username: pendingFriendUsername, pic: pendingFriendPic, isFriends: 0))
friendUserID += 1
}
}
}
}
}
}
}
...
func acceptFriendRequest(otherUserUID: String){
for var pendingFriend in pendingFriendUsers {
if pendingFriend.uid == otherUserUID {
pendingFriend.isFriends = 1
}
}
self.ref.collection("Users").document(self.uid).setData(
[ "friends": [
otherUserUID: true
] ]
, merge: true)
self.ref.collection("Users").document(otherUserUID).setData(
[ "friends": [
self.uid: true
] ]
, merge: true)
}
func declineFriendRequest(otherUserUID: String){
for var pendingFriend in pendingFriendUsers {
if pendingFriend.uid == otherUserUID {
pendingFriend.isFriends = 2
}
}
self.ref.collection("Users").document(self.uid).updateData([
"friends.\(otherUserUID)": FieldValue.delete(),
]) { err in
if let err = err {
print("Error updating document: \(err)")
} else {
print("Document successfully updated")
}
}
}
func checkFriendRequest(otherUserUID: String){
//0 if not found in friend list or friend request
//1 means theyre friends
//2 means that user sent self/me a friend request
var pendingFriendRequests: [String: Bool] = [:]
self.ref.collection("Users").document(self.uid).getDocument(){
(document, err) in
if let err = err {
print("Error getting documents \(err)")
} else {
pendingFriendRequests = document!.data()!["friends"] as! [String : Bool]
for key in pendingFriendRequests.keys {
if key == otherUserUID{
if pendingFriendRequests[key] == true {
self.isFriend = 1
}else if pendingFriendRequests[key] == false {
self.isFriend = 2
}
}
}
}
}
}
}
Edited with isFriends variable
A: Right now, you have this:
for var pendingFriend in pendingFriendUsers {
if pendingFriend.uid == otherUserUID {
pendingFriend.isFriends = 1
}
}
This doesn't actually modify anything in pendingFriendUsers, because pendingFriendUser is a struct, which gets passed by value in Swift, not by reference. So var pendingFriend is a copy of the version in pendingFriendUsers.
Instead, you could do something like this:
self.pendingFriendUsers = self.pendingFriendUsers.map { pendingFriend in
guard pendingFriend.uid == otherUserUID else { return pendingFriend } //return the original if the UID doesn't match
var modified = pendingFriend //make a mutable copy
modified.isFriends = 1
return modified //return the modified version
}
or:
guard let index = self.pendingFriendUsers.firstIndex(where: { $0.uid == otherUserUID }) else { return } //get the index of the matching UID
self.pendingFriendUsers[index].isFriends = 1 //modify the item at that index
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,199
|
package com.thoughtworks.go.config.update;
import com.thoughtworks.go.config.BasicCruiseConfig;
import com.thoughtworks.go.config.MailHost;
import org.junit.jupiter.api.Test;
import static org.assertj.core.api.Java6Assertions.assertThat;
class DeleteMailHostConfigCommandTest {
@Test
void shouldDeleteExistingBackupConfig() {
BasicCruiseConfig cruiseConfig = new BasicCruiseConfig();
MailHost mailho = new MailHost();
cruiseConfig.server().setMailHost(mailho);
DeleteMailHostCommand command = new DeleteMailHostCommand();
command.update(cruiseConfig);
assertThat(command.getPreprocessedEntityConfig()).isSameAs(mailho);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,353
|
Miss Pye, having grown up locally is looking forward to coming back to her roots. She completed her BA Hons degree in Costume with Performance Design at The Arts University Bournemouth, then attended the Bournemouth Poole and Dorset Secondary training partnership gaining her PGCE in secondary education specialising in Design Technology. Miss Pye was a textiles artist working on the costumes for the London 2012 Olympic Opening Ceremony and has had the good fortune to work on several film sets. From her family's catering background she developed a love of patisserie and baking and has spent many summers putting this into practice. In her spare time, she enjoys trampolining, sewing 1950s attire and working on charity events. With a dedicated pastoral approach, Miss Pye believes students should enjoy learning.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,319
|
Q: Use 64-bit library from 32-bit application My .NET application is 32-bit because I use TWAIN inside it. But I need to write 64-bit library or app and call its methods from my program.
I must create a 64-bit library to interact with WMI SMO, and its functions don't work properly if it compiled as 32-bit dll.
How can I resolve this issue? Is it possible to use a 64-bit library from a 32-bit application?
A: You cannot reference a 32-bit library from a 64-bit executable and viceversa. This is a technical limitation.
However, depending on your other constraints, you may decide to employ a service-based architecture to avoid issues like yours and be more agile in the future.
In fact, you could write "micro" services specialized to solve one single purpose (scanning, interacting with WMI, etc...) and then call them from your application.
This way, you are completely free to choose whichever technology/architecture you require to achieve every single task your application requires. Isolating your "features" in services, allows you to have a different evolutionary path per service, as each service is loosely coupled with each others.
This solution requires writing more code than using a library. First you need to create a windows service (msdn) to
wrap your code, then you need to implement the communication channel. Have a look at WCF and WebApi2 to have an idea of what you can achieve.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,237
|
Une étrange exploration dans la forêt africaine en l'an 2110... est une nouvelle de Jean Raspail qui parut pour la première fois dans Le Figaro du sous le titre d'Un Territoire inconnu.
Résumé
À cette époque-là, la société nouvelle, contrôlée par l'O.N.U., reposait sur l'émulation, la coopération et non plus sur la compétition. L'immense machine tourne toute seule, les hauts fonctionnaires du Secrétariat général s'ennuient et soudain on repère des fumées de campement au centre de l'Afrique... « S'il existe encore des sauvages sur cette Terre, il faut courir à leur secours ! »
L'expédition est composée selon la règle des couleurs, fondement de l'O.N.U. : Blancs,Noirs, Jaunes, Métis et Mulâtres dans la proportion croissante de un à trois dixièmes. Les charges, fonctions et spécialités sont réparties : ethnologues zambéziens, guides jivaros, topographes soudaniens, aviateurs carthaginois, médecins congo-bongolais, cuisiniers français, sociologues pékinois, aumôniers apaches, gardes américains, archéologues quéchuas, coiffeurs italiens, ingénieurs viêt-khmers, mécaniciens wallons, radio-électriciens bengalis, porteurs flamands, etc.
Éditions
1964 - Un Territoire inconnu dans Le Figaro journal quotidien, édition du , à Paris.
1971 - Une étrange exploration dans la forêt africaine en l'an 2081... dans Le Tam-Tam de Jonathan recueil de nouvelles, Éditions Robert Laffont, à Paris.
1977 - Un Territoire inconnu dans Boulevard Raspail recueil de chroniques, Société de production littéraire, à Paris.
2002 - In Le Son des tambours sur la neige, Éditions Robert Laffont à Paris.
Notes et références
Voir aussi
Jean Raspail
Le Son des tambours sur la neige, recueil dans lequel figure cette nouvelle.
Nouvelle française parue en 1964
Œuvre de Jean Raspail
Nouvelle de science-fiction
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,953
|
Q: is the sandbox broken - 500 error is the sandbox for Paypal broken ? I receive a 500 error.
https://www.sandbox.paypal.com/websc&cmd=_express-checkout&token=EC-78048295RW8002111
I wasnt sure if it needs cgi-bin in the url but that does not work either
https://www.sandbox.paypal.com/cgi-bin/websc&cmd=_express-checkout&token=EC-78048295RW8002111
It's from my prestashop plugin - which seems to be having issues in sandbox mode.
The docs seem to suggest that url should work:
https://developer.paypal.com/docs/classic/express-checkout/ht_ec-singleItemPayment-curl-etc/
thanks for any help ;)
A: Have just received this response from the support team:
We are already aware of the issue. Our engineers are working on it for a quick resolution.
I will let you know once the issue will be resolved.
Thank you in advance for you collaboration.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,751
|
'use strict'
/**
* Dependencies
* @ignore
*/
const socketio = require('socket.io')
/**
* Module Dependencies
* @ignore
*/
const MessageDispatcher = require('../../MessageDispatcher')
const UserSocketRegistry = require('./UserSocketRegistry')
const Logger = require('../../../logger')
/**
* Logger
* @ignore
*/
const log = Logger.bootLogger
/**
* Socket Manager
* @class
*/
class SocketManager {
/**
* constructor
*/
constructor (system) {
this.system = system
this.io = socketio(system.server)
this.socket = this.io.sockets
this.io.on('connect', socket => this.onServerConnection(socket))
Object.defineProperty(this, 'registry', { value: new UserSocketRegistry(), enumerable: true })
}
/**
* onServerConnection
*
* @description
* Callback for when a new socket connection is made to the server.
*
* @param {Socket} socket
*/
onServerConnection (socket) {
let { id } = socket
let { handshake: { session: { sub } } } = socket
if (sub) {
this.registry.register(sub, id)
}
log.debug({ user: sub, socket: id, registry: this.registry.registry }, `Socket session`)
socket.on('disconnect', reason => this.onSocketDisconnect(socket, reason))
socket.on('message', data => this.onSocketMessage(socket, data))
socket.on('error', error => this.onSocketError(socket, error))
}
/**
* onSocketDisconnect
*
* @description
* Callback for when a socket connection closes.
*
* @param {Socket} socket
*/
onSocketDisconnect (socket, reason) {
let { id } = socket
this.registry.deregister(id)
log.warn({ id: socket.id, reason, registry: this.registry.registry }, `Socket disconnect`)
}
/**
* onSocketMessage
*
* @description
* Callback for an incoming connection on a socket.
*
* @param {Socket} socket
* @param {Object} data
*/
onSocketMessage (socket, data) {
let { handshake: { session } } = socket
log.debug({ id: socket.id, data }, `Socket new message`)
MessageDispatcher.handle(data, socket, this.system)
.then(() => {
session.touch()
session.save()
log.trace({ id: socket.id }, 'message handling complete')
})
}
/**
* onSocketError
*
* @description
* Callback for when a socket fires an error event
*
* @param {Socket} socket
* @param {Error} error
*/
onSocketError (socket, error) {
log.warn({ id: socket.id, error }, `Socket error`)
socket.disconnect(true)
}
/**
* getSocket
*
* @description
* Retrieve a socket indexed by Socket or User ID
*
* @param {String} id
* @return {Socket}
*/
getSocket (id) {
let sid = this.registry.getSocketId(id) || id
let socket = this.socket.connected[sid]
return socket ? socket : undefined
}
}
/**
* Export
* @ignore
*/
module.exports = SocketManager
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,447
|
You are here: Home > Media Centre Posts > Why is Abiraterone so promising?
The launch of abiraterone acetate is encouraging news for patients with metastatic advanced prostate cancer and their families. Historically there have been few treatment options for advanced prostate cancer when it relapses after hormonal therapy and chemotherapy and so this new treatment has the potential to meet a significant and previously unmet need. Abiraterone is a tablet that has been shown to be well tolerated. In a large clinical trial it was found that men taking abiraterone in combination with a steroid had an average increase in survival of around 5 months compared to those men taking a placebo (a dummy pill) plus a steroid. Some men also achieved significant improvements in pain.
This is the latest in a range of new treatment options now available to men with progressive advanced prostate cancer such as new chemotherapy and immunotherapy techniques. In the last few years there has been significant progress made with continued investment into cancer research which bodes well for greater survival rates and a better quality of life for these men in the future.
NB: Dr Payne was involved in the clinical trials of abiraterone – comparing the drug with dummy pills. The National Institute for Health and Clinical Excellence (NICE) and the Scottish Medicines Consortium (SMC) are currently assessing whether to approve the medication for use on the NHS. A decision from NICE is expected in May 2012.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 259
|
\subsubsection{Anomalous Diagrams}
\label{ssub:anomalous_diagrams}
\begin{figure}[hb]
\centering
\begin{subfigure}{.4\textwidth}
\centering
\includegraphics[angle=270,scale=.5]{figures/2L_diagrams/ggZZ2L_anom1.eps}
\caption{}
\end{subfigure}
\hspace{1cm}
\begin{subfigure}{.4\textwidth}
\centering
\includegraphics[angle=270,scale=.5]{figures/2L_diagrams/ggZZ2L_anom2.eps}
\caption{}
\end{subfigure}
\caption{Two-loop \emph{anomaly style} diagrams for the production of $Z$ boson pairs.}
\label{fig:twotriangles}
\end{figure}
The two-loop $gg\to ZZ$ amplitude contains, in addition, two
topologies which consist of products of one-loop sub-diagrams. On the
one hand diagrams containing gluon self-energy contributions vanish
due to color conservation. The diagrams in
Fig.~\ref{fig:twotriangles}, on the other hand, give a finite mass
dependent contribution as long as both $Z$ bosons couple to distinct
fermion loops. These diagrams are proportional only to the axial
coupling of the $Z$ bosons to fermions; the vector component vanishes
due to $C$ invariance (\emph{Furry's theorem}). The diagrams have been
omitted in the previous section since they can be computed with their
full top mass dependence and, therefore, need no large-mass
expansion~\cite{Kniehl:1989qu,Campbell:2007ev}.
\begin{figure}[t]
\centering \includegraphics[angle=270,width=12cm]{figures/anomaly.eps}
\caption{Triangle diagrams representing the $Zgg$ form factor at lowest order.}
\label{fig:anomaly}
\end{figure}
In brevity we repeat the results from~\cite{Campbell:2007ev} and give the result in terms of our conventions. Let us denote the amplitude for a $Z$ coupling to two gluons by $T^{\mu \nu \rho}_{AB}$. We calculate the triangle shown in Fig.~\ref{fig:anomaly}, where all
momenta are outgoing $q_1+q_2+q_3=0$ and to begin with $q_1^2 \neq 0,
q_2^2 \neq 0$. The result for the two triangle diagrams (including
the minus sign for a fermion loop) is,
\begin{equation} \label{EqT}
T^{\mu \nu \rho}_{AB}(q_1,q_2) = i \frac{g_s^2}{16 \pi^2} \frac{1}{2}
\delta_{AB}
\Big(\frac{g_W}{2 \cos \theta_W} \Big) \tau_f\; \Gamma^{\mu \nu \rho}\,,
\end{equation}
where $\tau_f=\pm 1/2$ and,
\begin{equation}
\Gamma^{\mu \nu \rho}(q_1,q_2,m) = \frac{2}{i \pi^2} \int \; d^d l
\; \mbox{Tr}\Big\{ \gamma^\rho \gamma_5 \frac{1}{\slsh{l}-m}
\gamma^\mu \frac{1}{\slsh{l}+\slsh{q}_1-m}
\gamma^\nu \frac{1}{\slsh{l}+\slsh{q}_1+\slsh{q}_2-m} \Big\}\,.
\end{equation}
The most general form of $\Gamma$ consistent with QCD gauge invariance,
\begin{equation}
q_{1}^{\mu} \Gamma_{\mu \nu \rho}=
q_{2}^{\nu} \Gamma_{\mu \nu \rho} =0 \; ,
\end{equation}
can be written as,
\begin{eqnarray}
\Gamma^{\mu \nu \rho}&=& F_1(q_1,q_2,m) \;
\Big\{ \mbox{Tr}[\gamma^\rho \gamma^\nu \slsh{q_1} \slsh{q_2}\gamma_5 ] q_1^\mu
+\mbox{Tr}[\gamma^\rho \gamma^\mu \gamma^\nu \slsh{q_2}\gamma_5 ] q_1^2 \Big\}\nonumber \\
&+& F_2(q_1,q_2,m) \;
\Big\{\mbox{Tr}[\gamma^\rho\gamma^\mu\slsh{q_1}\slsh{q_2}\gamma_5 ] q_2^\nu
+\mbox{Tr}[\gamma^\rho\gamma^\mu\gamma^\nu\slsh{q_1}\gamma_5 ] q_2^2 \Big\}\nonumber \\
&+& F_3(q_1,q_2,m)\; (q_1^\rho+q_2^\rho)
\Big\{ \mbox{Tr}[ \gamma^\mu\gamma^\nu\slsh{q_1}\slsh{q_2}\gamma_5 ] \Big\}\nonumber \\
&+& F_4(q_1,q_2,m)\; (q_1^\rho-q_2^\rho)
\Big\{ \mbox{Tr}[\gamma^\mu\gamma^\nu\slsh{q_1}\slsh{q_2}\gamma_5 ] \Big\}\,.
\end{eqnarray}
By direct calculation it is found that $F_4=0$.
Contracting with the momentum of the
$Z$ boson we find that, $q_3=-q_1-q_2$
\begin{equation}
(q_3)_\rho \, \Gamma^{\mu \nu \rho}= \Big[ -q_1^2 \, F_1(q_1,q_2,m)
+q_2^2 \,F_2(q_1,q_2,m) -q_3^2 \, F_3(q_1,q_2,m) \Big]
\mbox{Tr}[ \gamma^\mu\gamma^\nu\slsh{q_1}\slsh{q_2}\gamma_5 ]\,.
\end{equation}
The divergence of the axial current is found by direct calculation to be,
\begin{equation}
(q_3)_\rho \, \Gamma^{\mu \nu \rho}=\Big[ 4 m^2 C_0(q_1,q_2;m,m,m) +2 \Big]
\mbox{Tr}[ \gamma^\mu\gamma^\nu\slsh{q_1}\slsh{q_2}\gamma_5 ]
\end{equation}
showing the contribution of the pseudoscalar current proportional to $m^2$ and the anomalous piece. Summation over one complete quark doublet ($\tau_f = \pm 1/2$) cancels the anomaly term and solely the piece proportional to the top mass remains.
For the particular case at hand we are
interested in on-shell $Z$'s and in
$q_2^2=\varepsilon_2 \cdot q_2=0,\, \varepsilon_3 \cdot q_3=0,\, q_3=-q_1-q_2$, so we get a contribution only from $F_1$. The result for $F_1$ is
\begin{eqnarray}
F_1(q_1,q_2,m) &=&\frac{1}{2 q_1 \cdot q_2}
\Bigg[2+4 m^2 C_0(q_1,q_2;m,m,m) \nonumber \\
&+&\Big(2 + \frac{q_1^2}{q_1 \cdot q_2}\Big)
\Big[B_0(q_1+q_2;m,m)-B_0(q_1;m,m)\Big]\Bigg]\,, \\
F_1(q_1,q_2,0) &=&\frac{2}{(q_3^2-q_1^2)}
\Bigg[1+\frac{q_3^2}{(q_3^2-q_1^2)} \log(\frac{q_1^2}{q_3^2})\Bigg]\,.
\end{eqnarray}
We further define a subtracted $F_1$ to take into account the contribution of the top and the bottom quarks,
\begin{equation}
{{\cal F}_1}(q_1,q_2,m)=\Big[ F_1(q_1,q_2,m)-F_1(q_1,q_2,0) \Big] \,.
\end{equation}
Analogous to \Eqno{eq:massiveggZZ_projected_amplitude} we define the projected matrix element for the anomaly piece
\begin{equation}
\label{eq:anomaly_2L_definition}
\ket{{\cal B}^0_\text{anom}(\alpha_S^0,m^0,\mu,\epsilon)} = \frac{\delta^{AB}}{N_A} \left(g^{\mu\nu}\,p_1p_2-p_2^\mu p_1^\nu \right) P_Z^{\alpha\rho'}(p_3)P_{Z,\rho'}^\beta(p_4) \ket{{\cal B}_{\text{anom},\mu\nu\alpha\beta}^{0,AB}(\alpha_S^0,m^0,\mu,\epsilon)}\,.
\end{equation}
The amplitude defined in \Eqno{eq:anomaly_2L_definition} is UV and IR finite and requires no renormalisation. Including the effect of both the $b$ quark (taken to be massless) and the $t$ quark we obtain (No statistical factor for identical $Z$ bosons is included).
\begin{align*}
\label{eq:anomaly_result_full}
\ket{{\cal B}_\text{anom}(\alpha_S^{(n_l)},m,\mu)} &= a_t^2 s^2 \mathcal{N} \cdot \left(\frac{\alpha_S^{(n_l)}}{4\pi}\right)^2 \numberthis \\
&\hspace{-2.5cm}\times \Bigg\{
\Big[(r_Z-x) \big(1+(r_Z-x) (1/r_Z-1/(2 r_Z^2))\big)\Big]
{\cal F}_1(p_1-p_3,-p_1,m) {\cal F}_1(p_3-p_1,-p_{2},m) \\
&\hspace{-2.5cm}+ \Big[(r_Z-1+x)\big(1+(r_Z-1+x) (1/r_Z-1/(2 r_Z^2))\big)\Big]
{\cal F}_1(p_1-p_4,-p_1,m) {\cal F}_1(p_4-p_1,-p_2,m)) \Bigg\}\,,
\end{align*}
where ${\cal N}$ is given in \Eqno{eq:Ndef}. Again we include the factors
$s^2$ to indicate the correct dimensionality of
${\cal F}_1(p_1,p_2,m)$. For completeness we also give the mass expansion
of \Eqno{eq:anomaly_result_full} in case only the top quark
contribution is considered, i.e. $ F_1(p_1,p_2,0)\to 0$. As expected
the expansion starts at $1/m^4$.
\begin{align*}
\label{eq:anomaly_result_top_exp}
\ket{{\cal B}_\text{anom,t}(\alpha_S^{(n_l)},m,\mu)} &= a_t^2 \mathcal{N} \cdot \left(\frac{\alpha_S^{(n_l)}}{4\pi}\right)^2 \;\Bigg\{
%
\frac{1}{r_t^2} \left[-\frac{1}{9}+\frac{r_Z}{9}+\frac{1-\tilde{x}}{18 r_Z}+\frac{-1+2 \tilde{x}}{72 r_Z^2}\right]
+ \frac{1}{r_t^3} \left[ -\frac{8 r_Z}{135}+\frac{2 r_Z^2}{45} \right.\\
&\hspace{-2cm}+ \left. \frac{(13-18 \tilde{x})}{270} + \frac{1-3 \tilde{x}}{270 r_Z^2}+\frac{-11+26 \tilde{x}}{540 r_Z} \right]
+ \frac{1}{r_t^4} \left[ -\frac{22 r_Z^2}{945}+\frac{22 r_Z^3}{1575}+\frac{r_Z (1511-2362 \tilde{x})}{56700} \right.\\
&\hspace{-2cm}+ \left. \frac{-3845+9892 \tilde{x}}{226800} - \frac{191 \left(1-4 \tilde{x}+2 \tilde{x}^2\right)}{226800 r_Z^2}+\frac{646-2129 \tilde{x}+382 \tilde{x}^2}{113400 r_Z} \right]
+ \frac{1}{r_t^5} \left[ -\frac{38 r_Z^3}{4725}+\frac{19 r_Z^4}{4725} \right. \\
&\hspace{-2cm}+ \frac{r_Z^2 (113-188 \tilde{x})}{9450} + \frac{r_Z (-783+2104 \tilde{x})}{75600}+\frac{-111+472 \tilde{x}-306 \tilde{x}^2}{75600 r_Z} + \frac{1-5 \tilde{x}+5 \tilde{x}^2}{5400 r_Z^2} \\
&\hspace{-2cm}+ \left. \frac{197-688 \tilde{x}+194 \tilde{x}^2}{37800} \right]
+ \frac{1}{r_t^6} \left[-\frac{1613 r_Z^4}{623700}+\frac{1613 r_Z^5}{1455300}+\frac{r_Z^3 (41432-71573 \tilde{x})}{8731800} \right. \numberthis \\
&\hspace{-2cm}+\frac{r_Z^2 (-457682+1261401 \tilde{x})}{87318000} + \frac{-1049213+4652126 \tilde{x}-3464248 \tilde{x}^2}{698544000} \\
&\hspace{-2cm}+ \frac{r_Z \left(622783-2250826 \tilde{x}+764954 \tilde{x}^2\right)}{174636000} + \frac{42658-222727 \tilde{x}+251038 \tilde{x}^2-18874 \tilde{x}^3}{116424000 r_Z} \\
&\hspace{-2cm}+ \left. \frac{9437 \left(-1+6 \tilde{x}-9 \tilde{x}^2+2 \tilde{x}^3\right)}{232848000 r_Z^2}\right] + {\cal O}\left(1/r_t^7,\epsilon\right)\Bigg\} \,.
\end{align*}
\section{Definition of Scalar Integrals}
\label{Intdef}
We work in the Bjorken-Drell metric so that
$l^2=l_0^2-l_1^2-l_2^2-l_3^2$.
The definition of the integrals is as follows
\begin{eqnarray}
&& A_0(m) =
\frac{\mu^{4-d}}{i \pi^{\frac{d}{2}}r_\Gamma}
\int d^d l \;
\frac{1}
{(l^2-m^2+i\varepsilon)}\,, \\
&& B_0(p_1;m,m) =
\frac{\mu^{4-d}}{i \pi^{\frac{d}{2}}r_\Gamma}
\int d^d l \;
\frac{1}
{(l^2-m^2+i\varepsilon)
((l+p_1)^2-m^2+i\varepsilon)}\,, \\
&& C_0(p_1,p_2;m,m,m) =
\frac{\mu^{4-d}}{i \pi^{\frac{d}{2}}r_\Gamma} \\
&& \times \int d^d l \;
\frac{1}
{(l^2-m^2+i\varepsilon)
((l+p_1)^2-m^2+i\varepsilon)
((l+p_1+p_2)^2-m^2+i\varepsilon)}\,,\nonumber \\
&&
D_0(p_1,p_2,p_3;m,m,m,m) =
\frac{\mu^{4-d}}{i \pi^{\frac{d}{2}}r_\Gamma} \\
\nonumber \\
&&
\times \int d^d l \;
\frac{1}
{(l^2-m^2+i\varepsilon)
((l+p_1)^2-m^2+i\varepsilon)
((l+p_1+p_2)^2-m^2+i\varepsilon)
((l+p_1+p_2+p_3)^2-m^2+i\varepsilon)}\,, \nonumber
\end{eqnarray}
We have removed the overall constant which occurs in $d$-dimensional integrals, ($d=4-2\epsilon$)
\begin{equation}
r_\Gamma\equiv\Gamma(1+\epsilon)=
1-\epsilon \gamma+\epsilon^2\Big[\frac{\gamma^2}{2}+\frac{\pi^2}{12}\Big]
\end{equation}
with the Euler-Mascheroni constant $\gamma=0.57721\ldots$. The large mass expansion of some of these integrals are
\input{src/LMEs/B0s.tex}
and
\input{src/LMEs/C0s.tex}
for $p_1^2=p_2^2=0,(p_1+p_2)^2=s$.
\section{Scale Dependence of the Finite Remainder}
\label{sec:scale_dependence_of_the_finite_remainder}
In this section we shortly summarise a convenient, and well-known, way to determine the dependence on the renormalisation scale $\mu=\mu_r$ of the one- and two-loop finite remainders used within this work, i.e. processes with a loop-induced leading-order matrix element. This determination is possible by exploiting the \emph{renormalisation group equation} (RGE) properties of the individual building block, e.g. $\alpha_S^{(n_f)}(\mu)$, as discussed below. Knowledge of this scale dependence, in return, offers a simple way to compute finite remainder results at arbitrary scales, provided the results at a starting scale $\mu_0$ are known. We mostly recycle our definitions from Sec.~\ref{sub:ggH_preliminaries}. In the following, however, we stick to a slightly more general notation when applicable. To this end we drop the amplitude specifications ${\cal A}$ and ${\cal B}$ from the finite remainder definition in Eq.~\eqref{eq:FinRem_def} and denote our previous amplitudes ${\cal A}$ and ${\cal B}$ simply by ${\cal M}$. We also replace our, to the $gg\to ZZ$ process specialised, IR constant $\Zop_{gg}^{(n_l)}$ from Eq.~\eqref{eq:Zop_as} by a more general IR constant $\Zop_{IR}$ following the notation in~\cite{Ferroglia:2009ii,Baernreuther:2013caa,Czakon:2014oma}. The finite remainder for $n_f$ quark flavours is thus defined by
\begin{align*}
\label{eq:scalecheck_FinRem_def}
\ket{{\cal F}(\alpha_S^{(n_f)},m,\mu)} &= \frac{1}{\Zop_{IR}} \,\ket{{\cal M}^r(\alpha_S^{(n_f)},m,\mu)} = \frac{Z_{UV}^{(n_f)}}{\Zop_{IR}} \left(\frac{N^\epsilon Z^{(n_f)}_{\alpha_S} \alpha_S^{(n_f)}(\mu)}{4\pi}\right) \Bigg[ \ket{{\cal M}^{(1),0}(m)} \\
&+ \left(\frac{N^\epsilon Z^{(n_f)}_{\alpha_S} \alpha_S^{(n_f)}(\mu)}{4\pi}\right) \ket{{\cal M}^{(2),0}(m)} \Bigg] + {\cal O}\left((\alpha_S^{(n_f)})^3\right) \,. \numberthis
\end{align*}
The mass dependence does not play any important role in the subsequent discussion and, hence, all results are valid for arbitrary masses $m$. $Z_{UV}^{(n_f)}$ denotes the process dependent $UV$ renormalisation constants and the mass renormalisation $m^0=Z_m m$ is again kept implicit. The strong coupling constants $\alpha_S$ is renormalised according to
\begin{equation}
\alpha_S^0 = N^\epsilon Z^{(n_f)}_{\alpha_S} \alpha_S^{(n_f)}(\mu) \quad \text{with} \quad N^\epsilon = \mu^{2\epsilon} \frac{e^{\epsilon\,\gamma_E}}{(4\pi)^\epsilon}\,,
\end{equation}
where the explicit $\mu$ dependence from the loop measure in Eq.~\eqref{eq:loopnorm_ggH} was shifted to $N^\epsilon$. The renormalisation constant $Z^{(n_f)}_{\alpha_S}$ and the coefficient of the beta function $\beta_0^{(n_f)}$ are given in Eq.~\eqref{eq:Zas_as}. The explicit scale and flavour dependence of $\alpha_S=\alpha_S^{(n_f)}(\mu)$ is neglected in the following for simplicity.
Equivalently to Eq.~\eqref{eq:scalecheck_FinRem_def} we define the perturbative expansion of the finite remainder as
\begin{equation}
\label{eq:scalecheck_FinRem_expansion}
\ket{{\cal F}(\alpha_S,m,\mu)} = \frac{\alpha_S}{4\pi} \ket{{\cal F}^{(1)}(m,\mu)} + \left(\frac{\alpha_S}{4\pi}\right)^2 \ket{{\cal F}^{(2)}(m,\mu)} + {\cal O}\left(\alpha_S^3\right) \,.
\end{equation}
Taking the derivative with respect to $\mu^2$ of Eq.~\eqref{eq:scalecheck_FinRem_def} and Eq.~\eqref{eq:scalecheck_FinRem_expansion} leads to
\begin{align*}
\label{eq:scalecheck_FinRem_der}
\mu^2 \frac{d\,}{d\mu^2} \ket{{\cal F}(\alpha_S,m,\mu)} &= \left(\mu^2 \frac{d}{d\mu^2}\left(\frac{\alpha_S}{4\pi}\right)\right) \ket{{\cal F}^{(1)}(m,\mu)} + \frac{\alpha_S}{4\pi} \;\mu^2 \frac{d\,}{d\mu^2} \ket{{\cal F}^{(1)}(m,\mu)} \\
&+2 \left(\frac{\alpha_S}{4\pi}\right) \left(\mu^2 \frac{d\,}{d\mu^2} \left(\frac{\alpha_S}{4\pi}\right)\right) \ket{{\cal F}^{(2)}(m,\mu)} + \left(\frac{\alpha_S}{4\pi}\right)^2\; \mu^2 \frac{d\,}{d\mu^2} \ket{{\cal F}^{(2)}(m,\mu)} \numberthis\\
&= \mu^2 \frac{d\,}{d\mu^2} \Bigg\{ \frac{Z_{UV}^{(n_f)}}{\Zop_{IR}} \left(\frac{N^\epsilon Z^{(n_f)}_{\alpha_S} \alpha_S^{(n_f)}(\mu)}{4\pi}\right) \Bigg[ \ket{{\cal M}^{(1),0}(m)} \\
&+ \left(\frac{N^\epsilon Z^{(n_f)}_{\alpha_S} \alpha_S^{(n_f)}(\mu)}{4\pi}\right) \ket{{\cal M}^{(2),0}(m)} \Bigg] \Bigg\} \,.
\end{align*}
The derivatives of $Z_{UV}^{(n_f)}$ and $Z_m$ vanish because these renormalisation constants are defined in the on-shell scheme. The explicit $\mu$ dependence within these expressions cancels against the $\alpha_S$ scale dependence. The derivative of $\Zop_{IR}$ with respect to $\mu$ is given by its RGE~\cite{Ferroglia:2009ii,Baernreuther:2013caa,Czakon:2014oma} and therefore
\begin{equation}
\mu^2 \frac{d\,}{d\mu^2} \frac{1}{\Zop_{IR}} = -\frac{1}{\Zop_{IR}^2} \frac{1}{2}\; \underbrace{\frac{d}{d\log\mu}\Zop_{IR}}_{-\hat\Gamma \cdot \Zop_{IR}} = \frac{\alpha_S}{4\pi} \frac{\hat\Gamma^{(1)}}{2}\cdot \frac{1}{\Zop_{IR}} + {\cal O}\left(\alpha_S^2\right)\,.
\end{equation}
The anomalous dimension operator $\hat\Gamma$ can be taken from~\cite{Czakon:2014oma} and references therein. For our $gg\to ZZ$ processes $\hat{\Gamma}$ simplifies to
\begin{align*}
\hat\Gamma = \frac{\alpha_S}{4\pi}\, \hat\Gamma^{(1)} + {\cal O}\left(\alpha_S^2\right) &= \frac{\alpha_S}{4\pi} \left(-4C_A\log\left(\frac{\mu^2}{-s-i\epsilon}\right) -2\beta_0^{(n_f)} \right) + {\cal O}\left(\alpha_S^2\right) \numberthis\\
&= \frac{\alpha_S}{4\pi} \left( \hat{K}^{(1)} + \hat{D}^{(1)}\cdot \log\left(\frac{\mu^2}{\mu_0^2} \right)\right)+ {\cal O}\left(\alpha_S^2\right)
\end{align*}
with
\begin{equation}
\label{eq:scalecheck_KD_def}
\hat{K}^{(1)} = -4C_A \,\log\left(\frac{\mu_0^2}{-s-i\epsilon}\right) - 2\beta_0^{(n_f)} \quad \text{and} \quad \hat{D}^{(1)} = -4C_A\,.
\end{equation}
The remaining derivatives up to ${\cal O}\left(\alpha_S^2\right)$
\begin{equation}
\mu^2 \frac{d}{d\mu^2} \left(\frac{g_s^2}{4\pi}\right) = \alpha_S \left(-\epsilon- \beta_0^{(n_f)}\,\frac{\alpha_S}{4\pi}\right),\quad \mu^2 \frac{d\,}{d\mu^2}\; N^\epsilon = \epsilon\;N^\epsilon \quad \text{and}\quad \mu^2 \frac{d\,}{d\mu^2}\; Z_{\alpha_S}^{(n_f)} = Z_{\alpha_S}^{(n_f)}\;\beta_0^{(n_f)} \frac{\alpha_S}{4\pi}
\end{equation}
combine to
\begin{equation}
\mu^2 \frac{d\,}{d\mu^2}\; \left(\frac{Z_{UV}^{(n_f)}}{\Zop_{IR}} \;\frac{N^\epsilon Z_{\alpha_S}\alpha_S}{4\pi}\right) = \frac{Z_{UV}^{(n_f)}}{\Zop_{IR}}\;\frac{N^\epsilon Z_{\alpha_S}\alpha_S}{4\pi} \left[ \frac{\alpha_S}{4\pi} \frac{\hat\Gamma^{(1)}}{2} \right]\,.
\end{equation}
Using the shorthand notation $\mu^2 \frac{d\,}{d\mu^2} \ket{{\cal F}} = \frac{d}{d\log\mu^2} \ket{{\cal F}} = \ket{{\cal F}^{'}}$ Equation~\eqref{eq:scalecheck_FinRem_der} becomes
\begin{align*}
\mu^2 \frac{d\,}{d\mu^2}\; \ket{{\cal F}(\alpha_S,m,\mu)} &= \frac{\alpha_S}{4\pi} \left(-\epsilon- \beta_0^{(n_f)}\frac{\alpha_S}{4\pi} \right) \ket{{\cal F}^{(1)}(m,\mu)} + \frac{\alpha_S}{4\pi} \ket{{\cal F}^{(1)'}(m,\mu)} \\
&+ 2 \left(\frac{\alpha_S}{4\pi}\right)^2 \left(-\epsilon- \beta_0^{(n_f)}\frac{\alpha_S}{4\pi} \right)\ket{{\cal F}^{(2)}(m,\mu)} + \left(\frac{\alpha_S}{4\pi}\right)^2 \ket{{\cal F}^{(2)'}(m,\mu)} \numberthis\\
&=\left(\frac{\alpha_S}{4\pi}\right)^2 \frac{\hat\Gamma^{(1)}}{2} \ket{{\cal F}^{(1)}(m,\mu)} + {\cal O}\left(\alpha_S^3\right)\,.
\end{align*}
Comparing each order in $\alpha_S$ yields the system of differential equations
\begin{align}
\label{eq:scalecheck_diffeq_sys}
\Rightarrow& \left(\frac{\alpha_S}{4\pi}\right) \left(\ket{{\cal F}^{(1)'}(m,\mu)}-\epsilon \ket{{\cal F}^{(1)}(m,\mu)}\right) = 0 \\
\Rightarrow& \left(\frac{\alpha_S}{4\pi}\right)^2 \left(\ket{{\cal F}^{(2)'}(m,\mu)}-2\epsilon \ket{{\cal F}^{(2)}(m,\mu)}-\left(\beta_0^{(n_f)}+\frac{\hat\Gamma^{(1)}}{2}\right)\ket{{\cal F}^{(1)}(m,\mu)}\right) = 0\,.
\end{align}
Solving the homogeneous differential equations for the leading- and next-to-leading-order finite remainder results in
\begin{equation}
\label{eq:scalecheck_hom_eq}
\ket{{\cal F}^{(1)}(m,\mu)} = \left(\frac{\mu^2}{\mu_0^2}\right)^\epsilon \ket{{\cal F}^{(1)}(m,\mu_0)} \quad \text{and} \quad \ket{{\cal F}^{(2)}(m,\mu)}_h = \left(\frac{\mu^2}{\mu_0^2}\right)^{2\epsilon} \ket{{\cal F}^{(2)}(m,\mu_0)}\,.
\end{equation}
The inhomogeneous equation for the NLO finite remainder can easily be solved by \emph{variation of constants}. We make an ansatz for the solution of the inhomogeneous equation and write the homogeneous solution as
\begin{equation}
\ket{{\cal F}^{(2)}(m,\mu)} = C(\mu)\,e^{F(\log\mu^2)}\, \quad \text{with} \quad F(\log\mu^2) = \int\limits_{\log\mu_0^2}^{\log\mu^2} \; 2\epsilon \;d\log\mu^2\,.
\end{equation}
Reinsertion into Eq.~\eqref{eq:scalecheck_diffeq_sys} yields the differential equation for $C(\mu)$
\begin{equation}
\label{eq:scalecheck_Cdiff}
C^{'}(\mu) = e^{-F(\log\mu^2)}\cdot \left(\beta_0^{(n_f)} + \frac{\hat\Gamma^{(1)}}{2}\right) \ket{{\cal F}^{(1)}(m,\mu)} \overset{\eqref{eq:scalecheck_hom_eq}}{=} \left(\frac{\mu^2}{\mu_0^2}\right)^{-\epsilon} \left(\beta_0^{(n_f)}+\frac{\hat\Gamma^{(1)}}{2}\right) \ket{{\cal F}^{(1)}(m,\mu_0)}\,.
\end{equation}
Solving Eq.~\eqref{eq:scalecheck_Cdiff} by an elementary integration using the decomposition of $\hat\Gamma^{(1)}$ into $\hat{K}^{(1)}$ and $\hat{D}^{(1)}$ from Eq.~\eqref{eq:scalecheck_KD_def} and combining the particular solution with the homogeneous solution from Eq.~\eqref{eq:scalecheck_hom_eq} yields for the scale dependence of the one- and two-loop finite remainders
\begin{align*}
\ket{{\cal F}^{(1)}(m,\mu)} &\overset{\epsilon\to 0}{=} \ket{{\cal F}^{(1)}(m,\mu_0)} \qquad \text{and} \numberthis\\
\ket{{\cal F}^{(2)}(m,\mu)} &\overset{\epsilon\to 0}{=} \ket{{\cal F}^{(2)}(m,\mu_0)} + \left[\log\left(\frac{\mu^2}{\mu_0^2}\right) \left(\beta_0^{(n_f)} +\frac{\hat{K}^{(1)}}{2} \right) + \frac{\hat{D}^{(1)}}{4}\log^2\left(\frac{\mu^2}{\mu_0^2}\right)\right] \ket{{\cal F}^{(1)}(m,\mu_0)} \numberthis \\
&= \ket{{\cal F}^{(2)}(m,\mu_0)} -2 C_A \log\left(\frac{\mu^2}{\mu_0^2}\right) \left[ \log\left(\frac{\mu_0^2}{-s-i\epsilon}\right) + \frac{1}{2} \log\left(\frac{\mu^2}{\mu_0^2}\right) \right] \ket{{\cal F}^{1}(m,\mu_0)}\,.
\end{align*}
\section{Conclusions}
\label{sec:conclusions}
In this paper we have presented a calculation of on-shell $Z$-boson pair
production via gluon fusion at the two-loop level. This occurs both
through diagrams that are mediated by a Higgs boson, with $H \to ZZ$,
and by continuum contributions in which the $Z$ bosons couple through loops
of quarks. We have considered contributions up to the two-loop level, corresponding
to NLO corrections, for the Higgs diagrams alone and also for the interference
between the two sets of diagrams.
In the continuum contribution the two-loop corrections containing loops of
massless quarks are known and we have reproduced results from the
literature. Our treatment of the massive quark loops is based on a large-mass
expansion up to order $1/m^{12}$, that is extended to the high-mass
region by using a combination of conformal mapping and \emph{Pad\'e approximation}.
This procedure was shown to provide an excellent approximation of the
Higgs contribution alone, where the exact result is known.
Additionally, applying the large-mass expansion in combination
with the conformal mapping and the \emph{Pad\'e approximation} to the
$gg\to ZZ$ amplitudes is obviously not limited to the interference calculation
alone. The same procedure can also be applied to the virtual two-loop
$gg\to ZZ$ amplitude including its full tensor structure. It might be
desirable to apply the presented procedure also to the Higgs-boson
pair-production process, because the latter offers identical kinematics.
Comparing those results with the recently published results including the full
top mass effects~\cite{Borowka:2016ehy} could lead to interesting insights
concerning the error estimate of the used approximation. However, this is kept
as future work.
We have used our calculation to provide theoretical predictions for the impact
of the interference contribution on the invariant mass distribution of
$Z$-boson pairs at the $13$~TeV LHC. In the high-mass region we have shown
that the impact of the NLO corrections to the interference are practically
identical to those for Higgs production alone. This explicit calculation
justifies using a procedure for estimating the number of off-shell events
due to the interference by rescaling the LO prediction by the on-shell
$K$-factor.
\section{Higgs Production in Gluon Fusion and Decay to $ZZ$}
\label{sec:higgs_production_via_gluon_fusion}
In this section we give a detailed discussion of single Higgs boson
production at LO and NLO QCD and its subsequent decay to a pair of
on-shell $Z$ bosons. As mentioned earlier the LO and NLO amplitudes
for single Higgs boson production have been known for a long time;
either approximate results in terms of Taylor expansions in the
inverse of the top quark mass
$s/m^2$~\cite{Dawson:1990zj,Dawson:1993qf,Harlander:2002wh,Anastasiou:2002yz,Aglietti:2006tp,Harlander:2009bw,Pak:2009bx}
or results keeping the exact top mass
dependence~\cite{Djouadi:1991tka,Aglietti:2006tp}.
It is understood that, whenever feasible and available, the exact
results for LO and NLO amplitudes are used. However, we are mainly
interested in approximations to the interference contributions
${\rm Re}\bra{{\cal A}_\text{LO}}\ket{{\cal B}_\text{(N)LO}}$, where ${\cal A}$ denotes
the Higgs-mediated and ${\cal B}$ the SM continuum amplitude. Since no
exact results are available for ${\cal B}_\text{NLO}$ we will use the,
so-called, large-mass expansion~\cite{Smirnov:2002pj} as an
approximation of the SM continuum. Hence, for consistency, we also
perform the expansion of the Higgs-mediated amplitude ${\cal A}$ to high
powers in $s/m^2$. Expansion of the two-loop Higgs-mediated amplitude
${\cal A}_\text{NLO}$ and its comparison to available results from the
literature provides moreover a helpful check of our expansion routines
due to the general structure of the LME.
Furthermore, the large-mass expansion in powers of $s/m^2$ is formally only
valid below the threshold of top quark pair-production, as $m$ is
assumed to be much larger than any other scale in the problem,
e.g. $s\ll m^2$. As extensively discussed in literature the naive LME
can be drastically improved at (and even far above) threshold by
taking the next mass threshold into account, see Ref~\cite{Smirnov:2002pj} and
references within, or by rescaling
the approximated NLO result by the exact LO result, see e.g.
Refs~\cite{Pak:2009dg,Grigo:2013rya}. We will address
this issue in Sec.~\ref{ssub:ggH_comparison_lme_with_full_result} and try to
draw conclusions for the SM continuum.
\subsection{Preliminaries}
\label{sub:ggH_preliminaries}
The amplitudes for single Higgs boson production
\begin{equation}
g(p_1,\alpha,A) + g(p_2,\beta,B) \to H(p_1+p_2),\;\;s= (p_1+p_2)^2 \, ,
\end{equation}
are illustrated in Fig.~\ref{fig:ggH_NLO_amps} for the one-loop and two-loop case.
\begin{figure}[t]
\centering
\begin{subfigure}{.4\textwidth}
\includegraphics[angle=270,scale=.6]{figures/1L_diagrams/ggH1L_1.eps}
\caption{}
\label{sfig:ggH_LO_amp}
\end{subfigure}
\hspace{.5cm}
\begin{subfigure}{.4\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggH2L_1.eps}
\caption{}
\end{subfigure}
\vspace{.5cm}
\begin{subfigure}{.4\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggH2L_2.eps}
\caption{}
\end{subfigure}
\hspace{.5cm}
\begin{subfigure}{.4\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggH2L_3.eps}
\caption{}
\end{subfigure}
\caption{Representative diagrams for the LO+NLO virtual $gg\to H\to ZZ$ amplitude.}
\label{fig:ggH_NLO_amps}
\end{figure}
The largest contribution is due to the internal
massive top quark loop; in the following we will ignore the
contribution of other quarks for the Higgs production process.
The $gg\to H$ amplitude, with color (Lorentz) indices $A,B (\alpha,\beta)$ for the initial state gluons,
can be written as
\begin{equation}
\label{eq:ggH_amp_decomposition}
\ket{{\cal A}^{0,AB}_{\alpha\beta}(\alpha_S^0,m^0,\mu,\epsilon)} =- i
\delta^{AB}\, \frac{g_W}{2 m_W} \frac{4}{3}\;\left(g_{\alpha\beta}\, p_1 \cdot p_2 - p_{1,\beta}p_{2,\alpha} \right)\; \ket{{\cal A}^0(\alpha_S^0,m^0,\mu,\epsilon)}\,,
\end{equation}
such that the reduced matrix element
$\ket{{\cal A}^0(\alpha_S^0,m^0,\mu,\epsilon)}$ is dimensionless
and can be expressed as a function of $\mu^2/s$ and $r_t=m^2/s$.
The bare on-shell amplitudes admit the perturbative expansion
\begin{equation}
\label{eq:ggH_amp_bare_expansion}
\ket{{\cal A}^0(\alpha_S^0,m^0,\mu,\epsilon)} = \frac{\alpha_S^0}{4\pi} \ket{{\cal A}^{0,(1)}(m^0,\mu,\epsilon)} + \left(\frac{\alpha_S^0}{4\pi}\right)^2 \ket{{\cal A}^{0,(2)}(m^0,\mu,\epsilon)} + {\cal O}{\left((\alpha_S^0)^3\right)}\,,
\end{equation}
where we introduced the parameter $\epsilon$ from dimensional regularisation
in $d=4-2\epsilon$ space-time dimensions
and $\mu$ to keep the amplitudes dimensionless.
The calculation is performed in Conventional Dimensional Regularisation (CDR) and the following definition of the $d$-dimensional loop integral
measure
\begin{equation}
\label{eq:loopnorm_ggH}
\int{\frac{d^4p}{(2\pi)^4}} \longrightarrow \mu^{2\epsilon} \underbrace{\frac{e^{\epsilon \gamma_E}}{(4\pi)^\epsilon}}_{\equiv S_\epsilon} \cdot \int{\frac{d^dp}{(2\pi)^d}}
\end{equation}
is used in accordance with the $\overline{MS}$-scheme, to avoid the proliferation of
unnecessary $\gamma_E-\log(4\pi)$ terms.
The ultraviolet (UV) renormalised amplitudes are given by
\begin{equation}
\label{eq:ggH_renormalisation_definition}
\ket{{\cal A}^r(\alpha_S^{(n_f)}(\mu),m,\mu,\epsilon)} = Z_{m} Z_g \ket{{\cal A}^0(\alpha_S^0,m^0,\mu,\epsilon)}\,,
\end{equation}
where $Z_g$ denotes the on-shell gluon renormalisation
constant. The $Ht\bar{t}$ vertex is renormalised, according
to~\cite{Braaten:1980yq}, by $g_H^0 = Z_{m} \,g_H$ with $g_H$ being
the Yukawa coupling for the top quark. The bare top quark mass is related to the renormalised mass, $m$, by $m^0 = Z_m m$. The necessary on-shell
renormalisation constants are given by
\begin{align}
Z_g &= 1- \frac{\alpha_S^{(n_f)}}{4\pi} T_F \left(\frac{\mu^2}{m^2}\right)^\epsilon \cdot \frac{4}{3\epsilon} + {\cal O}\left((\alpha_S^{(n_f)})^2,\epsilon \right) \quad \text{and} \\
Z_m &= 1- \frac{\alpha_S^{(n_f)}}{4\pi} C_F \left(\frac{\mu^2}{m^2}\right)^\epsilon \left[ \frac{3}{\epsilon} +4 \right] + {\cal O}\left((\alpha_S^{(n_f)})^2,\epsilon \right)\,,
\end{align}
with $T_F=1/2$. See appendix A of~\cite{Baernreuther:2013caa} and
references therein for more information. The mass renormalisation
enters as an overall factor in \Eqno{eq:ggH_renormalisation_definition} because
of the renormalisation of the Yukawa coupling, and also implicitly in
the relationship between the bare and renormalised mass.
We will always present mass-renormalised results
in the following.
The strong coupling constant is renormalised in the $\overline{MS}$-scheme according to
\begin{equation}
\alpha_S^0 = Z_{\alpha_S}^{(n_f)} \alpha_S^{(n_f)}(\mu)\,,
\end{equation}
with~\cite{Baernreuther:2013caa}
\begin{equation}
\label{eq:Zas_as}
Z_{\alpha_S}^{(n_f)} = 1- \frac{\alpha_S^{(n_f)}}{4\pi} \frac{\beta_0^{(n_f)}}{\epsilon} + {\cal O}\left((\alpha_S^{(n_f)})^2\right) \quad \text{and}\quad \beta_0^{(n_f)} = \frac{11}{3}C_A-\frac{4}{3}T_F n_f\,,
\end{equation}
where $n_f=6$ denotes the number of fermions and $\beta_0^{(n_f)}$ the coefficient of the beta function. The explicit scale dependence of the renormalised strong coupling constant $\alpha_S^{(n_f)}(\mu)$ is dropped in the following to simplify our notation.
All of our quantities are computed in five-flavour $(n_l=5)$ QCD. Hence, we decouple the top quark from the QCD running via
\begin{equation}
\alpha_S^{(n_f)} = \xi_{\alpha_S} \alpha_S^{(n_l)} \quad \text{and}\quad \xi_{\alpha_S} = 1+ \frac{\alpha_S^{(n_l)}}{4\pi} T_F \left[ \frac{4}{3}\log\left(\frac{\mu^2}{m^2}\right)\right] + {\cal O}\left((\alpha_S^{(n_l)})^2,\epsilon \right) \,,
\end{equation}
with $n_l$ the number of light quarks.
After UV renormalisation the two-loop amplitude still contains divergences of infrared origin. The structure of these divergences is, however, completely understood at two-loop level. The finite remainder is defined by infrared (IR) renormalisation
\begin{equation}
\label{eq:FinRem_def}
\ket{{\cal F}_{{\cal A},{\cal B}}\left( \alpha_S^{(n_l)},m,\mu \right)} = \left(\Zop_{gg}^{(n_l)}\right)^{-1} \ket{{\cal M}_{{\cal A},{\cal B}}^r \left( \alpha_S^{(n_l)},m,\mu,\epsilon \right)} \,.
\end{equation}
Expanding \Eqno{eq:FinRem_def} in $\alpha_S^{(n_l)}/(4\pi)$ yields the explicit expressions for the LO and NLO finite remainders
\begin{align}
\label{eq:FinRem_1_def}
\ket{{\cal F}_{{\cal A},{\cal B}}^{(1)}(m,\mu)} &= \ket{{\cal M}_{{\cal A},{\cal B}}^{r,(1)} \left( m,\mu\right)} \quad \text{and} \\
\ket{{\cal F}_{{\cal A},{\cal B}}^{(2)}(m,\mu)} &= \ket{{\cal M}_{{\cal A},{\cal B}}^{r,(2)}(m,\mu,\epsilon)} - \Zop_{gg}^{(n_l,1)}\, \ket{{\cal M}_{{\cal A},{\cal B}}^{r,(1)}(m,\mu,\epsilon)} \, .
\label{eq:FinRem_2_def}
\end{align}
The infrared renormalisation matrix $\Zop_{gg}^{(n_l)}$ is taken from~\cite{Ferroglia:2009ii,Baernreuther:2013caa,Czakon:2014oma} and reads for the gluon-gluon initial state with colourless final state in terms of the renormalised strong coupling constant
\begin{equation}
\label{eq:Zop_as}
\Zop_{gg}^{(n_l)} = 1+\frac{\alpha_S^{(n_l)}}{4\pi} \, \Zop_{gg}^{(n_l,1)} = 1+\frac{\alpha_S^{(n_l)}}{4\pi}\left(\frac{-2C_A}{\epsilon^2}-\frac{2C_A \log(-\mu^2/s)+\beta_0^{(n_l)}}{\epsilon}\right) +{\cal O}\left( (\alpha_S^{(n_l)})^2\right) \,.
\end{equation}
In the end we are interested in the amplitude for the process
\begin{equation}
g(p_1)+g(p_2) \to H \to Z(p_3)+Z(p_4) \, ,
\end{equation}
and we set up momentum conservation as $p_1+p_2 = p_3+p_4$. For the calculation at hand we also need the decay amplitude $H \to ZZ$, see Fig.~\ref{sfig:ggH_LO_amp}, which is given by
\begin{eqnarray}
\ket{{\cal M}^{\rho\sigma}}_{H\to ZZ} &=&i g_W \frac{m_W}{\cos^2 \theta_W} g^{\rho \sigma} \; .
\label{decayamplitude}
\end{eqnarray}
Combining Eqs.~(\ref{eq:ggH_amp_decomposition},\ref{decayamplitude}) the full amplitude for production and decay is
\begin{equation}
\label{eq:ggHZZ_amplitude_structure}
\ket{{\cal A}^{\alpha\beta\rho\sigma,AB}_\text{ggHZZ}(\alpha_S^{(n_l)},m,\mu,\epsilon)} =
{\cal N}\, \delta^{AB} \, \frac{4}{3}\; \frac{s}{s-m_H^2} \,\ket{{\cal A}(\alpha_S^{(n_l)},m,\mu,\epsilon)} \cdot
\Bigg( g^{\alpha \beta} -\frac{p_2^{\alpha} p_1^{\beta}}{p_1\cdot p_2}\Bigg) g^{\rho \sigma} ,
\end{equation}
where we have defined an overall normalisation factor,
\begin{equation}
{\cal N}= i \left( \frac{g_W}{2 \cos \theta_W} \right)^2 \,.
\label{eq:Ndef}
\end{equation}
From this it is straightforward to square the amplitude to obtain the result for the
Higgs-mediated diagrams alone. The sum over the polarisations of the gluons and the $Z$ bosons of momentum $p$ can be
performed as usual with the projection operators,
\begin{equation}
P_g^{\mu \nu}=-g^{\mu \nu}, \hspace{1cm} P_Z^{\rho \beta}(p) = -g^{\rho \beta}+\frac{p^\rho p^\beta}{m_Z^2} \, .
\label{Zpolsum}
\end{equation}
Using these projectors we get the subsidiary result
\begin{eqnarray}
P_{Z}^{\rho\sigma}(p_3)P_{Z \, \rho \sigma}(p_4)&=&2
\Bigg[\frac{(d-2)}{2}+\frac{1}{8}\frac{(s-2 m_Z^2)^2}{m_Z^4}\Bigg] \, .
\end{eqnarray}
Including also the sum over colors yields the matrix element squared for the signal in this channel,
(The statistical factor for identical $Z$ bosons is not included).
\begin{align*}
\label{eq:MEsq_ggH}
\mathcal{S}_{gg} &\equiv \bra{{\cal A}^{\alpha\beta\rho\sigma,AB}_\text{ggHZZ}(\alpha_S^{(n_l)},m,\mu,\epsilon)}\ket{{\cal A}^{AB}_{ \text{ggHZZ},\alpha\beta\rho'\sigma'}(\alpha_S^{(n_l)},m,\mu,\epsilon)} P_{Z \rho'}^{\rho}(p_3) P_{Z\sigma'}^{\sigma}(p_4) \numberthis \\
&= |{\cal N}|^2 \frac{64 N_A}{9} \left(\frac{s}{s-m_H^2}\right)^2 \bra{{\cal A}(\alpha_S^{(n_l)},m,\mu,\epsilon)}\ket{{\cal A}(\alpha_S^{(n_l)},m,\mu,\epsilon)}\cdot (1-\epsilon) \bigg[ 1-\epsilon + \frac{1}{8}\bigg(\frac{1}{r_Z}-2\bigg)^2\bigg] \,,
\end{align*}
where we use the notation $r_Z=m_Z^2/s$ and $N_A= N_c^2-1 = 8$.
\subsection{Large-Mass Expansion and Improvements}
\label{sub:mass_expansion_and_its_improvements}
Using the aforementioned conventions we can compute the leading- and
next-to-leading-order amplitude $\ket{{\cal A}^{(1,2)}(m,\mu,\epsilon)}$ for
single Higgs boson production. Although we always work with the loop
measure $S_\epsilon = \exp(\epsilon \gamma_E)(4\pi)^{-\epsilon}$ we factor out
\begin{equation}
\label{eq:loop_measure_factors}
S_\epsilon c_\Gamma = \frac{e^{\epsilon \gamma_E}}{(4\pi^\epsilon)}\cdot \Gamma(1+\epsilon)(4\pi)^\epsilon = 1+\epsilon^2\frac{\pi^2}{12}
+{\cal O}(\epsilon^3) \, ,
\end{equation}
in the results presented below to keep factors of $\pi^2$ implicit. The dimensional dependent factor $c_\Gamma$ denotes the somewhat more natural loop measure, because it cancels exactly the $\Gamma(1+\epsilon)$ factor obtained by the loop integration.
The exactly known leading-order result in $d$-dimensions ($d=4-2\epsilon$) yields~\cite{Resnick:1973vg,Georgi:1977gs,Dawson:1993qf,Anastasiou:2006hc,Harlander:2009bw}
\begin{align*}
\label{eq:ME_ggH_LO_exact}
\ket{{\cal A}^{(1)}(m,\mu,\epsilon)} &= S_\epsilon c_\Gamma \cdot 3 r_t \numberthis\\
&\times \Bigg( \frac{2\epsilon}{1-\epsilon} B_0\left(p_1+p_2;m,m\right) - \left(1-\frac{4}{1-\epsilon} r_t\right)s\, C_0\left(p_1,p_2;m,m,m\right) \Bigg)\,,
\end{align*}
where $s=(p_1+p_2)^2$. The
definitions of the integrals $B_0$ and $C_0$ are given in
appendix~\ref{Intdef}.
The essential idea of the large-mass expansion based on the method of
\emph{expansion by regions}~\cite{Smirnov:2002pj} is that the
integration domain is divided into different regions where the loop
momenta are soft, $k_i \sim p_i \ll m$ or hard, $p_i \ll k_i\sim m$.
The external momenta $p_i \ll m$ are always assumed to be small. In
the expansion of one-loop integrals only the region of a hard loop
momentum $k_1\sim m$ exists, because all propagators are associated
with the large mass $m$. As a result the one-loop expansion consists
only of a naive Taylor expansion and its result is given in terms of
simple massive one-loop vacuum integrals.
The two-loop integral expansion is more involved since the hard as
well as the soft region must be considered. The first region results,
with the help of~\cite{Davydychev:1995nq}, in scalar massive two-loop
vacuum integrals. The soft region produces a product of massive
one-loop vacuum integrals and massless one-loop bubble and triangle
integrals. All occurring integrals are well known and, although, the
intermediate expressions become huge, the final results are remarkably
simple, as can be seen below. We use our own fully automatic
\emph{in-house} software to perform the large-mass expansion, relying
extensively on the features of \texttt{FORM}~\cite{Vermaseren:2000nd}
and \texttt{Mathematica}. For a similar approach to Higgs boson
pair-production, see e.g.~\cite{Dawson:1998py}.
Using the large-mass expansion for the $B_0$ and $C_0$ integral, given in
Sec.~\ref{Intdef}, the corresponding expansion of the full result for
$\ket{{\cal A}^{(1)}(m,\mu,\epsilon)}$ in $d$ dimensions is
\input{src/LMEs/ggH_1L.tex}
Similarly the two-loop result can be expressed in terms of the leading-order amplitude $\ket{\bar{{\cal A}}^{(1)}(m,\mu,\epsilon)} = (S_\epsilon c_\Gamma (\mu^2/m^2)^\epsilon)^{-1} \ket{{\cal A}^{(1)}(m,\mu,\epsilon)}$ and with only mass renormalisation included
\input{src/LMEs/ggH_2L.tex}
The first terms of \Eqno{eq:ggH_LME1L} and
\Eqno{eq:ggH_LME2L} fully agree with available results in the
literature~\cite{Harlander:2009bw,Pak:2009bx}. Especially the NLO
corrections presented in~\cite{Harlander:2009bw} cover terms in the
expansion up to ${\cal O}\left(1/r_t^2,\epsilon^2\right)$ and we find full
agreement with our results for the amplitudes as well as the cross
sections. The analytic results for the exact LO and NLO amplitude
${\cal A}$, keeping the full top mass dependence, can be taken
from~\cite{Anastasiou:2006hc,Beerli:2008zz}\footnote{The overall sign
of the NLO term differs between the published
paper~\cite{Anastasiou:2006hc} and the thesis of
Beerli~\cite{Beerli:2008zz}. We believe that the sign in the latter
is correct, which is also supported by the comparison with the
NLO results using the large-mass expansion~\cite{Harlander:2009bw,Pak:2009bx}.}.
The NLO results for the virtual amplitude have also
been checked by our own independent program, using
\texttt{GiNaC}~\cite{Bauer:2000cp} to evaluate the harmonic
polylogarithms. This serves as a further independent check of the
mass expansion results in \Eqno{eq:ggH_LME1L} and
\Eqno{eq:ggH_LME2L}. This agreement will be illustrated in
Sec.~\ref{ssub:ggH_comparison_lme_with_full_result}.
The radius of convergence of the large-mass expansion is
given by $s/(4m^2)\lesssim 1$. The polynomial
growth leads to an extremely good convergence below and close to
threshold of top quark pair-production, as shown later.
\subsubsection{Rescaling with Exact Leading-Order Result}
\label{ssub:rescaling_with_exact_leading_order_result}
Above threshold, however, naively no convergence with respect to the exact result can be
expected. At least two procedures exist which
lead to major improvements in terms of convergence of the expanded
result even above threshold\footnote{The region above threshold could
also be approximated by fitting a suitable ansatz to the high-energy
limit~\cite{Marzani:2008az,Harlander:2009mq,Pak:2009dg}. This,
however, would require additional knowledge of the high-energy
behaviour and is beyond the scope of this work.}.
We recall these procedures in this subsection and the next.
A well known method of extending the naive large-mass expansion of the NLO cross section beyond its range of validity relies on factoring out the LO cross section with exact top mass dependence,
\begin{equation}
\label{eq:LME_rescaled}
\sigma^\text{NLO}_{\text{imp},N} \equiv \sigma^\text{LO}_\text{exact} \cdot \frac{\sigma^\text{NLO}_\text{exp}}{\sigma^\text{LO}_\text{exp}} = \sigma^\text{LO}_\text{exact} \cdot \frac{\sum\limits_{n=0}^N \; c_n^\text{NLO} (1/r_t)^n}{\sum\limits_{n=0}^N \; c_n^\text{LO} (1/r_t)^n} \,.
\end{equation}
The numerator and denominator are expanded to the same order in
$1/r_t$. It was argued for single Higgs boson production
in~\cite{Pak:2009dg} and for Higgs boson pair-production
in~\cite{Grigo:2013rya} that varying $N$ in the above formula allows to
check for additional power corrections. Including sufficient orders in
the expansion should lead to stable approximations
$\sigma^\text{NLO}_{\text{imp},N}$.
The method relies on the expansion of numerator and denominator in \Eqno{eq:LME_rescaled}
and evidently, requires the knowledge of all of the ingredients in
terms of series expansions. Although this requirement usually does not
pose any problem \emph{per se} it might turn out to be disadvantageous
in certain cases. In our particular case at hand, we require the SM
continuum as well as the Higgs-mediated amplitude as large-mass
expansions. Certainly the Higgs-mediated amplitude is well known at LO
and NLO including its full top mass dependence. Any approximation of
this amplitude poses a potential threat of introducing unnecessary
uncertainties. We will discuss this point further in
Sec.~\ref{sub:visualisation_of_large_mass_expansion_results_for_gg_to_zz}
and see that the method introduced in the next section provides a way
to circumvent this issue.
\subsubsection{Conformal Mapping and Pad\'e Approximants}
\label{ssub:conformal_mapping_and_pade_approximants}
Having sufficiently many terms in the $1/m$ expansion at hand allows for a
more powerful resummation method, the \emph{Pad\'e approximation}~\cite{Fleischer:1994ef,Fleischer:1996ju,Harlander:2001sa,Smirnov:2002pj,Press:2007:NRE:1403886}. The \emph{univariate Pad\'e approximant} $[n/m]$ to a given \emph{Maclaurin series} with a non-zero radius of convergence $z_0$
\begin{equation}
\label{eq:general_series_expansion}
f(z) = \sum\limits_{n=0}^{\infty}\; a_n z^n
\end{equation}
is defined via the rational function
\begin{equation}
f_{[n/m]}(z) = \frac{b_0+b_1 z+b_2 z^2+\ldots + b_n z^n}{1+c_1 z+c_2 z^2+\ldots + c_m z^m}
\label{eq:Padedefinition}
\end{equation}
such that its Taylor expansion reproduces the first $n+m$ coefficients
of $f(z)$; the coefficients $b_i$ and $c_i$ are uniquely defined by
this expansion. The advantage of \emph{Pad\'e approximants} over
other techniques, e.g. \emph{Chebyshev approximation}, lies in the
fact that they can provide genuinely new information about the
underlying function $f(z)$, see~\cite{Press:2007:NRE:1403886} for more
information.
The downside of \emph{Pad\'e approximants} is their
uncontrollability. In general, there is no way to tell how accurate
the approximation is, nor how far the range $z$ can be
extended. Computing the \emph{Pad\'e approximants} $[n/n]$ or
$[n/n\pm1]$ for different orders $n$ allows, at least, checking the
stability of the approximation. We will refer to $[n/n]$ as diagonal
and to $[n/n\pm1]$ as non-diagonal \emph{Pad\'e approximants} in the
following.
Although the \emph{Pad\'e approximation} can be directly applied to
\Eqno{eq:general_series_expansion}, it is advantageous to apply a
\emph{conformal mapping}~\cite{Fleischer:1994ef}
\begin{equation}
\label{eq:conformal_mapping}
w(z) = \frac{1-\sqrt{1-z/z_0}}{1+\sqrt{1-z/z_0}}
\end{equation}
first. The amplitudes at hand, $gg(\to H)\to ZZ$, with $z=s/m^2$
develop a branch cut starting from $z_0=4$ and extending to $+\infty$ due to the top
quark pair-production threshold. Applying the mapping,
\Eqno{eq:conformal_mapping}, transforms the entire complex plane
into the unit circle of the $w$-plane, such that the upper (lower)
side of the cut corresponds to the upper (lower) semicircle and the
origin of the original $z$-plane is left unchanged.
The initial power series can now be transformed into a new series in $w$~\cite{Smirnov:2002pj}
\begin{equation}
f(w) = \sum\limits_{n=0}^\infty\; \Phi_n w^n \,,
\end{equation}
where
\begin{align}
\Phi_0 = a_0 \quad \text{and}\quad
\Phi_n = \sum\limits_{k=1}^n\; \frac{(n+k-1)!(-1)^{n-k}}{(2k-1)!(n-k)!}\, (4z_0)^k \,a_k\,, \quad \text{if } n \ge 1\,,
\end{align}
and, subsequently, its \emph{Pad\'e approximants} computed. We will
illustrate these features
using the example of single Higgs boson production in the next section.
\subsubsection{Comparison of LME with Full Result}
\label{ssub:ggH_comparison_lme_with_full_result}
Let us briefly compare the results from the large-mass expansions,
\Eqno{eq:ggH_LME1L} and \Eqno{eq:ggH_LME2L}, and their,
previously discussed, improvements to the known LO and NLO QCD result
with full top mass
dependence~\cite{Spira:1995rr,Harlander:2005rq,Anastasiou:2006hc,Aglietti:2006tp}. We
include the subsequent $H\to ZZ$ decay, as given in
\Eqno{decayamplitude}, perform the UV+IR renormalisation and
compute the phase space integral over \Eqno{eq:MEsq_ggH}
including all corresponding phase space factors and coupling
constants. The NLO contribution so obtained is not physical, since we
neglect the real-radiation contribution for now. Considering the
obtained finite parts of the LO and virtual NLO corrections alone, on
the other hand, allow to better verify the validity of our
approximations. To be specific, we set
\begin{equation}
\label{eq:ggH_sigma_defs_virt}
\sigma_\text{H}^\text{LO} \sim {\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal A}^{(1)}(m,\mu)} \quad \text{and} \quad \sigma_\text{virt,H}^\text{NLO} \sim 2 \, {\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal A}^{(2)}(m,\mu)} \,.
\end{equation}
We utilise the $CT10nlo$ PDF set~\cite{Lai:2010vv} within \texttt{LHAPDF}~\cite{Buckley:2014ana} to determine $\alpha_S(\mu_f)$ and use the input parameters
\begin{align*}
\label{eq:ggH_input_parameters}
\sqrt{S} &= \unit[13]{TeV}\,, & \mu_f &= \mu_r = \sqrt{s}\,, \\
m &= \unit[173.5]{GeV}\,, & m_Z &= \unit[91.1876]{GeV}\,, \numberthis \\
m_W &= \unit[80.385]{GeV}\,, & G_F &= \unit[1.16639\cdot 10^{-5}]{GeV^{-2}}\,,
\end{align*}
where $S$ and $s$ denote the hadronic and partonic center-of-mass energy, respectively.
\begin{figure}[h!]
\centering
\begin{subfigure}{0.49\textwidth}
\includegraphics[width=\textwidth]{figures/ggHZZ_LO_LME_Pade.eps}
\end{subfigure}
\hfill
\begin{subfigure}{0.49\textwidth}
\includegraphics[width=\textwidth]{figures/ggHZZ_NLO_LME_Pade.eps}
\end{subfigure}
\caption{\textbf{Left panel:} Leading-order $gg\to H\to ZZ$ cross section. 1.) LME up to $1/m^{20}$ (orange). 2.) Exact result (black), LME with conformal mapping (blue) and \emph{Pad\'e approximants} $[4/4],[4/5],[5/4],[5/5]$ (yellow, purple, green, brown) agree perfectly. \textbf{Right panel:} Virtual NLO corrections to $gg\to H\to ZZ$ cross section. See text for details. Color code as in left panel. The bottom plots show the relative deviations with respect to the exact (N)LO results. The vertical dashed line denotes the top quark pair-production threshold.}
\label{fig:ggHZZ_LO_NLO_LME_Pade}
\end{figure}
The orange curves in Fig.~\ref{fig:ggHZZ_LO_NLO_LME_Pade} depict the
large-mass expansion results of \Eqno{eq:ggH_sigma_defs_virt} for
the LO and the NLO case, where each\footnote{The ambiguity between
expanding the product $\bra{{\cal F}_{\cal A}^{(1)}}\ket{{\cal F}_{\cal A}^{(1,2)}}$ or
expanding each $\ket{{\cal F}_{\cal A}^{(1,2)}}$ separately, consists only of
power corrections which are numerically negligible. We checked that
the difference in $\Delta\sigma/\sigma$ at threshold of both approaches is
$\lesssim 1\%$. The same arguments hold for the series expansions
including the conformal mapping.} finite remainder ${\cal F}_{\cal A}^{(1,2)}$ is expanded up to $1/m^{20}$. A minimum cut $\sqrt{s}\ge 2 m_Z$ has been imposed and the threshold for top quark pair-production is given by $s/m^2 = 4$.
The relative deviation
\begin{equation}
\frac{\Delta \sigma}{\sigma} = 1 - \frac{\sigma^\text{(N)LO}_\text{approx}}{\sigma^\text{(N)LO}_\text{exact}}
\end{equation}
of the approximated results with respect to the exact result are shown
in the bottom plots. The large-mass expansion describes the exact LO
and virtual NLO results up to the top threshold very well, with only
$5\%$ deviation at LO and $7\%$ at NLO at $s=4m^2$. As expected
however the large-mass expansion diverges for values above this
threshold. Improvements to this naive approximation by means of the
conformal mapping, \Eqno{eq:conformal_mapping}, are shown in blue. On
top we compute the diagonal, $[5/5]$ (brown) and $[4/4]$ (yellow), and
non-diagonal, $[4/5]$ (purple) and $[5/4]$ (green), \emph{Pad\'e approximants}
at amplitude level for the mapped series expressions
of each finite remainder, i.e. ${\cal F}_{{\cal A},[n/m]}^{(i)}$. Both results,
using the \emph{Pad\'e approximants} or the mapped series alone,
excellently reproduce the exact results (black curve) even far above
threshold; with less than $1\%$ deviation from the exact result over
the considered range. As a result the \emph{Pad\'e approximant} $[5/5]$
overlays all other curves in
Fig.~\ref{fig:ggHZZ_LO_NLO_LME_Pade}, some of which are scarcely
visible.
\begin{figure}[h!]
\centering
\includegraphics[width=.7\textwidth]{figures/ggHZZ_NLO_LME_rescaled.eps}
\caption{Virtual NLO corrections to $gg\to H\to ZZ$ cross section with rescaling from \Eqno{eq:LME_rescaled}. See text for details. 1.) Exact NLO result (black). 2.) Varying orders of rescaled LMEs are indicated by shaded grey area. Its envelope is given by $\sigma_{\text{imp},1}^\text{NLO}$ (orange) and $\sigma_{\text{imp},10}^\text{NLO}$ (blue). The bottom plot shows the relative deviations with respect to the exact NLO results. The vertical dashed line denotes the top quark pair-production threshold.}
\label{fig:ggHZZ_NLO_melnikov}
\end{figure}
The second choice of improving the naive LME is given by the rescaling
from \Eqno{eq:LME_rescaled}. The results are shown in
Fig.~\ref{fig:ggHZZ_NLO_melnikov}. The exact virtual NLO result is
again shown in black. The rescaled LMEs are indicated by the shaded
grey area and its envelope is given by the expansions
$\sigma_{\text{imp},1}^\text{NLO}$ (orange) and
$\sigma_{\text{imp},10}^\text{NLO}$ (blue). Although the heavy-quark
approximation $\sigma_{\text{imp},1}^\text{NLO}$ gives a reasonable
estimate of the exact result above threshold it fails to describe the
threshold behaviour and peak structure of the exact result. At
threshold the deviation is $10\%$. Taking higher orders in the
expansion into account improves the threshold prescription, with $3\%$
deviation for $\sigma_{\text{imp},10}^\text{NLO}$ at threshold, but
worsens the trend for higher energies. In both cases we find more than
$20\%$ deviation for $s/m^2>20$.
We end this section by drawing our conclusions from the results
presented. We see that, at least in the single-scale Higgs boson
production and having a sufficient number of terms in the LME at hand,
applying the conformal mapping (and the \emph{Pad\'e approximation})
yields excellent prescriptions of the exact results. The conformal
mapping is imperative, whereas the additional \emph{Pad\'e
approximants} give only small improvements in terms of uncertainty
reduction and stability of the approximations. We conclude that we
should favour these approximations over the rescaling method.
One important point to notice, however, is that the kinematics change
when moving from the single Higgs boson production to the SM $Z$ boson
pair-production\footnote{Even if the $H\to ZZ$ decay is
included. Effectively, only the $2\to 1$ kinematics of the Higgs boson production matter.}.
Therefore, the results discussed here may
not necessarily transfer easily. Still, the comparisons within this
chapter should give an idea of the validity of the improved large-mass
expansions. We will discuss analogous considerations for the Z boson
pair-production in
Sec.~\ref{sub:visualisation_of_large_mass_expansion_results_for_gg_to_zz}.
\section{Introduction}
\label{sec:intro}
The production of four charged leptons is a process of great
importance at the LHC. It was one of the discovery channels of the
Higgs boson at the LHC. It also provides fundamental tests of the
gauge structure of the electroweak theory through the high energy
behaviour. Four charged leptons are predominantly produced by quark
anti-quark annihilation; the mediation is by photons or $Z$ bosons
dependent on the mass of the four leptons, $m_{4l}$.
A smaller contribution, which however grows with energy is provided by
gluon-gluon fusion. The Higgs boson is of course produced in this
channel; in the Standard Model (SM) this occurs predominantly through
the mediation of a loop of top quarks. As pointed out by Kauer and
Passarino~\cite{Kauer:2012hd}, despite the narrow width of the Higgs
boson, the Higgs-mediated diagram gives a significant contribution for
$m_{4l}> m_H$. If we examine the tail of the Higgs-mediated diagrams
there are three phenomena occurring:
\begin{itemize}
\item The opening of the threshold for the production of real on-shell $Z$ bosons, $m_{4l}> 2 m_Z$.
\item The region $m_{4l}=2 m$, ($m$ is the top quark mass) where the
loop diagrams develop an imaginary part.
\item The large $m_{4l}$ region, $ m_{4l}>2 m$, where the destructive interference between the
Higgs-mediated diagrams leading to $Z$ bosons and the continuum production
of on-shell $Z$ bosons is most important.
\end{itemize}
\begin{figure}[ht]
\begin{center}
\includegraphics[angle=270,scale=0.7]{figures/ZZprod.eps}
\caption{Representative diagrams for the $ZZ$ production. In the following we will
suppress the $Z$-decays to leptons.}
\label{ZZprod}
\end{center}
\end{figure}
\renewcommand{\baselinestretch}{1.5}
A feature of this tail is that it depends on the couplings of
the Higgs boson to the initial and final state particles but not on
the width of the Higgs boson. Assuming the couplings of the on- and
off-peak Higgs-mediated amplitudes are the same, it has been proposed
to use this property to derive upper bounds on the width of the Higgs
boson~\cite{Caola:2013yja}. Note that models with different
on- and off-peak couplings can be constructed~\cite{Englert:2014aca}.
In the following we shall refer to the production of the bosons
$V_1,V_2$. Gluon-gluon fusion first contributes to the cross section
for electroweak gauge boson production $pp\to V_1V_2$ as shown in
Fig.~\ref{ZZprod}(c)-(e) at ${\cal O}(\alpha_S^2)$, which is the
next-to-next-to-leading-order (NNLO) with respect to the leading-order
(LO) QCD process shown in Fig.~\ref{ZZprod}(a); no two-loop $gg\to V_1
V_2$ amplitudes participate in this order in perturbation theory.
In the context of the Higgs boson width, however, the interference
between the Higgs-mediated $Z$ boson pair-production and the Standard
Model continuum at next-to-leading-order (NLO) QCD already requires
knowledge of the one- and two-loop $gg\to (H\to) V_1 V_2$
amplitudes. The requirement for more precise estimates to the Higgs
boson width were emphasised
in~\cite{ATLAS-CONF-2014-042,Khachatryan:2014iha,Melnikov:2015laa}.
Signal-background interference effects beyond the leading order
have been considered in ref.~\cite{Bonvini:2013jha} for the process
$gg\to H \to W^+ W^-$ for the case of a heavy Higgs boson.
In this work we will limit ourselves to the $Z$ boson pair final state,
due to its importance at the LHC. At LO~\cite{Dawson:1993qf} and
NLO~\cite{Spira:1995rr,Harlander:2005rq,Anastasiou:2006hc,Aglietti:2006tp}
the amplitudes for single Higgs boson production have been known for
quite some time. At LO, the amplitude for the SM continuum $gg\to ZZ$
process occurs via massless and massive fermion loops and
results are available in each case~~\cite{Glover:1988rg,Kauer:2012ma,Campbell:2013una,Campbell:2014gua}.
The situation, however, is different for the NLO continuum process,
although vast progress in terms of two-loop amplitudes has been
made~\cite{Gehrmann:2014bfa,Cascioli:2014yka,Caola:2014iua,Gehrmann:2015ora,Caola:2015ila,vonManteuffel:2015msa}.
Recently two-loop $gg\to ZZ$ amplitudes\footnote{Actually, the results
in~\cite{Caola:2015ila} and~\cite{vonManteuffel:2015msa} allow for
arbitrary off-shell electroweak gauge bosons in the final state.}
via \emph{massless} quarks became
available~\cite{Caola:2015ila,vonManteuffel:2015msa}. The complete
computation of two-loop amplitudes with \emph{massive} internal quark
loops, on the other hand, is commonly assumed to be just beyond
present technical capabilities.
Although the contribution of the top quark loops to these diagrams is smaller
than the contribution of the light quarks in the region just above the $Z$-pair
threshold, in the high $m_{4l}$ region the amplitude is dominated by the
contributions of longitudinal $Z$ bosons that couple to the top quark loops.
Recently a first heavy top quark
approximation for the two-loop $gg\to ZZ$ amplitude with internal top
quarks was published~\cite{Melnikov:2015laa}. In that work only the
leading term in the $s/m^2$ expansion was considered. In that
approximation, the vector-coupling of the $Z$ boson to the top quark
does not contribute. In addition an approximate treatment of this process at higher orders, based on soft gluon resummation,
was presented in Ref.~\cite{Li:2015jva}.
In the present work we will push this analysis further. We start by
presenting our results for the LO and NLO Higgs-mediated $ZZ$
production in terms of the $s/m^2$ expansion in
Sec.~\ref{sec:higgs_production_via_gluon_fusion}, despite the fact
that the full result is known. This part
is required for the later interference with the SM
continuum. Furthermore, it is well suited to introduce our notation in
Sec.~\ref{sub:ggH_preliminaries} and to assess the validity of the
approximation methods with respect to the exact known (N)LO amplitudes
in Sec.~\ref{sub:mass_expansion_and_its_improvements}.
The results for the LO and virtual NLO contributions to the SM
continuum with massive quark loops will be given in
Sec.~\ref{sec:virtual_corrections_to_sm_zz_production_via_massive_quark_loops}
as a large-mass expansion (LME) with terms up to $(s/m^2)^6$.
We will limit our discussion to the interference between the
Higgs-mediated term and the continuum term. Similar to~\cite{Melnikov:2015laa} we will consider
\emph{on-shell} $Z$ bosons in the final state. A theoretical
predictions for off-shell $Z$ bosons would be optimal, but in order to
reduce the number of scales in the problem, we restrict ourselves to
on-shell $Z$ bosons. Since we are primarily interested in the high-mass
behaviour this is an appropriate approximation.
A limited number of scales is beneficial when we
consider the extension of our approach to a full calculation.
In Sec.~\ref{sec:real_corrections_to_sm_zz_production} we summarize our treatment of the real
radiation contribution, which makes use of results already presented in Ref.~\cite{Campbell:2014gua}.
The results of our calculation, including loops of both massless and massive quarks,
will be presented in Sec.~\ref{sec:results}. We will compare the effects of the NLO
corrections to the interference contribution with the corresponding corrections to the
Higgs diagrams alone. In addition, we will discuss
the impact of our results on analyses of the off-shell region that
aim to bound the Higgs boson width.
All expansion results from Sec.~\ref{sec:higgs_production_via_gluon_fusion} and Sec.~\ref{ssub:non_anomalous_diagrams} are provided via ancillary files on \texttt{arXiv} as \texttt{FORM} and \texttt{Mathematica} readable code.
\section{Results}
\label{sec:results}
The individual components of the calculation that have been extensively
discussed above have been included in the parton-level Monte Carlo
code MCFM~\cite{Campbell:1999ah,Campbell:2011bn,Campbell:2015qma}.
The bulk of the calculation is performed in a
straightforward manner using the normal operation of MCFM at NLO.
The exception is the finite contribution to the two-loop amplitude containing
a closed loop of massless quarks. Since these contributions are computationally
expensive to evaluate, we choose to include their effects by reweighting
an unweighted sample of LO events.
For the two-loop amplitudes containing
massive loops of quarks the approximations used are as follows. The Higgs
amplitude is evaluated using the $[5/5]$~\emph{Pad\'e approximant} to the LME
after conformal mapping. As demonstrated in
Sec.~\ref{sec:higgs_production_via_gluon_fusion}, this is virtually identical
to the exact result. The massive quark box contributions are computed
by factoring out the exact LO amplitude according to
\Eqno{eq:LME_rescaled_Pade}, with the \emph{Pad\'e approximant} corresponding
to $n = m = 3$ in the definition given in \Eqno{eq:Padedefinition}.
The anomalous diagrams of Sec.~\ref{ssub:anomalous_diagrams} are not
included in the discussion of the massive quark loops below, but instead are
accounted for only when the sum of all loops is considered.
For massless quarks circulating in the loop the calculation is simplified by
the fact that the entire amplitude is proportional to the combination of
couplings $(v_f^2 + a_f^2)$, i.e. in the decomposition given in
\Eqno{eq:massiveggZZ_normalisation} the quantities
$\ket{\TW{{\cal B}}_{VV}^0}$ and $\ket{\TW{{\cal B}}_{AA}^0}$ are equal.
The calculation requires the one-loop master integrals up to $\epsilon^2$,
for which all orders results are given in ref.~\cite{Smirnov:2006ry} for bubble integrals
and refs.~\cite{Bern:1994zx,Bern:1993kr,Brandhuber:2004yw,Brandhuber:2005kd,Cachazo:2004zb} for the \emph{easy box}
(two opposite off-shell legs). The necessary results for the \emph{three-mass} triangle with massless propagators
and the \emph{hard box} (two adjacent off-shell legs) can be taken from refs.~\cite{Chavez:2012kn}
and~\cite{Anastasiou:2014nha} respectively. We use the \emph{coproduct} formalism~\cite{Duhr:2012fh,Duhr:2014woa}
to \emph{analytical continue} the results to the physical phase space regions. All master integrals have been numerically
cross-checked with \texttt{SecDec}~\cite{Borowka:2015mxa}. The two-loop master integrals for $gg\to ZZ$ are taken
from ref.~\cite{Gehrmann:2014bfa} and \texttt{GiNaC} is used to evaluate the polylogarithms.
Our results for this contribution agree with the earlier calculation of ref.~\cite{vonManteuffel:2015msa}.
The parameters for the following results have already been specified
in Sec.~\ref{ssub:ggH_comparison_lme_with_full_result}. Here we make only one
change: our central scale corresponds to the choice $\mu_r = \mu_f = M_{ZZ}/2$, where $M_{ZZ}$ is the
invariant mass of the $ZZ$ pair. As an estimate of the theoretical uncertainty we consider variations
by a factor of two about this value. We also introduce an uncertainty that is based on our combination of LME
and \emph{Pad\'e approximants} in the calculation of the massive quark loops, that has already been explored in
Fig.~\ref{fig:ggZZ_NLO_melnikov} (right). In order to obtain a more conservative error estimate we multiply the
deviations of the extremal values in the grey area with respect to $\sigma_{\text{imp},[3/3]}^\text{NLO}$ by a factor of two.
The impact of this variation on the complete NLO prediction for the massive loop is shown in Fig.~\ref{fig:NLO_massless_LMEuncertainty}.
Even for this choice, the impact of the approximation is estimated to be less than $20$\% throughout the distribution.
For the remaining plots in this section we no longer show the impact of this uncertainty, but it will be explicitly
included in Tables~\ref{tab:xsecs} and~\ref{tab:xsecsabove300} later on.
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.5\textwidth}
\includegraphics[width=\textwidth]{figures/NLOtoponly_envelopeHalfMZZ.eps}
\end{subfigure}
\caption{The uncertainty on the calculation of the massive loop interference contribution
stemming from the use of the LME expansion and \emph{Pad\'e approximants}. The central result is
shown as a solid histogram, with the dashed lines indicating deviations that correspond
to the grey area in Fig.~\ref{fig:ggZZ_NLO_melnikov}, multiplied by a factor of
two. All curves are computed for the central scale choice, $\mu_r = \mu_f = M_{ZZ}/2$.}
\label{fig:NLO_massless_LMEuncertainty}
\end{figure}
Results for both the massless and massive quark contributions to the interference, including the effects of scale variation,
are shown in Fig.~\ref{fig:NLO_massless_massive}. The interference is negative for both
the massless and massive quark contributions and is shown in Fig.~\ref{fig:NLO_massless_massive}
reversed in sign. In both cases the $K$-factor
decreases as the invariant mass of the $Z$-boson pair increases. The $K$-factor at small invariant
masses is larger for the massless loops; as the invariant mass increases, the NLO corrections are more
important for the massive loop. The NLO corrections are larger for the top quark loops
and exhibit a stronger dependence on $M_{ZZ}$. In both cases the NLO result lies outside the
estimated LO uncertainty bands and the scale uncertainty is not significantly reduced at NLO.
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.49\textwidth}
\includegraphics[width=\textwidth]{figures/newNLOnotop.eps}
\end{subfigure}
\hfill
\begin{subfigure}{0.49\textwidth}
\includegraphics[width=\textwidth]{figures/newNLOtoponly.eps}
\end{subfigure}
\caption{\textbf{Left panel:} Interference of the Higgs amplitude and massless quark loops at LO and NLO,
with the scale uncertainty indicated by the dashed histograms. The ratio of the
NLO and LO results is shown in the lower panel.
\textbf{Right panel:} The equivalent results for the interference of the Higgs amplitude and the top quark loops.}
\label{fig:NLO_massless_massive}
\end{figure}
The relative importance of the massive and massless loops can be
better-assessed from the NLO predictions shown in
Fig.~\ref{fig:compareNLO}. At smaller invariant masses, below the
top-pair threshold, the massless loops are most important. Around the
top-pair threshold the two are of a similar size, but at high energies
the massless loops are insignificant. In contrast, the top quark loop
quickly becomes the dominant contribution beyond this threshold and
exhibits a long tail out to invariant masses of around one TeV.
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.6\textwidth}
\includegraphics[width=\textwidth]{figures/compareNLO.eps}
\end{subfigure}
\caption{Comparison of the effect of the massless (magenta) and massive (red) loops in the NLO interference.
Also shown is the sum (blue) and the corresponding result for the Higgs amplitude squared (black).
All curves are computed for the central scale choice, $\mu_r = \mu_f = M_{ZZ}/2$.}
\label{fig:compareNLO}
\end{figure}
The full prediction for the interference that is obtained by summing over
both massless and top quark loops, as well as the numerically-small
anomalous contribution discussed in Sec.~\ref{ssub:anomalous_diagrams},
is shown in Fig.~\ref{fig:NLO_full_higgs}. The
relative size of the massless and top quark loops discussed above
means that the behaviour of the $K$-factor for the sum of both
contributions interpolates between the massless-loop $K$-factor for
small $M_{ZZ}$ and the massive loop one for high $M_{ZZ}$. It
therefore decreases from around $3$ at the peak of the distribution to
approximately $1.8$ in the tail. This is to be contrasted with the
$K$-factor distribution for the pure Higgs amplitudes alone, shown in
the right panel of Fig.~\ref{fig:NLO_full_higgs}. In that case the
$K$-factor decreases slowly from around $2.2$ at small invariant
masses to around $1.8$ in the far tail. We note that the $K$-factor
for the Higgs amplitudes alone, and the one for the interference with
the top quark loops, is almost identical.
In the high-energy limit this is guaranteed to be the case, due to the cancellation between
these two processes. This behaviour is shown explicitly in Fig.~\ref{fig:Krat}.
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.49\textwidth}
\includegraphics[width=\textwidth]{figures/newNLOall.eps}
\end{subfigure}
\hfill
\begin{subfigure}{0.49\textwidth}
\includegraphics[width=\textwidth]{figures/newNLOhiggs.eps}
\end{subfigure}
\caption{\textbf{Left panel:} Interference of the Higgs amplitude and quark loops at LO and NLO,
with the scale uncertainty indicated by the dashed histograms. The ratio of the
NLO and LO results is shown in the lower panel.
\textbf{Right panel:} The equivalent results for the Higgs amplitude squared.}
\label{fig:NLO_full_higgs}
\end{figure}
\begin{figure}[ht!]
\centering
\begin{subfigure}{0.5\textwidth}
\includegraphics[angle=-90,width=\textwidth]{figures/Krat.eps}
\end{subfigure}
\caption{The ratio of the $K$-factors for the square of the Higgs diagrams alone ($K_{\mbox{higgs}}$)
and the one for the interference ($K_{\mbox{inter}}$). The lines are fits to the individual histogram
bins that are good to the level of a few percent and are shown for the central scale (blue) as well as
the scale variations (red, green).}
\label{fig:Krat}
\end{figure}
The integrated cross-sections for the interference contributions and the Higgs amplitude squared
are shown in Table~\ref{tab:xsecs}. Note that, in this table, the total interference differs from
the sum of the massive and massless loops by a small amount that is due to the anomalous contribution.
At this level the differences between the effects of the NLO
corrections on the various contributions is quite small, with all corresponding to a NLO enhancement
by close to a factor of two. The $K$-factor for the massless loops is slightly larger, which
is also reflected in the result for the total interference. In addition to the scale uncertainty, we have
also indicated our estimate of the residual uncertainty related to the LME expansion that is indicated
in Fig.~\ref{fig:NLO_massless_LMEuncertainty}. The impact of this uncertainty is relatively small, at the
level of around $5$\%, due to the fact that the integrated cross-section is dominated by the region
$M_{ZZ}^2 \lesssim 5 m^2$ where the LME is expected to work well.
\begin{table}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
Contribution & $\sigma_{LO}$ [fb] & $\sigma_{NLO}$ [fb] & $\sigma_{NLO}/\sigma_{LO}$\\
\hline
Higgs mediated diagrams & ~~~$56.3^{+15.3}_{-11.4}$ & ~~$111.0^{+20.1}_{-16.6}$ & 1.97 \\
interference (total) & $-113.5^{+22.2}_{-29.5}$ & $-237.8^{+36.4}_{-45.4}$(scale)${}^{+5.4}_{-0.4}$(LME) & 2.09 \\
\hline
interference (massless loops) & $-60.2^{+11.0}_{-14.2}$ & $-132.7^{+20.5}_{-26.3}$ & 2.20 \\
interference (massive loop) & $-53.3^{+11.2}_{-15.3}$ & $-104.2^{+15.8}_{-18.7}$(scale)${}^{+5.4}_{-0.4}$(LME) & 1.95 \\
\hline
\end{tabular}
\end{center}
\caption{Integrated cross-sections at $\sqrt{S}=13$~TeV, using the input parameters of
Sec.~\ref{ssub:ggH_comparison_lme_with_full_result} and $\mu=M_{ZZ}/2$. Uncertainties correspond to scale variation
as described in the text and, for NLO results that include massive quarks, an estimate of the limitations
of the LME. The $K$-factor is computed using only the central result.}
\label{tab:xsecs}
\end{table}
For obtaining a bound on the width of the Higgs boson it is useful to focus on a high-mass region where
backgrounds from the continuum processes, represented at tree-level by $q\bar q \to ZZ$, are small but
the effect of the interference is still significant~\cite{Caola:2013yja,Campbell:2013una}. To that
end, in Table~\ref{tab:xsecsabove300} we show the cross-sections after the application of the
cut $M_{ZZ} > 300$~GeV. We see that, as expected, the impact of the massive top loop on the interference
is much greater, compared to the massless loops. This also has the effect of ensuring that the $K$-factors
for the Higgs amplitude squared and the total interference are almost equal. To estimate the cross-section
after the decays of the $Z$-bosons into electrons and muons we can simply take these results and multiply by a factor
of $4 \times BR(Z \to e^- e^+)^2$, where $BR(Z \to e^- e^+) = 3.363 \times 10^{-2}$. Assuming that the on-shell
Higgs cross-section takes its Standard Model value and that the Higgs boson couplings and width are related accordingly,
we can write the predictions for the off-shell region as,
\begin{eqnarray}
\sigma_{4\ell}^{LO} (m_{4\ell} > 300~\mbox{GeV}) =
\left(0.190^{+0.055}_{-0.040}\right) \times \left( \frac{\Gamma_H}{\Gamma_H^{SM} }\right)
-\left(0.275^{+0.079}_{-0.058}\right) \times \sqrt{ \frac{\Gamma_H}{\Gamma_H^{SM} }} \; \mbox{fb} \,, \label{eq:offshellLO} \\
\sigma_{4\ell}^{NLO} (m_{4\ell} > 300~\mbox{GeV}) =
\left(0.365^{+0.064}_{-0.054}\right) \times \left( \frac{\Gamma_H}{\Gamma_H^{SM} }\right)
-\left(0.526^{+0.092}_{-0.103}\right) \times \sqrt{ \frac{\Gamma_H}{\Gamma_H^{SM} }} \; \mbox{fb} \,. \label{eq:offshellNLO}
\end{eqnarray}
The linear terms derive from the Higgs cross-sections in Table~\ref{tab:xsecsabove300} while the terms that
scale as the square-root of the modified width reflect the total interference contributions.
The uncertainties reflect those shown in Table~\ref{tab:xsecsabove300}, with the scale and LME uncertainties
added linearly. It is interesting to compare these results with the corresponding on-shell Higgs cross-sections.
These are given by,
\begin{eqnarray}
\sigma_{4\ell}^{LO} (m_{4\ell} < 130~\mbox{GeV}) = 1.654^{+0.249}_{-0.220} \; \mbox{fb} \;, \qquad
\sigma_{4\ell}^{NLO} (m_{4\ell} < 130~\mbox{GeV}) = 3.898^{+0.770}_{-0.560} \; \mbox{fb} \,, \label{eq:onshellLOandNLO}
\end{eqnarray}
where the uncertainties correspond to our usual scale variation procedure.
From the results in Eqs.~(\ref{eq:offshellLO}) and~(\ref{eq:offshellNLO}) it is clear that the absolute rate of off-shell events varies
considerably between LO and NLO. On the other hand, the cross-sections in Eq.~(\ref{eq:onshellLOandNLO}) imply that the ratio of the
number of events in the off-shell region compared to the peak region is much better predicted,
\begin{eqnarray}
\frac{\sigma_{4\ell}^{LO} (m_{4\ell} > 300~\mbox{GeV})}{\sigma_{4\ell}^{LO} (m_{4\ell} < 130~\mbox{GeV})} =
\left(0.115^{+0.014}_{-0.010}\right) \times \left( \frac{\Gamma_H}{\Gamma_H^{SM} }\right)
-\left(0.166^{+0.020}_{-0.015}\right) \times \sqrt{ \frac{\Gamma_H}{\Gamma_H^{SM} }} \,, \nonumber \\
\frac{\sigma_{4\ell}^{NLO} (m_{4\ell} > 300~\mbox{GeV})}{\sigma_{4\ell}^{NLO} (m_{4\ell} < 130~\mbox{GeV})} =
\left(0.094^{+0.000}_{-0.002}\right) \times \left( \frac{\Gamma_H}{\Gamma_H^{SM} }\right)
-\left(0.135^{+0.000}_{-0.008}\right) \times \sqrt{ \frac{\Gamma_H}{\Gamma_H^{SM} }} \,.
\end{eqnarray}
The uncertainties in this equation are obtained by using both the LME uncertainty estimate and the scale
variation, but ensuring that the cross-sections that appear in the numerator and denominator are evaluated
at the same scale.
\begin{table}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
Contribution & $\sigma_{LO}$ [fb] & $\sigma_{NLO}$ [fb] & $\sigma_{NLO}/\sigma_{LO}$\\
\hline
Higgs mediated diagrams & ~~$42.1^{+12.1}_{-8.8}$ & ~~~$80.7^{+14.2}_{-12.0}$ & 1.92 \\
interference (total) & $-60.7^{+12.8}_{-17.4}$ & $-116.3^{+17.5}_{-19.9}$(scale)${}^{+5.4}_{-0.4}$(LME) & 1.91 \\
\hline
interference (massless loops) & $-12.5^{+2.5}_{-3.4}$ & $-22.5^{+3.2}_{-3.2}$ & 1.80 \\
interference (massive loop) & $-48.2^{+10.3}_{-14.1}$ & $-93.0^{+14.0}_{-16.4}$(scale)${}^{+5.4}_{-0.4}$(LME) & 1.93 \\
\hline
\end{tabular}
\end{center}
\caption{Cross-sections at $\sqrt{S}=13$~TeV in the region defined by $M_{ZZ} > 300$~GeV,
using the input parameters of
Sec.~\ref{ssub:ggH_comparison_lme_with_full_result}. Uncertainties correspond to scale variation
as described in the text and, for NLO results that include massive quarks, an estimate of the limitations
of the LME. The $K$-factor is computed using only the central result.}
\label{tab:xsecsabove300}
\end{table}
\section{Real Corrections to SM $ZZ$ Production}
\label{sec:real_corrections_to_sm_zz_production}
\begin{figure}[ht]
\begin{center}
\includegraphics[angle=270,scale=0.8]{figures/ggHg.eps}
\caption{Representative diagrams for the $0 \to ggHg$ and the $0 \to gHq \bar{q}$ amplitudes.}
\label{fig:ggHg_amps}
\end{center}
\end{figure}
\begin{figure}[ht]
\begin{center}
\includegraphics[angle=270,scale=0.6]{figures/ggZZg.eps}
\caption{Representative diagrams for the $0 \to ggZZg$ and the $0 \to gZZq \bar{q}$ amplitudes.}
\label{fig:ggZZg_amps}
\end{center}
\end{figure}
Representative diagrams for the real radiation contributions to this process are shown in
Figs.~\ref{fig:ggHg_amps} and \ref{fig:ggZZg_amps}.
The Higgs-mediated diagrams have previously been computed in~\cite{Ellis:1987xu}. They
can easily be adapted to our calculation by combining those results with the decay amplitude given in
\Eqno{decayamplitude} and ${\cal N}$ from \Eqno{eq:Ndef}. This procedure, together with
the strategy for handling the amplitudes for diagrams without a Higgs boson, is described
in detail in~\cite{Campbell:2014gua}. We adopt this implementation here.
Our calculation of the pure-Higgs contribution involves the computation of
the square of the diagrams shown in Fig.~\ref{fig:ggHg_amps}, together with
all crossings of the quarks in Fig.~\ref{fig:ggHg_amps}~(right) into the
initial state. Similarly, the interference contribution includes all
crossings of the diagrams shown in Fig.~\ref{fig:ggZZg_amps}.
In principle another contribution to the interference occurs
at this order, between tree-level amplitudes for the process $qg \to ZZ q$ and the $qg$-initiated diagrams
shown in Fig.~\ref{fig:ggHg_amps}~(right) and \ref{fig:ggZZg_amps}~(bottom-left). However this contribution
is subleading~\cite{Campbell:2014gua}, particularly for high invariant masses of the $ZZ$ system, so we do not
consider it here.
The real radiation diagrams contain infrared singularities, of soft and collinear origin, that
must be isolated and combined with the corresponding poles in the two-loop amplitudes. This
is handled using the dipole subtraction procedure~\cite{Catani:1996vz}.
\section{Virtual Corrections to SM $ZZ$ Production via Massive Quark Loops}
\label{sec:virtual_corrections_to_sm_zz_production_via_massive_quark_loops}
\begin{figure}[ht]
\centering
\begin{subfigure}{.3\textwidth}
\includegraphics[angle=270,scale=.6]{figures/1L_diagrams/ggZZ1L_1.eps}
\caption{}
\label{sfig:ggZZ_LO_amp}
\end{subfigure}
\hspace{.5cm}
\begin{subfigure}{.3\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggZZ2L_1.eps}
\caption{}
\end{subfigure}
\hspace{.5cm}
\begin{subfigure}{.3\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggZZ2L_2.eps}
\caption{}
\end{subfigure}
\begin{subfigure}{.3\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggZZ2L_3.eps}
\caption{}
\end{subfigure}
\hspace{.5cm}
\begin{subfigure}{.3\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggZZ2L_4.eps}
\caption{}
\end{subfigure}
\hspace{.5cm}
\begin{subfigure}{.3\textwidth}
\includegraphics[angle=270,scale=.6]{figures/2L_diagrams/ggZZ2L_6.eps}
\caption{}
\end{subfigure}
\caption{Representative diagrams for the LO and virtual NLO $gg\to ZZ$ amplitude.}
\label{fig:ggZZ_NLO_amps}
\end{figure}
After we set the stage in the previous chapters, including derivation
of known results for the single Higgs amplitudes and extending their
expansion to higher orders, we can now tackle the unknown QCD
corrections to $Z$ boson pair-production via massive quark loops in the
SM. Representative diagrams for the leading-order contribution are
illustrated in Fig.~\ref{sfig:ggZZ_LO_amp} and for the virtual
next-to-leading-order diagrams in Fig.~\ref{fig:ggZZ_NLO_amps}(b)-(f)
and Fig.~\ref{fig:twotriangles}, respectively.
These amplitudes were first studied for on-shell $Z$ bosons in
Ref.~\cite{Glover:1988rg}; more recently, the $Z$ decay and off-shell
effects were also calculated at
leading-order~\cite{Campbell:2013una}. Virtual
two-loop contributions with massless internal quark loops (and
subsequent $Z$ boson decay) became
available only recently\cite{Gehrmann:2014bfa,Cascioli:2014yka,Caola:2014iua,Gehrmann:2015ora,Caola:2015ila,vonManteuffel:2015msa}. Due
to the complexity of the computation and present technical limitations
no full two-loop correction to the amplitudes with \emph{massive}
internal quarks is presently known. The authors
of~\cite{Melnikov:2015laa} made the first attempt in approximating the
virtual NLO corrections with internal top quarks. Their results,
however, includes only the first term of the $1/m$ expansion. At this
order contributions from the vector coupling of the $Z$ bosons to the
quarks are neglected completely. This is not necessarily troubling
since the vector coupling contribution is $a_f/v_f\sim 2.5$ times
smaller than the axial coupling contribution.
However, to fully incorporate the physics of the $Z$ boson
interactions and to give an estimate of power corrections $s/m^2$ we
compute the virtual two-loop corrections up to
${\cal O}\left(r_t^{-7}\right)$. We keep the $Z$ bosons on-shell, sum over
their polarisations and project onto the tensor structure of the
$gg\to H\to ZZ$ amplitude
(\Eqno{eq:ggHZZ_amplitude_structure}) since we are only
interested in the interference of both.
This chapter is structured as follows: In Sec.~\ref{sub:ggZZ_preliminaries} we give our definitions of the SM $ZZ$ amplitude, as far as the conventions differ from Sec.~\ref{sub:ggH_preliminaries}. The leading-order and next-to-leading-order results are presented in Sec.~\ref{sub:ggZZ_large_mass_expansion_at_one_loop} and Sec.~\ref{sub:ggZZ_large_mass_expansion_at_two_loop}, respectively. The latter is divided into two parts; the first consists of diagrams where both $Z$ bosons couple to one fermion line and the second handles \emph{anomaly style} diagrams where a single $Z$ boson is connected to one fermion string.
\subsection{Preliminaries}
\label{sub:ggZZ_preliminaries}
The on-shell $Z$ boson pair-production in gluon-gluon fusion
\begin{equation}
g(p_1,\mu,A)+g(p_2,\nu,B) \to Z(p_3,m_Z,\alpha)+Z(p_4,m_Z,\beta) \, ,
\end{equation}
via the heavy top quark loop can be completely expressed in terms of kinematical invariants
\begin{equation}
p_3^2=m_Z^2=p_4^2\,,\quad s = (p_1+p_2)^2\,,\quad t=(p_1-p_3)^2\,,\quad u=(p_2-p_3)^2 \text{ and }\; s+t+u=2m_Z^2 \, ,
\end{equation}
or equivalently, using the on-shellness condition, by the rescaled variables
\begin{align}
r_t &= \frac{m^2}{s}\,,\quad r_Z = \frac{m_Z^2}{s}\,,\quad
x = \frac{m_Z^2-t}{s} = \frac{p_1p_3}{p_1p_2}\;\;\text{and}\;\; \tilde{x} = (1-x)x\,.
\label{eq:massiveggZZ_x_definition}
\end{align}
The SM continuum amplitudes
$\ket{{\cal B}^{0,AB}_{\mu\nu\alpha\beta}(\alpha_S^0,m^0,\mu,\epsilon)}$ admit the same
perturbative expansion as given in \Eqno{eq:ggH_amp_bare_expansion}
for the Higgs-mediated process. The bare amplitudes are renormalized
in accordance with Eqs.~(\ref{eq:ggH_renormalisation_definition}-\ref{eq:Zop_as}),
omitting the superfluous Higgs vertex renormalisation. As mentioned
earlier we project onto the tensor and color structure of the
Higgs-mediated amplitude (\Eqno{eq:ggHZZ_amplitude_structure})
with
\begin{equation}
\label{eq:massiveggZZ_projected_amplitude}
\ket{{\cal B}^0(\alpha_S^0,m^0,\mu,\epsilon)} = \frac{\delta^{AB}}{N_A} (g^{\mu\nu}\, p_1p_2-p_2^\mu p_1^\nu)\; P_Z^{\alpha\rho'}(p_3) P_{Z,\rho'}^{\beta}(p_4) \ket{{\cal B}^{0,AB}_{\mu\nu\alpha\beta}(\alpha_S^0,m^0,\mu,\epsilon)}\,,
\end{equation}
where $N_A=N_c^2-1=8$ and $P_{Z,\alpha\beta}(p)$ from \Eqno{Zpolsum}.\\
We shall consider a single quark of flavor $f$ to be circulating in the quark loop. The Standard Model coupling of this fermion to a $Z$ boson is given by,
\begin{equation}
-i \frac{g_W}{2 \cos \theta_W} \gamma^\mu \left(v_f - a_f \gamma_5\right),\;\;v_{f} = \tau_f -2 Q_f \sin^2 \theta_W,\;\;a_{f} = \tau_f,\;\;\tau_f =\pm \frac{1}{2}\, .
\end{equation}
The superposition of vector and axial coupling allows to write the scattering amplitude as
\begin{equation}
\label{eq:massiveggZZ_normalisation}
\ket{{\cal B}^0(\alpha_S^0,m^0,\mu,\epsilon)} = {\cal N} \; \left( v_f^2 \ket{\TW{{\cal B}}_{VV}^0(\alpha_S^0,m^0,\mu,\epsilon)} + a_f^2 \ket{\TW{{\cal B}}_{AA}^0(\alpha_S^0,m^0,\mu,\epsilon)} \right)\,,
\end{equation}
where we factored out the normalisation factor from \Eqno{eq:Ndef}. The mixed coupling structure $v_f a_f$ vanishes due to charge parity conservation. With the amplitudes outlined above it is straightforward to compute the interference.
\begin{align*}
\label{eq:general_H_ZZ_interference}
{\cal B}_{gg} &= 2{\rm Re}\Bigl\{ \bra{{\cal A}_{\alpha\beta\rho'\sigma'}^{AB}(\alpha_S^{(n_l)},m,\mu,\epsilon)}\ket{{\cal B}^{AB,\alpha\beta\rho\sigma}(\alpha_S^{(n_l)},m,\mu,\epsilon)}\; P^{\rho'}_{Z,\rho}(p_3) P^{\sigma'}_{Z,\sigma}(p_4) \Bigl\} \\
&= 2{\rm Re}\Bigl\{ {\cal N}^* \frac{8}{3} \frac{s\, N_A}{s-m_H^2} \bra{{\cal A}(\alpha_S^{(n_l)},m,\mu,\epsilon)}\ket{{\cal B}(\alpha_S^{(n_l)},m,\mu,\epsilon)} \Bigl\} \numberthis \\
&= |{\cal N}|^2 \frac{16}{3} \frac{s\, N_A}{s-m_H^2} \; {\rm Re}\Bigl\{ \bra{{\cal A}(\alpha_S^{(n_l)},m,\mu,\epsilon)} \left[v_f^2 \ket{\TW{{\cal B}}_{VV}(\alpha_S^{(n_l)},m,\mu,\epsilon)} + a_f^2 \ket{\TW{{\cal B}}_{AA}(\alpha_S^{(n_l)},m,\mu,\epsilon)} \right] \Bigl\}\,.
\end{align*}
Writing \Eqno{eq:general_H_ZZ_interference} in this way establishes that ${\cal A}(\alpha_S^{(n_l)},m,\mu,\epsilon)$ and ${\cal B}(\alpha_S^{(n_l)},m,\mu,\epsilon)$ are dimensionless quantities, i.e. we compute ${\cal B}(\alpha_S^{(n_l)},m,\mu,\epsilon)$ for $s=1$ in the following.
\subsection{Projected Exact Result at One Loop}
\label{sub:projected_exact_result_at_one_loop}
The leading-order amplitude for the SM continuum production of two $Z$
bosons is known exactly in $d=4-2\epsilon$ dimensions. The usual
normalisation factor \Eqno{eq:loop_measure_factors} is chosen. We
split the result, according to \Eqno{eq:massiveggZZ_normalisation},
into vector-vector ($VV$) and axial-axial ($AA$) contribution.
\begin{align*}
\label{eq:ggZZ_LO_exact_VV}
\ket{\TW{{\cal B}}_{VV}^{(1)}(r_t,\mu,\epsilon)} &= S_\epsilon\, c_\Gamma \cdot 2\Big\{
%
4 \epsilon(1-\epsilon) B_{\{1,2\}} + 2\epsilon \left(B_{\{1,3\}} + B_{\{2,3\}} - 2 B_{\{3\}}\right) \numberthis \\
%
&\hspace{-1cm}+ s C_{\{1,2\}} \left[ 8 r_t + 2\epsilon (1-4 r_t)- 2 \epsilon^2 \right]
%
+ s C_{\{23,1\}} \left[2 (1-4 r_t-2 r_Z) (1-x) - 4\epsilon (1-r_Z) (1-x) \right] \\
%
&\hspace{-1cm}+ s C_{\{12,3\}} \left[ \epsilon \left(2 (1-4 r_t-2 r_Z) -2\epsilon (1-2 r_Z) \right) \right]
%
+ s C_{\{1,3\}} \left[ 2 (1-4 r_t-2 r_Z) x - 4 \epsilon (1-r_Z) x \right] \\
%
&\hspace{-1cm}+ s^2 D_{\{1,2,3\}} \left[ 4 r_t (1-2 r_t-r_Z) + \epsilon \left((1-4 r_t) (1-r_Z)-x \right) + \epsilon^2 \left(-1+r_Z+x \right) \right] \\
%
&\hspace{-1cm}+ s^2 D_{\{2,1,3\}} \left[4 r_t (1-2 r_t-r_Z) + \epsilon \left( 4 r_t (-1+r_Z)-r_Z+x \right) + \epsilon^2 \left( r_Z-x \right) \right] \\
%
&\hspace{-1cm}+ s^2 D_{\{1,3,2\}} \left[(1-4 r_t-2 r_Z) \left(2 r_t-r_Z+x-x^2\right) +\epsilon \left( 4 r_t (-1+r_Z)+(1-2 r_Z) (r_Z-(1-x) x) \right)\right. \\
&\hspace{-1cm}+\left. \epsilon^2 \left(r_Z-(1-x) x \right) \right]
%
\Big\}\,.
\end{align*}
\begin{align*}
\label{eq:ggZZ_LO_exact_AA}
\ket{\TW{{\cal B}}_{AA}^{(1)}(r_t,\mu,\epsilon)} &= \ket{\TW{{\cal B}}_{VV}^{(1)}(r_t,\mu,\epsilon)} + S_\epsilon\, c_\Gamma \cdot 2r_t \Big\{
%
s C_{\{1,2\}} \left[(2-4 r_Z)/r_Z^2 \right] \numberthis \\
%
&\hspace{-1cm}+ s C_{\{23,1\}} \left[ 4(1-6 r_Z)(-1+x)/r_Z - 16 \epsilon (1-x) \right]
%
+ s C_{\{12,3\}} \left[ \epsilon \left(24+(2-8 r_Z)/r_Z^2 -16 \epsilon \right) \right] \\
%
&\hspace{-1cm}+ s C_{\{1,3\}} \left[4 \left(6-1/r_Z\right) x -16 \epsilon x \right] \\
%
&\hspace{-1cm}+ s^2 D_{\{1,2,3\}} \left[ -4+24 r_t+ 2 r_t/r_Z^2- 8 r_t/r_Z + \epsilon \left( 10-16 r_t+(1-x)/r_Z^2 - (3-2 x)/r_Z \right) -4\epsilon^2 \right] \\
%
&\hspace{-1cm}+ s^2 D_{\{2,1,3\}} \left[-4+24 r_t+2 r_t/r_Z^2 - 8 r_t/r_Z + \epsilon \left(10-16 r_t - (1+2 x)/r_Z + x/r_Z^2 \right) - 4\epsilon^2 \right] \\
%
&\hspace{-1cm}+ s^2 D_{\{1,3,2\}} \left[ 2 r_t/r_Z^2 - 12 r_Z - (8 r_t+2(1-x) x)/r_Z+2 \
\left(1+12 r_t+6 x-6 x^2\right) \right. \\
&\hspace{-1cm}+\left. \epsilon \left( 8 r_Z - 2 \left(8 r_t-(1-2 x)^2\right)+ (1-x)x/r_Z^2 - (1-2 x+2 x^2)/r_Z \right) -4 \epsilon^2 \right]
%
\Big\}\,.
\end{align*}
The notation for the scalar integrals $B,C$ and $D$ is given in Table~\ref{tab:ggZZ_int_def}.
\begin{table}
\begin{center}
\begin{tabular}{|l|l||l|l||l|l|}
\hline
$D_{\{1,2,3\}}$ &$D_0(q_1,q_2,q_3;m,m,m,m)$ &$C_{\{1,2\}}$ &$C_0(q_1,q_2;m,m,m)$ & $B_{\{1,2\}}$ &$B_0(q_{12};m,m)$ \\
$D_{\{1,3,2\}}$ &$D_0(q_1,q_3,q_2;m,m,m,m)$ &$C_{\{1,3\}}$ &$C_0(q_1,q_3;m,m,m)$ & $B_{\{1,3\}}$ &$B_0(q_{13};m,m)$ \\
$D_{\{2,1,3\}}$ &$D_0(q_2,q_1,q_3;m,m,m,m)$ &$C_{\{12,3\}}$ &$C_0(q_{12},q_3;m,m,m)$ & $B_{\{2,3\}}$ &$B_0(q_{23};m,m)$\\
& &$C_{\{23,1\}}$ &$C_0(q_{23},q_1;m,m,m)$ & $B_{\{3\}}$ &$B_0(q_3;m,m)$\\
\hline
\end{tabular}
\end{center}
\caption{Scalar integrals occurring in full LO SM continuum $ZZ$ production. The momenta are defined as $q_1=p_1, q_2=p_2, q_3=-p_3$ and $q_{ij} = q_i+q_j$. The scalar integrals are defined in appendix~\ref{Intdef}.}
\label{tab:ggZZ_int_def}
\end{table}
We re-introduced factors of $s$ in \Eqno{eq:ggZZ_LO_exact_VV} and \Eqno{eq:ggZZ_LO_exact_AA} to indicate the correct dimensionality of the expressions.
We note that, in contrast to the case where the $Z$ bosons are off-shell and their decays included,
these formulae for the interference take a very simple form. Eq.~(\ref{eq:ggZZ_LO_exact_VV},\ref{eq:ggZZ_LO_exact_AA})
extend the results of Ref~\cite{Campbell:2014gua} to include the terms of order $\epsilon^1$ and $\epsilon^2$.
\subsection{Large-Mass Expansion at One Loop}
\label{sub:ggZZ_large_mass_expansion_at_one_loop}
Equivalently, \Eqno{eq:ggZZ_LO_exact_VV} and \Eqno{eq:ggZZ_LO_exact_AA} can be expressed by means of the large-mass expansion. The result for the vector-vector part yields
\input{src/LMEs/ggZZ_1L_VV.tex}
The result for the axial-axial part is
\input{src/LMEs/ggZZ_1L_AA.tex}
The leading term in the vector-vector expansion is sub-dominant with respect to the axial-axial part. The reason for this difference has been given in~\cite{Melnikov:2015laa}.
\subsection{Large-Mass Expansion at Two Loops}
\label{sub:ggZZ_large_mass_expansion_at_two_loop}
The two-loop SM continuum amplitude consists in total of $93+16$
non-zero diagrams. $93$ diagrams belong to topologies where both $Z$
bosons couple to the same fermion string, as illustrated in
Fig.~\ref{fig:ggZZ_NLO_amps}. Due to momentum conservation and
assuming an anti-commuting $\gamma_5$ in $d$-dimensions, no $\gamma_5$
contribution arises in the fermion traces of the respective
diagrams. The large-mass expansion results for the vector-vector and
axial-axial part of these diagrams are shown in
Sec.~\ref{ssub:non_anomalous_diagrams}.
The remaining $16$ \emph{anomaly style} diagrams belong to the
topology shown in Fig.~\ref{fig:twotriangles}, where the $Z$ bosons
couple to distinct fermion lines. These diagrams must, in principle,
be handled with care when using dimensional regularisation due to the
non-conservation of the axial-current. Furthermore, contributions from
each quark-doublet have to be considered simultaneously. Only the sum
over one quark-doublet leads to a gauge anomaly free theory. In case
of massless quark doublets all contributions vanish and we only have
to consider the third-generation quark doublet, i.e. top and bottom
quarks. Results for these diagrams are presented in
Sec.~\ref{ssub:anomalous_diagrams}.
\subsubsection{Non-Anomalous Diagrams}
\label{ssub:non_anomalous_diagrams}
In this section we give explicit formulae for the large-mass expansions for the sum of the $93$ anomaly free diagrams. Including again only mass renormalisation, setting $\ket{\bar{{\cal B}}_{XX}^{(1)}(r_t,\mu,\epsilon)} = \left(S_\epsilon c_\Gamma \,(\mu^2/m^2)^\epsilon \,\right)^{-1} \ket{\TW{{\cal B}}_{XX}^{(1)}(r_t,\mu,\epsilon)} $ and $\log(-r_t)=\log(m^2/(-s-i\epsilon))$, we can write the divergent two-loop $VV$ part as
\input{src/LMEs/ggZZ_2L_VV.tex}
And the $AA$ part as
\input{src/LMEs/ggZZ_2L_AA.tex}
The leading term in \Eqno{eq:LME_ggZZ_2L_AA} can be compared to the projected results of~\cite{Melnikov:2015laa}. We find agreement with
their formula\footnote{Both Eq.(5) and Eq.(7) of~\cite{Melnikov:2015laa} contain typographical errors.}. We also performed a consistency check of the renormalisation scale dependence of the presented two-loop expansions by means of the technique given in Sec.~\ref{sec:scale_dependence_of_the_finite_remainder}.
\input{src/anombit.tex}
\subsection{Visualisation of Large-Mass Expansion Results for $gg\to ZZ$}
\label{sub:visualisation_of_large_mass_expansion_results_for_gg_to_zz}
Let us turn towards the graphical representations of the large-mass
expansion results for the SM continuum, Eqs.~(\ref{eq:LME_ggZZ_1L_VV}-\ref{eq:LME_ggZZ_2L_AA}), and their
improvements. We proceed analogously to
Sec.~\ref{ssub:ggH_comparison_lme_with_full_result} and compute the
UV+IR renormalised version of \Eqno{eq:general_H_ZZ_interference} and
again integrate over the $ZZ$ phase space. The setup from
\Eqno{eq:ggH_input_parameters} is utilised. Since we focus our
discussion in this section mainly on the different improvements of the
large-mass expansions we, again, do not take into account the full NLO
correction. We merely focus on the unknown virtual massive two-loop
contribution of the SM continuum interfered with the Higgs-mediated
process. That is, we set
\begin{equation}
\label{eq:ggZZ_sigma_defs_virt}
\sigma_\text{int}^\text{LO} \sim 2\,{\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal B}^{(1)}(m,\mu)} \quad \text{and} \quad \sigma_\text{virt,int}^\text{NLO} \sim 2\,{\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal B}^{(2)}(m,\mu)} \,,
\end{equation}
which also excludes the \emph{anomaly style} contribution from eq~\eqref{eq:anomaly_result_full} since this part can be computed without the necessity of any approximation.
It is important to notice the following conventions for our
approximations using \emph{Pad\'e approximants} below. As in
Sec.~\ref{ssub:ggH_comparison_lme_with_full_result} the \emph{Pad\'e approximants}
are computed at amplitude level for each finite
remainder ${\cal F}_{{\cal A},{\cal B}}$, including the conformal
mapping\footnote{Computing the \emph{Pad\'e approximants} for the
expanded product $\bra{{\cal F}_{\cal A}^{(1)}}\ket{{\cal F}_{\cal B}^{(1,2)}}$ yield no
reasonable result above threshold. We have checked this by
explicitly computing the \emph{homogeneous bivariate Pad\'e Approximants}
$[2/2]$-$[3/3]$~\cite{Cuyt1979,Guillaume2000197} for
the LO interference ${\rm Re}\bra{{\cal F}_{\cal A}^{(1)}}\ket{{\cal F}_{\cal B}^{(1)}}$,
where we treated the mapped variable $w$,
\Eqno{eq:conformal_mapping}, and its complex conjugated $\bar{w}$ as
independent variables.}. We know from our previous discussion that
the best approximation of the LO as well as the virtual NLO
contribution of the Higgs-mediated process is given by
${\cal F}_{{\cal A},[5/5]}^{(1,2)}$. It is understood that we will always use
this approximant in the following considerations. In principle, we can
also substitute the approximated Higgs-mediated amplitude
${\cal F}_{{\cal A},[5/5]}^{(1)}$ with its exact LO result. Doing so would
remove any uncertainties from the Higgs-mediated contribution. On the
other hand the numerical difference between both approaches is
negligible as discussed in
Sec.~\ref{ssub:ggH_comparison_lme_with_full_result}.
The vector-vector part of the SM continuum gives only a minor contribution to the total cross section, $\sigma_{VV}/\sigma_{AA}\sim 10^{-3}$. This relies on the fact that the mass expansion of the $VV$ part starts only at $1/m^4$ whereas the $AA$ part starts at $1/m^2$ and additionally $a_t^2/v_t^2 \sim 7$.
The interference including the exact top mass dependence is only known
at leading-order, which is shown in the left panel of
Fig.~\ref{fig:ggZZ_LO_NLO_LME_Pade}. Comparing the exact result
(black) and its naive large-mass approximation up to $1/m^{12}$
(orange) shows excellent agreement up to $s\sim 3 m^2$, with
approximately $1\%$ deviation from the exact result. At threshold the
deviation rises to $12\%$. In contrast the \emph{Pad\'e approximant}
${\cal F}_{{\cal B},[3/3]}^{(1)}$ (blue) deviates from the exact result by $6\%$
at threshold. The shaded grey area indicates the variation from
computing the \emph{Pad\'e approximants} $[2/2],[2/3],[3/2],[3/3]$
with $3-8\%$ deviation at threshold. Due to the change of sign of
their derivatives we get a better approximation closely above
threshold, as can be seen in the bottom plot of
Fig.~\ref{fig:ggZZ_LO_NLO_LME_Pade}. Nevertheless, the peak of the
exact LO result at $s\sim 5.2 m^2$ is with $10-11\%$ deviation quite
poorly approximated. Ineptly this is the region of interest for our
later analysis of the Higgs boson width. Going to large values of $s$
the deviations inevitably become larger, but the contribution to the
cross section is small due to the suppression by the flux.
This situation seems to continue in case of the next-to-leading-order
large-mass expansion as shown in the right panel of
Fig.~\ref{fig:ggZZ_LO_NLO_LME_Pade}. Evidently no exact result is
available and we have to rely on the approximate results. All
\emph{Pad\'e approximants} $[2/2],[2/3],[3/2],[3/3]$ for
${\cal F}_{\cal B}^{(2)}$ show a stable trend over the entire $s/m^2$ range. The
deviations between the diagonal and non-diagonal \emph{Pad\'e
approximants} are again indicated by the shaded grey area and the
approximant ${\cal F}_{{\cal B},[3/3]}^{(2)}$ is shown in orange. The steeper
rise near the top threshold suggests a better description of the
actual threshold properties of the NLO result with exact top mass
dependence in contrast to the naive large-mass expansion
(black). Comparing the trend above threshold with its analogous LO
situation we can only guess that we have to expect comparable
deviations from our \emph{Pad\'e approximations} with respect to the
unknown exact NLO result.
\begin{figure}[h!]
\centering
\begin{subfigure}{0.495\textwidth}
\includegraphics[width=\textwidth]{figures/ggZZ_LO_LME_Pade.eps}
\end{subfigure}
\hfill
\begin{subfigure}{0.495\textwidth}
\includegraphics[width=\textwidth]{figures/ggZZ_NLO_LME_Pade.eps}
\end{subfigure}
\caption{\textbf{Left panel:} Leading-order interference ${\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal B}^{(1)}(m,\mu)}$. Exact result (black), LME up to $1/m^{12}$ (orange) and envelope of \emph{Pad\'e approximants} $[2/2],[2/3],[3/2]$ and $[3/3]$ (blue) as grey area. Bottom plot shows the relative deviation from the exact result. \textbf{Right panel:} Next-to-leading-order interference ${\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal B}^{(2)}(m,\mu)}$. LME up to $1/m^{12}$ (black) and envelope of \emph{Pad\'e approximants} $[2/2],[2/3],[3/2]$ and $[3/3]$ (orange) as grey area. The vertical dashed line denotes the top quark pair-production threshold. See text for details.}
\label{fig:ggZZ_LO_NLO_LME_Pade}
\end{figure}
\begin{figure}[h!]
\centering
\begin{subfigure}{0.495\textwidth}
\includegraphics[width=\textwidth]{figures/ggZZ_NLO_LME_rescaled.eps}
\end{subfigure}
\hfill
\begin{subfigure}{0.495\textwidth}
\includegraphics[width=\textwidth]{figures/ggZZ_NLO_Pade_rescaled.eps}
\end{subfigure}
\caption{Next-to-leading-order interference
${\rm Re}\bra{{\cal F}_{\cal A}^{(1)}(m,\mu)}\ket{{\cal F}_{\cal B}^{(2)}(m,\mu)}$.
\textbf{Left panel:} Interference by rescaling, \Eqno{eq:LME_rescaled}.
\emph{Pad\'e approximant} $[3/3]$ as comparison
(black). Envelope of $\sigma_{\text{imp},n}^\text{NLO}$ for $n=\{1,\ldots,6\}$
as grey area; $n=1$ (orange) and $n=6$ (blue) shown explicitly.
\textbf{Right panel:} Interference by alternative rescaling,
\Eqno{eq:LME_rescaled_Pade}. \emph{Pad\'e approximant}
$[3/3]$ as comparison
(black). Grey area given by envelope of $\sigma_{\text{imp},[n/m]}^\text{NLO}$
with $n,m=\{2,3\}$; $[3/3]$ shown explicitly (orange). The vertical dashed
line denotes the top quark pair-production threshold. See text for details.}
\label{fig:ggZZ_NLO_melnikov}
\end{figure}
We can also consider rescaling the NLO large-mass expansion as
described in \Eqno{eq:LME_rescaled}. The resulting curves are
shown in the left panel of Fig.~\ref{fig:ggZZ_NLO_melnikov}. To guide
the eye we also include ${\cal F}_{{\cal B},[3/3]}^{(2)}$ (black). The envelope
of the different orders $n$ in the expansion
$\sigma_{\text{imp},n}^\text{NLO}$ is shown as grey area. For $s\le
20m^2$ the envelope is determined from $n=\{1,\ldots,6\}$, whereas for
$s>20m^2$ we only use $n=\{1,\ldots,5\}$ due to the instabilities for
$n=6$ in the high energy regime. The most interesting curves, namely
the heavy-quark approximation $n=1$ and the highest order in the
expansion $n=6$, are shown in orange and blue, respectively. Factoring
out the exact LO result seems to give a more natural description of
the threshold behaviour and peak structure in comparison to the plain
use of the \emph{Pad\'e approximation}.
The origin of the numerical instabilities of the $n=6$ expansion is
probably due to delicate numerical cancellations in the $(s/m^2)^6$
coefficients. One could try to cure this problem by switching to a
higher numerical precision or by a proper \emph{economisation}~\cite{Press:2007:NRE:1403886} of the
power series. With the \emph{Pad\'e approximation} we already have an
excellent method at hand and we adopt the idea of factoring out the
exact LO interference,
\begin{equation}
\label{eq:LME_rescaled_Pade}
\sigma_{\text{imp},[n/m]}^\text{NLO} = \sigma_\text{exact}^\text{LO} \cdot \frac{\sigma_{[n/m]}^\text{NLO}}{\sigma_{[n/m]}^\text{LO}}\,.
\end{equation}
Keeping our usual definition in mind
$\sigma_{\text{imp},[n/m]}^\text{(N)LO}$ denotes the (virtual N)LO
contribution using ${\cal F}_{{\cal A},[5/5]}^{(1)}$ and
${\cal F}_{{\cal B},[n/m]}^{(1,2)}$. The result is shown in
Fig.~\ref{fig:ggZZ_NLO_melnikov}, right panel. We immediately see the
advantages of this approach. Firstly we also get a similar, more
natural behaviour at threshold and of the peak structure above
threshold. Secondly we get a stable result across the entire range of
$s/m^2$. The grey area is again given by the envelopes due to the
variation between the (non-)diagonal \emph{Pad\'e approximants}
$[2/2],[2/3],[3/2]$ and $[3/3]$(orange). Ultimately by using the
\emph{Pad\'e approximants} in contrast to \Eqno{eq:LME_rescaled}
we could entirely remove the uncertainty of having to use an
approximation for the involved Higgs-mediated amplitude and fall back
to using the exactly known result for ${\cal F}_{{\cal A}}^{(1)}$.
Some concluding remarks. In contrast to the purely Higgs-mediated case, Sec.~\ref{ssub:ggH_comparison_lme_with_full_result}, it turns out that we require the \emph{Pad\'e approximation} in the interference case. Using the conformal mapping alone without an additional \emph{Pad\'e approximant} on top gives no reasonable approximation for the quantities discussed above. On the other hand we have seen that we hugely benefit by using \emph{Pad\'e approximations} due to their stability and the possibility of removing any uncertainty besides the approximated virtual massive two-loop $gg\to ZZ$ amplitude.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,587
|
Laura Giordano (born 9 June 1979 in Palermo, Italy) is an Italian lyric soprano.
Life and career
Born in Palermo, Laura Giordano made her operatic debut at a very young age in Noye's Fludde by Britten. She went on to sing in numerous opera houses and festivals in Europe, Asia and North and South America, including Teatro alla Scala, the Salzburg Festival, Opéra National de Paris, the Rossini Opera Festival in Pesaro, and the Santa Fe Opera in New Mexico.
The core of Giordano's repertory is formed by roles in the operas of Donizetti, Verdi, Mozart and Puccini. Her Donizetti roles include Norina (Don Pasquale) performed with Riccardo Muti at the Ravenna Festival, at Théâtre des Champs-Elysées in Paris, in Liège, Cologne, Como, Moscow and Saint Petersburg, and in further performances in Santiago de Chile, Treviso, at Teatro Massimo Bellini in Catania, and in Ferrara; Adina (L'elisir d'amore) in Las Palmas, Tel Aviv, Maribor and Oderzo; Marie (La fille du régiment) at Teatro Massimo in Palermo, in Lecce, and on a tour with Teatro Comunale di Bologna at the Savonlinna Opera Festival; and Betly (Betly) at Konzerthaus in Berlin.
Her performances in Verdi's operas include Falstaff (Nannetta) in Milan conducted by Riccardo Muti, and in Strasbourg, Rome, Brussels, Lyon, at the Santa Fe Opera, at Teatro Regio di Torino, and at Berwaldhallen in Stockholm; Rigoletto (Gilda) at Seoul Arts Center, and in Fano and Ancona; Un ballo in maschera (Oscar) conducted by Antonio Pappano at Accademia Nazionale di Santa Cecilia, under the baton of Valery Gergiev at Teatro Regio di Parma, and in Nice and Trieste; and Don Carlo (Tebaldo) conducted by Zubin Mehta at Maggio Musicale Fiorentino.
Amongst her Mozart roles are Pamina (Die Zauberflöte) in Palermo; Donna Anna (Don Giovanni) in Palermo, in Fano, and at Piccolo Festival FVG; Susanna (Le nozze di Figaro) at Teatro Regio di Parma, Semperoper in Dresden Teatro Carlo Felice in Genoa, and in Reggio Emilia; Zerlina (Don Giovanni) at Opernhaus Zürich; Despina (Così fan tutte) in Palermo, Catania and Zürich; and Barbarina (Le nozze di Figaro) at Opéra Bastille in Paris.
She sang Musetta in Puccini's La bohème multiple times, including performances at Teatro alla Scala, Teatro Real in Madrid, in Catania, and at Opéra Royal de Wallonie. She has also sung Lauretta in Puccini's Gianni Schicchi in a performance in Lisbon.
Giordano's first Bellini role was Elvira in I puritani at Teatro Massimo Bellini in Catania.
Her other significant roles, several of them in rarely performed operas, have been: Livietta (Il ritorno di Don Calandrino) conducted by Riccardo Muti at the Salzburg Festival, in Ravenna, in Las Palmas in Pisa, and in Piacenza; Corinna (Il viaggio a Reims) in Pesaro, at Teatro Real de Madrid and at the Rossini in Wildbad opera festival; Coraline (Le toréador) in Palermo; Carolina (Il matrimonio segreto) at Théâtre des Champs-Elysées in Paris, the Barbican Centre in London, and in Palermo; Donna Fulvia (La pietra del paragone) at Théâtre du Châtelet in Paris; Elena (Il cappello di paglia di Firenze) at Maggio Musicale Fiorentino; Sylvie (La colombe) at the Accademia Musicale Chigiana; Vi (Blue Monday) at the Teatro Nacional de São Carlos in Lisbon; and Aminta (L'Olimpiade) at Festival de Beaune.
Giordano has performed as a soloist in Mozart's Requiem at St. John in the Lateran; Rossini's Stabat Mater in Palermo; Poulenc's Stabat Mater in Palermo; Schubert's Mass No. 6 at Festival de Saint-Denis; Mozart's Krönungsmesse at the Mozart Festival in La Coruña; Mahler's Symphony No. 4 in Kazan; and Orff's Carmina Burana at Teatro Fraschini in Pavia, in Ravenna, and at the Baths of Caracalla in Rome.
In 2010 Giordano appeared as a guest artist in Roberto Alagna's concert at dedicated to Luis Mariano. In 2014 she took part in the TV show hosted on Rai 1 by Enrico Brignano, performing excerpts from Rigoletto and La traviata, and also made an appearance in the show as a vocal coach of Karin Proia. Moreover, the same year Giordano made her debut as an actress in her first movie, Laurus Nobilis.
Discography
DVD
La bohème, Giacomo Puccini. Opus Arte, OA 0961, 2006
La pietra del paragone, Gioachino Rossini. Naive, V5089, 2007
Don Pasquale, Gaetano Donizetti. Arthaus Musik, 101303, 2007
Riccardo Muti - Lezioni concerto, 5. La lezione: Muti incontra Cimarosa. Gruppo Editoriale L'Espresso S.p.A., 2009
Roberto Alagna Live. Deutsche Grammophon, LC00173 - 0762797 - EDV17, 2010
CD
Rossini Discoveries, Gioachino Rossini. Decca, 4702982, 2002
L'Olimpiade, Antonio Vivaldi. Opus 111, OP30316, 2003
Don Pasquale, Gaetano Donizetti. RAI Trade, RTTI 0002, 2008
Il ritorno di Don Calandrino, Domenico Cimarosa. Gruppo Editoriale L'Espresso S.p.A., 2009
References
External links
Laura Giordano (official website)
Laura Giordano (official YouTube channel)
Atelier Musicale Artists Management (management agency)
Operabase (performance schedule)
Interviews
Repubblica Tv (2015-10-17), Il Flauto Magico di Mozart debutta al Teatro Massimo
Messaggero Veneto (2015-06-30), Mi garba cantare a casa del doge
Repubblica Tv (2015-04-16), Cavalleria e Toréador, un dittico al Massimo di Palermo
Sicilia in Rosa (2015-02-16), Laura Giordano: "La musica? È arrivata come uno schiaffo"
Rai 1, Sottovoce – Laura Giordano (2012-07-25)
Living people
1979 births
Italian operatic sopranos
Musicians from Palermo
21st-century Italian singers
21st-century Italian women singers
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,146
|
{"url":"https:\/\/co-design.pop-coe.eu\/metrics\/global_efficiency.html","text":"# Global Efficiency\n\nThe Global Efficiency (GE) measures the overall quality of the parallelisation.\n\nTypically, inefficiencies in parallel code have two main sources: 1) overheads imposed by the parallel nature of a code (i.e., the Parallel Efficiency: PE); and 2) poor scaling of computation with increasing numbers of processes (i.e., the Computation Scaling: CompS). The Global Efficiency metric is the product of these two sub-metrics:\n\n$GE = PE \\times CompS$\n\nIn order to fully understand the formulas, you may also visit the glossary of the metrics terms.\n\nRelated programs: FFTXlib \u00b7 juKKR kloop \u00b7","date":"2022-05-25 23:05:09","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\": 0.48058953881263733, \"perplexity\": 3785.011436916471}, \"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-2022-21\/segments\/1652662594414.79\/warc\/CC-MAIN-20220525213545-20220526003545-00463.warc.gz\"}"}
| null | null |
cask 'torbrowser' do
version '6.0.1'
sha256 '7c0918b00ab3594c29be03942e2eb324e7c5531a105ece6238a42e4a43402d04'
url "https://dist.torproject.org/torbrowser/#{version}/TorBrowser-#{version}-osx64_en-US.dmg"
name 'Tor Browser'
homepage 'https://www.torproject.org/projects/torbrowser.html'
license :oss
gpg "#{url}.asc",
key_id: 'ef6e286dda85ea2a4ba7de684e2c6e8793298290'
app 'TorBrowser.app'
zap delete: [
'~/Library/Application Support/com.apple.sharedfilelist/com.apple.LSSharedFileList.ApplicationRecentDocuments/org.mozilla.tor browser.sfl',
'~/Library/Preferences/org.mozilla.tor browser.plist',
]
end
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,330
|
{"url":"https:\/\/math.stackexchange.com\/questions\/717124\/another-functional-equation-fx-f2x-f4x-lfloor-7x-rfloor","text":"# Another functional equation: $f(x) + f(2x) + f(4x) = \\lfloor 7x \\rfloor$\n\nI would like to find all continuous functions $f \\, : \\, \\mathbb{R} \\, \\longrightarrow \\, \\mathbb{R}$ such that :\n\n$$\\forall x \\in \\mathbb{R}, \\; f(x) + f(2x) + f(4x) = \\lfloor 7x \\rfloor \\tag{1}$$\n\nFollow-up :\n\nNow that I know there are no continuous functions satisfying $(1)$, I would like to find all functions $f \\, : \\, \\mathbb{R} \\, \\longrightarrow \\, \\mathbb{R}$, continuous at $0$, such that :\n\n$$\\forall x \\in \\mathbb{R}, \\; f(x) + f(2x) + f(4x) = x \\varphi(x) \\tag{2}$$\n\nwhere $\\displaystyle \\forall x \\in \\mathbb{R}, \\; \\varphi(x) = \\begin{cases} 1 & \\text{if } x \\in \\mathbb{Q} \\\\ 0 & \\text{if } x \\notin \\mathbb{Q} \\end{cases}$. I feel like there exist no such functions (but I might be mistaken). Here, both RHS and LHS are continuous at $x=0$. My try :\n\nIt is clear that $f(0)=0$. Since $\\varphi(qx)=\\varphi(x)$ for all $x \\in \\mathbb{R}$ and all $q \\in \\mathbb{Q}$, I think the following is true :\n\n$$f(8x) - f(x) = x \\varphi(x)$$\n\nwhich leads to :\n\n$$f(x) - f \\Big( \\frac{x}{8} \\Big) = \\frac{x}{8} \\varphi(x)$$\n\nwhich would lead to\n\n$$f(x) - f \\Big( \\frac{x}{2^{3k}} \\Big) = \\frac{x}{2^{3k}} \\varphi(x)$$\n\nAm I on the right track or is there an easier way ?\n\n\u2022 Unless I'm missing something, there is no such function. From continuity of $f$ and the functional equation, it would follow that $x\\mapsto\\lfloor7x\\rfloor$ is also continuous, which is not the case. \u2013\u00a0Dejan Govc Mar 18 '14 at 18:59\n\u2022 $f(x)+f(2x)+f(4x)$ is continuous. $\\lfloor 7x\\rfloor$ is not continuous. Contradiction. \u2013\u00a0Guy Mar 18 '14 at 19:04\n\u2022 Thanks, you're right ! I was thinking too complicated... \u2013\u00a0Odile Mar 19 '14 at 11:17\n\u2022 @DejanGovc post as answer? \u2013\u00a0user574848 May 24 at 11:21\n\u2022 @user574848: Done. \u2013\u00a0Dejan Govc May 24 at 12:17\n\nThere is no such function. From continuity of $$f$$ and the functional equation, it would follow that $$x\\mapsto\\lfloor7x\\rfloor$$ is also continuous, which is not the case.","date":"2019-07-22 03:13:14","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\": 2, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8144369721412659, \"perplexity\": 162.97160195659015}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"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-2019-30\/segments\/1563195527474.85\/warc\/CC-MAIN-20190722030952-20190722052952-00279.warc.gz\"}"}
| null | null |
package android.support.database.strategy;
import android.support.database.Behaviour;
import android.support.database.behaviour.InsertBehaviour;
/**
* Created by yexiaokang on 2017/8/19.
*/
public enum Insert implements Behaviour {
/**
* Use the following when no conflict action is specified.
* This is the normal insert behaviour
*/
NONE,
/**
* When a UNIQUE constraint violation occurs, the pre-existing rows that
* are causing the constraint violation are removed prior to inserting
* or updating the current row. Thus the insert or update always occurs.
* The command continues executing normally. No error is returned.
* If a NOT NULL constraint violation occurs, the NULL value is replaced
* by the default value for that column. If the column has no default
* value, then the ABORT algorithm is used. If a CHECK constraint
* violation occurs then the IGNORE algorithm is used. When this conflict
* resolution strategy deletes rows in order to satisfy a constraint,
* it does not invoke delete triggers on those rows.
* This behavior might change in a future release.
*/
REPLACE,
/**
* Same as {@link Insert#REPLACE}.
* <p>When replace a record, it will exclude the autoincrement field
*
* @see InsertBehaviour#includeAutoincrement()
*/
REPLACE_EXCLUDE_AUTOINCREMENT
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,508
|
Popular MP3
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TYPES OF INDUSTRIAL PIPES
January 23, 2019 by Jeremiah Miller
Industrial pipes are often cylindrical in shape. Industries having different sections contain different types of works, each section have their own requirements of machines and goods. Just like that, different parts of industries uses industrial pipes for different purpose, some of the sections use the pipes for the drainage of any liquid while some of the sections use those pipes as a channel for transport solid material. Industrial pipes Australia are rigid and often used in construction of architects. Industrial pipes needs to hold good quality of material in it otherwise it would cause a big loss as it would be risky enough not to be trusted. Industrial pipes can bear to a great limit of the motion of pressure having tightly fitted joints and connections, but also it all depends on the pipe which is designed for a particular task to hold a limited amount of pressure in it. Industrial pipes also hold a property of having an ability to bear temperature of the particular range, but that range is always enough to perform the operation undergoing. Moreover, industrial pipes can be flexed to a great extent, they are flexible enough but to a particular limit, otherwise it might break. Apart from that, industrial pipes can also be installed underground. Each of the industrial pipes is custom designed to perform the particular task that it is made for, they are ordered by the industries that also mention the purpose for the particular industrial pipe. Since industrial pipes are heat resistance, they are used in the operations that carry flammable gases. They come in all sizes and material, having an outstanding durability and they give really great results. Industrial pipes come in different types while some of them are as follow:
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Work Compensation Lawyers
January 8, 2019 by Jeremiah Miller
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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\section{Introduction}
Multiple-input multiple-output (MIMO) radar has found wide application in the past decades. By means of waveform diversity, MIMO radar allows significant improvement of performance to be made as compared with the conventional phased-array radar \citep{LS08}. There exist in the literature abundant works to investigate algorithms for target localization or to evaluate their performances in MIMO radar contexts \citep{JLL09,hong1,tang1} mostly under the umbrella of Gaussian clutter. The validity of the Gaussian clutter assumption is rooted in the central limit theorem and is realistic in the case of sufficiently large number of independent and identically distributed (i.i.d.) elementary scatterers. In applications of high-resolution radars, the radar clutter exhibits non-stationarity, and a Gaussian modeling of the clutter, be it white or colored, deviates heavily from the real data and thus is inadequate \citep{mim5}.
\nl{Non-Gaussian clutter scenarios have been first studied through $\alpha$-stable distribution and mixture noise distributions. Nevertheless, the so-called spherically invariant random process (SIRP) has, thanks to its ability to describe different scales of the clutter roughness and to incorporate various non-Gaussian distributions, become a favorite distribution family in the radar context \citep{mim5,mim12}. The latter is a two-scale, compound Gaussian process which is a product of two components: the {texture} and the {speckle}. The texture, which accounts for \bm{local power changes}, is the square root of a positive scalar random process, whereas the speckle, which accounts for a local scattering, is a complex Gaussian process.
Though abounding works have been dedicated to the estimation algorithms in the SIRP clutter context with zero mean observations \citep{mim14,mim15,mim17,gini1}, there are, to the best of our knowledge, few works dealing with jointly parameterized mean and parameterized covariance matrix in the context of SIRP clutter \citep{xzh4,xzh5,BAJ16}. Among these few works, we can, \bm{first}, cite the robust MUSIC (MUSIC-Tyler) based on a robust fixed point Tyler estimate of the covariance matrix \citep{Tyl87}, the robust covariation-based MUSIC (ROC-MUSIC) \citep{TN96}, adapted for $\alpha$-stable distribution, in which MUSIC method is applied to the covariation matrix instead of the estimated covariance matrix and the RG-MUSIC \citep{Cou15} based on the random matrix theory (namely, it takes into account the Marcenko-Pastur distribution of the eigenvalues of the covariance matrix to rectify its estimation). \bm{On the other hand,} the $\ell_{p}$-MUSIC \citep{ZSH13} \bm{is} based on $\ell_{p}$ norm minimization with $p<2$ in order to take into account impulsive noise. \bm{Finally, some algorithms rely on robust mixtures noise} as \citep{BAJ16,KS00}, in which the authors proposed respectively, ML based method in the presence of a mixture of K-distributed and Gaussian noise \citep{BAJ16} and ML based method in non-Gaussian noise with Gaussian mixtures \citep{KS00}.}
In this short communication, we focus on the direction-of-departure (DOD) and/or direction-of-arrival (DOA) estimation problems in the presence of SIRP noise/clutter, under an array processing model and a MIMO radar model. In \citep{xzh4,xzh5}, the authors designed estimators based on the Iterative Conditional Maximum Likelihood Estimator (ICdMLE) and the Iterative Joint Maximum Likelihood Estimator (IJMLE), which are, based on the \emph{conditional} likelihood of the observations on the texture realizations, and the \emph{joint} likelihood between the two, respectively. As a consequence, these two estimators are both \textit{eo ipso} suboptimal.
To overcome the algorithm suboptimality and the model limitations in \citep{xzh4,xzh5} (as the existence of only one Coherent Pulse Interval (CPI), the fact that DOD and DOA are assumed to share same values), we propose in this short communication an iterative ML estimator that is based on the marginal (\emph{exact}) observation likelihood for a general MIMO radar model under SIRP clutter, named the Iterative Marginal ML Estimator (IMMLE). As our derivations will show, the MIMO model in this paper after matched-filtering can be transformed into the same structure as the array processing model considered in \citep{xzh4}, meaning that the proposed IMMLE is directly applicable to the latter model without any further generalization.
\section{Model setup}\label{sec2}
\subsection{Observation model}
Consider a MIMO radar system with linear and possibly non-uniform arrays both at the transmitter and the receiver. Further assume that $K$ targets are illuminated by the MIMO radar, all modeled as far-field, narrowband, point sources \citep{LS08}. The radar output for the \textit{l}th pulse in a CPI, and after matched filtering in the case of transmission of orthogonal waveforms \citep{HBC08}, reads:
\begin{align}
\boldsymbol{Z}(l)=\frac{1}{\sqrt{T}}\boldsymbol{Y}(l)\boldsymbol{S}^H
&=\sum_{k=1}^{K}\sqrt{T}\alpha_{k}e^{2j\pi f_k l}\boldsymbol{a}_{(\mathcal R)}\left(\theta^{(\mathcal R)}_{k}\right)\boldsymbol{a}_{(\mathcal T)}^{T}\left(\theta^{(\mathcal T)}_{k}\right)\nonumber\\
&\quad +\boldsymbol{N}(l), \ \ \ \mbox{for} \ \ \ l=0,\dots,L-1
\label{1c}
\end{align}
where $L$ denotes the number of radar pulses per CPI; $\alpha_{k}$ and $f_k$ denote a complex coefficient proportional to the radar cross section (RCS) and the normalized Doppler frequency of the \textit{k}th target, respectively; $T$ is the number of snapshots per pulse, $\theta^{(\mathcal T)}_{k}$ and $\theta^{(\mathcal R)}_{k}$ represent the DOD and DOA of the \textit{k}th target, respectively; the transmit and receive steering vectors are defined as
$\boldsymbol{a}_{(\mathcal T)}(\theta^{(\mathcal T)}_{k})=[e^{j\frac{2\pi \sin\left(\theta^{(\mathcal T)}_{k}\right)}{\lambda}d_{1}^{(\mathcal T)}},\dots, e^{j\frac{2\pi \sin\left(\theta^{(\mathcal T)}_{k}\right)}{\lambda}d_{M}^{(\mathcal T)}}]^{T}$ and $\boldsymbol{a}_{(\mathcal R)}(\theta^{(\mathcal R)}_{k})=[e^{j\frac{2\pi \sin\left(\theta^{(\mathcal R)}_{k}\right)}{\lambda}d_{1}^{(\mathcal R)}},\dots, e^{j\frac{2\pi \sin\left(\theta^{(\mathcal R)}_{k}\right)}{\lambda}d_{N}^{(\mathcal R)}}]^{T}$, in which $M$ and $N$ represent the number of sensors at the transmitter and the receiver, respectively; $d_{i}^{(\mathcal T)}$ and $d_{i}^{(\mathcal R)}$ denote the distance between the $i$th sensor and the reference sensor for the transmitter and the receiver, respectively; $\lambda$ stands for the wavelength; $\boldsymbol{N}(l)$ denotes the received clutter matrix at pulse $l$; and $(\cdot)^{T}$ denotes the transpose of a matrix.
By stacking the output in Eq.~(\ref{1c}) into an $MN\times1$ vector denoted by $\boldsymbol{z}(l)$, we further have:
\begin{equation}
\boldsymbol{z}(l)=\text{vec}\left\{\boldsymbol{Z}(l)\right\}=\boldsymbol{A}\left(\boldsymbol{\theta}\right)\boldsymbol{v}(l)
+\boldsymbol{n}(l),\quad l=0,\dots,L-1,
\end{equation}in which
$\boldsymbol{A}\left(\boldsymbol{\theta}\right)=\left[\boldsymbol{a}\left(\theta^{(\mathcal T)}_{1},\theta^{(\mathcal R)}_{1}\right),\dots,\boldsymbol{a}\left(\theta^{(\mathcal T)}_{K},\theta^{(\mathcal R)}_{K}\right)\right]$ denotes the steering matrix after matched filtering, where $\boldsymbol{\theta}=\left[\theta^{(\mathcal T)}_{1},\theta^{(\mathcal R)}_{1},\dots,\theta^{(\mathcal R)}_{K}\right]^T$ is a vector parameter introduced to incorporate all the unknown DODs and DOAs of the targets, and, $
\boldsymbol{a}\left(\theta^{(\mathcal T)}_{k},\theta^{(\mathcal R)}_{k}\right)=\text{vec}\left\{\boldsymbol{a}_{(\mathcal R)}\left(\theta^{(\mathcal R)}_{k}\right)\boldsymbol{a}_{(\mathcal T)}^{T}\left(\theta^{(\mathcal T)}_{k}\right)\right\}
=\left(\boldsymbol{I}_M\otimes\boldsymbol{a}_{(\mathcal R)}\left(\theta^{(\mathcal R)}_{k}\right)\right)\boldsymbol{a}_{(\mathcal T)}\left(\theta^{(\mathcal T)}_{k}\right),$ in which $\boldsymbol{I}_M$ stands for the identity matrix of size $M$, and $\otimes$ denotes the Kronecker product; $\boldsymbol{v}(l)=\left[\sqrt{T}\alpha_{1}e^{2j\pi f_1 l},\dots,\sqrt{T}\alpha_{K}e^{2j\pi f_K l}\right]^T$; $\boldsymbol{n}(l)=\text{vec}\left\{\boldsymbol{N}(l)\right\}$ denotes the clutter vector after matched filtering at pulse $l$; and $\text{vec}\{\cdot\}$ stands for the vectorization of a matrix.
\subsection{Observation statistics}\label{IIB}
We model the clutter vectors $\boldsymbol{n}(l),\ l=0,\dots,L-1$ as independent, identically distributed (i.i.d.) Spherically Invariant Random Vectors (SIRVs), which can be formulated as the product of two components statistically independent of each other:
$\boldsymbol{n}(l)=\sqrt{\tau(l)}\boldsymbol{x}(l),\,l=0,\dots, L-1$, in which the texture terms $\tau(l)$, are i.i.d. positive random variables; the speckle terms $\boldsymbol{x}(l)$ are i.i.d. $MN$-dimensional circular complex Gaussian vectors with zero mean and second-order moments $\text{E}\left\{\boldsymbol{x}(i)\boldsymbol{x}^H(j)\right\}=\delta_{ij}\boldsymbol{\Sigma}$ where $\boldsymbol{\Sigma}$ denotes the speckle covariance matrix, $\text{E}{\{\cdot\}}$ is the expectation operator, $\delta_{ij}$ is the Kronecker delta. To avoid the ambiguity in the model arising from the scaling effect between the texture and the speckle, we assume that $\text{tr}\{\boldsymbol{\Sigma}\}=MN$, in which $\text{tr}\{\cdot\}$ denotes the trace.
In this paper, we mainly focus on two kinds of SIRP clutters that are prevalent in the literature, namely, the K-distributed and the t-distributed clutters. In both cases the texture is characterized by two parameters, the \emph{shape parameter} $a$ and the \emph{scale parameter} $b$:
\begin{itemize}
\item \textbf{K-distributed clutter}, in which $\tau(l)$ follows a \emph{gamma distribution} (denoted by $\tau(l)\sim \text{Gamma}(a, b)$), namely, $p(\tau(l);a,b)=\frac{1}{\Gamma(a)b^a}\tau(l)^{a-1}e^{-\frac{\tau(l)}{b}},$
in which $\Gamma(\cdot)$ denotes the gamma function.
\item \textbf{t-distributed clutter}, in which $\tau(l)$ follows an \emph{inverse-gamma distribution} (denoted by $\tau(l)\sim \text{Inv-Gamma}(a, b)$), thus, $
p(\tau(l);a,b)=\frac{b^a}{\Gamma(a)}\tau(l)^{-a-1}e^{-\frac{b}{\tau(l)}}.$
\end{itemize}
\subsection{Unknown parameter vector and likelihood function}
Under the assumptions above, the unknown parameter vector of our problem is given by:
\begin{equation}\label{xi}
\boldsymbol{\xi}=\left[\boldsymbol{\theta}^T,\Re\left\{\boldsymbol{\alpha}\right\}^T,
\Im\left\{\boldsymbol{\alpha}\right\}^T,\boldsymbol{f}^T,\boldsymbol{\zeta}^T,a,b\right]^T,
\end{equation}in which $\boldsymbol{\alpha}=\left[\alpha_1,\dots,\alpha_K\right]^T$ is a complex vector parameter including the RCS coefficients of all $K$ targets, $\boldsymbol{f}=\left[f_1,\dots,f_K\right]^T$ contains the normalized Doppler frequencies of the targets, $\boldsymbol{\zeta}$ is a $M^2N^2$-element vector containing the real and imaginary parts of the entries of the lower triangular part of $\boldsymbol{\Sigma}$, $\Re\{\cdot\}$ and $\Im\{\cdot\}$ denote the real and the imaginary part, respectively.
Let $\boldsymbol{z}=\left[\boldsymbol{z}^T(1),...,\boldsymbol{z}^T(L-1)\right]^T$ denotes the full observation vector after matched filtering, and $\boldsymbol{\tau}=\left[\tau(0),\dots,\tau(L-1)\right]^T$ represents the vector of texture realizations at all pulses. The full observation likelihood conditioned on $\boldsymbol{\tau}$ can be written as:
\begin{equation}\label{n1}
p\left(\boldsymbol{z} \left\mid \boldsymbol{\tau}; \boldsymbol{\bar{\xi}}\right. \right)=\prod_{l=0}^{L-1}\frac{\exp\left(-\frac{\vecnm{\boldsymbol{\rho}(l)}^2}{\tau(l)}\right)}
{\left|\pi\boldsymbol{\Sigma}\right|\tau^{MN}(l)};
\end{equation}in which $\boldsymbol{\bar{\xi}}=\left[\boldsymbol{\theta}^T,\Re\left\{\boldsymbol{\alpha}\right\}^T,
\Im\left\{\boldsymbol{\alpha}\right\}^T,\boldsymbol{f}^T,\boldsymbol{\zeta}^T\right]^T$ is the unknown parameter vector that does not contain the texture parameters $a$ and $b$, $\vecnm{\cdot}$ denotes the norm of a vector, and
\begin{equation}\label{rho}
\boldsymbol{\rho}(l)=
\boldsymbol{\Sigma}^{-1/2}\left(\boldsymbol{z}(l)-\boldsymbol{A}\left(\boldsymbol{\theta}\right)\boldsymbol{v}(l)\right),
\end{equation} which represents the clutter realization at pulse $l$ with its speckle spatially whitened.
The conditional likelihood in Eq.~(\ref{n1}), multiplied by $p(\boldsymbol \tau;a,b)$, leads to the joint likelihood between $\boldsymbol{z}$ and $\boldsymbol{\tau}$:
{\small \begin{equation}\label{n1a}
p\left(\boldsymbol{z}, \boldsymbol{\tau}; \boldsymbol{\xi} \right)
=p\left(\boldsymbol{z} \left\mid \boldsymbol{\tau}; \boldsymbol{\bar{\xi}}\right.\right)p(\boldsymbol \tau;a,b)
=\prod_{l=0}^{L-1}\frac{\exp\left(-\frac{\vecnm{\boldsymbol{\rho}(l)}^2}{\tau(l)}
\right)}{\left|\pi\boldsymbol{\Sigma}\right|\tau^{MN}(l)}p(\tau(l);a,b).
\end{equation}}
Finally, the full observation marginal (exact) likelihood, w.r.t. $\boldsymbol{\xi}$, is obtained by integrating out $\boldsymbol{\tau}$ from the joint likelihood in Eq.~(\ref{n1a}), as:
\begin{align}
p\left(\boldsymbol z; \boldsymbol{\xi} \right)=
\int_0^{+\infty}p\left(\boldsymbol z, \boldsymbol{\tau}; \boldsymbol{\xi} \right)\text{d}{\boldsymbol \tau}
&=\prod_{l=0}^{L-1}\int_{0}^{+\infty}\frac{\exp\left(-\frac{\vecnm{\boldsymbol{\rho}(l)}^2}
{\tau(l)}\right)}{\left|\pi\boldsymbol{\Sigma}\right|\tau^{MN}(l)} \nonumber\\
& \quad\times p(\tau(l);a,b)\text{d}\tau(l).\label{n1b}
\end{align}
\section{Iterative marginal maximum likelihood estimator}
The derivation procedure of the IMMLE is presented in this section. To begin with, let $\Lambda$ denote the marginal Log-Likelihood (LL) function, which is obtained from Eq.~(\ref{n1b}), as:
{\small \begin{equation}
\Lambda=\ln p\left(\boldsymbol{z}; \boldsymbol{\xi} \right)
=-LMN\ln\pi-L\ln|\boldsymbol{\Sigma}|+\sum_{l=0}^{L-1}\ln g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right),
\end{equation}}%
in which
{\small \begin{align}
g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)
=\int_{0}^{+\infty}\frac{\exp\left(-\frac{\vecnm{\boldsymbol{\rho}(l)}^2}{\tau(l)}
\right)}{\tau^{MN}(l)}
p(\tau(l);a,b)\text{d}\tau(l) \nonumber\\
=\left\{
\begin{aligned}
&\frac{2\vecnm{\boldsymbol{\rho}(l)}^{a-MN}K_{a-MN}\left(2\vecnm{\boldsymbol{\rho}(l)}/
b^\frac{1}{2}\right)}{b^\frac{MN+a}{2}\Gamma(a)},\text{K-distr. clutter}, \\
&\frac{b^a\Gamma(MN+a)}{\Gamma(a)\left(\vecnm{\boldsymbol{\rho}(l)}^2+b\right)^{MN+a}},\quad \text{t-distributed clutter},
\end{aligned}
\right.
\label{g}
\end{align}}
where $K_n(\cdot)$ is the modified Bessel function of the second kind of order $n$ (cf. \citep{tunaley10} for more details).
To begin with, we look for the estimates of the clutter parameters, i.e., of the speckle covariance matrix $\boldsymbol{\Sigma}$, and the texture parameters $a$ and $b$. Let $\hat{\boldsymbol{\Sigma}}$ denote the estimate of $\boldsymbol{\Sigma}$ when all the other unknown parameters are fixed, which can be obtained by solving the equation $\partial\Lambda/\partial\boldsymbol{\Sigma}=0$, as \citep{gini1}:
{\small \begin{equation}\label{Sig_es}
\hat{\boldsymbol{\Sigma}}=\frac{1}{L}\sum_{l=0}^{L-1}h_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)
\cdot\left(\boldsymbol{z}(l)-\boldsymbol{A}\left(\boldsymbol{\theta}\right)\boldsymbol{v}(l)\right)
\left(\boldsymbol{z}(l)-\boldsymbol{A}\left(\boldsymbol{\theta}\right)\boldsymbol{v}(l)\right)^H,
\end{equation}}%
in which
{\small \begin{align}
&h_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)
=-\frac{\frac{\partial g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}{\partial \vecnm{\boldsymbol{\rho}(l)}^2}}
{g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)} \nonumber \\
&=\left\{
\begin{aligned}
&\frac{K_{a-MN-1}\left(2\vecnm{\boldsymbol{\rho}(l)}/
b^\frac{1}{2}\right)}{b^\frac{1}{2}\vecnm{\boldsymbol{\rho}(l)}
K_{a-MN}\left(2\vecnm{\boldsymbol{\rho}(l)}/
b^\frac{1}{2}\right)},\text{K-distributed clutter}, \\
&\frac{MN+a}{\vecnm{\boldsymbol{\rho}(l)}^2+b},\quad \text{t-distributed clutter}.
\end{aligned}
\right.
\label{h}
\end{align}}%
Note that $\hat{\boldsymbol{\Sigma}}$ in Eq.~(\ref{Sig_es}) has an iterative nature, as can be seen from the expression of $\boldsymbol{\rho}(l)$ in Eq.~(\ref{rho}).
We further need to normalize $\hat{\boldsymbol{\Sigma}}$ to fulfill the assumption that $\text{tr}\{\boldsymbol{\Sigma}\}=MN$. Let $\hat{\boldsymbol{\Sigma}}_\text{n}$ denote the normalized estimate $\hat{\boldsymbol{\Sigma}}$, which is:
\begin{equation}\label{Sig_es_n}
\hat{\boldsymbol{\Sigma}}_\text{n}=MN
\frac{\hat{\boldsymbol{\Sigma}}}{\text{tr}\left\{\hat{\boldsymbol{\Sigma}}\right\}}.
\end{equation}
Similarly, the estimates of $a$ and $b$ when other unknown parameters are fixed, denoted by $\hat{a}$ and $\hat{b}$, can be found by equating $\partial\Lambda/\partial a$ and $\partial\Lambda/\partial b$ to zero, respectively, i.e., by solving numerically:
\begin{equation}
\label{a_es}
\frac{\partial\Lambda}{\partial a}=\sum_{l=0}^{L-1}\frac{j_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}
{g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}=0
\end{equation}
and
\begin{equation}
\label{b_es}
\frac{\partial\Lambda}{\partial b}=\sum_{l=0}^{L-1}\frac{k_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}
{g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}=0,
\end{equation}
w.r.t. $a$ and $b$, respectively, in which
{\small \begin{align}
&j_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)=\frac{\partial g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}{\partial a} \nonumber \\
&= \left\{
\begin{aligned}
&-\frac{1}{b^a\Gamma(a)}\int_0^{+\infty}\exp\left(-\frac{\vecnm{\boldsymbol{\rho}(l)}^2}{\tau(l)}
-\frac{\tau(l)}{b}\right)\tau(l)^{-MN+a-1}\\
&\cdot\left(\ln\left(\frac{b}{\tau(l)}\right)+\Psi(a)\right)\text{d}\tau(l),
\quad \text{K-distributed clutter},\\
&-\frac{b^a\Gamma(MN+a)\left(\ln\left(\frac{\vecnm{\boldsymbol{\rho}(l)}^2}{b}+1\right)-\Psi(MN+a)+\Psi(a)\right)}
{\Gamma(MN)\left(\vecnm{\boldsymbol{\rho}(l)}^2+b\right)^{MN+a}},\\
&\text{t-distributed clutter},
\end{aligned}
\right.
\end{align}}
where $\Psi(\cdot)$ denotes the digamma function, and
{\small \begin{align}
&k_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)=\frac{\partial g_{MN}\left(\vecnm{\boldsymbol{\rho}(l)}^2,a,b\right)}{\partial b} \nonumber \\
&=\left\{
\begin{aligned}
&\frac{1}{b^{a+2}\Gamma(a)}\int_0^{+\infty}\exp\left(-\frac{\vecnm{\boldsymbol{\rho}(l)}^2}{\tau(l)}
-\frac{\tau(l)}{b}\right)\\
&\cdot\tau(l)^{-MN+a-1}\cdot\left(\tau(l)-ab\right)\text{d}\tau(l), \text{K-distributed clutter},\\
&-\frac{ab^{a-1}\Gamma(MN+a)\left(-a\vecnm{\boldsymbol{\rho}(l)}^2+MNb\right)}
{\Gamma(a+1)\left(\vecnm{\boldsymbol{\rho}(l)}^2+b\right)^{MN+a+1}},\text{t-distr. clutter}.
\end{aligned}
\right.
\end{align}}%
Next, we consider the estimate $\hat{\boldsymbol{v}}(l)$, by solving $\partial\Lambda/\partial \boldsymbol{v}(l)=0$, which reads
\begin{equation}\label{v_es}
\hat{\boldsymbol{v}}(l)=\left(\tilde{\boldsymbol{A}}^H\left(\boldsymbol{\theta}\right)
\tilde{\boldsymbol{A}}\left(\boldsymbol{\theta}\right)\right)^{-1}
\tilde{\boldsymbol{A}}^H\left(\boldsymbol{\theta}\right)
\tilde{\boldsymbol{z}}(l),
\end{equation}in which
$\tilde{\boldsymbol{A}}\left(\boldsymbol{\theta}\right)=\boldsymbol{\Sigma}^{-\frac{1}{2}}\boldsymbol{A}\left(\boldsymbol{\theta}\right),
$ and
$ \tilde{\boldsymbol{z}}(l)=\boldsymbol{\Sigma}^{-1/2}\boldsymbol{z}(l),$ representing the steering matrix and the observation at pulse $l$, both pre-whitened by the speckle covariance matrix $\boldsymbol{\Sigma}$, respectively.
As the expressions in Eqs.~(\ref{Sig_es}), (\ref{a_es}), (\ref{b_es}) and (\ref{v_es}) suggest, the estimation of each of the parameters $a$, $b$, $\boldsymbol{\Sigma}$ and $\boldsymbol{v}(l)$ requires the knowledge of all the others of them, and furthermore the knowledge of the parameter vector $\boldsymbol{\theta}$. This mutual dependence between the unknown parameters makes it impossible to concentrate the LL function \emph{analytically}, i.e., to obtain a closed-form expression for the LL function concentrated w.r.t. each of the aforementioned parameters that is independent of the other ones. Instead, we resort to the so-called \emph{stepwise numerical concentration} approach.
This approach consists in concentrating the LL function iteratively, by assuming that certain parameters are known from the previous iteration. For the task under consideration, we assume, at each iteration, that $\hat{\boldsymbol{\Sigma}}$, $\hat{a}$ and $\hat{b}$ are known and use them to compute $\hat{\boldsymbol{v}}(l)$, which is then used in turn to update the values of $\hat{\boldsymbol{\Sigma}}$ and $\hat{a}$ and $\hat{b}$ to be used in the next iteration. This sequential updating procedure is repeated until convergence or a maximum iteration number is reached.
Next, we turn to the estimation of $\boldsymbol{\theta}$. The approach explained above allows us to drop all the constant terms in the LL function (including those terms that contain only $\boldsymbol{\Sigma}$, $a$ and $b$ as unknown parameters, as these are assumed to be known at each iteration). Furthermore, by inserting the expression of $\hat{\boldsymbol{v}}(l)$ in Eq.~(\ref{v_es}) into what remains in the LL function, we obtain the estimate of $\boldsymbol{\theta}$, denoted by $\hat{\boldsymbol{\theta}}$, as:
\begin{align}
\hat{\boldsymbol{\theta}}=
\left\{
\begin{aligned}
&\arg\min_{\boldsymbol{\theta}}\Bigg\{\sum_{l=0}^{L-1}\Big((MN-a)\ln\left(\vecnm{
\boldsymbol{P}_{\tilde{\boldsymbol{A}}(\boldsymbol{\theta})}^{\bot}\tilde{\boldsymbol{z}}(l)}\right)\\
& -\ln K_{a-MN}\left(\left.2\vecnm{\boldsymbol{P}_{\tilde{\boldsymbol{A}}(\boldsymbol{\theta})}^{\bot}
\tilde{\boldsymbol{z}}(l)}\middle/
b^\frac{1}{2}\right.\right)\Big)\Bigg\}, \text{K-distr. clutter},\\
&\arg\min_{\boldsymbol{\theta}}\left\{\sum_{l=0}^{L-1}\ln\left(\vecnm{\boldsymbol{P}_{\tilde{\boldsymbol{A}}
(\boldsymbol{\theta})}^{\bot}\tilde{\boldsymbol{z}}(l)}^2+b\right)
\right\}, \text{t-distributed clutter}.
\end{aligned}
\right.
\label{theta_es}
\end{align}in which
$\boldsymbol{P}_{\tilde{\boldsymbol{A}}(\boldsymbol{\theta})}^{\bot}=\boldsymbol{I}_{MN}
-\tilde{\boldsymbol{A}}(\boldsymbol{\theta})\left(\tilde{\boldsymbol{A}}^H(\boldsymbol{\theta})
\tilde{\boldsymbol{A}}(\boldsymbol{\theta})\right)^{-1}\tilde{\boldsymbol{A}}^H(\boldsymbol{\theta})
$ is the orthogonal projection matrix onto the null space of $\tilde{\boldsymbol{A}}(\boldsymbol{\theta})$.
Finally, the whole procedure of the IMMLE is summarized in Table.~1.
\begin{table*}[h!]
\begin{center}
\begin{tabular}{ll}
\hline \\[-8pt]
\ & The IMMLE procedures
\\[2pt]
\hline \\[-8pt]
\ \textbf{Initialization} & $i=0$, set $\hat{a}^{(0)}$, $\hat{b}^{(0)}$ to be two arbitrary positive numbers and $\hat{\boldsymbol{\Sigma}}^{(0)}_\text{n}=\boldsymbol{I}_{MN}$ \\[2pt]
\ \textbf{Step 1} & Iteration $i$, calculate $\hat{\boldsymbol{\theta}}^{(i)}$ from Eq.~(\ref{theta_es}) using $\hat{a}^{(i)}$, $\hat{b}^{(i)}$ and $\hat{\boldsymbol{\Sigma}}^{(i)}_\text{n}$ \\[2pt]
\ & Calculate $\hat{\boldsymbol{v}}^{(i)}(l)$ from Eq.~(\ref{v_es}) using $\hat{\boldsymbol{\theta}}^{(i)}$, $\hat{a}^{(i)}$, $\hat{b}^{(i)}$ and $\hat{\boldsymbol{\Sigma}}^{(i)}_\text{n}$ \\[2pt]
\ \textbf{Step 2} & Update $\hat{a}^{(i+1)}$ from Eq.~(\ref{a_es}) using $\hat{\boldsymbol{\theta}}^{(i)}$, $\hat{\boldsymbol{v}}^{(i)}(l)$, $\hat{\boldsymbol{\Sigma}}^{(i)}_\text{n}$, and $\hat{b}^{(i)}$ \\[2pt]
\ & Update $\hat{b}^{(i+1)}$ from Eq.~(\ref{b_es}) using $\hat{\boldsymbol{\theta}}^{(i)}$, $\hat{\boldsymbol{v}}^{(i)}(l)$, $\hat{\boldsymbol{\Sigma}}^{(i)}_\text{n}$, and $\hat{a}^{(i+1)}$ \\[2pt]
\ & Update $\hat{\boldsymbol{\Sigma}}^{(i+1)}_\text{n}$ from Eqs.~(\ref{Sig_es}) and (\ref{Sig_es_n}) using
$\hat{\boldsymbol{\theta}}^{(i)}$, $\hat{\boldsymbol{v}}^{(i)}(l)$, $\hat{a}^{(i+1)}$ and $\hat{b}^{(i+1)}$\\[2pt]
\ & Set $i\leftarrow i+1$ \\[2pt]
\ \textbf{Step 3} & Repeat Step 1 and Step 2 until convergence \\[2pt]
\hline
\end{tabular}
\caption{Summarization of the proposed algorithm}
\end{center}
\end{table*}
\noindent\textbf{Remark 1}: Let us recall the expression of $\hat{\boldsymbol{\theta}}$ for the Conventional ML Estimator (CvMLE), which treats the clutter as uniform white Gaussian distributed, denoted by CvMLE-U,
\begin{equation}\label{conv_ml}
\hat{\boldsymbol{\theta}}_{\rm CvMLE-U}= \arg\min_{\boldsymbol{\theta}}\sum_{l=0}^{L-1}\vecnm{\boldsymbol{P}_{{\boldsymbol{A}}
(\boldsymbol{\theta})}^{\bot}{\boldsymbol{z}}(l)}^2,
\end{equation}
as well as for both of the ICdMLE and IJMLE that we proposed in \citep{xzh4,xzh5}, which, adapted to the model in question, has the following expression,
{\small \begin{equation}\label{ICdMLE_ml}
\hat{\boldsymbol{\theta}}_{\rm ICdMLE/IJMLE}=\arg \min_{\boldsymbol{\theta}}\sum_{l=0}^{L-1}\dfrac{1}{\hat{\tau}_{\rm ICdMLE/IJMLE}(l)}
\vecnm{\boldsymbol{P}_{\tilde{\boldsymbol{A}}(\boldsymbol{\theta})}^{\bot}\tilde{\boldsymbol{z}}(l)}^2.
\end{equation}}%
Expression of $\hat{\boldsymbol{\theta}}_{\rm CvMLE-U}$ shows that the CvMLE-U considers simply the \emph{sum} of $\vecnm{\boldsymbol{P}_{{\boldsymbol{A}}
(\boldsymbol{\theta})}^{\bot}{\boldsymbol{z}}(l)}^2$ (the square of the norm of the projection of the observation at pulse $l$ onto the null space of the steering matrix), while the ICdMLE and IJMLE, as the expression of $\hat{\boldsymbol{\theta}}_{\rm ICdMLE/IJMLE}$ shows, consider the \emph{modified sum} of these terms (pre-whitened by the speckle covariance matrix, and weighted by the inverse of the texture realization at each pulse). It is precisely because of this modification that the ICdMLE and IJMLE gain their advantages in performance over the CvMLE-U. An iterative version of the CvMLE-U with no assumption on the covariance matrix, denoted by ICvMLE, can be easily derived, for which, the aforementioned conclusions remain valid for the ICvMLE. On the other hand, we can see from Eq.~(\ref{theta_es}) that the proposed IMMLE considers, instead of direct or modified sum of the projections, the sum of their \emph{logarithms} (modified by some algebraic operations), which is equivalent to the \emph{product} of them. Since a sum is small only if all its terms are small, while a product can be small even if only very few of its terms are small enough, we can conclude that underlying this contrast between summation and multiplication is a difference \textit{in essentia}, that the CvMLE-U, ICvMLE, ICdMLE and IJMLE treat all the pulses ``equally'', whereas the IMMLE focuses only on the ``best'' pulses. Due to space limitation, refer to Table. 2 for a concise comparison between the IMMLE, IJMLE and ICdMLE.
\begin{table}[h!]
\begin{center}
{\small
\begin{tabular}{|m{3cm}|m{1.6cm}|m{1.3cm}|m{1cm}|}
\hline
\ & ICdMLE & IJMLE & IMMLE\\
\hline
Likelihood & Conditional & Joint & Marginal\\
\hline
Texture modeling & Deterministic & \multicolumn{2}{c|}{Stochastic}\\
\hline
Considers $\boldsymbol{\tau}$ & \multicolumn{2}{c|}{Yes} & No\\
\hline
Considers $a$ and $b$ & No & \multicolumn{2}{c|}{Yes}\\
\hline
Numerical solution of equations & {No} & \multicolumn{2}{c|}{{Yes}}\\
\hline
Numerical integration & \multicolumn{2}{c|}{{No}} & {Yes}\\
\hline
Computational complexity & {Lowest} & {Higher than ICdMLE} & {Highest}\\
\hline
Iteration(s) required & \multicolumn{2}{c|}{{Two}} & {One}\\
\hline
Requires texture distribution & {No} & \multicolumn{2}{c|}{{Yes}}\\
\hline
Can be used for texture parameters estimation & {No} & \multicolumn{2}{c|}{{Yes}} \\
\hline
\end{tabular}}%
\caption{Comparison between ICdMLE, IJMLE and IMMLE }
\end{center}
\end{table}
\noindent\textbf{Remark 2}: As is clear from the procedure above, our algorithm does not entail the estimation of the RCS coefficients $\alpha_k$, and the normalized Doppler frequencies $f_k$, of the targets, but rather only involves estimating the vectors $\boldsymbol{v}(l)$, which are functions of them. Indeed, in applications where the estimation of those parameters are of interest, one can naturally find the ML or LS estimates of them by respectively equating an adequate cost function to zero, and then complement our algorithm accordingly. This, however, deviates from our topic, i.e., the DOD/DOA estimation, and due to space limitation, it is not to be discussed in this paper.
\noindent\textbf{Remark 3}: The convergence of the LL function is guaranteed by the fact that the value of the objective function to calculate $\hat{\boldsymbol{\theta}}$, $\hat{\boldsymbol{\Sigma}}$, $\hat{a}$ and $\hat{b}$ at each step can either improve or maintain but cannot worsen. As the simulations will show, the convergence of the estimates of the unknown parameters in $\boldsymbol{\theta}$ can be obtained by few iterations (one to two).
\noindent\textbf{Remark 4}: IMMLE has a computational complexity sightly higher than ICvMLE, ML-GM, IJMLE and CdMLE. Indeed, all of them possess a highly non-convex minimization step over a $2K$-dimensional parameter space, which is the most time-consuming stage compared to the updating steps of the speckle covariance matrix, $\boldsymbol{\Sigma}$ and the vector $\boldsymbol{v}$ (both of them mainly based on analytical expressions) and the potential numerical solving. Generally, a MUSIC-based algorithm has a lower complexity than the ML-based one, except for $\ell_p$-MUSIC algorithm where the signal/noise subspaces construction is time-consuming due to the $\ell_p$ norm minimization.
\section{Cram\'{e}r-Rao bound expression}
The CRB w.r.t. target direction parameters in a MIMO radar context in the presence of SIRP clutter has been derived in our previous works \citep{xzh3}, where we used an element-wise approach to calculate the Fisher information matrix (FIM). For the model considered in this paper, where the size of the unknown signal parameter vector (hence the dimension of the resulting FIM) is much larger, a block-wise expression for the CRB w.r.t. the signal DODs and DOAs (denoted by $\text{CRB}\left(\boldsymbol{\theta}\right)$) is required, the result of which is presented below.
The $2K\times 2K$ CRB matrix w.r.t. $\boldsymbol{\theta}$ in the presence of SIRP clutter is given by:
\begin{align}
\text{CRB}\left(\boldsymbol{\theta}\right)&=\left(\frac{2\kappa}{MN}\Re\left\{\sum_{l=0}^{L-1}\boldsymbol{H}^H(l)\tilde{\boldsymbol{D}}^H
\boldsymbol{P}_{\tilde{\boldsymbol{A}}(\boldsymbol{\theta})}^{\bot}\tilde{\boldsymbol{D}}\boldsymbol{H}(l)\right\}\right)^{-1} \nonumber \\
&=\frac{MN}{2\kappa L}\left(\Re\left\{\left(\tilde{\boldsymbol{D}}^H\boldsymbol{P}_{\tilde{\boldsymbol{A}}(\boldsymbol{\theta})}^{\bot}
\tilde{\boldsymbol{D}}\right)\odot\hat{\boldsymbol{P}}^T\right\}\right)^{-1},
\label{crb}
\end{align}in which
\begin{equation}\label{kappa}
\kappa=\left\{
\begin{aligned}
&\frac{\int_0^{+\infty}x^{MN+a-1}\frac{K_{a-MN-1}^2(x)}
{K_{a-MN}(x)}\text{d}x}{2^{MN+a-2}b\Gamma(MN)\Gamma(a)}, \quad \text{K-distributed clutter}, \\
&\frac{MNa(a+MN)}{b(a+MN+1)},\quad \text{t-distributed clutter},
\end{aligned}
\right.
\end{equation}where $K_n(x)$ is the modified Bessel functions of the second kind of order $n$, $\boldsymbol{H}(l)=\boldsymbol{I}_2\otimes\text{diag}\left\{\left[\boldsymbol{v}(l)\right]_1,\dots,\left[\boldsymbol{v}(l)\right]_K\right\}$, $\boldsymbol{J}_2$ is the all-ones matrix of size $2$,
$\hat{\boldsymbol{P}}=\frac{1}{L}\boldsymbol{J}_2\otimes\sum_{l=0}^{L-1}\boldsymbol{v}(l)\boldsymbol{v}^H(l),$ and
{\small \begin{align*}
&\hspace*{2.5cm}\tilde{\boldsymbol{D}}=\boldsymbol{\Sigma}^{-\frac{1}{2}}\left[\boldsymbol{D}^{(\mathcal T)},\boldsymbol{D}^{(\mathcal R)}\right]\\
&\text{where}\,\left\lbrace\begin{array}{l}
\boldsymbol{D}^{(\mathcal T)}=\left[\left.\frac{\partial\boldsymbol{a}\left(\theta^{(\mathcal T)},\theta^{(\mathcal R)}\right)}
{\partial\theta^{(\mathcal T)}}\right|_{\theta^{(\mathcal T)}=\theta^{(\mathcal T)}_1,\theta^{(\mathcal R)}=\theta^{(\mathcal R)}_1},
\dots,\right.\\
\hspace*{2cm}\left.\left.\frac{\partial\boldsymbol{a}\left(\theta^{(\mathcal T)},\theta^{(\mathcal R)}\right)}
{\partial\theta^{(\mathcal T)}}\right|_{\theta^{(\mathcal T)}=\theta^{(\mathcal T)}_K,\theta^{(\mathcal R)}=\theta^{(\mathcal R)}_K}\right] \\
\\
\boldsymbol{D}^{(\mathcal R)}=\left[
\left.\frac{\partial\boldsymbol{a}\left(\theta^{(\mathcal T)},\theta^{(\mathcal R)}\right)}
{\partial\theta^{(\mathcal R)}}\right|_{\theta^{(\mathcal T)}=\theta^{(\mathcal T)}_1,\theta^{(\mathcal R)}=\theta^{(\mathcal R)}_1},
\dots,\right.\\
\hspace*{2cm}\left.\left.\frac{\partial\boldsymbol{a}\left(\theta^{(\mathcal T)},\theta^{(\mathcal R)}\right)}
{\partial\theta^{(\mathcal R)}}\right|_{\theta^{(\mathcal T)}=\theta^{(\mathcal T)}_K,\theta^{(\mathcal R)}=\theta^{(\mathcal R)}_K}\right]
\end{array}\right.
\end{align*}}
\section{Numerical simulations}
For simulations, we consider a MIMO radar comprising $M=3$ sensors at the transmitter and $N=4$ at the receiver, both with half-wave length inter-element spacing. The DOD and DOA of the first source are respectively $18^\circ$ and $20^\circ$, and of the second source are $45^\circ$ and $40^\circ$. The coefficients $\alpha_1$ and $\alpha_2$ are chosen to be $2+3j$ and $1-0.5j$, and the normalized Doppler frequencies $f_1$ and $f_2$ are $0.3$ and $0.8$. There are $L=15$ pulses per CPI, and each pulse contains $T=5$ snapshots. For K-distributed clutter, we choose $a=2$ and $b=10$; and for t-distributed clutter, $a=1.1$ and $b=2$. The entries of the speckle covariance matrix $\boldsymbol{\Sigma}$ are generated by $[\boldsymbol{\Sigma}]_{m,n}=\sigma^2 0.9^{|m-n|}e^{j\frac{\pi}{2}(m-n)}, \ m,n=1,\dots,MN$, in which $\sigma^2$ is a factor to adjust speckle power. Each point of the MSE in the figures is generated by averaging the results of $500$ Monte-Carlo trials. The signal-to-clutter ratio (SCR) \citep{Akcakaya1} is defined by
$
\text{SCR}=\frac{1}{L}\frac{\sum_{l=0}^{L-1}\left(\boldsymbol{A}\left(\boldsymbol{\theta}\right)\boldsymbol{v}(l)\right)^H
\left(\boldsymbol{A}\left(\boldsymbol{\theta}\right)\boldsymbol{v}(l)\right)}
{\text{E}\{\tau(l)\}\text{tr}\left\{\boldsymbol{\Sigma}\right\}},
$
in which $\text{E}\{\tau(l)\}$ is equal to $ab$ for a K-distributed clutter and $b/(a-1)$ for a t-distributed clutter (for $a>1$).
Figs.~\ref{MSE_vs_SCR_Kdist} and \ref{MSE_vs_L_Kdist} investigate the performance of the proposed IMMLE estimator compared to the classical MUSIC method based on the Sample Covariance Matrix (MUSIC-SCM) as well, its robust version based on the well-known Tyler estimate of the covariance matrix \citep{Tyl87} (MUSIC-Tyler). Others robust MUSIC based algorithms are also considered such as the RG-MUSIC \citep{Cou15}, the $\ell_p$-MUSIC \citep{ZSH13} and the ROC-MUSIC \citep{TN96} as well the MKG algorithm proposed in \citep{BAJ16} and the ML-GM \citep{KS00}. Finally, we consider the ICdMLE, the IJMLE and the ICvMLE, as well the derived CRB.
\begin{figure*}[h!]
\centering
\input{MSE_vs_SCR_Kdist_500MC.tex}
\caption{MSE vs. SCR under K-distributed clutter, L = $15$}
\label{MSE_vs_SCR_Kdist}
\end{figure*}
\begin{figure*}[h!]
\centering
\input{MSE_vs_L_Kdist_500MCbis.tex}
\caption{MSE vs. $L$ under K-distributed clutter, SCR = $15$ dB}
\label{MSE_vs_L_Kdist}
\end{figure*}
In Fig.~\ref{MSE_vs_SCR_Kdist} and \ref{MSE_vs_L_Kdist}, the MSEs are plotted versus SCR with a fixed $L$, respectively versus the pulse number $L$ with a fixed SCR. \newline
It can be noticed that MUSIC-based algorithms, even the robust versions, do not outperform the proposed algorithm due to a
small number of pulses, which is a typical scenario in radar application. The MKG algorithm assumes a mixture of K-distributed and Gaussian noise, both of them, with a covariance matrix equals to the identity, which explains its poor performance. Whereas, the robust ML-GM is based on, empirically defined number of, Gaussian mixture with identity covariance matrix assumptions. Since, it is a ML estimator (i.e., an estimator based on a parametric model), its accuracy deteriorates if we deviate from the assumed model distribution. Concerning the ICdMLE, IJMLE and the ICvMLE, their performances are below the proposed algorithm. Consequently, from Figures \ref{MSE_vs_SCR_Kdist} et \ref{MSE_vs_L_Kdist}, we can assess that the IMMLE outperforms the aforementioned algorithms. The same behavior is noticed under the t-distributed clutter whether it is MSE versus SCR with fixed $L$ or versus $L$ with fixed SCR.
Finally, the reader is referred to Table. 2 for a concise comparison between the IMMLE, IJMLE and ICdMLE.
\newline
\noindent \textbf{Remark 5:} It is worth mentioning that, the proposed IMMLE estimates are approximation of the true ML estimates due to the iterative stepwise procedure, in which we have to solve numerically at each step three equations for the update of the parameters of the texture distribution and the speckle covariance matrix. Thus these latter, are not exact solutions either. Furthermore, it is worth mentioning that a theoretical analysis of the efficiency of the estimator on $\boldsymbol{\theta}$ is beyond the scope of this paper. Nevertheless, from our extensive simulations, we believe that our proposed algorithm would not be statistically efficient (i.e., its MSE does not attain the CRB).
\section{Conclusion}
This paper is dedicated to the design of the exact ML DOD and DOA estimation for MIMO radar in the presence of SIRP clutter. Specifically, our proposed iterative estimator is based on the marginal likelihood for which its related cost function is solved using stepwise numerical concentration approach. Finally, interconnections with the existing based likelihood methods, namely, the conventional, the conditional and the joint likelihood based estimators are investigated theoretically and numerically.
\bibliographystyle{elsarticle-num}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 3,508
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Q: Use RegExp in Awk I have CSV file:
<iframe src="https://localhost/get/44bc40f3bc04f65b7a35"></iframe>|name_1
<iframe src="https://localhost/get/5db0d477d707121934ff"></iframe>|name_2
<iframe src="https://localhost/get/6c95bd2b32ed45989c61"></iframe>|name_3
<iframe src="https://localhost/get/0a9c4655800e8a7b9ea2"></iframe>|name_4
<iframe src="https://localhost/get/754953b57a32e2841bda"></iframe>|name_5
and want use RegExp and Awk (or Gawk) to make this CSV file like this:
44bc40f3bc04f65b7a35|name_1
5db0d477d707121934ff|name_2
6c95bd2b32ed45989c61|name_3
0a9c4655800e8a7b9ea2|name_4
754953b57a32e2841bda|name_5
I have worked RegExp in Grep
$ grep -Po "[A-Za-z]*+\d++\w++" example.txt
44bc40f3bc04f65b7a35
5db0d477d707121934ff
6c95bd2b32ed45989c61
0a9c4655800e8a7b9ea2
754953b57a32e2841bda
but this RegExp not work in Awk. I think i'm not correct use regexp in Awk or this type of RegExp not worked in Awk.
$ awk -F "|" 'match($1, /[A-Za-z]*+\d++\w++/, a) {print a[0]"|"$2}' example.txt
db0d477d707121934ff|name_2
bd2b32ed45989c61|name_3
bda|name_5
Just Awk work fine:
$ awk -F "|" '{print $1"|"$2}' example.txt
<iframe src="https://localhost/get/44bc40f3bc04f65b7a35"></iframe>|name_1
<iframe src="https://localhost/get/5db0d477d707121934ff"></iframe>|name_2
<iframe src="https://localhost/get/6c95bd2b32ed45989c61"></iframe>|name_3
<iframe src="https://localhost/get/0a9c4655800e8a7b9ea2"></iframe>|name_4
<iframe src="https://localhost/get/754953b57a32e2841bda"></iframe>|name_5
A: Try:
$ awk -F'<iframe src="https://localhost/get/|"></iframe>' '{print $2 $3}' file
44bc40f3bc04f65b7a35|name_1
5db0d477d707121934ff|name_2
6c95bd2b32ed45989c61|name_3
0a9c4655800e8a7b9ea2|name_4
754953b57a32e2841bda|name_5
This works by setting the input field separator to be either iframe src="https://localhost/get/ or "></iframe> in which case the output you want is the second field followed by the third field.
Alternative Method Using Match
$ awk -F "|" 'match($1, /[[:xdigit:]]{20}/, a) {print a[0]"|"$2}' file
44bc40f3bc04f65b7a35|name_1
5db0d477d707121934ff|name_2
6c95bd2b32ed45989c61|name_3
0a9c4655800e8a7b9ea2|name_4
754953b57a32e2841bda|name_5
Note that awk supports POSIX regular expressions. That means that it recognizes character classes like [[:digit:]] or [[:alnum:]] but not necessarily \d or \w. As a GNU-specific extension, gawk supports \w (but not \d). For portability, stick to the POSIX classes as shown in man 7 regex.
Yet another method
Your regex [A-Za-z]*+\d++\w++ can be translated into awk as follows:
$ awk -F "|" 'match($1, /[[:alpha:]]*[[:digit:]]+[[:alnum:]]+/, a) {print a[0]"|"$2}' file
44bc40f3bc04f65b7a35|name_1
5db0d477d707121934ff|name_2
6c95bd2b32ed45989c61|name_3
0a9c4655800e8a7b9ea2|name_4
754953b57a32e2841bda|name_5
Note that this method requires that the 20-character hex string contains at least one digit.
A: *
*The difference between awk and grep invocations in your example is -P option in grep, which stands for "Use Perl regexp". If you replace it with -E, it will work just like your awk run. Awk does not support Perl extension.
*Your regexp is better be fixed, I don't think you need these extra + signs, to begin with. If I can assume that you need all letters or digits after get/ then I'd rather write:
awk -F "|" 'match($1, /get/([A-Za-z0-9]+)/, a) {print a[1]"|"$2}' example.txt
Here we use [A-Za-z0-9]+ match any number of small or upper letters or digits that come after /get, use a[1] to print a matched group inside the parentheses instead of the whole matching pattern a[0] which includes get/
A: awk '{gsub(/<.*get\//,"")gsub(/".*e>/,"")}1' file
44bc40f3bc04f65b7a35|name_1
5db0d477d707121934ff|name_2
6c95bd2b32ed45989c61|name_3
0a9c4655800e8a7b9ea2|name_4
754953b57a32e2841bda|name_5
A: Here is another solution:
awk -F"[/\">|]" 'BEGIN{ OFS = "|" }{ print $6, $11 }' yourfile
With the -F option in the beginning the Field Separator can be /, ", > and |. After that is done you can just print the fields $6 and $11 which contain your desired output together with the output field separator.
Output:
44bc40f3bc04f65b7a35|name_1
5db0d477d707121934ff|name_2
6c95bd2b32ed45989c61|name_3
0a9c4655800e8a7b9ea2|name_4
754953b57a32e2841bda|name_5
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 144
|
Q: Access name of .rmd file and use in R I am knitting a markdown file called MyFile.rmd.
How can I access the string MyFile during the knitting and use it for:
*
*use in the title section of the YAML header?
*use in subsequent R chunk?
---
title: "`r rmarkdown::metadata$title`"
author: "My Name"
date: "10. Mai 2015"
output: beamer_presentation
---
## Slide 1
```{r}
rmarkdown::metadata$title
```
leads to...
... which is incorrect as the file I am knitting is named differently.
> sessionInfo()
R version 3.1.2 (2014-10-31)
Platform: x86_64-apple-darwin13.4.0 (64-bit)
locale:
[1] de_DE.UTF-8/de_DE.UTF-8/de_DE.UTF-8/C/de_DE.UTF-8/de_DE.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] digest_0.6.8 htmltools_0.2.6 rmarkdown_0.5.1 tools_3.1.2 yaml_2.1.13
A: rmarkdown::metadata gives you the list of the meta data of the R Markdown, e.g. rmarkdown::metadata$title will be the title of your document. An example:
---
title: "Beamer Presentation Title"
author: "My Name"
date: "10\. Mai 2015"
output: beamer_presentation
---
## Slide 1
Print the title in a code chunk.
```{r}
rmarkdown::metadata$title
```
## Slide 2
The title of the document is `r rmarkdown::metadata$title`.
To obtain the filename of the input document, use knitr::current_input().
A: You could use the yaml library, like so:
library(yaml)
# Read in the lines of your file
lines <- readLines("MyFile.rmd")
# Find the header portion contained between the --- lines.
header_line_nums <- which(lines == "---") + c(1, -1)
# Create a string of just that header portion
header <- paste(lines[seq(header_line_nums[1],
header_line_nums[2])],
collapse = "\n")
# parse it as yaml, which returns a list of property values
yaml.load(header)
If you save the list returned by yaml.load, you can use it in various chunks as needed. To get the title, you can do this:
properties <- yaml.load(header)
properties$title
A: Just summarize Yihui's answer:
---
title: "`r knitr::current_input()`"
author: "My Name"
date: "10. Mai 2015"
output: beamer_presentation
---
## Slide 1
```{r}
knitr::current_input()
```
which knitted does the job.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,377
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1 доллар США — американские банкноты или монеты достоинством 1 доллар США.
В виде банкнот изготавливаются:
1 доллар США.
В виде монет чеканятся или чеканились:
Первые однодолларовые монеты США — монеты 1794–1804 годов. Доллар 1804 года является одной из самых дорогих монет в мире. В настоящее время известно о существовании всего 15 монет.
Доллар Гобрехта — очень редкая монета, чеканившаяся в 1836, 1838 и 1839 годах.
Доллар с сидящей Свободой — монета, выпускавшаяся с 1840 по 1873 год. На аверсе изображена сидящая женщина, символизирующая Свободу.
Золотой доллар.
Моргановский доллар — монета, выпускавшаяся с 1878 по 1904 год и в 1921 году.
Мирный доллар — монета, выпускавшаяся в 1921–1935 и 1964 годах.
Доллар Эйзенхауэра — монета, выпускавшаяся в 1971–1978 годах. На аверсе изображён президент Эйзенхауэр.
Доллар Сьюзен Энтони — монета, выпускавшаяся в 1979–1981, 1999 годах.
Доллар Сакагавеи — монета, находящаяся в обороте. Выпускается с 2000 года.
Программа однодолларовых президентских монет — серия монет США номиналом в один доллар с портретами всех американских президентов на аверсе.
Программа однодолларовых инновационных монет — серия монет США номиналом в один доллар, посвящённая инновациям.
Номиналы в долларах США
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 487
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ScriptManager.clearTimer();
ScriptManager.clearEl();
ScriptManager.clearTrigger();
// </d>
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{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,012
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package gov.va.vinci.leo.tools;
import org.apache.commons.lang3.StringUtils;
import javax.tools.JavaCompiler;
import javax.tools.JavaFileObject;
import javax.tools.StandardJavaFileManager;
import javax.tools.ToolProvider;
import java.io.File;
import java.util.Arrays;
/**
* Compile Java files. Requires tool.jar (on Windows) and classes.jar (on Mac)
* from java lib in the classpath to use this class.
*
* @author Prafulla
* @author Thomas Ginter
*/
public class AutoCompile {
/**
* Compile the files in the list. Optionally provide a destination directory where the
* compiled files should be placed.
*
* @param files List of java files to be compiled
* @param outputDirectory Optional, output directory where the compiled files will be written
* @throws Exception If there are errors compiling the files or we are unable to get a compiler
*/
public static void compileFiles(File[] files, String outputDirectory) throws Exception {
//If the list is null there is nothing to do
if (files == null) {
return;
}
JavaCompiler compiler = ToolProvider.getSystemJavaCompiler();
if (compiler == null) {
throw new Exception("Unable to get a Compiler from the ToolProvider");
}
StandardJavaFileManager stdFileManager = compiler.getStandardFileManager(null, null, null);
Iterable<String> compilationOptions
= (StringUtils.isBlank(outputDirectory)) ? null : Arrays.asList(new String[]{"-d", outputDirectory});
try {
Iterable<? extends JavaFileObject> compilationUnits = stdFileManager.getJavaFileObjectsFromFiles(Arrays.asList(files));
compiler.getTask(null, stdFileManager, null, compilationOptions, null, compilationUnits).call();
} finally {
stdFileManager.close();
}//finally
}//compileFiles method
}//end AutocCompile Class
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,172
|
A device that controls a powerful Russian-made satellite weapon is stolen by Russian terrorists, who try to escape by flying out but are shot down. The device is now on top of K2. The device is still active and where it will strike is indeterminable. The good guys have to get to the device and deactivate it. Only trouble, it's winter and it'll be difficult to climb K2. A few of the best mountain climbers in the world are recruited to take techs who can deactivate the device to the top of K2. However, before the climbing party leaves, a team member dies, and it might not be natural causes. The group who stole the device might have someone on the team. But they still go on.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,125
|
Deze lijst van voetbalinterlands is een overzicht van alle officiële voetbalwedstrijden tussen de nationale teams van Egypte en Slowakije. De landen speelden tot op heden een keer tegen elkaar: een vriendschappelijke wedstrijd die werd gespeeld op 4 februari 1994 in Sharjah (Verenigde Arabische Emiraten).
Wedstrijden
Samenvatting
Wedstrijden bepaald door strafschoppen worden, in lijn met de FIFA, als een gelijkspel gerekend.
Details
Eerste ontmoeting
De eerste ontmoeting tussen de nationale voetbalploegen van Egypte en Slowakije vond plaats op 4 februari 1994. Het vriendschappelijke duel, bijgewoond door 5.000 toeschouwers, werd gespeeld in het Sharjah Stadion in Sharjah, en stond onder leiding van scheidsrechter Ahmed Ibrahim Hakim uit de Verenigde Arabische Emiraten. Het duel werd gespeeld in het kader van een vierlandentoernooi, en betekende de tweede officiële interland van Slowakije sinds de vreedzame ontmanteling van Tsjecho-Slowakije.
Slowakije
Egypte
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,259
|
When you think about going to a baseball game and the food you might eat that day, you probably think of hot dogs, peanuts or nachos. You probably don't think of falafel (a Middle Eastern fritter made from chickpeas), but for me, baseball and falafel are intertwined!
You see, right near the stadium where the Detroit Tigers play is a bar/restaurant that serves amazing falafel pita sandwiches. They also have chicken shawarma, but I prefer falafel because they're a little lighter than the shawarma. They don't weigh me down quite as much, and leave me with plenty of room to get a big ol' ice cold beer at the ballpark.
This bar with the falafel also serves more traditional sports fan fare like buffalo wings. Upon studying the menu, I came up with the idea to combine the falafel with buffalo flavors. I mean, why the heck not? I stuck with some classic Middle Eastern flavors for the falafel – chickpeas are the main component of the fritters, along with garlic, cumin and lemon zest. But after pan-frying the falafel, I tossed them in a generous bath of buffalo sauce.
Because I wanted a double-dose of buffalo flavor, I also stirred buffalo sauce into Sabra Classic Hummus. This buffalo hummus mixture was spread on submarine buns, then the falafel were piled in. The sandwich is finished with crispy Romaine lettuce, juicy tomato and a bit of crumbled blue cheese.
Be sure to connect with Sabra on Facebook, Twitter, Instagram and Pinterest, and look for other National Hummus Day recipes using the hashtag #HummusDay.
In bowl of food processor fitted with knife blade attachment, pulse chickpeas, garlic, onions, egg, flour, parsley, baking powder, cumin, lemon zest, salt, black pepper and cayenne until chickpeas are very finely chopped. You don't want the mixture to be a smooth paste, but you do want the chickpeas to be finely ground.
Place 1/2 cup buffalo sauce in large bowl. In separate small bowl, stir together hummus and remaining 2 tablespoons buffalo sauce.
In large pot or high-sided skillet, heat 2 inches of oil to 370 degrees F over medium-high heat. Form chickpea mixture into balls and slightly flatten each ball (each ball should be about 1-1/2 to 2 tablespoons mixture; you should have about 24 falafel). In batches, drop falafel into hot oil and fry 3 to 4 minutes or until brown and crisp, turning once. Transfer falafel to paper-towel lined plate and continue frying remaining falafel.
When all falafel are fried, transfer to large bowl with buffalo sauce and toss until well combined.
Spread inside of submarine buns with hummus mixture. Divide lettuce over hummus, then divide falafel between buns (5-6 falafel per bun). Top with tomatoes and blue cheese and serve immediatley.
Disclosure: This was a sponsored post written by me on behalf of Sabra. Sponsored posts help me pay for the costs associated with this blog (groceries…lots of groceries), and help support me as I pursue a career in recipe development and food photography. All opinions are 100% my own.
What a unique sub! Buffalo sauce and falafel sounds like a match made in heaven!
Girl – Yes yes yes!! Love your creativity!!! Do you deliver?
How generous and creative! Love this idea, Lori. Falafel in a sandwich with greens and blue cheese sound heavenly!
I now NEED Buffalo falafel subs in the worst way!
Get the heck outta town! I'm going to need at least two of these flavorful huggers. Yum!
A double dose of chickpeas! Epic! The flavor combo here sounds out of this world amazing.
This looks like one flavorful sandwich! I love that you used Sabra Hummus for the buffalo sauce!
Thanks so much for stopping by, Nicole!
Oh my goodness this sounds amazing!! I happen to have a massive crush on falafel. This is a must try.
Serious. These look incredible, delicious, and I want them for dinner. Good thing I have tubs and tubs of Sabra Hummus for HummusDay!
It's never a bad thing to have a fridge full of Sabra hummus. Thanks Fabiola!
Lori, I think you hit on just about everything I love, all in one recipe. Falafel, hummus, raw veggies, things in buns…what else is there? (until dessert, of course). I've never made falafel from scratch before — Cannot wait to make these!
Thanks Amelia! This was my first time making falafel from scratch, and they were actually way easier than I thought they would be. The food processor makes quick work of mixing everything up!
What a great way to use hummus! These look terrific!
I am in love with this idea. It's like meatball sub meets wings meets Mediterranean. And that up close shot of the falafel is killer. I think it's about time for second dinner.
I had falafel for lunch today and I kept thinking it was missing something, clearly it was buffalo sauce. Such a great idea and I'd rather go for a big falafel sandwich over ballpark food any day.
You're a genius for combining two of my favorites into one delicious sub. Maybe now I can finally get my dad to eat falafel (he's not a fan of Middle Eastern cuisine, which is funny because it's the total opposite of me).
Awwwww. My parents ARE fans of Middle Eastern cuisine, but they live in a really small town. M.E. restaurants keep trying to open there, but go out of business quickly due to lack of customers. It stinks!
I love these!! They used to make a Buffalo hummus and it was awesome, but I can't find it anymore. It never occurred to me to just do it myself. DUH. Pairing it with falafel is brilliant.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,459
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\section{Introduction}
Submersions can be thought as a dual concept to submanifolds. The study of the equations describing the geometry of a submanifold of a Riemannian manifold has a long tradition, having the Gauss and Codazzi equations between the most celebrated results in the theory. These equations have been generalized to the Finsler setup from the 1930s (see \cite{2022HuberJavaloyes} for a summary of the topic). The systematic study of Riemannian submersions is much more recent, having its origin in the foundational paper by O'Neill about the fundamental equations \cite{1966ONeill} (see also \cite{1967Gray}). It is very natural to ask oneself about the generalization to the Finsler realm of these fundamental equations, but the closest achievement until now is the pioneering paper \cite{2001AlDur}, where the authors obtain some results
using the osculating metric associated with a geodesic vector field. In this paper, we intend to fill this gap providing a complete set of fundamental equations of a pseudo-Finsler submanifold, which describe the curvature tensor of the total space in terms of the configuration O'Neill tensors $\mathbf T$ and $\mathbf A$ and the curvature tensors of the fibers and the base manifold. We recall that the fundamental equations have been very helpful to compute the curvature tensor in many different situations as for example for models of spacetimes and to find new Einstein manifolds (see \cite[Chapter 9]{Besse}). We hope the equations presented in this manuscript will play its role in the development of Finsler spacetimes to find solutions of the Finslerian Einstein equations (see for exampe \cite{HPV19,JSV21}) and that will motivate the development of the theory of Finslerian submersions as in the case of classical Riemannian submersions (see for example \cite{FaIaPa04}). They will also play an important role in the theory of Finsler foliations \cite{AAD19,AAJ19,AEi22} and isoparametric functions \cite{HCSR21,Xu18}, which have already deserved some attention. Finsler submersions can also be useful in the study of some types of homogeneous Finsler manifolds \cite{XuZh18}.
It is well-known that Finsler Geometry entails cumbersome computations in coordinates. To overcome this problem we have used the anisotropic tensor calculus recently developed in \cite{Jav19,Jav20}. The main idea is to
perform all the computations with the anisotropic connections and when necessary, to fix a local vector field $V$ and to work with an associated affine connection $\nabla^V$. There are some priveleged choices of $V$ which economize computations. Moreover, it turns out that there is no need to use expressions in coordinates. The paper is organized as follows. In Section \ref{pseudo-Finsler}, we introduce the concept of pseudo-Finsler metric and its main features and we also briefly describe how the anisotropic tensor calculus works. In Section \ref{pseudo-sub}, we introduce the definition of pseudo-Finsler submersion and prove several basic properties, as for example that when one fixes a horizontal vector $v$, then the differential map of $\sigma$ with the fundamental tensor $g_v$ has a structure of pseudo-Riemannian submersion (see Lemma \ref{Lemma1}). This result was ennunciated in \cite[Prop. 2.2]{2001AlDur}. We also prove that when the pseudo-Finsler metrics are defined in the whole tangent bundle, then there exists a unique horizontal lift of every non-zero vector in the base. Finally, we compute an expresion which relates the Cartan tensors of the total space and the base being the first one evaluated in at least two $g_v$-horizontal vectors. In Section \ref{fundTA}, we introduce the fundamental tensors $\mathbf T$ and $\mathbf A$ and study their properties. One of the main differences with the Riemmannian case is that now the horizontal part of any vector field is an anisotropic vector field (see Section \ref{pseudo-Finsler} for its definition), this means that for every admissible vector $v$ there is a different horizontal part and we will have to deal with fiber derivatives of the horizontal and vertical parts, which are computed in Proposition \ref{dotPartialTopProposition}. The properties of $\mathbf T$ and $\mathbf A$ are very similar to those of their classical versions, with some exceptions. For example, the tensor $\mathbf A$ has a much more involved expression in terms of the Lie Bracket (see Proposition \ref{TAProposition2}). To compute the Gauss formula and its dual, we need to introduce two anisotropic tensors $\hat Q$ and $\tilde Q$, because as it was known from the theory of submanifolds, the Chern connection of the fibers does not coincide with the tangent part of the connection of the ambient space. A similar situation happens with the horizontal part of the Chern connection applied to $g_V$-horizontal vector fields for some horizontal vector field $V$ (see Proposition \ref{VXYhor}). In the subsection \ref{covderTA}, the formulas for the covariant derivatives of $\mathbf T$ and $\mathbf A$ are obtained, this time matching the classical version except for the cyclic formula in \cite[Lemma 7]{1966ONeill}. In Section \ref{fundeq}, we finally obtain the fundamental equations for the curvature tensor. As a first step, we introduce the vertical and horizontal curvature tensors in Definition \ref{RTopBotDefinition}. Then we decompose the curvature tensor using these horizontal and vertical ones and the tensors $\mathbf T$ and $\mathbf A$ in Theorem \ref{Unified} and obtain more specific formulas computing all the involved terms when the curvature tensor is evaluated in vertical and $g_v$-horizontal vectors in Corollary \ref{fundeqgeneral}. When $v$ is vertical, we can obtain a generalized Gauss equation as in the case of submanifolds, and when $v$ is horizontal a Dual Gauss equation (see Theorem \ref{gaussanddual}). Remarkably, Codazzi and its dual hold for an arbitrary admissible $v$. As an application, we obtain formulas for the flag curvature with arbitrary flagpole of the ambient space in Corollary \ref{flagcurgen}. Moreover, we also recover the O'Neill formulas (see \cite[Corollary 1]{1966ONeill}) for the flag curvature when the flagpole is horizontal in Corollary \ref{horflagpole}, while the formula for a vertical flagpole is slightly different (see Corollary \ref{verflagpole}). In the last section \ref{totallygeo}, we study Finslerian submersions with totally geodesic fibers. It is well-known that in the Riemannian case, this condition implies that the fibers are isometric in a natural way. In the Finslerian setup we need to add a further condition, namely, the horizontal regularity in Definition \ref{horreg}, to obtain these isometries. Then under this condition, we prove that a Finsler submersion has totally geodesic fibers if and only if it is an associated fiber bundle of a principle bundle, Theorem \ref{totgeoth}.
\section{Pseudo-Finsler manifolds}\label{pseudo-Finsler}
Given a manifold $\mathcal M$, we will denote by $\pi\colon T\mathcal M\to\mathcal M$ the natural projection from its tangent bundle $T\mathcal M$.
\begin{definition}
Let $\mathcal{A}\subset T\mathcal M\setminus\lbrace 0\rbrace$ be an open subset invariant under positive scalar multiplication. We say that a smooth function $L\colon\mathcal{A}\to\mathds R$ is a pseudo-Finsler metric when it satisfies:
\begin{enumerate}[(i)]
\item positive homogeneity of degree $2$, namely, $L(\lambda v)=\lambda^2L(v)$ $\forall\lambda>0$ and $v\in \mathcal{A}$,
\item non-degeneracy of its associated fundamental tensor $g_{ v}$ defined for each $v\in\mathcal{A}$ and $e,h\in T_{\pi(v)}\mathcal M$ by
\begin{equation*}
g_v(e,h) = \tfrac{1}{2} \left. \tfrac{\partial^2 L(v+se+th)}{\partial s\partial t} \right\vert_{s,t=0}
\,.
\end{equation*}
\end{enumerate}
Moreover, when $\mathcal{A}=T\mathcal M\setminus \bf 0$ and $g_v$ is always positive definite, we will say that $L$ is a Finsler metric, and it is common to work with the positive homogeneous function of degree one $F=\sqrt{L}$ rather than with $L$.
\end{definition}
Given a pseudo-Finsler metric, we will say that a vector $v$ is admissible if $v\in \mathcal{A}$. Moreover, given an open subset $\Omega\subset \mathcal M$, we will say that a vector field $V\in\mathfrak{X}( \Omega)$ is admissible if $V_p\in \mathcal{A}$ for all $p\in\Omega$. We will say that $V$ is locally admissible, if it is admissible in a certain neighborhood $\Omega$ of a given point.
The Cartan tensor $C$ associated with $L$ for each $v\in\mathcal{A}$ and $b,e,h\in T_{\pi(v)}\mathcal M$ is defined by
\begin{equation*}
C_v(b,e,h) = \tfrac{1}{2} \left. \tfrac{\partial^2 g_{v+tb}(e,h)}{\partial t} \right\vert_{t=0} = \tfrac{1}{4} \left. \tfrac{\partial^3 L(v+rb+se+th)}{\partial r\partial s\partial t} \right\vert_{r,s,t=0}
\,.
\end{equation*}
Observe that the Cartan tensor is symmetric and, by homogeneity, $C_v(v,\cdot,\cdot)=0$ for every $v\in\mathcal{A}$.
Let us denote by $\mathfrak{X}(\mathcal M)$ and $\mathfrak{X}^*(\mathcal M)$ the set of (smooth) vector fields and one-forms on $\mathcal M$, respectively. We will also denote by $T^*\mathcal M$ the cotangent bundle of $\mathcal M$ and by $\overset{r)}\bigotimes T\mathcal M\otimes \overset{s)}\bigotimes T^* \mathcal M$ the classical tensor bundle over $\mathcal M$ of type $(r,s)$. Now denoting $\pi_\mathcal{A}:\mathcal{A}\rightarrow \mathcal M$ the restriction of the natural projection $\pi$, we will consider the pullbacked bundle $\pi_\mathcal{A}^*(\overset{r)}\bigotimes T\mathcal M\otimes \overset{s)}\bigotimes T^* \mathcal M)$. We will say that a smooth section of this bundle $T:\mathcal{A}\rightarrow \pi_\mathcal{A}^*(\overset{r)}\bigotimes T\mathcal M\otimes \overset{s)}\bigotimes T^* \mathcal M)$ is an $\mathcal{A}$-anisotropic tensor field. When the subset $\mathcal{A}$ is obvious from the context, we will say simply ``anisotropic tensor''. Observe that for every $v\in \mathcal{A}$, $T_v$ (the evaluation of $T$ at $v\in\mathcal{A}$) is a multilinear map
\[T_v:(T^*_{p}M)^r\times (T_{p}M)^s\rightarrow \mathds R,\]
with $p=\pi(v)$. In particular, the fundamental tensor $g$ and the Cartan tensor $C$ of a pseudo-Finsler metric are $\mathcal{A}$-anisotropic tensors of type $(0,2)$ and $(0,3)$, respectively. We will denote by $\mathcal{T}^r_s(\mathcal M_\mathcal{A})$ the space of $\mathcal{A}$-anisotropic tensor fields. As in the classical case, $\mathcal{T}^1_0(\mathcal M_\mathcal{A})$ can be identified with $\mathcal{A}$-anisotropic vector fields $\mathcal X$, $\mathcal{A}\ni v\rightarrow {\mathcal X}_v\in T_{\pi(v)}\mathcal M$. Moreover, if
${\mathcal F}(\mathcal{A})$ is the space of smooth real functions on $\mathcal{A}$,
$\mathcal{A}$-anisotropic tensor fields can be described equivalently as ${\mathcal F}(\mathcal{A})$-multilinear maps
\begin{equation*}
T\colon {\mathcal T}^0_1(\mathcal M_\mathcal{A})^r\times {\mathcal T}^1_0(\mathcal M_\mathcal{A})^s \rightarrow {\mathcal F}(\mathcal{A})
\end{equation*}
very much as in the classical case, a tensor field can be thought of as a ${\mathcal F}(\mathcal M)$-multilinear maps.
\begin{definition}
An ($\mathcal{A}$-)anisotropic connection is a map
\begin{equation*}
\nabla\colon \mathfrak X(\mathcal M)\times\mathfrak X(\mathcal M) \rightarrow \mathcal{T}^1_0(\mathcal M_\mathcal{A})\,,\quad (E,H)\rightarrow \nabla_EH
\end{equation*}
satisfying for all vector fields $B,E,H$ and smooth real function $f$ on $\mathcal M$
\begin{enumerate}[(i)]
\item $\nabla_E(H+B)=\nabla_EH+\nabla_EB$,
\item $\nabla_E(fH)=(E(f)\cdot Y)\circ\pi_\mathcal{A}+(f\circ \pi_\mathcal{A})\nabla_EH$,
\item $\nabla_{E+B}H=\nabla_EH+\nabla_BH$,
\item $\nabla_{fE}H=(f\circ\pi_\mathcal{A})\nabla_EH$.
\end{enumerate}
\end{definition}
We will use the notation $\nabla^v_EH:=(\nabla_EH)_v$. Moreover,
since the expression $\nabla^v_EH$ depends only on the vector $e=E_{\pi(v)}$ rather than on its extension $E$, we shall prefer to write $\nabla^v_eH$.
Given a vector field $V$ on an open subset $\Omega$ of $\mathcal M$ which is $\mathcal{A}$-admissible, namely, such that $V_p\in \mathcal{A}$ for all $p\in\Omega$, we can consider an affine connection $\nabla^V$ on $\Omega$ defined for each $E,H\in {\mathfrak X}(\mathcal M)$ by $(\nabla^V_EH)_p:=(\nabla_EH)_{V_p}\equiv \nabla^{V_p}_EH $. Associated with the anisotropic connection there is a covariant derivative along curves which depends on the choice of an admissible vector field along the curve (with values in $\mathcal{A}$ at every instant). This covariant derivative is constructed locally with the Christoffel symbols of the anisotropic connection and for a curve $\gamma:[a,b]\rightarrow \mathcal M$ and an admissible vector field $W\in {\mathfrak X}(\gamma)$, it will be denoted by $D^W_\gamma: {\mathfrak X}(\gamma)\rightarrow {\mathfrak X}(\gamma)$, $X\rightarrow D^W_\gamma X$. When the curve $\gamma$ is regular at $p=\gamma(t)$, it is given by $(\nabla^{\tilde W}_V\tilde X)_p$, considering vector fields $\tilde W,V,\tilde X$ which are local extensions of $W,\dot\gamma,X$, respectively (see \cite[Section 2.2]{Jav20} for more details). Given an $\mathcal{A}$-anisotropic tensor $T$, we can also define a classical tensor of the same type $T_V$ as $(T_V)_p=T_{V_p}$. Departing from an anisotropic connection it is possible to define the covariant derivative of any anisotropic tensor as it was shown in \cite{Jav19}. These covariant derivatives can be obtained by the classical product rule, and they can also be expressed in terms of $\nabla^V$ with the help of fiber derivatives. If $T\in {\mathcal T}^r_s(\mathcal M_\mathcal{A})$, then its fiber derivative $\dot\partial T\in {\mathcal T}^r_{s+1}(\mathcal M_\mathcal{A})$ is defined as
\begin{equation*}
(\dot\partial_e T)_v
=
\left.\tfrac{\partial T_{v+te}}{\partial t} \right\vert_{t=0}
\end{equation*}
where $e\in T_{\pi(v)}\mathcal M$ is the $s+1$-covariant component of $\dot\partial T$ and sometimes we will put it in its natural place.
A key observation is that anisotropic connections allow us to define derivations of functions $\mathcal F(\mathcal{A})$. Indeed, for any $v\in \mathcal{A}$ and any choice of local admissible extension $V$ of $v$, $\nabla_Ef$ is smooth function on $\mathcal{A}$ given by
\begin{equation}\label{nablaf}
(\nabla_E f)(v)=E_{\pi(v)}(f(V))-\dot\partial_{\nabla^v_EV} f.
\end{equation}
It turns out that last expression is independent of the choice of $V$ (see \cite[Lemma 9]{Jav19}).
Moreover, this implies that given an anisotropic tensor $T \in \mathcal{T}^r_s(\mathcal M_\mathcal{A})$, vector fields $X_1,\ldots,X_r\in {\mathfrak X}(M)$ and one-forms $\theta^1,\ldots,\theta^s\in {\mathfrak X}^*(M)$, the covariant derivative
of $T$ with respect to $E$, denoted $\nabla_ET,$ is given by
\begin{align}
\nabla_ET(\theta^1,\ldots,\theta^s,X_1,\ldots,X_s) = &\nabla_E(T(\theta^1,\ldots,\theta^r,X_1,\ldots,X_s))\nonumber\\&-
\sum_{i=1}^r (T(\theta^1,\ldots,\nabla^v_E\theta^i,\ldots,\theta^r,X_1,\ldots,X_s)
\nonumber\\&-\sum_{j=1}^s (T(\theta^1,\ldots,\theta^r,X_1,\ldots,\nabla^v_EX_j,\ldots,X_s)\label{covder}
\end{align}
while the first term on the right, which is the anisotropic covariant derivative of $v\rightarrow T_v(\theta^1,\ldots,\theta^r,X_1,\ldots,X_s)$, can be expressed as
\begin{multline}\label{fiberder}
\nabla_E (T(\theta^1,\ldots,\theta^r,X_1,\ldots,X_s))(v)=E_{\pi(v)}(T_V(\theta^1,\ldots,\theta^r,X_1,\ldots,X_s))\\-(\dot\partial_{\nabla^v_EV} T)_v(\theta^1,\ldots,\theta^r,X_1,\ldots,X_s)).
\end{multline}
Observe that the anisotropic connection can also be applied to anisotropic vector fields. Indeed, if ${\mathcal X}\in{\mathcal T}^1_0(\mathcal M_\mathcal{A})$, then
\begin{equation}\label{nablaP}
\nabla^v_e{\mathcal X}=\nabla^v_e {\mathcal X}_V-\dot\partial {\mathcal X}_v(\nabla^v_eV),
\end{equation}
for any choice of admissible local extension $V$ of $v$.
We may define then the associated curvature tensor $R_v:{\mathfrak X}(\mathcal M)\times {\mathfrak X}(\mathcal M)\times {\mathfrak X}(\mathcal M)\rightarrow T_{\pi(v)}\mathcal M $ by
\begin{equation*}
R_v(E,H)B
=
\nabla^v_E \nabla_HB
-
\nabla^v_H \nabla_EB
-
\nabla^v_{[E,H]}B.
\end{equation*}
Then $R$ determines an anisotropic tensor field of type $(1,3)$, with the standard identifications (see \cite{Jav19,Jav20}). Moreover, this tensor can be expressed in terms of the curvature tensor $R^V$ of the affine connection $\nabla^V$ for any local admissible extension $V$ of $v$:
\begin{equation*}
R_V(E,H)B
=
R^V(E,H)B
-
P_V(H,B,\nabla^V_EV)
+
P_V(E,B,\nabla^V_HV)
\,,
\end{equation*}
where $P$ is the fiber derivative of $\nabla$ defined as
\begin{equation}\label{nablader}
P_v(E,H,B)
=
\left. \tfrac{\partial}{\partial t} \nabla^{v+tB\vert_{\pi(v)}}_EH \right\vert_{t=0}
\,,
\end{equation}
for $E,H$ and $B$ arbitrary vector fields on $\mathcal M$ (see \cite[Prop. 2.5]{Jav20}).
Let us introduce a privileged anisotropic connection associated with a pseudo-Finsler manifold.
\begin{definition}
The Levi-Civita--Chern connection of $(\mathcal M,L)$ a pseudo-Finsler manifold is the unique torsion-free $g$-compatible anisotropic connection $\nabla$, satisfying namely for each admissible $v\in\mathcal{A}$ and vector fields $E,H\in\mathfrak{X}(\mathcal M)$
\begin{enumerate}[(i)]
\item $\nabla^v_EH-\nabla^v_HE-[E,H]\vert_{\pi(v)}=0$,
\item $\nabla g = 0$.
\end{enumerate}
\end{definition}
By our definition of covariant derivation of anisotropic tensors, the metric-preser\-ving condition $\nabla g = 0$ is also referred to as the almost-compatibility condition due to the equivalent identity for each $v\in\mathcal{A}$ with locally admissible extension $V$ and $e,h,b\in T_{\pi(v)}\mathcal M$ with respective extensions $E$, $H$ and $B$
\begin{equation*} e (g_V(H,B)) \vert_{\pi(v)}
=
g_v(\nabla^v_eH,B)
+
g_v(H,\nabla^v_eB)
+
2C_v(h,b,\nabla^v_eV)
\, ,
\end{equation*}
observing that the fiber derivative of $g$ is $2C$.
Uniqueness can be easily checked from the associated Koszul formula
\begin{multline}\label{koszul}
2g_v(\nabla^v_eH,B)
=
e(g_V(B,H))+h(g_V(E,B))-b (g_V(E,H))
\\
+
g_v([B,E],h)+g_v(e,[B,H])+g_v([E,H],B)
\\
-
2C_v(b,h,\nabla^v_eV)
-
2C_v(e,b,\nabla^v_hV)
+
2C_v(e,h,\nabla^v_bV)
\end{multline}
(see for example \cite[Th. 4]{JSV22}).
\begin{lemma}
Let $\mathcal{X}$ be an anisotropic vector field, $v\in\mathcal{A}$ and $e,h\in T_{\pi(v)}\mathcal M$. If $\nabla$ is an anisotropic connection and $P$ its vertical derivative defined in \eqref{nablader}, then
\begin{equation}\label{dotPartialNablaMathcalX}
(\dot\partial (\nabla_e \mathcal{X}))_v(h)=P_v(e,\mathcal{X}_v,h)+(\nabla_e(\dot\partial \mathcal{X}))_v(h)-(\dot\partial \mathcal{X})_v(P_v(e,v,h)).
\end{equation}
\end{lemma}
\begin{proof}
Consider a locally admissible extension $V$ of $v$ and respective extensions $E$ and $H$ of $e$ and $h$.
By the chain rule, recalling \eqref{nablaP} and observing that the second fiber derivative $\dot\partial^2 {\mathcal X}$ is symmetric,
\begin{align*}
(\dot\partial (\nabla_e \mathcal{X}))_v(h)
= &
\left.\tfrac{\partial}{\partial t} \nabla^{v+th}_e\mathcal{X}\right\vert_{t=0}
\\
= &
\left.\tfrac{\partial}{\partial t}\left( \nabla^{v+th}_e\mathcal{X}_{V+tH}
-
(\dot\partial\mathcal{X})_{v+th}(\nabla^{v+th}_e(V+tH))\right)\right\vert_{t=0}
\\
=
&
\left.\tfrac{\partial}{\partial t} \nabla^{v+th}_e\mathcal{X}_{V} \right\vert_{t=0}
+
\left.\tfrac{\partial}{\partial t} \nabla^v_e\mathcal{X}_{V+tH} \right\vert_{t=0}
-
\left.\tfrac{\partial}{\partial t} (\dot\partial\mathcal{X})_{v+th}(\nabla^v_eV) \right\vert_{t=0}
\\
& \quad
-
\left.\tfrac{\partial}{\partial t} (\dot\partial\mathcal{X})_v(\nabla^{v+th}_eV) \right\vert_{t=0}
-
\left.\tfrac{\partial}{\partial t} (\dot\partial\mathcal{X})_v(\nabla^v_eV+t\nabla^v_eH)) \right\vert_{t=0}
\\
=
&
P_v(e,\mathcal{X}_v,h)
+
\nabla^v_e((\dot\partial \mathcal{X})_V(H))
-
(\dot\partial^2\mathcal{X})_v(\nabla^v_eV,h)
\\
& \quad
-
(\dot\partial\mathcal{X})_v(P_v(e,v,h))
-
(\dot\partial\mathcal{X})_v(\nabla^v_eH)
\\
=
&
P_v(e,\mathcal{X}_v,h)
+
(\nabla_e(\dot\partial\mathcal{X}))_v(h)
-
(\dot\partial \mathcal{X})_v((P_v(e,v,h)).
\end{align*}
\end{proof}
\section{Pseudo-Finsler submersions}\label{pseudo-sub}
Let $ (\mathcal M,L) $ and $ (\mathcal B,\tilde{L}) $ be pseudo-Finsler manifolds with $L:\mathcal{A}\subset T\mathcal M\rightarrow\mathds R$ and $\tilde L:\tilde\mathcal{A}\subset T\mathcal B\rightarrow\mathds R$, and $ \sigma \colon \mathcal M \to \mathcal B $ a submersion, namely, a surjective map with surjective differential. In this situation, it is well-known that the fibers $\mathcal F_q=\sigma^{-1}(q)$ are submanifolds of $\mathcal M$ for every $q\in\mathcal B$ and we will denote by $ \mathcal{V}_p := \ker\mathrm d\sigma\vert_p $ the vertical subspace, namely, the set of vectors tangent to the fibers, which will be called {\em vertical vectors} from now on.
As a counterpart to the vertical vectors, we define the {\em horizontal vectors} in the following way: an admissible vector $v\in\mathcal{A}_p:=\mathcal{A}\cap T_p\mathcal M$ is horizontal if and only if it is $g_v$-orthogonal to $\mathcal{V}_p$, namely, $g_v(v,w)=0$ for each vertical vector $w\in\mathcal{V}_p$. The subset of horizontal vectors of $T_p\mathcal M$ will be denoted by $ \mathcal{H}_p $, and has the structure of a (not necessarily linear) submanifold.
\begin{lemma}\label{horsub}
Given a submersion $ \sigma \colon \mathcal M \to \mathcal B $ between pseudo-Finsler manifolds $ (\mathcal M,L) $ and $ (\mathcal B,\tilde{L}) $,
$\mathcal{H}_p$ is a submanifold of $T_p\mathcal M$ at each $p\in\mathcal M$. Moreover, $\mathcal{H}=\cup_{p\in\mathcal M} \mathcal{H}_p$ is also a submanifold of $T\mathcal M$.
\end{lemma}
\begin{proof}
Let $e_1,\ldots,e_r$ be a basis of $\mathcal{V}_p$ and define the map $f:\mathcal{A}_p\rightarrow \mathds R^r$, given by
$f(v)=(g_v(v,e_1),\ldots,g_v(v,e_r))$. It turns out that $\mathrm{d} f_v(u)=(g_v(u,e_1),\ldots,g_v(u,e_r))$, and then, the kernel of $\mathrm{d} f_v$ is given by the $g_v$-orthogonal vectors to the fiber $\mathcal{V}_p$. As $g_v$ is a scalar product, the dimension of ${\rm ker}(\mathrm{d} f_v)$ is constantly equal to $n-r$, being $n=\dim\mathcal M$ (see \cite[Lemma 2.22]{1983ONeill}). This implies that $f$ is a submersion, and $\mathcal{H}_p=f^{-1}(0)$ a smooth submanifold. To prove that $\mathcal{H}$ is a submanifold of $T\mathcal M$, consider an adapted chart to the submersion to obtain a frame $E_1,\ldots, E_r$ of vertical vector fields in a neighborhood $U$of $p$. Then define $f: \mathcal{A}\cap TU\rightarrow \mathds R$ as above using now the frame $E_1,\ldots, E_r$ instead of $e_1,\ldots,e_r$ and follow the same steps as before.
\end{proof}
\begin{definition}
A submersion $\sigma \colon (\mathcal M,L) \to ( \mathcal B,\tilde{L}) $ between pseudo-Finsler manifolds is called pseudo-Finsler if both of the following conditions are satisfied:
\begin{enumerate}[(i)]
\item the submersion fibers are non-degenerate pseudo-Finsler submanifolds of the total space $(\mathcal M,L)$, namely, the restriction of $L$ to the tangent space to the fibers provides a pseudo-Finsler metric on the fibers,
\item $\mathrm{d}\sigma$ preserves the length of horizontal vectors, namely, $L(v)=\tilde{L}(\mathrm{d}\sigma(v))$ holds for each horizontal $v$.
\end{enumerate}
\end{definition}
Let us see that a pseudo-Finsler submersion induces a pseudo-Riemannian submersion at each tangent space. In the following, we will use the notation $\tilde v=\mathrm{d}\sigma(v)$, $\tilde x=\mathrm{d}\sigma(x),\ldots$ and will use a tilde to denote the elements associated with $\tilde L$, namely, $\tilde g$ denotes its fundamental tensor, $\tilde C$ its Cartan tensor and so on. Moreover, we will denote by $\Sigma$, the indicatrix of $L$, namely, the admissible vectors $v\in \mathcal{A}$ with $L(v)=1$. The indicatrix of $\tilde{L}$ will be denoted by $\tilde\Sigma$.
\begin{lemma}\label{Lemma1}
For each horizontal $ v $, $ \mathrm{d}\sigma\vert_p:T_p\mathcal M\to T_{\sigma(p)}\mathcal B $ is a pseudo-Riemannian submersion from $ (T_p\mathcal M,g_v)$ to $(T_{\sigma(p)}\mathcal B,\tilde g_{\tilde v})$, namely,
\begin{equation}\label{gv1}
g_v(x,y) = \tilde g_{\tilde v}(\tilde x,\tilde y)
\end{equation}
for all $x,y$ $g_v$-orthogonal to the vertical vectors at $T_p\mathcal M$. In particular,
\begin{equation}\label{gv2}
g_v(x,e) = \tilde g_{\tilde v}(\tilde x,\tilde e)
\end{equation}
for every $e\in T_p\mathcal M$ with $\tilde{e} = \mathrm{d}\sigma(e)$.
\end{lemma}
\begin{proof}
As lightlike vectors are in the closure of $\{v\in\mathcal{A}: L(v)\neq0\}$, it suffices to prove \eqref{gv1} under the condition $L(v)\neq0$ and extend it by continuity to the lightcone. By $0$-homogeneity of $ g $, we can furthermore assume that $L(v)=\pm 1$. By definition of a pseudo-Finsler submersion, $g_v(v,v)=\tilde g_{\tilde v}(\tilde v,\tilde v)$ for every horizontal $v$, and the vectors tangent to the indicatrix $\Sigma$ are therefore precisely those $g_v$-orthogonal to $v$. As a consequence, it suffices to prove that for $x,y\in T_v\Sigma$
\begin{equation}\label{igualsenza}
g_v(x,y) = \tilde{g}_{\tilde{v}}(\tilde{x},\tilde{y})
\,.
\end{equation}
Since the result is local, considering an adapted chart, we can assume that the submersion is a canonical projection $\sigma:\mathds R^n\rightarrow \mathds R^m$. Consider now $\Lambda_p=\Sigma_p\cap\mathcal{H}_p$, submanifold of $\Sigma_p$ by Lemma \ref{horsub} as it is the intersection of two transversal smooth submanifolds. Observe that $\mathrm{d} \sigma_p(\Lambda_p)\subset \tilde \Sigma_{\sigma(p)}$, and therefore $\Lambda_p\subset\mathrm{d}\sigma\vert_p^{-1}(\tilde\Sigma_{\sigma(p)})$. Moreover, as $\mathrm{d}\sigma_p$ is a submersion whose fibers are affines subspaces parallel to $\mathcal{V}_p$, the second fundamental form with respect to the trivial connection (the metric connection of any Euclidean metric on $T_p\mathcal M$) of $\mathcal C_p=\mathrm{d}\sigma\vert_p^{-1}(\tilde\Sigma_{\sigma(p)})$ has in its kernel all the vertical vectors. Then, recalling \cite[Eq. (2.5)]{JavSan14}) for the relationship between the fundamental tensor and the second fundamental form of $\Sigma$, \eqref{igualsenza} follows from the following observations:
\begin{enumerate}[(i)]
\item the second fundamental form of $\Sigma_p$ at $v$ having as transversal vector the vector $-v$ coincides with the restriction of $g_v$ to $T_v\Sigma_p\times T_v\Sigma_p$,
\item similarly the second fundamental form of $\tilde\Sigma_{\sigma(p)}$ at $\tilde v$ having as transversal vector $-\tilde{v}$ coincides with the restriction of $\tilde g_{\tilde v}$ to $T_{\tilde v}\tilde\Sigma_{\sigma(p)}\times T_{\tilde v}\tilde\Sigma_{\sigma(p)}$,
\item $\tilde \Sigma_{\sigma(p)}$ can be identified with $\mathcal C_p\cap\mathds R^m$ (or, to be precise, $\mathcal C_p\cap(\lbrace 0 \rbrace^{n-m}\times\mathds R^m)$),
\item $\mathcal C_p$ is invariant under vertical translations,
\item $v-\tilde v$ is tangent to $\mathcal C_p$ (it is indeed vertical), and thus the two second fundamental forms of $\mathcal C_p$ along the fiber $(\mathrm{d}\sigma)_p^{-1}(\tilde v)$ having as respective transversal vectors $-v$ and $-\tilde v$ coincide.
\end{enumerate}
\end{proof}
Recall that the Legendre map $\mathcal{L}_p\colon T_p\mathcal M\to T_p\mathcal M^*$ of a pseudo-Finsler metric is defined on each admissible $v$ as
$\mathcal{L}(v)=g_v(v,\cdot)$. If the pseudo-Finsler metric is defined in the whole tangent space, then $\mathcal{L}$ is bijective whenever $\dim\mathcal M\geq 3$ (see \cite{Min15,RuSu15}).
\begin{lemma}\label{legendrelemma}
If the Legendre map $\mathcal{L}$ at $p\in\mathcal M$ is injective (bijective), then each $\tilde v\in \tilde\mathcal{A}\cap T_{\sigma(p)}\mathcal B$ admits at most one (exactly one) horizontal vector $v\in T_p\mathcal M$ satisfying $\mathrm{d}\sigma(v)=\tilde v$.
\end{lemma}
\begin{proof}
For any two horizontal vectors $v_1,v_2\in T_p\mathcal M$ projecting onto $\tilde v$, let us prove $\mathcal{L}_p(v_1)=\mathcal{L}_p(v_2)$. Observe that $\mathcal{L}_p(v_1)$ and $\mathcal{L}_p(v_2)$ have the same kernel $\mathcal{V}_p$, such that each $g_{v_1}$-horizontal $x_1$ admits a vertical vector $w$ such that $x_2=x_1+w$ is $g_{v_2}$-horizontal. Denoting by $\tilde{x}$ their shared projection, by Lemma \ref{Lemma1},
\begin{equation*}
\mathcal{L}_p(v_1)(x_1)=g_{v_1}(v_1,x_1)=\tilde g_{\tilde v}(\tilde v,\tilde x)=g_{v_2}(v_2,x_2)=g_{v_2}(v_2,x_1)=\mathcal{L}_p(v_2)(x_1)
\,,
\end{equation*}
or specifically $\mathcal{L}_p(v_1)=\mathcal{L}_p(v_2)$, which implies that $v_1=v_2$ by injectivity of $\mathcal{L}_p$. Let us now prove the existence of the horizontal lift when $\mathcal{L}_p$ is invertible. Fixing an admissible vector $\tilde v$ at $\sigma(p)$, and a co-vector $\omega$ at $p$ with kernel $\mathcal{V}_p$ that acts on a transversal subspace as $\tilde g_{\tilde v}(\tilde v,\cdot)$ acts on its projection, it is straightforward to check that $v=(\mathcal{L}_p)^{-1}(\omega)$ is horizontal and projects onto $\tilde v$.
\end{proof}
\begin{remark}
Throughout this paper, we will need to consider only a local horizontal extension $V$ of a horizontal vector $v$. As the Legendre transform is locally invertible, the existence of these local horizontal extensions is always guaranteed for every horizontal $v$ by a local application of Lemma \ref{legendrelemma}. Indeed, consider an extension $\tilde V$ of the projection $\tilde v$ and a neighborhood of $v$ in $\mathcal{A}$ where the Legendre transform is injective. Then the one-form constructed in the proof of the last lemma is on the image of the Legendre transform for a small enough neighborhood of $g_v(v,\cdot)$ and its inverse by $\mathcal{L}$ gives a horizontal vector field which projects onto $\tilde V$.
\end{remark}
\begin{corollary}
For each horizontal $v$, $g_v$-horizontal $x$ and $y$ and arbitrary $e$ (at $p$),
\begin{equation}
C_v(x,y,e) = \tilde C_{\tilde v}(\tilde x,\tilde y,\tilde e)
- \tfrac{1}{2}g_v(\mathrm{I\!I}^\mathcal{H}_v(x,y),e)
\,,
\end{equation}
where $\mathrm{I\!I}^\mathcal{H}_v$ is the second fundamental form of $\mathcal{H}_p$ as a submanifold of $(T_p \mathcal M,g_v) $.
\end{corollary}
\begin{proof}
Observe that the space tangent to $\mathcal{H}_p$ at $v$ is the set of $g_v$-horizontal vectors (see the proof of Lemma \ref{horsub} and observe that this space is the kernel of $\mathrm{d}_vf$). Moreover, consider any smooth curve $\gamma$ of $\mathcal{H}_p$ with $\gamma(0)=v$ and $\dot\gamma(0)=x$, and vector fields $Y$ and $Z$ along $\gamma$ tangent to $\mathcal{H}_p$ with $Y(0)=y$ and $Z(0)=e-w$ for some (constant) vertical $w$, such that $Y$ and $Z$ are $g_\gamma$-orthogonal to $\mathcal{V}_p$ at each point of $\gamma$. By \eqref{gv1} we have
\begin{equation*}
g_\gamma(Y,Z+w) = \tilde g_{\tilde\gamma}(\tilde Y,\tilde Z)
\,.
\end{equation*}
By differentiating both sides,
\begin{equation*}
2C_\gamma(Y,Z+w,\dot\gamma) + g_\gamma(\dot Y,Z+w) + g_\gamma(Y,\dot Z)
=
2\tilde C_{\tilde\gamma}(\tilde Y,\tilde Z,\dot{\tilde\gamma}) + \tilde g_{\tilde\gamma}(\dot{\tilde Y},\tilde Z) + \tilde g_{\tilde\gamma}(\tilde Y,\dot{\tilde Z}).
\end{equation*}
By \eqref{gv1}, and as $\mathrm{d}\sigma(\dot X)=\dot{\tilde X}$ and $\mathrm{d}\sigma(\dot Y)=\dot{\tilde Y}$ by linearity of $\mathrm{d}\sigma$, this cancels to
\begin{equation*}
2C_v(y,z+w,x) + g_v(\dot Y,w) = 2\tilde C_{\tilde v}(\tilde y,\tilde z,\tilde x)
\,.
\end{equation*}
To conclude, the second fundamental form $\mathrm{I\!I}^\mathcal{H}_v(x,y)$ is the vertical part of the covariant differentiation of $Y$ along $x$.
\end{proof}
\section{The fundamental tensors T and A}\label{fundTA}
To introduce the fundamental tensors of a submersion we will need first to define the vertical and horizontal parts of each vector in $T\mathcal M$. In the pseudo-Finsler case, we will need a further condition to do this decomposition, namely, the vertical space must be non-degenerate for all $v\in \mathcal{A}$. By the definition of pseudo-Finsler submersion, this will be true for vertical and horizontal vectors, and we can reduce the domain where $L$ is defined to ensure this condition, which does always hold when $g_v$ is positive definite. As a consequence $\mathcal{A}$ could be non-connected. Then if $e\in T_{\pi(v)}\mathcal M$, one has the $g_v$-decomposition
\[e= e^\top_v+e^\bot_v\]
with $e^\top_v$ vertical and $e^\bot_v$ $g_v$-orthogonal to $\mathcal{V}_p$. In particular, given a vector field $X\in{\mathfrak X}(\mathcal M)$, we can define $X^\top$ and $X^\bot$ as the anisotropic vector fields such that the evaluation at $v\in \mathcal{A}_p$ is given by $X^\top_v=(X_p)^\top_v$ and $X^\bot_v=(X_p)^\bot_v$, respectively.
\begin{lemma}\label{XT}
Given a pseudo-Finsler submersion and a vector field $X\in{\mathfrak X}(\mathcal M)$, then $X^\top,X^\bot\in {\mathcal T}^1_0(\mathcal M)$.
\end{lemma}
\begin{proof}
It suffices to prove the smoothness of $X^\top$. Consider in a neighborhood of $p\in\mathcal M$ coordinates $(x_1,\ldots,x_n)$ adapted to the submersion and $v\in \mathcal{A}_p$. Some constant coefficient combination of the induced vector fields $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$ produces a non-degenerate reference frame $e_1,\ldots,e_n$ such that the first vectors $e_1,\ldots,e_r$ form a reference frame for the vertical space in a neighborhood $U$ of $p$,
and for all $1\leq s\leq n$, $e_1,\ldots,e_s$ spans a $g_w$ non-degenerate subspace for $w$ in a neighborhood of $v$
(this can be done by perturbing the frame of partial vector fields by reducing the neighborhood if necessary). Under these conditions, we can apply the Gram-Schmidt process to obtain a $g_w$-orthonormal basis $e_1(w),\ldots,e_n(w)$ with a smooth dependence on $w$ in some neighborhood of $v$, such that $e_1(w),\ldots, e_r(w)$ is a basis of the vertical subspace. Locally, $X^\top_w = \sum_{i=1}^r \frac{g_w(e_i(w),X)}{g_w(e_i(w),e_i(w))} e_i(w)$, smooth in $w$.
\end{proof}
Observe that Lemma \ref{XT} can be applied to one single vector $e\in T_p\mathcal M$, obtaining a smooth map $e^\top:
\mathcal{A}_p\rightarrow T_{\pi(e)}\mathcal M$.
The letter $p$ will always denote a point of the manifold $\mathcal M$, the letters $s,u,w$ will always denote vertical vectors, namely elements of $\mathcal{V}_p$, $v$ an admissible vector, namely an element of $\mathcal{A}_p$, and $x,y,z$ $ g_v $-horizontal vectors of $T_p\mathcal M$. The capital letters $S,U,W$ will always denote vertical vector fields, $V$ a locally admissible extension of $v$, and $X,Y,Z$ locally $g_V$-horizontal vector fields, preferably projectable onto some vector fields $\tilde X,\tilde Y,\tilde Z$ of the base manifold in the sense that $\mathrm{d}\sigma\cdot X=\tilde X\circ\sigma$ and so on.
The covariant derivative $\nabla^v_h E$ of an arbitrary extension $E$ of $e$ along each vector $h$ at $p$ defines the $1$-form $\nabla^v E \colon \mapsto \nabla^v_h E$ on $T_p\mathcal M$ which splits as
\begin{multline*}
\nabla^v E
=
(\nabla^v (E^\top+E^\bot))^\top_v+(\nabla^v (E^\top+E^\bot))^\bot_v
\\
=
(\nabla^v E^\top)^\top_v
+
(\nabla^v E^\bot)^\top_v
+
(\nabla^v E^\top)^\bot_v
+
(\nabla^v E^\bot)^\bot_v
\,,
\end{multline*}
whose middle terms are independent of $E$ and define the tensorial expression
\begin{equation*}
\chi^v e
=
(\nabla^v E^\bot)^\top_v
+
(\nabla^v E^\top)^\bot_v
\,.
\end{equation*}
Recall by Lemma \ref{XT} that $ E^\top $ and $ E^\bot $ are anisotropic vector fields. Furthermore, the tensor $\chi$ is well-defined if we substitute $E$ for an anisotropic vector field $\mathcal{X}$, with the definition that $\mathcal{X}^\top \colon v \mapsto (\mathcal{X}_v)^\top_v$ and similarly $\mathcal{X}^\bot \colon v \mapsto (\mathcal{X}_v)^\bot_v$.
\begin{definition}\label{ATDefinition}
Assume that the vertical space is $g_v$-non-degenerate for all $v\in \mathcal{A}$. Then the fundamental anisotropic tensors of a pseudo-Finsler submersion are defined as the
$(1,2)$-anisotropic tensors $\mathbf T$ and $\mathbf A$ given by
\begin{equation*}
\mathbf T^v_b e = \chi^v_{b^\top_v}e, \quad \mathbf A^v_b e = \chi^v_{b^\perp_v}e,
\end{equation*}
for any $v\in\mathcal{A}_p$ and $e,b\in T_p\mathcal M$. Therefore $\chi=\mathbf T+\mathbf A$.
\end{definition}
We may check the following properties, which differ slightly from \cite{1966ONeill}. Note our use of the musical isomorphism to define $ C^\sharp $ as the symmetric type $ (1,2) $ tensor
determined for arbitrary vectors $ b $, $ e $ and $ h $ at $ p $ by
\begin{equation*}
g_v(b,C^\sharp_v(e,h))
=
C_v(b,e,h)
\,,
\end{equation*}
and by
homogeneity, $ C^\sharp_v(v,\cdot) = 0 $.
\begin{proposition}\label{dotPartialTopProposition}
For each admissible $ v $, vertical $ w $ and $ g_v $-horizontal $ x $ at $ p $,
\begin{equation*
(\dot\partial w^\top)_v = (\dot\partial w^\bot)_v = 0
\,, \quad
(\dot\partial x^\top)_v = 2C^\sharp_v(x,\cdot)^\top_v
\,, \quad
(\dot\partial x^\bot)_v = -2C^\sharp_v(x,\cdot)^\top_v
\,.
\end{equation*}
That is to say, for an arbitrary vector $e$ at $p$,
\begin{equation}\label{vertder}
(\dot\partial e^\top)_v= 2C^\sharp_v(e^\perp_v,\cdot)^\top_v, \quad (\dot\partial e^\perp)_v= -2C^\sharp_v(e^\perp_v,\cdot)^\top_v.
\end{equation}
\end{proposition}
\begin{proof}
The identities involving $w$ are due to $w^\top$ being identically equal to $w$ and $w^\bot$ being identically zero. As for the identities involving $x$, consider an arbitrary $ e \in T_p \mathcal M $ along with a smooth one real parameter family of $ g_{v+te} $-orthonormal bases $ (u_1(t),\ldots,u_{ r}(t)) $ of $ \mathcal{V}_p $ and denote by $ (\dot{u}_1(t),\ldots,\dot{u}_r(t)) $ their derivatives with respect to $t$. Observe that by continuity, $\varepsilon_i:=g_{v+te}(u_i(t),u_i(t))$ is constantly equal to $1$ or $-1$ and it does not depend on $t$. Then
\begin{equation*}
x^\top_{v+te}=\sum_{i=1}^r \varepsilon_i g_{v+te}(u_i(t),x) u_i(t)
\,,
\end{equation*}
which differentiates at $t=0$ to
\begin{equation*}
(\dot\partial x^\top)_v(e)
=
\sum_{i=1}^r \varepsilon_i \left( 2C_v(u_i(0),x,e) u_i(0)
+
g_v(\dot u_i(0),x) u_i(0)
+
g_v(u_i(0),x) \dot{u}_i(0) \right)
\,.
\end{equation*}
As $u_i(0)$ and $\dot u_i(0)$ are vertical, the last two terms are zero and the results follow straightforwardly. The last identities \eqref{vertder} follow from the first part of the proposition and the fact that $0=\dot\partial e=\dot\partial (e^\top)_v+\dot\partial (e^\perp)_v$, and then
$\dot\partial (e^\perp)_v=-\dot\partial (e^\top)_v$.
\end{proof}
\begin{lemma}\label{Lemma3}
For each admissible $ v $, vertical $ w $ and $ g_v $-horizontal $ x $ at $ p $ and a vertical vector field $ U $ with $ u = U_p $, any locally admissible extension $ V $ of $v$ and locally $ g_V $-horizontal vector field $ Y $ with $ y = Y_p $,
\begin{align}\label{UTA}
\nabla^v_w U &= \mathbf T^v_w u + (\nabla^v_w U)^\top_v,& \nabla^v_x U &= \mathbf A^v_x u + (\nabla^v_x U)^\top_v,
\\ \label{YTA}
\nabla^v_w Y &= (\nabla^v_w Y)^\bot_v + \mathbf T^v_w y - 2C^\sharp_v(y,\nabla^v_w V)^\top_v, & \nabla^v_x Y &= (\nabla^v_x Y)^\bot_v + \mathbf A^v_x y - 2C^\sharp_v(y,\nabla^v_x V)^\top_v.
\end{align}
Furthermore, if $ Y $ is projectable,
\begin{equation}\label{Lieconversion}
(\nabla^v_w Y)^\bot_v = \mathbf A^v_y w.
\end{equation}
\end{lemma}
\begin{proof}
From the decomposition of the covariant derivative of an arbitrary vector field $ E $ with $ e = E_p $ and by definition of the tensors $ \mathbf T $ and $ \mathbf A $
\begin{equation*}
\nabla^v E
=
(\nabla^v E^\top)^\top_v
+
(\mathbf T+\mathbf A)^v e
+
(\nabla^v E^\bot)^\bot_v
\,,
\end{equation*}
we obtain for a vertical vector field $ U $ with $ u = U_p $
\begin{equation}\label{nablaU}
\nabla^v U
=
(\nabla^v U)^\top_v
+
(\mathbf T+\mathbf A)^v u,
\end{equation}
which implies \eqref{UTA}.
Moreover, from the above decomposition, taking into account that $Y=Y^\bot_V$ and $Y^\top_V=0$, and using \eqref{nablaP} and Proposition \ref{dotPartialTopProposition},
\begin{multline}\label{nablaY}
\nabla^v Y
=
(\mathbf T+\mathbf A)^v y
+
(\nabla^v Y)^\bot_v
-
(\dot\partial Y^\top)_v(\nabla^v V)^\top_v
-
(\dot\partial Y^\bot)_v(\nabla^v V)^\bot_v
\\
=
(\mathbf T+\mathbf A)^v y
+
(\nabla^v Y)^\bot_v
-
2C^\sharp_v(y,\nabla^v V)^\top_v
\,,
\end{multline}
which is equivalent to \eqref{YTA}. For the last identity \eqref{Lieconversion} observe that the Lie bracket of a vertical vector field and a projectable vector field is vertical, and then one has that $(\nabla^v_w Y)^\bot_v=(\nabla^v_y W)^\bot_v=\mathbf A^v_yw$, being $W$ a vertical extension of $w$, where in the last equality we have used \eqref{UTA}.
\end{proof}
\begin{proposition}\label{TAProposition}
For each admissible $ v $, arbitrary $e$, $ g_v $-horizontal $ x $ and vertical $ u $ and $ w $ at $p$,
\begin{enumerate}[(i)]
\item $\mathbf T^v_e$ and $\mathbf A^v_e$ are skew-symmetric on $ T_p \mathcal M $, and map $g_v$-horizontal vectors into vertical ones and vice versa,
\item if $v$ is vertical, then $\mathbf T^v$ satisfies $\mathbf T^v_u w = \mathbf T^v_w u$,
\item if $ v $ is horizontal, then $\mathbf A^v$ satisfies $\mathbf A^v_x v = - \mathbf A^v_v x$ and in particular $\mathbf A^v_vv = 0$.
\end{enumerate}
\end{proposition}
\begin{proof}
Consider a $ g_v $-horizontal vector $ y $ at $ p $.
By \eqref{nablaY} and \eqref{nablaU}, for any local choice of an admissible extension $ V $ of $v$, vertical extension $ U $ of $u$ and $ g_V $-horizontal extension $ Y $ of $y$, it follows that
\begin{multline*}
g_v((\mathbf T+\mathbf A)^v y,u)
=
g_v(\nabla^v Y,u) + 2C_v(y,u,\nabla^v V)
=
- g_v(y,\nabla^v U)
\\
=
- g_v(y,(\mathbf T+\mathbf A)^v u)
\,,
\end{multline*}
where we have also used in the second equality that $ g_V(Y,U)=0$, and therefore $0=e(g_V(Y,U))=g_v(\nabla^v_e Y,u)+ g_v(y,\nabla^v_e U) + 2C_v(y,u,\nabla^v_e V)$ for any $e\in T_p\mathcal M$. This proves the skew-symmetry. The last statement of $(i)$ is a direct consequence of Lemma \ref{Lemma3}.
The symmetry of $ \mathbf T^v $ for vertical vectors is because it coincides with the second fundamental form of the submersion fibers (see for example \cite[\S 3.1]{2022HuberJavaloyes}), while the antisymmetry $ \mathbf A^v_x v = - \mathbf A^v_v x $ under the assumption that $ v \in \mathcal{H}_p $ can be obtained from the Koszul formula \eqref{koszul} under the following form, where $ W $ is a local vertical extension of $ w $, $ V $ a locally horizontal extension of $ v $ and $ X $ a locally $ g_V $-horizontal extension of $ x $,
\begin{equation}\label{vwxY}
g_v(w,\nabla^v_x V)
=
\tfrac{1}{2} g_v([X,V],w)
-C_v(x,w,\nabla^v_v V)
\,,
\end{equation}
(recall that the Lie bracket of a projectable vector field with another vertical one is vertical and $C_v(v,\cdot,\cdot)=0$ by homogeneity).
By \eqref{YTA} and \eqref{vwxY}, we obtain
\begin{equation}\label{vwxy}
g_v(w,\mathbf A^v_x v)
=
g_v(w,\nabla^v_x V)
=
\tfrac{1}{2} g_v([X,V],w)
- C_v(x,w,\nabla^v_{v} V)
\,,
\end{equation}
where, using again \eqref{YTA}, we can compute
\begin{multline*}
g_v([X,V],w)
=
g_v(\nabla^v_xV,w)-g_v(\nabla^v_vX,w)
\\
=
g_v(\mathbf A^v_xv,w)-g_v(\mathbf A^v_vx,w)
+2C_v(x,\nabla^v_vV,w)
\,,
\end{multline*}
such that \eqref{vwxy} simplifies to
\begin{equation*}
g_v(w,\mathbf A^v_x v)
=
\tfrac{1}{2} g_v([X,V],w)
-C_v(x,w,\nabla^v_v V)
=
\tfrac{1}{2}g_v(\mathbf A^v_x v - \mathbf A^v_v x,w)
\,.
\end{equation*}
Conclude by non-degeneracy.
\end{proof}
While we have used the tilde $ \tilde{\text{ }} $ to designate features of the base manifold $\mathcal B$ of the submersion, let us use the caret $ \hat{\text{ }} $ for the submersion fibers as submanifolds of $\mathcal M$.
\begin{definition}
Let $ \hat{Q} $ be the anisotropic tensor given for each admissible $ v $ and $ g_v $-horizontal $ x $ by $ \hat{Q}^v_x = \hat{Q}^v x = 0 $ and for each vertical $ u,w $ by
\begin{multline*}
\hat{Q}^v_u w
=
-
\left(
\mathbf T^v_v C^\sharp_v(u,w)
+
C^\sharp_v(\mathbf T^v_v u,w)
+
C^\sharp_v(u,\mathbf T^v_v w)
\right. \\ \left.
+
C^\sharp_v(C^\sharp_v(u,w)^{\top}_v,\mathbf T^v_v v)
-
C^\sharp_v(C^\sharp_v(u,\mathbf T^v_v v)^{ \top}_v,w)
-
C^\sharp_v(u,C^\sharp_v(w,\mathbf T^v_v v)^{\top}_v)
\right)^\top_v
\,.
\end{multline*}
\end{definition}
\begin{proposition}[Gauss formula] \label{GaussFormulaVert}
For any vertical $ u $, vertical admissible $ v $, and vertical $ w $ at $ p $ and any vertical extension $ W $ of $ w $,
\begin{equation*}
(\nabla^v_u W)^\top_v
=
\hat\nabla^v_u W
+
\hat{Q}^v_u w
\,.
\end{equation*}
\end{proposition}
\begin{proof}
Completely analogous to the Gauss formula for pseudo-Finsler submanifolds (see for example \cite[Eq. (17) and Lemma 2]{2022HuberJavaloyes} for a reference using the same approach).
\end{proof}
\begin{definition}\label{tildeQ}
Let $ \tilde{Q} $ be the anisotropic tensor given for each admissible $ v $ and vertical $ w $ by $ \tilde{Q}^v_w = \tilde{Q}^v w = 0 $ and for each $ g_v $-horizontal $ x,y $ by
\begin{equation*}
\tilde{Q}^v_x y
=
\left(\ \mathbf A^v_v C^\sharp_v(x,y) + C^\sharp_v(\mathbf A^v_v x,y) + C^\sharp_v(x,\mathbf A^v_v y) \right)^\bot_v
\,.
\end{equation*}
\end{definition}
\begin{proposition}[Dual Gauss formula]\label{VXYhor}
For each projectable and horizontal $ V $ defined in a neighborhood of $ p $, and projectable $ X $ and $ Y $ locally $ g_V $-horizontal,
\begin{equation*}
(\nabla^V_X Y)^\bot_V
=
(\tilde\nabla^{\tilde{V}}_{\tilde{X}} \tilde{Y})^\ast_V
+
\tilde{Q}^V_X Y.
\end{equation*}
where $(\cdot)^*_V$ denotes the $g_V$-horizontal lift.
\end{proposition}
\begin{proof}
Let $ v = V_p $, $ x = X_p $ and $ y = Y_p $. Using the Koszul formula \eqref{koszul} on $ \mathcal B $
\begin{multline*}
\tilde{g}_{\tilde{v}}(\tilde{x},\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{Z})
=
\tilde{y}(\tilde{g}_{\tilde{V}}(\tilde{X},\tilde{Z}))
+\tilde{z}(\tilde{g}_{\tilde{V}}(\tilde{Y},\tilde{X}))
-\tilde{x}(\tilde{g}_{\tilde{V}}(\tilde{Y},\tilde{Z}))
\\
+\tilde{g}_{\tilde{v}}([\tilde{X},\tilde{Y}],\tilde{z})+\tilde{g}_{\tilde{v}}(\tilde{y},[\tilde{X},\tilde{Z}])+\tilde{g}_{\tilde{v}}(\tilde{x},[\tilde{y},\tilde{z}])
\\
-2\tilde{C}_{\tilde{v}}(\tilde{x},\tilde{z},\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{V})
-2\tilde{C}_{\tilde{v}}(\tilde{y},\tilde{x},\tilde\nabla^{\tilde{v}}_{\tilde{z}} \tilde{V})
+2\tilde{C}_{\tilde{v}}(\tilde{y},\tilde{z},\tilde\nabla^{\tilde{v}}_{\tilde{x}} \tilde{V})
\end{multline*}
can be lifted to
\begin{multline*}
2g_v(x,(\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{Z})^\ast_v)
=
y(g_V(X,Z))+z(g_V(Y,X))-x(g_V(Y,Z))
\\
+g_v([X,Y],z)+g_v(y,[X,Z])+g_v(x,[Y,Z])
\\
-2C_v(x,z,(\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{V})^\ast_v)
-2C_v(y,x,(\tilde\nabla^{\tilde{v}}_{\tilde{z}} \tilde{V})^\ast_v)
+2C_v(y,z,\tilde\nabla^{\tilde{v}}_{\tilde{x}} \tilde{V})
\,.
\end{multline*}
This is because by Proposition \ref{dotPartialTopProposition}, $\tilde{y}(\tilde{g}_{\tilde{V}}(\tilde{X},\tilde{Z}))=\tilde y(g_V(X,Z)\circ\sigma)=y(g_V(X,Z))$, and so on for the similar terms, and also because \[\tilde{g}_{\tilde{v}}([\tilde{X},\tilde{Y}],\tilde{z})=g_v([\tilde{X},\tilde{Y}]^*_v,z))
=g_v([X,Y]^\perp_v,z)=g_v([X,Y],z).\]
Making use of the Koszul formula \eqref{koszul} for $\mathcal M$, it follows that
\begin{multline}\label{koszulmedia}
2g_v(x,\nabla^v_y Z - (\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{Z})^\ast_v)
=
-2C_v(x,z,\nabla^v_y V - (\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{V})^\ast_v)
-2C_v(y,x,\nabla^v_z V - (\tilde\nabla^{\tilde{v}}_{\tilde{z}} \tilde{V})^\ast_v)
\\
+2C_v(y,z,\nabla^v_x V - (\tilde\nabla^{\tilde{v}}_{\tilde{x}} \tilde{V})^*_v),
\end{multline}
where, by our assumption that $ V $ is horizontal, we may substitute $ Z $ and $ Y $ for $ V $ to furthermore obtain
$
g_v(x,\nabla^v_v V - (\tilde\nabla^{\tilde{v}}_{\tilde{v}} \tilde{V})^\ast_v)=0.$ As $(\nabla^v_v V)^\top=\mathbf A^v_vv=0$ by part $(iii)$ of Proposition \ref{TAProposition}, we conclude that $\nabla^v_v V = (\tilde\nabla^{\tilde{v}}_{\tilde{v}} \tilde{V})^\ast_v$,
and, replacing now only $Z$ by $V$ in \eqref{koszulmedia}, we get
\begin{equation*}
g_v(x,\nabla^v_y V - (\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{V})^\ast_v)
=
0
\,,
\end{equation*}
which, using \eqref{YTA} under the form $ (\nabla^v_y V)^\top_v = \mathbf A^v_y v $, we find that $\nabla^v_y V - (\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{V})^\ast_v=\mathbf A^v_yv$, so using this (and similar formulas replacing $y$ with $x$ and $z$)
in \eqref{koszulmedia}, we obtain that
\begin{equation*}
g_v(x,\nabla^v_y Z - (\tilde\nabla^{\tilde{v}}_{\tilde{y}} \tilde{Z})^\ast_v)
=
-g_v(x,C^\sharp_v(z,\mathbf A^v_y v))
-g_v(x,C^\sharp_v(y,\mathbf A^v_z v))
+g_v(\mathbf A^v_x v,C^\sharp_v(y,z))
\,.
\end{equation*}
Conclude the result at $p$ by Proposition \ref{TAProposition}, namely, using that $\mathbf A^v_xv=-\mathbf A^v_vx$ and then the skew-symmetry of $\mathbf A^v_v$, and the non-degeneracy of $ g_v $. The same process completes the proof on the neighbourhood where $V$ is horizontal and $X$ and $Y$ are $g_V$-horizontal.
\end{proof}
\begin{proposition}\label{TAProposition2}
For each horizontal $ v $ at $p$ with a local horizontal projectable extension $V$ with its projection $\tilde V$ satisfying that $\tilde{\nabla}^{\tilde{v}}\tilde{V} = 0$, and $ g_v $-horizontal vectors $ x,y $ with $g_V$-horizontal extensions $ X $ and $ Y $,
\begin{align}\label{Axy}
\mathbf A^v_x y
=&
\tfrac{1}{2}[X,Y]^\top_v
+
(\mathbf T+\mathbf A)^v_{C^\sharp_v(x,y)} v
+
C^\sharp_v(\mathbf A^v_x v,y)^\top_v
-
C^\sharp_v(x,\mathbf A^v_y v)^\top_v
\,,\\ \label{Axv}
\mathbf A^v_x v
=&
\tfrac{1}{2}[X,V]^\top_v\,,
\end{align}
and $ \mathbf A^v $ satisfies the almost-antisymmetry $\mathbf A^v_x y + \mathbf A^v_y x = 2(\mathbf T+\mathbf A)^v_{C^\sharp_v(x,y)} v$.
\end{proposition}
\begin{proof}
From the Koszul formula \eqref{koszul} where $ W $ is a local vertical extension of a vertical $ w $,
\begin{equation}\label{vwxY2}
g_v(w,\nabla^v_x Y)
=
\tfrac{1}{2} g_v([X,Y],w)
-C_v(w,y,\nabla^v_x V)
-C_v(x,w,\nabla^v_y V)
+C_v(x,y,\nabla^v_w V)
\,,
\end{equation}
and using \eqref{Lieconversion} with $Y=V$ and applying \eqref{UTA} and \eqref{YTA}, it follows that
\begin{equation*}
\nabla^v_w V
=
(\nabla^v_w V)^\top_v
+
(\nabla^v_v W)^\bot_v
=
\mathbf T^v_w v
+
\mathbf A^v_v w
\,.
\end{equation*}
Replacing the above identity in the last term of \eqref{vwxY2}, we obtain two terms, being the first one
\begin{equation*}
C_v(x,y,\mathbf T^v_w v)
=
g_v(C^\sharp_v(x,y)^\top_v,\mathbf T^v_w v)
=
-g_v(\mathbf T^v_{C^\sharp_v(x,y)^\top_v} w,v)
=
g_v(w,\mathbf T^v_{C^\sharp_v(x,y)^\top_v} v),
\end{equation*}
where we have applied parts $(i)$ and $(ii)$ of Proposition \ref{TAProposition}, and the second one,
\begin{equation*}
C_v(x,y,\mathbf A^v_v w)
=
g_v(C^\sharp_v(x,y),\mathbf A^v_v w)
=
-g_v(\mathbf A^v_v C^\sharp_v(x,y),w)
\,,
\end{equation*}
where we have applied again part $(i)$ of Proposition \ref{TAProposition}. Using \eqref{YTA} and \eqref{vwxY2},
\begin{align}\nonumber
g_v(w,\mathbf A^v_x y)
&=
g_v(w,\nabla^v_x Y) + 2C_v(y,\nabla^v_x V,w)\\\nonumber
&= \tfrac{1}{2} g_v([X,Y],w)
+C_v(w,y,\nabla^v_x V)
-C_v(x,w,\nabla^v_y V)
+C_v(x,y,\nabla^v_w V)\\
&=\tfrac{1}{2} g_v([X,Y],w)
+C_v(w,y,\nabla^v_x V)
-C_v(x,w,\nabla^v_y V)
+g_v(w,\mathbf T^v_{C^\sharp_v(x,y)^\top_v} v)\nonumber \\
&\quad\quad\quad\quad-g_v(\mathbf A^v_v C^\sharp_v(x,y),w).\label{vwxy2}
\end{align}
Conclude \eqref{Axy} taking into account that
\begin{equation*}
\mathbf A^v_v C^\sharp_v(x,y)^\bot_v = -\mathbf A^v_{C^\sharp_v(x,y)} v
\,,
\end{equation*}
and using $\nabla^v_x V = \mathbf A^v_x v$ and $\nabla^v_y V = \mathbf A^v_y v$ by Proposition \ref{VXYhor}.
Finally, \eqref{Axv} and the almost-antisymmetry of $\mathbf A$ are straightforward consequences of \eqref{Axy}.
\end{proof}
The dual Gauss formula can be extended from Proposition \ref{VXYhor} to the following formula \eqref{GaussFormula}.
\begin{corollary}
For each horizontal $ v $, arbitrary vector $ e $ at $ p $ and arbitrary projectable vector field $ H $ with $ h = H_p $,
\begin{equation}\label{GaussFormula}
(\nabla^v_e H)^\bot_v
=
(\tilde\nabla^{\tilde{v}}_{\tilde{e}} \tilde{H})^\ast_v
+
\tilde{Q}^v_e h
+
(\mathbf T^v_e h + \mathbf A^v_e h + \mathbf A^v_h e)^\bot_v
\,.
\end{equation}
\end{corollary}
\begin{proof}
Consider a locally projectable horizontal extension $ V $ of $ v $.
Now by definition
\begin{equation}\label{nablaHfirst}
(\nabla^v_e H)^\perp_v=(\nabla^v_e (H^\perp))^\perp_v+(\nabla^v_e (H^\top))^\perp_v
=(\nabla^v_e (H^\perp))^\perp_v+\mathbf T^v_e (h^\top_v)+\mathbf A^v_e (h^\top_v).
\end{equation}
Consider a $g_V$-horizontal projectable extension $X$ of $e^\perp_v$ and a vertical extension $U$ of $e^\top_v$. Then,
using \eqref{vertder} to cancel the fiber derivative in the first equality, Proposition \ref{VXYhor}, the fact that the Lie bracket of projectable vectors is vertical when one of them is vertical and \eqref{UTA},
\begin{align*}
(\nabla^v_e (H^\perp))^\perp_v&=(\nabla^v_e (H^\perp_V))^\perp_v
=(\nabla^v_X (H^\perp_V))^\perp_v+(\nabla^v_U (H^\perp_V))^\perp_v\\
&=
(\tilde\nabla^{\tilde v}_{\tilde e}\tilde H)^*_v +\tilde Q^v_e h+(\nabla^v_{H^\perp_V}U)^\perp_v
=(\tilde\nabla^{\tilde v}_{\tilde e}\tilde H)^*_v +\tilde Q^v_e h+\mathbf A^v_h (e^\top_v).
\end{align*}
Replacing the last identity in \eqref{nablaHfirst}, and taking into account that
\[\mathbf T^v_e (h^\top_v)+\mathbf A^v_e (h^\top_v)+\mathbf A^v_h (e^\top_v)=(\mathbf T^v_e h + \mathbf A^v_e h + \mathbf A^v_h e)^\bot_v\]
as a consequence of part $(i)$ of Proposition \ref{TAProposition}, we obtain \eqref{GaussFormula}.
\end{proof}
We can thus remove the choice of an extension of $ v $ from the hypotheses.
\begin{proposition}[Dual Gauss formula]\label{GaussProposition}
For each horizontal $ v $, $ g_v $-horizontal $ x $ and $ y $ at $ p $ and any projectable extension $ Y $ of $ y $,
\begin{equation*}
(\nabla^v_x Y)^\bot_v
=
(\tilde\nabla^{\tilde{v}}_{\tilde{x}} \tilde{Y})^\ast_v
+
\tilde{Q}^v_x y
\,.
\end{equation*}
\end{proposition}
One very nicely behaved consequence of the dual Gauss formula is the relationship between geodesics of the base and horizontal geodesics.
\begin{corollary}\label{horizontalGeodesicCorollary}
A horizontal curve is a geodesic of $(\mathcal M,L)$ if and only if it is the horizontal lift of a geodesic of $(\mathcal B,\tilde L)$.
\end{corollary}
\begin{proof}
By Proposition \ref{GaussProposition}, the projection of a horizontal curve $ \gamma $ satisfying $ D^{\dot\gamma}_\gamma \dot\gamma = 0 $ satisfies the geodesic equation (consider for example a horizontal projectable vector field $V$ that locally extends $\dot\gamma$). Conversely, the same proposition allows us to deduce $ (D^{\dot\gamma}_{\gamma} \dot\gamma)^\bot_{\dot\gamma} = 0 $ for the horizontal lift of a geodesic, while by Lemma \ref{Lemma3}
\begin{equation*}
(D^{\dot\gamma}_{\gamma} \dot\gamma)^\top_{\dot\gamma}
=
\mathbf A^{\dot\gamma}_{\dot\gamma} \dot\gamma
-
2C^\sharp_{\dot\gamma}(\dot\gamma,D^{\dot\gamma}_{\gamma} \dot\gamma)^\top_{\dot\gamma}
\,,
\end{equation*}
where the last term is zero by homogeneity, and the first term to the right is zero by part $(iii)$ of Proposition \ref{TAProposition}.
\end{proof}
\begin{corollary}\label{horizontalEverywhereCorollary}
If a geodesic of $(\mathcal M,L)$ is horizontal at one instant, then it is horizontal everywhere.
\end{corollary}
\begin{proof}
Assume that $\gamma$ is horizontal at $t_0\in [a,b]$. Then consider the lift $\beta$ of the geodesic with initial velocity
$\mathrm{d}\sigma(\dot\gamma(t_0))$ (recall Corollary \ref{horizontalGeodesicCorollary}). As both $\gamma$ and $\beta$ have the same initial velocity, they must coincide and $\gamma$ is horizontal everywhere.
\end{proof}
The last two corollaries were obtained in \cite[Theorems 3.1 and 5.3]{2001AlDur} for Finsler metrics using two different methods, the minimizing property of geodesics (which holds only in the positive definite case) and symplectic reduction, which can be also applied to pseudo-Finsler submersions.
\subsection{Covariant derivatives of T and A}\label{covderTA}
Note that skew-symmetry from Proposition \ref{TAProposition} implies that $ \nabla \mathbf T $ and $ \nabla \mathbf A $ are skew-symmetric.
Algebraicity of $ \nabla \mathbf T $ and $ \nabla \mathbf A $ can be imported intact from \cite{1966ONeill} as follows.
\begin{lemma}\label{Lemma4}
For each admissible $ v $, vertical $ u,w $ and $ g_v $-horizontal $ x,y $ and arbitrary $ e $ at $ p $,
\begin{align*}
(\nabla_w \mathbf A)^v_ue
&=
-\mathbf A^v_{T^v_w u}e
\,, &\quad
(\nabla_x \mathbf A)^v_we
&=
-\mathbf A^v_{\mathbf A^v_x w}e
\,, \\
(\nabla_w \mathbf T)^v_ye
&=
-\mathbf T^v_{\mathbf T^v_w y}e
\,, &\quad
(\nabla_x \mathbf T)^v_ye
&=
-\mathbf T^v_{\mathbf A^v_x y}e
\,.
\end{align*}
More succinctly, for every $h$ at $p$
\begin{equation*}
(\nabla_h \mathbf A)^v_w e = -\mathbf A^v_{(\mathbf T+\mathbf A)^v_hw}e
\,, \quad
(\nabla_h \mathbf T)^v_x e = -\mathbf T^v_{(\mathbf T+\mathbf A)^v_hx}e
\,.
\end{equation*}
\end{lemma}
\begin{proof}
For a locally admissible extension $ V $ of $ v $ satisfying $ \nabla^v V = 0 $, a vertical extension $U$ of $u$ and a $g_V$-horizontal extension $Y$ of $y$, all the identities are a direct consequence of definitions taking into account that $\mathbf A^v_u=0$ for any vertical $u$ and $\mathbf T^v_x=0$ for any $g_v$-horizontal $x$ and the identities of Lemma \ref{Lemma3}.
\end{proof}
\begin{lemma}\label{Lemma5}
For each admissible $ v $, vertical $ u,w $ and $ g_v $-horizontal $ x $ and arbitrary $e$ and $h$ at $ p $,
\begin{align*}
&
((\nabla_w \mathbf A)^v_e u)^\top_v
=
(\mathbf T^v_w \mathbf A^v_e - \mathbf A^v_e \mathbf T^v_w) u
\,, \quad
((\nabla_x \mathbf A)^v_e u)^\top_v
=
(\mathbf A^v_x \mathbf A^v_e - \mathbf A^v_e \mathbf A^v_x) u
\,, \\
&
((\nabla_w \mathbf T)^v_e u)^\top_v
=
(\mathbf T^v_w \mathbf T^v_e - \mathbf T^v_e \mathbf T^v_w) u
\,, \quad
((\nabla_x \mathbf T)^v_e u)^\top_v
=
(\mathbf A^v_x \mathbf T^v_e - \mathbf T^v_e \mathbf A^v_x) u
\,,
\end{align*}
and for each $ g_v $-horizontal $ y $ at $ p $
\begin{align*}
&
((\nabla_w \mathbf A)^v_e y)^\bot_v
=
(\mathbf T^v_w \mathbf A^v_e - \mathbf A^v_e \mathbf T^v_w) y
\,, \quad
((\nabla_x \mathbf A)^v_e y)^\bot_v
=
(\mathbf A^v_x \mathbf A^v_e -\mathbf A^v_e \mathbf A^v_x) y
\,, \\
&
((\nabla_w \mathbf T)^v_e y)^\bot_v
=
(\mathbf T^v_w \mathbf T^v_e - \mathbf T^v_e \mathbf T^v_w) y
\,, \quad
((\nabla_x \mathbf T)^v_e y)^\bot_v
=
(\mathbf A^v_x \mathbf T^v_e - \mathbf T^v_e \mathbf A^v_x) y
\,.
\end{align*}
More synthetically,
\begin{align}\label{nablaT+Au}
((\nabla_e (\mathbf T+\mathbf A))^v_h u)^\top_v
& =
(\mathbf T+\mathbf A)^v_e (\mathbf T+\mathbf A)^v_h u - (\mathbf T+\mathbf A)^v_h (\mathbf T+\mathbf A)^v_e u
\,, \\
\label{nablaT+Ay}((\nabla_e (\mathbf T+\mathbf A))^v_h y)^\bot_v
& =
(\mathbf T+\mathbf A)^v_e (\mathbf T+\mathbf A)^v_h y - (\mathbf T+\mathbf A)^v_h (\mathbf T+\mathbf A)^v_e y
\,.
\end{align}
\end{lemma}
\begin{proof}
Consider a locally admissible extension $ V $ of $ v $ satisfying $ \nabla^v V = 0 $, an arbitrary extension $ E $ of $ e $ and any vertical extension $ U $ of $ u $. Then
\begin{equation*}
(\nabla_h \mathbf T)^v_e u
=
\nabla^v_h (\mathbf T^V_E U)
-
\mathbf T^v_{\nabla^v_h E} u
-
\mathbf T^v_e \nabla^v_h U
\,,
\end{equation*}
whose second term will vanish when taking the vertical part, while the vertical part of the third term is by part $(i)$ of Proposition \ref{TAProposition} and \eqref{UTA},
\begin{equation*}
-(\mathbf T^v_e \nabla^v_h U)^\top_v=-\mathbf T^v_e (\nabla^v_h U)^\perp_v
=
-\mathbf T^v_e (\mathbf T+\mathbf A)^v_h U
\,.
\end{equation*}
The vertical part of the first term may also be computed into a similar form as follows. For any vertical extension $ W $ of $ w $, by Lemma \ref{Lemma3}, noting that the product $ g_V((\mathbf T+\mathbf A)^V_E U,W) $ is zero by properties of $ \mathbf T $ and $ \mathbf A $, and using also \eqref{UTA} and skew-symmetry of $\mathbf T$ and $\mathbf A$,
\begin{multline*}
g_v(\nabla^v_h (\mathbf T^V_E U),w)
=
-
g_v( \mathbf T^v_e u,\nabla^v_h W)
-
2C_v(\mathbf T^v_e u,w,\nabla^v_h V)
\\
=
-g_v((\mathbf T^v_e u,(\mathbf T+\mathbf A)^v_h w)
=
g_v((\mathbf T+\mathbf A)^v_h \mathbf T^v_e u,w)
\,.
\end{multline*}
It follows then, putting together the last three identities, that
\[((\nabla_h \mathbf T)^v_e u)^\top_v=-\mathbf T^v_e (\mathbf T+\mathbf A)^v_h u+(\mathbf T+\mathbf A)^v_h \mathbf T^v_e u.\]
With analogous computations, we obtain other three identities
\begin{align*}
((\nabla_h \mathbf A)^v_e u)^\top_v&=-\mathbf A^v_e (\mathbf T+\mathbf A)^v_h u+(\mathbf T+\mathbf A)^v_h \mathbf A^v_e u,\\
((\nabla_h \mathbf T)^v_e y)^\bot_v&=-\mathbf T^v_e (\mathbf T+\mathbf A)^v_h y+(\mathbf T+\mathbf A)^v_h \mathbf T^v_e y,\\
((\nabla_h \mathbf A)^v_e y)^\bot_v&=-\mathbf A^v_e (\mathbf T+\mathbf A)^v_h y+(\mathbf T+\mathbf A)^v_h \mathbf A^v_e y.
\end{align*}
All the identities in Lemma \ref{Lemma5} follow from the above four, by taking into account that $\mathbf T_x=\mathbf A_w=0$.
\end{proof}
\section{Fundamental equations}\label{fundeq}
Our aim is to generalise the fundamental equations $ \lbrace 0 \rbrace $ to $ \lbrace 4 \rbrace $ in \cite{1966ONeill}. To this end, let us compute the curvature tensor in terms of the following curvature tensors.
\begin{definition}\label{RTopBotDefinition}
Let $ R^\top $ and $ R^\bot $ be given for each admissible $ v $ and tangent vectors $ b,e,h $ at $p$ with extensions $ B,E,H $ by
\begin{equation*}
R^\top_v(b,e)w
=
\left(
\nabla^v_b (\nabla_E H^\top)^\top
-
\nabla^v_e (\nabla_B H^\top)^\top
-
\nabla^v_{[B,E]} H^\top
\right)^\top_v
\,,
\end{equation*}
and replacing the vertical operator $ \text{ }^\top $ by the horizontal operator $ \text{ }^\bot $
\begin{equation*}
R^\bot_v(b,e)h
=
\left(
\nabla^v_b (\nabla_E H^\bot)^\bot
-
\nabla^v_e (\nabla_B H^\bot)^\bot
-
\nabla^v_{[B,E]} H^\bot
\right)^\bot_v
\,.
\end{equation*}
\end{definition}
One can easily checked that $R^\top$ and $R^\bot$ are anisotropic tensors by the same method as the proof that the curvature tensor of a Riemannian manifold is indeed a tensor.
\begin{proposition}\label{RTopRBotProposition}
With the same notation, for $v$ admissible with a locally admissible extension $V$ and $w$ vertical with vertical extension $W$,
\begin{multline*}
R^\top_v(b,e) w
=
\left(
\nabla^v_b (\nabla^V_E W)^\top_V
-
\nabla^v_e (\nabla^V_B W)^\top_V
-
\nabla^v_{[B,E]} W
\right)^\top_v
\\
-
2C^\sharp_v(\mathbf T^v_ew,\nabla^v_bV)^\top_v
+
2C^\sharp_v(\mathbf T^v_bw,\nabla^v_eV)^\top_v
\\
-
P_v(e,w,\nabla^v_bV)^\top_v
+
P_v(b,w,\nabla^v_eV)^\top_v
\,,
\end{multline*}
and
\begin{multline*}
R^\bot_v(b,e)h
=
\left(
\nabla^v_b (\nabla^V_E H^\bot_V)^\bot_V
-
\nabla^v_e (\nabla^V_B H^\bot_V)^\bot_V
-
\nabla^v_{[B,E]} H^\bot_V
\right)^\bot_v
\\
-
P_v(e,h^\bot_v,\nabla^v_bV)^\bot_v
+
P_v(b,h^\bot_v,\nabla^v_eV)^\bot_v
\\
+
2 (\mathbf T^v_e+\mathbf A^v_e) C^\sharp_v(h^\bot_v,\nabla^v_bV)^\top_v
-
2 (\mathbf T^v_b+\mathbf A^v_b) C^\sharp_v(h^\bot_v,\nabla^v_eV)^\top_v
\,.
\end{multline*}
\end{proposition}
\begin{proof}
In the first case,
as $W^\top=W$, we only have to apply the chain rule and \eqref{vertder} to compute,
\begin{align*}
\nabla^v_b (\nabla_E W)^\top
=&
\nabla^v_b (\nabla^V_E W)^\top_V
-
(\dot\partial (\nabla_e W)^\top)_v(\nabla^v_bV)
\\=&
\nabla^v_b (\nabla^V_E W)^\top_V
-2C^\sharp_v(\mathbf T^v_ew,\nabla^v_bV)^\top_v
-
P_v(e,w,\nabla^v_bV)^\top_v
\,.
\end{align*}
Putting together the above identity and an analogous one commuting the role of $e$ and $b$, one concludes the expression for $R^\top$.
In the second case,
by \eqref{vertder}
\begin{equation}\label{nablabrac}
\nabla^v_{[B,E]} H^\bot
=
\nabla^v_{[B,E]} H^\bot_V
-
(\dot\partial (H^\bot))_v(\nabla^v_{[B,E]} V)
=
\nabla^v_{[B,E]} H^\bot_V
+
2C^\sharp_v(h^\bot_v,\nabla^v_{[B,E]}V)^\top_v
\,,
\end{equation}
whose last term will vanish under $(\cdot)^\bot_v$, while
\begin{equation*}
(\nabla^v_b (\nabla_E H^\bot)^\bot)^\bot_v
=
(\nabla^v_b (\nabla^V_E H^\bot)^\bot_V)^\bot_v
-
((\dot\partial (\nabla_e H^\bot)^\bot)_v(\nabla^v_bV))^\bot_v
\,.
\end{equation*}
Using the chain rule and \eqref{dotPartialNablaMathcalX},
\begin{align*}
((\dot\partial (\nabla_e H^\bot)^\bot)_v(\nabla^v_bV))^\bot_v
=&
((\dot\partial (\nabla^v_e H^\bot)^\bot)_v(\nabla^v_bV))^\bot_v
+
(((\dot\partial (\nabla_e H^\bot))_v(\nabla^v_bV))^\bot_v)^\bot_v
\\
=&
((\dot\partial (\nabla_e H^\bot))_v(\nabla^v_bV))^\bot_v
\\=&
P_v(e,h^\bot_v,\nabla^v_bV)^\bot_v
+
((\nabla_e(\dot\partial H^\bot))_v(\nabla^v_bV))^\bot_v
\,,
\end{align*}
because by \eqref{vertder}, the first term to the right and the term coming from the third one in \eqref{dotPartialNablaMathcalX} are the $g_v$-horizontal part of a vertical vector.
Lastly, again by \eqref{vertder},
\begin{multline*}
(\nabla^V_E H^\bot)^\bot_V
=
(\nabla^V_E H^\bot_V)^\bot_V
-
((\dot\partial (H^\bot))_V(\nabla^V_EV))^\bot_V
\\
=
(\nabla^V_E H^\bot_V)^\bot_V
+
2(C^\sharp_V(H^\bot_V,\nabla^V_EV)^\top_V)^\bot_V
=
(\nabla^V_E H^\bot_V)^\bot_V
\,.
\end{multline*}
All that remains is to compute $((\nabla_e(\dot\partial (H^\bot)))_v(\nabla^v_bV))^\bot_v$, which we may obtain from \eqref{vertder} under the form
\begin{equation*}
(\nabla_e (\dot\partial (H^\bot))_v(\nabla^v_bV)
=
-2(\nabla_e (C^\sharp(H^\bot,\cdot)^\top))_v(\nabla^v_bV)
\end{equation*}
and by extending the definition of the tensors $\mathbf T$ and $\mathbf A$ to anisotropic vector fields, we obtain
\begin{align*}
-2((\nabla_e (C^\sharp(H^\bot,\cdot)^\top))_v(\nabla^v_bV))^\bot_v
=&
-2(\nabla^v_eC^\sharp(H^\bot,\nabla_BV)^\top)^\bot_v
\\=&
-2(\mathbf T^v_e + \mathbf A^v_e)C^\sharp_v(h^\bot_v,\nabla^v_bV)^\top_v
\,.
\end{align*}
Putting together all the last identities, we conclude that
\begin{multline*}
(\nabla^v_b (\nabla_E H^\bot)^\bot)^\bot_v=(\nabla^v_b (\nabla^V_E H^\bot_V)^\bot_V)^\bot_v-
P_v(e,h^\bot_v,\nabla^v_bV)^\bot_v
\\
+2(\mathbf T^v_e + \mathbf A^v_e)C^\sharp_v(h^\bot_v,\nabla^v_bV)^\top_v
\,.
\end{multline*}
Taking into account the expression obtained by commuting $e$ and $b$ and \eqref{nablabrac}, one deduces the expression for $R^\bot$.
\end{proof}
\begin{proposition}\label{RTopProposition}
For each admissible $ v $, arbitrary $ e $ and $ h $ at $ p $, vertical $w$ and $ g_v $-horizontal $ x $,
\begin{equation*}
R^\top_v(e,h)x
=
-
2 C^\sharp_v(R_v(e,h)v,x)^\top_v
\,,
\quad
R^\bot_v(e,h)w
=
0
\,.
\end{equation*}
\end{proposition}
\begin{proof}
The second identity is merely a consequence of the fact that for any vertical extension $W$ of $w$, the anisotropic vector field $W^\top$ is constant and identical to $W$ on each tangent space while $W^\bot$ is identically equal to zero. As for the first identity,
consider a locally admissible extension $ V $ of $ v $ satisfying that $ \nabla^v V = 0 $, extensions $ E $ of $e$ and $ H $ of $ h $ whose Lie bracket $[E,H]$ vanishes at $ p $ and a local $ g_V $-horizontal extension $ X $ of $ x $.
Note how by Proposition \ref{dotPartialTopProposition}
\begin{equation*}
(\nabla^V_H X^\top)^\top_V
=
(\nabla^V_H X^\top_V)^\top_V
-
((\dot\partial X^\top)(\nabla^V_H V))^\top_V
=
-2C^\sharp_V(X,\nabla^V_H V)^\top_V
\,.
\end{equation*}
Differentiating $ \Xi_{v}(\cdot,\cdot) = -2(C^\sharp_v(\cdot,\cdot))^\top_v $, with $\nabla^v_HV=\nabla^v_eV=0$ for our choice of $V$,
\begin{multline*}
\nabla^v_e ((\nabla^V_H X^\top)^\top_V)
=
\nabla^v_e (\Xi_V(X,\nabla^V_H V))
=
\Xi_v(x, \nabla^v_e \nabla^V_H V)
=
-2C^\sharp_v(x,\nabla^v_e \nabla^V_H V)^\top_v
\,.
\end{multline*}
Reinserting into
\begin{equation*}
(\nabla^v_e (\nabla_H X^\top)^\top)^\top_v
=
(\nabla^v_e (\nabla^V_H X^\top)^\top_V)^\top_v
=
-2 C^\sharp_v(x,\nabla^v_e \nabla^V_H V)^\top_v
=
-2 C^\sharp_v(x,\nabla^v_e \nabla_H V)^\top_v
\end{equation*}
concludes the first identity up to commuting $ e $ and $ h $.
\end{proof}
Observe how, if $ v $ is vertical, then we may relate $ R^\top_v $ with the intrinsic curvature tensor $ \hat{R} $ of the submersion fibers. In analogy to the vertical derivative $P$ of the Chern connection $\nabla$ of the ambient pseudo-Finsler manifold $(\mathcal M,L)$, let $\hat P$ designate that of $\hat\nabla$ and $\tilde P$ that of $\tilde\nabla$.
\begin{proposition}\label{RTopVProposition}
For each vertical admissible $ v $ and vertical $ s $, $ u $ and $ w $ at $ p $,
\begin{multline*}
R^\top_v(u,w)s
=
\hat R_v(u,w)s
+
(\hat\nabla_u\hat Q)^v_ws - (\hat\nabla_w\hat Q)^v_us
+
\hat Q^v_u\hat Q^v_ws - \hat Q^v_w\hat Q^v_us
\\
-
P_v(w,s,\hat Q^v_uv+\mathbf T^v_uv)^\top_v
+
P_v(u,s,\hat Q^v_wv+\mathbf T^v_wv)^\top_v
\\
-
2C^\sharp_v(\mathbf T^v_ws,\hat Q^v_uv+\mathbf T^v_uv)^\top_v
+
2C^\sharp_v(\mathbf T^v_us,\hat Q^v_wv+\mathbf T^v_wv)^\top_v
\,.
\end{multline*}
\end{proposition}
\begin{proof}
It follows from \cite[Theorem 5]{2022HuberJavaloyes}, taking into account that
we may differentiate the Gauss formula in Proposition \ref{GaussFormulaVert} with respect to the fiber derivative in the direction of $\hat Q^v_wv$ to obtain
\begin{equation*}
P_v(u,s,\hat Q^v_wv)^\top_v
+
2C^\sharp_v(\mathbf T^v_us,\hat Q^v_wv)^\top_v
=
\hat P_v(u,s,\hat Q^v_wv)
+
(\dot\partial\hat Q)^v_us(\hat Q^v_wv)
\,.
\end{equation*}
Here we have used the chain rule and \eqref{vertder}.
\end{proof}
Similarly, if $v$ is horizontal, then we may relate $R^\bot_v$ with the horizontal lift $\tilde{R}^\ast_v$ of the curvature tensor of the base manifold.
\begin{proposition}\label{RBotVProposition}
For each horizontal $ v $ at $ p $ and $ g_v $-horizontal $ x $, $ y $ and $ z $,
\begin{align*}
& R^\bot_v(x,y)z
\\
& \quad =
\tilde{R}^\ast_v(\tilde{x},\tilde{y})\tilde{z}
+
\mathbf A^v_z\mathbf A^v_y x
-
\mathbf A^v_z\mathbf A^v_x y
+
((\nabla_x \tilde{Q})^v_y z - (\nabla_y \tilde{Q})^v_x z)^\bot_v
+
\tilde{Q}^v_y \tilde{Q}^v_x z - \tilde{Q}^v_x \tilde{Q}^v_y z
\\ & \qquad
+
2 \mathbf A^v_z C^\sharp_v(y,\mathbf A^v_xv)^\top_v
+
2 \mathbf A^v_y C^\sharp_v(z,\mathbf A^v_xv)^\top_v
-
2 \mathbf A^v_z C^\sharp_v(x,\mathbf A^v_yv)^\top_v
-
2 \mathbf A^v_x C^\sharp_v(z,\mathbf A^v_yv)^\top_v
\\ & \qquad
-
P_v(y,z,\mathbf A^v_xv)^\bot_v
+
P_v(x,z,\mathbf A^v_yv)^\bot_v
+
((\dot\partial\tilde Q)^v_yz(\mathbf A^v_xv))^\bot_v
-
((\dot\partial\tilde Q)^v_xz(\mathbf A^v_yv))^\bot_v
\,.
\end{align*}
\end{proposition}
\begin{proof}
Consider an extension $ \tilde{V} $ of $ \tilde{v} $ satisfying $ \tilde\nabla^{\tilde{v}} \tilde{V} = 0 $, and let $ V $ be, locally, the horizontal lift of $ \tilde{V} $. In particular, by the dual Gauss formula $ (\nabla^v V)^\bot_v $ vanishes on $g_v$-horizontal vectors, while, by Lemma \ref{Lemma3}, $\nabla^v_x V = \mathbf A^v_x v$ and $\nabla^v_y V = \mathbf A^v_y v$. Let $ \tilde{X} $, $ \tilde{Y} $ and $ \tilde{Z} $ be extensions of $ \tilde{x} $, $ \tilde{y} $ and $ \tilde{z} $ with mutual Lie brackets that vanish at $ p $. Let $ X $, $ Y $ and $ Z $ be, locally, their respective $ g_V $-horizontal lifts. In particular, $ [X,Y]^\bot_v = [Y,Z]^\bot_v = [Z,X]^\bot_v = 0 $.
From Proposition \ref{RTopRBotProposition}, it suffices to compute $(\nabla^v_{[X,Y]} Z)^\bot_v$ (recall that $Z^\bot_V=Z$) and by Proposition \ref{GaussProposition}
\begin{equation*}
(\nabla^v_x(\nabla^V_YZ)^\bot_V)^\bot_v
=
(\nabla^v_x (\tilde\nabla^{\tilde V}_{\tilde y} \tilde Z)^\ast_V)^\bot_v
+
(\nabla^v_x \tilde Q^V_Y Z)^\bot_v
\,.
\end{equation*}
Observe that we may use the almost-antisymmetry of $\mathbf A$ to rewrite \eqref{Axy} under the form
\begin{equation}\label{AtvxyLieAtvyx}
\tfrac{1}{2}
\mathbf A^v_x y
=
\tfrac{1}{2}[X,Y]^\top_v
+
\tfrac{1}{2}
\mathbf A^v_y x
+
C^\sharp_v(\mathbf A^v_x v,y)^\top_v
-
C^\sharp_v(x,\mathbf A^v_y v)^\top_v
\,.
\end{equation}
Then, using the above identity, \eqref{Lieconversion} (recall that $[X,Y]$ is vertical by assumption)
and part $(iii)$ of Proposition \ref{TAProposition}, we get
\begin{equation*}
(\nabla^v_{[X,Y]} Z)^\bot_v
=
\mathbf A^v_z [X,Y]
=
\mathbf A^v_z \mathbf A^v_x y
-
\mathbf A^v_z \mathbf A^v_y x
+
2 \mathbf A^v_z C^\sharp(\mathbf A^v_v x,y)^\top_v
-
2 \mathbf A^v_z C^\sharp_v(x,\mathbf A^v_v y)^\top_v
\,.
\end{equation*}
By the dual Gauss formula, and since for our choice of extension $\tilde\nabla^{\tilde v}\tilde V=0$,
\begin{multline*}
(\nabla^v_x (\tilde\nabla^{\tilde V}_{\tilde y} \tilde Z)^\ast_V)^\bot_v
+
(\nabla^v_x \tilde Q^V_Y Z)^\bot_v
=
(\tilde\nabla^{\tilde v}_{\tilde x} \tilde\nabla_{\tilde Y} \tilde Z)^\ast_v
+
\tilde Q^v_x(\tilde\nabla^{\tilde v}_{\tilde y} \tilde Z)^\ast_v
\\
+
((\nabla_x\tilde Q)^v_yz)^\bot_v
+
\tilde Q^v_{\nabla^v_xY}z
+
\tilde Q^v_y(\tilde\nabla^{\tilde v}_{\tilde x}\tilde Z)^\ast_v
+
\tilde Q^v_y\tilde Q^v_xz
+
((\dot\partial\tilde Q)^v_yz(\mathbf A^v_xv))^\bot_v
\,.
\end{multline*}
When subtracting to the formula we have obtained the formula commuted in $x$ and $y$, the pairs of $\tilde Q^v(\tilde\nabla^{\tilde v}\tilde Z)^\ast_v$ terms cancel each other out, as well as the pair of $\tilde Q^v_{\nabla^v_xY}z$ and $\tilde Q^v_{\nabla^v_yX}z$ by our assumption that $[X,Y]^\bot_v=0$.
\end{proof}
\begin{theorem}[Unified fundamental equation of a pseudo-Finsler submersion]
For each admissible $ v $ and arbitrary vectors $ e $ and $ h $ at $ p $,
\begin{multline}\label{Unified}
R_v(e,h)
=
R^\top_v(e,h)
+
R^\bot_v(e,h)
+
(\nabla_e (\mathbf T+\mathbf A))^v_h - (\nabla_h (\mathbf T+\mathbf A))^v_e
\\
+
(\mathbf T+\mathbf A)^v_h(\mathbf T+\mathbf A)^v_e - (\mathbf T+\mathbf A)^v_e(\mathbf T+\mathbf A)^v_h
\,,
\end{multline}
where $R$ stands for the curvature tensor of the Chern connection of the ambient manifold, $R^\top$ and $R^\bot$ are as defined in Definition \ref{RTopBotDefinition} and $\mathbf T,\mathbf A$ are the O'Neill tensors of Definition \ref{ATDefinition}.
\end{theorem}
\begin{proof}
For a choice of extensions whose Lie brackets vanish at $ p $,
\begin{equation}
R_v(e,h)
=
\nabla^v_e \nabla_H
-
\nabla^v_h \nabla_E
\,.
\end{equation}
Consider any vector field $ B $ with $ b = B_p $. By definition of the tensors $ \mathbf T $ and $ \mathbf A $,
\begin{equation*}
\nabla^v_e \nabla_H B
=
\nabla^v_e (\nabla_H B^\top)^\top
+
\nabla^v_e ((\mathbf T+\mathbf A)_H B)
+
\nabla^v_e (\nabla_H B^\bot)^\bot
\,,
\end{equation*}
where we may further expand the first and last terms as
\begin{align*}
\nabla^v_e (\nabla_H B^\top)^\top
&=
(\nabla^v_e (\nabla_H B^\top)^\top)^\top_v
+
(\mathbf T+\mathbf A)^v_e (\nabla^v_h B^\top)^\top_v,\\
\nabla^v_e (\nabla_H B^\perp)^\perp
&=
(\nabla^v_e (\nabla_H B^\perp)^\perp)^\perp_v
+
(\mathbf T+\mathbf A)^v_e (\nabla^v_h B^\perp)^\perp_v,
\end{align*}
and the middle term as
\begin{align*}
\nabla^v_e ((\mathbf T+\mathbf A)_H B)
=&(\nabla_e (\mathbf T+\mathbf A))^v_h b
+(\mathbf T+\mathbf A)^v_{\nabla^v_e H} b+(\mathbf T+\mathbf A)^v_{h}{\nabla^v_e B} \\
=&(\nabla_e (\mathbf T+\mathbf A))^v_h b
+
(\mathbf T+\mathbf A)^v_{\nabla^v_e H} b
\\
&+
(\mathbf T+\mathbf A)^v_h (\nabla^v_e B^\top)^\top_v
+
(\mathbf T+\mathbf A)^v_h (\mathbf T+\mathbf A)^v_e b
+
(\mathbf T+\mathbf A)^v_h (\nabla^v_e B^\bot)^\bot_v
\end{align*}
to obtain a long expression, to which we may substract the corresponding expressions commuted in $e$ and $h$. Pairs of $(\mathbf T+\mathbf A)^v(\nabla^vB^\top)^\top_v$ and $(\mathbf T+\mathbf A)^v(\nabla^vB^\bot)^\bot_v$ terms cancel each other out. Cancel $(\mathbf T+\mathbf A)^v_{\nabla^v_eH}b$ with $-(\mathbf T+\mathbf A)^v_{\nabla^v_hE}b$ by our assumption that $[E,H]_p=0$ to conclude.
\end{proof}
\begin{corollary}[Fundamental equations of a pseudo-Finsler submersion]\label{fundeqgeneral}
For each admissible $v$, vertical $s',s,u,w$ and $g_v$-horizontal $x,y,z,z'$
\begin{align*}\tag*{\{0\}}\label{0}
&
g_v(R_v(w,u)s,s')
=
g_v(R^\top_v(w,u)s,s')
+
g_v(\mathbf T^v_ws,\mathbf T^v_us')
-
g_v(\mathbf T^v_us,\mathbf T^v_ws')
\,,
\\ & \tag*{\{1\}}\label{1}
g_v(R_v(w,u)s,z)
=
g_v((\nabla_w \mathbf T)^v_us,z) - g_v((\nabla_u \mathbf T)^v_ws,z)
\,,
\\ & \tag*{\{1'\}}\label{1p}
g_v(R_v(x,u)s,w)
=
g_v(R^\top_v(x,u)s,w)
+
g_v(\mathbf T^v_us,\mathbf A^v_xw)
-
g_v(\mathbf A^v_xs,\mathbf T^v_uw)
\,,
\\ & \tag*{\{2\}}\label{2}
g_v(R_v(x,u)s,z)
=
g_v((\nabla_x \mathbf T)^v_us,z)
-
g_v((\nabla_u \mathbf A)^v_xs,z)
-
g_v(\mathbf A^v_{\mathbf A^v_xu}s,z)
-
g_v(\mathbf T^v_ux,\mathbf T^v_sz)
\,,
\\ & \tag*{\{2'\}}\label{2p}
g_v(R_v(x,y)s,w)
=
g_v(R^\top_v(x,y)s,w)
+
g_v(\mathbf A^v_ys,\mathbf A^v_xw)
-
g_v(\mathbf A^v_xs,\mathbf A^v_yw)
\,,
\\ & \tag*{\{3\}}\label{3}
g_v(R_v(x,y)s,z)
=
g_v((\nabla_x \mathbf A)^v_ys,z) - g_v((\nabla_y \mathbf A)^v_xs,z)
+
g_v(\mathbf A^v_yx,\mathbf T^v_sz)
-
g_v(\mathbf A^v_xy,\mathbf T^v_sz)
\,,
\\ & \tag*{\{4\}}\label{4}
g_v(R_v(x,y)z,z')
=
g_v(R^\bot_v(x,y)z,z')
+
g_v(\mathbf A^v_xz,\mathbf A^v_yz')
-
g_v(\mathbf A^v_yz,\mathbf A^v_xz')
\,.
\end{align*}
\end{corollary}
\begin{proof}
Recall that, by Proposition \ref{RTopProposition}, $ R^\bot_v(w,u)s = R^\bot_v(x,u)s = R^\bot_v(x,y)s = 0 $.
Moreover, using \eqref{nablaT+Au} and being $e,h$ arbitrary vectors at $p$,
\begin{multline}\label{NaTA}
(\nabla_e (\mathbf T+\mathbf A))^v_hs - (\nabla_h (\mathbf T+\mathbf A))^v_es
\\
+
(\mathbf T+\mathbf A)^v_h(\mathbf T+\mathbf A)^v_es - (\mathbf T+\mathbf A)^v_e(\mathbf T+\mathbf A)^v_hs
\\
=((\nabla_e (\mathbf T+\mathbf A))^v_hs - (\nabla_h (\mathbf T+\mathbf A))^v_es)^\bot_v
\\
+
(\mathbf T+\mathbf A)^v_e(\mathbf T+\mathbf A)^v_hs - (\mathbf T+\mathbf A)^v_h(\mathbf T+\mathbf A)^v_es\,.
\end{multline}
Then the six first identities are obtained from \eqref{Unified} using the last identity together with the observation about $R^\bot$, an application of Lemma \ref{Lemma4} and the properties of $\mathbf T$ and $\mathbf A$ in Proposition \ref{TAProposition}. In particular,
\begin{equation*}
((\nabla_w \mathbf A)^v_u s - (\nabla_u \mathbf A)^v_w s)^\bot_v=- \mathbf A^v_{\mathbf T^v_wu} s + \mathbf A^v_{\mathbf T^v_uw} s = 0
\end{equation*}
by Lemma \ref{Lemma4} and the symmetry of $\mathbf T$ to get $\{0\}$ and $\{1\}$,
\[
((\nabla_x \mathbf A)^v_u s - (\nabla_u \mathbf T)^v_x s)^\bot_v
=-
\mathbf A^v_{\mathbf A^v_xu} s
+
\mathbf T^v_s\mathbf T^v_ux
\]
where we have used $\mathbf T^v_{\mathbf T^v_ux}s=\mathbf T^v_s\mathbf T^v_ux$, to get $\{1'\}$ and $\{2\}$,
\[ ((\nabla_x \mathbf T)^v_y s - (\nabla_y \mathbf T)^v_x s)^\bot_v=
-
\mathbf T^v_s\mathbf A^v_xy
+
\mathbf T^v_s\mathbf A^v_yx
\]
where we have used $ \mathbf T^v_{\mathbf A^v_xy} s = \mathbf T^v_s \mathbf A^v_xy $, to get $\{2'\}$ and $\{3\}$.
For the remaining identity $\{4\}$, we proceed analogously. Using \eqref{nablaT+Ay}
\begin{multline*}
(\nabla_e (\mathbf T+\mathbf A))^v_hz - (\nabla_h (\mathbf T+\mathbf A))^v_ez
\\
+
(\mathbf T+\mathbf A)^v_h(\mathbf T+\mathbf A)^v_ez - (\mathbf T+\mathbf A)^v_e(\mathbf T+\mathbf A)^v_hz
\\
=((\nabla_e (\mathbf T+\mathbf A))^v_hz - (\nabla_h (\mathbf T+\mathbf A))^v_ez)^\top_v
\\
+
(\mathbf T+\mathbf A)^v_e(\mathbf T+\mathbf A)^v_hz - (\mathbf T+\mathbf A)^v_h(\mathbf T+\mathbf A)^v_ez
\end{multline*}
and by Lemma \ref{Lemma4},
\[ ((\nabla_x \mathbf T)^v_y z - (\nabla_y \mathbf T)^v_x z)^\top_v
=-
\mathbf T^v_{\mathbf A^v_xy} z
+
\mathbf T^v_{\mathbf A^v_yx} z\,.\]
\end{proof}
Let us recall one of the most important geometric invariants of a pseudo-Finsler manifold and let us introduce the analogous definitions for the vertical and horizontal parts of the curvature tensor presented in Definition \ref{RTopBotDefinition}.
\begin{definition}
For a flagpole $v\in\mathcal{A}$ and flag $e\in T_{\pi(v)}\mathcal M$ such that $L(v)g_v(e,e)\neq g_v(v,e)^2$, denote by $K_v(e)$ the flag curvature of $(\mathcal M,L)$ with the following definition:
\begin{equation*}
K_v(e)=\frac{g_v(R_v(v,e)e,v)}{L(v)g_v(e,e)-g_v(v,e)^2}
\,,
\end{equation*}
while the {\em vertical flag curvature} $K_v^\top(e)$ and the {\em horizontal flag curvature} $K_v^\bot(e)$ are defined in the same way replacing $R$ by $R^\top$ and $R^\bot$, respectively.
\end{definition}
\begin{corollary}\label{flagcurgen}
For each vertical $w$, admissible $v$ and $g_v$-horizontal $x$,
\begin{multline*}
K_v(w)
=
K^\top_v(w)+\frac{g_v((\nabla_v\mathbf T)^v_ww-(\nabla_w(\mathbf T+\mathbf A))^v_vw,v)}{L(v)g_v(w,w)-g_v(v,w)^2}
\\
+
\frac{-g_v(\mathbf A^v_{(\mathbf T+\mathbf A)^v_vw}w,v)+g_v(\mathbf T^v_ww,(\mathbf T+\mathbf A)^v_vv)-g_v((\mathbf T+\mathbf A)^v_vw,\mathbf T^v_wv)}{L(v)g_v(w,w)-g_v(v,w)^2}
\,,
\end{multline*}
and
\begin{multline*}
K_v(x)
= K^\top_v(x)+K^\bot_v(x)+
\frac{g_v((\nabla_v\mathbf A)^v_xx-(\nabla_x(\mathbf T+\mathbf A))^v_vx,v)}{L(v)g_v(x,x)-g_v(v,x)^2}
\\
+
\frac{-g_v(\mathbf T^v_{(\mathbf T+\mathbf A)^v_vx}x,v)+g_v(\mathbf A^v_xx,(\mathbf T+\mathbf A)^v_vv)-g_v((\mathbf T+\mathbf A)^v_vx,\mathbf A^v_xv)}{L(v)g_v(x,x)-g_v(v,x)^2}
\,.
\end{multline*}
\end{corollary}
\begin{proof}
By \eqref{Unified},
recalling that $R^\bot_v(\cdot,\cdot)w=0$
\begin{multline*}
g_v(R_v(v,w)w,v)
=
g_v(R^\top_v(v,w)w+(\nabla_v(\mathbf T+\mathbf A))^v_ww-(\nabla_w(\mathbf T+\mathbf A))^v_vw,v)
\\
+g_v(\mathbf T^v_ww,(\mathbf T+\mathbf A)^v_vv)
-
g_v((\mathbf T+\mathbf A)^v_vw,\mathbf T^v_wv)
\,,
\end{multline*}
and similarly
\begin{multline*}
g_v(R_v(v,x)x,v)
=
g_v((R^\top+R^\bot)_v(v,x)x+(\nabla_v(\mathbf T+\mathbf A))^v_xx-(\nabla_x(\mathbf T+\mathbf A))^v_vx,v)
\\
+
g_v(\mathbf A^v_xx,(\mathbf T+\mathbf A)^v_vv)
-
g_v((\mathbf T+\mathbf A)^v_vx,\mathbf A^v_xv)
\,.
\end{multline*}
To conclude, observe that by Lemma \ref{Lemma4}
\begin{equation*}
(\nabla_v \mathbf A)^v_w = -\mathbf A^v_{(\mathbf T+\mathbf A)^v_vw}
\end{equation*}
and
\begin{equation*}
(\nabla_v \mathbf T)^v_x = -\mathbf T^v_{(\mathbf T+\mathbf A)^v_vx}
\,.
\end{equation*}
\end{proof}
\begin{theorem}[Generalised Gauss equation and dual Gauss equation]\label{gaussanddual}
For each vertical admissible $ v $ and vertical vectors $ s $, $ u $ and $ w $ at $ p $,
\begin{multline}\tag*{\{0'\}}\label{0p}
(R_v(w,u)s)^\top_v
=
\hat{R}_v(w,u)s
+
\mathbf T^v_w \mathbf T^v_u s
-
\mathbf T^v_u \mathbf T^v_w s
\\
+
(\hat\nabla_u\hat Q)^v_ws - (\hat\nabla_w\hat Q)^v_us
+
\hat Q^v_u\hat Q^v_ws - \hat Q^v_w\hat Q^v_us
\\
-
P_v(w,s,\hat Q^v_uv+\mathbf T^v_uv)^\top_v
+
P_v(u,s,\hat Q^v_wv+\mathbf T^v_wv)^\top_v
\\
-
2C^\sharp_v(\mathbf T^v_ws,\hat Q^v_uv+\mathbf T^v_uv)^\top_v
+
2C^\sharp_v(\mathbf T^v_us,\hat Q^v_wv+\mathbf T^v_wv)^\top_v
\,.
\end{multline}
For each horizontal $v$ and $g_v$-horizontal vectors $x$, $y$ and $z$,
\begin{multline}\tag*{\{4'\}}\label{4p}
(R_v(x,y)z)^\bot_v
=
\tilde{R}^\ast_v(\tilde{x},\tilde{y})\tilde{z}
+
\mathbf A^v_x \mathbf A^v_y z
-
\mathbf A^v_y \mathbf A^v_x z
+
\mathbf A^v_z\mathbf A^v_y x
-
\mathbf A^v_z\mathbf A^v_x y
\\
+
2 \mathbf A^v_z C^\sharp(y,\mathbf A^v_xv)^\top_v
+
2 \mathbf A^v_y C^\sharp_v(z,\mathbf A^v_xv)^\top_v
-
2 \mathbf A^v_z C^\sharp_v(x,\mathbf A^v_yv)^\top_v
-
2 \mathbf A^v_x C^\sharp_v(z,\mathbf A^v_yv)^\top_v
\\
+
((\nabla_x \tilde{Q})^v_y z - (\nabla_y \tilde{Q})^v_x z)^\bot_v
+
\tilde{Q}^v_y \tilde{Q}^v_x z - \tilde{Q}^v_x \tilde{Q}^v_y z
+
((\dot\partial\tilde Q)^v_yz(\mathbf A^v_xv))^\bot_v
-
((\dot\partial\tilde Q)^v_xz(\mathbf A^v_yv))^\bot_v
\\
-
P_v(y,z,\mathbf A^v_xv)^\bot_v
+
P_v(x,z,\mathbf A^v_yv)^\bot_v
\,.
\end{multline}
\end{theorem}
\begin{proof}
From \ref{0} and \ref{4} together with Propositions \ref{RTopVProposition} and \ref{RBotVProposition}.
\end{proof}
\begin{corollary}\label{verflagpole}
For $v\in\mathcal{V}$ and $w$ vertical, assuming that the plane $\{v,w\}$ is non-degenerate,
\begin{multline*}
K_v(w)
=
\hat K_v(w)
-
\frac{g_v(\mathbf T^v_ww,\mathbf T^v_vv)-g_v(\mathbf T^v_vw,\mathbf T^v_vw)}{L(v)g_v(w,w)-g_v(v,w)^2}
\\
-
\frac{g_v( P_v (w,w,\mathbf T^v_vv)-(\hat\nabla_v\hat Q)^v_ww,v)+C_v(w,\hat Q^v_vu,\mathbf T^v_vv)}{L(v)g_v(w,w)-g_v(v,w)^2}
\end{multline*}
where $\hat K$ denotes the flag curvature intrinsic to the submersion fibers.
\end{corollary}
\begin{proof}
The proof is completely analogous to \cite[Corollary 6]{2022HuberJavaloyes}.
\end{proof}
\begin{corollary}\label{horflagpole}
For $v\in\mathcal{H}$ and $x$ $g_v$-horizontal, denoting by $\tilde v$ and $\tilde x$ their projections, and $w$ vertical, assuming that the planes $\{v,w\}$ and $\{v,x\}$ are non-degenerate,
\begin{align*}
K_v(w)&= \frac{g_v((\nabla_v\mathbf T)^v_ww,v)+g_v(\mathbf A^v_vw,\mathbf A^v_vw)-g_v(\mathbf T^v_wv,\mathbf T^v_wv))}{L(v)g_v(w,w)-g_v(v,w)^2},\\
K_v(x)&=\tilde K_{\tilde v}(\tilde x)
- \frac{3g_v(\mathbf A^v_x v,\mathbf A^v_x v)}{L(v)g_v(x,x)-g_v(v,x)^2}
\,.
\end{align*}
\end{corollary}
\begin{proof}
The expression for $K_v(w)$ follows from $\{2\}$ in Corollary \ref{fundeqgeneral} taking into account that
\[g_v(\mathbf A^v_{\mathbf A^v_vw}w,v)=-g_v(\mathbf A^v_{\mathbf A^v_vw}v,w)=g_v(\mathbf A^v_v{\mathbf A^v_vw},w)=-g_v(\mathbf A^v_vw,\mathbf A^v_vw)\]
by skew-symmetry and part $(iii)$ of Proposition \ref{TAProposition}, and $g_v(\nabla_w\mathbf A)^v_vw,w)=0$. To check this take a horizontal extension $V$ with $\nabla^v_vV=0$, and $W$ a vertical extension of $w$, then
\begin{equation}\label{depaso}
g_v(\nabla_w\mathbf A)^v_vw,w)=g_v(\nabla^v_w(\mathbf A^V_VW),v)-g_v(\mathbf A^v_{\nabla^v_wV}w,v)-g_v(\mathbf A^v_v(\nabla^v_wW),v).
\end{equation}
Using again skew-symmetry and part $(iii)$ of Proposition \ref{TAProposition},
\[g_v(\mathbf A_v(\nabla^v_wW),v)=-g_v(\mathbf A_vv,\nabla^v_wW)=0,\]
\[g_v(\nabla^v_w(\mathbf A^V_VW),v)=w(g_V(\mathbf A^V_VW,V))-g_v(\mathbf A^v_vw,\nabla^v_wV)=-g_v(\mathbf A^v_vw,\mathbf A^v_vw),\]
since $g_V(\mathbf A^V_VW,V)=-g_V(W,\mathbf A^V_VV)=0$ and $(\nabla^v_wV)^\bot_v=\mathbf A^v_vw$ by \eqref{Lieconversion}. Finally,
\begin{align*}
g_v(\mathbf A^v_{\nabla^v_wV}w,v)&=-g_v(\mathbf A^v_{\nabla^v_wV}v,w)=g_v(\mathbf A^v_v{(\nabla^v_wV)},w)=-g_v(\mathbf A^v_vw,{\nabla^v_wV})\\
&=
-g_v(\mathbf A^v_vw,\mathbf A^v_vw).
\end{align*}
Applying the last three identities in \eqref{depaso}, we conclude that $g_v(\nabla_w\mathbf A)^v_vw,w)=0$.
From \ref{4p}, by properties of the Cartan tensor, part $(iii)$ of Proposition \ref{TAProposition},
the identity $\tilde{Q}^v_v=\tilde{Q}^vv=0$ for $v$ horizontal, and taking into account that $g_v(\mathbf A^v_v\mathbf A^v_xx,v)$ and $g_v(\mathbf A^v_v C^\sharp_v(x,\mathbf A^v_xv)^\top_v,v)$ also vanish by skew-symmetry, we get
\begin{align}\nonumber
g_v(R_v(v,x)x,v)
& =
g_v(\tilde{R}^\ast_v(\tilde{v},\tilde{x})\tilde{x}
+ 3 \mathbf A^v_x \mathbf A^v_x v
+ (\nabla_v \tilde{Q})^v_x x - (\nabla_x \tilde{Q})^v_v x,v)
\\ & \qquad
- g_v((\dot\partial\tilde Q)^v_vx(\mathbf A^v_xv)+ P_v(v,x,\mathbf A^v_xv),v)
\,.\label{eqrefer}
\end{align}
By definition of $\nabla\tilde Q$, for a locally horizontal admissible extension $V$ of $v$ and locally $g_V$-horizontal extension $X$ of $x$ satisfying $\nabla^v_xV=\mathbf A^v_xv$ and $\nabla^v_vV=0$
\begin{equation*}
(\nabla_v\tilde Q)^v_x x
=
\nabla^v_v\tilde Q^V_X X
-
\tilde Q^v_{\nabla^v_vX}x
-
\tilde Q^v_x\nabla^v_vX
\end{equation*}
such that, using that $g_v(\tilde Q^v_yz,v)=0$ for $v$ horizontal and arbitrary $y$ and $z$ by definition of $\tilde Q$ and properties of the Cartan tensor, and, in particular $g_V(\tilde Q^V_XX,V)=0$,
\begin{equation*}
g_v((\nabla_v\tilde Q)^v_x x,v)
=
v(g_V(Q^V_XX,V))
-
g_v(\tilde Q^v_xx,\nabla^v_vV)
-
2C_v(\tilde Q^v_xx,v,\nabla^v_vV)
=
0
\,,
\end{equation*}
furthermore, since $Q^V_VX=0$,
\begin{multline*}
(\nabla_x\tilde Q)^v_vx
=
\nabla^v_x\tilde Q^V_V X
-
\tilde Q^v_{\nabla^v_xV}x
-
\tilde Q^v_v\nabla^v_xX
-
(\dot\partial\tilde Q)^v_vx(\mathbf A^v_xv)
=
-
\tilde Q^v_{\mathbf A^v_xv}x
-
(\dot\partial\tilde Q)^v_vx(\mathbf A^v_xv)
\\
=
-
(\dot\partial\tilde Q)^v_vx(\mathbf A^v_xv).
\end{multline*}
Finally, from \cite[Eq. (56)]{Jav20},
\begin{equation*}
g_v(P_v(v,x,\mathbf A^v_xv),v) = 0.
\end{equation*}
All the above identities imply that in \eqref{eqrefer} all the terms to the right are zero but the first two. Taking into account that by skew-symmetry $g_v(\mathbf A^v_x \mathbf A^v_x v,v)=-g_v(\mathbf A^v_x v,\mathbf A^v_x v)$, we conclude.
\end{proof}
\begin{remark}
The expressions for the flag curvatures in the last corollary were previously obtained in \cite[Theorem 5.1]{Vitorio2010} (see also \cite[Theorem 5.12]{Vitorio2017} for the second one). In the first identity, the covariant derivative of $\mathbf T$ in this reference is slightly different from ours.
On the other hand, in the positive definite case, $g_v(\mathbf A^v_xv,\mathbf A^v_xv)$ is non-negative. As a consequence of this corollary, we have just shown that Finsler submersions never increase the flag curvature along horizontal flags, in the sense that
\begin{equation}
K_v(x) \leq \tilde K_{\tilde v}(\tilde x)
\end{equation}
along each horizontal $v$ and for every $g_v$-horizontal $x$. This fact has been previously derived by a different method in \cite[Theorem 6.1]{2001AlDur}.
\end{remark}
\section{Submersions whose fibers are totally geodesic}\label{totallygeo}
Let us study pseudo-Finsler submersions having fibers which are totally geodesic, namely, the geodesics of their fibers are also geodesics of the total space. In the classical case of Riemannian submersions, this corresponds with the submersions having $\mathbf T=0$, and they can be characterized by the fact that all the fibers are isometric. We will see that the Finslerian case is more complex. Observe that the fibers of a pseudo-Finsler submersion are totally geodesic if and only if $\mathbf T^v_vv=0$ for all admissible vertical $v$. This turns out to be equivalent to $\mathbf T^v_vu=0$ for all admissible vertical $v$ and arbitrary $u$ (see \cite[Prop. 4]{2022HuberJavaloyes} and the basic properties of $\mathbf T$).
\begin{proposition}
Given two pseudo-Finsler submersions $\sigma_1:(\mathcal M_1,L_1)\rightarrow (\mathcal M_2,L_2)$ and $\sigma_2:(\mathcal M_2,L_2)\rightarrow (\mathcal M_3,L_3)$, it holds that
\begin{enumerate}[(i)]
\item their composition is also a pseudo-Finsler submersion,
\item if this composition has totally geodesic fibers, then so does the submersion onto the final base.
\end{enumerate}
\end{proposition}
\begin{proof}
To check $(i)$, let $g^1$ and $g^2$ be the fundamental tensors of $L_1$ and $L_2$, respectively. By hypothesis every $v$ horizontal with respect to the initial submersion satisfies $L_1(v)=L_2(\mathrm{d}\sigma_1(v))$. We have a similar identity for the final submersion. By definition, a vector is horizontal if and only if $g_v(v,\cdot)$ vanishes on the vertical subspace; because the vertical subspace of the composition contains the vertical subspace of the initial submersion, a horizontal vector with respect to the composition is in particular a horizontal vector for the initial submersion, such that $L_1(v)=L_2(\mathrm{d}\sigma_1(v))$ is satisfied for that initial submersion, whereby \eqref{gv1} yields that $g^1_v(v,e)=0$ if and only if $ g^2_{\mathrm{d}\sigma_1(v)}(\mathrm{d}\sigma_1(v),\mathrm{d}\sigma_1(e))=0$ for each vector $e\in T_{\pi(v)}\mathcal M_1$. Now observe that $\mathrm{d}\sigma_1(v)$ is horizontal with respect to the final submersion, which concludes, since then $L_1(v)=L_2(\mathrm{d}\sigma_1(v))=L_3(\mathrm{d}\sigma_2(\mathrm{d}\sigma_1(v)))$. This is because if $\tilde e$ is $\sigma_2$-vertical, then as $\mathrm{d}\sigma_1$ is surjective, $\tilde e=\mathrm{d}\sigma_1(e)$ for some $e\in T\mathcal M_1$. Moreover, $e$ is $\sigma_1\circ\sigma_2$-vertical and as $v$ is $\sigma_1\circ\sigma_2$-horizontal, it follows that
$g^1_v(v,e)=0$. Applying \eqref{gv2}, we conclude that $g^2_{\mathrm{d}\sigma_1(v)}(\mathrm{d}\sigma_1(v),\tilde e)=0$, and then $\mathrm{d}\sigma_1(v)$ is $\sigma_2$-horizontal as required.
To check $(ii)$, let $\gamma_2$ be a geodesic of a fiber of $\sigma_2$ and $\gamma_1$ one of its horizontal lifts by
$\sigma_1$. If $\nabla$ and $\tilde\nabla$ are the Chern connections of $(\mathcal M_1,L_1)$ and $(\mathcal M_2,L_2)$, respectively, and $D_{\gamma_1}$ and $\tilde D_{\gamma_2}$ their associated covariant derivatives alogn $\gamma_1 $ and $\gamma_2$, respectively, then
\begin{equation}\label{dsigma1}
\mathrm{d}\sigma_1(D^{\dot\gamma_1}_{\gamma_1}\dot\gamma_1)=\tilde D^{\dot\gamma_2}_{\gamma_2}\dot\gamma_2
\end{equation}
by the dual Gauss formula (see Proposition \ref{GaussProposition} and observe that $\tilde Q^v_vv=0$ for all $v$ horizontal). Moreover, it is easy to check that $\gamma_1$ lies in a fiber of $\sigma_2\circ\sigma_1$, and $D^{\dot\gamma_1}_{\gamma_1}\dot\gamma_1$ is $\sigma_2\circ\sigma_1$-vertical because $\sigma_2\circ\sigma_1$ has totally geodesic fibers. Summing up, from the last claim and \eqref{dsigma1},
\[\mathrm{d}\sigma_2(\tilde D^{\dot\gamma_2}_{\gamma_2}\dot\gamma_2)=\mathrm{d}\sigma_1( \mathrm{d}\sigma_1(D^{\dot\gamma_1}_{\gamma_1}\dot\gamma_1))=0,\]
and therefore $\tilde D^{\dot\gamma_2}_{\gamma_2}\dot\gamma_2$ is $\sigma_2$-vertical, which implies that $\mathbf T^{\dot\gamma_2}_{\dot\gamma_2}\dot\gamma_2=0$, with $\mathbf T$ the corresponding O'Neill tensor of $\sigma_2$, and therefore $\sigma_2$ has totally geodesic fibers.
\end{proof}
Let $\sigma:(\mathcal M,L)\rightarrow (\mathcal B,\tilde L)$ be a pseudo-Finsler submersion. Assume that the domains of $L$ and $\tilde L$ are the slit tangent bundle and $\dim\mathcal M\geq 3$. For each piecewise smooth path $ \tilde\gamma $ of the base manifold joining $\tilde p$ and $\tilde q$, let $ F_{\tilde\gamma} :\mathcal F_{\tilde p}\rightarrow \mathcal F_{\tilde q}$ be the map defined for every $p\in \mathcal F_{\sigma(p)}$ as follows. Let $\gamma$ be the horizontal lift of $\tilde \gamma$ with initial point at $p$, which there exists and it is unique by Lemma \ref{legendrelemma}. Then $F_{\tilde\gamma}(p)$ is defined as the endpoint of $\gamma$. As horizontal lifts are determined by ODE's, the map $F_{\tilde\gamma}$ will be smooth because of the smooth dependence of the solutions of these EDO's from initial values. Moreover, it is a diffeomorphism because if $\tilde\beta$ is the reverse curve of $\tilde \gamma$, then $F_{\tilde \beta}$ is the smooth inverse of $F_{\tilde\gamma}$.
Due to the relationship between (ambient) horizontal geodesics and geodesics of the base manifold, we may prove that, under the right conditions, the map $F_{\tilde\gamma}$ is an isometry.
\begin{definition}\label{horreg}
A pseudo-Finsler submersion is said to be horizontally regular when for each admissible vertical $ v $ and projectable horizontal $ X $
\begin{equation*}
(\nabla^v_v (X^\top))^\top_v = 0.
\end{equation*}
\end{definition}
\begin{proposition}\label{isometry}
Let $\sigma:(\mathcal M,F)\rightarrow (\mathcal B,\tilde F)$ be a horizontally regular Finsler submersion between complete Finsler manifolds. Then the map $ F_{\tilde\gamma} $ is an isometry for all $\tilde \gamma$ if and only if the fibers are totally geodesic.
\end{proposition}
\begin{proof}
Assume first that the fibers are totally geodesic. To show that $ F_{\tilde\gamma} $ is an isometry, assume that $ \tilde\gamma $ is parametrised in $[0,1]$ such that $ \tilde\gamma(0) = \tilde{p} $ and $ \tilde \gamma(1) = \tilde{q} $. Consider a smooth admissible curve
$\alpha_0:[0,1]\rightarrow \mathcal F_{\tilde p}$ and then define $\lambda:[0,1]\times [0,1]\rightarrow \mathcal M$ such for every $t\in[0,1]$, $s\rightarrow \lambda(s,t)$ is the horizontal lift of $\tilde \gamma$ with initial point at $\alpha_{ 0}(t)$.
Denote $\alpha_s$ the curve in the fiber $\mathcal F_{\tilde\gamma(s)}$ defined as $\alpha_s(t)=\lambda(s,t)$ and $\beta_t$ the horizontal curve given by $\beta_t(s)=\lambda(s,t)$. Let $X_s(t)=\frac{d}{d s}\lambda(s,t)=\dot\beta_t(s)$ be one of the variational vector fields of $\lambda$. Observe that $X_s$ is always horizontal. Then, recalling the commutation of the covariant derivative in a two-parameter map (see \cite[Prop. 3.2]{Jav14}),
\begin{equation}\label{previous}
(D^{\dot\alpha_s}_{\beta_t} \dot\alpha_s)^\top_{\dot\alpha_s}
=
(D^{\dot\alpha_s}_{\alpha_s} X_s)^\top_{\dot\alpha_s}
=
(D^{\dot\alpha_s}_{\alpha_s} X_s^\top)^\top_{\dot\alpha_s}
+
\mathbf T^{\dot\alpha_s}_{\dot\alpha_s} (X_s)^\bot_{\dot\alpha_s}
=
\mathbf T^{\dot\alpha_s}_{\dot\alpha_s} {(X_s)}^\bot_{\dot\alpha_s}
=
0
\end{equation}
by horizontal regularity and total geodesicity of the fibers, therefore
\begin{equation*}
g_{\dot\alpha_s}(D^{\dot\alpha_s}_{\beta_t} \dot\alpha_s,\dot\alpha_s)
=
0
\,.
\end{equation*}
Let us now show that $ \alpha_1 = F_{\tilde\gamma} \circ \alpha_0 $ has the same length as $ \alpha_0 $. Assuming that $ F(\dot\alpha_s)^2 = g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s) $ is non-negative, and by the Leibniz integral rule, the variation
\begin{equation*}
\tfrac{\partial}{\partial s} \int^1_0 \sqrt{g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)} \mathrm{d} t
=
\int^1_0 \tfrac{\partial}{\partial s} \sqrt{g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)} \mathrm{d} t
=
\int^1_0 \tfrac{\frac{\partial}{\partial s} g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)}{\sqrt{g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)}} \mathrm{d} t
\end{equation*}
of the length of $ \alpha_s $ is zero. Indeed,
\begin{equation*}
\tfrac{\partial}{\partial s} g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)
=
2 g_{\dot\alpha_s}(D^{\dot\alpha_s}_{\beta_t} \dot\alpha_s,\dot\alpha_s)
+
2 C_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s,D^{\dot\alpha_s}_{\beta_t} \dot\alpha_s)
=
0.
\end{equation*}
Consequently, the image by $ F_{\tilde\gamma} $ of any vertical geodesic between two fixed points $ p $ and $ p' $ has the same length. A geodesic that attains the minimal length exists by completeness,
therefore, given $p_1,p_2\in \mathcal F_{\tilde p} $, it follows that $d_{\mathcal F_{\tilde p}}(p_1,p_2)\geq d_{\mathcal F_{\tilde q}}(F_{\tilde\gamma} (p_1),F_{\tilde\gamma} (p_2))$ and reasoning with the inverse $F_{\tilde \beta}$, one obtains the other inequality, concluding that $F_{\tilde\gamma} $ is an isometry.
Conversely, assume that the maps $F_{\tilde\gamma}$ are always isometries. Since by the first two identities in \eqref{previous} and skew-symmetry
\begin{equation*}
g_{\dot\alpha_s}(D^{\dot\alpha_s}_{\beta_t} \dot\alpha_s,\dot\alpha_s)
=
g_{\dot\alpha_s}(\mathbf T^{\dot\alpha_s}_{\dot\alpha_s} X_s,\dot\alpha_s)
=
-
g_{\dot\alpha_s}(X_s,\mathbf T^{\dot\alpha_s}_{\dot\alpha_s} \dot\alpha_s)
\,,
\end{equation*}
we have that if
\begin{equation*}
\int^1_0 \tfrac{\frac{\partial}{\partial s} g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)}{\sqrt{g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)}} \mathrm{d} t
=
\tfrac{\partial}{\partial s} \int^1_0 \sqrt{g_{\dot\alpha_s}(\dot\alpha_s,\dot\alpha_s)} \mathrm{d} t
=
0
\end{equation*}
then, by the mean value theorem, for some $t_1\in[0,1]$
\begin{equation*}
g_{\dot\alpha_s(t_1)}(X_s(t_1),\mathbf T^{\dot\alpha_s(t_1)}_{\dot\alpha_s(t_1)} \dot\alpha_s(t_1))
=
\tfrac{\partial}{\partial s} g_{\dot\alpha_s(t_1)}(\dot\alpha_s(t_1),\dot\alpha_s(t_1))
=
0
\,.
\end{equation*}
By the same argument we obtain a sequence of values $t_n\in[0,\frac{1}{2^n}]$ converging to $0$ for which $g_{\dot\alpha_s(t_n)}(X_s(t_n),\mathbf T^{\dot\alpha_s(t_n)}_{\dot\alpha_s(t_n)} \dot\alpha_s(t_n))$ is zero, and by continuity
\begin{equation*}
g_{\dot\alpha_s(0)}(X_s(0),\mathbf T^{\dot\alpha_s(0)}_{\dot\alpha_s(0)} \dot\alpha_s(0))
=
0
\,.
\end{equation*}
Conclude by non-degeneracy.
\end{proof}
\begin{corollary}
For a horizontally regular Finsler submersion with totally geodesic fibers, the flow of a projectable horizontal geodesic vector field gives rise to an isometry between the fibers.
\end{corollary}
\begin{proof}
Consider on some neighbourhood a projectable vector field satisfying the geodesic equation. Provided it is horizontal, each of the geodesic arcs described by its flow is the horizontal lift of some geodesic arc $ \tilde\gamma $ described by the flow of its projection, which satisfies the geodesic equation by the dual Gauss formula; by construction $ F_{\tilde\gamma} $ is that horizontal flow, which we have just shown to be an isometry between the fibers in Proposition \ref{isometry}.
\end{proof}
\begin{theorem}\label{totgeoth}
If the ambient manifold of a Finsler submersion is connected and geodesically complete, then its base manifold is geodesically complete. If additionally the submersion is horizontally regular and has totally geodesic fibers, then it is the projection of a bundle associated with a principal fiber bundle whose structure group is the Lie group of isometries of the fiber.
\end{theorem}
\begin{proof}
First recall that the isometries of a Finsler manifold form a Lie group (see \cite{2002DengHou} for further details).
For completeness, observe that each geodesic arc is extended by the projection of the extension of its horizontal lift. For the second part of the theorem, fix a point $ \tilde{p} $ in the base manifold $ \mathcal B $ and denote by $ \mathbf{F} $ its submersion fiber at some point $\tilde p$, namely, $\mathbf{F}=\mathcal F_{\tilde p}$. Let us futhermore denote by $ \mathbf{G} $ the Lie group of isometries of $ \mathbf{F} $, by $ \mathbf{G}_{\tilde{q}} $ the set of isometries from $ \mathbf{F} $ to $ \sigma^{-1}(\lbrace \tilde{q} \rbrace) $, and by $ \mathbf{E} $ their union. Note how $ \mathbf{G} $ acts (diffeomorphically) and freely on $ \mathbf{E} $ as
\begin{equation*}
\begin{array}{ccc}
\mathbf{G} \times \mathbf{E} & \to & \mathbf{E}
\\
(\mathbf{g},\mathbf{e}) & \mapsto & \mathbf{e} \circ \mathbf{g}
\mathrlap{\,.}
\end{array}
\end{equation*}
Let us show that $ \mathbf{E} $ is a fiber bundle over $ \mathcal B $ for the map $ \mu \colon \mathbf{E} \to \mathcal B $ sending each $ \mathbf{G}_{\tilde{q}} $ onto $ \tilde{q} $, with structure group $ \mathbf{G} $, equipped with the differentiable structure obtained as follows.
For the choice of a point $ \tilde{q}_i $ in each set of some open covering of $ \mathcal B $ by geodesically convex sets $ \mathbf{U}_i $, and the choice of a smooth path $ \tilde\gamma_i $ joining $ \tilde{p} $ to $ \tilde{q}_i $, there exists, at each point $ \tilde{q} $ and for each $ \mathbf{U}_i $ that contains it, a unique geodesic arc $ \tilde\gamma_{i,\tilde{q}} $ from $ \tilde{q}_i $ to $ \tilde{q} $. By the previous proposition, both $ \tilde\gamma_i $ and $ \tilde\gamma_{i,\tilde{q}} $ induce an isometry between the corresponding submersion fibers, respectively $ F_{\tilde\gamma_i} $ and $ F_{\tilde\gamma_{i,\tilde{q}}} $. Let us define a local section $ \mathbf{e}_i $ on each $ \mathbf{U}_i \ni \tilde{q} $ as the unique isometry
\begin{equation*}
\mathbf{e}_i(\tilde{q}) = F_{\tilde\gamma_{i,\tilde{q}}} \circ F_{\tilde\gamma_i} \in \mathbf{G}_{\tilde{q}}
\,.
\end{equation*}
When $ \mathbf{U}_i \cap \mathbf{U}_j \ni \tilde{q} $, we can define the transition functions
\begin{equation*}
\mathbf{g}_{i\!j}(\tilde{q}) = \mathbf{e}^{-1}_j(\tilde{q}) \circ \mathbf{e}_i(\tilde{q}) \in \mathbf{G}
\,,
\end{equation*}
satisfying
\begin{equation*}
\mathbf{e}_i(\tilde{q}) = \mathbf{e}_j(\tilde{q}) \circ \mathbf{g}_{i\!j}(\tilde{q})
\end{equation*}
and the cocycle condition with respect to the group action
\begin{equation*}
\mathbf{g}_{ik}(\tilde{q}) = \mathbf{g}_{jk}(\tilde{q}) \circ \mathbf{g}_{i\!j}(\tilde{q})
\,.
\end{equation*}
By the fiber bundle construction theorem, $ \mathbf{E} $ is a fiber bundle over $ \mathcal{B} $ for the map $ \mu $.
Denote by $ \mathbf{E}' $ the quotient of $ \mathbf{E} \times \mathbf{F} $ by the (diffeomorphic) action
\begin{equation*}
\begin{array}{ccc}
\mathbf{G} \times \mathbf{E} \times \mathbf{F} & \to & \mathbf{E} \times \mathbf{F}
\\
(\mathbf{g},(\mathbf{e},p)) & \mapsto & (\mathbf{e} \circ \mathbf{g},\mathbf{g}^{-1}(p))
\mathrlap{\,.}
\end{array}
\end{equation*}
By the definition of the action, the map
\begin{equation*}
\begin{array}{ccc}
\mathbf{E} \times \mathbf{F} & \to & \mathcal B
\\
(\mathbf{e},p) & \mapsto & \mu(\mathbf{e})
\end{array}
\end{equation*}
is identically equal to $\tilde{q}$ on the whole equivalence class of $(\mathbf{e},p)$ in $ \mathbf{E}' $, and induces a well-defined projection $ \mu' \colon \mathbf{E}' \to \mathcal B $, defining a fiber bundle associated with the principal fiber bundle $ \mu \colon \mathbf{E} \to \mathcal B $. Let us now show that the submersion coincides with this associated fiber bundle.
Indeed, the action on $ \mathbf{E} \times \mathbf{F} $ is free, by freedom of the action of $ \mathbf{G} $ on $ \mathbf{E} $ and proper, because the action of $G$ on $\mathbf F$ is proper (recall that the isometries of a Finsler manifold is a closed subgroup of the isometries of the Riemannian metric obtained by averaging \cite{Torrome}).
The quotient $ \mathbf{E}' $ of $ \mathbf{E} \times \mathbf{F} $ by the free and proper action of $ \mathbf{G} $ is a manifold by the quotient manifold theorem (see \cite[Th. 7.10]{2000Lee}). Finally, the map
\[ \phi: \mathbf{E}\times \mathbf{F}\rightarrow\mathcal M\]
defined as $\phi(\mathbf{e},p)=\mathbf{e}(p)$ induces a map $\phi'$ from the quotient $\mathbf{E}'$, which is in fact a diffeomorphism and maps fibers of $\mu'$ to fibers of $\sigma$. To check that $\phi'$ is surjective observe that fixing $\mathbf{e}\in\mathbf{E}$, $\phi'(\mathbf{e},\cdot)$ maps $\mathbf{F}$ to a fiber of $\mathcal M$ diffeomorphically. Moreover, if
$\phi(\mathbf{e},p)=\phi(\mathbf{e}',p')$, then $\mathbf{e}^{-1}\circ\mathbf{e}'\in \mathbf{G}$ and
$(\mathbf{e},p)$ and $(\mathbf{e}', p')$ are representatives of the same class under the action of $\mathbf{G}$.
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,458
|
from cookiecutter.main import cookiecutter
class Runner:
def __init__(self, settings):
self.settings = settings
def run(self):
cookiecutter(
self.settings.template,
self.settings.checkout,
self.settings.no_input,
self.settings.extra_context,
self.settings.replay,
self.settings.overwrite_if_exists,
self.settings.output_dir,
self.settings.config_file
)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,790
|
Graus è un comune spagnolo di 3.605 abitanti situato nella comunità autonoma dell'Aragona, nella provincia di Huesca.
Il comune è anche il capoluogo della comarca della Ribagorza e dista 20 km da Huesca.
Il capoluogo del comune è un piccolo pittoresco paese pirenaico a 469 m s.l.m., base di partenza per escursioni in montagna, caratterizzato da strette e tortuose vie sulle quali si affacciano edifici medievali. Dista 98 km dal confine francese, 90 da Saragozza e 122 da Jaca.
Origini del nome
Secondo alcuni studiosi il toponimo deriva dal termine ibero-basco gara-us che significa "rupe brulla", secondo altri dal latino Gradus "valico stretto", "gola".
Storia
Nei territori pirenaici è certa la presenza umana già nel Paleolitico come attestano i numerosi giacimenti archeologici dei dolmen e di resti trovati all'interno di grotte con pitture rupestri. I popoli che abitarono la zona furono diversi: Celti, Iberi, Baschi, che opposero molta resistenza ai Romani che dovettero combattere oltre 20 anni prima di potersi insediare su queste terre nel III secolo a.C. Augusto nel I secolo modernizzò l'organizzazione amministrativa e la dominazione romana continuò fino alla discesa dei popoli barbari che li sostituirono con i Visigoti fino agli inizi dell'VIII secolo quando la Spagna fu invasa dagli Arabo Berberi.
I Visigoti che, inizialmente, avevano fondato un unico regno, si suddivisero poi in vari piccoli regni che opposero molta resistenza ai Musulmani ma senza coordinarsi e Graus fu uno dei territori più settentrionali raggiunti dagli Arabi. Con l'appoggio del conte di Tolosa i cristiani riuscirono ad organizzarsi e a liberarsi della dominazione araba, nacque così nell'XI secolo la contea di Ribagorza, che assieme a quelle di Aragona e di Sobrarbe era sotto il dominio del re pamplonese Sancho III Garcés detto El Mayor. Alla morte di questo re nel 1035 il figlio Ramiro I ereditò la contea di Aragona e il figlio Gonzalo le contee di Sobrarbe e Ribagorza ma morì assassinato e tutto passò al fratello Ramiro primo re del regno di Aragona che si formò unendo le tre contee. Da allora Graus seguì le sorti grandiose e gloriose del regno di Aragona senza alcun protagonismo. Ramiro I fortificò i vari paesi e combatté i musulmani sottraendo loro nuove terre. Morì nel 1064 e il figlio Sancho Ramirez continuò l'azione politica e militare del padre e intrattenne buoni rapporti con la chiesa di Roma, accettò la riforma gregoriana e cluniacense e instaurò il rito romano al posto di quello mozarabico, allargò i confini del regno e portò la capitale a Jaca, che poi il figlio Pietro I succedutogli alla sua morte nel 1094 portò a Huesca. Il successore di Pietro I nel 1134 morì e il regno di Aragona passò al fratello Alfonso I detto "Il Battagliero" (El Batallador) per le numerose battaglie combattute contro i Mori che gli fecero conquistare anche Saragozza.
Il XV secolo fu un secolo di forti tensioni sociali e politiche di rivalità fra nobili e monarchi. Il matrimonio nel 1469 fra Ferdinando II di Aragona e Isabella di Castiglia aprì una nuova epoca e l'unione personale dei due regni dieci anni dopo iniziò la nuova monarchia cattolica di Spagna. (Il titolo di Cattolici fu dato al re Ferdinando e alla regina Isabella dal papa Alessandro VI Borgia).
I secoli dal XVI al XVIII furono secoli di progresso economico per la Spagna che era divenuta una delle grandi potenze mondiali e così fu anche per Graus e si costruirono diversi nuovi edifici religiosi e civili. Durante la guerra d'indipendenza subì dei danni da parte delle truppe napoleoniche.
Oggi è un attivo centro turistico che utilizza il suo pittoresco centro storico medievale, la sua posizione per cui è detta "Porta dei Pirenei", i corsi d'acqua e le bellezze naturali dei dintorni per attirare i turisti, gli sportivi e gli amanti della montagna e della villeggiatura. Ha il titolo di muy noble y muy antigua villa. Vi si parla un dialetto aragonese chiamato grausino.
Dintorni
A 9,9 km Secastilla con il Santuario di Torreciudad fondato nel 1976 dall'Opus Dei.
A 9 km Panillo con il monastero Buddhista fondato nel 1984.
A 27 km Roda de Isabena paese che fu una delle più antiche sedi vescovili dei Pirenei con una cattedrale romanica del 1067 ricca di opere d'arte che fa parte dell'antica fortezza.
A 44 km Seira con le pittoresche gole dell'Ésera dette El Congosto de Ventamilla.
Monumenti e luoghi d'interesse
Basilica de la Virgen de la Peña, santuario del 1538 cui si accede percorrendo una lunga galleria. È dichiarata patrimonio de interés historico y artistico.
Iglesia Parroquial de San Miguel, romanica del XIII secolo.
Ermita de San Gregorio, del XII secolo.
Puente de Abajo, ponte del XII secolo su fondamenta romane rifatto nel XVI secolo.
Casa consistorial del XVIII secolo.
Ayuntamiento, sede municipale del XVI secolo.
Casa de los Mur del XVI secolo.
Casa de los Solanos.
Casa de Bardaxi, coeve.
Parque Natural.
Feste
Di interesse turistico nazionale sono le feste che si svolgono dal 12 al 15 settembre in onore di Cristo e di San Vincenzo Ferreri che comprendono oltre ai riti religiosi la cosiddetta Espera de los Gaileros con musica, danze e la popolare Mejiganga, rappresentazione popolare di tono satirico-umoristico.
Durante luglio e agosto si hanno i concerti di musica classica dei festivales de Ribagorza, sempre in luglio si svolge la festa della longaniza y el comercio con la preparazione del salsicciotto più grande del mondo. La longaniza è una specialità gastronomica del paese, un insaccato di suino protetto, con la sigla "C" che indica la qualità certificata.
Le feste patronali si svolgono dal 12 al 15 del mese di settembre, inoltre ogni lunedì si svolge in modo particolare il mercado de la trufa cioè del tartufo.
Altri progetti
Collegamenti esterni
*
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,731
|
Xiaodongzhuang (kinesiska: 小东庄) är en ort i Kina. Den ligger i storstadsområdet Tianjin, i den norra delen av landet, omkring 20 kilometer öster om stadens centrum. Antalet invånare är .
Runt Xiaodongzhuang är det mycket tätbefolkat, med invånare per kvadratkilometer. Närmaste större samhälle är Junliangcheng, km öster om Xiaodongzhuang. Trakten runt Xiaodongzhuang består till största delen av jordbruksmark.
Genomsnittlig årsnederbörd är millimeter. Den regnigaste månaden är juli, med i genomsnitt mm nederbörd, och den torraste är januari, med mm nederbörd.
Källor
Orter i Tianjin
WP:Projekt Kina
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,452
|
Q: Npm test cannot find socket.io.js file I am trying to make tests using Karma and i am following Angulars own testing guide. Currently i am using the most basic test class but having issues running it.
The test class:
describe('1st test',()=>{
it('true is true', ()=> expect(true).toBe(true)); });
When i run it using npm test i get the following error in the command prompt:
[1] 24 04 2017 11:27:24.957:WARN [web-server]: 404: /base/node_modules/socket.io-client/dist/socket.io.js
[1] Chrome 57.0.2987 (Windows 10 0.0.0) ERROR
[1] {
[1] "__zone_symbol__error": {
[1] "originalStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "zoneAwareStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js"
[1] },
[1] "stack": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "name": "Error",
[1] "message": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "originalStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "zoneAwareStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "__zone_symbol__stack": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "originalErr": {
[1] "__zone_symbol__error": {
[1] "originalStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "zoneAwareStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js"
[1] },
[1] "stack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "name": "Error",
[1] "message": "XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "originalStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "zoneAwareStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__currentTask": {
[1] "type": "microTask",
[1] "state": "notScheduled",
[1] "source": "Promise.then",
[1] "zone": "<root>",
[1] "cancelFn": null,
[1] "runCount": 0
[1] },
[1] "__zone_symbol__stack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__message": "XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__name": "Error",
[1] "__zone_symbol__originalStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__zoneAwareStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js"
[1] },
[1] "__zone_symbol__currentTask": {
[1] "type": "microTask",
[1] "state": "notScheduled",
[1] "source": "Promise.then",
[1] "zone": "<root>",
[1] "cancelFn": null,
[1] "runCount": 0
[1] },
[1] "__zone_symbol__name": "Error",
[1] "__zone_symbol__message": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "__zone_symbol__originalStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "__zone_symbol__zoneAwareStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js"
[1] }
[1]
[1] Chrome 57.0.2987 (Windows 10 0.0.0) ERROR
[1] {
[1] "__zone_symbol__error": {
[1] "originalStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "zoneAwareStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js"
[1] },
[1] "stack": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "name": "Error",
[1] "message": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "originalStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "zoneAwareStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "__zone_symbol__stack": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "originalErr": {
[1] "__zone_symbol__error": {
[1] "originalStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "zoneAwareStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js"
[1] },
[1] "stack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "name": "Error",
[1] "message": "XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "originalStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "zoneAwareStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__currentTask": {
[1] "type": "microTask",
[1] "state": "notScheduled",
[1] "source": "Promise.then",
[1] "zone": "<root>",
[1] "cancelFn": null,
[1] "runCount": 0
[1] },
[1] "__zone_symbol__stack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__message": "XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__name": "Error",
[1] "__zone_symbol__originalStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js",
[1] "__zone_symbol__zoneAwareStack": "Error: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js"
[1] },
[1] "__zone_symbol__currentTask": {
[1] "type": "microTask",
[1] "state": "notScheduled",
[1] "source": "Promise.then",
[1] "zone": "<root>",
[1] "cancelFn": null,
[1] "runCount": 0
[1] },
[1] "__zone_symbol__name": "Error",
[1] "__zone_symbol__message": "(SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "__zone_symbol__originalStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js",
[1] "__zone_symbol__zoneAwareStack": "Error: (SystemJS) XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError: XHR error (404 Not Found) loading src/node_modules/socket.io-client/dist/socket.io.js\n\tError loading src/node_modules/socket.io-client/dist/socket.io.js as \"socket.io-client\" from src/app/global/socket.service.js"
[1] }
I have checked the file that it cannot find and it exists in the directory stated by the error so i have no clue what to do.
A: Those are 404 errors for XHR requests. It doesn't mean that the file does not exist in the file system but that it is not served by accessing that particular path. It is unlikely that your server serves the entire file system so you shouldn't be worried if the file exists there or not but if it is being served from a particular URL that your program is trying to access it from.
See the code examples in this answer for an example:
*
*Getting data from/writing data to localhost with Express
Here the socket.io.js file is referenced in HTML as:
<script src="/socket.io/socket.io.js"></script>
not as an absolute path in the file system, and not with a path including node_modules. Paths in the HTTP requests and paths in the file system are a different thing and 404 errors are HTTP-specific.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,204
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Marble coating blade with wheat straw handle knives - Full extend design.
Safe to use on all kitchen cooking foods, meats, fruit, vegetables, etc.
The professional kitchen knife set suitable for gift, best home kitchen use.
Q1: What is the material of modern kitchen knife set?
A1: High quality stainless steel blade and new plastic material handle with wheat straw material.
Q2: Can you make OEM/ODM service?
A2: Yes, we can accept OEM service. Also we have our own designer team so it's also welcome to choose our ODM products.
Q3: Can be the handle change the other color?
A3: Yes, the hand color can be changed to your request.
Q4: Can I get the sample?
A4: Yes, we are honored to offer you samples.
Q5: How do you charge the sample fees?
A5: The sample is free to base on our current color handle, the freight cost charge by your side.
Q6: How long for sampling?
A6: Normally, it need 3-7days.
Q7: What about your factory production capacity?
A7: 100000pcs/per month, but this according to the different products. We do our best for your require.
Q8: If i place an order, do you inspect the product? How?
A8: Yes, each step of production and finished products will be carried out inspection by QC before shipping.
Q9: Can you accept the small order?
A9: Yes, we can accept the trial orders, it is the starting of the business, but according to the different product, the price may be higher as normal quantity.
Q10: Can you give me a discount?
A10: Sure, contact us to discuss the further details.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 31
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"Confident" este un cântec a cântăreaței americane Demi Lovato. Lansat ca al doilea single de pe al cincilea album de studio Confident (2015). Piesa a fost lansată pe data de 18 septembrie 2015, de casa de discuri Hollywood și Island Records ca descărcare instantanee cu comandarea albumului pe iTunes, precum și o descărcare pe site-urile digitale, cum ar fi Google Play și 7digital. piesa a fost scrisă de Lovato, Max Martin, Ilya Salmanzadeh, și Savan Kotecha.
Lista pieselor
Descărcare digitală
"Confident" – 3:25
Clasamente
Datele lansărilor
Referințe
Discuri single din 2015
Cântece din 2015
Pop rock
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 80
|
Contents
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Chapter 25
Chapter 26
Chapter 27
Chapter 28
Chapter 29
Chapter 30
Chapter 31
Chapter 32
Chapter 33
Chapter 34
Chapter 35
Chapter 36
Chapter 37
Chapter 38
Chapter 39
Chapter 40
Chapter 41
Chapter 42
Chapter 43
Chapter 44
Chapter 45
Chapter 46
Chapter 47
Chapter 48
Chapter 49
Chapter 50
Chapter 51
Chapter 52
Chapter 53
Chapter 54
Chapter 55
Chapter 56
Chapter 57
Chapter 58
Chapter 59
Chapter 60
A Note on the Author
By the Same Author
To my brother, Brian Whitehouse
1932–2013
Chapter 1
The sun was slanting through the high-vaulted roof of the souk, throwing down shafts of light in which dust motes and thin drifts of cigarette smoke swirled lazily. Miles Brookhaven began to relax as he walked down the long central avenue, breathing in smells of powdery piles of spices, reaching over to touch the shiny purple skins of aubergines, and exchanging a shouted greeting with the stallholder.
He stopped at a food stall on the corner of one of the side aisles where the same old man who'd been there since God knows when had a juicing machine. As he usually did when he took this route, Miles stopped for a glass of fresh orange juice. Against the wall behind the counter a shawarma of meat the size of a tree trunk rotated on a long sharp pole. Miles propped one hip on a stool in the corner, from where he could look down the main aisle, the way he'd come, but his eye was drawn to the spit. There was something different. Usually, as he drank, he would watch a short balding man called Afiz, his apron stained by the spattering juices, wielding a long knife of incredible sharpness, peeling shavings of meat off the shawarma like strips of wallpaper.
He and Afiz had established a friendly unspoken ritual – Afiz would turn and gesture to Miles with his knife, as if to ask You want some? Miles would shake his head and hold up his glass to show that was what he'd come for. Afiz would laugh and turn back to the shawarma.
But it wasn't Afiz who tended the spit today. Instead a young man held the long knife. He was tall with a prominent Adam's apple and long black hair tied back into a knot, and he stared at Miles with dark indifferent eyes, then turned away to serve a customer. He had none of Afiz's practised delicacy; instead he just hacked at the meat, which fell in chunks instead of paper-thin slices. That seemed odd, Miles thought as he sipped his juice. Holding the glass in one hand he reached into his pocket for some coins to pay, and it was then he sensed movement, looked up and saw the young man coming towards him, holding the knife in one hand, his eyes glazed and hostile.
Not pausing to think, Miles tilted his glass of juice and hurled its contents straight into the eyes of his attacker. The long-haired youth was caught by surprise, blinking furiously, trying to get the juice out of his eyes. Miles took a step back, and as the young man lunged forward, swiping hard with the knife, he threw the empty glass at his face.
It hit the youth square in the eye. He yelled in pain and dropped the knife, which fell onto the tiled floor of the stall and bounced from its point, erratically, before landing at last, like an offering, at Miles's feet. As Miles bent down and grabbed it, the young man ran out of the far side of the stall.
Miles stared at the fleeing figure and when he turned back he saw that the juice man had fled as well. The commotion was drawing a crowd. Miles understood from the jabber of Arabic that they were wondering what this Westerner was doing, holding that knife. He put it down on the counter and without looking around strode quickly down the aisle towards the exit from the souk. The last thing he or his colleagues needed was the attention of the police.
By the time he'd reached the modern end of the souk, no one in the crowd of shoppers seemed to be taking special notice of him. He slowed to a stroll, forced himself to breathe normally and began to review what had happened. Was the young man just another extremist who hated Westerners? He didn't think so. The fact that he'd been working at the stall where Miles regularly stopped – and that the juice man had fled as well – made it seem more likely that he'd been targeted.
Perhaps the group he'd been working with had been penetrated – but by whom? That was what made it hard to deal with the rebels. Too many conflicting interests; too many irons in the fire. Your enemy's enemy wasn't necessarily your friend. Whatever the explanation, he couldn't go on using the same cover. It was time to move on.
Outside in the bright sun, Miles realised that his hand felt sticky, raised it and found it covered in blood. More blood was running down the sleeve of his jacket, and moving his shoulder made him wince in pain. That swipe with the knife must have connected.
He'd begun to feel faint; best get back to the office fast. He heard a gasp and looked up to see a young woman staring in horror at his jacket. Behind her a little man with a bushy black moustache was pointing at him. The blood was flowing fast down his arm now, dripping from the cuffs of his shirt and jacket onto the paving stones. His vision was blurring, and he'd started to sway as he walked. Seeing him stagger, the little man put his arm round him, waving with his other arm at a taxi. 'Hospital, hospital,' he shouted at the driver, and as he bundled him into the back of the car, Miles passed out.
Chapter 2
Liz Carlyle was sitting at her desk in Thames House, the London headquarters of Britain's MI5, frowning at the pile of papers neatly stacked in the centre of her desk. She'd just got back from a three-week holiday walking in the Pyrenees and was wishing she'd stayed there. A spectacled head poked round the door, followed by the rest of Peggy Kinsolving, Liz's long-standing research assistant and now her deputy in the Counter-Terrorist section that Liz ran.
'Welcome back,' said Peggy. 'Did you have a good time? You must be fit as a flea. It's never stopped raining here since you went away.' She waved a hand at the pile of paper. 'Don't worry about that lot. I've read it all and it's just background stuff. The top one is the only important one – I've summarised where we've got to in all the current investigations. You've got a meeting with the Home Secretary on Friday to bring her up to date. If you like I'll come with you.'
Peggy stopped to draw breath and Liz smiled fondly at her younger colleague. 'It is actually great to be back, though I didn't feel that when I woke up this morning. We had a wonderful time. Walked miles, ate too much, drank some great wine. Martin is fine, though he's still wondering whether to leave the DGSE and go into private security work. He fancies getting out of Paris and living in the South – his family home was near Toulouse. But it's a big step to leave government service and go private and there's a lot of competition in the private security field – just like here. Anyway, how are you? And how's Tim?'
Tim was Peggy's boyfriend, a lecturer in English at King's College, London University, a very bright lad if a bit of a sensitive soul. Peggy said, 'I'm fine, and so is Tim, thanks. He's still doing the vegetarian cooking course – advanced level now. I hadn't realised it could be so tasty. I'm quite converted.' They both smiled and Peggy went on, 'There's one thing you won't be too pleased about. We've been given an extra responsibility. I was only told about it on Friday. It's a "watching brief" – whatever that is – for under-the-counter arms supplies to the Arab Spring rebels.'
Liz knew all too well what a watching brief was. It meant extra responsibility with no additional resources. Then if anything bad happened you were to blame. She sighed. 'Is there any intelligence that arms are going from dealers in this country to the rebels?'
'Not that I've heard. It's not so much the rebels per se that anyone's worried about; it's the jihadis who've infiltrated them. The Foreign Secretary went to a meeting in Geneva last week and this was on the agenda. There's a lot of concern about al-Qaeda-type groups leaking into the Arab Spring countries. There were some gruesome pictures on TV while you were away of what they were doing to their captives.'
'I saw them on French TV. But I would have thought they could get arms quite easily from the countries who support them.'
'I know that seems more likely. But the conference decided that each country should put measures in place to ensure that no undercover supplies from the EU countries get to these jihadis. It seems to be more of a matter for Eastern Europe than us, but DG told me on Friday that it's been decided that we were to have the "watching brief".'
'Great. But what about Six? I wonder what they have on this.'
'Quite a lot, I imagine. But guess who's running their part of the show – your favourite officer, Bruno Mackay. Bruno rang me on Friday to welcome us on board. Said he'd like to come over to see you when you were back.'
Liz put her head in her hands and groaned. 'Did I just say I was glad to be back?'
Peggy grinned. 'Bruno told me something quite interesting. Do you remember Miles Brookhaven, who used to be in the CIA station here? Andy Bokus's deputy?' When Liz nodded she went on, 'Apparently he was nearly killed a few months ago. He was under cover in an aid charity the Agency had set up in Syria, running a source in a rebel group, and he was attacked in the souk. They aren't sure if his cover had been blown, or if it was just an opportunist attack, but from what Bruno said, it sounded planned to me. Miles needed a series of blood transfusions – they had to get him out of there pretty quickly.'
'Poor Miles. He was a bit naïve when he was here. He tried to recruit me once – he took me on the London Eye in a private pod and plied me with champagne. It was fun, and I enjoyed watching him waste the Agency's money. I wonder if he's grown up.'
'I hope so.' Peggy got up to go. 'I'll leave you to catch up.'
But as Peggy walked out of the door she bumped into someone coming in. 'Whoops. Sorry, Geoffrey. I was just going. Liz will be delighted to see you!'
The tall, heron-like figure of Geoffrey Fane, a senior MI6 officer, sauntered into the room. 'Good morning Elizabeth. Delighted to see you looking so fit and well. Been on holiday I hear. I hope you had a wonderful time and that our friend Seurat is in good form.'
One of Geoffrey Fane's characteristics, which drove Liz mad, was his inquisitive interest in her private life and his delight in showing how much he knew about it. She would much have preferred him not to know that she was seeing Martin Seurat of the French Military Intelligence Service, the DGSE. But he did know, and she suspected that he had learned about it from Bruno Mackay, who had been the deputy head of MI6's Paris Station when she'd first met Martin.
What she didn't like to acknowledge, though everyone else knew it, was that Geoffrey Fane himself would have liked to be in Martin Seurat's shoes. He was divorced, a lonely man and evidently deeply attracted to Liz, who though she admired and respected his professional skill, couldn't disguise the fact that on a personal level she found him pompous and patronising. What she couldn't – or wouldn't let herself see – was that much of his manner towards her was a cover for his feelings. So he went on calling her 'Elizabeth', though he knew that she preferred to be called 'Liz', and she went on grinding her teeth at the sight of him while everyone marvelled that they seemed to have such a successful working relationship.
Now Fane folded himself elegantly into the chair that Peggy had just left and crossed one long, tailored leg over the other, showing a length of subtly striped sock and a highly polished black brogue. 'I was delighted to learn we'll be working together on the arms supply front,' he said.
'Yes. I've just heard about it from Peggy. I gather we've been given a watching brief and no extra resources, so I don't suppose we'll be doing much active investigation. Anyway, Peggy told me that there's no specific intelligence about any UK arms dealers being involved, and we certainly haven't the time to go looking.'
Fane leant forward in his chair. 'That might have been true last week, but things are moving on. I had a call from Andy Bokus over the weekend. They've just posted a new man into Yemen. An old friend of yours if I'm not mistaken. Miles Brookhaven. I'm sure you remember him from when he was here at Grosvenor with Bokus. I gather he was quite smitten.'
Liz gazed at the languid figure in the chair, clenched her jaw and said nothing. Fane smiled slightly and went on, 'He had a bit of a rough time on his previous posting but he's recovered now. The Agency have sent him to Sana'a to pick up a rather promising contact of the Embassy. Bokus seems to think there may be something interesting to come out for us as well as for the Agency.'
Chapter 3
Miles Brookhaven's shoulder still ached at night if he turned over awkwardly in bed, and now driving the heavy SUV he felt a twinge of pain whenever the car bounced over a rut in the road or he had to turn the wheel suddenly to avoid a pothole. The long knife had slashed the tendons at the top of his arm and the doctors at the military hospital in Germany to which he had been evacuated had told him that they would never properly heal. But, under pressure from CIA headquarters in Langley, the doctors had finally authorised his posting. There was a shortage of experienced case officers with fluent Arabic, and as the Arab Spring spread and jihadis joined the rebels in one country after another, the need for intelligence both from the front line of the fighting and from the countries on the periphery had become urgent.
From NSA at Fort Meade and CIA at Langley to GCHQ in Cheltenham, the eyes and ears of the West were focused on the movement of weapons around the world, and in particular to the countries where rebel groups were fighting governments. It was clear that supplies of some of the most sophisticated weapons were getting through to jihadis.
Yemen was a special focus of attention. Overhead surveillance had shown piles of what appeared to be weapons crates stacked on the dockside at Aden. The photographs had landed on the desk of an analyst in Langley at the same time as a report from the Commercial Counsellor at the US Embassy in Sana'a. The diplomat's report described his meetings with the Minister of Trade, one of his main contacts. The Minister seemed uninterested in developing trade with the US except in one area – weapons. The Minister explained this as the need for his government to be able to protect both its own citizens and foreigners against increasingly aggressive jihadi groups. But the Embassy report pointed out that the same message was coming from the Interior and Defence Ministers – departments where issues of weapons supply seemed a more natural fit.
The other notable feature of the diplomat's meetings with the Minister of Trade was the frequency of his references to 'his' Foundation, set up he said to help the homeless and in desperate need of funds. But research by the Embassy had thrown up no sign of such a charity. It was this last point that had brought the diplomatic cable onto the desk in Langley and which had resulted a few months later in the diplomat being moved to a senior post in another country and Miles Brookhaven turning up in the Sana'a Embassy as the new Commercial Counsellor. He had one brief – to recruit the Minister of Trade.
As he drove, Miles could see the mountains in the west, ranged in a rough semicircle around the city. His health not fully restored, it had taken him a few days to get used to the thin air of Sana'a. This was to be his first meeting with Jamaal Baakrime, the Trade Minister, his recruitment target, and he was feeling rather nervous. In normal circumstances he would have taken months to get to know a target, to assess weaknesses and vulnerabilities before he made his pitch, but he had been told not to hang about with this one but to go straight in and offer him cash for information. If the approach failed he would be quickly withdrawn and posted somewhere else where he could be useful.
He parked on a wide street and walked, turning into a much narrower side street lined by squat government buildings, concrete blocks mostly, put up in the Sixties with US aid money. They were interspersed with a few more recent constructions built as Yemen began to develop its oil resources. Not that Yemen nowadays showed much sign of being oil- or gas-rich. On the streets, even here in the capital, poverty was rife, and as he walked along Miles reflected that if the Minister's charity existed there was plenty for it to do.
Inside the Trade Ministry, a guard with a holstered pistol was sitting in a chair in one corner of the entrance hall reading a magazine. He raised his eyes lazily as Miles came in, then resumed reading. A young uniformed woman behind the front desk took his name, consulted a sheet of paper, then waved Miles to follow her. She led him up the stairs to the first floor, into a large open-plan office where a dozen men and women sat typing, and on into a long corridor with dark little offices on either side, occupied by men sitting behind desks covered with piles of papers.
At the end of the corridor she knocked on a large, closed door. A loud voice boomed out in Arabic and the woman opened the door and ushered Miles inside.
Baakrime's office could have been in a different world. It was roughly forty feet long, lined by picture windows with fabulous views of the mountains. The floors were polished mahogany boards overlaid by a rich sprinkling of fine Persian carpets. Gaudy oil paintings hung on the walls, scenes from the Arabian Nights, featuring scantily draped female figures.
Baakrime came out from behind a large antique desk, his hand extended. He was a diminutive square-shouldered man, with short black hair brushed back in a lacquered wave, and a thick Groucho Marx moustache. 'It is delightful to meet you, Mr Brookhaven. Your predecessor and I had an excellent relationship,' he declared. 'Come, let us make ourselves comfortable.' He gestured towards a sitting area, where two sofas were adorned by soft cushions covered in coloured damask.
They sat down at right angles to each other. 'Coffee is coming,' Baakrime said hospitably. He wore a light grey suit and a white shirt with a canary-coloured tie. A triangle of paisley silk handkerchief peeked out from the breast pocket of his jacket.
'It's good of you to see me,' said Miles. 'I know you have a full schedule.'
'Nonsense. I always have time for my friends,' said Baakrime. He shrugged his shoulders. 'And you know what they say – if you want to get something done, ask a busy man.'
They chatted for a few minutes inconsequentially. Miles was accustomed to the Arab insistence that all business, however pressing, was prefaced by small talk. The coffee arrived, brought in by a young woman, smartly dressed in Western clothes – a noticeably short skirt and a blouse unbuttoned at the top. Baakrime ogled her legs with unconcealed pleasure and, as she put the tray on the low table and bent down to pour out the coffee, his eyes moved to her cleavage.
When she had left, Baakrime continued chatting idly, asking after the welfare of Miles's family. When Miles explained that he was unmarried, Baakrime asked after his parents. He moved on to describe the location, ambience and menu of a new restaurant that Miles must try, and recommended two holiday resorts on the Egyptian coast along the Red Sea.
When finally Baakrime paused to sip his coffee, Miles said, 'I understand that your Department has a role in the import of arms to your country.'
Baakrime stopped sipping but continued to hold his cup to his mouth. He said nothing for a moment, then put the cup on the table, looking all the time at Miles. He said, 'That is true. It is a trade that interests me greatly. We have, as you know, many threats to our country, both internal and external.'
'Yes indeed,' replied Miles. Then, with the instruction from Langley to 'get on with it' at the front of his mind, he said, 'It is also a subject that greatly interests those who sent me to your country.'
Baakrime didn't reply. Miles hoped that he had picked up the hint he had been offered about who Miles worked for. Then, his eyes slipping away from Miles, Baakrime remarked, 'These affairs can be a little complicated, but I might be able to help you get started on your work.'
Miles's heart gave a lurch. Baakrime had recognised the bait. It was time to see if he would swallow it. He said, 'That would be very much appreciated by my government. You know, data is freely available these days – we in the West with all our computers are positively awash with information. But knowledge is scarce, and can be expensive to find. Don't you agree?'
Baakrime smiled and nodded. 'How true that is, my friend.'
Miles ploughed on. 'My colleague also told me that another interest of yours is the Foundation you have set up to help the homeless in your country. That is such an excellent cause that I am authorised to offer you substantial and regular contributions to help in its work. In fact,' he said, reaching into his pocket, 'not knowing what the bank account details for the Foundation are, I have brought our first contribution of ten thousand dollars with me.' And he put a thick white envelope on the table, thinking that if Langley had got this wrong he was going to look awfully stupid.
But Baakrime rapidly swept up the envelope and stuffed it into his pocket. 'That is so very kind and much appreciated. The Foundation is helping many people, I am glad to say. But the recent upheavals in my country mean more people are suffering than ever before, and we cannot keep up. We find it is better not to operate through the banks. They are not so reliable always. This' – he patted his pocket – 'would be the best way to make your contribution in future. I will ensure the cash gets to where it can best be used.'
You bet, thought Miles, but he merely smiled and nodded.
Baakrime said, 'In return for your generosity to my Foundation, you must tell me how I can best help you.'
Miles decided to strike while the iron was hot. 'We know that Yemen is one of the countries through which weapons are reaching rebel groups. And not just legitimate rebels – but others fighting with them, outsiders. Jihadis, extremists, al-Qaeda supporters.'
Baakrime smiled and shrugged but said nothing.
'What we want to know is the sources of those weapons and in particular any sources in Europe or the United States.'
Baakrime's manner changed from the wily to the businesslike. 'These young men. They think they are all Osama Bin Ladens. They are crude and cruel and defame the name of Islam. They are indeed a threat to us all. I will do what I can to help you, my friend. Come back in one week and I will see what I can find out.'
Chapter 4
A raw day. Viewed from the window of Liz's office, the Thames looked battleship-grey, sprinkled with the frothy white lines of waves stirred up by the October wind. To Liz, her skin still brown from her holiday in the Pyrenees, the sun was a faraway memory.
She turned back to the pile of forms on her desk. The Service was blessedly free of much of the bureaucracy that affected the Civil Service, but it strongly believed in annual appraisals of staff, and now that Liz was responsible for managing a team of people, she had to write their performance assessments. She took the task seriously, knowing how important it was to the careers of her team, as well as to the Service itself as a tool for getting the right people in the right jobs. But it was not her favourite pastime. Even though she was now a manager, Liz was still an operational officer at heart. Too much time spent sitting behind her desk made her restless and irritable.
'That looks like fun.' Peggy Kinsolving was standing in the doorway.
Liz looked up. 'I thought you were at the conference.'
'I am. It's the lunch break, so I nipped back to check how that surveillance operation is going on.' Peggy was running an investigation into a group of young men in Camden Town who had just come back from Pakistan.
'Anything happening?'
'No. No movement at all so far. I think they're all still in bed.'
Liz nodded. Peggy had transferred to MI5 from MI6 several years ago. She had been a diffident, shy girl but a genius at research. She would follow a lead like a bloodhound but if you'd asked her to go out and interview someone she would have panicked and frozen with nerves. But over the years, under Liz's guidance, she had grown in confidence and now she was running her own operations, and directing a small team. Peggy had become a skilled interviewer, and had discovered a talent for finding out what made people tick, getting underneath their reserves and breaking down their defences.
But though her personality had developed, her appearance had hardly changed from her days as a librarian. She was a little short of medium height, with long brown hair she tied back in a wispy ponytail. Her spectacles, round and brown, seemed to be too big for her face and were forever slipping down her nose. The sight of Peggy pushing back her spectacles was often the preface to a remark that would begin the unravelling of some knotty problem.
'What's going on at the conference? Any good?' Liz asked. It was a Home Office-run conference aimed mainly at regional police forces, and designed to draw their attention to a nationwide growth in gun crime. Little of the agenda had much direct connection with the work of Liz's team, but she had thought it worthwhile to send someone to register an interest and demonstrate that they were taking their watching brief seriously.
Peggy said, 'Actually it's not been too bad. This afternoon might be quite interesting.'
'Really? What's happening?'
Peggy seemed to be struggling not to laugh. 'Well, it was meant to be a keynote address from the Foreign Office. You remember Henry Pennington?'
Liz groaned. She'd crossed swords with Henry Pennington several times over the years. A long lean man with a large nose that dominated his thin face, he was a panicker. Any indication that something might be going wrong caused him to begin rubbing his hands together in a washing motion and breathing heavily. At such times he was liable to make sudden decisions, which on one or two occasions had landed Liz in difficult situations. She never forgot the time he had volunteered her services as an undercover protection officer for a Russian oligarch, almost succeeding in getting her killed in the process.
'But sadly,' Peggy went on, 'Henry's indisposed. So they've put together a panel instead. Some senior officers from the North and the Midlands are going to be talking about their experience of the arms trade. I thought you might be interested.'
Liz thought about this. Her interest was in illegal arms shipments abroad, but there might be something worth hearing and the alternative was the pile of assessment forms on her desk. 'I think I'll come along.'
When they arrived at the conference room in the Queen Elizabeth Conference Centre in Parliament Square the session had already started. The room was three-quarters full and they slipped into seats at one side of a back row. There were three people on the stage, sitting in a semicircle so that Liz could only see two of them clearly. They were discussing the impact of Britain's gun laws, and Liz recognised one of the speakers – a senior policewoman from Derbyshire, notorious for her impatience with junior officers. The man next to her, who was obviously from the Home Office, was praising the government's tough stance on firearms as if one of his political masters were in the audience. He contrasted the UK's ban on handguns with America, where more often than not there didn't seem to be any gun laws at all. The policewoman from Derbyshire agreed with him that the total ban on handguns in the UK was a great thing.
Suddenly the third member of the panel, who Liz couldn't see properly, interrupted. 'Make no mistake, this country has a gun culture too – it's just invisible to most of us. All the government has really managed to do is drive gun sales further underground. We only hear about them when some drug dealer gets shot in Merseyside. Things have got worse in the last ten years, not better. We need to remember that when we congratulate ourselves on not being like the Americans.'
The bluntness of his remarks would have seemed out of place if the delivery had not been so self-assured, and as it was there was a murmur of assent round the room. The Home Office man looked uncomfortable. Liz sat up and leaned over to try to catch a glimpse of the man's face. There was something in the voice that was familiar.
Peggy noticed. 'What is it?' But Liz put a finger to her lips. The man she couldn't see properly was still talking.
It couldn't be, Liz told herself. She could see that the man was dressed in a suit, not a uniform, and from what she could see of him he looked pretty smart for a policeman. The man she was thinking of had always been a bit of a clothes horse.
Then he shifted in his chair and she could see his profile. She recognised the sharp nose and rugged chin. The hair now was thinner than before, but well cut, with only a few flecks of grey. He was still good-looking; whatever you thought of him you had to give him that.
'You look as if you've seen a ghost,' whispered Peggy.
Liz sighed, leaning back in her chair as the Derbyshire woman started up again. 'It's not a ghost,' she said at last. 'Just somebody I used to know. Though it was a long time ago.'
Chapter 5
Liz had been in MI5 just eighteen months. She had applied on the spur of the moment, in her last year at Bristol University. She had been thinking vaguely and without much enthusiasm that she might stay on at the university to do research when the chance remark of a visiting lecturer had coincided with an intriguing newspaper advertisement for logical, level-headed and decisive people to do important work in the national interest. She had sent in her cv, such as it was, without much hope of any response, and had been amazed to be called for an interview. After that the recruitment process had ground slowly on until, at the end of it all, she'd found herself a member of MI5, Britain's Security Service.
Although she was still on probation and in the training period, Liz felt settled and comfortable in the Service. Each morning when she left the flat in Holloway that she shared with four other Bristol graduates to take the underground to Thames House, she looked forward to the day.
Even though she'd been at university in a city, she wasn't really a city girl. She had grown up in the Wiltshire countryside where her father had been the land agent for a large estate. He was dead now and the estate had been broken up after the death of the last owner without an heir, but her mother still lived in the octagonal Gatehouse where Liz had been brought up. Susan Carlyle managed the flourishing garden centre that now occupied the old kitchen gardens of the estate.
Liz was enjoying living in London and felt guilty that she didn't go down to Wiltshire more often, as she knew her mother was lonely. Susan Carlyle didn't disguise the fact that she would like Liz to abandon what she thought of as a 'dangerous job' and marry a nice young man, a solicitor or a doctor or something safe. Liz couldn't think of anything she wanted to do less.
Between them Liz and her flatmates had a fairly wide circle of friends. There was a faint shadow over Liz's social life in that she couldn't join in enthusiastically when everyone else was talking about their work, but she had taken those she lived with into her confidence and told them that she worked for one of the intelligence agencies, so they protected her and didn't question it when they heard her telling casual acquaintances that she worked for a PR agency.
The secondment to Merseyside police came as a considerable jolt. Liz knew that at some stage, as part of the training programme, she would be sent off on attachment to learn how a provincial police force and its Special Branch worked, but she wasn't expecting it so soon. And Liverpool was alien territory to her – she had never been further north than Nottingham.
It was the period before the Peace Process had taken hold in Northern Ireland and she was one of a team collating intelligence on the threat from the Provisional IRA. Liverpool had an established community of Irish expats, many with nationalist sympathies and a few with actual links with the Provos. The Special Branch had some sources that from time to time provided useful intelligence, so she'd already had some dealings with Merseyside Special Branch officers and she had not much liked them. As she'd travelled up on the train to Liverpool that gloomy, showery day she was feeling nervous.
As it turned out she had good reason to be, but not because of the IRA. In the police headquarters' rectangular red office block near the docks, a gloomy middle-aged sergeant with a pencil behind his ear had sent her upstairs with a grunt and a jerk of his thumb. One floor up she found a large open-plan room with a dozen or so desks in untidy rows, about half of them occupied by men, some young, some middle-aged, some in shirtsleeves, some in leather jackets, some typing, some talking on the phone. Cigarette smoke hung in the air in a blue uncirculating haze.
Every man looked up as Liz came into the room. She asked where Detective Inspector Avery could be found, and one of them pointed to the back of the room where a small office had been partitioned off with opaque glass. As Liz walked through the rows of desks, someone gave a low wolf whistle. Liz tried not to react, but she felt herself blush.
She knocked on the door, and a gruff voice said, Come in. Opening the door, she found a wide-shouldered man in shirtsleeves, with a tie pulled down an inch or two from his collar. He looked close to retirement age, and had greying hair cut very short, though he had let his sideboards grow in some misguided youthful impulse.
Avery looked annoyed by her interruption. 'What can I do for you, miss?'
'DI Avery?' The man nodded. 'I'm here from Box 500,' said Liz, using the acronym by which the police referred to MI5. 'My name's Liz Carlyle.'
He stared at her. 'You're Carlyle?' He sounded astonished. 'I was expecting a George Carlyle, or a John Carlyle, or even a Seamus Carlyle. But nobody said anything about a Liz Carlyle.' He was looking at her with distaste; Liz didn't know what to say. Avery suddenly added, 'I suppose you're a graduate.'
'Yes.' Never had she felt less proud of it.
'Good. You'll be used to reading then.' He pointed to three stacks of papers on a side table. 'You can start with them. I've got more important things to do than read bumf from the Home Office all day. Come back in the morning and you can tell me what's in it.'
After this welcome, Liz reckoned things would have to get better. She was wrong. By her third day she had acquired a nickname – Mata Hari – but not much else in the form of contact with her new workmates, whose initial curiosity was swiftly followed by the hazing rituals of an American college fraternity. The first morning when Liz went to the desk she had been allocated, she found a large cigar lying on the desk top. An hour later when she came back with a cup of muddy coffee from the vending machine in the hall, she found that someone had moved the cigar suggestively to the seat of her chair. While the men around her watched surreptitiously she broke the cigar in half and threw it in the wastepaper basket.
The next morning another cigar was in place. Again Liz threw it away, and this time she said loudly, without looking around, 'I hope you boys can put cigars on expenses. If this goes on, it's going to cost you a fortune.'
All week she ate lunch alone and saw no one after work. The only other woman in the office, the typist for DI Avery, was a middle-aged woman called Nellie who came in at exactly nine in the morning and left at precisely five at night. She had clearly never read Germaine Greer or heard of sister-solidarity; she made a point of ignoring Liz.
Not all the men joined in the harassment. Some just ignored her and one in particular was quite polite – McManus, a tall, sharp-featured detective sergeant who dressed better than the others.
The work itself was dull, a relentless progress through mind-bogglingly dense papers from the Home Office. Liz was desperate to get her teeth into something real; otherwise she would finish her secondment without knowing any more about how a police force ran than she had when she came. She resented Avery's using her as an intellectual dogsbody, covering his back in case some civil servant expected a response to one of the documents sent seemingly by the truckload from Whitehall and Scotland Yard.
The harassment persisted, though not any longer with cigars. Purvis, a tall man with a dimple in his chin, seemed particularly intent on making Liz feel unwelcome. 'Ask our new graduate colleague,' he would say when someone had a question at the weekly briefing meeting.
Liz ignored this as best she could, but it made for stressful working hours, and she wasn't sure how long she could put up with it in silence. Part of her was determined not to let these bastards get to her; another part wanted to run back to London. Then one morning she arrived to find a bundle of dirty shirts on her desk, with a note pinned to them. Washed, ironed and folded by Thursday please. She felt the eyes of the room upon her as she stood by her desk. Suddenly furious, she picked up the shirts, walked over to the open window and dumped them out into the alley below.
And then things changed – whether because she'd shown she'd had enough or because some of the men had begun to feel embarrassed, she never knew. As her third week in Liverpool was drawing to a close, she was sitting looking at what seemed an undiminished stack of typescript pages when McManus stopped beside her desk. 'That looks interesting,' he said, pointing at a Home Office circular on top of the pile.
She looked up at him warily. 'It's absolutely entrancing,' she said dryly. 'I'd be happy to tell you all about it.'
'No, thanks.' McManus paused for a moment, and she watched him as he seemed to be making up his mind about what to say next. He was a good-looking man – and he seemed to know it. Not my type, Liz told herself; her last boyfriend had been a gentle guitar player at Bristol. Besides, McManus must have been fifteen years older than she was. There was no way she was interested in him.
'Fancy joining us on a little mission?' he said lightly. 'Or are you chained to your desk?'
'I'm just following orders,' she said, nodding towards Avery's office.
'Boss is in Manchester today, so why don't you come along?'
She hesitated, but anything was better than reading any more bumf. 'OK. What is it?'
'I'll explain in the car.'
Outside they joined two detective constables, Cardew and Purvis, who looked surprised when McManus explained Liz would be joining them. He added, 'You boys go ahead. We'll see you there.'
Cardew, who Liz suspected had been the wolf whistler on her first appearance in the office, rolled his eyes. McManus gave him a look and he and Purvis stomped off to their car.
McManus drove her in his black Range Rover away from the Docks and towards the eastern suburbs of the city. It was unseasonably warm and Liz put her window down as the evening turned from dusk to dark. The smoky yellow of sodium lights lined the streets in glowing dots. They climbed a bit and were going past a series of large institutional-looking buildings, a few modern but mostly Victorian. 'Where are we?' asked Liz.
'The university,' said McManus. After a pause, he added, 'I was there.'
'Really? What did you read?'
'Business studies. It seemed the practical thing to do. I'm a local lad – my dad was first-generation Paddy, and worked on the docks till they closed. I didn't know what I wanted to do; I just wanted to get out from his way of life.'
'Why did you join the police then?'
'Because I was bored by business.' He turned his head and gave a wry smile. 'If I'd stuck with it I'd have gone mad before I was thirty.'
Liz laughed, and McManus said, 'Where'd you grow up?'
She explained, and he said, 'Sounds very posh. Your dad a grandee?'
'No, he just worked for one.' This time McManus laughed. They were in the suburbs now, tree-lined streets with large detached houses. 'This is what I was aiming for,' he said, gesturing around them.
'Aren't you still?'
He shook his head. Liz said, 'What's changed?'
'It's called maintenance,' he said with a trace of bitterness. But then his tone changed and he was all business. 'I'm meeting an informant. He's just over from Belfast; the RUC's passed him on to us.'
'You sound doubtful.'
McManus nodded. 'I am. He's a tricky little sod.'
'How so?'
'His RUC handler said they were never sure how reliable his information was. They had doubts about his real allegiances – nothing they could put their finger on. He seemed to provide just enough to keep them interested but not enough to be really useful. They were pretty sure he could have given more if he'd wanted to. They've sent him over here to see what we make of him. He's supposed to be getting alongside the Provo sympathisers here.'
He pulled the car over and parked at the top of a rise. Down the street a little below them was a small precinct of shops. At the corner there was one still open; it had a retro neon sign Liz could just make out – it said Café Noir. McManus pointed at it. 'That wine bar's where we're going to meet. I'm going to stand by the door smoking and chummy will come past me and go in. I'll have a look round to check he's clean then join him – there's a little room in the back where we can talk. Purvis and Cardew are parked further along that street, watching my back from there. You can watch it as well from up here. I'm not expecting any bother. I'll leave the keys in case you need to move the car, but whatever you do, don't drive down the hill.'
'I thought—' Liz began to say, but McManus had already opened the door and was halfway out of the car.
He said, 'Won't be long.' And he slammed the door and began striding quickly down the hill towards the wine bar.
Liz sat there, fuming. The paper work in the office was bad enough, but having the promise of something real to do, only to have it snatched away, was worse. Why had McManus brought her here if it was only to leave her in the car while he met this informant? He already had two detectives watching his back – though it seemed a bit unprofessional to have them both in the same place – so he didn't need her as well. And what if she did see anything? She had no way of contacting him to warn him. Maybe he'd brought her along to find out more about her so that he could pass it on to the others. Yet he didn't seem that kind of man. So what exactly was he doing?
McManus had almost reached the wine bar now, and he stopped and casually lit a cigarette. He lounged by the entrance, studying the menu in the window. There was no sign of his contact, or anyone else – the street was deserted.
Then she heard footsteps on the pavement behind the car. Two sets. She grabbed her bag and rummaged through it, keeping her head down in case she was noticed sitting alone in a parked car and someone got the wrong idea. The footsteps had reached the car now, but thankfully they didn't slow down, just went on past. Slowly she lifted her head and saw two men wearing short leather jackets, jeans and trainers. They looked young and fit. She wondered if they were plainclothes policemen – but these two weren't Purvis and Cardew, and McManus hadn't mentioned any other backup.
A car came up towards her from down the hill, and as its headlights swept across the pavement she saw the two men suddenly stop and tuck themselves into some bushes growing at the front of someone's garden. The car went past and the two men continued down the hill. They didn't want to be seen; Liz wondered why. Unless they weren't police at all.
The two men stopped again and exchanged a few words. They were still only about forty yards ahead of Liz, and she watched as one of them crossed the street. The other one waited for a moment; he was out of the direct light of the street lamp but she could see him clearly enough. He had his hand behind his back and as he brought it round something glinted momentarily, and she caught on: it was a handgun. He held it for a moment then tucked it away under his jacket.
She hesitated. Was it a gun? Could they be plain-clothes police? If it was and they weren't, there was no time to waste. The two men were now halfway between her and McManus, still outside the wine bar. They would reach him in a couple of minutes.
Liz slid over behind the wheel and turned the keys in the ignition. The engine responded right away. She turned on the side lights and pulled out into the street, coasting down the hill. As she passed the two men, one on each side of the street, she tensed, half expecting them to fire at her. They were striding quickly now and the one on the right-hand pavement had pulled his gun and was carrying it in his hand openly.
As she passed them she suddenly switched on the headlights full beam, blinding a van coming up the other way, and as she picked up speed she began hitting the horn so it sounded loudly in short warning beeps.
When she reached the wine bar she braked hard, coming to a sudden halt just in front of the entrance. She reckoned she was seventy or eighty yards in front of the men. McManus was looking startled. She pushed the button and the window on the passenger side came down. She yelled, 'Get in quick.'
'What the hell—?' he said.
'There's a couple of gunmen just behind me. For God's sake, get in.'
McManus looked over the top of the car back up the hill. The two men had stopped; they must have been uncertain what was happening. By now Purvis and Cardew had seen the commotion and came roaring up from the other direction.
'What's going on, Guv?'
McManus was shouting into the car radio, calling out an armed team. He broke off to yell at the two men, 'Up the hill. Two of them. Get up there now. See if you can follow them but hang back – one of them's got a gun, possibly both. Armed response is on the way. Keep in touch.'
'Park up there,' he said to Liz, pointing to a space in front of the line of shops.
'Shouldn't we help go after them?'
He shook his head. 'No. A gun fight's no place for you. Anyway, odds are they'll be gone. I need to wait here in case chummy shows. Though that seems a bit unlikely now.'
'You think he set you up?'
He nodded. 'Must have. Unless he's blown and they were after him. If he doesn't turn up we'll know which it is. Either way, he's not going to be any more use. I'd give you odds he's safely back in Ireland by now, thinking he's helped assassinate a Special Branch officer.'
They sat in silence then, too shaken to talk, McManus keeping his eyes on the street ahead, while Liz kept a lookout behind through the rear-view mirror. It must have been ten minutes before a car pulled up in front, and Cardew and Purvis got out.
Cardew came over and spoke through the open window of the passenger seat. 'No trace. The boys are out combing the streets but it looks as if they've got clean away. We don't know the car and we've got no description so there's not much chance. Jesus, Guv, we were wondering what the hell Mata Hari was doing, driving down like that.'
McManus stared at him. 'She was saving my life, Officer, while you were sitting picking your toes. And don't call her Mata Hari. Her name is Liz.'
Chapter 6
Word spread quickly in the Special Branch office that Liz had saved McManus's skin and the atmosphere got a lot more friendly; even Nellie the typist began to talk to her. When Avery stopped offloading Whitehall's paperwork onto her and started asking her to analyse the intelligence reports coming in from Belfast to see if they threw up any leads to local activity, she felt that at last she'd been accepted as someone who might have something useful to contribute.
That wasn't all that changed. Looking back on it now, she supposed it had been inevitable that after their run-in with the IRA she and McManus would be drawn together. Their shared danger formed a bond which at first made them friends, and then, as if it were the most natural thing in the world, something more than friends.
It didn't happen right away. McManus was cautious about getting involved with the Spook, the woman from MI5, and at first he was just cordial. Three days after the drama of what they now all accepted had been an assassination attempt, he casually asked her to join him for a drink – but when she walked into the pub she saw that Purvis and Cardew were sitting at the table with him. A week later he asked her again, this time on his own, but before he'd even got her a drink, he was called on his mobile and had to go – an informant had been arrested for benefits fraud and he had to sort things out.
A few days later she had left her car at a local garage for its MOT on her way to work and to her annoyance the garage had rung late in the afternoon to say the car wouldn't be ready until the next day. She was waiting for a bus down the street from the office when a man's voice called out, 'Want a lift?'
She turned, ready to tell the man to buzz off, when she saw it was McManus at the wheel of a smart Audi. He lifted his hands in mock-surrender. 'Don't shoot. It's only me.'
She laughed. 'What are you doing here?'
'Looking for damsels in distress. Hop in.'
'What happened to the Range Rover?' she asked as she got in.
'Strictly for operations,' he said, as he accelerated away. 'This one's mine. Now where are you going?'
When she told him he gave a little groan. 'That's a very respectable address.'
'Well, of course,' she replied with a grin. 'I'm a very respectable person. The lady who owns the house is the widow of some former contact of the Service. I don't know the details. I've got a couple of rooms on the top floor.'
'I bet she watches you like a hawk. That can't help your social life.'
Liz suppressed the temptation to ask, What social life?
'Tell you what,' said McManus, glancing sideways at Liz, 'why don't you come back to my place for a drink? Then I'll run you home,' he added quickly, as if he didn't want to scare her off.
He accelerated past a dawdling queue of cars, his eyes straight ahead. Liz considered what to say. She sensed her answer was going to make a big difference to her relationship with McManus, and she wasn't sure it was a step she wanted to take. But then she thought of what otherwise awaited her that evening in her flat – a quick glass of mediocre wine, a shallow bath (the hot water tank was minute), followed by a solitary microwaved supper, a little television, a couple of chapters of the disappointing thriller she was reading, and lights out. Not a very exciting prospect.
So she said, 'OK. Thanks.'
Looking back, she supposed the whole affair wasn't surprising. McManus was an attractive figure to a young woman. Good-looking, confident, mature – he could see Liz was pretty inexperienced and hadn't been around much and he enjoyed showing her the town. He knew Liverpool like the back of his hand: from the industrial wastelands to the newly fashionable dockland; from the gentility of its grandest suburbs to clubs so rough that even the bouncers were scared of the clientele; from fancy French restaurants where the city's famous footballers spent £1,500 on a bottle of wine they couldn't pronounce to the bingo hall where he said his mother had been a habituée. Wherever they went the proprietor knew the Special Branch detective, and treated him with respect.
Liz was less certain what McManus saw in her. She sometimes wondered if in other circumstances he would have given her a second look. Observing the admiring glances he attracted from women of all sorts, from restaurant cloakroom girls to the chic owner of an upmarket boutique, she knew that he could have had his choice of women. But circumstances were what they were, and the simple fact remained that she had probably saved his life. If his interest in her arose out of gratitude, Liz couldn't really object, since she was also grateful to him.
It was an intense affair, and for all the excitement of their social life, what really kept the two together was a mutual commitment to their work. Liz had already discovered a capacity for immersion in the job, and now that Avery had given her something substantial to do, she was interested and intent on doing it well. But she was nothing like McManus. As she quickly discovered, life for him was filtered through work. In the pubs and restaurants they visited, his conversations with the manager were information-gathering exercises. Even when they were most relaxed – a walk on the beach, a quiet meal in a country pub where no villain had ever set foot – McManus was alert, noticing anything out of the ordinary, any behaviour in the least bit strange. This was the first time Liz had experienced something that she later encountered often in her colleagues and indeed learned to practise herself, the acute awareness of one's surroundings of the true intelligence officer.
But she soon discovered that McManus's almost forensic attentiveness was focused not so much on intelligence gathering as on a righteous passion to sniff out wrongdoing and see it punished. He was a zero-tolerance police officer, openly disdainful of the way so many of the criminals he had hunted down wriggled free in their progress from arrest to the jury's verdict. The only time Liz saw McManus lose his temper was when the Crown Prosecution Service refused to prosecute the leader of a drug ring, a man called Pears whom McManus had pursued for years, because in their view there was insufficient evidence to secure a conviction.
If Liz sometimes found McManus's crusading spirit unsettling, she also admired it. Where some of his colleagues appeared quite happy to accept the odd freebie – drinks in a pub, a taxi ride home, free admission to a club – McManus wasn't: when one evening the owner of a local restaurant brought them two brandies at the end of their meal and said they were 'on the house', McManus insisted they be added to the bill. But with Liz he was relaxed; she found him caring, loving and warm. To her surprise he seemed happy to be open about their relationship, and made no effort to disguise it from their colleagues. She was startled but flattered when quite early on he asked her to think about moving into his flat, and though she didn't take that step she did find herself wondering how she could get her secondment to Liverpool extended.
They had been together for two months when things went suddenly wrong. They were in McManus's flat, an elegant one-bedroom pad high enough up in a new block to give a spectacular view over the Mersey. McManus was in a jubilant mood, and over a glass of wine he explained that Pears, the drug dealer, had been arrested again and this time the Crown Prosecution Service were going to prosecute.
'What changed?' asked Liz.
'New evidence,' said McManus.
'Really, what sort of evidence?' She was curious to know, since the CPS had previously complained that the available evidence was too circumstantial.
'A witness has come forward. He's prepared to say he saw Pears make a big sale.'
'That's excellent,' said Liz. 'Why did he come forward now? It must be a bit risky for him. Are you going to have to protect him?'
McManus shrugged. 'Maybe it was my appeal to his better nature – not that this particular little runt has one.' He paused and looked at Liz with a grin. 'Maybe it had something to do with letting him off another charge if he came good in this case.'
'A deal, in other words,' said Liz, starting to understand.
'If you want to call it that.'
'What else should I call it? The little runt, as you call him, has decided he's seen something because that way he gets off.'
'It may be a rough kind of justice, but believe me it's still justice. He would have seen Pears do other deals plenty of times.'
'But not this one?'
Again McManus shrugged, this time in acknowledgement. His jubilation was gone. He said defensively, 'What the hell. I didn't say it was ideal. But this way we'll get a result.'
Liz said, 'It's wrong. You know that.'
He looked at her and shook his head. 'Forget about it. More wine?'
'No, thanks. You haven't answered my question.'
'I didn't hear any question.' He'd got up and was pouring himself a glass of Chianti.
Liz said, 'You know what I mean. I know what you've done, and it's wrong.'
'Says who?' His voice was sharp now. 'Says Liz Carlyle, twenty-something trainee spook from London. The same Liz Carlyle who's never walked a beat, never made an arrest, never looked down the barrel of a gun held by some scumbag who'd as soon pull the trigger as sneeze. A Liz Carlyle who might be just a little out of her depth here.'
He had never spoken like this to her before. She said as calmly as she could, 'It's not right, Jimmy. Not because little Liz Carlyle says so. It's not right because it just isn't. You can't go round making up evidence just because you're convinced someone is guilty. You can't be judge and jury; that's not your job.'
'Nice speech, Liz, but if we can't rely on the legal system, what else can we do? If I have to bend the rules to get this bastard, I will. It's the results that matter. Getting Pears off the streets and locked up where he belongs.'
'It's not some minor rules you're bending, it's the law. Here you are saying Pears can't stand above the law, but then where are you standing?'
McManus made a show of looking at his watch. 'Time's up,' he announced. 'Our booking's ten minutes from now. You better get your coat.'
The flippancy in this dismissal enraged Liz. 'I'll get my coat,' she snapped. 'And see myself out.'
They didn't speak for three days, each locked into their conviction that they were right. Finally Liz decided it was ridiculous to behave this way – she was never going to agree with what he'd done, and her whole view of the man had changed. But even if they weren't going to be lovers any more, it seemed ridiculous not to be on speaking terms, so towards the end of the day, when McManus came into the office and sat down at his desk, she went over.
'Fancy a drink?' she said lightly. Purvis at the desk next to them was pretending not to listen.
'Got a lot on,' McManus said tersely, without lifting his head from the papers he was reading.
'OK,' said Liz. The rebuff couldn't have been clearer.
She gave it a week, then tried again, and received the same short shrift. After that, they ignored each other, which made for a certain tension in the office, though nothing like it had been when she first arrived. She went back to spending the evenings boringly alone, now looking forward to the end of her Liverpool posting. She missed McManus – or she missed the man she had thought he was, though it gave her a sliver of comfort to know that that man did not exist.
When McManus left Liverpool on promotion to Greater Manchester, she barely noticed, so accustomed by then was she to not having him in her life. She was not invited to his leaving do, and he did not even bother to say goodbye. So she could only imagine his reaction when the drug dealer Pears was convicted and given eight years.
Then one morning she heard Purvis complaining that he'd paid more than he could afford for a second-hand Audi he'd bought from McManus when he'd left for Manchester. Liz's car was once again in the garage and suddenly she found herself offering to buy the Audi off Purvis for the same price he'd paid McManus. Purvis accepted with alacrity. Since she was never going to see or hear from McManus again, Liz reckoned this would be the legacy of their affair.
Chapter 7
The sky was black over the mountains as Miles drove his SUV along the sandy road into the countryside. The Trade Minister, Baakrime, had said that he would have something to tell Miles in a week, and the previous day an invitation had arrived at the US Embassy inviting him to lunch at the Minister's farm in the hills outside Sana'a.
Miles's colleagues in Langley were waiting impatiently for the payback on the cash that Baakrime had been given, information they were sure the Minister was holding about the sources of arms that were getting into the hands of jihadis through Yemen.
But Miles was uncomfortable, nervous about this journey away from comparative safety in the bustle of the town. Minister Baakrime had fallen for his recruitment pitch suspiciously quickly, taking the envelope of cash and promising information. But had he really agreed, or was this invitation a trap either to kill Miles or expose him as a foreign spy? He had consulted Langley overnight but they were keen to get the information and prepared to take a risk, so he was instructed to go to the meeting – and wear a tracking device that would be monitored by a drone far overhead. It wouldn't help if he was killed but might if he was kidnapped – small consolation.
Miles glanced uneasily at the darkening sky. The climate in Yemen, normally so hot and dry, could produce sudden short but heavy cloudbursts, and it looked as if that was exactly what was on the way.
Seconds later it arrived. Rain beat thunderously on the top of the car; the wipers sweeping at top speed from side to side of the windscreen had no effect and the glass ran with a stream of water thick with sand raised by the sudden wind and the force of the rain. Miles could see nothing. The fields of arable crops and fruit orchards that bordered the road disappeared from sight and he stopped the car where he was, in the middle of the single-track road, hoping no vehicle was coming the other way. If there was, he wouldn't see it and it wouldn't see him until it was upon him.
He sat sweating with heat and tension until suddenly the rain stopped, the wipers cleared away the sand and he could see the road again. It was more of a small river now and his wheels threw up a fountain of water on each side of the car as he drove slowly on. As the sun came out again, he saw in the distance the red walls of what he took to be his objective, the Minister's farm.
The carved wooden gates of the compound were open as Miles drove up. A young man in a white robe and a Western-style sports jacket saluted and waved him in through the gates, then walked across to open the car door as Miles parked against the wall beside a wet and muddy silver Mercedes with a two-digit licence plate – 12.
'Salaam aleikum. Come this way, sir.'
Miles followed the young man into a lofty hall. Sunlight glanced though small windows set high up in the walls, but below the room was in shade and at first Miles, coming in from the bright sunlight, could see little. As his eyes got used to the dim light he saw the rough stone walls, the red-tiled floor covered with rugs in subdued colours, and around the room ottomans and chairs covered with cushions and throws of bright silks. This was a very luxurious farmhouse.
'Sit down, sir. The Minister will be here shortly,' said the young man in unaccented English. He clicked his fingers and a servant appeared with a tray of glasses of fruit juice.
Miles sat on the edge of an ottoman, sipping a glass of pomegranate juice. His sense of unease grew as he waited, wondering what would happen next.
'My friend.' A loud voice echoed across the hall as Baakrime in a long white robe strode towards Miles, his hand held out. 'It is delightful to see you here. I must apologise for our weather. These rain storms blow up at this time of the year, but they are soon over. Unlike your hurricanes, they do little damage.' He pumped Miles's hand enthusiastically, setting up a sharp twingeing pain in his shoulder.
'I thought it best to meet here. It is safe and away from prying eyes. Everyone here is family or old servants of my family. The road you came along is watched by my people and the young man who met you is one of my sons. He is my secretary. He was at school in England and at Cambridge University. Do you know England?'
'Yes,' said Miles. 'I have worked in London.'
'London. I love that city.' The Minister rubbed his hands together. 'We go there every year. My wife enjoys the shopping. Oxford Street, Harrods. I come back a poor man.' He smiled and Miles smiled back. Baakrime's poverty was not to be taken seriously.
'But let us eat while we talk,' and the clapping of his hands produced two servants with trays of little dishes and jugs of more fruit juice. One tray was placed on a brass-topped table beside Miles and the other beside Baakrime. Glasses were filled and the servants withdrew.
'I am honoured to see you in my house,' Baakrime began. 'Sana'a is not a safe place for me to meet you. This is a dangerous time in my country. There is much discontent; the people are unsettled, the country is fragile. There are elements in Sana'a who want to overthrow the government and would like to be able to show we were pawns of the United States. In our desert regions, the jihadi groups have established strongholds. They are in league with groups in other countries and they want to kill us all.'
Miles nodded. 'Yes. These are disturbed times in the whole of this region.'
'You asked about weapons,' the Minister went on. 'They are everywhere. Where are they coming from? Iran, Pakistan – all the places you would expect.' He paused to eat some things from the dishes.
'What do you know of arms from the United States or Europe?' Miles was anxious to come to the point and get away. In spite of the Minister's assurances of security, he felt exposed in this place with no backup.
'My friend, with the arms trade it is always difficult to know the origin. These people are masters at deceit – false invoices, changing documentation while a cargo is en route, and of course there are corrupt officials in every port and so much money to be made. The people who run this trade are very rich indeed – unlike in my country, where the poor are everywhere.'
He sipped his juice and looked at Miles over the rim of his glass.
There was silence. Miles ate some food. He felt sure there was more information to come, but it seemed to need some assistance. 'Yes. The poor. I hope our contribution helped a little.'
'Yes indeed, my friend. We are so grateful. But there is so much need.'
Miles felt in his pocket and produced an envelope. 'Let's hope this will satisfy some of it,' he said, placing the envelope carefully on the tray of dishes beside him.
Baakrime began to talk again as though there had been no interruption. 'Yes. The people who run this trade are very clever at disguising themselves. But I have heard that there is a main middleman for deals from Europe. They call him Calibre. His real name is never used. I hear that he is meeting the leader of a group of jihadis or rebels – I don't know what group or exactly who they are, though I understand they are being funded by al-Qaeda. The meeting is in Paris in the next week or so. It is to arrange a shipment. The delivery will come through Yemeni ports, I hear.'
Miles nodded and waited. His face was calm but he was excited. At last he had something for his money, though it was pretty vague and probably not anything that could be acted on.
But Baakrime had not finished. 'I will try to find out more about this meeting and if I do my secretary will get a message to you.' He stopped for a sip of juice. 'There is one more thing. It is generally thought that the arms that come via this route are for use in Arab countries, and that may be so, but I have heard that the man behind those deals, this Calibre, is using someone from England to help with this latest deal. The arms trade is a very tight-knit network, almost like a club, but it seems someone British is applying for membership.'
Chapter 8
It was eight o'clock in the evening and Liz was tidying up the kitchen after her supper. Unusually for her she'd been cooking. Martin was convinced that only French women knew how to cook and she had promised herself that next time he came to London for the weekend she was going to surprise him by producing the perfect soufflé. So she had been practising on herself and this evening she reckoned she'd cracked it. She had just eaten what she considered to be a masterly example – cheese and spinach soufflé à la mode de Carlyle. She was just wondering what to do with the half that remained, asking herself if it would be OK if she heated it up again for tomorrow night, when the phone rang. It was the Duty Officer.
'Evening, Liz. The Six Duty Officer has just rung with a message for you from Bruno Mackay,' he said. 'Would you join him and Geoffrey Fane at Grosvenor tomorrow morning at half past eight for a meeting with Mr Bokus? Apparently something urgent has just come in from Langley. He said you should bring an overnight bag.'
'Oh thanks,' said Liz. 'And did he say what I should put in it? Jeans and a T-shirt, a fur coat or a long black garment suitable for interviewing Arab sheiks?'
''Fraid that's all the message said.'
'OK. Thanks. I suppose I'll just have to use my initiative.'
'Good night then,' said the Duty Officer cheerily, and rang off.
At quarter past eight the following morning she was walking across Grosvenor Square towards the American Embassy, carrying an overnight bag, when she spotted Geoffrey Fane and Bruno Mackay getting out of a taxi. It was uncanny how similar they looked. Fane, his tall, slim, pinstriped figure, nowadays with a slight stoop that made him look even more heron-like than when he was younger. Bruno, equally tall and slim, equally elegantly clad, though his suit was finely checked rather than pinstriped and the colour lighter than Fane's navy blue. Bruno's shock of fair hair and deeply tanned face contrasted with Fane's pale skin and black hair, but they might have been, if not father and son, at least related. They certainly came out of the same mould.
'Good morning, Elizabeth,' said Fane as they all reached the steps up to the Embassy front door together. 'Glad to see you've come prepared,' he added, glancing at her bag.
'Good morning,' she replied, her heart sinking as she noticed that Bruno was carrying a black leather valise. It looked as though wherever she was going, he was going too.
In Andy Bokus's office in the CIA suite of rooms behind the locked and alarmed steel door in the Embassy, a plate of oversized bagels and cream cheese was set out on the table. 'Help yourselves to breakfast,' said Andy, waving his hand at the plate. 'Coffee's over there.'
Fane shuddered slightly at the sight of the bagels, and from the corner of her eye Liz caught Bokus's grin. Liz enjoyed watching Bokus and Fane playing a game with each other. It was a game that neither acknowledged but she suspected both understood. In Fane's presence Bokus played up his roots as a son of humble immigrants – his grandfather had been a coalminer in the Ukraine and his father had landed on Ellis Island at the age of sixteen with nothing but the clothes he stood up in. Bokus senior had ended up running a gas station in Ohio and making enough to put Andy through college. Andy was bright, or he wouldn't be where he now was, heading the CIA station in London. But he didn't like London and he didn't like most of the Britishers he met. And in particular he didn't like Fane, who struck him as snobbish, self-satisfied and devious. So to Fane, Bokus presented himself as rather stupid and very uncouth, hence the enormous bagels. Fane responded by shooting his cuffs and adopting an exaggeratedly public school drawl and a patronising manner.
How much of all this psychological drama Bruno was following Liz didn't know. He was contentedly munching a bagel, seemingly oblivious. But she knew that you could never tell with Bruno.
'Well, I've got things you folks need to know,' said Bokus. 'We'll go down to the Bubble.' The Bubble was the secure room in the bowels of the basement, purpose-built to foil any attempt at eavesdropping. It always struck Liz as strange and illogical that, as the main threat of eavesdropping in London must come from the British intelligence services, the Agency conducted its most sensitive conversations with the British in their most secure room.
The door of the windowless room closed with a pneumatic hiss behind them and they sat down on padded benches around a central table. The faint hum of the high-frequency-wave baffler had a rather soporific effect on Liz and she hoped that the hastily convened meeting was going to produce something worthwhile.
'Geoffrey, you and Bruno know something of what I'm going to say, but I'll just recap for Liz here. We recently sent an officer to Sana'a. He had one objective, to make a quick pitch to a highly placed official who'd been making it pretty obvious to the Commercial Counsellor – a State Department man – both that he was in the arms business and that he could be bought. So we sent young Miles Brookhaven. You all know him from his time here.' He grinned at Liz; she pretended not to notice. 'He made a quick pitch and it came good. The guy is now signed up. He's going to give us stuff on arms supplies going through Yemen, to rebels and jihadis. As you know we're particularly looking for anything coming out of Europe and the States.'
Fane shifted in his seat, unwrapping his long legs and crossing an ankle over a knee. He clearly found the narrow benches uncomfortable but, more importantly, he couldn't bear to let Andy Bokus talk for more than a few minutes without interrupting. 'I mentioned to Elizabeth that you thought young Brookhaven was making progress,' he said.
'Yes. He's done quite well.' He looked at Liz, 'You heard he had a rough time in his last posting? Quite badly injured.'
'Yes. I heard.' Liz was wondering when he was going to get on to whatever had brought them here.
'So,' said Bokus. 'What he's got from this new source – we're calling him Donation – is that there is a European arms dealer who is arranging supplies from somewhere in Eastern Europe, the old Soviet Union probably.' He paused for effect. No one spoke; they all knew there was more to come. 'We don't know what nationality the arms dealer is. They call him Calibre. But Donation says that he's using someone to help him ship the arms – a transport expert, I guess. And this expert is a Brit.'
'Are you sure this Donation isn't just telling Miles what he thinks he wants to hear?' said Liz after a moment. 'It all sounds a bit too pat.'
'Wait till you hear the rest,' said Bokus. 'He says that there's a meeting arranged between Calibre and a jihadi leader, tomorrow in Paris.'
'Big city, Paris,' said Bruno dreamily.
'In the Luxembourg Gardens,' Bokus went on. 'At twelve noon.'
'So that's where we're going,' said Liz, turning to Bruno.
'That's where you're going,' he replied with a grin. 'I'm going to Sana'a.'
Chapter 9
Jean Perlue was excited. He had been in the DCRI, the French equivalent of MI5, for just eighteen months and this was his first real surveillance operation. At the briefing the team had been told that an international arms dealer was going to meet a dangerous jihadi in the Luxembourg Gardens, and that it was vital that both of them were photographed and followed.
So at the age of only twenty-four Jean was engaged in counter-terrorist work of international importance. His instructions were to hang around, just inside the park gates on the Boulevard St Michel, looking natural and merging into the surroundings until he heard the controller, speaking through his earpiece, telling him to move.
But it was a frosty morning and his feet were very cold and he had been in place for more than an hour. He was running out of things to do to look natural. He'd bought a crêpe from the mobile cart parked outside the gates and eaten it slowly; he'd read a copy of Le Monde standing up until he knew the front page virtually by heart. Now he was stamping his feet and looking angrily at his watch, for the fourth time, as if he was waiting for a girlfriend who was late. Just as he was wondering what to do next, conscious that the crêpe seller was staring at him, he heard a voice in his ear.
'We have a possible for Numéro Un. Male, Caucasian, about forty-five to fifty, one hundred eighty centimetres, grey/brown hair, brown leather knee-length jacket.' The voice was Gustave Dolet's. Jean Perlue knew he was sitting with Michel Vallon in a Renault car parked in the Rue Gay-Lussac with a view of one of the other gates of the Gardens. 'He's heading into the Gardens now. Michel is going with him.'
Jean felt his throat constrict as his excitement rose. His cold feet were forgotten, as was the cover of waiting for a girlfriend. He peered eagerly in the direction of the action, hoping desperately that the target would come his way. This was better than a training exercise.
The thought had hardly entered his head when he heard, 'We have an Arab at the gates. He's nervous, looking around. I think he may be Numéro Deux. Thin, about twenty-five to thirty, white trainers, jeans, navy-blue sweater. He's in the Gardens now, heading west.'
'I have him,' came the hoarse voice of Rabinac. Rabinac had been one of his teachers on the course and had been known by all the students as Mr Croaky. 'They're walking towards each other and slowing down. I'll have to pass them.'
At that moment Jean Perlue, from his post by the Boulevard St Michel gates, saw them. They were walking together now, approaching a park bench. He saw Rabinac walk past as they sat down. Rabinac came on towards Jean and, without giving any sign of recognition, went through the gates, passed the crêpe seller and walked off along the Boulevard.
'I have eyeball,' said Perlue into his microphone, his voice rising in excitement. 'They're talking. The Arab has a piece of paper and the European is shaking his head. They're too far away for my camera to get a clear picture. I'll go closer.'
'Stay where you are,' came the urgent voice of the controller. 'Michel has photographed the targets.'
'I might be able to see the paper he's holding if I walk slowly.'
'On no account approach them,' ordered the controller, but Perlue was already on the move, heading diagonally across the grass towards the bench where the two men were sitting.
His headphones were silent now, but even if anyone had spoken he wouldn't have heard. He was sure he could get a perfect photograph of the men and the paper that was still in the Arab's hand. As he approached the bench, he was making a show of consulting his wristwatch, muttering, tapping it and shaking his wrist. When he reached the bench he smiled at the two men and leaning over them asked them if they had the right time. The European replied and Perlue thanked him and continued on down the path.
When he was a little distance away from the bench he heard the voice of Rabinac in his ear. 'You've blown it Perlue, you idiot. They're leaving.'
Suddenly the Luxembourg Gardens were full of movement and the headphones were busy as the surveillance team tried to get into position to follow the two men as they went off quickly in different directions. Now that it was pretty certain that Perlue had blown the surveillance operation, the controller was less concerned with the team being seen, more with keeping in contact with the targets.
Rabinac got up from the bench where he had been sitting, stretching and yawning as if still dozy from a good nap. The Arab was a good fifty metres past him now, heading down the avenue of trees towards the palace. Marcel Laperrière would be waiting there, ready to chuck away his newspaper and walk as a front tail, while Rabinac followed from behind.
But what about the other man? Perlue knew that if he had stayed at his post the European would be passing right in front of him now, but instead he was far away on the wrong side of the Gardens. He felt mortified at what he had done; he knew that in all likelihood he would be back on the training course the next day. That is if he wasn't sacked. He prayed that Gustave and Michel had had time to move their car closer to the entrance gate.
'Get back to your position, Perlue,' came the instruction from Control, and he walked quickly towards the Boulevard Saint Michel, seeing as he approached the gates that a large crowd had gathered on the pavement just outside the gardens. There must have been close to 200 people there. What was this about? Surely nothing he'd done had caused this. Then he saw that some kind of performance was under way just outside the gates. A juggler perhaps, or a mime. Someone good enough to capture the attention of a large audience.
Perlue was at the gate now, puffing a little. Breathlessly, he started to offer excuses for what he'd done, but the controller cut him short. There would be time for that later but now he wanted only to find the European. Perlue stared at the crowd, hoping that Gustave and Michel were across the street, also watching for him.
Control asked tersely, 'Anyone got sight of Numéro Un?'
'Can't spot him,' Gustave replied.
'Negative,' said Michel.
Perlue went out of the gates into the boulevard and saw that the performance was by a couple of mimes, one male, one female. Numéro Un must be somewhere in the crowd. It was a motley mix of tourists, families, local residents, small children with their minders, and businessmen stopping to see what was going on. Jean Perlue looked for anyone whose face was turned away from the mimes, watching for a figure who was not interested in the performance but merely using it as cover.
He was desperate now to make up for his mistake and wanted above everything to be the one who found Numéro Un. He must be here somewhere – he couldn't have got past Gustave and Michel, could he? But everyone he could see had their eyes fixed intently on the two performers, though there were so many spectators that he couldn't properly inspect even half of them. It would have been easy for Numéro Un to insinuate himself into the middle of the crowd, and put himself out of sight of any of the watchers.
After five minutes, movement began in the crowd. Some of those on the edges started to drift off. The performance was coming to an end. The mimes came out among the spectators, each holding out a hat, bowing exaggeratedly when anyone dropped money in. They moved quickly, trying to catch people before they left. Perlue followed on behind them into the middle of the onlookers, but there was still no sign of Numéro Un.
On the other side of the boulevard he could see Gustave scanning the dispersing crowd; there was a strained look on his face, and it was obvious he was getting nowhere. But then neither was anyone else.
He looked behind him, in case Numéro Un had somehow slipped back into the park, but there was only a woman holding the hand of a small child, who was holding in his other hand the string of a fat pink pig balloon, which bobbed in the air above his head.
Jean Perlue turned back and saw that the crowd was getting smaller and smaller. He stared at each departing spectator, hoping against hope that he'd find Numéro Un among them. Some of them stared back, clearly wondering what was wrong with the young man with the drawn and anxious face. Like the sand seeping through an hourglass, his chances were inexorably running out, and finally only three or four people remained, chatting idly as the mimes picked up their props and pooled the money they had collected.
Suddenly the radio silence was broken and in his ear he heard the voice of Rabinac. 'We have Numéro Deux, just ahead of us. He's leaving the park. Do we pick him up?'
There was a pause. Control was consulting. 'No. Keep with him but if you think there's any danger of losing him, then pick him up. Gustave and Michel, get over there and help Rabinac and Marcel. There's nothing more to be done where you are. Numéro Un has given us the slip.' Then came the words Jean Perlue did not want to hear. 'Perlue. You come straight back to base.'
Chapter 10
There was silence in the Control Room in the headquarters of the DCRI where Liz was sitting with her opposite number Isabelle Florian. They had just heard that not only had Numéro Un disappeared but Rabinac, Marcel and the others had also lost Numéro Deux in the crowd.
Isabelle ran her hands through her hair. 'I'm sorry, Liz,' she said. Her English was fluent. 'We should never have had that young man on the team.'
The control officer broke in: 'The trouble was that we had too many operations going on today and this one came in at short notice. Perlue passed all the training courses but it looks as though his temperament let him down in the excitement. I shall be sending him for retraining.'
'Never mind,' said Liz. 'It happens to us all sometimes.' Thank God Bruno wasn't here, she thought. He'd certainly have made some scathing remark that would have ruined Anglo/French cooperation for good.
Isabelle sighed and said, 'Well, let's hope we've got some decent photographs. They're just being printed up; let's go back to my office where we can have a cup of coffee and look at them.'
Liz had been working with Isabelle Florian on and off for several years now. When she had first heard that her opposite number in the French Service was a woman, she had expected to encounter an epitome of Parisian style. She had been pleasantly surprised to find that Isabelle, a woman a little older than Liz, was more given to wearing jeans, a sweater and flat shoes than high heels and an elegant black number. Her pleasantly weathered face was normally bare of make-up and her hair was usually tied back in a ponytail.
But as they walked back to Isabelle's office Liz couldn't help remarking on the change in Isabelle's appearance. Today she looked far more as Liz had originally imagined her. The jeans and sweater had been replaced by a black skirt and tights and a silk blouse the colour of ripe cherries. The ponytail had gone and her hair had been cut stylishly short.
When she complimented Isabelle, the Frenchwoman said, 'I never feel quite comfortable dressed up like this, but I've been promoted and they told me I had to dress the part. I have to go to more meetings and talk to government ministers and my bosses thought I looked too workmanlike.'
'Well. It suits you. Not that the other didn't,' added Liz hastily.
Isabelle smiled. 'And you, Liz. You look flourishing. How is our friend Martin?'
'Well, thank you. We've just been on holiday. The curious yellow shade of my face is the remains of a tan.'
Liz had first met Martin Seurat when she had been working with Isabelle on the case of a dissident Irish Republican group. The leader of the group had kidnapped one of Liz's colleagues, Dave Armstrong, and taken him to the South of France, where Martin Seurat had been instrumental in saving his life.
Liz now stood by the window in Isabelle's office, admiring the glimpse of the Eiffel Tower, which was just visible from the corner of the window. A girl came in clutching a sheaf of A4-sized photographs that she put down on Isabelle's desk, saying cheerfully, 'I think you'll be pleased with these.'
As she went out Isabelle said, 'Come and have a look, Liz. Let's hope they are some use.'
The two women leaned over the desk, their heads close together, looking at the picture on top of the pile. It was of Numéro Un, the European, as he walked towards the rendezvous with the Arab. At the same moment, Isabelle exclaimed, 'It can't be,' and Liz said, 'Isn't that . . .'
They were both staring at the picture in astonishment.
Isabelle nodded. 'Yes, it's Antoine Milraud.' A former officer of the DGSE, and a former friend and colleague of Martin Seurat, Milraud had been dismissed from the DGSE after an operation had gone disastrously wrong. Milraud was suspected of taking money that had gone missing from an arms deal, but he had disappeared before he could be prosecuted.
Martin Seurat had made it his mission to capture Milraud; he blamed him for having betrayed both their friendship and the Service they both worked for. It later became apparent that Milraud had used the money he'd stolen to launch his own career as an arms dealer, where he skirted the border of legality until he crossed it with a vengeance. The Irish Republican who had kidnapped Dave Armstrong had been one of his customers and Milraud had assisted in the kidnap.
That was several years ago, and Milraud hadn't been seen in France since – though there had been a host of rumoured sightings, including one of his wife, Annette. Reliable reports had come in that Milraud had continued acting as a middleman for arms sales; he had been linked to major transactions in a range of conflict-torn territories from Central Africa to Chechnya.
'Why would he resurface in Paris now?' asked Liz. 'He's taking a hell of a risk.'
Isabelle pursed her lips, and started to push her hair back on one side, until she remembered that she no longer had long hair. Her hairdresser had told her that the style was chic for a woman of a certain age. Isabelle had liked the result, though she had bristled at being called 'a woman of a certain age'. She said to Liz, 'It must mean this is a big transaction. Only a lot of money would get Milraud to take such a risk.'
'Mmm,' said Liz, unconvinced. 'It still seems very strange to choose Paris when they could have met in any city in the world.'
Isabelle looked at Liz. She found her English colleague's habit of looking for hidden meanings unsettling. She added, 'I'll need to tell Martin.'
'Of course,' said Liz, though there was resignation in her voice.
Isabelle said hesitantly, 'Is he still so . . . obsessed with Milraud?'
Liz sighed, and Isabelle added gently, 'It's understandable, Liz. The two of them worked closely together. That must make Milraud's betrayal very painful.'
'I know, but I had hoped he was getting over it. There's been no real sign of Milraud for several years. Just rumours and false leads. Martin used to jump at each one, but the last time there'd been a possible sighting he didn't seem to feel the need to go rushing off after it. I thought that was a good sign.'
'This is different, alas.' They looked through the sheaf of photographs. 'I'm afraid there can be no doubt. It is Milraud. Which makes it especially galling that he got away.'
Liz shrugged. 'These things happen.'
Isabelle admired her equanimity. Had their roles been reversed, she liked to think she would have stayed equally calm. But she wouldn't have bet on it. 'Anyway,' she replied. 'we will do our very best to find him. I'll get these photographs out straightaway. We'll check the airlines, the railway stations, the hotels. But I'm afraid he'll be long gone by now.'
Liz nodded. 'Unless you think there's anything I can do here, I need to be getting back to London. I want to send the pictures out to Bruno Mackay. He's gone out to Sana'a to join the CIA man there whose source gave us this lead. I'll send the pictures of Numéro Deux too. Maybe someone out there can identify him, though it's pretty unlikely. He could be absolutely anybody.' Then, seeming to sense Isabelle's gloom, Liz added, 'Cheer up, Isabelle. You may get a break. If Milraud was stupid enough to show up in the Luxembourg Gardens, he may have made some other mistakes as well.'
Chapter 11
Three hours later Isabelle was still in the office, Liz having long gone. Isabelle would have liked her to stay longer, though she knew that there was nothing she could do by sticking around. She liked her English colleague, not least because she was a woman who seemed comfortable with herself. She was intelligent and very focused but she was also attractive and easy to get on with. Too many of Isabelle's female colleagues seemed so intent on proving to their male colleagues that they were their equals that they lost all femininity.
It also pleased her to see Liz so happy in her relationship with Martin Seurat, even if inevitably it made her a little jealous. Isabelle was divorced. Her former husband was a diplomat; their two careers just hadn't fitted together and Isabelle had not been prepared to give up hers for her marriage. And nowadays she worked such long and irregular hours that there didn't seem much prospect that she'd find a successor to him.
She was married to her work, she thought to herself, imagining her own obituary. How ghoulish – she decided to stop feeling sorry for herself and get on with finding Milraud.
Ten minutes later, as she was wishing for the hundredth time she hadn't given up her beloved Gitanes Blondes, there was a knock on her door.
'Entrez,' said Isabelle mildly, thinking it was time she went home. Her young son was at her mother's apartment; he often spent the night there when Isabelle was working late. So often in fact that Isabelle sometimes wondered guiltily if he would grow up thinking he had two mothers. But it wasn't too late to collect him now.
Her assistant Madeline came in, looking unusually excited. 'I think we've found something. They have been checking the hotels of the inner arrondissements and they've discovered where Milraud was staying.'
'Was?'
'Yes. He checked out two hours ago. A place on the Rue Jacob. He must have gone back there when we lost him. He got the receptionist to call him a taxi.'
'Where was he going?'
'The taxi company can't reach the driver.' She saw the disappointment on Isabelle's face. 'There's more. We know the alias he's using. It's Pigot.'
'Pigot?'
'Yes.'
'I don't believe it.' It was almost the exact name of Milraud's Irish Republican customer – who had been gunned down attempting to escape from their hideout off the south coast of France. Calling himself after his dead colleague seemed a bad joke, unless Milraud was thumbing his nose at his pursuers.
Isabelle shook her head, trying to focus on what needed to be done. 'I want the airlines contacted, and we need to check car rental agencies and the train stations.'
Madeline said mildly, 'It's all under way.'
'Good,' said Isabelle. 'Could you ring my mother please? Ask her if she'll keep Jean-Claude tonight. I'll be here a while yet.'
Five minutes later Madeline came in again. 'A Monsieur Pigot made a reservation on an Air France flight to Berlin. Business Class.'
'That's him all right,' said Isabelle. Milraud had always liked the best; Seurat had once told her that his expenses had been legendary in the DGSE. 'I want him arrested at the gate, and held at the airport until I get out there.'
'Too late. The flight took off from Charles De Gaulle twenty minutes ago.'
Damn. Another tantalisingly close miss. But this time she knew exactly where Milraud was. 'Get me the BfV on the phone – I want the Germans to be waiting for the plane when Milraud lands.'
'Anything else?'
'Yes. Book me on the first flight to Berlin in the morning.' She paused for a moment, thinking of something. 'Book two seats while you're at it.'
She examined her options. What should she ask the Germans to do? Arrest Milraud? Martin Seurat would be delighted to lay hands on him but Liz would be worried that the trail to her case would go cold as a result. Milraud would be sure to have some plausible story about his meeting in the Luxembourg Gardens. So put him under surveillance instead? But did she dare risk losing him again?
Minutes later she was on the phone to her opposite number in the BfV, Germany's security service, asking him to set up surveillance on an international arms dealer travelling under the name Pigot, who would land at Tegel in one hour. Photographs of the man were on their way. He was a former intelligence officer and highly surveillance-conscious.
Then she rang Martin Seurat.
Chapter 12
Hans Anspach of the BfV stifled a yawn as the flight information line on the board at Tegel airport flipped over. Air France 1134 from Paris had landed. Anspach signalled to his colleague, Pieter Dimitz, who was coming back from the terminal's Starbucks with two cardboard cups of coffee in his hands. 'You'd better dump those,' he said.
The junior officer groaned. 'Don't tell me the flight's on time.'
'Yes. It's just landed. And I bet our man will be one of the first through. Control has just told me that the French say he has no checked baggage on board.'
Anspach had been halfway home when the call had come, telling him to go to Tegel airport where a French arms dealer called Pigot would land at ten minutes past nine. Anspach and his hastily put together team were to follow Pigot wherever he went and stay with him till they were told to stand down. No reason was given at this stage, though according to the French he was likely to be alert for surveillance.
That probably means they screwed up and he saw them, thought Anspach grumpily. He was missing seeing his son's school play and he was going to get hell from his wife when he eventually got home.
Sitting inconspicuously in a small interview room just behind the passport control desks, Gunter Beckerman was waiting for a buzzer to alert him that Pigot was at the passport desk. He would send a warning to Anspach's phone, before following Pigot through Customs and into the Arrivals Hall.
There Anspach and Dimitz were taking up their positions. Dimitz wore a dark blue suit and had now put on a peaked cap. Holding a sign reading Herr Rossbach, he went to stand alongside the waiting chauffeurs next to the exit point from Customs.
Anspach stood further back at a newsstand, idly examining a copy of Der Spiegel. As he turned the pages he kept a deceptively casual eye on everyone emerging into the Hall. He wasn't relying on Beckerman's call to tell him the suspect was coming through, for it was perfectly possible that Monsieur Pigot might now have a different name, and a different passport, from those he had used to board in Paris.
His phone vibrated and he glanced down at its screen. Coming now. Brown leather coat, read the message attached to a photograph of a man in a leather coat and roll-neck sweater, carrying a laptop bag on one shoulder.
And then, not thirty seconds later, he spotted him.
Pigot was medium height, broad-shouldered, dressed in the smart casual clothes of a businessman. But, unlike a visiting businessman, he wasn't carrying a suit bag, only the laptop case hanging from a shoulder strap. He was walking quickly – though not so quickly as to call attention to himself – and heading towards the far exit, under the sign for taxis. Anspach followed, knowing both Dimitz and Beckerman were behind him.
Outside, the sky was pitch-black, but the pavement was eerily illuminated by the series of sodium lights lining the front of the terminal's façade. Anspach saw his quarry standing in the taxi queue, which was short this late at night. He waited until Dimitz passed him, no longer wearing his peaked chauffeur's hat. Then both men got into the back of a Mercedes saloon parked by the kerb in which the final member of the team had been sitting in the driver's seat. He'd prevented vigilant security and parking staff from having it towed away by waving his security pass at them.
From the car they watched Pigot enter a taxi. When it drove off they followed. Beckerman, having joined the taxi queue two behind Pigot, was in another taxi, not far behind. The convoy headed off on Route 11 for the centre of Berlin.
Ten minutes later a message flashed up on Anspach's phone. 'Booking in name of Pigot made two days ago at Westin Grand Hotel, Unter den Linden. 3 nights, arriving yesterday. Await further inquiries.'
'What do they mean, "Await further inquiries"?' muttered Anspach. 'How can we await anything? We're right behind the guy,' and he tapped furiously on his phone.
Twenty minutes later as they drove, still in convoy, into the centre of Berlin, Anspach's phone vibrated again. 'A Madame Pigot checked out of Westin Grand this pm. No forwarding address. No trace so far of any other booking in central hotels in name of Pigot.'
'Don't lose that cab,' said Anspach to the driver. 'We don't know where the hell we're going now.'
'Well, we'll be on Unter den Linden in a minute,' he replied. 'So perhaps he's rebooked.'
The lights were bright along the pavements of Unter den Linden, traditionally Berlin's most glamorous avenue, but the atmosphere was marred by a darkened construction site running all the way down the street's centre, where work was going on to connect the subway between the former west and east sectors of the city. The beautiful trees were virtually invisible behind the boards and railings; what could be seen of them was covered in white dust that each day's excavations threw up.
'If he gets out here and crosses the road we'll lose him behind the hoardings,' said Dimitz.
'It depends if he's spotted us,' grunted Anspach.
But the taxi containing Pigot didn't stop. It drove on at a stately pace down Unter den Linden until it turned off into a small quiet square and drew up in front of the Hotel Schmitzkopf, an ornate six-storey building with little balconies and flower boxes, an oasis of nineteenth-century solidity amid the city's East German decrepitude and its obsession with new build. This was a hotel designed for comfort rather than style. We have seen it all before, the hotel's stone façade seemed to say. Fads come and go, but the Hotel Schmitzkopf remains the same.
Warned by a text from Beckerman in his taxi, Anspach and Dimitz had stopped their Mercedes further up Unter den Linden, behind a skip that was half full of broken asphalt. They waited fifteen minutes, then Anspach got out. He turned into the square, climbed the steps to the glass and oak door of the hotel and went into the ground-floor lobby, where at this late hour the soft sofas and chintz-covered armchairs were unoccupied.
At the reception desk a young blonde woman in a smart black suit gave a welcoming smile from behind a large bowl of wrapped sweets. Her face fell slightly when Anspach produced a card identifying him as a government official and asked for the manager.
'He's on his break,' the girl said hesitantly. 'Do you want me to fetch him for you, sir?'
'That won't be necessary. Tell me – a man came in a few minutes ago and checked in. A Herr Pigot, if I'm not mistaken.'
'No, sir,' said the girl. 'That was Herr Pliska. He's a Polish gentleman. He and his wife are in Room 403. She arrived this afternoon. We have no guest called Pigot and no booking in that name.'
'Oh,' said Anspach. Then, after a pause while he absorbed the new information he said, 'I must have made a mistake. Got the wrong hotel. Please don't mention to anyone that I was inquiring for Herr Pigot. It's a matter of national security,' he added solemnly.
'Certainly not, sir,' the girl replied, wide-eyed. 'Shall I let you know if Herr Pigot turns up?'
'Please do,' replied Anspach. 'Here is a card with a number to ring,' and he handed her an official-looking card with no name on it and a telephone number that didn't exist.
Chapter 13
'What the hell's going on?' Annette Milraud got up from the chair where she had been watching TV as soon as Milraud walked into the hotel room. 'Why did you tell me to change hotels and names? Did something happen in Paris?'
She paused as Milraud dropped his bag onto the floor and sat down on the bed. He was tired. Tired of life on the run. He'd always known that things would be difficult when he cheated his old employer, the DGSE, and went underground. But he'd thought that eventually some sort of steady state would emerge, allowing him to live without constantly looking over his shoulder.
He'd been wrong. His old employer had not forgotten him. And everywhere he went he'd been conscious that somewhere in the shadows they were there, waiting to pounce if he gave them the smallest chance. There had been financial rewards greater than anything he had ever enjoyed before, but with them went a total lack of peace of mind.
Annette was always angry these days, always nagging. He was taking too many risks, she said, but he had tried to explain that it was only because he took risks that he could make the kind of money he did and she could live in the style she demanded. Risk and money were linked like uneasy soulmates, bonded as unhappily as . . . Milraud and Annette.
They had been together seventeen years, married for fifteen of them. At first, they had been very happy. He enjoyed his work at the DGSE and she was content with their life in the prosperous Parisian suburbs, such a far cry from her humble origins in Toulon in the south of France. He realised later that her single goal then had been to have children, and that compared with this nothing else mattered.
It was when, after every kind of test, the doctors had finally told them that having a family simply wasn't going to happen, that Annette's dissatisfaction had begun. It was as if money had replaced children as her objective, and making the kind of money she had in mind was no more likely for Milraud as an officer in the DGSE than having children with his wife.
Then an operation to bust an arms deal had gone wrong, through an untimely intervention by the Swiss authorities. For a few hours the money at the heart of the deal had floated in a kind of no man's land between the dealer and the buyer. It was there for the asking, and before anyone had thought to reclaim it, Milraud had seen his chance – and taken it.
And since then money had led to the pursuit of more money – and more trouble. He had left the Service under a cloud that soon turned into a criminal investigation and a warrant for his arrest. He had fled France, escaping by the skin of his teeth, with Interpol fast on his heels. In the years that followed, his new business dealing in arms had become global. He had set up shop in Venezuela, where he had made certain arrangements that he felt confident would keep him safely out of the reach of Interpol and the European and American intelligence services. From there he ventured forth carefully, using a multitude of different passports, and usually to countries where there was no danger of extradition – certain Central European countries, the Middle East, parts of Asia, other South American countries. This trip to Western Europe was an exception and, as he was now realising, a mistake.
Now Annette was looking at him with irritation. 'Go and get a shower and change. I've hung your clothes up in the wardrobe. Let's get out of this stuffy old hotel and get some dinner. I've booked a table at a restaurant round the corner. You can tell me what happened while we eat.'
When he came out of the bathroom, Annette was getting dressed. She had put on a chic, tight-fitting black dress, and was trying on necklaces. He recognised one of them, a heavy silver chain he had bought for her in Geneva. The others she had bought for herself, with his money.
'Which one?' she asked as he came out of the bathroom.
'Which one what?'
'Necklace, you idiot,' she said, half crossly, half affectionately. He noticed the small chicken wing flaps of skin under her arms. Annette was growing older and he couldn't offer her the certainty of a secure retirement.
She settled on a simple affair of thin gold strings braided together and turned for his approval. He nodded without looking at the necklace. 'I'm a bit tired,' he said.
'Of course you are, chéri.' She looked as if she would give him a hug for a moment, but the damp towel he'd wrapped around his middle put her off. 'I think some supper would be just the thing. I've been cooped up all day waiting for you and worrying. Go on, darling, put some clothes on.'
He shook his head and she stared at him. He said, 'I don't think we should go out tonight. In fact, I know we shouldn't go out tonight.'
'Why?' she demanded. 'This is the first time in months that I've been out of that violent, uncultured dump where you make us live, and now you say I have to stay in our hotel room?'
Milraud's shoulders slumped. Annette looked at him despairingly. 'I'm not asking to go dancing, Antoine; just a decent meal in a restaurant where the food isn't Spanish. I thought that was the whole point of my joining you here in Germany.'
'It was.'
'Then what's changed?'
He sighed. 'I think they may be onto me.'
Annette looked at him, disbelieving. 'Who's they?' she demanded.
He shrugged his shoulders. 'Does it matter? The French – it could be our old friend Martin Seurat. Or the English. Or any number of countries. It doesn't really matter. This is Western Europe, not South America. Countries here cooperate.'
'What makes you think they've spotted you?'
'I can't be sure, but I had a meeting in Paris, in the Luxembourg Gardens. Somebody interrupted us – a young man. There was something odd about him, so we broke off the meeting. I haven't been able to make contact since.'
'Why didn't you call me when it happened? I wouldn't have come.'
'It was too late by then and you were already en route. That's why I told you to change the hotel and use the other passport. If they did spot me, they'll have a picture, and it won't have taken them too long to trace where I was staying in Paris, to get the name I was using and from that discover the flight I took to Berlin. As long as you left no trace at the last hotel of where you were going, then we should be OK for twenty-four hours or so – long enough for the meeting tomorrow. It's critical I attend that; if it goes well there will be other larger deals, and then we'll have enough money to retire somewhere nice and not to have to go on taking all these risks.'
Annette was shaking her head. 'I told you not to go to Paris. But would you listen? Of course not. You seemed to get pleasure from thumbing your nose at your old colleagues – even if it meant both of us ending up in prison. How could you?'
She looked on the verge of angry tears, but Milraud had seen this display often enough before to feel unmoved. It was a bit rich of Annette to complain about their being forced to live in Venezuela in one breath, then in the next to moan about the risks.
He said patiently, 'I am doing my best, Annette. And if it comes good . . . Believe me, there's a lot of money at stake or I wouldn't have taken these risks.'
'But what if they are already onto you? What if they picked you up when you got here?'
It was a possibility he didn't want to face and certainly one he didn't want to discuss with Annette. 'I'm sure we've got at least twenty-four hours. Time for me to make this deal and get us out of here.'
'To go where?' Annette said, in a whine, her voice like a distant but approaching siren.
'Just in case they're watching the airports, I think we should take the train to Poland. If we don't hang about there, we should be safe to fly back home.'
'Home? You call Caracas home?' The sirens in her voice were at full blast now.
'It's home for now, Annette, and at least it's safe. The point is, if things go well tomorrow, then we can start to think about living somewhere else.' He raised a hand to stave her off before she could get started. 'No, not Paris, that's true. But somewhere better than Caracas. A place where you can feel you're back in civilisation.'
'Like where? You said yourself all the Western services are on the same side.'
He sometimes forgot how quick his wife was. He said, slightly faltering, 'I thought we might try South Africa.'
She stared at him, then laughed disdainfully. 'Cape Town here we come, eh? Well I can't see that's much better than where we are now. Believe me, if that's the only choice on offer, you'd better just buy one ticket. I'll come back to Europe and take my chances.'
He didn't respond to this; after all, it was not a new threat. He inched along the bed and reached for the phone on the bedside table. 'So what do you want from room service?'
Chapter 14
'What are we waiting for?' Martin Seurat demanded.
Isabelle sighed. Martin had seemed edgy throughout the flight from Paris. Normally a calm man, he had barely sat still, crossing and uncrossing his legs, folding and unfolding a copy of Le Monde. Isabelle had tried to divert him by asking about his daughter, now studying at the Sorbonne and the apple of her father's eye, but he had cut off the conversation and stared moodily out of the plane window.
Now sitting in the BfV conference room with Isabelle and the German investigating officer, his tension was even more obvious. She sensed that Martin's excitement at the prospect of finally getting his hands on Antoine Milraud was dwarfed by his fear of letting the man slip through his fingers.
Seurat went on, 'We know where Milraud is, so why don't we arrest him at once? If we hang around he'll disappear again. We don't know whether he saw the surveillance last night but he obviously suspected something in Paris. It sounds as if his wife, Annette, has joined him here; he must have contacted her, told her to change hotels, and to use a different name. He's clearly thinking we're not far behind him. There's no problem with the warrant, so let's get on with it before they vanish.'
The German said mildly, 'We can do that, of course. If that's what you want.' He looked pointedly at Isabelle, as if to say, This is your problem, not mine.
'Martin, you know as well as I do that we are not the only people interested in Milraud,' said Isabelle. 'And we've only found him at all because of the information we got from the British. At the very least, we need to consult them before making an arrest.'
Seurat was shaking his head, more from frustration than disagreement. He looked at the German. 'Is that what you think?'
The German frowned and shrugged his shoulders. He was a youthful-looking man, a classic German with light blonde hair and pinkish skin which was turning red as he tried to follow the argument between his two French visitors. 'Well, as I said, it's really up to you. We have no information against the man but the warrant is outstanding, and the request to help came from your country.' He paused. 'The matter is complicated by the fact that he is not alone.'
Seurat said impatiently. 'There's a warrant out for his wife as well. She's helped that bastard every step of the way.'
The German acknowledged this with a nod, but said firmly, 'Nevertheless, since other countries' services are involved, I would feel more comfortable knowing they agreed with the action we decide to take.'
Seurat looked exasperated, but when he turned to Isabelle there was resignation in his eyes. 'All right. Ring London. Let's talk to Liz.'
Liz Carlyle was at her desk in Thames House when the call came through from Berlin. She had heard about the German surveillance from Isabelle the previous evening, when she'd arrived back from Paris. So she knew that Milraud and Annette had changed names and hotels and that they must suspect that their pursuers were not far behind.
Now Isabelle explained the dilemma. 'Martin is keen to go in now and arrest the pair of them, but I felt we must consult you first. If we do arrest him he's most unlikely to talk about what he's doing here and you'll lose your lead to his contacts. What do you think, Liz?'
Martin Seurat was tapping his fingers on the top of the conference room table and he suddenly leant forward and spoke into the speakerphone. 'Liz, you know that Milraud has broken French law in too many ways to list. Larceny, kidnapping, conspiracy to murder. These aren't trivial offences. We at the DGSE want to see him extradited and put on trial, and who can blame us?'
Isabelle said, without looking at Martin, 'He's a big fish all right. And of course we have our national priorities. But perhaps we need to take a wider view.'
Isabelle sensed that Martin was bristling. He ignored her raised placatory hand, and said, with indignation in his voice, 'What you call our national priorities ignores the fact that Milraud has been involved in arms deals all over Europe. Indirectly, he's killed people on at least three continents. It also ignores the fact that Milraud was instrumental in kidnapping an MI5 officer in Northern Ireland, and bringing him to the south of France. If we hadn't moved in, I doubt very much that that officer would still be alive.'
Isabelle said calmly, 'But there may be other sharks swimming with him that we can catch. That's what you think, Liz, as I understand it. And the Americans too. Is that not correct?'
To Isabelle's relief, Liz Carlyle broke in, her tone brisk but conciliatory. 'I have something to propose. But first let me ask our German colleague, are you confident of keeping Milraud under surveillance?'
Isabelle thought, can a fish swim? No intelligence officer worth his salt would say no to that question. Where was Liz going with this?
The German replied stiffly, 'Of course.'
'Good,' said Liz, 'then I advise the following: we keep tabs on Milraud, and obviously his wife as well. But if he goes off to meet anyone connected with his activities in Paris, it seems to me very unlikely that Annette will go with him. He wouldn't want to involve her or expose her to the risk. I'm sure she knows exactly how he makes his living, and we know that when he escaped from us in the south of France several years ago, it was with her help. But I can't believe she's actively involved in his deals, whatever they are.'
'So?' asked Seurat impatiently.
Liz said patiently, 'So, if he goes out and leaves her in the hotel, then that should be the time for you, Martin, to go in. After all, you know the woman well, don't you?'
'I do.'
'So you can work on her. You can explain that if she tells us who Milraud is working with, and helps us move up the ladder of this deal, then we can see that things don't go too hard on her. Or her husband.'
'I'm not prepared to promise that. I want things to go hard for the bastard.'
'Martin. It's up to you what you say. We all know you won't have any influence over what happens to them when they're arrested.'
There was silence while all the participants considered this. At last Seurat stirred. Leaning towards the speaker on the table he said, 'All right, Liz. You win – you, and the Americans. But let's not lose him, OK? Nothing personal, but he's caused me a lot of trouble. I couldn't bear it if he had the chance to cause any more.'
Chapter 15
The Schweiber Mansion at the eastern end of Unter den Linden had once housed the private collection of Ernst Schweiber, a German manufacturer who became fabulously wealthy in the late nineteenth century. He and his sons after him had used their wealth to amass an eclectic collection of paintings, furniture and objets d'art from all over the world, which they had housed in their grand baroque mansion. But the mansion had had the misfortune to be in the path of the Red Army when it arrived in Berlin in 1945.
The Schweiber family had by then been long dispersed, some to other parts of Europe, some to their deaths in concentration camps. By the time the Russians arrived, part of the collection had already been removed by the Nazis. What remained was taken as booty by the conquerors, some of it to find its way eventually into galleries and museums in Moscow and Leningrad.
After the Berlin Wall went up, the Schweiber Mansion found itself in East Berlin, no longer grand but grimy, broken-windowed and pocked by shell holes. The house became home to a department of the Stasi and was feared and avoided as far as possible by East Berliners. As part of the restoration of East Berlin, the building had fairly recently been renovated to something of its former grandeur. But now, instead of sitting in an avenue of equally grand mansions, it rested uneasily between two glass-fronted office blocks, surprising the tourists who came to see what was at least a part of the Schweiber Collection, gathered together again from around the world and returned to Berlin after much diplomacy and haggling.
If Hans Anspach had known about the diplomacy and haggling, he would probably not have thought it worthwhile. He was gazing at a rather gruesome painting of someone being flayed alive. But in any case his mind was not on the art, for though the headphones he was wearing looked like those the museum supplied to visitors who wanted a commentary as they toured its collection, what he was hearing through them had nothing to do with art.
'Still here,' came from Beckerman, who was a few rooms away. Taking a couple of casual-looking steps, Anspach could see, through an arch, the back of Antoine Milraud's head. The Frenchman was standing with half a dozen other visitors in front of a Corot which had lately made the news – to the embarrassment of the German authorities, it was now thought not to have been part of the Schweiber collection at all but to have been plundered by Field Marshal Göring from a French aristocrat in Burgundy, whose descendants were threatening to sue for its return. Beckerman added, 'No movement.'
It had been easy enough to follow Milraud to the gallery. He had left the hotel half an hour ago, dressed in a white roll-neck sweater and a grey tweed jacket. He had walked, without looking around, straight down Unter den Linden, then fifty metres along a side street to the Schweiber Collection. With two teams of three on his tail, there had been no chance of losing him, and with the museum busy but not too crowded, it was simple enough to keep tabs on the man as he wandered through the ground-floor rooms.
He had been in the building over half an hour now, and there had seemed no particular rhyme or reason to his progress. He had looked at paintings and porcelain and classical sculptures. To Anspach's experienced eye, he seemed to be killing time rather than appreciating the objects.
But the Corot was holding his attention far longer than anything else had. Was he waiting for someone? Was this the meeting point? Anspach edged into the next gallery, from where he could get a wider view of the room where Milraud stood.
He noticed the black man as soon as he walked into Milraud's gallery. Berlin was full of students from Africa, but this man was no student – he was tall, slim and beautifully dressed in a tailored grey wool suit, a cream silk shirt, and a tie. The fact that he was probably the only man in the gallery wearing a tie would have made it remarkable enough, but this was clearly a designer tie, broad, silky, with a brightly coloured pattern. His figure was elegant but his height and broad shoulders suggested there was strength behind the smooth façade.
The man didn't glance in Milraud's direction; he moved towards the far wall, where a group of young Chinese tourists stood giggling in front of a large nude. As Anspach watched, he saw Beckerman stroll in from the other gallery; he had joined the back of a tour group that gathered briefly at the Corot before moving in Anspach's direction. The group, with Beckerman in tow, walked into the gallery where Anspach stood and gathered at another picture.
Anspach glanced again in Milraud's direction. The Chinese had moved on from the nude, but where was the black man? Then he spotted him; he had been hidden by another group listening to an English-speaking guide. Now he walked up to the Corot and stood next to Milraud, with only a foot or two of space between them. Both were examining the Corot as if they were experts, and when the Frenchman turned his head slightly Anspach could see the two men were talking.
He drew back until he was out of sight, then looking down he said softly into his microphone: 'We have contact. Newcomer. Black male. One hundred eighty-five centimetres tall, slim, smart grey suit.'
The two men stayed standing, side by side, until suddenly the black man turned and walked out of the room. Milraud waited a few minutes then left the room too, going quickly towards the museum's exit. As he left the building he headed off in the direction of his hotel, watched by Anspach, who had joined Dimitz in an unmarked car parked in the car park outside the building. Three spaces away a second car was parked containing two more officers of the BfV; a third team member was busy buying a newspaper from a kiosk outside the entrance of the museum.
'We'll take the Newcomer; you take the main man,' ordered Anspach on the car's radio. 'When you've housed him at the hotel, come and help us. If he goes somewhere else, stick with him.'
Anspach settled down to wait for the black man he'd labelled Newcomer, and a few minutes later he emerged, with Beckerman fifteen metres behind him, examining a map of Berlin with apparent concentration. Anspach waited until Newcomer had walked a couple of hundred metres away from Unter den Linden, towards a shopping district, busy on a Saturday morning. When it got hard to see him in the crowd on the pavements, Anspach nodded at Dimitz, who started up the car and drove in the direction their target had gone. They could see Beckerman, struggling to keep up with Newcomer, who was striding quickly past the shops as if on his way to keep an urgent appointment.
They drove past both men, and Dimitz pulled up, just short of a pedestrian-only area. Anspach hopped out, waved to Dimitz as if to thank him for the lift, and walked swiftly into Nadelhoff's department store, an old-fashioned emporium that was adjusting badly to its new concrete and glass quarters. Inside he loitered on the ground floor, looking at men's shirts near the front windows, waiting for Newcomer to walk past. When he did, Anspach abandoned the shirts and left Nadelhoff's, just in time to see his target disappear through the swing doors of a shopping centre – six stories of small independent shops known collectively as the Boutique Mall. Whoever this elegant black man was, he seemed to know his way around this part of Berlin.
Anspach spoke into the mike under his lapel. 'He's gone into the Boutique Mall. I'll try and keep with him in there; park the car and come round to cover the rear entrance. Beckerman, watch the front. Control: get the other team over here as soon as they've seen their target home.'
Anspach spotted his target easily enough as soon as he went into the Mall. He was in a record shop on the ground floor, leafing through CDs. Anspach walked past and went into a shop opposite; from there he could see the door of the record shop.
He was beginning to feel desperately exposed but he didn't want to call in either of the other two to take his place for fear of leaving the exits unmonitored. The black man was taking his time – or was he killing time? He had twice looked at his watch but he went on flipping through CDs.
Then he moved, suddenly and quickly, heading straight for the atrium in the centre of the Mall. If he had clocked Anspach he didn't show it; he walked fast, looking straight ahead, and by the time Anspach was out of the shop, he had crossed the atrium and was striding down the aisle leading to the rear exit.
'Dimitz, target coming your way. He's yours,' he said into his mike. He was hanging back now to avoid detection if his target should look back. He gave it a good sixty seconds, then said into his mike, 'Have you got him?'
The reply was a grunt.
'Which way is he going?'
'He's not "going". The bastard's just standing on the kerb.'
'Any cabs around?'
'No. If he wanted one there's a taxi rank that he walked right past.'
So what was he doing? Waiting to see if he was being tailed? Possibly, but there were better ways to shake off surveillance, or even just to see if it was there. Waiting in one spot wouldn't do the trick, since the watchers didn't have to show themselves.
Anspach decided he should risk a closer look. He had reached the rear entrance and could see the black man now, across the street, staring into the window of a women's shoe shop. It seemed contrived, unnatural. Was he using the window to spot surveillance? Anspach had a premonition. 'Dimitz, quick get the car.'
'I'm in it already. Just round the corner.'
'Come and pick me up.'
But it was too late. There was a whoosh of an approaching car – a black Mercedes limousine, with tinted windows – the screech of brakes, and in an instant the black man had disappeared into the back seat, slamming the door behind him. The Mercedes executed a three-point turn at the expert hands of an invisible driver, then accelerated away down the street.
By then Anspach had his phone in his hand, and its camera snapped and snapped again. 'Dimitz, where the hell are you?' he shouted into his mike, not caring now if he was overheard.
'I'm stuck. There's a rubbish truck in front of me and I can't get round.'
Chapter 16
Anspach strode round the corner into a cacophony of car horns. Dimitz was sitting at the wheel of his car, at the head of a line of stationary cars, all blowing their horns at the garbage truck blocking their way.
'What in heaven's name is going on?' shouted Anspach.
'The driver's in that café and he won't come out.'
'I'll get him out fast enough,' said Anspach, and headed into the café waving a card identifying him as on special government business. In seconds he was out again, shouting at a couple of men in yellow jackets who had come out of the café. They got into the rubbish truck and drove it off up the street. By then Beckerman had joined his colleagues in the car.
'I've passed the registration number of the Mercedes to Control and he's asking the traffic police to look out for it,' said Anspach. 'We've got no reason to stop him, unless Traffic can get them for exceeding the speed limit, but at least we should get a fix on where he's going.'
'I got some good photographs of both of them in the gallery,' said Beckerman. 'I swear that was no chance meeting. They were discussing something. It was an RV.'
'Yeah. And I think that black fellow clocked us, at least by the end. That was a very smooth getaway,' added Anspach.
'Where to, boss?' asked Dimitz.
Anspach snorted. 'God knows. We'll join up with the other team and hope Traffic get lucky.'
And they did – up to a point. Ten minutes later a report came in that a traffic patrol car had spotted the Mercedes heading north on the E26 near Westend. Unfortunately the patrol car had been going in the other direction.
A quick conference with Control sent both teams off to Tegel airport, where the BfV officers and the police and immigration officials were all alerted to look out for a tall, elegantly dressed black man, and to note his passport details and where he was heading.
Tegel was crowded when Anspach and his team arrived. They had to push past long queues of passengers at the departure desks in the hexagonal International Terminal A to reach the office where the airport team had their base. There was no news of their target. He had not been observed going through security or passport checks at Departure and no sightings of the Mercedes had been reported by police outside the terminal.
But Anspach wasn't going to give up; he found a ticketing supervisor, and with the man by his side, slowly worked his way along the lines of check-in desks – British Airways, Lufthansa, Delta and all the other airlines running international flights from the airport, showing the desk clerks the clearest picture he had of the dozens taken by Beckerman in the Schweiber Museum. As each desk clerk peered at the tiny image on the mobile phone's screen they responded with a shake of the head.
As he was working his way along the desks he heard Beckerman's voice through his earpiece say, 'They've seen the car at Terminal D.'
'What's Terminal D?' he asked the supervisor, who was still with him.
'It's Air Berlin – domestic flights.'
He looked at the man, puzzled. 'Domestic flights?' Surely their target wasn't going somewhere else in Germany.
The man added, 'Private jets use it as well.'
'How do I get there?'
'Not easily. There's a bus . . .' the man started to explain, but Anspach was already racing for the terminal doors, shouting into his mike for Beckerman to pick him up.
At Terminal D they were directed to the far end of the departures hall. There they found one small counter manned by a middle-aged woman in a blue suit and forage cap who greeted them with a smile.
'Guten Tag,' she said, 'and how may I help you today?'
'Have you seen this man?' asked Anspach, thrusting the mobile phone in front of her face.
Taken aback, the woman paused. 'Our clients expect confidentiality, Herr . . .?'
'Anspach.' He brought out his card – official-looking, special government business, it breathed authority.
The woman's eyes widened. 'Yes, I have seen this gentleman.'
'Where is he flying to?'
'Rotterdam.'
'When is he leaving?'
She looked at Anspach with mild surprise. 'His plane took off ten minutes ago.'
Chapter 17
Martin Seurat knew he had to work fast. He'd been waiting, hidden from view by a pillar in the lobby of the hotel, and as soon as he'd seen Milraud leave he had come upstairs. But if Milraud had only gone out for a paper or to get some fresh air, then he would be back soon, and before that Seurat had to make his pitch.
He had no idea how Annette would react when he turned up at the door of her room. Once, they had known each other very well. In the DGSE, officers worked in small teams, often abroad and in stressful circumstances, and they got to know each other intimately. He and Milraud had worked together on and off for over a decade, and despite some fundamental differences in personality – Martin was quieter, more analytical, focused on getting the job done; Milraud was flamboyant, sometimes inspired, sometimes simply erratic – they had grown to trust each other. Whenever they could, they liked to socialise together and to include their wives, who could easily feel ignored and left out because of the secret nature of the work their husbands did.
Annette Milraud had been a lively young woman then, apparently carefree, without any children. She loved the good things of life: the Milraud apartment was beautifully furnished, her clothes expensive and stylish – enough so that Seurat's wife used to wonder enviously how she could afford it all on the salary of a DGSE officer. Once a week Annette ran a little market stall in the Marais where she sold jewellery, some antique, some that she'd made herself and some just rather pretty junk that she had picked up for practically nothing. She was always wearing three or four of the more flamboyant rings from her stock when they met. It was difficult to believe that the stall brought in enough extra money to finance her lifestyle.
When the four of them got together for an evening, Annette drank more than any of them, smoked incessantly, and liked to dance. She had the kind of loud, extravagant joie de vivre that hinted at dissatisfaction or even desperation lying not far beneath. Seurat's wife had got on with her well enough, though she'd never trusted her in the way her husband trusted Antoine Milraud, and she had made it clear that she didn't want to see the Milrauds too often.
Now Seurat knocked on the door of Room 403, taking care that he could not be seen through the spyhole. He heard nothing at first, then there were steps inside the room. 'Oui?' a woman's voice called out.
He replied in accented English, hoping she would think he was a concierge. 'I have a message for you, Madame.'
'A message?' She sounded suspicious. 'Put it under the door.'
He sighed – it had never been easy to hoodwink Annette. He said quite loudly, in French now, 'Come on, Annette, open the door.'
'Who is that?' He could hear the surprise in her voice.
'It's Martin Seurat.'
There was no reply for a moment. Then the door opened a crack, held on its chain. Annette stared out at him, surprise replaced by hostility.
'What the hell do you want?' she demanded.
'If you let me in, I'll explain.' When she hesitated he added, 'I need to talk to you alone, Annette. Before Monsieur Pliska gets back.' He could see her flinch at the name. 'We know what name you're using, and the one Antoine used in Paris – Pigot. If need be we can even find out the one you used before that.'
'If you know so much, why do you want to talk to me?'
'Because you can help us. And help Antoine. You don't need me to tell you how much trouble he's in.'
Annette stared at him, as if considering what to do, then she suddenly closed the door. For a moment Seurat thought that would be it. But the door opened again, and she stood there, looking angry. 'I suppose you'd better come in,' she said.
Annette had obviously been packing. Two suitcases lay open on the floor and a smaller Vuitton bag was on the bed.
'Going already?' asked Seurat. 'You've only just arrived.'
Annette shrugged. 'That was the plan,' she said.
'Mind if I sit down?' said Seurat, taking one of the two armchairs. 'Whether you go and where you go is going to be up to you, Annette. If you help me you at least may be able to go wherever you like. Don't cooperate and you'll be seeing the inside of a French prison before long.'
'The Germans may have something to say about that.'
Seurat shook his head. 'I don't think so. They'll accept a European arrest warrant, and there's more than enough evidence behind it. You have aided and abetted your husband, a man who's facing charges on everything from illegal arms dealing to kidnapping.'
'What do you want me to do? You know I would never betray Antoine,' she said defiantly, gesturing to emphasise the point. Seurat noticed that nowadays she limited herself to two rings – but both looked a good deal more valuable than in the days of the market stall.
'I'm not asking you to betray him; I'm asking you to help him. And you can do that by helping me.'
She looked at him sardonically. 'That sounds unlikely. How does it work?'
'He listens to you, Annette. You know he does. He thinks all the rest of us are fools and you and he are the only clever ones.'
Annette grimaced. 'I'm not sure he'd include me, not these days. He'd tell you I'm always whining. Anyway, I don't see how I can help you. I don't know the details of Antoine's business. I never have. He's an old-fashioned Frenchman that way.'
Seurat eyed her sceptically but she returned his look with a stare of her own, as if daring him not to believe her. He was confident she knew more than she was letting on, but her true value lay in her influence over Milraud, not in any information she might have about his activities. He said, 'I believe you. But a judge might not – you're in this up to your neck, as I'm sure you know. But if you cooperate – and more importantly if you get Antoine to cooperate – there's still a chance you can lead a normal life again.'
It was her turn to look sceptical, so he went on: 'I mean it. I'm not saying Antoine won't have to serve time in prison, he will – and you may too – but perhaps for less time than otherwise. To be quite clear, what I'm saying is that Antoine can help himself by cooperating and you can help him and yourself by persuading him. Life in prison won't be pleasant, but I can't imagine life on the run is much fun either.'
'It's had its moments.'
'Where do you call home these days?'
She shrugged. He said, 'Come on, Annette, we're already checking with the airlines for passengers called Pliska. I'll know all your recent movements soon enough.'
She hesitated, then said sourly, 'Caracas. We have a flat there.'
'Good God. I don't imagine that's the safest place for a woman left on her own a lot of the time, as I imagine you are.'
'Venezuela's a very beautiful country,' she replied defensively.
'I should think it needs to be – to compensate for all the other disadvantages.'
Annette laughed out loud. A good sign, thought Seurat; he had always been able to make her laugh in the past. He went on, 'The good news is that you won't be going back there for a while. The bad news is that it could be a long, long while. That's up to you.'
'So what do you want me to do?'
'Encourage Antoine to work with us. That's all.'
She was still thinking about this when there was the sound of footsteps in the corridor, then a card was inserted in the lock and the door swung open. Antoine Milraud walked into the room. When he saw Seurat he didn't seem surprised.
'So it's you on my tail, is it? I wondered who had stirred up the Germans. I knew we'd meet again one day.'
Seurat had to admire the man's sangfroid: Milraud had always been nerveless, even in the most hair-raising situations. But then Seurat supposed you had to be if you were going to live on the run. He looked at his former colleague, the man who had been his trusted friend and had become his nemesis, haunting his dreams, filling his head with thoughts of revenge, and said, 'I doubt this is how you envisaged our meeting.'
Milraud shrugged, and sat down heavily. 'Some days life is a bowl of cherries; some days the bowl holds only a few stones. I knew someone was onto me, but I congratulate you on your efficiency. I was hoping I was a few hours ahead.'
He started to reach into his jacket pocket, but Seurat put up a warning hand. 'Don't even think of doing something stupid. I'm armed and downstairs in the lobby there are two members of the local police and an officer of the BfV.'
'I was going for a cigarette actually,' said Milraud, bringing out a pack of Disque Bleus and a gold lighter. He inhaled greedily, then blew out a long funnelling plume of smoke. 'So, what happens now?'
Seurat outlined the position. If Milraud cooperated Seurat would do everything he could to get a reduction in his sentence. There was no point in pretending that Milraud wouldn't be serving time, and some hefty time at that, but equally, his assistance, if it led to other convictions, would be taken into account by the court. If he didn't cooperate, then he could expect the maximum sentence. Seurat said softly, 'I think we're talking twenty years.'
Milraud nodded and stubbed out his cigarette. 'That was very well put, Martin. You haven't lost your touch for clarity. But I have to say I doubt there's much really that you will be able to do for me. I've rubbed too many noses in the dirt. Even if your offer is sincere – and I have no reason to doubt that it is,' he said with a wry smile, 'I have to question your ability to see it through. I'm cooked, as the Americans like to say, though if I take my punishment like a man I will have a chance of breathing free air again some day. If I squeal, then I have very little chance at all.'
'So you won't cooperate?'
'Regretfully, no. Believe me, the sort of people I work with are not the kind one wishes to annoy.'
'What a pity,' said Seurat. It was clear to him that Milraud was far more scared of his arms-dealing associates than he was of the French authorities. Seurat decided it was time to play his trump card, and hoped that Annette would play her role. He said firmly, 'In that case you leave me no choice. I will have you placed under arrest . . . and Annette as well.'
'Annette?' Milraud's voice rose in alarm. 'Why Annette? She's done nothing.'
'On the contrary, she's helped you virtually every step of the way. Beginning with your escape from France. It's a serious offence and she will do serious time.' He paused to let this sink in, then added, 'I think I can guarantee ten years minimum.'
Milraud stared at him, his eyes widening in shock. There was a loud gasp. Annette had her hand over her mouth and she was shaking her head almost theatrically in disbelief.
Whether the appalled look on her face was genuine or not, it was doing the trick. Milraud stood up and rushed to her, throwing a comforting arm around her shoulders. 'It's all right, chérie.'
Annette started to cry, tears the size of raindrops rolling down both cheeks, her sobs growing louder despite her husband's efforts to console her. 'Ten years,' she wailed, as Seurat watched, mentally giving her performance five stars.
His arm still around Annette, Milraud looked at Seurat with undisguised hatred. 'I tell you, she has nothing to do with my affairs, and I don't believe you have any evidence that she has. So leave her out of this.'
'It's too late for that, Antoine. As for evidence, don't worry: we have a strong case – for starters, just travelling on a false passport will get her behind bars. You should have thought of that before you had her fly from Caracas.'
Annette had moved away from her husband's embrace and sat down on the bed, where she began to rock backwards and forwards, still sobbing heavily. When Milraud moved towards her she pushed him away, and Milraud's face fell.
He turned to Seurat angrily. 'What would it take for you to drop charges against her?'
'The truth. All of it. Let's start with why you came to Berlin.'
Chapter 18
Back in Thames House in London Liz Carlyle was feeling out of sorts. An investigation with which she was vitally concerned was unravelling without her and she didn't like it. She was happiest when she was at the centre of events; watching other people take the decisions and viewing the action from far away was not how she liked things to be.
She had been unimpressed by the surveillance efforts of the French, and from what she had heard of the Berlin operation it hadn't been much better. 'A4 could knock that lot into a cocked hat,' she'd said to Peggy Kinsolving.
Geoffrey Fane had been unusually quiet too, and there had been no response from Bruno Mackay in Sana'a, after she'd sent him the French surveillance pictures of the young Arab in the Luxembourg Gardens. She wondered if MI6 were doing something they were not telling her about.
To make matters worse, she was anxious about Martin Seurat. She knew how obsessed he was with Antoine Milraud. She knew how personally he had taken the betrayal, and she was worried that he might not be able to keep his cool when faced with Milraud again.
In spite of all her frustration, some progress had been made in London. Peggy had managed to put together details of the route taken by the private jet boarded by Milraud's contact, the elegant black man, at Tegel airport in Berlin. The Germans had not asked for any special monitoring of the flight as they had no case against the passenger, so Peggy had worked from the records, something she loved doing.
'Smart plane,' she observed. 'Pilatus PC-12, registered in Russia. Even hiring that costs a bomb. We're dealing with real money here.'
The plane had landed in Rotterdam to refuel and taken off straight away, heading for Prestwick Airport. No one had disembarked at Rotterdam. Twenty minutes into the flight, the pilot had requested permission to divert to a small private airfield in North Wales. Inquiries at the airfield afterwards by the local Special Branch had established that the plane had indeed landed there; that one passenger had got out and had been picked up by a private car. No one had asked to see his passport. The duty desk clerk had been confused and had thought that the plane had flown from Prestwick, and yes, the manager of the airport thought that the passenger might have been a tall black man, and no, nobody had noted the registration number of the car.
'I can't believe the sloppiness at that airport,' said Peggy. 'Special Branch is reporting them to the Civil Aviation Authority. I hope they lose their licence.'
'If they've got one,' observed Liz.
Peggy had also circulated the photographs taken in the Schweiber Museum to Special Branches across the UK, with a request for any information about the black man. No replies had been received so far.
The arrival of a detailed report from Martin Seurat of his interview with the Milrauds in the Berlin hotel room gave Liz something to focus on. It was clear that her concerns that Martin might be hindered by his personal animosity towards Milraud had not been realised. In fact, it had probably helped him in being quite ruthless in using Annette's fear of prison to get Milraud to agree to cooperate.
The report was followed by a phone call from Martin, who reported that the Germans had agreed to release the Milrauds into his custody and he was about to leave Berlin for Paris in the company of a small posse of French and German police and security officers.
'Where are you going to take them?'
'I've arranged a safe house in Montreuil.'
'Montreuil? I thought that was a fashionable holiday resort. Why so posh?'
'Not that one. This one's a suburb of Paris. Not posh at all.'
'Good. I wouldn't like to think of the Milrauds living it up at your taxpayers' expense.'
'Can you come over tomorrow? I think we should hook him in firmly to what you want him to do without giving him too much time to think about it.'
'All right. I'll let you know if there's any problem; otherwise you can expect me by lunchtime.'
Liz put the phone down thoughtfully. She and Martin had been close now for more than two years. So close that he had made it very clear the previous year that he would have liked her to give up her job and come to Paris to live with him. From her reaction he had realised that that had been a mistake. Her job was an essential part of Liz's being and without it she could never be happy, even with him. Since then he had been talking about leaving his own job in the DGSE and looking for some other form of work. Liz wondered whether, now that he had finally caught Antoine Milraud, the man who had obsessed him for so long, he might actually decide to do it. She wasn't sure that would be a good idea for him or, she had to admit, for her either.
She was sometimes concerned that there could be a conflict of interest when they found themselves working closely together on a case, as now, but so far they had managed to keep their personal relationship and their work in separate compartments. She knew that her bosses had their eye on the situation and that she needed to be scrupulously professional if she were not to be moved onto other work. Yet now here she was, going to Paris to take over the handling of Milraud. She planned to use him to flush out whatever was going on, and she had to take the lead, because the intelligence so far, such as it was, was pointing to the UK.
Martin met her at the Gare du Nord and they drove to Montreuil, where they found Antoine Milraud waiting in the living room of a nondescript stucco bungalow. He had a young DGSE officer called Thibault and a couple of tough-looking guards for company. Little had been spent on the décor of the bungalow – a few battered-looking chairs, a much-used coffee table and a frayed sofa furnished the sitting room and the walls were enlivened by gaudy reproductions of Impressionist paintings. I wonder who buys this sort of stuff in the first place thought Liz, hoping that the kitchen was equipped rather more expansively – with perhaps a pâté and a bottle of Chablis in the fridge.
Milraud had dressed up for the occasion. In his smart wool jacket, open-necked shirt and polished loafers, he looked more like a successful film director or the owner of a trendy gallery than a renegade intelligence officer.
Liz gazed coldly at the elegant figure who stood up to greet her. She ignored the hand he held out and did not respond to his 'Good morning, Madame'. This was a man who had not only destroyed the peace of mind of the man she loved but had worked alongside a psychopathic killer to kidnap her colleague Dave Armstrong and hold him for days in a damp cellar; a man she had last seen as a shadowy figure in the darkness on a French holiday island, escaping capture and leaving his partner to die in a hail of bullets. She was glad to see him confined in this charmless room, but she would rather have seen him in prison. In the present circumstances she had no option but to work with him, but she certainly wasn't going to treat him with any warmth.
The young officer came in with a cafetière of coffee and three small cups, then quickly withdrew. Milraud fidgeted nervously on the sofa. As soon as Martin had handed out the cups, Liz said abruptly, 'Let's get started. As you know I'm from MI5 and I'm going to ask you some questions about your recent activities. Firstly I'd like to know why you were in Berlin.'
She knew from Martin that Milraud's English was excellent, almost idiomatic, yet she wondered whether he would pretend not to understand her, and he was giving her a questioning look. But then he said, 'Has Martin not told you what I've said already? I'm surprised; I thought you two were close,' he added.
Liz ignored this. 'Of course he has. But I want to hear it from the horse's mouth.'
He nodded. 'As you wish. As I told Martin, I was in Berlin to act as a liaison – the venue was not of my own choosing. I know you are aware of my earlier meeting in the Luxembourg Gardens – this was to follow up my instructions from that.'
'Whose instructions?'
Milraud put both hands up and shrugged, in the universal sign for who knows? He said, 'He is Arab, he is young, he is anti-establishment. To me that means the rebels of the Arab Spring. Other than that, I have no idea.'
'Do you often do business with anonymous contacts?'
Milraud gave her a patronising smile. 'Of course. I doubt I know the real name of any of my customers, Madame.'
'Why did this mysterious person send you to Berlin?'
Milraud looked at her as if checking to see that she was minimally intelligent.
'To meet the black man in the Schweiber Museum?'
'Yes.'
'And what is his name?'
'Smith,' said Milraud without batting an eye. Then he added, 'Perhaps it was Jones.'
Liz sighed. 'All right, let's forget about names for the moment. What was he going to do for you?'
Milraud was silent. 'I don't know,' he said at last.
Liz looked at Milraud, waiting. Then he said, 'I've never seen him before. I don't know him.'
'Then what were you talking about?'
'He was setting up another meeting. For tomorrow. He wanted to make sure we weren't being watched yesterday. I didn't think I was, then I went back to the hotel and found Martin talking with my wife. So I was wrong and the black man was right to be suspicious.'
Liz sat back in her chair. This didn't make any sense. Milraud seemed to be claiming that all these meetings were only to set up further meetings.
He saw her scepticism. 'I could still go to the next meeting tomorrow,' he offered. 'Then I might find out what it's all about.'
Not a big help, thought Liz, since the black man had left Germany and was presumably now somewhere in Britain. She wondered if Milraud knew this and was lying – if she accepted his offer, he would simply invent another rendezvous point, dutifully go there, then profess regretful surprise when the black man didn't show up.
Liz switched tack. 'What about the man from the Arab Spring? How were things left with him?'
Milraud looked at her blandly. 'He was not forthcoming. He only wanted assurance I could supply the weapons he wanted.'
'What kind of weapons?'
'Automatic rifles. In time he said they'd want more sophisticated armoury – RPGs, SAM missiles, you name it. But you have to walk before you can run, and at the moment it's basic infantry armaments he's looking for. Rifles, ammunition.'
'So what was agreed?'
'Nothing. I named a price, and he made a counter-proposal. As is usually the case, we settled somewhere in between. Then we aborted the meeting because we were being watched. The next step is for him to confirm the order – which in practice means a down payment of twenty-five per cent. Not unreasonable,' Milraud said with a hint of pride, and Liz realised he was enjoying talking about his trade. He wouldn't have had many opportunities to do so in recent years.
'How is he going to contact you?'
'If he contacts me at all, after that clumsy surveillance in Paris, it will be by email. Third party – a dummy box I've created. He'll ask for a meeting, though it won't read that way – on the surface, it will look like a misplaced request for a booking at a restaurant. The name of the restaurant will contain a link to another site – that site will contain coordinates which when I apply them to a pre-existing grid will give me the location, time and date of the meeting.'
'You told Martin this will be in the UK.'
'Yes. That is what the man told me in Paris before we aborted our meeting.'
'He didn't give you any sense of where in the UK?'
'No.' Milraud looked at her impatiently. 'I have already told Martin all this.'
She ignored him. 'When are you expecting an email?'
Milraud shrugged. 'I am working to my client's schedule, not my own. When it arrives, it arrives.'
Martin interjected. 'My colleague Thibault has taken charge of his laptop and will monitor all emails.'
'This better be right, Antoine,' said Liz, 'or any reassurance you have received, about how Annette will be treated will no longer be valid.'
Milraud looked at her wide-eyed. 'Do I take it you are in charge then?' He seemed surprised.
'As far as you're concerned I am.'
Milraud turned to Seurat, as if expecting him to dissent, but Seurat said simply, 'She's right, Antoine. She will be directing what you are to do.'
Milraud looked confused as he tried to take this in. At last he nodded again, and gave an ironic shrug. 'I am accustomed to it. Annette wears the trousers in my family too.'
Chapter 19
'Donation will see us this evening,' said Miles Brookhaven, putting down the phone. 'I told his son I was bringing a colleague from the British Embassy and he didn't ask any questions. His son seems to be a sort of secretary for this so-called charity he runs. Well, he calls it a charity, but as far as I can see it's a kind of private fund-raising operation. God knows what shady deals they're doing. Anyway, we're to go out to his farm this evening.'
'A farm?' Bruno Mackay raised his eyebrows. 'How far away is it? I don't fancy a long drive in the dark in this place. I won't be at all popular with Geoffrey Fane if I end up as a kidnap victim.'
'It's not that far. About ten miles or so along a fairly decent road. It'll be dark when we come back, but Donation seems to have some sort of security operation set up to control who goes along that road, so it should be OK.'
Bruno Mackay was sitting in Miles Brookhaven's office in the American Embassy in Sana'a. The surveillance pictures from Paris were spread out on the table in front of them.
'I hope it's worth the journey,' said Bruno. 'I can't imagine they'll make much out of these photographs. I don't know why Liz Carlyle bothered sending them. The guy looks like thousands of young men you might meet anywhere from Algiers to Afghanistan.'
'Maybe he does, to you, but Donation or his son may recognise him – or know someone who will.'
'Let's hope so. Our French colleagues certainly seem to have messed up thoroughly in Paris. First they blew the surveillance and then they lost both of their targets.'
'I don't think it's a complete disaster. I've heard from Andy Bokus that they know the European who met this guy in Paris. He's called Milraud, a DGSE officer who left the Service and turned rogue.'
'Oh him. The French have been looking for him for years. He stole a lot of cash and set himself up as an arms dealer. If he's reappeared it will have set the cat among the pigeons. He used to work with Liz Carlyle's boyfriend Martin Seurat; Seurat's sworn to get him.'
'Well, apparently they have got him. They pinned him down in a hotel in Berlin and they're hoping to find out why he was meeting this guy in Paris' – he waved at the photograph of the young Arab – 'and what he went to Berlin for.'
Five hours later Miles Brookhaven was driving the Embassy SUV along the road through fields and small apple orchards. The sun was setting over the line of hills in a clear pink and red sky.
'No clouds tonight, thank God,' remarked Miles. 'Last time I came along here there was a downpour. I couldn't see a thing. Had to stop dead in the middle of the road.'
'Hmm,' said Bruno, who was sitting uneasily sideways in his seat, keeping an eye on the road behind them and looking from side to side.
'Don't worry,' said Miles. 'I'm sure it's OK. I told you, he's got this road monitored. It feels safe to me.'
'Hmm,' said Bruno again.
Miles drove on another few miles and then Bruno, who was peering out of the front windscreen, said, 'I thought you said there were no clouds tonight. What's that then?' He pointed to what looked like a small black cloud low in the sky ahead of them.
'It looks like smoke. It's just about where Donation's compound is. Perhaps they're burning rubbish.' But as they got nearer the cloud seemed to separate itself and gradually it became a moving mass of birds.
'Vultures,' said Bruno. 'Something's died.'
'Probably a cow or a buffalo. We'll soon find out. We're less than a mile away from the farm now.'
As they came up to the walls of the compound, another cloud of flapping vultures rose up to join those circling in the sky. Miles turned the car to go under the arch and then slammed on the brakes.
'My God,' shouted Bruno. 'What the hell's that?' A body clad in what had been white robes was swinging in the arch, dangling from a rope round its neck. Its face was a raw mass of bloodied flesh and its eyes had been pecked out. The legs, swinging in mid-air, ended in shiny black leather shoes.
'It's Donation's son.' Miles's voice shook.
'Turn round,' yelled Bruno. 'Let's get out of here.'
'Donation may be inside. He may need our help.'
'If he's in there,' said Bruno, 'he's long past our help. Can't you see? It's a warning. Go on, get out or we'll be next.'
Suddenly Miles jerked into action. With squealing tyres throwing up sand and stones he turned the car and drove off, back down the road they had come along.
Bruno was leaning forward now, holding on to the dashboard. 'I thought you said they had security on this road.'
'That's what Donation told me and I believed him. I thought they knew what they were doing. It all seemed very casual but I figured they were the best judge of what was safe. I bet it's that bloody French surveillance operation that's blown it. The guy in Paris knew he was being followed, so he knew there'd been a leak and they've traced it back to Donation and his son.'
They fell silent, each thinking over the implications of what had happened. Miles drove fast, bouncing the heavy car over the ruts in the road, while Bruno kept a sharp eye on the fields to each side. The light was fading now as Bruno turned to look over his shoulder at the road behind them.
'How much further?'
'About six miles.'
'Well, get a move on. There's company behind us.'
'I know. It came out of a field track just back there.'
A battered-looking pickup truck was approaching at high speed. As it got nearer two men in black balaclavas stood up in the back, each waving a heavy weapon in one hand.
Miles had his foot on the floor but the pickup truck was gaining on them. 'Hold on,' shouted Miles, 'I'm gonna knock them off,' and as the pickup drew alongside them, he turned the wheel of the SUV hard to the left.
But the truck driver had anticipated the manoeuvre and with a burst of acceleration managed to block their sideways move. There was a loud bang as metal hit metal, and the two vehicles each did a sweeping one-eighty and came to a halt side by side, slewed along the road.
The two armed men leapt down and pulled open Miles's door.
'Get out. Both of you,' said one in an accent straight from the streets of south London.
The two climbed out of the SUV, and the man with the London accent motioned with his rifle for them to move away from the car. 'Get down on your knees,' he ordered, and when Bruno hesitated he pointed the gun at his head. 'Get down, I said.'
As they knelt on the sandy road, Miles glanced at Bruno. He had clasped his hands behind his head and was staring straight ahead. Miles knew he was waiting for the shot. Then they'll shoot me, he thought. There was silence for a moment. A breeze had picked up, bringing a faint smell of petrol from the pickup truck. Behind them one of the men moved close; Miles could hear him breathing, noisily and fast. This is it, thought Miles, trying to come up with something meaningful for his final thought.
But then the Londoner spoke again. 'This is a warning. Keep out of our business and go home or you'll end up like that corpse at the farm. Now get back in that car and bugger off.'
And as Miles got slowly to his feet, he saw the man and his colleague leap back into their pickup truck. The engine started, the truck turned in a cloud of dust and drove back along the road the way it had come.
Miles stood with Bruno in the road for a moment, looking after the rapidly disappearing truck.
'What on earth was that all about?' said Bruno, his voice shaking very slightly. 'Why did they let us go?'
'Are you complaining?' asked Miles with a tremulous laugh. 'Perhaps they've got too much going on to want two dead diplomats on their hands.'
Bruno said, 'Maybe that's it. We've been lucky this time. Let's get the hell back to your Embassy.'
Chapter 20
'I need a drink,' said Miles as he parked the dusty SUV in the car park underneath the US Embassy. 'Come on up. I've got a bottle of Scotch in my cupboard.'
As he was getting the bottle and glasses out, Miles's eye fell on a piece of paper propped up on his desk. He read its message out loud: The Ambassador would like to see you in his office as soon as you get back.
Looking at Bruno he said, 'Something must have happened. I have a regular meeting with him on Monday mornings and he never asks to see me otherwise.'
'Surely he won't still be in his office at this hour,' said Bruno. 'Sit down and drink up. You've deserved it.'
But Ambassador Thomas B. Rodgers III, not a man to leave his post when there was still business to do, was at his desk.
'Come in, young man,' he called out as Miles appeared in his outer office. 'I've had a complaint about you.'
Ambassador Rodgers was a State Department professional. Sana'a was a tough posting, potentially dangerous, requiring diplomatic skills; not the sort of plum Embassy that presidents gave as a reward to their business friends and supporters. Thomas B. Rodgers had been round the block a few times, served in more junior posts in some tough places, and now in his mid-fifties had made it to Ambassador. He was used to dealing with the CIA.
'I'm sorry to hear that, sir.' Miles's voice was calm but his heart lurched. He hadn't yet made up his mind what, if anything, he was going to say about the events of this evening. He knew for certain that if the Ambassador found out that not only had he nearly got himself kidnapped or killed, but that he had led a British colleague into the same danger, there'd be a request to Langley for his withdrawal. Yet surely the news couldn't have got back to the Embassy so quickly.
'It concerns Minister Baakrime. You told me that you were hoping to use him as a source of information on arms supplies. Well, you should know that your contact with him has been noticed by the Yemenis, and I've been warned that we should steer clear of him. Other members of the government don't trust him. He's been making too much money on the side.' He waved an exasperated hand. 'I know, I know, most of them are at it in one way or another, but he's been making more than other people.'
'I see,' said Miles, wondering what else the Ambassador had been told.
'I don't know how much you know about him, but apparently he's working with the Russians.'
'With the Russians?' Miles was taken off guard and his surprise showed. 'No. I didn't know that. What's he doing for them?'
'I wasn't told. But probably much the same as you were hoping he'd do for you. Whatever it is, he's visited the Caucasus twice in the past year. Dagestan apparently. God knows what for, but whatever it is it seems to be making Minister Baakrime a lot of money. I'd be grateful if you'd steer clear of him from now on. I think he may shortly find himself in prison.'
If he's not already dead, thought Miles, remembering the hideous sight of the Minister's son, dangling in the entrance to the farm.
Chapter 21
It was hard work trying to extract any useful information from Milraud. It had needed frequent reminders from Martin that Annette's treatment depended on his cooperation to get him to fill in any of the details; even then he could only be described as a reluctant witness.
Eventually Liz had got him to admit that the Arab had got in touch with him via a contact in Yemen – a man who had put business his way before. He did not know his identity, he'd said, or who the Arab was – he never asked such questions. The request had been for comparatively small arms, as he'd said at the beginning, and he had been told these were for use by rebel groups in the Arab Spring countries. He had assumed this meant Syria, but he had not asked. It was not his concern. The Arab had said that the arms were to be delivered to Dagestan, one of the former Soviet republics, from where they would be moved on to their destination. He'd quoted an inflated price for the deal and there had been a bit of haggling, but he was very pleased with the final bargain they'd struck.
When Liz asked if he was not surprised that the delivery was to be to Dagestan, he'd said that nothing surprised him. He had both delivered arms to Dagestan before and bought arms there. When she asked more about the black man he'd met in Berlin, all he would say was that the Arab had asked him to meet the man – who he guessed must be arranging the onward shipment, though he couldn't be sure of this as the man was so jumpy they had had no significant conversation.
As Martin drove her to the Gare du Nord to catch the last Eurostar to London, Liz was mulling over all this.
'You know,' she said after a while, 'I don't believe a quarter of what Milraud said. The trouble is, I'm so tired I can't work it out.'
'I can't say I'm surprised to hear you say that. Milraud's not one to give up easily. It sounded unlikely to me too; I'm sure some vital parts are missing. I just don't believe he wanders around the world having meetings with people he doesn't know anything about. He wouldn't have lasted as long as he has, with me on his tail, if that's how he did business.'
'I know. And I can't understand why the Arab Spring rebels would want to buy small weapons at a high price from someone like him. Surely they are getting all they need from Iran and Hezbollah and the like.'
'Why don't you stay the night and we can talk about it in the morning?'
'I'd love to, but I can't. Peggy rang to say there was some new information about the black man. One of the Special Branches think they know who he is.'
'Let Peggy deal with it,' he said, as he stopped the car at the station.
She touched his hand on the wheel. 'No. I want to do it myself. I want to be sure Monsieur Milraud isn't going to get away with anything now we've got him. For your sake, as well as my own.'
She kissed him on the cheek, jumped out of the car and was gone into the station before he could say anything.
Liz got up early in the morning and was at work by eight. Peggy Kinsolving, another early riser, was already there at her desk in the open-plan office.
'Here's the number to call,' Peggy said, handing Liz a piece of paper. 'It's DS Halliday from Cheshire Special Branch. He said he's fairly sure he knows the black man.'
Halliday wasn't in his office until ten, but when he answered the phone he sounded cheerful and eager to help. 'I've had your photo. I'm pretty certain I know your guy. It looks like Lester Jackson, who owns a club in Wilmslow. I'll send you one of our pictures of him, so you can see what you think. He's well known to me and my colleagues.'
'Tell me more.'
'He's a tried and true bad guy, involved in trafficking drugs and women. But the frustrating thing is we've never managed to pin anything on him – not a single thing. The only trouble he's been in that I know of was years ago. Some teenage scrapes, and one arrest for burglary – but he was underage, and I don't think he even saw the inside of a young offenders' institution. He's never done time as an adult.'
'You say he owns a club. What sort of club?'
'It's called Slim's. In Wilmslow, which is in my bailiwick here in Cheshire. He gets quite a lot of the football fraternity in the restaurant and there's gambling and girls, and drugs, of course. Sometimes it gets a bit wild at the weekends but nothing too bad, just some young footballer drinking too much or snorting too much coke and getting involved with the paparazzi.
'There's an upstairs operation as well, with girls providing special services, as you might say, but we've had no complaints and we've never bothered them up to now. Recently Immigration have been sniffing around. They've a strong suspicion that some of the girls may have been trafficked, probably from Eastern Europe, and they think he may be selling women on, because his own upstairs operation isn't very big. Between you and me they're planning a raid pretty soon and I'm helping them. I've got my eye on one of the girls as a possible inside source. The club's in Cheshire, like I said, just inside our border, but Jackson lives in Greater Manchester's area. You should talk to them; they know him pretty well. How's he come across your radar anyway?'
Liz said cautiously, 'We're investigating a dodgy-looking arms deal on the Continent and it's possible he may be involved.'
'Guns? Jackson's crooked as a dog's hind leg, but as far as I know he's never sold weapons. Still, there's always a first time – he's not somebody who would turn down an opportunity.'
'If I wanted the Manchester angle who should I contact?'
'You should probably call the Deputy Head of Special Branch there.' His voice sounded unenthusiastic.
'Not the Head then?'
'No, he's new. It's his deputy who knows Jackson. He says he's been helpful in the past.'
'What. You mean he's a source?'
'I wouldn't go that far. But you're better off getting the story from him.'
Halliday sounded oddly wary and Liz decided not to press the point. 'OK, the Deputy Head it is. What's his name?'
'McManus. Do you want me to ring him first?'
'Not Jimmy McManus?' said Liz before she could stop herself.
'Yes. That's him. Do you know him?'
'No, not really,' she said, trying to recover from the surprise. 'I met him quite a time ago. I'll ring him myself,' she added, though her heart was sinking at the prospect.
When the photographs came through Liz looked at them carefully, trying not to jump to conclusions. Some had been taken in the street, some in what looked like a restaurant but was probably the club. But there wasn't any doubt – it was the same man. The same handsome face, with wide-set thin eyes, a sharp chin made sharper by the width of the high cheekbones. Afro-Caribbean, almost African but lighter-skinned, just the dark side of café au lait. Hair neatly cropped and, in all the pictures, very smartly dressed.
'What do you think?' asked Peggy, looking over Liz's shoulder, unwilling to hope for too much. 'Could it be our chap?'
'"Could be" is the understatement of all time. He's our man all right.'
'But do we have any real evidence he's one of the bad guys? Maybe he's just a respectable businessman holidaying in Berlin.'
'No. Milraud admitted he had a rendezvous with him and that the mysterious Arab set it up. What he hasn't told us is why he met him and what they said to each other – nothing, according to him, except to arrange another meeting, but I don't believe it. That's just one of the things he's holding back. So far we don't have anything on Mr Jackson, and the Germans couldn't hold him just for standing in front of a picture in a gallery, but I'm convinced he's in it up to his neck. A Mercedes that comes out of nowhere, a private jet that diverts to God knows where, and most of all the contact with Milraud – that's enough for me. And Halliday says he's a tried and true bad guy.'
She looked at Peggy, who seemed convinced. 'Now,' said Liz, looking pointedly at her phone, 'I've got someone else to ring to try and find out more.' And Peggy took the hint and left Liz alone to make the call.
Chapter 22
'Special Branch. McManus speaking.'
The voice was familiar, even after all these years, but it was more subdued, as if its owner had lost some vitality. Liz said brightly, 'Hello there, it's Liz Carlyle from MI5. I'm assuming I don't have to say "remember me?"'
There was a long pause, followed by the quick sharp laugh she remembered well. 'You can say that again. Hello, Liz. I take it this is a business call.'
You bet it is, she thought firmly. 'I sent round a photograph recently asking for information. I'm surprised I didn't hear from you. It's been identified as one Lester Jackson. Apparently you know the man.'
There was another, shorter pause.
'Yes, I do. I didn't see your photograph. What's he gone and done now?'
'I was hoping you'd tell me. Has he got form?'
'Strictly speaking no. But this isn't Little Lord Fauntleroy you're asking about. Why are you looking at him?'
'He's cropped up possibly in contact with someone we're investigating on the Continent,' she said cautiously. 'We're trying to work out what role he might be playing.'
There was another pause, then McManus said, 'I would have thought the Continent was a step too far for our friend Jackson.'
'Oh really. Why's that?'
'Frankly, this guy is not the sharpest knife in the box. He's home-grown and strictly a small-time villain. On his own patch he does OK, and most of his business is legit – his club has its dodgy angles but the restaurant's not bad. To tell you the truth, there're a few shenanigans that go on upstairs, but nothing to get excited about. I'm surprised to find him showing up on your radar.'
'Your colleagues over in Cheshire seem to take a different view.'
'You must mean Halliday.' McManus gave a derisory snort. 'He's a young man who gets a bit over-excited. Not much goes on in Cheshire and he's got a bee in his bonnet about the club. He's cross that he's never managed to get anything on Jackson.'
'He said Jackson was a source of yours.'
'Is that what he called him?' McManus laughed, but there was nothing amused about its tone. 'Listen, the guy's helped me out on a few occasions, pointed me the right way when I was bringing down the coke traffickers in this town. He's done enough for us that we leave him alone.'
I get it, thought Liz angrily. Let Jackson traffic in women in return for helping out once in a while with drugs. Drugs got the headlines, while prostitution was just seen as a necessary evil – however many lives it ruined, however many women it kept in a kind of slavery. 'So why was he in Berlin then?' she asked. Immediately the words were out of her mouth she wished she hadn't been so specific.
'I haven't a clue. But believe me, if he's got himself tangled up in something big-time, Jackson is not playing a large role in it. He's small beer, Liz. Honestly.'
'OK. Thanks for letting me know.'
She paused for a second, feeling awkward. Then McManus said, his voice softening, 'It's been a long time. So how goes life for you?'
'Good, thanks. Same employer, as you can see.'
McManus laughed. 'I always had you down as a lifer. You had the talent, and the commitment. I wouldn't be surprised if you end up running the whole shebang one day.'
'Don't count on it.' McManus had always been a charmer when he wanted to be. 'But what about you? You must like Manchester if you're still there.'
'Like? I don't know about that.' His voice was flatter now. 'It's a living. I can't complain.'
'Oh.' It wasn't the answer she'd expected. 'Well, I'd better get moving; we've got our weekly brief in a minute. Thanks for the info.'
'Any time.'
She didn't like leaving it like this. She said, wanting to give the conversation a better ending, 'I may have to come up to your part of the world. If I do, I'll drop in and say hello.'
'You do that. It would be good to see you again.' Then he added, 'Just don't make a special trip on account of Lester Jackson. Take my word for it, the guy's nothing for you to worry about.'
Putting the phone down, Liz felt troubled by the conversation. She stood up and went over to the window, looking down as a small tug chugged along the river. The Thames was lifeless-looking and grey under the overcast sky of late autumn. His account of Lester Jackson just hadn't rung true. 'Small beer; not the sharpest knife in the box' did not describe the elegantly dressed man who, according to the Germans, had strolled into the Schweiber Museum, had conducted quite clever counter-surveillance, had been whisked off the street by a Mercedes, picked up by a private plane and collected by yet another limousine from a private airfield.
Why had McManus tried to downplay Jackson's importance? Come to that, why had he not responded to the photograph Peggy had circulated to all Special Branches? He must have received it and known very well who it was.
She thought back to the McManus she had known years ago and the reason she had split up with him. Then he had been prepared to bend the rules in his pursuit of criminals who he was convinced were guilty, even when he couldn't prove it. Was he now bending the rules in pursuit of something else? His own interests perhaps?
And McManus had exhibited all the verbal tics of the practised liar – 'honestly', 'believe me', 'to tell you the truth', and 'frankly'. She realised that she didn't believe a word he'd said about Lester Jackson, and now she was worried that in talking to him she had given too much away.
Chapter 23
Katya knew all about the police in her country – they were armed and violent and sometimes if you paid them enough they would go away – but she didn't know about the British police. People said they were different, but those were normal people, people who were in the country legally, people with the right stamps in their passports, people who had genuine passports. Not people like her for whom the smallest brush with the authorities could mean disaster.
So when a young man knocked at the door of the house where she rented a room and said he was a policeman, an icy panic gripped her. He flashed an identity card so quickly that she couldn't have seen it even if her eyes had been working properly. She'd been woken by his knock and was still half asleep as well as scared.
'Detective Sergeant Halliday,' he said. 'Can I come in?' and before she could say anything he pushed past her and went into the lounge. The three other girls who lived in the house had all gone out to work. They had nine-to-five jobs, but Katya got home at four o'clock in the morning and usually slept till the early afternoon.
The lounge was in a mess. One of the girls slept there on the sofa and she'd left her clothes and underwear scattered on the floor. Halliday sat down on the one chair while Katya stood in the doorway in her nightclothes and nervously waited for him to say something.
'I expect you know what this is about, love,' he said with a smile that was only superficially friendly. He seemed young to be a detective; his hair was spiky and shiny with wax, like the kids she saw sometimes on her way home, coming out of the clubs.
She didn't say anything and he laughed. 'Come on, Katya. Speak to me.'
'Just tell me what you want,' she said, not that she had much doubt. He must know she was there illegally, without proper papers, and she feared the worst – deportation back to Dagestan, the country she had been so happy to leave. But if he'd come to arrest her, why was he on his own? It seemed odd.
'I'm interested in your place of work.'
'Slim's?'
'That's right, love. You work upstairs, don't you?'
'Yes.'
'Funny kind of place, Slim's. I mean, it's a club downstairs, full of respectable citizens having dinner and a drink or two and a dance. But if someone wants a special dessert they can get it upstairs.'
Katya said nothing, wondering what he was getting at. She didn't know whether what went on upstairs in Slim's was legal or not, all she knew was that neither she nor any of the girls who worked there had the right papers. But if he was inquiring into what went on in the club, why had he come to her? She didn't run the place. Whatever he wanted, she wished he'd get on with it. But his next remark gave her a shock. 'How well do you know Mr Jackson?'
She shrugged. 'He is there most nights, but he doesn't often speak to the staff.'
Halliday sneered. 'Oh, so he's too grand to talk to the people who help make him rich.'
She didn't reply; the less she said the better. She must do nothing to rouse his interest and then he might go away. She knew Jackson, of course, but as a daunting presence rather than as someone you could talk to. He was the owner of the club, with the power to hire and fire. But it was more than that – he owned them, the girls, and she had no doubt that he was behind the operation that brought them into the country.
The girls in the upstairs room at Slim's were stunners – the prettiest girls their home country had to offer. Katya was proud of this, since part of her job was selecting the girls who got brought over. For that, she had to travel to Dagestan from time to time, and when she did she used a false passport that was given to her for the journey, then taken away. It said she was Bulgarian. The girls she recruited came to the UK in a lorry; she knew that from talking to them when they got here. The other part of her job was managing the girls once they arrived.
There was an air of menace about the man Jackson; behind his stylish clothes and cool manner she sensed a brutality that scared her. The other girls saw it too, though as far as she knew he had never hurt any of them.
There was another strange thing about him. In Katya's experience any owner would have occasionally sampled the goods; that was a right that came with the territory. But not Jackson; he never talked to any of the girls, let alone touched them, and he only occasionally had a word with Katya, just to check that the customers were happy and that there had been no complaints. There never had been and he seemed satisfied, but she still found him frightening.
Halliday's breezy manner had changed. His voice sounded ominous when he said, 'Your employer is about to find himself brought down a peg or two.'
'Oh?' said Katya.
'Yes. And you're going to help me do it.'
Chapter 24
The two men sat in a dimly lit alcove on the raised dais at the back of the dining room. Slim's, named after Joe Slim, the Manchester United footballer who'd started the club eight years earlier, was in Wilmslow, ten miles or so south of Manchester. It was said that the Aston Martin dealership in Wilmslow sold the highest number of Aston Martins in the UK, so affluent was the local lifestyle. The room was crowded this evening, loud with music and the raised voices of a group of young men and girls at a long table. One of the two men looked around and smiled in satisfaction at the packed tables.
He was the owner, a tall black man known as Jackson. No one at the club ever used his first name. Jackson had acquired the club after Joe Slim was found, early one morning, face down in the Manchester Ship Canal. It was generally assumed that he had fallen in from the towpath while he was drunk, but no one seemed to know why he was down there and no witnesses had ever come forward.
Jackson dressed as smartly as his well-heeled clients, and tonight he wore an elegant blue suit, a cream-coloured woven shirt, and a subtly patterned Hermès tie. His companion was less flashy but his suit looked equally pricey; he had the air of a successful self-made businessman – the kind of man who paid in cash from a roll of banknotes held by a silver money clip.
'Good trip then?' asked the man who looked like a businessman.
Jackson gave him a quizzical look, then seemed to decide the question was innocent enough. 'Not bad. Though I had a spot of bother with the locals. I don't know what sparked them off but they seemed to be wondering what this uppity nigger was doing over there.' Jackson chuckled. 'They didn't find out though.'
'What were you doing over there? Was it business?'
Jackson laughed sarcastically. 'I wasn't in Berlin for my health, man. I was chasing up a new opportunity.'
'German girls?'
Jackson shook his head. 'I'm getting tired of that line of work – too many hassles. I'm thinking of branching out a bit.'
When he didn't elaborate, the other man said, 'Well, it must have been important if you took a chance like that.'
'What chance?'
The other man shrugged. 'You don't want to get European police forces on your tail. They can be a bit nasty. Just watch out if you're up to something dodgy over there.'
Jackson said nothing at first. Then, 'I don't know if it was the police. I didn't see any uniforms.'
The other man said, 'But you got out all right?'
Jackson looked amused. 'I'm here, aren't I?'
'It looks that way to me,' said the other man. His role there was hard to place. He didn't act like a customer; he was too self-confident to be a dependant; yet the black man didn't seem the type to have friends.
'Anyway,' said Jackson, 'when are they coming?'
The businessman looked at his watch. 'Any time now.'
And as if in response, the maître d' came up to their table, looking agitated. 'Mr Jackson,' he said breathlessly. 'There are Immigration officers outside the back door. They're asking for you.'
Jackson raised his eyes but didn't seem surprised. 'Thank you, Émile.'
The maître d' went on, 'They have police officers with them. They say they want to check the papers of the girls.'
Jackson looked at his companion, who also didn't seem surprised by Émile's news. Jackson said to him, 'You better excuse me. I like to leave by the front door of the places I own.' He turned to Émile. 'This gentleman's my guest, so put our dinner on the house tab.'
'Of course, Mr Jackson. But what should I tell the police?'
'Tell them if they want to see me they need to make an appointment. Like my guest here,' he added with a smile. And then, without any show of haste, Jackson was out of the front door of the club in ten seconds, leaving Émile to deal with the officers of the law. Jackson's guest remained seated at the table, and after a moment signalled for a waiter and calmly ordered a large cognac.
Chapter 25
She had seen Halliday twice and each time he had pumped her about the upstairs operation at Slim's. She'd explained that she didn't know any details of the business; once a week Khoury, the accountant, showed up, and sat in the little room next to the cloakroom, where he went through all the tabs the girls had handed in. But he didn't talk to Katya about the business, and she certainly didn't know the turnover figures. She only knew that none of the girls, even the desperate ones, dared to try and skim any of the money. They handed it all over. That's how scared they were of Mr Jackson.
At their second meeting, Halliday had told her about the coming raid. 'Not a word to anyone,' he'd said. 'They'll be picking up the girls, and that'll include you. But don't worry – I'll see you right.'
And he had been true to his word – too true for Katya's liking. The police and Immigration officers had come in the back entrance, quite politely. This had seemed curious to her – she'd expected something like the movies, with armed officers breaking down the door, waving guns and shouting as they forced their way in. But instead they had waited outside, only four of them, in plain clothes, while the bouncer had called Émile, the maitre d' of the restaurant, who had come back and let them in.
It had all been tidily done, and despite the initial panic of the upstairs customers – that early in the evening, there had only been one group of stag-night revellers and a sad-looking man who said his wife had recently died – it was soon clear that the police were only interested in the accountant's room, where one of them had gone right away, and in the employees. None of them was English, of course, and none of them had papers as far as Katya knew – not a National Insurance card, a driving licence, or anything at all.
There had been seven girls working that night, including Katya (though her job was to supervise the goings on, not participate), and they had all been escorted out to the police van while Émile had wrung his hands and promised that he would have them out as soon as he could locate the club's lawyer.
It hadn't been quite that easy – by three in the morning there had still been no sign of the lawyer, and until the preliminary hearing scheduled for noon there didn't seem much chance of any of them getting released. It didn't matter much, since they were in a long row of adjacent cells, and the girls – however frightened they really were – took things in good part, calling out to each other, whistling to keep their spirits up, even briefly having a sing-along, until the duty sergeant came along and told them to pipe down. After which they curled up on their respective bunks and settled in for the night.
Which made Katya's release so noticeable, since alone among the seven of them she got called, by the same duty sergeant, and brought out from the cell. 'You're free to go,' the officer said grumpily, giving her back the small handbag she'd been made to hand over when they'd booked her and the other girls.
'What about my friends?'
'What about them?'
'Aren't they getting out too?'
The duty sergeant shook his head. 'Not as far as I know. What's the problem – do you want to go back to your cell?'
It would have been better if she had, Katya reflected, as she opened the front door to her home. Her housemates were still asleep and she crept in quietly. It wasn't that much later – or earlier depending on your time frame – than her usual return home after the club's closing.
She was worried about being the only one released. It must have been Halliday's doing, she decided. Hadn't he told her he would look after her? But she wished he hadn't done it this way. The other girls were bound to wonder what set her apart. They were loyal to her, but only up to a point. She did her best to look after them, and she could protect them from the drunks or abusers or the ones who didn't want to pay. But there was no disguising the nature of the operation upstairs in Slim's, and any girl who thought her duties stopped at serving drinks didn't last long – which meant an ignominious return to Dagestan, since none of them had papers that would enable them to find a proper job.
Katya knew this, since she had come to the UK originally by the same route and found herself in the same position. Back in Dagestan the offer had seemed irresistible: Come and be a hostess in a deluxe club in glamorous Manchester. Earn five times the wage you are earning now. Meet interesting powerful people and live a life of Western luxury.
She knew the come-on lines by heart now, since she used them herself when she was sent home to recruit new girls. Occasionally when she was interviewing a girl who seemed particularly sweet and likeable, she tried to hint that perhaps the job wouldn't be quite what it seemed; that maybe the girl should have a long think about what was on offer before accepting. But Dagestan was dire; no one under the age of thirty could see their future in a positive light, and the girls she saw didn't want to know that the West was not the land of milk and honey; that men could be just as exploitative in Manchester as in Makhachkala; that the money on offer would go mainly to pay the rent charged for their squalid shared rooms, or in 'fees' for nebulous services to the owner of Slim's.
Now she hesitated for a moment in the hall, wondering if she should have a cup of tea, then decided to go upstairs. Normally she would head straight to bed, but she felt grimy after her time in the cells, and went and ran a bath, closing the bathroom door so as not to wake the other girls in the house. She was pleasantly surprised to see that Michele, the French girl who always had a bath late at night, had for once cleaned the bath after using it.
Katya lay and soaked for a while, wondering if by now the other girls had been let out. She hoped so, as otherwise she knew they would be wondering why Katya had been released. And if word got round it would be certain to be picked up by Émile – he was cat-like, that man, always lurking nearby, avid for gossip. And Émile would never keep news like that to himself, which meant he would tell . . . Katya shuddered, and quickly got out of the bath.
She managed to fall asleep, half waking when the other residents of the little house got up and went to work, then falling asleep again. When she finally rose it was just past noon.
Downstairs the kitchen was in its usual post-breakfast disarray – used cereal bowls and half-drunk mugs of tea and coffee. She opened the back door to air the place and started tidying up – her housemates were younger than Katya, and, just as at work, she looked after them. Maybe someday she'd have her own children to look after, but in the meantime she did not mind looking after girls younger and less worldly than herself. Michele in particular seemed in need of sisterly advice, especially when she expressed interest in ditching her boring secretarial job and coming to work at Slim's, something Katya had so far managed to steer her away from.
She had just finished with the leftover dishes when she heard the postman push the mail through the letter box slot. No point rushing to see it; bills and more bills would be lying on the mat. But a minute later she thought she heard a tap at the front door, so this time she left the kitchen and went out into the hall. She opened the front door, but there was nobody there – she must have imagined the noise. Then she bent down and picked up the post, examining it as she walked back to the kitchen. There was one envelope for her, which she was opening when she came into the kitchen again, not paying much attention. It took a moment to notice the man now sitting at the kitchen table, and she jumped when she saw him.
'You startled me,' she said, feeling flustered at first, then fearful as she realised who it was.
'Did I now?' said Lester Jackson mildly. 'Maybe you were expecting the police instead of me.'
'Why would I be expecting them?' Katya managed to say.
Jackson shrugged. 'There has to be someone in the force you're friends with. Seeing as you were the first one sprung last night. Why don't you sit down and tell me who your friend is?'
He gestured at the chair next to him, and Katya stiffened. 'I would, Mr Jackson, but I have to go out now . . .'
Jackson was smiling as he shook his head. 'I don't think so. Sit down, Katya,' and there was something so steely in his voice that reluctantly she did.
'Now,' Jackson continued, his voice mild again. 'Was it DCI Lansley or DCI Robertson? Or is it your friend from Special Branch – maybe Detective Halliday?'
When the French girl Michele came back from work later that day she was surprised to find the front door of the house unlocked. She went in and called out for Katya, who was usually at home at this time, getting dressed for work at the club. Michele didn't care what Katya said; Slim's sounded fun, and a thousand times more exciting than her own job, typing the correspondence of a fat and unsuccessful property developer. She was going to tackle Katya again about it; Michele knew she was attractive enough to work at Slim's – it was only the older woman who was standing in her way.
'Katya,' she called out, but there was no reply. Funny that, thought Michele, as she walked towards the kitchen at the back of the house. She couldn't remember the last time Katya wasn't at home when she came back from work.
And sure enough, Katya was at home – and in the kitchen too. It was when Michele found her that the screaming started.
Chapter 26
Dinner alone. God knows, Martin was used to it, but it seemed strange to have seen Liz only so briefly in Paris, considering how close they had become. He knew that she found it awkward to be working so closely with him, and particularly to be the cause of delaying what she knew Martin had been working for years to achieve, the trial and conviction of Antoine Milraud. He saw that she had been relieved to go straight back to London the other evening after seeing Milraud, and though he understood the reason why, it made him sad.
He hoped the slight chill in their relationship was only temporary. Even in the fog that seemed to distort everything connected with Milraud, he knew that Liz promised a happy life ahead and Milraud only represented the past.
He was annoyed when the phone rang on the table in his study and broke into his reverie. It turned out to be his young colleague from the safe house in Montreuil, Jacques Thibault.
'Yes? What is it?' he asked sharply.
'He's had an email.'
Seurat was alert now. 'What did it say?'
'It's calling him to another meeting – in London. It's encoded in the form he described when your British colleague was here. He says it's from the Arab.'
So Liz was right, and the UK connection was proving key. 'When's the meeting and where?'
'Two days from now. We're working out exactly where but I wanted to tell you straightaway. The instructions are in the form of coordinates disguised as sports scores. As soon as we've unzipped it, I'll let you know.'
'Do you know the time?'
'Four o'clock in the afternoon.'
Dusk at this time of year, which would make surveillance of the meeting more difficult. 'I'll let London know. Contact me as soon as you've worked out all the details.'
'OK. I'll get back to you shortly.'
Young Thibault was a computer genius, a real geek, thought Martin. Let's hope he can get more out of that message than the time and place of the meeting.
Chapter 27
When the all-clear came through from the A4 team looking for counter-surveillance, Milraud was let out of the car. He walked along Regent's Park Road, and turned left through the open gate of Primrose Hill Park. Eight pairs of eyes watched him go.
The light was fading now after a bright late-autumn day. It was 3.45 in the afternoon and it would be practically dark by 4.30 at this time of the year. Maureen Hayes, sitting in an apparently closed up and deserted park-keepers' shed, was observing Milraud's progress across the park. His light-coloured raincoat made him easy to spot as he sat down on a bench at the top of the hill. She didn't envy him sitting out there on this chilly evening.
His was the only bench occupied; the wind was getting up and everyone else in the park seemed to be hurrying home. A woman in a fake fur coat was dawdling along, holding a little plastic bag in one hand and apparently urging the terrier she had on an extending lead to do his business so they could leave. Three small boys in school uniform went out of the gate chattering, one holding a football under his arm. A faint aroma of burning leaves seeped through the wooden slats of Maureen's hut.
For a second the setting sun caught a window of one of the tall glass buildings somewhere in the City to the south, and a flash of brilliance lit up Milraud's figure, sitting alone at the top of the hill, and momentarily blinded Maureen as she peered at him through her binoculars.
When she could see again, she noticed several people were walking into the park through the same gate Milraud had used. Perhaps an underground train had just come in or maybe they'd got off a bus. Then, as she watched, a young man separated from the others and turned up the path that led to the seat where Milraud still waited.
Could this be Zara, as Milraud's Arab contact was now codenamed? She had been told to expect a tall, thin young Arab, dressed scruffily like a student. But this young man was wearing a dark business suit and carrying a briefcase and a rolled-up copy of the Evening Standard. He was tall and thin all right, and dark-skinned, but he looked more like a City worker returning to his flat in this expensive part of London than a student or a jihadi.
The man was passing Milraud without a glance, when he suddenly stopped, and seemed to be admiring the view. To Maureen's practised eye he was looking for signs of surveillance. Then he stepped behind Milraud's bench and seemed to be rubbing his hands up and down the Frenchman's back. Maureen stared at them through her binoculars, thinking that in other circumstances this would look like some kind of gay encounter.
The newcomer slowly circled around the bench and sat down at the far end from Milraud. There was a pause and the two men seemed to be talking. Then Milraud got up and took off his raincoat, folding it and laying it on the bench. Again the smart young man appeared to be stroking Milraud's body, his chest this time. Whatever was going on? After a short time, Milraud got up, put his raincoat on again and the two men conversed, apparently calmly. After a further ten minutes, the young man stood up and walked away down the hill, in the opposite direction from which he had come, and Milraud retraced his steps to the waiting car.
As Zara headed for the far gate, Maureen alerted the Ops Room, and as he left the park six of Maureen's colleagues were on his tail.
Chapter 28
Liz had had a bit of a struggle persuading the Home Office that she had enough on Lester Jackson to justify a warrant to intercept his communications. On the face of it, a small-time Manchester club owner with no criminal record, let alone any proved involvement in terrorist-related activity, did not present any threat to national security. She had argued strongly that his covert contact with Milraud, a man well known to the French as an arms supplier, in an apparent plot to supply weapons to a group of jihadis, justified the warrant. Eventually she had won the day, but the warrant was to be reviewed after two weeks and if by then no information indicating a national security threat had emerged, it would be cancelled. She had come away from the meeting in Whitehall feeling disgruntled. Two weeks was a very short time in which to prove anything.
She was reading the first transcripts when her phone buzzed – an internal call. She picked it up, impatient at the interruption.
'Liz, you'd better come down.' It was Wally Woods in the A4 Operations room.
'What's happened?'
'Your Zara operation. The meeting took place and we've got the Frenchman back safely in our custody. That's all OK, but we're following Zara and I need to know how you want us to handle it.'
'Give me five minutes?'
'Make it three.'
She rang off and looked at the transcripts again. At 16:45 the day before, Jackson had taken a call on his mobile. The caller had been located two thousand miles away, though they still hadn't tracked the signal down specifically. The conversation had been in English, with the caller speaking fluently but with what sounded like a Russian accent. The transcript read:
Caller: It's Tag here.
Jackson: What's the state of play?
Caller: It's ready to go.
Jackson: There may be some more to come. But for now, have you got everything?
Caller: Yeah, all of it.
Jackson: Twenty pieces?
Caller: (Impatiently) Yes, yes. They all look good to me, though I'm no expert.
Jackson: Can you confirm the route?
Caller: Same as last time.
Jackson: Why not a different port?
Caller: That's up to me, my friend. Once I deliver, the shipment's all yours. Until then it's my worry.
Jackson: Have you got a date?
Caller: Not yet, but it won't be long now. We have some snow so it is hard to be more specific than that.
Jackson: I need 12 hours' warning.
Caller: I can do better than that – I'll give you 24.
Jackson: OK, I'll hold you to that.
Liz shook her head, trying to make sense of it, then got up and walked to the lifts in the centre of Thames House. As she went, she thought about the transcript. Given Jackson's background, it would be fair to assume the conversation was about human trafficking – the goods being East European women shipped over on a lorry for service in places like Slim's.
But something was wrong with that – Liz simply didn't believe twenty would be coming in one shipment, one lorry perhaps. Not to work at Slim's at any rate, where Halliday had explained only half a dozen women were on the game upstairs in the club. And even if Jackson was involved in trafficking women for other places, twenty pieces seemed an improbably large number at one time and an odd term ('pieces') to use, even if the caller was not speaking in his native language and was trying to be discreet. And wasn't it rather strange to say he was no expert, if he was talking about women?
So what on earth was Jackson importing? If it was guns, why only twenty, if they were then going to be re-sent to . . . God knows where? Was this what Milraud had been talking to him about?
She pondered all this as she walked along to the A4 Ops Room. Inside Wally Woods and two colleagues sat, headphones on, in front of a row of TV monitors. Wally was talking into the microphone on the desk and waved her to the battered old leather sofa just inside the door that was kept specially for visiting case officers. The Ops Room was Wally's domain and no one was welcome when an operation was going on except by invitation.
'Which side of Pentonville Road?' he asked the microphone.
Over the speaker a voice Liz recognised as Daley, a veteran surveillance officer, replied, 'South side and walking fast.'
'I have him,' said another voice, more muffled.
Wally kept his eye on the screen but spoke to Liz. 'This Zara's led us a pretty dance. He walked all the way to Great Portland Street station and went into the Tube. We had to rush in there, but then the bugger came out again and caught a bus.'
'Do you think he saw you?'
Wally shook his head. 'Don't think so. You told us to take extra care and we have. I just think he's been trained, and he's being extra careful too.'
'Where did he get off?'
'In Euston Road, by the British Library. He hung about for a bit – I think he wanted to see who else got off the bus. None of us was on it – I've got three cars on this so he was easy enough to follow. It must be the only time in my life I've been grateful for the traffic on the Euston Road.'
As they spoke, video pictures appeared on one of the TV screens of Zara walking up the Pentonville Road, just past King's Cross. It was a hazy picture, taken through the window of one of the surveillance cars, but Liz could clearly see the tall, dark-suited figure striding along the pavement. She watched as it turned and moved towards the entrance of a large building set back from the road. A group of young people were talking by the front door.
Maureen Hayes's voice came through the speaker. 'Zara entering a building. It looks like some sort of college. Groups of young people outside.'
Wally replied, 'Send Tia up to check it out.'
And as Liz watched, a young woman in a hooded jacket and headscarf walked to the front of the building. She threaded her way through the groups of chattering young people, went up the steps and inside.
She was gone about five minutes, and when she came out she said, 'It's called Dinwiddy House.'
Wally turned to Liz, who shrugged. Tia was saying, 'It's a hostel for students at London University. Most of them are at SOAS – School of African and Oriental Studies.'
It made sense. Zara was young, Middle Eastern, like any number of SOAS students.
'Any sign of Zara?' asked Wally.
'No. There's a common room and bar on the ground floor but I couldn't see him in there, though it was pretty crowded and I might have missed him. But I think he went upstairs. That's where their rooms are.'
Wally turned his swivel chair to face Liz. 'You want us to ask around a bit? Try and find out if he lives there?'
Liz shook her head. 'Too risky, especially if he comes downstairs again when you're asking questions. But I'd like an eye kept overnight, just in case he's only visiting. Maybe he's got a girlfriend there. Can you do that?'
Wally nodded. 'It'll be another team but I'll make sure they're well briefed. What do you want us to do if he leaves? Follow him?'
Liz nodded. 'Yes, please. And keep Peggy posted. I've got to go out now to debrief Milraud.' She stood up. 'Thanks, Wally. That's a great help. Now I've got some chance of finding out who this Zara is.'
Two hours later, after Peggy had made a series of urgent phone calls resulting in a senior university administrator being rooted out of his home to consult the file in his office, Liz knew. Zara did indeed live in the hostel known as Dinwiddy House, and was studying for a Masters degree in International Relations at SOAS. He was a Yemeni called Samara and was in the UK on a temporary students' visa. The address given on his visa application and supplied to the college was in Sana'a, the capital of Yemen. He hadn't drawn himself to the attention of the college authorities in any way and a search of the records in MI5 and MI6 came up 'No Trace'. But then, thought Liz ruefully, if this guy was any good, that's what you'd expect.
Chapter 29
The Royal Standard Hotel was in an undistinguished street between Victoria Station and Buckingham Palace. Though it billed itself as 'situated in the shadow of Buckingham Palace', it was in fact much nearer to Victoria Station. An anonymous sort of place, part of a small chain, it provided everything a mid-level businessman or official visiting London might require: wi-fi, cable TV with 'adult' films, in-room tea and coffee, minibar and even an ironing board and iron. All of its 361 rooms were furnished identically, and carpeted and upholstered in variations on the colour theme of beige and maroon.
All in all it was the sort of place where people could come and go without anyone taking much notice. Which is why, a few years ago, Liz Carlyle's colleagues had identified it as perfect for the sort of rendezvous they occasionally needed to conduct. The manager had been recruited as what was called a 'facilities agent', to provide a room or rooms as required, without asking any questions about who might occupy them or what might go on in them. In return he received a present at Christmas and the satisfaction of knowing that he was helping Her Majesty's Government.
On this occasion, two pairs of interconnecting rooms had been booked on different floors. In one of the pair on the eighth floor, Liz Carlyle was sitting, waiting for Dicky Soames, the burly A4 officer and member of the team 'minding' Milraud while he was in London, to produce his charge, so she could find out what had happened at the meeting on Primrose Hill.
There was a light tap on the door and a deep cockney voice said, 'Here we are. OK to come in?' Milraud entered followed closely by Soames, who closed the door firmly behind him and put the lock on.
'I'll be next door, if you want anything,' said Soames, and he went into the other room leaving the intercommunicating door slightly ajar.
Liz motioned the Frenchman to one of the two chairs. She thought how tired and strained he looked. Much more so than when she'd first met him in the safe house in Montreuil.
'Would you like something to drink?' she asked. 'There's tea or coffee, or a drink from the minibar if you'd prefer.'
Milraud shook his head. 'Non, merci,' he said shortly. He had kept on his mackintosh, and he looked chilled, even though it was warm in the room.
Liz switched the kettle on, and as she waited for it to boil, she pointed out of the window at the coloured lights strung across the street. 'Christmas starts earlier every year,' she said cheerfully. Milraud glanced out and nodded, but he seemed a million miles away. Liz took her time making coffee for herself and chatting inconsequentially, hoping to relax the man a little.
She sipped her coffee and winced. 'You made the right decision,' she said, but Milraud's smile was perfunctory. He was clearly impatient for the debrief to begin.
'So how did it go?' asked Liz, sitting down at last.
Milraud shrugged. 'Much as expected.'
'Was he concerned about security? I mean, since your Paris meeting was aborted because of the surveillance.'
Milraud sat up. 'Yes. He was worried that I might have been followed and he checked me out for a microphone. I assured him he need not worry; that I was once an intelligence officer and I know about these things.' He gave a wry smile. 'I explained that I had gone to Paris and Berlin under different passports and I was at least twelve hours ahead of anyone hunting me.' He grimaced; they both knew Milraud had thought this himself.
'Do you think he suspects you?'
'In his position I certainly would – I never trust my customers, so why should they trust me? But when he pressed me about being spotted in Paris, I told him I had as much right to worry about him as he had about me. That shut him up.'
'So after that, what did you discuss? He called the meeting, didn't he? What did he want to say to you?'
'He wanted to add to his order. That was for firearms, as you know.'
'What else does he want?'
'It's a bit surprising. He wants grenades – two dozen of them.'
'Really?' Liz was astonished. The whole business seemed surprising, as it was generally assumed that the jihadi groups fighting in the Arab Spring countries had no difficulty acquiring weapons from their supporters, but this requirement was even more unexpected.
'That's right. And then the oddest thing of all – he wants more ammunition for the weapons he'd ordered. Not more weapons; just more ammunition. Twenty thousand rounds.'
'Twenty thousand?' Liz could not contain her astonishment. It sounded as if Zara was equipping an infantry battalion. And why so much ammunition for only twenty weapons?
'I agree it doesn't make sense, unless he already has a lot of weapons at his disposal. But I didn't have that impression from our first meeting. It's quite peculiar.'
Milraud looked uneasy; Liz sensed there was something on his mind. She waited, but he said no more. Eventually she asked, 'Let's come back again to this black man you met in Berlin. What did he want?'
'I was asked to meet him. I was told he wanted to see who was involved in the deal. I was told he has not done this type of business before.'
'What did he say?'
'Almost nothing. He just asked about my business – how long I'd been supplying, what parts of the world I supplied, that sort of thing.'
'What did you tell him?'
'Very little, but it seemed to satisfy him. Then he rushed off. He was very jumpy.'
He still looked uncomfortable. Then he shrugged and returned to the subject of the meeting with Zara. 'Anyway, I wasn't sure how you wanted me to play it today. So I told him that I would check if I could get the goods he wanted in time and get back to him. He pressed me, so I had to promise to let him know tomorrow.'
'How are you to do that?'
'By email.'
She knew from Seurat that the French were in control of the email traffic.
Milraud asked, 'What do you want me to say?'
'Can you supply the extra things he wants in time?'
'Yes. I only have to email my supplier.'
'Where is he?'
'In Bulgaria.'
Liz didn't hesitate. 'Do it then and tell him you can fulfil the supplementary order. But also tell him you need to know precisely where and how it should be delivered. Press him for details.'
She looked at Milraud intently. He might have been surprised by Zara's request, but she was certain he was holding something back. It didn't make sense that he knew nothing about Jackson. Milraud was acting as if Jackson was Zara's contact and he had nothing to do with him, but she was sure that wasn't the case. Maybe if young Thibault over in Paris could hack into their back email exchanges the full truth would emerge – and a lot sooner than if she waited for Milraud to come clean.
Chapter 30
It was almost eight when Liz left the hotel. Milraud would be spending the night there in the other pair of interconnecting rooms, under the watchful eye of Dicky Soames and his colleagues, before returning to Paris with them as close escorts. There was no way Liz was going to be responsible for losing the man whom Martin Seurat had spent so many years hunting.
In the dark, Thames House looked like a lit-up half-filled egg box: unoccupied offices were dark, but enough officers worked late hours to dot the heavy masonry façade with the lights of their midnight oil. In her office Liz found a handwritten slip from Peggy: Halliday rang. Said call him any time. He has news.
When she reached Halliday there was the background noise of a raucous party going on. 'Hang on a minute,' he shouted. Gradually the noise subsided, until she could hear only traffic whizzing past in the background, tyres wet from rain. Halliday must have stepped outside from whatever club he was visiting. 'Sorry about that,' he said.
'It's Liz Carlyle; I got a message to ring you. But I don't want to interrupt the party.'
'I'm working, believe it or not. I'm drinking vodka and tonic without the vodka, and waiting for the barman to offer to sell me three grams of coke. I thought I'd better take your call outside. I've got some news for you. Not good, I'm afraid.'
'What's happened?'
'We raided Slim's with Immigration – that's the club owned by Lester Jackson. We arrested half a dozen girls working upstairs – they were "hostesses" but they were doing more than serving drinks. All from somewhere in Eastern Europe most likely but they didn't have a set of papers between them.
'Normally that would have been enough to close the place down, and maybe let me squeeze our high-flying friend Mr Jackson a bit. But he wasn't there and he didn't seem to care, and I now know why. He had a leading brief go to the lock-up by breakfast time, and bob's your uncle, it turned out all the girls had proper papers and valid passports – the solicitor claimed he'd been holding them on the girls' behalf.'
'What sort of passports?'
'Bulgarian – every one. And now that it's in the EU that means they can work here, come and go as they please. Not that I believe for a minute their papers were kosher. None of those girls speaks Bulgarian.'
'How do you know? Do you speak it?'
Halliday laughed. 'No. But one of the cleaners at the police station is from Sofia. She said the girls couldn't understand a word she said.'
'But you had to let them go anyway?'
'Yes. No choice. They're all living in Manchester, so it's not up to me. I would have tried to work the prostitution angle, but Manchester SB couldn't be bothered. These days it's hard to convict unless you show the girls involved are either under duress or illegal immigrants. None of the girls would make a complaint so we couldn't do either.'
'Too bad,' said Liz, though she wasn't very surprised. Jackson seemed unlikely to jeopardise his club by laying himself open to a single police raid.
Halliday paused and Liz heard the sound of a bus passing. As it died down Halliday went on, 'That isn't good news, but there's worse to come. I had a source in the club – an older woman who functioned as a kind of "mother" to the working girls. Name of Katya.'
'You "had" a source?'
'Katya was found strangled in the kitchen of her digs two mornings ago. The uniform thought it was a burglary gone wrong but it doesn't ring true to me. There was no sign of forced entry, nothing taken. One of her flatmates found her when she came home from work.'
'Do you see a connection with the club?'
'Yes I do, not that I can prove it.' He hesitated, then finally said, 'The thing is, when we arrested the girls we took Katya in, too. But she was released hours before the others were. I don't know why – she was the only one sprung early. It would have looked peculiar. I didn't ask for her to be let go, that's for sure.'
Liz sensed he was very upset by this. She said encouragingly, 'Maybe Forensics will find something.'
'I don't think so. The killer was very careful. Her place was in the Greater Manchester area and the CID guys there have made it a low priority.'
'Why's that?'
'Either because they reckon it's a one-off and won't lead anywhere, or because they know where it leads and have been warned off.'
'What does that mean?' She didn't like the sound of it at all.
'Ask your friend in Manchester Special Branch.'
He's not my friend, thought Liz, but there was no point in saying this. She asked, 'This woman Katya, did she have a Bulgarian passport too?'
'I don't know what passport she had, but I know she wasn't from Bulgaria.'
'Then where was she from?'
'One of those funny ex-Soviet countries – the ones that end in "stan". Hers was called Dagestan. At least that's what she told me. Never heard of it myself. Have you?'
'Yes,' said Liz flatly. She had heard of it quite recently. 'Listen, I wonder if you can help me with something.'
'Just say the word,' said Halliday so breezily that Liz wondered whether perhaps there had been some vodka in his tonic after all.
'You remember I told you that we'd learned that Jackson was connected to an arms dealer.'
'Yes, I do.'
'Well, we've now had confirmation that he's involved.' She hesitated, then decided she had to trust him – so far at least, he had been completely straight with her, unlike her old friend McManus. 'I think there might be a connection between his role in this arms deal we're investigating and his usual business at the club – bringing in the women, I mean.'
'What kind of connection?'
'Not sure yet.' Liz was working largely on intuition now; she couldn't give Halliday any specifics because she didn't have any. She went on, 'That's where you could be of help. Can you keep an even closer eye than usual on what goes on at Slim's?'
'Yeah, I can do that. But what am I looking for?'
'I know it sounds rather pathetic but I can't actually tell you. Anything that looks stranger than usual. It's about bringing stuff into the country. Importing stuff that could be arms but it probably wouldn't look like that.'
'If you seriously think he's into weaponry, it would probably be wise to run it by Manchester SB, just to be diplomatic.'
'Do you have to? I thought you said Slim's was on your patch?'
There was a pause, then Halliday said, 'No, I don't have to if you're not going to.' He gave a short laugh. 'I see you don't trust McManus either.'
The overnight team outside Dinwiddy House had had a busier time than expected. At twenty past seven in the evening Zara had emerged, dressed now in a black hoody and jeans and carrying a small backpack. He had walked to Euston Station and after collecting a ticket from a pre-paid ticket machine, boarded a train for Manchester Piccadilly. Two of the team had accompanied him, while the Ops Room had dispatched another team to Manchester to be ready to meet the train, in case he stayed on all the way to Manchester.
Which he did. At Manchester the original team handed him over to the new team, which went with him, first on the metro to Manchester Victoria station and then on a local train, from which he got off at Eccles.
By this time it was past eleven o'clock and Liz in bed was on a conference call link to the Ops Room in Thames House. 'Eccles,' she said. 'What on earth can he be going there for? Does anyone know anything about Eccles?'
Peggy, in her flat in Muswell Hill, a few miles further north from Liz, was in on the call and also searching the internet. 'Eccles is part of Salford, about four miles from Manchester. The interesting thing is that it has quite a large Yemeni community. There have been Yemenis in Eccles since the 1940s,' she read out from a website. 'Large numbers came in in the 1950s. There's a Yemeni Community Association. Perhaps he has friends there.'
Meanwhile the team in Manchester was reporting that they had followed Zara to a small terraced house, No. 31 Ashby Road. The door had been opened by a lady, probably in her late sixties, in traditional Muslim dress, who had kissed Zara and welcomed him into the house. They hoped Liz did not require overnight watch on the house, as it was a very quiet neighbourhood and therefore it would be difficult to remain unobserved. Liz had agreed that they could stand down for the night; it seemed most unlikely that anything was imminent. She and Peggy would meet in Thames House at seven in the morning and decide what to do next about Zara.
Chapter 31
Miles woke up slightly hungover, the after-effect of a long evening at the French Embassy, and discovered that his mobile phone was ringing. 'Hello,' he said tentatively; the screen read 'number unknown'.
'Ah, the croaky voice of a man who's had a good night out.'
It was Bruno Mackay. At the best of times, Miles felt a mild antipathy towards his British Intelligence counterpart, and right now there was a jauntiness about the man he could do without.
'What can I do for you, Bruno?' he said shortly.
'I've had a communiqué from London. It seems there's been some progress. Better if we talk face to face, old man? I'll see you at Sharim's café in an hour.'
Miles made it in fifty minutes, feeling slightly revived after a long shower and a shave. He drove cautiously into the old city, keeping an eye on his rear-view mirror; after their experience on the road from Donation's farm, he felt that his car might be a marked vehicle. Parking in a Diplomatic parking bay, under the eye of a policeman, he walked along the pavement until he saw the wide awning of Sharim's – and Bruno, in a white cotton jacket and pink tie, sitting at an outside table.
Miles joined him. Bruno gave a commanding wave and a waiter scurried over with a fresh pot of coffee and a cup for Miles, who watched while the man poured out the syrupy local brew. Miles added two sugar cubes from the little clay pot on the table. As he stirred them in with a tiny wooden spoon, he said to Bruno, 'So what's the news?'
'London's identified the guy they sent the photographs of. The one at the meet in the Luxembourg Gardens that we were going to ask Donation about. His name is Samara and he's Yemeni. He's doing a Master's degree at London University, the School of Oriental and African Studies, SOAS we call it. On the surface he looks perfectly legit. Only quite obviously he's not. I've been asked to check out his credentials here, and I thought you might be able to help me.'
Why? wondered Miles, but then Bruno said, 'You're a bit better placed to ask, I think. If you get my drift.'
And Miles now understood. Official Yemeni–American relations were blossoming. A cynic might say that the United States was propping up a weak local government to further its own interests, but for whatever reason, a request for help from the American Embassy was likely to get a quicker, more favourable reaction than if the Brits had asked.
'It may take me a little while,' Miles said.
'Not a problem, old boy. We've got a couple of hours on London as it is. They'll still be fast asleep.'
Miles's contact was a middle-level officer in the Yemeni Intelligence Service called Arack, who had been a graduate student at the University of Southern California. It was never entirely clear what he had studied there, and he seemed to know the beaches north of Santa Monica rather better than the classrooms of USC. But he was a useful contact, since the Yemeni bureaucracy was both legendarily cumbersome and unreceptive to foreign approaches, and Arack was always willing to help the Americans, provided the request was relatively easy to fulfil and his reward readily forthcoming. He was known to Miles and his colleagues, semi-derisorily, as 'Sweet Tooth' because of his love of sugary cakes and desserts, which made payment for his services unusually easy.
Miles and Arack met now for coffee and a baklava-like concoction in a café near the Yemeni Ministry of Defence. Arack listened sympathetically while Miles explained what he was looking for. 'We just want confirmation that the personal details we have for this student are correct and that he is known to your authorities and is in London legitimately.'
'Is there any reason to think he is not?' asked Arack mildly.
'No,' said Miles, though it didn't take a genius to realise there had to be a question about the 'student', or else Miles wouldn't be checking him out. 'It's just a formality.'
Arack nodded, happy to hear that this was not something he would have to call to the attention of his superiors. 'Naturally births and deaths are registered here, as they are in the United States, and there is a department for that purpose. But you might find its office difficult to navigate. Let me make a few calls and get back to you. Give me the details please, and I would be grateful if you could ask the waiter to come over.'
Arack rang Miles just before dinner. There was a shortage of eligible Western women in Sana'a and Miles was about to have dinner with one of them – a new shapely secretary called, appropriately, Marilyn, who had come out to work in the Embassy the month before. He waited impatiently as Arack went through the standard Middle Eastern formalities, applied rigorously even to a phone call. How was Miles? As if they hadn't met five hours before. Was not the weather good this day, and would it not be fine throughout the evening? At last Arack came to the point, though even then he spoke elliptically. 'I am afraid I have surprising news for you, my friend.'
'Really?'
'We have no record of this man, you see.'
'What do you mean?'
'Just what I have said. There is no birth certificate, no record of an education and no passport.'
'Could the name be spelled differently?'
'I have pursued all possible variants. More important, the residential address you say this fellow gave in Sana'a does exist but . . . it is a bicycle shop. I can assure you, there is no citizen with the particulars you supplied.'
Miles mind was no longer on his date with Marilyn. 'OK. Thank you for checking this for me.'
'My pleasure. I wish you luck finding this gentleman. But I can assure you, it will not be in the Yemen.'
Damn, thought Miles as he put down the phone, then picked it up to cancel his date. He hoped Marilyn wouldn't be too disappointed – though he was, especially since he realised there would be a further call to make. It looked like he would be having dinner with Bruno Mackay instead.
Chapter 32
Since Peggy Kinsolving had joined MI5, and particularly since she had been working with Liz Carlyle, she had found out a lot of things about herself that she didn't know. At school and university she had been a quiet, studious, and rather shy girl. She loved acquiring information, categorising it, sorting it out so she could access it and apply her considerable intelligence and her almost photographic memory to it.
These were the qualities that had taken her from her grammar school in the north of England to Oxford, where as predicted she had obtained a good 2:1 degree. No one, including Peggy herself, had ever thought she had the intellectual confidence and verve that makes a first-class scholar.
Her social life at university had followed the same cautious pattern. She had joined a few societies of the intellectual type and one day, showing much daring, she had gone with a friend to a meeting of the college dramatic society, who were looking for backstage staff. Everyone in the society, it seemed, wanted to be on the stage and in the limelight, and no one was prepared to do the behind-the-scenes work. Peggy thought that job would suit her very well, and it did. She brought her formidable information-sorting skills to organising the props, the scenery, the sound effects; eventually she became completely indispensable to any performance.
She would stand in the wings, noticing every detail, knowing everyone's part better than they did themselves and making sure that at least from a technical point of view the performance was perfect. She loved the drama but only from behind the scenes. She could never be persuaded to take even the smallest role on the stage. The thought of appearing before an audience petrified her.
Satisfied with her 2:1 and pursuing what she thought was her métier, Peggy had taken a job in a small research library, working on sorting and cataloguing the papers of an obscure female Victorian novelist. But after a couple of years she had begun to find the work dull and unsatisfying, and her social life in a small town where she knew no one was practically non-existent.
So when she saw an advertisement for a research post in a government department in London, with some hesitation she applied and found herself working as a research assistant in MI6. A chance secondment to MI5 a few years later led to her working with Liz Carlyle. At first her work had been purely research, but Liz had seen something in her young assistant that made her think there was more to Peggy than met the eye, and she had gradually encouraged her to take on a more upfront role.
At first Liz had given her some simple interviews to do, then she had moved her on to situations where Peggy had to play a role, to pretend to be someone other than an MI5 officer.
This was when they both realised that Peggy had a penchant for acting a part. Though she would still rather die than go on stage and act before an audience, put in a one-to-one situation she could convincingly present herself as anything from a housewife to a hedge-fund manager – and enjoy doing it.
Today she was an electoral registration officer. She'd dressed primly: a mid-length blue skirt, matching tights, sensible shoes, and dark paisley shirt under a navy blue blazer. She carried a clipboard and pen, and with her glasses firmly in place on her nose looked entirely like the local authority bureaucrat she was pretending to be.
At two o'clock that afternoon she knocked on the door of 29 Ashby Road. Most of the area seemed to be lived in by Muslim families, but she knew from the electoral register that this house was occupied by a Mrs Margaret Donovan. The door was opened by a large red-faced woman whom she guessed to be in her early seventies.
'What can I do for you, luv?'
'Mrs Donovan, is it?' asked Peggy, and she explained that she was from the electoral office, confirming the names of the occupants of voting age in each house along the street.
'Wasn't there someone here a few months ago about that?' the woman asked.
Peggy sighed. 'Probably. There seems to be a lot of duplication in this job. I've only been at it three weeks, but you're not the first one to tell me it's all been done before.'
The woman smiled sympathetically, and just then the phone in the hall rang. 'I'd better get that,' she said. Peggy started to make her excuses but Mrs Donovan waved her in. 'Come inside and close the door before you catch your death.' While she went to the phone, Peggy waited patiently in the hall. The woman wasn't long. 'Bloody tele sales,' she announced, coming back into the hall.
'They are a nuisance,' said Peggy, shivering slightly. It was a raw day outside, and in her anxiety to look authentic she had not put enough clothes on. The weather had been hovering between autumn and winter for several days, but today for the first time you could sense the months of real cold ahead.
'You look like you're freezing, dearie. Come into the kitchen and have a cuppa and warm yourself.'
Peggy didn't even pretend to protest, foreseeing a golden opportunity to gossip about the neighbours. As the kettle warmed on the gas hob, she looked around the room, which had family photographs all along the top of a sideboard. 'Your children?' she offered.
'All five of them. Grown up now,' the woman added sadly, 'and my poor Leonard gone ten years now. Still, mustn't grumble.'
'Have you been in this house a long time then?'
Mrs Donovan gave a little laugh. 'Each one of my children was born and raised here. It will be forty years come October.'
'Gosh,' said Peggy appreciatively. 'I suppose the neighbourhood's changed a bit since then.'
The woman gave Peggy a sideways look. 'Not for the worse,' she said firmly.
'Not at all,' said Peggy. 'I can see that. It looks a fine street to me.'
The woman relaxed. 'It's just that so many left when the Asians came. Not me, mind; I wasn't going anywhere. I always said, there's good and bad – white, black, and all the in-betweens. Why should I up sticks if people treat me right? Who cares what colour they are?'
'Who indeed?' said Peggy, taken aback by the old woman's almost aggressive tolerance and the implication that she – Peggy – might not agree.
The old lady went to the stove and poured boiling water from the kettle into two waiting mugs. 'Milk?' she asked and Peggy nodded. When she brought the mugs over she pushed the sugar bowl along the kitchen table, and Peggy shook her head and pushed it back.
They sipped in quiet contentment for a moment. Then Peggy said casually, 'You've got good neighbours then?'
'The best,' Mrs Donovan declared. 'The Desais live on that side,' she said, and proceeded to talk about the Hindu family next door. Peggy nodded as the old woman took her through three generations of Desai family tree, little realising that her listener was entirely uninterested in them, and was only waiting for her to talk about her neighbours on the other side.
Peggy's cup had been refilled by the time she felt able to ask about the other neighbouring family. 'Mrs Atiyah,' Mrs Donovan said, and her face seemed to light up. 'Isn't it a lovely name?'
'Very pretty. What kind of name is it?'
'The family was from Yemen, luv. What we used to call Aden before they went and got themselves independent. Though then there was a lot of trouble, and that's when the Atiyahs moved here.'
This time Peggy paid close attention while Mrs Donovan went through the generations of Atiyahs. Mr Atiyah senior had passed away several years before, leaving Mrs A solitary in the house, though she had two daughters (and five grandchildren) living nearby, and almost every day one of them paid a visit on their mother. 'She's seventy-two next March, not that she looks a day over seventy if you ask me yourself.'
'It's nice she's got daughters to look after her,' said Peggy, resisting the temptation to finish her tea, since she couldn't be sure she would be offered another refill. 'Though I suppose she would have liked a son as well.'
'Oh she's got a son, all right. He's the youngest child and the apple of her eye. And Mrs A spoils him rotten. You'd think he was still a schoolboy from the way he lets his mum take care of him – I've seen him lug his laundry home for her to do, and him living all the way down in London.'
'He's got no family then?'
Mrs Donovan shook her head. 'No, he's still a student. If you ask me, it's all very well everyone going to university these days, but sometimes they carry it on too long. Mika is twenty-six if he's a day. By that age my Leonard had been working for ten years, yet this lad's still at his books.' She shook her head uncomprehendingly. 'My nephew Arnold—' she started to say, but Peggy cut in quickly to impede the diversion. 'Do you reckon his mum minds? I mean, his being a student and all?'
For a moment the old lady looked confused, as if her nephew Arnold was being discussed, then she realised Peggy was talking about the Atiyah boy and she shook her head decisively. 'No, his mum thinks the sun shines out of that boy's eyes. Even when it's grey and overcast outside.' She gave a little chuckle.
'They say Middle Eastern lads are very dutiful sons.'
The woman gave a little harrumph, and Peggy realised she didn't like her neighbour's son much. She said nothing but waited patiently, and sure enough there was more to come. 'Like I say, the boy's been spoilt. Why, last year he said he wanted to go back to his homeland – he meant Yemen – and his mum coughed up the air fare. What was the point, I ask you? He's born and bred British just like you and me, so why start pretending you're not? Never go backwards, that's my motto.'
'Maybe he wanted to explore his roots. Like that programme on the TV.'
'I can't see him sobbing over his great-grandmother like what's-his-name did. He's a hard little bugger, our Mika.'
'Did he like it in Yemen?'
Mrs Donovan shrugged. 'I didn't think it was my business to ask. Mrs A knows I don't approve of the boy – he's not polite, at least not to the likes of an old lady like me.'
'Really?' said Peggy, trying to sound indignant.
'Not since he went to the Middle East. He hardly says hello when he sees me.'
'Are they a very religious family?'
Mrs Donovan paused, as if she had never thought about this before, and said reflectively, 'The old man was, but not Mrs A. Since he died I don't think she goes to the mosque much. And when one of her daughters married an English bloke, she didn't bat an eye.'
'And Mika?'
She shrugged, and looked at the mugs on the table. Peggy realised she was in danger of outstaying her welcome; the old lady liked to talk, but on her own terms, and that didn't seem to include answering too many of a stranger's questions. Peggy got up from her chair. 'Golly, what you've said has been so interesting I could stay and listen all day. But duty calls, and I have to get back to work. Thanks so much for the tea, Mrs Donovan.'
'Call me Maggie, dear.'
'Right, Maggie. You've been very helpful.'
'Have I?' asked Maggie, and her face was suddenly cheerful again. 'That's kind of you to say, luv, though I don't see how.'
'I'll just leave you this,' said Peggy, putting a small printed leaflet on the kitchen table. 'It explains about the Electoral Registration process and it's got my phone number on it in case there's anything you want to inquire about.'
'Thanks, luv,' said Mrs Donovan, picking up the leaflet and putting it on the sideboard beside the photographs.
Chapter 33
Martin Seurat looked moodily out of the window of his office in the headquarters of the DGSE, France's external intelligence service. He occupied a small room in a corner of one of the white stone buildings just off the Boulevard Mortier on the outskirts of Paris. Outside, the gravelled courtyard had darkened to the colour of slate from the rain that had come down in a short heavy burst earlier that morning. The sky had stayed overcast, with no hint of sun, and now the wind was picking up. It all seemed like a plot by winter to hurry things along, thought Seurat, who every year wanted to hibernate at this season and wake up only when the clocks changed in spring.
He couldn't quite understand why he was feeling so low. After all, he had achieved his ambition of capturing his old colleague Antoine Milraud, the man who had betrayed his friendship and his trust. Why wasn't he feeling elated?
He supposed the trouble was that he did not yet have the pleasure of seeing the man in court, answering for his crimes. That pleasure had to wait until the operation in Britain was concluded. But he wasn't directing that operation; he was having to leave that to Liz Carlyle, since it was happening on her turf. So at present he had only a minor role to play, keeping Annette sweet and monitoring the arrangements at the safe house in Montreuil.
He could hear the noise of workmen moving furniture around across the passage. He'd left his door open, and occasionally he saw one of the workmen passing by, carrying a chair or a cupboard. A colleague had returned from a posting in Taiwan and was moving into the vacant office. Funnily enough, that very room across the passage used to be Antoine Milraud's office. Seurat had spent many an hour there, talking with his old friend and colleague, sometimes cracking open a bottle of Bordeaux if they had stayed working late enough to deserve a glass or two, talking quietly until the phone would ring and – Seurat could hear her voice from the other side of the room – Annette would demand to know when Antoine was coming home and did he really expect his dinner to be waiting when he did?
Annette was not so chirpy now, living with a guard in the small flat the Service kept in the Fifth Arrondissement, while her husband twiddled his thumbs in the Montreuil bungalow not far away from this office. Seurat had talked to Liz that morning and heard her account of her debriefing of Milraud in London. Both had agreed that he was still holding something back, and only superficially cooperating. Whatever it was the man was not saying was bound to be important, or why keep it secret? Maybe it was something that reflected badly on him. But why would he bother, considering the mess he was already in? Liz thought it most likely to concern Lester Jackson's role in the whole affair, and Martin did not disagree.
The problem was there didn't seem any obvious way to prise more information out of Milraud. He'd already been threatened with the prosecution of Annette, and had folded accordingly. They could always threaten him again, but to what end? Putting Annette in prison wasn't going to tell them anything more about Lester Jackson or the young Arab whom Seurat still thought of as Zara. And in any case, after a while repeated threats failed to frighten, as if the ferocious dog barking from inside a house turned out, when the front door was opened, to be a chihuahua.
'Monsieur?'
The voice was gentle but Seurat was startled none the less. Looking up, he found a young man in the doorway. At first he thought he must be one of the moving men, but no, this fellow had longish hair and wore a cotton jacket and chinos. He looked like a student rather than a workman.
'What is it?'
'Forgive me, Monsieur. I am Jacques Thibault. I have been helping out with Antoine Milraud.'
Seurat stared at Thibault; he seemed very young to be guarding his ex-colleague. Then he remembered. 'Ah, of course. You are the computer genius.'
Thibault gave a modest shrug. 'You are too kind.'
'How goes it? Anything more to report?'
'In fact, yes. As you know, I have control of Monsieur Milraud's laptop and I read all his emails. That includes the recent communiqué asking him to come to London. He claims he wiped all the earlier emails on security grounds. What he doesn't realise is that I have been working hard to find them nonetheless.'
Seurat saw the importance of this immediately. 'And have you?' he asked eagerly.
'Only up to a point. I am sure you are familiar with reverse engineering.'
'I think so. You go backwards to reconstruct a trail. It's especially useful to see how something began, isn't it?'
'In a sense.' Thibault had lost his air of diffidence and had come into the office, sitting down when Seurat pointed to the empty leather chair across from his desk. 'But I would argue that it is most valuable when something has been destroyed rather than built.'
'Really?' said Seurat, trying to be patient.
Thibault nodded vigorously. 'Suppose you are confronted with a brick house and want to see how it came to that state: through reverse engineering you gradually work your way back until the walls have come down and the first foundations are about to be poured – the bricks for the walls may not even have been delivered. Now that is a beautiful process in its own way, but it doesn't tell you much if what you want to learn about is the finished house.'
Seurat nodded politely at this elaborate metaphor, but privately he wondered what point Thibault was trying to make. If Thibault sensed his doubts he gave no sign of it, and continued: 'Think about it this way – what if this finished brick house is destroyed? Accidentally or on purpose, it doesn't matter. Either way all the information you want is lost, irretrievably. Unless' – and he started to smile – 'you could reverse-engineer the act of destruction, slowly work your way back from the present position of crumbled walls and masonry dust to the house in its former glory.'
'You can do this with Milraud's emails?'
'Yes.' Thibault was sure of his ground, his voice entirely confident. 'Not the whole house at first, more like one of its rooms. But in time I am certain it can all be reconstructed.'
'Do we know anything yet?'
'We do, but I don't know how much of it is of value. The first exchange occurred when Milraud was in South America. I am not quite sure where.'
Caracas, thought Seurat, and motioned for Thibault to continue. He said, 'I can be more precise about the sender of the email, however. His message came from Yemen. Not far outside the city of Sana'a.'
'Hang on a minute. Was the sender the same person in England who's contacted him recently?'
'Yes, well, at least it is the same email address.'
Liz had told Seurat about Peggy Kinsolving's fact-finding mission to the northern town of Eccles. Atiyah, apparently the real name of the young Arab they'd been calling Zara, had visited Yemen within the last year according to the neighbour. It all fitted. 'What about more recent communications?' he asked Thibault.
The young boffin shook his head. 'Not yet. There is a large gap to be filled between these first exchanges and the last email, which we've already seen. I am confident of filling in this gap, but it will take some time.'
'Oh,' said Seurat, sounding disappointed. Thibault was obviously elated to have recovered even a small part of what had been deleted, but if it didn't actually tell them much then there didn't seem any reason to get excited.
Thibault said, 'I will send you the transcripts of what I have managed to disinter.'
Seurat wondered if there was any point in passing this on to Liz. Probably not; the 'breakthrough' hadn't amounted to much.
Thibault shifted in his armchair, ready to depart. Then he said, 'There is one other thing that may be of interest. It's a reference in the very first email from Yemen to the man who made the introduction to Milraud.'
Seurat was suddenly alert. 'Does it give his name?'
'No, because he was careful not to use it. But does sound as if the man is a senior person in the government. Possibly a minister, I can't tell exactly.'
Chapter 34
Miles Brookhaven was used to working in the Middle East, but each year as winter hove onto the horizon he thought fondly of home. He had just been reading a letter from his mother, describing the Thanksgiving dinner she was planning in the small town in upstate New York where she lived. It would soon be snowing there, with icy winds coming in over the Great Lakes from Canada.
It all seemed a very long way away from this café in Sana'a where he was sitting at an outside table under a hot Yemeni sun, watching as Arack used his fork to attack a second helping of pistachio and syrup-laden pastry. CIA HQ back in Langley must have seen some odd expense claims over the years, everything from 'booze and babes' to a legendary purchase of a racehorse, but Miles couldn't imagine what they would think of the three-figure bill he was planning to submit for the purchase of Yemeni patisserie.
Not that he had much to show for it in return: Arack seemed as mystified by the disappearance of Baakrime as Miles was. Arack said between mouthfuls, 'No one knows anything. I made up an excuse to ring his office at the Ministry. It's quite obvious that his secretary doesn't have a clue where he is. My own Minister's secretary was trying to contact him to find out if he was coming to a meeting just two days ago, but no one could say whether he was coming or not.'
He looked contemplatively at his fork. Noticing a vestigial smear of cream on one of its tines, he licked it clean, then said, 'His son has disappeared as well, you know. That is also a mystery. He worked for the Minister's charity but no one there seems to know where he is either.'
'That's interesting. What do you think it's all about?' asked Miles. He couldn't prevent his mind recalling the bloodied, white-robed corpse, whose shiny black shoes seemed to make the mental image even more gruesome.
Arack shrugged, and looked over at the waiter, who fortunately for Miles's expense account was occupied serving another table. He said, 'Someone must know something. You have to remember what kind of man Baakrime is.'
'How do you mean?'
'He's rich, he's powerful; inevitably, he has enemies. Maybe something has frightened him enough to go into hiding. And that would also frighten anyone who knows his whereabouts. All one can do is keep asking, though it requires care in order not to make people suspicious.' Arack's face suddenly broadened into a smile. He had caught the waiter's eye.
Two nights later, Miles left the embassy after attending a reception given by Ambassador Rodgers for what he called 'a visiting fireman' – in this case, a natural gas producer from Monroe, Louisiana. It was a three-line whip for most of the Embassy staff, and the trade representatives of other missions and embassies had also been invited. The party was never going to be a barrel of laughs, but Miles didn't mind – among the other conscripts was the voluptuous Marilyn.
But when Marilyn cut him dead in the embassy foyer and went instead to talk to the other secretaries, Miles realised she was still cross with him for cancelling their dinner date. His disappointment turned to annoyance when a little while later he saw her with Bruno Mackay, who seemed to be putting something into his phone – probably her number, thought Miles jealously.
Miles compensated for Marilyn's snub by having three glasses of wine, which meant he left the party feeling mellow. He lived in an apartment in the old quarter of the city, and he drove there in the dark with extra caution. Since the events on the road from Baakrime's farm he had changed the car he drove, but he still felt uneasy driving at night, even in the city.
He parked in the underground car park of his apartment block. As he slammed and locked the car door his eye caught a movement behind a car two spaces along the row. His stomach lurched and he stood still, keeping the car between himself and whoever was there, alert for what might happen next.
Then a figure emerged into the light, and after a second or two Miles recognised Minister Baakrime. Standing just fifteen feet away under the yellowish glow of the car park lights he was a changed figure. The finery of his office apparel – silk ties, handmade shoes from London – was all gone. He wore a brown canvas jacket with side pockets, and unpressed cotton trousers; he looked utterly nondescript, which must have been his intention, though no doubt it galled him.
'What are you doing—' Miles began to ask, but Baakrime put an urgent finger to his lips, then beckoned him into the shadows in the corner of the car park. 'Quiet, my friend. I do not want to be seen by anyone.'
'I've been looking for you,' whispered Miles.
Baakrime gave a melancholy smile. 'You are not the only one.'
'Why have you disappeared?'
'I had no choice.' His voice was hoarse. 'My son was murdered. That was a warning to me that I would be next.'
'I am sorry for your loss,' said Miles. There was no point in saying that he had seen the body of the young man. 'Who did it? Who's after you?'
'I know their identity, and it will be of great interest to your country and the British. But I cannot stay in Yemen much longer; I want to emigrate to America. I think I have information to earn that right; I have been of considerable assistance to you in the past and I have more to tell.'
Miles said nothing while he digested this. His estimate of Baakrime was that he was a wily old crook who had probably got what he deserved, though what had happened to his son could not be wished on anyone. It was not as if the Minister had been helping the United States out of the goodness of his heart: each time he'd imparted any information he had been well paid for it. There was no question of giving him free passage into America and, as he no doubt hoped, a pension to live on when he got there. Those who held the purse strings in Washington would never authorise it unless he had a lot more to offer.
Miles spoke carefully now. 'You have been a valuable friend to my country, but I think you will admit your efforts have always been rewarded. In the end we are all in business, even if our goods are information. I need to know what more you have to offer before I can put your proposal to my superiors. I need to know a lot more about these people. Why are they hunting for you, where do they come from, what are their plans? If you can tell me that, then maybe I can help you.'
Baakrime didn't reply at first. He exhaled noisily, wiped at his thinning hair impatiently, then looked around. He was less agitated now, but his shoulders were slumped; clearly Miles had not told him what he had been hoping to hear. At last he spoke, 'Very well. I see the position, even if I regret it. I will tell you what I know, though I want guarantees of protection – until you arrange for me to go to America. Is that agreed?'
'There's only so much I can do in Yemen. Once you have left this country I can guarantee your safety. Until then, I can only give advice.'
Baakrime thought about this, his lips pursed. 'It will have to do for now,' he said grudgingly. Then he added reluctantly, 'The man attending the meeting in Paris is not working for the rebels of the Arab Spring. He is working for jihadis. Ones that are based here in Yemen.'
He looked at Miles as if he had handed him a gift of unexpected value, but Miles shook his head, to show that he would have to do better than this. 'We already know that. We have learned a lot more since you alerted us to the meeting in Paris.'
If the Yemeni was disappointed he didn't show it, but like a magician whose rabbit has failed to impress, simply produced another one. 'Of course, but I doubt you know much about these jihadis. You see, they are not Yemeni, they are British.
Miles remembered the British voice in the armed gang who had forced him and Bruno Mackay off the road. 'Believe it or not, I did know that too. But tell me more.'
'Do you know their plans?'
'Which plans?'
'Ah,' said Baakrime, 'I thought not. These British men are here only temporarily. Soon they will be returning to their country, with a small stop in Paris. I don't think they're going there to see the Eiffel Tower.' He laughed. 'And when they get to England I don't think they'll be training to become lawyers.'
'What are they going to do?' Miles demanded. Baakrime didn't reply. Miles insisted: 'What are these men planning to do?'
'It's not entirely clear,' said Baakrime, which to Miles meant either he didn't know or he was holding the information as a bargaining chip. The latter seemed most likely when he said, 'I believe I can find out.' He paused, then added, 'If . . .'
'If you can tell me when these men are going to Paris and what they are planning to do in England, then I think I may be able to help you. It will take forty-eight hours and the information will need to be checked.'
Baakrime said slowly, 'Forty-eight hours is a long time in my position.'
'You've made it this far; what's a few more days? Get me the information I want and then I'll set everything up. How long will it take you?'
'I will meet you tomorrow at this time, but not here.'
'I have a small place I keep as a safe house,' said Miles and he gave him an address in the old city. 'Come there tomorrow but be sure you are not followed.'
'Trust me, my friend. I still have a few friends who look after me.'
Chapter 35
At eight o'clock the following morning Liz Carlyle, Geoffrey Fane and Andy Bokus were sitting in the basement secure room in the Grosvenor Square Embassy. Each had in front of them a copy of the message that had come in overnight from Miles Brookhaven in Sana'a, describing his meeting with Baakrime in the car park.
'Well,' said Bokus, looking at his two British visitors, 'I've been in touch with Langley overnight. We don't want this Donation guy, so it's up to you. Are you prepared to have him?'
'Come now, Andy,' said Fane in his most patronising tone. 'I know it's early in the morning and you may well have been up half the night, but let's just talk about this for a minute. As I read what Baakrime said to Brookhaven, it's the US he wants to go to. He made no mention of the UK.'
'He wants to get out of there before someone tops him, and I don't suppose for a minute he's going to turn down a passage to London. It seems to me that it's you who stand to benefit from whatever he has to say. He's talking about British jihadis, not American, so it's your side who should bear the cost. That's what I've advised Langley and they agree.'
There was silence for a moment. Geoffrey Fane was leaning forward on the bench with his elbows on the table and his fingertips together. Liz Carlyle knew that any meeting between these two had to begin with some sort of ritual sparring match, and she was used to biding her time until the first bout was over. It looked as though it was, so she said, 'I think you'll agree, Andy, that it's crucial that we find out what Donation knows about these British jihadis he's talking about. It seems to me that Miles has asked him all the right questions. What we don't yet know is whether he can answer them. It's far too early to consider giving him asylum, let alone accepting him as a defector.'
'It's easy for you to say that, sitting here in London,' replied Bokus testily. 'The guy wants an answer and he's expecting Miles to give him one. Can he get out of the country or not? That's what he wants to know. Miles may not be able to get him to spill his guts if he can't give him the assurance he wants.'
'I hear what you say,' said Fane, 'but you and Langley seem to have made your mind up that the answer's No. It's just as much in your interests as ours to find out what these jihadis are planning to do. They may be British, but how do we know they're not planning an attack on a US target? Maybe it's the Embassy here. You won't look so clever if your colleagues get blown up.'
'Give it a break, Geoffrey. Our security is better than a bunch of home-grown jihadis can breach, and you know it.'
Round two over, thought Liz as Fane turned to her. And asked, 'Is your Service prepared to sponsor this character at the Defector subcommittee?'
Liz knew that doing that would mean making a case that Donation was likely to have information, or had already given information so valuable to the UK that he should be accepted as a defector with all the expenditure of cash and resource that that implied.
'Not as things stand now,' she replied. 'They'd never accept it. I'm afraid I think we will have to rely on Miles to extract whatever information Donation has, while making no promises about his future.'
'He's going to love that as a brief,' grunted Bokus.
'Have you any better idea?' asked Fane.
'I'm sure he'll do it perfectly,' said Liz with a charming smile. I hope so, she thought to herself. If not, all we're left with is the Jackson end of this puzzle and whatever we can get out of Antoine Milraud.
'OK,' said Bokus with a shrug. 'I'll let Miles have the good news.'
Chapter 36
It was an unprepossessing kind of place – just a small room with a little kitchen and a lavatory on the first floor above a minimarket in the old city. But it was safe. The minimarket was owned and run by the father of a longstanding and trusted CIA contact who lived in Virginia. Access to the upstairs was through the shop and under the watchful eye of its owner.
Miles and Bruno Mackay sat on a scruffy sofa gloomily contemplating a bottle of scotch and three glasses lined up on a low table in front of them. They'd read the instructions that had come in from Bokus earlier in the day.
'God knows how they expect us to get the story out of him when we've got nothing to offer in return,' Miles had said angrily.
'I know. I sometimes wonder if our lords and masters have forgotten what it's like at the sharp end, dealing with real people. String him along, they say, cheerfully, till he's told you all he knows, then we'll think about whether it's good enough and if not we'll throw him back to whoever's hunting him.'
They'd made their plan: who was to start the conversation, who was to say what and when, and now they sat in silence waiting for the concealed buzzer that would indicate that their visitor was in the shop. Silence; just the sound of shopping going on downstairs and the ring of the till as purchases were made.
Time passed. Miles looked at his watch for the third time. The Yemeni was now half an hour late. They both knew not to expect punctuality in this part of the world, but how long should they wait?
'I'm having a drink,' said Bruno suddenly. He unscrewed the cap on the whisky bottle and poured out two generous slugs, slopping in some water from a jug.
Miles was brooding over the fact that Marilyn had sent him an email, asking if he would be her guest at a small chamber concert hosted by the Ambassador's wife that evening. Though he wasn't especially interested in classical music, he was still interested in Marilyn, but he'd had to decline the invitation because of this meeting. Not being able to tell her the truth about his evening plans, he'd had to give her a vague excuse, and from her reaction he'd sensed that that had been his last chance. If Baakrime wasn't going to turn up it would be all the more galling.
'He's not coming,' said Bruno, another half-hour and another drink later. 'Let's pack up and go and get some dinner.'
'OK but we'd better let London know first.'
'Yes. Liz Carlyle is going to be pretty fed up that we haven't got any information about these British jihadis.'
Miles slept badly, dreaming of a sailing expedition from his childhood when they had run aground off Nantucket. In real life no one had been hurt; in his dream, inexplicably someone had drowned, lost in the shoals after the boat overturned in the incoming tide. He woke in a sweat at three in the morning, then turned on the BBC World Service, which eventually lulled him back to sleep shortly before dawn.
At the Embassy he found a message from the Ambassador's secretary, summoning him to see Rodgers. He went along anxiously, thinking he must have been spotted meeting Baakrime two days before, and wondering how to explain this violation of the Ambassador's orders to stay well away from the Minister. But he found the Ambassador unaccountably good-humoured, honouring Miles with a beneficent smile as he entered his office. 'Miles, Miles, how good to see you. All going well?'
'Yes, sir,' Miles said cautiously, wondering what was coming next.
'I've got some news. You remember our conversation about Minister Baakrime?'
'Yes,' said Miles.
'Well, you don't have to worry about him any more.'
'Oh. What's happened?' He felt a sense of dread. Had Baakrime been right to fear for his safety? He should have done more to protect him.
The Ambassador didn't answer him directly. 'Yes, you won't have to avoid that gentleman any more.'
Miles stared at Rodgers, unable to pretend he was anything but horrified. 'Is he—'
Rodgers nodded. 'Yep. The Yemeni government has informed me this morning that Mr Baakrime is currently a resident of Moscow, courtesy of an Aeroflot flight he caught yesterday in Istanbul. Fine by me, I have to say, though the Yemenis are not at all amused. They reckon he took twenty-five million bucks of government money with him. I bet the Russians won't let him keep a dime of it. What do you know about that?'
Far more than you, thought Miles, wondering what Bokus was going to say when he learned that Baakrime had not needed any of the US government's money – he was perfectly capable of paying his own way.
Chapter 37
There had been no further emails to Milraud from the young jihadi Zara in the UK. When Seurat pressed him Milraud merely shrugged, and said that when he'd met the young Arab both in Paris and Primrose Hill he had not been given a schedule for his next communication. Milraud's insouciance infuriated Seurat, but he did his best not to show it – he didn't want to give his former colleague the satisfaction of seeing him get angry, when getting angry wouldn't do any good.
But he needed to move things forward. Liz had told him that the Americans' source in Yemen, the government Minister who had started this whole operation going, had now fled the country; she had also told him of MI5's discovery that Zara, far from being the Yemeni student he claimed, was a native-born Briton. It was quite possible that whatever Zara was plotting could be well advanced, which made it crucial to find out what else Milraud knew. He clearly wasn't going to volunteer information, so Seurat had to find some way to lever it out of him. He knew Milraud well enough to know that threats and confrontation would get him nowhere, so he fell back on his ace card – Annette.
He arranged to meet her at a café near the Seine, a few streets from the Musée d'Orsay. The café straddled an intersection of two streets that met at right angles; its outside tables allowed a clear view of both the pedestrians and the cars that drove past. When Annette arrived, accompanied by her two guards, Milraud had been sitting for fifteen minutes, and was satisfied that he would recognise any returning cars or pedestrians that might indicate surveillance. The coast seemed to be clear, as he expected it to be.
As Annette sat down, her two guards took up positions at a nearby table. The waiter came over and she ordered a large Campari and soda before asking Seurat, 'To what do I owe the privilege of being let out of my cage?'
'I thought you might enjoy a little outing.' He knew that Annette was allowed out once a day for a stroll, but only when accompanied by her armed escorts. Meeting Seurat here, she could at least enjoy the pretence of being an ordinary Parisian.
'Come, come, Martin,' she said. 'We both know your concern for my welfare is strictly professional. You never cared a damn about me.'
'That's not true at all—' Martin protested.
Annette dismissed this with a curt wave of her hand. 'Even if you did regard me as a friend back then, you are not going to let that affect you now. So tell me what you want from this tête-à-tête.'
Seurat said nothing while the waiter was putting Annette's drink on the table. The two guards, alert and watchful, weren't even pretending to talk to each other; they were scanning the comings and goings at the café tables and in the street. When the waiter had left, Seurat said quietly, 'Antoine is holding back on us, Annette. I don't know if he's actually lied to us, but he certainly hasn't told us the whole truth.'
Annette lifted her drink and took a long swallow. Putting the glass down, she pursed her lips, as though considering what to say.
Seurat sighed. 'I haven't got time for games, Annette. If Antoine is concealing information, it will come out sooner or later, and then things will go very hard for him. And you.'
'You've already made that clear.' She reached into her bag and brought out a packet of cigarettes – Russian Sobranies. She lit one with a wafer-thin gold lighter from Cartier – he remembered Milraud showing it to him after he had bought it for Annette's Christmas present years ago. A reminder of more innocent times.
He said, 'Yes, but what I haven't told you is what Antoine has got himself involved in. This isn't a normal kind of arms deal we are talking about.'
'No?' Annette said neutrally, but she was tapping the fingers of one hand on the Formica tabletop, and Martin sensed her curiosity.
'No, it's much worse than that. Your husband would like to think he's supplying arms to freedom fighters in the Arab Spring, but that's not the real situation and I think he knows it. He's helping to arm terrorists – al-Qaeda supporters.'
Annette frowned and shook her head. 'You've been listening to the Americans too much, Martin. They think anyone who doesn't agree with them is a terrorist – and that all Muslims are al-Qaeda supporters.'
'Don't pretend to be simple-minded. What I'm telling you is true. I can't be sure yet exactly how the weapons Antoine has agreed to supply to these people will be used, but it's not for any struggle against dictatorship, I can tell you. Antoine's buyer is a radical jihadi, whose sole purpose is to kill anyone who fits his distorted idea of an enemy of Islam. His mission is likely to be to murder as many people as possible. Innocent people, by any civilised standard.'
He was staring at Annette but her eyes avoided his face, gazing past him to the street outside. She took a deep drag of her cigarette, then slowly blew it out in a white trail that hung in a plume over the table. She tapped her milky pink nails on the table. 'Antoine is many things, Martin, most of them good. You may not approve of his life now, but he is as human as you are, in every essential way. I am sure he would never sell weapons to anyone like the man you are describing.'
'He may not have known at first, I grant you that. But I think he's guessed now and he's doing it just the same.'
'Can you prove it?'
'No. Not yet. But everything is pointing to the truth of what I'm saying.' He judged that it was better to be up-front with Annette; if he misled her she would press him until that became clear. 'What I do know without a doubt is that his customer is English, even if he's ethnically Arab. And why would an English citizen want twenty thousand rounds of ammunition – and it is even looking possible that it is to be delivered to England – unless he was planning a terrorist attack of some kind? It simply doesn't make sense if he's a "freedom fighter" in Yemen, does it?'
He could see she was taking this in, and beginning to waver from her previous defiance, so he turned the screw further. 'We don't know what his plans are, but we need to find out before there's a bloodbath. You wouldn't want to have that blood on Antoine's hands, would you?' He added more gently, 'Or on your own.'
'I'd like another drink,' Annette said loudly, and Seurat signalled to the waiter. Annette sighed. 'You were always a persuasive bastard, Martin. Antoine used to come home and describe how the two of you had interrogated someone. You know my husband – he'd have been direct and aggressive. But he admired your method; he said you could charm the birds out of the trees.'
Seurat gave a non-committal shrug. Annette laughed. 'Still the modest one. That was something else Antoine admired.'
'There was a lot I admired in Antoine too,' said Martin.
'Yes, perhaps there was.' She sounded wistful. 'But not any more. I can see that in your eyes.'
'No. Not any more. Not after what he did. I took that as a personal betrayal.'
'Really?' She looked at him thoughtfully. 'I don't think I'd realised that – though I suppose I should have done. You were always so upright; nothing tempted you off the path of duty.' Her face looked sad and drawn as she sat quietly while the waiter brought her drink. When he had gone she sat up straight as though she had resolved something. 'So back to the beginning – what is it you want me to do?' she asked.
'Talk to Antoine. If you believe what I've told you about his client, and I think you do, then make him believe it too. Forget about jail sentences or clemency or anything like that; I'm not bargaining right now. I just don't believe Antoine would want to see dozens, maybe hundreds of innocent people massacred because he'd helped their killers.'
Chapter 38
Milraud watched as Annette got up from the bed, dressed in a silk slip and nothing else. She took a cigarette from the packet on the bedside table, lit it with the Cartier lighter he had given her years ago, and then went to the window, where she stood staring down at the narrow street that snaked along until, just out of sight, it reached the Seine.
He sat up in the bed, so that his back was cushioned by the pillows that he'd propped against the headboard. He said softly, 'Chérie, it is good to be with you.'
'Yes, my darling,' she said, but there was a hint of sadness in her voice and she didn't turn round.
He said, 'Martin is no fool, you know.'
Now she did turn round, and looked at him, her eyes filling with tears.
He went on, 'He let me come to see you because he knew how much I wanted to. Enough to tell him what he wants, in the hope that he will let us stay together.'
'Yes,' acknowledged Annette. 'But better this time together than no time at all.' She had been surprised, sitting in the flat, reading an old paperback novel she had found on a shelf and trying to ignore the guard who was making tea in the kitchen, when Antoine had arrived. He told her that he had suddenly been told to grab his coat and go for a drive; he'd had no idea that he was being taken into Paris to see his wife. In a rare tactful act, the now-combined force of armed escorts had left them alone, though they were hovering nearby – in the hall outside the flat, on the ground floor with the concierge, and outside by the parked Mercedes that had chauffeured Antoine from Montreuil.
Milraud looked at his wife, still as attractive to him as she'd been when they'd first met some twenty years before. He tried not to think of what prison would do to her figure, and to her spirited approach to life. It would do the same to him, no doubt, but he had already resigned himself to a long spell behind bars.
'Are there important things you haven't told Martin?' she asked.
Milraud raised his eyes towards the ceiling. He assumed the flat was bugged, especially if they'd let him see Annette here. She understood, and came back to the bed, stopping to turn on the radio on the bedside table. The station was playing Edith Piaf and they both laughed as they heard the song in mid-flow – 'Je Ne Regrette Rien'.
Annette lay down next to Antoine and whispered, 'So are there?'
'Of course. But why are you asking now? Has Seurat put you up to this?' He only slightly lowered his voice; he didn't care if the microphone picked this up over the radio; he was angry that they were being manipulated.
She didn't waver, whispering right away, 'He says the people you are supplying are much worse than you realise. They're not rebels fighting in the Middle East. He said they're al-Qaeda or their equivalent, and they're planning a terrorist attack.'
Milraud shifted uneasily on the bed, moving an inch or two away from his wife. 'How does he know?' He realised he had not spent any time questioning the intentions of the young Arab he had first met in the Luxembourg Gardens. His initial introduction to the man had come from Minister Baakrime, whom he had dealt with often before. He had simply assumed that the Minister had either been bribed by Yemen's insurgents to help them get arms, or was actually a secret sympathiser with the rebels.
He realised now that he had been naïve, but what did it matter? He had never made judgements about his clients, and he had helped arm revolutionaries across most of the world. There was no telling which side was right and which wrong, and if someone in his trade tried to make those sort of judgements they'd soon go mad or out of business. These affairs often ended in a place no one had foreseen. Look at Iraq now, or Libya, or Syria.
He was about to say as much to Annette, when she put a firm finger to his lips. 'Listen to me, Antoine. Naturally, Martin wanted me to talk to you; of course he wants me to persuade you to tell him everything you know. I would never hide what he said from you. I don't think we have any choice. If you know more about what's going on, then you should tell me and I will tell Martin.'
'But then I have nothing left to bargain with.'
'We are in no position to bargain, chéri. But even if we were, I have to tell you that if Martin is telling the truth – and I think he is – then I don't want you to help these people. They are killers; they kill children and their mothers. They have no just cause, only hate.'
Milraud lay back, his head against the pillow, and stared at the ceiling while he thought about this. Had Annette gone soft on him? It seemed improbable – if anything she had always been the tougher of the two of them, more businesslike, never very concerned about the morality of his trade. He knew she was scared of going to prison, but he also knew that she was very loyal to him – and her concern about what this young Yemeni, if that's what he was, was going to do with the weapons he was supplying was genuine. And he had to admit it did alarm him too – the thought of this character and his followers or colleagues killing dozens of innocents in Western Europe was appalling.
'OK,' he said at last, though he didn't look at his wife, but kept his eyes on the ceiling, as if addressing a deity or, he was pretty certain, the listening ears of his former colleagues in the DGSE. 'I'll tell him what I know. But it's not much.'
'I expect anything will help,' said Annette lightly.
Milraud turned on the bed and looked at her at last. 'The originator of the contact in Paris was a Yemeni minister. That's why I thought this was legit.' Legit struck him as a funny way to describe the transaction, but he knew Annette would understand what he meant – he had thought he was simply supplying one side of the innumerable civil wars that seemed to be proliferating all over the region.
'I understand. But now?'
'What Martin has told you could make sense. I haven't told him everything I know. I haven't told him exactly where the shipment is being assembled, though he knows the country. He doesn't know anything about the onward shipping arrangements. He knows there's a British person involved but he doesn't know that the order is now to be delivered to England. And he doesn't know that originally it was to go somewhere else.'
'Where?'
'Here,' he said simply. 'Paris.'
Annette looked shocked. 'So what changed?'
Milraud shrugged. 'I don't know. But if Martin's right about these people, it means the target's changed. Now it must be in Britain.'
Chapter 39
This time they were to meet in Fane's office in the MI6 building at Vauxhall Cross. As she walked across Vauxhall Bridge from Thames House, leaning into the gusty wind that was blowing off the river, Liz recalled the email exchanges between Grosvenor Square and Vauxhall that had preceded this meeting. Their tone suggested that the encounter between Fane and Bokus was going to be as rough as the weather, and she was not looking forward to playing the role of peacemaker.
Fane's office was a spacious room, high up in one of the semicircular protuberances at the front of the building. Its two large windows had a commanding view of the Thames – to the right Parliament and the MI5 building on the north bank, and to the left across to Kensington and Chelsea and upriver to Hammersmith. Somehow Fane had managed to acquire the sort of antique official furniture usually only to be found in the Foreign Office, and he had added some oriental rugs and a table that he had inherited from his grandmother. The whole effect was of a country gentleman's study, and about as far as you could get from the bleak, functional office that Bokus inhabited in the American Embassy in Grosvenor Square.
Liz knew that Bokus never felt comfortable in Fane's office, and when she arrived he was standing by the windows, looking stiff and awkward. Fane's secretary, Daisy, followed her into the room with a pot of coffee on a silver tray with china cups and saucers. Bokus waved her away when she offered him a cup and sat down heavily in one of the chairs round the table.
'Let's get on with it,' he said as soon as Daisy had left the room.
Fane took the chair at the head of the table and gestured to Liz to sit down opposite Bokus. He took his time sipping his coffee before saying, 'Thank you for the email, Andy; I think we are all sorry to learn that your source Donation has left Yemen. And very surprised to learn that he has gone to Moscow. I for one was not aware that he was in touch with the Russians. Were you, Elizabeth?'
Liz did not reply, and Bokus broke in, 'Not Moscow. Our latest information is that he's gone to Dagestan. We don't know why. He may have arms-dealing contacts there, or maybe the Russians have shipped him off there to get him out of Russia. But it seems that somehow he's got himself mixed up with jihadis – and got on their bad side; I told you his son was murdered. This is a man used to playing both sides from the middle, only suddenly he was squeezed from either end. The Yemeni government was growing fed up with him; now the jihadis have as well. So he's done a runner. But instead of running our way, as he would have done if you'd been a bit quicker on your feet, he's gone in the other direction.'
Fane shook his head and said, 'You were handling Donation – we weren't. If you'd been prepared to take a risk and be a little more generous, then maybe we would have got something back. Instead, the bird's flown the coop and taken his information and his money with him.'
Liz was about to intervene, but as she drew in her breath to speak, Bokus snapped, 'You can blame us all you like, but it isn't the United States that's at risk from this arms deal he was telling us about. It's you, and you weren't willing to do anything to keep him sweet and find out what he knew.' Bokus looked angry enough to spit. 'As always, you expect us to bail you out, and if we don't, you scream bloody murder and say it's all our fault. But you can't pin this one on the Agency.'
Bokus sat back in his chair, his face red and his arms crossed over his stomach. Liz could see that Fane was taken aback by the American's aggression. She had long suspected that Bokus's usual front was a pose. The bluff, rough Yank who spoke in monosyllables was, she had always been pretty sure, put on for Fane's benefit – a kind of defence mechanism against the smooth English gentleman. A tirade like this from Bokus was unprecedented, and unique for its articulate delivery, which meant that it came from the heart and what they were seeing was the real Bokus behind the taciturn façade.
Since Fane looked as if he was gathering himself for a counter-offensive, Liz decided to intervene before things got totally out of hand. She said calmly, 'I think we need to move on. Donation's gone, and we won't get any more from him, wherever he is. We need to focus now on what we've learned.'
'OK,' said Bokus. 'Donation was only the middleman. The coalface is this guy Atiyah. He's the one you've got to worry about, and he's been operating right under your noses. He's a Brit, and you didn't know anything about him.'
'For God's sake,' broke in Fane, 'how is that supposed to be helpful? We've got a British citizen gone bad – is that a unique situation? You want to tell me how the American Somalis slipped through your nets? Or the Boston bombers? Two can play at that blame game, you know.'
Liz broke in, 'Or we can accept that we both face the same difficulties and work together to sort them out.'
Fane was silent and Bokus gave her a long stare, but her words seemed to have a calming effect. Bokus threw both hands up in a parody of surrender. 'OK. But I didn't start this.'
'Oh no?' Fane said, ready to dive in again, until Liz gave him a look that could freeze stone. She continued quickly, 'Why don't we start with what we know?' Before either man could say anything at all, she added, 'Antoine Milraud the French arms dealer has decided to be a little more forthcoming. I'm not sure he's telling us everything he knows, but it's more than he was telling us before.'
'How'd you manage that?' asked Bokus. 'Feminine charm?' Liz was relieved to see him grin.
'It was the French, actually, who got him to talk.'
'Monsieur Seurat?' asked Fane.
Give it a rest, Geoffrey, thought Liz, doing her best to ignore him. 'The man Milraud met in Berlin, the black man in the museum, will be receiving a delivery of guns and ammunition in the next ten days or so, somewhere here in the UK. Originally the delivery was going to be in Paris.'
'So what's changed?' asked Bokus. Liz rather liked the way he was always happy to ask the obvious questions – whereas Fane would hold back, unwilling to admit there were things he didn't understand.
She said, 'It's looking increasingly likely that the arms aren't for use in the Middle East – why bring them all the way to France or Britain if they were? We don't know why at first it was Paris, but I'm now afraid they're intended for a terrorist attack and that it's going to take place here in Britain.' She noticed that both Bokus and Fane's eyes widened at this.
Fane said, 'You say "all the way to France or Britain" – where do we think these arms are coming from?'
'Milraud says it's Dagestan.'
'Where our friend Donation – Baakrime – is right now,' said Bokus.
Liz nodded. 'I doubt it's a coincidence.'
Bokus said, 'But he's unreachable there – for us and for you. Neither of us has any permanent post in Dagestan and we'd never get anyone in there in time to find out anything useful. If we're going to crack this open it's not going to be through Dagestan or Baakrime.'
'That's right,' Liz said firmly, determined that the question of who was to blame for Baakrime's flight from Yemen should not be reopened. 'But we still need Miles Brookhaven in Yemen on the case. If he can find out the identities of the British youths who went out to Yemen – the ones Baakrime said were planning on returning home for some purpose – then we can keep tabs on them if and when they come back into Europe.' Liz didn't really think Miles would be able to find out anything useful, but she felt it was important to keep the Americans on board. Which meant providing at least a pretence of a job for Miles Brookhaven to do in Yemen.
'That's if they haven't got new false documents,' said Bokus doubtfully.
She went on, 'We've got two potential sources of information here: the young man Atiyah, who's been the contact with Antoine Milraud – we've got twenty-four-hour surveillance on him. And this man Lester Jackson.'
'That's the black man from Berlin?' Fane asked.
'Yes. He owns a club just outside Manchester. He's well known to the local Special Branch, but it's all standard criminal stuff – drugs and the white slave trade. Since Jackson's shipped women, he'll know how to ship arms, I imagine. I bet he's being employed for his expertise in trafficking.'
'Trafficking from where?' asked Bokus.
'I wonder,' said Fane caustically.
Liz gave a resigned smile. 'Dagestan, of course,' she said. 'We know that at least one of the women in his club came from there. Anyway, we can't do much besides watch Atiyah for now – if we brought him in, it could blow the case without our finding out what he's planning to do. If he's been terrorist-trained in Yemen, he's not going to crack under questioning. We just have to hope he makes some kind of a mistake in the next week – you know, phone someone or send an incriminating email.'
'Why don't you turn him over to the Yemenis?' asked Bokus, and Fane laughed.
Liz shook her head in regret. 'I wish we could. But there's the small matter of his being a British citizen. So we'll watch him all right, but I think Jackson's a better bet. He's got no reason to think we suspect him – as far as he knows he got away in Berlin – but we know more about him than he realises.'
Fane said, 'Who's going to direct the operation to put the squeeze on this chap? Local Special Branch?' He sounded sceptical.
'No,' said Liz. 'It will be us. I'm going up to Manchester tomorrow.'
'Rather you than me,' said Fane, looking at the rain now lashing the windows and sounding pleased for the first time that day.
Chapter 40
Liz had done her homework before she'd made an appointment to see the Chief Constable of Greater Manchester Police. She already knew that Greater Manchester was one of the largest police forces in Britain and she was expecting to find that the Chief Constable was one of the old school, a man who had risen through the ranks, man and boy a policeman, near to retirement and fiercely loyal to his colleagues and to the old style of policing.
What she found was that Chief Constable Richard Pearson was forty-seven years old, the youngest Chief Constable of any of the larger forces in England. He had degrees from Nottingham and Edinburgh Universities and a D.Phil. from Oxford and had been a part-time officer in the Territorial Army for a number of years. He had risen fast through the police ranks and had been in his present post only six months, appointed as part of a push from the Home Office for a new image for policing. He had previously spent two years as Chief Constable of the Cheshire force. With all this information filed away in her mind, Liz set off in quite cheerful mood on the train to Manchester, thinking that perhaps the interview with the Chief was going to be less tricky than she'd thought.
The Police HQ was a three-mile taxi ride from the station, on a smart new industrial estate, full of brick-and glass buildings dominated by the service industries. After signing the register in the large atrium and receiving a visitor's pass to hang round her neck, Liz waited, sitting on a sofa in front of a low table spread with newspapers and assorted police leaflets, one of which was covered with photographs of Manchester's most wanted criminals. She flicked through these with interest and was not in the least surprised to find that Lester Jackson was not among them. She was musing that he was probably worse than any of those whose mugshots were on display when a young female constable arrived to escort her to the Chief's office on the top floor.
A tall, lean man with a shock of blonde hair got up from the desk as she was shown into the room and came forward with a big smile and a hand held out. As she shook it Liz said, 'Thank you for seeing me at such short notice.'
'It sounded important.'
'It is,' she said as they sat down in easy chairs in a corner of the room. 'We're working on a counter-terrorism case that involves a young British man of Yemeni origin. His mother lives in Eccles.'
'That's part of our Salford West Division,' said the Chief.
'He doesn't live there. He's a student at SOAS in London – though he's told the college he's from Yemen and he's given them false identity details. But he's Eccles born and bred and he comes home periodically.'
Pearson gave a half-smile. 'I suppose even terrorists have mothers. Do you have a name for this chap?'
'We do. His name is Atiyah. We call him Zara and we'll be briefing your counter-terrorist team as things develop. But Zara is not the reason I'm here to see you.'
Liz paused but Pearson said nothing. She went on, 'We believe that Zara is due to receive a shipment of guns and ammunition from abroad – it's coming from one of the ex-Soviet republics – Dagestan. We're not sure exactly when but believe it will arrive in the next week or so – as part of a delivery to a middleman in Manchester, who we think regularly receives shipments through this route as part of a criminal business.'
'Does the middleman have a name?'
'Lester Jackson. He owns a club called Slim's in Wilmslow – it's a combination of flash restaurant, small-time casino and brothel.'
The Chief Constable nodded. 'I know all about Jackson and Slim's. I was Chief in Cheshire before I came here. But I wouldn't have associated him with terrorism.'
'No, I understand. And I don't think he is directly involved in terrorism. I think he's just making some extra cash by adding the weapons to one of his regular deliveries. I very much doubt if he realises all the consequences of what he's involved in.'
'How did he get drawn into this?'
'I think it must be through his transport contact in Dagestan. It's complicated, but the whole affair seems to have originated with a corrupt government minister in Sana'a in Yemen. We think he has been buying weapons in Dagestan and selling them on at a huge profit to whoever will buy them. He seems to have agreed to supply a bunch of jihadis – and Zara is one of them. At first we thought they were going to use them in the Arab Spring countries, but it's beginning to look as if they plan to use them here.'
'OK,' said the Chief slowly. 'I know you'll be keeping our counter-terrorist team up to date with all this, but what did you particularly want to see me about? I get the feeling there's something else you haven't told me yet.'
Liz smiled. 'You're right – there is something else. As you said, Lester Jackson is well known to the Manchester and Cheshire forces. I gather he's never been arrested, and I'm told he's been helpful on more than one occasion. He's not actually a source, but one of your officers knows him pretty well.'
'Meaning?'
Liz paused, then said, 'This is where it gets difficult. What I mean is that your officer may know him too well. So I can't approach this officer in the normal way and ask for help with Jackson. In fact, I did speak to him on the phone to see what he knew about Jackson, and he gave me a cock-and-bull story – that Jackson was small beer of no possible interest to my Service. He was so dismissive that I didn't tell him that Jackson was suspected of being an accomplice to terrorists trying to bring arms into this country. To be quite honest with you, Mr Pearson, I just didn't trust the officer.'
'Are you trying to say that he is involved in some way with Lester Jackson?'
'Yes. I'm afraid I am,' said Liz. She could see Pearson starting to bristle. 'I should tell you,' she went on, now feeling very uncomfortable, 'that the officer in question is someone I used to know. It was a long time ago – in Liverpool where I was seconded to the Special Branch when I first joined the Service.' When Pearson looked at her curiously she said, 'The officer's name is James McManus. He's Deputy Head of your Special Branch.'
To her surprise the defensiveness she had sensed building in the Chief Constable almost instantaneously disappeared. He nodded and his face grew friendly again, and though it wouldn't have been right to say he was smiling, he somehow seemed relieved. 'Are you telling me you think McManus is involved with this arms delivery?' he asked.
'No. But I think he may be involved with Jackson in a way that isn't altogether . . . healthy.'
Pearson gave a snort. 'That's a generous way of putting it. What I think you really mean is that McManus is in Jackson's pocket, so you can't trust him to help you investigate.'
Liz said nothing and the Chief waved a dismissive hand. 'Don't worry about offending me. Forgive me if I seemed a bit chilly a moment ago – nobody in my job wants to hear that one of his senior officers isn't trusted. But in this case . . . you should know that you aren't the only one with questions about McManus.'
'Really?' She found herself half relieved and half upset, but all attention.
'Absolutely. McManus is one of our most senior officers, as you know, very experienced, with a record anyone would envy. He's popular, and said to be charming with the ladies,' he added, smiling at Liz.
She was too old to blush, and she looked straight back at Pearson, expressionless. He went on, 'But a few years ago, when I was still in Cheshire, I got the feeling that something wasn't right with McManus's relationship with Jackson. He wasn't my officer then, of course, but I did mention it to the Chief here in those days, Sir Charles Worthington. He was a very senior Chief Constable, not long to go to retirement, and frankly he was more interested in international policing and making trips all over the world than in running this Force. Anyway, for whatever reason he did nothing about it.
'But when I came here I realised that there were certain no-go areas for McManus – certain select villains he didn't want to pursue. One of them was certainly Lester Jackson. It is true that we do get the odd titbit supplied by Jackson, and McManus uses that to justify his relationship. But nothing ever substantial enough to make it right that Jackson is allowed to operate so freely from that club. And from what you're telling me he's now involving himself in even more serious crime.'
Liz listened intently as Pearson went on: 'For the last three months my Professional Standards Unit has been covertly investigating McManus. They've put together quite a significant file and we are just about ready to confront him with the evidence. We know now he's been taking money from Jackson – and others – to warn them of criminal investigations and keep the CID off their backs. His usual way of doing that is to claim they're acting as valuable Special Branch sources.'
Liz nodded, though part of her was deeply dismayed. 'Thank you for being so candid with me. As I told you, I knew Jimmy McManus more than ten years ago when I was very junior in the Service. He was kind to me when some of his colleagues were bullying me; for a time he and I were close. But our relationship ended when I thought he was being dishonest. He was convinced that a man he had been investigating was guilty of drug dealing. When the man was acquitted, Jimmy fitted him up by getting someone to give false evidence against him in another case. I've no doubt the man was a drug dealer, and I suppose you could say that in one sense Jimmy was acting on the side of the angels in those days – certainly he thought he was, and he liked to call it "conviction policing". But I thought it was corrupt, and we fell out. Still, I never thought that he'd go over to the criminals' side and take money to protect them.'
The Chief nodded. 'In my experience, conviction policemen are dangerous people to have in a force. They can easily get disillusioned and cynical, and if they have a shaky moral compass in the first place they can become thoroughly crooked. I don't know enough about McManus's personal life to understand precisely what turned him bad, but something certainly has.'
Liz said, 'I'm afraid that what you've told me just makes my problem worse. McManus is in a unique position to give me an inside view of Jackson, but not if I can't trust him. The only other possible source of information I have on Jackson is a young DI in Cheshire called Halliday.'
'Oh yes,' interrupted Pearson. 'He's a good lad, though a bit green.'
'Yes. And he's nothing like as close to Jackson as McManus is. My problem is that we've seen enough to know that Jackson's very alert to surveillance, so I can't rely on that to find out when this delivery's due or where it's going to come into the country. But if I tell McManus what I know, he'll most likely leak it all to Jackson.'
The Chief Constable said, 'Yes. I can see the problem. We need to put pressure on McManus to get him to help, but we've got to do it in such a way that he doesn't tell Jackson what's going on. In other words, we've got to scare him rigid.'
The Chief thought for a moment, then said, 'Here's what I propose. I said we were just about ready to confront him with what we've learned in our investigation. Well, we'll bring that confrontation forward and we'll do it tomorrow morning. I'll make it quite clear to him that we have enough on him to prosecute him for corruption, and if he's convicted he's likely to get a good stretch in prison, which he'll know anyway.
'If you agree, I'd like to add that we have now learned that he may be involved in acts preparatory to the commission of terrorism. He won't know what I'm talking about so I'll tell him that you will be meeting him in the afternoon. You'll be seeking his help and what you'll be telling him is Top Secret. If he doesn't fully support you or if he leaks what you say, we'll throw the book at him and he'll be in prison for the rest of his life.'
'Right,' said Liz, her breath taken away by the Chief Constable's decisive response. 'That should sort it.'
'We'll have him in at eleven tomorrow. I'll tell him to be ready to see you at two pm, but I don't think I'll tell him who he'll be seeing – unless you want me to. Surprise sometimes helps on this sort of occasion. What do you think?'
'I agree,' said Liz weakly. The Chief Constable was well into his stride now.
'If you care to come here at about one o'clock I'll brief you on how it went before you see him. Does that suit you?'
'Yes. Thank you,' said Liz.
Chapter 41
At one o'clock the following day Liz was standing in the Chief Constable's office looking out of the window. Pearson had not appeared and Liz was wondering whether that meant the interview with McManus was going well or badly.
She was feeling nervous, uncertain how McManus would react when he found out that it was his old flame Liz Carlyle who had come to put pressure on him. It wouldn't take him long to work out that she would know all about the accusations of corruption against him. Would that make him more or less willing to cooperate?
As she was mulling this over, the door suddenly swung open and the Chief Constable strode into the room. 'Good afternoon,' he said cheerfully, shaking Liz's hand with a firm grip. 'Sorry to keep you waiting. You must have been wondering what was going on. Well, I'm pleased to say that we've put the fear of God into him. He obviously had no idea that we knew what he's been up to. I suppose he's got away with it for so long under the last regime that he hadn't noticed things have changed.
'He denied everything at first, of course, but when he saw the amount of evidence my team has collected, he went silent. I think he's scared enough now that you can be confident he'll cooperate with you. When I told him that he was at risk of a charge under the Terrorism Act if he leaked anything you were going to tell him, he went pale.
'So he's all yours now. He's been taken to get some lunch with one of my investigating officers; he'll be back at two. Is there anything you'd like to ask me?'
'I don't think so. Not at this stage. I take it he still doesn't know who he'll be meeting this afternoon?'
'Correct. I just told him it was someone from the Security Service.'
'Have you any free time later on so I could look in and tell you how it went?'
'I'll be in the building all afternoon, though I've got various appointments. But Constable Symes will find me if you come back here when you've finished.'
An hour later Liz was sitting at one end of a long table in a conference room two floors down in the Police HQ, waiting for McManus to arrive. She had resolved to keep the conversation strictly on the subject of Jackson and the arms delivery and not to get drawn into any reminiscing.
When McManus walked into the room his appearance shocked her. His face was gaunt and pale and he seemed to have shrunk since she last saw him on the stage at the conference in London. His eyes were cast down as he walked into the room, but when he looked up and recognised her, they flared with anger.
'So, it's you,' he said. 'I should have guessed it would be, after your phone call. I suppose you're the one that's been investigating me.'
'Sit down, Jimmy,' she said. 'Of course I haven't been investigating you. That's nonsense, as you know perfectly well. Whatever trouble you're in has nothing to do with me and I don't even know the details of what's gone on. I'm here for one reason only, and that is to get your help with a counter-terrorist investigation.'
'That's what the Chief said, but I don't know anything about any terrorists. I just deal with ordinary villains.'
'Deal' is the right word, thought Liz. But she went on, 'I want to ask you some questions about Lester Jackson. When I spoke to you on the phone, you told me he was just a small-time crook, not one of the real bad guys. If that ever was true, it's not true now. What I'm going to tell you is strictly Top Secret and you must keep it to yourself.'
'The Chief told me that already. I don't need a lecture from someone I bedded years ago.'
'There's no point in insulting me, Jimmy. It won't help you or the investigation. Let's get the business done without getting personal and then I can go away and leave you alone.'
'To my fate,' he responded bitterly.
'So, Lester Jackson. Our information is that he's got himself involved with a shipment of weapons and ammunition coming into the country for terrorist purposes.'
'I don't believe it. That sort of thing wouldn't interest him. He's not an extremist; he's not religious. Why would he want to get involved with terrorists?'
'For money, I should think. I'm not suggesting he is going to carry out a terrorist attack himself. I believe he regularly brings merchandise into the country covertly – girls, drugs, maybe other contraband as well. I also believe he has contacts in Dagestan, the ex-Soviet republic in Central Asia where weapons are easily available. What do you know about any of that?'
There was a pause while McManus shifted uncomfortably in his chair and looked out of the window. Then he said, 'I suppose this conversation's being recorded?'
'Not as far as I know,' Liz replied.
'That doesn't mean a thing. I know that if Jackson finds out I've told you stuff about his business he'll kill me.'
'And if you don't tell me, the Chief will throw the book at you. So get on with it, Jimmy.'
McManus sighed and shuffled his feet. Then he said in a low voice, 'He has a contact with a man in Dagestan. He gets girls there and brings them in, hidden in lorries. He gets other stuff as well, drugs and legit stuff – cheap clothing that gets sold in the markets here. Some of the girls are for the club and some he sells on. It's not like proper trafficking,' he said defensively. 'Most of them know what they're coming here for, and they're so glad to get out of that hellhole that they don't mind. The ones who work at his club, Slim's, are treated fairly well. If they don't like it, they can go home, the way they came in. But most of them stay.'
'How did he get to know anyone in Dagestan?' asked Liz. 'It seems an unlikely place for him to have a contact.'
'It's someone who used to live in Manchester. I think he came in on political asylum from Chechnya when the war was on there. But I don't know much about him.'
'So he provides the goods and the transport for Jackson. What do you know about the way it's done? I mean the route, the name on the lorries, that kind of thing.'
'He told me once that they come through Turkey. I don't know what registration they have when they set off, but by the time they get here they have Bulgarian number plates and documents. I think they come into this country through different ports. I can't tell you more than that.'
'Come on, Jimmy. Do the lorries have a name on the side? How do we recognise them? We're talking about weapons coming in for terrorists to use. You don't want to have the deaths of innocent people on your conscience, do you?'
'Along with all the other things, you mean.' He put his head in his hands and muttered, 'They're Mercedes, rigid sides like a box with double doors at the back. Not huge – medium-sized, long-distance lorries, I'd say. They're blue – a sort of royal blue, with a white stripe low down on the side, and they've got DSA written on the side in white capital letters with a big white flash in front of it – shaped like a tick in a kid's schoolbook. I don't know what the letters stand for, if you're about to ask.'
Liz wondered how he could give such a good description. What exactly was his relationship with Jackson's trucking operations? But she would leave that to the Chief Constable and his team to find out.
'So can you tell me where the lorries unload their cargoes?'
'He'll know it's me, if I tell you that. Who's going to protect me when he finds out? He'll get me – even if I'm in prison. I've had it now: if Jackson doesn't get me, Mr Clean, the new Chief Constable, will.' McManus got up from the table and went over to the window; he rested his forehead on the glass and rocked backwards and forwards, gently banging his head on the pane. 'I want protection,' he said, without turning round.
'It goes without saying, that as far as my Service is concerned your help will remain entirely confidential, but I can't give any undertakings about what will happen if Jackson is prosecuted. That's for the Chief Constable.'
McManus snorted, and Liz continued, 'I'll tell the Chief how helpful you've been and I'm sure he'll arrange to have you looked after.'
'He'd be pleased if I was killed. It would save him the embarrassment.'
'Come on, Jimmy. You know that's not true. Just tell me what else you know and then we can get this over with.'
McManus sat down again, at the other end of the table from Liz and started to talk freely for the first time. It was as though something had clicked into position in his head.
'He's got four lockups – warehouses – on industrial estates. They're all off the ring road round the south side of Manchester, the M60. I could point them out on a map. There's one on the south-east side, near Denton, one near Stockport, the other two are up the south-west side, one near Sale and the other near Eccles. He reckons if one of them gets busted, he's still got the others. They're all in different names. You can get to all of them straight off the M60, without going anywhere near central Manchester. That's the attraction for him. The lorries always come in from ports on the south or the east coasts.'
'Thank you. That's just what I needed to know. Is there anything else you can tell me that will be helpful – we're trying to stop these weapons from getting into the hands of terrorists.'
He shook his head. He was knocking his wrist on the edge of the table. 'When I first knew you, Liz, I was determined to get the villains, whatever it took. The system just wasn't capable of punishing all the bastards I came across – too many were getting away with it. You said it wasn't the right way to go about it and I should have listened to you.' He added bitterly, 'I never thought I'd end up on the side of the bastards.'
He looked so drawn and despairing that Liz couldn't help feeling sorry for him. But she said nothing. She rang the bell attached to the underneath of the table to indicate that the interview was over. A uniformed sergeant came in to take McManus away. Liz didn't shake his hand, but just said, 'Goodbye, Jimmy. Thanks for your help.'
Chapter 42
By half past five Peggy, in Thames House, was talking to her contact in the Border Agency, passing on the description of the lorry that might contain the weapons. She asked for all ports to be alerted but stressed that it was most likely to turn up on the east or south coasts. 'Please let us know as soon as it's sighted, but we don't want it searched, or anything done to make the driver think he's under suspicion. We need it to get to its destination, because what we are most interested in are the people who will meet it there.'
'How are you going to keep tabs on it if we don't delay it?' asked the contact.
'I was just coming on to that,' Peggy replied. 'We want your people to put a marker on it, and that'll help us pick it up even if we miss it at the port. We'll be able to keep our distance as we follow it. Have they got the equipment at all the likely ports?'
'Yes.' But he sounded doubtful. 'That is provided it doesn't turn up somewhere very small.'
'No. Our expectation is that it'll be on a normal freight route.'
'What about the tunnel?'
'I suppose that's a possibility.'
'OK. I'll alert our people there as well.'
Peggy went to see Wally Woods, the chief A4 controller, responsible for the implementation of all surveillance requests. 'Just giving you early warning,' she said. 'It'll most likely be in the next week or so, and we may not know till it arrives at a port. The Border Agency will try to put a marker on. It'll be going up to the outskirts of Manchester. One of four possible destinations. I'll come down tomorrow and give you as much detail as we've got.'
Wally Woods grunted. 'I don't know where we'll find the manpower. We're chock-a-block already.' Wally liked Peggy but he reserved the right to be grumpy with case officers.
'Oh, Wally. Please do your best. Liz says it's really important,' said Peggy, knowing perfectly well that Wally would die in a ditch rather than let Liz Carlyle down.
Peggy's next call was to Ted Poyser. Known to everyone in the Service as Technical Ted, he was the head of its eavesdropping operations. Ted had joined MI5 from the army after a legendary career in some of the most dangerous spots in Northern Ireland at the height of the Troubles. He was getting on now and due to retire in a couple of years, so he left most of the sharp-end work to his younger colleagues, spending most of his time on planning and research.
Peggy found him at his bench in the basement workshops, surrounded, as he nearly always was, with strange-looking bits of electrical kit, wires and laptops, their screens showing changing patterns of wavy lines. Ted seemed to like to work in a clutter. Once a compulsive smoker, the smoking ban had turned him into a compulsive sweet-eater instead, and his bench was always littered with discarded sweet papers and mugs containing cold coffee dregs. As he never seemed to put on weight and no one had ever seen him eating a meal, it was widely rumoured that he lived on a diet of Werther's Originals washed down with coffee. Some of the younger intelligence officers called him 'Grandad' after the Werther's advertisements, but Peggy always addressed him as Ted. Even though Ted was nearly sixty now, he wore his hair, which was very black (unnaturally black, some said), in a ponytail, and he rode a flashy Harley-Davidson motorbike while wearing the latest in leather gear. No one, except probably Personnel, knew anything about his private life, and no one had ever met a partner, or even a friend.
Ted still liked to play an active part in the occasional, particularly interesting operation, and his eyes lit up when Peggy told him that she needed eavesdropping and cameras planted in four warehouses on the outskirts of Manchester, as part of an operation to prevent a group of jihadis taking delivery of guns and ammunition. By now one of the police officers from the Chief Constable's inquiry team had sent down the map coordinates for the warehouses, and also a description and approximate dimensions, all of which they must have got from McManus. By the time Peggy left him, Ted had summoned a Planning Team for first thing the following morning and was contentedly poring over maps.
Peggy went back to her office just in time to pick up a call from Liz, who had returned from Manchester and was back in her flat.
'How was it?' Peggy asked. 'It can't have been easy.'
'It wasn't – at first. He tried to embarrass me by making it personal. But in the end I found it rather sad. It seems such a pity that he's got himself into such a mess. And he really has. I don't know all the details, but it looks as though he's in pretty deep with some very unsavoury characters, and not just the one we're interested in. He's facing a long stretch in prison according to the Chief Constable. Whom I liked, by the way. He's young and seems very straight.'
'Thank goodness for that,' said Peggy.
'How have you got on?' asked Liz.
'I've alerted the Border Agency, Wally and Technical Ted. I got the coordinates through from Manchester, and Ted and his team are going up tomorrow.
'I've warned Wally and Borders that it could come in at any south or east coast port. Borders mentioned the tunnel and I asked them to warn them as well, though it's probably unlikely – they're much more alert there because of all the illegal immigrants. I'll get on to the French tomorrow to ask them to keep a lookout. It would be great if we could know when it's boarding the ferry on their side. The ferries to Harwich come from the Hook of Holland, so I'll get on to the Dutch as well tomorrow first thing.' She suddenly stopped, breathless.
'Go home, Peggy,' said Liz. 'You've done all you can for one day. You sound exhausted.'
So Peggy went home, but later that night she dreamed of lorryloads of guns arriving at darkened ports all round the coast – ports that she hadn't thought of – and driving off unwatched into the night.
Chapter 43
Peggy was always early for appointments. She wished she wasn't, because it often meant standing around for ages with cold feet, especially at railway stations where there were never any empty seats in the waiting areas. But she knew herself well enough to know that she would always be the same. She just seemed to have a chronic fear of being late. Today was no exception. She was waiting at Paddington Station for Jacques Thibault, the young computer wizard from the DGSE who was responsible for monitoring Antoine Milraud's computer.
With access to Milraud's computer and his password, Thibault had been able to follow current communications easily enough. He was also working hard to reconstruct the archive of Milraud's previous exchanges with the Yemeni middleman who had first introduced him to the young Arab now known as Zara.
But it was taking time and the Yemeni middleman was no longer responding to emails, as Thibault had found out when he initiated messages purporting to come from Milraud. Expand the net – that had been Seurat's instruction to Thibault, and Thibault had done his best. But efforts to work his way into Zara's system had been balked by a sophisticated firewall that Thibault quickly realised would take months, possibly years to break. And there was no question of that; Seurat had made clear that he wanted results yesterday. He had instructed Thibault to seek the help of the British, who were in any case the senior partners in this operation. Peggy in turn had passed the problem to her colleagues in GCHQ.
The day before, her contact in GCHQ had rung to say that they had got something and asked Peggy and her French colleague to come urgently for a meeting. So Peggy, early as ever, was stamping her feet and waiting for Thibault, who had caught an early Eurostar from Paris, to turn up at Paddington so they could catch the 10.15 train to Cheltenham.
Not for the first time she wondered why GCHQ had put itself in Cheltenham. It was such an awkward place to get to from London. It took at least two hours by train, and you had to change and get on an uncomfortable little local train for the last part of the journey. It was no better by car, she thought crossly as she looked at her watch again. It was still only five past ten, so there was nothing to worry about, she told herself, but she was still relieved to see the tall, slim figure of Jacques Thibault walking with long strides across the concourse towards her. She waved and he smiled back with a schoolboy grin that made Peggy feel quite motherly. With his longish, wavy hair, anorak and laptop bag over his shoulder, he looked about eighteen, though Peggy guessed he was probably more or less the same age as she was.
Thankfully for Peggy, whose French was not very advanced, Jacques Thibault had one English grandmother, which meant that he spoke fluent English, honed by annual visits as a boy to Granny Fairfax in her crumbling rectory in the Norfolk Broads. The train was crowded so they didn't talk much in any case. Peggy read the Guardian on her iPad while Thibault opened his laptop, plugged earphones into his ears and tapped away on the keyboard.
After they had changed to the local train for the final part of the journey, Peggy explained that they would be met at the station by Charlie Simmons, who had been working on Zara's communications. They would have a sandwich lunch in his office so not to waste any time. 'He said he had something urgent for us,' said Peggy. 'And he particularly wanted you to be here.'
'It's about Zara,' said Charlie as they sat down in his office, overlooking the walkway known as The Street, which ran round the GCHQ headquarters building, whose shape made it inevitable that it would be called 'The Doughnut'. 'We've been following his chat. There's a lot of it – he seems to have contacts all over the globe.'
'Yes,' said Thibault non-committally, munching a sandwich, his long body slouched in his chair.
'I can't say we've got very far,' Charlie Simmons went on. 'Most of the messages are encrypted; a few are not. They're pretty humdrum – Facebook messages to his friends, that sort of thing.'
'And the encrypted ones?' asked Thibault, in a voice without much hope. 'I know that even with supercomputers, it can take a lot of time to crack the latest kinds of coding.'
But Simmons surprised them now. 'Oh we've cracked that easily enough. Only Level Two. I'm about to send you the results, but I'm sure you'll agree they're disappointing. It's just a lot of jihadi chat-room stuff – nothing firm. It's as if they're egging each other on, but in the most general ways.'
Peggy was familiar enough with these kinds of jihadi online discussions and she was sure that Thibault was too. Death to the West; death to the Jews; death to the Infidels. A kind of OCD with Death, but rarely much specific detail about how to bring these deaths about.
'Could he have inserted more secret information in these emails?' Thibault asked. 'I'm thinking of the odd coordinates in the emails Zara sent to Milraud. I wondered if it was something a bit like those old codes, the book codes and the one-time pads that needed some sort of external reference to translate them.'
'I don't think so,' Simmons said. He sounded cheerful, and Peggy wondered why, and why he had invited them urgently to come to Cheltenham. Maybe he was one of those oddballs who were happiest with bad news.
Thibault was obviously thinking the same thing. He asked bluntly, 'So nothing to report then after all?'
'On the contrary, I've got plenty to tell you. It just came as a bit of surprise. You see, it seems our Zara is something of a mother's boy.'
'So?' said Thibault, his impatience now undisguised.
Charlie Simmons wasn't going to be rushed. 'He goes home practically every other weekend. All the way up to Manchester.' He paused, then went on. 'And while he's home he's often online – like most students these days. He takes his laptop home with him, and engages in the usual correspondence. But then something else happens, and here's the funny thing.'
'What's that?'
'His mother goes online as well.'
'His mother?'
'So it seems. It's a Gmail account in her name, and the recipients – on the surface at least – are other ladies who appear to be of Middle Eastern origin and of a certain age.'
'I don't get it.' Thibault was sitting up now.
'Neither did I. But then I had a closer look. The PC his mother apparently uses only comes to life when Zara's at home. The rest of the time it's in deep hibernation mode. I mean deep – I bet the old lady doesn't even know how to turn it on. Not surprising; it would be odder if the old lady were actually internet-savvy. I think it's pretty clear she's not. Zara's using her machine, and the people he's talking to are doing the same thing – using some unlikely dummy as the supposed sender of the emails.'
'That sounds clever,' said Peggy.
'And simple,' added Thibault.
'Yes. So simple I almost overlooked it. We could have wasted half the firepower of GCHQ on this and got absolutely nowhere, when the answer was staring us in the face. Though if you read the emails you'd be none the wiser. A recipe for tabbouleh. A discussion of how best to cook lamb shanks, with an awful lot of talk about whether it should be four and a half hours or three days. Food is the usual topic, which means numbers – one hundred and fifty grams of couscous, ninety minutes simmering etc.'
'So have you broken this food code?' Peggy was awestruck by the almost basic ingenuity of this. A circle of middle-aged Middle Eastern women, babbling about cooking techniques and recipes and food shops – perfect cover for what she assumed were in fact lethal instructions and commands.
'Pretty much. I'll spare you the details, but basically, every time numbers get used they have to be prefixed by something to indicate what they're referring to – is it time, or quantities or the geographical coordinates of a place?'
'Can't the prefix be in the numbers themselves?' asked Thibault, leaning forward, his elbows on the table, his hands supporting his chin.
'They could be, but then too often they would be the same. The repetition would be suspicious. Anyway, I've made enough progress to want to let you know.'
Oh gosh, thought Peggy. Simmons has made a breakthrough, but it's still only conceptual. He's brought us all the way here to tell us that he's cracked the code, but he doesn't know what the decoded material actually means. It was the classic folly of cryptanalysts the world over – fantastic excitement when they cracked a code, as if that were the be-all and end-all. If code breakers had run Bletchley rather than worked in it, the Germans would have won the War.
'I congratulate you,' Thibault said gravely. 'You have done remarkable work. Please keep me posted with any results that come from it.' He reached for his coat. 'I need to be getting back now.'
'What?' Simmons suddenly was almost shouting. 'It's the results I've brought you here about. Don't you want to know them? You should. There are five conspirators heading for Paris – they're going to meet up with an associate of Zara's called Michel Ramdani. He lives in Paris. He's going to send the five men on to England – it's not clear how they'll be travelling but he's responsible for the arrangements.'
'When are they due to arrive in Paris?' asked Peggy, reaching for her notebook.
'The day after tomorrow. I'd better tell you Ramdani's address. It seems that's where this little conclave is supposed to meet.'
Chapter 44
It was half past eleven, dark, windy and pouring with rain, when a small convoy set off from Greater Manchester police headquarters. There were six black Range Rovers with tinted windows. In one was Technical Ted and two colleagues, in a second, three more from Thames House. Both of their vehicles had an assortment of oddly shaped bags and holdalls in the back. In the other four were eight police officers, two in each vehicle, but only one of each pair was recognisable from the word 'POLICE' on the front and back of his black pullover. All the other men wore anonymous dark clothes.
The cars stayed in convoy as they joined the southbound M60, Manchester's ring road. Some miles on, at a junction marked Denton, one of the Thames House cars and two of the police cars peeled off, while the other three kept on the M60, circling the south of Manchester until they reached the turning for Eccles, where they too left the motorway and at a small roundabout headed into an industrial estate, led by one of the police cars. Ted, who was in the passenger seat of the Thames House vehicle, was talking to one of his colleagues in the other convoy.
'All's going fine here,' he was hearing. 'No problem with the alarm. It's just the usual Chubb as we'd been told. The whole place is quiet as the grave. There's nothing in here but empty wine crates and cardboard cartons, doesn't look as though it's been used for ages. We've put in three mikes and we're doing two cameras; Frankie's just working on the first one now. Then we'll be testing it back to the Ops Room and we should be off to the next place in less than an hour.'
'Sounds good,' said Ted. 'We've just arrived at our first stop, so let's hope it's as easy.'
The three cars pulled up on a square of tarmac outside a large metal warehouse which stood on its own, separated by at least fifty metres of grass and weeds from the next building on the narrow road. The wind was rattling the structure, making it reverberate like a drum. Dim lights on tall concrete lamp posts weakly illuminated the road and the front of the warehouse. One of the police officers came across to Ted's car.
'There's resident security on this estate. They'll be holed up in their hut on the other side. We've warned them we're doing a search here and told them to keep away. If they come out, leave them to us.'
Ted nodded. 'Suits us. We'll be inside and we'll stay there unless you alert us to get out.'
The policeman nodded, and as he did they both saw the lights of a car across the estate.
'Looks as though they're out of their box,' said Ted. 'Over to you.' And as he turned away, one of the police cars drove off in the direction of the headlights.
By this time the small door to the side of the roller door was open and Ted's two colleagues were inside. They had rigged up a couple of lights which showed that the interior of the warehouse was partitioned along one side, forming what seemed, judging by the doors, to be three separate rooms and leaving a large open space in which a lorry or several cars could be parked. It was not what Ted had been expecting. He opened one of the doors and found a room with four bunk beds in a row, very close to each other. The next room was a very small shower room with a lavatory and wash basin, and in the final room, which was a primitive kitchen, there was a pile of boxes, some open, some taped up, all of which seemed to be full of bedding – duvets, pillows and towels.
'Looks like he's expecting visitors,' said Ted.
'Or maybe he's had visitors,' replied Ted's colleague Alfie, who had come in behind him, clutching a drill. 'Some of this stuff has been used.'
'We're going to need six cameras to cover this lot,' said Alfie, 'so we'd better get going. We need to fit four mikes as well.'
'OK. While you do that I'll get onto the others and see how they're getting on. This is going to take longer than we thought.'
The other team had just arrived at their second target, the warehouse on the industrial estate near Stockport, and reported back over a mobile phone. 'Looks as though he uses this one as a store for his club. It feels quite used, as though people have been in recently. There's restaurant-type tables and chairs, boxes of glasses and china and crates of wine and beer.'
'Yeah. Well, that makes sense. It's the one nearest his club. Stick in a couple of mikes and cameras and make sure you leave it as found. Then get out asap, just in case anyone turns up. We've got him under control but he must have staff who go there to get stuff, though probably not in the middle of the night – let's hope not anyway. Then let me know when you're finished, as it may be best for you to do the last one. This one's a bit complicated and we're going to be here some time.'
As he finished speaking, one of the policemen came in.
'We've just had the alert that Jackson's leaving the club. It's about the time he usually leaves so I don't think there's anything to be worried about. We've got a static surveillance near his house and they'll report in when he gets there. And we've got a team trailing him just in case he comes this way.'
'Thanks. We'll be at least another couple of hours, so let's hope he goes comfortably to bed. Do the others know? They're nearer him than we are.'
'Yes. Everyone's been warned.'
It was four thirty by the time Ted and his colleagues were ready to leave the warehouse near Eccles, having fixed and tested enough mikes and cameras to provide comprehensive coverage of all the rooms, including the bathroom and the open garage space. There had been no more interest from the security guards, who had been told firmly by the police officers that they would never work in the security business again if they spoke a word in the wrong place. Jackson had gone straight home and apparently gone to bed; his lights were out.
It was still pitch-dark and raining as the little convoy left the industrial estate. The other team had taken on the fourth target, the warehouse near Sale, but were finding it less straightforward than their other two and they were still there.
'There's something not right with this lock,' they had reported when they'd arrived. 'It's wired up to something. Could be some sort of a remote alarm.'
'Well, for God's sake go carefully,' Ted had replied. 'Liz Carlyle and that little Peggy'll kill us if we cock it up. Send us a photograph.'
And with advice from Ted, sitting on the edge of one of the bunk beds and working from a greatly enlarged photograph on his laptop screen, they had managed to disable what was indeed a remote alarm that would have triggered an alert somewhere, possibly in Jackson's bedroom, if they hadn't noticed it. Once safely inside they had found that this building too had been partitioned down one side, to make what was in fact a set of offices. The three rooms contained desks and chairs and carpets and heaters and a number of large locked filing cabinets.
'Do you want us to open them?'
'No.' Ted made the decision without consulting anyone. 'Leave them alone. We can look at them another time when we've got someone with us who can make sense of what's in them. Just do the mikes and cameras and then get the hell out. It's getting late.'
Chapter 45
'You look tired.' Liz was watching Martin Seurat closely as they sat in the restaurant.
He started to deny it but then smiled, 'I am a bit,' he acknowledged.
'Small wonder,' she said, and signalled to the waiter to come and take their order.
It had been a long day, especially for Martin – he would have got up in the dark to catch the first Eurostar from Paris, arriving at St Pancras as most people were on their way to work. He'd taken the tube to Westminster and joined the hordes of civil servants heading for their desks in the government offices around Whitehall. Liz had given him coffee in the Thames House canteen, then they'd gone upstairs for the first of the day's meetings, a catch-up with Peggy. The three of them had sat in Liz's office while Peggy pulled together the different strands of the investigation so far. She described what had been found at Jackson's four lockup warehouses the previous night.
'It looks as though he's been using one of them to store his most confidential papers,' she said. 'That was the one with the tamper alarm on the lock and all the locked filing cabinets. The police are going to want to have a look at them when this bit of the operation is over. The only other interesting one is the one near Eccles. That looked as though it had been used for sleeping in, presumably for some of the girls he brings in. But there is space in any of them for a lorry to be parked, so if the guns are coming in concealed in one of his deliveries, they could arrive at any of the four warehouses. We've fitted them all with mikes and cameras so we should be able to see and hear what's going on. We just have to hope that we get enough warning to be able to do something about it.'
'What about the lorry that's supposed to be coming soon from Dagestan?' asked Seurat. 'Any more news on that?'
'Well, we've got the description from McManus of the type of lorry we're looking for, its colour and the name on the side. So if it's the same as usual, we should get warning from the port when it arrives. I'm hoping we might hear from across the Channel – I've alerted all the likely ports in Holland, Belgium and France.'
'It's possible we may hear something on Jackson's phone, but it's been very quiet,' Liz added. 'They're too cunning to risk phone chatter.'
'You seem to have that side of things pretty well covered,' said Martin. 'Well done.'
Peggy smiled, looking pleased.
Then they'd moved on to what Thibault and GCHQ had discovered about the jihadis. Martin said, 'It seems fairly clear that a group of Yemeni-based, English-born terrorists are heading towards England, stopping in Paris to rendezvous.' He explained that the flat of the Parisian radical Ramdani, which was going to be the meeting place, was already under surveillance by Isabelle Florian's people.
Martin went on to say that they hadn't been able to get eavesdropping inside the flat because it was in a tenement building occupied by a mixture of immigrant families and old people who had been there for years. No one was going to be able to enter or leave the flat without being observed.
At this point he paused and looked at Liz. 'We need to settle the key issue.'
'What's that?' asked Liz.
'Are we going to arrest these people when they arrive at this flat in Paris, or are we going to keep them under surveillance and let them come on to you?'
'I've discussed this with DG and he's talked it through with the Home Secretary and the Chief Constable in Manchester. The Home Secretary wanted us to ask your colleagues to make arrests. She said that we couldn't take the risk of allowing a gang of jihadis into the country when we might not be able to keep them under our control. But DG pointed out that there may be nothing for your colleagues to hold them on, particularly if they carry no weapons. They may well have perfectly valid documents. So she's agreed that you should just follow and watch and hand them on to us. We need to know what they're planning to do before we act.'
Martin nodded. 'I was hoping you'd say that. That is the view of Isabelle and the Interior Ministry, and my own Service agrees. But we do have to remember that there's always the chance, however good the surveillance, that they could give us the slip between Paris and Britain.'
'We just have to take that chance. If we detain them now, we have nothing to charge them with – even in France, they'll be out within days. Besides, there's every chance that others are joining them in the UK – not just Zara. If we grab this bunch the others may find out, and then we'll never locate them.'
Martin was smiling now. 'Clear, as ever. Let's hope the others think so too.'
'Frankly,' said Liz, 'it doesn't much matter if they don't, now we have the Home Secretary's agreement.'
'The others' had been Geoffrey Fane and the CIA Head of Station Andy Bokus. Bokus was already in Fane's office when Liz and Seurat arrived, and judging from the chilly silence they were not enjoying each other's company.
When Liz introduced Seurat, Bokus merely grunted and looked grumpily out of the window, as if he wished he were somewhere else.
'Cheer up, Andy,' said Fane. 'You'll find life south of the river isn't all that bad' – a reference to the impending move of the US Embassy from Grosvenor Square to a new, more isolated but thought to be safer, location in Wandsworth.
Liz noticed that the CIA man was losing weight, though not much – his suit was a little looser at the shoulders than it once would have been, but his buttoned-up jacket did his bulging midriff no favours.
They'd all sat down and waited awkwardly while Daisy brought in a tray of coffee.
'Don't bother, Daisy,' said Liz. 'I'll pour it out.' As she reached forward to pour out the coffee, she'd noticed that Bokus was already drumming his thumbs on the arms of his chair impatiently.
When the coffee was poured, Fane said, 'Elizabeth, why don't you bring us all up to date?'
Liz had been startled by how rude the two men were being to Martin. Bokus hadn't even acknowledged his presence when she'd introduced him and now Fane was behaving as if he wasn't there. But she made no comment and proceeded to summarise the situation. When she finished there was a heavy silence.
Bokus said gruffly, 'You mean to tell me, you got five bad guys – I mean really bad guys – right within your sights, and you want to let them come on here to do God knows what?' He was staring at Liz and sounded incredulous.
'We don't have any intention of letting them do anything. Nor do the French.'
'No. We certainly do not,' said Martin Seurat.
Bokus ignored him – it was Liz he was going for. He said in the folksy voice Liz had always been wary of, 'Listen, I'm just a country boy from Ohio. Sometimes I get a little lost if anything gets too complicated. But we used to say back home that a bird in the hand beats two birds in the bush any old day.'
'Did you really say that?' Seurat asked with feigned innocence, and Liz just managed not to laugh. She noted that Fane was staying quiet.
For a brief moment Bokus's eyes flashed, but he stuck to his Huck Finn persona. 'We sure did,' he said, still looking only at Liz. 'And I'm thinking it applies here pretty well. Why risk losing these guys if we can pick 'em up easier than a bird dog grabs a grouse?'
'Why indeed?' muttered Fane.
Liz was about to reply when Seurat broke in. He said simply, 'Here is why.' He looked at Bokus with a steeliness Liz had never seen before. 'The initial information in this case came from you, the Americans. Believe me, we are all grateful for that. And then, the focus shifted to here in the United Kingdom – this man Jackson appeared, and we learned that these British Yemenis are on their way to this country, almost certainly to commit an atrocity.
'But the fact remains, they are meeting first in Paris. And we believe they were originally considering Paris as the target of their operation – whatever this operation is.'
'Not any more—' Bokus started to say. Seurat held up a hand and the American stopped.
'Hear me out, Monsieur. My point is that Paris has already featured in this case – this is where Zara and the arms dealer Milraud met, and where I fear the other side first suspected they had been observed.'
'Whose fault was that?' Bokus demanded.
'Ours. Not all of us share the American infallibility. In any case, Paris is now again the focal point of this operation and of our cooperation.' He looked around at them all. 'Naturally, we need to respect each other's point of view and to take dissenting opinions into account. But you will appreciate that since this part of the operation is taking place on French soil, then we – the French – must make the final decisions about it. So, since you are asking' – which, thought Liz, no one was – 'I must tell you that I agree with our colleagues here. We will not arrest the jihadis who are meeting in this apartment, and instead we will follow them to their exit point which we all believe will be the UK border.'
Seurat took a deep breath. 'I am sorry if you are not in accord with this, Mr Bokus. And I know that you think this will be the weak decision of another one of those cheese-eating surrender monkeys. But it is the monkeys' decision nonetheless.'
This speech had produced a startled silence in the room. Even Bokus had looked embarrassed in the face of Martin's eloquence. When Liz seized the opportunity to say that the Home Secretary, the DG and the Chief Constable of Greater Manchester police had all agreed to let the operation run to the UK, no one had anything more to say and the meeting had broken up in a chilly atmosphere of recrimination.
Now the waiter arrived and Liz said, 'So what do you want to eat, my cheese-eating friend?'
Seurat laughed. 'I'll just have a starter, I think. They will feed me on the train.'
'Somehow after an hour with Andy Bokus, I don't feel very hungry either – just a starter will do me too. But I need a glass of wine.'
When the waiter had left, Seurat sat back and sighed. 'You OK?' asked Liz.
He smiled. 'Yes. That was just a sigh of relief. A day I am glad is over. Though I will be happier when tomorrow is over as well.'
'Are you worried about it?'
Seurat shrugged. 'No more than I would be normally. Isabelle and her people are in charge, and I have every confidence in them. Thibault seems quite sure that what GCHQ have told him is right. He says it all makes perfect sense. It should be fine, and with any luck they will all be in the UK the day after tomorrow. Then it's your problem,' he said, with a smile.
'Thanks a lot,' said Liz with an affectionate grin. Martin seemed more like his old self now, and she was relieved to see it. His put-down of Bokus hadn't bothered her one bit – in fact, she'd loved it. It was such a change from the catlike way Geoffrey Fane danced around their American colleague. Though it had been direct, it had also been controlled, with no sign of the irritability Martin had been showing recently about Milraud.
Their food arrived, and they ate quickly, talking now of anything but work. Liz told him how her mother, whom he had met several times, had thought about giving up work at the nursery garden she ran, and how her partner Edward had dissuaded her since he rightly sensed she'd go mad if she didn't have enough to do. And Martin talked about his daughter; he was worried about what she'd do after she graduated from the Sorbonne.
It was funny, thought Liz, that when things had been tense between them they had not talked about personal affairs at all; now she felt they were back on their old intimate footing again and it made her happy.
She said, a little reluctantly, 'Tomorrow, will you be there?'
Seurat raised his eyebrows. 'At Ramdani's flat? No. Only the surveillance will be there. I will be with Isabelle and we'll be sitting safe and sound in the DCRI HQ. Nothing to worry about.'
'Good,' she said, forcing a smile. She wished she felt less worried about this operation. She was used to the mix of apprehension and excitement that came just before the action, but somehow this time it felt different. She reached across the table and held Martin's hand. 'There's a train at the crack of dawn, you know.'
He tilted his head back and smiled. 'And how tempting it is. But I should go back tonight.' He shook his head. 'I'd never forgive myself if something went wrong tomorrow and I wasn't there.'
'But you said there was nothing to worry about.' Liz kicked herself for letting her concern show.
Martin put one of his hands on top of hers and looked into her eyes. 'There isn't. But I just feel I need to be there. You'd feel the same, wouldn't you?'
'Of course I would. You're quite right.'
Martin looked at her. 'It'll soon be over.'
'I hope so.'
'And when it is, I was thinking . . .'
'Yes?' asked Liz.
Martin was smiling. 'You remember the hotel in the hills near Toulon?'
'How could I forget?' They had begun their affair there. She remembered the flowers in the garden of the small auberge where they had stayed as spring arrived.
'I thought a few days there would not go amiss.'
'D'accord,' said Liz. 'I'd like that very much.'
'Good,' said Martin. 'I'd like it too. Because I love you very much, Miss Liz Carlyle.' And then, as if embarrassed by his display of emotion, he signalled furiously to the waiter for the bill.
Chapter 46
'It's at moments like this,' Isabelle Florian declared, 'that I miss cigarettes the most.'
She gave Seurat a wry smile, and he nodded. 'I know. Anything is better than the waiting.'
Not for the first time, Seurat thought how fortunate he was to have Isabelle as his counterpart at the domestic intelligence Service. Relations between the DGSE and the DCRI were almost always tense, fuelled by the same kind of competition that seemed to affect domestic and external intelligence services the world over. But whereas Liz had to put up with the know-it-all patronising of Geoffrey Fane, Seurat had long ago established an excellent relationship with Isabelle, one based on mutual respect and by now a genuine liking for each other.
They sat in the operations room in the building that housed the DCRI. It was a windowless and low-ceilinged space, with a series of desk consoles ranged in a half-circle at one end to face a row of large screens that hung from the wall. At the moment just two of them were active. One screen showed a distant shot of the entrance to a tall grim-looking tower block, and the other, its picture obviously coming from a concealed fixed camera, showed the length of a passageway with one open side. You could clearly see the doors of individual flats that ran off the passage; the one in the centre of the picture belonged to the flat of the suspect, Ramdani. But there was no sign of anyone moving in either of these camera views.
At the centre console Alex Carnier, a veteran DCRI Operations control officer, struggled to suppress a yawn. He had a headphone set dangling loosely around his neck, and on the desk in front of him a microphone sat on its stand. He was directing the surveillance operation, but seemed happy enough to have Isabelle and Martin Seurat watching him work.
He turned his head to Isabelle and said, 'They're late.'
She shrugged. 'You said that five minutes ago. They could turn up at any time. What do you want me to do? Ask them to hurry up?'
Carnier gave a grin full of yellow teeth; unlike Isabelle, he pretty clearly hadn't given up cigarettes. The new regulations banning smoking in public buildings applied here too; Seurat thought it must have been hell for a twenty-a-day man.
'It's only been half an hour,' said Seurat mildly. 'There may be some reason for the delay.' Though Thibault had been very specific: the latest decoded email had said four o'clock sharp for the rendezvous at the flat.
Carnier brushed his greying hair back with one hand, then leaned forward and spoke into the microphone. 'Team Three, anything alive out there?' he asked, more in hope than expectation.
There was the crackle of a car radio, then a voice replied dully, 'Rien.'
Isabelle had explained to Martin that there were six teams, each of three people, on the operation. Most of them were in cars, parked safely out of sight, though there would also be a few surveillance officers on foot around Ramdani's tower block, which along with half a dozen other relics of some bright city planner's 'vision' in the 1970s sat on the edge of Seine-Saint-Denis, one of the constellation towns just north-east of Paris. It was all public housing, now inhabited overwhelmingly by first- and second-generation immigrants. Seurat often wondered whose idea it had been to create these hellholes so far away from the rest of the city's life. Or perhaps that had been the rationale: to deposit the North Africans who'd flocked to France in the aftermath of the Algerian war out of sight of the public face of the city, known the world over for its elegance.
Isabelle said, 'Martin, if you want some coffee there's a machine down the hall.'
'It's broken,' said Carnier. 'But there's a café on the corner.'
'I'm fine,' said Martin. Sod's law said that if he went out now the jihadis would show up.
But three hours later there was still no sign of them. Not that there had been any sign of Ramdani either. The surveillance had begun the day before, with only one team, but Ramdani hadn't left his flat in that time. A light in the front room suggested that he was there, but it had stayed on all night and that, taken with the failure of the others to arrive, meant that his presence in the flat was now open to question. Seurat said as much to Isabelle.
'I know,' she said. 'That's worrying me too. What if they changed plans and are meeting somewhere else?'
'It doesn't seem likely or we'd have seen Ramdani leave his flat. Thibault says GCHQ will notify him immediately if there's any change of plan.'
'Still, I'd like to make sure. Alex,' she said, turning to Carnier, 'I'd like to establish if Ramdani is actually in the flat. Any ideas?'
Twenty minutes later they watched on the screen a young man walking along the corridor of the tower block. He wore a parka and trainers, and carried a sheaf of flyers advertising a local takeaway pizza joint. Carnier said his name was Philippe, and that he had been with the DCRI for less than a year. 'But he's good,' Carnier said. 'He wanted to be an actor but he got tired of waiting at tables to pay his rent.'
They watched as Philippe began at the far end, ringing the buzzer of each flat one by one. Most of the doors were answered, sometimes by small children, always with the chain on, and Philippe would give them one of the pamphlets, then move on to the next apartment.
When he got to Ramdani's door he paused, and looked around. The corridor was empty as he rang the buzzer. He waited a good thirty seconds but no one answered, so he buzzed again. Still nothing. Philippe knelt down and looked through the letter box, then he stood up and moved over to peer through the small window to the side of the door of the flat. His voice came over the speaker on Carnier's desk, saying quietly, 'Nothing doing. And no sign of him through the window. I can see into the living room.'
Carnier said, 'Are you sure the buzzer's working? Maybe you should knock.'
'I can hear the buzzer from outside. The walls of this place are paper-thin.'
'Maybe he's in the shower – or asleep. Try knocking.'
So this time Philippe knocked on the door as well as pushing the buzzer again. 'That's enough,' said Isabelle. 'He'll alert the neighbours and they'll think it strange he's so persistent.'
But it was too late, the door to the next apartment opened and an old lady with a walking stick came out, remonstrating furiously. As Philippe beat a hasty retreat, Carnier gave a laugh. The old lady was still shouting at him as he reached the far exit of the corridor by the lifts.
'Well,' said Isabelle. 'At least we've learned his neighbour's nosy. Could be useful.'
'She was as mad as a hornet,' Philippe said when he got downstairs and left the building. 'She told me that if her neighbour wanted a pizza he would have answered the door. I asked if she'd seen him recently.'
'And?'
'She said . . .' and he hesitated.
Seurat and Isabelle leaned forward to hear. Carnier snapped impatiently, 'What did she say?'
'She said I should piss off.'
Carnier looked at Isabelle. 'What do you think?'
'I don't think he's in there.'
She looked at Seurat. He thought she was probably right. He knew what he wanted to happen next, but it was risky. Liz would never forgive them if they blew the whole operation. He waited for Isabelle to speak – this was her operation after all, even if it came at his instigation. In one sense he was only a privileged guest here.
Isabelle said, 'I think we should go in and have a look.'
Carnier said, 'You sure? What happens if Ramdani comes back while we're in there?'
'We'll put people outside to stop him coming in. We tell him that a smell of gas has been reported and we're required to check it out for safety reasons. So no one is allowed into the building while we do that. If he goes away your people will follow him, Alex.'
'And if the others turn up?'
'We tell them the same thing. They have to wait till the all-clear is given. I hope none of that happens, but I think it's a chance we have to take. It doesn't make any sense that the meet didn't take place. Ramdani wouldn't have had much time to change the plans, so it would have to have been a change from the other end. But whatever's happened, we're now out of touch with the jihadis – that's worrying, to say the least. I'm hoping there may be something in the flat that indicates what they're up to.'
As Carnier digested this, he looked at Seurat, who stayed silent – Isabelle had just decided to do what he'd been hoping she would, but it was certainly risky and he wondered what Liz would have done.
Finally, Carnier shrugged. 'OK. I'll set it up.'
Isabelle nodded. 'We'll need the locksmith – I don't want any doors broken down. The idea of this is to keep it as quiet as possible. A discreet entry, a quick search by officers with concealed weapons to make sure he's not there, then I'll go in.'
'I'll come with you,' said Seurat immediately. Carnier looked at him dubiously. Isabelle said quickly, 'It's all right, Alex. Martin knows more than anyone else what we're looking for.'
Chapter 47
The sky was pitch-black now, and the corridor was only dimly lit by the few bulbs that were working. Seurat could just make out the two armed officers who were to go in first in case Ramdani was there. They were wearing fluorescent yellow jackets with 'GAZ' written on the back. One of them carried a bag, its contents not usual for a gas fitter.
Their instructions were to get inside the flat as quickly and quietly as they could and to avoid, if at all possible, attracting the attention of the other inhabitants in the block. The previous summer had seen three nights of rioting at this estate, which had started when the arrest of a drug dealer had gone wrong and a child had been shot. The last thing Isabelle wanted was anything to happen that might make trouble flare up again, alert Ramdani and the jihadis and send them off on another tack.
As the 'gas men' approached the door of Ramdani's flat, Seurat and Isabelle waited at the end of the corridor. It was dinner time and quiet except for the sound of television coming from the flats. At each end of the corridor two more men in yellow GAZ jackets stood ready to detain anyone who tried to come along the corridor.
The first policeman rang the buzzer and knocked on the door. They all waited tensely in silence, and suddenly a door opened. But it was not the one to Ramdani's flat. An old lady came out of the next apartment; Isabelle groaned.
'Who the hell are you?' the old lady was demanding. She had her stick with her and waved it threateningly at the men standing by Ramdani's door.
'It's perfectly all right, Madame,' said the senior of the men politely. 'Just go back inside, if you don't mind. We've had a report of a smell of gas coming from this flat and we need to go inside to investigate. It is probably nothing serious but we need to check. Do you know whether your neighbour is in?'
'I've no idea where he is,' she said. 'It's not my business to keep track of my neighbours. But if he's not answering the door, I suppose he's out. What are you going to do – break his door down?'
'No, Madame, there'll be no need for that,' replied the 'gas man', soothingly. 'Now if you'd just like to go inside out of the cold, we'll check it out. I don't think there's any problem but better safe than sorry.' And he ushered her gently back inside her flat and waited until she closed the door.
There was a sigh of relief from the end of the corridor as Isabelle let out her pent-up breath. 'He did well,' said Seurat. 'That old lady reminds me of my grandmother. Terribly nosy, absolutely fearless, and won't take any nonsense from anyone.'
The man with the bag started work on the lock. It took him only seconds to have the door open and the two officers went inside. Isabelle and Seurat went up to the front door but they waited outside by the door as the armed officers went through the apartment. After a few minutes they came back to the external corridor. 'The place is empty.'
'All right, thank you very much,' Isabelle said. 'Stay here please. I don't think we'll be very long, and then you can go and tell the old lady that everything's fine. Hopefully she'll forget about us.' In her grey parka and jeans Isabelle cut an unremarkable figure, but she was clearly in charge. 'All right, Martin? Time to have a look around, eh?'
They started in the living room at the front of the flat. There were thin curtains hanging at the window but they were not drawn. An unshaded bulb hanging from the centre of the ceiling was switched on. The room was tidy but minimally furnished with a threadbare sofa, one grubby armchair that had a rip in the fabric on its back, and a low table marked by the rings of mugs and glasses, on which was a two-day-old copy of Libération and a newspaper in Arabic.
The centre of the floor was covered by a faded carpet, the floorboards visible round the edges, and there was an electric fan heater in one corner that wasn't plugged in. There were no cupboards, desks or anything else that might contain papers, and no laptop or other electronic device. A new-looking television set on a stand in one corner of the room was the only thing that looked as if it had cost any significant sum.
Down the corridor, back towards the front door, was a small kitchen. There were cupboards above and below a worktop on one side of the room; the officers had left them open. Seurat peered in at their small collection of tins, cereal boxes, small bags of flour and sugar, a carton of salt, and an old jam jar full of couscous. The fridge was almost empty – wilted stalks of celery, three eggs in a little rack, a half-full milk carton and a chunk of hard cheese that looked as though it had been there for a good long time.
'If this guy was expecting visitors he hasn't exactly stocked up to feed them,' Seurat said.
Isabelle was examining the oven and grill. 'Thank God it's gas,' she said. 'I suddenly wondered if these flats only had electricity.'
Opposite the kitchen was a small bathroom. There was a bath with a shower over it and a plastic shower curtain. It was bone-dry. No one had taken a shower or a bath for a long time. The sink was streaked with the detritus of Ramdani's last shave – little black hairs that studded the basin like steel filings. On the porcelain top a razor lay carelessly askew, next to a can of shaving foam, its cap off. Seurat opened the mirrored front of the small bathroom cabinet and saw one stick of roll-on deodorant, a box of plasters, a pair of tweezers and an opened pack of razor blades.
'Looks as though he doesn't have a beard,' was Seurat's comment.
Behind him Isabelle was pulling at the wooden slats on the side of the bath, but they wouldn't give. She said, 'I don't think he's hiding under there. Not if he ever planned on coming out.'
Seurat snorted. 'I have to say, this all seems unnecessarily grim. I don't have any clear picture of Ramdani, but this flat barely feels lived in.'
'I know what you mean. After three days here any sane man would jump off the balcony outside. I bet he hasn't been here long. And wasn't planning to stay much longer.'
Seurat nodded. 'But where has he gone? And why isn't he here now? I don't get it.'
'Come on. Don't let's hang about. There's still the bedroom to check.'
The bedroom did nothing to lift their spirits. It held a double bed, and a cheap-looking desk with four drawers, containing a few pens and pencils, some rubber bands, paper clips and a lot of dust but no paper. A metal chair, a small bedside table on which sat a little lamp and nothing else, and a built-in cupboard that contained one empty suitcase completed the furnishings.
'Honestly,' said Seurat, gazing at the paltry contents of the room, 'this could be a doss house. Do we know anything about this guy?'
'Not much,' admitted Isabelle. 'We're working on it, but for now we've only official records. We couldn't exactly ask around here if we didn't want him scared off.'
'So what do we know?'
'He's a native Parisian – Yemeni mother, French father. The father is not in evidence, and the mother is dead. Ramdani grew up half a mile from here. He's twenty-five, and when he left school he went and lived in Yemen for a few years – at least that's what Immigration say; if he went anywhere else it didn't get stamped in his passport. He's on benefits now, but used to work in a little bar down the street – again, we didn't want to ask questions; I got this from the Office of Employment.' She added a little defensively, 'We can find out a lot more than that, of course. But at this stage discretion seemed to be preferable to a lot of inquiries.'
Seurat sighed, looking around the dismal room. 'I'm just frustrated the jihadis didn't arrive. And this guy seems to have disappeared and left virtually no sign of himself. Your people are sure he couldn't have got past them somehow?'
'Absolutely. We've had the camera up for thirty-six hours. No one's been out of this flat – or in. You've seen for yourself there's no other way out. And if he were hiding in here, we'd have found him by now. There isn't room to swing a cat in this place. He must have left before we started watching. That's the only explanation, isn't it?' she added, since she wasn't sure Seurat was paying attention to what she was saying.
He wasn't. He had gone over to one corner of the room and was looking up at the ceiling, where there was a square metal grating.
'What is it?' asked Isabelle, slightly annoyed. Martin was always so inquisitive, she thought, even when it just wasted time.
'It must be some kind of central heating system. There are no radiators in this place and it's not freezing cold,' he said, purposely misunderstanding her.
'What about it? The officers will have checked that out.'
'Maybe. But maybe not.' He picked up the metal chair and put it under the grating. 'There are no screws in this. How do you think it stays in place?'
'Heaven knows. It probably rests on a ledge. Can't you see?'
As he stood on the chair the cover was only inches above his upraised hand. He reached up and gently pushed. One corner of the grating lifted and then dropped back into place. He pushed again with both hands and the entire square cover lifted up and he was able to move it over to rest on the inside of the ceiling. He stared up into the hole above his head. He poked one arm through until it disappeared into the gap, and felt around in the blackness. Then he climbed down.
'Find anything?' asked Isabelle sarcastically. She shared Seurat's frustration, but couldn't see the point of what he was doing now.
'Not yet.' He reached into his pocket and took out a small metal torch. Then he repositioned the chair to place it directly below the opening in the ceiling and climbed back onto it again.
'You're not thinking of going up there, are you? Let the officers do it. They're younger than you.'
'Don't worry – if anyone's going to have to crawl along a shaft it won't be me. I just want to take a look to see where it goes.'
He hoisted both arms up into the gap, holding his torch in his teeth, and leaned his elbows on the ceiling. Then before Isabelle could protest Seurat pulled himself straight up into the air until his head disappeared into the opening. He's strong, thought Isabelle admiringly despite herself, for inwardly she thought this was all a waste of time.
'Come on down, Martin,' she said, staring at his legs hanging in the air. He must have replied, but his voice was muffled by the surrounding walls of the shaft.
'What did you say?' Isabelle half shouted, and just then he dropped back down again, missing the chair and falling onto the floor with a thud.
'Are you all right?' Isabelle was standing over him and held out a hand. But he sat up, shook his head and said, 'I'm fine. My arms suddenly gave way.'
'I told you it was a job for a younger man,' she said unsympathetically. 'What's up there anyway?'
'There's a duct, quite wide. You could crawl along it if you were slim – and young,' he added with a smile. 'It must run along the top of all the flats in this corridor. I can't swear to it, but I thought I heard something moving up there.'
'It was probably rats. Or the pipes heating up.'
'Mmm. Perhaps. We need to find out exactly where it goes. There may be an exit he could have used.'
'Why would he want to go out that way?'
'He may have spotted the surveillance and put two and two together.'
'Yes, and he may have invisible powers too,' she replied caustically.
'But don't you see? That would explain why the others didn't show. He may have warned them.'
'I don't believe it,' said Isabelle. 'We've been very careful.'
Martin shrugged. 'Well, whatever. But the duct's certainly a possible way out – and in as well. Do you think we can get a plan of where it goes?'
'I'm sure we can from the building management company. But I can't see much point in doing it now. There's nothing here to help us learn what the jihadis are planning to do, or where they are. And if you're right and Ramdani saw the surveillance, he's probably not coming back. If we stay here much longer we're going to have the old lady next door coming out to find out what's going on. I think we ought to leave now and get the "gas men" to tell the old lady that everything's fine. Then we can get the plans tomorrow and see whether it's worth sending someone to explore the ducting.'
Although he hated leaving the job unfinished, Seurat couldn't think of any reason to object to Isabelle's plan, so they put everything back as it was in the flat and went towards the front door where the two officers were waiting in the passage.
They closed the door behind them and Isabelle and Martin Seurat began to walk off down the corridor as the officer rang the old lady's bell. Nothing happened. So he rang again and put his ear to the door, listening for her. Then the officer called out to Isabelle, 'I think you should come.'
'What is it?' she said as she and Seurat walked back.
'Listen.'
Isabelle bent down and opened the letter box. She could hear a gasping, choking sound.
She said, 'I think she's ill. Sounds like a heart attack. Open the door.'
The lock was no more difficult than the one on the flat next door and within seconds the door was open. Martin elbowed Isabelle out of the way and went in first. He's acting as if the old bird is his grandmother, Isabelle thought with amusement.
The flat had exactly the same layout as the one next door and the sounds were coming from behind the closed door to the living room in front. Martin pushed the door open and saw the old lady standing up, held on her feet by a thin, dark young man. He had one arm round her neck and with his other hand he was pushing a revolver hard into the side of her throat. The old lady's eyes were open but only the whites were showing; her mouth was slack and saliva was dribbling out and down her chin. Her skin was a bluish white and there was a raw, rattling noise coming from her open mouth.
'Let her go,' shouted Martin. 'Can't you see? You're suffocating her.'
The young man, whom Isabelle recognised from the photos as Ramdani, tightened his grip on the old lady's throat, and pointed his pistol at Martin. He didn't look much more than twenty years old, thought Isabelle, and he looked frantic.
'Stop it!' Martin commanded. 'She's choking. She can't breathe.'
Isabelle added, trying to sound calm, 'Put the gun down. We don't want anyone to get hurt. And let the lady go.'
The man stared at Isabelle, and for a moment she thought that her words had got through to him. Martin must have thought so too, for he took a step forward and extended his hand. 'Just give me the gun.'
Ramdani relaxed his grip on the old lady's throat, but instead of handing over the gun, he held his arm straight out and fired.
Isabelle watched in horror as the shot hit Martin square in the chest, its force knocking him to his knees. Immediately one of the armed officers behind her raised his own weapon and fired back.
Ramdani's face creased in agonised surprise. He dropped the gun as his legs gave way, and he knocked down the old lady as he fell.
There were three bodies on the floor now, but only one of them was moving. The old lady was gasping and shuddering, the other two were still. One of the officers was on his phone calling for backup and medical assistance, Isabelle was kneeling on the floor, holding Martin's head up, his blood running over her hands and down his jacket. She was shouting, 'Martin, Martin,' but he didn't respond and she knew that he was dead.
Later Isabelle could only dimly recall the sequence of events that followed the shootings. Looking back she realised that she had acted automatically to try to prevent a public furore. She had sent the officers in their GAZ jackets outside to explain to interested bystanders, attracted by the ambulance and the presence of the police, that there had been a gas emergency, and that an old lady had had a heart attack, but the emergency was over and the old lady was still alive and was going to hospital.
She had insisted that the bodies of Martin and the young man, presumably Ramdani, once they had been formally declared dead, be left where they were until the middle of the night, when they could be taken out secretly.
She had stayed, at first sitting on the floor beside Martin, tears running down her cheeks, then sitting in the kitchen making the dreadful but necessary phone calls. Throughout this, some of her colleagues had thoroughly searched the old lady's flat. It was obvious how Ramdani had got in. The grating in the bedroom was off and there was a gaping hole in the ceiling. Why he had chosen her flat no one could explain, unless he had gone into the ducting when he heard the knocking on his door and thought he could hide there. Or perhaps he had heard Seurat come up into the crawl space, and panicked. He might have thought it would be safe to hide in the old lady's flat, or possibly he'd thought he could escape through her front door, until he'd realised that the officers were outside in the corridor. When they'd broken in, he must have intended to use her as a shield for his escape.
By the time the medical team returned to remove the bodies, Isabelle was back in the living room sitting beside Martin. Before Ramdani was taken out, she ordered a policeman to search his body thoroughly. She was glad she did – in a trouser pocket they found a folded train schedule. It was for the Eurostar from Paris's Gare du Nord to London.
Watching as Martin was zipped into a bag and taken away, Isabelle thought how unnecessary his death was, and cursed herself for letting him push ahead of her as they came into the flat. Like the grandmother he had been telling her about, and like the old lady who had now been taken despite her protests to hospital, Martin had been absolutely fearless. And curious, fatally curious.
The only good thing to come out of this whole dreadful night, she told herself, was that it was now pretty certain that the terrorists were heading for England.
Chapter 48
At the safe flat in Paris, Annette Milraud was in the kitchen making a late supper. Her husband Antoine was with her. Martin Seurat had decided to move Antoine from the Montreuil house to share the flat, judging that he was likely to cooperate more if he was with his wife than if they were kept separated. As well as the guards, Jacques Thibault was there this evening too. He was monitoring Milraud's laptop and phone for any messages from Zara or the contact in Dagestan – any communication at all that might throw light on what might happen next. If need be, he could immediately ask Milraud to explain.
Annette poked her head round the sitting-room door. 'Would you like to join us for supper?'
'No, thanks,' said Thibault. 'I'll stay here.'
As well as Milraud's laptop, he kept checking his own for any news of the operation at Ramdani's flat. From the kitchen he could hear the low murmur of the Milrauds' conversation. Once Annette gave out a loud groan, and he heard Antoine say, 'It will be all right, I promise.'
It was about eleven o'clock when the landline phone rang. Thibault picked it up, thinking with relief it must be Isabelle at last. But it was a man's voice. He identified himself as a senior police officer. 'Am I speaking to Monsieur Thibault?'
'Yes,' said Jacques, warily, wondering why on earth a police officer had his name and this number.
'I have been asked to ring you by Madame Isabelle Florian.'
'Is she all right?' asked Thibault.
'Yes. But she wished me to tell you that there has been some shooting at a flat in Seine-Saint-Denis. The occupant of an apartment has been shot dead.'
The policeman seemed to hesitate and Thibault sensed that there was more to come. 'Is he the only casualty?'
The policeman said slowly, 'One other person was shot. He is also dead, alas.'
'I'm sorry,' said Thibault, thinking it must be some poor policeman who had gone first into the flat. Thibault barely registered what the caller said next – 'A Monsieur Martin Seurat from your Service, I believe' – but then the words sank in.
'Martin Seurat? Are you sure?'
'Positive, Monsieur. He was dead when he reached the hospital. I am so very sorry.'
In the background Thibault heard Annette clearing the table in the kitchen. He thanked the policeman for calling and hung up. He would learn the details later on; right now, he was too stunned to take in much more than the death of a senior officer of the DGSE.
'What's wrong?' Milraud was in the doorway to the kitchen, eyeing him suspiciously.
Thibault stared back at him. 'There's been a shooting.'
'Where?' Milraud asked, bewildered. Milraud had not been told anything about Ramdani or the anticipated arrival in Paris of the group of jihadis, but that didn't stop Thibault's mounting anger.
'In a tower block The wrong man got shot. Martin Seurat is dead.'
'What?'
'I said Seurat is dead.'
A plate shattered on the floor in the kitchen. A moment later Annette appeared in the doorway. 'What did you say?'
'I think you heard me.'
She looked at Thibault with disbelief, her arms outstretched. For once Antoine didn't try to comfort her but sat down heavily in one of the sitting-room chairs. He was clearly stunned, one hand on his forehead, his head bowed.
'But why?' asked Annette, as tears began to trickle from her eyes.
Thibault sensed that she must have had feelings for Seurat. He said, 'I don't know the details. Obviously something went badly wrong.' He stared angrily at Milraud.
Annette was crying openly now. 'But this is too dreadful.'
'I know,' said Thibault in a cold voice.
Milraud looked up. 'How can that have happened? I never imagined anything like this.'
'Oh no?' said Thibault. 'What did you think was going to happen when you met that Arab in the Luxembourg Gardens? What did you think would result from your meeting in Berlin? Did you think it was all just a harmless game?'
Milraud said, 'Martin was my colleague for years. Whatever our later differences, he and I were once very close.'
Thibault looked at him incredulously. 'You talk as if you were old pals who sadly no longer saw each other. We all know your story – they use you as a case history of betrayal in the Ethics lecture when we join the Service. So don't try to whitewash your past; it just dirties the name of a man who was widely admired. One who died trying to prevent the harm you were encouraging.'
Milraud sat up. 'You are blaming me for his death? I've told you everything I know.'
'No doubt.' Thibault shook his head in contempt. 'What a pity you couldn't have done it earlier.'
Ten minutes later Thibault sat gazing at the screen of his laptop but not seeing it. He could not have tolerated any more talking with either Milraud, but thankfully they had withdrawn to their bedroom. There was no one for him to phone: Isabelle would be busy for hours now, or she wouldn't have asked a police officer to break the news.
Then his mobile phone bleeped and the screen lit up. It was a text message from Peggy in London:
Charlie has just unzipped message: expected party in Paris cancelled. Group delayed leaving Yemen, now going straight on to UK. Ramdani to make own way and join them there. Sorry so late in letting you know. Problem with decoding. Peggy.
He stared blankly at the screen now, trying to still a surge of nausea. Perhaps if there hadn't been a decoding problem and the message had come through earlier, Martin Seurat would still be alive.
Chapter 49
Liz was lying on her bed in her Kentish Town flat, shoes off but still fully dressed. Isabelle had promised to let her know as soon as there was news of the group of jihadis, due to arrive at four o'clock at the flat in Paris. But she had heard nothing before she left work at seven and still nothing three hours later, by which time she had stretched out on her bed, with both her phones beside her. She was half asleep when her landline started ringing. She sat up and grabbed the handset.
'Hello.'
'Liz, it's Peggy.'
'Oh, Peggy. I thought you might be Isabelle. Have you heard anything from Paris?'
'No. But it will have been a no-show. That's probably why they haven't rung. I've just heard from Charlie Simmons. There's been a message in the cooking code. It was sent this morning but it's taken him ages to decrypt because it was full of mistakes. He thinks whoever sent it didn't properly understand the rules and it was badly encoded. Anyway he's managed to get into it and apparently it says that they're not going to Paris after all. They're coming straight on to Britain. I'm just about to text Jacques Thibault. They must all have been wondering why no one turned up at the flat. They were probably hanging on, hoping they were just late.'
'Yes, but I'm surprised they didn't let us know that no one had appeared. I wonder what they've been doing. I'm going to ring Isabelle now to see what's going on.'
'OK. While you do that I'll text Jacques. Then I think we need to warn A4 that Zara might be on the move soon. Because if his friends are on the way here, they may arrive tonight, and he's the only angle we've got on them.'
'Yes, and when I've spoken to Isabelle, or Martin if I can't get her, I'll warn the Manchester Counter-Terrorist Group that we may have some action for them soon. Our friends may well be heading for one of those warehouses.'
Liz put the phone down and was just about to pick it up again to ring Isabelle when her mobile suddenly came to life.
'Liz, it's Isabelle.'
'Hello. We've been wondering where you were. I gather no one turned up. You've must have had a rather boring evening.'
There was a pause. Then Isabelle said, 'Well, actually that's not quite true.'
'Why? What happened?'
'It's true the people we expected didn't show – but we were puzzled why and we decided to search the flat to see if there were any clues to what was going on, and our man didn't seem to be there. But he was there – he was hiding in the next-door flat and, Liz, I'm so sorry . . .' Her voice crumpled.
'What is it? What's happened?'
She could hear Isabelle sucking her breath in, trying to pull herself together. Then she managed to say, 'Martin, he was shot.'
'Shot?'
'Liz, I am so sorry. Martin is dead.'
Liz went ice-cold. She didn't want to believe what she'd heard. She took a deep breath, trying to control herself, and said as calmly as she could, 'What happened?'
While Isabelle explained, Liz tried to focus, to listen. But the words ran like noisy flowing water in the background while one brutal fact kept occupying the foreground – Martin was dead. Isabelle was explaining that when the jihadis hadn't shown up, she and Martin had taken a gamble and gone in, hoping to find evidence of what was being plotted. Martin had been curious, Isabelle explained – and Liz thought, damn Martin, he was always curious.
And it was here Liz completely tuned out, not wanting to hear the details of the death of the man she loved. Isabelle was still talking as a thousand images flashed through Liz's head: of her first meeting with Martin at the DGSE's old-fashioned headquarters on the outskirts of Paris; of Martin down at Bowerbridge, her family home, and the way he had taken to the place – so quintessentially English, he'd said; and of how Martin had chuckled when he'd seen the childhood relics Liz still kept in her bedroom there – the rosettes from gymkhanas, the watercolours she had liked to paint as a girl, and the photograph taken by her father as she stood gap-toothed and beaming and no more than nine years old, holding perhaps the titchiest fish ever to be yanked (and that with some grownup help) from the waters of the river Nadder.
All this came at her in a concentrated rush, which made her smile momentarily – though each time she had a loving image of him the grim news of his death intervened, and her memories fell away like waves hitting an unexpected reef.
She became aware that Isabelle was no longer talking. Liz did her best to pull herself together. She said mechanically, 'Thanks, Isabelle, for letting me know.'
'Liz, did you hear what I said? I said I thought you would want to come over.'
'Of course. Should I be arranging things?'
There was an awkward pause, and Liz suddenly realised that she had no real position in this. She hadn't been Martin's wife, not even legally his partner; officially, she had no real status in Martin's life at all.
She asked Isabelle, 'Have you told Mimi?' Martin's daughter.
'Not yet.'
'Or Claudette?' Martin's ex-wife, who lived in the Brittany countryside. It had not been a happy divorce – she had left Martin for an old boyfriend – but lately they had re-established speaking terms and could discuss their daughter civilly enough. Martin's bitterness at his wife's desertion had obviously been intense, but she remembered now how once as they were having coffee after dinner, he'd explained that since Liz had come into his life, his anger with his ex-wife had evaporated.
'No, I haven't called her yet. Listen, Liz, give me half an hour and I will phone you back. But remember one thing. You were the most important person in Martin's life.'
'It's kind of you to say that, Isabelle.' She was doing her very best not to sob but her eyes filled with tears.
'I'm not just saying it to be kind – he told me often enough.'
It was long after midnight when Isabelle called again. In the intervening time Liz had got up and made coffee, checked her diary for appointments the next day, rung Peggy and told her the news and that she'd be in Paris tomorrow, then asked her to tell DG about Martin. She went online and booked a ticket on the Eurostar, then put a few things in an overnight bag, just in case. Finally, having run out of diversions, she collapsed in an armchair in her sitting room. She sat still for several minutes, slowly composing herself. She didn't actually want to think any more about Martin just now – it was too painful. But quite unbidden, the memory of their last meeting came back to her, and she thought of his words – Because I love you very much, Miss Liz Carlyle. And suddenly she started to cry, then cried and cried until she could cry no more.
When her tears were utterly exhausted, she got up and went to the bathroom and washed her face. As she dried it the phone rang.
It was Isabelle again. 'I have reached Mimi and Claudette, Liz.'
'I hope they are all right.' She had little sense of Claudette. Early on in their relationship it had been clear that Martin didn't want to talk about his ex in any detail, something that Liz had always been grateful for, since it meant there were no shadows hanging over them.
'Well, Claudette was shocked, of course. I don't know if you ever met her.'
'No, I didn't.'
'She likes to control things in her life, Liz, so the unexpected tends to throw her at first, then she reasserts control, if you understand.'
'Yes,' said Liz, but she wasn't sure what Isabelle was getting at.
'At first she decided there must be a funeral right away. I explained that couldn't happen. Because of the circumstances there will have to be a post-mortem and there may be an inquiry, though it will be secret of course. Everything is being done to make sure there is no publicity – at present anyway – as we don't want to alert Zara and his friends. And I told Claudette you should be consulted.'
'Thank you,' Liz said mechanically. She didn't really feel able to cope with all this at present.
'She didn't like that – not because it was you, Liz; she has no axe to grind, but because she always wants to decide everything herself. But she did say she would be happy to have you attend the service.'
'That's big of her,' said Liz. Then she took a deep breath and forced herself to focus. 'I don't think there's much family, Isabelle. Martin's parents are both dead and he was an only child. My real concern is what Mimi wants. It's her wishes we should follow here.'
Liz had only met the girl once. Martin's relations with his daughter had been strained after the divorce; living with her mother, Mimi had not surprisingly sided with her in what had been an angry parental split. But since coming to Paris to attend the Sorbonne, she had begun to see her father regularly, and relations had improved immeasurably. When Liz had met her, not in Martin's flat but on the neutral ground of a café, conversation had been polite but strained at first.
Then Martin had excused himself to make a phone call and Liz had admired Mimi's new pair of boots, and suddenly they had begun to talk freely about all sorts of things – clothes, films, and why they hated cigarette smoke and were glad Martin had given up, and whether Paris was rainier than London – and their conversation was so spontaneous and friendly that when Martin had come back from making his call, he felt (as he said affectionately to Liz later that evening) that he was almost surplus to requirements.
Now Isabelle said, 'Actually, I have just come off the phone to Mimi – that's why I am so late ringing you again. Her mother broke the news to her, and of course she is distraught. I'd given Claudette my number and Mimi must have got it from her. At first, she wanted all the details of her father's death. To tell you the truth, I ducked that, Liz. I hope you think that was the right thing to do.'
'Yes,' said Liz, thinking that she didn't know the details either. She hadn't been listening when Isabelle was telling her what had happened. 'She'll learn all about it in due course,' she said, thinking, So will I.
'She wanted to make sure you'd been told, but she didn't have your number. I think she was relieved to learn that I'd already been in touch. She said she hoped you would come over right away. She'll take this very hard but I'm sure your presence here would be a great comfort.'
'I will,' said Liz. 'I'll be on the Eurostar that gets in at quarter past ten. But I don't really know Mimi at all.'
'I'll send a car to meet you and take you to the flat. Right now you are the one link to her father. She said that the last time she saw Martin he told her he hoped to marry you. He told her everyone has a true love in their life but not everyone is lucky enough to find them. He said he was one of the lucky ones.'
Chapter 50
Peggy Kinsolving liked to wake early – one of the best things in life was having a job she was eager to get to. In her earlier incarnation as a librarian, there had been mornings when she could barely get out of bed, especially in the dark winter months, but ever since she'd joined MI5 there had never been any problem about getting up.
This morning, however, she was fast asleep when her alarm rang at 6.30. After Liz's phone call telling her the dreadful news from Paris she had just sat in a chair for half an hour, everything spinning in her head. She hadn't been able to make up her mind whether she should ring DG straightaway or wait until morning. Should she ring Geoffrey Fane? She seemed to be immobilised, as if all the stuffing had been knocked out of her.
Then suddenly she had pulled herself together. What would Liz do in my shoes? she'd thought. Well, she wouldn't be sitting here like this. Peggy had long ago observed that the worse the situation, the more calmly Liz behaved, and she had drawn strength from Liz's cool efficiency. Well now, she said to herself, I must do the same. So she'd grabbed the phone, dialled DG's PA and passed on the news. 'He'll want to know now,' was the advice, so Peggy had rung him. DG had asked for an update on the operation and had told her that she must be the main liaison with Manchester Police until Liz was able to take over again. She'd then rung the Duty Officer at Vauxhall Cross and given him the barest account of what had happened to pass on to Geoffrey Fane. She had decided to leave informing Andy Bokus until morning. Having done all that, she began to feel better about herself and got into bed. But it was past two o'clock and her mind was racing. She was thinking what she must do in the morning; how awful Liz must be feeling; whatever could have happened in Paris – and so it went on until she fell asleep at about five o'clock, only to be woken an hour and a half later.
When she got to Thames House, Peggy found that word had already spread about Martin Seurat's death. A few colleagues asked her what had happened, but she didn't know any more than they did. As more people arrived for work, they were also greeted with the news. Soon an almost palpable gloom settled over the open-plan office where Peggy had her desk. Liz was a very popular colleague, much admired by the younger officers. It was widely known that her partner was a DGSE officer whom she'd met when she was posted to Northern Ireland, and that an operation there had ended violently in the South of France. Some people knew that Martin Seurat had saved the life of Dave Armstrong, one of their colleagues, who had been kidnapped. So Seurat was something of a hero in the Counter-Terrorist branch, even though not many people had met him.
When everyone had arrived for work Peggy told them all that Martin had been killed in Paris in the course of the operation they were working on; that Liz had gone to Paris and would be getting a full briefing but at present she could not tell them exactly what had happened. DG had asked her, she explained, to stand in for Liz until she was back. She was going to move into Liz's office for the time being and she'd asked for her calls to be put through to Liz's extension. If anything relevant came through to any of them they were to come in and tell her immediately. She might well be going up to Manchester very soon. Then she went into Liz's office, closed the door and set about trying to get to grips with what was going on.
An hour later she felt like a salesman who'd made the rounds but come back with an empty order book. She had begun by calling Charlie Simmons at GCHQ. He'd had the news of Seurat's death by now, and sounded very subdued. There had been no further email traffic to or from Zara since the email had come in announcing that the meeting in Paris had been cancelled. He explained again that the reason why it had taken so long to unzip that message was that it seemed to have been sent by someone who was not familiar with the code. 'This may mean that those who usually send the messages are not there,' he said.
'That makes sense,' said Peggy. 'Presumably the messages are usually sent by the people who are on their way here. But they must be communicating somehow or how are they going to meet up with Zara?'
'However they are doing it, we're not onto it. Perhaps they made all the arrangements in advance and don't need to communicate.'
'Maybe so,' said Peggy, but she was sceptical. The silence seemed ominous.
Next she checked in with A4 and was told that Zara was acting like a model student, attending lectures, working in the library. 'Completely normal,' said the Duty Controller. Too normal, thought Peggy, sceptical again.
Finally she checked with her contact at the Border Agency. He was in constant contact with their counterparts on the Continent and no vehicle of the description she had given him had been reported crossing the Channel or the North Sea. Peggy asked, 'What if the vehicle were coming in a roundabout way?'
'How do you mean?'
'Say from much further away than the usual Channel ports. Like Scandinavia – ferries from Norway come here, don't they?'
'Sure, though that wouldn't help them escape detection. There isn't a port within five hundred miles east of here that hasn't been given the details of the vehicle we're looking for. And just to be safe, we circulated them to Ireland too. In fact, unless this lorry's coming from Brazil you don't have anything to worry about. If it sails, we'll know.'
'All right,' said Peggy, tempted to ask him to cover Brazil as well, but even she could see that was absurd. 'Thank you,' she added, realising that perhaps she had been a bit rude. She was getting very tense. It wasn't just the aftermath of Seurat's death and the absence of sleep, it was the absence of developments. No news was usually good news, but right now Peggy wanted something to happen.
Chapter 51
It had been a really tedious few days. Maureen Hayes had wanted to take the week as holiday because her son was home on leave from Afghanistan, but she'd been told she had to work. Wally Woods, her A4 controller, had said that they needed all the resource they could muster to cover what was thought to be a developing terrorist plot.
But so far nothing had happened. The target Maureen and her team were covering, Zara, had gone reliably as clockwork every day of the week from his hostel, Dinwiddy House, to SOAS, where he had attended lectures, sat working in the library and drunk coffee in the snack bar with other students. He did not seem to have any close friends whom he met regularly but he chatted in a friendly enough way to whoever was around. She and her team had been unable to get near enough to overhear any of his conversations but everything looked perfectly natural. Then at about five or six in the evening, he had left the university area and gone back to Dinwiddy House, where, according to her overnight shift colleagues, he had stayed until the following morning. If he was plotting a terrorist outrage, thought Maureen, he must be doing it from his room, as there were no outward signs of a conspiracy.
Today was Friday, and at the early morning briefing before they took over the surveillance, she and her team had been told to be extra-vigilant. Something that had happened the previous day in Paris had led the desk officers to think that a group of possibly up to six people would be arriving in Britain, if they were not already here, and Zara would be meeting up with them. They were thought to be intending to carry out some form of terrorist attack, but what, where and when was not known. It was vital, they had been told, that if Zara broke his routine or met a group of people who had not been seen before, they reported at once; and above all that they did not lose him.
So Maureen and her team were very alert this morning, and rather disappointed when Zara came out at the usual time and headed off to SOAS just the same as on all the previous days. Marcus Washington went into the building and reported that Zara was in a lecture. After the lecture he went to the library, where he was reported by Marcus, by then sitting two places away from him, to be concentrating on a large book from which he was taking notes. Just before twelve noon, he looked at his watch, packed up his things, returned the book to the desk and came out of the library.
'On the move,' said Marcus quietly into his microphone as Zara left the library to be picked up by Maureen and her partner, Duff Wells, as he came down the steps.
'Having an early lunch,' reported Maureen to the Control Room. But instead of heading off to the snack bar where he usually went at lunchtime, Zara walked quickly out into Tottenham Court Road, ran straight across, narrowly avoiding being run over by a bus, and headed fast towards Goodge Street underground station.
'He's doing anti-surveillance,' reported Maureen as Wells, who had anticipated the move and was already on the other side of the road, went into the station ahead of Zara. Maureen caught up, arriving at the station as Zara and Wells with a small group of passengers were waiting for the lift to take them to the platforms. Maureen, Wells and Zara, with about fifteen other people, piled into the ancient lift, which creaked its way down and juddered to a halt at platform level. Zara was first out, hurrying along the tunnel to the southbound platform.
'Doesn't look as though he's going to see his mum,' reported Maureen. 'Euston is north.'
Then began a short tour of the underground system as Zara, with Maureen and Wells accompanying him, went south on the Northern Line to Tottenham Court Road, west on the Central Line to Oxford Circus and finally back north on the Victoria Line to Euston, where he took the exit for the mainline station. Each time he changed trains he hung back and tried to be the last onto the train, but Maureen and Duff Wells knew all about that anti-surveillance ploy and, helped by the crowded platforms, one or other of them managed to board the train after him without drawing attention to themselves.
Half an hour later, as Maureen, now ahead of Zara, emerged up the stairs from the underground onto the concourse of Euston Station, she was relieved to see another colleague, Fred Watson, standing in the crowd in front of the departure board.
As she followed Zara towards the booking hall and watched from the door as he collected a ticket from the fast ticket machine, she heard Fred talking to the Control Room. 'There's a Manchester train at one o'clock; we'll go with him if he catches it. Gets there at seven minutes past three.'
'OK,' came back from Wally Woods. 'I'm alerting the police to meet the train at all the stops. I'll get a team out to meet you in Manchester. Keep us posted.'
Back in the main concourse Zara joined the crowd in front of the departure board, where he stood waiting, watched from different directions by the three pairs of eyes of the A4 team.
As soon as the platform for the 13.00 train to Manchester Piccadilly flashed up on the board, Duff Wells moved fast, ahead of the crowd, towards Platform 5 and Fred Watson followed, more casually. Maureen stayed in the concourse waiting for Zara to move too. But Zara didn't move. Maureen muttered into her microphone, 'Watch out for a last-minute rush. He's still here and he's very alert for surveillance.' At 12.55 Zara was still on the concourse.
Then suddenly he moved fast, out of the concourse, towards the platforms. 'On the move,' said Maureen. She was trying to keep up with him, but she lost sight of him in the crowd of people now rushing to get seats on another train. 'Control lost,' she shouted as she ran towards the platforms.
Fred and Duff were still waiting at the top of the ramp leading down to platform 5, but there was just a trickle of latecomers now and Zara was not among them.
'Pretty sure we haven't missed him.' It was Duff Wells. 'Fred got here before anyone else. Between us we've clocked everyone who got on.'
As Maureen ran up to join them, Wally Woods said, 'Try the next train', over their headphones. 'Thirteen-oh-three, Platform seven, for Birmingham.'
'I'll wait here till this train leaves in case he's just delaying,' panted Maureen as Duff and Fred set off running to Platform 7 where the stragglers were still boarding. Duff waited at the end while Fred sprinted along the platform, scanning the passengers without much hope of seeing his target, but then near the far end of the platform he spotted Zara, just about to get onto the train.
'Got him,' he shouted. 'Second carriage. I'm boarding now.' Duff joined a chattering group of grey-haired men dressed in walking clothes who were getting into a carriage in the middle of the train. Last to arrive was Maureen, clambering into the final carriage, just before the doors were locked and the guard signalled the driver to go. She stood leaning on the door, gasping for breath, her heart pumping at twice its normal speed. I'm getting too old for this, she thought to herself.
'Phew,' she heard Fred say. 'That was a close one. But we're still with him. I've got eyeball. He's just three rows in front of me.'
'OK,' said Wally from Thames House Control Room. 'Well done.'
'The train stops at Rugby, Coventry and Birmingham International; Birmingham New Street is the last stop,' continued Fred.
'Get off where he does, but I'll try to get the police to be at the stops along the way – I'm hoping they'll be able to take him on if he gets off before New Street. I'll get our teams to meet you at New Street in case that's where he's going.'
Rugby and Coventry came and went and it wasn't until Birmingham International was announced that Zara got up and joined the line of passengers waiting to get off the train.
What on earth is he up to? wondered Maureen. Don't say he's going to a conference – not after all this trouble.
But it wasn't to the Conference Centre he was heading. As soon as he left the train, he made a beeline for the Skyrail to the airport and got on the first train that came in, with the A4 team in hot pursuit.
'What do you want us to do if he checks in for a flight?' asked Maureen.
Wally replied, 'You'll have to let him go. But get all the details.'
But at the airport Zara didn't go to the departure hall; he went instead to the arrivals hall, and straight to the Hertz car-hire desk.
'He's hiring a car. We've got no wheels so we'll have to let him go or hire one ourselves.'
'Get the number and make of the car and we'll pick him up on the road. There's a police team coming out now to join you.'
As the A4 team watched, Zara hired a dark blue Ford S Max and drove off, heading for the airport exit.
While Wally Woods in London passed the target to the police surveillance teams, Maureen and her colleagues went off in different directions to get some lunch in the airport cafés. By the time she had finished a not very enticing salad, Maureen heard over her headphones that Zara's car had been picked up by the cameras, heading towards the M6 Toll. That was a silly choice if he's trying to avoid surveillance, she thought. He'll be on camera all the way.
Chapter 52
Peggy had been staring out of the window, feeling as sluggish as the Thames at low tide, when the phone on Liz's desk rang.
'Hi, Border Agency here. I think we have something for you.'
'Where?'
'Hook of Holland. They called five minutes ago. There's a Stena Line ferry leaving for Harwich at fourteen thirty their time; that's half past one here, so fairly soon. Scheduled arrival time at Harwich is twenty hundred hours, British time. The lorry came in just before the deadline – they have to be quayside sixty minutes before sailing. It's got the markings you're looking for, though it's carrying Turkish registration plates.' He read out the registration number. 'Just one driver, Turkish passport, name of Deniz Keskin, date of birth thirtieth October 1963.'
'I bet that's a false passport. If that's our lorry it's come from Dagestan and he's not Turkish. What's it carrying?'
'Mattresses. Lots of mattresses, according to the manifest.'
Plus a few other things, thought Peggy. And she asked, 'Has anyone looked inside?'
'No. The Dutch are giving it a bit of space – as we requested. You said don't scare them off.'
'That's right.'
'It was weighed – all the vehicles are, so that can't have aroused suspicion; it was apparently normal weight for its declared load. But it's hard to tell much without looking inside. We can have Customs search it when it arrives if you want. Easy enough to do.'
'No, thanks. We don't want to risk alerting them at this stage. But please ask them to try and put the marker on as it goes through.'
As Peggy put the phone down she was hoping she'd taken the right decision. It was a big risk to allow into the country a lorry that she was pretty sure was carrying weapons, detonators and heaven knew what else, intended for a group of jihadis who had gone off the map and could be anywhere in the country. But she didn't have much time to worry about it. As soon as she put the phone down, she picked it up again and rang Wally Woods in the A4 Control room.
'Hi, Liz.'
'No, it's Peggy. Liz is out today.'
'Oh?'
Obviously the news from Paris hadn't percolated to the A4 control room. Peggy said, 'I'm running the op until Liz gets back. I've just heard news of our lorry from the Border Agency. It's on board the Stena ferry at the Hook of Holland coming to Harwich.' She passed on the details she'd been given. 'They're going to get the marker on at Harwich.'
'OK. We'll be there. You still reckon it's headed for one of those warehouses?'
'Yes. But we don't know which one. If I learn anything else I'll let you know. Anything new on Zara?'
'Yes. He's made a move. I was just about to pick up the phone to tell Liz when you rang. Is she OK by the way? It's not like her to leave her post just as things start hotting up.'
'Yes. She's fine but someone close to her has died.' She hoped she'd said enough and not too much.
'Oh. I'm sorry to hear that,' said Wally and went on, 'Zara took a train to Birmingham.'
'Birmingham?'
'Yeah. He's doing anti-surveillance but not all that cleverly. He took the Skyrail from the train to the airport and now he's in a hire car. Last seen heading towards the M6.'
'Oh God. Have you lost him?' asked Peggy, thinking of the lorryload of weapons she had just agreed to let into the country.
'No. Not as you might say "lost". We're not with him at the moment but we know roughly where he is and what car he's in, so the police teams will be behind him soon. He'll be on the cameras, and if he takes the M6 or the Toll, he'll be snapped every few hundred yards. And we can always stop him at the Toll gate if we need to. The paying system can break down for a bit.'
He must be heading towards Manchester, thought Peggy. Nothing else makes sense.
'Our team is ready to join in in case he goes off the M6 up a minor road,' went on Wally. 'Don't worry, Peggy. I think we can cope with little Mr Zara whatever he does.'
'He may be picking up some others somewhere.'
'Yeah. That occurred to us. He's hired a big enough car.'
'You've got all the addresses he might be going to, haven't you?' asked Peggy anxiously. 'The four warehouses and his mother's house.'
'Relax, Peggy. We've got it all in the brief. And we're in touch with Manchester CT Unit.'
'OK, Wally, thanks. Keep me posted please. I'll be on my mobile.'
'You going somewhere?'
'Yes. I'm going up to Manchester to liaise with the police. I'll be in the Ops Room up there.'
There was no point in hanging around in London. Not with both Zara and the lorry apparently heading for Manchester. So Peggy went back to the open-plan office and told the others where she would be, then headed out of Thames House and hailed a passing taxi. As she leaned back in her seat, she pulled out her mobile. The last thing Liz probably needed now was a phone call, but knowing Liz she would be wondering what was going on and, after all, she had asked Peggy to keep her informed. So Peggy sent her a text:
Off to Manchester – lorry and Z on their way.
The other package's whereabouts still unknown.
Hope you are all right. PK
She hoped Liz wouldn't be away too long. She wasn't at all sure that she could fill her shoes.
Chapter 53
Peggy made it to Euston with just enough time to buy a ticket. The train to Manchester was packed but she managed to find a seat that wasn't booked, though she had to ask a rude young man to move his coat and briefcase so that she could sit down. As the train pulled out of the station she closed her eyes and rehearsed in her mind everything that had happened and what she thought was about to happen. She was worried that they had seen no trace of the jihadis. Where were they? Were they travelling together or separately? Perhaps they were on the train. Perhaps they didn't exist. Had they all misinterpreted the intelligence? And if they did exist and were on their way to meet Zara, what was it they were planning to do?
She was relieved that Manchester Police had set up an Ops Room. The responsibility to prevent whatever was planned no longer lay entirely on her shoulders. The police were now in charge of the action and she was their adviser.
Her thoughts drifted to Paris and to Liz. She wondered what she was doing and how she was getting on. What had happened in Paris the previous evening and why had Martin been shot? She tried to imagine the chain of events but she couldn't make any sense of it.
When the refreshment trolley came through the carriage she realised she was starving. She had had no lunch and hardly any sleep the night before. She bought a sandwich and a black coffee and began to feel a bit better. She tried to relax, watching the reflections in the window and the bright lights of occasional stations. She had a feeling she wouldn't be relaxing again for a while.
There was a long queue for taxis at Manchester Piccadilly Station, and when Peggy eventually got to the front and the cab drove off, she remarked to the driver how busy the place seemed. He laughed. 'It's the pop concert.'
'Who's playing?'
He named a boy band Peggy had only vaguely heard of and added, 'It'll be worse tomorrow. There's another performance and the match – United's playing City at Old Trafford. There'll be gridlock, so I think I'll stay at home.'
At Police HQ Peggy signed in at the front desk. 'Third floor,' she was told. 'They're expecting you.'
When the lift doors opened she found a tall, youngish-looking police officer waiting for her. It took a minute before she realised who it was.
'I'm Richard Pearson, the Chief Constable. You must be Peggy.'
'Yes,' replied Peggy, rather breathlessly. 'Good evening.'
'I wanted to meet you to say how pleased we are to have you with us – but also how sorry I was about the sad events in Paris. I don't know exactly what happened but I understand that Liz Carlyle has lost someone close to her. Please pass on my sympathy when you see her.'
'Thank you,' said Peggy, very surprised. 'None of us knows the details, but Liz has gone over there and I expect she'll have heard the full story by now. It seems that the group of jihadis changed their plans. They seem to have bypassed Paris and now we think they're coming straight here. Your people will be more up to date than me – I've been on the train for the last couple of hours.'
'Yes,' replied the Chief Constable. 'There have been some developments. Let me take you into the Ops Room and introduce you. The officer in charge is Chief Superintendent George Lazarus, Head of our Counter-Terrorist Unit. He'll brief you on what's going on.'
He led her down a corridor and into a large, brightly lit room. A big square table with chairs around it filled one end, and at the other a line of eight or ten desks, each with a computer, a phone and headphones, faced a wall of screens. A large digital clock on the wall showed 8.27 pm.
The desks were all occupied; there was a mix of men and women, some in uniform, some in plain clothes, some talking on the phone, some tapping on keyboards, some sitting back in their chairs. The atmosphere seemed busy but calm.
The Chief Constable introduced Peggy to Chief Superintendent Lazarus. Then with a quick, 'Let me know, George, as soon as anything starts to happen,' he left.
'Come and sit down and have a cup of coffee and a bun and I'll tell you what's going on. Then I'll introduce you to the team,' said Lazarus, shepherding Peggy to the table. He was a big man, with large hands and feet. He quite dwarfed Peggy. As they sat down he picked up a paper from the table. 'There was a call for you from Thames House Duty Officer about half an hour ago. He said that someone rang on one of your agent lines and asked you to ring back. Here's the number.'
'OK, thanks,' said Peggy, taking the slip of paper and glancing at it before putting it in her pocket. 'I'll ring them later.'
'Right then,' said Lazarus. 'The situation at present is that the Stena ferry carrying the lorry should be just about in to Harwich. The lorry will be allowed through Customs with no fuss, as you requested, and a marker will be put on covertly as it goes through. We have surveillance waiting to go with it wherever it goes. If it comes up here, as we expect, it should arrive any time from two o'clock onwards, provided it doesn't stop or get lost. Zara in his hire car has arrived at his mum's house in Eccles. We have three teams of A4 there, but they're having to stand off a bit as the area is difficult for surveillance. They are doing drive-bys and Zara's rental car is still there outside Mum's house. If he goes out they should pick him up. My only worry is if he leaves over the garden wall, but that's unlikely if he's going to make contact with the lorry. He didn't seem aware of surveillance. I gather he led your lot a bit of a dance on the way here, so he probably thinks he's clean now.
'I'm sure you're briefed on McManus,' he went on. 'Well, he's working with us now. He's got no choice,' and he smiled grimly. 'He's been told that if he doesn't hear from Jackson, he's to drop by the club at about twelve thirty and try and find him. If Jackson's going to meet the lorry he should make a move any time from one o'clock onwards. We've got an armed team standing by and we're going to conceal a couple of surveillance officers by the entrances to each of the industrial estates to warn us of who's coming in. We've got all the comms and the cameras coming in to us here, so this is Mission Control,' he said with a grin. 'But what I want to know from you is what's happened to your band of terrorists. I gather they didn't turn up in Paris.'
'No,' said Peggy 'but we're pretty sure they're out there somewhere and intending to meet up with Zara. What we don't know is what they're planning to do.'
'Well, let's hope we find out before they do it,' said Lazarus, sucking his breath in through his teeth with a faint hiss. 'Now come and meet the people.'
They walked side by side across to the desks. At the first desk was Lazarus's deputy, a balding man with a pate that gleamed in the bleaching glare of the overhead strip lighting. His headphones were hanging round his neck. Lazarus said, 'Andy's got all the surveillance comms on his desk. What's happening at the moment, Andy?'
'Not a lot,' was the reply. 'The ship's just docked.' Andy turned a knob and the sound of the A4 teams at Harwich, talking to each other and to A4 control, floated into the room.
Peggy and Lazarus moved along the line of desks meeting all the officers. A young woman Detective Sergeant, Emily something, was monitoring the cameras that Technical Ted and his team had placed at the warehouses. 'Do we know yet which warehouse they'll be going to?' she asked Peggy.
'No. 'Fraid not,' Peggy replied. 'Could be any of them. The one in Denton seems to hold all the paperwork of Lester Jackson's club, but the one in Eccles has beds.'
'Let's have a look, Emily,' said Lazarus. She leaned forward and clicked her mouse. Suddenly the screens on the bank of monitors on the wall cleared, replaced a moment later by views of the warehouses. Two were old brick buildings that looked pretty run-down; the Denton facility was a long, hangar-like building and the Eccles one was a large aluminium shed that was indistinguishable from those dotting the outskirts of every town in England.
Technical Ted and his team had put cameras inside and outside each warehouse, and Emily panned through the pictures from each.
A curly-haired man called Ames who had his headphones on sat up quickly and raised his hand.
'Yes?' said Lazarus.
'McManus has heard from Lester Jackson. Jackson wants to meet him at Slim's at midnight. McManus wants to know if he should go.'
No one said anything for a moment. To Peggy's surprise she saw they were all looking at her. Yes, she thought, it was a question for her to answer.
'He should go. Definitely,' she said. 'I don't know why Jackson wants McManus there, but it gives us an opportunity to know where Jackson is while we wait for the lorry.'
Ames said, 'Jackson may suspect McManus.'
'Good point.' It was Lazarus now, giving his view. 'But we'll have to take that risk. It would only create more suspicion if McManus refused to meet him.'
'But what if Jackson doesn't go to the warehouse?' asked Ames.
'If Jackson leaves the club, McManus should tell us right away.'
Ames asked, 'Should he follow him?'
Lazarus turned to Peggy again. 'No,' she said. 'Far too risky. But at least we'll know from McManus when Jackson's on the move. Probably just as the lorry arrives. Or at least we hope so.'
Chapter 54
Slim's Club was full. On a week night the customers were beginning to drift home by midnight, but this was Friday, and people seemed happy to stay out late. McManus found a space at the very end of the car park and walked back towards the club's entrance. He could hear the loud music from the dance floor while he was still fifty yards away. He nodded at the bouncers standing by the front door and went through into the restaurant, surprised to find that most of the tables were still occupied.
Lester Jackson, in an elegant dark suit with cream shirt and crimson tie, was sitting in his usual place against the back wall. He nodded almost imperceptibly when he saw McManus, who walked over and joined him, sliding in behind the table to sit on the banquette next to his host.
'Bang on time,' said Jackson without looking at McManus.
'Have you ever known me to be late?' The waiter came to the table. McManus saw that Jackson was drinking his usual fizzy water with a slice of lime. 'Whisky and soda,' McManus said, thinking it would look odd if he ordered anything other than his usual. 'So what's happening?' he asked casually.
Jackson didn't answer at once. He was looking around the room, as if counting heads – or the money the heads would bring in. He took a small sip of his water and said, 'I got a nice little deal proceeding.'
'That's good,' said McManus, as if it had nothing to do with him.
'Big delivery. From abroad.'
'Girls?'
Jackson shrugged and pulled one of his cuffs. 'And then some. I could use a little help with this one.'
McManus said nothing. The waiter came back with his drink, and he took a large swallow, then put his glass down. 'I've been meaning to speak to you.'
'Oh yeah?'
'I'm going to be retiring soon.'
'Retiring? You ain't that old, man.' Jackson's voice had suddenly lost its polish.
'My pension says I am.' McManus tried a smile. 'Things are going to change.'
'How's that?'
'Well, once I'm no longer working I'm not going to be much use to you, am I? It's not like I'll know what's going on.'
Jackson looked amused. 'You'll still know plenty as far as I'm concerned. And you'll know how to find out what you don't know. Your buddies will still be working in the department, won't they?'
McManus didn't say anything. He sensed this was not the time to push the story of his retirement.
Jackson said, 'You're gonna help me tonight, aren't you? Or you getting cold feet in your old age? Looking for your bus pass maybe.'
'I'm OK,' said McManus resolutely. 'What is it you need me for?'
'I got a dude collecting something from me, only I haven't done business with him before. I want backup – in case he gets some odd idea of lifting one over me. I just need you to be there. Right?'
'Since when did you need extra firepower? I know you're carrying.' He gestured at Jackson's jacket. 'I'm not. What use am I going to be if things get rough? Or are you expecting me to arrest him?'
'It's not about shooting – or arresting. I just want you there. OK?' It was not really a question; the expression on Jackson's face was telling McManus it had better be OK.
'Where are we going?'
'Not far.'
'How far? I haven't got much petrol in my car. I'll need to fill up.'
Jackson gave him a thoughtful look. 'You won't need it. I'll drive you.'
'When do you want to leave?'
'Now is not too soon.'
McManus nodded and stood up. 'OK, let me have a slash first and then we can go.'
'Do it later.'
'What do you mean?'
Jackson stared at him expressionlessly. 'I said, do it later.'
'Can't a man go to the bog?'
'Sure you can,' said Jackson, relenting. 'But leave your phone behind.'
'Why?'
'Why do you think?'
'What's the matter? Don't you trust me?' McManus demanded, trying to put outrage in his voice.
Jackson looked amused. 'I trust you, Jimmy, as much as I trust anyone.' He paused. 'Which means I don't trust you at all.'
McManus shrugged. 'OK then. I can wait. Let's go.'
Outside it was suddenly cold; frost was settling on the bonnets of the vehicles in the car park. McManus said, 'If it's not far I'll follow you. Then I can go straight home after.' He started to head for his car, but Jackson put a hand on his arm.
'Whoa. You're coming with me.' He pointed to the sleek silver Audi coupé he kept in a special slot reserved for him.
'How do I get home then?'
'I give you a lift or drive you back here for your car. But I need you with me.'
By now McManus was scared. It was clear from the way Jackson was behaving that he didn't trust him, so why did he want McManus to go with him? It didn't make any sense unless he wanted to use him as cover for whatever he was up to. They'd told him at headquarters, when they'd accused him of corruption, that the only way of avoiding a very long stretch was to help them get Jackson behind bars. They'd said that if he didn't cooperate he'd find himself charged with abetting terrorism, because Jackson had got himself involved with a bunch of jihadis. They'd said they were expecting something to go down tonight and he was supposed to warn them if Jackson moved out of the club, but with Jackson being so suspicious, he wasn't going to be able to do that. His only hope was that when they got wherever they were going he might get a chance to send a text to say where he was.
'Here,' said Jackson curtly, handing him the car keys, 'You drive.' He took out his phone. 'I'm turning this off for safety's sake. Give me yours and I'll turn that off too.'
Chapter 55
Andy, the bald man, yawned loudly. It was almost one o'clock. On the table was a litter of paper plates covered with crumbs, curling sandwiches, sausage rolls and other delicacies provided by the canteen, together with several Thermos jugs of coffee. They had monitored the lorry's progress for more than four hours as it had worked its way across country from the east coast, come up the M1, then, as if drawn by a magnet, moved west towards Manchester. It had been tailed the whole way by A4 teams.
'Any news of McManus?' The Chief Constable had been looking in from time to time during the evening, but now he'd sat down at the table, looking as if he had come to stay. He had been told earlier in the evening about McManus's text message.
Lazarus shook his head. 'No, sir. And his phone's switched off. As is Jackson's. They may still be at the club, but we don't know for sure.'
'Something coming through now,' Emily, the Detective Sergeant, announced. 'It's the Eccles estate.'
On one of the monitors a misty picture came up, showing a stretch of road, some bushes and the outline of a car in the distance, coming towards the camera. Officers Fielding and Pierce from Manchester Police's CT unit were lying hidden in a shallow ditch that ran along the edge of the estate on the east side. A couple of their colleagues were in similar positions at the west entrance. While Pierce kept a lookout, Fielding lay on his belly and watched through the special nightscope of a videocam recorder he had perched on a low tripod. The feed from Fielding's camera, displayed on the screen in the Ops Room, showed an Audi coupé slowing as it turned off the approach road into the estate, then driving away from the camera on one of the estate's narrow roads.
In the Ops Room, Emily said, 'That's Jackson's car.'
'But where's McManus?' asked Peggy.
There was no sign of any other vehicle. Andy was talking into his microphone, and he suddenly held up a hand. He flicked a switch and his conversation was audible on the speaker.
'Picture's clear enough,' said Andy. 'How many in the car?'
Pierce spoke from the ditch at the estate. 'Two guys. A black guy – I think it was Jackson. And a white driver. Mid-fifties maybe. Clean-cut.'
'Thanks.'
The Chief Constable asked, 'You reckon that's McManus?'
'Has to be,' said Lazarus. 'Otherwise he would have called us.'
'Jackson's no fool,' said Emily. 'He'll be keeping a close eye on everyone around him, and being extra-careful. I'm sure that's why his phone's off and probably why McManus's is off as well.'
They watched the monitor anxiously. From the perimeter where Fielding and Pierce were hidden, you couldn't see the Jackson warehouse, and the cameras in the warehouse – one on its exterior, the others inside – had so far shown no movement.
Suddenly the camera outside the warehouse came to life as a light went on, and the vast front door of the warehouse began to lift up slowly. Two figures were visible, standing just outside the building.
'Is that McManus?' Peggy asked.
'Yes,' said a new voice in the room. 'That's him all right.'
All heads turned to the door. It was Liz Carlyle, standing just inside the room, wearing her overcoat. Peggy leapt up, knocking her chair over. 'Liz.' The relief in her voice was clear. 'I didn't expect you back tonight. How are you?'
It had been a hard day by any measure, and it wasn't going to be over any time soon. But at least she would be concentrating now on something that didn't drain her emotions, something that called on her professional skills rather than her feelings.
She'd had plenty of time on the journey back to mull over her day's hurried trip to Paris: Isabelle meeting her at the Gare du Nord; the conversation and the tears on the drive out to Martin's flat; the realisation, when she stood in the sitting room and looked out of the window at the Square opposite, the trees bare of leaves now on this raw day, of just how much of her life, her emotions and, as she had thought, her future lay there.
Foolishly Liz had imagined she could collect all her belongings in a suitcase and take them back with her, but it took less than five minutes in the flat to recognise just how many clothes, books and odds and ends she had accumulated over the few years of her relationship with Martin. After the flat there had been a brief meeting with Claudette, Martin's ex-wife, who had been civil, if not exactly cordial. And finally a tearful hour with Mimi, Martin's adored daughter.
There had been no reason to stay longer, since she would be coming back again soon – for the funeral, for the gathering of her possessions, and (this she had promised the girl) to spend some more time with Mimi. So she had headed back to the station and caught a late afternoon Eurostar back to London. She'd gone to her flat, planning to leave the operation in Manchester to the police, but after an hour at home she'd felt so desolate and restless that when eventually she'd checked her mobile and seen the text from Peggy announcing that she was leaving for Manchester, she had decided that she would go to join her.
She'd managed to get what must have been about the last seat on the packed train by travelling first class; she'd fallen briefly asleep, waited in a long queue for a taxi, and now here she was, slightly dazzled in the brightly lit Ops Room but relieved to be able to focus on something that had nothing to do with Martin Seurat and the grief that flooded through her in unpredictable waves.
Peggy said, 'Zara's been at his mother's house in Eccles. We've just heard from the A4 team there that he's gone out. He's in the car he rented in Birmingham.'
'Are they still with him?'
'Yes.'
'Here's the lorry,' Andy announced as the grey, wavy picture from the night-vision camera at the gate appeared on the screen again. The images from the camera outside the warehouse were showing on another screen. The lorry drove into the picture, made an enormous 180-degree turn, and stopped, facing out on the hard standing where Jackson and McManus stood waiting. Jackson waved it backwards and the lorry reversed slowly into the warehouse, gave a belch of exhaust, and stopped.
Jackson and McManus went in and attention in the Ops Room switched to the pictures from the cameras inside the warehouse. After a moment the driver jumped down from the cab. He was a short, stocky man in a thick dark pea jacket.
'You made it at last,' said Jackson, his voice clearly audible in the Ops Room.
'Ya. That was one good long hell of a drive.' His English was heavily accented and quite difficult to make out on the microphones. 'We had to stop a lot for fuel.'
'I bet you did,' said Jackson knowingly. 'Everything all right with the cargo?'
'Yeah. You want to see?'
The man made to go for the rear of the lorry, but Jackson put up a hand. 'Wait a minute. Tell me about the journey. Any problems?'
'The journey? It was difficult, especially in Germany. Snow has come early this year.'
'I'm not asking about the weather. I meant, when you got to Harwich. Were you stopped at Customs? Have they been through the cargo?'
'No. I expected them to open the doors, but they didn't.'
Jackson turned to McManus, who was standing beside him, before turning back to the driver and asking, 'Do they usually look inside?'
'Always. In my experience. But not tonight.'
'I don't like the sound of that.'
McManus said, 'Could just be shortage of staff, weight of traffic, Christmas spirit – anything. I wouldn't read much into it. And he got here, didn't he?'
Jackson's eyes stayed on McManus. 'Yeah, but what I'm wondering is if anyone else came along for the ride.'
The three men in the warehouse now moved outside to the tarmac forecourt, and all the microphone could pick up was the faint sound of voices.
'What are they waiting for?' asked Andy. He sounded cross.
Before anyone could answer, the three men started walking back into the warehouse. The driver was gesturing at the back of the lorry. 'I should open it up now?' His voice came through loudly.
'Not yet,' said Jackson curtly.
The driver was insistent. 'I have done. Let me unload and then I can be gone. I have mattresses to go to Glasgow by tomorrow. And there is a breathing cargo here that needs some air.'
Jackson laughed harshly. '"A breathing cargo". I like that. Don't you, Jimmy?'
McManus shrugged. 'I hope you haven't dragged me out here for a bunch of tarts.'
'You'll see soon enough.' And Jackson walked to the front of the warehouse again, while McManus stood still, half in shadow, and the lorry driver lit a cigarette.
Back in the Ops Room, Emily asked, 'Why doesn't he want the lorry opened up?'
'Because the main customer hasn't arrived,' said Liz.
'If he ever does,' said Andy.
'He will. Zara's come all the way from London,' Liz said. 'I don't think it's just to see his mum.'
Fielding's camera had picked up another car coming into the estate, a dark Ford S Max. Peggy looked at Liz. 'That's the car Zara hired.'
Thirty seconds later, as the S Max appeared on the monitor parking on the tarmac outside Jackson's warehouse, Lazarus was on the radio to the armed police team. 'Target has arrived.' Turning to Liz, he said, 'Time to go in?'
'I think we should wait a bit.'
'You sure? If the guns are in the lorry we'll find them. We can strip the bloody thing down to nuts and bolts if we have to.'
'We still don't know where the others are.'
'You think they're coming to the warehouse?'
'Possibly. I'd like to hear what Zara and Jackson say to each other.'
'I make Jackson as just the middleman.'
'I think you're right, but don't we need to hear them make the transaction if we're going to be sure of a successful prosecution? Otherwise we haven't got much to stick on Zara. He can say he's come to collect mattresses, and without more evidence a jury might give him the benefit of the doubt.'
The Chief Constable broke in. 'We'll take the risk, George. Bring the armed team forward but hold off going in for a bit.'
Lazarus nodded and radioed some orders.
Chapter 56
What the hell is happening now, thought McManus as he stood at the door of the warehouse and watched the dark-coloured car pull up. He hadn't believed Liz Carlyle when she'd told him that Jackson had got himself involved with a bunch of jihadis, but there was something going on here that was out of the run of Jackson's usual style. Who was this 'customer' who'd arrived and what was he collecting?
He wondered how much Liz Carlyle and the team back at HQ really knew about the situation. If they'd known tonight would be dangerous, they should have warned him. When Jackson had asked him to meet up at Slim's, there had been no reason for him to think there could be trouble brewing; it was only when Jackson had insisted on taking away his mobile phone that he'd grown worried. Without his phone, and without a gun, he felt doubly exposed.
They should have issued him with a weapon if they were putting him into a potentially violent situation, McManus thought angrily. But he knew the Chief Constable would never have authorised that, given the accusations against him. Not that the 'customer' who had just arrived looked very menacing. OK, he was Middle Eastern-looking, but he was slightly built, not much more than five feet nine, and looked more like a student, in his jeans, trainers and roll-neck sweater underneath a parka, than a jihadi terrorist. McManus hadn't wanted to believe Liz Carlyle's claims of a jihadi threat, and part of him still didn't. And even if this guy was a fanatic, intent on slaughtering innocents, there wasn't much he could do about it. Not when Jackson had his phone – and a gun.
Now Jackson signalled for McManus to follow him. They walked out to the tarmac in front, where the new arrival stood by his car, watching warily as Jackson and McManus approached.
'Good timing,' said Jackson. He gestured at the lorry. 'Your goods have just shown up.'
'Who's this guy?' the man demanded, pointing at McManus.
'My business associate,' said Jackson. McManus took a step back and kept his hands loose by his side. If he was supposed to be the heavy then he'd better act like one.
'You were supposed to come alone.' For all his youthful appearance, the man spoke with authority and without any signs of nerves. He's been trained, thought McManus.
Jackson seemed to sense this too. 'I'm sorry, man, but I didn't think it mattered.'
The young guy shook his head. 'I can see you're new to this. Rule Number One: no surprises. Understood?'
Jackson nodded reluctantly. It was clear now to McManus who was running the show, and it wasn't Jackson.
They walked together into the building, where the lorry driver was stubbing out a cigarette with the heel of his shoe. 'OK please to open up?' the driver asked.
Jackson shook his head. 'Not yet.' He turned to McManus and gestured at the new arrival. 'I'll be back in a minute. Give our friend a coffee, will you? There's a machine in the kitchen over there.' He pointed towards the end door in the partition at the side of the warehouse.
The Middle Eastern-looking guy said sharply, 'I don't want coffee. What's the hold-up?'
'Don't worry: I just want to have a look around outside,' said Jackson. 'Can't be too careful, can we? Then we'll get down to business.' And Jackson walked out of the warehouse before anyone could object.
McManus turned towards the other man. 'What's your name, mate?'
'Whatever,' the man said impatiently, his eyes following Jackson.
'All right, "Whatever" – are you sure you don't want coffee?'
In the Ops Room Peggy asked, 'What's Jackson doing?'
Lazarus looked at Andy, who said, 'Can't see him. He's out of camera range.'
'Perhaps he's gone to have a pee,' said Emily.
Nobody laughed. The atmosphere in the room had tautened with Jackson's sudden disappearance from view.
Lazarus said, 'Andy, get me Team Three.'
A moment later Andy said, 'On the line now.'
'Yes?' a disembodied voice came over the loudspeaker.
'Jackson's outside the warehouse. Don't know where he is – out of camera range. Hold your position until we know where he is.'
There was a pause. 'Do my best. But I've got two men closing in now.'
The Chief Constable looked at Liz and winced.
The lorry driver was growing agitated, which didn't improve his English. 'Doors to open,' he was insisting.
McManus shook his head. 'Not yet. The man will be back any time now.' The Middle Eastern guy was standing by the front of the warehouse, looking out. McManus had given up efforts to make conversation.
'Not waiting,' the driver said, going to the back of the lorry.
McManus took three strides and caught up with him as the driver was reaching for the steel handles of the twin back doors. He put a hand on the man's shoulder. 'The boss will be back in a minute,' he said firmly. 'So cool it.'
The driver stepped back from the lorry door. He shook his head. 'I am not liking this.'
'You'll survive,' said McManus. Out of the corner of his eye he saw something move outside, and then Jackson came back inside the warehouse, a tense expression on his face.
'He wants to open the lorry.' It was McManus speaking.
'Yeah, well, we got bigger problems. There's a car down the road that wasn't there before.'
'So? Lots of people must come in and out of here.'
'At three in the morning? I don't think so.' He stared at McManus. 'You wouldn't know anything about it, would you, Jimmy?'
'Me? Why would I?'
'You tell me. First you say you're retiring, then you try to duck out of driving over here with me. And you didn't like it when I took your phone. Who are you working for tonight?'
'I didn't realise I was working. You said could I lend a hand, and here I am. What's this about anyway?' He pointed over at the Middle Eastern customer who was watching them from the front of the warehouse.
'Never mind him,' Jackson said curtly. He seemed to have made up his mind. 'Here's what we're going to do. We'll open the doors and let the cargo out. I want you to take them into that room – I've got beds in there, and they can spend the rest of the night here. While you doss them down I'll finish up with my customer here. Got that?
'OK.' McManus was thinking hard about his options, which seemed dismayingly limited. If there were police outside and they raided now, how was it going to look? They'd never believe he'd been forced into giving up his phone; they'd assume he'd been trying to double-cross them. Yet it was equally clear Jackson wasn't going to let him out of his sight – not long enough to get away, at any rate – and Jackson had a gun . . .
Jackson turned to the driver, 'Go on. Open up.' Then he looked back at McManus. 'Just try something now,' he said, his voice full of menace, 'and it will be the last thing you ever do try.'
'What on earth?' asked Peggy as they watched the monitor. The back doors of the lorry had been opened, and a pile of mattresses dragged out by the driver and chucked onto the warehouse floor. Now down a step at the back of the HGV came one, two, three, and finally a fourth woman.
They were all bedraggled, thin with matted blonde hair, and each clutched a suitcase. In spite of their winter coats and trousers, they looked frozen and they screwed up their eyes, dazzled by the light. They looked to be in their twenties – except for the last one, whom Liz watched with a growing sense of outrage: the girl could not have been more than sixteen years old.
Once out of the lorry, they huddled together in a little circle, clearly apprehensive about their new surroundings. The youngest was shivering uncontrollably, and one of the other women put an arm around her shoulder.
Jackson stepped forward. 'Welcome to England and the Jackson Hotel. You'll be spending the rest of the night here. My associate Mr McManus will show you to your quarters.'
The oldest-looking of the women stepped forward. 'We have not eaten for twelve hours,' she said. 'We're hungry.'
Jackson was unfazed. 'You'll have to wait till breakfast.' He made a show of looking at his watch. 'That won't be long now. So why don't you all get some sleep?'
McManus ushered the women towards the side of the warehouse, a plan starting to form in his mind. As he led the women along the partition towards the door into the so-called bedroom, he looked over his shoulder and saw Jackson and the driver conferring at the back of the lorry, while the young Middle Eastern guy stood by looking impatient. It wasn't going to take them long to locate the cargo in the lorry and bring it out; McManus would probably have less than a minute. But it might be time enough.
When they reached the first door in the partition, the girls stopped and looked back at him for directions. He nodded and indicated that they should open the door. He then stood in the doorway and watched as the girls put down their suitcases in the small spaces between the bunk beds. One of them opened the door into the tiny bathroom next door. He felt sorry for them in this comfortless place after their long journey in the back of the lorry.
'There's a kitchen next door,' he said. 'You can make some coffee.' From his position at the door in the partition, he looked back at the lorry. There was no sign of Jackson or the other two. They must all be inside the vehicle.
McManus walked fast back towards the front of the warehouse. As he passed by the lorry, he paused, listening carefully, then he set off, running fast towards the warehouse door.
McManus had gone out of sight of the internal camera as he'd taken the women towards the bedroom, and the attention of the watchers in the Ops Room had focused on Jackson as he clambered into the back of the lorry with the driver and Zara. As they all watched there was silence in the room. Liz now thought it was improbable that the other jihadis would be appearing, and she was willing Zara to get on and retrieve his 'goods' from the lorry, so they could send the armed team in to arrest him and Jackson.
Suddenly at the bottom of the screen a figure appeared, walking quickly towards the front of the warehouse. 'It's McManus,' Lazarus exclaimed just as the figure broke into a run, his shoes slapping noisily on the warehouse's concrete floor.
The outside camera took over, showing McManus as he reached the tarmac forecourt and ran out into the road. He was raising his arms and shouting, so loudly that in the Ops Room his voice came through clearly. 'Don't shoot, don't shoot,' he yelled. 'I'm police. DI McManus.' Fifty yards or so ahead of him an armed policeman had emerged from the undergrowth, holding an assault rifle aimed at McManus.
There was the flat crack of a gunshot and McManus half turned, clutching his stomach with both hands, stumbled, and fell. He lay motionless on his side. In the glare of the outside security lights the camera showed a small pool forming next to the inert figure; like a leak from a broken pipe the little pool gradually got bigger and began to trickle along the road.
'Oh God,' said Peggy as the policeman in a bullet-proof vest came slowly forward, his rifle still held high.
Liz looked on in disbelief. Had this policeman shot McManus, an unarmed man? Then at the side of the picture, she saw Jackson standing just outside the warehouse, a gun in his right hand.
Jackson must have seen the policeman at that moment, because he lifted his arm, aimed and fired. The same flat crack split the air, but now almost simultaneously there was a second noise – this time a burst of metallic-sounding gunfire. Jackson spun around, tottered for two steps and fell to his knees. One hand was still clutching his gun, but the other was pressed against his gut. He tried to stand again, but could only make it into a low crouch. He lifted the hand from his stomach and stared at it with a mixed expression of awe and disbelief; it was coated in blood.
He turned awkwardly on his heels to face the approaching policeman, who was shouting at him to drop the gun and stay where he was. But Jackson paid no attention; defiantly he managed to struggle to his feet and point his gun in the policeman's direction. There was the sound of another burst of fire, then silence. This time Jackson dropped for good.
Chapter 57
People always said old people went to bed early, and Mrs Donovan wouldn't argue with that. Ever since the nine o'clock news on TV had moved to ten she'd never watched it. Nowadays she went to bed at half past nine and listened to the ten o'clock news on the radio.
But what people didn't understand was that just because you went to bed early, it didn't mean you slept. Every night she woke up, uncertain and hazy, lifting her head off the pillow to see the bright red illuminated numbers on the clock on her bedside table. They might say 12:30, or 2:17, or – when she was lucky – 5:45. Rare was the night she managed as much as four hours' continuous sleep; rarer still those where she slept through until dawn.
Tonight was no different. It was four o'clock and she was sitting at the kitchen table with a cup of milky tea, a biscuit and a copy of the Sun, which her grandson Michael had left behind.
She had put the telephone on the table within arm's reach, but it remained defiantly silent. She wasn't going to use it herself, because though she was accustomed to being up at this ungodly hour (the experts all said it was better to get up than lie in bed twisting and turning), she knew that not many other people were. Old as she was, Mrs Donovan hadn't lost many of her marbles – she might keep odd hours, but she knew what was and wasn't usual.
She'd tried to tell them about it earlier in the evening. Someone had said they'd ring her back, but they hadn't. She'd never believed for a moment that that girl who had shown up really was from Electoral Registration. But she had liked the look of her and she'd kept her leaflet with the phone number, behind the little cactus plant that Michael had given to her.
She couldn't have said quite who that girl did work for, but she was sure it was something to do with those thrillers she liked to watch when they were on the TV early enough in the evening. 'Spooks', that's what they were called. She was one of them, Mrs Donovan was sure. She knew she was right because the number on the leaflet was a London number. Why would the electoral registration office for Eccles have a London telephone number?
She wouldn't have noticed any of this – or rung the number – if things hadn't suddenly grown very peculiar next door. Mrs Atiyah had come round three days before, to say that she was going to visit her sister down in Croydon. Would Mrs D mind keeping an eye out for her cat Domingo? He was a fat tabby with a scrunched ear from a long-ago fight who liked to sleep in Mrs Atiyah's porch. He wasn't actually the Atiyah cat – Domingo made it clear he belonged to nobody – but the Yemeni woman was soft-hearted and treated the animal like a favourite child. There was always food for Domingo when he deigned to visit.
That was all fine, and Mrs Atiyah had gone off – Mrs D had seen the minicab arrive two days before – but then this morning the peculiar thing had happened. Just as she was putting some Go-Cat in the bowl in her neighbour's porch, with Domingo purring and rubbing himself against her legs, the front door had opened. She'd looked up, startled, expecting to see either Mrs Atiyah, back early, or one of her children. Instead a young man had stood there, Middle Eastern and bearded. He'd been just as startled as she was.
Mrs Donovan had stood up smiling, ready to introduce herself, pointing at Domingo to explain her presence. But the young man hadn't smiled or said a word, just gone back inside and firmly closed the door. Rude, Mrs Donovan had thought, but then later, back in her house, she had thought it also very odd. In that household, only Mrs Atiyah's son Mika was capable of that kind of behaviour, and it wasn't Mika who'd opened the door. So who was this stranger?
All day the question gnawed at her, competing with her usual instinct not to get involved, to leave things be, not to make a fuss. But she had been increasingly aware of something going on next door; of people – not just one surly young man, but others: someone playing the radio in the kitchen, while somebody else ran a bath, and someone came thumping down the stairs. You wouldn't have known, from the street, that anyone was there, since the curtains in front, both upstairs and down, were tightly drawn. It was only that the walls in these terrace houses were so thin that you always knew if there was anyone in.
If they were burgling the place in Mrs Atiyah's absence, it seemed a funny way of going about it; on the other hand, Mrs Atiyah would have mentioned it if she had invited people to use her house when she was away. And why would she have asked Mrs Donovan to feed Domingo if she had guests staying there?
Mrs Donovan was afraid of sticking her nose where it didn't belong. But what if these people were not in fact burglars, but something worse? Mrs Donovan was no coward, but neither was she a fool; she didn't think it would be sensible to go and knock on the door and ask who they were and what was going on. There wouldn't be much she could do if the strangers suddenly bundled her inside and . . . she didn't even want to think about it.
Then she had seen Mika, Mrs A's son, arrive. He'd parked outside and run into his mother's house, carrying a bag. Before Mrs Donovan could get to the door and go out to intercept him, he had gone inside, slamming the door.
He was driving a brand-new car from the look of it, a big one too, which struck Mrs Donovan as a bit much. These students were meant to be paying their own fees these days – weren't they always complaining about that? So how could Mika afford such a flashy car?
Finally Mrs Donovan decided that she needed to do something. She wondered again about whether she should knock on the door now Mika was back and ask him what was going on and whether his mother knew about all these people being there. But again she thought that wouldn't be wise. From the way he had rushed into the house, she didn't think she'd be welcome; the thought of the hostile young man she had seen that morning put her off the idea completely.
It was then she remembered her recent visitor who'd said she was from the electoral registration office. Whoever she really was, perhaps she could help. It had been evening by then, after six o'clock, so she wasn't sure if she'd still be there. But nowadays people seemed to work long hours and they all had these mobile phones, so she thought it worth giving it a try. She took the card down from the sideboard and dialled the number.
A woman's voice had answered, and thinking it was the girl she'd met, Mrs Donovan began to explain – until the woman interrupted. Once the confusion was sorted out, and Mrs Donovan had explained who she was trying to reach and why, the woman had promised to pass the message on. She'd said she'd be rung back right away. But nothing had happened.
It was nearly ten hours now since she'd rung. She'd seen Mika go out in his car, but the other people were still next door. She could hear them moving about even though it was the middle of the night. Mika had not come back; the car wasn't there. She'd been tempted to ring the number again but there probably wasn't any point.
Perhaps she was just being a silly old woman. Part of Mrs Donovan hoped she was, and that she was wrong in her suspicions of the people next door. I think I'll just forget about it, she thought, taking a sip of her now-tepid tea. I expect there's some innocent explanation and Mrs Atiyah will sort it out when she gets back. She yawned and stood up to go back to bed. Not that she would sleep.
Chapter 58
It took eighteen minutes to reach the warehouse from Police HQ. They went in convoy, three cars in all. Chief Superintendent Lazarus stayed behind in the Ops Room to coordinate the operation. Liz drove with Chief Constable Pearson in his BMW; his driver, Tom, had turned the heater on high to melt the frost on the windows – the car had been standing outside waiting for the call to move and was cold inside as well as out.
At first there was no conversation in the car. They were listening to the radio transmissions as police cars converged on the industrial estate. Two ambulances were not far behind.
Then Pearson broke the silence. 'I wasn't expecting to see you here tonight.'
'Well—' she began, then found herself with nothing more to say. She hadn't expected to find herself here either. It seemed unreal. But she was grateful for the almost frantic sequence of events, since it kept her from thinking of the terrible happenings in Paris the night before. The night before? Incredibly it was only last night, even though it seemed to be days since she had first heard the news of Martin's murder.
Pearson said, 'I'm delighted that you're here. Don't get me wrong, I think young Peggy is extremely good. But I know she was glad when you showed up.' He paused to listen to a burst of radio transmission then said, 'I think you're pretty remarkable, frankly, after the twenty-four hours you've had.'
'I wanted to see things through,' Liz said.
'Of course. But listen, if this gets too much for you at any point, just let me know. Tom will drive you back to Police HQ and sort you out with one of our guest rooms. Then you can pick things up again tomorrow.'
The driver nodded. 'I'll be with the car. Just let me know if you want to go.'
'That's kind of you, but honestly—'
Pearson lifted a hand to interrupt. 'Understood. Just remember if you change your mind, the offer holds.'
As they approached the trading estate they could see a ghoulish glow created by the dim sodium lights that lined the narrow strips of road and trailed off into the industrial enclave. Tom drove quickly, following the other two cars, turning right then left into a kind of cul de sac, at the end of which was a tarmac apron in front of a large metal warehouse. Scrubby grass and undergrowth filled the spaces between the warehouse and the adjacent buildings, derelict-looking brick and concrete workshops.
A lone policeman stood at the front of the tarmac, waving a flashlight to steer them around a small area which was marked by traffic cones. Behind the cones something lay on the ground covered by a tarpaulin sheet. With a jolt, Liz realised she was looking at McManus's dead body. They were now part of the drama that they'd been watching on the screens in the Ops Room. She felt as if she had stepped from the audience onto the stage.
Three police vans and an ambulance were already neatly parked and Tom pulled up beside them. Another two cars bringing Peggy and some more uniformed officers had just arrived. Liz and Pearson followed the policemen into the warehouse, stepping gingerly over Jackson's body, which was still lying in the entrance, also under a sheet.
Two members of the armed team were inside. One stood guard over Zara, who was handcuffed, sitting on a wooden crate. He was staring vacantly into space, pointedly ignoring the people around him. The other armed officer was trying to calm down the women, who had emerged from their tiny bedroom compartment at the side of the warehouse. The youngest was still shaking but now she was screaming too and tears were running down her face. Another, who seemed to be the oldest, was clawing at the arm of the policeman and shouting, 'Not to shoot.'
The policeman was trying to unhook her hands and saying, 'I'm not going to shoot you. You are quite safe here.'
But he was having no effect. The women were all clearly terrified and Liz couldn't blame them; two men had been shot dead nearby less than half an hour after their arrival. This was not what they thought they'd come to England for.
'Where's the lorry driver?' asked Liz, suddenly realising that someone was missing.
The armed policeman pointed to the cab. 'He locked himself in when the shooting started. I've been trying to coax him out, but he's scared to death.'
'At least we know where he is. We'll get to him in a minute,' said Chief Constable Pearson. 'First I want these women out of the way. Put them somewhere until we decide what to do with them.'
Peggy, who had come in behind Liz, stepped forward and touched the arm of the woman who was clutching at the policeman.
'Come with me,' she said in a gentle voice. 'No one's going to hurt you. Let's go and see if we can make some coffee. Then I'll ask someone to get you something to eat.'
The woman let go of the policeman and grasped Peggy's hand. She looked at Peggy's face with frightened, anxious eyes, then after a moment she turned to the others and said something. It seemed to calm them, and then, like a mother hen, Peggy rounded up the little group and ushered them back towards the bedroom.
The armed policeman, the Chief Constable and Liz all watched in silence. A silence that was broken when one of the policemen came up to the group and asked, 'When we search the lorry, what are we looking for, sir?'
Pearson looked at Liz. She said, 'Guns and grenades. The firearms are probably a mix of assault rifles and handguns. And a lot of ammunition – they asked for twenty thousand rounds. That will take up a fair amount of space.'
Pearson said, 'I'm sure the driver knows where the cargo is hidden, so we should talk to him first. But whatever he says, take it slowly. I don't want anything going off because someone gets impatient.'
The other officer had joined them. 'I've frisked the suspect, sir,' he said, pointing to Zara. 'He wasn't armed.'
Liz asked, 'Was he carrying any ID?'
'No.'
'How about valuables? Any cash?'
'He only had a few quid in his pocket, but he had something else worth a hell of a lot of money. A ticket for the derby tomorrow, at Old Trafford.' He handed the ticket to Liz. As she studied it, he added, 'They're like gold dust.'
Liz handed the ticket to Pearson, and said, 'We've dug pretty deep into young Atiyah's past but I've never seen anything in the file about a love of football. Nor that he had the money to fund this sort of expense.' She turned to the policeman. 'Do you ever go to Old Trafford?'
'I've been known to attend a match,' he admitted.
'Do they search you when you go through the gates?'
'No. It wouldn't be practical. You've got sixty thousand people going in in a short time. The queues would go back for miles if they searched everyone. They tried it for the Olympics and it caused chaos.'
Pearson was looking on with growing apprehension. He said, 'It's cold enough now for everyone in the crowd to be bundled up. You could smuggle a weapon or a grenade in under an overcoat easily enough if there's no proper searching.'
'Exactly,' said Liz. 'And if you had six people in different parts of the stadium, then even if one got spotted you'd have five others who might not have been.'
The policeman said, 'To do what? Shoot Wayne Rooney?' He gave a weak laugh. 'And why do you say "six people"? The suspect only had the one ticket.'
Pearson didn't bother to explain. He saw what Liz was driving at, and he said, 'So the other jihadis must already have their tickets. Which means—'
'It means they've arrived and are holed up somewhere nearby.' Liz pointed towards the solitary figure of Atiyah, sitting in handcuffs on the wooden crate, then asked the policeman, 'Are you sure he didn't have anything else on him? Anything at all – a crumpled bus ticket, or a pocket comb. Anything.'
The officer shook his head. 'No, and he didn't say a word – he wouldn't even tell me his name. I don't think you'll get much out of him.'
Pearson said to Liz, 'Do you want to have a go here or wait until we take him back to headquarters?'
'Here please.' It was critical to try to get Atiyah to talk before he had time to collect his thoughts and invent a story – or just clam up and ask for a lawyer.
As Liz started to walk over to Atiyah, Peggy, who had come back from tending the Dagestan women, intercepted her. 'Could I have a word, Liz?' She held up her mobile phone. 'I've just had a message relayed from Thames House. It could be important.'
'Give me two minutes, Peggy. I need to talk to Zara urgently.' And she strode over and stood in front of Atiyah. He ignored her, continuing to stare out towards the parked cars on the hard standing in front of the warehouse.
Liz said, 'You all right? You didn't get hurt in the shooting?'
He didn't reply. His eyes remained focused on the distance, trance-like. For a moment Liz wondered if he was drugged, but then she remembered him from the video feed – he had been perfectly lively then, even aggressive. She said, 'Tell me if you got hurt; there are paramedics here now.' When he still didn't reply, Liz said softly, 'Mika, we know who you are.'
This time he blinked. For a moment Liz thought he was going to say something, but he didn't. She went on, 'We know the lorry has brought other things into the country, besides the women and the mattresses. We're going to start searching it in a minute or two. When we find what we're looking for, you'll be arrested.
'But that's not all we're searching for. I think you know that. At least five of your colleagues have entered the country from Yemen; I think they're supposed to meet you once you've collected the guns that are in the lorry. I didn't realise you were interested in football – are your colleagues going to the match too?'
He flinched slightly, then pursed his lips. Liz went on, 'I'm certain we'll be able to find them, especially if they show up at the match tomorrow.' She was watching him carefully. Without these guns, his comrades shouldn't be able to do much even if they made it inside the stadium next day – unless . . . And Liz shuddered at the thought. Unless they already had other weapons.
The only way to be sure was to find them. She suddenly hated the idea that Martin might have died for nothing; that despite the sacrifice of his life, and all their efforts, these terrorists might still manage to launch an attack. If only Zara could be made to talk. But looking at him she realised he was determined to give nothing away – he had adopted the same vacant stare again, as if transfixed.
Liz said, her voice hardening, 'Your colleagues will go down all right. But the big loser is going to be you, Mika, because you're the one we can tie to the guns we're about to find. We clocked you a long time ago, and you've been followed ever since. We watched your meeting in Primrose Hill, and the dealer you saw there is in custody in France. He's told us everything we need to put you away. I reckon you're facing thirty years. You might get out in time for the 2040 Olympics. Just think how old you'll be then.'
Liz gave a sigh. 'It's not as if you will have helped your cause very much, either. But there is a way you can help yourself, a way you could be out of prison in just a few years – you'd still have a life left. But, Mika, you have to tell us where the others are.'
Atiyah continued to sit impassively and Liz realised she was hitting a brick wall. He was obviously a fully signed up jihadi. This was his martyrdom and if she had said two hundred years in prison instead of thirty, he would have been pleased. She made one last try: 'We're going to catch your colleagues anyway; it's just going to speed things up if you tell us where they are. Think about what I'm saying; soon it'll be too late for me to help you.'
Atiyah turned his head very slowly, and for the first time Liz felt hopeful that he might reply. His eyes met hers, and he held her gaze as his lips began to move. Then his mouth opened and he spat in her face.
Liz jerked back in surprise. She tried to collect herself, and wiped the spittle from her cheek with the sleeve of her coat. She was determined not to show her shock, or anger. She said calmly, 'If you help us, I promise you I will do everything in my power to help you.' A thought came suddenly into her head. She added, 'I'll also make sure your mother doesn't get dragged into this.'
Atiyah's eyes flared for an instant, and for a moment Liz thought he would spit at her again. But then he regained control, and his eyes resumed their opaque stare.
Liz turned round and saw that Pearson was waiting, standing halfway between her and the lorry. She shrugged as she walked towards him, leaving Atiyah in the care of his armed guard. Peggy was there too, waiting for her, and Liz remembered that she had something to tell her.
In the background, behind Pearson, three policemen had approached the lorry, gesturing to the driver to come out of the cab. One of them went round to the driver's side and climbed up on the step next to the door of the cab. He knocked on the glass and shouted through the window, 'Open up. We want to talk to you.'
The Chief Constable and Peggy turned round and Liz stopped and watched as the policeman, losing patience, shouted, 'Open up, or we'll have to smash the window.'
The driver was looking frightened – though suddenly Liz wondered if that was an act. She was about to shout a warning when she saw the man slide across the front seat of the cab to the passenger side. Opening the door, he leapt down just as two of the policemen came round the front of the lorry.
They were less than ten feet away when from the pocket of his pea jacket the driver drew out a small grenade. With his free hand he prised the pin off, then chucked the grenade underhand, like a child playing rounders.
The nearest policeman to him flinched and turned away with his arms holding his head. The grenade landed on the cement floor, just missing the policeman, then bounced high in the air, angled towards . . . towards Liz. She tensed, waiting for it to explode. There was nowhere to go and nothing she could do.
Then an outstretched arm, black-clad, with silver on its shoulder, reached out and grabbed the grenade as it started to come down. In one quick motion the arm then threw the grenade straight out of the open front of the warehouse.
It travelled twenty yards and hit with a sharp thump on the tarmac forecourt, where it promptly exploded. As dirt-coloured shards burst through the air, the noise of the explosion was astonishingly small, almost muffled. But it was followed by a series of sharp pings – the shrapnel was hitting the sides of the parked police cars.
Pearson ran to the front of the warehouse. 'Who's hit?' he shouted. But the two ambulance attendants had been shielded from the blast by one of the police cars. They looked dazed but unhurt. Tom, the Chief's driver, was the sole policeman outside, and he'd been in his car on the radio. He held a hand up to show he was OK.
Visibly relieved, Pearson came back into the warehouse, where the driver had been wrestled to the ground and handcuffed. As an armed policeman watched him, Atiyah at last showed some emotion – he was smiling broadly.
'Are you all right, Liz?' asked Pearson.
She nodded. 'Just surprised to be breathing. For a moment I was sure that was it.' She looked at Pearson. 'This is the first time I've had to thank anyone for saving my life. Thank you very much.'
'Pure instinct,' he said. 'I was in the Territorial Army and sometimes it seemed half our training was about dealing with incendiary devices and grenades. Never had to use it then.' He shook his head. 'And never thought I'd have to use it here. There must have been something wrong with that grenade, but thank God there was.'
Behind them they heard a quiet groan. Liz turned and saw Peggy squatting down against the side wall of the warehouse. She was holding her arm, which was bleeding badly just below the elbow.
'Were you hit?' Liz asked.
Peggy grimaced and slumped down, her back against the wall and her legs splayed out in front of her, flat on the floor. As Liz rushed to her, Pearson said, 'I'll get a paramedic.'
Liz crouched down next to Peggy. She saw at once that the wound was bad; shrapnel had ripped through the layers of sweater and shirt Peggy wore; there was a deep jagged cut in her forearm, which was bleeding profusely.
She saw Peggy's eyes glaze and start to shut. The girl was going into shock. 'Peggy!' Liz shouted, and the eyes fluttered open, stared vaguely at Liz, then shut again.
The paramedic had arrived and Liz stood up to get out of his way. As he examined Peggy's arm, she moaned in pain, and he took a syringe and vial out of his pack and injected something into Peggy's other arm. Morphine, Liz guessed; the pain of the shrapnel piece must be excruciating.
Two more paramedics arrived, carrying a stretcher between them. They carefully lifted Peggy onto it, then carried her towards one of the ambulances parked on the tarmac outside.
The medic who had injected Peggy looked at Liz. 'She should be fine, but we need to get her seen to properly right away – that's a nasty wound she's got. Do you want to come with us in the ambulance?'
Liz hesitated. She wanted to be with Peggy, but there was still everything to play for. She shook her head. 'I'm still needed here. But please keep me posted.'
Pearson was on his phone, but he rang off when he saw Liz. He said, 'We're going to have to make a decision about the match.'
She nodded. 'I know. It's your call of course, but I'm worried about these other jihadis. We just have no idea where they are or whether they have any weapons. I would hazard a guess that they haven't, but there are no guarantees. They might have access to some cache somewhere.'
'Yes. You're right. And if they do turn up at the match armed or carrying explosives of some kind, we can't be sure we'll be able to stop them getting in. We've got the seat number of Atiyah's ticket, but even if we searched everyone with a seat in the same block we might not catch them. They could have seats in any part of the ground.'
The Chief Constable was frowning. 'I'm beginning to think we have no option but to cancel the match. We can't take the risk. But if we do it's going to cause an immense furore. There'll be chaos on the streets, the media will have a field day, the Home Secretary will get drawn in and all of us including your Service will come in for a load of criticism. I need to speak to the Home Office before we do anything and you'll want to talk to your management too.'
Liz looked at her watch. It was now a quarter past five. 'I'll get on to the Duty Officer. DG will certainly want to be informed.'
Pearson looked round the warehouse. 'We're not needed here any more. We'll go back to HQ and set up a conference call and then everyone can have their say and get themselves prepared for the shit storm we're going to face. We can start the ball rolling while we drive back.'
Chapter 59
At six o'clock they were in Pearson's corner office, joined now by Lazarus and several of Pearson's senior colleagues, called in to help plan what would be a major operation whatever was decided about cancelling the derby match. Outside it was still pitch-dark, and whenever Liz looked towards the windows she saw, reflected against the black sky, the dark-uniformed figures sitting round the conference table in the middle of the room.
Liz had rung the hospital from the car and learned that Peggy's condition was stable, but that they would soon be operating on the arm to remove the shrapnel, which had fragmented into a number of small pieces. Confident there was nothing she could do for Peggy at the moment, Liz was focused on the decision Pearson was going to have to make.
Reports had been called for from all police divisions for any sightings of a group of men acting suspiciously, but in the absence of any descriptions, no one was surprised that no reports had been received. Liz's colleagues in Thames House had been in touch with GCHQ, the DCRI in France and the UK Border Agency, but no new information was forthcoming. A4 and police surveillance teams were still out in the area – they all knew the urgency of the situation and would instantly have communicated news of any sightings.
From the speakerphone in the middle of the table an automated voice suddenly announced, 'Your call is ready to begin. All participants are now signed in.' Pearson took a deep breath and said, 'Good morning, everyone, I apologise for the uncivilised hour but we have an urgent decision to take in connection with the Zara Operation on which I think you are all briefed. Will you all please introduce yourselves?'
With the preliminaries over, Pearson outlined the situation, calmly summarising the dramatic events of the previous few hours. He concluded, 'We are confident that the target of this jihadi group is the derby match between Manchester United and Manchester City at Old Trafford this afternoon. We will of course continue to question Zara, but so far we've got nothing out of him, and I don't believe that will change. He's already asking for a lawyer. We have interdicted the arms imported for use in that attack, but the whereabouts of the group other than Zara is unknown and we do not know if they are armed or have access to arms. So a risk exists that they may attempt to proceed with the attack and that there may be casualties – possibly many.'
The gravelly tones of DG came through now. 'What security measures can you take at the ground that might help detect these people if they turn up? I'm assuming searching all the fans is impossible.'
'Yes. It would take too long and we don't have the manpower,' agreed Pearson. 'What I can do is double the number of officers patrolling the gates and the stands, and I can insert more plain-clothes officers into the crowd. Obviously we'll be closely monitoring the CCTV cameras too, but we have no descriptions of these jihadis and it's very likely we won't spot them until they start something.'
'The Home Secretary will wish to know what the law and order implications are if the match is cancelled at this short notice.' It was the Head of Counter-Terrorism at the Home Office.
'It won't surprise you, or her, that cancelling the match at this late hour will create plenty of problems on the street. Even if we announce cancellation now there will still be hundreds arriving in a short space of time, and we may have some violence when they find out the match is cancelled. The later we leave it to announce cancellation, the worse it will be. That's why need to decide now, I'm afraid.'
DG spoke again: 'Liz, what's your view of things? Do you have a recommendation?'
Liz was drawing in her breath to say that she thought the only safe option was to cancel the match, when Pearson's office door opened and a young uniformed policeman came in, looking nervous. Without saying a word he handed Liz a slip of paper. It read, Most urgent call for you. She started to shake her head, but something in the young officer's eyes made her change her mind.
'Excuse me,' she said, leaning towards the speakerphone. 'I've just been told there's a very urgent call for me. It may be relevant so I think I'd better take it.'
Liz was gone less than five minutes. When she came back into the room she walked straight up to the table and, still standing, leant towards the phone. 'I apologise for the interruption,' she said. 'You were asking my view a moment ago and I was about to recommend cancellation. But as a result of the phone conversation I've just had, I can now confidently recommend that we should let the match go ahead.'
As she paused to catch her breath, Pearson broke in. 'What's happened?'
'I've just been speaking to the lady who lives next door to Zara's mother. Zara was at the house early last night, before he went to the warehouse to collect the weapons.'
'Go on.' It was DG's voice.
'The neighbour told me that Mrs Atiyah is away, but when Mrs Atiyah told her she'd be gone for a few days she didn't say anything about anyone coming to stay. Yet the neighbour swears there are people in the house – she says she can hear them through the wall. And she actually saw one of them yesterday. It was a young man.'
Liz paused. No one spoke. 'I think they've got to be the jihadis. And they're still there now.'
Chapter 60
Liz had always looked forward to Christmas and enjoyed it, but not this year. She and Martin had planned to spend it in Paris, but after the funeral service and the day spent removing her belongings from Martin's flat, Paris was the last place she wanted to spend any more time, let alone the holidays.
She'd gone instead to Bowerbridge, her childhood home. Her mother and Edward Treglown, her mother's partner, had been doing their best to help her come to terms with Martin's death and she didn't want to hurt their feelings by refusing to join them for Christmas.
She had arrived on Christmas Eve afternoon and they had gone to the midnight service at the village church. Liz had been brought up an Anglican but she no longer believed in any sort of God, though she knew that the moral principles she tried to live by were firmly rooted in her Anglican upbringing. And in fact, the church service with its familiar words, its music and the carols she knew so well had proved oddly soothing.
Afterwards they had walked back through the estate that her father had managed for the then owners. She'd played there with the children of the big house when she was quite young and she knew every track and path. It was a clear starry night and frost was beginning to settle on the fields. So bright was the moon that they hardly needed their torches. As she walked, she thought about everything that had happened to her since the previous Christmas and wondered, not for the first time, whether the time had come to quit and find another job.
She had stayed on for Christmas Day, worried that the sadness she couldn't disguise might spoil the day, but by the evening she had eaten enough and drunk enough of Edward's favourite burgundy to dull the pain and they had spent a pleasant evening dozing in front of the fire and the TV. Then on Boxing Day they had joined a big lunch party at some neighbour's, where blessedly no one present had known anything about Liz's relationship with Martin or about his death, so all she had to do was dodge the usual questions about her job.
She left the next day, promising to try and come down for New Year's Eve, though she knew that both her mother and Edward understood that she wasn't going to make it. Wherever she was, Liz knew she was going to be in pain, and she thought it best to suffer alone – why spoil any more of the holidays for everybody else?
After an aimless evening in her flat, she went back to work the next day. She was used to the sense of anticlimax that came at the end of an operation, whatever the outcome, but this time grief redoubled her deflation. Yet she knew that immersion in work would be the one thing to get her through the coming days, so she was glad to be back at her desk.
The office was quiet, virtually empty of staff on her floor, though in pockets round the building people were working as hard as ever. She rang the A4 Control Room to say thank you for all their efforts in following Zara on his anti-surveillance route to Eccles and the lorry from Harwich. Wally Woods answered.
'Happy New Year,' she said. 'Don't you ever have a holiday?'
'My work is my leisure,' he responded with a snort. 'We're in the middle of another drama now.'
On her return from Manchester, Liz had filed a brief report but knew she would have to supplement it. Not that there was much to add.
On that morning in Manchester when a half-conscious Peggy, just before surgery, had through sheer stubbornness insisted word must get through to Liz that Mrs Donovan had phoned the Thames House switchboard, it hadn't taken Einstein to see the link with Atiyah. Liz had called Mrs Donovan straight away.
She had expected a call that early would wake the old lady up, but quite the opposite proved to be the case. 'About time someone rang,' Maggie Donovan had said irritably. 'I haven't been able to sleep. I've been waiting for hours.' Then the old lady, still sharp as a tack, had told Liz about the visitors next door, and how Mrs Atiyah was away and hadn't said anything about people coming to stay.
Within half an hour, still in the moonless dark, a dozen armed police had filled the terraced street, blocking each end, climbing over garden walls and finally simultaneously breaking down the front and back doors and charging into the house. Inside they'd found the five associates of Mika Atiyah asleep on inflatable mattresses on the sitting room and bedroom floors. Caught off guard, they had been taken into custody without resistance. When they were questioned, they had all protested, claiming that they had come to Manchester to go to the football match with their friends. It hadn't taken much of a search of the house to find the little pile of tickets for the match, lying on top of a cupboard. But when they were asked to explain why the tickets were all for different parts of the ground, they became less vocal.
After it was discovered that each of them carried a Yemeni passport, even though they spoke in a variety of British regional accents and not one of them could understand the interpreter who was summoned to speak to them in Arabic, they'd refused to say anything at all. Not that that had helped them for long in concealing their identities, once the detailed inquiries were completed. They were all now in prison in Manchester, as was Mika Atiyah, on a variety of charges under the Counter-Terrorism Act.
The media had got hold of the fact that armed police had made arrests at a house in Eccles that were thought to be connected in some way with a shooting incident at a warehouse on an industrial estate off the M60, but so far it hadn't leaked out that a terrorist plot had been disrupted, or what the intended target had been. Jackson's death had gone unmourned in the Manchester metropolitan area, while the newspapers had been spare with the usual effusive eulogies for a murdered policeman – word seemed to have got round about McManus's less savoury activities.
And any sympathy Liz might have felt for her former lover disappeared when Halliday rang her. 'You know,' he said after congratulating Liz on the arrest of the terrorist suspects, 'I felt guilty that I was somehow responsible for the woman Katya's death – by tipping off Jackson accidentally somehow. But I've discovered it wasn't anything I'd done. McManus was in the police station the night Katya and the other girls were brought in. The desk sergeant told me it was McManus who got her released first – I guess to alert Jackson that she was an informer.'
On that night – really, very early morning – when the Atiyah house was being entered by armed police, a search party had been busy back in the warehouse. It had taken them almost three hours to find the weapons hidden in the lorry, and it would have taken longer than that but for a stray remark from one of the Dagestani women the police had begun to interview. She complained about how long the journey had taken. She said that the driver was forever stopping to fill up – yet he wouldn't let them out at the petrol stations to stretch their legs or go to the toilet. Since lorries of that size had enormous petrol tanks, this continual stopping for fuel seemed peculiar.
It was then that they checked the fuel tank itself, where they soon found that half of it had been fitted with a metal partition. In the newly created compartment they found twenty AK-47s wrapped in oilskins, grenades in metal containers, and box after box of ammunition clad in bubble wrap. It was an ingenious hiding place, and a stupid one, since a stray spark and some leaked petrol fumes could have set off the ammunition and blown the lorry sky-high.
Liz remained concerned about Peggy Kinsolving, whom the doctors had told to take six weeks' sick leave. Peggy had been through the mill, with fragments of shrapnel embedded deep in her arm. Some had chipped the bone, and it had required two bouts of surgery to remove them all and repair the bone. She'd been in Manchester Royal Infirmary for more than a week as they monitored her for shock and infection.
Liz had visited her just hours after the tumultuous events at the warehouse had concluded with the arrest of the jihadis at the Atiyah house. She had found Peggy not long out of a first operation on her arm, propped up in bed and still looking dazed and shaken. A TV set on the wall of her room was showing the game between City and United. Liz sat down and they watched together in silence. As the camera panned around the stadium, which was packed to the rafters with noisy fans, waving, cheering and singing, they looked at each other. Liz voiced what they were both thinking.
'Look at them,' she said. 'Think what that would look like if Zara and his friends had got through. If they'd got those guns and grenades in there, into different parts of the stadium, they could have killed hundreds of people before anyone stopped them.'
The two were silent; wild cheering filled the room as a man in red scored a goal. 'I can't forgive myself for not checking the message Mrs Donovan left with the Thames House switchboard when I first got it. We could have picked up the terrorists hours earlier and arrested Zara before he ever came to the warehouse.'
'I'm not sure about that. We needed to have Zara go where the weapons were to have a good chance of prosecution. Anyway there's no point in beating yourself up. As it's turned out they were stopped. Thanks to you and everybody else working on this case, it didn't happen.'
'Yes,' replied Peggy, reaching out with her good arm for Liz's hand. 'And that includes Martin.'
Liz nodded, her eyes filling with tears.
Now, weeks later, there was still a lot more investigation to do both for the police and Liz's team before any trials could take place. Research into the young Atiyah's finances had unearthed a recent series of deposits into his bank account, totalling £177,000 – deposits which to Liz's fury, the particular branch had never thought to question, as if it were entirely normal for a student from Eccles to have that kind of money at his disposal. It had proved possible to trace the money to a Lebanese bank, which had so far been stubbornly slow to assist with British efforts to uncover the money's original source.
Following the spider web of connections from Atiyah back to his controllers in the Middle East was challenging and time-consuming, but Liz consoled herself that there was already ample evidence to prosecute Atiyah and his cohorts. Antoine Milraud, appalled by what his young customer had been planning to do, was cooperating fully with Isabelle Florian in Paris, and had agreed to give evidence in court.
Martin would have been pleased by this, Liz thought, as she stood up and went over to the window of her office. There had been a fall of snow the night before, but it was melting now, leaving a thin layer of slush on the pavement along the Embankment. The Thames was a dull grey and restless, with choppy waves stirred up by the winter wind. Martin had liked to tease her that the Seine was the superior river, and today she would have agreed.
Would she ever stop missing Martin? Even now she could only feel heartbreakingly alone in a world without him. His death had served some purpose, she knew. Had he not succeeded in flushing out Ramdani the terrorist would have warned his colleagues bound for England that they were blown. They would have melted away and Liz would still be searching for half a dozen lethal men. She couldn't make room for the thought that this was any kind of compensation for Martin's death – it wasn't – but at least it gave some meaning to it. He had been dedicated and professional to the end, and Martin would have been the first to scoff at any suggestion that he should have hesitated to act because of possible danger. He knew, just as Liz knew, that risk came with the job.
A knock on the open door of her office shook Liz from her reverie. 'Come in,' she said.
It was Geoffrey Fane, and for once he actually looked friendly, almost shy. 'Elizabeth,' he said awkwardly.
Liz smiled to herself. There was no point in getting cross; he really couldn't help it. 'Hello, Geoffrey,' she said. 'I actually do prefer Liz, you know.'
'Of course,' he said, coming into the room. Liz went back to her desk and sat down, motioning Fane to a chair. But he shook his head; unusually for him, he seemed to understand that his presence might not be entirely welcome. 'I just wanted to say how very sorry I was to hear about Martin Seurat. I know you two were close.' He paused, as if hearing his words and how lame they sounded.
'Thank you,' she said simply.
He gave a little cough. 'I gather you did stellar work up in Manchester.'
'It's kind of you to say that. A lot of things didn't go right.'
'Possibly, but when do they ever? And you did prevent the very worst happening. Well done.'
Is this why Fane had come? Liz wondered. Gentle commiseration followed by a pat on the back? She'd known him long enough to know there had to be some other agenda.
And so there was. Fane came right into the room now, sat down, straightened his long back and crossed a languid leg over one knee. This was the Geoffrey Fane she knew. She watched him warily, waiting for what was to come. He said, 'I've got a bit of news actually.'
'Really?' She tried to look surprised.
'I had a meeting with our friend Andy Bokus yesterday. Not an altogether happy encounter, you could say. I pointed out that there was a missing link in this case, one that would have helped us a lot.'
'Baakrime.'
'That's right. The Minister,' Fane said, with the mild surprise he always showed when he found that Liz had got there too. 'He was both the instigator and the linchpin of this whole affair.'
'But currently unavailable.'
'So it would seem. Thanks to American cack-handedness. When they shilly-shallied he must have panicked and decided to throw in his lot with the Russians.'
'You said that to Bokus again?'
'In so many words.'
'That couldn't have gone down well.'
Fane gave a sly smile. Liz could see he was enjoying himself now.
'Actually, he had bigger things to worry about.'
'Oh?'
'Yes. It seems he's being moved on. Back to Langley.'
'I'd have thought he'd be pleased. Bokus never liked it here.'
'That's true. Or at least he never liked us – or to be even more precise, me.' Fane's grin now could only be described as wicked. 'But from his account, it sounded as if he was leaving under something of a cloud. No trumpets at the Langley gates when Andy reappears.'
'But what's he done wrong?'
'He's being blamed for Baakrime's disappearance.'
'Really? It was Miles Brookhaven out in Sana'a who was running Baakrime.'
'Ah, but it was Bokus who was giving him the line to take and Bokus who was pushing Miles to squeeze Baakrime.'
'I suppose so,' said Liz dubiously.
'And that's what provides the delicious irony – and what I suppose must gall Bokus the most.'
He paused, savouring his position as the fount of high-end gossip. Go on, spill the beans, Liz thought to herself, but she hesitated, knowing that Fane was longing to be asked. Finally curiosity prevailed. 'What delicious irony, Geoffrey?'
Pleased to be asked at last, Fane said, 'You see, they've already named the new Station Head for London. Usually, there's just the slightest lag – out of courtesy to the departing Head. Not this time.'
'Who is it?' But Fane was now laughing too hard to reply. 'Come on, Geoffrey, what's so funny?'
And at last he managed to croak, 'Miles Brookhaven.'
Liz stared at Fane, wondering if he was pulling her leg. It seemed too improbable to credit, until one looked at its natural symmetry. It was Miles who had first relayed the tip that arms were being sent to the UK, and Miles who had triggered the convoluted sequence of events that had ended – thank God – in a failed conspiracy to kill countless numbers of people.
So Miles's return to the UK somehow seemed entirely fitting. It was this – as well as the thought that she quite liked Miles, and was curious to learn what he was like after several years away – that meant Liz was glad to hear the news. Glad enough in fact to join Geoffrey Fane and find herself laughing too.
A Note on the Author
Dame Stella Rimington joined the Security Service (MI5) in 1968. During her career she worked in all the main fields of the Service: counter-subversion, counter-espionage and counter-terrorism. She was appointed Director–General in 1992, the first woman to hold the post. She has written her autobiography and eight Liz Carlyle novels. She lives in London and Norfolk.
By the Same Author
The Liz Carlyle series
At Risk
Secret Asset
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Present Danger
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Open Secret: The Autobiography of the
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Rip Tide
A Liz Carlyle Novel
When pirates attack a cargo ship off the Somalian coast and one of them is found to be a British-born Pakistani, alarm bells start ringing in London. MI5 Intelligence Officer Liz Carlyle is brought in to establish how and why a young British Muslim could go missing from his family in Birmingham and end up onboard a pirate skiff in the Indian Ocean, armed with a Kalashnikov. After an undercover operative connected to the case turns up dead in Athens it looks like piracy may be the least of the Service's problems. Liz and her team must unravel the connections between Pakistan, Greece and Somalia. And they don't have long, as trouble is brewing closer to home: the kind of explosive trouble that MI5 could do without...
'Rimington's best work demonstrates a flair for narrative, with a sense of authenticity and an insider's grasp on the pressing issues of the day'
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The Geneva Trap
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Geneva, 2012. When a Russian intelligence officer approaches MI5 with vital information about the imminent cyber-sabotage of an Anglo–American Defence programme, he refuses to talk to anyone but Liz Carlyle. But who is he, and what is his connection to the British agent?
At a tracking station in Nevada, US Navy officers watch in horror as one of their unmanned drones plummets out of the sky, and panic spreads through the British and American Intelligence services. Is this a Russian plot to disable the West's defences? Or is the threat coming from elsewhere?
As Liz and her team hunt for a mole inside the MOD, the trail leads them from Geneva, to Marseilles and into a labyrinth of international intrigue, in a race against time to stop the Cold War heating up once again...
'Rimington's best work demonstrates a flair for narrative, with a sense of authenticity and an insider's grasp on the pressing issues of the day' Washington Post
'For a pacy page-turner, she's a safe bet ... Rimington is particularly strong in her accounts of procedure, unsurprisingly, given her past role as Head of MI5' Independent
'Liz Carlyle is an MI5 agent with the traditional thriller-heroine mix of dysfunctional personal life and steely ambition' Daily Telegraph
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First published in Great Britain 2014
This electronic edition published in 2014 by Bloomsbury Publishing Plc
Copyright © 2014 by Stella Rimington
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{"url":"http:\/\/conal.net\/blog\/posts\/deriving-parallel-tree-scans","text":"## Deriving parallel tree scans\n\nThe post Deriving list scans explored folds and scans on lists and showed how the usual, efficient scan implementations can be derived from simpler specifications.\n\nLet\u2019s see now how to apply the same techniques to scans over trees.\n\nThis new post is one of a series leading toward algorithms optimized for execution on massively parallel, consumer hardware, using CUDA or OpenCL.\n\nEdits:\n\n\u2022 2011-03-01: Added clarification about \"`\u2205`\" and \"`(\u2295)`\".\n\u2022 2011-03-23: corrected \"linear-time\" to \"linear-work\" in two places.\n\n### Trees\n\nOur trees will be non-empty and binary:\n\n``data T a = Leaf a | Branch (T a) (T a)instance Show a \u21d2 Show (T a) where show (Leaf a) = show a show (Branch s t) = \"(\"++show s++\",\"++show t++\")\"``\n\nNothing surprising in the instances:\n\n``instance Functor T where fmap f (Leaf a) = Leaf (f a) fmap f (Branch s t) = Branch (fmap f s) (fmap f t)instance Foldable T where fold (Leaf a) = a fold (Branch s t) = fold s \u2295 fold tinstance Traversable T where sequenceA (Leaf a) = fmap Leaf a sequenceA (Branch s t) = liftA2 Branch (sequenceA s) (sequenceA t)``\n\nBTW, my type-setting software uses \"`\u2205`\" and \"`(\u2295)`\" for Haskell\u2019s \"mempty\" and \"mappend\".\n\nAlso handy will be extracting the first and last (i.e., leftmost and rightmost) leaves in a tree:\n\n``headT \u2237 T a \u2192 aheadT (Leaf a) = aheadT (s `Branch` _) = headT slastT \u2237 T a \u2192 alastT (Leaf a) = alastT (_ `Branch` t) = lastT t``\n\nExercise: Prove that\n\n``headT \u2218 fmap f \u2261 f \u2218 headTlastT \u2218 fmap f \u2261 f \u2218 lastT``\n\nConsider the `Leaf` and `Branch` cases separately:\n\n`` headT (fmap f (Leaf a))\u2261 {- fmap on T -} headT (Leaf (f a))\u2261 {- headT def -} f a\u2261 {- headT def -} f (headT (Leaf a))``\n`` headT (fmap f (Branch s t))\u2261 {- fmap on T -} headT (Branch (fmap f s) (fmap f t))\u2261 {- headT def -} headT (fmap f s)\u2261 {- induction -} f (headT s)\u2261 {- headT def -} f (headT (Branch s t))``\n\nSimilarly for `lastT`.\n\n### From lists to trees and back\n\nWe can flatten trees into lists:\n\n``flatten \u2237 T a \u2192 [a]flatten = fold \u2218 fmap (:[])``\n\nEquivalently, using `foldMap`:\n\n``flatten = foldMap (:[])``\n\nAlternatively, we could define `fold` via `flatten`:\n\n``instance Foldable T where fold = fold \u2218 flatten``\n``flatten \u2237 T a \u2192 [a]flatten (Leaf a) = [a]flatten (Branch s t) = flatten s ++ flatten t``\n\nWe can also \"unflatten\" lists into balanced trees:\n\n``unflatten \u2237 [a] \u2192 T aunflatten [] = error \"unflatten: Oops! Empty list\"unflatten [a] = Leaf aunflatten xs = Branch (unflatten prefix) (unflatten suffix) where (prefix,suffix) = splitAt (length xs `div` 2) xs``\n\nBoth `flatten` and `unflatten` can be implemented more efficiently.\n\nFor instance,\n\n``t1,t2 \u2237 T Intt1 = unflatten [1\u20253]t2 = unflatten [1\u202516]``\n``*T> t1(1,(2,3))*T> t2((((1,2),(3,4)),((5,6),(7,8))),(((9,10),(11,12)),((13,14),(15,16))))``\n\n### Specifying tree scans\n\n#### Prefixes and suffixes\n\nThe post Deriving list scans gave specifications for list scanning in terms of `inits` and `tails`. One consequence of this specification is that the output of scanning has one more element than the input. Alternatively, we could use non-empty variants of `inits` and `tails`, so that the input & output are in one-to-one correspondence.\n\n``inits' \u2237 [a] \u2192 [[a]]inits' [] = []inits' (x:xs) = map (x:) ([] : inits' xs)``\n\nThe cons case can also be written as\n\n``inits' (x:xs) = [x] : map (x:) (inits' xs)``\n``tails' \u2237 [a] \u2192 [[a]]tails' [] = []tails' xs@(_:xs') = xs : tails' xs'``\n\nFor instance,\n\n``*T> inits' \"abcd\"[\"a\",\"ab\",\"abc\",\"abcd\"]*T> tails' \"abcd\"[\"abcd\",\"bcd\",\"cd\",\"d\"]``\n\nOur tree functor has a symmetric definition, so we get more symmetry in the counterparts to `inits'` and `tails'`:\n\n``initTs \u2237 T a \u2192 T (T a)initTs (Leaf a) = Leaf (Leaf a)initTs (s `Branch` t) = Branch (initTs s) (fmap (s `Branch`) (initTs t))tailTs \u2237 T a \u2192 T (T a)tailTs (Leaf a) = Leaf (Leaf a)tailTs (s `Branch` t) = Branch (fmap (`Branch` t) (tailTs s)) (tailTs t)``\n\nTry it:\n\n``*T> t1(1,(2,3))*T> initTs t1(1,((1,2),(1,(2,3))))*T> tailTs t1((1,(2,3)),((2,3),3))*T> unflatten [1\u20255]((1,2),(3,(4,5)))*T> initTs (unflatten [1\u20255])((1,(1,2)),(((1,2),3),(((1,2),(3,4)),((1,2),(3,(4,5))))))*T> tailTs (unflatten [1\u20255])((((1,2),(3,(4,5))),(2,(3,(4,5)))),((3,(4,5)),((4,5),5)))``\n\nExercise: Prove that\n\n``lastT \u2218 initTs \u2261 idheadT \u2218 tailTs \u2261 id``\n\n`` lastT (initTs (Leaf a))\u2261 {- initTs def -} lastT (Leaf (Leaf a))\u2261 {- lastT def -} Leaf a lastT (initTs (s `Branch` t))\u2261 {- initTs def -} lastT (Branch (\u22ef) (fmap (s `Branch`) (initTs t)))\u2261 {- lastT def -} lastT (fmap (s `Branch`) (initTs t))\u2261 {- lastT \u2218 fmap f -} (s `Branch`) (lastT (initTs t))\u2261 {- trivial -} s `Branch` lastT (initTs t)\u2261 {- induction -} s `Branch` t``\n\n#### Scan specification\n\nNow we can specify prefix & suffix scanning:\n\n``scanlT, scanrT \u2237 Monoid a \u21d2 T a \u2192 T ascanlT = fmap fold \u2218 initTsscanrT = fmap fold \u2218 tailTs``\n\nTry it out:\n\n``t3 \u2237 T Stringt3 = fmap (:[]) (unflatten \"abcde\")``\n``*T> t3((\"a\",\"b\"),(\"c\",(\"d\",\"e\")))*T> scanlT t3((\"a\",\"ab\"),(\"abc\",(\"abcd\",\"abcde\")))*T> scanrT t3((\"abcde\",\"bcde\"),(\"cde\",(\"de\",\"e\")))``\n\nTo test on numbers, I\u2019ll use a handy notation from Matt Hellige to add pre- and post-processing:\n\n``(\u219d) \u2237 (a' \u2192 a) \u2192 (b \u2192 b') \u2192 ((a \u2192 b) \u2192 (a' \u2192 b'))(f \u219d h) g = h \u2218 g \u2218 f``\n\nAnd a version specialized to functors:\n\n``(\u219d*) \u2237 Functor f \u21d2 (a' \u2192 a) \u2192 (b \u2192 b') \u2192 (f a \u2192 f b) \u2192 (f a' \u2192 f b')f \u219d* g = fmap f \u219d fmap g``\n``t4 \u2237 T Integert4 = unflatten [1\u20256]t5 \u2237 T Integert5 = (Sum \u219d* getSum) scanlT t4``\n\nTry it:\n\n``*T> t4((1,(2,3)),(4,(5,6)))*T> initTs t4((1,((1,2),(1,(2,3)))),(((1,(2,3)),4),(((1,(2,3)),(4,5)),((1,(2,3)),(4,(5,6))))))*T> t5((1,(3,6)),(10,(15,21)))``\n\nExercise: Prove that we have properties similar to the ones relating `fold`, `scanlT`, and `scanrT` on list:\n\n``fold \u2261 lastT \u2218 scanlTfold \u2261 headT \u2218 scanrT``\n\n`` lastT \u2218 scanlT\u2261 {- scanlT spec -} lastT \u2218 fmap fold \u2218 initTs\u2261 {- lastT \u2218 fmap f -} fold \u2218 lastT \u2218 initTs\u2261 {- lastT \u2218 initTs -} fold headT \u2218 scanrT \u2261 {- scanrT def -} headT \u2218 fmap fold \u2218 tailTs\u2261 {- headT \u2218 fmap f -} fold \u2218 headT \u2218 tailTs\u2261 {- headT \u2218 tailTs -} fold``\n\nFor instance,\n\n``*T> fold t3\"abcde\"*T> (lastT \u2218 scanlT) t3\"abcde\"*T> (headT \u2218 scanrT) t3\"abcde\"``\n\n### Deriving faster scans\n\nRecall the specifications:\n\n``scanlT = fmap fold \u2218 initTsscanrT = fmap fold \u2218 tailTs``\n\nTo derive more efficient implementations, proceed as in Deriving list scans. Start with prefix scan (`scanlT`), and consider the `Leaf` and `Branch` cases separately.\n\n`` scanlT (Leaf a)\u2261 {- scanlT spec -} fmap fold (initTs (Leaf a))\u2261 {- initTs def -} fmap fold (Leaf (Leaf a))\u2261 {- fmap def -} Leaf (fold (Leaf a))\u2261 {- fold def -} Leaf a scanlT (s `Branch` t)\u2261 {- scanlT spec -} fmap fold (initTs (s `Branch` t))\u2261 {- initTs def -} fmap fold (Branch (initTs s) (fmap (s `Branch`) (initTs t)))\u2261 {- fmap def -} Branch (fmap fold (initTs s)) (fmap fold (fmap (s `Branch`) (initTs t)))\u2261 {- scanlT spec -} Branch (scanlT s) (fmap fold (fmap (s `Branch`) (initTs t)))\u2261 {- functor law -} Branch (scanlT s) (fmap (fold \u2218 (s `Branch`)) (initTs t))\u2261 {- rework as \u03bb -} Branch (scanlT s) (fmap (\u03bb t' \u2192 fold (s `Branch` t')) (initTs t))\u2261 {- fold def -} Branch (scanlT s) (fmap (\u03bb t' \u2192 fold s \u2295 fold t')) (initTs t))\u2261 {- rework \u03bb -} Branch (scanlT s) (fmap ((fold s \u2295) \u2218 fold) (initTs t))\u2261 {- functor law -} Branch (scanlT s) (fmap (fold s \u2295) (fmap fold (initTs t)))\u2261 {- scanlT spec -} Branch (scanlT s) (fmap (fold s \u2295) (scanlT t))\u2261 {- lastT \u2218 scanlT \u2261 fold -} Branch (scanlT s) (fmap (lastT (scanlT s) \u2295) (scanlT t))\u2261 {- factor out defs -} Branch s' (fmap (lastT s' \u2295) t') where s' = scanlT s t' = scanlT t``\n\nSuffix scan has a similar derivation.\n\n`` scanrT (Leaf a)\u2261 {- scanrT def -} fmap fold (tailTs (Leaf a))\u2261 {- tailTs def -} fmap fold (Leaf (Leaf a))\u2261 {- fmap on T -} Leaf (fold (Leaf a))\u2261 {- fold def -} Leaf a scanrT (s `Branch` t)\u2261 {- scanrT spec -} fmap fold (tailTs (s `Branch` t))\u2261 {- tailTs def -} fmap fold (Branch (fmap (`Branch` t) (tailTs s)) (tailTs t))\u2261 {- fmap def -} Branch (fmap fold (fmap (`Branch` t) (tailTs s))) (fmap fold (tailTs t))\u2261 {- scanrT spec -} Branch (fmap fold (fmap (`Branch` t) (tailTs s))) (scanrT t)\u2261 {- functor law -} Branch (fmap (fold \u2218 (`Branch` t)) (tailTs s)) (scanrT t)\u2261 {- rework as \u03bb -} Branch (fmap (\u03bb s' \u2192 fold (s' `Branch` t)) (tailTs s)) (scanrT t)\u2261 {- functor law -} Branch (fmap (\u03bb s' \u2192 fold s' \u2295 fold t) (tailTs s)) (scanrT t)\u2261 {- rework \u03bb -} Branch (fmap ((\u2295 fold t) \u2218 fold) (tailTs s)) (scanrT t)\u2261 {- scanrT spec -} Branch (fmap (\u2295 fold t) (scanrT s)) (scanrT t)\u2261 {- headT \u2218 scanrT -} Branch (fmap (\u2295 headT (scanrT t)) (scanrT s)) (scanrT t)\u2261 {- factor out defs -} Branch (fmap (\u2295 headT t') s') t' where s' = scanrT s t' = scanrT t``\n\nExtract code from these derivations:\n\n``scanlT' \u2237 Monoid a \u21d2 T a \u2192 T ascanlT' (Leaf a) = Leaf ascanlT' (s `Branch` t) = Branch s' (fmap (lastT s' \u2295) t') where s' = scanlT' s t' = scanlT' tscanrT' \u2237 Monoid a \u21d2 T a \u2192 T ascanrT' (Leaf a) = Leaf ascanrT' (s `Branch` t) = Branch (fmap (\u2295 headT t') s') t' where s' = scanrT' s t' = scanrT' t``\n\nTry it:\n\n``*T> t3((\"a\",\"b\"),(\"c\",(\"d\",\"e\")))*T> scanlT' t3((\"a\",\"ab\"),(\"abc\",(\"abcd\",\"abcde\")))*T> scanrT' t3((\"abcde\",\"bcde\"),(\"cde\",(\"de\",\"e\")))``\n\n### Efficiency\n\nAlthough I was just following my nose, without trying to get anywhere in particular, this result is exactly the algorithm I first thought of when considering how to parallelize tree scanning.\n\nLet\u2019s now consider the running time of this algorithm. Assume that the tree is balanced, to maximize parallelism. (I think balancing is optimal for parallelism here, but I\u2019m not certain.)\n\nFor a tree with $n$ leaves, the work $W\\phantom{\\rule{0.167em}{0ex}}n$ will be constant when $n=1$ and $2\\cdot W\\phantom{\\rule{0.167em}{0ex}}\\left(n\/2\\right)+n$ when $n>1$. Using the Master Theorem (explained more here), $W\\phantom{\\rule{0.167em}{0ex}}n=\\Theta \\phantom{\\rule{0.167em}{0ex}}\\left(n\\phantom{\\rule{0.167em}{0ex}}\\mathrm{log}n\\right)$.\n\nThis result is disappointing, since scanning can be done with linear work by threading a single accumulator while traversing the input tree and building up the output tree.\n\nI\u2019m using the term \"work\" instead of \"time\" here, since I\u2019m not assuming sequential execution.\n\nWe have a parallel algorithm that performs $n\\phantom{\\rule{0.167em}{0ex}}\\mathrm{log}\\phantom{\\rule{0.167em}{0ex}}n$ work, and a sequential program that performs linear work. Can we construct a linear-parallel algorithm?\n\nYes. Guy Blelloch came up with a clever linear-work parallel algorithm, which I\u2019ll derive in another post.\n\n### Generalizing `head` and `last`\n\nCan we replace the ad hoc (tree-specific) `headT` and `lastT` functions with general versions that work on all foldables? I\u2019d want the generalization to also generalize the list functions `head` and `last` or, rather, to total variants (ones that cannot error due to empty list). For totality, provide a default value for when there are no elements.\n\n``headF, lastF \u2237 Foldable f \u21d2 a \u2192 f a \u2192 a``\n\nI also want these functions to be as efficient on lists as `head` and `last` and as efficient on trees as `headT` and `lastT`.\n\nThe `First` and `Last` monoids provide left-biased and right-biased choice. They\u2019re implemented as `newtype` wrappers around `Maybe`:\n\n``newtype First a = First { getFirst \u2237 Maybe a }instance Monoid (First a) where \u2205 = First Nothing r@(First (Just _)) \u2295 _ = r First Nothing \u2295 r = r``\n``newtype Last a = Last { getLast \u2237 Maybe a }instance Monoid (Last a) where \u2205 = Last Nothing _ \u2295 r@(Last (Just _)) = r r \u2295 Last Nothing = r``\n\nFor `headF`, embed all of the elements into the `First` monoid (via `First \u2218 Just`), fold over the result, and extract the result, using the provided default value in case there are no elements. Similarly for `lastF`.\n\n``headF dflt = fromMaybe dflt \u2218 getFirst \u2218 foldMap (First \u2218 Just)lastF dflt = fromMaybe dflt \u2218 getLast \u2218 foldMap (Last \u2218 Just)``\n\nFor instance,\n\n``*T> headF 3 [1,2,4,8]1*T> headF 3 []3``\n\nWhen our elements belong to a monoid, we can use `\u2205` as the default:\n\n``headFM \u2237 (Foldable f, Monoid m) \u21d2 f m \u2192 mheadFM = headF \u2205lastFM \u2237 (Foldable f, Monoid m) \u21d2 f m \u2192 mlastFM = headF \u2205``\n\nFor instance,\n\n``*T> lastFM ([] \u2237 [String])\"\"``\n\nUsing `headFM` and `lastFM` in place of `headT` and `lastT`, we can easily handle addition of an `Empty` case to our tree functor in this post. The key choice is that `fold Empty \u2261 \u2205` and `fmap _ Empty \u2261 Empty`. Then `headFM` will choose the first leaf, and `lastT`\n\nWhat about efficiency? Because `headF` and `lastF` are defined via `foldMap`, which is a composition of `fold` and `fmap`, one might think that we have to traverse the entire structure when used with functors like `[]` or `T`.\n\nLaziness saves us, however, and we can even extract the head of an infinite list or a partially defined one. For instance,\n\n`` foldMap (First \u2218 Just) [5 \u2025]\u2261 foldMap (First \u2218 Just) (5 : [6 \u2025])\u2261 First (Just 5) \u2295 foldMap (First \u2218 Just) [6 \u2025]\u2261 First (Just 5)``\n\nSo\n\n`` headF d [5 \u2025]\u2261 fromMaybe d (getFirst (foldMap (First \u2218 Just) [5 \u2025]))\u2261 fromMaybe d (getFirst (First (Just 5)))\u2261 fromMaybe d (Just 5)\u2261 5``\n\nAnd, sure enough,\n\n``*T> foldMap (First \u2218 Just) [5 \u2025]First {getFirst = Just 5}*T> headF \u22a5 [5 \u2025]5``\n\n### Where to go from here?\n\n\u2022 As mentioned above, the derived scanning implementations perform asymtotically more work than necessary. Future posts explore how to derive parallel-friendly, linear-work algorithms. Then we\u2019ll see how to transform the parallel-friendly algorithms so that they work destructively, overwriting their input as they go, and hence suitably for execution entirely in CUDA or OpenCL.\n\u2022 The functions `initTs` and `tailTs` are still tree-specific. To generalize the specification and derivation of list and tree scanning, find a way to generalize these two functions. The types of `initTs` and `tailTs` fit with the `duplicate` method on comonads. Moreover, `tails` is the usual definition of `duplicate` on lists, and I think `inits` would be `extend` for \"snoc lists\". For trees, however, I don\u2019t think the correspondence holds. Am I missing something?\n\u2022 In particular, I want to extend the derivation to depth-typed, perfectly balanced trees, of the sort I played with in A trie for length-typed vectors and From tries to trees. The functions `initTs` and `tailTs` make unbalanced trees out of balanced ones, so I don\u2019t know how to adapt the specifications given here to the setting of depth-typed balanced trees. Maybe I could just fill up the to-be-ignored elements with `\u2205`.\n\n1. #### Jake McArthur:\n\nThe correspondence of tailsTs with duplicate comes up if you change your tree from a free monad to a cofree comonad.\n\n2. #### Conal Elliott \u00bb Blog Archive \u00bb Composable parallel scanning:\n\n[\u2026] About \u00ab Deriving parallel tree scans [\u2026]\n\n3. #### Russell O'Connor:\n\nYour theorem `headT \u2218 fmap f \u2261 f \u2218 headT` is the free theorem for `headT` (and `lastT`), so you don\u2019t even have to do anything to prove it. It follows from the type.\n\n4. #### conal:\n\nThanks for the tip, Russell! I\u2019ll keep an eye out for these free theorems.\n\n5. #### Conal Elliott \u00bb Blog Archive \u00bb Parallel tree scanning by composition:\n\n[\u2026] final form is as in Deriving parallel tree scans, changed for the new scan interface. 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Es befanden sich 1690–1801 sechzehn Schweizer Truppen in österreichischen Diensten, darunter eine Schweizergarde.
Sie dienten, teilweise fremdfinanziert, den Habsburgern, neben zahlreichen Schweizer Söldnern, im Spanischen, im Polnischen und im Österreichischen Erbfolgekrieg sowie in deren schier endlosen Auseinandersetzungen mit Frankreich.
Schweizer Truppen in fremden Diensten hiess der von Behörden der Schweizer Eidgenossenschaft mit Staatsverträgen geregelte Solddienst von geführten, ganzen Truppenkörpern im Ausland.
Diese Verträge enthielten ein Kapitel, das die militärischen Angelegenheiten regelte: die sogenannte Kapitulation (oder Privatkapitulation, wenn einer der Vertragspartner ein privater Militärunternehmer war).
Übersicht der Schweizer Truppen in österreichischen Diensten
Verdrängung aus den alten Stammlanden
1474 beendete Herzog Sigismund von Tirol einen langen Konflikt mit den Eidgenossen.
Er vereinbarte mit den Acht Orten und ihren Zugewandten ein Bündnis, die «Ewige Richtung». Darin verzichtete er auf die habsburgischen Stammlande in der Eidgenossenschaft und sicherte sich deren Beistand gegen Sold.
Die Kapitulation der Eidgenossen von 1480 mit Frankreich, und besonders dessen finanzielle Möglichkeiten, verhinderten jedoch längere Zeit eine militärische Annäherung.
1557 wurde die «Ewige Richtung» von Kaiser Ferdinand I. mit den nun Dreizehn Orten erneuert (bestätigt alle 10 Jahre).
Zwar hatten unter den Landsknechten Maximilians I. auch eine grosse Anzahl Schweizer Söldner in Burgund, in den Niederlanden und in Italien gekämpft: 1496 hatten ihm Bern und Uri die Werbung von 4'000 und 1516 alle Dreizehn Orte von 15'000 Mann bewilligt.
Es dauerte aber bis nach dem Angriff Frankreichs Ludwig XIV. 1672 auf die Niederlande und dem Stimmungsumschwung in der Eidgenossenschaft, bis der streng katholische Kaiser Leopold I. mit den Eidgenossen eine Kapitulation mit Truppenaushebung erreichte.
Spanischer Erbfolgekrieg
1690 waren die Dreizehn Orte endlich bereit, mit dem Kaiser eine Kapitulation für eine Schweizer Truppe in österreichischen Diensten abzuschliessen. 1702, nach Ausbruch des Spanischen Erbfolgekrieges, doppelten sie nach. 1704 folgte Graubünden.
Die Truppen der Eidgenossenschaft konnten nur defensiv in Vorderösterreich verwendet werden. Graubünden hingegen bewilligte 1704 Einheiten, die auch ausserhalb der Waldstädte und offensiv eingesetzt werden durften.
Nachdem im Rastatter Friede 1714, der den Spanischen Erbfolgekrieg beendete, die Lombardei wieder dem österreichischen Habsburger Kaiser Karl VI. zugesprochen worden war, gelang es diesem 1726, die vom spanischen Habsburger Philipp IV. 1639 mit Graubünden abgeschlossene Mailänder Konvention zu erneuern.
Aus dem im Dom zu Mailand beidseitig feierlich beschworenen Bündnis folgten die Aushebungen in Graubünden von 1734 und 1743.
Polnischer Thronfolgekrieg
Der Polnische Thronfolgekrieg wurde in Polen, am Rhein und in Italien, wo Österreich seine Position zu behaupten hatte, ausgetragen. Der Kaiser verstärkte sein Heer auch mit Schweizer Truppen. Spätestens nach dem Friedensschluss von Wien 1738 wurden sie wieder entlassen.
Österreichischer Erbfolgekrieg
Als 1740 auch der letzte österreichische Habsburger, Kaiser Karl VI., 1740 ohne männlichen Erben starb, hatte sich seine gemäss der Pragmatischen Sanktion vorgesehene Nachfolgerin und Tochter, Maria Theresia, gegen gleich drei weitere Bewerber um sein Erbe zu behaupten. Im Frieden von Aachen 1748 musste Österreich jedoch schliesslich einzig Schlesien Preussen überlassen und Maria Theresia hatte sich als Erbin durchgesetzt. Sie wurde dabei auch von einem Bündner Regiment unterstützt.
Bereits 1736 hatte Maria Theresia den Herzog Franz Stephan von Lothringen geheiratet. Dieser brachte dabei seine lothringische Schweizergarde, mit Zustimmung der katholischen Orte, nach Wien mit.
Franz Stephan hatte zu Gunsten der Heirat mit Maria Theresia auf das Herzogtum Lothringen und Bar verzichtet.
Er erhielt als Apanage dabei die Anwartschaft auf das Grossherzogtum Toskana, das er ein Jahr später, nach dem Tod des letzten Medici, in habsburgischer Sekundogenitur übernahm. Ab 1739 lebte das Paar in Wien; das Grossherzogtum Toskana wurde von Beamten als eine seiner Finanzquellen verwaltet.
Als Karl VI. 1740 starb, ging der Kaiserthron an den kurbayrischen Wittelsbacher Karl VII.
Franz Stephan, Grossherzog der Toskana, wurde zum Mitregenten von Maria Theresia. Als Erbtochter Karls VI., war sie nun Erzherzogin von Österreich sowie Königin von Ungarn und Böhmen und damit zur Herrscherin über die Habsburgischen Erblande geworden.
Damit nicht genug: Als 1745 Karl VII. bereits starb, wurde Franz Stephan in Frankfurt am Main zum Kaiser des Heiligen Römischen Reiches Deutscher Nation gekrönt. Er war also in nur neun Jahren vom einfachen Herzog zum mächtigsten Kaiser Europas und Begründer der Dynastie Habsburg-Lothringen aufgestiegen.
Koalitionskriege
1797 beendete der Frieden von Campo Formio den Ersten Koalitionskrieg des revolutionären Frankreich gegen das militärisch geschlagene Österreich.
Da entstand 1798 unter der Leitung des österreichischen leitenden Ministers Franz von Thugut am kaiserlichen Hof in Wien ein Plan, der schliesslich zur Finanzierung von mehreren Schweizer Emigranten-Regimentern im österreichischen Heer in englischem Solde führte.
Diese Emigranteneinheiten waren die letzten Schweizer Truppen in österreichischen Diensten.
Anmerkungen
Literatur
Beat Emmanuel May (von Romainmôtier): Histoire Militaire de la Suisse et celle des Suisses dans les differents services de l'Europe. Tome VIII, J. P. Heubach et Comp., Lausanne 1788, .
Felix Burckhardt: Die schweizerische Emigration 1798-1801, Dissertation Universität Basel, Verlag von Helbing & Lichtenhahn, Basel 1908.
Moritz von Wattenwil: Die Schweizer in fremden Kriegsdiensten. Separatdruck aus dem Berner Tagblatt, Bern 1930, .
Ferdinand Schramm-Schiessl von Perstorff: Die Schweizer und Bündner Regimenter in kaiserlich-österreichischen Diensten von 1691–1750. Artikel, Bündnerisches Monatsblatt: Zeitschrift für bündnerische Geschichte, Landes- und Volkskunde 1691–1699, Heft 1, 1937.
Paul de Vallière, Henry Guisan, Ulrich Wille: Treue und Ehre, Geschichte der Schweizer in fremden Diensten (übersetzt von Walter Sandoz). Les éditions d'art ancien, Lausanne 1940, .
René Chartrand: Émigré and Foreign Troops in British Service (1): 1793–1802, Osprey Publishing, ISBN 978-1-85532-766-5, Oxford 1999.
Marc Höchner: Selbstzeugnisse von Schweizer Söldneroffizieren im 18. Jahrhundert (= Herrschaft und soziale Systeme in der frühen Neuzeit, Band 18). V & R Unipress, Göttingen 2015, ISBN 978-3-8471-0321-9 (Diss. Universität Fribourg, 2013, 284 S.).
Hubert Foerster: Erfolg trotz Misserfolg: Der Aufstand und der Befreiungskrieg 1799 in der Schweiz; Ein Beitrag zu den antihelvetischen Bewegungen. Schriftenreihe der Eidgenössischen Militärbibliothek und des Historischen Dienstes, Nr. 48, Bern 2012, ISBN 3-906969-47-9.
Siehe auch
Schweizer Truppen in fremden Diensten
Schweizer Truppen in lothringischen Diensten
Schweizer Truppen in englischen und britischen Diensten
Weblinks
Die Welt der Habsburger
Stammbaum der Herzöge von Lothringen (englisch)
Oberst Ferdinand-Isaac de Rovéréa (englisch)
Einzelnachweise
oesterreichischen Diensten
Österreichisch-schweizerische Beziehungen
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Q: Exchanging data from web I am creating a program which calculates the solar radiations received on earth. The program is complete and working. I have made it specifically for my city, but I want to generalize it for the whole world. Only one thing is standing between me and my goal:
double Hm[] = {
4.38, 5.18, 5.93, 6.65, 6.67, 6.40, 5.44, 5.27, 5.62, 5.24, 4.5, 4.11
};
This data is the only thing I want to generalize. These values are for my city only. I want to take Longitude and Latitude as an input and then this input goes to the NASA website, extract only the data I want and return those values in an array as shown above. How can I do that?
A: You can use LocationManager for getting location.
You can send Lat Lon you got to your web service you can take responce in something like JSON
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 2,133
|
package de.katzenpapst.amunra.mothership;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Random;
import cpw.mods.fml.relauncher.Side;
import cpw.mods.fml.relauncher.SideOnly;
import de.katzenpapst.amunra.AmunRa;
import de.katzenpapst.amunra.helper.PlayerID;
import de.katzenpapst.amunra.network.packet.PacketSimpleAR;
import micdoodle8.mods.galacticraft.api.galaxies.CelestialBody;
import micdoodle8.mods.galacticraft.api.galaxies.CelestialBody.ScalableDistance;
import micdoodle8.mods.galacticraft.api.galaxies.GalaxyRegistry;
import micdoodle8.mods.galacticraft.api.galaxies.Moon;
import micdoodle8.mods.galacticraft.api.galaxies.Planet;
import micdoodle8.mods.galacticraft.api.galaxies.Satellite;
import net.minecraft.entity.player.EntityPlayer;
import net.minecraft.nbt.NBTTagCompound;
import net.minecraft.nbt.NBTTagList;
import net.minecraft.server.MinecraftServer;
import net.minecraft.world.WorldSavedData;
import net.minecraftforge.common.DimensionManager;
public class MothershipWorldData extends WorldSavedData {
public static final String saveDataID = "ARMothershipData";
private NBTTagCompound dataCompound;
// orbit distances should stay the same
private HashMap<CelestialBody, Float> orbitDistances;
private int highestId = 0;
private int numTicksWithoutSave = 0;
// https://github.com/Questology/Questology/blob/d125a9359e50a84ccee0c5100f04464a0d13e072/src/main/java/demonmodders/questology/handlers/event/GenericEventHandler.java
protected HashMap<Integer, Mothership> mothershipIdList;
protected HashMap<Integer, Mothership> mothershipsByDimension;
public MothershipWorldData(String id) {
super(id);
mothershipIdList = new HashMap<Integer, Mothership>();
mothershipsByDimension = new HashMap<Integer, Mothership>();
orbitDistances = new HashMap<CelestialBody, Float>();
}
public HashMap<Integer, Mothership> getMotherships() {
return (HashMap<Integer, Mothership>) mothershipIdList.clone();
}
protected void updateAllOrbits() {
HashMap<CelestialBody, Integer> bodies = this.getBodiesWithShips();
for(CelestialBody b: bodies.keySet()) {
this.updateOrbitsFor(b);
}
}
protected void updateOrbitsFor(CelestialBody parent) {
if(parent == null) return;
List<Mothership> list = getMothershipsForParent(parent);
int numShips = list.size();
float twoPi = (float) Math.PI * 2;
float angle = (twoPi / numShips);
Random rand = new Random(parent.getName().hashCode());
float phaseOffset = rand.nextFloat()*twoPi;
float orbitDistance = getMothershipOrbitDistanceFor(parent);
for(Mothership ms: list) {
if(phaseOffset > twoPi) {
phaseOffset -= twoPi;
}
ms.setPhaseShift(phaseOffset);
ms.setRelativeDistanceFromCenter(new ScalableDistance(orbitDistance, orbitDistance));
phaseOffset += angle;
}
this.markDirty();
}
public float getMothershipOrbitDistanceFor(CelestialBody parent) {
if(orbitDistances.get(parent) != null) {
return orbitDistances.get(parent);
}
// recalc
float orbitSize = -1;
if(parent instanceof Planet) {
// now try to find out what the closest thing here is
for (Moon moon : GalaxyRegistry.getRegisteredMoons().values()) {
if(moon.getParentPlanet() != parent)
continue;
if(orbitSize == -1 || orbitSize > moon.getRelativeDistanceFromCenter().unScaledDistance) {
orbitSize = moon.getRelativeDistanceFromCenter().unScaledDistance;
}
}
for (Satellite satellite : GalaxyRegistry.getRegisteredSatellites().values()) {
if(satellite.getParentPlanet() != parent) {
continue;
}
if(orbitSize == -1 || orbitSize > satellite.getRelativeDistanceFromCenter().unScaledDistance) {
orbitSize = satellite.getRelativeDistanceFromCenter().unScaledDistance;
}
}
if(orbitSize == -1) {
orbitSize = 10.0F;
} else {
orbitSize -= 1.0F;
}
} else {
// todo figure out
orbitSize = 5.0F;
}
orbitDistances.put(parent, orbitSize);
return orbitSize;
}
/**
* Creates new mothership for given player and given parentBody, sends it to all clients and returns it.
*
* @param player
* @param currentParent
* @return
*/
// @SideOnly(Side.SERVER)
public Mothership registerNewMothership(EntityPlayer player, CelestialBody currentParent) {
int newId = ++highestId;
// failsafe
if(mothershipIdList.get(newId) != null) {
throw new RuntimeException("Somehow highestID is already used");
}
// find dimension ID
int newDimensionID = DimensionManager.getNextFreeDimId();
DimensionManager.registerDimension(newDimensionID, AmunRa.config.mothershipProviderID);
Mothership ship = new Mothership(newId, new PlayerID(player));
ship.setParent(currentParent);
ship.setDimensionInfo(newDimensionID);
mothershipIdList.put(newId, ship);
mothershipsByDimension.put(newDimensionID, ship);
this.updateOrbitsFor(currentParent);// Do I even need this on server side?
this.markDirty();
NBTTagCompound data = new NBTTagCompound();
ship.writeToNBT(data);
AmunRa.packetPipeline.sendToAll(new PacketSimpleAR(PacketSimpleAR.EnumSimplePacket.C_NEW_MOTHERSHIP_CREATED, new Object[]{
data
}));
return ship;
}
/**
* Add an existing mothership object, usually one which the server sent here
*
* @param ship
* @return the definite mothership object as it should be used and stuff
*/
@SideOnly(Side.CLIENT)
public Mothership addMothership(Mothership ship) {
if(MinecraftServer.getServer() != null && !MinecraftServer.getServer().isDedicatedServer()) {
// don't do this on an integrated SSP server, because for these, the list is up to date already
this.updateOrbitsFor(ship.getParent());
// here we have a stupid case where the ship we get is a duplicate of one in the list
return this.getByMothershipId(ship.getID());
}
// probably got from server
if(ship.getID() > highestId) {
highestId = ship.getID();
}
if(mothershipIdList.get(ship.getID()) != null) {
throw new RuntimeException("Mothership "+ship.getID()+" is already registered, this shouldn't happen...");
}
maybeRegisterDimension(ship.getDimensionID());
//DimensionManager.registerDimension(ship.getDimensionID(), AmunRa.instance.confMothershipProviderID);
mothershipIdList.put(ship.getID(), ship);
mothershipsByDimension.put(ship.getDimensionID(), ship);
this.updateOrbitsFor(ship.getParent());
// this.markDirty();// not sure if needed. does the client even save this?
return ship;
}
/**
* Should only be used if only the number of ships around a body is required, otherwise just get the full list
*
* @param parent
* @return
*/
public int getNumMothershipsForParent(CelestialBody parent) {
int result = 0;
Iterator it = mothershipIdList.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pair = (Map.Entry)it.next();
Mothership curM = (Mothership) pair.getValue();
CelestialBody curParent = curM.getParent();
if(curParent != null && curParent.equals(parent)) {
result++;
}
}
return result;
}
public boolean hasMothershipsInOrbit(CelestialBody parent) {
Iterator it = mothershipIdList.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pair = (Map.Entry)it.next();
Mothership curM = (Mothership) pair.getValue();
if(curM.getParent() == parent) return true;
}
return false;
}
/**
* Get all motherships for a certain parent
* @param parent
* @return
*/
public List<Mothership> getMothershipsForParent(CelestialBody parent) {
LinkedList<Mothership> result = new LinkedList<Mothership> ();
Iterator it = mothershipIdList.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pair = (Map.Entry)it.next();
Mothership curM = (Mothership) pair.getValue();
CelestialBody curParent = curM.getParent();
if(curParent != null && curParent.equals(parent)) {
result.add(curM);
}
}
return result;
}
/**
* Get all motherships owned by a certain player
* @param player
* @return
*/
public int getNumMothershipsForPlayer(PlayerID player) {
int num = 0;
Iterator it = mothershipIdList.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pair = (Map.Entry)it.next();
Mothership curM = (Mothership) pair.getValue();
if(curM.isPlayerOwner(player)) {
num++;
}
}
return num;
}
public int getNumMothershipsForPlayer(EntityPlayer player) {
return getNumMothershipsForPlayer(new PlayerID(player));
}
/**
* Gets a list of CelestialBodies which have motherships.
* @return a map where the key is the celestial body and the value is the number of motherships around it
*/
public HashMap<CelestialBody, Integer> getBodiesWithShips() {
HashMap<CelestialBody, Integer> result = new HashMap<CelestialBody, Integer>();
Iterator it = mothershipIdList.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pair = (Map.Entry)it.next();
Mothership curM = (Mothership) pair.getValue();
CelestialBody parent = curM.getParent();
if(parent == null) continue;
if(result.get(parent) == null) {
result.put(parent, 1);
} else {
result.put(parent, result.get(parent)+1);
}
}
return result;
}
public Mothership getByDimensionId(int dimId) {
return mothershipsByDimension.get(dimId);
}
public Mothership getByMothershipId(int id) {
return mothershipIdList.get(id);
}
public Mothership getByName(String name) {
Iterator it = mothershipIdList.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pair = (Map.Entry)it.next();
Mothership curM = (Mothership) pair.getValue();
if(curM.getName().equals(name)) {
return curM;
}
}
return null;
}
/**
* This should only ever be called when the save is loaded initially
*/
@Override
public void readFromNBT(NBTTagCompound data) {
NBTTagList tagList = data.getTagList("MothershipList", 10);
mothershipIdList.clear();
mothershipsByDimension.clear();
for (int i = 0; i < tagList.tagCount(); i++)
{
NBTTagCompound nbt2 = tagList.getCompoundTagAt(i); // I think I have to unregister them on player logout.
Mothership m = Mothership.createFromNBT(nbt2);
if(highestId < m.getID()) {
highestId = m.getID();
}
if(DimensionManager.isDimensionRegistered(m.getDimensionID())) {
if(DimensionManager.getProviderType(m.getDimensionID()) != AmunRa.config.mothershipProviderID) {
// now that shouldn't happen
throw new RuntimeException("Dimension "+m.getDimensionID()+" should be registered for an AmunRa Mothership, registered for "+DimensionManager.getProviderType(m.getDimensionID())+" instead");
}
// it's fine otherwise
} else {
DimensionManager.registerDimension(m.getDimensionID(), AmunRa.config.mothershipProviderID);
}
mothershipIdList.put(m.getID(), m);
mothershipsByDimension.put(m.getDimensionID(), m);
}
this.updateAllOrbits();
}
/**
* This should only be called on the client if the server has sent some generic change data
* @param data
*/
/*public void updateFromNBT(NBTTagCompound data) {
NBTTagList tagList = data.getTagList("MothershipList", 10);
for (int i = 0; i < tagList.tagCount(); i++)
{
NBTTagCompound mothershipNBT = tagList.getCompoundTagAt(i);
int id = mothershipNBT.getInteger("id");
Mothership m = this.getByMothershipId(id);
if(m != null) {
m.updateFromNBT(mothershipNBT);
} else {
m = Mothership.createFromNBT(mothershipNBT);
if(highestId < id) {
highestId = id;
}
if(DimensionManager.isDimensionRegistered(m.getDimensionID())) {
if(DimensionManager.getProviderType(m.getDimensionID()) != AmunRa.config.mothershipProviderID) {
// now that shouldn't happen
throw new RuntimeException("Dimension "+m.getDimensionID()+" should be registered for an AmunRa Mothership, registered for "+DimensionManager.getProviderType(m.getDimensionID())+" instead");
}
// it's fine otherwise
} else {
DimensionManager.registerDimension(m.getDimensionID(), AmunRa.config.mothershipProviderID);
}
mothershipIdList.put(m.getID(), m);
mothershipsByDimension.put(m.getDimensionID(), m);
}
}
}*/
/**
* Hack for client-side dimension registration
*
* @param dimId
*/
protected void maybeRegisterDimension(int dimId) {
if(!DimensionManager.isDimensionRegistered(dimId)) {
DimensionManager.registerDimension(dimId, AmunRa.config.mothershipProviderID);
} else {
// just check if it's registered the right way
int type = DimensionManager.getProviderType(dimId);
if(type != AmunRa.config.mothershipProviderID) {
throw new RuntimeException("Dimension "+dimId+" could not be registered for mothership because it's already taken");
}
}
}
@Override
public void writeToNBT(NBTTagCompound data) {
NBTTagList tagList = new NBTTagList();
// HashMap<Integer, Mothership> mothershipIdList
for (Mothership m : mothershipIdList.values())
{
NBTTagCompound nbt2 = new NBTTagCompound();
m.writeToNBT(nbt2);
tagList.appendTag(nbt2);
}
data.setTag("MothershipList", tagList);
}
public void tickAllMotherships() {
boolean hasChanged = false;
for (Mothership m : mothershipIdList.values())
{
if(!m.isInTransit()) {
continue;
}
numTicksWithoutSave++;
hasChanged = true;
if(m.modRemainingTravelTime(-1) <= 0) {
// arrived
// we will need the worldprovider here
m.getWorldProviderServer().endTransit();
AmunRa.packetPipeline.sendToAll(new PacketSimpleAR(PacketSimpleAR.EnumSimplePacket.C_MOTHERSHIP_TRANSIT_ENDED, m.getID()));
}
}
if(hasChanged || (!hasChanged && numTicksWithoutSave > 0)) {
// if no changes, but still unsaved changes
if(numTicksWithoutSave >= 1200) {
numTicksWithoutSave = 0;
this.markDirty(); //
}
/*if(hasChanged) {
NBTTagCompound data = new NBTTagCompound ();
TickHandlerServer.mothershipData.writeToNBT(data);
AmunRa.packetPipeline.sendToAll(new PacketSimpleAR(PacketSimpleAR.EnumSimplePacket.C_UPDATE_MOTHERSHIP_LIST, data));
}*/
}
}
/**
* This is just so that the progress bar is updated on client
*/
public void tickAllMothershipsClient() {
for (Mothership m : mothershipIdList.values())
{
if(!m.isInTransit()) {
continue;
}
if(m.getRemainingTravelTime() > 0) {
m.modRemainingTravelTime(-1);
}
}
}
public void unregisterAllMotherships() {
for (Integer dimID : mothershipsByDimension.keySet())
{
DimensionManager.unregisterDimension(dimID);
}
mothershipIdList.clear();
mothershipsByDimension.clear();
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,692
|
\section{Introduction}
\label{sec:introduction}
Let $n \geq 2$ be the dimension
and let $\Omega \subsetneq \Rd$
be a domain (open, connected) with
a non-empty compact Lipschitz boundary
($\partial \Omega \in C^{0,1}$).
For given $\ensuremath{\varepsilon} > 0$,
we shall consider the following Hele-Shaw-type problem
on a parabolic cylinder $Q := \Omega \times (0, T]$ for some $T > 0$:
find $u^\ensuremath{\varepsilon}: Q \to [0, \infty)$ that formally satisfies
\begin{equation}
\label{HSt}
\begin{cases}
- \Delta u(x,t) = 0 & \text{in $\set{u > 0}$},\\
V_\nu(x,t) = g\pth{\frac{x}{\ensuremath{\varepsilon}},\frac{t}{\ensuremath{\varepsilon}}} \abs{Du^+(x,t)}
& \text{on $\partial \set{u>0}$},
\end{cases}
\end{equation}
with some boundary data to be specified later
(Theorem~\ref{th:well-posedness}),
\begin{figure}
\fig{Fig1}{4.5in}
\caption{The Hele-Shaw problem in a plane}
\label{fig:setting}
\end{figure}
where $D = D_x$ and $\Delta = \Delta_x$ are respectively the gradient and the Laplace operator in the $x$ variable,
and $V_\nu$ is the normal velocity of
the \emph{free boundary} $\partial \set{u > 0}$.
Assuming that the solution is sufficiently smooth,
the free boundary $\partial \set{u > 0}$ is a level set of $u$
and its normal velocity can be expressed as
\begin{align*}
V_\nu = \frac{u_t^+}{\abs{Du^+}}.
\end{align*}
Here $u_t^+$ and $Du^+$ formally represent
the limits of the derivatives from the positive set of $u$.
The well-posedness of problem \eqref{HSt}
requires some basic regularity of $g$.
In the following we shall assume that $g$ satisfies
\begin{align}
\label{g-bound}
0 < m \leq g \leq M
\end{align}
for some positive constants $m, M$,
and that $g$ is $L$-Lipschitz both in $x$ and $t$, that is,
\begin{align}
\label{g-Lipschitz}
\abs{g(x,t) - g(y,s)} \leq L(\abs{x-y} + \abs{t-s})
\qquad
\text{for all $(x,t), (y,s) \in \Rd\times\ensuremath{\mathbb{R}}$.}
\end{align}
We are interested in the behavior of the solutions of \eqref{HSt}
in the \emph{homogenization limit} $\ensuremath{\varepsilon} \to 0+$.
Since we want to observe an averaging behavior,
we further assume that $g$ is a $\ensuremath{\mathbb{Z}}^{n+1}$-periodic function, i.e.,
\begin{align}
\label{g-periodic}
g(x + k, t +l) &= g(x, t)
&& \text{for all $(x,t) \in \Rd \times \ensuremath{\mathbb{R}},\ (k,l) \in \ensuremath{\mathbb{Z}}^n \times \ensuremath{\mathbb{Z}}$.}
\end{align}
In the following we define
\begin{align*}
g^\ensuremath{\varepsilon}(x,t) := g\pth{\frac{x}{\ensuremath{\varepsilon}}, \frac{t}{\ensuremath{\varepsilon}}}.
\end{align*}
Note that $g^\ensuremath{\varepsilon}$ is an $\ensuremath{\varepsilon} \ensuremath{\mathbb{Z}}^{n+1}$-periodic $\frac{L}{\ensuremath{\varepsilon}}$-Lipschitz function
which satisfies \eqref{g-bound}.
Problem \eqref{HSt} with $g \equiv const$ is the standard
Hele-Shaw problem with no surface tension.
In two dimensions, it was introduced in \cite{HS} as a model
of a slow movement of a viscous fluid injected
in between two nearby parallel plates
that form the so-called Hele-Shaw cell.
This problem
naturally generalizes to all dimensions $n \geq 1$.
In particular, in three dimensions it serves as a model
of a pressure-driven flow of an incompressible fluid through a porous medium.
Following this motivation,
problem \eqref{HSt} with general $g$ describes a pressure-driven flow of
an incompressible fluid in an inhomogeneous, time-dependent medium.
Free boundary problems with similar velocity laws
have various applications in the plastics industry
and other fields \cites{PR,Richardson,Steinbach}.
The Hele-Shaw-type problem \eqref{HSt}
can also be viewed as a quasi-stationary limit
of the one-phase Stefan problem with a latent heat of phase transition
depending on position and time
\cite{AGM,LR,Primicerio,Rou}.
\subsection*{Homogenization overview}
There is a large amount of literature on homogenization
which is beyond the scope of this discussion
and thus we refer the reader to \cite{JKO,Tartar}
and the references therein.
In the context of viscosity solutions for fully nonlinear problems
of the first and second order,
the now standard approach to homogenization
using correctors that are the solutions of an appropriate cell problem
was pioneered by Lions, Papanicolau \& Varadhan \cite{LPV}
for first order equations,
and later by Evans \cite{Evans92} for second order equations.
Unfortunately, this approach does not apply to problems
like \eqref{HSt} because the zero level of solutions has a special
significance and the perturbation of a test function
by a global periodic solution
of some cell problem, i.e., the corrector, does not seem feasible.
The idea of using obstacle problems to recover the homogenized operator
without the need for a cell problem
was developed by Caffarelli, Souganidis \& Wang \cite{CSW}
for the stochastic homogenization of fully-nonlinear second order
elliptic equations.
It was then applied to the homogenization of the Hele-Shaw problem
with spatial periodic inhomogeneity
by Kim \cites{K07},
using purely the methods of the theory of viscosity solutions.
This approach was later extended to a model of contact angle dynamics \cite{K08},
and an algebraic rate of the convergence of free boundaries
was obtained \cite{K09}.
Kim \& Mellet \cites{KM09,KM10} later succeeded at applying
a combination of viscosity and variational approaches
and obtained a homogenization result in the setting of
spatial stationary ergodic random media.
A related question of long-time asymptotics of
the spatially inhomogeneous Hele-Shaw problem was addressed by the author
\cite{P11}.
This technique relies on the special structure of
problem \eqref{HSt} with time-independent $g$
which allows us to rewrite
the problem as a variational inequality of a certain obstacle problem.
We refer the reader to \cite{Rodrigues} and the references therein
for an exposition of obstacle problems and their homogenization.
The homogenization results for spatial media do not necessarily
translate directly to the spatiotemporal homogenization.
In the context of Hamilton-Jacobi equations, for instance,
a lack of uniform estimates in the spatiotemporal case was
encountered by Schwab \cite{Schwab} who
proved the homogenization
in spatiotemporal stationary ergodic random media, which was
before established by Souganidis \cite{Soug99} for spatial random media.
The situation seems to be more extreme in the case of the Hele-Shaw problem since the homogenization in spatiotemporal media
is qualitatively different even in the periodic case.
This difference can be already observed in the homogenization of a single
ordinary differential equation
of the type $x_\ensuremath{\varepsilon}'(t) = f(x_\ensuremath{\varepsilon}(t)/\ensuremath{\varepsilon}, t/\ensuremath{\varepsilon}, x_\ensuremath{\varepsilon}(t))$.
In fact, the Hele-Shaw problem \eqref{HSt} reduces
to this type of ODE in one dimension; see Section~\ref{sec:homogenizationin1D}
for further discussion.
It is known that, under some assumptions on $f$, $x_\ensuremath{\varepsilon} \rightrightarrows \bar x$ locally uniformly,
where $\bar x$ is the solution of a homogenized ODE
\cite{Piccinini1,IM10}.
However, the form of the homogenized problem depends on $f$.
If, for illustration, $f(y,s,x) = f(y)$ is a $\ensuremath{\mathbb{Z}}$-periodic
Lipschitz function,
then the homogenized problem has the form $\bar x'(t) = \bar f$
where $\bar f$ is the constant
\begin{equation}
\label{hom-const}
\bar f = \pth{\int_0^1 (f(y))^{-1} \;dy}^{-1}.
\end{equation}
On the other hand,
if $f(y,s,x) = f(y,s)$ has a nontrivial dependence on both $y$ and $s$,
then no such explicit formula exists and, in fact, the right-hand side
of the homogenized problem, while still a constant,
can possibly attain any value in the range $[\min f, \max f]$.
Another explanation of the qualitative difference,
slightly more intuitive, is the interpretation
of the Hele-Shaw problem \eqref{HSt} as the quasi-stationary
limit of the Stefan problem.
Then the quantity $(g(x/\ensuremath{\varepsilon},t/\ensuremath{\varepsilon}))^{-1}$ can be interpreted
as the \emph{latent heat of phase transition}, that is,
the ``energy'' necessary
to change a unit volume of the dry region into the wet region
and advance the free boundary \cite{P11}.
The ``energy flux'' is proportional to the gradient of the pressure.
If $g$ does not depend on time, the homogenized latent heat
is simply the average, i.e., $\int_{[0,1]^n} (g(x))^{-1} \;dx$,
giving the homogenized velocity recovered in \cite{KM09}
(of the form \eqref{hom-const}).
This formula was rigorously justified in \cite{KM09}
using the variational formulation via an obstacle
problem,
which allows one to solve for the shape of the wet region
at a given time without the need to solve the problem
at previous times.
The variational formulation was introduced in \cite{EJ}
using a transformation due to \cite{B72}.
In a sense, for the evolution overall,
the free boundary feels the same influence of the medium no matter how it passes through it. Or, in other words, the amount of the energy required
to fill a given region is always the same.
This drastically changes when the latent heat depends on
both space and time.
In this case, the energy required to advance the boundary depends
on the specific history of the motion of the free boundary through the space-time.
The variational formulation does not apply anymore.
What is more, it is no longer obvious that the problem should homogenize.
As the one-dimensional situation indicates---Section~\ref{sec:homogenizationin1D}---the homogenized velocity
might have a complicated dependence on the gradient,
with velocity pinning and directional dependence.
Some of these features appear in the spatial homogenization
of non-monotone problems \cite{K08,K09}.
Our approach can be characterized as geometric, relying on
maximum principle arguments.
A similar approach to homogenization,
albeit of (local) geometric motions,
was recently pursued by Caffarelli \& Monneau \cite{CM12}.
The geometric approach to the Hele-Shaw problem
is complicated by the nonlocal nature of the problem.
Indeed, \eqref{HSt} can be interpreted as a geometric motion
of the free boundary with the velocity given
by a nonlocal operator based on the Dirichlet-to-Neumann map.
Thus the domain is of crucial importance.
\subsection*{Main results}
We present new well-posedness and homogenization results
for the Hele-Shaw-type problem \eqref{HSt}.
The well-posedness result is
a generalization and an improvement of the previous results
in \cites{K03,K07}.
In full generality, we consider the Hele-Shaw type problem
\begin{align}
\label{hs-f}
\begin{cases}
-\Delta u(x,t) = 0 & \text{in } \set{u > 0} \cap Q,\\
u^+_t(x,t) = f(x,t,Du)\abs{Du^+(x,t)}^2 & \text{on } \pset{u > 0} \cap Q.
\end{cases}
\end{align}
Let us introduce a number of assumptions
on the function $f(x,t,p): Q \times \Rd \to \ensuremath{\mathbb{R}}$.
In the following, we use the semi-continuous envelopes $f_*$ and $f^*$
of $f$ on $Q \times \Rd$ (see \eqref{usclsc} for definition).
Let us point out that we do not require continuity of $f(x,t,p)$ in $p$.
We shall need this generality to
handle the homogenized problem.
\begin{enumerate}[\bfseries ({A}1)]
\item (non-degeneracy)
There exist constants $m$ and $M$
such that $0 < m \leq f(x,t,p) \leq M$
for all $(x,t,p) \in Q \times \Rd$.
\item (Lipschitz continuity)
There exists a constant $L > 0$
such that $f$ is $L$-Lipschitz in $x$ and $t$ for all $p$.
\item (monotonicity)
$f^*(x,t, a_1 p) \abs{a_1 p} \leq f_*(x,t, a_2 p) \abs{a_2 p}$
for any $(x, t, p) \in Q \times \Rd$
and $0 < a_1 < a_2$.
\end{enumerate}
The assumption (A3) implies that the free boundary
velocity is monotone with respect to the gradient,
while allowing for certain jumps.
We have the following well-posedness theorem
for viscosity solutions that are introduced in Section~\ref{sec:viscosity-solutions}.
\begin{theorem}[Well-posedness]
\label{th:well-posedness}
Let $Q := \Omega \times (0,T]$ where $\Omega$
is a domain that satisfies the assumptions above
and $T > 0$.
Assume that either $f(x,t,p) = f(x,t)$
or $f(x,t,p) = f(p)$
and that $f$ satisfies (A1)--(A3).
Then for any positive function $\psi \in C(\partial \Omega \times [0,T])$
strictly increasing in time,
and for any open set $\Omega_0 \subset \Rd$
with smooth boundary, $\partial \Omega_0 \in C^{1,1}$,
such that $\Omega^c \subset \Omega_0$ and $\Omega_0 \cap \Omega$
is bounded (see Figure~\ref{fig:setting}),
there exists a unique bounded viscosity solution
$u: \cl Q \to [0, \infty)$ of the Hele-Shaw-type problem \eqref{hs-f}
such that $u^* = u_* = u$ on $\partial_P Q := \cl Q \setminus Q$
with boundary data $u(x,t) = \psi(x,t)$ on $\partial \Omega \times [0,T]$
and initial data $u(\cdot, 0) > 0$ in $\Omega \cap \Omega_0$
and $u(\cdot, 0) = 0$ in $\Omega \setminus \Omega_0$.
The solution is unique in the sense that
$u_* = v_*$ and $u^* = v^*$ for any two viscosity solutions
$u$, $v$ with the given boundary data.
\end{theorem}
In the context of a flow in porous media,
$\Omega^c$ represents the source of a liquid
with the prescribed pressure $\psi$
on its boundary, and $\Omega_0$ is the initial wet region.
The situation is depicted in Figure~\ref{fig:setting}.
\medskip
However, the main result of this paper concerns the homogenization of \eqref{HSt}.
Note that because the solution of the homogenized problem
might be discontinuous,
we cannot expect uniform convergence of solutions in general.
Furthermore, we do not know
if the homogenized velocity
($r$ in Theorem~\ref{th:intro-homogenization} below) is continuous.
\begin{theorem}[Homogenization]
\label{th:intro-homogenization}
Suppose that $g \in C(\Rd \times \ensuremath{\mathbb{R}})$
satisfies \eqref{g-bound}--\eqref{g-periodic}.
Then there exists a function $r: \Rd \to \ensuremath{\mathbb{R}}$
such that
\[f(x,t,p) := \frac{r(p)}{\abs p}\] satisfies (A1)--(A3),
and,
for any $Q$ and initial and boundary data $\Omega_0, \psi$
that satisfy the assumptions in Theorem~\ref{th:well-posedness}
the following results hold:
\begin{enumerate}
\item
the unique solutions $u^\ensuremath{\varepsilon}$ of \eqref{HSt}
with data $\Omega_0, \psi$
converge in the sense of half-relaxed limits as $\ensuremath{\varepsilon}\to0$
(Definition~\ref{def:half-relaxed}) to
the unique solution $u$ of
\eqref{hs-f} with $f(x,t,p) = r(p)/\abs{p}$
and the same data $\Omega_0, \psi$;
\item
if $u$ is also continuous on a compact set
$K \subset \cl Q$ then $u^\ensuremath{\varepsilon} \rightrightarrows u$
converge uniformly on $K$;
\item
the free boundaries $\partial \set{(u^\ensuremath{\varepsilon})_* > 0}$
converge uniformly
to the free boundary $\partial \set{u_* > 0}$
with respect to the Hausdorff distance
(Definition~\ref{def:Hausdorff-convergence}).
\end{enumerate}
\end{theorem}
\subsection*{Sketch of the proof}
Let us give an overview of the main ideas in the paper.
Since the variational formulation via an obstacle problem,
discussed above, is not available,
we have to rely solely on the technically heavy tools of the viscosity theory. The time-dependence of $g$ poses significant new challenges which require a rather nontrivial extension of the previous results.
The first step is the identification of the homogenized problem.
Since the solutions of \eqref{HSt} are harmonic in space in their positive sets on any scale $\ensuremath{\varepsilon}$,
their limit as $\ensuremath{\varepsilon} \to 0$ should also be harmonic in space.
The free boundary velocity of the homogenized problem is, however,
unknown.
Following the ideas from \cite{CSW,K07},
we identify the correct homogenized free boundary velocity
by solving an obstacle problem and study its behavior as $\ensuremath{\varepsilon}\to0$.
To motivate our approach, suppose that the solutions $u^\ensuremath{\varepsilon}$ in the limit $\ensuremath{\varepsilon} \to 0$ converge
to a solution $u$ of the homogenized Hele-Shaw problem with
the free boundary law given as
\begin{align*}
V_\nu &= r(Du) & &\text{on } \pset{u > 0},
\end{align*}
where $a \mapsto r(a q)$ is an increasing function on $\ensuremath{\mathbb{R}}_+$ for any $q \in \Rd$.
The crucial observation is that this problem has traveling wave solutions;
these are the planar solutions of the form
$P_{q,r}(x,t) = (\abs{q}r t + x \cdot q)_+$,
with the particular choice $r = r(q)$.
Here $(\cdot)_+$ stands for the positive part, i.e.,
$s_+ := \max(s, 0)$.
If $r > r(q)$ then $P_{q,r}$ is a supersolution
of the homogenized problem,
and if $r < r(q)$ it is a subsolution.
This observation allows us to identify the correct velocity $r(q)$
by solving the $\ensuremath{\varepsilon}$-problem \eqref{HSt} for each $\ensuremath{\varepsilon} > 0$ and comparing
the solution with the planar solution $P_{q,r}$ for given $r$ and $q$.
Loosely speaking, if $r$ is too large for a given $q$,
the solutions $u^\ensuremath{\varepsilon}$ should evolve slower than $P_{q,r}$
for small $\ensuremath{\varepsilon}$
and, similarly, if $r$ is too small then $u^\ensuremath{\varepsilon}$ should evolve faster
than $P_{q,r}$ for small $\ensuremath{\varepsilon}$.
To make this idea rigorous,
we solve the following obstacle problems
for each fixed $\ensuremath{\varepsilon}$:
given a fixed domain $Q$, $r$ and $q$,
find the largest subsolution $\lu_{\e;q,r}$ of the $\ensuremath{\varepsilon}$-problem
that stays under $P_{q,r}$ in $Q$
and the smallest supersolution $\uu_{\e;q,r}$ of the $\ensuremath{\varepsilon}$-problem
that stays above $P_{q,r}$ in $Q$.
Following this reasoning,
we find two candidates for the correct homogenized velocity $r(q)$
by ``properly measuring'' how much contact there is as $\ensuremath{\varepsilon} \to 0$
between the obstacle $P_{q,r}$
and the largest subsolution $\lu_{\e;q,r}$,
yielding $\lr(q)$,
and the smallest supersolution $\uu_{\e;q,r}$, yielding $\ur(q)$.
Since the free boundary velocity law is nonlocal,
given by a Dirichlet-to-Neumann map,
a good choice of the domain $Q$ for this procedure is important.
For homogenization to occur, it is necessary that both candidates
$\lr(q)$ and $\ur(q)$
yield the same limit problem \eqref{hs-f}.
We need to give a proper meaning to the intuitive idea of
evolving slower or faster than the obstacle $P_{q,r}$.
The selection of a quantity
that not only ``properly measures'' the amount of contact
between the obstacle $P_{q,r}$ and the solutions of the obstacle problem
in the homogenization limit $\ensuremath{\varepsilon}\to0$
but is also convenient to work with is far from obvious.
In particular, we want to take advantage
of the natural monotonicity (Birkhoff property)
of the solutions of the obstacle problem,
which is a consequence of the periodicity of the medium.
In \cite{CSW} as well as in \cite{K07,K08,K09},
the authors consider the coincidence set
of the solutions of the obstacle problem
and the obstacle $P_{q,r}$ as their quantity of choice.
This choice is motivated by the fact that
if there is a contact on a sufficiently large set,
then the solution must
be close to the obstacle everywhere.
This kind of estimate can be established using the
Alexandroff-Bakelman-Pucci estimate
in the case of fully nonlinear elliptic problems in \cite{CSW}.
Since ABP-type estimates are not available for
the Hele-Shaw problem,
the closeness to the obstacle must be derived by other means
\cite{K07}.
The disadvantage of this approach in the context of free boundary problems,
in particular for the homogenization of \eqref{HSt},
stems from the restriction that it imposes on
the directions in which one can translate
the solutions of the obstacle problem
to take advantage of their natural monotonicity property,
and to recover the monotonicity
of the contact set.
This creates technical difficulties and
requires a separate treatment of rational and irrational directions.
It seems that, to overcome these difficulties,
it is necessary to scale the solutions in time
for the arguments to work.
However, a scaling in time is not available
in the time-dependent medium.
The main new idea in this paper is
the introduction of \emph{flatness}
and its critical value $\ensuremath{\varepsilon}^\ensuremath{\beta}$ (Section~\ref{sec:flatness}).
More specifically, to recover the correct homogenized
boundary velocity for a given gradient $q$
we test
if,
for some fixed $\ensuremath{\beta} \in (0,1)$,
the free boundary of the solution of the obstacle problem
stays $\ensuremath{\varepsilon}^\ensuremath{\beta}$-close to the obstacle for a unit time
on an arbitrary small scale $\ensuremath{\varepsilon}$.
This choice is motivated by the new cone flatness property
(Proposition~\ref{pr:cone-flatness}):
the boundary of the solution of the obstacle problem
stays between two cones that are $\sim \ensuremath{\varepsilon} \abs{\ln\ensuremath{\varepsilon}}^{1/2}$ apart.
Therefore if, for $\ensuremath{\varepsilon} \ll 1$, the boundary is farther than $\ensuremath{\varepsilon}^\ensuremath{\beta} \gg \ensuremath{\varepsilon} \abs{\ln\ensuremath{\varepsilon}}^{1/2}$,
it will be detached from the obstacle on a large set.
This is formulated in the \emph{detachment lemma} (Lemma~\ref{le:detachment-ball}).
Moreover, the improvement of the \emph{local comparison principle}
(Theorem~\ref{th:localComparison})
for $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat solutions to allow for $\ensuremath{\beta} \in (4/5,1)$
is a necessary ingredient to close the argument.
This approach allows us to prove that
the candidates $\lr(q)$ and $\ur(q)$
almost coincide, i.e., $\lr = \ur^*$,
and that $f(x,t,p) = \lr(p)/\abs p$ satisfies
the assumptions (A1)--(A3).
Moreover, if the medium is time-independent, that is, $g(x,t) = g(x)$,
we are able to show that $\lr = \ur$
are continuous and one-homogeneous
(Section~\ref{sec:time-independent}),
essentially recovering the result of \cite{K07}.
Once the homogenized velocity has been identified,
we can prove that the half-relaxed limits
satisfy the homogenized problem (Section~\ref{sec:convergence}).
The perturbed test function method cannot be applied to free boundary
problems and a different, more geometric argument
based on a comparison of the solutions
of the $\ensuremath{\varepsilon}$-problem with rescaled translations of the solutions of the obstacle problems must
be engaged.
The detachment lemma (Lemma~\ref{le:detachment-ball}) plays a key role in this argument.
Finally, since the comparison with barriers guarantees that
the limits have the correct boundary data,
the comparison principle for the limit problem establishes
that the upper and lower half-relaxed limits coincide with the
unique solution of the homogenized problem.
\subsection*{Open problems}
Let us conclude the introduction by mentioning some of the open problems.
We have been only able to show that the homogenized velocity
$r$ is semi-continuous.
However, it seems quite reasonable
to expect continuity, or even H\"older continuity.
As the one-dimensional case suggests in Section~\ref{sec:homogenizationin1D},
this is the highest regularity one may hope for
in a general situation.
Another open problem is the question of a convergence rate of
free boundaries in the homogenization limit.
An algebraic rate was obtained in the periodic spatial homogenization case
\cite{K09}.
These issues will be addressed in future work.
A related open problem is the homogenization in
random media when the free boundary problem does not have a variational
structure, unlike \cite{KM09,KM10}.
This problem has not yet been solved since there is no obvious
subadditive quantity to which the subadditive ergodic theorem
can be applied to overcome the lack of compactness of a general
probability space.
Finally, in the current paper, we use the monotone propagation property
of the free boundary---that is, that the wet region cannot recede---to
obtain some of the important estimates.
The presented method, however, seems robust enough to handle
non-monotone problems such as the model of contact angle dynamics
as in \cite{K08,K09}. This will also be a subject of future work.
\subsection*{Outline}
The exposition of the proof of the homogenization result
was split in a number
of steps.
First we give the necessary definitions of solutions
in Section~\ref{sec:viscosity-solutions},
together with some preliminary results including the comparison principle
and a well-posedness result.
These are used in Section~\ref{sec:limit-problem}, where we identify a candidate for the limit velocity.
Finally, Section~\ref{sec:convergence} is devoted
to showing that the solutions converge in the homogenization limit.
In addition, the appendices contain some auxiliary results
used throughout the text.
\section{Viscosity solutions}
\label{sec:viscosity-solutions}
In this section, we briefly revisit the theory of viscosity solutions
of the Hele-Shaw problem \eqref{HSt}.
We generalize the definitions introduced in \cites{K03,K07},
and outline the proof of the comparison principle in our settings, Section~\ref{sec:comparison-principle},
with an aim at establishing Theorem~\ref{th:well-posedness} in Section~\ref{sec:well-posedness}.
Let us mention that viscosity solutions
seem to be the natural class of solutions of \eqref{HSt}
since the problem has a maximum principle structure.
Moreover, due to the possible topological changes
and merging of free boundaries, the solutions
might be discontinuous \cite{K03}.
The standard notion of solutions due to \cite{EJ}
does not apply when $g$ in \eqref{HSt} depends on time.
Since viscosity solutions are the only weak notion of solutions used
throughout the paper,
we will often refer to them simply as \emph{solutions}.
Before we give the definitions of viscosity solutions,
we need to introduce some notation.
For given radius $\rho > 0$ and center $(x,t) \in \Rd\times\ensuremath{\mathbb{R}}$,
we define the open balls
\begin{align*}
B_\rho(x,t) &:= \set{(y,s) \in \Rd\times\ensuremath{\mathbb{R}}:
\abs{y-x}^2 + \abs{s-t}^2 < \rho^2},\\
B_\rho(x) &:= \set{y \in \Rd : \abs{y-x} < \rho}.
\end{align*}
Let $E \subset \ensuremath{\mathbb{R}}^d$ for some $d \geq 1$.
Then $USC(E)$ and $LSC(E)$ are respectively
the sets of all upper semi-continuous
and lower semi-continuous functions on $E$.
For a locally bounded function $u$ on $E$
we define the semi-continuous envelopes
$u^{*, E} \in USC(\ensuremath{\mathbb{R}}^d)$ and $u_{*,E} \in LSC(\ensuremath{\mathbb{R}}^d)$
as
\begin{align}\label{usclsc}
\begin{aligned}
u^{*,E} &:=
\inf \set{v \in USC(\ensuremath{\mathbb{R}}^d): v \geq u \text{ on } E},\\
u_{*, E} &:= \sup \set{v \in LSC(\ensuremath{\mathbb{R}}^d): v \leq u \text{ on } E}.
\end{aligned}
\end{align}
Note that $u^{*,E} : \ensuremath{\mathbb{R}}^d \to [-\infty, \infty)$
and $u_{*, E} : \ensuremath{\mathbb{R}}^d \to (-\infty, \infty]$ are finite on $\cl E$.
We simply write $u^*$ and $u_*$
if the set $E$ is understood from the context.
The envelopes can be also expressed as
\[
u^{*,E}(x) = \lim_{\ensuremath{\delta}\to0} \sup \set{u(y) : y \in E,\ \abs{y - x} < \ensuremath{\delta}}
\quad \text{for } x \in \cl E,
\qquad u_{*,E} = -(-u)^{*,E}.
\]
It will be also useful to use a shorthand notation for
the set of positive values of a given function $u: E \to [0,\infty)$,
defined on a set $E \subset \Rd\times \ensuremath{\mathbb{R}}$,
\begin{equation*}
\Omega(u; E) := \set{(x,t) \in E: u(x,t) > 0}, \qquad
\Omega^c(u; E) := \set{(x,t) \in E: u(x,t) = 0},
\end{equation*}
and the closure $\cl\Omega(u;E) := \cl{\Omega(u;E)}$.
For $t \in \ensuremath{\mathbb{R}}$, the time-slices $\cl\Omega_t(u; E)$,
$\Omega_t(u; E)$ and $\Omega^c_t(u; E)$
are defined in the obvious way, i.e.,
\begin{align*}
\cl\Omega_t(u;E) = \set{x : (x,t) \in \cl\Omega(u;E)}, \qquad \text{etc.}
\end{align*}
We shall call the boundary of the positive set in $E$ the \emph{free
boundary} of $u$ and denote it $\Gamma(u;E)$, i.e.,
\begin{align*}
\Gamma(u; E) = (\partial \Omega(u; E)) \cap E.
\end{align*}
If the set $E$ is understood from the context, we shall simply write
$\Omega(u)$, etc.
For given constant $\tau \in \ensuremath{\mathbb{R}}$ we will often abbreviate
\begin{align*}
\set{t \leq \tau} := \set{(x,t) \in \Rd\times\ensuremath{\mathbb{R}}: t \leq \tau},
\qquad \text{etc.}
\end{align*}
It will be convenient to define viscosity solutions
on general parabolic neighborhoods;
we refer the reader to \cite{WangI} for a more general definition.
\begin{definition}[Parabolic neighborhood and boundary]
\label{def:parabolic-nbd}
\ \\A nonempty set $E \subset \Rd\times \ensuremath{\mathbb{R}}$ is called a \emph{parabolic neighborhood} if $E = U \cap \set{t \leq \tau}$ for some open set $U \subset \Rd\times \ensuremath{\mathbb{R}}$ and some $\tau \in \ensuremath{\mathbb{R}}$.
We say that $E$ is a parabolic neighborhood of $(x,t) \in \Rd\times\ensuremath{\mathbb{R}}$
if $(x,t) \in E$.
Let us define $\partial_P E := \cl{E} \setminus E$, the \emph{parabolic boundary} of $E$.
\end{definition}
\begin{remark}
Since $\operatorname{int} E = U \cap \set{t < \tau}$,
one can observe that
$\partial_P E = \partial E \setminus (U \cap \set{t = \tau})$.
If $E = U$ then $\partial_P E = \partial E$. Note that we do not require that $E \neq U$, which is usually assumed, simply because it is unnecessary in this paper.
Adding this requirement does not change any of the results presented.
\end{remark}
Following \cites{K03,K07},
we define the viscosity solutions for \eqref{hs-f}.
In the following definitions,
$Q \subset \Rd\times\ensuremath{\mathbb{R}}$ is an arbitrary
parabolic neighborhood in the sense of
Definition~\ref{def:parabolic-nbd}
and
$f(x,t,p): Q \times \Rd \to \ensuremath{\mathbb{R}}$ satisfies assumptions (A1)--(A3)
(Section~\ref{sec:introduction}).
We do \emph{not} assume that $f$ is continuous in $p$.
\begin{definition}
\label{def:visc-test-sub}
We say that a locally bounded, non-negative
function $u: Q \to [0,\infty)$ is
a \emph{viscosity subsolution} of \eqref{hs-f} on $Q$ if
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(}\roman{enumi}\textup{)}}
\item \emph{(continuous expansion)}
\[
\cl\Omega(u; Q) \cap Q \cap \set{t \leq \tau}
\subset \cl{\Omega(u;Q) \cap \set{t < \tau}}
\quad \text{for every $\tau > 0$},
\]
\item \emph{(maximum principle)}\\
for any $\phi \in C^{2,1}$
such that $u^* - \phi$ has a local maximum at
$(x_0, t_0) \in Q \cap \cl\Omega(u;Q)$
in $\cl\Omega(u;Q) \cap \set{t \leq t_0}$, we have
\begin{enumerate}[({ii}-1)]
\item if $u^*(x_0, t_0) > 0$ then $-\Delta \phi(x_0, t_0) \leq 0$,
\item if $u^*(x_0, t_0) = 0$ then either
$-\Delta \phi(x_0, t_0) \leq 0$ or
$D \phi(x_0, t_0) = 0$\\
or
$[\phi_t - f^*(x_0,t_0, D\phi(x_0, t_0)) \abs{D\phi}^2] (x_0, t_0) \leq 0$.
\end{enumerate}
\end{enumerate}
\end{definition}
\begin{remark}
The condition (i) in Definition~\ref{def:visc-test-sub} is necessary
to prevent a scenario where a `bubble' closes instantly;
more precisely, a subsolution cannot become instantly
positive on an open set surrounded by a positive phase,
or cannot fill the whole space instantly.
\end{remark}
The definition of a viscosity supersolution is similar.
\begin{definition}
\label{def:visc-test-super}
We say that a locally bounded, non-negative
function $u: Q \to [0,\infty)$ is
a \emph{viscosity supersolution} of \eqref{hs-f} on $Q$ if
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(}\roman{enumi}\textup{)}}
\item (monotonicity of support)\\
if $(\xi,\tau) \in \Omega(u_*; Q)$ then
$(\xi, t) \in \Omega(u_*;Q)$
for all $(\xi,t) \in Q$, $t \geq \tau$.
\item (maximum principle)\\
for any $\phi \in C^{2,1}$
such that $u_* - \phi$ has a local minimum at $(x_0, t_0) \in Q$
in $\set{t \leq t_0}$, we have
\begin{enumerate}[({ii}-1)]
\item if $u_*(x_0, t_0) > 0$ then $-\Delta \phi(x_0, t_0) \geq 0$,
\item if $u_*(x_0, t_0) = 0$ then either
$-\Delta \phi(x_0, t_0) \geq 0$ or
$D \phi(x_0, t_0) = 0$
\\or
$[\phi_t - f_*(x_0,t_0,D\phi(x_0, t_0)) \abs{D\phi}^2] (x_0, t_0)
\geq 0$.
\end{enumerate}
\end{enumerate}
\end{definition}
\begin{remark}
The condition (i) in Definition~\ref{def:visc-test-super}
states that the support of a supersolution is nondecreasing.
It is a purely technical assumption which simplifies
the proof of the comparison theorem
by preventing an instantaneous disappearance of
a component of a positive phase.
Indeed, it can be shown that all solutions of the Hele-Shaw problem
have nondecreasing support.
This assumption can be removed by using the tools developed in
\cite{KP12}.
\end{remark}
Finally, we define a viscosity solution by combining the two previous definitions.
\begin{definition}
\label{def:visc-sol}
A function $u: Q \to [0,\infty)$ is a \emph{viscosity solution}
of \eqref{hs-f} on $Q$ if $u$ is both a viscosity subsolution
and a viscosity supersolution on $Q$.
\end{definition}
It will be useful to state an equivalent definition of viscosity solutions
via a comparison with barriers,
which in our case are strict classical sub- and supersolutions.
This definition seems more natural,
as it follows a more common pattern appearing
in the treatment of free boundary problems;
see \cites{ACSI,BS98,CS,CV,KP11,KP12}.
\begin{definition}
For a given closed set $K \in \Rd\times\ensuremath{\mathbb{R}}$,
we say that $\phi \in C^{2,1}_{x,t}(K)$
if there exists open set $U \supset K$
and $\psi \in C^{2,1}_{x,t}(U)$,
i.e., a function twice continuously differentiable in space
and once in time,
such that $\phi = \psi$ on $K$.
\end{definition}
\begin{definition}
\label{def:classical-sub}
Given a nonempty open set $U \subset \Rd\times\ensuremath{\mathbb{R}}$,
a function $\phi \in C(\cl U) \cap C^{2,1}_{x,t} (\cl\Omega(\phi; U))$
is called a \emph{subbarrier} of \eqref{hs-f} in $U$
if there exists a positive constant $c$ such that
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(}\roman{enumi}\textup{)}}
\item
$- \Delta \phi < -c$ on $\Omega(\phi; U)$, and
\item
$\abs{D\phi^+} > c$ and
$\phi^+_t - f_*(\cdot, \cdot, D \phi^+) \abs{D \phi^+}^2 < -c$
on $\Gamma(\phi; U)$.
\end{enumerate}
Here $D\phi^+$ and $\phi_t^+$ denote the limits of $D\phi$ and $\phi_t$, respectively, on $\Gamma(\phi;U)$ from $\Omega(\phi;U)$.
\end{definition}
\begin{definition}
\label{def:classical-super}
Given a nonempty open set $U \subset \Rd\times\ensuremath{\mathbb{R}}$,
a function $\phi \in C(\cl U) \cap C^{2,1}_{x,t} (\cl\Omega(\phi;U))$
is called a \emph{superbarrier} of \eqref{hs-f} in $U$
if there exists a positive constant $c$ such that
\begin{enumerate}
\renewcommand{\labelenumi}{\textup{(}\roman{enumi}\textup{)}}
\item
$- \Delta \phi > c$ on $\Omega(\phi; U)$, and
\item
$\abs{D\phi^+} > c$ and $\phi^+_t
- f^*(\cdot, \cdot, D \phi^+) \abs{D \phi^+}^2 > c$
on $\Gamma(\phi; U)$.
\end{enumerate}
\end{definition}
A notion of \emph{strict separation} is a crucial concept in the theory
and will be used multiple times
(our definition differs slightly from the one introduced in \cite{K03}).
\begin{definition}[Strict separation]
Let $E \subset \Rd \times \ensuremath{\mathbb{R}}$ be a parabolic neighborhood,
and $u, v : E \to \ensuremath{\mathbb{R}}$ be
bounded non-negative functions on $E$,
and let $K \subset \cl E$.
We say that $u$ and $v$ are strictly separated
on $K$ with respect to $E$,
and we write $u \prec v$ in $K$ w.r.t. $E$,
if
\begin{align*}
u^{*,E} &< v_{*,E} &&\text{in }&
&K \cap \cl\Omega(u;E).
\end{align*}
\end{definition}
A definition of viscosity solutions using barriers follows.
\begin{definition}
\label{def:visc-barrier}
We say that a locally bounded, non-negative
function $u: Q \to [0,\infty)$ is
a \emph{viscosity subsolution} of \eqref{hs-f} on $Q$ if
for every bounded parabolic neighborhood $E = U \cap \set{t \leq \tau}$,
$E \subset Q$,
and every superbarrier $\phi$ on $U$
such that $u \prec \phi$ on $\partial_P E$ w.r.t. $E$,
we also have $u \prec \phi$ on $\cl E$ w.r.t. $E$.
Similarly, a locally bounded, non-negative
function $u: Q \to [0,\infty)$ is
a \emph{viscosity supersolution} if
Definition~\ref{def:visc-test-super}(i) holds and
for any
subbarrier $\phi$ on $U$
such that $\phi \prec u$ on $\partial_P E$ w.r.t. $E$,
we also have $\phi \prec u$ on $\cl E$ w.r.t. $E$.
Finally, $u$ is a viscosity solution if it is both
a viscosity subsolution and a viscosity supersolution.
\end{definition}
\bigskip
We finish this section by stating the equivalence
of the two definitions of viscosity solutions.
\begin{proposition}
\label{pr:visc-equivalency}
The definitions of viscosity subsolutions (resp. supersolutions)
in Definition~\ref{def:visc-test-sub} (resp. \ref{def:visc-test-super})
and in Definition~\ref{def:visc-barrier}
are equivalent.
\end{proposition}
Before proceeding with the proof,
let us state a useful property of a strict separation
on time dependent sets.
\begin{lemma}
\label{le:ordering-open}
Suppose that $E$ is a bounded parabolic neighborhood
and $u$, $v$ are non-negative locally bounded functions on $E$.
The set
\begin{align}
\label{ordering-set}
\Theta_{u,v;E} := \set{\tau: u \prec v
\text{ in $\cl E \cap \set{t \leq \tau}$
w.r.t. $E$}}
\end{align}
is open
and $\Theta_{u,v;E} = (-\infty, T)$ for some $T \in (-\infty, \infty]$.
\end{lemma}
\begin{proof}
By replacing $u$ by $u^{*,E}$ and $v$ by $v_{*, E}$, we may assume that $u$ is upper semi-continuous and $v$ is lower semi-continuous.
Let us choose $\tau \in \Theta_{u,v;E}$.
Clearly $s \in \Theta_{u,v;E}$ for all $s \leq \tau$.
Therefore, we only need to show that there exists $s > \tau$ such that
$s \in \Theta_{u,v;E}$.
We can assume that
$\set{t \leq \tau} \cap \cl E \neq \emptyset$,
otherwise the claim is trivial since $\cl E$ is compact.
Suppose that $u \not\prec v$
on $\cl E \cap \set{t \leq s}$ w.r.t. $E$ for all $s > \tau$.
Hence
there is a sequence $(x_k, t_k) \in \cl\Omega(u;E)$ such that
$t_k \searrow \tau$ and
$u(x_k, t_k) \geq v(x_k, t_k)$.
By compactness of $\cl E$,
we can assume that $(x_k, t_k) \to (\xi, \tau)$,
and we see that
$(\xi, \tau) \in \cl\Omega(u;E) \cap \set{t \leq \tau}$.
By semi-continuity we have $u(\xi, \tau) - v(\xi, \tau) \geq 0$,
a contradiction with $u \prec v$ in $\cl E \cap \set{t \leq \tau}$.
\qedhere\end{proof}
\begin{proof}[Proof of Proposition~\ref{pr:visc-equivalency}]
Let us prove the statement for subsolutions.
Suppose that $u$ is a subsolution on some parabolic neighborhood $Q$
in the sense
of Definition~\ref{def:visc-test-sub}.
By taking the semi-continuous envelope,
we can assume that $u \in USC(\cl Q)$.
Let
$E = U \cap \set{t \leq \tau} \subset Q$ be a bounded
parabolic neighborhood
and let $\phi$ be a superbarrier on $U$
such that $u \prec \phi$ on $\partial_P E$ w.r.t. $E$.
Consider the set $\Theta_{u, \phi, E}$,
introduced in \eqref{ordering-set}.
Let us define $\hat t = \sup \Theta_{u, \phi, E}$.
Since $E$ is bounded, we have $\hat t > -\infty$.
To show that $u$ is a subsolution
in the sense of Definition~\ref{def:visc-barrier},
it is enough to show that
$u \prec \phi$ in $\cl E$,
which is equivalent to
$\hat t = +\infty$.
Hence suppose that $\hat t < +\infty$.
Lemma~\ref{le:ordering-open} implies that
\begin{compactenum}[(a)]
\item
$u \not\prec \phi$ in $\cl E \cap \set{t \leq \hat t}$ w.r.t. $E$,
\item
$u \prec \phi$ in $\cl E \cap \set{t \leq s}$ for all $s < \hat t$,
and, finally,
\item
$\cl\Omega(u; E) \cap \set{t \leq \hat t} \subset \cl\Omega(\phi;E)$
due to (b) and the continuous expansion in
Definition~\ref{def:visc-test-sub}(i).
\end{compactenum}
By semi-continuity and compactness, $u - \phi$ has a local maximum
at some point $(\xi, \ensuremath{\sigma})$ in $\cl\Omega(u;E) \cap \set{t \leq \hat t}$.
Then (a) implies that $u(\xi, \ensuremath{\sigma}) - \phi(\xi, \ensuremath{\sigma}) \geq 0$.
Consequently, since (b) holds,
we must have $\ensuremath{\sigma} = \hat t$,
and $u \prec \phi$ on $\partial_P E$ implies
$(\xi, \hat t) \in E$.
Finally, using (c) we conclude that
$(\xi, \hat t) \in E \cap \cl\Omega(u; E) \cap \cl\Omega(\phi;E)$.
Therefore either:
\begin{compactitem}
\item
$u(\xi,\hat t) > 0$,
and that is a contradiction with
Definition~\ref{def:visc-test-sub}(ii-1); or
\item $u(\xi, \hat t) = 0$,
but then $\phi(\xi,\hat t) = 0$ and consequently
$(\xi, \hat t) \in \Gamma(\phi;E)$,
which yields a contradiction with
Definition~\ref{def:visc-test-sub}(ii-2).
\end{compactitem}
The proof for supersolutions is analogous,
but more straightforward since the support of $u_*$ cannot
decrease due to Definition~\ref{def:visc-test-super}(i).
\bigskip
The direction from Definition~\ref{def:visc-barrier}
to Definitions~\ref{def:visc-test-sub} and \ref{def:visc-test-super}
is quite standard.
In particular,
Definition~\ref{def:visc-test-sub}(i)
can be shown by a comparison with barriers
as in the proof of Corollary~\ref{co:HS-expansion-speed}.
Furthermore,
any test function $\phi$ such that $u - \phi$ has a local maximum
at $(\hat x, \hat t)$
and violates either of
Definition~\ref{def:visc-test-sub}(ii)
can be used for a construction of a superbarrier
in a parabolic neighborhood of $(\hat x, \hat t)$
by considering
$\tilde \phi(x,t)
= (\phi(x,t) + |x - \hat x|^4 + |t - \hat t|^2 - \phi(\hat x, \hat t))_+$.
Once again,
the proof for supersolutions is similar.
\qedhere\end{proof}
\begin{definition}
For a given function $f$
and a nonempty parabolic neighborhood $Q \subset \ensuremath{{\mathbb{R}^{\dimension}}} \times \ensuremath{\mathbb{R}}$,
we define the following classes of functions:
\begin{itemize}
\item $\overline{\mathcal{S}}(f,Q)$,
the set of all viscosity supersolutions of the Hele-Shaw problem
\eqref{hs-f} on $Q$;
\item $\underline{\mathcal{S}}(f,Q)$,
the set of all viscosity subsolutions of \eqref{hs-f} on $Q$;
\item $\mathcal{S}(f,Q) = \overline{\mathcal{S}}(f,Q) \cap \underline{\mathcal{S}}(f,Q)$,
the set of all viscosity solutions of \eqref{hs-f} on $Q$.
\end{itemize}
\end{definition}
We have the following obvious inclusions:
\begin{lemma}
Let $Q \subset \Rd \times \ensuremath{\mathbb{R}}$ be a nonempty parabolic neighborhood.
If $f \leq g$ for given functions $f,g: Q \times \Rd \to (0,\infty)$ then
\begin{align*}
\overline{\mathcal{S}}(f, Q) &\supset \overline{\mathcal{S}}(g, Q) &&\text{and}&
\underline{\mathcal{S}}(f,Q) &\subset \underline{\mathcal{S}}(g, Q).
\end{align*}
\end{lemma}
Finally, we observe that subsolutions and supersolutions can be
``stitched'' together.
\begin{lemma}
\label{le:stitch}
Let
$Q_1, Q_2 \subset \Rd\times\ensuremath{\mathbb{R}}$ be two parabolic neighborhoods
such that $Q_1 \subset Q_2$
and let $u_i \in \underline{\mathcal{S}}(f, Q_i)$, $i = 1,2$,
for some $f(x,t,p) : Q_2 \times \Rd \to (0,\infty)$.
We define
\begin{align*}
u =
\begin{cases}
\max (u_1, u_2) & \text{in } Q_1,\\
u_2 & \text{in } Q_2 \setminus Q_1.
\end{cases}
\end{align*}
If $u_1^{*,Q_1} \leq u_2^{*,Q_2}$ on $(\partial_P Q_1) \cap Q_2$
and
\begin{equation}
\label{incl-on-pb}
\cl\Omega(u_1; Q_1) \cap \partial_P Q_1 \subset \cl\Omega(u_2;Q_2)
\end{equation}
then $u \in \underline{\mathcal{S}}(f, Q_2)$.
Let now $v_i \in \overline{\mathcal{S}}(f, Q_i)$, $i = 1,2$,
and we define
\begin{align*}
v =
\begin{cases}
\min (v_1, v_2) & \text{in } Q_1,\\
v_2 & \text{in } Q_2 \setminus Q_1.
\end{cases}
\end{align*}
If $(v_1)_{*,Q_1} \geq (v_2)_{*,Q_2}$ on $(\partial_P Q_1) \cap Q_2$,
then $v \in \overline{\mathcal{S}}(f, Q_1)$.
\end{lemma}
\begin{proof}
Let us show the proof for subsolutions.
We use Definition~\ref{def:visc-test-sub}.
Clearly $\Omega(u;Q_2) = \Omega(u_1;Q_1) \cup \Omega(u_2;Q_2)$
and therefore, by \eqref{incl-on-pb},
we have
\begin{align*}
\cl\Omega(u;Q_2) &= \cl\Omega(u_1;Q_1) \cup \cl\Omega(u_2;Q_2)
\\&= (\cl\Omega(u_1;Q_1) \cap Q_1) \cup
(\cl\Omega(u_1;Q_1) \cap \partial_P Q_1)
\cup \cl\Omega(u_2;Q_2)
\\&= (\cl\Omega(u_1;Q_1) \cap Q_1) \cup \cl\Omega(u_2;Q_2).
\end{align*}
Thus Definition~\ref{def:visc-test-sub}(i) is clear.
To prove (ii),
suppose that $u^* - \phi$ has a local maximum at
$(\hat x,\hat t) \in \cl\Omega(u; Q_2)$
in $\cl\Omega(u; Q_2) \cap \set{t \leq \hat t}$.
We observe that, by hypothesis,
$u^* = \max(u_1^{*,Q_1}, u_2^{*,Q_2})$ on $Q_1$
and $u^* = u_2^{*,Q_2}$ on $Q_2 \setminus Q_1$.
Consequently, if $(\hat x, \hat t) \in Q_1$ then $u_1^* - \phi$
and/or $u_2^* - \phi$ have a local maximum
in the appropriate set,
and if $(\hat x, \hat t) \in Q_2 \setminus Q_1$ then $u_2^* - \phi$
has a local maximum. Definition~\ref{def:visc-test-sub}(ii)
follows from this observation.
The proof for supersolutions is straightforward
since there is no restriction on their supports.
\qedhere\end{proof}
\subsection{Comparison principle}
\label{sec:comparison-principle}
A crucial result for a successful theory of viscosity solutions
is a comparison principle.
In this section we establish the comparison principle
between strictly separated viscosity solutions
on an arbitrary parabolic neighborhood.
We only give an outline of the proof
and point out the differences from previous results
in \cites{K03,K07,KP12}.
\begin{theorem}
\label{th:comparison}
Let $Q$ be a bounded parabolic neighborhood.
Suppose that $f : Q \times \Rd \to \ensuremath{\mathbb{R}}$ is a given function
that satisfies the assumptions (A1)--(A3) and is of the form either
$f(x,t,p) = g(x,t)$
or
$f(x,t,p) = f(p)$.
If $u \in \underline{\mathcal{S}}(f, Q)$ and $v\in \overline{\mathcal{S}}(f, Q)$
such that $u \prec v$ on $\partial_P Q$ w.r.t. $Q$,
then $u \prec v$ in $\cl Q$ w.r.t. $Q$.
\end{theorem}
\begin{proof}
\emph{Step 1.} We can assume that $u \in USC(\cl Q)$ and $v \in LSC(\cl Q)$.
Similarly to the proof of Proposition~\ref{pr:visc-equivalency},
we consider the set $\Theta_{u,v; Q}$,
defined in \eqref{ordering-set},
and introduce $\hat t_0 := \sup \Theta_{u,v; Q} > -\infty$.
Then $u \prec v$ on $\cl Q$ w.r.t. $Q$
is equivalent to $\hat t_0 = +\infty$.
Therefore suppose that $\hat t_0 < +\infty $.
Let us introduce the sup- and inf-convolutions
\begin{align*}
Z(x,t) &= \pth{1+ \frac{3 L r}{m}} \sup_{\cl \Xi_r(x,t)}u,\\
W(x,t) &= \pth{1 - \frac{3 L r}{m}} \inf_{\cl \Xi_{r - \ensuremath{\delta} t}(x,t)} v(x,t),
\end{align*}
with $0 < r \ll \frac{m}{3L}$ and $\ensuremath{\delta} = \frac{r}{2T}$
where $T := \inf \set{\tau: Q \subset \set{t \leq \tau}}$.
The open set $\Xi_r(x,t)$
is defined as
\begin{align*}
\Xi_r(x,t)
= \set{(y,s) : (\abs{y - x}-r)_+^2 + \abs{s - t}^2 < r^2}.
\end{align*}
We refer the reader to \cites{K03,KP12}
for a detailed discussion of the properties of $Z$ and $W$.
In particular,
$Z$ and $W$ are well-defined on the closure
of the parabolic neighborhood $Q_r$,
\begin{align*}
Q_r := \set{(x,t): \cl\Xi_r(x,t) \subset Q},
\end{align*}
and $Z \in USC(\cl Q_r)$
and $W \in LSC(\cl Q_r)$.
Moreover,
$Z \in \underline{\mathcal{S}}(f, Q_r)$ and $W \in \overline{\mathcal{S}}(f, Q_r)$.
This is straightforward if $f(x,t,p) = f(p)$.
However, if $f$ depends on $(x,t)$ as in $f(x,t,p) = g(x,t)$
we have to use Proposition~\ref{pr:nonuniform-perturbation}
with factors $a = (1 \pm 3Lr/m)$, yielding the same conclusion.
Finally, and this is the main motivation for
the choice of the convolutions,
both $Z$ and $W$ have an important
interior/exterior ball property of their level sets
both in space for each time and in space-time;
see \cite{KP12}.
\emph{Step 2.} Since $u \prec v$ on $\partial_P Q$ w.r.t. $Q$,
we claim that
if $r$ is chosen small enough
we have $Z \prec W$ on $\partial_P Q_r$ w.r.t. $Q_r$.
Suppose that this is not true for any $r > 0$.
Let $r_k \to 0$ as $k \to \infty$.
There exists a sequence of $(x_k,t_k) \in \partial_P Q_{r_k}
\cap \cl\Omega(Z_{r_k};Q_{r_k})$
such that $Z_{r_k}(x_k,t_k) \geq W_{r_k}(x_k,t_k)$.
By definition of $Z$ and $W$ and semi-continuity,
there exist $(\xi_k, \ensuremath{\sigma}_k), (\zeta_k, \tau_k) \in \cl\Xi_{r_k}(x_k,t_k)$
such that $(\xi_k, \ensuremath{\sigma}_k) \in \cl\Omega(u;Q)$
and $u(\xi_k, \ensuremath{\sigma}_k) \geq v(\zeta_k, \tau_k)$.
By compactness of $\cl\Omega(u;Q)$,
for a subsequence, still denoted by $k$,
$(\xi_k,\ensuremath{\sigma}_k)$ converges to a point
$(\xi, \ensuremath{\sigma})$, which must lie in $\partial_P Q \cap \cl\Omega(u; Q)$.
A simple observation $(\zeta_k, \tau_k) \in \cl\Xi_{2r_k}(\xi_k, \sigma_k)$ then implies that $(\zeta_k, \tau_k) \to (\xi, \ensuremath{\sigma})$.
This is a contradiction with the semi-continuity of $u$ and $v$
and the fact that $u(\xi, \ensuremath{\sigma}) < v(\xi, \ensuremath{\sigma})$.
\emph{Step 3.} We claim that $W$ is a supersolution of \eqref{hs-f} on $Q_r$ with the free boundary velocity increased by $\delta$, that is,
with $f^\ensuremath{\delta}(x,t,p) = f(x,t,p) + \ensuremath{\delta} \abs{p}^{-1}$.
As we already know that $W \in \overline{\mathcal{S}}(f, Q_r)$, we only need to verify Definition~\ref{def:visc-test-super}(ii-2).
Thus suppose that $\phi$ is a $C^{2,1}$-function
and $W - \phi$ has a strict local minimum $0$ at $(\xi, \ensuremath{\sigma})$
in $\set{t \leq \ensuremath{\sigma}}$, and
$W(\xi, \ensuremath{\sigma}) = 0$, $-\Delta \phi(\xi, \ensuremath{\sigma}) < 0$ and $\abs{D \phi(\xi, \ensuremath{\sigma})} > 0$.
By semi-continuity there exists $(y, s) \in \cl\Xi_{r - \ensuremath{\delta} \ensuremath{\sigma}}(0,0)$
with
\[\pth{1 - \frac{3Lr}m}v(\xi + y, \ensuremath{\sigma} + s) = W(\xi, \ensuremath{\sigma}).\]
Moreover, the definition of $\phi$ and $W$ implies that
\begin{align}
\label{phi-v-with-shift}
\phi(x,t) \leq W(x,t) \leq \pth{1 - \frac{3Lr}m}v(x + y + z, t + s),
\end{align}
for any $(y + z, s) \in \cl \Xi_{r - \ensuremath{\delta} t}(0,0)$
and $(x,t)$ in a neighborhood of $(\xi, \sigma)$, $t \leq \sigma$.
Since the set $\cl \Xi_{r -\delta t}(0,0)$ is decreasing in $t$,
we can take any $z \in \cl B_{\ensuremath{\delta} (\ensuremath{\sigma} - t)}(0)$.
Let us define $\nu = -\frac{D\phi}{\abs{D\phi}} (\xi, \ensuremath{\sigma})$,
the unit outer normal to $\Omega_\ensuremath{\sigma}(\phi)$ at $\xi$.
With the particular choice $z = -\ensuremath{\delta} (\ensuremath{\sigma} - t)\nu$,
we can rewrite \eqref{phi-v-with-shift} after a change of variables as
\begin{align*}
\psi(x,t) := \pth{1 - \frac{3Lr}m}^{-1}\phi(x - y + \ensuremath{\delta} (\ensuremath{\sigma} + s - t) \nu, t - s) \leq
v(x,t),
\end{align*}
for $(x,t)$ close to $(\zeta, \mu) := (\xi + y, \ensuremath{\sigma} + s)$, $t \leq \sigma + s$.
Additionally, the equality holds at $(\zeta, \mu)$
by the choice of $(y,s)$.
Thus $\psi$ is a test function for $v$ at $(\zeta, \mu)$ for Definition~\ref{def:visc-test-super}(ii-2),
and we have
\begin{align*}
[\phi_t - \ensuremath{\delta} \abs{D\phi}](\xi,\ensuremath{\sigma}) &= \pth{1 - \frac{3Lr}m}\psi_t(\zeta, \mu)\\
& \geq
\pth{1 - \frac{3Lr}m} f_*\pth{\zeta, \mu, D\psi(\zeta, \mu)} \abs{D \psi(\zeta, \mu)}^2\\
&= \pth{1 - \frac{3Lr}m}^{-1} f_*\pth{\zeta, \mu, \pth{1 - \frac{3Lr}m}^{-1} D\phi(\xi, \sigma)} \abs{D \phi(\xi, \sigma)}^2.
\end{align*}
If $f(x,t,p) = g(x,t)$,
then we use the assumptions (A1)--(A2) that imply
$g(\zeta,\mu) \geq \pth{1 - \frac{3Lr}{m}}g(\xi, \ensuremath{\sigma})$,
and if $f(x,t,p) = f(p)$ we use the assumption (A3)
to conclude that
\begin{align*}
[\phi_t - \ensuremath{\delta} \abs{D\phi}](\xi,\ensuremath{\sigma}) \geq f_*(\xi, \ensuremath{\sigma}, D\phi(\xi,\sigma))
\abs{D\phi(\xi,\ensuremath{\sigma})}^2.
\end{align*}
Therefore $W \in \overline{\mathcal{S}}(f^\ensuremath{\delta}, Q_r)$.
\bigskip
\emph{Step 4.} We proceed with the proof of the comparison theorem.
Let us set \[\hat t = \sup \Theta_{Z, W; Q_r}\]
and observe that $\hat t \leq \hat t_0$.
As in the proof of Proposition~\ref{pr:visc-equivalency},
we have $\cl\Omega(Z) \cap \set{t \leq \hat t}
\subset \cl{\Omega(W) \cap \set{t < \hat t}}$,
and there exists
$(\xi, \hat t) \in Q_r \cap \cl\Omega(Z) \cap \cl\Omega(W)$
such that $(\xi, \hat t)$ is a point of maximum of $Z - W$
in $\cl\Omega(Z) \cap \set{t \leq \hat t}$,
and $Z(\xi, \hat t) - W(\xi, \hat t) \geq 0$.
Since the set $\Omega^c_{\hat t}(W)$ has the interior
ball property of radius $r/2$,
we can show by a barrier argument
that in fact $Z = 0$ on $\Omega^c(W) \cap \set{t \leq \hat t}$,
see \cites{K03,KP12} for details.
Most of the arguments are analogous since $Z \in \underline{\mathcal{S}}(M, Q_r)$,
$W \in \overline{\mathcal{S}}(m, Q_r)$.
This together with the assumption that $\Omega(W)$ cannot decrease
in time due to Definition~\ref{def:visc-test-super}(i)
yields
$\Omega(Z) \cap \set{t \leq \hat t} \subset \Omega(W)$.
Suppose that $Z(\xi, \hat t) > 0$.
Let us denote by $U$ the connected component of $\Omega_{\hat t}(W)$
that contains $(\xi, \hat t)$.
Since $Z(\cdot, \hat t)$ is subharmonic in the (open) set
$U$,
$Z(\cdot, \hat t)$ cannot be identically zero on $\partial U$.
But the conclusion of the previous paragraph implies that
$Z \equiv 0$ on $(\partial U \times \set{\hat t}) \cap Q_r$.
Hence $(\partial U \times \set{\hat t}) \cap \partial_P Q_r \neq \emptyset$.
We have $0 < Z < W$ at the points of $\partial_P Q_r$ where $Z > 0$ by the assumption of strict separation.
Consequently, a straightforward argument of
adding a small harmonic function to
$Z$, using the comparison principle for subharmonic
and superharmonic functions,
yields $Z(\xi, \hat t) < W(\xi, \hat t)$, a contradiction with the previous paragraph.
Therefore $Z(\xi, \hat t) = W(\xi, \hat t) = 0$
and $Z \leq W$ on $\set{t \leq \hat t}$.
We arrive at the last step of the comparison theorem,
where we need to find some weak ordering of gradients
and the corresponding velocities, as in \cite{KP12}.
\emph{Step 5.} Hopf's lemma implies that the ``gradients'' of $Z$ and $W$ are
strictly ordered.
Indeed, let $\nu$ be the unique unit outer normal of
$\Omega_{\hat t}(Z)$ and $\Omega_{\hat t}(W)$ at $\xi$.
Since $Z(\cdot, \hat t)$ is subharmonic in $U$
and $W$ is superharmonic in
$U$,
and there is an interior ball of $\Omega_{\hat t}(Z)$ at $\xi$,
and $W(\cdot, \hat t) \not\equiv Z(\cdot, \hat t)$ in $U$,
we can apply the Hopf's lemma in $U$ at $\xi \in \partial U$
and conclude that
\begin{align}
\label{gradient-order}
\alpha := \limsup_{h\to 0} \frac{Z(\xi - h\nu, \hat t)}{h}
< \liminf_{h\to 0} \frac{W(\xi - h\nu, \hat t)}{h} =: \beta.
\end{align}
The assumption on $f$ implies
\begin{align}
\label{speed-order}
f^*\pth{\xi, \hat t, -\ensuremath{\alpha} \nu} \abs{-\ensuremath{\alpha} \nu}
\leq f_*\pth{\xi, \hat t, -\beta \nu} \abs{-\ensuremath{\beta} \nu}.
\end{align}
It can be showed as in \cite{K03} (see also \cite{KP11,KP12})
that the sets $\cl\Omega(Z)$ and $\Omega^c(W)$
have interior space-time balls at $(\xi, \hat t)$
with space-time slopes that represent the velocities
of the free boundaries at $(\xi, \hat t)$
which we denote as $m_Z$ and $m_W$.
A barrier argument yields $0 < \ensuremath{\delta} \leq m_W \leq m_Z < \infty$,
using the result of Step 3.
\emph{Step 6.} Now we proceed as in \cite{KP12}:
There are points $(\xi_u, t_u)$
and $(\xi_v, t_v)$
on the free boundaries of $u$ and $v$, respectively,
such that $(\xi_u, t_u), (\xi_v, t_v) \in \partial \Xi_r(\xi, \hat t)$.
Since $Z \leq W$ for $t\leq \hat t$ and $Z > 0$ on $\Xi_r(\xi_u, t_u)$,
we can deduce that $v > 0$ on $\bigcup \cl \Xi_{r-\delta t} (x,t)$,
where the union is over all points $(x,t) \in \Xi_r(\xi_u, t_u)$, $t \leq \hat t$.
Therefore as in \cite[Lemma~3.23]{KP12}, given arbitrary $\eta > 0$ small
we can put a radial test function under $v$ in a neighborhood of $(\xi_v, t_v)$
that touches $v$ from below at $(\xi_v, t_v)$ with free boundary velocity $m_Z - \delta + \eta$
and gradient $-\pth{1 -\frac{3Lr}m}^{-1} (\beta - \eta)\nu$ at $(\xi_v, t_v)$.
The definition of supersolution then implies
\begin{align*}
m_Z - \delta + \eta&\geq f_*\pth{\xi_v, t_v, -\pth{1 -\frac{3Lr}m}^{-1} (\beta - \eta) \nu} \pth{1 -\frac{3Lr}m}^{-1}(\beta - \eta),
\intertext{which can be estimated using (A1)--(A2) or (A3), depending on the form of $f$ as in Step 3, as}
&\geq f_*(\xi, \hat t, -(\beta - \eta)\nu) (\beta - \eta).
\end{align*}
Sending $\eta\to0$, the lower semi-continuity of $f_*$ implies
\begin{align*}
m_Z - \delta \geq f_*(\xi, \hat t, - \beta \nu) \beta.
\end{align*}
A similar, but more direct proof using the fact that $Z = 0$ on $\cl \Xi_r(\xi, \hat t)$ establishes
\begin{align*}
m_Z \leq f^*(\xi, \hat t, -\alpha \nu) \alpha.
\end{align*}
When we combine these two inequalities through \eqref{speed-order}, we arrive at
\begin{align*}
m_Z \leq f^*\pth{\xi, \hat t, - \ensuremath{\alpha} \nu}\ensuremath{\alpha} \leq f_*\pth{\xi, \hat t, -\ensuremath{\beta} \nu}\ensuremath{\beta} \leq m_Z - \ensuremath{\delta},
\end{align*}
a contradiction.
Therefore $Z$ and $W$ cannot cross, which in turn implies
that $u$ and $v$ cannot cross and hence
$u \prec v$ in $\cl Q$ w.r.t. $Q$.
\qedhere\end{proof}
\subsection{Well-posedness}
\label{sec:well-posedness}
We will use the comparison principle, Theorem~\ref{th:comparison},
to establish the well-posedness
of the Hele-Shaw-type problem \eqref{hs-f}, Theorem~\ref{th:well-posedness}.
Since Theorem~\ref{th:comparison} requires
strictly separated boundary data,
it does not provide uniqueness of solutions
for general boundary data.
However,
it is possible to prove uniqueness
when the boundary data features some special structure;
see \cites{K03,K07}.
We show the well-posedness theorem for boundary data
strictly increasing in time.
\begin{proof}[Proof of Theorem~\ref{th:well-posedness}]
In this proof we return to the parabolic cylinder
$Q = \Omega \times (0,T]$
introduced in Section~\ref{sec:introduction}.
\medskip
\noindent
\emph{Bounded support.}
First observe that exactly one of $\Omega$ and $\Omega_0$
is bounded and that both $\partial \Omega$
and $\partial \Omega_0$ are bounded.
Let $u$ be a bounded viscosity solution on $Q$
with the correct boundary data.
We will show that
$\cl\Omega(u;Q) \subset B_R(0) \times [0,T]$
for some large $R$.
If $\Omega$ is bounded, this is obvious.
Therefore assume that $\Omega$ is unbounded and therefore
$\Omega_0$ is bounded.
Let $d = \operatorname{diam} \Omega_0 > 0$,
set $K := \sup_Q u < \infty$ (by assumption)
and recall that $M$ is the upper bound on $f$.
We define $\mu := 2 \sqrt{2 n K M T}$
and $R := d + \mu$.
The claim now follows from Corollary~\ref{co:HS-expansion-speed}
applied with $E = \cl{\Omega_0}^c$.
We set $\Sigma := B_R(0) \cap \Omega$, which is clearly bounded.
Note that $\partial \Omega_0 \subset B_R(0)$.
\medskip
\noindent
\emph{Barriers.}
Before proceeding with the proof of existence,
let us first construct barriers for the Perron's method.
We write $\Omega_0^\rho := \Omega_0 + B_\rho(0)$, $\Omega_0^0 := \Omega_0$.
Since $\partial \Omega_0$ is $C^{1,1}$ and compact,
it is a set of a positive reach
and there exists $\rho_0 > 0$ such that $\partial \Omega_0^\rho \in C^{1,1}$
and $\partial \Omega_0^\rho \subset B_R(0)$ for $\rho \in [0, \rho_0]$.
For $\rho \in [0, \rho_0]$, $t \in [0,T]$,
let us define $u_t^\rho \in C(\Omega)$,
such that
$\Delta u_t^\rho = 0$ in $\Omega_0^\rho \cap \Omega$
with boundary data
$u_t^\rho(x) = \psi(x, t)$ on $\partial \Omega$,
$u_t^\rho = 0$ in $\Omega \setminus \Omega_0^\rho$.
Since $\partial \Omega \subset B_R(0)$ and
$\partial \Omega \in C^{0,1}$,
we can extend $\psi$ to $C(\Rd \times [0,T])$
as $\Delta \psi = 0$ in $(B_R(0) \setminus \partial \Omega) \times [0,T]$
and $\psi = 0$ in $B_R^c(0) \times [0,T]$.
Observe that $\psi \in \mathcal{S}(M, \Sigma \times [0, T])$.
Now we introduce the barriers at the initial time $t = 0$.
By the Hopf's lemma and comparison for harmonic functions,
and the fact that $\psi$ is bounded from above and from zero
on $\partial \Omega \times [0, T]$,
there exists a constant $c > 0$ such that
$0 < c^{-1} \leq \abs{D(u_t^\rho)^+} \leq c$
on
$\partial \Omega_0^\rho$ for all $\rho \in [0, \rho_0/2]$, $t \in [0,T]$.
Therefore we can find $\omega > 0$ large enough
so that $U(x,t) := u_t^{\ensuremath{\omega}^{-1} t}(x)$
is a subsolution $U \in \underline{\mathcal{S}}(m/2, \Omega \times (0, \ensuremath{\delta}])$,
and $V(x,t) := u_t^{\ensuremath{\omega} t}(x)$ is a
supersolution $V \in \overline{\mathcal{S}}(M, \Omega \times (0, \ensuremath{\delta}])$
for some $\ensuremath{\delta} \in (0, \rho_0 /\omega)$.
Moreover, $U, V \in C(\Omega \times [0,\ensuremath{\delta}])$.
Let us extend $U$ and $V$ by defining $U(x,t) = u_t^{\ensuremath{\omega}^{-1} \ensuremath{\delta}}(x)$
and $V(x,t) = \psi(x,t)$ for $t \in (\ensuremath{\delta}, T]$ (recall the extension
of $\psi$ above).
It is straightforward to show
that $U \in \underline{\mathcal{S}}(m/2, Q)$, $V \in \overline{\mathcal{S}}(M, Q)$
and that $U$ and $V$
satisfy the boundary condition on $\partial_P Q$
in the correct sense.
That is, $U^* = U_* = V^* = V_* = \psi$ on $\partial \Omega \times [0,T]$
and $U^* = U_* = V^* = V_* = u_0^0$.
\medskip
\noindent
\emph{Uniqueness.}
Let $u$ and $v$ be two solutions with the correct
boundary data.
We first check the value of $u$ and $v$ at $t = 0$.
In general, for any solution $u$,
we have $u^*(\cdot, t)$ is subharmonic in $\Omega$
and $u_*(\cdot, t)$ is superharmonic in $\Omega_t(u_*; Q)$
(see \cite{KP12} for a detailed proof).
Thus a simple argument,
using the facts that a support of a subsolution must expand continuously,
and that a support of a supersolution cannot decrease,
shows that
$u_*(\cdot, 0) = u^*(\cdot, 0) = v_*(\cdot, 0) = v^*(\cdot, 0) = u_0^0$
on $\Omega$.
Now we observe that the support of a supersolution $u$
is strictly increasing
at $t = 0$, i.e.,
$\cl\Omega_0(u_*;\cl Q) = \cl{\Omega \cap \Omega_0}
\subset \Omega_t(u_*;\cl Q)$ for any $t > 0$.
This follows from the comparison with a perturbation of $U$.
Specifically, fix a point $\zeta \in B_R(0) \setminus \cl\Omega$
and define $U_\kappa(x,t) = (U(x,t) + \kappa \abs{x - \zeta}^2
- 4 \kappa R^2)_+$.
Clearly $U_\kappa \rightrightarrows U$ uniformly on $\cl Q$
as $\kappa \to 0$ and $U_\kappa$ is a subbarrier
on $\Omega \times (0,\ensuremath{\delta}]$ for $\kappa > 0$
small enough (recall that $U \in \underline{\mathcal{S}}(m/2, Q)$ with $m/2< m$).
Moreover, the support of $U$ strictly increases in the above sense.
Since $U_\kappa \prec u$ on $\partial_P Q$ for any $\kappa > 0$,
we conclude that the support of $u_*$ must strictly increase
by sending $\kappa \to 0$.
Because the comparison principle requires strictly separated boundary
data, we have to perturb the solutions
and create this separation.
We use a perturbation in time.
Such perturbation is more involved
because of the time-dependence of $f(x,t,p)$,
and in that case we will use the nonlinear
perturbation from Appendix~\ref{sec:nonlinear-perturbation}.
Thus let
\begin{compactitem}
\item
$\ensuremath{\theta}_\eta(t) = t - \eta$ if $f(x,t,p) = f(p)$, or let
\item
$\ensuremath{\theta}_\eta(t)$ be the function constructed
in Section~\ref{sec:nonlinear-scaling-sub} with $\tau = - \eta$
and $\rho = 0$ if $f(x,t,p) = f(x,t)$.
\end{compactitem}
Observe that, in either case, $\ensuremath{\theta}_\eta$
is well-defined on $[0,\infty)$, $\ensuremath{\theta}_\eta' \in (0,1]$
and therefore it is invertible,
$\ensuremath{\theta}_\eta(0) = -\eta$ and $\ensuremath{\theta}_\eta(t) \to t$ as $\eta \to 0$,
locally uniformly in $t$.
We define $w(x,t) = u(x,\ensuremath{\theta}_\eta(t))$.
As we showed above, the support of $v_*$
is strictly increasing at $t = 0$.
Furthermore, the boundary data $\psi$ on $\partial \Omega_0$
is strictly increasing in time
and $\ensuremath{\theta}_\eta(t) < t$ for all $t \geq 0$.
Since $\ensuremath{\theta}_\eta^{-1}(0) > 0$,
let us define
$Q_\eta := \Sigma \times (\ensuremath{\theta}_\eta^{-1}(0), T]$,
where $\Sigma$ was introduced above,
and we have for all small $\eta > 0$
\begin{align*}
w \prec v\quad \text{on } \partial_P Q_\eta.
\end{align*}
The comparison theorem~\ref{th:comparison} then implies
$w \prec v$ on $\cl Q_\eta$,
and thus $u^*(x,\ensuremath{\theta}_\eta(t)) \leq v_*(x,t)$.
By sending $\eta \to 0$, we conclude
$u_* \leq (u^*)_* \leq v_*$ on $Q$.
Futhermore, clearly
\begin{align*}
u^*(x,t) \leq v_*(x, \ensuremath{\theta}_\eta^{-1}(t))\quad
\text{on } \cl Q \times [0, \ensuremath{\theta}_\eta^{-1}(T)],
\end{align*}
which yields $u^* \leq (v_*)^* \leq v^*$ on $Q$ after sending $\eta \to 0$.
The argument repeated with $u$ and $v$ swapped
implies that the solution is unique.
\medskip
\noindent
\emph{Existence.}
We use the classical Perron-Ishii method \cite{Ishii87}; see also \cites{CIL,K03}.
Let
\begin{align*}
u &= \sup \set{w \in \underline{\mathcal{S}}(f, Q): w \leq V}, &
v &= \inf \set{w \in \overline{\mathcal{S}}(f,Q): w \geq U}.
\end{align*}
Clearly $u \geq U$ and $v \leq V$.
Since $(V^*)_* = V$
and $(U_*)^* = U$,
a standard argument yields that $u, u^*,v, v_* \in \mathcal{S}(f, Q)$.
Moreover,
because of the continuity of $U$ and $V$ at $\partial_P Q$
we conclude that all these solutions have the correct boundary data.
Therefore, by uniqueness,
if $w \in \mathcal{S}(f, Q)$ with the correct boundary data, we have
$w_*=u_* = v_* = (u^*)_*$ and $w^* = u^* = v^* = (v_*)^*$, and thus
\[
(w^*)_* = w_*, \qquad (w_*)^* = w^*,
\]
which completes the proof.
\qedhere\end{proof}
The last step of the previous proof provides the following regularity.
\begin{corollary}
\label{co:regularity}
Let $u$ be the unique solution of the problem \eqref{hs-f}
with data satisfying the assumptions of Theorem~\ref{th:well-posedness}.
Then
\begin{equation}
\label{u-upper-lower}
(u_*)^* = u^* \qquad \text{and} \qquad (u^*)_* = u_*.
\end{equation}
and
\begin{equation}
\cl\Omega(u_*;Q) = \cl\Omega(u;Q)=\cl\Omega(u^*;Q).
\end{equation}
\end{corollary}
\begin{proof}
As was indicated above, \eqref{u-upper-lower}
was shown in the last step of the proof of Theorem~\ref{th:well-posedness}.
Let us point out that $\cl\Omega(u) = \cl\Omega(u^*)$ is always true.
It is enough to prove that $\cl\Omega(u^*) \subset \cl\Omega(u_*)$
since the other direction is obvious.
Suppose that $(\xi, \ensuremath{\sigma}) \in \cl\Omega(u^*) \setminus \cl\Omega(u_*)$.
Then there exists $\ensuremath{\delta} > 0$ such that
$B_\ensuremath{\delta}(\xi,\ensuremath{\sigma}) \cap \cl\Omega(u_*) = \emptyset$
and therefore $u_* = 0$ on $B_\ensuremath{\delta}(\xi,\ensuremath{\sigma})$.
However, \eqref{u-upper-lower} implies that
$u^* = (u_*)^* = 0$ on $B_\ensuremath{\delta}(\xi, \ensuremath{\sigma})$,
a contradiction with the choice of $(\xi, \ensuremath{\sigma})$.
\qedhere\end{proof}
\section{Identification of the limit problem}
\label{sec:limit-problem}
This section is devoted to the key step in the proof of Theorem~\ref{th:intro-homogenization},
the identification of a viable candidate for the homogenized velocity $r(Du)$.
The results are quite technical, an unfortunate consequence of the employed arguments, and therefore we first give a brief overview of the content of this section.
We motivate the approach by considering the one-dimensional case in Section~\ref{sec:homogenizationin1D}.
The numerical results there suggest that the form of the homogenized velocity $r(Du)$ is qualitatively different from the time-independent case.
In particular, it indicates that a simple closed expression for $r(Du)$ is not available.
Therefore we proceed with the identification of $r(Du)$ by other means.
We follow the idea from \cite{CSW,K07} of using a class of certain obstacle problems.
They are introduced in Section~\ref{sec:obstacle-problem},
where for each scale $\ensuremath{\varepsilon} >0$, gradient $q$ and candidate $r$ of the free boundary velocity we set up a domain, an obstacle and define a subsolution and a supersolution of an obstacle problem.
The main feature of the solutions of the obstacle problems is their natural monotonicity with respect to a family of scalings and translations, Section~\ref{sec:monotonicity-obstacle}.
To quantify the viability of the homogenization velocity candidate, we introduce a new quantity, called \emph{flatness}, in Section~\ref{sec:flatness}, and derive its basic properties.
It measures how far the obstacle problem solutions detach from the obstacle, that is, how ``flat'' they are.
The following three sections then cover three important results that build on the definitions above.
First, the \emph{local comparison principle}, Section~\ref{sec:local-comparison-principle}, improves on the theorem of the same name from \cite{K08}, and allows us to locally compare the subsolutions of the obstacle problems with the supersolutions, away from the domain boundary, even after the boundary values are no longer ordered, provided that the flatness of those solutions is bounded with a certain rate.
Section~\ref{sec:cone-flatness} then presents the new \emph{cone flatness} property, which is a consequence of our particular choice of the domain for the obstacle problems, and which guarantees that the free boundary of a given obstacle problem solution lies in between two nearby cones whose distance can be estimated.
Finally, the \emph{detachment lemma} of Section~\ref{sec:detachment}, a direct consequence of the cone flatness property, guarantees that if the flatness of a given obstacle problem solution exceeds a certain value, the solution is separated from a significant portion of the obstacle.
These three results motivate the introduction of the two candidates for the homogenized velocity in Section~\ref{sec:homogenized-velocity}.
We then deduce the semi-continuity of the candidates and their coincidence up to the points of discontinuity, as explained in Section~\ref{sec:semi-continuity}.
The main tools are the natural monotonicity of the obstacle problem, the local comparison principle and the detachment lemma.
In Section~\ref{sec:time-independent} we make a brief detour to show that if the scaling in time is available, as is the case in the time-independent medium, we can prove that the homogenized velocity is in fact continuous and positively one-homogeneous, which recovers the results of \cite{K08} via our methods.
\subsection{Homogenization in one dimension}
\label{sec:homogenizationin1D}
Let us first explore the homogenization of \eqref{HSt} in one dimension.
For simplicity,
we take $\Omega = (0, \infty)$ and $\Omega_0 = (-\infty, x_0)$
for some $x_0 > 0$.
Furthermore, let $g(x,t): \ensuremath{\mathbb{R}} \times \ensuremath{\mathbb{R}} \to [m,M]$
be a positive, $L$-Lipschitz, $\ensuremath{\mathbb{Z}}^2$-periodic function.
Finally, we assume that the boundary data on $\partial \Omega = \set{0}$
are given by a function $\psi(0,t) = \psi(t) > 0$,
$\psi \in C([0,\infty))$.
Let $u$ be the unique solution of \eqref{HSt} with the above data.
We observe that, in this simple setting, $\Omega(u; \Omega) = (0, x(t))$,
where $x(t)$ is the position of the free boundary
$\partial \Omega(u)$ at time $t$,
i.e., $\partial \Omega(u) = \set{(x(t),t) : t \geq 0}$.
Therefore the normal velocity is simply $V_\nu = x'$.
Moreover, the harmonic function $u(\cdot, t)$
must be linear on $(0,x(t))$ with the slope $\abs{Du(x(t), t)} = \psi(t)/x(t)$.
Therefore, for every fixed $\ensuremath{\varepsilon} > 0$,
the free boundary condition in \eqref{HSt}
is equivalent to a simple ODE of the form
\begin{align}
\label{eq:boundaryode}
\begin{cases}
x_\ensuremath{\varepsilon}'(t) = g\pth{\frac{x_\ensuremath{\varepsilon}(t)}{\ensuremath{\varepsilon}}, \frac{t}{\ensuremath{\varepsilon}}} \frac{\psi(t)}{x_\ensuremath{\varepsilon}(t)},\\
x_\ensuremath{\varepsilon}(0) = x_0.
\end{cases}
\end{align}
Due to the ODE homogenization result of \cite{IM10} (see also \cite{Piccinini1}),
$x_\ensuremath{\varepsilon} \to x$ locally uniformly,
where $x(t)$ is the solution of
\begin{align*}
\left\{
\begin{aligned}
x'(t) &= f (x(t), t),\\
x(0) &= x_0.
\end{aligned}
\right.
\end{align*}
The arguments in this paper then imply that $f$ is of the form
\[
f(x(t), t) = r(\psi(t)/x(t)).
\]
The function $r(q)$ can be found numerically by scaling \eqref{eq:boundaryode} and thus solving
\begin{align}\label{eq:boundaryodescaled}
\left\{
\begin{aligned}
x'(t) &= q g\pth{x(t), t},\\
x(0) &= x_0,
\end{aligned}
\right.
\end{align}
in some $t \in [0, T]$ for $T$ large. Then $r(q)$ can be approximated by $x(T)/T$. The limit $\lim_{T \to \infty} \frac{x(T)}{T}$ is independent of $x_0$ since we can squeeze any solution for $x_0 \neq 0$ between $x(t)$ and $x(t) + 1$ using the comparison principle, and use the periodicity of $g$.
Consider the medium
\begin{align*}
g(x,t) = \sin^2 ( \pi(x-t)) + 1.
\end{align*}
Then for $q \in [1/2, 1]$,
$x(t) = t + \pi^{-1} \arccos \sqrt{q^{-1} - 1}$
is a solution of \eqref{eq:boundaryodescaled}.
Thus we see that $r(q) = 1$ for $q \in [1/2, 1]$.
If the boundary propagates in the opposite direction,
i.e., when $g = \sin^2(\pi(-x-t)) + 1$,
this effect does not appear and $r(q)$ is strictly increasing.
The situation can be even more complicated;
see Figure~\ref{fig:pinningintervalmultiple} for an example of this phenomenon.
\begin{figure}[t]
\centering
\fig{Fig2}{4.25in}
\caption{The graph of $r(q)$ for $g(x,t) = \sin(2 \pi (x-3t)) \sin(2 \pi (2t + x)) + 11/10$}
\label{fig:pinningintervalmultiple}
\end{figure}
\subsection{Obstacle problem}
\label{sec:obstacle-problem}
This section is devoted to the introduction of the class of obstacle problems whose solutions are the key tool for the identification of the homogenized free boundary velocity $r(Du)$ in Theorem~\ref{th:intro-homogenization}.
The obstacle problems lead to the following class of solutions:
for every scale $\ensuremath{\varepsilon} >0$, gradient $q \in \Rd \setminus \set0$ and candidate $r > 0$ of the homogenized velocity,
we define
the solutions of an obstacle problem on $\Rd\times\ensuremath{\mathbb{R}}$ as
\begin{subequations}
\label{obstaclesolution}
\begin{align}
\label{eq:obstaclesupersolution}
\uu_{\e;q,r} &= \pth{\inf\set{
u :
u \geq P_{q,r} \text{ on $\Rd\times\ensuremath{\mathbb{R}}$},
u \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon},Q_q)
}}_*,\\
\overline{u}_{\e;q,r} &= \pth{\sup\set{
u :
u \leq P_{q,r} \text{ on $\Rd\times\ensuremath{\mathbb{R}}$, }
u \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)
}}^*.
\end{align}
\end{subequations}
As a mnemonics, notice that $u$ is \emph{above} the bar in $\uu_{\e;q,r}$ because $\uu_{\e;q,r}$ is \emph{above} the obstacle.
The obstacles $P_{q,r}$ are the planar traveling wave solutions of the homogenized Hele-Shaw problem, and they are introduced in Section~\ref{sec:planar-solutions} below.
The domain of the obstacle problem, $Q_q$, requires a special attention.
It is constructed in Section~\ref{sec:domain-geometry}.
After the definition has been clarified, the basic properties of the solutions $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ are discussed in Section~\ref{sec:obstaclesolutions-properties}.
\subsubsection{Planar solutions}
\label{sec:planar-solutions}
The homogenized Hele-Shaw problem with boundary velocity law
$V_\nu = r(Du)$
admits simple traveling wave solutions.
We call them planar solutions since each
is a positive part of a linear function
with gradient $q \in \Rd\setminus\set0$
moving in the direction $\nu := -q/\abs q$
with the normal velocity $V_\nu = r(q)$.
These solutions serve as the obstacles for the obstacle
problem \eqref{obstaclesolution} that will be used to find a candidate of
the homogenized free boundary velocity $r(q)$.
For any $q \in \Rd\setminus\set0$ and $r > 0$ we define
\begin{align*}
P_{q,r}(x,t) := (\abs{q} r t + x \cdot q)_+
= \abs q \pth{rt - x\cdot \nu}_+,
\qquad \nu := \frac{-q}{\abs{q}},
\end{align*}
where $(s)_+ := \max(s, 0)$ is the positive part.
We interpret $r$ as the velocity of the free boundary
\begin{align*}
\Gamma(P_{q,r}) = \set{(x,t) : rt = x\cdot \nu}
\end{align*}
and $q \neq 0$ as the gradient $q = D P_{q,r}$.
Thus $\nu$ is the unit outer normal
vector to the free boundary at every time.
The following observation is a
trivial consequence of the nondegeneracy assumption
\eqref{g-bound}.
\begin{proposition}
\label{pr:planar-solution-range}
$P_{q,r} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon})$ if
\begin{align*}
r &\leq m \abs{q}
\intertext{and $P_{q,r} \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon})$ if}
r &\geq M \abs{q}.
\end{align*}
\end{proposition}
\begin{proof}
Use \eqref{g-bound}.
\qedhere\end{proof}
In view of Proposition~\ref{pr:planar-solution-range},
we will often impose the following restriction on the values
of $r$ and $q$ throughout the paper:
\begin{align}
\label{rqrestriction}
q \neq 0, \qquad m \leq \frac{r}{\abs{q}} \leq M.
\end{align}
\begin{proposition}
\label{pr:obstacle-ordering}
For given $q \neq 0$ and $r > 0$ we have
\begin{align*}
P_{q,r} (x - y, t - \tau) &\leq P_{q,r}(x,t) \text{ for all $x, t$}
&\quad& \text{if and only if}\quad y \cdot \nu \leq r \tau,
\intertext{and}
P_{q,r} (x - y, t - \tau) &\geq P_{q,r}(x,t) \text{ for all $x, t$}
&&\text{if and only if}\quad y \cdot \nu \geq r \tau.
\end{align*}
\end{proposition}
\begin{proof}
The statement is obvious from the identity
\begin{align*}
\abs{q}(r (t - \tau) - (x-y) \cdot \nu)
= \abs q(r t - x \cdot \nu) + \abs{q} (y \cdot \nu - r \tau),
\end{align*}
and from the definition of $P_{q,r}$.
To prove the only-if direction, consider $x = 0$ and $t$ sufficiently large so that the first two terms above are positive.
\qedhere\end{proof}
\subsubsection{Domain}
\label{sec:domain-geometry}
\begin{figure}
\centering
\fig{Fig3}{4.5in}
\caption{The cone $\Omega_q$ and the support of
the sub/supersolution boundaries at a time $t$}
\label{fig:obstacledomain}
\end{figure}
In this section
we construct a class of space-time cylinders $Q_q$
that serve as the domains for the obstacle problem
introduced in \eqref{obstaclesolution}.
The base of the cylinder is chosen to be a cone with a prescribed opening angle and an axis in the direction that coincides with the normal of the free boundary of the obstacle.
This particular choice of geometry will let us control
how fast the solution of the obstacle problem detaches
from the obstacle at the lateral boundary of the domain $Q_q$.
It is achieved by a comparison with the class of planar subsolutions and supersolutions $R^\pm_\xi$
constructed below.
In fact, this is the main motivation for this technical construction.
With these barriers at our disposal, we can show that the free boundaries of the solutions $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ must lie inside an intersection of certain cones; see Proposition~\ref{pr:obstacle-sols}.
The ability to control the boundary behavior
will be necessary for the extension of the monotonicity property
of the obstacle problem beyond the straightforward
inclusion of domains.
One of the main consequences of this extra monotonicity is
the cone flatness property, Proposition~\ref{pr:cone-flatness}.
Throughout the rest of this section,
we fix $q\in\Rd\setminus\set0$ and $r>0$
and let $\nu = \frac{-q}{\abs{q}}$.
For simplicity, we shall denote
\begin{align*}
\Gamma_t = \Gamma_t(P_{q,r}).
\end{align*}
Since the homogenization result is trivial
when $m = M$,
we may assume throughout the rest of the paper that $0 < m < M$.
We define angles $\ensuremath{\theta}, \ensuremath{\theta}^+ \in (0,\frac\pi2)$,
and $\ensuremath{\theta}^- \in (\ensuremath{\theta}, \frac\pi2)$ as
\begin{subequations}
\label{eq:choiceoftas}
\begin{align}
\label{eq:choiceofta}
\ensuremath{\theta} &= \arccos \sqrt{\frac{m}{M}}\\
\label{eq:choiceoftaplus}
\ensuremath{\theta}^+ &= \ha\pi - \ensuremath{\theta}\\
\label{eq:choiceoftaminus}
\ensuremath{\theta}^-&= \ha\pi + \ensuremath{\theta} - \ensuremath{\varphi}^-, && \text{where } \ensuremath{\varphi}^- := \arccos\frac{m}{M}.
\end{align}
\end{subequations}
Note that indeed $\ensuremath{\theta}^- \in (\ensuremath{\theta}, \frac\pi2)$
since $\ensuremath{\varphi}^- \in (\ensuremath{\theta}, \frac\pi2)$
by $\cos \ensuremath{\varphi}^- = \frac{m}{M} < \sqrt{\frac{m}{M}} = \cos \ensuremath{\theta}$.
Let ${\rm Cone}_{p, \ensuremath{\theta}}(x)$ be the open cone with axis
in the direction of $p\in\Rd$, opening angle $2\ensuremath{\theta}$
and vertex $x$, that is,
\begin{align*}
{\rm Cone}_{p, \ensuremath{\theta}}(x)
&:= \set{y : (y - x) \cdot p > \abs{y - x} \abs{p} \cos \ensuremath{\theta}}.
\end{align*}
Angles $\ensuremath{\theta}$, $\ensuremath{\theta}^+$ and $\ensuremath{\theta}^-$ then define cones $\Omega_q$,
$C_t^+$ and $C_t^-$ for every $t \geq 0$,
\begin{align*}
\Omega_q &:= {\rm Cone}_{\nu, \ensuremath{\theta}}(V),\\
C_t^- &:= {\rm Cone}_{\nu, \ensuremath{\theta}^-}(V^-_t),\\
C_t^+ &:= {\rm Cone}_{-\nu, \ensuremath{\theta}^+}(V^+_t),
\end{align*}
where $V = -\nu$ and the vertices $V^\pm_t$ are uniquely determined by
the equality
\begin{align}
\label{same-base}
\Omega_q \cap \Gamma_t = C_t^- \cap \Gamma_t = C_t^+ \cap \Gamma_t,
\end{align}
see Figure~\ref{fig:obstacledomain}.
Finally, we define the space-time cylinder
\begin{align*}
Q_q := \Omega_q \times (0, \infty).
\end{align*}
\begin{remark}
\label{re:speedVplust}
Since the positions of the vertices $V^\pm_t$ of the cones $C^\pm_t$
clearly depend linearly on time,
we can extend their definition to all $t \in \ensuremath{\mathbb{R}}$
as
\[
V^\pm_t = V^\pm_0 + r^\pm_V t \nu \qquad t \in \ensuremath{\mathbb{R}},
\]
where $r^\pm_V$ are constants and we call them the velocities of $V^\pm_t$.
Note that $V = V^+_{-r^{-1}} = V^-_{-r^{-1}} = - \nu$
due to \eqref{same-base} and the fact that $V = -\nu \in \Gamma_t(P_{q,r})$
at $t = -\ov r$.
The velocities of $r^\pm_V$ can be found explicitly.
Since the cones share a base, \eqref{same-base},
we infer that (see Figure~\ref{fig:obstacledomain})
\[(r_V^+ - r)\tan \ensuremath{\theta}^+ = r \tan \ensuremath{\theta} = (r - r_V^-) \tan \ensuremath{\theta}^-.\]
Since $\ensuremath{\theta}^+ = \frac{\pi}{2} - \ensuremath{\theta}$,
$\tan \ensuremath{\theta}^+ = \frac{1}{\tan \ensuremath{\theta}}$ holds and we have
\begin{align*}
r_V^+ = r(1 + \tan^2 \ensuremath{\theta}) = \frac{r}{\cos^2 \ensuremath{\theta}} = \frac{M}{m}r > r.
\end{align*}
We can similarly express the velocity $r^-_V$ as
\begin{align*}
r_V^- = \pth{1 - \frac{\tan \ensuremath{\theta}}{\tan \ensuremath{\theta}^-}} r = c_{\frac Mm} r \in (0, r).
\end{align*}
\end{remark}
The reason for this choice of domain $Q_q$ is the result in Lemma~\ref{le:matchingplanarsolutions} below.
But we begin with the following geometrical result.
Let us introduce the set of all directions
of rays in $\partial \Omega_q$:
\begin{align}\label{eq:Xi}
\Xi = \set{\xi \in \ensuremath{{\mathbb{R}^{\dimension}}} : \quad \abs{\xi} = 1, \quad \xi \cdot \nu = \cos \ensuremath{\theta}}.
\end{align}
\begin{proposition}
\label{pr:cone-planes}
For any given $x \in \partial \Omega_q$, $x \neq V$,
there exists unique $\xi \in \Xi$
such that $x \in L_\xi \subset \partial \Omega_q$,
where $L_\xi$ is the ray
\begin{align}\label{eq:Lxi}
L_\xi := \set{\ensuremath{\sigma} \xi +V: \ensuremath{\sigma} \geq 0}.
\end{align}
Furthermore, there exist unique $\xi^-$ and $\xi^+$
such that $\abs{\xi^\pm} = 1$,
\begin{align*}
L^\pm_{\xi,t}
:= \set{\ensuremath{\sigma} \xi^\pm + V^\pm_t: \ensuremath{\sigma} \geq 0} \subset \partial C_t^\pm
\end{align*}
and
\begin{align*}
L^\pm_{\xi,t} \cap L_\xi = \set{I_{\xi,t}} \qquad \text{for all $t \geq 0$}
\end{align*}
for some point $I_{\xi,t}$.
Let $\eta^\pm_\xi$ be the unit normal to $\partial C_t^\pm$
on $L^\pm_{\xi,t}$ with $\eta^\pm \cdot \nu > 0$.
There exist unique constants $\mu^\pm$, $r^\pm$ and $T^\pm$
depending only on $(q, r, m, M)$,
and independent of $\xi$, such that
\begin{align}
\label{Rxi-touches-Ct}
\Gamma_t(R^\pm_\xi) \cap \partial C^\pm_t &= L^\pm_{\xi,t}
&& \text{for all $t \geq 0$,}\\
\label{eq:planarsolutionsmatch}
R^\pm_\xi &= P_{q,r} && \text{on $L_\xi \times [0, \infty)$}
\end{align}
where
\begin{align}
\label{eq:Rpmxi}
R^\pm_\xi (x,t):= P_{-\mu^\pm \eta_\xi^\pm, r^\pm}(x, t - T^\pm).
\end{align}
\end{proposition}
\begin{proof}
The existence of $\xi, \xi^+, \xi^-$ is
straightforward.
$\mu^\pm$ and $r^\pm$ can be expressed using the following geometric
considerations.
By definition,
\begin{align*}
\cos \ensuremath{\theta} = \xi \cdot \nu.
\end{align*}
We introduce $\ensuremath{\varphi}^\pm \in (0,\pi/2)$ via $\cos \ensuremath{\varphi}^\pm = \xi \cdot \eta^\pm$ and note that
\begin{align}
\label{eq:taandvp}
\ensuremath{\theta}^+ &= \ha\pi + \ensuremath{\varphi}^+ - \ensuremath{\theta}, & \ensuremath{\theta}^- &= \ha{\pi} + \ensuremath{\theta} - \ensuremath{\varphi}^-.
\end{align}
Observe that the point $I_{\xi,t} \in L_\xi \cap \Gamma_t$ moves in the direction $\xi$ along $L_\xi$ with the velocity $r_I$ given as
\begin{align*}
r_I = \frac{r}{\cos \ensuremath{\theta}}.
\end{align*}
Since $R^\pm_\xi$ propagates in the direction $\eta^\pm$,
$\abs{\eta^\pm} = 1$,
and
\[
\set{I_{\xi,t}} = L^\pm_{\xi,t} \cap L_\xi
= \Gamma_t(R^\pm_\xi) \cap \partial C_t^\pm \cap L^\xi
\]
by condition \eqref{Rxi-touches-Ct},
we must have $I_{\xi,t} \in \Gamma_t(R^\pm_\xi)$
for all $t \geq 0$
and therefore
\begin{align}
\label{eq:rpm}
r^\pm = r_I (\xi \cdot \eta^\pm)
=r_I \cos \ensuremath{\varphi}^\pm = r \frac{\cos \ensuremath{\varphi}^\pm}{\cos \ensuremath{\theta}}.
\end{align}
The slope $\mu_L$ of $P_{q,r}$ on $L_\xi$ is
\begin{align*}
\mu_L = \abs{q} \cos \ensuremath{\theta}.
\end{align*}
In \eqref{eq:planarsolutionsmatch}
we require that this is also the slope of $R^\pm_\xi$ on $L_\xi$, i.e.,
\begin{align}
\label{eq:mupm}
\mu^\pm = \frac{\mu_L}{\cos \ensuremath{\varphi}^\pm} = \frac{\abs{q} \cos \ensuremath{\theta}}{\cos \ensuremath{\varphi}^\pm}.
\end{align}
Finally, the constants $T^\pm$ are then fixed
by requiring $V^\pm_{T^\pm} = 0$,
and therefore, by Remark~\ref{re:speedVplust},
\[
T^\pm = \ov{r^\pm_V} - \ov r.
\]
It is straightforward to check that
$R^\pm_\xi$ defined with such unique choice of
$\mu^\pm$, $r^\pm$ and $T^\pm$
satisfy \eqref{Rxi-touches-Ct} and \eqref{eq:planarsolutionsmatch}.
\qedhere\end{proof}
We complete the family of $R_\xi^\pm$ defined in \eqref{eq:Rpmxi}
by introducing two special planar functions:
\begin{align}\label{eq:Rpm0}
\begin{aligned}
R^+_0(x,t) &= P_{q,\max(M\abs{q},r)}(x,t),\\
R^-_0(x,t) &= P_{q,\min(m\abs{q},r)}(x,t).
\end{aligned}
\end{align}
\begin{corollary}
\label{co:values-Rxi}
We have
\begin{align*}
R^-_\xi &\leq P_{q,r} \leq R^+_\xi &
&\text{on $Q_q$ for all $\xi \in \Xi \cup \set0$},
\end{align*}
with equality
\begin{align*}
R^\pm_\xi &= P_{q,r} &
&\text{on $L_\xi \times [0,\infty)\subset \partial Q_q$
for $\xi \in \Xi$},\\
R^\pm_0 &= P_{q,r} && \text{at $t = 0$},
\end{align*}
where $L_\xi$ was introduced in \eqref{eq:Lxi}.
Moreover,
\begin{align}
\label{Rxi-outside}
\sup_{\xi\in\Xi} R^-_\xi \geq P_{q,r} \geq \inf_{\xi\in\Xi} R^+_\xi
\qquad \text{on $\Omega_q^c \times [0,\infty)$.}
\end{align}
Finally,
\begin{align}
\label{Rxi-cone}
\bigcap_{\xi\in\Xi} \Omega_t(R_\xi^+) &= C^+_t, &
\bigcap_{\xi \in\Xi} \Omega_t^c(R_\xi^-) &= \overline{C^-_t}
\qquad \text{for all $t\geq 0$.}
\end{align}
\end{corollary}
\begin{proof}
Since the statement is obvious for $\xi = 0$ by the definition
of $R^\pm_0$ in \eqref{eq:Rpm0},
we fix $\xi \in \Xi$.
Let $\eta$ be the unit normal vector to $\partial \Omega_q$
on $L_\xi \subset \partial \Omega_q$.
Clearly $\eta \cdot \xi = 0$.
Define the half-space $H:= \set{x : (x - V) \cdot \eta > 0}$
and observe that $\Omega_q \subset H$
and $\partial \Omega_q \cap \partial H = L_\xi$.
Let us first show that $P_{q,r} = R^\pm_\xi$ on $\partial H$.
We note that if $x \in \partial H$ we can express
$x - V= \ensuremath{\sigma} \xi + y$, where $\ensuremath{\sigma} \in \ensuremath{\mathbb{R}}$ and
$y \cdot \xi = y \cdot \eta = 0$.
Since $\set{\nu, \eta^+, \eta^-} \subset \operatorname{span} \set{\xi, \eta}$,
and $P_{q,r} = R^\pm_\xi$ on $L_\xi$ by Proposition~\ref{pr:cone-planes},
linearity implies
\begin{align*}
P_{q,r}(x,t) &= P_{q,r}(V + \ensuremath{\sigma}\xi + y,t) = P_{q,r}(V+ \ensuremath{\sigma}\xi,t)
\\&=
R^\pm_\xi(V+\ensuremath{\sigma}\xi,t) = R^\pm_\xi(x,t) \qquad (x,t)\in \partial H \times \ensuremath{\mathbb{R}}.
\end{align*}
Additionally, by linearity, since $V^\pm_t \in H$ for all $t\geq 0$,
and
\[V^+_t \in \cl\Omega_t(R^+_\xi) \setminus \cl\Omega_t(P_{q,r}),
\qquad
V^-_t \in \cl\Omega_t(P_{q,r}) \setminus \cl\Omega_t(R^-_\xi),
\]
we must have
\[
R^-_\xi \leq P_{q,r} \leq R^+_\xi \qquad
\text{on $H \times [0,\infty) \supset Q_q$}
\]
and
\[
R^-_\xi \geq P_{q,r} \geq R^+_\xi \qquad
\text{on $H^c \times [0,\infty)$}.
\]
Since for every $(x,t) \in \Omega_q^c \times [0,\infty)$
there exists $\xi \in \Xi$ such that $x \in H^c$,
\eqref{Rxi-outside} is also proved.
To show one inclusion ($\supset$) of \eqref{Rxi-cone},
we recall that by \eqref{Rxi-touches-Ct} and the choice $\eta^\pm \cdot \nu > 0$,
we clearly have
\[
C^+_t \subset \Omega_t(R^+_\xi), \qquad
\cl{C^-_t} \subset \Omega^c_t(R^-_\xi), \qquad t \geq 0.
\]
To show the other inclusion ($\subset$) of \eqref{Rxi-cone},
we only need to show that for any $x \in \Rd$
there exists $\xi \in \Xi$ that generates
$\xi^\pm, \eta^\pm$ as in Proposition~\ref{pr:cone-planes}
and $x$
can be expressed as
\begin{align}
\label{express-x}
x = \ensuremath{\alpha}^\pm \xi^\pm + \ensuremath{\beta}^\pm\eta^\pm + V^\pm_t,
\end{align}
and therefore clearly for such $\xi$ we have
\begin{align*}
x \notin C^+_t \quad \Leftrightarrow \quad \ensuremath{\beta}^+ \geq 0
\quad \Leftrightarrow \quad R^+_\xi(x,t) = 0
\quad \Leftrightarrow \quad x \notin \Omega_t(R^+_\xi),\\
x \notin \cl{C^-_t} \quad \Leftrightarrow \quad \ensuremath{\beta}^- < 0
\quad \Leftrightarrow \quad R^-_\xi(x,t) > 0
\quad \Leftrightarrow \quad x \notin \Omega^c_t(R^-_\xi).
\end{align*}
To find such $\xi$ for a given $x \in \Rd$,
first observe that there exists $y \in \partial \Omega_q$
such that $x = y + a \nu$ for some $a \in \Rd$.
Since $y \in \partial \Omega_q$, there exists $\xi \in \Xi$
such that $y \in L_\xi$
and therefore $x = \ensuremath{\sigma} \xi + a \nu + V$.
Finally, we can write $x$ as \eqref{express-x}
since $\set{\xi, \nu, V^\pm_t - V} \subset \operatorname{span} \set{\xi^\pm, \eta^\pm}$.
This concludes the proof.
\qedhere\end{proof}
The following geometrical observation,
which shows a connection between the cones $C^\pm_t$
and the planar functions $R^\pm_\xi$,
will be useful later
in deriving monotonicity properties of the solutions
of the obstacle problems.
\begin{proposition}
\label{pr:cone-plus-generator}
For fixed $a > 0$, the following are equivalent
(all with superscript either $+$ or $-$):
\begin{enumerate}
\item
$\pm a R_\xi^\pm(a^{-1} x,a^{-1} t)
\leq \pm R_\xi^\pm(x - y, t - \tau)$ for all $\xi \in \Xi$,
$(x, t) \in \Rd \times \ensuremath{\mathbb{R}}$;
\item
$a C^\pm_{a^{-1} t} \subset C^\pm_{t - \tau} + y$ for some $t \in \ensuremath{\mathbb{R}}$;
\item
$y \in \cl{{\mathrm{Cone}}_{\pm\nu,\ensuremath{\theta}^\pm}(r_V^\pm \tau \nu + (a-1) V_0^\pm)}$
where $r_V^\pm$ were introduced in Remark~\ref{re:speedVplust}.
\end{enumerate}
\end{proposition}
\begin{proof}
Clearly $a R^\pm_\xi(a^{-1}x, a^{-1}t)$ is just a translation
of $R^\pm_\xi(x,t)$ by scaling invariance,
therefore we can compare the planar solutions
in (a)
by comparing their supports.
In fact,
we have
\[aV^\pm_{a^{-1}t} \in \Gamma_t(a R^\pm_\xi(a^{-1}\cdot, a^{-1}\cdot))\]
by construction
and
\[
\Omega_t(R^\pm_\xi(\cdot - y, \cdot - \tau)) = \Omega_{t-\tau}(R^\pm_\xi) + y,
\]
for all $\xi \in \Xi$.
Therefore by Corollary~\ref{co:values-Rxi}
it is straightforward that (a) is equivalent to
\begin{align*}
aV^\pm_{a^{-1}t} \in \cl{C^\pm_{t-\tau}} + y \qquad \text{for some $t \in \ensuremath{\mathbb{R}}$.}
\end{align*}
This is clearly equivalent to (b)
since $aV^\pm_{a^{-1}t}$ is the vertex of $aC^\pm_{a^{-1}t}$.
And it is also equivalent to (c) if we take $t = 0$
and compute
\begin{align}
\label{vertex-inclusion}
a V_0^\pm \in \cl{C^\pm_{-\tau}} + y = \cl{{\mathrm{Cone}}_{\mp\nu,\ensuremath{\theta}^\pm}(V^\pm_{-\tau} + y)}.
\end{align}
Since $V^\pm_{-\tau} = V_0^\pm - r_V^\pm \tau \nu$
by Remark~\ref{re:speedVplust},
we can rewrite \eqref{vertex-inclusion} as (c).
\qedhere\end{proof}
The main motivation for such an involved choice of the domain
for the obstacle problem is the following observation
on the properties of functions $R^\pm_\xi$.
\begin{lemma}
\label{le:matchingplanarsolutions}
Suppose that $q$ and $r$ satisfy the condition \eqref{rqrestriction}.
Then the choice of the angles $\ensuremath{\theta}, \ensuremath{\theta}^\pm$ in \eqref{eq:choiceoftas},
depending only on $\frac{m}{M}$,
guarantees that
\begin{align*}
R^+_\xi &\in \overline{\mathcal{S}}(M) \subset \overline{\mathcal{S}}(g^\ensuremath{\varepsilon})
&&\text{and} & R^-_\xi &\in \underline{\mathcal{S}}(m)\subset\underline{\mathcal{S}}(g^\ensuremath{\varepsilon}) &
&\text{for all $\xi \in \Xi \cup \set0$},
\end{align*}
where $R^\pm_\xi$ were introduced in \eqref{eq:Rpmxi}
and \eqref{eq:Rpm0}.
\end{lemma}
\begin{proof}
The statement is obvious for $R_0^\pm$, by definition.
For any $\xi \in \Xi$, by Proposition~\ref{pr:planar-solution-range},
we only need to verify $r^+ \geq M \mu^+$ and $r_- \leq m \mu^-$.
Recall that $q$ and $r$ satisfy \eqref{rqrestriction}.
We first observe that $\cos \ensuremath{\varphi}^+ = 1$ by \eqref{eq:taandvp} and
\eqref{eq:choiceoftaplus}.
Then, in order,
\eqref{eq:rpm}, \eqref{eq:mupm}, \eqref{eq:choiceofta},
and finally \eqref{rqrestriction} lead to
\begin{align*}
\frac{r^+}{\mu^+}
= \frac{r}{\abs{q}} \frac{\cos^2 \ensuremath{\varphi}^+}{\cos^2 \ensuremath{\theta}}
= \frac{r}{\abs{q}} \frac{M}{m} \geq M,
\end{align*}
which verifies that $R_\xi^+ \in \overline{\mathcal{S}}(M)$.
Similarly, \eqref{eq:rpm}, \eqref{eq:mupm}, \eqref{eq:choiceofta}, \eqref{eq:choiceoftaminus}
and \eqref{rqrestriction} yield
\begin{align*}
\frac{r^-}{\mu^-}
= \frac{r}{\abs{q}} \frac{\cos^2 \ensuremath{\varphi}^-}{\cos^2 \ensuremath{\theta}}
= \frac{r}{\abs{q}} \frac{m}{M} \leq m,
\end{align*}
which verifies that $R_\xi^- \in \underline{\mathcal{S}}(m)$.
\qedhere\end{proof}
\subsubsection{Properties of the obstacle solutions}
\label{sec:obstaclesolutions-properties}
The basic properties of the solutions $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ of the obstacle problem introduced in \eqref{obstaclesolution} are
gathered in the following proposition.
\begin{proposition}
\label{pr:obstacle-sols}
For any $q \in \ensuremath{{\mathbb{R}^{\dimension}}}\setminus\set0$, $r > 0$ satisfying \eqref{rqrestriction} and any $\ensuremath{\varepsilon} > 0$, the following statements apply.
\begin{enumerate}
\item
$\uu_{\e;q,r} \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$
and
$\uu_{\e;q,r} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q \setminus (\Gamma(\uu_{\e;q,r})\cap \Gamma_{q,r}))$;
\item
$\overline{u}_{\e;q,r} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$
and
$\overline{u}_{\e;q,r} \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q \setminus (\Gamma(\overline{u}_{\e;q,r})\cap \Gamma_{q,r}))$;
\item
$(\uu_{\e;q,r})^* \geq \uu_{\e;q,r} \geq P_{q,r}$
and
$(\overline{u}_{\e;q,r})_* \leq \overline{u}_{\e;q,r} \leq P_{q,r}$ in $\Rd\times\ensuremath{\mathbb{R}}$,
with equality on $(\Rd\times\ensuremath{\mathbb{R}})\setminus Q_q$;
\item
$\Omega_t(\uu_{\e;q,r}) \subset C^+_t \cup \Omega_t(P_{q,r})$
and $\Omega_t^c(\overline{u}_{\e;q,r}) \subset \cl{C^-_t} \cup \Omega^c_t(P_{q,r})$;
\item
$\uu_{\e;q,r} \in \underline{\mathcal{S}}(M, \Rd\times\ensuremath{\mathbb{R}})$
and
$\overline{u}_{\e;q,r} \in \overline{\mathcal{S}}(m,\Rd\times\ensuremath{\mathbb{R}})$;
\item
$\uu_{\e;q,r} \leq \inf_{\xi\in\Xi\cup\set0} R^+_\xi$
and $\lu_{\e;q,r} \geq \sup_{\xi\in\Xi\cup\set0} R^-_\xi$
on $Q_q$.
\end{enumerate}
\end{proposition}
\begin{proof}
(a) and (b) are standard; see \cite[Lemma~4]{K07}.
To prove (c), we recall the definition of $R_\xi^\pm$
in \eqref{eq:Rpmxi} and \eqref{eq:Rpm0}.
Lemma~\ref{le:matchingplanarsolutions}
and Corollary~\ref{co:values-Rxi}
yield that $\max(P_{q,r}, R_\xi^+) \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$
and $\min(P_{q,r}, R_\xi^-) \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$ and thus, by definition,
\begin{align}
\label{Rxi-bounds-u}
\uu_{\e;q,r} \leq \max(P_{q,r}, R_\xi^+), \qquad
\lu_{\e;q,r} \geq \min(P_{q,r}, R_\xi^-)
\qquad \text{for all $\xi \in \Xi \cup \set0$,}
\end{align}
which proves (f).
Therefore Corollary~\ref{co:values-Rxi}
implies that
\begin{align*}
(\uu_{\e;q,r})^* = \uu_{\e;q,r} = (\overline{u}_{\e;q,r})_* = \overline{u}_{\e;q,r} = P_{q,r}
\qquad \text{on $(\Rd\times\ensuremath{\mathbb{R}})\setminus Q_q$},
\end{align*}
and (c) is proved.
The result of (d) follows from \eqref{Rxi-bounds-u},
using the support property of $R^\pm_\xi$ in \eqref{Rxi-cone}.
Since $q$ and $r$ satisfy \eqref{rqrestriction},
(e) follows from (a)--(b) and Proposition~\ref{pr:planar-solution-range}.
\qedhere\end{proof}
\begin{remark}
\label{rem:extension}
Fix $T > 0$ and a pair $q,r$ that satisfies \eqref{rqrestriction}.
We shall show that $\uu_{\e;q,r}$ coincides with the function defined on $Q^T = Q_q \cap \set{t < T}$,
\begin{align*}
v = \pth{\inf \set{u \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q^T):
u \geq P_{q,r} \text{ on $Q^T$}}}_{*,\cl{Q^T}},
\end{align*}
which is the solution of the obstacle problem considered in \cite{K07}.
An analogous result holds for $\lu_{\e;q,r}$.
Indeed, $v \leq \uu_{\e;q,r}$ trivially, but also,
since $v \leq R_\xi^+$ on $\cl Q^T$,
we can define for any $s< T$ the function
\begin{align*}
w(x,t) =
\begin{cases}
v(x,t) & t\leq s,\\
R_0^+(x,t) & t > s.
\end{cases}
\end{align*}
Clearly $w \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon},Q_q)$
and hence, by definition,
$\uu_{\e;q,r} \leq w = v$ for $t \leq s$ for any $s < T$.
\end{remark}
\subsection{Monotonicity of the obstacle problem}
\label{sec:monotonicity-obstacle}
The solutions $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ of the obstacle problem introduced in \eqref{obstaclesolution} admit a natural monotonicity property with respect to the hyperbolic scaling and a change of scale $\ensuremath{\varepsilon}$.
Moreover, the periodicity of the medium also yields the monotonicity with respect to translations by multiplies of $\ensuremath{\varepsilon}$, both in space and time.
Since we have a control on how fast the free boundaries of $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ detach from the obstacle at the parabolic boundary of the domain $Q_q$, we can extend the allowed range of translations.
This is the main result of this section.
The second result is the monotonicity of $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ in time.
\begin{proposition}[Monotonicity]
\label{pr:monotonicity}
Let $q$ and $r$ satisfy \eqref{rqrestriction},
and let $a > 0$.
Then
\begin{align*}
\uu_{\e;q,r}(x,t) \leq a^{-1}\uu_{a\ensuremath{\varepsilon};q,r}(a(x - y), a(t - \tau))
\end{align*}
for all $(x,t) \in \Rd\times\ensuremath{\mathbb{R}}$
and for all $(y, \tau) \in \ensuremath{\varepsilon}\ensuremath{\mathbb{Z}}^{n+1}$ such that
\begin{align}
\label{y-in-taplus-cone}
y\cdot \nu &\geq \max (r\tau, M\abs{q} \tau), &
&\text{and}&
y &\in \cl {{\mathrm{Cone}}_{\nu,\ensuremath{\theta}^+}(r^+_V \tau \nu + (a - 1) V^+_0)},
\end{align}
where $r_V^\pm$, $V_0^\pm$ and $\ensuremath{\theta}^\pm$ were defined
in Section~\ref{sec:domain-geometry}.
Similarly,
\begin{align*}
\overline{u}_{\e;q,r}(x, t) \geq a^{-1}\overline{u}_{a\ensuremath{\varepsilon};q,r}(a (x-y),a (t-\tau))
\end{align*}
for all $(x,t) \in \Rd\times\ensuremath{\mathbb{R}}$
and for all $(y, \tau) \in \ensuremath{\varepsilon}\ensuremath{\mathbb{Z}}^{n+1}$ such that
\begin{align}
\label{y-in-taminus-cone}
y\cdot \nu &\leq \min (r\tau, m\abs{q} \tau), &
&\text{and}&
y &\in \cl {{\mathrm{Cone}}_{-\nu,\ensuremath{\theta}^-}(r^-_V \tau \nu + (a - 1) V^-_0)}.
\end{align}
\end{proposition}
\begin{proof}
We shall show the ordering for $\overline{u}_{\e;q,r}$,
the proof for $\uu_{\e;q,r}$ is analogous.
Fix $a > 0$ and $(y, \tau)$
that satisfies the hypothesis \eqref{y-in-taplus-cone}
and define the cylinder \[\Sigma = Q_q \cap (a^{-1}Q_q + (y, \tau)).\]
Let us, for simplicity,
denote \[v^a(x,t) := a^{-1}\overline{u}_{a\ensuremath{\varepsilon};q,r}(a(x-y), a(t-\tau)).\]
Due to the $\ensuremath{\varepsilon}\ensuremath{\mathbb{Z}}^{n+1}$-periodicity of $g^\ensuremath{\varepsilon}$,
we observe that $v^a \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, a^{-1}Q_q +(y,\tau))$.
We immediately have
\begin{align}
\label{obs-va-order}
P_{q,r}(x,t) \geq P_{q,r}(x-y,t-\tau) = a^{-1} P_{q,r}(a(x-y),a(t-\tau))
\geq v^a(x,t)
\end{align}
by \eqref{y-in-taminus-cone},
the scale invariance and Proposition~\ref{pr:obstacle-ordering}.
Our goal is to show that
\begin{align}
\label{uorder-complement}
\text{$\lu_{\e;q,r} \geq v^a$ on $\Sigma^c\quad$
and $\quad\cl\Omega(v^a)\cap \Sigma^c \subset \cl\Omega(\lu_{\e;q,r})$}
\end{align}
since then we can apply Lemma~\ref{le:stitch}
with $Q_1 = \Sigma$, $Q_2 = Q_q$,
$u_1 = v^a$ and $u_2 = \lu_{\e;q,r}$
to conclude that $u \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$,
where $u$ is defined in Lemma~\ref{le:stitch}.
Because clearly $u \leq P_{q,r}$ by \eqref{obs-va-order},
we have by definition of $\lu_{\e;q,r}$ that
$\lu_{\e;q,r} \geq u \geq v^a$ on $\Sigma$
and therefore $\lu_{\e;q,r} \geq v^a$ on $\Rd\times\ensuremath{\mathbb{R}}$.
Let us prove \eqref{uorder-complement}.
We can write $\Sigma^c = A_1 \cup A_2 \cup A_3$
where
$A^1 := (Q_q)^c$,
$A_2 := Q_q \cap ((a^{-1} \Omega_q + y)^c \times [\tau,\infty))$
and
$A_3 := Q_q \cap \set{t < \tau}$.
We therefore prove $\lu_{\e;q,r}(x,t) \geq v^a(x,t)$
for all $(x,t) \in \Sigma^c$ by considering the following three cases:
\begin{itemize}
\item
$(x,t) \in A_1$:
This case is very simple since
Proposition~\ref{pr:obstacle-sols}(c)
and \eqref{obs-va-order},
yield
\begin{align*}
\lu_{\e;q,r}(x,t) = P_{q,r}(x,t) \geq v^a(x,t).
\end{align*}
\item
$(x,t) \in A_2$:
Observe that \eqref{y-in-taminus-cone} implies
Proposition~\ref{pr:cone-plus-generator}(c)
and therefore Proposition~\ref{pr:cone-plus-generator}(a)
holds.
We use
Proposition~\ref{pr:obstacle-sols}(f),
Proposition~\ref{pr:cone-plus-generator}(a)
and \eqref{Rxi-outside}, all properly rescaled, and estimate
\begin{align*}
\lu_{\e;q,r}(x,t) &\geq \sup_{\xi\in\Xi} R^-_\xi(x,t)
\geq \sup_{\xi\in\Xi} a^{-1}R^-_\xi(a(x-y), a(t -\tau))
\\&\geq a^{-1}P_{q,r}(a(x-y), a(t-\tau)) \geq v^a(x,t).
\end{align*}
\item
$(x,t) \in A_3$:
Since $0 \leq t \leq \tau$,
Proposition~\ref{pr:obstacle-sols}(f)
and \eqref{y-in-taminus-cone}
yield
\begin{align*}
\lu_{\e;q,r}(x,t) &\geq R^-_0(x,t) = \abs{q} (m\abs{q} t - x\cdot\nu)_+
\\&\geq \abs{q}(r(t- \tau) - (x - y)\cdot \nu)_+
= a^{-1}P_{q,r}(a(x-y), a(t-\tau))
\\&\geq v^a(x,t).
\end{align*}
\end{itemize}
Therefore we have shown the first part of \eqref{uorder-complement},
and the second part can be inferred from the above estimates as well.
Consequently, the argument using Lemma~\ref{le:stitch} described above
applies and we can conclude that
$\lu_{\e;q,r} \geq v^a$ on $\Rd\times\ensuremath{\mathbb{R}}$.
\qedhere\end{proof}
The second important result of this section is the almost obvious fact
that the solutions of the obstacle problems are increasing in time.
However, in contrast to \cite{K07},
Proposition~\ref{pr:monotonicity} only implies
the monotonicity in time for multiplies of $\ensuremath{\varepsilon}$
due to the time-dependence of $g$.
Here we present an argument using a nonlinear scaling of the solutions,
which also provides a lower bound
on the speed of the free boundary.
\begin{lemma}[Monotonicity in time]\label{le:monotonicity-in-time}
Let $r$ and $q$ satisfy \eqref{rqrestriction}
and let $\ensuremath{\varepsilon} > 0$. Then
\begin{align*}
\uu_{\e;q,r}(x - y, t_1) &\leq \uu_{\e;q,r}(x,t_2),\\
\overline{u}_{\e;q,r}(x - y, t_1) &\leq \overline{u}_{\e;q,r}(x,t_2)
\end{align*}
for any $t_1 < t_2$ and $x \in \Rd$,
and $y \in \Rd$ such that $\abs{y} \leq \rho$
for some positive constant $\rho$ depending only on $m, L, \ensuremath{\varepsilon}, t_1, t_2$.
\end{lemma}
\begin{proof}
Let us fix $r$, $q$ and $\ensuremath{\varepsilon}$.
We first prove the result for $\uu := \uu_{\e;q,r}$.
Since we have to compare solutions
that are not necessarily translated by a multiply of $\ensuremath{\varepsilon}$,
we cannot use the monotonicity of
Proposition~\ref{pr:monotonicity} directly.
Instead,
we will compare $\uu$
with its nonuniform perturbation
using Proposition~\ref{pr:nonuniform-perturbation}.
First, assume that $0 < t_1 < t_2 < t_1 + \ensuremath{\gamma}$
where $\gamma := \frac mL$.
We shall define
\begin{align}
\label{def-of-v-mont}
v(x,t) = \inf_{y \in \cl B_\rho(0)} \uu_{\e;q,r}(x - y, \ensuremath{\theta}(t)),
\end{align}
where $\rho$ will be a positive constant
and $\ensuremath{\theta}(t) := \ensuremath{\theta}(t;\bar\ensuremath{\lambda}) = \bar\ensuremath{\lambda} + f(t;\bar\ensuremath{\lambda})$
for some $\bar\ensuremath{\lambda}$ determined below.
Function $f(t;\ensuremath{\lambda})$ is the function constructed in
Section~\ref{sec:nonlinear-scaling-super}
for given parameters $\gamma$ (defined above), $a = 1$,
and $\ensuremath{\lambda} > 0$. Because $D\rho \equiv 0$ we set $\ensuremath{\alpha} = a = 1$.
With these parameters, the expression for $\ensuremath{\theta}(t;\ensuremath{\lambda})$
is simply
\begin{align*}
\ensuremath{\theta}(t;\ensuremath{\lambda}) =
t - \gamma W\pth{-\frac{\ensuremath{\lambda}}{\ensuremath{\gamma}} e^{-\frac \ensuremath{\lambda} \ensuremath{\gamma}} e^{\frac t \ensuremath{\gamma}}},
\end{align*}
where $W$ is the Lambert W function.
Therefore $\ensuremath{\theta}(t; \ensuremath{\lambda})$ is well-defined on $t \in [0, t_\ensuremath{\lambda}]$
and $\ensuremath{\lambda} \in [0,\gamma]$,
where $t_\ensuremath{\lambda} := \log \frac \ensuremath{\gamma} \ensuremath{\lambda} + \frac \ensuremath{\lambda}\ensuremath{\gamma} -1$
(we set $t_0 = +\infty$).
Observe that $t_\ensuremath{\lambda}$ is strictly decreasing in $\ensuremath{\lambda}$
and positive for $\ensuremath{\lambda} \in (0, \ensuremath{\gamma}]$ with
$t_\ensuremath{\lambda} = 0$ at $\ensuremath{\lambda} = \ensuremath{\gamma}$ and $t_\ensuremath{\lambda} \to +\infty$
as $\ensuremath{\lambda} \to 0+$.
Therefore for every $t > 0$ there exists a unique $\ensuremath{\lambda} =: \ensuremath{\lambda}_t$
such that $t_\ensuremath{\lambda} = t$.
From the properties of $W$, we observe that $\ensuremath{\theta}(t; 0) = t$
and $\ensuremath{\theta}(t; \ensuremath{\lambda}_t) = t + \ensuremath{\gamma}$ for any $t > 0$.
By continuity and the assumption $t_2 \in (t_1, t_1 + \ensuremath{\gamma})$,
there exists $\bar \ensuremath{\lambda} \in (0, \ensuremath{\lambda}_{t_1})$
such that $\ensuremath{\theta}(t_1; \bar\ensuremath{\lambda}) = t_2$.
Since $\bar\ensuremath{\lambda} < \ensuremath{\lambda}_{t_1}$ we must have $t_{\bar\ensuremath{\lambda}} > t_1$.
Let us
thus take $\ensuremath{\theta}(t) = \ensuremath{\theta}(t;\bar \ensuremath{\lambda})$
and consider $\uu_{\e;q,r}(x-y, \ensuremath{\theta}(t))$
for some $y \in \Rd$ and $(x,t) \in \set{t \leq t_{\bar\ensuremath{\lambda}}}$.
It was shown in the proof of Proposition~\ref{pr:monotonicity}
that
$\uu_{\e;q,r}(x,t) \leq \uu_{\e;q,r}(x - y, t - \tau)$
on $(Q_q \cap (Q_q + (y,\tau))^c$
for all $(y,\tau)$ satisfying \eqref{y-in-taplus-cone}.
Let us set $\tau = - \bar\ensuremath{\lambda}$. There exists $\rho>0$
such that $(y,\tau)$ satisfies \eqref{y-in-taplus-cone}
for all $y \in \cl B_\rho(0)$.
Since $t + \bar \ensuremath{\lambda}\leq \ensuremath{\theta}(t)$ for all $t \in [0, t_{\bar\ensuremath{\lambda}}]$,
the same argument yields
$\uu_{\e;q,r}(x, t) \leq \uu_{\e;q,r}(x-y, \ensuremath{\theta}(t))$
on $(Q_q \cap (Q_q + (y,\tau))^c \cap \set{t \leq t_{\bar\ensuremath{\lambda}}}$
for $(y,\tau) \in \cl B_\rho(0) \times \set{-\bar\ensuremath{\lambda}}$.
Hence the argument using Lemma~\ref{le:stitch}
as in the proof of Proposition~\ref{pr:monotonicity}
(with Remark~\ref{rem:extension}
for extension of $v$ for $t > t_{\bar\ensuremath{\lambda}}$)
yields that $\uu_{\e;q,r}(x,t) \leq \uu_{\e;q,r}(x-y,\ensuremath{\theta}(t))$
for $(x,t) \in \set{t < t_{\bar\ensuremath{\lambda}}}$.
In particular $u(x,t_1) \leq u(x - y, \ensuremath{\theta}(t_1)) = u(x - y, t_2)$
for all $y \in \cl B_\rho(0)$,
and that is what we wanted to prove.
The proof for general $0 < t_1 < t_2$ can be done iteratively
by splitting $[t_1, t_2]$ into subintervals of length smaller than
$\ensuremath{\gamma}$.
We skip the proof for $\overline{u}_{\e;q,r}$, which follows the same idea,
but is simpler.
In particular, it is unnecessary to restrict
$t_2 < t_1 + \ensuremath{\gamma}$
since $f$ constructed in Section~\ref{sec:nonlinear-scaling-sub}
can handle arbitrary tranlations.
\qedhere\end{proof}
For completeness, and as a consequence of the monotonicity in time in the previous lemma,
we also show that the solutions are harmonic in their positive sets.
\begin{proposition}
\label{pr:harmonic}
Let $r$ and $q$ satisfy \eqref{rqrestriction}
and let $\ensuremath{\varepsilon} > 0$.
The functions $\uu_{\e;q,r}(\cdot, t)$ and
$\pth{\overline{u}_{\e;q,r}}_*(\cdot, t)$ are harmonic
in $\Omega_t(\uu_{\e;q,r})\cap\Omega_q$
and $\Omega_t((\overline{u}_{\e;q,r})_*) \cap \Omega_q$,
respectively, for every $t$.
\end{proposition}
\begin{proof}
Since the claim is trivially true for $t \leq 0$,
we fix $\tau > 0$.
We first show that
$\uu_{\e;q,r}(\cdot, \tau)$ is harmonic
in the open set $E := \Omega_\tau(\uu_{\e;q,r})\cap \Omega_q$.
For any $w \in LSC(\cl{E})$,
superharmonic in $E$, such that $w \geq \uu_{\e;q,r}(\cdot, \tau)$
on $\partial E$, we define a new function on $Q_q \cap \set{t \leq \tau}$
\begin{align*}
v(x, t) =
\begin{cases}
\min \pth{\uu_{\e;q,r}(x,t), w(x)} & x \in E,\\
\uu_{\e;q,r}(x, t) & \text{otherwise}.
\end{cases}
\end{align*}
By definition of $w$, we observe
that $v \in LSC(Q_q \cap \set{t \leq \tau})$
and $v \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q \cap \set{t \leq \tau})$.
Therefore $\uu_{\e;q,r}(x, \tau) \leq v(x, \tau) \leq w(x)$ in $E$
by definition of $\uu_{\e;q,r}$ and Remark~\ref{rem:extension}.
We recall that $\uu_{\e;q,r}(\cdot, \tau)$ is superharmonic in
$E$, and therefore it is the smallest superharmonic function,
i.e., it must be harmonic.
Now we show that $(\overline{u}_{\e;q,r})_*(\cdot, \tau)$ is harmonic
in the open set $E = \Omega_\tau((\overline{u}_{\e;q,r})_*)\cap\Omega_q$.
First we recall that $(\overline{u}_{\e;q,r})_*(\cdot, \tau)$
is superharmonic since $\overline{u}_{\e;q,r} \in \overline{\mathcal{S}}(m, \Rd\times\ensuremath{\mathbb{R}})$
by Proposition~\ref{pr:obstacle-sols}.
Let $w \in LSC(\cl E)$ be a superharmonic function in $E$,
such that $w \geq (\overline{u}_{\e;q,r})_*(\cdot, \tau)$ on $\partial E$.
By the monotonicity in time, Lemma~\ref{le:monotonicity-in-time},
and in particular the positive speed of expansion,
we observe that $\overline{u}_{\e;q,r}(x, t) = 0$ for
$x \in \Gamma_\tau((\overline{u}_{\e;q,r})_*)$
and $t < \tau$.
Since $\overline{u}_{\e;q,r}(\cdot, t)$ is subharmonic in $\Omega_q$ for all $t$,
we must have $\overline{u}_{\e;q,r}(x,t) \leq w(x)$ for $t < \tau$.
Therefore, $(\overline{u}_{\e;q,r})_*(x, \tau) \leq w(x)$.
Again, $(\overline{u}_{\e;q,r})_*(\cdot, \tau)$ is the smallest superharmonic function
and thus harmonic.
\qedhere\end{proof}
\subsection{Flatness}
\label{sec:flatness}
In this section we introduce \emph{flatness},
the quantity that will be a measure
of how good our guess of the homogenized
velocity $r(q)$ is.
More specifically, it measures how far, for given $\ensuremath{\varepsilon}$, $q$ and $r$, the free boundaries of the obstacle problem solutions $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$,
introduced in \eqref{obstaclesolution}, detach from the free boundary of the obstacle $P_{q,r}$ up to the given time.
For $\eta \in \ensuremath{\mathbb{R}}$, denote
\begin{align*}
P_{q,r}^\eta(x,t) := P_{q,r} (x - \eta \nu, t).
\end{align*}
Note that $P_{q,r}^\eta$ represents a translation of $P_{q,r}$
in the direction $\nu$ by distance $\eta$.
\begin{definition}
\label{def:flatness}
For $q \neq 0$, $r > 0$ and $\ensuremath{\varepsilon} > 0$,
we define the \emph{upper flatness} $\underline\Phi_{\e;q,r}(\tau)$,
i.e., the flatness of $\uu_{\e;q,r}$ up to time $\tau \in \ensuremath{\mathbb{R}}$
as
\begin{align*}
\underline\Phi_{\e;q,r}(\tau) =
\inf \set{\eta:
\uu_{\e;q,r} \leq P^\eta_{q,r} \text{ in } \set{t \leq \tau}}.
\end{align*}
Similarly, the \emph{lower flatness} $\overline\Phi_{\e;q,r}(\tau)$
is defined as the flatness of $\overline{u}_{\e;q,r}$ up to time $\tau \in \ensuremath{\mathbb{R}}$,
\begin{align*}
\overline\Phi_{\e;q,r}(\tau) =
\inf \set{\eta:
\overline{u}_{\e;q,r} \geq P^{-\eta}_{q,r} \text{ in } \set{t \leq \tau}}.
\end{align*}
We say that $\uu_{\e;q,r}$ (resp. $\overline{u}_{\e;q,r}$) is $\eta$-flat up to time $\tau$
if $\underline\Phi_{\e;q,r}(\tau) \leq \eta$ (resp. $\overline\Phi_{\e;q,r}(\tau) \leq \eta$).
\end{definition}
The flatness is Lipschitz and monotone:
\begin{proposition}
\label{pr:phi-lipschitz}
Let $q \in \Rd \setminus \set0$, $r > 0$ and $\ensuremath{\varepsilon} > 0$.
Then for any $t \in \ensuremath{\mathbb{R}}$, $h > 0$,
\begin{align*}
\underline\Phi_{\e;q,r}(t) &\leq \underline\Phi_{\e;q,r}(t + h)
\leq \underline\Phi_{\e;q,r}(t) + h(M\abs{q} - r)_+,\\
\overline\Phi_{\e;q,r}(t) &\leq \overline\Phi_{\e;q,r}(t + h)
\leq \overline\Phi_{\e;q,r}(t) + h(r - m\abs{q})_+.
\end{align*}
In particular, $\overline\Phi_{\e;q,r}$ and $\underline\Phi_{\e;q,r}$ are
nondecreasing and Lipschitz continuous.
\end{proposition}
\begin{proof}
$\overline\Phi_{\e;q,r}$ and $\underline\Phi_{\e;q,r}$ are clearly nondecreasing
from their definition.
To show the upper bound for $\underline\Phi_{\e;q,r}(\tau + h)$ for any $\tau \in \ensuremath{\mathbb{R}}$,
$h > 0$,
we may assume that $r < M \abs q$, otherwise $\uu_{\e;q,r} = P_{q,r}$
and so $\underline\Phi_{\e;q,r}(\tau + h) = \underline\Phi_{\e;q,r}(\tau) = 0$.
We only need to compare $\uu_{\e;q,r}$ with an appropriate translation
of $R^+_0=P_{q,M\abs{q}}$, using the fact that
$\uu_{\e;q,r}(\cdot, \tau) \leq P_{q,r}^\eta(\cdot, \tau)$
and the identity
$P_{q, M\abs q}(\cdot, h) = P_{q,r}^{(M\abs q - r)h}(\cdot, h)$.
Similarly, the upper bound for $\overline\Phi_{\e;q,r}(\tau+h)$ follows
from comparison with a translation of $R^-_0 = P_{q,m\abs{q}}$.
\qedhere\end{proof}
Finally, we can always find a point on the free boundary of $\uu_{\e;q,r}$ or $\overline{u}_{\e;q,r}$
whose spatial distance from the free boundary of the obstacle is exactly equal to the flatness at the given time.
\begin{lemma}
\label{le:furthest-point}
Let $q \in \Rd \setminus \set0$, $r > 0$ and $\ensuremath{\varepsilon} > 0$.
Let $\tau \in \ensuremath{\mathbb{R}}$.
Then for any $\eta \in (0, \underline\Phi_{\e;q,r}(\tau)]$ there exists
$(\zeta, \sigma) \in \Gamma(\uu_{\e;q,r})$
such that
$\zeta \cdot \nu = r \sigma + \eta$
and $\underline\Phi_{\e;q,r}(t) \leq \eta$ for all $t \leq \ensuremath{\sigma}$.
Similarly,
for any $\eta \in (0, \overline\Phi_{\e;q,r}(\tau)]$ there exists
$(\zeta, \sigma) \in \Gamma(\overline{u}_{\e;q,r})$
such that
$\zeta \cdot \nu = r \sigma - \eta$
and $\overline\Phi_{\e;q,r}(t) \leq \eta$ for all $t \leq \ensuremath{\sigma}$.
\end{lemma}
\begin{proof}
We will show the proof for $\uu_{\e;q,r}$,
the proof for $\lu_{\e;q,r}$ is similar.
Let us define $T = \sup\set{t \leq \tau: \underline\Phi_{\e;q,r}(t) \leq \eta}$.
Clearly $T \in (0, \tau]$. Moreover, by continuity,
$\underline\Phi_{\e;q,r}(T) = \eta$.
We claim that we can find
$(\zeta, \sigma) \in \Omega^c(P_{q,r}^\eta) \cap \cl\Omega(\uu_{\e;q,r})
\cap \set{t \leq T}$.
If such point does not exist,
we must have
$\operatorname{dist}(\Omega^c(P_{q,r}^\eta) \cap \set{0 \leq t \leq T},
\cl\Omega(\uu_{\e;q,r})) > 0$
and therefore there exists $\ensuremath{\delta} \in (0, \eta)$
such that
$\Omega^c(P_{q,r}^{\eta-\ensuremath{\delta}}) \cap \cl\Omega(\uu_{\e;q,r})
\cap \set{t \leq T} = \emptyset$.
Since $P_{q,r}^{\eta-\ensuremath{\delta}}$ is harmonic in its positive set,
it follows that $\uu_{\e;q,r} \leq P_{q,r}^{\eta-\ensuremath{\delta}}$ in $\set{t \leq T}$.
But that is a contradiction with
$\underline\Phi_{\e;q,r}(T) = \eta > \eta - \ensuremath{\delta}$.
Therefore $(\zeta, \ensuremath{\sigma})$ exists and
we observe that
$(\zeta, \ensuremath{\sigma}) \in \Gamma(\uu_{\e;q,r}) \cap \Gamma(P_{q,r}^\eta)
\cap \set{0 < t \leq T}$.
\qedhere\end{proof}
\subsection{Local comparison principle}
\label{sec:local-comparison-principle}
An important tool in the analysis of the behavior in the homogenization
limit is the \emph{local comparison principle}.
In contrast to the standard comparison principle, which requires
ordering on the parabolic boundary to guarantee ordering
in the whole cylinder,
the local comparison relies on the extra information about flatness
of solutions in the sense of Definition~\ref{def:flatness},
and therefore allows for solutions to cross on the lateral boundary
but still guarantees the ordering of free boundaries
for a short time
in a small region far from the boundary.
This is a generalization of a result that appeared in
\cite{K07,K08} (with $\ensuremath{\beta} =1$).
\begin{theorem}[Local comparison principle]
\label{th:localComparison}
Let $\ensuremath{\beta} \in (\frac45, 1)$, $r_1 > r_2 > 0$ and $q_1, q_2 \in \Rd$
such that $a^3 q_1 = q_2$ for some $a > 1$.\\
Then there exists
$\ensuremath{\varepsilon}_0 = \ensuremath{\varepsilon}_0(r_1,r_2, \abs{q_1}, \abs{q_2}, m, M, n, \ensuremath{\beta})$ such that
\[
\max \pth{\overline\Phi_{\ensuremath{\varepsilon}; q_1, r_1}(1), \underline\Phi_{\ensuremath{\varepsilon}; q_2, r_2}(1)} > \ensuremath{\varepsilon}^\ensuremath{\beta}
\qquad \text{for } \ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0.
\]
In other words,
$\overline{u}_{\ensuremath{\varepsilon}; q_1, r_1}$ and $\uu_{\ensuremath{\varepsilon}; q_2, r_2}$ cannot be both $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat up to time $t = 1$ for any $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$.
\end{theorem}
Before proceeding with the proof of Theorem~\ref{th:localComparison},
we first present a simple result on the nondegeneracy of $\uu_{\e;q,r}$.
This is an improvement of
Corollary~\ref{co:hs-nondegeneracy}
when we also know that $\uu_{\e;q,r}$ is $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat.
\begin{lemma}
\label{le:nondegeneracyObstacleSupersolution}
For given $q \in \Rd \setminus \set0, r>0$,
there exists a positive constant $c = c(r, \abs{q}, M, n)$ such that if
$\underline\Phi_{\e;q,r}(T) \leq \eta$, for some $\ensuremath{\varepsilon}, \eta, T > 0$, then
\begin{align*}
\uu_{\e;q,r}(x,t) \geq c \frac{\rho^2}{\eta}
\end{align*}
for every $0 < \rho \leq \eta$ and $(x,t)$ such that $B_\rho(x) \subset \Omega_t(\uu_{\e;q,r}) \cap \Omega_q$, $t \leq T$.
\end{lemma}
\begin{proof}
Let $c_H = c_H(n)$ be the constant from Harnack inequality such that
if $\psi \geq 0$, harmonic in $B_1(0)$ then
$\inf_{\cl{B}_{1/2}(0)} \psi \geq c_H \sup_{\cl{B}_{1/2}(0)} \psi$.
Let us denote $u = \uu_{\e;q,r}$. Fix $(\xi, \tau)$ such that $B_\rho(\xi) \subset \Omega_{\tau}(u) \cap \Omega_q$.
Then we have two possible scenarios:
\begin{enumerate}
\item $\xi \cdot \nu \leq r \tau + \frac{\rho}{4}$: We estimate
\begin{align*}
u(\xi - \frac{\rho}{2} \nu, \tau) \geq P_{q,r}(\xi - \frac{\rho}{2}\nu, \tau) \geq \abs{q} \frac{\rho}{4} \geq \frac{\abs{q}}{4} \frac{\rho^2}{\eta}.
\end{align*}
\item $\xi \cdot \nu > r \tau + \frac{\rho}{4}$:
In this case necessarily $r < M \abs{q}$,
otherwise $u = P_{q,r}$, a contradiction.
We have $u \in \underline{\mathcal{S}}(M, \Rd\times \ensuremath{\mathbb{R}})$
by Proposition~\ref{pr:obstacle-sols}
and $u$ is nondecreasing in time by Lemma~\ref{le:monotonicity-in-time}.
Due to the flatness assumption $\underline\Phi_{\e;q,r}(\tau) \leq \eta$,
$u(\cdot, t) = 0$ on $\cl{B}_{\rho/4}(\xi)$ for
$t \leq \tau - \frac{\eta}{r}$.
Consequently,
Corollary~\ref{co:hs-nondegeneracy} implies
\begin{align*}
\sup_{B_{\rho/4}(\xi)} u(\cdot, \tau) \geq \frac{\rho^2}{32n M \frac{\eta}{r}} = \frac{r}{32n M} \frac{\rho^2}{\eta}.
\end{align*}
\end{enumerate}
Since $u(\cdot, \tau)$ is harmonic in
$\Omega_\tau(u) \cap \Omega_q \supset B_{\rho}(\xi)$,
the Harnack inequality, properly rescaled, yields
\[
u(\xi, \tau) \geq c_H \min \pth{\frac{\abs{q}}{4}, \frac{r}{32nM}} \frac{\rho^2}{\eta}.\qedhere
\]
\qedhere\end{proof}
An important tool in the proof of the local comparison principle
is the nonuniform perturbation in space,
see Section~\ref{sec:nonlinear-perturbation}.
We will construct a particular radius $\rho(x)$
in the following lemma.
\begin{lemma}
\label{le:radial-nonlin-perturbation}
There exists $R > 0$ and a smooth,
radially symmetric function $\rho : \cl B_{2 R}(0) \setminus B_R(0)
\to [1,6]$
such that
$\rho \Delta \rho \geq (n-1) \abs{D \rho}^2$
in $B_{2 R}(0) \setminus \cl B_R(0)$,
$\rho = 1$ on $\partial B_{2R}(0)$ and
$\rho > 5$ on $\cl B_{R+1}(0) \setminus B_R(0)$.
\end{lemma}
\begin{proof}
Let $\ensuremath{\varphi}$ be the radially symmetric smooth positive solution
of
\begin{align*}
\begin{cases}
\Delta \ensuremath{\varphi}^{2-n} = 0, & \text{ in } B_{2} \setminus \cl B_1\\
\ensuremath{\varphi} = 1 & \text{ on } \partial B_{2},\\
\ensuremath{\varphi} = 6 & \text{ on } \partial B_{1}.\\
\end{cases}
\end{align*}
It was observed in \cite[p. 148]{C87}
that
$\ensuremath{\varphi}$ satisfies $\ensuremath{\varphi} \Delta \ensuremath{\varphi} \geq (n-1) \abs{D\ensuremath{\varphi}}^2$.
Moreover,
note that also the rescaled function $a \ensuremath{\varphi}(b x)$ satisfies the same inequality for any $a, b$ positive.
By continuity, there exists $\ensuremath{\delta} > 0$ such that $\ensuremath{\varphi} > 5$
on $\cl B_{1 + \ensuremath{\delta}} \setminus B_1$.
Define $R = 1/\ensuremath{\delta}$ and
$\rho(x) = \ensuremath{\varphi}(\ensuremath{\delta} x)$
on $\cl B_{2R} \setminus B_R$.
It is straightforward to verify that $\rho$ satisfies
the properties asserted in the statement and the proof is finished.
\qedhere\end{proof}
The following technical lemma
shows
that the closure of a time-slice of $\Omega(\lu_{\e;q,r})$
is the time-slice of the closure,
a fact
that will be used throughout the proof of the local comparison principle
and
that allows us to work with simple time-slices of
a subsolution with nondecreasing support.
\begin{lemma}
Let $u \in \underline{\mathcal{S}}(M, Q)$
be a bounded subsolution on $Q$
for some $M > 0$ and $Q = E \times (t_1,t_2)$,
$E \subset \Rd$ open,
and assume that $\Omega_t(u^*) \subset \Omega_s(u^*)$ for all
$t, s$ such that
$t_1 < t \leq s < t_2$.
Then for any $(x,t) \in Q$ we have
\begin{align*}
x &\in \cl{\Omega_t(u^*;Q)}&&\text{if and only if}
& (x,t) \in \cl\Omega(u;Q),
\end{align*}
or, equivalently,
\begin{align*}
\cl{\Omega_t(u^*;Q)} = \cl\Omega_t(u;Q).
\end{align*}
\end{lemma}
\begin{proof}
Since $\cl\Omega(u) = \cl\Omega(u^*)$,
we can WLOG assume that $u = u^*$.
Let $(\xi, \tau) \in Q$.
It is clear that if $\xi \in \cl{\Omega_\tau(u)}$
then $(\xi, \tau) \in \cl \Omega(u)$.
Now suppose that $(\xi, \tau) \in \cl\Omega(u) \cap Q$.
By Definition~\ref{def:visc-test-sub}(i)
and the monotonicity assumption,
$(\xi, \tau) \in \cl{\Omega(u) \cap \set{t < \tau}}
\subset \cl{\Omega_\tau(u)} \times \set{t \leq \tau}$.
\qedhere\end{proof}
\begin{proof}[Proof of Theorem~\ref{th:localComparison}]
The proof proceeds in a number of technical steps.
Suppose that such $\ensuremath{\varepsilon}_0$ does not exist and
both $\overline{u}_{\ensuremath{\varepsilon}; q_1, r_1}$ and $\uu_{\ensuremath{\varepsilon}; q_2, r_2}$
are $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat up to time $t = 1$
along some subsequence of $\ensuremath{\varepsilon}$ that converges to 0.
We will show that this leads to a contradiction
with the comparison principle for the Hele-Shaw problem
for $\ensuremath{\varepsilon}$ sufficiently small.
\parahead{(1) Setup}
Let $\ensuremath{\alpha} := \frac{4}{5}$, and observe that, since $\ensuremath{\beta} \in (4/5,1)$,
\begin{align}
\label{alpha-beta-relation}
2 - 2 \ensuremath{\beta} \leq 4 - 4 \ensuremath{\beta} < \ensuremath{\alpha} < \ensuremath{\beta}.
\end{align}
First,
for fixed $\ensuremath{\varepsilon} > 0$,
we introduce notation that will be used throughout the proof.
Symbol $c$ with a subscript will denote quantities independent of $\ensuremath{\varepsilon}$.
We shall use multiple perturbations of
the subsolution $\overline{u}_{\ensuremath{\varepsilon},q_1,r_1}$.
Throughout the proof,
$\nu = \frac{-q_1}{\abs{q_1}} = \frac{-q_2}{\abs{q_2}}$.
Let us denote $\Lambda := 6$ and define the constant
\begin{align}
\label{choice-gamma}
\ensuremath{\gamma} = \frac{1}{\Lambda}\min\pth{\frac{m}{L}(a^{1/2} - 1), \ov2}
\end{align}
and a translation
\begin{align*}
\xi_0 \in \operatorname*{arg\,min}_{\substack{\xi \in \ensuremath{\varepsilon} Z^n\\
\xi \cdot \nu \leq - \ensuremath{\varepsilon}}}
\abs{\xi}.
\end{align*}
Clearly $\abs{\xi_0} \leq n \ensuremath{\varepsilon}$.
For $\ensuremath{\lambda} \in [0,\Lambda]$,
we define the perturbations
\begin{align*}
u^\ensuremath{\lambda}(x,t) :=
\sup_{y \in \cl B_{\ensuremath{\lambda} \ensuremath{\gamma} \ensuremath{\varepsilon}}(x)} a\lu_{\ensuremath{\varepsilon},q_1,r_1}(y - \xi_0,t),
\end{align*}
where $\cl B_0(x) := \set{x}$ for simplicity.
The choice of $\gamma$ and $\xi_0$,
and Proposition~\ref{pr:nonuniform-perturbation}
imply that $u^\ensuremath{\lambda} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, \tilde Q)$,
where $\tilde Q := \set{(x,t) : B_{(n+1)\ensuremath{\varepsilon}}(x) \times \set{t} \subset Q_q}$.
Furthermore,
since $\ensuremath{\gamma} \leq \frac1{2\Lambda}$ we note that
$u^\ensuremath{\lambda} \leq P_{aq_1, r_1}$ for all $\ensuremath{\lambda} \in [0, \Lambda]$.
Next, we introduce the time
\begin{align*}
T := \frac{3 \ensuremath{\varepsilon}^\ensuremath{\beta}}{r_1 -r_2} = c_T \ensuremath{\varepsilon}^\ensuremath{\beta} > 0.
\end{align*}
If $\ensuremath{\varepsilon}$ is small enough so that
$T \leq 1$ then $u^\ensuremath{\lambda}$ and $v$ will be
$(\ensuremath{\varepsilon}^\ensuremath{\beta} + n\ensuremath{\varepsilon})$-flat and $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat,
respectively, up to time $t = T$,
and if also
$\ensuremath{\varepsilon}^\ensuremath{\beta} \geq n\ensuremath{\varepsilon} \geq \abs{\xi_0 \cdot \nu}$
then the free boundary of $u^\ensuremath{\lambda}$
must completely overtake the free boundary of $v$
by time $T$.
\newcommand{\operatorname{Cyl}}{\operatorname{Cyl}}
Finally, we choose the constants $\ensuremath{\sigma}_1, \ensuremath{\sigma}_2$ as
\begin{align*}
\ensuremath{\sigma}_1 &:= \frac{3a \abs{q_1} + 1}{\abs{q_2} - a \abs{q_1}} \ensuremath{\varepsilon}^\ensuremath{\beta}
= c_{\ensuremath{\sigma}_1} \ensuremath{\varepsilon}^\ensuremath{\beta},\\
\ensuremath{\sigma}_2 &:= T r_1 = c_{\ensuremath{\sigma}_2} \ensuremath{\varepsilon}^\ensuremath{\beta}.
\end{align*}
Let us also denote $\ensuremath{\sigma} := \ensuremath{\sigma}_1 + \ensuremath{\sigma}_2$ and remark
that $\ensuremath{\sigma} = c_\ensuremath{\sigma} \ensuremath{\varepsilon}^\ensuremath{\beta}$.
The motivation behind this particular choice of $\ensuremath{\sigma}_1$
is the fact that
\begin{align}
\label{eq:boundsigma1}
v(x,t) - u^\ensuremath{\lambda}(x,t) \geq P_{q_2, r_2}(x,t) -P_{aq_1,r_1}(x,t) \geq \ensuremath{\varepsilon}^\ensuremath{\beta}
\end{align}
for $x \cdot \nu \leq - \ensuremath{\sigma}_1$, $t \in [0,T]$, $\ensuremath{\lambda} \leq \Lambda$.
Similarly,
$\ensuremath{\sigma}_2$ is motivated by
\begin{align}
\label{obstacle-reach}
{\cl\Omega_t(P_{a q_1,r_1}) \cup \cl\Omega_t(P^{\ensuremath{\varepsilon}^\ensuremath{\beta}}_{q_2,r_2})}
\subset \set{x : x \cdot \nu \leq \ensuremath{\sigma}_2}
\end{align}
for $t \in [0,T]$,
i.e. the free boundaries of $\overline{u}_{\ensuremath{\varepsilon};q_1,r_1}$ and
$\uu_{\ensuremath{\varepsilon};q_2,r_2}$ will be contained in
$\set{x : 0 \leq x \cdot \nu \leq \ensuremath{\sigma}_2}$ for $t \in [0,T]$.
\parahead{(2) Fast-shrinking domain}
For a point $x \in \Rd$,
we define $x^\perp$ to be the
component of $x$ orthogonal to $\nu$,
\begin{align*}
x^\perp = x - (x \cdot \nu) \nu = (I - \nu \otimes \nu) x.
\end{align*}
By symbol $\operatorname{Cyl}(\rho; \eta_1, \eta_2)$ we shall
denote the open space cylinder of radius $\rho$
with axis parallel to $\nu$,
between the hyperplanes $\set{x : x \cdot \nu = \eta_1}$ and
$\set{x: x \cdot \nu = \eta_2}$,
\begin{align*}
\operatorname{Cyl}(\rho; \eta_1, \eta_2)
= \set{x : \abs{x^\perp} < \rho,
\ \eta_1 < x \cdot \nu < \eta_2}.
\end{align*}
Let us define a ``fast-shrinking'' domain $\Sigma$,
whose each time-slice $\Sigma_t$ is the cylinder
\begin{align*}
\Sigma_t :=
\begin{cases}
\operatorname{Cyl}(\rho_\Sigma(t); -\ensuremath{\sigma}_1, \ensuremath{\sigma}_2) & t \geq 0,\\
\emptyset & t < 0.
\end{cases}
\end{align*}
with radius
\begin{align*}
\rho_\Sigma(t) = \frac{\tan\ensuremath{\theta}}2 - \ensuremath{\varepsilon}^{-\ensuremath{\alpha}} t,
\end{align*}
where $\ensuremath{\theta}$ is the opening of the cone $\Omega_q$ defining $Q_q$,
introduced in \eqref{eq:choiceofta}.
In particular, the radius $\rho_\Sigma(t)$ of the cylinder $\Sigma_t$
shrinks with velocity
$\ensuremath{\varepsilon}^{-\ensuremath{\alpha}}$.
Furthermore,
if $\ensuremath{\varepsilon}$ is small enough,
we have $\Sigma + (\cl B_{(n+1)\ensuremath{\varepsilon}} \times \set0) \subset Q_q$,
which also implies
$u^\ensuremath{\lambda} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, \Sigma)$ for $\ensuremath{\lambda} \in [0,\Lambda]$.
Additionally,
$\Sigma_t \neq \emptyset$ for all $t \in [0, T]$
since, as $\ensuremath{\alpha} < \ensuremath{\beta}$ by \eqref{alpha-beta-relation},
\begin{align*}
\rho_\Sigma(T) &= \frac{\tan\ensuremath{\theta}}2 - \ensuremath{\varepsilon}^{-\ensuremath{\alpha}} T
= \frac{\tan\ensuremath{\theta}}2 - c_T \ensuremath{\varepsilon}^{\ensuremath{\beta} - \ensuremath{\alpha}} > 0
&&\text{for $\ensuremath{\varepsilon}>0$ small.}
\end{align*}
Finally, we need to fix a small time step $\tau$,
\begin{align}
\label{lc-def-of-tau}
\tau := \frac{\ensuremath{\gamma}^2 \ensuremath{\varepsilon}^2}{4n a \ensuremath{\sigma} \abs{q_1} M} \quad
= c_\tau \ensuremath{\varepsilon}^{2-\ensuremath{\beta}}.
\end{align}
Let us explain how such choice of $\tau$ guarantees
that, for $\ensuremath{\lambda} \in [0,\Lambda]$,
the support of $u^\ensuremath{\lambda}$
cannot expand by more than $\ensuremath{\gamma} \ensuremath{\varepsilon}$
from time $\hat t - \tau$ to time $\hat t$
for any $\hat t \in [\tau, T]$ in the sense
\begin{align}
\label{expand-by-gae}
\cl\Omega_s(u^\ensuremath{\lambda})
&\subset \cl\Omega_{\hat t - \tau}(u^\ensuremath{\lambda}) + \cl B_{\ensuremath{\gamma}\ensuremath{\varepsilon}}
&&s \in [\hat t-\tau, \hat t].
\end{align}
This follows from a straightforward application of
Corollary~\ref{co:HS-expansion-speed},
on the parabolic cylinder
\[E = \set{(x,t) : -\ensuremath{\sigma}_1 < x \cdot \nu,\ t > \hat t - \tau},\]
taking advantage of the bound
\begin{align}
\label{eq:boundulambda}
u^\ensuremath{\lambda}(x,t) &\leq P_{aq_1,r_1}(x,t) \leq a \abs{q_1} \ensuremath{\sigma} \quad =: c_K \ensuremath{\varepsilon}^\ensuremath{\beta},
\end{align}
valid for $x\cdot \nu \geq -\ensuremath{\sigma}_1$ and $t \in [0, T]$
due to \eqref{obstacle-reach}.
\parahead{(3) Boundary crossing point}
With these definitions at hand,
we proceed with the proof.
As we observed above,
since $n \ensuremath{\varepsilon} \leq \ensuremath{\varepsilon}^\ensuremath{\beta}$ if $\ensuremath{\varepsilon}$ is small enough,
$u^\ensuremath{\lambda}$ is $(\ensuremath{\varepsilon}^\ensuremath{\beta} + n\ensuremath{\varepsilon})$-flat
for any $\ensuremath{\lambda} \in [0, \Lambda]$,
and $v$ is $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat,
the choice of $\xi_0$ and $T$ guarantees
that the supports of $u^\ensuremath{\lambda}$ and $v$ are originally strictly ordered,
but
will cross each other
before the time $T$ in the domain $\Sigma$.
So let $\hat t$ be the first time that
the boundaries of $u^5$ and $v$ touch in $\Sigma$,
i.e.,
\begin{align*}
\hat t := \sup \set{s : \cl\Omega(u^5) \cap
\Omega^c(v) \cap \cl\Sigma \cap \set{t \leq s} = \emptyset} < T.
\end{align*}
Observe that $\hat t \geq \tau$ since
$\cl\Omega_t(u^5)$ cannot advance more than $\ensuremath{\gamma}\ensuremath{\varepsilon}$
in time $\tau$ by \eqref{expand-by-gae}.
We claim that the set $\cl\Omega_{\hat t}(u^5) \cap
\Omega^c_{\hat t}(v) \cap \cl\Sigma_{\hat t}$
is nonempty.
Indeed, if that were not the case,
$\operatorname{dist}(\cl\Omega_{\hat t}(u^5),
\Omega^c_{\hat t}(v) \cap \cl\Sigma_{\hat t}) > 0$
because the first set is closed and the second one is compact.
Therefore, there exists $\ensuremath{\delta} > 0$ such that
the distance is positive for $t < \hat t + \ensuremath{\delta}$ by
Corollary~\ref{co:HS-expansion-speed}
and the monotonicity of $\Omega^c_t(v) \cap \cl\Sigma_t$,
which yields a contradiction with the definition of $\hat t$.
Let us thus choose
\begin{align*}
\hat x \in \cl\Omega_{\hat t}(u^5) \cap
\Omega^c_{\hat t}(v) \cap \cl\Sigma_{\hat t}.
\end{align*}
We will show that the existence of such point leads to a contradiction
with the comparison principle for the Hele-Shaw problem
since $u^5$ and $v$ were strictly separated on a much larger
domain $\cl \Sigma_{\hat t - \tau}$ at time $t = \hat t - \tau$.
To make this idea rigorous, we introduce the cylinders
\begin{align*}
S_i &= \operatorname{Cyl}(\rho_i; -\ensuremath{\sigma}_1, \ensuremath{\sigma}_2) & &\text{for } i=1,2,
\end{align*}
(see Figure~\ref{fig:localcomparison2})
\begin{figure}
\centering
\fig{Fig4}{4.5in}
\caption{Arrangement of cylinders}
\label{fig:localcomparison2}
\end{figure}
with radii
\begin{align*}
\rho_1 &:= \rho_\Sigma(\hat t - \tau) - 2 \ensuremath{\gamma} \ensuremath{\varepsilon}, &
\rho_2 &:= \rho_1 - \pth{\frac{c_K}{c_N}\ensuremath{\varepsilon}^{2\ensuremath{\beta} - 2} + 2}c_\ensuremath{\sigma} \ensuremath{\varepsilon}^\ensuremath{\beta},
\end{align*}
where constant $c_N$ is specified below in \eqref{cN}.
We point out that if $\ensuremath{\varepsilon}$ is small enough, we have
\begin{align}
\label{rho2-rhoSigma}
\rho_2 > \rho_\Sigma(\hat t) + 2 R \ensuremath{\sigma},
\end{align}
where $R$ is the constant from
Lemma~\ref{le:radial-nonlin-perturbation},
since
\[
\rho_\Sigma(\hat t - \tau) - \rho_\Sigma(\hat t) = \tau \ensuremath{\varepsilon}^{-\alpha}
= c_\tau \ensuremath{\varepsilon}^{2-\ensuremath{\alpha}-\ensuremath{\beta}} \gg \rho_{\Sigma}(\hat t - \tau) -\rho_2 + 2R\sigma
\qquad \text{if } \ensuremath{\varepsilon} \ll 1
\]
due to \eqref{alpha-beta-relation}.
In particular, we have
\begin{align*}
\Sigma_{\hat t - \tau} \supset S_1 \supset S_2 \supset \Sigma_{\hat t},
\end{align*}
see Figure~\ref{fig:localcomparison2}.
Now we consider function $u^2$.
First, clearly $\cl\Omega_t(u^5) = \cl\Omega_t(u^2) + \cl B_{3\ensuremath{\gamma}\ensuremath{\varepsilon}}(0)$.
The definition of $\hat t$ yields
$\cl\Omega_t (u^5) \cap
\Omega^c_t(v) \cap \cl\Sigma_t = \emptyset$
for $t < \hat t$,
and thus, in particular,
\begin{align*}
\pth{\cl\Omega_{\hat t - \tau} (u^2) + \cl B_{3\ensuremath{\gamma} \ensuremath{\varepsilon}}(0)}
\cap \Omega^c_{\hat t - \tau}(v)
\cap \cl\Sigma_{\hat t - \tau} = \emptyset.
\end{align*}
As $\Omega^c_t(v)$ cannot expand
(due to the monotonicity, Lemma~\ref{le:monotonicity-in-time}),
and the positive set of $u^2$ cannot expand by more than $\ensuremath{\gamma} \ensuremath{\varepsilon}$
in time $\tau$,
in the sense of \eqref{expand-by-gae},
we conclude that
\begin{align}
\label{u1-v-separated}
(\cl\Omega_t(u^2) + \cl B_{2\ensuremath{\gamma} \ensuremath{\varepsilon}}(0))
\cap \Omega^c_t (v) \cap \cl S_1 &= \emptyset
&&t \in [0, \hat t].
\end{align}
Let us define the closed sets
\begin{align*}
A_t &:= \pth{\cl\Omega_{t}(u^2) + \cl B_{\ensuremath{\gamma}\ensuremath{\varepsilon}}(0)}
\cap \cl S_1
&&t \in [0, \hat t].
\end{align*}
Due to the nondegeneracy of $v$ (Lemma~\ref{le:nondegeneracyObstacleSupersolution} applied with $\eta = \ensuremath{\varepsilon}^\ensuremath{\beta}$ and $\rho = \ensuremath{\gamma} \ensuremath{\varepsilon}$), we observe that
\begin{align}
\label{cN}
v(x,t) \geq c(r_2, \abs{q_2}, M, n) \ensuremath{\gamma}^2 \ensuremath{\varepsilon}^{2-\ensuremath{\beta}} \quad = c_N\ensuremath{\varepsilon}^{2-\ensuremath{\beta}} \qquad x \in A_t, \ t \in [0, \hat t],
\end{align}
where $c(r_2, \abs{q_2}, M, n)$ is the constant
from Lemma~\ref{le:nondegeneracyObstacleSupersolution}.
Consider the function
$w^t(x) = \frac{u^2(x,t) - v(x,t)}{c_N \ensuremath{\varepsilon}^{2-\ensuremath{\beta}}}$
defined on $A_t$.
Function $w^t$ is subharmonic on $A_t$ for all $t \in [0, \hat t]$
since $v(\cdot, t)$ is superharmonic in $\operatorname{int} A_t$ as $A^t \subset \Omega_t(v) \cap \Omega_q$,
and $u^2(\cdot, t)$ is clearly subharmonic in $\operatorname{int} A_t$.
Our goal is to apply Lemma~\ref{le:subharmonicOnThinDomain}
to function $w^t$.
Since $\ensuremath{\varepsilon}$ is small,
we have $\ensuremath{\varepsilon}^{\ensuremath{\beta}} \geq c_N \ensuremath{\varepsilon}^{2-\ensuremath{\beta}}$
and thus $w^t \leq -1$ on $\partial A_t \cap \set{x : x \cdot \nu = -\ensuremath{\sigma}_1}$
by \eqref{eq:boundsigma1}.
Additionally,
$u^2 = 0$ on $\partial A_t \cap S_1$ by definition of $A_t$
and therefore
$w^t \leq -1$ on $\partial A_t \cap S_1$.
Finally, by \eqref{eq:boundulambda},
\begin{align*}
w^t \leq
\frac{u^2}{c_N \ensuremath{\varepsilon}^{2-\ensuremath{\beta}}}
\leq \frac{c_K}{c_N} \ensuremath{\varepsilon}^{2\ensuremath{\beta} -2} \qquad \text{on } A_t.
\end{align*}
Therefore Lemma~\ref{le:subharmonicOnThinDomain}
implies that $w^t < 0$
on the set $S_1 \cap \set{x : \abs{x^\perp} \leq \rho_1'}$,
where $\rho_1'$ is given by
\begin{align*}
\rho_1' = \rho_1 - c_T
\pth{\frac{c_K}{c_N}\ensuremath{\varepsilon}^{2\ensuremath{\beta} - 2} + 2}c_\ensuremath{\sigma} \ensuremath{\varepsilon}^\ensuremath{\beta},
\end{align*}
and $\rho' \geq \rho_2$ by the choice of $\rho_1$ and $\rho_2$.
We conclude that $w^t < 0$ in $\cl{S_2} \cap A_t$
for all $t \in [0, \hat t]$,
and therefore,
with \eqref{u1-v-separated},
\begin{align}
\label{u2-prec-v}
u^2 &\prec v &&\text{ in $\cl S_2 \times [0, \hat t]$}.
\end{align}
Let $R$ and $\rho(x)$ be from
Lemma~\ref{le:radial-nonlin-perturbation} and define
\begin{align*}
x_\ensuremath{\varphi} := \hat x^\perp + (\ensuremath{\sigma}_2 + R\ensuremath{\sigma}) \nu
\end{align*}
and
\begin{align*}
\ensuremath{\varphi}(x) := \ensuremath{\gamma} \ensuremath{\varepsilon} \rho \pth{\frac{x - x_\ensuremath{\varphi}}{\ensuremath{\sigma}}}
\qquad \text{on } A_\ensuremath{\varphi} := B_{2R\ensuremath{\sigma}}(x_\ensuremath{\varphi}) \setminus \cl B_{R\ensuremath{\sigma}}(x_\ensuremath{\varphi}).
\end{align*}
Let us set
$c_\rho = \max_{\cl B_{2R}(0) \setminus B_R(0)} \abs{D\rho} < \infty$.
Clearly
\begin{align*}
\max_{\cl A_\ensuremath{\varphi}} \abs{D\ensuremath{\varphi}}
= \frac{\ensuremath{\gamma} \ensuremath{\varepsilon}}{\ensuremath{\sigma}} c_\rho
= \frac{c_\rho}{c_\ensuremath{\sigma}} \ensuremath{\gamma} \ensuremath{\varepsilon}^{1-\ensuremath{\beta}}.
\end{align*}
Hence for $\ensuremath{\varepsilon}$ small enough we have
\begin{align}
\label{smallness-Dvp}
(1 - \max_{\cl A_\ensuremath{\varphi}} \abs{D\ensuremath{\varphi}})^{-2} \leq a^{1/2}.
\end{align}
Moreover, $\ensuremath{\sigma}_2 < \ensuremath{\sigma}$ yields
$\hat x \in \cl B_{(R+1) \ensuremath{\sigma}}(x_\ensuremath{\varphi}) \setminus B_{R\ensuremath{\sigma}}(x_\ensuremath{\varphi})$
and hence,
by the definition of $\rho$,
we also have $\ensuremath{\varphi}(\hat{x}) > 5\ensuremath{\gamma}\ensuremath{\varepsilon}$.
\parahead{(4) Non-uniform perturbation}
In the final step, we consider the nonuniform perturbation $u^\ensuremath{\varphi}$
of $\overline{u}_{\e;q,r}$,
\begin{align*}
u^\ensuremath{\varphi}(x,t) := \sup_{y \in \cl B_{\ensuremath{\varphi}(x)}(x)}
a \overline{u}_{\ensuremath{\varepsilon};aq_1, r_1}(y - \xi_0, t)
\qquad (x,t) \in A_\ensuremath{\varphi} \times (0, \hat t].
\end{align*}
This function is a subsolution $u^\ensuremath{\varphi} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_\ensuremath{\varphi})$
on $Q_\ensuremath{\varphi} := A_\ensuremath{\varphi} \times (0,\hat t]$
for all $\ensuremath{\varepsilon}$ small enough
so that \eqref{smallness-Dvp} holds,
due to Proposition~\ref{pr:nonuniform-perturbation}
and the choice of $\gamma$ in \eqref{choice-gamma}.
We want to show that $u^\ensuremath{\varphi} \prec v$ on
the parabolic boundary
$\partial_P (A_\ensuremath{\varphi} \times (0, \hat t])$
to be able to use the comparison theorem for the Hele-Shaw problem:
\begin{compactitem}
\item
$u^\ensuremath{\varphi} \prec v$ on $A_\ensuremath{\varphi} \times \set{0}$
by the choice of $\xi_0$,
which implies $u^\Lambda \prec v$ on $A_\ensuremath{\varphi} \times \set{0}$,
and $u^\ensuremath{\varphi} \leq u^\Lambda$.
\item
$\cl\Omega(u^\ensuremath{\varphi}; Q_\ensuremath{\varphi})
\cap \pth{\partial B_{R\ensuremath{\sigma}}(x_\ensuremath{\varphi}) \times [0,\hat t]} = \emptyset$
since $\cl B_{R \ensuremath{\sigma}} (x_\ensuremath{\varphi}) \subset
\set{x: x\cdot\nu\geq \ensuremath{\sigma}_2}$, recalling \eqref{eq:boundulambda}
and \eqref{obstacle-reach}.
\item
To show $u^\ensuremath{\varphi} \prec v$ on
$\partial B_{2R\ensuremath{\sigma}}(x_\ensuremath{\varphi}) \times [0, \hat t]$ w.r.t. $Q_\ensuremath{\varphi}$,
we split $\partial B_{2R\ensuremath{\sigma}}(x_\ensuremath{\varphi})$ into three parts.
By construction of $S_2$
and \eqref{rho2-rhoSigma},
we have
\[
\partial B_{2R\ensuremath{\sigma}}(x_\ensuremath{\varphi}) \subset \set{x : x\cdot \nu \leq - \ensuremath{\sigma}_1}
\cup S_2 \cup \set{x: x\cdot\nu \geq \ensuremath{\sigma}_2}.
\]
But the strict separation $u^\ensuremath{\varphi} \prec v$
follows from $u^\ensuremath{\varphi} = u^1$ on $\partial B_{2R\ensuremath{\sigma}(x_\ensuremath{\varphi})} \times [0,\hat t]$,
and then \eqref{eq:boundsigma1} for the first part,
\eqref{u2-prec-v} for the second part,
and \eqref{eq:boundulambda} with \eqref{obstacle-reach}
for the third part.
\end{compactitem}
Therefore $u^\ensuremath{\varphi} \prec v$ on $\cl A_\ensuremath{\varphi} \times [0,\hat t]$
by the comparison theorem, Theorem~\ref{th:comparison}.
This is however a contradiction with
\begin{align*}
\hat x \in \cl \Omega_{\hat t}(u^\ensuremath{\varphi}) \cap \Omega^c_{\hat t}(v),
\end{align*}
since $u^\ensuremath{\varphi}(\hat x, \hat t) > 0$
due to $\hat x \in \cl \Omega_{\hat t}(u^5)$
and $\ensuremath{\varphi}(\hat x) > 5 \ensuremath{\gamma} \ensuremath{\varepsilon}$.
This finishes the proof of the local comparison principle.
\qedhere\end{proof}
\subsection{Cone flatness}
\label{sec:cone-flatness}
In this part
we show that the positive and the zero sets of the obstacle solutions $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$
cannot form long, thin fingers in the direction orthogonal to the obstacle, and, on the contrary, that their free boundaries are in fact in between
two cones that are $\sim \ensuremath{\varepsilon} \abs{\ln \ensuremath{\varepsilon}}^{1/2}$ apart.
We refer to this property as \emph{cone flatness}.
It is a consequence of the monotonicity of the obstacle problem and our particular choice of the domain $Q_q$
for the obstacle problem that allows us to control how fast the free boundaries of $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ detach from the obstacle at the boundary of the domain $Q_q$; see also the discussion in previous sections.
We have the following result:
\begin{proposition}[Cone flatness]
\label{pr:cone-flatness}
Let $q$ and $r$ satisfy \eqref{rqrestriction}
and let $T > 0$.
There exist positive constants $K$ and $\ensuremath{\varepsilon}_0$,
both depending only on $(n, m, M, \abs q, T)$,
such that
for any $(\zeta, \ensuremath{\sigma}) \in \Omega^c(\uu_{\e;q,r})$, $\ensuremath{\sigma} \leq T$ and $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
\begin{align*}
{\mathrm{Cone}}_{\nu, \ensuremath{\theta}^+}(\zeta + K \ensuremath{\varepsilon} \abs{\ln \ensuremath{\varepsilon}}^{1/2} \nu)
\times (-\infty, \ensuremath{\sigma}] \subset \Omega^c(\uu_{\e;q,r}),
\end{align*}
and similarly for any $(\zeta,\sigma) \in \Omega^c(\overline{u}_{\e;q,r})$
and $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
\begin{align*}
{\mathrm{Cone}}_{\nu, \ensuremath{\theta}^-}(\zeta + K \ensuremath{\varepsilon} \abs{\ln \ensuremath{\varepsilon}}^{1/2} \nu)
\times (-\infty, \ensuremath{\sigma}]
\subset \Omega^c(\overline{u}_{\e;q,r}).
\end{align*}
(Note the different opening angles $\ensuremath{\theta}^+$ and $\ensuremath{\theta}^-$, introduced in Section~\ref{sec:domain-geometry}.)
\end{proposition}
\begin{proof}
Before proceeding with the detailed proof let us give its brief outline. We use $u$ to denote either $\uu_{\e;q,r}$ or $(\overline{u}_{\e;q,r})_*$.
\begin{compactenum}[Step 1.]
\item A supersolution $u \in \overline{\mathcal{S}}(m, Q_q)$, harmonic in its positive set,
can be bounded from above using Proposition~\ref{pr:solUpperBoundZeroSet}.
\item The monotonicity property,
Proposition~\ref{pr:monotonicity}, implies that $u$
is in fact small in the whole cone.
\item A barrier prevents $u$ from becoming positive
too far from the boundary of the cone.
\end{compactenum}
\medskip
\emph{Step 1.} For the sake of brevity, let us denote $\uu = \uu_{\e;q,r}$.
By Remark~\ref{rem:grid-approx-cone}
there exists $\ensuremath{\lambda} = \ensuremath{\lambda}(n, \ensuremath{\theta}^+) > 0$
such that
\begin{align}
\label{cone-ball-cover}
{{\mathrm{Cone}}_{\nu,\ensuremath{\theta}^+}(x)} \subset
\pth{{\mathrm{Cone}}_{\nu,\theta^+}(x)}^{{::}\ensuremath{\varepsilon}} + B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(0)
\qquad \text{for any $x \in \Rd$, $\ensuremath{\varepsilon} > 0$,}
\end{align}
where the notation ${::}\ensuremath{\varepsilon}$
is introduced in Definition~\ref{def:grid}.
Pick $(\zeta, \ensuremath{\sigma}) \in \Omega^c(\uu) \cap \set{t \leq T}$.
Note that $\uu$ is in $\overline{\mathcal{S}}(m, Q_q)$,
nondecreasing in time by Lemma~\ref{le:monotonicity-in-time}
and harmonic in $\Omega_t(\uu; Q_q)$ by
Proposition~\ref{pr:harmonic}.
Therefore
Proposition~\ref{pr:solUpperBoundZeroSet} yields
\begin{align}
\label{rational-bound}
\uu(z,t) &\leq \frac Cm \frac{\abs{z - \zeta}^2}{\ensuremath{\sigma} - t}
\leq \frac Cm \frac{\ensuremath{\lambda}^2 \ensuremath{\varepsilon}^2}{\ensuremath{\sigma} - t}
&& \text{for all } z \in \cl B_{\ensuremath{\lambda} \ensuremath{\varepsilon}} (\zeta),\ t < \ensuremath{\sigma}.
\end{align}
\emph{Step 2.} By the monotonicity result of Proposition~\ref{pr:monotonicity},
we have
\begin{align}
\label{epsilon-monotonicity}
\uu(z + y,t) \leq \uu(z, t - \ensuremath{\varepsilon})
\end{align}
for all $(z,t) \in \Rd\times\ensuremath{\mathbb{R}}$ and
\begin{align}
\label{y-in-cone}
y \in
\pth{{{\mathrm{Cone}}_{q,\theta^+}} (c\ensuremath{\varepsilon} \nu)}^{::\e},
\end{align}
where $c = \max (M\abs q, r_V^+)$.
\eqref{cone-ball-cover} implies
${{\mathrm{Cone}}_{\nu,\theta^+}} (c\ensuremath{\varepsilon} \nu + \zeta)
\subset \zeta + \pth{{{\mathrm{Cone}}_{\nu,\theta^+}} (c\ensuremath{\varepsilon} \nu)}^{::\e} + B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(0)$
and therefore for any
$x \in {\mathrm{Cone}}_{\nu,\theta^+} (c\ensuremath{\varepsilon} \nu + \zeta)$
there exist $z \in B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta)$ and $y$ satisfying \eqref{y-in-cone},
and,
combining \eqref{epsilon-monotonicity} and \eqref{rational-bound},
we obtain
\begin{align}
\label{rational-bound-in-cone}
\uu(x,t) = \uu(z + y, t) \leq
\uu(z,t - \ensuremath{\varepsilon})
\leq \frac{C \ensuremath{\lambda}^2}{m} \frac{\ensuremath{\varepsilon}^2}{\ensuremath{\varepsilon} + \ensuremath{\sigma} - t}.
\end{align}
\emph{Step 3.} Let us now choose $\mu$ satisfying
\begin{align*}
\frac{m\mu^2}{2nMC\ensuremath{\lambda}^2\ensuremath{\varepsilon}^2} = -2 \ln \ensuremath{\varepsilon},
\end{align*}
i.e.,
\begin{align*}
\mu = \tilde K \abs{\ln \ensuremath{\varepsilon}}^{1/2} \ensuremath{\varepsilon}.
\end{align*}
Hence, provided that $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0 = \ensuremath{\varepsilon}_0(T)$,
we have
\begin{align}
\label{eps-and-T}
\ensuremath{\varepsilon} \pth{e^{\frac{m \mu^2}{2nM C\ensuremath{\lambda}^2\ensuremath{\varepsilon}^2}} - 1} =
\ensuremath{\varepsilon} (\ensuremath{\varepsilon}^{-2} - 1) > T.
\end{align}
We choose $K$ such that for all $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
\begin{align*}
\operatorname{dist}\pth{{\mathrm{Cone}}_{\nu,\ensuremath{\theta}^+}(K \abs{\ln\ensuremath{\varepsilon}}^{1/2}\ensuremath{\varepsilon} \nu),
\partial {\mathrm{Cone}}_{\nu,\ensuremath{\theta}^+}(c\ensuremath{\varepsilon}\nu)} > \mu.
\end{align*}
Consequently,
whenever $\xi \in {\mathrm{Cone}}_{\nu, \ensuremath{\theta}^+}(\zeta + K \ensuremath{\varepsilon} \abs{\ln \ensuremath{\varepsilon}}^{1/2} \nu)$
we also have
$\cl B_\mu(\xi) \subset {\mathrm{Cone}}_{\nu, \ensuremath{\theta}^+}(\zeta + c\ensuremath{\varepsilon}\nu)$
and therefore the bound \eqref{rational-bound-in-cone}
for
$\uu$
holds in $\cl B_\mu(\xi) \times [0, \ensuremath{\sigma}]$.
Since $\ensuremath{\sigma} \leq T$ and \eqref{eps-and-T} holds,
we set $A = \frac{C\ensuremath{\lambda}^2\ensuremath{\varepsilon}^2}{m}$ and
apply Corollary~\ref{co:rational-contract-bound}
to conclude that $\uu(\xi, t) = 0$ for $t \in [0,\ensuremath{\sigma}]$.
This finishes the proof for $\uu_{\e;q,r}$.
The proof for $\overline{u} = \overline{u}_{\e;q,r}$ is analogous since
$\overline{u} \in \overline{\mathcal{S}}(m, Q_q)$ due to Proposition~\ref{pr:obstacle-sols}(e),
$(\overline{u}_{\e;q,r})_*$ is harmonic in the positive phase
by Proposition~\ref{pr:harmonic},
and the correct monotonicity holds due to
Proposition~\ref{pr:monotonicity}
with a cone ${{\mathrm{Cone}}_{\nu,\theta^-}}$.
\qedhere\end{proof}
\begin{remark}
\label{rem:reversed-cone-flatness}
The statement of Proposition~\ref{pr:cone-flatness}
can be reversed in the following sense:
For $q$, $r$, $T$, $K$ and $\ensuremath{\varepsilon}_0$ from Proposition~\ref{pr:cone-flatness}
we have that
if
$(\zeta,\sigma) \in \Omega(\uu_{\e;q,r})$, $\ensuremath{\sigma}\leq T$
and $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
then
\begin{align}
\label{reversed-flatness}
{\mathrm{Cone}}_{-\nu, \ensuremath{\theta}^+}(\zeta - K \ensuremath{\varepsilon} \abs{\ln\ensuremath{\varepsilon}}^{\frac12}\nu) \times
[\ensuremath{\sigma}, +\infty) \subset \Omega(\uu_{\e;q,r}).
\end{align}
The proof is very simple:
suppose that the inclusion \eqref{reversed-flatness} is violated,
i.e., we have $(\zeta, \ensuremath{\sigma}) \in \Omega(\uu_{\e;q,r})$
for which there exists $(y,s)$
that belongs to the left-hand side of \eqref{reversed-flatness}
but belongs also to $\Omega^c(\uu_{\e;q,r})$.
Since $s \geq \ensuremath{\sigma}$, the monotonicity in time, Lemma~\ref{le:monotonicity-in-time},
implies that $(y, \ensuremath{\sigma}) \in \Omega^c(\uu_{\e;q,r})$
and we can apply Proposition~\ref{pr:cone-flatness}
and conclude that
$(\xi, \sigma) \in \Omega^c(\uu_{\e;q,r})$
since $\xi\in
{\mathrm{Cone}}_{\nu, \ensuremath{\theta}^+}(y + K \ensuremath{\varepsilon} \abs{\ln \ensuremath{\varepsilon}}^{1/2} \nu)$.
But that is obviously a contradiction.
A similar reflection argument applies for $\lu_{\e;q,r}$ as well.
\end{remark}
Since it is easier to work with a ball instead of a cone,
we also formulate the following consequence of
Proposition~\ref{pr:cone-flatness}.
\begin{corollary}
\label{co:ball-outlier}
Let $q$ and $r$ satisfy \eqref{rqrestriction}
and let $\ensuremath{\lambda} > 0$
and $\ensuremath{\beta} \in [0, 1)$.
There exists $\ensuremath{\varepsilon}_0 = \ensuremath{\varepsilon}_0(n, m, M, q, \ensuremath{\lambda},\ensuremath{\beta})$
such that for all $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
the following statements hold:
\begin{itemize}
\item
if $\underline\Phi_{\e;q,r}(1) > \frac34 \ensuremath{\varepsilon}^\ensuremath{\beta}$
then there exists $(\zeta,\ensuremath{\sigma})$ such that
$\zeta \cdot \nu = r \ensuremath{\sigma} + \frac12 \ensuremath{\varepsilon}^\ensuremath{\beta}$,
$\underline\Phi_{\e;q,r}(\ensuremath{\sigma} + \ensuremath{\lambda} \ensuremath{\varepsilon}) \leq \ensuremath{\varepsilon}^\ensuremath{\beta}$
and
\begin{align*}
B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma}) \subset
\Omega(\uu_{\e;q,r}) \cap
Q_q \cap C^+ \cap \set{t \leq 1};
\end{align*}
\item
if $\overline\Phi_{\e;q,r}(1) > \frac34 \ensuremath{\varepsilon}^\ensuremath{\beta}$
then there exists $(\zeta,\ensuremath{\sigma})$ such that
$\zeta \cdot \nu = r\ensuremath{\sigma} - \frac12 \ensuremath{\varepsilon}^\ensuremath{\beta}$,
$\overline\Phi_{\e;q,r}(\ensuremath{\sigma} + \ensuremath{\lambda} \ensuremath{\varepsilon}) \leq \ensuremath{\varepsilon}^\ensuremath{\beta}$
and
\begin{align*}
B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma}) \subset
\Omega^c(\lu_{\e;q,r}) \cap
Q_q \cap C^+ \cap \set{t \leq 1}.
\end{align*}
\end{itemize}
\end{corollary}
\begin{proof}
Fix $q$ and $r$.
For $T = 1$
we have constants $\ensuremath{\varepsilon}_0$ and $K$ from Proposition~\ref{pr:cone-flatness}.
Let us define
\begin{align}
\label{def-eta}
\eta := \inf \set{s : B_\ensuremath{\lambda}(s \nu) \subset {\mathrm{Cone}}_{\nu,\ensuremath{\theta}^+}(0)}
= \frac{\ensuremath{\lambda}}{\sin \ensuremath{\theta}^+}.
\end{align}
Because $\ensuremath{\beta} <1$,
we may assume, making $\ensuremath{\varepsilon}_0$ smaller if necessary,
that for all $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
\begin{align}
\label{shazam}
\max \set{r\ensuremath{\lambda}\ensuremath{\varepsilon} + \eta \ensuremath{\varepsilon} + K \ensuremath{\varepsilon} \abs{\ln\ensuremath{\varepsilon}}^{\frac12},
2\ensuremath{\lambda} \ensuremath{\varepsilon} (M-m)\abs q}
< \frac18 \ensuremath{\varepsilon}^\ensuremath{\beta}.
\end{align}
Lemma~\ref{le:furthest-point}
provides a point $(\hat \zeta,\hat \ensuremath{\sigma}) \in \partial \Omega(\uu_{\e;q,r})$
such that $\hat \zeta = r \hat \sigma + \frac58 \ensuremath{\varepsilon}^\ensuremath{\beta}$
and $\underline\Phi_{\e;q,r}(\hat\ensuremath{\sigma}) = \frac58 \ensuremath{\varepsilon}^\ensuremath{\beta}$.
We define
\begin{align*}
\zeta &= \hat \zeta + (r\ensuremath{\lambda}\ensuremath{\varepsilon} - \frac18 \ensuremath{\varepsilon}^\ensuremath{\beta}) \nu, &
\sigma &= \hat \sigma + \ensuremath{\lambda} \ensuremath{\varepsilon}.
\end{align*}
Clearly $\zeta \cdot \nu = r \sigma + \frac12 \ensuremath{\varepsilon}^\ensuremath{\beta}$.
Furthermore,
due to \eqref{shazam}, the definition of $\eta$ in
\eqref{def-eta}
and Remark~\ref{rem:reversed-cone-flatness},
we have
\begin{align*}
B_{\ensuremath{\lambda} \ensuremath{\varepsilon}} (\zeta, \ensuremath{\sigma})
\in {\mathrm{Cone}}_{-\nu,\ensuremath{\theta}^+} (\hat \zeta - K \ensuremath{\varepsilon} \abs{\ln\ensuremath{\varepsilon}}^{\frac12}\nu)
\times [\hat \ensuremath{\sigma}, +\infty)
\subset \Omega(\uu_{\e;q,r}),
\end{align*}
see Figure~\ref{fig:ball-outlier}.
\begin{figure}
\centering
\fig{Fig5}{4.5in}
\caption{Construction of the outlying ball in Corollary~\ref{co:ball-outlier}}
\label{fig:ball-outlier}
\end{figure}
Since $B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta,\ensuremath{\sigma}) \subset \Omega^c(P_{q,r})$,
Proposition~\ref{pr:obstacle-sols}(d)
implies $B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta,\sigma) \subset Q_q \cap C^+$.
Finally,
Proposition~\ref{pr:phi-lipschitz}
and \eqref{shazam}
imply that
\begin{align*}
\underline\Phi_{\e;q,r}(\ensuremath{\sigma}+\ensuremath{\lambda}\ensuremath{\varepsilon}) =
\underline\Phi_{\e;q,r}(\hat\ensuremath{\sigma}+2\ensuremath{\lambda}\ensuremath{\varepsilon})
&\leq \underline\Phi_{\e;q,r}(\hat\ensuremath{\sigma}) + 2\ensuremath{\lambda}\ensuremath{\varepsilon}(M\abs q - r)_+
\\&\leq \frac58\ensuremath{\varepsilon}^\ensuremath{\beta} + 2\ensuremath{\lambda}\ensuremath{\varepsilon}(M-m)\abs q < \frac34 \ensuremath{\varepsilon}^\ensuremath{\beta}.
\end{align*}
In particular, $\underline\Phi_{\e;q,r}(\ensuremath{\sigma} + \ensuremath{\lambda}\ensuremath{\varepsilon}) \leq \ensuremath{\varepsilon}^\ensuremath{\beta}$
and $\ensuremath{\sigma} + \ensuremath{\lambda} \ensuremath{\varepsilon} \leq 1$, that is,
$B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta,\sigma) \subset \set{t\leq 1}$.
This finishes the proof for $\uu_{\e;q,r}$.
The proof for $\lu_{\e;q,r}$ is analogous.
\qedhere\end{proof}
\subsection{Detachment lemma}
\label{sec:detachment}
The main consequence of the cone flatness property of the previous section (Proposition~\ref{pr:cone-flatness}), combined
with the monotonicity of solutions
across scales (Proposition~\ref{pr:monotonicity}),
is the detachment lemma, presented in this section.
Heuristically speaking, we can conclude that if
the solution of the obstacle problem is not $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat on some scale,
the boundary of the solution on a smaller scale is completely detached
from the obstacle
by a fraction of $\ensuremath{\varepsilon}^\ensuremath{\beta}$
on a substantial portion of $Q_q$.
\begin{lemma}[Detachment]
\label{le:detachment-ball}
For every $q$ and $r$ satisfying \eqref{rqrestriction}
and $\ensuremath{\beta} \in (0,1)$,
there exist $\ensuremath{\gamma} \in (0,1)$ and $\ensuremath{\varepsilon}_0 > 0$ such that
for $\mu := \frac{1}{4\sqrt{1 +r^2}}$:
\begin{itemize}
\item
if
$\overline\Phi_{\e;q,r}(1) > \frac34\ensuremath{\varepsilon}^\ensuremath{\beta}$ for some $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
then
\begin{align}
\label{detachment-expansion-set-lu}
\lu_{\ensuremath{\gamma} \ensuremath{\varepsilon}; q,r} &= 0
&\text{in} \quad &
B_{\ensuremath{\gamma}\mu \ensuremath{\varepsilon}^\ensuremath{\beta}} (r \nu, 1);
\end{align}
\item
if
$\underline\Phi_{\e;q,r}(1) > \frac34\ensuremath{\varepsilon}^\ensuremath{\beta}$ for some $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$
then
\begin{align}
\label{detachment-expansion-set-uu}
\uu_{\ensuremath{\gamma} \ensuremath{\varepsilon}; q,r} &> 0
&\text{in} \quad &
B_{\ensuremath{\gamma}\mu\ensuremath{\varepsilon}^\ensuremath{\beta}} (r \nu, 1).
\end{align}
\end{itemize}
\end{lemma}
\begin{proof}
We give the proof for $\uu_{\e;q,r}$.
The proof for $\overline{u}_{\e;q,r}$ is analogous.
The basic idea of the proof is quite straightforward.
If $\uu_{\e;q,r}$ is not $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat for sufficiently small $\ensuremath{\varepsilon}$,
the cone flatness proposition implies
that there must be a ball of sufficiently large radius $\ensuremath{\lambda} \ensuremath{\varepsilon}$
with a center
still $\frac{1}{2}\ensuremath{\varepsilon}^\ensuremath{\beta}$ away from the obstacle in the direction of $\nu$,
and $\uu_{\e;q,r}$ is positive on this ball.
Then by looking at a smaller scale $\ensuremath{\gamma} \ensuremath{\varepsilon}$,
$\ensuremath{\gamma} < 1$,
and using the monotonicity property,
we can cover the ball given in \eqref{detachment-expansion-set-uu}
by the translates of the rescaled ball of radius $\ensuremath{\gamma} \ensuremath{\lambda} \ensuremath{\varepsilon}$
and conclude that $\uu_{\ensuremath{\gamma} \ensuremath{\varepsilon}; q, r}$ must be positive in that set.
These translates cover the ball $B_{\gamma\mu\ensuremath{\varepsilon}^\ensuremath{\beta}}(r\nu,1)$.
For any $\ensuremath{\gamma} \in (0,1)$, let us define the sets
\begin{align*}
E &:= \set{(x,t) : x\cdot\nu = rt + \frac12 \ensuremath{\varepsilon}^\ensuremath{\beta}} \cap
Q_q \cap \set{t \leq 1}, \\
H &:= \set{(x,t) : x\cdot \nu \leq r t},\qquad
S_\ensuremath{\gamma} := \bra{(1- \gamma) \cl C^+} \cap \set{t \geq 0}.
\end{align*}
Recall that if $-(y,\tau) \in H \cap S_\ensuremath{\gamma}$
then $(y,\tau)$ satisfies \eqref{y-in-taplus-cone} (with $a = \ensuremath{\gamma}$).
We set $\ensuremath{\lambda} = \sqrt{n + 1}$ and
note that the choice of $\ensuremath{\lambda}$
guarantees $\ensuremath{\mathbb{Z}}^{n+1} + B_\ensuremath{\lambda} = \ensuremath{\mathbb{R}}^{n+1}$.
We set $\mu = \frac1{4\sqrt{1 + r^2}}$
and observe that
\begin{align*}
B_{\ensuremath{\gamma}\mu\rho}(r\nu,1) \subset
B_{\ensuremath{\gamma}\mu\rho}(r\nu,1) + H \subset H + \frac{\gamma\rho}4 (\nu,0)
\end{align*}
for any $\rho > 0$.
Take $\ensuremath{\varepsilon}_0$ from Corollary~\ref{co:ball-outlier}.
By making $\ensuremath{\varepsilon}_0$ smaller if necessary,
we can assume that
\begin{align*}
\ensuremath{\lambda} \ensuremath{\varepsilon} \leq \mu \ensuremath{\varepsilon}^\ensuremath{\beta} \qquad \text{for } \ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0.
\end{align*}
This guarantees,
since $-\gamma E \subset H - \frac\gamma2\ensuremath{\varepsilon}^\ensuremath{\beta}(\nu,1)$,
that
\begin{align}
\label{gamma-cons-1}
B_{\ensuremath{\gamma}\mu \ensuremath{\varepsilon}^\ensuremath{\beta}}(r\nu, 1) + B_{\ensuremath{\gamma} \ensuremath{\lambda} \ensuremath{\varepsilon}}(0,0)
- \ensuremath{\gamma} E
=B_{\ensuremath{\gamma}\mu \ensuremath{\varepsilon}^\ensuremath{\beta} + \ensuremath{\gamma} \ensuremath{\lambda} \ensuremath{\varepsilon}}(r\nu, 1)
- \ensuremath{\gamma} E
\subset H
\end{align}
for all $\ensuremath{\gamma} > 0$, $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$.
We claim that we can find $\ensuremath{\gamma} \in (0,1)$ such that,
for all $\ensuremath{\varepsilon} \in (0,\ensuremath{\varepsilon}_0)$,
\begin{align}
\label{gamma-cons-2}
B_{\ensuremath{\gamma}\mu \ensuremath{\varepsilon}^\ensuremath{\beta}}(r\nu, 1) + B_{\ensuremath{\gamma} \ensuremath{\lambda} \ensuremath{\varepsilon}}(0,0)
- \ensuremath{\gamma} E
\subset S_\ensuremath{\gamma}.
\end{align}
This is indeed possible by choosing $\ensuremath{\gamma}$ small,
because
the set on the left-hand side is bounded,
contained in a ball $B_{\ensuremath{\gamma} R}(r\nu, 1)$ for some $R > 0$,
independent of $\ensuremath{\varepsilon} < 1$,
and $(r \nu, 1)$ is in the interior of $S_\ensuremath{\gamma}$ for all $\ensuremath{\gamma} < 1$.
Such choice of $\ensuremath{\gamma}$ and $\ensuremath{\varepsilon}_0$
that provides \eqref{gamma-cons-1} and \eqref{gamma-cons-2}
then assures by Lemma~\ref{le:grid-cover} that
\begin{align}
\label{choice-gamma-cons}
B_{\ensuremath{\gamma} \mu \ensuremath{\varepsilon}^\ensuremath{\beta}} (r\nu, 1) &\subset (S_\gamma \cap H)^{{::}\ensuremath{\gamma}\ensuremath{\varepsilon}}
+ B_{\ensuremath{\gamma}\ensuremath{\lambda}\ensuremath{\varepsilon}}(0,0)
+\ensuremath{\gamma}(\zeta,\ensuremath{\sigma})
\end{align}
for any $\ensuremath{\varepsilon}< \ensuremath{\varepsilon}_0$ and $(\zeta, \ensuremath{\sigma}) \in E$,
where the notation ${::}\ensuremath{\gamma}\ensuremath{\varepsilon}$ is introduced in
Definition~\ref{def:grid}.
By Corollary~\ref{co:ball-outlier} we can find $(\zeta,\ensuremath{\sigma}) \in E$ such that
\begin{align}
\label{z-sigma}
\uu_{\e;q,r} &> 0 \text{ in }
B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta,\ensuremath{\sigma}).
\end{align}
Additionally, any $(y,\tau) \in -(S_\ensuremath{\gamma} \cap H)^{{::}\ensuremath{\gamma}\ensuremath{\varepsilon}}$
satisfies the condition \eqref{y-in-taplus-cone}
with $a = \ensuremath{\gamma}$
and also $(y,\tau) \in \ensuremath{\gamma}\ensuremath{\varepsilon} \ensuremath{\mathbb{Z}}^{n+1}$.
Hence, by the monotonicity proposition~\ref{pr:monotonicity},
\begin{align*}
\uu_{\ensuremath{\gamma} \ensuremath{\varepsilon}; q, r} (x,t)
\geq \ensuremath{\gamma}\uu_{\e;q,r}\pth{\frac{x + y}{\ensuremath{\gamma}}, \frac{t + \tau}{\ensuremath{\gamma}}}
> 0
\end{align*}
for all $(x,t)$ such that
$\ensuremath{\gamma}^{-1}\pth{x + y, t + \tau} \in B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma})$;
this can be rewritten as
\begin{align*}
(x,t) \in (S_\gamma \cap H)^{{::}\ensuremath{\gamma}\ensuremath{\varepsilon}} + B_{\ensuremath{\gamma} \ensuremath{\lambda} \ensuremath{\varepsilon}}(0,0)
+ \ensuremath{\gamma} (\zeta,\ensuremath{\sigma}).
\end{align*}
Therefore $\uu_{\ensuremath{\gamma}\ensuremath{\varepsilon};q,r} > 0$ in $B_{\ensuremath{\gamma}\mu\ensuremath{\varepsilon}^\ensuremath{\beta}}(r\nu,1)$
by \eqref{choice-gamma-cons}.
\qedhere\end{proof}
\subsection{Homogenized velocity}
\label{sec:homogenized-velocity}
By now we have laid enough groundwork to be able to introduce reasonable candidates for
the homogenized velocity $r(Du)$.
The two main items in our toolbox, the local comparison principle (Theorem~\ref{th:localComparison})
and the detachment lemma (Lemma~\ref{le:detachment-ball}),
both work at the critical flatness rate $\ensuremath{\varepsilon}^\beta$ for $\beta$ in the interval $(\frac45, 1)$.
We therefore fix an exponent $\ensuremath{\beta} \in (\frac45, 1)$ in the rest of the paper
and make the educated guess that if $r$ is a good candidate for the homogenized velocity at
a given gradient $q \in \ensuremath{\mathbb{R}}^n \setminus \set0$,
then the flatness of both $\uu_{\e;q,r}$ and $\overline{u}_{\e;q,r}$ decreases \emph{faster} than $\ensuremath{\varepsilon}^\beta$ as $\ensuremath{\varepsilon} \to 0+$.
This consideration leads us to the two candidates for the homogenized velocity:
given $q \in \Rd \setminus \set0$,
we define the lower and upper homogenized velocities
\begin{subequations}
\label{homogenized-velocities}
\begin{align}
\lr(q) &:= \sup \set{r > 0 :
\limsup_{\ensuremath{\varepsilon}\to0} \ensuremath{\varepsilon}^{-\ensuremath{\beta}} \overline\Phi_{\e;q,r}(1) < 1},\\
\ur(q) &:= \inf \set{r > 0 :
\limsup_{\ensuremath{\varepsilon}\to0} \ensuremath{\varepsilon}^{-\ensuremath{\beta}} \underline\Phi_{\e;q,r}(1) < 1}.
\end{align}
\end{subequations}
If $q = 0$ we set $\lr(0) = \ur(0) = 0$.
Heuristically speaking,
if $r > \lr(q)$ then
$\overline{u}_{\e;q,r}$ does not have $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat free boundary
up to time $1$ for a subsequence of $\ensuremath{\varepsilon}$ converging to $0$.
Similarly, if $r < \ur(q)$ then
$\uu_{\e;q,r}$ does not have $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat free boundary up to time $1$
for a subsequence of $\ensuremath{\varepsilon}$ converging to $0$.
In this and in the following subsection
we show that the two candidates $\lr$ and $\ur$ give rise to the same homogenized velocity $r$
(in the sense of Proposition~\ref{pr:r-semicontinuity}).
In this section we show that they are well defined and satisfy some basic monotonicity properties.
\begin{proposition}
\label{pr:bound-on-r}
For all $q \in \Rd$, the homogenized velocities
$\overline{r}(q)$ and $\underline{r}(q)$
introduced in \eqref{homogenized-velocities}
are well defined.
Moreover,
\begin{align*}
\overline{r}(q) &\in [m\abs{q}, M\abs{q}],
& \underline{r}(q) &\in [m\abs{q}, M\abs{q}].
\end{align*}
\end{proposition}
\begin{proof}
We will show the statement for $\lr(q)$,
the proof for $\ur(q)$ is similar.
First, we consider $r \in (0, m\abs{q}]$.
Due to Proposition~\ref{pr:planar-solution-range},
$P_{q,r} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$
and therefore $\lu_{\e;q,r} = P_{q,r}$, by definition.
In particular, $\overline\Phi_{\e;q,r} \equiv 0$
and we conclude that $\lr(q) \geq m\abs{q}$.
On the other hand, if $r > M \abs{q}$,
we can find $r_2$, $q_2$ such that
$q_2 = a^3 q$ for some $a > 1$ and $M \abs{q_2} = r_2 < r_1$.
Since $P_{q_2,r_2} \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$,
we now have $\uu_{\e;q,r} = P_{q_2, r_2}$, which is $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat
for all $\ensuremath{\varepsilon} > 0$.
Therefore, by the local comparison, Theorem~\ref{th:localComparison},
there exists $\ensuremath{\varepsilon}_0$ such that
$\overline{u}_{\e;q,r}$ cannot be $\ensuremath{\varepsilon}^\ensuremath{\beta}$-flat for any $\ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0$.
We conclude that $\lr(q) \leq M \abs{q}$.
\qedhere\end{proof}
$\lr$ and $\ur$ also enjoy the following monotonicity property.
\begin{proposition}
For any $a > 1$ and $q \in \Rd$ we have
\begin{align*}
\lr(q) &\leq \lr(a q) & \ur(q) &\leq \ur(aq).
\end{align*}
\end{proposition}
\begin{proof}
The statement is trivial for $q = 0$.
Hence we assume that $q \neq 0$
and $a > 1$.
Suppose that $\lr(q) > \lr(aq)$.
By definition of $\lr$, there exists $r$ with
$\lr(q) \geq r > \lr(aq)$ and $\ensuremath{\varepsilon}_0 > 0$ such that
$\limsup_{\ensuremath{\varepsilon}\to0}\ensuremath{\varepsilon}^{-\ensuremath{\beta}}\overline\Phi_{\e;q,r}(1) < \ensuremath{\theta} < 1$
for some $\ensuremath{\theta}$.
Observe that due to Proposition~\ref{pr:nonuniform-perturbation}
we have $a \overline{u}_{\e;q,r} \in \underline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$,
and clearly $a \overline{u}_{\e;q,r} \leq a P_{q,r} = P_{aq,r}$.
Therefore, by definition of $\overline{u}_{\ensuremath{\varepsilon};aq,r}$ and flatness of $\overline{u}_{\e;q,r}$,
\[
P_{aq, r}^{-\ensuremath{\varepsilon}^\ensuremath{\beta} \ensuremath{\theta}} = a P_{q,r}^{-\ensuremath{\varepsilon}^\ensuremath{\beta}\ensuremath{\theta}} \leq a \overline{u}_{\e;q,r}
\leq \overline{u}_{\ensuremath{\varepsilon};aq, r} \qquad \text{in } \set{t \leq 1},\ \ensuremath{\varepsilon} \ll 1,
\]
and we conclude that
$\limsup_{\ensuremath{\varepsilon}\to0}\ensuremath{\varepsilon}^{-\ensuremath{\beta}}\overline\Phi_{\ensuremath{\varepsilon};aq,r}(1) \leq \ensuremath{\theta} < 1$ and therefore $r \leq \lr(aq)$, a contradiction.
The proof for $\ur$ is analogous.
\qedhere\end{proof}
However,
the local comparison principle,
Theorem~\ref{th:localComparison},
also provides some ordering between $\ur$ and $\lr$.
\begin{corollary}
\label{co:mixed-monotone}
For any $q \in \Rd$ and $a > 1$ we have
\begin{align*}
\lr(q) \leq \ur(aq).
\end{align*}
\end{corollary}
\begin{proof}
Again, the ordering is obvious if $q = 0$.
Therefore let us assume that $q \neq 0$ and $a > 1$.
Suppose that the ordering does not hold, i.e.,
$\ur(aq) < \lr(q)$.
Set $q_1 = q$ and $q_2 = a q$.
By the definition of $\lr$ and $\ur$,
there are $r_1$ and $r_2$ with
\begin{align*}
\ur(a q) < r_2 < r_1 < \lr(q)
\end{align*}
such that
\[
\limsup_{\ensuremath{\varepsilon}\to0} \ensuremath{\varepsilon}^{-\ensuremath{\beta}} \max \pth{\overline\Phi_{\ensuremath{\varepsilon}; q_1, r_1}(1), \underline\Phi_{\ensuremath{\varepsilon}; q_2, r_2}(1)} < 1.
\]
But that is in contradiction with the conclusion of
Theorem~\ref{th:localComparison}.
\qedhere\end{proof}
\subsection{Semi-continuity}
\label{sec:semi-continuity}
In this section we prove that
$\ur(q) \leq \lr(q)$,
and that in fact $\lr$ and $\ur$ are semi-continuous.
\begin{proposition}
\label{pr:r-semicontinuity}
The functions $\lr$ and $\ur$ are semi-continuous on $\Rd$ and
\begin{align*}
\lr = \ur^* &\in USC(\Rd), & \ur = \lr_* &\in LSC(\Rd).
\end{align*}
Furthermore, $a \mapsto \lr(a q)$ and $a \mapsto \ur(aq)$ are nondecreasing
on $(0, \infty)$ for any $q \in \Rd$.
(Recall that we set $\lr(0) = \ur(0) = 0$.)
\end{proposition}
To prove this proposition, we start with the following geometric result.
\begin{lemma}
\label{le:halfContinuity}
For every $q \in \Rd$, $\eta > 0$ there exists $\ensuremath{\delta} > 0$ such that
\begin{align*}
\underline{r}(p)& \leq \overline{r}(q) + \eta, &
\underline{r}(q)& \leq \overline{r}(p) + \eta,
\end{align*}
for all $p \in \Rd$, $\abs{p - q} < \ensuremath{\delta}$.
\end{lemma}
\begin{proof}
By the continuity of $\lr$ and $\ur$ at $0$ (Proposition~\ref{pr:bound-on-r}),
we can assume that $q \neq 0$.
We will prove the second inequality.
The proof of the first one is analogous.
Let us further assume that $\ur(q) - \eta > 0$.
We set $r = \ur(q) - \eta/2$ and $\hat r = \ur(q) - \eta = r - \eta/2 < r$.
Let $\ensuremath{\lambda} := \sqrt{n+1}$.
By the definition of $\ur(q)$,
there exists $\ensuremath{\varepsilon} >0$ small enough
such that Corollary~\ref{co:ball-outlier} applies
and there exists $(\zeta, \ensuremath{\sigma})$
such that $B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma}) \subset \Omega(\uu_{\e;q,r}) \cap \set{0 \leq t \leq 1}$
and $\zeta \cdot \nu = r \ensuremath{\sigma} + \ensuremath{\varepsilon}^\ensuremath{\beta}/2$.
Moreover, $\underline\Phi_{\e;q,r}(\ensuremath{\sigma} + \ensuremath{\lambda} \ensuremath{\varepsilon}) \leq \ensuremath{\varepsilon}^\ensuremath{\beta}$.
Suppose that the conclusion does not hold and there exists
a sequence $\set{q_k}_k \subset \Rd$ such that $q_k \to q$
as $k \to \infty$ and $\ur(q) > \lr(q_k) + \eta$.
By the definition of $\lr$, there exists
$\ensuremath{\varepsilon}_k \in (0, \ensuremath{\varepsilon})$, $\ensuremath{\varepsilon}_k \to 0$ such that
$\overline\Phi_{\ensuremath{\varepsilon}_k;q_k, \hat r} \geq \frac34 \ensuremath{\varepsilon}_k^\ensuremath{\beta}$.
We will show that this leads to a contradiction with the comparison principle
for $k$ large enough.
Recall that we set $\nu = -q/\abs q$ and $\nu_k = -q_k/\abs{q_k}$.
By Lemma~\ref{le:furthest-point},
there exists $(\zeta_k,\ensuremath{\sigma}_k) \in \Gamma(\overline{u}_{\ensuremath{\varepsilon}_k;q_k,\hat r})$
such that $\zeta_k \cdot \nu_k = \hat r \ensuremath{\sigma}_k - \frac12\ensuremath{\varepsilon}_k^\ensuremath{\beta}$
and $\overline\Phi_{\ensuremath{\varepsilon}_k;q_k,\hat r}(\ensuremath{\sigma}_k) = \frac12 \ensuremath{\varepsilon}_k^\ensuremath{\beta}$.
We introduce the scaling factor $\ensuremath{\gamma}_k := \ensuremath{\varepsilon}_k/\ensuremath{\varepsilon}$.
By the choice of $\ensuremath{\lambda}$ we can find $(\xi_k, \tau_k) \in \ensuremath{\varepsilon} \ensuremath{\mathbb{Z}}^{n+1}$
such that
\begin{align*}
(X_k, T_k) := (\xi_k, \tau_k) + \ensuremath{\gamma}_k^{-1} (\zeta_k, \ensuremath{\sigma}_k)
\in B_{\ensuremath{\lambda}\ensuremath{\varepsilon}} (\zeta, \ensuremath{\sigma}).
\end{align*}
By compactness and taking a subsequence, we can assume that $(X_k, T_k) \to (X,T) \in \cl B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta, \sigma)$.
Let $v_k$ be the rescaled solution
\begin{align*}
v_k(x,t) = \ensuremath{\gamma}_k^{-1} \overline{u}_{\ensuremath{\varepsilon}_k;q_k,\hat r} (\ensuremath{\gamma}_k(x - \xi_k), \ensuremath{\gamma}_k(t - \tau_k)).
\end{align*}
Observe that $(X_k, T_k) \in \Gamma(v_k)$
and$(X_k, T_k) \in B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma}) \subset \Omega(\uu_{\e;q,r})$.
If we prove that $\uu_{\e;q,r} \leq v_k$ in $B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma})$
for large $k$,
we get a contradiction.
To that end, we will use the monotonicity of the obstacle problem
and the fact that $\lu_{\ensuremath{\varepsilon}_k;q_k,\hat r}$
is a solution away from the boundary of the obstacle.
Let $\Sigma_k := \ensuremath{\gamma}_k^{-1} Q_{q_k} + (\xi_k, \tau_k)$.
Furthermore, let us define the translated planar solutions
\begin{align*}
P_k(x,t) &:= P_{q_k, \hat r}(x - \xi_k, t - \tau_k)&
P^{-\rho_k}_k &:= P^{-\rho_k}_{q_k, \hat r}(x - \xi_k, t - \tau_k),
\end{align*}
where $\rho_k := \frac12 \ensuremath{\gamma}_k^{-1} \ensuremath{\varepsilon}_k^\ensuremath{\beta} = \frac12 \ensuremath{\varepsilon}_k^{\ensuremath{\beta} -1} \ensuremath{\varepsilon}$.
Note that $\rho_k \to \infty$ since $\ensuremath{\varepsilon}_k \to 0$ and $\ensuremath{\beta} < 1$.
We observe that
$v_k \in \mathcal{S}(g^\ensuremath{\varepsilon}, \Sigma_k \setminus \Gamma(P_k))$.
Therefore it is enough to show that
\begin{compactenum}
\item
$v_k \geq P_{q,r}$
in $Q_q \cap \set{t \leq T_k}$,
\item
$\cl\Omega(\uu_{\e;q,r})\cap \set{t \leq T_k} \cap \Gamma(P_k) = \emptyset$, and
\item
$\uu_{\e;q,r} \leq (v_k)_*$
on $\partial_P (Q_q \cap \Sigma_k \cap \set{t \leq T_k})$.
\end{compactenum}
By the properties of the obstacle problem,
Proposition~\ref{pr:obstacle-sols},
we have
\begin{align*}
\uu_{\e;q,r} &= P_{q,r} \quad\text{on } \partial Q_q &
(v_k)_* = P_k \quad\text{on } \partial \Sigma_k.
\end{align*}
Moreover, the flatness assumptions provide the bounds
\begin{align*}
\uu_{\e;q,r} &\leq P_{q,r}^{\ensuremath{\varepsilon}^\ensuremath{\beta}} & v_k&\geq P_k^{-\rho_k},
\quad \text{ both in }
\set{t \leq T_k}.
\end{align*}
Therefore, since
$\partial_P (Q_q \cap \Sigma_k \cap \set{t \leq T_k})
\subset (\partial Q_q \cup \partial \Sigma_k) \cap \cl Q_q$,
we only have to show that, for large $k$,
\begin{align}
\label{order1}
P_k > P_{q,r}^{\ensuremath{\varepsilon}^\ensuremath{\beta}} \quad\text{in } \cl Q_q \cap \set{t\leq T_k}
\cap \cl\Omega(P_{q,r}^{\ensuremath{\varepsilon}^\ensuremath{\beta}})
\end{align}
and
\begin{align}
\label{order2}
P_k^{-\rho_k} > P_{q,r} \quad\text{in } \cl Q_q \cap \set{t\leq T_k}
\cap \cl\Omega(P_{q,r}),
\end{align}
and (a)--(c) shall follow.
A straightforward calculation yields
\[
P_k^{-\rho_k}(x,t) = P_{q_k, \hat r}(x - X_k, t - T_k)
\to P_{q,\hat r}(x - X, t - T), \quad {k \to\infty}
\]
where the convergence is locally uniform in $\Rd \times \ensuremath{\mathbb{R}}$.
A similar calculation yields
\[
P_k(x,t) = P^{\rho_k}_{q_k, \hat r}(x - X_k, t - T_k) \to \infty,
\quad k \to \infty
\]
locally uniformly, since $\rho_k \to \infty$.
Finally, we realize that
\[
P_{q,\hat r}(x - X, t - T) > P_{q,r} \quad\text{in } \cl Q_q \cap \set{t\leq \ensuremath{\sigma} + \ensuremath{\lambda}\ensuremath{\varepsilon}}
\cap \cl\Omega(P_{q,r}),
\]
due to $\hat r < r$
and $X \cdot \nu \geq r T + \ensuremath{\varepsilon}^\ensuremath{\beta}/4$,
and
we conclude that the orderings \eqref{order1} and \eqref{order2}
hold for sufficiently large $k$.
Therefore by the stitching lemma, Lemma~\ref{le:stitch},
and the obstacle monotonicity
we have that $\uu_{\e;q,r} \leq v_k$ in $B_{\ensuremath{\lambda} \ensuremath{\varepsilon}}(\zeta, \ensuremath{\sigma})$
for large $k$ but that is a contradiction with the facts
that $\uu_{\e;q,r} > 0$ on $B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta,\ensuremath{\sigma})$
and $(X_k, T_k) \in B_{\ensuremath{\lambda}\ensuremath{\varepsilon}}(\zeta,\ensuremath{\sigma}) \cap \Gamma(v_k)$.
The proof of the lemma is finished.
\qedhere\end{proof}
We can now proceed with the proof of semi-continuity.
\begin{proof}[Proof of Proposition~\ref{pr:r-semicontinuity}]
Let us fix $q \in \ensuremath{\mathbb{R}}$ and $\eta > 0$,
and let $\ensuremath{\delta} > 0$ be from Lemma~\ref{le:halfContinuity}.
We will show that $\lr$ is USC at $q$.
First, we observe that whenever $p \in \Rd$ and $\abs{p - q} < \ensuremath{\delta}$,
there exists $a > 1$ such that $\abs{a p - q} < \ensuremath{\delta}$.
Therefore, for such $p$ and $a$,
Corollary~\ref{co:mixed-monotone}
and Lemma~\ref{le:halfContinuity} yield
\[
\lr(p) - \eta \leq \ur(a p) -\eta \leq \lr(q).
\]
Since $\ensuremath{\delta} > 0$ exists for arbitrary $\eta > 0$, we conclude that $\lr$
is $USC$ at $q$ for all $q \in \Rd$, and thus $\lr \in USC(\Rd)$.
We can similarly show that $\ur \in LSC(\Rd)$.
Furthermore, we have $\ur \leq \lr$ on $\Rd$ by Lemma~\ref{le:halfContinuity}
which together with the semi-continuity implies
\[
\ur \leq \lr_*, \qquad\qquad \ur^* \leq \lr.
\]
On the other hand, Corollary~\ref{co:mixed-monotone} yields
$\lr(a^{-1} q) \leq \ur(q) \leq \lr(q) \leq \ur(aq)$
for any $a > 1$ and therefore
\[
\lr_*(q) \leq \liminf_{a \to1+} \lr(a^{-1} q) \leq \ur(q)
\leq \lr(q) \leq \limsup_{a \to1+} \ur(aq) \leq \ur^*(q)
\qquad q \in \Rd.
\]
Finally, the monotonicity of $\lr(aq)$ and $\ur(aq)$
in $a$ is a simple consequence of
\[
\ur(a_1 q) \leq \lr(a_1 q) \leq \ur(a_2 q) \leq \lr(a_2 q)
\qquad 0 < a_1 < a_2,\ q \in \Rd.
\]
This finishes the proof.
\qedhere\end{proof}
\subsection{Time-independent medium}
\label{sec:time-independent}
In this section, we show that in a time-independent medium,
i.e., when $g(x,t) = g(x)$,
we can in fact prove that $\lr$ and $\ur$ are continuous and therefore coincide, and that they are positively one-homogeneous.
We thus obtain the same results as in \cite{K07}.
We are able to achieve this because of the key feature
of a time-independent medium, that is,
the availability of scaling in time
independently of space.
Indeed,
it is trivial to check that
if $u \in \mathcal{S}(g^\ensuremath{\varepsilon})$ then also $[(x,t) \mapsto a u(x, a t)] \in \mathcal{S}(g^\ensuremath{\varepsilon})$
for any $\ensuremath{\varepsilon} > 0$ and $a > 0$.
The main result of this section is the following proposition.
\begin{proposition}
\label{pr:timeind}
Suppose that $g(x,t) = g(x)$.
Then
\begin{align}
\label{lrur-coincide}
\lr = \ur \in C(\Rd).
\end{align}
and
\begin{align*}
\lr(q) = \ur(q) = \abs{q} \ur\pth{\frac{q}{\abs{q}}}
\qquad \text{for all $q \in \Rd \setminus \set0$.}
\end{align*}
\end{proposition}
Let us first show the continuity.
We will need the following scaling result.
\begin{lemma}
\label{le:flatness-scaling}
Suppose that $g(x,t) = g(x)$.
Then for any $q \in \Rd$ and $\ensuremath{\varepsilon}, r > 0$
we have
\begin{align}
\label{flatness-scaling}
\underline\Phi_{\ensuremath{\varepsilon}; aq, ar} (t) \leq a\underline\Phi_{a^{-1}\ensuremath{\varepsilon};q,r}(t)
\qquad \text{for all $a>1$ and $t\in\ensuremath{\mathbb{R}}$.}
\end{align}
\end{lemma}
\begin{proof}
Since the statement is trivial for $q = 0$,
we assume $q \neq 0$.
Fix $a > 1$ and define
\[
v(x,t) = a^2 \uu_{a^{-1} \ensuremath{\varepsilon}; q, r}(a^{-1} x, t).
\]
Since $\uu_{a^{-1} \ensuremath{\varepsilon}; q, r} \in \overline{\mathcal{S}}(g^{a^{-1} \ensuremath{\varepsilon}}, Q_q)$,
the scaling of the problem and the independence of $g$ on time
yield
that $v \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon},a Q_q)$.
Moreover,
\[
v(x,t) \geq a^2 P_{q,r}(a^{-1}x,t) = P_{aq,ar}(x,t).
\]
Therefore, by the definition of $\uu_{\ensuremath{\varepsilon}; aq, ar}$
and $Q_q \subset a Q_q$, we get
\[
\uu_{\ensuremath{\varepsilon}; aq, ar} \leq v \qquad \text{in $\Rd\times\ensuremath{\mathbb{R}}$.}
\]
By the definition of flatness, we have
for any $\tau \in \ensuremath{\mathbb{R}}$ and $t \leq \tau$
\begin{align*}
\uu_{\ensuremath{\varepsilon}; aq, ar}(x,t) \leq
v(x,t) &\leq a^2P_{q,r}(a^{-1} x - \underline\Phi_{a^{-1}\ensuremath{\varepsilon};q,r}(\tau)\nu, t)
\\&= P_{aq,ar}(x - a\underline\Phi_{a^{-1}\ensuremath{\varepsilon};q,r}(\tau)\nu, t).
\end{align*}
Thus \eqref{flatness-scaling} follows.
\qedhere\end{proof}
\begin{proof}[Proof of Proposition~\ref{pr:timeind}, continuity]
Fix $q \in \Rd$.
Since $\lr(0) = \ur(0)$ by definition,
we shall assume that $q \neq 0$.
By Lemma~\ref{le:halfContinuity} and
Corollary~\ref{co:mixed-monotone}
we infer
\[
\ur(q) \leq \lr(q) \leq \ur(a q) \qquad \text{for all } a > 1.
\]
Now fix $\ensuremath{\delta} > 0$.
By definition of $\ur(q)$, there exists
$r \in [\ur(q), \ur(q) + \ensuremath{\delta}/2)$
such that $\limsup_{\ensuremath{\varepsilon}\to0} \ensuremath{\varepsilon}^{-\ensuremath{\beta}} \underline\Phi_{\e;q,r}(1) = \ensuremath{\theta} < 1$.
Now Lemma~\ref{le:flatness-scaling} for any $a \in (1, \ensuremath{\theta}^{1/(\ensuremath{\beta}-1)})$
yields
\begin{align*}
\limsup_{\ensuremath{\varepsilon}\to0} \ensuremath{\varepsilon}^{-\ensuremath{\beta}} \underline\Phi_{\ensuremath{\varepsilon}; aq, ar}(1)
\leq \limsup_{\ensuremath{\varepsilon}\to0} a^{1-\ensuremath{\beta}}(a^{-1} \ensuremath{\varepsilon})^{-\ensuremath{\beta}} \underline\Phi_{a^{-1}\ensuremath{\varepsilon};q,r}(1)
= a^{1-\ensuremath{\beta}}\ensuremath{\theta} < 1.
\end{align*}
Therefore if also $a < 1 + \ensuremath{\delta}/2r$,
the definition of $\ur(aq)$ implies
\begin{align*}
\ur(aq) \leq ar \leq r+\ensuremath{\delta}/2 \leq \ur(q) + \ensuremath{\delta}.
\end{align*}
Since $\ensuremath{\delta} > 0$ was arbitrary,
we have $\lr(q) = \ur(q)$.
The semi-continuity result of Proposition~\ref{pr:r-semicontinuity}
implies $\lr = \ur \in C(\Rd)$.
\qedhere\end{proof}
For the proof of the one-homogeneity,
we will use the following scaling result.
\begin{lemma}
\label{le:phi-scaling}
Suppose that $g(x,t) = g(t)$.
Then for any $q \in \Rd$ and $\ensuremath{\varepsilon}, r > 0$ we have
\begin{align}
\label{uPhi-scaling}
\underline\Phi_{\e;q,r}(t) = \underline\Phi_{\ensuremath{\varepsilon}; aq, ar}(a^{-1} t) \qquad
\text{for all $a > 0$, $t \in\ensuremath{\mathbb{R}}$.}
\end{align}
The same holds for $\overline\Phi$.
\end{lemma}
\begin{proof}
Let us first observe that a simple calculation yields
\begin{align}
\label{pqr-scaling}
a P_{q,r}^\eta(x, at) = P_{aq,ar}^\eta(x,t)
\qquad \text{for all $q \in \Rd$, $a,r>0$.}
\end{align}
Fix $a > 0$ and consider $v(x,t) = a \uu_{\e;q,r}(x,at)$.
Since $g$ is time-independent, we have
$v \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon}, Q_q)$
and \eqref{pqr-scaling} yields
$v \geq P_{aq,ar}$. Therefore,
by definition of $\uu_{\ensuremath{\varepsilon};aq,ar}$,
we obtain
\begin{align}
\label{uu-1}
\uu_{\ensuremath{\varepsilon};aq,ar}(x,t) \leq v(x,t) = a \uu_{\e;q,r}(x,at).
\end{align}
Similarly, we set $w(x,t) = a^{-1} \uu_{\ensuremath{\varepsilon}; aq, ar} (x,a^{-1} t)$
and observe that $w \in \overline{\mathcal{S}}(g^\ensuremath{\varepsilon},Q_q)$
and $w \geq P_{q,r}$.
Therefore
\begin{align}
\label{uu-2}
\uu_{\e;q,r} (x,t) \leq w(x,t) = a^{-1} \uu_{\ensuremath{\varepsilon}; aq, ar} (x,a^{-1} t).
\end{align}
Since $a > 0$ was arbitrary, \eqref{uu-1} and \eqref{uu-2} yield together
\begin{align*}
\uu_{\e;q,r}(x,t) = a^{-1} \uu_{\ensuremath{\varepsilon}; aq, ar} (x,a^{-1} t).
\end{align*}
The scaling of the planar solutions in \eqref{pqr-scaling}
and the definition of flatness thus yield \eqref{uPhi-scaling}.
The same statement for $\overline\Phi$ can be obtained analogously.
\qedhere\end{proof}
We can now finish the proof of the main result of this section.
\begin{proof}[Proof of Proposition~\ref{pr:timeind}, homogeneity]
Let us fix $q \in \Rd \setminus\set0$
and $a \in (0,1)$.
Then Proposition~\ref{pr:phi-lipschitz} and Lemma~\ref{le:phi-scaling}
yield
\[
\overline\Phi_{\ensuremath{\varepsilon};aq,ar}(1) \leq \overline\Phi_{\ensuremath{\varepsilon};aq,ar}(a^{-1})
\leq \overline\Phi_{\e;q,r}(1).
\]
Thus the definition of $\lr$ implies
\[
\lr(aq) \geq a \lr(q),
\]
and an analogous argument with $\underline\Phi$ yields
\[
\ur(aq) \leq a \ur(q).
\]
Since $\ur = \lr$ by \eqref{lrur-coincide},
we can conclude that
\begin{align}
\label{urlr-scale}
\ur(aq) = \lr(aq) = a\ur(q) = a \lr(q)\qquad\text{for all $a \in (0,1)$.}
\end{align}
Since $q \in \Rd\setminus\set0$ was arbitrary,
we can define $q' = aq$ and show that
\[
a^{-1}\ur(q') = a^{-1}\lr(q') = \ur(a^{-1}q') = \lr(a^{-1}q')\qquad\text{for all $a \in (0,1)$.}
\]
Therefore \eqref{urlr-scale} holds for all $q \in \Rd$ and $a > 0$.
In particular, if $q \neq 0$, for $a = 1/\abs{q}$ which concludes the proof.
\qedhere\end{proof}
\section{Convergence in the homogenization limit}
\label{sec:convergence}
In the view of Proposition~\ref{pr:r-semicontinuity},
we define the homogenized Hele-Shaw problem
\begin{align}
\label{hs-homogenized}
\begin{cases}
-\Delta u = 0, & \text{in } \set{u > 0} \cap Q,\\
V = \overline{r}(D u), & \text{on } \partial \set{u > 0} \cap Q,
\end{cases}
\end{align}
with the appropriate domain $Q:= \Omega \times (0,T]$
and the initial and boundary data $\Omega_0, \psi$
as in Theorem~\ref{th:well-posedness}.
The viscosity solutions of this problem were introduced in
Section~\ref{sec:viscosity-solutions}.
We show that the solutions of the inhomogeneous Hele-Shaw problem \eqref{HSt}
converge as $\ensuremath{\varepsilon} \to 0$
to the unique solution of the homogenized problem \eqref{hs-homogenized}
with the same initial and boundary data,
in the sense of half-relaxed limits (denoted $\operatorname*{\star-liminf}$ and $\operatorname*{\star-limsup}$ here, Definition~\ref{def:half-relaxed} below).
Since the solution of the homogenized problem might be discontinuous,
a half-relaxed limit is the standard mode of convergence
in such setting.
In particular, we prove the following homogenization result.
\begin{theorem}[Homogenization]
\label{th:homogenization}
Suppose that the domain $Q := \Omega \times (0,T]$
and the initial and boundary data satisfy the assumptions of
Theorem~\ref{th:well-posedness}.
Let $u^\ensuremath{\varepsilon} \in \mathcal{S}(g^\ensuremath{\varepsilon}, Q)$ be the unique solution of
problem \eqref{HSt} for $\ensuremath{\varepsilon}>0$
and let $u \in \mathcal{S}(\lr(p)/\abs{p}, Q)$ be the unique solution
of \eqref{hs-homogenized}.
Then $u_* = \operatorname*{\star-liminf}_{\ensuremath{\varepsilon}\to0} u^\ensuremath{\varepsilon}$
and $u^* = \operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0} u^\ensuremath{\varepsilon}$.
Moreover, $\partial \Omega((u^\ensuremath{\varepsilon})_*;Q)$ converge to
$\partial \Omega(u_*;Q)$ uniformly in Hausdorff distance.
\end{theorem}
Before proceeding with the proof of Theorem~\ref{th:homogenization},
we recall the notion of half-relaxed limits and the notion of convergence in
Hausdorff distance.
\begin{definition}
\label{def:half-relaxed}
Suppose that $\set{v_\ensuremath{\varepsilon}}_{\ensuremath{\varepsilon} > 0}$
is a family of functions
$v_\ensuremath{\varepsilon} : E \to \ensuremath{\mathbb{R}}$ on some set $E \subset \Rd\times\ensuremath{\mathbb{R}}$.
For $(x,t) \in \cl E$ we define
the \emph{upper and lower half-relaxed limits}
\begin{align*}
\operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0} v_\ensuremath{\varepsilon}(x,t) &:=
\lim_{\eta\to0}
\sup \set{v_\ensuremath{\varepsilon}(y,s): (y,s) \in E,\ \abs{y - x} + \abs{s - t} + \ensuremath{\varepsilon} \leq \eta}
\\
\operatorname*{\star-liminf}_{\ensuremath{\varepsilon}\to0} v_\ensuremath{\varepsilon}(x,t) &:=
\lim_{\eta\to0}
\inf \set{v_\ensuremath{\varepsilon}(y,s): (y,s) \in E,\ \abs{y - x} + \abs{s - t} + \ensuremath{\varepsilon} \leq \eta}.
\end{align*}
\end{definition}
Let us recall that $v_\ensuremath{\varepsilon} \rightrightarrows v$ uniformly as $\ensuremath{\varepsilon}\to0$
on a compact set $K \subset E$, where $v \in C(K)$, if
$ \operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0} v_\ensuremath{\varepsilon} = \operatorname*{\star-liminf}_{\ensuremath{\varepsilon}\to0} v_\ensuremath{\varepsilon} = v$ on $K$.
\begin{definition}
\label{def:Hausdorff-convergence}
Let $E_\ensuremath{\varepsilon}, E \subset \Rd\times\ensuremath{\mathbb{R}}$ for $\ensuremath{\varepsilon} > 0$
be compact sets.
We say that $E_\ensuremath{\varepsilon}$ converges to $E$ uniformly in Hausdorff distance
as $\ensuremath{\varepsilon} \to 0$
if for every $\ensuremath{\delta} > 0$ there exists $\ensuremath{\varepsilon}_0 > 0$ such that
\[
E \subset E_\ensuremath{\varepsilon} + B_\ensuremath{\delta}(0,0), \quad E_\ensuremath{\varepsilon} \subset E + B_\ensuremath{\delta}(0,0)
\quad \text{for } \ensuremath{\varepsilon} < \ensuremath{\varepsilon}_0.
\]
\end{definition}
The following characterization is well known \cite{CC06}.
\begin{proposition}
\label{pr:hausdorff}
Compact sets $E_\ensuremath{\varepsilon} \subset \Rd\times\ensuremath{\mathbb{R}}$
converge to a compact set $E\subset\Rd\times\ensuremath{\mathbb{R}}$ uniformly in Hausdorff
distance if and only if the following two statements are true:
\begin{compactenum}[(i)]
\item
if $(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon}) \in E_\ensuremath{\varepsilon}$ for all $\ensuremath{\varepsilon} > 0$, then every limit point of
$\set{(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon})}_{\ensuremath{\varepsilon}>0}$ lies in $E$;
\item for every $(x,t) \in E$ there exists $(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon}) \in E_\ensuremath{\varepsilon}$
such that $(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon}) \to (x,t)$ as $\ensuremath{\varepsilon}\to0$.
\end{compactenum}
\end{proposition}
To prove Theorem~\ref{th:homogenization},
we first address the convergence of the supports,
which is important
for the stability of subsolutions of one-phase problems
such as \eqref{hs-f}.
Nondecreasing-in-time subsolutions
in fact satisfy $\cl\Omega(\overline u) = \limsup_{\ensuremath{\varepsilon}\to0} \cl\Omega(u^\ensuremath{\varepsilon})$,
see \cite[Lemma 4.1]{K08},
but we will only need the following simpler statement
(we give the proof here for completeness).
\begin{lemma}
\label{le:convergence-of-supports}
Let $v^\ensuremath{\varepsilon}\in\underline{\mathcal{S}}(M, Q)$
be a locally uniformly bounded sequence
of subsolutions on the space-time cylinder
$Q = \Omega \times (0, T]$
for open domain $\Omega$ and $T > 0$,
and let $\overline v = \operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0} v^\ensuremath{\varepsilon}$.
Moreover,
suppose that $(v^\ensuremath{\varepsilon})^*$ are nondecreasing in time
and
$\cl\Omega_0(v^\ensuremath{\varepsilon};Q) \subset \cl\Omega_0(\overline v;Q)$
for all $\ensuremath{\varepsilon} > 0$.
If $\ensuremath{\varepsilon}_k \to 0$ as $k\to\infty$,
$(x_k, t_k) \in \cl\Omega(v_{\ensuremath{\varepsilon}_k}; Q)$ for all $k$
and $(x_k, t_k) \to (\hat x, \hat t)$
then $(\hat x, \hat t) \in \cl\Omega(\overline v; Q) \cup \partial_P Q$.
\end{lemma}
\begin{proof}
Suppose that $\ensuremath{\varepsilon}_k \to 0$ as $k \to \infty$ and
$(x_k, t_k) \in \cl\Omega(v_{\ensuremath{\varepsilon}_k})$ such that
$(x_k, t_k) \to (\hat x, \hat t) \in \cl Q$.
We argue by contradiction and assume that
$(\hat x, \hat t) \notin \cl\Omega(\overline v) \cup \partial_P Q$.
Let us denote $H_\rho(\xi,\ensuremath{\sigma}) = B_\rho(\xi,\ensuremath{\sigma}) \cap \set{t \leq T}$ in the following.
In particular, there exists $\ensuremath{\delta} > 0$ such that
$\cl H_{2\ensuremath{\delta}}(\hat x, \hat t) \subset Q \setminus \cl \Omega(\overline v)$.
Since $(v_\ensuremath{\varepsilon})^*$ is nondecreasing in time,
so is $\overline v$ and thus
\[
\cl B_{\ensuremath{\delta}}(x_k) \cap \cl\Omega_0(v_\ensuremath{\varepsilon}) \subset
\cl B_{2\ensuremath{\delta}}(\hat x) \cap \cl\Omega_0(\overline v) = \emptyset
\]
for $k$ large
so that
$\cl B_\ensuremath{\delta}(x_k) \subset B_{2\ensuremath{\delta}}(\hat x)$.
Hence the nondegeneracy property of subsolutions,
Corollary~\ref{co:hs-nondegeneracy}, applies and,
used with the monotonicity in time,
yields the estimate
\[
\sup_{H_{2\ensuremath{\delta}}(\hat x, \hat t)} v_{\ensuremath{\varepsilon}_k} \geq
\sup_{H_{\ensuremath{\delta}}(x_k, t_k)} v_{\ensuremath{\varepsilon}_k} \geq \frac{\ensuremath{\delta}^2}{2nM t_k}
\to \frac{\ensuremath{\delta}^2}{2nM\hat t} > 0 \quad \text{as } k \to \infty.
\]
But this is a contradiction with the property of $\operatorname*{\star-limsup}$,
namely that $v_{\ensuremath{\varepsilon}_k}$ converges to $0$ uniformly on
$\cl H_{2\ensuremath{\delta}}(\hat x, \hat t)$ as $k \to \infty$.
The proof is complete.
\qedhere\end{proof}
\begin{corollary}
\label{co:finite-speed-stable}
Let $v_\ensuremath{\varepsilon}$ be a sequence of subsolutions
as in Lemma~\ref{le:convergence-of-supports}.
Then $\overline v = \operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0}v_\ensuremath{\varepsilon}$
satisfies Definition~\ref{def:visc-test-sub}(i).
\end{corollary}
\begin{proof}
Suppose that $\overline v$ does not satisfy
Definition~\ref{def:visc-test-sub}(i)
and there exists
\[(\xi, \tau) \in (\cl\Omega(\overline v) \cap Q)
\setminus \cl{\Omega(\overline v) \cap \set{t < \tau}}.\]
Then for some $\delta > 0$ we have
\begin{align}
\label{ball-outside-support}
\cl B_{2\ensuremath{\delta}}(\xi,\tau) \cap \set{t\leq T} \subset
Q \setminus \cl{\Omega(\overline v) \cap \set{t < \tau}}.
\end{align}
Let $K = \sup_{\Omega\times[0,\tau]} v_\ensuremath{\varepsilon} < \infty$
and find $h \in (0, \ensuremath{\delta})$ such that
\[
\sqrt{2nKM h} < \ensuremath{\delta}.
\]
By the definition of $\operatorname*{\star-limsup}$ there exists
sequences $\ensuremath{\varepsilon}_k \to0$ and $(\xi_k, \tau_k) \to (\xi, \tau)$
as $k \to \infty$ such that
$(\xi_k, \tau_k) \in \Omega(v_{\ensuremath{\varepsilon}_k})$.
Therefore by Corollary~\ref{co:HS-expansion-speed}
we conclude that there exist $(x_k, t_k) \in \cl\Omega(v_{\ensuremath{\varepsilon}_k}) \cap \cl B_{2\ensuremath{\delta}}(\xi,\tau) \cap \set{t \leq \tau - h}$.
By compactness, $(x_k, t_k) \to (\hat x, \hat t)
\subset \cl B_{2\ensuremath{\delta}}(\xi,\tau) \cap \set{t \leq \tau- h}$
as $k \to \infty$
up to a subsequence,
and Lemma~\ref{le:convergence-of-supports} implies
that $(\hat x, \hat t) \in \cl \Omega(\overline v)$.
But that is a contradiction with \eqref{ball-outside-support}.
\qedhere\end{proof}
In the following proposition we show that the half-relaxed limits of the solutions of the inhomogeneous
Hele-Shaw problem \eqref{HSt} are solutions
of the homogeneous problem \eqref{hs-homogenized}.
\begin{proposition}
\label{pr:homogenization-interior}
Let $v_\ensuremath{\varepsilon}$ be a locally uniformly bounded sequence
of subsolutions of \eqref{HSt}
on the space-time cylinder $Q = \Omega \times (0,T]$
satisfying also the hypothesis of Lemma~\ref{le:convergence-of-supports}.
Then the upper half-relaxed limit
\begin{align*}
\overline{v} = \operatorname*{\star-limsup}_{\ensuremath{\varepsilon} \to 0} v_\ensuremath{\varepsilon}
\end{align*}
is a viscosity subsolution of the homogenized problem
\eqref{hs-homogenized} on $Q$.
Similarly, if $v_\ensuremath{\varepsilon}$ is a locally uniformly bounded sequence of supersolutions of \eqref{HSt} on $Q$ then the lower half-relaxed limit
\begin{align*}
\underline{v} = \operatorname*{\star-liminf}_{\ensuremath{\varepsilon} \to 0} v_\ensuremath{\varepsilon}
\end{align*}
is a supersolution of \eqref{hs-homogenized} on $Q$.
\end{proposition}
\begin{proof}
We will prove that $\overline{v}$
is a subsolution of \eqref{hs-homogenized}.
A parallel, simpler argument yields that $\underline{v}$ is a supersolution;
in this case there is no need to study the convergence of supports.
We shall use Definition~\ref{def:visc-test-sub}.
Its part (i) follows from Corollary~\ref{co:finite-speed-stable}.
To prove that $\overline v$ satisfies Definition~\ref{def:visc-test-sub}(ii) we shall argue by contradiction:
if $\overline{v}$ fails to be a subsolution at some point $P$ of
its free boundary,
it is possible to compare $v_\ensuremath{\varepsilon}$ with a \emph{rescaled translation} of the subsolution
of the obstacle problem
$\overline{u}_{\hat\ensuremath{\varepsilon};q,r}$ for some small $\ensuremath{\varepsilon},\hat\ensuremath{\varepsilon}$ and appropriate $q$, $r$,
in the neighborhood of $P$
and it can be shown that this subsolution
must be detached enough from its obstacle to prevent the free boundary
from reaching $P$ in the limit $\ensuremath{\varepsilon}\to0$.
We accomplish this by translating the rescaled $v_\ensuremath{\varepsilon}$ into $w_\ensuremath{\varepsilon}$ in \eqref{w-epsilon} in such a way that the point $P$ is moved into the neighborhood of point $(r\nu,1)$ where the function $\overline{u}_{\hat\ensuremath{\varepsilon};q,r}$ is 0 due to the detachment lemma (Lemma~\ref{le:detachment-ball}).
First, we can assume that $v_\ensuremath{\varepsilon}$ are $USC$
since that does not change $\operatorname*{\star-limsup}$.
Let us therefore suppose that $\overline{v}-\phi$ has a local maximum at
$(\hat x, \hat t)$
in
$\cl\Omega(\overline v) \cap \set{t \leq \hat t}$
for some $\phi \in C^{2,1}_{x,t}$.
If $\overline{v}(\hat x,\hat t) > 0$ then a standard
stability argument
yields that $-\Delta \phi(\hat x, \hat t) \leq 0$
and thus $\overline{v}$
satisfies Definition~\ref{def:visc-test-sub}(ii-1).
Hence suppose that
$\overline{v}$ does not satisfy Definition~\ref{def:visc-test-sub}(ii-2)
and
$\overline{v}(\hat x, \hat t)=0$,
$-\Delta \phi(\hat x, \hat t) > 0$,
$q := D \phi(\hat x, \hat t) \neq 0$
and $\phi_t > \overline{r}(D\phi) \abs{D\phi}$ at $(\hat x, \hat t)$.
In this case we can find $r \in \ensuremath{\mathbb{R}}$
such that
\begin{align*}
\frac{\phi_t}{\abs{D\phi}} > r > \overline{r}(q)
= \overline{r}(D\phi) \qquad \text{at } (\hat x, \hat t).
\end{align*}
Let us first observe that we can assume
that $\phi \in C^\infty$.
In fact,
let us take
\begin{align*}
\psi(x,t) = \abs{q} r (t - \hat t) + q \cdot (x - \hat x)
+ \ov2 A (x - \hat x) \cdot (x - \hat x),
\end{align*}
where $A = D^2\phi(\hat x, \hat t) + \eta I$,
with $\eta > 0$ chosen small so that $-\Delta \psi = -\operatorname{tr} A
= -\Delta \phi(\hat x, \hat t) - n \eta > 0$.
By the Taylor's theorem, there exists $\ensuremath{\delta} > 0$
such that
$\phi(x,t) - \phi(\hat x, \hat t) \leq \psi(x,t)$
for $\abs{x - \hat x} \leq \ensuremath{\delta}$, $\hat t - \ensuremath{\delta} \leq t \leq \hat t$,
and the equality holds only at $(\hat x,\hat t)$.
Therefore $\overline{v} - \psi$ has
a strict maximum $0$ at $(\hat x, \hat t)$
in $\cl\Omega(\overline{v}) \cap \set{t \leq \hat t}
\cap \cl B_\ensuremath{\delta}(\hat x,\hat t)$.
Consequently,
along a subsequence of $\ensuremath{\varepsilon}\to0$,
using Lemma~\ref{le:convergence-of-supports},
there exist $(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon})$ such that
$(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}) \to (\hat x, \hat t)$ as $\ensuremath{\varepsilon}\to0$
and $v_\ensuremath{\varepsilon} - \psi$ has a local maximum at $(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})$
in $\cl\Omega(v_\ensuremath{\varepsilon}) \cap \set{t \leq t_\ensuremath{\varepsilon}} \cap \cl B_\ensuremath{\delta}(\hat x, \hat t)$.
To simplify the notation in the rest of the proof,
$\ensuremath{\varepsilon}$ only takes on the values from this subsequence.
Since $-\Delta \psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}) > 0$,
we must have $v_\ensuremath{\varepsilon}(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon}) = 0$ and therefore
\begin{align}
\label{max-estimate}
v_\ensuremath{\varepsilon}(x,t) \leq \bra{\psi(x,t)- \psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})}_+
\qquad \text{on $\cl B_\ensuremath{\delta}(\hat x, \hat t) \cap \set{t \leq t_\ensuremath{\varepsilon}}$.}
\end{align}
By the definition of $\lr(q)$ and the choice of $r > \lr(q)$,
there exists arbitrarily small $\ensuremath{\varepsilon}'$ such that
$\overline\Phi_{\ensuremath{\varepsilon}';q,r}(1) > \frac34 \ensuremath{\varepsilon}'^\ensuremath{\beta}$.
Hence Lemma~\ref{le:detachment-ball} implies that
there exists arbitrarily small $\hat\ensuremath{\varepsilon} = \ensuremath{\gamma} \ensuremath{\varepsilon}'$ such that
$\overline{u}_{\hat \ensuremath{\varepsilon}; q, r} = 0$ in
$B_{c \hat \ensuremath{\varepsilon}^{\ensuremath{\beta}}}(r\nu, 1)$, where $c$ is a positive constant
independent of $\hat \ensuremath{\varepsilon}$.
Let us fix one such $\hat \ensuremath{\varepsilon}$ small enough so that
\begin{align}
\label{choice-of-hate}
c \hat \ensuremath{\varepsilon}^{\ensuremath{\beta} - 1} > r + 2 \sqrt{n} + 1,
\end{align}
and set $h := \ensuremath{\varepsilon}/\hat \ensuremath{\varepsilon}$.
Denoting $q_\ensuremath{\varepsilon} = D\psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})$,
we can estimate
\begin{align}
\label{psi-taylor}
\begin{aligned}
\psi(x,t) - \psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}) &=
\psi_t(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})(t - t_\ensuremath{\varepsilon})
+ D\psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}) \cdot (x - x_\ensuremath{\varepsilon})\\
&\quad + \ov2 D^2\psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})(x -x_\ensuremath{\varepsilon}) \cdot (x - x_\ensuremath{\varepsilon})\\
&\leq \abs{q} r (t - t_\ensuremath{\varepsilon})
+ q \cdot (x - x_\ensuremath{\varepsilon})
\\&\quad+ \abs{q_\ensuremath{\varepsilon} - q} \abs{x - x_\ensuremath{\varepsilon}}
+ \norm{A} \abs{x - x_\ensuremath{\varepsilon}}^2.
\end{aligned}
\end{align}
We introduce $E:= Q_q \cap C^+ \cap \set{t \leq 2}$
and
its diameter $d := \operatorname{diam} E$,
and set (recall that $\nu := -q/\abs{q}$)
\begin{align}
\label{ye-taue}
\begin{aligned}
\tilde y_\ensuremath{\varepsilon} &= x_\ensuremath{\varepsilon} - h r \nu
+ \norm{A} h^2 d^2 \abs{q}^{-1} \nu
+ \abs{q_\ensuremath{\varepsilon} - q} h d \abs{q}^{-1}\nu
+ (r + \sqrt{n}) \ensuremath{\varepsilon} \nu,\\
\tilde \tau_\ensuremath{\varepsilon} &= t_\ensuremath{\varepsilon} - h.
\end{aligned}
\end{align}
Then we choose $(y_\ensuremath{\varepsilon}, \tau_\ensuremath{\varepsilon}) \in \ensuremath{\varepsilon}\ensuremath{\mathbb{Z}}^{n+1}$ such that
\begin{align*}
(y_\ensuremath{\varepsilon}, \tau_\ensuremath{\varepsilon}) \in \operatorname*{arg\,min}_{(x,t) \in \ensuremath{\varepsilon} \ensuremath{\mathbb{Z}}^{n+1}}
\bra{\abs{x - \tilde y_\ensuremath{\varepsilon}} +
\abs{t - \tilde \tau_\ensuremath{\varepsilon}}}.
\end{align*}
Clearly
\begin{align}
\label{grid-error}
\abs{y_\ensuremath{\varepsilon} - \tilde y_\ensuremath{\varepsilon}} &\leq \sqrt{n} \ensuremath{\varepsilon}, &
\abs{\tau_\ensuremath{\varepsilon} - \tilde \tau_\ensuremath{\varepsilon}} &\leq \ensuremath{\varepsilon}
\end{align}
and therefore, recalling \eqref{choice-of-hate}
and $h = \ensuremath{\varepsilon}/\hat\ensuremath{\varepsilon}$,
we have for all $\ensuremath{\varepsilon}$ sufficiently
small
\begin{align*}
\operatorname{dist}((y_\ensuremath{\varepsilon} + h r \nu, \tau_\ensuremath{\varepsilon} + h), (x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}))
&\leq (r + 2\sqrt{n} + 1) \ensuremath{\varepsilon}
\\&\quad + \abs{q_\ensuremath{\varepsilon} - q} h d \abs{q}^{-1}
+ \norm{A} h^2 d^2 \abs{q}^{-1}
\\&\leq c \hat \ensuremath{\varepsilon}^{\ensuremath{\beta} - 1} \ensuremath{\varepsilon} = c h \hat\ensuremath{\varepsilon}^\ensuremath{\beta}.
\end{align*}
since $q_\ensuremath{\varepsilon} \to q$ as $\ensuremath{\varepsilon} \to 0$.
In particular,
\begin{align}
\label{xe-te-in-ball}
(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}) \in B_{c h \hat\ensuremath{\varepsilon}^\beta} (y_\ensuremath{\varepsilon} + h r\nu, \tau_\ensuremath{\varepsilon} + h)
\subset h E + (y_\ensuremath{\varepsilon}, \tau_\ensuremath{\varepsilon})
\end{align}
for sufficiently small $\ensuremath{\varepsilon}$.
From this and the definition of $d = \operatorname{diam} E$,
we infer that $\operatorname{dist}((x,t), (x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})) \leq hd$
for all $(x,t) \in hE + (y_\ensuremath{\varepsilon}, \tau_\ensuremath{\varepsilon})$.
Thus we estimate,
using \eqref{psi-taylor}, \eqref{ye-taue} and \eqref{grid-error},
\begin{align*}
\abs{q} (r(t - \tau_\ensuremath{\varepsilon}) - (x - y_\ensuremath{\varepsilon})\cdot\nu)
&= \abs{q} r \pth{t - t_\ensuremath{\varepsilon} + t_\ensuremath{\varepsilon} - \tilde \tau_\ensuremath{\varepsilon}
+ \tilde \tau_\ensuremath{\varepsilon} - \tau_\ensuremath{\varepsilon}}\\
&\quad + q \cdot \pth{x - x_\ensuremath{\varepsilon} + x_\ensuremath{\varepsilon} - \tilde y_\ensuremath{\varepsilon}
+ \tilde y_\ensuremath{\varepsilon} - y_\ensuremath{\varepsilon}}\\
&\geq \psi(x,t) - \psi(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon})
- \norm{A} \abs{x - x_\ensuremath{\varepsilon}}^2 - \abs{q_\ensuremath{\varepsilon} - q} \abs{x - x_\ensuremath{\varepsilon}}\\
&\quad + \abs{q} r (-h) + hr \abs{q} \\
&\quad + \norm{A} h^2 d^2
+ \abs{q_\ensuremath{\varepsilon} - q} h d + \pth{r + \sqrt{n}}\ensuremath{\varepsilon} \abs{q}\\
&\quad - (\abs{q} r + \abs{q}\sqrt{n}) \ensuremath{\varepsilon} \\
&\geq \psi(x,t)- \psi(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon})
\qquad \text{for all } (x,t) \in hE + (y_\ensuremath{\varepsilon}, \tau_e).
\end{align*}
And therefore \eqref{max-estimate} yields
\begin{align*}
v_\ensuremath{\varepsilon}(x,t)
&\leq \bra{\psi(x,t)-\psi(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon})}_+
\leq P_{q,r}(x - y_\ensuremath{\varepsilon}, t - \tau_\ensuremath{\varepsilon}) &
&\text{on } \pth{h E + (y_\ensuremath{\varepsilon}, \tau_\ensuremath{\varepsilon})} \cap \set{t \leq t_\ensuremath{\varepsilon}}.
\end{align*}
Let us define
\begin{align}
\label{w-epsilon}
w_\ensuremath{\varepsilon}(x,t) =
\begin{cases}
h^{-1} v_\ensuremath{\varepsilon}
\pth{h x + y_\ensuremath{\varepsilon}, h t + \tau_\ensuremath{\varepsilon})}
& (x,t) \in Q_q \cap C^+,\\
0 & (x,t) \in Q_q \setminus C^+.
\end{cases}
\end{align}
We observe that $w_\ensuremath{\varepsilon} \in \underline{\mathcal{S}}(g^{\ensuremath{\varepsilon}/h}, Q_q)$
and $w_\ensuremath{\varepsilon} \leq P_{q,r}$ on
$Q_q \cap \set{t \leq 1 + h^{-1}(\tilde\tau_\ensuremath{\varepsilon} - \tau_\ensuremath{\varepsilon})}$.
Therefore, by definition, $w_\ensuremath{\varepsilon} \leq \lu_{\ensuremath{\varepsilon}/h;q,r} = \lu_{\hat\ensuremath{\varepsilon};q,r}$
in $Q_q \cap \set{t \leq 1 + h^{-1}(\tilde\tau_\ensuremath{\varepsilon} - \tau_\ensuremath{\varepsilon})}$.
But by the choice of $\hat\ensuremath{\varepsilon}$ we have
that $\lu_{\hat\ensuremath{\varepsilon};q,r} = 0$ in $B_{c\hat\ensuremath{\varepsilon}^\ensuremath{\beta}}(r\nu, 1)$,
and by scaling back, we have
\begin{align*}
v_\ensuremath{\varepsilon} = 0 \qquad \text{in $B_{ch\hat\ensuremath{\varepsilon}^\ensuremath{\beta}}(y_\ensuremath{\varepsilon} + hr\nu, \tau_\ensuremath{\varepsilon} + h)
\cap \set{t \leq t_\ensuremath{\varepsilon}}$}.
\end{align*}
When we apply Definition~\ref{def:visc-test-sub}(i)
and \eqref{xe-te-in-ball},
we obtain $(x_\ensuremath{\varepsilon}, t_\ensuremath{\varepsilon}) \notin \cl\Omega(v_\ensuremath{\varepsilon})$
and that is a contradiction with the choice of $(x_\ensuremath{\varepsilon},t_\ensuremath{\varepsilon})$.
\qedhere\end{proof}
In particular, if the solution of the homogenized problem
\eqref{hs-homogenized} is unique,
then the limit is the unique solution.
Now we have enough information to finish the proof of the main homogenization result.
\begin{proof}[Proof of Theorem~\ref{th:homogenization}]
We shall denote
$\overline{u} = \operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0} u^\ensuremath{\varepsilon}$
and $\uu = \operatorname*{\star-liminf}_{\ensuremath{\varepsilon}\to0} u^\ensuremath{\varepsilon}$.
Let $U \in \underline{\mathcal{S}}(m/2, Q)$ and $V \in \overline{\mathcal{S}}(M, Q)$
be the boundary barriers constructed
in the proof of Theorem~\ref{th:well-posedness}.
From that proof we know that
$U_* \leq u^\ensuremath{\varepsilon} \leq V^*$ for all $\ensuremath{\varepsilon} > 0$ and
hence $U_* \leq \uu \leq \overline{u} \leq V^*$ on $Q$.
This implies that $\uu$ and $\lu$ have the correct boundary
data in the sense of Theorem~\ref{th:well-posedness}.
Since $\overline{u} \in \underline{\mathcal{S}}(\lr(p)/\abs{p}, Q)$
and $\uu \in \overline{\mathcal{S}}(\lr(p)/\abs{p}, Q)$
by Proposition~\ref{pr:homogenization-interior},
the proof of uniqueness in Theorem~\ref{th:well-posedness} yields
$u_* \leq \uu \leq \overline{u} \leq u^*$,
which implies $\uu = u_*$, $\overline{u} = u^*$ using Corollary~\ref{co:regularity}.
Let us proceed with the proof of the
convergence of free boundaries.
Denote
$E_\ensuremath{\varepsilon} = \partial \Omega((u^\ensuremath{\varepsilon})_*;Q)$
and $E = \partial \Omega(u_*;Q)$.
Clearly $E,E_\ensuremath{\varepsilon}$ are compact subsets of $\cl Q$
(see also the proof of Theorem~\ref{th:well-posedness}).
We point out that due to the choice of initial and boundary data,
we have that
\begin{align}
\label{boundary-decomp}
E = \Gamma(u_*;Q)
\cup (\partial \Omega \times [0,T])
\cup ((\Omega \cap \Omega_0) \times \set{0})
\subset \Gamma(u_*;Q) \cup \partial_P Q,
\end{align}
and similarly for $E_\ensuremath{\varepsilon}$
(see the proof of Theorem~\ref{th:well-posedness}).
Let us also recall that
$\cl\Omega(u_*) = \cl\Omega(u)$
by Corollary~\ref{co:regularity}.
Therefore $\partial \Omega(u_*) = \cl\Omega(u) \setminus
\Omega(u_*)$.
The same is true for $u^\ensuremath{\varepsilon}$.
We shall establish
the uniform convergence $E_\ensuremath{\varepsilon} \to E$ in
Hausdorff distance
using the characterization in Proposition~\ref{pr:hausdorff}.
Let us denote
\begin{align*}
H_\rho(\xi,\ensuremath{\sigma}) := B_\rho(\xi,\ensuremath{\sigma}) \cap \set{t\leq T},
\qquad (\xi,\ensuremath{\sigma}) \in \Rd \times \ensuremath{\mathbb{R}}, \rho > 0,
\end{align*}
in the following.
\textbf{(i)}
Suppose that $\ensuremath{\varepsilon}_k \to 0$ as $k \to\infty$
and $(x_k,t_k) \in E_{\ensuremath{\varepsilon}_k}$ such that $(x_k, t_k) \to (\hat x, \hat t)$.
Clearly $\hat t \leq T$.
Lemma~\ref{le:convergence-of-supports}
yields that $(\hat x,\hat t) \in \cl\Omega(u)$.
Furthermore, due to the behavior of $u^\ensuremath{\varepsilon}$
at the parabolic boundary of $\partial_P Q$,
we have $\hat x \in \Omega$,
and $(\hat x, \hat t) \in E$ if also $\hat t = 0$.
Let us therefore assume that $\hat t > 0$ and
$(\hat x, \hat t) \in \Omega(u_*)$.
There exists $\ensuremath{\delta} > 0$ such that $\cl H_\ensuremath{\delta}(\hat x, \hat t)
\subset \Omega(u_*)$.
Semi-continuity yields
$\eta = \min_{\cl H_\ensuremath{\delta}(\hat x, \hat t)} u_* > 0$.
The property of $\operatorname*{\star-liminf}$ then implies
that $\min_{\cl H_\ensuremath{\delta}(\hat x, \hat t)} (u^{\ensuremath{\varepsilon}_k})_* > \eta/2 > 0$
for $k$ sufficiently large.
But this is a contradiction with
$(x_k, t_k) \in H_\ensuremath{\delta}(\hat x, \hat t) \cap \partial \Omega((u^\ensuremath{\varepsilon})_*)$
for large $k$.
\textbf{(ii)}
Now suppose that there exist $(\hat x, \hat t) \in E$
and $\ensuremath{\delta} > 0$ such that
for some $\ensuremath{\varepsilon}_k \to 0$ as $k \to \infty$
we have
$E_{\ensuremath{\varepsilon}_k} \cap \cl H_\ensuremath{\delta}(\hat x,\hat t) = \emptyset$
for all $k$.
Again, since $u^\ensuremath{\varepsilon}$ and $u$ satisfy the same boundary conditions,
we must have $(\hat x, \hat t) \in Q$
by \eqref{boundary-decomp}.
Thus we can assume that $\cl H_\ensuremath{\delta}(\hat x, \hat t) \subset Q$.
Suppose that $\cl H_\ensuremath{\delta}(\hat x,\hat t) \cap
\cl \Omega(u^{\ensuremath{\varepsilon}_k}) = \emptyset$
for infinitely many $k$.
In this case $u = 0$ in $\cl H_\ensuremath{\delta}(\hat x, \hat t)$
and we get a contradiction with the choice of $(\hat x, \hat t)
\in \partial \Omega(u_*)$.
Hence we must have $\cl H_\ensuremath{\delta}(\hat x,\hat t) \subset \Omega((u^{\ensuremath{\varepsilon}_k})_*)$
for all but finitely many $k$.
We can assume for simplicity that this inclusion holds for all $k$.
We recall that the comparison with the barrier $U$
in the proof of Theorem~\ref{th:well-posedness}
yields that the support of $u$ strictly expands at $t = 0$.
Consequently, $\hat x \notin \cl\Omega_0(u; Q)$.
By taking $\ensuremath{\delta}$ smaller if necessary,
we can assume that $\cl B_\ensuremath{\delta}(\hat x) \cap \cl\Omega_0(u; Q) = \emptyset$.
Therefore $\cl B_\ensuremath{\delta}(\hat x) \cap \cl\Omega_0(u^\ensuremath{\varepsilon}; Q) = \emptyset$
and $u^\ensuremath{\varepsilon}$ is nondegenerate in the sense of
Corollary~\ref{co:hs-nondegeneracy}.
Since $(u^\ensuremath{\varepsilon})_* = ((u^\ensuremath{\varepsilon})^*)_*$ is nondecreasing in time
and $(\hat x, \hat t- \ensuremath{\delta}/2) \in
\cl\Omega((u^{\ensuremath{\varepsilon}_k})_*) = \cl\Omega(u^{\ensuremath{\varepsilon}_k})$,
we have
\[
\sup_{x \in B_{\ensuremath{\delta}/2}(\hat x)} (u^{\ensuremath{\varepsilon}_k})_*(x,t) \geq 2nM
\frac{\ensuremath{\delta}^2}{\hat t - \ensuremath{\delta}/2}
\qquad
\text{for } t \in I := [\hat t -\ensuremath{\delta}/2, \hat t + \ensuremath{\delta}/2] \cap (0,T].
\]
Furthermore, we observe that
$B_{\sqrt3 \ensuremath{\delta}/2}(\hat x) \in \Omega_t((u^{\ensuremath{\varepsilon}_k})_*)$
for $t \in I$
and, since $(u^\ensuremath{\varepsilon})_*$ is harmonic in $\Omega_t((u^\ensuremath{\varepsilon})_*)$
due the same argument as in the proof of Proposition~\ref{pr:harmonic},
the Harnack inequality yields
$\inf_{x \in B_{\ensuremath{\delta}/2}(\hat x)} (u^{\ensuremath{\varepsilon}_k})_*(x,t) \geq \eta > 0$
for $t \in I$,
in particular $\inf_{H_{\ensuremath{\delta}/2}(\hat x, \hat t)} u^{\ensuremath{\varepsilon}_k} \geq \eta$
with $\eta$ independent of $k$.
Therefore, by regularity, Corollary~\ref{co:regularity},
we have
\[
u_* = (u^*)_* = \pth{\operatorname*{\star-limsup}_{\ensuremath{\varepsilon}\to0}u^\ensuremath{\varepsilon}}_* \geq \eta > 0
\qquad \text{in } H_{\ensuremath{\delta}/2}(\hat x, \hat t).
\]
But that is a contradiction with the choice of $(\hat x, \hat t)
\in \partial \Omega(u_*)$.
The proof of uniform convergence in Hausdorff distance is finished.
\qedhere\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 9,568
|
The bibliographic databases that are held here can no longer be searched automatically (since 2012). But they can still be accessed as text files. Here they are listed.
This is a collection of several thousand bibliographic entries in syntax, semantics and computational linguistics covering the mid-1990s to the early part of the present century.
You can access it in various formats: bibtex, endnote (aka refer), html
It includes systematic coverage of conferences such as COLING, ACL, and EACL, and the Journal Computational Linguistics during this period. but these are more completely covered by the ACL bibliography (described next).
This was the bibliography I was constructing as part of the Digital Archive of Research Papers in Computational Linguistics constructed by the ACL.
It has been absorbed into the ACL anthology: See http://aclweb.org/anthology/.
HPSG
This is the bibliography of work in Head Driven Phrase Structure Grammar.
You can access it here: http://hpsg.fu-berlin.de/HPSG-Bib/
This is the bibliography of work in Lexical Functional Grammar compiled by Peter Austin (Department of English, University of Hong Kong; pkaustin@hkuxa.hku.hk) and Mary Dalrymple (Xerox PARC; dalrymple@parc.xerox.com), and others.
You will find links to it here: http://www.essex.ac.uk/linguistics/external/lfg/FAQ/Bibliography.html
TSNLP
This bibliography was developed as part of the Test Suites for Natural Language Processing Project, which aimed at developing guidelines for the creation of test suites for evaluation and testing of NL systems, as well at constructing actual test suites in three languages: English, French and German. For more information check the TSNLP Home Page.
This bibliography is no longer being developed or maintained.
You can access it in various formats: bibtex, endnote (aka refer), html.
CL/MT
This is a collection of about 6,000 bibliographic entries in syntax, semantics and computational linguistics, written by members of the CL/MT group at Essex during the 1980s and 1990s. It includes relatively systematic coverage of conferences such as COLING, ACL, EACL, TMI (Theoretical & Metholdological Issues in Machine Translation) and journals such as Machine Translation, Computational Linguistics, and Linguistics & Philosophy. It includes unsystematic coverage of a whole lot of other things we were interested in at various time in the past.
This bibliography is no longer being developed or maintained. See the CLgroup bibliography for something similar and more up-to-date.
This is the bibliography of computational linguistics and Natural Language Processing in the 198Os developed at the University of Sussex, and published as: Gerald Gazdar, Alex Franz, Karen Osborne and Roger Evans Natural Language Processing in the 1980s , CSLI Publications, Stanford, Ca., CSLI Lecture Notes, 12. It contains over 2,000 entries, and gives relatively systematic coverage of a number of conferences and journals throughout the 1980s, including relevant papers from AAAI, ACL, AISB, COLING, EACL, and IJCAI.
You can access it one format: endnote (aka refer) format
Please also read the README file
CSLI
This is the "lingbib.csli" a bibliography constructed at CSLI, based on the bibliography files of Andras Kornai with significant contributions by Bill Poser, Paul Kiparsky, and Gerald Gazdar. It is version is 1.0 (as of July 1992 -- so far as we know, this is the only version that has ever been distributed). It contains over 3,000 entries from over 400 conferences and journals in mainstream linguistics, including: AAAI, ACL Proceedings, Acta Linguistica, Berkeley Linguistics Society, CLS, Cognition, Cognitive Psychology, Cognitive Science, Computer Speech and Language, IJCAI, IULC, Journal of Linguistics, Journal of Phonetics, Language, Lingua, Linguistic Analysis, Linguistic Inquiry, Linguistics, NELS, Natural Language and Linguistic Theory, Ohio State Working Papers in Linguistics, Studies in African Linguistics, Syntax and Semantics, Synthese, Theoretical Linguistics.
This biblography is Copyright (C) 1992 Center for the Study of Language and Information and is subject to the terms and conditions set forth in the CSLI General Public License.
You can access it one format:
endnote (aka refer) format
Maintained by Doug Arnold (doug#essex.ac.uk) 19th August 1995
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 1,496
|
Nouakchott (district), een regio van Mauritanië
Nouakchott (stad), de hoofdstad van de gelijknamige regio
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,872
|
<!---
For community contributors -- Please fill out Part 1 of the following template. This will help our team collaborate with you and give us an opportunity to provide valuable feedback that could inform your development process. Sections in Part 2 are not mandatory to get the conversation started, but will help our team understand your vision better and allow us to give better feedback.
--->
**_Part 1 - Required information_**
# Feature title
## :memo: Summary
One paragraph explanation of the feature.
## :checkered_flag: Motivation
Why are we doing this? What use cases does it support? What is the expected outcome?
## 🤯 Explanation
Explain the proposal as if it was already implemented in the GitHub package and you were describing it to an Atom user. That generally means:
- Introducing new named concepts.
- Explaining the feature largely in terms of examples.
- Explaining any changes to existing workflows.
- Design mock-ups or diagrams depicting any new UI that will be introduced.
**_Part 2 - Additional information_**
## :anchor: Drawbacks
Why should we *not* do this?
## :thinking: Rationale and alternatives
- Why is this approach the best in the space of possible approaches?
- What other approaches have been considered and what is the rationale for not choosing them?
- What is the impact of not doing this?
## :question: Unresolved questions
- What unresolved questions do you expect to resolve through the Feature Request process before this gets merged?
- What unresolved questions do you expect to resolve through the implementation of this feature before it is released in a new version of the package?
## :warning: Out of Scope
- What related issues do you consider out of scope for this Feature Request that could be addressed in the future independently of the solution that comes out of this Feature Request?
## :construction: Implementation phases
- Can this functionality be introduced in multiple, distinct, self-contained pull requests?
- A specification for when the feature is considered "done."
## :white_check_mark: Feature description for Atom release blog post
- When this feature is shipped, what would we like to say or show in our Atom release blog post (example: http://blog.atom.io/2018/07/31/atom-1-29.html)
- Feel free to drop ideas and gifs here during development
- Once development is complete, write a blurb for the release coordinator to copy/paste into the Atom release blog
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,696
|
Q: Why is PyQt executing my actions three times? I'm still kind of new to PyQt but I really have no idea why this is happening.
I have a Mainwindow that I create like this:
class MainWindow(QtGui.QMainWindow):
#initialize
def __init__(self):
#Call parent constructor
super(MainWindow, self).__init__()
#Load the interface
self.ui = uic.loadUi(r"Form Files/rsleditor.ui")
#Show the ui
self.ui.show()
and when I wanted to override the close event with:
def closeEvent(self, event):
quit_msg = "Are you sure you want to exit the program?"
reply = QtGui.QMessageBox.question(self, 'Message',
quit_msg, QtGui.QMessageBox.Yes, QtGui.QMessageBox.No)
if reply == QtGui.QMessageBox.Yes:
self.saveSettings()
event.accept()
else:
event.ignore()
I read I had to change the uic.loadUI call to:
self.ui = uic.loadUi(r"Form Files/rsleditor.ui", self)
But when I do this, all of my actions start being set off three times. I'm pretty sure I'm setting up the signals and slots correctly as they were working before adding this. Any help?
The pyuic file:
RSLEditorClass.setIconSize(QtCore.QSize(24, 24))
self.centralWidget = QtGui.QWidget(RSLEditorClass)
sizePolicy = QtGui.QSizePolicy(QtGui.QSizePolicy.Preferred, QtGui.QSizePolicy.Preferred)
sizePolicy.setHorizontalStretch(0)
sizePolicy.setVerticalStretch(0)
sizePolicy.setHeightForWidth(self.centralWidget.sizePolicy().hasHeightForWidth())
self.centralWidget.setSizePolicy(sizePolicy)
self.centralWidget.setObjectName(_fromUtf8("centralWidget"))
self.vLayMain = QtGui.QVBoxLayout(self.centralWidget)
self.vLayMain.setObjectName(_fromUtf8("vLayMain"))
self.hLayFilePath = QtGui.QHBoxLayout()
self.hLayFilePath.setObjectName(_fromUtf8("hLayFilePath"))
self.lbFilePath = QtGui.QLabel(self.centralWidget)
self.lbFilePath.setObjectName(_fromUtf8("lbFilePath"))
self.hLayFilePath.addWidget(self.lbFilePath)
self.txtFilePath = QtGui.QLineEdit(self.centralWidget)
self.txtFilePath.setObjectName(_fromUtf8("txtFilePath"))
self.hLayFilePath.addWidget(self.txtFilePath)
self.vLayMain.addLayout(self.hLayFilePath)
self.splitTxt = QtGui.QSplitter(self.centralWidget)
sizePolicy = QtGui.QSizePolicy(QtGui.QSizePolicy.Preferred, QtGui.QSizePolicy.Expanding)
sizePolicy.setHorizontalStretch(0)
sizePolicy.setVerticalStretch(1)
sizePolicy.setHeightForWidth(self.splitTxt.sizePolicy().hasHeightForWidth())
self.splitTxt.setSizePolicy(sizePolicy)
self.splitTxt.setLineWidth(0)
self.splitTxt.setMidLineWidth(0)
self.splitTxt.setOrientation(QtCore.Qt.Vertical)
self.splitTxt.setHandleWidth(10)
self.splitTxt.setObjectName(_fromUtf8("splitTxt"))
self.tabMain = QtGui.QTabWidget(self.splitTxt)
self.tabMain.setBaseSize(QtCore.QSize(0, 0))
self.tabMain.setElideMode(QtCore.Qt.ElideNone)
self.tabMain.setTabsClosable(True)
self.tabMain.setMovable(False)
self.tabMain.setObjectName(_fromUtf8("tabMain"))
self.tabOutput = QtGui.QTabWidget(self.splitTxt)
self.tabOutput.setObjectName(_fromUtf8("tabOutput"))
self.tabRSLOut = QtGui.QWidget()
self.tabRSLOut.setObjectName(_fromUtf8("tabRSLOut"))
self.vLayRSL = QtGui.QVBoxLayout(self.tabRSLOut)
self.vLayRSL.setObjectName(_fromUtf8("vLayRSL"))
self.txtRSLOut = QtGui.QTextEdit(self.tabRSLOut)
self.txtRSLOut.setTextInteractionFlags(QtCore.Qt.TextEditable)
self.txtRSLOut.setObjectName(_fromUtf8("txtRSLOut"))
self.vLayRSL.addWidget(self.txtRSLOut)
self.tabOutput.addTab(self.tabRSLOut, _fromUtf8(""))
self.tabRIBOut = QtGui.QWidget()
self.tabRIBOut.setObjectName(_fromUtf8("tabRIBOut"))
self.vLayRib = QtGui.QVBoxLayout(self.tabRIBOut)
self.vLayRib.setObjectName(_fromUtf8("vLayRib"))
self.txtRIBOut = QtGui.QTextEdit(self.tabRIBOut)
self.txtRIBOut.setTextInteractionFlags(QtCore.Qt.TextEditable)
self.txtRIBOut.setObjectName(_fromUtf8("txtRIBOut"))
self.vLayRib.addWidget(self.txtRIBOut)
self.tabOutput.addTab(self.tabRIBOut, _fromUtf8(""))
self.vLayMain.addWidget(self.splitTxt)
RSLEditorClass.setCentralWidget(self.centralWidget)
self.menuBar = QtGui.QMenuBar(RSLEditorClass)
self.menuBar.setGeometry(QtCore.QRect(0, 0, 750, 21))
self.menuBar.setObjectName(_fromUtf8("menuBar"))
self.menuFile = QtGui.QMenu(self.menuBar)
self.menuFile.setObjectName(_fromUtf8("menuFile"))
self.menuRecent_Files = QtGui.QMenu(self.menuFile)
self.menuRecent_Files.setObjectName(_fromUtf8("menuRecent_Files"))
self.menuEdit = QtGui.QMenu(self.menuBar)
self.menuEdit.setObjectName(_fromUtf8("menuEdit"))
self.menuSearch_Options = QtGui.QMenu(self.menuEdit)
self.menuSearch_Options.setObjectName(_fromUtf8("menuSearch_Options"))
self.menuView = QtGui.QMenu(self.menuBar)
self.menuView.setObjectName(_fromUtf8("menuView"))
self.menuSearch = QtGui.QMenu(self.menuBar)
self.menuSearch.setObjectName(_fromUtf8("menuSearch"))
self.menuDebug = QtGui.QMenu(self.menuBar)
self.menuDebug.setObjectName(_fromUtf8("menuDebug"))
self.menuHelp = QtGui.QMenu(self.menuBar)
self.menuHelp.setObjectName(_fromUtf8("menuHelp"))
RSLEditorClass.setMenuBar(self.menuBar)
self.mainToolBar = QtGui.QToolBar(RSLEditorClass)
self.mainToolBar.setObjectName(_fromUtf8("mainToolBar"))
RSLEditorClass.addToolBar(QtCore.Qt.TopToolBarArea, self.mainToolBar)
self.actionNew = QtGui.QAction(RSLEditorClass)
icon1 = QtGui.QIcon()
icon1.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/new.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionNew.setIcon(icon1)
self.actionNew.setObjectName(_fromUtf8("actionNew"))
self.actionOpen = QtGui.QAction(RSLEditorClass)
icon2 = QtGui.QIcon()
icon2.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/open.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionOpen.setIcon(icon2)
self.actionOpen.setObjectName(_fromUtf8("actionOpen"))
self.actionSave = QtGui.QAction(RSLEditorClass)
icon3 = QtGui.QIcon()
icon3.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/save.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionSave.setIcon(icon3)
self.actionSave.setObjectName(_fromUtf8("actionSave"))
self.actionSave_As = QtGui.QAction(RSLEditorClass)
icon4 = QtGui.QIcon()
icon4.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/save as.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionSave_As.setIcon(icon4)
self.actionSave_As.setObjectName(_fromUtf8("actionSave_As"))
self.actionUndo = QtGui.QAction(RSLEditorClass)
icon5 = QtGui.QIcon()
icon5.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/undo.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionUndo.setIcon(icon5)
self.actionUndo.setObjectName(_fromUtf8("actionUndo"))
self.actionRedo = QtGui.QAction(RSLEditorClass)
icon6 = QtGui.QIcon()
icon6.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/redo.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionRedo.setIcon(icon6)
self.actionRedo.setObjectName(_fromUtf8("actionRedo"))
self.actionCut = QtGui.QAction(RSLEditorClass)
icon7 = QtGui.QIcon()
icon7.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/cut.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionCut.setIcon(icon7)
self.actionCut.setObjectName(_fromUtf8("actionCut"))
self.actionCopy = QtGui.QAction(RSLEditorClass)
icon8 = QtGui.QIcon()
icon8.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/copy.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionCopy.setIcon(icon8)
self.actionCopy.setObjectName(_fromUtf8("actionCopy"))
self.actionPaste = QtGui.QAction(RSLEditorClass)
icon9 = QtGui.QIcon()
icon9.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/paste.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionPaste.setIcon(icon9)
self.actionPaste.setObjectName(_fromUtf8("actionPaste"))
self.actionSelect_All = QtGui.QAction(RSLEditorClass)
self.actionSelect_All.setObjectName(_fromUtf8("actionSelect_All"))
self.actionCompile = QtGui.QAction(RSLEditorClass)
icon10 = QtGui.QIcon()
icon10.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/compile.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionCompile.setIcon(icon10)
self.actionCompile.setObjectName(_fromUtf8("actionCompile"))
self.actionFind_Replace = QtGui.QAction(RSLEditorClass)
self.actionFind_Replace.setObjectName(_fromUtf8("actionFind_Replace"))
self.actionPreferences = QtGui.QAction(RSLEditorClass)
icon11 = QtGui.QIcon()
icon11.addPixmap(QtGui.QPixmap(_fromUtf8(":/icons/prefs.png")), QtGui.QIcon.Normal, QtGui.QIcon.Off)
self.actionPreferences.setIcon(icon11)
self.actionPreferences.setObjectName(_fromUtf8("actionPreferences"))
self.actionAdd_Shader_Path = QtGui.QAction(RSLEditorClass)
self.actionAdd_Shader_Path.setObjectName(_fromUtf8("actionAdd_Shader_Path"))
self.actionAdd_Textures_Path = QtGui.QAction(RSLEditorClass)
self.actionAdd_Textures_Path.setObjectName(_fromUtf8("actionAdd_Textures_Path"))
self.actionAdd_Archives_Path = QtGui.QAction(RSLEditorClass)
self.actionAdd_Archives_Path.setObjectName(_fromUtf8("actionAdd_Archives_Path"))
self.actionAdd_PointCloud_Path = QtGui.QAction(RSLEditorClass)
self.actionAdd_PointCloud_Path.setObjectName(_fromUtf8("actionAdd_PointCloud_Path"))
self.actionAdd_BrickMap_Path = QtGui.QAction(RSLEditorClass)
self.actionAdd_BrickMap_Path.setObjectName(_fromUtf8("actionAdd_BrickMap_Path"))
self.actionAdd_PhotonMap_Path = QtGui.QAction(RSLEditorClass)
self.actionAdd_PhotonMap_Path.setObjectName(_fromUtf8("actionAdd_PhotonMap_Path"))
self.actionEditor_Docs = QtGui.QAction(RSLEditorClass)
self.actionEditor_Docs.setObjectName(_fromUtf8("actionEditor_Docs"))
self.actionDelight_Docs = QtGui.QAction(RSLEditorClass)
self.actionDelight_Docs.setObjectName(_fromUtf8("actionDelight_Docs"))
self.actionPRMan_Docs = QtGui.QAction(RSLEditorClass)
self.actionPRMan_Docs.setObjectName(_fromUtf8("actionPRMan_Docs"))
self.actionFundza = QtGui.QAction(RSLEditorClass)
self.actionFundza.setObjectName(_fromUtf8("actionFundza"))
self.actionColor_Picker = QtGui.QAction(RSLEditorClass)
self.actionColor_Picker.setObjectName(_fromUtf8("actionColor_Picker"))
self.actionClose_Tab = QtGui.QAction(RSLEditorClass)
self.actionClose_Tab.setObjectName(_fromUtf8("actionClose_Tab"))
self.actionExit = QtGui.QAction(RSLEditorClass)
self.actionExit.setObjectName(_fromUtf8("actionExit"))
self.menuRecent_Files.addSeparator()
self.menuFile.addAction(self.actionNew)
self.menuFile.addAction(self.actionOpen)
self.menuFile.addAction(self.menuRecent_Files.menuAction())
self.menuFile.addSeparator()
self.menuFile.addAction(self.actionSave)
self.menuFile.addAction(self.actionSave_As)
self.menuFile.addSeparator()
self.menuFile.addAction(self.actionClose_Tab)
self.menuFile.addAction(self.actionExit)
self.menuSearch_Options.addAction(self.actionAdd_Shader_Path)
self.menuSearch_Options.addAction(self.actionAdd_Textures_Path)
self.menuSearch_Options.addAction(self.actionAdd_Archives_Path)
self.menuSearch_Options.addAction(self.actionAdd_PointCloud_Path)
self.menuSearch_Options.addAction(self.actionAdd_BrickMap_Path)
self.menuSearch_Options.addAction(self.actionAdd_PhotonMap_Path)
self.menuEdit.addAction(self.actionUndo)
self.menuEdit.addAction(self.actionRedo)
self.menuEdit.addSeparator()
self.menuEdit.addAction(self.actionCut)
self.menuEdit.addAction(self.actionCopy)
self.menuEdit.addAction(self.actionPaste)
self.menuEdit.addSeparator()
self.menuEdit.addAction(self.actionSelect_All)
self.menuEdit.addSeparator()
self.menuEdit.addAction(self.menuSearch_Options.menuAction())
self.menuEdit.addAction(self.actionColor_Picker)
self.menuSearch.addAction(self.actionFind_Replace)
self.menuDebug.addAction(self.actionCompile)
self.menuDebug.addSeparator()
self.menuDebug.addAction(self.actionPreferences)
self.menuHelp.addAction(self.actionEditor_Docs)
self.menuHelp.addSeparator()
self.menuHelp.addAction(self.actionDelight_Docs)
self.menuHelp.addSeparator()
self.menuHelp.addAction(self.actionPRMan_Docs)
self.menuHelp.addSeparator()
self.menuHelp.addAction(self.actionFundza)
self.menuBar.addAction(self.menuFile.menuAction())
self.menuBar.addAction(self.menuEdit.menuAction())
self.menuBar.addAction(self.menuView.menuAction())
self.menuBar.addAction(self.menuSearch.menuAction())
self.menuBar.addAction(self.menuDebug.menuAction())
self.menuBar.addAction(self.menuHelp.menuAction())
self.mainToolBar.addAction(self.actionNew)
self.mainToolBar.addAction(self.actionOpen)
self.mainToolBar.addAction(self.actionSave)
self.mainToolBar.addAction(self.actionSave_As)
self.mainToolBar.addAction(self.actionUndo)
self.mainToolBar.addAction(self.actionRedo)
self.mainToolBar.addAction(self.actionCut)
self.mainToolBar.addAction(self.actionCopy)
self.mainToolBar.addAction(self.actionPaste)
self.mainToolBar.addAction(self.actionCompile)
self.mainToolBar.addAction(self.actionPreferences)
self.retranslateUi(RSLEditorClass)
self.tabMain.setCurrentIndex(-1)
self.tabOutput.setCurrentIndex(0)
QtCore.QMetaObject.connectSlotsByName(RSLEditorClass)
def retranslateUi(self, RSLEditorClass):
RSLEditorClass.setWindowTitle(_translate("RSLEditorClass", "RSLEditor", None))
self.lbFilePath.setText(_translate("RSLEditorClass", "File Path:", None))
self.tabOutput.setTabText(self.tabOutput.indexOf(self.tabRSLOut), _translate("RSLEditorClass", "RSL Output", None))
self.tabOutput.setTabText(self.tabOutput.indexOf(self.tabRIBOut), _translate("RSLEditorClass", "RIB Output", None))
self.menuFile.setTitle(_translate("RSLEditorClass", "File", None))
self.menuRecent_Files.setTitle(_translate("RSLEditorClass", "Recent Files", None))
self.menuEdit.setTitle(_translate("RSLEditorClass", "Edit", None))
self.menuSearch_Options.setTitle(_translate("RSLEditorClass", "Search Options", None))
self.menuView.setTitle(_translate("RSLEditorClass", "View", None))
self.menuSearch.setTitle(_translate("RSLEditorClass", "Search", None))
self.menuDebug.setTitle(_translate("RSLEditorClass", "Debug", None))
self.menuHelp.setTitle(_translate("RSLEditorClass", "Help", None))
self.actionNew.setText(_translate("RSLEditorClass", "New", None))
self.actionNew.setToolTip(_translate("RSLEditorClass", "Create a new file", None))
self.actionNew.setShortcut(_translate("RSLEditorClass", "Ctrl+N", None))
self.actionOpen.setText(_translate("RSLEditorClass", "Open", None))
self.actionOpen.setToolTip(_translate("RSLEditorClass", "Open a file", None))
self.actionOpen.setShortcut(_translate("RSLEditorClass", "Ctrl+O", None))
self.actionSave.setText(_translate("RSLEditorClass", "Save", None))
self.actionSave.setToolTip(_translate("RSLEditorClass", "Save the current file", None))
self.actionSave.setShortcut(_translate("RSLEditorClass", "Ctrl+S", None))
self.actionSave_As.setText(_translate("RSLEditorClass", "Save As", None))
self.actionSave_As.setToolTip(_translate("RSLEditorClass", "Save as a new file", None))
self.actionSave_As.setShortcut(_translate("RSLEditorClass", "Ctrl+Alt+S", None))
self.actionUndo.setText(_translate("RSLEditorClass", "Undo", None))
self.actionUndo.setToolTip(_translate("RSLEditorClass", "Undo last action", None))
self.actionUndo.setShortcut(_translate("RSLEditorClass", "Ctrl+Z", None))
self.actionRedo.setText(_translate("RSLEditorClass", "Redo", None))
self.actionRedo.setToolTip(_translate("RSLEditorClass", "Redo last action", None))
self.actionRedo.setShortcut(_translate("RSLEditorClass", "Shift+Z", None))
self.actionCut.setText(_translate("RSLEditorClass", "Cut", None))
self.actionCut.setToolTip(_translate("RSLEditorClass", "Cut selected text", None))
self.actionCut.setShortcut(_translate("RSLEditorClass", "Ctrl+X", None))
self.actionCopy.setText(_translate("RSLEditorClass", "Copy", None))
self.actionCopy.setToolTip(_translate("RSLEditorClass", "Copy selected text", None))
self.actionCopy.setShortcut(_translate("RSLEditorClass", "Ctrl+C", None))
self.actionPaste.setText(_translate("RSLEditorClass", "Paste", None))
self.actionPaste.setToolTip(_translate("RSLEditorClass", "Paste text", None))
self.actionPaste.setShortcut(_translate("RSLEditorClass", "Ctrl+V", None))
self.actionSelect_All.setText(_translate("RSLEditorClass", "Select All", None))
self.actionSelect_All.setToolTip(_translate("RSLEditorClass", "Select all text", None))
self.actionSelect_All.setShortcut(_translate("RSLEditorClass", "Ctrl+A", None))
self.actionCompile.setText(_translate("RSLEditorClass", "Compile", None))
self.actionCompile.setToolTip(_translate("RSLEditorClass", "Compile current document", None))
self.actionCompile.setShortcut(_translate("RSLEditorClass", "Alt+E", None))
self.actionFind_Replace.setText(_translate("RSLEditorClass", "Find/Replace", None))
self.actionFind_Replace.setToolTip(_translate("RSLEditorClass", "Open find/replace dialog", None))
self.actionFind_Replace.setShortcut(_translate("RSLEditorClass", "Ctrl+F", None))
self.actionPreferences.setText(_translate("RSLEditorClass", "Preferences", None))
self.actionPreferences.setToolTip(_translate("RSLEditorClass", "Open the preferences dialog", None))
self.actionPreferences.setShortcut(_translate("RSLEditorClass", "Ctrl+P", None))
self.actionAdd_Shader_Path.setText(_translate("RSLEditorClass", "Add Shader Path", None))
self.actionAdd_Textures_Path.setText(_translate("RSLEditorClass", "Add Textures Path", None))
self.actionAdd_Archives_Path.setText(_translate("RSLEditorClass", "Add Archives Path", None))
self.actionAdd_PointCloud_Path.setText(_translate("RSLEditorClass", "Add PointCloud Path", None))
self.actionAdd_BrickMap_Path.setText(_translate("RSLEditorClass", "Add BrickMap Path", None))
self.actionAdd_PhotonMap_Path.setText(_translate("RSLEditorClass", "Add PhotonMap Path", None))
self.actionEditor_Docs.setText(_translate("RSLEditorClass", "Editor Docs", None))
self.actionDelight_Docs.setText(_translate("RSLEditorClass", "3Delight Docs", None))
self.actionPRMan_Docs.setText(_translate("RSLEditorClass", "PRMan Docs", None))
self.actionFundza.setText(_translate("RSLEditorClass", "Fundza", None))
self.actionColor_Picker.setText(_translate("RSLEditorClass", "Color Picker", None))
self.actionColor_Picker.setShortcut(_translate("RSLEditorClass", "Ctrl+I", None))
self.actionClose_Tab.setText(_translate("RSLEditorClass", "Close Tab", None))
self.actionExit.setText(_translate("RSLEditorClass", "Exit", None))
import rsleditor_rc
A: If you scroll down to the bottom of the setupUi method in the ui module, you'll see this line:
QtCore.QMetaObject.connectSlotsByName(RSLEditorClass)
What this does, is to automatically connect signals to slots based on the format of the slot names. The format of the slot name is:
on_[object name]_[signal name]
So looking at the way you're connecting your actions:
self.ui.actionNew.triggered.connect(self.on_actionNew_triggered)
it should be clear that you're using this format when naming your slots, and that that is where the problem lies. But why does this cause the slot to be called three times? Well, in PyQt, the triggered signal has two overloads: one which sends the default checked parameter, and one which doesn't. If you don't specify which one of these you want to connect to, both overloads will get connected. So in your case, what happens is that on_actionNew_triggered gets connected twice by connectSlotsByName and once by your own explicit connection, making three in all. And presumably, a similar story can be told for the other actions.
To fix this, you can either rename your slots so they don't use the auto-connection format:
self.ui.actionNew.triggered.connect(self.handleActionNew)
...
def handleActionNew(self):
...
or get rid of the explicit connection, and use the pyqtSlot decorator to select the right overload for auto-connection:
@QtCore.pyqtSlot()
def on_actionNew_triggered(self):
...
or:
@QtCore.pyqtSlot(bool)
def on_actionNew_triggered(self, checked):
...
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 6,264
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Kremselen is een buurtschap in de gemeente Meierijstad in de Nederlandse provincie Noord-Brabant. Het ligt drie kilometer ten zuidwesten van het dorp Sint-Oedenrode.
Buurtschap in Noord-Brabant
Geografie van Meierijstad
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,246
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{"url":"https:\/\/chesterrep.openrepository.com\/handle\/10034\/621199\/browse?view=list&rpp=20&etal=-1&sort_by=1&type=title&starts_with=G&order=ASC","text":"Now showing items 787-806 of 1928\n\n\u2022 #### Galerkin finite element approximation of a stochastic semilinear fractional subdiffusion with fractionally integrated additive noise\n\nAbstract A Galerkin finite element method is applied to approximate the solution of a semilinear stochastic space and time fractional subdiffusion problem with the Caputo fractional derivative of the order $\\alpha \\in (0, 1)$, driven by fractionally integrated additive noise. After discussing the existence, uniqueness and regularity results, we approximate the noise with the piecewise constant function in time, in order to obtain a regularized stochastic fractional subdiffusion problem. The regularized problem is then approximated by using the finite element method in spatial direction. The mean squared errors are proved based on the sharp estimates of the various Mittag\u2013Leffler functions involved in the integrals. Numerical experiments are conducted to show that the numerical results are consistent with the theoretical findings.\n\u2022 #### Gas decomposition and electrode degradation characteristics of a 20% C 3 F 7 CN and 80% CO 2 gas mixture for high voltage accelerators\n\nAbstract: Sulphur hexafluoride (SF6) is a potent greenhouse gas used in high voltage accelerators. As a promising alternative to SF6, the C3F7CN\/CO2 gas mixture and its by\u2010products are of great interest to ensure the safe operation of accelerators that will adopt any SF6\u2010free solution. This work experimentally examines the electrical ageing characteristics of a 20% C3F7CN\/80% CO2 gas mixture tested using spark gaps under a pressure of 7.2 bar (abs.). Gas samples were collected after 1000 DC breakdowns and analysed using gas chromatography mass spectrometry (GC\u2010MS) with an estimated toxicity value of 54,459 ppmv, which indicates the aged mixture to be non\u2010toxic. Subsequent investigation was conducted on the gas\u2010solid interface after 500 breakdowns for both SF6 and the 20% C3F7CN\/80% CO2 gas mixture. Aged electrodes were analysed using X\u2010ray photoelectron spectroscopy (XPS) and X\u2010ray diffraction (XRD) techniques. Electrode surface analysis revealed the formation of metal fluorides on the electrode surface tested using the 20% C3F7CN\/80% CO2 mixture, whereas metal fluorides and sulphides were detected for electrodes tested with SF6. The findings provide a reference on the toxicity and gas\u2010solid interaction of the electrically aged 20% C3F7CN\/80% CO2 gas mixture for potential retro\u2010fill application in high voltage accelerators.\n\u2022 #### Gene editing enables rapid engineering of complex antibiotic assembly lines\n\nAbstract: Re-engineering biosynthetic assembly lines, including nonribosomal peptide synthetases (NRPS) and related megasynthase enzymes, is a powerful route to new antibiotics and other bioactive natural products that are too complex for chemical synthesis. However, engineering megasynthases is very challenging using current methods. Here, we describe how CRISPR-Cas9 gene editing can be exploited to rapidly engineer one of the most complex megasynthase assembly lines in nature, the 2.0 MDa NRPS enzymes that deliver the lipopeptide antibiotic enduracidin. Gene editing was used to exchange subdomains within the NRPS, altering substrate selectivity, leading to ten new lipopeptide variants in good yields. In contrast, attempts to engineer the same NRPS using a conventional homologous recombination-mediated gene knockout and complementation approach resulted in only traces of new enduracidin variants. In addition to exchanging subdomains within the enduracidin NRPS, subdomains from a range of NRPS enzymes of diverse bacterial origins were also successfully utilized.\n\u2022 #### Gene expression signatures predict response to therapy with growth hormone\n\nAbstract: Recombinant human growth hormone (r-hGH) is used as a therapeutic agent for disorders of growth including growth hormone deficiency (GHD) and Turner syndrome (TS). Treatment is costly and current methods to model response are inexact. GHD (n = 71) and TS patients (n = 43) were recruited to study response to r-hGH over 5 years. Analysis was performed using 1219 genetic markers and baseline (pre-treatment) blood transcriptome. Random forest was used to determine predictive value of transcriptomic data associated with growth response. No genetic marker passed the stringency criteria for prediction. However, we identified an identical set of genes in both GHD and TS whose expression could be used to classify therapeutic response to r-hGH with a high accuracy (AUC > 0.9). Combining transcriptomic markers with clinical phenotype was shown to significantly reduce predictive error. This work could be translated into a single genomic test linked to a prediction algorithm to improve clinical management. Trial registration numbers: NCT00256126 and NCT00699855.\n\u2022 #### Gene expression signatures predict response to therapy with growth hormone.\n\nRecombinant human growth hormone (r-hGH) is used as a therapeutic agent for disorders of growth including growth hormone deficiency (GHD) and Turner syndrome (TS). Treatment is costly and current methods to model response are inexact. GHD (n\u2009=\u200971) and TS patients (n\u2009=\u200943) were recruited to study response to r-hGH over 5 years. Analysis was performed using 1219 genetic markers and baseline (pre-treatment) blood transcriptome. Random forest was used to determine predictive value of transcriptomic data associated with growth response. No genetic marker passed the stringency criteria for prediction. However, we identified an identical set of genes in both GHD and TS whose expression could be used to classify therapeutic response to r-hGH with a high accuracy (AUC\u2009>\u20090.9). Combining transcriptomic markers with clinical phenotype was shown to significantly reduce predictive error. This work could be translated into a single genomic test linked to a prediction algorithm to improve clinical management. Trial registration numbers: NCT00256126 and NCT00699855.\n\u2022 #### Gene Panel Testing for Breast Cancer Reveals Differential Effect of Prior BRCA1\/2 Probability\n\nWhilst panel testing of an extended group of genes including BRCA1\/2 is commonplace, these studies have not been subdivided by histiotype or by a priori BRCA1\/2 probability. Patients with a breast cancer diagnosis undergoing extended panel testing were assessed for frequency of actionable variants in breast cancer genes other than BRCA1\/2 by histiotype and Manchester score (MS) to reflect a priori BRCA1\/2 likelihood. Rates were adjusted by prior testing for BRCA1\/2 in an extended series. 95\/1398 (6.3%) who underwent panel testing were found to be positive for actionable non-BRCA1\/2 breast\/ovarian cancer genes (ATM, BARD1, CDH1, CHEK2, PALB2, PTEN, RAD51C, RAD51D, TP53). As expected, PALB2, CHEK2 and ATM were predominant with 80-(5.3%). The highest rate occurred in Grade-3 ER+\/HER2\u2212 breast cancers-(9.6%). Rates of non-BRCA actionable genes was fairly constant over all likelihoods of BRCA1\/2 but adjusted rates were three times higher with MS 9 (BRCA1\/2 = 1.5%, other = 4.7%), but was only 1.6% compared to 79.3% with MS \u2265 40. Although rates of detection of non-BRCA actionable genes are relatively constant across BRCA1\/2 likelihoods this disguises an overall adjusted low frequency in high-likelihood families which have been heavily pre-tested for BRCA1\/2. Any loss of detection sensitivity for BRCA1\/2 actionable variants in breast cancer panels should lead to bespoke BRCA1\/2 testing being conducted first.\n\u2022 #### Gene-Environment Interactions Relevant to Estrogen and Risk of Breast Cancer: Can Gene-Environment Interactions Be Detected Only among Candidate SNPs from Genome-Wide Association Studies?\n\nIn this study we aim to examine gene\u2013environment interactions (GxEs) between genes involved with estrogen metabolism and environmental factors related to estrogen exposure. GxE analyses were conducted with 1970 Korean breast cancer cases and 2052 controls in the case-control study, the Seoul Breast Cancer Study (SEBCS). A total of 11,555 SNPs from the 137 candidate genes were included in the GxE analyses with eight established environmental factors. A replication test was conducted by using an independent population from the Breast Cancer Association Consortium (BCAC), with 62,485 Europeans and 9047 Asians. The GxE tests were performed by using two-step methods in GxEScan software. Two interactions were found in the SEBCS. The first interaction was shown between rs13035764 of NCOA1 and age at menarche in the GE|2df model (p-2df = 1.2 \u00d7 10\u22123). The age at menarche before 14 years old was associated with the high risk of breast cancer, and the risk was higher when subjects had homozygous minor allele G. The second GxE was shown between rs851998 near ESR1 and height in the GE|2df model (p-2df = 1.1 \u00d7 10\u22124). Height taller than 160 cm was associated with a high risk of breast cancer, and the risk increased when the minor allele was added. The findings were not replicated in the BCAC. These results would suggest specificity in Koreans for breast cancer risk.\n\u2022 #### Generation and Characterization of the Drosophila melanogaster paralytic Gene Knock-Out as a Model for Dravet Syndrome\n\nDravet syndrome is a severe rare epileptic disease caused by mutations in the SCN1A gene coding for the Nav1.1 protein, a voltage-gated sodium channel alpha subunit. We have made a knock-out of the paralytic gene, the single Drosophila melanogaster gene encoding this type of protein, by homologous recombination. These flies showed a heat-induced seizing phenotype, and sudden death in long term seizures. In addition to seizures, neuromuscular alterations were observed in climbing, flight, and walking tests. Moreover, they also manifested some cognitive alterations, such as anxiety and problems in learning. Electrophysiological analyses from larval motor neurons showed a decrease in cell capacitance and membrane excitability, while persistent sodium current increased. To detect alterations in metabolism, we performed an NMR metabolomic profiling of heads, which revealed higher levels in some amino acids, succinate, and lactate; and also an increase in the abundance of GABA, which is the main neurotransmitter implicated in Dravet syndrome. All these changes in the paralytic knock-out flies indicate that this is a good model for epilepsy and specifically for Dravet syndrome. This model could be a new tool to understand the pathophysiology of the disease and to find biomarkers, genetic modifiers and new treatments.\n\u2022 #### Generation of Alkalinity by Stimulation of Microbial Iron Reduction in Acid Rock Drainage Systems: Impact of Natural Organic Matter Types\n\nAbstract: To determine the role of organic matter in the attenuation of acid rock drainage (ARD), microcosm-based experiments were performed using ARD stimulated with plants and manures. Initial mineralogical, organic geochemical and microbial analyses indicated a predominance of goethite, a substantial amount of organic carbon originating from local sources, and a bacterial community comparable with those detected in a range of ARD sites worldwide. After 100 days of incubation, changes in the mineralogical, organic and microbiological composition of the ARD demonstrated that the plant additions stimulate microbes with the potential to degrade this organic matter but do not necessarily cause substantial Fe(III) reduction. Conversely, the greatest observed stimulation of Fe(III) reduction, associated with an increase in pH to near-neutral values, was observed using manure additions. These results demonstrate that the use of the optimal natural carbon source is important and can promote the metabolism of microorganisms potentially fuelling a range of geomicrobial processes, including iron and sulfate reduction.\n\n\u2022 #### Genetic and process engineering strategies for enhanced recombinant N -glycoprotein production in bacteria\n\nAbstract: Background: The production of N-linked glycoproteins in genetically amenable bacterial hosts offers great potential for reduced cost, faster\/simpler bioprocesses, greater customisation, and utility for distributed manufacturing of glycoconjugate vaccines and glycoprotein therapeutics. Efforts to optimize production hosts have included heterologous expression of glycosylation enzymes, metabolic engineering, use of alternative secretion pathways, and attenuation of gene expression. However, a major bottleneck to enhance glycosylation efficiency, which limits the utility of the other improvements, is the impact of target protein sequon accessibility during glycosylation. Results: Here, we explore a series of genetic and process engineering strategies to increase recombinant N-linked glycosylation, mediated by the Campylobacter-derived PglB oligosaccharyltransferase in Escherichia coli. Strategies include increasing membrane residency time of the target protein by modifying the cleavage site of its secretion signal, and modulating protein folding in the periplasm by use of oxygen limitation or strains with compromised oxidoreductase or disulphide-bond isomerase activity. These approaches achieve up to twofold improvement in glycosylation efficiency. Furthermore, we also demonstrate that supplementation with the chemical oxidant cystine enhances the titre of glycoprotein in an oxidoreductase knockout strain by improving total protein production and cell fitness, while at the same time maintaining higher levels of glycosylation efficiency. Conclusions: In this study, we demonstrate that improved protein glycosylation in the heterologous host could be achieved by mimicking the coordination between protein translocation, folding and glycosylation observed in native host such as Campylobacter jejuni and mammalian cells. Furthermore, it provides insight into strain engineering and bioprocess strategies, to improve glycoprotein yield and titre, and to avoid physiological burden of unfolded protein stress upon cell growth. The process and genetic strategies identified herein will inform further optimisation and scale-up of heterologous recombinant N-glycoprotein production.\n\u2022 #### Genetic and process engineering strategies for enhanced recombinant N -glycoprotein production in bacteria\n\nAbstract: Background: The production of N-linked glycoproteins in genetically amenable bacterial hosts offers great potential for reduced cost, faster\/simpler bioprocesses, greater customisation, and utility for distributed manufacturing of glycoconjugate vaccines and glycoprotein therapeutics. Efforts to optimize production hosts have included heterologous expression of glycosylation enzymes, metabolic engineering, use of alternative secretion pathways, and attenuation of gene expression. However, a major bottleneck to enhance glycosylation efficiency, which limits the utility of the other improvements, is the impact of target protein sequon accessibility during glycosylation. Results: Here, we explore a series of genetic and process engineering strategies to increase recombinant N-linked glycosylation, mediated by the Campylobacter-derived PglB oligosaccharyltransferase in Escherichia coli. Strategies include increasing membrane residency time of the target protein by modifying the cleavage site of its secretion signal, and modulating protein folding in the periplasm by use of oxygen limitation or strains with compromised oxidoreductase or disulphide-bond isomerase activity. These approaches achieve up to twofold improvement in glycosylation efficiency. Furthermore, we also demonstrate that supplementation with the chemical oxidant cystine enhances the titre of glycoprotein in an oxidoreductase knockout strain by improving total protein production and cell fitness, while at the same time maintaining higher levels of glycosylation efficiency. Conclusions: In this study, we demonstrate that improved protein glycosylation in the heterologous host could be achieved by mimicking the coordination between protein translocation, folding and glycosylation observed in native host such as Campylobacter jejuni and mammalian cells. Furthermore, it provides insight into strain engineering and bioprocess strategies, to improve glycoprotein yield and titre, and to avoid physiological burden of unfolded protein stress upon cell growth. The process and genetic strategies identified herein will inform further optimisation and scale-up of heterologous recombinant N-glycoprotein production.\n\u2022 #### Genetic and process engineering strategies for enhanced recombinant N-glycoprotein production in bacteria.\n\n<h4>Background<\/h4>The production of N-linked glycoproteins in genetically amenable bacterial hosts offers great potential for reduced cost, faster\/simpler bioprocesses, greater customisation, and utility for distributed manufacturing of glycoconjugate vaccines and glycoprotein therapeutics. Efforts to optimize production hosts have included heterologous expression of glycosylation enzymes, metabolic engineering, use of alternative secretion pathways, and attenuation of gene expression. However, a major bottleneck to enhance glycosylation efficiency, which limits the utility of the other improvements, is the impact of target protein sequon accessibility during glycosylation.<h4>Results<\/h4>Here, we explore a series of genetic and process engineering strategies to increase recombinant N-linked glycosylation, mediated by the Campylobacter-derived PglB oligosaccharyltransferase in Escherichia coli. Strategies include increasing membrane residency time of the target protein by modifying the cleavage site of its secretion signal, and modulating protein folding in the periplasm by use of oxygen limitation or strains with compromised oxidoreductase or disulphide-bond isomerase activity. These approaches achieve up to twofold improvement in glycosylation efficiency. Furthermore, we also demonstrate that supplementation with the chemical oxidant cystine enhances the titre of glycoprotein in an oxidoreductase knockout strain by improving total protein production and cell fitness, while at the same time maintaining higher levels of glycosylation efficiency.<h4>Conclusions<\/h4>In this study, we demonstrate that improved protein glycosylation in the heterologous host could be achieved by mimicking the coordination between protein translocation, folding and glycosylation observed in native host such as Campylobacter jejuni and mammalian cells. Furthermore, it provides insight into strain engineering and bioprocess strategies, to improve glycoprotein yield and titre, and to avoid physiological burden of unfolded protein stress upon cell growth. The process and genetic strategies identified herein will inform further optimisation and scale-up of heterologous recombinant N-glycoprotein production.\n\n\u2022 #### Genetically defined favourable adiposity is not associated with a clinically meaningful difference in clinical course in people with type 2 diabetes but does associate with a favourable metabolic profile\n\nAbstract: Aims: Change in weight, HbA1c, lipids, blood pressure and cardiometabolic events over time is variable in individuals with type 2 diabetes. We hypothesised that people with a genetic predisposition to a more favourable adiposity distribution could have a less severe clinical course\/progression. Methods: We involved people with type 2 diabetes from two UK\u2010based cohorts: 11,914 individuals with GP follow\u2010up data from the UK Biobank and 723 from Salford. We generated a \u2018favourable adiposity\u2019 genetic score and conducted cross\u2010sectional and longitudinal studies to test its association with weight, BMI, lipids, blood pressure, medication use and risk of myocardial infarction and stroke using 15 follow\u2010up time points with 1\u2010year intervals. Results: The \u2018favourable adiposity\u2019 genetic score was cross\u2010sectionally associated with higher weight (effect size per 1 standard deviation higher genetic score: 0.91 kg [0.59,1.23]) and BMI (0.30 kg\/m2 [0.19,0.40]), but higher high\u2010density lipoprotein (0.02 mmol\/L [0.01,0.02]) and lower triglycerides (\u22120.04 mmol\/L [\u22120.07, \u22120.02]) in the UK Biobank at baseline, and this pattern of association was consistent across follow\u2010up. There was a trend for participants with higher \u2018favourable adiposity\u2019 genetic score to have lower risk of myocardial infarction and\/or stroke (odds ratio 0.79 [0.62, 1.00]) compared to those with lower score. A one standard deviation higher score was associated with lower odds of using lipid\u2010lowering (0.91 [0.86, 0.97]) and anti\u2010hypertensive medication (0.95 [0.91, 0.99]). Conclusions: In individuals with type 2 diabetes, having more \u2018favourable adiposity\u2019 alleles is associated with a marginally better lipid profile long\u2010term and having lower odds of requiring lipid\u2010lowering or anti\u2010hypertensive medication in spite of relatively higher adiposity.\n\u2022 #### Geochemical compositional controls on DNA strand breaks induced in in vitro cell-free assays by crushed rock powders from the Panasqueira mine area, Portugal\n\nAbstract: DNA strand breaks are a common form of DNA damage that can contribute to chromosomal instability or gene mutations. Such strand breaks may be caused by exposure to heavy metals. The aim of this study was to assess the level of DNA strand breaks caused by \u00b5m-scale solid particles of known chemical composition with elevated heavy metals\/metalloids, notably arsenic, using an in vitro cell-free DNA plasmid scission assay. These samples were incubated with and without H2O2 to see whether damage occurs directly or indirectly through the Fenton reaction. Levels of DNA damage in the absence of H2O2 were < 10%, but in the presence of H2O2, all samples showed higher levels of damage ranging from 10 to 100% suggesting that damage was being incurred through the Fenton reaction. Using bivariate correlation analysis and multiple linear regression, manganese oxide (MnO), sulphur (S), copper (Cu), and zinc (Zn) concentrations in the particulates were found to be the most significant predictors of DNA damage. The mechanism of this DNA damage formation has yet to be thoroughly investigated but is hypothesised to be due to reactive oxygen species formation. Further work is required to assess the extent of contribution of reactive oxygen species to this DNA damage, but this study highlights the potential role of chemistry and\/or mineralogy to the extent and\/or nature of DNA damage caused by particulates.\n\n\u2022 #### Geostatistical model of the spatial distribution of arsenic in groundwaters in Gujarat State, India\n\nAbstract: Geogenic arsenic contamination in groundwaters poses a severe health risk to hundreds of millions of people globally. Notwithstanding the particular risks to exposed populations in the Indian sub-continent, at the time of writing, there was a paucity of geostatistically based models of the spatial distribution of groundwater hazard in India. In this study, we used logistic regression models of secondary groundwater arsenic data with research-informed secondary soil, climate and topographic variables as principal predictors generate hazard and risk maps of groundwater arsenic at a resolution of 1 km across Gujarat State. By combining models based on different arsenic concentrations, we have generated a pseudo-contour map of groundwater arsenic concentrations, which indicates greater arsenic hazard (> 10 \u03bcg\/L) in the northwest, northeast and south-east parts of Kachchh District as well as northwest and southwest Banas Kantha District. The total number of people living in areas in Gujarat with groundwater arsenic concentration exceeding 10 \u03bcg\/L is estimated to be around 122,000, of which we estimate approximately 49,000 people consume groundwater exceeding 10 \u00b5g\/L. Using simple previously published dose\u2013response relationships, this is estimated to have given rise to 700 (prevalence) cases of skin cancer and around 10 cases of premature avoidable mortality\/annum from internal (lung, liver, bladder) cancers\u2014that latter value is on the order of just 0.001% of internal cancers in Gujarat, reflecting the relative low groundwater arsenic hazard in Gujarat State.\n\u2022 #### Gertrude Elles: the pioneering graptolite geologist in a woolly hat. Her career, achievements and personal reflections from her family and colleagues\n\nAbstractGertrude Elles gained worldwide renown for her seminal work with Ethel Wood on A Monograph of British Graptolites, which is still used today. She gained the MBE, pioneered female geological education, became the first female reader in Cambridge University and one of the first tranche of female Fellows of the Geological Society in 1919. An eccentric with a vast array of hats, PhD students and lodgers, she was a stalwart member of the Sedgwick Club and life member of the British Federation of University Women. She wrote obituaries for colleagues describing their achievements with humour and good nature. Her family describe her as \u2018a fabulous woman\u2019 with a huge range of interests including archaeology, botany and music. She related her geological and botanical knowledge in showing a nephew that plants growing along the Moine Thrust reflected change in the underlying rocks. Cambridge colleagues recall her as a \u2018marvellous and well-respected figure\u2019 who caused some amusement by her big old cluttered table from which she swept away material making room for new samples (and work for technicians). She died in 1960 in her beloved Scotland. However, her legacy survives in the classification of a group of fossils extinct for nearly 400 myr.","date":"2021-12-09 10:33:18","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\": 0.3572723865509033, \"perplexity\": 8185.073509398891}, \"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\/1637964363791.16\/warc\/CC-MAIN-20211209091917-20211209121917-00098.warc.gz\"}"}
| null | null |
Q: How to determine the ANOVA table for split-plot with repeated measures experiment? I have an experiment where observations are one of eight different treatment combinations and are then either one of two different heights. There are 4 blocks, with 8 plots for each block, and the two heights are thought of as split-plots.
I.e. the data table resembles:
I don't know how to set up the anova table, i.e. the sources of variation. Would somebody mind running through the set up of the sources of variation with me as I understand that in some cases split plot analysis can be used for repeated measures experiment. In this case when there are only two repeated measures, so split plot methods can be used I just am at a loss when it comes to the analysis of variance table.
As i am a new user I cannot post the data table. The treatments in the experiment consist of 8 combinations of factor C (1,2,3,4) and factor D (1,2). There are 4 blocks each with the 8 plots per block, the different heights are split-plots. So there are 64 observations, i.e 32 per height (ground level and 10m)
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\chapter{ Introduction}}
In this contribution I will report on certain {\it non-linear} and {\it
non-local} extensions of the conformal or Virasoro algebra
(with central extension). These algebras will be
called $V$-algebras.
The procedure will be to
go from physics to mathematics. In particular, I will start with a certain
$1+1$
dimensional field theory (non-abelian Toda field theory)
defined by its action functional. The
corresponding equations of motion admit three conserved ``left-moving" and
three conserved ``right-moving" quantities (spin-two currents),
expressed in terms of the
interacting fields. I will define canonical Poisson
brackets from the action functional, which provide the phase space with
the canonical symplectic structure. Then I compute the Poisson brackets
of the conserved quantities. This algebra closes in a non-linear
and non-local way, reminiscent of $W$-algebras (which are however local),
so it seemed appropriate to call this algebra a $V$-algebra. The explicit
classical solutions of the equations of motion induce a ``chiral"
realization of the conserved quantities (i.e. of the generators of the
$V$-algebra) in terms of free fields. This will be the contents of section 2.
Then, in the third section, I relate the previously found results to
standard mathematical structures. I show that exactly the same
algebra is obtained as the second Gelfand-Dikii symplectic structure based
on a second-order differential operator, that is a $2\times 2$-matrix
Schr\"odinger operator $L=-\partial^2+U$. Applying a straightforward matrix
generalization of the classical work on the resolvent by Gelfand and Dikii,
this provides us with an infinity of Hamiltonians in involution, hence an
infinity of conserved charges for the non-abelian Toda field theory I
started with. It also immediately implies the connection with matrix
KdV-hierarchies.
In the fourth section, I generalize all these developments to
$n\times n$-matrix $m^{\rm th}$-order differential operators
$L=-\partial^m +U_2\partial^{m-2}+U_3\partial^{m-3}+\ldots +U_m$ and
so-called $V_{n,m}$-algebras. The latter are $n\times n$-matrix
generalizations of the $W_m$-algebras. They are non-local due to the reduction
to
$U_1=0$ and the
non-commutativity of matrices. This fourth section is mathematically
self-contained,
(except that the reader is referred to my original paper for the proofs). Of
course,
taking this section just by itself, misses the whole point of the present
contribution (at least in my physicist's point of view), which is to relate
abstract
mathematical structures to certain physical
theories.
\chapter{The non-abelian Toda field theory}
\section{The action and equations of motion\ }
The one plus one dimensional field theory I will consider is defined by its
action functional
$$ S\equiv S[r,t,\varphi]= {2\over \gamma^2}\int \sqrt{2}\sigma \sqrt{2}\tau
\left( \partial_u r\partial_v r+{\rm th}^2 r\, \partial_u t\partial_v t +\partial_u \varphi\partial_v\varphi + {\rm ch} 2r\, e^{2\varphi}
\right) \ ,
\eqn\uiii$$
where $\tau$ is the time and $\sigma$ the space-coordinate and
$u=\tau+\sigma$ and $v=\tau-\sigma$.
The three fields are $r(\tau,\sigma), t(\tau,\sigma)$ and $\varphi(\tau,\sigma)$. This action
has the following physical interpretation.
The first two, $\varphi$-independent terms, constitute a sigma-model
describing a string on a two-dimensional black hole background,
while the last two, $\varphi$-dependent terms
correspond to an ``internal" field $\varphi$ (or a flat third dimension) and
a tachyon potential ${\rm ch} 2r\, e^{2\varphi}$. However, this interpretation need
not concern us here. The constant $\gamma^2$ plays the role
of the Planck constant and will be seen later to control the central
charge of the conformal algebra. As discussed below, this model is obtained
by gauging a {\it nilpotent} subalgebra of the Lie algebra $B_2$.
The theory defined by the above action is known as the non-abelian Toda
theory \REF\LS{A.N. Leznov and M.V. Saveliev, {\it Two-dimensional exactly and
completely integrable dynamical systems},
\CMP 89 1983 59 .} [\LS]
associated with the Lie algebra $B_2$
\REF\GS{J.-L. Gervais and M.V. Saveliev, {\it Black holes from non-abelian
Toda theories}, \PL B286 1992 271 .}
[\GS].
The general solution of the
equations of motion is in principle contained in ref. \GS\ where it is shown
how the solutions for an equivalent system of equations can be obtained from
the general scheme of ref. \LS. However, it is non-trivial to actually spell
out the solution and put it in a compact and useful form. This was done
in
\REF\NAT{A. Bilal, {\it Non-abelian Toda theory: a completely integrable
model for strings on a black hole background}, \NP B422 1994 258 .}
ref. \NAT.
The equations of motion\ obtained from the action \uiii\ read
$$\eqalign{
\partial_u\partial_v r&={{\rm sh} r\over {\rm ch}^3r}\partial_u t\partial_v t +{\rm sh} 2r\, e^{2\varphi}\cr
\partial_u\partial_v t&=-{1\over {\rm sh} r\, {\rm ch} r}\left( \partial_u r\partial_v t+\partial_u t\partial_v r\right)\cr
\partial_u\partial_v\varphi&={\rm ch} 2r\, e^{2\varphi} \ .}
\eqn\di$$
Using these equations of motion it is completely straightforward to show
that the following three quantities are conserved [\GS]:
$$\eqalign{
T\equiv T_{++}&=(\partial_u r)^2+{\rm th}^2r\, (\partial_u t)^2 +(\partial_u \varphi)^2-\partial_u^2\varphi\cr
V^\pm\equiv
V^\pm_{++}&={1\over \sqrt{2}} \left( 2\partial_u\varphi-\partial_u\right)
\left[e^{\pm i\nu}\left(\partial_u r\pm i {\rm th} r\, \partial_u t\right)\right]\cr }
\eqn\dii$$
i.e.
$$\partial_v T=\partial_v V^\pm=0\ .
\eqn\diii$$
Here $\nu$ is defined by
$$\partial_v\nu={\rm ch}^{-2}r\, \partial_v t\quad , \qquad \partial_u\nu=(1+{\rm th}^2r)\partial_u t
\eqn\div$$
where the integrability condition is fulfilled due to the equations of motion\ \di.
\section{The symplectic structure and the constraint algebra}
Now I will define the symplectic structure using canonical Poisson brackets.
Recall that $u=\tau+\sigma,\ v=\tau-\sigma$. The action \uiii\ can then equivalently
be written as (as usual, a dot denotes $\partial_\tau$
while a prime denotes $\partial_\sigma$)
$$ S= {1\over \gamma^2}\int \sqrt{2}\tau \sqrt{2}\sigma
\left[ {1\over 2} (\dot r^2-r'^2)+{1\over 2}{\rm th}^2 r\, (\dot t^2-t'^2)
+{1\over 2} (\dot \varphi^2-\varphi'^2) + 2\, {\rm ch} 2r\, e^{2\varphi} \right] \ .
\eqn\ti$$
The constant $\gamma^2$ can be viewed as the Planck constant $2\pi \hbar$,
already included ino the classical action, or merely as a coupling
constant. The canonical momenta then are
$$\Pi_r=\gamma^{-2}\, \dot r\quad , \quad \Pi_t=\gamma^{-2} {\rm th}^2 r\, \dot t\quad , \quad
\Pi_\f=\gamma^{-2}\, \dot\varphi
\eqn\tii$$
and the canonical (equal $\tau$) Poisson brackets are
$$\eqalign{
\{r(\tau,\sigma)\, ,\, \Pi_r(\tau,\sigma')\} &=
\{t(\tau,\sigma)\, ,\, \Pi_t(\tau,\sigma')\} =
\{\varphi(\tau,\sigma)\, ,\, \Pi_\f(\tau,\sigma')\} = \delta(\s-\s')\cr
\{r(\tau,\sigma)\, ,\, r(\tau,\sigma')\} &=
\{r(\tau,\sigma)\, ,\, t(\tau,\sigma')\} = \ldots = 0\cr
\{\Pi_i(\tau,\sigma)\, ,\, \Pi_j(\tau,\sigma')\}&=0\ . \cr
}
\eqn\tiii$$
It follows that
the only non-zero equal $\tau$ Poisson brackets are
$$\eqalign{
\{r(\tau,\sigma)\, ,\, \dot r(\tau,\sigma')\} =\gamma^2 \delta(\s-\s')\quad &, \quad
\{t(\tau,\sigma)\, ,\, \dot t(\tau,\sigma')\} ={\gamma^2 \over {\rm th}^2 r}\delta(\s-\s')\ ,\cr
\{\varphi(\tau,\sigma)\, ,\, \dot \varphi(\tau,\sigma')\} =\gamma^2 \delta(\s-\s')\quad &, \quad
\{\dot r(\tau,\sigma)\, ,\, \dot t(\tau,\sigma')\} ={2\gamma^2 \over {\rm sh} r\, {\rm ch} r}\, \dot t
\, \delta(\s-\s')\ ,\cr }
\eqn\tiv$$
and those derived from them by applying $\partial_\sigma^n \partial_{\sigma'}^m$.
Before one can compute the Poisson bracket algebra of the $T, V^\pm$ one
has to rewrite them in terms of the fields and their momenta. This means
in particular that second (and higher) $\tau$-derivatives have to be
eliminated first, using the equations of motion. One might object that one
is not allowed to use the equations of motion\ in a canonical formulation. However, the
conserved quantities given above are only defined up to
terms that vanish on solutions of the equations of motion. So the correct
starting point for a canonical formulation are the expression where all
higher $\tau$-derivatives are eliminated, while the expressions given in
eq. \dii\ are merely derived from the canonical ones by use of the
equations of motion.
One has, for example, using the $\varphi$-equation of motion \di\
$$\partial_u^2\varphi={1\over 4}(\ddot\varphi+2\dot\varphi'+\varphi'')={1\over 2}(\varphi''+\dot\varphi')+{\rm ch} 2r
\, e^{2\varphi}=\partial_\sigma\partial_u\varphi+{\rm ch} 2r\, e^{2\varphi}
\eqn\tv$$
where in the canonical formalism
$$\partial_u r= {1\over 2}(\gamma^2 \Pi_r+ r')\ ,\
\partial_u t= {1\over 2}(\gamma^2 {\rm th}^{-2} r\, \Pi_t+ t')\ ,\
\partial_u \varphi= {1\over 2}(\gamma^2 \Pi_\f+ \varphi')\ .
\eqn\tva$$
The canonical expressions for the $++$ components of the constraints are
$$\eqalign{
T&=(\partial_u r)^2+{\rm th}^2 r\, (\partial_u t)^2 + (\partial_u\varphi)^2 -(\partial_u\varphi)' -{\rm ch} 2r
e^{2\varphi}\cr
V^\pm&={1\over \sqrt{2}} e^{\pm i\nu}\Big[ 2\partial_u\varphi\partial_u r\pm 2i {\rm th} r\partial_u\varphi\partial_u t
+ 2{\rm th}^3 r(\partial_u t)^2 \mp 2i {\rm th}^2 r\partial_u r\partial_u t\cr
&\phantom{=e^{\pm i\nu}\Big[}+{{\rm sh} r\over {\rm ch}^3 r}\partial_u t\, t'\mp i {\partial_u
r\, t'+\partial_u t\, r'\over {\rm ch}^2 r}-(\partial_u r)'\mp i{\rm th} r(\partial_u t)'-{\rm sh} 2r\,
e^{2\varphi}\Big]\ \cr }
\eqn\tvb$$
where the substitutions \tva\ are understood.
For the Poisson bracket of $T$ with itself one then obtains
$$\gamma^{-2} \{T(\sigma)\, ,\, T(\sigma')\} =
(\partial_\sigma-\partial_{\sigma'})\left[ T(\sigma') \delta(\s-\s')\right]-{1\over 2} \delta'''(\s-\s') \ .
\eqn\tvii$$
${\bar T}$ satisfies the same algebra (with $\sigma\rightarrow -\sigma, \sigma'\rightarrow -\sigma'$)
while
$\{T(\sigma)\, ,\, {\bar T}(\sigma')\} =0$.
These are just two copies of the conformal algebra. If $\sigma$
takes values on the unit circle one can define the modes
$$
L_n=\gamma^{-2} \int_{-\pi}^{\pi}\sqrt{2}\sigma\left[ T(\tau,\sigma)+{1\over 4}\right]
e^{in(\tau+\sigma)}
\eqn\tx$$
and similarly for $\bar L_n$.
Then the bracket \tvii\ becomes a Virasoro algebra
$$
i\{L_n\, ,\, L_m\} =(n-m)L_{n+m}+{c\over 12} (n^3-n)\delta_{n+m,0}
\eqn\txi$$
and idem for $\{\bar L_n\, ,\, \bar L_m\}$,
while $\{L_n\, ,\, \bar L_m\}=0$. Here $c$ is the central charge given by
$$c={12 \pi\over \gamma^2}\ .
\eqn\txii$$
The occurrence of a central charge already at the classical, Poisson bracket
level is due to the $\partial^2\varphi$-term in $T$. This is reminiscent of the
well-known Liouville theory. The factor $i$ on the left hand side of
equations \txi\ may seem strange at first sight, but one should remember
that quantization replaces $i$ times the canonical Poisson bracket by the
commutator. Hence \txi\ is indeed the Poisson bracket version of the
(commutator) Virasoro algebra.
In order to compute Poisson brackets involving $V^\pm$ one needs the
Poisson brackets involving the field $\nu$. Now, $\nu$ is only defined
through its partial derivatives, and thus only up to a constant. This
constant, however, may have a non-trivial Poisson bracket with certain modes
of $\nu$ and/or of the other fields. From \div\ one has using \tii
$$\nu'=\gamma^2\Pi_t+t'
\eqn\txiii$$
whereas one does not need $\dot\nu$ explicitly. Equation \txiii\ implies
$$
\partial_\sigma\partial_{\sigma'}\{\nu(\sigma)\, ,\, \nu(\sigma')\}=\{\nu'(\sigma)\, ,\, \nu'(\sigma')\} = 2\gamma^2
\delta'(\s-\s')\ .
\eqn\txiv$$
This is integrated to yield $\{\nu(\sigma)\, ,\, \nu(\sigma')\}=-\gamma^2 \epsilon(\s-\s') +h(\sigma)-
h(\sigma')$ where I already used the antisymmetry of the Poisson bracket.
$\epsilon(\s-\s')$ is defined to be $+1$ if $\sigma>\sigma'$, $-1$ if $\sigma<\sigma'$ and $0$ if
$\sigma=\sigma'$. The freedom to choose the function $h$ corresponds to the
above-mentioned freedom to add a constant $\nu_0$ to $\nu$ with $\{\nu(\sigma)\,
,\, \nu_0\}=h(\sigma)$. However, if one imposes invariance under translations
$\sigma\rightarrow\sigma+ a,\ \sigma'\rightarrow\sigma' +a$ then $h$ can be only linear and one arrives at
$$\{\nu(\sigma)\, ,\, \nu(\sigma')\}=-\gamma^2 \epsilon_\alpha(\sigma-\sigma')\equiv -\gamma^2\left[
\epsilon(\s-\s')-{\alpha\over\pi} (\sigma-\sigma')\right]\ .
\eqn\txv$$
There is only one parameter $\alpha$ left, related to the zero-mode of $\nu$.
Before doing the actual calculation it is helpful to show how the result is
constrained by dimensional and symmetry considerations. First consider
$\{V^+(\sigma)\, ,\, V^+(\sigma')\}$. Each $V^+$ contains a factor
$e^{i\nu}$. The Poisson bracket of $e^{i\nu(\sigma)}$ with $e^{i\nu(\sigma')}$
leads to a term $\gamma^2 \epsilon(\s-\s') V^+(\sigma) V^+(\sigma')$. All other terms are
local, i.e. involve $\delta(\s-\s')$ or derivatives of $\delta(\s-\s')$. On dimensional
grounds\foot{
The naive dimensional counting assigns dimension 1 to each derivative,
dimension 0 to all fields $r,t,\varphi, \nu$ and functions thereof, except for
functions of $\varphi$. $e^{2\varphi}$ has dimension 2 as seen from the action,
while $\delta(\s-\s')$ has dimension 1.}
$\delta(\s-\s')$ must be multiplied by a dimension 3 object (3 derivatives) and $\delta'(\s-\s')$
by a dimension 2 object. Furthermore, these objects must contain an
overall factor $e^{2i\nu}$. If the $T, V^+$ and $V^-$ form a closed
algebra, there are no such objects. The same reasoning applies to
$\{V^-(\sigma)\, ,\, V^-(\sigma')\}$. Thus one expects
$$\gamma^{-2}\{V^\pm(\sigma)\, ,\, V^\pm(\sigma')\}=\epsilon(\s-\s')
V^\pm(\sigma)V^\pm(\sigma')\ .
\eqn\txvii$$
It is a bit tedious, but otherwise straightforward to verify that this is
indeed correct. Note that it is enough to do the computation for $V^+$
since $V^-$ is the complex conjugate of $V^+$ (treating the fields $r,t,\varphi$
and $\nu$ and their derivatives as real):
$$V^-=(V^+)^*\ .
\eqn\txviii$$
What can one say about $\{V^+(\sigma)\, ,\, V^-(\sigma')\}$? Using
\txviii\ one sees that $\{V^+(\sigma)\, ,\, V^-(\sigma')\}^* =
-\{V^+(\sigma)\, ,\, V^-(\sigma')\}\vert_{\sigma\leftrightarrow \sigma'}$. This
fact, together with the same type of arguments as used above, implies
$$\eqalign{
\gamma^{-2} \{V^+(\sigma)\, ,\, V^-(\sigma')\}=&-\epsilon(\s-\s')
V^+(\sigma)V^-(\sigma') +(\partial_\sigma-\partial_{\sigma'})[a\delta(\s-\s')] \cr
&+i b \delta(\s-\s') +i(\partial_\sigma^2+\partial_{\sigma'}^2)[d\delta(\s-\s')] +\tilde c \delta'''(\s-\s')\cr
}
\eqn\txix$$
where $a,b,d,\tilde c$ are real and have (naive) dimensions 2,3,1 and 0.
Hence $\tilde c$ is a c-number. Also, $b,a,d$ cannot contain a factor
$e^{\pm i \nu}$. If one assumes closure of the algebra one must have $b=d=0$
and $a\sim T_{++}$. After a really lengthy computation one indeed finds
equation \txix\ with $b=d=0$, $a=T_{++}$ and $\tilde c=-{1\over 2}$.
Finally the Poisson bracket of $T$ with $V^\pm$ simply shows that
$V^\pm$ are conformally primary fields of weight (conformal dimension)
2. The complete algebra thus is
$$\eqalign{
\gamma^{-2} \{T(\sigma)\, ,\, T(\sigma')\} &=
(\partial_\sigma-\partial_{\sigma'})\left[ T(\sigma') \delta(\s-\s')\right]-{1\over 2} \delta'''(\s-\s') \cr
\gamma^{-2} \{T(\sigma)\, ,\, V^\pm\sigma')\} &=
(\partial_\sigma-\partial_{\sigma'})\left[ V^\pm(\sigma') \delta(\s-\s')\right]\cr
\gamma^{-2} \{V^\pm(\sigma)\, ,\, V^\pm(\sigma')\}&=\epsilon(\s-\s')
V^\pm(\sigma)V^\pm(\sigma')\cr
\gamma^{-2} \{V^\pm(\sigma)\, ,\, V^\mp(\sigma')\}&=-\epsilon(\s-\s')
V^\pm(\sigma)V^\mp(\sigma')\cr
&\phantom{=}+(\partial_\sigma-\partial_{\sigma'})\left[ T(\sigma') \delta(\s-\s')\right] -{1\over 2}
\delta'''(\s-\s') \ .\cr }
\eqn\txx$$
The algebra of the $--$ components ${\bar T}, {\bar V}^\pm$
looks exactly the same except for the replacements $\sigma\rightarrow
-\sigma,\ \sigma'\rightarrow -\sigma'$ and hence $\partial\rightarrow -\partial,\ \epsilon(\s-\s')\rightarrow -\epsilon(\s-\s')$.
The algebra \txx\ is the correct algebra for $\sigma\in {\bf R}$. If $\sigma\in
S^1$, one must replace $\epsilon(\s-\s')\rightarrow \epsilon_1(\sigma-\sigma')$ which is a
periodic function (cf. eq. \txv). Also $\delta(\s-\s')\rightarrow {1\over 2}\partial_\sigma
\epsilon_1(\sigma-\sigma')=\delta(\s-\s')-{1\over
2\pi}$ while $\delta'(\s-\s')$ remains unchanged. But since the right hand sides of
\txx\ can be written using only $\epsilon(\s-\s'),\, \delta'(\s-\s')$ and $\delta'''(\s-\s')$, only the
replacement $\epsilon(\s-\s')\rightarrow \epsilon_1(\sigma-\sigma')$ is relevant. One then defines the modes
$$
V^\pm_n= \gamma^{-2} \int_{-\pi}^\pi \sqrt{2}\sigma V^\pm(\tau,\sigma)\, e^{in(\tau+\sigma)}
\eqn\txxi$$
and similarly for $\bar V^\pm_n$.
The mode algebra is\foot{As a further consistency check one can verify that
the Jacobi identities are satisfied.}
$$\eqalign{
i\{L_n\, ,\, L_m\} &=(n-m)L_{n+m}+{c\over 12} (n^3-n)\delta_{n+m,0}\cr
i\{L_n\, ,\, V^\pm_m\} &=(n-m)V^\pm_{n+m}\cr
i\{V^\pm_n\, ,\, V^\pm_m\} &={12\over c}\sum_{k\ne 0} {1\over k}
V^\pm_{n+k}V^\pm_{m-k}\cr
i\{V^\pm_n\, ,\, V^\mp_m\} &=- {12\over c}\sum_{k\ne 0} {1\over k}
V^\pm_{n+k}V^\mp_{m-k}+ (n-m)L_{n+m}+{c\over 12} (n^3-n)\delta_{n+m,0}\ .\cr
}
\eqn\txxii$$
This is a non-linear algebra, reminiscent of the $W$-algebras. What is new
here are the non-local terms involving $\epsilon(\s-\s')$, or the ${1\over k}$ in the
mode algebra.
\section{The classical solutions of the equations of motion\ }
The essential point for solving the equations of motion\ \di\ is to realize the underlying
Lie algebraic structure. Following Gervais and Saveliev
[\GS], one first introduces fields $a_1, a_2, a_+$ and $a_-$
subject to the following equations of motion\
$$
\eqalign{
&\partial_u\partial_v a_1=-2(1+2a_+a_-)e^{-a_1}\cr
&\partial_u\partial_v(2a_2-a_1)+2\partial_u(\dva_+ a_-)=0\cr
&\partial_u\left(e^{a_1-2a_2}\dva_+\right)=2a_+ e^{-2a_2}\cr
&\partial_u\left[e^{-a_1+2a_2}\left(\dva_--a_-^2\dva_+\right)\right]=2 a_-
(1+a_+a_-) e^{-2a_1+2a_2}\ .\cr }
\eqn\dvii$$
It is then straightforward, although a bit lengthy, to show that we can
identify
$$\eqalign{
\varphi&=-{1\over 2}a_1\cr
{\rm sh}^2 r&=a_+a_-\cr
\partial_u t&={1\over 2i}\left[(1+2a_+a_-){\dua_-\over a_-}- {\dua_+\over a_+}
-2(1+a_+a_-)\partial_u(a_1-2a_2)\right]\cr
\partial_v t&={i\over 2}\left[(1+2a_+a_-){\dva_+\over a_+}- {\dva_-\over a_-}
\right]\cr}
\eqn\dviii$$
i.e. with these definitions the equations \dvii\ and \di\ are equivalent.
The integrability condition for solving for $t$ is given by the
equations \dvii.
The advantage of equations \dvii\ over equations \di\ is that they follow
from a Lie algebraic formulation. The relevant algebra is $B_2$ with
generators $h_1, h_2$ (Cartan subalgebra) and $E_{e_1}, E_{e_2}, E_{e_1-e_2},
E_{e_1+e_2}$ and their conjugates $E_\alpha^+=E_{-\alpha}$. Then $H=2h_1+h_2,
J_+=E_{e_1}$ and $J_-=E_{-e_1}$ span an $A_1$ subalgebra. $H$ induces a
gradation on $B_2$. The gradation 0 part ${\cal G}_0$ is spanned by $h_1,
h_2, E_{e_2}$ and $E_{-e_2}=E_{e_2}^+$. The corresponding group elements
$g_0\in G_0$ can be parametrized as
$$g_0=\exp(a_+ E_{e_2})\exp(a_- E_{e_2}^+)\exp(a_1 h_1 +a_2 h_2)\ .
\eqn\dx$$
One can then show that equations \dvii\ are equivalent to
$$\partial_u( g_0^{-1}\partial_v g_0)=[J_-\, ,\, g_0^{-1}J_+ g_0]\ .
\eqn\dix$$
If one now defines $g_+$ and $g_-$ as solutions of the ordinary differential
equations
$$\eqalign{
\partial_u g_+^{-1}&=g_+^{-1} (g_0^{-1} J_+ g_0)\cr
\partial_v g_-&=g_- (g_0 J_- g_0^{-1})
\ .\cr}
\eqn\dxii$$
and $g=g_-g_0g_+$ then one can show [\GS,\NAT] that eq. \dix\ is in turn
equivalent to
$$\partial_u (g^{-1}\partial_v g)=0 \quad {\rm and} \quad \partial_v (\partial_u g g^{-1})=0 \ .
\eqn\dxiv$$
The general solution to these equations is well-known:
$g=g_L(u) g_R(v)$. But each group element $g_L$ and $g_R$ has again a Gauss
decomposition $g_L(u)=g_{L-}(u)g_{L0}(u)g_{L+}(u)$ and
$g_R(v)=g_{R-}(v)g_{R0}(v)g_{R+}(v)$ so that
$g=g_{L-}(u)g_{L0}(u)g_{L+}(u)g_{R-}(v)g_{R0}(v)g_{R+}(v)$.
On the other hand we also have the decomposition $g=g_-g_0g_+$ where $g_+$ and
$g_-$
must obey the differential
equations \dxii.
The latter translate into
$$\eqalign{
\partial_u g_{L+}(u)=-{\cal F}_L(u) g_{L+}(u) \quad &,
\quad {\cal F}_L(u) =g_{L0}^{-1}(u) J_+ g_{L0}(u)\ ,\cr
\partial_v g_{R-}(v)= g_{R-}(v){\cal F}_R(v) \quad &,
\quad {\cal F}_R(v) =g_{R0}(v) J_- g_{R0}^{-1}(v)\ .\cr
}
\eqn\dxx$$
The strategy then is
\pointbegin
Pick some arbitrary $g_{L0}(u),\ g_{R0}(v)\in G_0$.
\point
Compute the solutions $g_{L+}(u)$ and $g_{R-}(v)$ from the first order
ordinary differential equations \dxx.
\point
Let
$$
\Gamma=g_{L0}(u) g_{L+}(u) g_{R-}(v) g_{R0}(v)
\eqn\dxxi$$
and choose a basis $\langle$ of states annihilated by ${\cal G}_+$ (i.e. by $E_{e_1},
E_{e_1-e_2}$ and $E_{e_1+e_2}$. Then using the different decompositions of $g$
we have
$$G_{\alpha\beta}\equiv \langle \l_\b \vert g_0 \langle=\langle \l_\b \vert \Gamma \langle\ .
\eqn\dxxii$$
This yields all matrix elements of $g_0$, solution of equation \dix, which
in turn, as shown above, yields the solution for the $a_1,a_2,a_+$ and
$a_-$, parametrized in terms of the arbitrary $g_{L0}(u)$ and $g_{R0}(v)$.
\par
I will now display the resulting solutions. For details of the derivation, see
ref. \NAT. One introduces three arbitrary (``left-moving") functions $f_1(u),
f_+(u)$ and $f_-(u)$ of {\it one} variable (parametrizing $g_{L0}(u)$),
and three arbitrary
(``right-moving") functions $g_1(v),
g_+(v)$ and $g_-(v)$ of {\it one} variable (parametrizing $g_{R0}(v)$).
To write the results in a more compact way,
introduce the functions of one variable
$$\eqalign{
F_1(u)&=-\int^u e^{f_1} (1+2f_+f_-)\cr
F_2(u)&=2\int^u e^{f_1} f_-\cr
F_3(u)&=-2\int^u e^{f_1}f_+(1+f_+f_-) \cr}
\eqn\dxxxvi$$
as well as
$$F_+=F_1+f_+ F_2\quad , \quad F_-=F_3-f_+ F_1
\eqn\dxxxvii$$
and similarly for $G_1(v)$ etc, with $f_i(u)\rightarrow g_i(v)$. Then one introduces
the quantities $X,Y,Z$ and $V,W$ that depend on both variables $u$ and $v$:
$$\eqalign{
X&=1+f_+g_++F_+G_++F_-G_-\cr
Y&=(1+f_+f_-)(1+g_+g_-)+f_-g_-+(F_1-f_-F_-)(G_1-g_-G_-)\cr
&\phantom{=}+(F_2+f_-F_+)(G_2+g_-G_-)\cr
Z&=1+2F_1G_1 +F_2G_2+F_3G_3+(F_1^2+F_2F_3)(G_1^2+G_2G_3)\cr
V&=-g_--f_+-g_+g_-f_+-g_-F_+G_+-g_-F_-G_-+F_-G_1-F_+G_2\cr
W&=-f_--g_+-f_+f_-g_+-f_-F_+G_+-f_-F_-G_-+F_1G_--F_2G_+\ .\cr
}
\eqn\dxxxxiv$$
They are sums of products of left-moving times right-moving quantities.
The complete solution then is
$$\eqalign{
e^{a_1}&= e^{-f_1-g_1}Z\cr
e^{-a_2}&=\pm e^{f_2+g_2}{Y\over Z}\cr
a_+&=e^{f_1-2f_2}{V\over Y}\cr
a_-&=e^{2f_2-f_1}{YW\over Z}\cr
}
\eqn\dxxxxvii$$
while the fifth equation is the relation
$XY-Z=VW$.
{}From equations \dviii\ one immediately finds
$$\eqalign{
\varphi&={1\over 2}\left( f_1+g_1-\log Z\right)\cr
{\rm sh}^2 r&={VW\over Z}={XY\over Z}-1\cr
t&=t_0+i\int^u f_-f_+'-i\int^v g_-g_+' +{i\over 2} \log {V\over W} \ .
\cr }
\eqn\dxxxxxiii$$
which is the general solution of the equations of motion\ .
Using the equations of motion\ for $\varphi,\, r$ and $t$ it was shown above that the
quantities $T\equiv T_{++}$ and $V^\pm\equiv V^\pm_{++}$ are conserved, i.e.
can only depend on $u$. This means that they must be expressible entirely in
terms of the $f_i(u)$'s. Given the complexity of the solutions
\dxxxxxiii\ and \dxxxxiv\ this is highly non-trivial and
constitutes a severe consistency check. The same considerations
apply to ${\bar T}\equiv T_{--}$ and ${\bar V}^\pm\equiv V^\pm_{--}$.
One can indeed show [\NAT] that
$$\eqalign{
T_{++}&={1\over 4}(f_1')^2-{1\over 2}f_1''+ f_+'(f_-'-f_-^2f_+') \cr
V^+&={1\over \sqrt{2} } (f_1'-\partial_u) \, \left[
\exp\left( -2\int^u f_-f_+'\right) (f_-'-f_-^2f_+') \right]\cr
V^-&={1\over \sqrt{2} } (f_1'-\partial_u) \, \left[
\exp\left( +2\int^u f_-f_+'\right) f_+' \right]\ . \cr}
\eqn\ddxiii$$
It is then natural to set
$$\eqalign{
f_-&=e^{\sqrt{2}\varphi_1}\cr
f_+'&={1\over \sqrt{2}}e^{-\sqrt{2}\varphi_1} (\partial_\sigma\varphi_1+i\partial_\sigma\varphi_2)\cr
f_1&=\sqrt{2}\varphi_3\cr
}
\eqn\txxxviii$$
(and with a similar relation between the $g_i$ and $\bar\varphi_i$).
The conserved quantities $T, V^\pm$ are easily expressed in terms of the
$\varphi_j$ as
$$\eqalign{
T&={1\over 2} \sum_{j=1}^3 (\partial_\sigma \varphi_j)^2 -{1\over \sqrt{2}}\partial_\sigma^2\varphi_3\cr
V^\pm&={1\over 2}(\sqrt{2}\partial_\sigma\varphi_3-\partial_\sigma)\left[ e^{\mp i\sqrt{2}\varphi_2}
(\partial_\sigma\varphi_1\mp i\partial_\sigma\varphi_2)\right]\ .\cr
}
\eqn\txxxxiii$$
Thus in terms of the $\varphi_j$, $T$ has the standard form of a stress-energy
tensor
in a conformal field theory with a background charge. The $V^\pm$ are
{\it local} expressions of the
fields $\varphi_j$, analogous to standard vertex operators. (Of course, their
Poisson
brackets exhibit the non-local $\epsilon(\s-\s')$-function.)
\section{Canonical transformation to free fields}
In principle one could now deduce the Poisson brackets
of the $f_i$ and $g_i$ through the transformation induced by the classical
solution \dxxxxxiii\ and \txxxviii.
More precisely, one would have to allow formally that the $f_i$ and $g_i$,
respectively the $\varphi_i$ and $\bar\varphi_i$,
depend both on $u$ and $v$ since one has to consider the full phase space
and not only the manifold of solutions to the equations of motion.
Nevertheless, equations \dxxxxxiii\ and \txxxviii, as well as their time
derivatives, constitute a phase space transformation from $r(\tau,\sigma),\,
t(\tau,\sigma),\, \varphi(\tau,\sigma)$ and their momenta $\Pi_r(\tau,\sigma),\, \Pi_t(\tau,\sigma),\,
\Pi_\f(\tau,\sigma)$ to new phase space variables $f_i(\tau,\sigma),\ g_i(\tau,\sigma)$ and to
$\varphi_i(\tau,\sigma), \bar\varphi_i(\tau,\sigma)$. (Of
course, the equations of motion still imply $\partial_v f_i=\partial_u
g_i=\partial_v\varphi_i=\partial_u\bar\varphi_i=0$.) In
practice, this would be very complicated to implement. It is much simpler
to use the Poisson brackets of the $T$ and $V^\pm$ derived before, and then
consider the $T$ and $V^\pm$ (or ${\bar T}$ and ${\bar V}^\pm$) as given in terms of
the $\varphi_i$ only (or $\bar\varphi_i$ only). Thus one does the phase space
transformation in two steps: $r(\tau,\sigma),\,
t(\tau,\sigma),\, \varphi(\tau,\sigma),\, \Pi_r(\tau,\sigma),\, \Pi_t(\tau,\sigma),\,
\Pi_\f(\tau,\sigma)\ \rightarrow\ T_{\pm\pm}(\tau,\sigma),\, V^+_{\pm\pm}(\tau,\sigma),\,
V^-_{\pm\pm}(\tau,\sigma)\ \rightarrow\ \varphi_i(\tau,\sigma),\, \bar\varphi_i(\tau,\sigma)$.
This yields the Poisson brackets of the $\varphi_i$ and of the $\bar\varphi_i$ in a
relatively easy way. These Poisson brackets are simple, standard
harmonic oscillator Poisson brackets:
$$\{\partial_\sigma \varphi_i(\sigma)\, ,\, \partial_{\sigma'} \varphi_j(\sigma')\}=\gamma^2 \delta_{ij} \delta'(\s-\s')
\eqn\txxxix$$
or
$$\{\varphi_i(\sigma)\, ,\, \varphi_j(\sigma')\}=-{\gamma^2\over 2} \delta_{ij} \epsilon(\s-\s')\ .
\eqn\txxxx$$
If one considers $\sigma\in S^1$, the mode expansion\foot{
Note that this is a Fourier expansion in $\sigma$. The factor $e^{-in\tau}$ is
only extracted from the $\varphi_n^j$ for convenience. The $\varphi_n^j$ are still
functions of $\tau$. It is only if one imposes the equations of motion that
the $\varphi_n^j$ are constant.}
$$\partial_\sigma \varphi_j(\tau,\sigma)={\gamma\over \sqrt{2\pi}} \sum_n
\varphi^j_n e^{-in(\tau+\sigma)} \eqn\txxxxi$$ %
leads to
$$i\{\varphi^j_n\, ,\, \varphi^k_m\}=n \delta^{ij} \delta_{n+m,0}
\eqn\txxxxii$$
as appropriate for three sets of harmonic oscillators. The Poisson brackets of
the
$\bar\varphi_i$ are analogous, while $\{\varphi_i(\sigma)\, ,\, \bar\varphi_j(\sigma')\}=0$.
\section{Associated linear differential equation}
{}From experience with integrable models, in particular the (non-affine,
conformally invariant) Toda models
\REF\BG{A. Bilal and J.-L. Gervais, {\it Systematic approach to
conformal systems with extended Virasoro symmetries}, Phys. Lett.
{\bf B206} (1988) 412; {\it Systematic construction of $c=\infty$
conformal systems from classical Toda field theories}, Nucl. Phys.
{\bf B314} (1989) 646; {\it Systematic construction of conformal
theories with higher-spin Virasoro symmetries}, Nucl. Phys. {\bf B318}
(1989) 579.}
[\BG] one expects that the conserved quantities appear as coefficients of
an ordinary linear differential equation, e.g. for the $A_{m-1}$ Toda model
$$\left[\partial_u^m-\sum_{k=2}^m W^{(k)}(u) \partial_u^{m-k}\right]\psi(u) =0\ .
\eqn\qi$$
The $ W^{(k)}(u), k=2,\ldots m$ are the conserved quantities which form the
$W_m$-algebra, while the solutions $\psi_j(u)$ of this equation, together
with solutions $\chi_j(v)$ of a similar equation in $v$, are the
building blocks of the general solution to the Toda equations of motion.
The $ W^{(k)}(u)$ have (naive) dimension $k$.
In the present theory all conserved quantities have dimension 2,
so one might expect a second-order differential
equation of the form $(-\partial^2+U)\psi=0$. To fit three conserved quantities into
$U$, it needs to be at least a $2\times 2$-matrix. In ref. \NAT, I guessed the
following
linear differential equation
$$\left[ \partial_u^2-\pmatrix{ \alpha T(u)& \beta_+ V^+(u)\cr \beta_- V^-(u) & \delta T(u)
\cr } \right] \Psi(u)=0\ .
\eqn\qii$$
Actually, if one inserts the free-field representation \txxxxiii\ of $T$ and
$V^\pm$ one
can see that this equation admits the very simple solution
$$\eqalign{
\psi_1&=\exp(a\varphi_1 +ib\varphi_2 +d\varphi_3)\cr
\psi_2&=\exp(a\varphi_1 -ib\varphi_2 +d\varphi_3)\ .\cr
}
\eqn\qqi$$
if and only if
$$\eqalign{
a={1\over \sqrt{2}}\quad &, \quad b=d= -{1\over \sqrt{2}}\cr
\alpha=\delta=1 \quad &, \beta_+=\beta_-=-\sqrt{2}\ .\cr
}
\eqn\qqii$$
The existence of this very simple solution already is an indication that the
differential equation \qii\ is probably the correct generalization of \qi\
for $W$-algebras to the present $V$-algebra. That this is indeed so was shown
in
\REF\MCK{A. Bilal, {\it Multi-component KdV hierarchy, $V$-algebra and
non-abelian Toda theory}, Lett. Math. Phys. {\bf 32} (1994) 103.}
ref. \MCK, and will be reviewed in the next section.
\chapter{The $V$-algebra as second Gelfand-Dikii bracket,\break
the resolvent and matrix KdV-hierarchy}
\def\partial{\partial}
\def\partial_\sigma{\partial_\sigma}
\def\partial_{\sigma'}{\partial_{\sigma'}}
\def\partial_u{\partial_u}
\def\partial_v{\partial_v}
\def\varphi{\varphi}
\def\chi{\chi}
\def\sigma{\sigma}
\def\sigma'{\sigma'}
\def\lambda{\lambda}
\def\tau{\tau}
\def\int {\rm d} \sigma\, {\int {\rm d} \sigma\, }
\def\int {\rm d} \sigma'{\int {\rm d} \sigma'}
\def\delta(\s-\s'){\delta(\sigma-\sigma')}
\def\delta'(\s-\s'){\delta'(\sigma-\sigma')}
\def\delta''(\s-\s'){\delta''(\sigma-\sigma')}
\def\delta'''(\s-\s'){\delta'''(\sigma-\sigma')}
\def\epsilon(\s-\s'){\epsilon(\sigma-\sigma')}
\def\omega{\omega}
\def\Omega{\Omega}
\def\alpha{\alpha}
\def\beta{\beta}
\def\gamma{\gamma}
\def\gamma^2{\gamma^2}
\def\gamma^{-2}{\gamma^{-2}}
\def{\rm d}{{\rm d}}
\def\sqrt{2}{\sqrt{2}}
\def\langle{\langle}
\def\rangle{\rangle}
\def\vert 0\rangle{\vert 0\rangle}
\def{V^+}{{V^+}}
\def{V^-}{{V^-}}
\defGelfand-Dikii\ {Gelfand-Dikii\ }
\def{\cal P} {\Psi}
\def\psi_1{\psi_1}
\def\psi_2{\psi_2}
\def\dd #1 #2{{\delta #1\over \delta #2}}
\def{\rm tr}\ {{\rm tr}\ }
\def\lambda{\lambda}
\def\widetilde{\widetilde}
\REF\BAK{I. Bakas, {\it Higher spin fields and the Gelfand-Dikii\ algebras},
Commun. Math. Phys. {\bf 123} (1989) 627.}
In this section, following ref. \MCK, I will relate the $V$-algebra obtained
above
to well-known
mathematical structures. In the case of the standard $W_m$-algebras it was
shown [\BAK]
that their classical version coincides with second Gelfand-Dikii symplectic
structure
\REF\GELDA{I.M. Gel'fand and L.A. Dikii, {\it Fractional powers of
operators and hamiltonian systems},
Funct. Anal. Applic. {\bf 10} (1976) 259;
{\it The resolvent and
hamiltonian systems}, Funct. Anal. Applic. {\bf 11} (1977) 93.}
\REF\ADLER{M. Adler, {\it On a trace functional for
pseudo-differential operators and the symplectic structure of the
Korteweg-de Vries type equations}, Invent. math. {\bf 50} (1979)
219.}
\REF\LEBMAN{D.R. Lebedev and Yu.I. Manin, {\it Gel'fand-Dikii
Hamiltonian operator and the coadjoint representation of the Volterra
group}, Funct. Anal. Applic. {\bf 13} (1979) 268.}
\REF\GELDB{I.M. Gel'fand and L.A. Dikii, {\it A family of
Hamiltonian structures connected with integrable non-linear
differential equations}, Inst. Appl. Math. Acad. Sci. USSR preprint
n$^{\rm o}$ 136, 1978 (in Russian).}
\REF\KW{B.A. Kupershmidt and G. Wilson, {\it Modifying Lax equations
and the second Hamiltonian structure}, Invent. Math. {\bf 62} (1981)
403.}
\REF\DIK{L.A. Dikii, {\it A short proof of a Kupershmidt-Wilson
theorem}, Commun. Math. Phys. {\bf 87} (1982) 127.}
[\GELDA-\DIK] on the space of ordinary differential operators of order $m$ (as
in eq.
\qi). What I will show here in the remainder of this contribution,
is that a straightforward matrix generalization
leads to a whole family of $V$-algebras. In this section I will obtain
the algebra of the previous section
from the $2\times 2$-matrix second-order differential operator
$$ L=\partial^2-U\ ,\quad
U=\pmatrix{ T& -\sqrt{2} V^+\cr -\sqrt{2} V^- & T \cr }\ .
\eqn\uiii$$
The general case of $n\times n$-matrix $m^{\rm th}$-order differential
operators
leading to $V_{n,m}$-algebras will be treated in the next section.
If not indicated otherwise, $\partial\equiv \partial_\sigma$, and $U$
depends on $\sigma$. $U$ may also depend on other parameters $t_1, t_2, \ldots$.
In the previous section $U$ depended on $\tau$ and $\sigma$, but, upon imposition of
the
equations of motion\ , only through
the combination $\sigma+\tau$. Actually, as usual (see below),
this is the first flow of the matrix KdV hierarchy: ${\partial\over \partial t_1}
U=\partial_\sigma U$, so that $\tau$ is identified with $t_1$.
Let $f$ and $g$ be differential polynomial functionals on the space of
second-order differential operators $L$, i.e. polynomial functionals
of $U$ (and its derivatives). One defines the pseudo-differential
operator %
$$X_f=\partial^{-1} X_1 +\partial^{-2} X_2\quad , \quad X_1=\dd f U
\eqn\di$$
where\foot{
On the circle $S^1$ e.g., $\partial^{-1}$ is well-defined on functions
without constant Fourier mode, i.e. $f(\sigma)=\sum_{m\ne 0}
f_m e^{-im\sigma}$. One easily sees that $(\partial^{-1}f)(\sigma)=\int {\rm d}\sigma'\
{1\over 2} \epsilon(\s-\s') f(\sigma')$ with $\epsilon(\s-\s') = {1\over \pi i}\sum_{m\ne 0} {1\over
m} e^{im(\sigma-\sigma')}$.
}
$\partial^{-1}\partial=\partial\d^{-1}=1$ and ${\delta\over \delta U}$ is defined
as
$${\delta\over \delta U}=\pmatrix{
{1\over 2} {\delta\over \delta T}&
-{1\over \sqrt{2}} {\delta\over \delta {V^-}}\cr
-{1\over \sqrt{2}} {\delta\over \delta {V^+}}&
{1\over 2} {\delta\over \delta T}\cr }
\eqn\dii$$
so that ${\delta\over \delta U} \int {\rm tr}\ U^n =n U^{n-1}$, and $X_2$
is determined (cf. e.g. [\ADLER,\LEBMAN]) by requiring\foot{
This condition is necessary since the coefficient of $\partial$ in $L$
vanishes. I will discuss this condition in more detail in the
next section.} ${\rm res} [L,X_f]=0$. As usual, the residue of a
pseudo-differential operator, denoted ${\rm res}$, is the coefficient
of $\partial^{-1}$. One then has
$$X_2'={1\over 2}\left( \dd f U \right)'' +{1\over 2}
\left[ U, \dd f U \right]\ .
\eqn\diii$$
Integrating this equation yields $X_2$. Here, one observes a new
feature as compared to the scalar case: since in general $\left[ U,
\dd f U \right]\ne 0$, $X_2$ will be given by a non-local
expression involving an integral. This is the origin of the
non-local terms ($\sim\epsilon(\s-\s')$) in the $V$-algebra \txx.
In analogy with
the standard procedure [\GELDA-\DIK], I define the second Gelfand-Dikii\ bracket
in the matrix case as follows
$$\{f,g\}_{\rm GD2}=a\int {\rm d} \sigma\, {\rm tr}\ \ {\rm res} \left( L(X_f L)_+ X_g
-(L X_f)_+ L X_g \right) \ .
\eqn\dvi$$
Using the definitions of $L$ and $X_f, X_g$ it
is straightforward to obtain
$$\eqalign{
\{f,g\}_{\rm GD2}=a\int {\rm d} \sigma\, {\rm tr}\ \Bigg( &{1\over 2} \dd f U \partial^3 \dd g U
+{1\over 2} \left[ U, \dd g U \right] \left( \partial^{-1}
\left[ U, \dd f U \right] \right)\cr
&-\dd f U (U\partial+{1\over 2} U') \dd g U
+ \dd g U (U\partial+{1\over 2} U') \dd f U
\Bigg)\cr }
\eqn\dvii$$
where the $\partial^{-1}$ is meant to act only on $\left[ U, \dd f U
\right]$.
Inserting the definitions of the $2\times 2$-matrices $U$ and ${\delta
\over \delta U}$ one obtains
$$\eqalign{
\{f,g\}_{\rm GD2}=-{a\over 2} \int {\rm d} \sigma\, \Bigg[ &
-{1\over 2} \dd f T \partial^3 \dd g T
-{1\over 2} \dd f {V^+} \partial^3 \dd g {V^-}
-{1\over 2} \dd f {V^-} \partial^3 \dd g {V^+} \cr
&+T\left(\dd f T \partial \dd g T + \dd f {V^+} \partial \dd g {V^-}
+ \dd f {V^-} \partial \dd g {V^+} -(f \leftrightarrow g) \right) \cr
&+{V^+}\left(\dd f T \partial \dd g {V^+} + \dd f {V^+} \partial \dd g T
-(f \leftrightarrow g) \right) \cr
&+{V^-}\left(\dd f T \partial \dd g {V^-} + \dd f {V^-} \partial \dd g T
-(f \leftrightarrow g) \right) \Bigg]\cr
-{a\over 2} \int \int {\rm d}\sigma{\rm d}\sigma'
& \epsilon(\s-\s')\left( {V^+} \dd f {V^+} - {V^-} \dd f {V^-} \right)(\sigma)
\left( {V^+} \dd g {V^+} - {V^-} \dd g {V^-} \right)(\sigma') \ .\cr }
\eqn\dix$$
For completeness, let me note that the first Gelfand-Dikii bracket
defined by
$\{f,g\}_{\rm GD1} = a \int {\rm d} \sigma\, {\rm tr}\ \ {\rm res} ([L,X_f]_+ X_g)$
is simply
$$\{f,g\}_{\rm GD1}=-a\int {\rm d} \sigma\, \left( \dd f T \partial \dd g T +
\dd f {V^+} \partial \dd g {V^-} + \dd f {V^-} \partial \dd g {V^+} \right) \ .
\eqn\dviii$$
Taking $f,g$ to be $T, {V^+}$ or ${V^-}$ one concludes that
the
second Gelfand-Dikii\ bracket \dix\ coincides with the $V$-algebra \txx\
provided
one chooses
$$ a=-2\gamma^{-2}\ .
\eqn\dx$$
Henceforth I will adopt this choice and the only Poisson bracket used
is the second Gelfand-Dikii\ bracket, unless otherwise stated.
Let me note that the free field representation of $T, V^\pm$ in terms of the
$\varphi_i$
obtained in the previous section constitutes a Miura transformation since it
maps the
second Gelfand-Dikii\ symplectic structure to the much simpler
free-field Poisson brackets. I will discuss the Miura transformation in more
detail
in the next section.
\REF\GELD{I.M. Gel'fand and L.A. Dikii, {\it Asymptotic behaviour of the
resolvent of Sturm-Liouville equations and the algebra of the Korteveg-de Vries
equations}, Russ. Math. Surv. {\bf 30}
(1975) 77.}
Let me now turn to some results that are a bit more specific to second-order
differential operators, generalizing the classical work of ref. \GELD.
For a
hermitian $n\times n$-matrix $U$ define the $n\times n$-matrix resolvent $R$ as
$$ R(x,y;\xi)=\langle
x\vert (-\partial^2+U+\xi)^{-1}\vert y\rangle \eqn\cci$$ %
which is a solution of
$$(-\partial_x^2 +U(x)+\xi)R(x,y;\xi)=\delta(x-y)\ .
\eqn\ccii$$
Just as in the scalar case, $n=1$, the restriction of the resolvent to the
diagonal, $R(x;\xi)\equiv R(x,x;\xi)$ has an asymptotic expansion for
$\xi\rightarrow\infty$ of the form
$$R(x;\xi)=\sum_{n=0}^\infty {R_n[u]\over \xi^{n+1/2}}\ .
\eqn\cciii$$
This equation is to be understood as an equality of the asymptotic
expansions in half-integer powers of $1/\xi$, disregarding any terms
that vanish exponentially fast as $\xi\rightarrow\infty$.
{}From the defining differential equation for $R(x,y;\xi)$ one easily
establishes that
$R\equiv
R(x;\xi)=R(x,x;\xi)$ satisfies
$$R'''-2(UR'+R'U)-(U'R+RU')+[U,\partial^{-1}[U,R]]=4\xi R'
\eqn\ccxaa$$
(where $R'\equiv \partial_x R(x;\xi)$ etc.) and hence that the coefficients $R_n$ of
the asymptotic expansion \cciii\ satisfy
$$ 4R'_{n+1}=R_n'''-2(UR'_n+R'_nU)-(U'R_n+R_nU')+[U,\partial^{-1}[U,R_n]]\ .
\eqn\ccxa$$
This allows us to determine the $R_n$ recursively:
$$\eqalign{
R_0=&{1\over 2}\cr
R_1=&-{1\over 4}U\cr
R_2=&{1\over 16}(3U^2-U'')\cr
R_3=&-{1\over 64}(10U^3-5UU''-5U''U-5{U'}^2+U^{(4)})\cr
R_4=&{1\over 256}(35U^4-21U^2U''-21U''U-28UU''U
-28{U'}^2U-28U{U'}^2-14U'UU'\cr
&\phantom{{1\over 256}(} +7UU^{(4)}+7U^{(4)}U+14U'U'''+14U'''U'
+21{U''}^2+U^{(6)})\ . \cr }
\eqn\ccxb$$
\REF\OMG{E. Olmedilla, L. Martinez Alonso and F. Guil, {\it
Infinite-dimensional Hamiltonian systems associated with matrix
Schr\"odinger operators}, Nuovo Cim. {\bf 61B} (1981) 49.}
One of the reasons why one is interested in the coefficients $R_n$ is the
following:
if one defines an infinite family of Hamiltonians as
$$H_n={(-4)^n\over 2(2n-1)}\int {\rm d} x\ {\rm tr}\ R_n(x)
\eqn\cccia$$
one can show [\MCK] that all $H_n$ are in involution\foot{
Such a family of $H_n$ in involution for the matrix Schr\"odinger operator $L$
was already obtained a long time ago in ref.
\OMG.}:
$$\{ H_n,H_m\}=0\ .
\eqn\cccii$$
The proof uses the fact that the recursion relation between the $H_n$
(inherited from
the one between the $R_n$) relates the first and second Gelfand-Dikii\ brackets:
$\{H_n,H_m\}_2\sim \{H_n,H_{m+1}\}_1=-\{H_{m+1},H_n\}_1\sim
-\{H_{m+1},H_{n-1}\}_2
= \{H_{n-1},H_{m+1}\}_2$, so
that by iteration one arrives at $\{H_n,H_m\}_2=\{H_0,H_{n+m}\}_2=0$.
The first few $H_n$ are
$$\eqalign{
H_1&={1\over 2}\int{\rm tr}\ \, U = \int T \cr
H_2&={1\over 2}\int{\rm tr}\ \, U^2 = \int (T^2+2{V^+}{V^-}) \cr
H_3&={1\over 2}\int{\rm tr}\ \, (2U^3+U'^2)
= \int (2T^3+12T{V^+}{V^-}+T'^2+2{V^+}'{V^-}') \cr
H_4&={1\over 2}\int{\rm tr}\ \, (5U^4+10UU'^2+U''^2)\cr
&= \int (5T^4+20{V^+}^2{V^-}^2+60T^2{V^+}{V^-}+10TT'^2\cr
&\phantom{= \int ( } +20T{V^+}'{V^-}'+20T'{V^+}{V^-}'+20T'{V^+}'{V^-} +T''^2
+2{V^+}''{V^-}'') \ .\cr}
\eqn\cccvi$$
Note that within our non-abelian Toda field theory,
$H_1$ is the integral of the $(++)$-component of the energy-momentum
tensor $T\equiv T_{++}$. Thus the Hamiltonian of the theory is $H=L_0+\bar L_0=
H_1+\bar H_1$. The other $H_n$ are
all in involution with $H_1$ (and obviously also with $\bar H_1$)
and thus are conserved under the conformal evolution of
the non-abelian Toda field theory.
\REF\CD{F. Calogero and A. Degasperis, {\it Nonlinear evolution equations
solvable by the inverse spectral transform associated with the multichannel
Schr\"odinger problem, and properties of their solution}, Lett. Nuovo Cim. {\bf
15}
(1976) 65;
}
As usual one can define an infinite hierarchy of flows by\foot{
As before, $\{\cdot,\cdot\}$ is meant to be the second Gelfand-Dikii\ bracket.
But since $4 {\partial U\over \partial t_r}=4\gamma^{-2} \{U,H_r\}_{\rm GD2} =
\gamma^{-2} \{U,H_{r+1}\}_{\rm GD1}$, one can also use the first Gelfand-Dikii\ bracket
and the next higher Hamiltonian instead.}
$${\partial U\over \partial t_r}=\gamma^{-2} \{U,H_r\}\ .
\eqn\si$$
Since $\{H_r,H_s\}=0$ it follows from the Jacobi identity that all
flows commute.
The flow in $t_2$ gives the matrix
generalisation of the KdV equation:\foot{
Matrix KdV flows were already discussed a long time ago in ref. \CD.}
$${\partial U\over \partial t_2}=(3U^2-U'')'
\eqn\sii$$
or in components
$${\partial T\over \partial t_2}=(3T^2-T''+6{V^+}{V^-})'\quad , \quad
{\partial V^\pm\over \partial t_2}=(6TV^\pm-{V^\pm}'')'
\eqn\siii$$
Just as the Virasoro algebra is a subalgebra of the $V$-algebra \txx,
the KdV equation is simply obtained by setting $V^\pm=0$.
Note that since all $H_n$ are symmetric in ${V^+}$ and ${V^-}$ any
non-local term ($\sim\epsilon(\s-\s')$) that might appear cancels in all flow
equations, and the latter are always partial {\it differential}
equations.\foot{This also follows from the equivalence with the first
Gelfand-Dikii\ bracket which is local, see above.}
\chapter{The Gelfand-Dikii\ symplectic structure for general $n\times n$-matrix $m^{\rm
th}$-order differential operators and the $V_{n,m}$-algebras}
\def\partial{\partial}
\def\partial_\sigma{\partial_\sigma}
\def\partial_{\sigma'}{\partial_{\sigma'}}
\def\varphi{\varphi}
\def\sigma{\sigma}
\def\sigma'{\sigma'}
\def\lambda{\lambda}
\def\int {\rm d} \sigma\, {\int {\rm d} \sigma\, }
\def\int {\rm d} \sigma'{\int {\rm d} \sigma'}
\def\delta(\s-\s'){\delta(\sigma-\sigma')}
\def\delta'(\s-\s'){\delta'(\sigma-\sigma')}
\def\delta''(\s-\s'){\delta''(\sigma-\sigma')}
\def\delta'''(\s-\s'){\delta'''(\sigma-\sigma')}
\def\epsilon(\s-\s'){\epsilon(\sigma-\sigma')}
\def\epsilon{\epsilon}
\def\alpha{\alpha}
\def\beta{\beta}
\def\gamma{\gamma}
\def\gamma^2{\gamma^2}
\def\gamma^{-2}{\gamma^{-2}}
\def{\rm d}{{\rm d}}
\def\sqrt{2}{\sqrt{2}}
\def{V^+}{{V^+}}
\def{V^-}{{V^-}}
\defGelfand-Dikii\ {Gelfand-Dikii\ }
\def{\cal P} {\Psi}
\def\dd #1 #2{{\delta #1\over \delta #2}}
\def\bin #1 #2{{ #1\choose #2}}
\def{\rm tr}\ {{\rm tr}\ }
\def{\rm res}\ {{\rm res}\ }
\def\widetilde{\widetilde}
\def{\bf 1}{{\bf 1}}
\def\Lemma #1 {\noindent{\bf Lemma #1 :\ }}
\def\noindent{\it Proof :\ }{\noindent{\it Proof :\ }}
\def{\cal V}(f){{\cal V}(f)}
\def{\tilde{\cal V}}(f){{\tilde{\cal V}}(f)}
\def{\tilde f } {{\tilde f } }
\def{\tilde g } {{\tilde g } }
\def{\cal P} {{\cal P} }
\def{\hat \d} {{\hat \partial} }
\REF\MW{A. Bilal, {\it Non-local matrix generalizations of $W$-algebras},
Princeton
University preprint PUPT-1452 (March 1994), hep-th@xxx/9403197, Comm. Math.
Phys. in
press.}
In this section, following ref. \MW, I will compute the
second Gelfand-Dikii\ bracket of
two functionals $f$ and $g$ of the $n\times n$ matrix coefficient functions
$U_k(\sigma)$ of the linear $m^{\rm th}$-order differential operator\foot{
Throughout the rest of this paper, $m$ will denote the order of $L$ which is a
positive integer.}
$$L=-\partial^m+\sum_{k=1}^mU_k\partial^{m-k}\equiv \sum_{k=0}^mU_k\partial^{m-k}\ .
\eqn\di$$
where $\partial={d\over d\sigma}$.
To make subsequent formula more compact, I formally introduced
$U_0=-{\bf 1}$. The fuctionals $f$ and $g$ one considers are of the form
$f=\int{\rm tr}\ P(U_k)$, where $P$ is some polynomial in the $U_k,\,
k=1,\ldots m$, and their derivatives (i.e. a differential polynomial
in the $U_k$). $P$ may also contain other constant or non-constant
numerical matrices so that these functionals are fairly general. (Under
suitable boundary conditions, {\it any} functional of the $U_k$ and their
derivatives can be approximated to arbitrary ``accuracy" by an $f$ of
the type considered.) The integral can either be defined in a formal
sense as assigning to any function an equivalence class by considering
functions only up to total derivatives (see e.g. section 1 of the second ref.
\GELDA), or in the standard way if one restricts the integrand, i.e. the
$U_k$, to the class of e.g. periodic functions or sufficiently fast
decreasing functions on ${\bf R}$, etc. All that matters is that the
integral of a total derivative vanishes and that one can freely
integrate by parts.
To define the Gelfand-Dikii\ brackets, it is standard to use pseudo-differential
operators
[\ADLER,\LEBMAN] involving integer powers of $\partial^{-1}$, as already encountered
in the
previous section. Again, $\partial^{-1}$
can be defined in a formal sense by $\partial \partial^{-1}=\partial^{-1}\partial=1$, but one can
also give a concrete definition on appropriate classes of functions.
For example for $C^\infty$-functions $h$ on ${\bf R}$ decreasing
exponentially fast as $\sigma\rightarrow\pm\infty$ one can simply define
$(\partial^{-1}h)(\sigma)=\int_{-\infty}^\infty {\rm d} \sigma'
{1\over 2}\epsilon(\s-\s') h(\sigma')$.
Throughout this section, I will only state the results. The reader is referred
to ref.
\MW\ for all proofs.
\section{The Gelfand-Dikii\ brackets for general $U_1, \ldots U_m$}
In analogy with the scalar case (i.e. $n=1$)
[\ADLER,\LEBMAN,\GELDB,\DIK], I define the second Gelfand-Dikii\ bracket
associated with the $n\times n$-matrix $m^{\rm th}$-order differential
operator $L$ as
$$
\{f,g\}_{(2)}=a\int{\rm tr}\ \ {\rm res}\ \left( L(X_f L)_+ X_g -(L X_f)_+ L X_g
\right)
\eqn\partial_v$$
where $a$ is an arbitrary scale factor and $X_f, X_g$ are the
pseudo-differential operators
$$\eqalign{
X_f=\sum_{l=1}^m \partial^{-l}X_l\quad &, \quad X_g=\sum_{l=1}^m
\partial^{-l}Y_l\cr
X_l= \dd f {U_{m+1-l}} \quad &, \quad Y_l= \dd g {U_{m+1-l}} \ .\cr}
\eqn\dvi$$
The functional derivative of $f=\int {\rm d} \sigma\, {\rm tr}\ P(U)$ is defined as usual by
$$\left( \dd f {U_k}(\sigma) \right)_{ij} =
\sum_{r=0}^\infty \left( -{d\over d\sigma}\right)^r
\left( {\partial {\rm tr}\ P(u)(\sigma)\over \partial (U_k^{(r)})_{ji} }\right)
\eqn\dvii$$
where $(U_k^{(r)})_{ji}$ denotes the $(j,i)$ matrix element of the $r^{\rm
th}$ derivative of $U_k$. It is easily seen, that for $n=1$, equations
\partial_v-\dvii\ reduce to the standard
definitions of the Gelfand-Dikii\ brackets [\ADLER,\LEBMAN,\GELDB,\DIK].
For $m=2,\, n=2$ and
with the extra restrictions $U_1=0,\ {\rm tr}\ \sigma_3U_2=0$, equation
\partial_v\ was shown in the previous section to reproduce the
original $V$-algebra \txx\
(with $a=-2\gamma^2$).
Working through the algebra (see [\MW] for details) and defining
for $l\ge 1$
$$
S_{r,l}^{q,j}=\sum_{s=\max(0,r)}^{\min(q,j)} (-)^{s-r}
\bin s-r+l-1 {l-1} \bin q s \ ,
\eqn\dxiv$$
with
$S_{r,l}^{q,j}=0$ if $\max(0,r)>\min(q,j)$, one obtains the
\noindent
{\bf Proposition 1 :} The second Gelfand-Dikii\ bracket associated with the
$n\times n$-matrix $m^{\rm th}$-order differential operator $L$ as
defined by eqs. \di, \partial_v\ and \dvi\ equals
$$\eqalign{
\{f,g\}_{(2)}&=a\int{\rm tr}\ \sum_{j=0}^{m-1} {\cal V}(f)_j Y_{j+1} \cr
{\cal V}(f)_j&=\sum_{l=1}^m \sum_{p=0}^{2m-j-l}
\sum_{q=\max(0,p+j+l-m)}^{\min(m,p+j+l)} \left( S_{q-p,l}^{q,j} -
\bin q-l p \right)
U_{m-q}(X_lU_{m-p-j-l+q})^{(p)}\ .\cr}
\eqn\dxxiii$$
It is not obvious that \dxxiii\ satisfies antisymmetry or the Jacobi identity,
but
this will follow from the Miura transformation discussed below.
\section{The Gelfand-Dikii\ brackets reduced to $U_1=0$}
The problem of consistently restricting a given symplectic manifold
(phase space) to a symplectic submanifold by imposing certain constraints
$\phi_i=0$ has been much studied in the literature. The basic point is
that for a given phase space one cannot set a coordinate to a given
value (or function) without also eliminating the corresponding
momentum. More generally, to impose a constraint $\phi=0$
consistently, one has to make sure that for any functional $f$ the
bracket $\{\phi,f\}$ vanishes if the constraint $\phi=0$ is imposed
{\it after} computing the bracket. In general this results in a
modification of the original Poisson bracket.
Here, I want to impose $\{f, U_1\}\vert_{U_1=0}=0$ for all $f$. Since
$Y_m=\dd g {U_1}$, one sees from \dxxiii\ that this requires
${\cal V}(f)_{m-1}\vert_{U_1=0}=0$. In practice this determines $X_m$ which
otherwise would be undefined if one starts with $U_1=0$. In the
scalar case it is known [\ADLER,\LEBMAN] that $X_m$ should be determined by
${\rm res}\ [L,X_f]=0$. The following Lemma shows that this is still
true in the matrix case.
\Lemma 2 One has
${\cal V}(f)_{m-1}=-{\rm res}\ [L,X_f]$,
and for
$U_1=0$, ${\rm res}\ [L,X_f]=0$ is equivalent to
$$X_m={1\over m} \sum_{l=1}^{m-1}\left(
\partial^{-1}[U_{m+1-l},X_l]
+\sum_{k=l}^m (-)^{k-l}\bin k {l-1}
(X_lU_{m-k})^{(k-l)}\right) \ .
\eqn\dxxix$$
Note the commutator term which is a new feature of the present matrix
case as opposed to the scalar case.
One of the main results then is the following
\noindent
{\bf Theorem 3 :} The second Gelfand-Dikii\ bracket for $n\times n$-matrix
$m^{\rm th}$-order differential operators $L$ with vanishing $U_1$ is
given by
$$\eqalign{
\{f,g\}_{(2)}=&a\int{\rm tr}\ \sum_{j=0}^{m-2} {\tilde{\cal V}}(f)_j Y_{j+1}\ , \cr
{\tilde{\cal V}}(f)_j=&{1\over m} \sum_{l=1}^{m-1} [U_{m-j},\partial^{-1} [X_l,U_{m-l+1}]]
\cr
+&{1\over m} \sum_{l=1}^{m-1}\Big\{ \sum_{k=0}^{m-l} (-)^k
\bin k+l {l-1} (X_lU_{m-k-l})^{(k)} U_{m-j} \cr
&\phantom{{1\over m} \sum_{l=1}^{m-1}} -\sum_{k=0}^{m-j-1}
\bin k+j+1 j U_{m-k-j-1} (U_{m-l+1}X_l)^{(k)} \Big\} \cr
+&\sum_{l=1}^{m-1}\, \sum_{p=0}^{2m-j-l}
\sum_{q=\max(0,p+j+l-m)}^{\min(m,p+j+l)} C_{q-p,l}^{q,j}
U_{m-q}(X_lU_{m-p-j-l+q})^{(p)}\ , \cr
C_{q-p,l}^{q,j}=&S_{q-p,l}^{q,j} -
\bin q-l p -{1\over m}(-)^{q-p+j}\bin q j \bin p-q+j+l {l-1}\cr }
\eqn\dxxx$$
where the $S_{q-p,l}^{q,j}$ are defined by eq. \dxiv, and it is
understood that $U_0=-1$ and $U_1=0$.
\noindent
{\bf Remark 4 :} If one takes $m=2$, $L=-\partial^2+U$, so that
$U_2\equiv U$ and $X_1\equiv X$, only ${\tilde{\cal V}}(f)_0$ is non-vanishing:
$${\tilde{\cal V}}(f)_0=-{1\over 2} [U,\partial^{-1}[U,X]]+{1\over 2}(XU+UX)'
+{1\over 2}(X'U+UX')-{1\over 2}X'''
\eqn\dxxxiii$$
and with $X=\dd f U$ and $Y=\dd g U$ one obtains (using $\int x
\partial^{-1} y=-\int (\partial^{-1}x) y$)
$$\{f,g\}_{(2)}=a\int{\rm tr}\ \left( -{1\over 2} [U,X]\partial^{-1}[U,Y]
+{1\over 2}(X'Y+YX'-XY'-Y'X)U-{1\over 2}YX'''\right)
\eqn\dxxxiv$$
which obviously is a generalization of the original $V$-algebra \txx\
to arbitrary $n\times n$-matrices $U\equiv U_2$.
To appreciate the structure of the non-local terms, I explicitly
write this algebra in the simplest case for $n=2$ (but {\it without}
the restriction
${\rm tr}\ \sigma_3U=0$ which is present for \txx). Let
$$U=\pmatrix{ T+V_3&-\sqrt{2}{V^+}\cr -\sqrt{2}{V^-}& T-V_3\cr} \ .
\eqn\dxxxv$$
Then one obtains from \dxxxiv\ (with $a=-2\gamma^2$) the algebra
$$\eqalign{
\gamma^{-2} \{T(\sigma)\, ,\, T(\sigma')\} &=
(\partial_\sigma-\partial_{\sigma'})\left[ T(\sigma') \delta(\s-\s')\right]-{1\over 2} \delta'''(\s-\s') \cr
\gamma^{-2} \{T(\sigma)\, ,\, V^\pm(\sigma')\} &=
(\partial_\sigma-\partial_{\sigma'})\left[ V^\pm(\sigma') \delta(\s-\s')\right]\cr
\gamma^{-2} \{T(\sigma)\, ,\, V_3(\sigma')\} &=
(\partial_\sigma-\partial_{\sigma'})\left[ V_3(\sigma') \delta(\s-\s')\right]\cr
\gamma^{-2} \{V^\pm(\sigma)\, ,\, V^\pm(\sigma')\}&=\epsilon(\s-\s')
V^\pm(\sigma)V^\pm(\sigma')\cr
\gamma^{-2} \{V^\pm(\sigma)\, ,\, V^\mp(\sigma')\}&=-\epsilon(\s-\s')
(V^\pm(\sigma)V^\mp(\sigma')+V_3(\sigma)V_3(\sigma'))\cr
&\phantom{=}+(\partial_\sigma-\partial_{\sigma'})\left[ T(\sigma') \delta(\s-\s')\right] -{1\over 2}
\delta'''(\s-\s') \cr
\gamma^{-2} \{V_3(\sigma)\, ,\, V^\pm(\sigma')\}&=\epsilon(\s-\s')
V^\pm(\sigma)V_3(\sigma')\cr
\gamma^{-2} \{V_3(\sigma)\, ,\, V_3(\sigma')\}&=\epsilon(\s-\s')
(V^+(\sigma)V^-(\sigma')+ V^-(\sigma)V^+(\sigma'))\cr
&\phantom{=}+(\partial_\sigma-\partial_{\sigma'})\left[ T(\sigma') \delta(\s-\s')\right] -{1\over 2}
\delta'''(\s-\s') \ .\cr
}
\eqn\dxxxvi$$
Once again, one sees that $T$ generates the conformal algebra with a
central charge, while $V^+$, $V^-$ and $V_3$ are conformally primary
fields of weight (spin) two. It is easy to check on the example \dxxxiv,
that antisymmetry and the Jacobi identity are satisfied. For general $m$
this follows from the Miura transformation to which I now turn.
\section{The Miura transformation : The case of general $U_1, \ldots U_m$}
\noindent
{\bf Definition and Lemma 5 :} Introduce the $n\times
n$-matrix-valued functions $P_j(\sigma),\, j=1, \ldots m$. Then for
functionals $f,g$ (integrals of traces of differential polynomials)
of the $P_j$ the following Poisson bracket is well-defined %
$$\{f,g\}=a\sum_{i=1}^m \int{\rm tr}\ \left( \left( \dd f {P_i} \right)'
\dd g {P_i} - \left[ \dd f {P_i} , \dd g {P_i} \right] P_i \right)
\eqn\ti$$
or equivalently for $n\times n$-matrix-valued (numerical) test-functions
$F$ and $G$
$$\{\int{\rm tr}\ FP_i,\int{\rm tr}\ GP_j\}=a\, \delta_{ij} \int{\rm tr}\ \left( F'G
-[F,G]P_i \right) \ .
\eqn\tii$$
\noindent
Note that due to the $\delta_{ij}$ in \tii\ one has $m$ decoupled
Poisson brackets. In the scalar case ($n=1$), \tii\ simply gives
$\{P_i(\sigma),P_j(\sigma')\}=(-a)\delta_{ij}\delta'(\s-\s')$. These are $m$ free fields
or $m$ $U(1)$ current algebras. In the matrix case, one easily sees that
each $P_j$ actually gives a $gl(n)$
current algebra. So one has no longer free fields but $m$ completely
decoupled current algebras. This is still much simpler than the
bracket \dxxiii. To connect both brackets one starts with the
following obvious
\Lemma 6 Let $P_j, j=1, \ldots m$ be as in Lemma 5. Then
$$L=-(\partial-P_1)(\partial-P_2)\ldots (\partial-P_m)
\eqn\tiii$$
is a $m^{\rm th}$-order $n\times n$-matrix linear differential
operator and can be written $L=\sum_{k=0}^mU_{m-k}\partial^k$ with $U_0=-1$
as before. This identification gives all $U_k, k=1,\ldots m$ as
$k^{\rm th}$-order differential polynomials in the $P_j$, i.e. it
provides an embedding of the algebra of differential polynomials in
the $U_k$ into the algebra of differential polynomials in the $P_j$.
One has in particular
$$U_1=\sum_{j=1}^m P_j\quad , \quad
U_2=-\sum_{i<j}^m P_iP_j+\sum_{j=2}^m (j-1)P_j'\ .
\eqn\tiv$$
\noindent
I will call the embedding given by \tiii\ a (matrix) Miura
transformation. The most important property of this Miura
transformation is given by the following matrix-generalization of a
well-known theorem [\KW, \DIK].
\noindent
{\bf Theorem 7 :} Let $f(U)$ and $g(U)$ be functionals of the $U_k,\
k=1,\ldots m$. By Lemma 6 they are also functionals of the $P_j,\
j=1,\ldots m$: $f(U)={\tilde f } (P)$, $g(U)={\tilde g } (P)$ where ${\tilde f } (P)=f(U(P))$
etc. One then has
$$\{{\tilde f } (P),{\tilde g } (P)\}=\{f(U),g(U)\}_{(2)}
\eqn\tv$$
where the bracket on the l.h.s. is the Poisson bracket \ti\ and the
bracket on the r.h.s. is the second Gelfand-Dikii\ bracket \dxxiii.
\noindent
The previous Theorem states that one can either compute
$\{U_k,U_l\}$ using the complicated formula \dxxiii\ or using the
simple Poisson bracket \ti\ for more or less complicated functionals
$U_k(P)$ and $U_l(P)$. In particular Lemma 5 implies the
\noindent
{\bf Corollary 8 :} The second Gelfand-Dikii\ bracket \dxxiii\ obeys
antisymmetry and the Jacobi identity. Bilinearity in $f$ and $g$
being evident, it is a well-defined Poisson bracket.
\section{The Miura transformation : The case $U_1=0$}
As seen from \tiv, $U_1=0$ corresponds to $\sum_{i=1}^m P_i=0$. In
order to describe the reduction to $\sum_i P_i=0$ it is convenient to
go from the $P_i, i=1,\ldots m$ to a new ``basis": $Q=\sum_{i=1}^m
P_i$ and ${\cal P} _a, a=1,\ldots m-1$ where all ${\cal P} _a$ lie in the
hyperplane $Q=0$. Of course, $Q$ and each ${\cal P} _a$ are still $n\times
n$-matrices. More precisely:
\noindent
{\bf Definition and Lemma 9 :} Consider a $(m-1)$-dimensional
vector space, and choose an overcomplete basis of $m$ vectors $h_j,\
j=1,\ldots m$. Denote the components of each $h_j$ by $h_j^a,\ a=1,
\ldots m-1$. Choose the $h_j$ such that
$$\eqalign{
\sum_{j=1}^m h_j &=0\ ,\cr
h_i\cdot h_j &= \delta_{ij}-{1\over m}\ ,\cr
\sum_{i=1}^m h_i^a h_i^b &=\delta_{ab} \cr}
\eqn\txv$$
and define the completely symmetric rank-3 tensor $D_{abc}$ by
$$D_{abc}=\sum_{i=1}^m h_i^a h_i^b h_i^c \ .
\eqn\txvi$$
Define $Q$ and ${\cal P} _a,\ a=1,\ldots m-1$ to be the following linear
combinations of the $P_j$'s
$${\cal P} _a=\sum_{j=1}^m h_j^a P_j \quad , \quad Q=\sum_{j=1}^m P_j\ .
\eqn\txvii$$
If one considers the $P_j$ as an orthogonal basis in a $m$-dimensional
vector-space, then the ${\cal P} _a$ are an orthogonal basis in a
$(m-1)$-dimensional hyperplane orthogonal to the line spanned by $Q$.
\noindent
{\bf Proposition 10 :} The Poisson bracket \ti\ can be reduced to the
symplectic submanifold with $Q\equiv \sum_{j=1}^m P_j =0$. The reduced
Poisson bracket is
$$\{f,g\}=a\int{\rm tr}\ \Big( W_bV_b'
-[V_b,W_c]D_{bcd}{\cal P} _d -{1\over m} [V_b,{\cal P} _b]\partial^{-1}[W_c,{\cal P} _c]\Big)
\eqn\txxi$$
where $V_b$ and $W_b$ denote
$V_a= \dd f {{\cal P} _a} , \ V_0= \dd f Q , \
W_a= \dd g {{\cal P} _a}$ and $W_0= \dd g Q $.
Equivalently one has
$$\{\int{\rm tr}\ F{\cal P} _b,\int{\rm tr}\ G{\cal P} _c\}=a\int{\rm tr}\ \Big( GF'\delta_{bc}
-[F,G]D_{bcd}{\cal P} _d -{1\over m} [F,{\cal P} _b]\partial^{-1}[G,{\cal P} _c]\Big) \ .
\eqn\txxii$$
\noindent
{\bf Corollary 11 :} The Poisson bracket \tii\ when reduced to
$Q=\sum_{j=1}^mP_j=0$ can be equivalently written as
$$\eqalign{
\{f,g\}=a\int{\rm tr}\ &\Big( GF'(\delta_{ij} -{1\over m}) -
[F,G] (\delta_{ij} -{2\over m}){1\over 2}(P_i+P_j)\cr
&-{1\over m} [F,P_i]\partial^{-1}[G,P_j]\Big) \ . \cr}
\eqn\txxv$$
\noindent
{\bf Theorem 12 :} Let $U_1=0$ and hence $Q=\sum_{j=1}^mP_j=0$.
By the Miura transformation of Lemma 6 any functionals $f(U),\, g(U)$
of the $U_k$ only $(k=2,\ldots m$) are also functionals
${\tilde f } ({\cal P} )=f(U({\cal P} )),\, {\tilde g } ({\cal P} )=g(U({\cal P} ))$ of the ${\cal P} _a, a=1, \ldots m$
only. The reduced second Gelfand-Dikii\ bracket \dxxx\ of $f$ and $g$ equals the
reduced Poisson bracket \txxi\ of ${\tilde f } $ and ${\tilde g } $.
\noindent
{\bf Corollary 13 :} The second Gelfand-Dikii\ bracket \dxxx\ obeys
antisymmetry and the Jacobi identity. Bilinearity in $f$ and $g$ being
evident, it is a well-defined Poisson bracket.
\section{The conformal properties}
\REF\MAT{P. Mathieu, {\it Extended classical conformal algebras and
the second Hamiltonian structure of Lax equations}, Phys. Lett. {\bf
B208} (1988) 101.}
\REF\DIZ{P. Di Francesco, C. Itzykson and J.-B.
Zuber, {\it Classical $W$-algebras}, Commun. Math. Phys.
{\bf 140} (1991) 543 .}
In the scalar case, i.e. for $n=1$, the second Gelfand-Dikii\ bracket (with
$U_1=0$) gives the $W_m$-algebras [\BAK,\MAT,\DIZ]. The interest in the
$W$-algebras stems from the fact that they are extensions of the
conformal Virasoro algebra, i.e. they contain the Virasoro algebra as a
subalgebra. Furthermore, in the scalar case, it is known that certain
combinations of the $U_k$ and their derivatives yield primary fields of
integer spins $3,4,\ldots m$. It is the purpose of this section to
establish the same results for the matrix case, $n>1$. From now on,
I only consider the second Gelfand-Dikii\ bracket \dxxx\ for the case
$U_1=0$. I will simply write $\{f,g\}$ instead of $\{f,g\}_{(2)}$.
Also, it is often more convenient to replace the scale factor $a$ by
$\gamma^2$ related to $a$ by
$$a=-2\gamma^2\ .
\eqn\qi$$
(Note that $\gamma^2$ need not be positive.)
\subsection{The Virasoro subalgebra}
For the original $V$-algebra \txx\ (corresponding to
$m=2,\, n=2$ and an additional constraint ${\rm tr}\ \sigma_3U_2=0$) one sees that
$T={1\over 2}{\rm tr}\ U_2$ generates the conformal algebra. I will now show
that for general $m, n$, the generator of the conformal algebra is
still given by this formula.
\Lemma 14 For arbitrary $m\ge 2$ one has
$$\eqalign{
\{\int{\rm tr}\ FU_2,\int{\rm tr}\ GU_2\}&
= a\int{\rm tr}\ \Big(
-{1\over m} [F,U_2]\partial^{-1}[G,U_2]-[F,G](U_3-{m-2\over 2}U_2')\cr
&+{1\over 2}(F'G-G'F+GF'-FG')U_2 -{1\over 2} \bin m+1 3 G F''' \Big) \
. \cr}
\eqn\qii$$
Note that for $m=2$ one has to set $U_3=0$.
\noindent
{\bf Proposition 15 :} Let $T(\sigma)={1\over 2} {\rm tr}\ U_2(\sigma)$. Then
$$\gamma^{-2} \{T(\sigma_1),T(\sigma_2)\}=(\partial_{\sigma_1}-\partial_{\sigma_2})
\left( T(\sigma_2)\delta(\sigma_1-\sigma_2)\right)
-{n\over 4}\bin m+1 3 \delta'''(\sigma_1-\sigma_2) \ .
\eqn\qiv$$
Equivalently, if, for $\sigma\in S^1$, one defines for integer $r$
$$L_r=\gamma^{-2}\int_{-\pi}^\pi {\rm d}\sigma\, T(\sigma)e^{ir\sigma} +{c\over
24}\, \delta_{r,0}
\eqn\qv$$
where
$$c={6\pi\over \gamma^2} \, n\, \bin m+1 3 = {12\pi\over (-a)} n\, \bin m+1 3
\eqn\qvi$$
then the $L_r$ form a Poisson bracket version of the Virasoro algebra
with (classical) central charge $c$ :
$$i\{L_r,L_s\}=(r-s)L_{r+s}+{c\over 12} (r^3-r) \delta_{r+s,0} \ .
\eqn\qvii$$
Also, if $\{A_\mu\}_{\mu=1,\ldots n^2-1}$ is a basis for the traceless
$n\times n$-matrices, then each $S_\mu (\sigma)={\rm tr}\ A_\mu U_2(\sigma),
\mu=1,\ldots n^2-1$ is a conformally primary field of conformal
dimension (spin) 2: %
$$\gamma^{-2} \{T(\sigma_1),S_\mu(\sigma_2)\}=(\partial_{\sigma_1}-\partial_{\sigma_2})
\left( S_\mu(\sigma_2)\delta(\sigma_1-\sigma_2)\right)
\eqn\qviii$$
or for the modes $(S_\mu)_r=\gamma^{-2}\int_{-\pi}^\pi {\rm d}\sigma\,
S_\mu(\sigma)e^{ir\sigma}$ one has
$i\{L_r,(S_\mu)_s\}=(r-s)(S_\mu)_{r+s}$. Equations \qiv\ and \qviii\
can be written in matrix notation as (${\bf 1}$ denotes the $n\times n$
unit matrix)
$$\gamma^{-2} \{T(\sigma_1),U_2(\sigma_2)\}=(\partial_{\sigma_1}-\partial_{\sigma_2})
\left( U_2(\sigma_2)\delta(\sigma_1-\sigma_2)\right)
-{1\over 2}\bin m+1 3 {\bf 1}\, \delta'''(\sigma_1-\sigma_2) \ .
\eqn\qix$$
\subsection{The conformal properties of the $U_k$ for $k\ge 3$}
Having computed the conformal properties of
the matrix elements of $U_2$, I will now give
those of all the other $U_k$, i.e. compute $\{T(\sigma_1),U_k(\sigma_2)\}$ or
equivalently, for any (test-) function $\epsilon(\sigma)$, compute $\{\int \epsilon T,
U_k(\sigma_2)\}$
for all $k\ge 3$. I will find that this Poisson bracket is linear in the
$U_l$ and their derivatives and is formally identical to the result of
the scalar case. It then follows that appropriately symmetrized
combinations $W_k$ can be formed that are $n\times n$-matrices, each
matrix element of $W_k$ being a conformal primary field of dimension
(spin) $k$.
\noindent
{\bf Proposition 16 :}
The conformal properties of all matrix elements of all $U_k, k=2,\ldots
m$ are given by
$$\eqalign{
\gamma^{-2} \{\int \epsilon T, U_k\} &= -\epsilon U_k'-k\epsilon' U_k +{k-1\over 2} \bin m+1 {k+1}
\epsilon^{(k+1)}\cr
& +\sum_{l=2}^{k-1}\left[ \bin m-l {k+1-l} -{m-1\over 2} \bin m-l {k-l}
\right] \epsilon^{(k-l+1)} U_l \cr }
\eqn\qxv$$
which is formally the same equation as in the scalar case $n=1$ [\DIZ].
Since the conformal properties \qxv\ are formally the same as in the
scalar case, and in the latter case it was possible to form
combinations $W_k$ that are spin-$k$ conformally primary fields [\DIZ], one
expects a similar result to hold in the matrix case. Indeed, one has the
\noindent
{\bf Theorem 17 :} For matrices $A_1, A_2, \ldots A_r$ denote by
$S[A_1,A_2,\ldots,A_r]$ the completely symmetrized product normalized
to equal $A^r$ if $A_s=A$ for all $s=1,\ldots r$. Let
$$\eqalign{
W_k=&\sum_{l=2}^k B_{kl\, }U_l^{(k-l)} +\sum_{0\le p_1\le \ldots \le p_r
\atop \sum p_i+2r=k} (-)^{r-1} C_{p_1\ldots p_r}\, S[U_2^{(p_1)},
\ldots, U_2^{(p_r)}]\cr
&+\sum_{0\le p_1\le \ldots \le p_r
\atop s\le l\le k-\sum p_i-2r} (-)^{r} D_{p_1\ldots p_r,l}\,
S[U_2^{(p_1)}, \ldots, U_2^{(p_r)},U_l^{(k-l-\sum p_i-2r)}]\cr }
\eqn\qxix$$
where the coefficiets $B_{kl}, C_{p_1\ldots p_r}$ and $D_{p_1\ldots
p_r,l}$ are the same as those given in ref. \DIZ\ for the scalar case, in
particular
$$B_{kl}=(-) ^{k-l}\ { \bin k-1 {k-l} \bin m-l {k-l} \over
\bin 2k-2 {k-l} }\ .
\eqn\qxx$$
Then the $W_k$ are spin-$k$ conformally primary
$n\times n$-matrix-valued fields, i.e.
$$\gamma^{-2} \{\int \epsilon T, W_k\}=-\epsilon W_k' -k\epsilon' W_k \ .
\eqn\qxxi$$
For $\sigma\in S^1$ one can define the modes
$(W_k)_s=\gamma^{-2}\int_{-\pi}^\pi {\rm d}\sigma\, W_k(\sigma) e^{is\sigma}$ and the Virasoro
generators $L_r$ as in \qv. Then one has equivalently
$$i\{L_r,(W_k)_s\}=\big( (k-1)r-s\big) (W_k)_{r+s}
\eqn\qxxia$$
where each $(W_k)_s$ is a $n\times n$-matrix.
\noindent
{\bf Examples :} From the previous Theorem and the results of ref.
\DIZ\ (their Table I) one has explicitly:
$$\eqalign{
W_3&= U_3-{m-2\over 2} U_2' \cr
W_4&= U_4-{m-3\over 2} U_3'+{(m-2)(m-3)\over 10} U_2''
+{(5m+7)(m-2)(m-3)\over 10 m (m^2-1)} U_2^2 \cr
W_5&= U_5-{m-4\over 2} U_4'+{3(m-3)(m-4)\over 28} U_3''
-{(m-2)(m-3)(m-4)\over 84} U_2'''\cr
&+{(7m+13)(m-3)(m-4)\over 14 m (m^2-1)} (U_2W_3+W_3U_2) \ . \cr}
\eqn\qxxiii$$
\section{Example of the $V_{n,3}$-algebra}
{}From the previous subsection one might have gotten the impression that
the matrix case is not very different from the scalar case. This is
however not true. In the previous subsection only the conformal
properties, i.e. the Poisson brackets with $T={1\over 2} {\rm tr}\ {\bf 1} U_2$
were studied, and since the unit-matrix ${\bf 1}$ always commutes, most of
the new features due to the non-commutativity of matrices were not
seen. Technically speaking, only ${\rm tr}\ {\tilde{\cal V}}(f)$ was needed, not ${\tilde{\cal V}}(f)$
itself (cf. eq. \dxxx). In this subsection, I will give the Poisson brackets,
for the (more
interesting) reduction to $U_1=0$, of any two matrix elements of $U_2$
or $U_3$, or equivalently of $U_2$ or $W_3$, for $m=3$.
This is the complete algebra, giving a matrix generalization
\REF\ZAM{A.B. Zamolodchikov, {\it Infinite additional symmetries
in two-dimensional conformal field theory}, Theor. Math. Phys. {\bf 65}
(1985) 1205. }
of Zamolodchikov's $W_3$-algebra [\ZAM]. (Recall that $F$ and $G$ are $n\times
n$-matrices of
test-functions.)
$$\eqalign{
\{\int{\rm tr}\ FU_2,\int{\rm tr}\ GU_2\}=a \int{\rm tr}\ \Big(
&-{1\over 3}[F,U_2]\partial^{-1}[G,U_2] -[F,G]W_3\cr
&+{1\over 2}(F'G+GF'-FG'-G'F)U_2-2GF'''\Big)\ ,
\cr}
\eqn\uiii$$
$$\eqalign{
\{\int{\rm tr}\ FU_2,\int{\rm tr}\ GW_3\}=a \int{\rm tr}\ \Big(
&-{1\over 3}[F,U_2]\partial^{-1}[G,W_3] -{1\over 6}[F,G]U_2^2\cr
&+(-{1\over 4}[F',G']+{1\over 2}[F'',G]+{1\over 12}[F,G'']) U_2\cr
&+(F'G+GF'-{1\over 2}FG'-{1\over 2}G'F)W_3 \Big)\ ,
\cr}
\eqn\uiv$$
$$\eqalign{
\{\int{\rm tr}\ F&W_3,\int{\rm tr}\ GW_3\}
=a \int{\rm tr}\ \Big(
-{1\over 3}[F,W_3]\partial^{-1}[G,W_3]\cr
&-{1\over 6}[F,G](W_3U_2+U_2W_3)
+{2\over 3}(FU_2GW_3-GU_2FW_3)\cr
&+{5\over 12}(F'U_2GU_2-G'U_2FU_2)+{1\over 12}(FG'-GF')U_2^2
+{1\over 12}[F,G]U_2'U_2\cr
&+{7\over 12}[F',G']W_3-{1\over 6}[F,G]''W_3
+{1\over 12}(FG'''+G'''F-F'''G-GF''')U_2\cr
&+{1\over 8}(F''G'+G'F''-F'G''-G''F')U_2+{1\over 6}GF^{(5)} \Big)\ .
\cr}
\eqn\uv$$
One remarks that in the scalar case ($n=1$) this reduces to the
Poisson bracket version of Zamolodchikov's $W_3$-algebra [\ZAM], as it
should. In the matrix case however, even if $F=f{\bf 1},\ G=g{\bf 1}$ (with
scalar $f,g$) this is a different algebra, i.e.
$\{\int{\rm tr}\ W_3(\sigma),\int{\rm tr}\ W_3(\sigma')\}$ does not reduce to the
$W_3$-algebra, since the r.h.s. contains the non-linear terms and
${\rm tr}\ U_2^2\ne ({\rm tr}\ U_2)^2$. In other words, the scalar ($n=1$)
$W_m$-algebras are not subalgebras of the matrix $V_{n,m}$-algebras.
The only exception is $m=2$, since one always has a Virasoro subalgebra.
\refout
\end
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,436
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CyberCom Chief Alexander Lays Down Cyber Red Line; Destroy A Network, Risk War
From: Breaking Defense
By Colin Clark
CAPITOL HILL: On the day that China's president took personal charge of his country's new cyber body, pledging to make the People's Republic of China a "cyber power," the outgoing head of America's Cyber Command laid out a clear red line that, if crossed, could lead to war.
"If it destroys government or other networks, I think it would cross that line," Army Gen. Keith Alexander, head of both Cyber Command and the National Security Agency, told the Senate Armed Services Committee today when asked what level of cyber "attack" would potentially cause America to go to war.
Hackers Threaten Brazil's World Cup
From: Voice of America
SAO PAULO — Brazilian hackers are threatening to disrupt the World Cup with attacks ranging from jamming websites to data theft, adding cyber warfare to the list of challenges for a competition already marred by protests, delays and overspending.
In a country with rampant online crime, a challenging telecommunications infrastructure and little experience with cyber attacks, authorities are rushing to protect government websites and those of FIFA, soccer's governing body.
360 million newly stolen credentials on black market: cybersecurity firm
From: Reuters
BOSTON: A cybersecurity firm said on Tuesday that it uncovered stolen credentials from some 360 million accounts that are available for sale on cyber black markets, though it is unsure where they came from or what they can be used to access.
The discovery could represent more of a risk to consumers and companies than stolen credit card data because of the chance the sets of user names and passwords could open the door to online bank accounts, corporate networks, health records and virtually any other type of computer system.
BlackBerry announces $200 smartphone, but emphasises security
From: The Guardian
BlackBerry chief executive brings the company back to its business roots, focusing on secure messaging and a new phone that features the 'classic' trademark trackpad and keyboard
Samuel Gibbs in Barcelona
BlackBerry is working with Chinese manufacturing company Foxconn to produce a new, cheaper smartphone that will sell for under $200, it was announced on Tuesday.
The chief executive John Chen told the Mobile World Congress in Barcelona that the Z3 will launch in Indonesia in April, and that it was built in just three months instead of the usual 12.
Russia, China the most dangerous countries for mobile cyber attacks
Feb 21st
From: The Sydney Morning Herald
Jordan Robertson
In news accounts of cyber attacks plaguing computer networks around the world, the bad actors are almost always the same – faceless adversaries hailing from shadowy regions of Asia and Eastern Europe.
But what if the issue was examined from the other direction, from the perspective of people living in countries identified as launching pads for the world's hacking attacks?
New research by security firm Lookout shows that when it comes to cyber threats, countries reap what countries sow.
Read Complete Article
NIST Unveils Crypto Standards Proposal
Editor's Note: To read about the Center for Regulatory Effectiveness's work enhancing federal cyber security transparency, see the Internet Architecture Board's (IAB) comments to NIST in the matter of the NIST Special Publication 800-90A (Recommendation for Random Number Generation Using Deterministic Random Bit Generators) review proceeding here and CRE's comments on NIST Special Publication 800-137 Information Security Continuous Monitoring for Federal Information Systems and Organizations here.
From: GovInfoSecurity.com
Feedback Sought on Development Process
Because of concerns of possible National Security Agency meddling with its cryptographic standards, the National Institute of Standards and Technology has issued a draft report proposing revisions in how it develops cryptographic standards.
Nursing Homes Are Exposed to Hacker Attacks
From: The Wall Street Journal
Cybersecurity Experts Find Trove of Information on File-Sharing Web Site
By Rachael King
Computer-security researchers have discovered on a website documents that could allow hackers easily to obtain electronic medical records and payment information from health-care providers.
The documents—found by two cybersecurity firms on a site commonly used by hackers—detail the type of equipment used in computer networks, the Internet addresses for computers and other devices, and the passwords to network firewalls run by health-care providers such as nursing homes, doctors' offices and hospitals.
Workshop: Low Intensity Cyber Operations – The International Legal Regime
From: NATO Cooperative Cyber Defence Centre of Excellence
Time: 17-18 February
Place: Tartu, Estonia
Organisers: NATO CCD COE and the University of Tartu, in cooperation with the Baltic Yearbook of International Law, Estonian Foreign Policy Institute and Stockton Center for the Study of International Law at the US Naval War College
Participation: Only for invited individuals
Workshop proceedings will be published in the 2014 Baltic Yearbook of International Law.
How to reduce data breach and cyber security risk
From: Out-Law.com
FOCUS: Most companies will have an information security breach in most years, so businesses should focus on preparing for incidents, because they are inevitable.
The increase in the volume of data that businesses now store; the growing use of mobile devices, and the trend of users connecting their own devices to corporate networks are factors making data breaches more likely. And proposed changes to EU law mean that organisations will no longer be able to keep breaches a secret.
Iran Supreme Leader Khamenei Tells Students to Get Ready for Cyber War
From: The Algemeiner
Iran's Supreme Leader, Ayatollah Ali Khamenei, sent a message to the country's university students to prepare for cyber war, according to Iran's semi-official government news agency Mehr , on Wednesday. The radical cleric also referred to cyber-war as "a demanding field," according to the report.
"You are the cyber-war agents and such a war requires Ammar-like insight and Malik Ashtar-like (two Prophet's Companions in early Islamic history) resistance; get yourselves ready for such war wholeheartedly," his message said.
FISMA Focus
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 5,379
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Q: How do I calculate conversion rate using Google Analytics? I have created a goal in Google Analytics that is met when the user completes the sign up process. The page that they end up at is
http://my_url.com/?just_signed_up=true
In order to calculate my conversation rate, I need to do this calculation:
goal (sign ups) / new visitors
I know that in order to use new visitor numbers, I can either set up a profile with a new visitor filter or just apply a new user advanced segment on my normal profile.
My problem is that the Google Analytics tracking code is on every page in my site, including my landing page. The sign up process goes like this:
Landing page -> Sign up form -> http://my_url.com/?just_signed_up=true
When the user ends up at the URL above, am I correct in thinking that Google Analytics will no longer consider them a new user, as they would have the cookie from when they landed on the landing page at the start of the sign up process? If so, then there will never be a new visitor that meets the goal. How would I calculate my conversion rate in this scenario?
A: A returning visitor is someone who starts an additional Google Analytics session while using your site. To start a new session they need to close their browser or stop using the site for a period of at least 30 minutes. Browsing around your site clicking links isn't considered stating a new session. This means that, in most cases, signups from new users will be from people (or at least browsers with a set of cookies) who have not visited your site before.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,537
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The 2020 Ooredoo Ride of Champions' Corporate Challenge Provides a Friendly Match Between Qatar's Fittest Firms 16 December 2020
Champions and Public Continue to Win with the 2020 Ooredoo Ride of Champions 8 December 2020
The Qatar Cyclists Center organised 2020 Ooredoo Ride of Champions Believed to Be The Most Accessible Mass Participation Event in Qatar's History, Featuring Best Buddies, Accessible Qatar, and More! 19 November 2020
Safety First to Enable Fun Was the Key Theme at the 2020 Ooredoo Ride of Champions, Enabling Thousands to Have Fun, Get Exercise and look after their Mental Health in 2020 14 November 2020
Inmates Could be Freed to Ease Virus Pressure on Jails due to Coronavirus 25 March 2020
Coronavirus to be Tracked Using its Genetic Code 22 March 2020
Japan PM Abe says Postponing Tokyo Olympics an Option 22 March 2020
China Reports 39 New Confirmed Cases of Coronavirus, All Imported 22 March 2020
The Best and Wworst of F1's Esports Weekend 22 March 2020
Italy Sees 651 New Virus Deaths, Toll Nearly 5,500 22 March 2020
UAE's Arabian Gulf Super Cup to Provide Early Treat
Sports 22 August 2019 Arabia Day Newsdesk 0
The Arabian Gulf Super Cup, the traditional season-opener to the UAE's football season is still a month away and so is the start of...
The Arabian Gulf Super Cup, the traditional season-opener to the UAE's football season is still a month away and so is the start of the Arabian Gulf League, but fans who live and breathe the sport in the country will have their hunger sated with an early treat.
The Arabian Gulf Cup, which runs side-by-side to the UAE's top flight, kicks off over the weekend with a couple of interesting fixtures.
And it is Al Jazira who will get the season into gear by welcoming promoted side Khor Fakkan in a Group A game at the Mohammed bin Zayed Stadium on Thursday evening.
At about the same time, another promoted Hatta are away to Ajman, while Bani Yas play host to League champions Sharjah, later at night.
And Al Jazira's new coach Jurgen Streppel is hoping to start his reign with a win following a camp in his native Netherlands.
"There is no doubt that all players will make every effort throughout the game to ensure the best result," Streppel said in his pre-match briefing.
And despite having some great individual talent, the Dutchman focused on the collective.
"We have some of the best players in the UAE and the region and have the expertise and skills. But individual effort is not enough to win because football is a collective game and requires everyone to work together to win matches," added the 50-year-old, who previously worked with 22-year-old Barcelona midfielder Frenkie De Jong.
Al Jazira were quite active in the transfer market, signing Sultan Al Ghaferi and Mourad Batna from city rivals Al Wahda, Amer Abdulrahman from Al Ain, Serbian defender Milos Kosanovic and Keno, to add to the big acquisition of Omar Abdulrahman.
Meanwhile, club captain and goalkeeper Ali Khaseif is looking to bring the pride back to Abu Dhabi.
"We trained well and have made all the preparations. We are trying to provide a level that is according to the standard of Al Jazira and we will do our best to bring Abu Dhabi's pride to the forefront," said Khaseif.
Meanwhile, Bani Yas coach Winfried Schafer said his motto is to build a team for the future. "A tough game awaits us. We must fight hard to get a positive result against Sharjah, who are a very strong team," said Schafer.
"Our goal is to build a team for the future. We have a clear vision and that is to create a competitive team. We now have the youngest team in the competition with the average age being 23," added the former Al Ain manager.
On Friday night, holders Shabab Al Ahli begin their title defence with a tough Group A game against Al Ain at the iconic Hazza bin Zayed Stadium.
FIXTURES:
Al Jazira vs Khor Fakkan, Mohammed bin Zayed Stadium, 7.15 p.m
Ajman vs Hatta, Rashid bin Saeed Stadium, 7.15 p.m
Bani Yas vs Sharjah, Bani Yas Stadium, 9.30 p.m
Al Nasr vs Ittihad Kalba, Al Maktoum Stadium, 7.15 p.m
Al Wahda vs Fujairah, Al Nahyan Stadium, 7.15 p.m
Al Ain vs Shabab Al Ahli, Hazza bin Zayed Stadium, 9.30 p.m
© ArabiaDay 2020. All Rights Reserved. Site by www.arabiaday.com
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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Q: Custom Tooltip for input text i am trying to implement a tooltip like div using css.
i am refering
https://www.w3schools.com/howto/tryit.asp?filename=tryhow_css_tooltip
above example shows a top tool tip.
can any one help me change this to bottom tool tip. I am very weak in CSS and design.
A: Here you go this could help you.
<!DOCTYPE html>
<html>
<style>
.tooltip {
position: relative;
display: inline-block;
border-bottom: 1px dotted black;
}
.tooltip .tooltiptext {
visibility: hidden;
width: 120px;
background-color: #555;
color: #fff;
text-align: center;
border-radius: 6px;
padding: 5px 0;
position: absolute;
z-index: 1;
bottom: -40px;
left: 50%;
margin-left: -60px;
opacity: 0;
transition: opacity 0.3s;
}
.tooltip .tooltiptext::after {
content: "";
position: absolute;
bottom: 100%;
left: 50%;
margin-left: -5px;
border-width: 0px 10px 10px 10px;
border-style: solid;
border-color: #555 transparent;
}
.tooltip:hover .tooltiptext {
visibility: visible;
opacity: 1;
}
</style>
<body style="text-align:center;">
<h2>Tooltip</h2>
<p>Move the mouse over the text below:</p>
<div class="tooltip">Hover over me
<span class="tooltiptext">Tooltip text</span>
</div>
</body>
</html>
A: Here it is...
.tooltip {
position: relative;
display: inline-block;
border-bottom: 1px dotted black;
}
.tooltip .tooltiptext {
visibility: hidden;
width: 120px;
background-color: black;
color: #fff;
text-align: center;
border-radius: 6px;
padding: 5px 0;
/* Position the tooltip */
position: absolute;
z-index: 1;
top: 100%;
left: 50%;
margin-left: -60px;
}
.tooltip:hover .tooltiptext {
visibility: visible;
}
<!DOCTYPE html>
<html>
<body style="text-align:center;">
<div class="tooltip">Hover over me
<span class="tooltiptext">Tooltip text</span>
</div>
</body>
</html>
A: Here is how you can do it with some changes in CSS.
.tooltip {
position: relative;
display: inline-block;
border-bottom: 1px dotted black;
}
.tooltip .tooltiptext {
visibility: hidden;
width: 120px;
background-color: #555;
color: #fff;
text-align: center;
border-radius: 6px;
padding: 5px 0;
position: absolute;
z-index: 1;
bottom: -215%;
left: 50%;
margin-left: -60px;
opacity: 0;
transition: opacity 0.3s;
}
.tooltip .tooltiptext::after {
bottom: 100%;
left: 50%;
border: solid transparent;
content: " ";
height: 0;
width: 0;
position: absolute;
pointer-events: none;
border-color: rgba(136, 183, 213, 0);
border-bottom-color: #555;
border-width: 6px;
margin-left: -6px;
}
.tooltip:hover .tooltiptext {
visibility: visible;
opacity: 1;
}
<div style="text-align:center;">
<h2>Tooltip</h2>
<p>Move the mouse over the text below:</p>
<div class="tooltip">Hover over me
<span class="tooltiptext">Tooltip text</span>
</div>
</div>
A: Try this...!!!
<!DOCTYPE html>
<html>
<style>
.tooltip {
position: relative;
display: inline-block;
border-bottom: 1px solid black;
}
.tooltip .tooltiptext {
visibility: hidden;
width: 120px;
background-color: black;
color: #fff;
text-align: center;
border-radius: 6px;
padding: 5px 0;
/* Position the tooltip */
position: absolute;
z-index: 1;
top: 100%;
left: 50%;
margin-left: -60px;
}
.tooltip:hover .tooltiptext {
visibility: visible;
}
</style>
<body style="text-align:center;">
<div class="tooltip">Hover Me
<span class="tooltiptext">bottom tooltip</span>
</div>
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 80
|
from redis.connection import ConnectionPool, UnixDomainSocketConnection
try:
from redis.connection import SSLConnection
except ImportError:
SSLConnection is None
from threading import Lock
from rb.router import PartitionRouter
from rb.clients import RoutingClient, LocalClient
class HostInfo(object):
def __init__(self, host_id, host, port, unix_socket_path=None, db=0,
password=None, ssl=False, ssl_options=None):
self.host_id = host_id
self.host = host
self.unix_socket_path = unix_socket_path
self.port = port
self.db = db
self.password = password
self.ssl = ssl
self.ssl_options = ssl_options
def __eq__(self, other):
if self.__class__ is not other.__class__:
return NotImplemented
return self.host_id == other.host_id
def __ne__(self, other):
rv = self.__eq__(other)
if rv is NotImplemented:
return NotImplemented
return rv
def __hash__(self):
return self.host_id
def __repr__(self):
return '<%s %s>' % (
self.__class__.__name__,
' '.join('%s=%r' % x for x in sorted(self.__dict__.items())),
)
def _iter_hosts(iterable):
if isinstance(iterable, dict):
iterable = iterable.iteritems()
for item in iterable:
if isinstance(item, tuple):
host_id, cfg = item
cfg = dict(cfg)
cfg['host_id'] = host_id
else:
cfg = item
yield cfg
class Cluster(object):
"""The cluster is the core object behind rb. It holds the connection
pools to the individual nodes and can be shared for the duration of
the application in a central location.
Basic example of a cluster over four redis instances with the default
router::
cluster = Cluster(hosts={
0: {'port': 6379},
1: {'port': 6380},
2: {'port': 6381},
3: {'port': 6382},
}, host_defaults={
'host': '127.0.0.1',
})
`hosts` is a dictionary of hosts which maps the number host IDs to
configuration parameters. The parameters correspond to the signature
of the :meth:`add_host` function. The defaults for these parameters
are pulled from `host_defaults`. To override the pool class the
`pool_cls` and `pool_options` parameters can be used. The same
applies to `router_cls` and `router_options` for the router. The pool
options are useful for setting socket timeouts and similar parameters.
"""
def __init__(self, hosts, host_defaults=None, pool_cls=None,
pool_options=None, router_cls=None, router_options=None):
if pool_cls is None:
pool_cls = ConnectionPool
if router_cls is None:
router_cls = PartitionRouter
self._lock = Lock()
self.pool_cls = pool_cls
self.pool_options = pool_options
self.router_cls = router_cls
self.router_options = router_options
self._pools = {}
self._router = None
self.hosts = {}
self._hosts_age = 0
self.host_defaults = host_defaults or {}
for host_config in _iter_hosts(hosts):
self.add_host(**host_config)
def add_host(self, host_id=None, host='localhost', port=6379,
unix_socket_path=None, db=0, password=None,
ssl=False, ssl_options=None):
"""Adds a new host to the cluster. This is only really useful for
unittests as normally hosts are added through the constructor and
changes after the cluster has been used for the first time are
unlikely to make sense.
"""
if host_id is None:
raise RuntimeError('Host ID is required')
elif not isinstance(host_id, (int, long)):
raise ValueError('The host ID has to be an integer')
host_id = int(host_id)
with self._lock:
if host_id in self.hosts:
raise TypeError('Two hosts share the same host id (%r)' %
(host_id,))
self.hosts[host_id] = HostInfo(host_id=host_id, host=host,
port=port, db=db,
unix_socket_path=unix_socket_path,
password=password, ssl=ssl,
ssl_options=ssl_options)
self._hosts_age += 1
def remove_host(self, host_id):
"""Removes a host from the client. This only really useful for
unittests.
"""
with self._lock:
rv = self._hosts.pop(host_id, None) is not None
pool = self._pools.pop(host_id, None)
if pool is not None:
pool.disconnect()
self._hosts_age += 1
return rv
def disconnect_pools(self):
"""Disconnects all connections from the internal pools."""
with self._lock:
for pool in self._pools.itervalues():
pool.disconnect()
self._pools.clear()
def get_router(self):
"""Returns the router for the cluster. If the cluster reconfigures
the router will be recreated. Usually you do not need to interface
with the router yourself as the cluster's routing client does that
automatically.
This returns an instance of :class:`BaseRouter`.
"""
cached_router = self._router
ref_age = self._hosts_age
if cached_router is not None:
router, router_age = cached_router
if router_age == ref_age:
return router
with self._lock:
router = self.router_cls(self, **(self.router_options or {}))
self._router = (router, ref_age)
return router
def get_pool_for_host(self, host_id):
"""Returns the connection pool for the given host.
This connection pool is used by the redis clients to make sure
that it does not have to reconnect constantly. If you want to use
a custom redis client you can pass this in as connection pool
manually.
"""
if isinstance(host_id, HostInfo):
host_info = host_id
host_id = host_info.host_id
else:
host_info = self.hosts.get(host_id)
if host_info is None:
raise LookupError('Host %r does not exist' % (host_id,))
rv = self._pools.get(host_id)
if rv is not None:
return rv
with self._lock:
rv = self._pools.get(host_id)
if rv is None:
opts = dict(self.pool_options or ())
opts['db'] = host_info.db
opts['password'] = host_info.password
if host_info.unix_socket_path is not None:
opts['path'] = host_info.unix_socket_path
opts['connection_class'] = UnixDomainSocketConnection
if host_info.ssl:
raise TypeError('SSL is not supported for unix '
'domain sockets.')
else:
opts['host'] = host_info.host
opts['port'] = host_info.port
if host_info.ssl:
if SSLConnection is None:
raise TypeError('This version of py-redis does '
'not support SSL connections.')
opts['connection_class'] = SSLConnection
opts.update(('ssl_' + k, v) for k, v in
(host_info.ssl_options or {}).iteritems())
rv = self.pool_cls(**opts)
self._pools[host_id] = rv
return rv
def get_local_client(self, host_id):
"""Returns a localized client for a specific host ID. This client
works like a regular Python redis client and returns results
immediately.
"""
return LocalClient(
self, connection_pool=self.get_pool_for_host(host_id))
def get_local_client_for_key(self, key):
"""Similar to :meth:`get_local_client_for_key` but returns the
client based on what the router says the key destination is.
"""
return self.get_local_client(self.get_router().get_host_for_key(key))
def get_routing_client(self):
"""Returns a routing client. This client is able to automatically
route the requests to the individual hosts. It's thread safe and
can be used similar to the host local client but it will refused
to execute commands that cannot be directly routed to an
individual node.
See :class:`RoutingClient` for more information.
"""
return RoutingClient(self)
def map(self, timeout=None, max_concurrency=64):
"""Shortcut context manager for getting a routing client, beginning
a map operation and joining over the result.
In the context manager the client available is a
:class:`MappingClient`. Example usage::
results = {}
with cluster.map() as client:
for key in keys_to_fetch:
results[key] = client.get(key)
for key, promise in results.iteritems():
print '%s => %s' % (key, promise.value)
"""
return self.get_routing_client().map(
timeout=timeout, max_concurrency=max_concurrency)
def fanout(self, hosts=None, timeout=None, max_concurrency=64):
"""Shortcut context manager for getting a routing client, beginning
a fanout operation and joining over the result.
In the context manager the client available is a
:class:`FanoutClient`. Example usage::
with cluster.fanout(hosts='all') as client:
client.flushdb()
"""
return self.get_routing_client().fanout(
hosts=hosts, timeout=timeout, max_concurrency=max_concurrency)
def all(self, timeout=None, max_concurrency=64):
"""Fanout to all hosts. Works otherwise exactly like :meth:`fanout`.
Example::
with cluster.all() as client:
client.flushdb()
"""
return self.fanout('all', timeout=timeout,
max_concurrency=max_concurrency)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,978
|
Today I attended the CIPR's first PR Show in London. And what an experience! I've organised events (in my previous incarnation as a PR practitioner) all around the world, for audiences in the tens of thousands, with keynote speakers including Ministers (even "Prime" Ministers!). But I rarely get the opportunity to attend one as a delegate, so this was a welcome opportunity to see things from the other side.
I have no doubt that there will be lots of discussions about how the event can be improved. But one thing stuck out in my mind. I met face-to-face a number of people that I had only ever "known" before on social media. And it was fascinatingly BETTER.
I had a brief twenty minute conversation with Stephen Waddington- the CIPR president elect for 2014, and Ketchum's Digital guru about an event we are collaborating on next March. I came away enthused by his enthusiasm. And no amount of exclamation marks or emoticons can do the same thing- make a real, genuine commitment on a person to person basis.
When I first arrived at the event I was surprised at how quiet it was. The stands were manned, and there were plenty of visitors- but a surprising number of people were standing around looking intently at their phones- hooked into twitter, and e mails and phone calls with clients. It took a while…but by 3pm the place was positively humming with people talking, exchanging ideas, making connections.
And how stimulating that is! I came away from the event buzzing with ideas about new things to explore and try, new contacts to see where we can work together. And I was also there to help my twenty or so PR students, both MA and BA undergraduates experience an event not only as a participant but also take the opportunity to do some critical analysis and reflect on the power of personal contact in a world that is increasingly intermediated by social media.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,535
|
const requireDir = require('require-dir')
requireDir('./tasks', { recurse: true })
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 2,233
|
Leteča špagetna pošast (v angleščini Flying Sphagetti Monster, pogosto okrajšano kot FSM) je božanstvo parodične religije, cerkve leteče špagetne pošasti ali pastafarianizma (po ital. pasta = testenine, z aluzijo na rastafarijanstvo). Letečo špagetno pošast je prvič omenil Bobby Henderson leta 2005 v javnem pismu kansaškemu odboru za šolstvo, ki je dovolil poučevanje inteligentnega načrta namesto evolucije. Henderson je inteligentnemu načrtu nasprotoval s trditvijo, da je stvarnik sveta Leteča špagetna pošast in izzval kogarkoli, naj dokaže nasprotno. Zahteval je enak obseg poučevanja o nauku Leteče špagetne pošasti in drugih svetovnih nazorih.
Nehote je Henderson ustvaril pošast: v nekaj mesecih je njegova spletna stran dosegla čez milijon zadetkov. Henderson je izdal knjigo, Sveto Pismo Leteče Špagetne Pošasti. Privrženci Cerkve LSP organizirajo letna srečanja.
Novejši dogodek s pastafarijanstvom je zahteva avstrijskega ateista Nika Alma, da mu izdajo vozniško dovoljenje s fotografijo, na kateri ima na glavi cedilo za špagete. Prvič je zahteval tako vozniško dovoljenje leta 2008, uspel je leta 2011. Poleg izpričevanja svojih dvomov v ustaljena verstva, je Alm protestiral proti predpisu, ki prepoveduje uradne slike s pokivali, razen, če gre za versko obeležje. Trdi, da je tak predpis diskriminatoren, saj dovoljuje uradne fotografije s pokrivalom le pripadnikom verskih skupnosti ne pa tudi pripadnikom drugih prepričanj. Aprila 2016 je na Novi Zelandiji potekala prva uradno priznana pastafarianska cerkvena poroka.
Sklici in opombe
Viri
Zunanje povezave
Parodične religije
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 577
|
Automating Opportunities and Contracts with EchoSign, Amazon puts music in the cloud, Salesforce.com buys Radian6 and gets some reaction. App picks of the week.
With the winter season before us and technology manufacturers pumping out new products, what do we want, and how does the cloud make it possible?
A review of the official mobile apps of Salesforce.com.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,857
|
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE reference PUBLIC "-//IXIA//DTD IXIA DITA Composite//EN" "../../system/dtd/ixia/IxiaDitabase.dtd">
<reference id="blackbery_io_filetransfer_upload" xml:lang="en-us">
<title><apiname>upload()</apiname></title>
<shortdesc><i>Uploads a file from the device to a remote server using an HTTP multi-part POST request. Both HTTP and HTTPS protocols are supported. The function is an asynchronous call and will not block execution.</i></shortdesc>
<refbody>
<section>
<title>Synopsis:</title>
<pre scale="80">upload(filePath, server, successCallback, errorCallback, options)</pre>
</section>
<section>
<title>Parameters:</title>
<dl>
<dlentry>
<dt><varname>filePath</varname> {String}</dt>
<dd><p>Full path of the file on the device.</p></dd>
</dlentry>
<dlentry>
<dt><varname>server</varname> {String}</dt>
<dd><p>URL of the remote server that will receive the file.</p></dd>
</dlentry>
<dlentry>
<dt><varname>successCallback</varname> {function}</dt>
<dd><p>Callback function that will be invoked if the file upload is successful. </p></dd>
<dl>
<dlentry>
<dt><varname>successCallback.info</varname> {Object}</dt>
<dd><p>Object containing information about the upload.</p></dd>
<dl>
<dlentry>
<dt><varname>successCallback.info.bytesSent</varname>{Number}</dt>
<dd><p>Total number of bytes sent to the server.</p></dd>
</dlentry>
<dlentry>
<dt><varname>successCallback.info.responseCode</varname>{Number}</dt>
<dd><p>HTTP response code returned by the server.</p></dd>
</dlentry>
<dlentry>
<dt><varname>successCallback.info.response</varname>{String}</dt>
<dd><p>Response returned by the server.</p></dd>
</dlentry>
</dl>
</dlentry>
</dl>
</dlentry>
<dlentry>
<dt><varname>errorCallback</varname> {function}</dt>
<dd><p>Callback function that will be invoked if the file upload fails.</p></dd>
<dl>
<dlentry>
<dt><varname>errorCallback.info</varname> {Object}</dt>
<dd><p>Object containing information about the failed upload request.</p></dd>
<dl>
<dlentry>
<dt><varname>errorCallback.info.code</varname>{Number}</dt>
<dd><p>Error code indicating the type of error that occurred.</p></dd>
</dlentry>
<dlentry>
<dt><varname>errorCallback.info.source</varname>{String}</dt>
<dd><p>Path of the original file.</p></dd>
</dlentry>
<dlentry>
<dt><varname>errorCallback.info.target</varname>{String}</dt>
<dd><p>URL of the remote server.</p></dd>
</dlentry>
<dlentry>
<dt><varname>errorCallback.info.http_status</varname>{Number}</dt>
<dd><p>HTTP status code. This attribute is only available when a response code is received from the HTTP connection.</p></dd>
</dlentry>
</dl>
</dlentry>
</dl>
</dlentry>
<dlentry>
<dt><varname>options</varname>{Object}</dt>
<dd><p>Optional Object literal that allows the user to customize the file key, file name, MIME type, parameters, and chunked mode of the upload request. It is not required to provide all parameters, and these do not have to be specified in any particular order.</p></dd>
<dl>
<dlentry>
<dt><varname>options.fileKey</varname>{String}</dt>
<dd><p>Name of the form element. If not set, this defaults to "file".</p></dd>
</dlentry>
<dlentry>
<dt><varname>options.fileName</varname>{String}</dt>
<dd><p>Name that the file will be saved as on the remote server. If not set, this defaults to "image.jpg".</p></dd>
</dlentry>
<dlentry>
<dt><varname>options.mimeType</varname>{String}</dt>
<dd><p>MIME type of the data being uploaded. If not set, this defaults to "image/jpeg".</p></dd>
</dlentry>
<dlentry>
<dt><varname>options.params</varname>{Object}</dt>
<dd><p>A set of optional key/value pairs to be passed along in the HTTP request.</p></dd>
</dlentry>
<dlentry>
<dt><varname>options.chunkedMode</varname>{Boolean}</dt>
<dd><p>Specifies whether the data should be uploaded in chunked streaming mode. If not set, this defaults to true.</p></dd>
</dlentry>
<dlentry>
<dt><varname>options.chunkSize</varname>{Number}</dt>
<dd><p>Specifies the size of each chunk when chunkedMode is true. If not set, this defaults to 1024 bytes.</p></dd>
</dlentry>
</dl>
</dlentry>
</dl>
</section>
<example>
<title>Example:</title>
<p><pre scale="80">
<script type="text/javascript">
function uploadSuccess(result) {
alert("Upload was successful");
console.log("Bytes sent: " + result.bytesSent);
console.log("Response code: " + result.responseCode);
console.log("Response: " + result.response);
}
function uploadError(result) {
alert("Upload failed");
console.log("Error code: " + result.code);
console.log("Source: " + result.source);
console.log("Target: " + result.target);
console.log("HTTP Status: " + result.https_status);
}
function fileUpload() {
var parameters, options;
try {
parameters = { app : "webworks" };
options = {
fileKey : "file",
fileName : "blackberry.jpg",
mimeType : "image/jpeg",
params : parameters,
chunkedMode : true,
chunkSize : 1024
};
blackberry.io.filetransfer.upload("/accounts/1000/shared/camera/image_123.jpg", "http://www.blackberry.com/upload", uploadSuccess, uploadError, options);
} catch(e) {
alert("Exception in fileUpload: " + e);
}
}
</script>
</pre></p>
</example>
<example>
<title>Example:</title>
<p><pre scale="80">
/*
FILENAME: upload.php
A sample PHP upload server script which will save files into an "upload" directory. Make sure to create and set 777 permissions for the "upload" folder in the same directory as upload.php.
*/
<?php
if (move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"])) {
echo "Success";
} else {
echo "Error";
}
?>
</pre></p>
</example>
</refbody>
</reference>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,760
|
Q: How to program several sizes of screens in uwp? Hi i am Developing a app on universal windows 10 app platform, and when i put some Objects on the XAML window and when change the screen size it's stay how it was before so it's fit the new screen size and if i fix it it's fit the old screen size and i need the app to be universal.
my xaml code:
<Page
x:Class="App17.MainPage"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:local="using:App17"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
mc:Ignorable="d">
<Grid Loaded="Grid_Loaded">
<Grid.Background>
<ImageBrush Stretch="UniformToFill" ImageSource="Assets/backPHONE.jpg"/>
</Grid.Background>
<Grid.RowDefinitions>
<RowDefinition Height="55*"/>
<RowDefinition Height="9*"/>
</Grid.RowDefinitions>
<RelativePanel>
<TextBlock x:Name="textBlock" HorizontalAlignment="Left" Margin="41,155,0,0" TextWrapping="Wrap" VerticalAlignment="Top" Height="128" Width="360" Text="test 123" FontFamily="Century Gothic" FontStretch="UltraExpanded" FontStyle="Italic" FontSize="26" Tapped="textBlock_Tapped" TextAlignment="Center"/>
<ComboBox x:Name="comboBox" HorizontalAlignment="Left" Margin="84,700,0,-666.8" VerticalAlignment="Top" Width="264" SelectionChanged="comboBox_SelectionChanged" Height="32" Grid.Row="1">
<ComboBoxItem Content="test" IsSelected="True"/>
</ComboBox>
<TextBlock x:Name="textBlock2" HorizontalAlignment="Left" Margin="164,675,-2.8,-665" TextWrapping="Wrap" Text="test" VerticalAlignment="Top" Height="20" Width="120" Grid.Row="1" SelectionHighlightColor="#FFF10000" FontFamily="Century Gothic" FontSize="12" FontStretch="UltraExpanded" FontStyle="Italic" Foreground="White"/>
</RelativePanel>
</Grid>
thanks
A: Let me introduce two possible ways to handle various screen sizes related with Universal Windows 10 Apps.
1. Adaptive Triggers - you can use them to adjut your layout to different device families - for instance you can define how your Page will look on the smartphone and while app is launched on the PC. Below I tried to add some short instructions how you can start using them.
a) When you have Visual Studio opened, please right click on the MainPage.xaml and select "Design in Blend...":
b) Go to "States" tab and select icon (marked red rectangle below) called "Add state group":
c) Now you have to add state, so click the icon next to "Visual State Group" as show below:
d) Now add two state: "Mobile" and "Desktop" as shown below:
e) Last step is to set minimum window width to detect when screen size is changed:
f) Now you are able to set minimum window width for the "Mobile" state:
Please type 320 as below (remember that these are effective pixels not physical)
Once you click "OK" Visual State for "Mobile" will be configured.
Do the same above steps but for "Desktop" state and type 1024 for minimum window width:
Now I can switch between two states and adjust the design for each of them:
(Note that currently selected state is marked with red rectange):
1. Device-Family Folders - you can specify two separate views for Pages but with one code behind. This can help if you have to craft your design specially for mobile devices and for instance fot the PCs.
a) Right click on the project and add new falder called "DeviveFamily-Mobile":
b) Add new Xaml View called the same like your previous page - in my case this is "MainPage":
c) Now you have one code behind class but two separate views that will be applied accordingly to device family:
I hope that it will help you to start.
You can also watch very good Channel9 video or visit my blog where I am trying to present a lot of helpful samples related with UWP.
A: you must use AdaptiveTriggers
http://www.wintellect.com/devcenter/jprosise/using-adaptivetrigger-to-build-adaptive-uis-in-windows-10
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 1,488
|
En förtrollad jul () är en amerikansk-kanadensisk julfilm från 1985 i regi av Phillip Borsos. I huvudrollerna ses Mary Steenburgen och Harry Dean Stanton.
Rollista i urval
Mary Steenburgen - Ginny Grainger
Gary Basaraba - Jack Grainger
Harry Dean Stanton - Gideon
Arthur Hill - Caleb Grainger
Robbie Magwood - Cal Grainger
Elisabeth Harnois - Abbie Grainger
Michelle Meyrink - Betty
Elias Koteas - Eddie
Wayne Robson - Harry Dickens
Jan Rubes - Jultomten
Sarah Polley - Molly Monaghan
Graham Jarvis - Frank Crump
Timothy Webber - Herbie Conklin
Joy Thompson - Mrs. Monaghan
John Friesen - Mr. Noonan
Externa länkar
Filmer 1985
Engelskspråkiga filmer
Amerikanska julfilmer
Kanadensiska julfilmer
Amerikanska fantasyfilmer
Kanadensiska fantasyfilmer
Filmer från Walt Disney Pictures
Filmer inspelade i Ontario
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,412
|
\section{%
\if@noskipsec\leavevmode\fi
\par
\setlength\@tempskipa{-3.5ex\@plus-1ex\@minus-.2ex}
\@afterindenttrue
\ifdim\@tempskipa<\z@
\@tempskipa-\@tempskipa\@afterindentfalse
\fi
\if@nobreak
\everypar{}
\else
\addpenalty\@secpenalty\addvspace\@tempskipa
\fi
\@ifstar
{\unnumberedsection}
{\@dblarg{\numberedsection}}}
\def\numberedsection[#1]#2{%
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\section{Introduction}
Nanoindentation methods have become extremely important in the last
couple of decades due to the increasing interest in the study of the
viscoelastic properties of nano-systems. Viscoelasticity is typical of a number of materials: polymers, plastics,
composites, metals and alloys, building materials, and
biological tissues. Studying viscoelasticity in micro and nano scale
is crucial to shed new light onto the understanding of a wide range
of practical problems like drug delivery by nanoparticle carriers
\cite{fiel2011}, biomechanics of living cells and their response to
external forces \cite{hoffman2009,bremmel2010,rebelo2014}, and the possibility
of diagnosing diseases at early stages
\cite{darling2006,darling2007,ketene2011, fritsch2010, rebelo2013}.
Nanoindentation with the atomic force microscope (AFM) is one the
most popular methods to probe soft samples. Conventional AFM force
curves can be fitted with an appropriate model to
extract the materials properties, whereas the models from Hertz
\cite{hertz,sneddon1965} and Oliver-Pharr
\cite{oliverpharr1992,oliverpharr2004} are the most used ones. The
former is mainly used to study thin soft films and biological
samples (tissues and cells). The latter is usually applied to hard
materials like metals, glass and plastic. Both models consider that
samples can be described by a purely elastic solid. On the other
hand, viscoelastic materials are usually described in terms of the generalized Maxwell model, which is composed of associations of springs and dashpots, where each
spring element has a spring constant $E_i$, and each dashpot has a
relaxation time $\tau_i$ \cite{fung}. This model comprises four
particular cases: the elastic case (represented a single spring
element), Kelvin-Voigt model (represented by a spring and dashpot
connected in parallel), Maxwell model (represented by a spring and
dashpot connected in series) and the standard linear solid (SLS)
model (represented by the combination of a spring element connected
in parallel with a Maxwell arrangement of spring and dashpot).
The Maxwell model does not describe creep or recovery, and the
Kelvin-Voigt model does not describe stress relaxation. The SLS model is
the simplest model that predicts both phenomena, but it fails to
describe materials with more than one relaxation time scale. One can progressively add more Maxwell elements in order to describe materials with multiple relaxation times. The aforementioned models are able to describe materials exhibiting exponential shear relaxation in time $G(t) \propto e^{-t/\tau}$.
However, there are many classes of materials (e.g. living cells, hydrogels, cross-linked polymers, colloidal suspensions and foams) whose viscoelastic properties are described by a power-law relaxation $G(t)\propto t^{-n}$, which cannot be modelled by an association of spring and dashpots (unless a very large number of elements is added) \cite{fabry2001,djordjevic2003,jaishankar2012}. The natural framework to model power law rheology is the fractional calculus \cite{podlubny1999}. Analogously to the viscoelastic elements (springs and dashpots), one defines a fractional element (Scott-Blair element) whose constitutive stress-strain equation is $\sigma(t) \propto d^n \epsilon (t)/ d t^n$, where $0 \le n \le 1$ and $d^n / d t^n$ is the fractional derivative operator. The fractional element interpolates between two responses: for $n = 0$ ($n=1$) one obtains the Hookean elastic spring (Newtonian dashpot) constitutive equation \cite{scottblair1947, schiessel1995, jaishankar2012}.
Several methods have been proposed to obtain viscoelastic properties of samples with the AFM. Darling \textit{et al.} modified the closed-loop feedback control of the z-axis movement to perform stress relaxation tests in their samples \cite{darling2007,ketene2011}. This approach allows the determination of intrinsic relaxation times of viscoelastic materials. Some groups modified the AFM to perform force-modulated dynamic rheology \cite{rebelo2014, fabry2001,mahaffy2004,alcaraz2002,nalam2015}. This allows the determination of the exponents of the power law response of viscoelastic materials in the frequency domain $G(\omega) \propto \omega^n$, but demands complicated modifications in the AFM apparatus as well. Within the framework of conventional AFM force curves, most of the studies in the literature only address the instantaneous elasticity modulus of the materials, disregarding viscoelastic effects because of the lack of simple models to extract the viscoelastic properties from the force curves. Ren \textit{et al.} measured the frequency-dependent instantaneous elasticity moduli $E_0$ of cancer cells subjected to anticancer drugs by changing the loading speeds $v_L$ \cite{ren2015}. Although their forces curves clearly show viscoelastic effects, they focused on the analysis of the power law relationship $E_0 \propto f_z^m$ ($v_L \propto f_z$) to classify the action mechanisms of the anticancer drugs. Some of the authors made use of an empirical model to determine the apparent viscosity of living cells and asphalt binder directly from force curves \cite{rebelo2013, rebelo2014b}. The major limitation in the investigation of viscolelastic effects by means of force curves is the loading speed of the cantilever. High loading frequencies (typically above 10-30 Hz) induces cantilever oscillations that reduces the accuracy of the curve fitting \cite{ren2015}. In this regard, Chyasnavichyus \textit{et al.} proposed the application of the known frequency-temperature superposition to surpass the limited range of AFM loading rates, and constructed the master curves of PnBMA polymers by fitting AFM forces curves measured at different temperature and loading rates with a viscoelastic force model based on the SLS model \cite{tsukruk2014}. None of the aforementioned works considered power law relaxation mechanisms in the viscoelastic modeling of the materials.
In this work, we present an analytical model for the AFM force curves of viscoelastic samples indented by axisymmetric indenters that accounts for power law viscoelastic relaxation. We validate our model numerically with Finite Elements Modeling using a computational two-body indentation model described elsewhere (see Supplementary Material) \cite{santos2012}, and apply it in the AFM study of polyacrylamide gels. Polycrylamide gels are often considered as standard to test viscoelastic models of soft samples, with many works reporting their properties \cite{nalam2015,gavara2010,abidine2015}. We demonstrate that the measured properties of the gels are in good agreement with other studies in the literature, and that our model is able to extract viscoelastic properties that could only be previously determined with dynamic rheology methods.
\begin{figure*}[ht]
\center \includegraphics[scale=0.7]{fig1.pdf}
\caption{(a) Schematics of the indentation of soft samples by conical and spherical indenters. The sample thickness $h$ is much larger than the maximum indentation $\delta_{max}$ in order to avoid finite thickness effects. (b) Schematics of the regular (left) and fractional (right) SLS model where the dashpot is replaced by a fractional element \cite{scottblair1947, schiessel1995, jaishankar2012}. (c) Representation of the fractional element that interpolates between Hookean spring and Newtonian dashpot. (d) Relaxation function of the fractional-like SLS model given by Eq. \ref{eq:SLSrelax}. (e) Frequency dependence of the storage and shear moduli given by Eq. \ref{eq:glin}.}
\label{fig:modelAFM}
\end{figure*}
\section{Theoretical modeling}
The analysis of force-indentation $F(\delta)$ curves are usually performed wtihin the framework of the Hertz contact theory \cite{hertz,sneddon1965}:
\begin{equation}
\label{eq:Hertz}
F_H(\delta)= \Gamma E^* \delta^\lambda.
\end{equation}
\noindent The subscript H stands for Hertz model, $\Gamma$ and $\lambda$ are geometry-dependent parameters. $E^* = E/(1-\nu^2)$ is the reduced elasticity modulus, and $\nu$ is the Poisson ratio ($\nu=0.5$ for incompressible materials). The elasticity modulus $E$ is related to the shear modulus $G$ as $2 G = E/(1+\nu)$. For pyramidal indenters, one has $\Gamma = 2\tan \theta/\pi$ ($\theta$ being the half-opening angle of the indenter) and $\lambda=2$. For spherical indenters, one has $\Gamma = 4\sqrt{R}/3$ ($R$ being the indenter radius) and $\lambda=3/2$. The schematics of the indentation of soft samples is shown in Figure \ref{fig:modelAFM}.
Hertz theory is based on the following major assumptions: (i) the sample is assumed as a purely elastic half-space, (ii) the stress-strain response is linear, (iii) the elasticity modulus is constant. Therefore, Hertz model is not appropriate to describe viscoelastic materials. Despite of that, several groups have proposed modifications to Hertz model in order to include viscoelastic effects \cite{rebelo2014,darling2006, darling2007, ketene2011, mahaffy2004, alcaraz2002}. For example, Darling \textit{et al.} and Ketene \textit{et al.} studied the viscoelastic properties of living cells using a modified AFM apparatus to perform stress relaxation experiments: they produced an initial indentation in the cells, and recorded the stress relaxation in the cell by monitoring the cantilever deflection as a function of time \cite{darling2006,darling2007,ketene2011}. To explain their measurements, they assumed that the cells could be described by the SLS viscoelastic model, and used the functional method originally proposed by Lee and Radok to obtain the force load history of samples subjected to an instantaneous step indentation (as performed in stress relaxation tests) \cite{leeradok1960}. For this specific case, the force load history essentially keeps the mathematical form of Hertz model, but replacing the fixed value of $E$ by the SLS relaxation function. For other indentation histories, their corresponding force histories must be determined.
Here we employ the functional method to determine the relaxation properties of soft viscoelastic samples subjected to an indentation history $\delta(t)$ similar to the loading conditions in typical AFM force curves. We derived analytical formulae to extract the viscoelastic parameters directly from the force curves, without any modification in the AFM apparatus to impose a prescribed indentation or load history like in previous reports \cite{rebelo2014,darling2006, darling2007, ketene2011, mahaffy2004, alcaraz2002}. It is known that the functional method is valid for the cases where the contact area increase monotonically \cite{leeradok1960}. This restriction has been removed by the challenging formulation proposed by Ting \cite{ting1966}. However, Vandamme \textit{et al.} have shown that the functional method not only works well under during the loading phase where the indentation depth $\delta(t)$ is a monotonically increasing function, but it also remains valid remains valid to calculate the initial unloading phase \cite{vandamme2006}.
In time domain, the elastic Hertz-like model has the form $F(t)=\Gamma E^* \delta^\lambda(t)$. We assume $F(t) = F_{max} \bar{F}(t)$ and $\delta(t) = \delta_{max} \bar{\delta}(t)$, where $F_{max}$ and $\delta_{max}$ represent the maximum load and indentation depth, respectively. $\bar{F}(t)$ and $\bar{\delta}(t)$ represent the load and indentation histories, respectively. In Laplace domain, the associated Hertz-like elastic problem becomes:
\begin{equation}
\label{eq:hertzlaplace}
\tilde{F}(s) = \frac{\Gamma \delta_{max}^\lambda}{(1-\nu^2)F_{max}} \frac{\tilde{E}(s)}{s} s\tilde{\delta^\lambda}(s).
\end{equation}
\noindent From the constitutive equation of a given viscoelastic model one determines $\bar{E}(s)$, the relaxation function $\tilde{R}(s)=\tilde{E}(s)/s$ and the creep compliance function $\tilde{J}^{-1}(s)=s\tilde{E}(s)$ in Laplace space. Applying the convolution property of the Laplace transform, we obtain:
\begin{equation}
\label{eq:functionaleqgeneral1}
\bar{F}(t) = \frac{F(t)}{F_0} = E_0 \int_0^t \bar{R}(t-t^\prime) \frac{d \bar{\delta}^{\lambda}(t^\prime)}{d t^\prime} d t^\prime,
\end{equation}
\noindent where
$F_0 = \Gamma \delta_{max}^\lambda/(1-\nu^2)$ is a force normalisation factor, and $\bar{R}(t) = R(t)/E_0$ with $E_0$ being the instantaneous elasticity modulus. An alternative formulation of Eq.~\ref{eq:functionaleqgeneral1} is obtained by providing a force history $F(t)$ to determine $\delta(t)$:
\begin{equation}
\label{eq:functionaleqgeneral2}
\delta^\lambda(t) = \frac{(1-\nu^2)}{\Gamma} \int_0^t J(t-t^\prime) \frac{d F(t^\prime)}{d t^\prime} d t^\prime,
\end{equation}
\noindent where $J(t)$ is the creep compliance function. This approach was adopted by Chyasnavichyus \textit{et al.}, who assumed a constant rate force ramp and the creep compliance function of the SLS model to measure the dynamic properties of PnBMA polymers \cite{tsukruk2014}.
The choice between formulations depends on the working principle of the indenter. Regular nanoindenters apply the load vertically in the same axis of the indenter, aiding the precise control on the load history $F(t)$. In this case, the formulation given by Eq. \ref{eq:functionaleqgeneral2} is more appropriate. In a typical AFM force curve (see Fig.~\ref{fig:modelAFM}), the load is applied indirectly by extending the piezo, and making contact between the cantilever and the sample. The cantilever deflects upward as the piezo extends until a maximum deflection (force) is achieved. Then, the piezo retracts ending the contact between the cantilever and sample. One has precise control in the rate of expansion/retraction of the piezo, but neither in the force nor the indentation histories.
The indentation depth in AFM force curves are computed as $\delta(t) = [z(t)-z_0] - [d(t)-d_0]$, where $z$ is the piezo displacement, $d$ is the cantilever deflection, and $(z_0,d_0)$ represents the contact point in the force curve. If the sample is infinitely hard, no indentation occur, and the rate of the piezo extension is equal to the rate of cantilever deflection $\dot{z}(t) = \dot{d}(t)$. If the sample is infinitely soft, there is no cantilever deflection, and the indentation rate becomes equal to rate of piezo extension $\dot{z}(t) = \dot{\delta}(t)$. The intermediate cases comprehend soft samples in general, for which we have $\dot{\delta}(t) = \dot{z}(t)- \dot{d}(t)$. It is known that $\dot{z}(t) \approx \pm 2 L_z f_z$ (for load and unload phases, respectively), where $f_z$ is the vertical scan rate, and $L_z$ is the amplitude of the piezo extension.
Predicting the indentation history that samples undergo during a force curve measurement is difficult. As a first approximation, one can assume a linear indentation history $\delta(t) \propto z(t) = v t$. The deviation from this behavior can be modeled by a second order contribution, which has small effects on the overall measurement (See Supplementary Material). The linear indentation history approximation leads to simple analytical formulae, which can be easily incorporated to current software packages for the analysis of AFM force curves. One important aspect of indentation experiments with AFM force curves is the detection of the contact point \cite{rudoy2010,benitez2013,roy2014}. Here, we have employed the bi-domain polynomial (BDP) Method of Roy \textit{et al.} which allows a quick method to detect the contact point in the force curves \cite{roy2014}.
\subsection{Viscoelastic modeling}
We adopted the fractional SLS viscoelastic model that comprises either a single relaxation time or a power law relaxation (see Figure \ref{fig:modelAFM}). The choice between relaxation types depends upon a single parameter $0\leq n \leq 1$. The shear relaxation function of the fractional SLS viscoelastic model is given by (See Supplementary Material):
\begin{equation}
\label{eq:SLSrelax}
R(t) = E_0\left[(1 -\alpha) + \alpha E_{n,1}\left[-\left(\frac{t}{\tau}\right)^n\right]\right].
\end{equation}
\noindent The shear modulus relaxes from the instantaneous $G(0) =G_0$ to the relaxed modulus $G(t \rightarrow \infty) =G_\infty$, where the amplitude of relaxation is $G_1 = G_0-G_\infty$. $E_{n,1}(z)$ is the generalized Mittag-Leffler function. For $n=1$ one has $E_{1,1}(z)=\exp (z)$ which results in the shear relaxation function of the conventional SLS model \cite{schiessel1995,shukla2007,jaishankar2012}. $\alpha$ is defined such that $G_1 = \alpha G_0$ and $G_\infty =(1-\alpha) G_0$. It can be regarded as a parameter that describes the \emph{viscoelastic relaxation amplitude} of the material, whereas $\alpha = 0$ represents the elastic limit, and $\alpha = 1$ represents the viscoelastic limit corresponding to the Kelvin-Voigt model. Finally, $\tau$ is a relaxation time. The complex shear modulus is given by $G^*(\omega)= i\omega\int_0^\infty R(t) \exp (-i \omega t ) dt$, and the storage and loss moduli are, respectively:
\begin{eqnarray}
\label{eq:glin}
G'(\omega) = Re[G^*(\omega)]=G_0\left[ (1-\alpha) + \alpha \frac{ (\omega\tau)^{2n}}{1+(\omega\tau)^{2n}} \right] \\
G''(\omega)= Im[G^*(\omega)]=G_0\alpha\left[ \frac{ (\omega \tau)^n}{1+(\omega\tau)^{2n}} \right]. \nonumber
\end{eqnarray}
\noindent The storage modulus $G'(\omega)$ suggests that the instantaneous elasticity modulus measured by an AFM force curve should exhibit a frequency-dependent behavior of the form $E_0(\omega) = E_0(1 - \alpha) + E_0\alpha f(\omega)$, such that $f(\omega) \rightarrow 0$ for $\omega \rightarrow 0$ (in the the limit of very slow piezo extension), and $f(\omega ) \rightarrow 1$ for $\omega\tau >> 1$ (in the limit very fast piezo extension). There are two crossovers between $G'(\omega)$ and $G''(\omega)$. For $n=1$ and $\alpha \ge 0.82$, they are located at $\omega_{1,2} =(\alpha/ 2 \tau) [1\pm \sqrt{1 - 4(1-\alpha)/\alpha^2}]$. Figures \ref{fig:modelAFM}(d)-(e) show the general behavior of $R(t)$ and $G^*(\omega)$ for different values of $n$. The exponent $n$ governs the dynamics of $G''(\omega)$, and $\tau = \omega_0^{-1}$ is the inverse frequency for which $G''(\omega)$ reaches its maximum value. The dynamic viscosity is determined by
\begin{equation}
\label{eq:eta}
\eta= \frac{G''(\omega)}{\omega} = G_0 \alpha\frac{ \omega^{n-1}\tau^n}{1+(\omega\tau)^{2n}}.
\end{equation}
\begin{figure*}[ht]
\center \includegraphics[scale=0.6]{fig2.pdf}
\caption{Generic force curves exhibiting the role of the parameters $E_0$, $\alpha$, $\beta_L$ and $n$ in the loading curves. The curves were generated using $\alpha = 0.9$, $\beta_L= 0.05$ (blue), $0.5$ (orange), $1.0$ (green), and $n = 1$ (solid), $0.8$ (dashed), $0.6$ (dotted).}
\label{fig:diagramforce}
\end{figure*}
\subsection{Force curve model}
The linear indentation history during a force curve measurement is given by:
\begin{eqnarray}
\label{eq:deltahistory}
\bar{\delta}_L(t) = t/\tau_L ~~~(t\leq \tau_L) \\ \nonumber
\bar{\delta}_U(t) = 1 - (t-\tau_L)/\tau_U ~~~(\tau_L \leq t \leq \tau_L+\tau_U),
\end{eqnarray}
\noindent where $\tau_L$ and $\tau_U$ represent the duration of the load (approach) and unload (retract) phases of the force curve. Replacing $\bar{\delta}_{L}(t)$ in Eq. \ref{eq:functionaleqgeneral1}, the resulting integral can be solved analytically, and the load curves of conical and spherical indenters can be cast in the following expression:
\begin{equation}
\label{eq:forceconeapp}
\frac{\bar{F}_L(\bar{\delta})}{E_0} = \bar{\delta}^\lambda\left[ 1 - \alpha +\alpha \Gamma(\lambda+1) E_{n,\lambda+1}\left[-\left(\frac{\bar{\delta}}{\beta_L}\right)^n \right] \right],
\end{equation}
\noindent where $\Gamma(z)$ is the Gamma function, and $E_{n,m}(z)$ is the generalised Mitta-Leffler function \cite{schiessel1995,shukla2007,jaishankar2012}. We simplified notation by making $\bar{\delta}_{L}(t) \rightarrow \bar{\delta}$, and $\beta_{L}=\tau/\tau_{L}$. Although the validity of the functional method for the unload curve is not completely understood except near $\bar{\delta}=1$ \cite{vandamme2006}, we provide analytical formulas for the unload curves for the specific case of $n=1$ in the Supplementary Material.
The force curves $d~versus~z$ must be transformed into the form $\bar{F}~versus~\bar{\delta}^\lambda$, where $\bar{F} = k_c(d-d_0)/F_0$, $F_0 = \Gamma \delta_{max}^\lambda/(1-\nu^2)$, and $\bar{\delta}=\delta/\delta_{max}$. We must also compute $\tau_L$ directly from the force curves. This results in universal curves as the ones shown in Figure \ref{fig:diagramforce}. In the elastic limit (by elastic we mean either a truly elastic material or a viscoelastic material whose relaxation time is much longer than $\tau_L$), the force curve is a straight line whose slope is $E_0$. The slope of the force curve is $E_0$ for $\bar{\delta} \rightarrow 0$ and $E_0[1-\alpha(1-f(\beta_L,n,\lambda))]$ for $\bar{\delta} \rightarrow 1$. The function $f(\beta_L,n,\lambda) = \Gamma (\lambda+1) E_{n,\lambda+1}(-\beta_L^{-n})$ has a weak dependence on the indenter geometry and exhibits the following behavior: (i) $f(\beta_L,n,\lambda)\rightarrow 0$ for $\beta_L\rightarrow 0$, and (ii) $f(\beta_L,n,\lambda)\rightarrow 1$ for $\beta_L\rightarrow \infty$. At the opposite end of the elastic limit, one has the full instantaneous relaxation case obtained when $\alpha \rightarrow 1$ and $\beta_L \rightarrow 0$. For fixed values of $\alpha$ and $n$, increasing values of $\beta_L$ make the curves to move towards the elastic limit case. For a fixed values of $\beta_L$ and $n$, increasing values of $\alpha$ make the curves to move towards the full instantaneous relaxation case. We remark that the aforementioned behavior are general trends of force curves taken in fractional SLS viscoelastic materials (see Figure \ref{fig:diagramforce}), and are valid even for nonlinear indentation profiles.
\begin{figure*}[ht]
\center \includegraphics[scale=0.70]{fig3.pdf}
\caption{(a) Comparison of AFM force curves measured in polyacrylamide gels with different $C_{bis}$. The vertical scan frequency is $f_z = 1$Hz. A maximum deflection of $d_{trigger}=100$ nm was imposed in all measurements. From those curves, we constructed (b) their respective indentation histories $\delta(t)$.
(c) Normalised force curves of polyacrylamide gels. Solide lines represent experimental curves, dashed lines represent the fitting with Hertz model, and symbols represent the fitting with different relaxation exponents. The parameters extracted from those curves are shown in Table \ref{tab:poliac1hz}.}
\label{fig:poliac1hz}
\end{figure*}
\begin{table*}[ht]
\caption{List of viscoelastic parameters extracted from the force curves of Figure \ref{fig:poliac1hz}(c). The lines lacking data of the parameters $\alpha$, $\tau$, $\beta_L$ and $n$ represent the data obtained with the conventional Hertz fit.}
\begin{center}
\begin{tabular}{@{} crrrrrrrrr @{}}
\hline
\hline
$C_{bis}$ & $f_z$( Hz) & $E_{0}$ (kPa) & $\alpha$ & $\tau$(s)& $\beta_L$ & $r^2$ & RMSE & $n$ \\
\hline
0.2\% & 1 & 7.778 & & & & 0.9952 & 0.1575 & \\
0.4\% & 1 & 15.402 & & & & 0.9960 & 0.2880 & \\
0.8\% & 1 & 39.404 & & & & 0.9921 & 1.0189 & \\ \hline
0.2\% & 1 & 19.154 & 0.677 & 0.005 & 0.043 & 0.9997 & 0.0393 & 0.91\\
0.4\% & 1 & 21.751 & 0.931 & 0.072 & 0.863 & 0.9998 & 0.0681 & 0.79\\
0.8\% & 1 & 73.645 & 0.638 & 0.009 & 0.129 & 0.9999 & 0.1375 & 1\\
\hline
\hline
\end{tabular}
\end{center}
\label{tab:poliac1hz}
\end{table*}%
\section{Experimental validation}
\subsection{Experimental details}
Polyacrylamide gels were prepared from the stock solution of 30\% acrylamide (by weight) in three different concentrations of N,N-bis acrylamide (0.2\%, 0.4\% and 0.8\% in volume). The final gels have a concentration of 0.375 M Tris-HCL, with pH 8.8. The gels were polymerized chemically by addition of 10$\mu$L of tetramethylethylenediamine (Temed) and 0.1 mL of 15\% ammonium persulfate solution/10mL of gel solution
An AFM (MFP-3D, Asylum-Research, Santa Barbara, CA, USA) was used to measure conventional force curves. Soft cantilevers (Microlever, MLCT-AUHW, Veeco, USA) with a spring constant of 0.015 N/m were used to probe the gels, verified by the thermal method \cite{radmacher2007}. A maximum force trigger of $d_{trigger}=100$nm were imposed to avoid excessive gel indentation. The used AFM tips have pyramidal shapes with half-opening angles of $\theta= 38^\circ$. To reduce adhesion effects in the cantilever, the force measurements were performed in distilled water. Different vertical scan frequencies (0.1Hz, 0.5Hz, 1Hz, 5Hz and 10Hz) were applied to samples to investigate the effect of varying load rates in the viscoelastic properties. The frequency-dependent data were averaged over different ($n=5$) locations for each gel.
\begin{figure*}[ht]
\center \includegraphics[scale=0.6]{fig4.pdf}
\caption{(left) Frequency dependence of $E_0$, $\alpha$, $\beta_L$ and $\tau$. Square (circle) symbols represent the fitting with fractional (regular) SLS viscoelastic model. Colors represent different $C_{bis}$ concentrations: 0.2\% (black), 0.4\% (blue), 0.8\% (red). The dash-dotted lines in panel (a) represent the fitting go $E_0$ with Hertz model. (right) Frequency dependence of the relaxation exponents. The horizontal lines represent the exponents (0.83 and 0.87) extracted from the polyacrylamide gels measured by Abidine \textit{et al.} \cite{abidine2015} (See Table \ref{tab:abidine}).}
\label{fig:rheology}
\end{figure*}
\subsection{Study of polyacrylamide gels}
Figure \ref{fig:poliac1hz}(a) shows AFM force curves (with their respective contact points aligned) of polyacrylamide gels measured with $f_z=1$Hz, and their respective time-dependent indentation profiles $\delta(t)$ are shown in Figure \ref{fig:poliac1hz}(b). The comparison of the indentation depths clearly shows that the increase of the bisacrylamide concentration ($C_{bis}$) enhances the stiffness of the gels. The indentation profile is not linear, and the time to reach the maximum indentation are also different, suggesting distinct viscoelastic relaxation properties among samples. We tested the validity of the linear indentation approximation in all measurements in this work. We obtained that all force curves exhibited very small deviation from linear behavior, lying within a quasi-linear indentation regime. Therefore, we can use Eq. \ref{eq:forceconeapp} to extract the viscoelastic parameters of the polyacrylamide gels at the expense of very small errors. The error analysis of the linear indentation approximation is discussed in the Supplementary Material.
Figure \ref{fig:poliac1hz}(c) shows the $\bar{F}~versus~\bar{\delta}^2$ curves for different $C_{bis}$ measured with $f_z = 1$ Hz. The comparison of $E_0$ slopes shows that the increase of $C_{bis}$ enhances the gel stiffness. The relaxation times can be qualitatively compared by how close to $\bar{\delta}=0$ the force curve deviates from the $E_0$ slope. Therefore, the largest relaxation time must be observed in the 0.4\% sample, while samples 0.2 and 0.8\% should exhibit comparable values of $\tau$. The same trend is observed in the value of $\beta_L$. All viscoelastic parameters fitted from Figure \ref{fig:poliac1hz}(c) are listed in Table \ref{tab:poliac1hz}. For these individual force curves, we obtained exponents $n$ varying between 0.79 and 1.0.
The dynamic behavior of the gels is shown in Figure \ref{fig:rheology}. Here we adopted two fitting strategies. First, we assumed that gels can be described by the conventional SLS model (exponent $n=1$). Second, we assumed that the gels can be described by the fractional SLS model where $n$ is also a fitting parameter. In both cases, the stiffness of the gels are proportional to $C_{bis}$, and do not exhibit any appreciable frequency dependence. The Hertz fitting of the force curves provided lower values $E_H < E_0$ compared to our model. This is due to viscoelastic relaxation that reduces the instantaneous elasticity modulus during the course of the force curve measurement. This can be better seen in Figure \ref{fig:poliac1hz}(c) that shows that the slope of the force curves near $\bar{\delta}\rightarrow 0$ is always larger than the slope of the Hertz fit. In the limit of very slow piezo extension, one obtains $E_H \approx E_\infty$. For very fast piezo extension, Hertz fitting provides $E_H \approx E_0$. The nearly constant difference $E_0-E_H$ suggests that the relaxation times of the gels are much shorter than 0.1 s (only accessible for frequencies above 10 Hz). This figure also shows that The relaxation times are inversely proportional to $C_{bis}$, exhibiting power law decay, while $\alpha$ and $\beta_L$ are nearly frequency independent.
The comparison of both fitting strategies show that $E_0$, $\alpha$ and $\tau$ are inversely proportional no the exponent $n$. This is reasonably simple to understand with the help of the relaxation functions in Figure \ref{fig:modelAFM}(d). Although similar, those curves are not exponential decaying functions, except for the case $n=1$. The determination of $E_0$ is governed by the derivative of the force curve $dF_L/d\bar{\delta^\lambda}$ at $t \rightarrow 0$, which is roughly proportional to $R_{n}(t)$ (note that $R(t)$ depends $E_{n,1}[-(t/\tau)^n]$, while $F_L(t)$ depends on $E_{n,\lambda+1}[-(t/\tau)^n]$, but the shape of those function are qualitatively similar to each other). An attempt to fit the relaxation function $R_{n<1}(t)$ with an exponential decay leads to overestimated values near $t = 0$ in order to reduce the total fitting error, resulting in an underestimated value of $E_0$. The functions $R_{n<1}(t)$ seem to converge to values $R_{n<1}(t\rightarrow\infty) > 1 - \alpha_{n<1}$, this will lead to underestimated values of $\alpha$. The overall adjustment of the curves $R_{n<1}(t)$ with a exponential decay demands an steeper descend behavior near $t=0$, that leads to underestimated values of $\tau$.
The fitted relaxation exponents (for the cases where $n$ was also a fitting parameter) are shown in Figure \ref{fig:rheology}. Despite of the fluctuations, the relaxation exponents are nearly frequency independent between 0.1 Hz and 10 Hz. The average exponents for the 0.2\%, 0.4\% and 0.8\% samples are 0.796, 0.681 and 0.793, respectively. The average exponent for all $C_{bis}$ cases is 0.756. The fluctuations and error bars are attributed to spatial non-homogeneities in the gel composition, and other complicated effects which are not included in our model like hydrodynamic interaction of the cantilever with liquid (all measurements were performed in liquid to minimize the adhesiveness of the gels) and residual adhesive effects.
\section{Discussion}
Abidine \textit{et al.} investigated the dynamic rheology of polyacrylamide gels in a wide range of frequencies combining conventional shear rheometry (for low frequencies ranging between 10$^{-3}$ Hz and 1 Hz), and AFM based dynamic indentation (for high frequencies ranging from 1 Hz to 300 Hz) \cite{abidine2015}. They obtained elasticity moduli varying from 1 kPa to 30 kPa for bisacrylamide concentrations (in weight) ranging between 5\% and 15\%. They have shown that the storage modulus exhibits a constant plateau between 10$^{-3}$ Hz and 50 Hz, for all bisacrylamide concentrations. The storage shear modulus $G'(\omega)$ is one order of magnitude higher than the loss modulus $G''(\omega)$ from frequencies up to 100 Hz, above which there is a crossover between $G'(\omega)$ and $G''(\omega)$. It is instructive to fit their rheological measurements with the fractional SLS relaxation model in order to extract quantities that can be compared to ours.
\begin{table*}[ht]
\caption{SLS parameters extracted from the dynamic measurements of the gels measured by Abidine \textit{et al.} \cite{abidine2015}.}
\begin{center}
\begin{tabular}{crrrrrrrr}
\hline
\hline
$C_{bis}$ & $E_{0}$(kPa) & $E_{\infty}$(kPa) & $E_{1}$(kPa) & $\alpha$ & $\tau$ (s) & $r^2_{G'}$ & $r^2_{G''}$ & $n$\\
\hline
\multicolumn{9}{c}{conventional SLS model ($n = 1$)} \\
\hline
5\% & 15.816 & 1.290 & 14.525 & 0.918 & 3.66 $\times 10^{-4}$ & 0.9945 & 0.9970 & 1\\
7.5\% & 19.996 & 2.086 & 17.910 & 0.895 & 5.48 $\times 10^{-4}$ & 0.9974 & 0.9972 & 1\\
\hline
\multicolumn{9}{c}{fractional SLS model ($n$ is a fitting parameter)} \\
\hline
5\% & 28.056 & 1.274 & 26.781 & 0.954 & 1.36 $\times 10^{-4}$ & 0.9963 & 0.9998 & 0.87 \\
7.5\% & 29.734 & 2.079 & 27.654 & 0.930 & 2.29 $\times 10^{-4}$ & 0.9984 & 0.9994 & 0.83 \\
\hline
\hline
\end{tabular}
\end{center}
\label{tab:abidine}
\end{table*}%
Figure \ref{fig:abidine} shows an example of the rheological data of Abidine \textit{et al.} Both regular and fractional SLS models are able to describe well the storage modulus in the whole frequency range, with the fractional model exhibiting a slightly improved accuracy. The conventional SLS model fails to describe $G''(\omega)$ for frequencies up to 10 Hz, which is precisely the largest frequency that the forces curves can be measured in our AFM. On the other hand, the fractional model (with $n = 0.83$) describes $G''(\omega)$ very well between 1 Hz and 300 Hz. Abidine's data suggests that an exponent a little smaller than 0.83 between 1 Hz and 10 Hz. The viscoelastic parameters fitted from Abidine's measurements are shown in Table \ref{tab:abidine}. We focused our comparison in Abidine's gels with lowest $C_{bis}$ (5.0\% and 7.5\%) because the cut-off frequency $\omega_0$ is out of the measured frequency range for gels with higher $C_{bis}$.
The fitted values of $\omega_0^{-1}$ of the order of $10^{-4}$ s confirm that our measurements were performed much below $\omega_0$, in a frequency range in which the instantaneous elasticity modulus must be almost fully relaxed. The $E_0$ values in both works are in good agreement, and in both studies $E_0$ is proportional to $C_{bis}$, while $\alpha$ and $\tau$ are inversely proportional to $C_{bis}$. The fitted values of $\alpha$ are compatible ($\alpha > 0.82$) with the values for which there is a double crossover between $G'(\omega)$ and $G''(\omega)$ (see Figure \ref{fig:abidine}). The parameters $E_0$, $\alpha$ from Abidine's measurements are in good agreement with the values estimated from our measurements. The $\tau$ values obtained estimated from our model are of the order of $10^{-3}$ s, nearly one order of magnitude higher than the $\omega_0^{-1}$ values estimated from Abidine's measurements. One possible reason for this is that the time resolution of the AFM force measurements is limited by the reading frequency of the AFM controller of 2 kHz.
We also obtained a very good agreement between the exponents from the frequency dependent AFM force curves (ranging between 0.681 and 0.796) and the exponents from Abidine's measurements (ranging between 0.83 and 0.87). We remind that our exponents were fitted from rheological data acquired between 0.1 Hz and 10 Hz, while Abidine's exponents were fitted from data in the range 1 Hz to 300 Hz. However, a quick look in Abidine's $G''(\omega)$ curves suggests that slightly lower exponents would be obtained data between 1 Hz and 10 Hz, becoming even closer to our values.
Finally, Eq. \ref{eq:eta} suggests that the dynamic viscosity can be written in form $\eta(\omega) = G_0(\omega) \alpha(\omega) \tau(\omega)$, where $E_0 = 2(1+\nu)G_0$, $\alpha$ and $\tau$ are the frequency-dependent parameters extracted from the force curves. Figure \ref{fig:abidine}(b) compares our measured effective viscosities with Abidine's data. Both studies exhibit very good qualitative agreement, with similar slopes. The quantitative discrepancy is attributed to different gel composition between both studies.
\begin{figure*}[ht]
\center \includegraphics[scale=0.6]{fig5.pdf}
\caption{(a) AFM based dynamic rheology of polyacrylamide gels (7.5\% of bisacrylamide) measured by Abidine \textit{et al} \cite{abidine2015}. Solid lines represent the fitting of the data with the complex shear modulus of the SLS relaxation function (power law relaxation exponent $n=1$). Dashed lines represent the fitting of the data with complex shear modulus of the fractional-like SLS relaxation function (power law relaxation exponent $n=0.83$). Table \ref{tab:abidine} shows the fitted parameters from Abidine's gels. (b) Frequency dependence of the apparent viscosity $\eta=E_0 \alpha \tau$. Colors represent different $C_{bis}$ values: 0.2\% (black), 0.4\% (blue), 0.8\% (red). Square (circle) symbols represent the fitting with fractional (regular) SLS viscoelastic model. Pink symbols represent the dynamic viscosity of Abidine's gels calculated as $\eta(\omega) = G''(\omega)/\omega$: square (5\%), triangle (7.5\%), diamond (15\%). }
\label{fig:abidine}
\end{figure*}
\section{Conclusions}
We derived an analytical force-indentation model to describe viscoelastic materials with power law relaxation, that can be easily incorporated in the analysis of AFM forces. The major approximation to derive this model is that the indentation profile is linear. This approximation is valid as long as the degree of nonlinearity of the indentation profile is small. The force model is in excellent agreement with FEM simulations of viscoelastic materials indented by conical and spherical indenters (see Supplementary Material). Experimentally, we tested the model by measuring the viscoelastic properties of polyacrylamide gels with AFM force curves with varying load load frequencies at room temperature. The viscoelastic properties of the gels exhibited very good agreement with the results of Abidine \textit{et al.} \cite{abidine2015}. A very important characteristic of the model is that it was able to reproduce the viscoelastic properties of the gels, regardless the measurement method. For example, we have used simple AFM force curves, while Abidine \textit{et al.} used a dynamic indentation experimental method based on a custom modification of the AFM apparatus to measure $G'(\omega)$ and $G''(\omega)$. Most strikingly, the model is able to determine the viscoelastic relaxation exponent without a direct measurement of $G''(\omega)$. Our exponent values ranging 0.681 and 0.796 (measured between 0.1 Hz and 10 Hz) are in very good agreement with the exponents ranging between 0.83 and 0.87 from Abidine's data (measured between 1 Hz and 300 Hz).
In principle, force curve based rheology is limited to low loading frequencies (up to 30 Hz) to avoid strong cantilever oscillations during approach and retract motions, but one can use the time-temperature superposition principle by performing force measurements at different temperatures to study the viscoelastic response of the materials in a much wider range of frequencies. This method was recently demonstrated by Chyasnavichyus \textit{et al.} \cite{tsukruk2014}. Finally, the proposed model is simple enough to be easily incorporated in AFM data analysis softwares.
\noindent \textbf{Acknowledgements.} The authors acknowledge the financial support from the Brazilian National Research Council (CNPq).
\pagebreak
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,011
|
Q: Typecsript and NextJS how to check if "document" exists NextJS renders part of the code server-side. I can handle that. But I would need to check if cookies are set and here I run into problems. I tried:
!!document && !!document.cookie
and
document !== undefined && !!document.cookie
Each time I get the same error:
ReferenceError: document is not defined
I just want to make this clear, I do not want to render the component exclusively client side. The only thing I want to check is if document.cookie is defined or not :)
I am thankful for every answer that would help me solve this problem
UPDATE:
I accepted the first answer since it was the correct one. However, I realised my approach was wrong and solved it by explicitly checking the type of document:
typeof document !== 'undefined' && !!document.cookie
I guess I did not have enough coffee yet and I missed something pretty obvious.
A: Next.js uses pre-rendering, so you should wrap any use of document in the useEffect hook as document is not available on the server. That way the particular code gets executed just client-side.
https://nextjs.org/docs/migrating/from-create-react-app#safely-accessing-web-apis
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,492
|
<?php
set_include_path(get_include_path().PATH_SEPARATOR.
realpath("../src"));
require_once 'com/mikebevz/xsd2php/Wsdl.php';
require_once 'data/expected/ContactPersonWsdl/services/NavService.php';
use dk\nordsign\application\services;
use com\mikebevz\xsd2php\wsdl;
class WsdlTest extends PHPUnit_Framework_TestCase
{
/**
* XSD to PHP convertor class
* @var com\mikebevz\xsd2php\AbstractWsdl
*/
private $tclass;
private $expDir = "data/expected/ContactPersonWsdl";
private $genDir = "data/generated/ContactPersonWsdl";
protected function setUp ()
{
//$this->xsd = dirname(__FILE__)."/../resources/ubl2.0/maindoc/UBL-Order-2.0.xsd";
$this->tclass = new wsdl\WsdlFactory();
}
protected function tearDown ()
{
$this->tclass = null;
}
public function testWsdlNavService() {
if (file_exists($this->genDir)) {
rmdir_recursive($this->genDir);
}
mkdir($this->genDir."/public/schemas", 0777, true);
$service = new services\NavService();
$this->tclass->setService($service);
$this->tclass = $this->tclass->getImplementation($service);
$this->tclass->setLocation("http://mylocation.com/soap/");
$this->tclass->setSchemasPath("../resources/ContactWsdl");
$this->tclass->setPublicPath(realpath($this->genDir."/public/schemas"));
$this->tclass->setPublicUrl("/schemas");
$this->tclass->setDebug(true);
$schemas = array("CodeList_CurrencyCode_ISO_7_04.xsd",
"CodeList_LanguageCode_ISO_7_04.xsd",
"CodeList_MIMEMediaTypeCode_IANA_7_04.xsd",
"CodeList_UnitCode_UNECE_7_04.xsd",
"ContactCompany.xsd",
"ContactPerson.xsd",
"UBL-CommonAggregateComponents-2.0.xsd",
"UBL-CommonBasicComponents-2.0.xsd",
"UBL-QualifiedDatatypes-2.0.xsd",
"UnqualifiedDataTypeSchemaModule-2.0.xsd");
$wsdl = $this->tclass->toXml();
//print_r($wsdl);
//file_put_contents($this->expDir."/NavService.wsdl", $wsdl);
$expected = file_get_contents($this->expDir."/NavService.wsdl");
$this->assertEquals($expected, $wsdl);
foreach ($schemas as $schema) {
$exp = file_get_contents($this->expDir."/public/schemas/".$schema);
$act = file_get_contents($this->genDir."/public/schemas/".$schema);
$this->assertEquals($exp, $act);
}
if (file_exists($this->genDir)) {
rmdir_recursive($this->genDir);
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,304
|
Hackman Hacks Benyiwa Doe's Claims
By Ghanaian Chronicle
Mr. Hackman Owusu Agyeman the acting minister of interior has denied as absolutely untrue a statement made by Mrs. Ama Benyiwa Doe the defeated Member of Parliament for Gomoa-west and sections of the media that sought to create the impression that Ivorian rebels had lost confidence in President John Agyepong Kufuor.
Speaking on the Peace FM newspaper review morning show yesterday, Benyiwa Doe alleged that the president's inability to bring peace to the country was due to the warring faction in the neighboring Cote D' Ivoire's loss of confidence in President Kufour.
She said Hackman at a point in time had asked that the rebels be crushed because there was no way they could hold the country to ransom. Reacting to the allegation, Hackman said not only was Doe's information untrue; it was very bad for the peace process and mediation in our part of Africa since strenuous efforts were still being made by not only the Economic Community of West African States (ECOWAS) but the United Nation (UN) which was fashioning ways to bring the long standing hostility to a halt.
According to the minister, the President had rather made immense contributions to the peace process in Cote d'Ivoire, which had at least minimized the tension in the war- torn country.
He noted further that Thabo Mbeki, the South African president was made to replace President Kufuor in the mediation process upon realizing that he (Kufuor) had only a couple of months to campaign in his re-election bid and could not have looked on with the seat slipping to his political opponents who were then briskly campaigning.
He asserted that President Mbeki taking over from President Kufuor could not be represented as meaning the Ivorians had lost confidence in him.
He further indicated that Kofi Annan was brought into the process in his capacity as the UN secretary general which made it incumbent on him to react to all threatening security situations the world over.
Hackman stressed that he never asked anybody anywhere to 'crush' the rebels in the country as purported by Benyiwa Doe, knowing well that without them, the much expected peace could not prevail. He concluded that Benyiwa Doe had gotten all her claimed facts upside down since the picture she was painting did not represent or reflect the real situation.
In furtherance of his speech, the minister said President Kufuor's mediation efforts played a significant role in his second term bid as the chairman of ECOWAS, stressing that this and other things reflected in the trust reposed in him not only in his country but in the West African sub-region.
Meanwhile, most callers and text messages to the programme condemned the former legislator in no uncertain terms with some branding her as too politically inclined because she had on series of occasions stated categorically that if offered an appointment by the NPP government, she would refuse it.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,090
|
About BSA
For Students & Caregivers
Academic & Supportive Programs
Caregiver Involvement Association
How to Enroll at BSA
NWEA Assessments
Broome Street Academy's robust extracurricular program is rooted in the belief that important learning experiences are not limited to the classroom.
BSA Sports and Clubs
BSA offers opportunities to participate in numerous athletics and service clubs, including basketball teams, soccer teams, track and field teams, the culinary club, student council, and many more. Students are encouraged to express their interests and start new clubs, too!
Every BSA student has access to The Door after school. The Door hosts daily activities including recreation, health and wellness, LGBTQ programs, and performing and visual arts with expert teaching artists.
Broome Street Academy has also forged exciting relationships beyond the walls of the building for students to explore their creativity and self-expression both inside and outside of the school.
BSA has also become a chapter for the MOUSE Squad. MOUSE Squad gives our students hands-on experiences with computers and other technology that are not available during a traditional school day.
Broome Street Academy has also been recognized as a chapter in the National Honor's Society. The NHS is the nation's premier organization to recognize outstanding high school students. Memberships are more than just honor roll students – NHS services to honor those students who have demonstrated leadership, scholarship, service, and character. Students that embody our PRIDE principles are invited to join and serve.
© 2020 Broome Street Academy | Enroll | Contact | Email Sign-up | Privacy Policy
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,598
|
<?php
declare(strict_types=1);
namespace CrudJsonApi\Listener;
use Cake\Event\EventInterface;
use Cake\ORM\Table;
use Crud\Listener\BaseListener;
use RuntimeException;
class SearchListener extends BaseListener
{
/**
* Settings
*
* @var array
*/
protected $_defaultConfig = [
'enabled' => [
'Crud.beforeLookup',
'Crud.beforePaginate',
],
'collection' => 'default',
];
/**
* Returns a list of all events that will fire in the controller during its lifecycle.
* You can override this function to add your own listener callbacks
*
* @return array<string, mixed>
*/
public function implementedEvents(): array
{
return [
'Crud.beforeLookup' => ['callable' => 'injectSearch'],
'Crud.beforePaginate' => ['callable' => 'injectSearch'],
];
}
/**
* Inject search conditions into the query object.
*
* @param \Cake\Event\EventInterface $event Event
* @return void
*/
public function injectSearch(EventInterface $event): void
{
if (!in_array($event->getName(), $this->getConfig('enabled'))) {
return;
}
$repository = $this->_table();
if ($repository instanceof Table && !$repository->behaviors()->has('Search')) {
throw new RuntimeException(
sprintf(
'Missing Search.Search behavior on %s',
get_class($repository)
)
);
}
$filterParams = ['search' => $this->_request()->getQuery('filter', [])];
$filterParams['collection'] = $this->getConfig('collection');
$event->getSubject()->query->find('search', $filterParams);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 107
|
HNP - An inspection delegation of the Party Central Committee on April 12 had a working session with the Hanoi Party Committee on reviewing the five-year implementation of Resolution No. 33-NQ/TW (11th tenure) on "Building and developing Vietnamese culture and people to meet the requirement of the country's sustainable development" in the city.
Secretary of the Hanoi Party Committee Hoang Trung Hai speaks at the working session.
The event was co-chaired by Vo Van thuong, Politburo member, Secretary of the Party Central Committee (PCC), Head of the PCC's Commission for Communication and Education, Head of the Steering Committee; and Hoang Trung Hai, Politburo member, Secretary of the Hanoi Party Committee.
In attendance were Le Manh Hung, Deputy Head of the PCC's Commission for Communication and Education, Deputy Head of the Steering Committee; Nguyen Thi Bich Ngoc, Deputy Secretary of the Hanoi Party Committee, Chairwoman of the Municipal People's Council, among others.
At the session, Nguyen Van Phong, Head of the Municipal Party Committee's Commission for Communication and Education, gave a report on the outcomes of the five-year implementation of the resolution in the city.
Accordingly, considering the cultural development as the city's immediate and long-term mission, Hanoi has always renewed the leadership and worked out a big program on developing culture and people.
During the period of 2015-2020, the Hanoi Party Committee has focused on directing the effective implementation of Program 04-CTr/TU on socio-economic development, improving the qualifications of human resources, and building elegant and civilized Hanoians in the period of 2016-2020.
The report also highlighted the city's outstanding outcomes of executing the resolution, including strengthening exchanges and international integration, contributing to promoting the images of the capital's culture and people in the eyes of international friends.
Mass movements have been launched citywide associated with political mission in localities and new-style rural building program which attracted large participation oflocal residents.
The city has issued many policies and concrete measures for cultural development, building elegant, civilized Hanoians, and gained remarkable achievements.
At the session, delegates focused on exchanging on some outstanding outcomes as well as existing problems in carrying out the resolution.
Some of them suggested the city to further promote the role of all-level Party Committees and authorities in culture; and strengthen the popularization of the resolution.
Making conclusion at the session, Vo Van Thuong, Head of the PCC's Commission for Communication and Education, appreciated the achievements of Hanoi in carrying out Resolution No. 33-NQ/TW.
Head of the PCC's Commission for Communication and Education Vo Van Thuong makes conclusion at the session.
To better implement the resolution in the coming time, Vo Van Thuong asked the Municipal Party Committee to focus on reviewing the implementation of the resolution to draw experiences and promptly remove existing problems and weaknesses; fulfilling tasks stipulated in the resolution; enhancing the cooperation with the Central agencies, sectors and levels in carrying out the tasks, among others.
On behalf of the city leaders, Hoang Trung Hai, Secretary of the Municipal Party Committee stressed the importance of cultural development.
Thus, the Municipal Party Committee issued Program 04 on socio-cultural development, improving the qualifications of human resources, and building elegant, civilized Hanoians for the period of 2016-2020, he added.
Acknowledging all the delegation's concerns, the Secretary requested the municipal entities to elaborate guiding documents on developing culture and people to meet the requirement of sustainable development of the capital city and the country in the context of globalization and international integration today.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,119
|
When maintaining peak productivity is everything in a business perspective, why let something as resolvable as lock and key troubles get in the way of smooth business transactions? At All County Locksmith Store we give priority to your distress calls as your most reliable commercial locksmith operating in Parlin, NJ area. We deal with all kind of emergencies that might hamper your operations like commercial lockouts.
Caught up in an unfortunate commercial lockout? Call All County Locksmith Store on 732-475-3007 for immediate assistance!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,913
|
\section{Introduction}
While a lot of effort is put into the search for new physics at large accelerators like the LHC at CERN, another approach is to instead search for new physics at very low energy scales and small couplings.
In this context, the axion is one of the best known candidates: it was introduced in 1977 by Peccei and Quinn to solve the strong CP problem \cite{pecceiQuinn}, but despite all effort, it has not been found yet, and both its mass and its coupling to photons are still unknown.
In addition to the axion, several extensions of the standard model predict similar particles, so called ``axion-like particles'' (ALPs) (see, e.g., \cite{ringwald:2010} for a review).
But while the axion must satisfy a certain relation between its mass and its coupling to photons in order to solve the strong CP problem,
in general there is no relation between mass and coupling constants for ALPs.
These ALPs have been suggested to explain several physical phenomena \cite{ringwald:2014}, such as the anomalous gamma-ray transparency \cite{DeAngelis:2007, Horns:2012}, the soft X-ray excess from the Coma cluster \cite{Conlon:2014}, or dark matter \cite{Preskill:1983, Abbott:1983, Dine:1983, Arias:2012}.
Interestingly, both the anomalous gamma-ray transparency and the soft X-ray excess from the Coma cluster may be explained with ALPs in a similar parameter region: the gamma-ray transparency can be resolved with a photon-ALP coupling constant $g \gtrsim 10^{-(10-11)}$ GeV$^{-1}$ and an ALP mass $m_\phi \lesssim 10^{-7}$ eV \cite{Meyer:2013}, while the soft X-ray excess can be explained with $g \gtrsim 10^{-13}$ GeV$^{-1}$ and $m_\phi \lesssim 10^{-12}$ eV \cite{Angus:2014}.
\\
In this work, we propose a method to improve current limits in the mass region of $10^{-14}$ eV $\lesssim m_\phi \lesssim 10^{-12}$ eV, therefore reducing the parameter space available for explaining the soft X-ray excess and the gamma-ray transparency with ALPs:
for a suitable ALP mass, CMB photons crossing a galaxy cluster can undergo resonant photon-ALP conversion inside the cluster's magnetic field, therefore distorting the black-body spectrum of the CMB.
Galaxy clusters are one of the few large scale astrophysical objects with known magnetic fields which allows to derive constraints not only on the combination $gB$ from distortions of the CMB \cite{resPhotonAxion}, but also on $g$ itself.
In particular, we use observations of the thermal Sunyaev-Zeldovich effect by OVRA, WMAP, MITO, and the \textit{Planck} satellite
and use these to obtain limits on the coupling between ALPs and photons.
While current data leads to limits only slightly better than the ones obtained from SN1987A, a future PRISM-like experiment might significantly improve current limits.
The currently strongest limits in this mass-region are derived from the absence of a gamma-ray flash at the time of the SN1987A and limits the coupling to photons to $g\lesssim 5.3 \cdot 10^{-12}$ GeV$^{-1}$ \cite{sn1987:3}.
\\
This paper is structured as follows: in section 2 we introduce the framework of resonant photon-ALP conversion inside galaxy clusters
and describe the cluster models used.
The resulting constraints are presented in section 3.
This section also contains an estimate of the expected sensitivity of a PRISM-like experiment.
In section 4 we discuss the results and how they may be further strengthened.
In the appendix we present in detail how the multiple level crossing has been calculated.
\\
Throughout the paper, we set $c = \hbar = k_B = 1$. We denote spatial vectors with bold face symbols.
\section{Framework of photon-ALP oscillations}
In this chapter, we derive an expression for the conversion probability from photons to ALPs and compare the corresponding temperature change with observations of the thermal Sunyaev-Zeldovich effect.
Using suitable models for the profile of galaxy clusters, we then determine upper limits for the coupling constant between photons and ALPs.
\subsection{Resonant photon-ALP conversion and its effect on the CMB temperature}
For the deduction of the conversion probability we closely follow \cite{resPhotonAxion}.
Axion-like particles (ALPs), in this work denoted by $\phi$, are pseudoscalar bosonic particles, that couple to photons through the interaction Lagrangian \cite{photonALPLagrangian}
\begin{equation}
\mathcal{L} = - \frac{1}{2}g F_{\mu \nu} \tilde{F}^{\mu \nu} \phi= g \mathbf{B} \cdot \mathbf{E} \phi,
\end{equation}
where $\tilde{F}^{\mu \nu} = \epsilon^{\mu \nu \alpha \beta} F_{\alpha \beta}/2$ is the dual of the electromagnetic field strength tensor, $\mathbf{B}$ and $\mathbf{E}$ are the magnetic and electric field, respectively, and $g$ denotes the axion-photon coupling constant.
\\
In an external magnetic field, this interaction Lagrangian is well known to produce effective mass-mixing between photons and ALPs. The new propagation eigenstates are then rotated with respect to the interaction eigenstates by an angle $\theta$ given by \cite{raffelt:1988}
\begin{equation}
\label{eq:thetaNoPlasma}
\sin 2\theta = \frac{2gB\omega}{\sqrt{m_{\phi}^4 + (2gB\omega)^2}}, \qquad
\cos 2\theta = \frac{m_{\phi}^2}{\sqrt{m_{\phi}^4 + (2gB\omega)^2}}.
\end{equation}
Here, $\omega$ denotes the energy of the photon, $B$ is the component of the magnetic field perpendicular to the propagation direction of the photons, $m_{\phi}$ is the ALP mass and $g$ is the coupling constant as before.
From here on, we will refer to $\theta$ as magnetic mixing angle.
\\
Using typical values for the parameters in our study, the relevant dimensionless parameter in these expressions reads
\begin{equation}
\frac{2gB\omega}{m^2_\phi} \simeq 1.38 \cdot 10^{-6} \frac{g}{10^{-12} \mbox{ GeV}^{-1}} \frac{B}{\mu \mbox{G}} \frac{\omega}{10^{-4} \mbox{ eV}} \left( \frac{10^{-13} \mbox{ eV}}{m_\phi} \right)^2.
\end{equation}
For the parameter ranges considered here this will always be much smaller than unity.
The misalignment between interaction eigenstates and propagation eigenstates will produce photon-ALP oscillations with a wavenumber given by \cite{raffelt:1988}
\begin{equation}
\label{eq:k}
k = \frac{\sqrt{m_\phi^4 + (2gB \omega)^2}}{2 \omega}.
\end{equation}
Inside a plasma, photon-ALP mixing will be modified: the refractive properties of the plasma lead to a non-trivial dispersion relation, which can be parametrized by an effective photon mass $m_\gamma$.
In this case, for a given magnetic field the effective mixing angle in the plasma $\tilde{\theta}$ is related to the mixing angle $\theta$ at vanishing charge density by \cite{raffelt:1996}
\begin{equation}
\label{eq:angleInPlasma}
\sin 2\tilde{\theta} = \frac{\sin 2 \theta}{[\sin^2 2 \theta + (\cos 2 \theta - \xi)^2 ]^{1/2}},
\end{equation}
\begin{equation}
\cos 2\tilde{\theta} = \frac{\cos 2 \theta - \xi}{[\sin^2 2 \theta + (\cos 2 \theta - \xi)^2 ]^{1/2}},
\end{equation}
where $\xi$ is defined as
\begin{equation}
\xi \equiv \cos 2 \theta \left( \frac{m_\gamma}{m_\phi} \right)^2.
\end{equation}
If in some region in space the resonance condition
\begin{equation}
\label{eq:resonanceCondition}
m_\gamma = m_\phi
\end{equation}
is satisfied, one has $\tilde{\theta} \rightarrow \pi/4$ and resonant photon-ALP conversion is possible.
\\
In this study, we will consider such resonant photon-ALP conversion occurring in galaxy clusters, where one has both
external magnetic fields and a nonzero free electron density.
\\
Due to the high ionization fraction in the intra-cluster medium, contributions from the scattering off neutral atoms can be neglected and the effective photon mass is given by the plasma frequency \cite{born:1980}
\begin{equation}
\label{eq:m_gamma}
m_\gamma^2 \simeq \omega_P^2 = \frac{4 \pi \alpha}{m_e} n_e,
\end{equation}
where $\alpha$ is the fine structure constant, $m_e$ is the electron mass and $n_e$ is the free electron density.
Using this relation, one can rewrite the resonance condition (\ref{eq:resonanceCondition}) to
\begin{equation}
m_\phi = 3.72 \left( \frac{n_e}{\mbox{m}^{-3}} \right)^{1/2} \cdot 10^{-14}~\mbox{eV}.
\end{equation}
As the free electron density cannot reach arbitrary values inside a galaxy cluster, the resonance condition can only be satisfied
for a certain range of ALP masses $m_\phi$.
Assuming for example a minimal density of $0.3$ m$^{-3}$ and a maximal density of $10^{4}$ m$^{-3}$,
resonant photon-ALP conversion will only occur for $2 \cdot 10^{-14}$ eV $\lesssim m_\phi \lesssim 3.7 \cdot 10^{-12}$ eV.
The first value is the average baryon density in the universe at redshift zero and the second value is a typical density in the core of a galaxy cluster.
\\
The distance between the photon production and the resonance as well as the distance between the resonance and the detection is much larger than the oscillation length causing an incoherent superposition of the oscillation patterns.
In this case, the transition probability is given by \cite{parke:1986}
\begin{equation}
\label{eq:generalConversionFormula}
P_{\gamma \rightarrow \phi} \simeq \frac{1}{2} + \left( p - \frac{1}{2} \right) \cos 2\tilde{\theta}_0 \cos 2 \tilde{\theta}_D,
\end{equation}
where $\tilde{\theta}_0$ is the effective mixing angle at production, $\tilde{\theta}_D$ is the effective mixing angle at detection and $p$ is the level crossing probability. Therefore, a transition from a medium dominated to the vacuum state corresponds to
$\cos2\tilde\theta_0\simeq-1$ and $\cos2\tilde\theta_D\simeq\cos2\theta$ or \textit{vice versa} and for $\theta\ll1$ one obtains
a conversion probability $P_{\gamma \rightarrow \phi}$ close to unity for $p\ll1$, corresponding to an adiabatic transition.
\\
As it was argued in \cite{resPhotonAxion}, the high plasma density at the time of the creation of the CMB photons leads to a value of $\tilde{\theta}_i$ close to $\pi/2$.
Using typical values of the free electron density in the solar system, one can see that also $\tilde{\theta}_D$ is very close to $\pi/2$.
A more detailed discussion of this point can be found in the appendix.
\\
For our range of ALP masses, there is not only one, but several resonances:
the first resonance occurs when the free electron density decreases due to the cosmic expansion.
Inside the transversed cluster, there are several resonances: additionally to the two resonances due to the increasing (decreasing) electron density when entering (leaving) the cluster, density fluctuations enable even more resonances.
Finally, there are also resonances when the photons enter the Milky Way.
\\
The level crossing probability $p_i$ for a single resonance $i$ is given by the Landau-Zener expression \cite{kuo:1989}
\begin{equation}
\label{eq:LandauZener}
p_i \simeq \exp(-2\pi R k \sin^2\theta_{\text{res}}),
\end{equation}
where $k$ is the oscillation wavenumber given in eq. (\ref{eq:k}), $\theta_{\text{res}}$ is the magnetic mixing angle given by eq. (\ref{eq:thetaNoPlasma}) at this resonance and $R$ is the scale parameter defined as
\begin{equation}
\label{formula:scaleParameter}
R = \left| \frac{d \ln m^2_\gamma(t)}{dt} \right|^{-1}_{t=t_{\text{res}}}.
\end{equation}
The level crossing probability $p_i$ takes into account the deviation from adiabaticity of the photon-ALP conversion in the resonance region. One has $p_i \simeq 0$ for a completely adiabatic transition and $p_i = 1$ for an extremely non-adiabatic one.
\\
The Landau-Zener expression (\ref{eq:LandauZener}) only holds for the case when the free electron density varies linearly during the resonance.
For $\theta_{\rm res}$ the resonance half-width is, according to equation (\ref{eq:angleInPlasma}), $\delta \xi \simeq \sin 2 \theta_{\text{res}}$, corresponding to a resonance width in the density scale of
\begin{equation}
\Delta n_e^R \simeq n_e^R \sin(2\theta_\text{res}) \simeq n_e^R \frac{2gB\omega}{m_\phi^2}
\simeq 10^{-6} n_e^R,
\end{equation}
where $n_e^R$ is the resonance density defined by eq. (\ref{eq:resonanceCondition}).
Due to this extremely narrow resonance region, the approximation of a linear density change is very well fulfilled.
\\
Expanding the sine in eq. (\ref{eq:LandauZener}) and approximating $k \simeq m_\phi^2/(2\omega)$, the exponent becomes
\begin{equation}
2\pi Rk \sin^2(\theta_\text{res}) \simeq \frac{g^2B^2_\text{res} R \omega \pi}{m_\phi^2}
\simeq \mathcal{O}(10^{-6}),
\end{equation}
for typical values used in this work.
As derived in Appendix A, for such small exponents, the total level crossing probability becomes
\begin{equation}
p = \prod_i p_i \simeq 1 - \sum_i \frac{g^2 B^2_i R_i \omega \pi}{m_\phi^2},
\end{equation}
where the sum is over all resonances $i$.
Together with the obtained expressions for $\cos 2 \tilde{\theta}_0$, and $\cos 2 \theta_D$, the conversion probability then is
\begin{equation}
\label{formula:convProb}
P_{\gamma \rightarrow \phi} \simeq \sum_i \frac{g^2 B^2_i R_i \omega \pi}{m_\phi^2} = \frac{g^2\omega \pi}{m_\phi^2} \sum_i B_i^2 R_i.
\end{equation}
In the present study we will neglect the first resonance because of the unknown magnetic field, and the resonances in the Milky Way because of the small scale parameter $R$.
These approximations are conservative, since they tend to underestimate the actual conversion probability, as it will be discussed in Appendix A.
\\
The conversion of photons into ALPs will always reduce the number of photons in the beam, causing an apparent temperature decrease.
The intensity for a given photon energy and temperature $I(\omega, T)$ can be related with the apparent temperature change over
\begin{equation}
\label{formula:deltaT}
I' = \frac{dI(\omega, T)}{dT} \simeq \frac{\Delta I}{\Delta T} ~\rightarrow~ \Delta T(\omega) \simeq \frac{\Delta I(\omega, T)}{I'(\omega, T)}.
\end{equation}
The intensity of the photon beam after crossing the galaxy cluster is reduced by a factor of $1-P_{\gamma \rightarrow \phi}$, giving $\Delta I(\omega, T) = -P_{\gamma \rightarrow \phi} I_0(\omega, T)$.
With $I(\omega, T) \propto \left[ \exp(\omega / T_\text{CMB} ) -1 \right]^{-1}$
one arrives at
\begin{equation}
\label{formula:DeltaTFromConversion}
\Delta T_{\gamma \rightarrow \phi}(\omega) = - P_{\gamma \rightarrow \phi} \frac{T^2_{CMB}}{\omega} \left[ 1 - \exp \left( -\frac{\omega}{T_{\text{CMB}}} \right) \right]
\end{equation}
for the apparent temperature change in dependence of the photon energy due to resonant photon-ALP conversion.
\subsection{Comparing the conversion probability with the tSZ-parameter}
\label{section:comparingTSZandPAC}
From the \textit{Planck} 2015 data \cite{planck2015SZ}, we have information about the temperature differences in the directions of galaxy clusters.
These temperature differences depend on the frequency observed and are related to the (by definition frequency-independent) thermal Sunyaev-Zeldovich (tSZ) Compton parameter $y$ by the expression \cite{sunyaev:1972}
\begin{equation}
\label{formula:tSZ}
\Delta T_{\text{tSZ}}(\omega) = f(\omega)~ T_{\text{CMB}} ~ y,
\end{equation}
with $f(\omega) = (\omega/T_{\text{CMB}})\cdot \coth [ \omega/(2 T_{\text{CMB}}) ] - 4$.
The function $f(\omega)$ is negative for values of $\omega$ smaller than $3.83\,T_{\text{CMB}}$, otherwise the function is positive.
\begin{figure}[t]
\centering
{
\includegraphics[width=10cm]{figure1.pdf}
}
\caption{Comparison of the spectral profile of the tSZ effect (with y = $10^{-5}$) and the photon-axion conversion.
The $x-$axis shows the photon energy, the $y-$axis shows the relative distortion of the effective CMB temperature,
$\Delta T/T_{\text{CMB}}$ in units of $10^{-5}$.
For this plot, the parameters $g = 5\cdot 10^{-13}$ GeV$^{-1}$, $B$ = 2 $\mu$G, $R$ = 0.5 Mpc, $m_{\phi} = 10^{-13}$ eV were used.
Different values for these parameters change the normalization of the effect but not its dependence on the photon frequency.
The filled circles refer to the centers of the frequency bands of the \textit{Planck}-mission.}
\label{tSZandConversion}
\end{figure}
In figure (\ref{tSZandConversion}) the relative temperature change due to the thermal SZ effect as well as due to photon-ALP conversion is shown.
As mentioned above, the relative temperature change due to the photon-ALP conversion is always negative, because the effect always removes photons from the beam.
The tSZ-effect, in contrast, creates a negative temperature change for small energies ($\omega < 3.83~ T_{\text{CMB}}$) and a positive temperature change at higher photon energies.
\\
For the Coma cluster, detailed measurements of its thermal SZ-effect for photon energies in the range of $10^{-4}$ eV $ \lesssim \omega \lesssim 1.1 \cdot 10^{-3}$ eV exist \cite{battistelli:2003}.
These values are presented in table (\ref{table:tSZdata}).
\begin{table}[t]
\begin{center}
\begin{tabular}
{|l|c|r|} \hline Experiment & $\omega$ [10$^{-4}$ eV] & $\Delta$ T$_{tSZ}$ [$\mu$K]\\
\hline \hline
OVRA & 1.32 & -520 $\pm$ 83 \\ \hline
WMAP & 2.51 & -240 $\pm$ 180 \\ \hline
WMAP & 3.87 & -340 $\pm$ 180 \\ \hline
MITO & 5.91 & -184 $\pm$ 39 \\ \hline
MITO & 8.85 & -32 $\pm$ 79 \\ \hline
MITO & 11.25 & 172 $\pm$ 36 \\ \hline
\end{tabular}
\end{center}
\caption{Measurements of the thermal SZ-effect in the Coma cluster.}
\label{table:tSZdata}
\end{table}
In this case, one can build the reduced $\chi^2$-function
\begin{equation}
\label{formula:chiSqTest}
\chi^2(y, g) = \frac{1}{N-2} \sum_{i}^{N} \left[ \frac{\Delta T_i^{\text{exp}} - \Delta T_i^{\text{theo}}(y, g)}{\sigma_i^{\text{exp}}} \right]^2,
\end{equation}
where $\Delta T_i^{\text{exp}}$ are the observed temperature changes at photon energy $\omega_i$, $\sigma_i$ are their standard errors and
\begin{equation}
\Delta T_i^{\text{theo}}(y, g) = \Delta T_{\gamma \rightarrow \phi}(\omega_i, g) + \Delta T_{\text{tSZ}}(\omega_i, y),
\end{equation}
is the prediction by the theory, with $\Delta T_{\gamma \rightarrow \phi}$ and $\Delta T_{\text{tSZ}}$ given by eq. (\ref{formula:DeltaTFromConversion}) and eq. (\ref{formula:tSZ}), respectively.
For each ALP mass $m_{\phi}$, we calculate the $\chi^2$-function in the ($y, g$)-parameter space and determine the limit on $g$ as the largest value of it still inside the respective confidence interval.
\\
Often, e.g. in the case of the \textit{Planck}-mission \cite{planck2015SZ}, only the $y$-parameters of galaxy clusters are easily available, but not the temperature changes due to the thermal SZ-effect at different frequencies.
In such cases, a simple and conservative bound can be obtained when assuming that the magnitude of the temperature change due to the photon-ALP conversion must be smaller than the magnitude of the temperature change due to the tSZ-effect:
\begin{equation}
\label{formula:temperatureComparison}
| \Delta T_{\gamma \rightarrow \phi} | \lesssim | \Delta T_{\text{tSZ}} |
\end{equation}
Using equations (\ref{formula:convProb}), (\ref{formula:DeltaTFromConversion}), (\ref{formula:tSZ}), and solving for the coupling constant $g$, one arrives at
\begin{equation}
\label{formula:gBound}
g \lesssim m_\phi \left( \frac{y}{\sum_i B_i^2 R_i} \right)^{1/2} h(\omega),
\end{equation}
where we defined
\begin{equation}
\label{formula:h}
h(\omega) \equiv \left| \frac{4-x\cdot \coth(x/2)}{T_\text{CMB} \pi [1 - \exp(-x)]} \right| ^{1/2}, ~~ x \equiv \frac{\omega}{T_{\text{CMB}} }.
\end{equation}
Note that this approach does not take any non-resonant conversion into account, and therefore conservatively overestimates $g$.
\subsection{Galaxy cluster model and multiple resonances}
Both the magnetic field strength as well the scale parameter at resonance depend on the free electron density inside the considered galaxy cluster.
On top of the smooth, large-scale electron density profile, galaxy clusters exhibit smaller turbulent contributions in the electron density.
For example, in the Coma cluster, density fluctuations of $\sim$5\% on scales of $\sim$30 kpc and $\sim$(7-10)\% on scales of $\sim$500 kpc have been observed \cite{churazov:2012}.
As estimated before, in terms of the electron density variation the resonance is extremely narrow,
$\Delta n_e/n_\text{res} \approx 10^{-6}$, such that small turbulent contributions can already cause several distinct resonances.
We therefore have to evaluate the expression $\sum_i B_i^2 R_i$ from eq. (\ref{formula:convProb}), where the index $i$ labels the individual resonances.
\\
We assume that the magnetic field strength follows the free electron density, such that
\begin{equation}
\label{eq:magneticFieldWithDensity}
\left< | \textbf{B}(r) | \right> \simeq B_{\text{max}} [ n_e(r) / n_{\text{max}} ]^\eta,
\end{equation}
where $\eta \lesssim \mathcal{O}(1)$ and $B_\text{max}$ will be specified later.
Due to the narrowness of the resonance region, one can approximate the magnetic field as being constant during the resonance.
Furthermore, the magnetic field during the resonance is completely determined by the resonance density $n_\text{res}$, such that all resonances will experience the same field strength.
Averaging over several resonances, one therefore obtains
\begin{equation}
\sum_i B_i^2 R_i \simeq \frac{2}{3} B_\text{res}^2\sum_i R_i,
\end{equation}
where the factor of 2/3 accounts for the fact that only the two transverse components of the magnetic field enter the conversion probability (\ref{formula:convProb}).
\\
To investigate the consequences of turbulent density contributions, we will consider an electron density with a dominant smooth component $n_\text{s}$ and a spectrum of modulations with wavenumber $k$, amplitude $\delta(k)$, and phase $\phi_k$, such that
\begin{equation}
n_e(r) = n_\text{s}(r) \cdot \prod_k [1+\delta(k) \sin(kr+\phi_k) ].
\end{equation}
We will assume some upper cutoff $k_\text{max}$ induced by viscous damping and $\delta(k) \ll 1$ as indicated by observation.
As $\Delta n_\text{res} / n_\text{res} \ll 1$, one can neglect ``incomplete'' resonances (where the density has an extremum), but always assume the electron density to vary linearly during the resonance.
The scale parameter $R_i$, defined by eq. (\ref{formula:scaleParameter}), of an individual resonance $i$ at radius $r_i$ then is
\begin{equation}
\label{eq:scaleParameterSpectrum}
R_i = \Big| \frac{d \ln[n_\text{s}(r)]}{dr} + \sum_k \frac{\delta(k) k \cos(kr+\phi_k)}{1+ \delta(k) \sin(kr + \phi_k)} \Big|_{r = r_i}^{-1}.
\end{equation}
Using $\delta($500 kpc) $\sim(7-10)\%$, $\delta($30 kpc) $\sim5\%$ from the Coma cluster, and assuming a power law $\delta(k) \propto k^{-\xi}$ provides $\xi\simeq 0.12 ... 0.25$, implying $\delta(k) k \propto k^{0.75...0.88}$.
The scale parameter will therefore be dominated by the contribution from the largest wavenumber not affected by damping,
while, due to the random phases $\phi_k$, the contributions from larger scales approximately average out and can be neglected.
With $\delta(k) \ll 1$, and denoting the dominating wavenumber as $k_\text{dom}$, one thus has
\begin{equation}
\label{eq:RdominantWavenumber}
R_i \simeq \Big| \frac{d \ln[n_s(r)]}{dr} + \delta(k_\text{dom}) k_\text{dom} \cos(k_\text{dom}r_i + \phi_k) \Big|^{-1}.
\end{equation}
The spatial width $\Delta r$ containing all the resonances can be estimated by the width, within which the smooth profile varies by a factor of ($1 \pm \delta_\text{max}$), where $\delta_\text{max}$ is the largest modulation amplitude.
Explicitly, $\Delta r \simeq 2 \delta_\text{max} n_\text{res} / |n_\text{s}'|$, where $n_\text{s}' = dn_\text{s}(r)/dr|_{\text{res}}$ is the derivative of the smooth profile, evaluated at resonance.
As $\delta(k_\text{dom}) \leq \delta(k_\text{max})$, one can conservatively set
\begin{equation}
\label{formula:DeltaR}
\Delta r \simeq 2 \delta(k_\text{dom}) n_\text{res} / |n_\text{s}'|,
\end{equation}
neglecting a factor of order unity.
Including additional modes would enable resonances within an even larger region, making this statement only more conservative.
\\
The number of resonances can then be estimated by
\begin{equation}
\label{eq:numberofresonances}
N \simeq 2 \frac{k_\text{dom} \Delta r}{2 \pi} \simeq \frac{2\delta(k_\text{dom}) k_\text{dom} n_\text{res}}{\pi |n_\text{s}'|},
\end{equation}
where the factor of two in the first equality arises because there are two resonances per complete period.
Note that there must always be at least one resonance (if $n_\text{res} < n_\text{max}$) for radially incoming (outgoing) photons due to the increasing (decreasing) electron density.
\\
As $N$ has to be a natural number, this formula is only a good approximation for $N \gg 1$, while, for $N \approx 1$, we expect an uncertainty of order unity.
The assumption $N \gg 1$ implies certain conditions on the density fluctuations:
density fluctuations of $\sim 5\%$ on scales of 30 kpc have been detected in the Coma cluster, providing $N \gtrsim 2$, while projection effects preclude strong limits for smaller scales \cite{churazov:2012}.
Rotation measures, however, indicate that the Coma cluster contains magnetic fields with coherence lengths down to $\sim2$ kpc \cite{bonafede:2010}.
Due to turbulence, one expects density fluctuations on similar length scales.
Again using $\delta(k) \propto k^{-\xi}$ with $\xi\simeq 0.12 ... 0.25$, one obtains $\delta($2 kpc) $\simeq(2.5...3.6)\%$, while $N \gg 1$ implies $\delta$(2 kpc) $\gg 0.2\%$.
In the last inequality, we assumed the $\beta$-model introduced below as a smooth profile and the parameters of the Coma cluster.
In the same cluster, and $\delta($2 kpc) $\sim 3\%$, one obtains $N \gtrsim 15$.
For such a high number of resonances, the second term in (\ref{eq:RdominantWavenumber}) dominates, and one can approximate the sum over all resonances by averaging the trigonometric functions, leading to
\begin{equation}
\sum_i R_i \simeq \sum_i \Big| \delta(k_\text{dom}) k_\text{dom} \cos(k_\text{dom}r_i + \phi_k) \Big|^{-1} \simeq N\frac{\pi}{2\delta(k_\text{dom}) k_\text{dom}} \simeq \frac{n_\text{res}}{|n_\text{s}'|} = R_\text{s},
\end{equation}
where $\langle | \cos(x) | \rangle = 2/\pi$ and eq. (\ref{eq:numberofresonances}) for $N$ has been used.
$R_\text{s}$ is the scale parameter one obtains from the smooth profile without density modulation.
\\
If $N$ is $\mathcal{O}(1)$, the scale parameter (\ref{eq:scaleParameterSpectrum}) becomes $R_i \approx R_\text{s} / N$.
One could then still approximate $\sum_i R_i \simeq R_\text{s}$, inducing a relative error of $\lesssim \mathcal{O}(1)$.
\\
As we have seen, the exact dominating wavenumber $k_\text{dom}$ and corresponding amplitude $\delta(k_\text{dom})$ do not influence the transition probability as long as $N \gg 1$.
We will therefore not specify them in any more detail and approximate
\begin{equation}
\label{eq:finalConvProb}
\sum_i B_i^2 R_i \simeq \frac{2}{3} B_\text{res}^2 R_\text{s}.
\end{equation}
Independent of the multiple resonances due to the turbulent structure, there is one region of resonance when the photons enter the galaxy cluster, and one region of resonances when the photons leave the cluster.
Thus, an additional factor of 2 has to be included in the conversion probability.
In total, one therefore obtains
\begin{equation}
\label{eq:convProb2}
P_{\gamma \rightarrow \phi} \simeq \frac{4g^2 B_\text{res}^2 R_\text{s} \omega \pi}{3m_\phi^2}.
\end{equation}
For numerical calculations, we will consider a $\beta$-profile as the dominant smooth profile
\begin{equation}
\label{formula:betaProfile}
n_\text{s}(r) = \max \left( n_0,~ n_{\text{max}} [ 1 + (r / r_c)^2]^{-3 \beta / 2} \right),
\end{equation}
where $\beta$ is $\mathcal{O}$(1), $n_{\text{max}}$ is the free electron density in the cluster center, and $r_c$ is the core radius of the cluster.
We include the average cosmological electron density $n_0$ as a lower boundary for the electron density, and therefore, due to eq. (\ref{eq:resonanceCondition}), a lower cutoff for the ALP-masses $m_\phi$ able to undergo resonance.
\\
We furthermore will focus on two clusters: the Coma cluster and the Hydra A cluster.
In \cite{bonafede:2010}, rotation measure images have been used to determine the Coma clusters magnetic field strength as well as the parameter $\eta$, defined by eq. (\ref{eq:magneticFieldWithDensity}).
In this analysis, degeneracy between $B_\text{max}$ and $\eta$ has been found: a larger $B_\text{max}$ implies a larger $\eta$ and \textit{vice versa}.
The best fit gave $B_\text{max} = 4.7\mu$G, $\eta = 0.5$, while $B_\text{max} = 3.9\mu$G, $\eta = 0.4$ and $B_\text{max} = 5.4\mu$G, $\eta = 0.7$ are still within 1 $\sigma$.
To illustrate the dependence of our approach on $B_\text{max}$ and $\eta$, we will perform our analysis with these three pairs of values.
We furthermore adopt the values $r_c = (291 \pm 17)$ kpc, $n_\text{max} = (3.44 \pm 0.04)\cdot 10^3$ m$^{-3}$, and $\beta = 0.75 \pm 0.03$ from the same work.
\\
In contrast, the Hydra A cluster is a cool-core cluster and exhibits magnetic field strengths of $6~\mu$G coherent on scales of 100 kpc and magnetic field strengths of 30 $\mu$G on scales of 4 kpc \cite{taylor:1993}.
The electron density in the cluster center is $n_\text{max} \simeq 10^{4}$ m$^{-3}$ and the core radius is $r_c \simeq 130$ kpc \cite{davis:2009}.
We also adopt the value $\beta = 1$ from \cite{davis:2009}, while different values of $\eta$ have been used in the same work: mostly, $\eta = 0.9$ has been used, but also $\eta = 1/2, \eta = 2/3,$ and $\eta = 1$ have been considered.
Due to its strong influence on the possible limits on $g$, we will work with $\eta = 0.9$ as well as with $\eta = 2/3$.
The former value is suggested by observations of Abell 119 \cite{dolag:2001}, while the latter is predicted by flux conservation and is closer to the value observed in the Coma cluster.
\\
For both clusters, we will assume $y \simeq 10^{-5}$, a typical value of the Compton parameter observed in galaxy clusters \cite{planck2015SZ}.
\section{Results}
\subsection{Coupling constant constraints}
\begin{figure}[t]
\centering
{
\includegraphics[width=15cm]{figure2.pdf}
}
\caption{
Obtained limits for different pairs of values for $B_\text{max}, \eta$ obtained by the $\chi^2$-test with the data from the Coma cluster.
The black solid line shows the limits at 95\% C.L. when assuming the best-fit model from \cite{bonafede:2010}.
Using the upper and lower limits on $B_\text{max}, \eta$ at $1\sigma$, one obtains the red dashed-dotted line and the green dashed line, respectively.
The limits at 99\% C.L. for the different values are indicated by the dotted lines.
While the limits on the coupling constant $g$ for high ALP masses (corresponding to resonances near the center of the cluster) are quite similar, the difference increases for smaller ALP masses (corresponding to resonances in the outer regions of the cluster).
The shaded area is excluded by limits obtained from SN1987A.}
\label{fig:etaComparison}
\end{figure}
In figure (\ref{fig:etaComparison}), the limits obtained from the $\chi^2$-analysis of the Coma cluster are presented.
Due to the strong influence of $\eta$ on the magnetic field strength in the outer regions of the cluster, the limits from three different pairs of values for ($B_\text{max}$, $\eta$) are shown.
These three pairs of values are the ones already presented in the previous section: (4.7$\mu$G, 0.5) provides the best fit to the observed rotation measures, while (3.9$\mu$G, 0.4) and (5.4$\mu$G, 0.7) are the lower and upper limits, respectively, at $1 \sigma$ confidence level.
\\
For the best-fit model from \cite{bonafede:2010}, the obtained limits are up to a factor of 5 stronger than the limits derived from SN1987A.
For $B_\text{max} = 3.9\mu$G, $\eta = 0.4$, the obtained limits are even stronger for small ALP masses, while $B_\text{max} = 4.7 \mu$G, $\eta = 0.7$ produces limits slightly weaker than the ones from SN1987A.
\\
In this context, a warning is necessary: the $\beta$-model of the free electron density as used in \cite{bonafede:2010} and adapted here is based on measurements of the Coma cluster's X-ray emission \cite{briel:1992}.
In this work, the largest distance from the cluster center probed is $\sim$15 $r_c$, as the signal becomes undetectable in the noise for larger distances.
This maximal tested radius corresponds to $n_e \simeq 8$ m$^{-3}$ and $m_\phi \simeq 1.0 \cdot 10^{-13}$ eV.
Although the density obviously has to decrease to the average cosmological density, the exact profile is not known.
We will assume that the $\beta$-profile holds down to the average cosmological density, but one should keep in mind that for $m_\phi \lesssim 10^{-13}$ eV, the limits are obtained under this assumption.
\begin{figure}[t]
\centering
{
\includegraphics[width=15cm]{figure3.pdf}
}
\caption{
Comparison of the limits at 95\% C.L. obtained by the $\chi^2$-test (\ref{formula:chiSqTest}) and the direct temperature comparison (\ref{formula:temperatureComparison}) with different photon energies.
For the photon energy $\omega$ = 8.97$\cdot 10^{-4}$ eV (green solid line), the function $h(\omega)$ in eq. (\ref{formula:h}) becomes almost zero. But the rapid change of the function causes a high uncertainty, effectively even weakening the obtained limits.
For all other (center-) frequencies measured by \textit{Planck}, the obtained limits lie inside the red/light gray area, where $\omega = 35.44\cdot 10^{-4}$ eV gives the weakest bounds and $\omega = 5.91 \cdot 10^{-4}$ eV gives the strongest bounds.
These bounds are approximately 20\% weaker than the limits obtained by the $\chi^2$-test (black solid line), proving that the direct temperature comparison provides realistic results.
The blue/dark gray shaded area is excluded by limits obtained from SN1987A.
For this plot, the data of the Coma cluster and $\eta$ = 1/2 has been used.
}
\label{fig:gBoundsDifferentOmega}
\end{figure}
\\
In figure (\ref{fig:gBoundsDifferentOmega}), a comparison of the limits obtained by the $\chi^2$-test (\ref{formula:chiSqTest}) and the limits obtained by the temperature comparison (\ref{formula:temperatureComparison}) is presented.
For the limits by the temperature comparison, the value of $g$ was determined according to eq. (\ref{eq:finalConvProb}).
Additionally, we included Gaussian error propagation to estimate the uncertainty and to obtain limits at different confidence levels.
We adopted the uncertainties given by \cite{bonafede:2010}, i.e. $\sigma_{B_\text{max}}/B_\text{max} = 16\%$, $\sigma_{n_\text{max}}/n_\text{max} = 1.1\%$, $\sigma_{\beta}/\beta = 4\%$, $\sigma_{r_c}/r_c = 6\%$, as well as $\sigma_\omega/\omega = 20\%$(low frequency instrument)/33\%(high frequency instrument) from \cite{planck2010HFI}.
Although the total conversion probability (\ref{eq:finalConvProb}) does not dependent explicitly on $N$ anymore, the multiple resonances induce an additional uncertainty.
When assuming $N \gtrsim 40$ (see above), and $\sigma_N \propto \sqrt{N}$, one obtains $\sigma_N/N \simeq 16\%$.
As $\sum_i R_i \propto N \propto R_\beta \propto r_c$, we absorb this uncertainty into $\sigma_{r_c}/r_c$ and conservatively set $\sigma_{r_c}/r_c \simeq 25\%$.
Finally, we conservatively set $\sigma_y/y = 50\%$.
The error is usually dominated by the uncertainty of the photon energy $\omega$; only for very low and for very high photon energies, the uncertainty is dominated by $\sigma_y$.
\\
In order to avoid the singular behavior of $R_\beta$ for $n_\text{res} \rightarrow n_c$, we exclude the innermost region with $n_\text{res} / n_c \gtrsim 10\%$, and, due to its large influence, we fix $\eta = 1/2$.
This plot should therefore demonstrate the robustness of the obtained limits with respect to astrophysical uncertainties.
\\
The used photon energy determines the numerical value of the function $h(\omega)$, see eq. (\ref{formula:h}). This function is usually of order unity, but reaches zero for $\omega = 3.83\,T_{\text{CMB}}$.
Naively, one could expect to obtain arbitrarily strong limits when simply using a photon energy close to this value.
This is, however, unphysical:
the \textit{Planck} high-frequency channels have bandwidths of $\Delta \nu / \nu \approx 0.33$ \cite{planck2010HFI}, meaning that every frequency map is actually an average over a range of frequencies.
One therefore also would have to take an appropriate average over the function $h(\omega)$, preventing arbitrarily small limits.
\begin{figure}[ht]
\centering
{
\includegraphics[width=15cm]{figure4.pdf}
}
\caption{
Comparison of the obtained limits at 95\% C.L. for two different clusters: the Coma cluster and the Hydra A cluster.
Additionally to each labeled line, the limits at 99\% C.L. are shown as dotted lines.
The higher central density of the Hydra A cluster allows limits for slightly higher masses, while the strength of the limits strongly depends on the assumed values of $\eta$.
For this plot, $\omega = 5.9\cdot 10^{-4}$ eV has been used.
}
\label{fig:ComaHydra}
\end{figure}
\\
A comparison of the limits obtained from the Coma cluster and from the Hydra A cluster using eq. (\ref{eq:finalConvProb}) is shown in figure (\ref{fig:ComaHydra}).
For the Hydra A cluster, we have assumed 25\% uncertainties for all cluster parameters, i.e. $r_c$, $n_\text{max}$, $B_\text{max}$, $\beta$, and kept $\sigma_y = 50\%$ as before.
The higher central electron density in the Hydra A cluster allows higher ALP masses to undergo resonant conversion, therefore slightly expanding the mass range accessible for the method presented here.
The higher value of $\eta$ leads to a faster decrease of the magnetic field strength with increasing radius, such that the obtained limits become weaker for smaller ALP masses.
For illustration, we also show the limits for the Hydra A cluster with $\eta = 2/3$ and the Coma cluster with $\eta = 0.7$: in this case, the radial decrease of the magnetic field is very similar, while the higher value of the central magnetic field in the Hydra A cluster leads to a higher conversion probability and therefore slightly stronger limits.
\subsection{Future perspective}
In a more detailed study, one would have to simultaneously fit several contributions to the recorded data, e.g. the tSZ effect, thermal dust and synchrotron radiation.
Including photon-ALP conversion in this procedure, one would then obtain limits on the coupling constant $g$.
To estimate the possible limits with this approach, we restrict ourselves to a simpler approach:
we simulate a tSZ-signal according to eq. (\ref{formula:tSZ}), where we use the uncertainties of the \textit{Planck}-experiment, multiplied with a factor referred to as ``error penalty''.
This error penalty parametrizes the additional uncertainty induced by subtracting the foreground emission.
In a second step, we perform a $\chi^2$-analysis, according to eq. (\ref{formula:chiSqTest}), where we fit both a tSZ-signal and photon-ALP conversion to the simulated signal.
We thus obtain an upper limit on the sensitivity for the coupling constant $g$.
\\
We use the temperature sensitivities described in \cite{planck2010LFI} and \cite{planck2010HFI}, where the sensitivities range between $\Delta T/T \simeq 2.2\cdot 10^{-6}$ for $\nu = 143$ GHz and $\Delta T/T \simeq 6 \cdot 10^{-3}$ for $\nu = 857$ GHz.
In figure (\ref{fig:Future}), we show the possible limits with error penalties of 1, 5, and 10, where the parameters of the Coma cluster have been used and we averaged the limits obtained from ten different simulated realizations.
\\
One also can extent this approach to proposed future experiments for highly sensitive CMB observation like PRISM \cite{PRISM} or PIXIE \cite{PIXIE}.
PRISM is proposed to have 32 broad-range frequency channels as well as 300 narrow frequency channels covering the range from 30 GHz to 6000 GHz.
The simulated 4-year sensitivities range from
$\delta I_\nu$ = 3.6$\times10^{-27}$ Wm$^{-2}$Hz$^{-1}$sr$^{-1}$ for frequencies between 30 GHz and 180 GHz
to $\delta I_\nu$ = 1.6$\times10^{-26}$ Wm$^{-2}$Hz$^{-1}$sr$^{-1}$ for frequencies greater than 3 THz.
Performing the same analysis as before, we arrive at sensitivities of $\gtrsim 10^{-14}$GeV$^{-1}$, two orders of magnitude below the currently strongest limits for this mass range.
This result is also displayed in figure (\ref{fig:Future}).
PIXIE is proposed to have 400 channels covering the same frequency range as PRISM, reaching slightly worse sensitivities than PRISM.
We therefore expect PIXIE to be sensitive to values of $g$ very similar to the ones presented for PRISM.
\begin{figure}[t]
\centering
{
\includegraphics[width=15cm]{figure5.pdf}
}
\caption{
Projected sensitivity of the presented approach when using the full \textit{Planck} data or data from a future, PRISM-like experiment.
The sensitivity of \textit{Planck} with no error penalty is shown with the solid red line, while the dashed black line corresponds to a PRISM-like experiment.
The dashed-dotted lines are obtained when an error penalty of 5 is used, the dotted lines for error penalties of 10.
The shaded area is excluded by limits obtained from SN1987A.
For this plot, the parameters of the Coma cluster have been used.
}
\label{fig:Future}
\end{figure}
\section{Discussion and Conclusions}
In this study, we have shown that \textit{Planck's} recent measurements of the tSZ Compton-parameter $y$ \cite{planck2015SZ} can be used to constrain the coupling constant $g$ of pseudoscalar ALPs to photons.
To this end, we compared the temperature change due to the tSZ-effect of the CMB photons reaching us from galaxy clusters with the temperature change due to resonant photon-ALP conversion.
Photon-ALP conversion is most effective at resonance where it leads to the strongest limits on the coupling constant.
On the other hand, resonant photon-ALP conversion is only possible for a limited range of ALP masses, typically of the order of 10$^{-13}$ eV; this range depends on the effective photon mass in the galaxy clusters, and therefore on the free electron density.
\\
The strength of the obtained limits depends both on the density profile in the galaxy cluster as well as on the magnetic field.
In our study, we used a $\beta$-model, extended with density modulations, for these profiles, and typical values for the magnetic field strength, electron densities and the observed $y$-parameter for galaxy clusters.
Under these assumptions, we can derive limits on the photon-ALP coupling constant $g$, which are slightly stronger than the existing bounds in this mass region from SN1987A \cite{sn1987:3}, and furthermore provide an independent constraint.
\\
The scaling of the magnetic field with the electron density has a strong influence on the limits for smaller ALP masses.
Further investigations of the magnetic field strength in the outer regions of galaxy clusters or the selection of suitable clusters will be necessary to solve this problem and might provide stronger limits.
Another approach might be to consider the magnetized jets emitted by AGNs.
\\
We estimated the parameter region a PRISM-like experiment would be sensitive to and found that for the mass range given by the resonance condition, sensitivities down to $g \gtrsim 10^{-14}$ GeV$^{-1}$ are realistic.
This is especially interesting, as this is part of the parameter space has been invoked to explain the soft X-ray excess of the Coma cluster \cite{Angus:2014} or the anomalous gamma-ray transparency \cite{Meyer:2013}.
\section*{Acknowledgements}
We would like to thank Anne-Christine Davis, Alexandre Payez and Andreas Ringwald for useful comments.
This work was supported by the Deutsche Forschungsgemeinschaft (DFG) through the Collaborative Research Centre SFB 676 ``Particles, Strings and the Early Universe''.
Furthermore, we acknowledge support from the Helmholtz Alliance for Astroparticle Physics (HAP) funded by the Initiative and Networking Fund of the Helmholtz Association.
|
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"redpajama_set_name": "RedPajamaArXiv"
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| 2,794
|
Product building
We know product building.
A product is a product and a project is a project, and ne'er the twain shall meet.
If a software system is not designed from Day 1 as a product, you can't retro-fit features and engineering to it to make it one. Most of the custom software applications built for Wall Street giants are just in-house projects. They will never be products.
We know how to make a product (i) extensible and maintainable, (ii) scalable for high loads, (iii) work across multiple platforms, and (iv) look good. And we know how to manage the nail-biting criticality of release management whenever a new version is released.
We can build your product for you, run it, and maintain it.
We built this product for our internal business division, with goals of uber-scalability and high reliability. This product can send millions of emails a day. Its architecture permits scale-out scalability and one cluster can have 50+ servers, all managed from a single control, pushing out 100,000+ emails per hour.
We built this in a mixture of Perl and Java, and extensively used email sub-systems already available on Unix. The core MTA is Sendmail. Broadside operates as a SaaS service out of data centres in Singapore.
Kvadrato
This was a start-up, building a messaging and collaboration app with some nice new wrinkles. A sort-of more business-like and structured upgrade from WhatsApp.
We built the app using Ionic and Cordova, and it works on iPhones and Android from the first release. We used CouchDB and PouchDB for the data synchronisation and message passing, thus eliminating a lot of the overheads of writing and using web service calls. The server-side code therefore has become extremely low maintenance.
People are now setting up tasks, notices, sharing documents, all collaboratively. The world is about to become a more friendly place. Such collaboration!
Bitlogg
This was built initially for one of our enterprise clients, a diversified financial services group. This product runs on Linux servers and provides a scalable, high throughput log aggregation and forensic analysis platform. It receives log messages from all the custom business applications of the client, across various units, and allows inspection at a centralised repository.
The product was built primarily in PHP, on Linux. It uses the LKE stack, with LogStash accepting log messages from various components in their data centre, massaging the format to convert to the one true internal structure. It is then stored in ElasticSearch, and a modified version of Kibana allows access controlled browsing of the repository.
This product was for a start-up based in Singapore and the US. Their aim was to pay retail users for customer acquisition by passing special offers to them. The user would accept the offer for a trial, for a test drive, for a free sample, and would get paid hard cash from the advertiser.
This product was mobile-first, and used a browser interface only for administration and back-office workers. We built the app using Ionic, and the server-side code was in PHP, with a PostgreSQL database. There were stringent internationalisation requirements, which we met. Sweet!
Keurix
We built this for a Dutch client, who are in the business of inspections of civil constructions. Their site inspectors carry large blueprints and checklists into a construction site, make detailed inspections, make notes on a notepad (with a pen!) and submit reports to the construction company.
We helped them automate the entire on-site activity onto an iPad. Huge PDF blueprints of 1GB+ in size are supported, as are totally configurable checklists. Inspectors can tap a specific point in a blueprint and raise a red flag. Issues are later tabulated and tracked for closure. It all works wonderfully. Buildings in the Netherlands are now safer and defect-free thanks to Keurix.
ECS for KRCL
Konkan Railway Corporation is a govt owned railway service, with 70+ railway stations and 736 km of tracks along Indian western coast. They needed an office productivity and collaboration suite.
We built the Enterprise Collaboration Suite for them, based on our internal Servaya platform. This platform provided email services, controlled and secure web browsing, malware filtering, an Intranet, file servers, and various other features. It ran on multiple servers spread over three data centres, all controlled from one administrative console. It serves 3,000 users reliably and has scalability to grow with KRCL.
|
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"redpajama_set_name": "RedPajamaCommonCrawl"
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| 6,436
|
{"url":"https:\/\/mathoverflow.net\/questions\/313600\/explicit-left-adjoint-to-forgetful-functor-from-cartesian-to-symmetric-monoidal","text":"# Explicit Left Adjoint to Forgetful Functor from Cartesian to Symmetric Monoidal Categories\n\nThere is a forgetful functor from the category of (small) Cartesian monoidal categories (a symmetric monoidal category in which the tensor product is given by the categorical product) to the category of symmetric monoidal categories. This functor is known to have a left adjoint and right adjoint. The right adjoint is easy to describe: it takes a symmetric monoidal category to its category of cocommutative coalgebras. I am looking for an explicit description of the left adjoint though. Does anyone know what it is?\n\nThere is a MO question about this, but it does not give an explicit description: Cartesian envelope of a symmetric monoidal category\n\nI am also interested in the left adjoint of the forgetful functor with domain semi-Cartesian monoidal categories, which are symmetric monoidal categories in which the tensor unit is terminal.\n\nAnother way to look at this is from the point of view of the \"commutative algebra of categories,\" as described (in the quasicategorical case) here: https:\/\/arxiv.org\/abs\/1606.05606. Using that machinery, i.e. knowing that Cartesian monoidal categories are precisely the symmetric monoidal categories which are modules over the solid semi-ring category $$Fin^{op}$$, it would suffice to have an explicit description of the tensor product on (presentable) symmetric monoidal categories (that recovers the structure described in the above reference). So maybe this is known?\n\n\u2022 Wouldn\u2019t it be \u201cinduction\u201d to the semiring $Fin^{op}$? I.e. tensor with $Fin^{op}$? Oct 24, 2018 at 0:52\n\u2022 @NoahSnyder maybe so! I guess I don't know what induction would be here. To me, I'd think it'd be something like $C\\otimes_{Fin^{iso}}Fin^{op}$? But I'm not sure how this kind of thing works. Oct 24, 2018 at 5:05\n\u2022 @NoahSnyder Ah I think maybe I misread your comment. Yeah, I agree that it's tensor with $Fin^{op}$, but what is that tensor? I don't think it's just the usual tensor of categories is it? It's some complicated tensor product of $Fin^{iso}$-modules? Oct 24, 2018 at 16:24\n\u2022 Where did the word \"presentable\" sneak in? Are you asking about locally presentable categories or about arbitrary ones? It seems to me that the answers might be different. In particular, I believe the tensor product of locally presentable categories does have a somewhat explicit description, but I don't know of one in the general case. Oct 25, 2018 at 20:18\n\u2022 @MikeShulman I suppose I was just going off of the fact that \"presentability\" (by which I mean local presentability, since the term has been sadly mucked up) is one of the requirements on the categories addressed in the Berman paper cited in the question. Oct 25, 2018 at 22:15\n\nI don't know the answer, although I've thought about this quite a bit. I think the best that can be said in general is that the free cartesian monoidal category on $$\\mathcal{C}^\\otimes$$ is a full subcategory of $$\\text{Fun}^\\otimes(\\mathcal{C}^\\otimes,\\text{Set}^\\times)^\\text{op}$$, which is the category of symmetric monoidal functors. That is Theorem 3.10 of https:\/\/arxiv.org\/abs\/1606.05606.\n\nThere is some hope of giving a more explicit description if we know something about maps into tensor products. Here are three examples. (I would love to understand what they have in common.)\n\nBaby example: If $$\\mathcal{C}^\\times$$ is already cartesian monoidal, then $$\\mathcal{C}_{\/Y\\times Z}\\cong\\mathcal{C}_{\/X}\\times_\\mathcal{C}\\mathcal{C}_{\/Y}$$. That is, a map $$X\\rightarrow Y\\times Z$$ corresponds to two maps $$X\\rightarrow Y$$ and $$X\\rightarrow Z$$. In this case, the free cartesian monoidal category on $$\\mathcal{C}$$ is $$\\mathcal{C}$$ itself.\n\nExample 2: If $$\\mathcal{C}^\\amalg$$ is cocartesian monoidal, we may ask whether the coproduct behaves like a disjoint union (that is, $$\\mathcal{C}$$ is disjunctive) in the following sense: any map $$X\\rightarrow Y\\amalg Z$$ decomposes canonically as the coproduct of two maps $$X_Y\\rightarrow Y$$ and $$X_Z\\rightarrow Z$$. That is, $$\\mathcal{C}_{\/Y\\amalg Z}\\cong\\mathcal{C}_{\/Y}\\times\\mathcal{C}_{\/Z}$$. In this case, I believe that the free cartesian monoidal category can be described as a category of spans, something like the following:\n\nObjects are just objects of $$\\mathcal{C}$$. Morphisms from $$X$$ to $$Y$$ are spans $$X\\leftarrow T\\rightarrow Y$$, such that $$T$$ admits a decomposition $$T_1\\amalg\\cdots\\amalg T_n$$ with each $$T_i\\rightarrow X$$ an inclusion of a direct summand; that is, $$T_i\\amalg T_i^\\prime\\cong X$$.\n\nFor example, when $$\\mathcal{C}=\\text{Fin}$$, the category of finite sets, the free cartesian monoidal category is spans of finite sets. The $$\\mathcal{C}=\\text{Fin}$$ case is a main result of the paper cited above (at the level of $$\\infty$$-categories). The general case is not written down anywhere, as far as I know.\n\nExample 3: Given an operad $$\\mathcal{O}$$, there is a category $$[\\mathcal{O}^\\otimes]$$, whose objects are finite sets. A map from $$S$$ to $$T$$ is a function $$f:S\\rightarrow T$$ and an $$|f^{-1}(t)|$$-ary operation for each $$t\\in T$$. Concatenation gives $$[\\mathcal{O}^\\otimes]$$ a symmetric monoidal structure.\n\nNote that $$[\\mathcal{O}^\\otimes]$$ is disjunctive (in the sense of Example 2) but not cocartesian monoidal. In this case, the free cartesian monoidal category is the full subcategory of $$\\text{Fun}^\\otimes([\\mathcal{O}^\\otimes],\\text{Set}^\\times)^\\text{op}=\\text{Alg}_\\mathcal{O}(\\text{Set})^\\text{op}$$ spanned by the free $$\\mathcal{O}$$-algebras. Classically, the free $$\\mathcal{O}$$-algebra on $$n$$ generators is $$\\coprod_{k\\geq 0}\\mathcal{O}(k)\\times_{\\Sigma_k}n^k.$$ That effectively computes the free cartesian monoidal category on $$[\\mathcal{O}^\\otimes]$$ as a type of span construction, similar to Example 2. See the paper above, Remark 5.1 (also Example 3.16). The span construction can be made precise without much difficulty for ordinary categories, but it is an open problem for $$\\infty$$-categories.","date":"2023-02-06 03:15:10","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\": 43, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9603413939476013, \"perplexity\": 174.8089195719994}, \"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-2023-06\/segments\/1674764500303.56\/warc\/CC-MAIN-20230206015710-20230206045710-00825.warc.gz\"}"}
| null | null |
Where Will Amazon Move its New Headquarters (HQ2)?
By Carmello on December 20, 2017
In September, Amazon announced plans to create a second headquarters, to be located outside the Seattle region which hosts the current headquarters. The Amazon site states that the company is looking for a locale with strong local talent and " . . . a stable and business-friendly environment to continue hiring and innovating on behalf of our customers."
At stake are five billion dollars in direct investments and over fifty thousand high-paying jobs. Amazon estimates that it has indirectly contributed $55 billion to the Seattle economy with forty thousand plus employees; the host city of HQ2 could benefit even more. No wonder that cities as diverse and prominent as Atlanta, Pittsburgh, and Toronto are leading the pack of 238 proposals.
The International Factor
The company web site has specified only that the new headquarters will be located in North America. That leaves open the possibility that the location will be in Canada. In favor of that: A non-US headquarters would signal that Amazon intends to move more prominently onto the global stage.
But doesn't it already have a global footprint? Yes, but the ratio of Amazon's North American revenue to World revenue increased from 1.51 in 2014 to 1.82 in 2016. Granted that this isn't really a problem, considering that overall revenue increased from $89 billion to $136 billion over the same period, but it hints that Amazon is missing global opportunities for parallel expansion. The question then is whether global opportunities are being lost because of a domestic focus. To expand globally one must think globally, and perhaps the thinking of Amazon's top executive team is that an international perspective might be better gained by having the new headquarters located outside of the US.
The National Factor
Meanwhile, what Amazon needs now is to grapple with its rapidly expanding US operations. Seattle is already bursting at the seams with traffic congestion and skyrocketing housing prices. For domestic operations, Amazon needs to expand from Seattle but still stay in the US.
A central location within the country would make geographical sense. Atlanta might then be selected, as it is an airline hub with direct connections to most cities in the US.
The Texas Factor
Texas is more centrally located, however, and has plenty of room for a company to expand. Dallas, within the fourth largest Metropolitan Statistical Area in the US, has a national reputation for the 'business-friendly environment' that Amazon seeks.
Texas may also benefit from a personal factor. In his youth, Amazon CEO Jeff Bezos lived in Houston, spent summers on his grandfather's ranch in South Texas, and the test bed facility for Blue Origin, his rocket company, is located in West Texas.
The Negotiation Factor
For the beneficence of bringing fifty thousand jobs and tens of billions of dollars in investments to the host city, Amazon just might appreciate something in return. Therefore, the city that becomes the new headquarters will do so by making the best offer. Tax breaks and supporting infrastructure development are likely to be among the negotiated factors that will affect Amazon's final decision.
Whatever other factors are in its favor as a potential location for the new headquarters, a city will stand above the pack by its willingness to work with Amazon.
|
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| 5,140
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Q: Homeomorphisms between circles and rectangles In our topology class we learn that in $\mathbb{R}^2,$ circles and rectangles are homeomorphic to each others.
I can understand the underline idea intuitively.
But can we find an explicit homeomorphic between them?
If so how?
Also our professor said that, "we can describe any point in the rectangle $[0,1]\times[0,1]$ using a single coordinate."
I wonder how such thing is possible.
As I think, for this we need a bijection between $[0,1]\times[0,1]$ and some (closed?) interval in $\mathbb{R}.$
Can some one explain this phenomena?
A: One very visual way of seeing such a homeomorphism is to place one inside the other, choose a point that's inside both (say point $P$), and let the image of a point $Q$ in the circle be where the line emanating from $P$ and through $Q$ intersects the rectangle.
EDIT: Added an example image:
A: consider $S^1$ parametrized by $(\cos(t),\sin(t))$. Extend this vector from the origin to its first intersection with the square. You can explicate the first quadrant and then just argue by symmetry. For $0 \leq t \leq \pi/4$, you're going to intersect with the $x=1$ line segment, and for $\pi/4 \leq t \leq \pi/2$, the vector will intersect $y=1$ at $\cot(t)$.
However, the real idea is given in the answer by Fimpellizieri.
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Q: ServiceName is getting ignored while connecting to HA enabled HDFS from Java using RPC Here is my build config method
private Configuration buildConfiguration() {
Configuration conf = new Configuration();
if (connectivityDetail.isSecureMode()) {
conf.set("hadoop.security.authentication", "kerberos");
conf.set("hadoop.http.authentication.type", "kerberos");
conf.set("dfs.namenode.kerberos.principal", connectivityDetail.getHdfsServicePrincipal());
}
if (isHAEnabled()) {
String hdfsServiceName = connectivityDetail.getHdfsServiceName();
conf.set("fs.defaultFS", "hdfs://" + hdfsServiceName);
conf.set("dfs.ha.namenodes." + hdfsServiceName, "nn0,nn1");
conf.set("dfs.nameservices", hdfsServiceName);
conf.set("dfs.client.failover.proxy.provider." + hdfsServiceName,
"org.apache.hadoop.hdfs.server.namenode.ha.ConfiguredFailoverProxyProvider");
conf.set("dfs.namenode.rpc-address." + hdfsServiceName + ".nn1",
connectivityDetail.getNameNodeUris().get(0));
conf.set("dfs.namenode.rpc-address." + hdfsServiceName + ".nn0",
connectivityDetail.getNameNodeUris().get(1));
}
return conf;
}
hdfsService name is incorrectly set into the configuration object but I am able to get FileSystem and everything is working correctly. I am not sure why it is not using service name?
This is how I am creating Path
public static Path getHdfsResourceLocation(String resourceLocation) throws Exception {
String[] hdfsURIs = OrchestrationConfigUtil.getHdfsUri();
Path hdfsResourceLoc = null;
if (isHAEnabled()) {
hdfsResourceLoc = new Path(resourceLocation);
} else {
hdfsResourceLoc = FileContext.getFileContext(new URI(hdfsURIs[0])).makeQualified(new Path(resourceLocation));
}
return hdfsResourceLoc;
}
Everything is working fine with wrong service name and I am not sure why?
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,798
|
"This is no dry academic treatise. Coupled with superb photography by Bill Wassman, Cummings's learned but accessible book enriches our knowledge far beyond such a facile purpose."
Buddhist Stupas in Asia: The Shape of Perfection is a lavish exploration of one of the oldest and most persistent religious and architectural icons in the world. Transcending architecture and archaeology, the stupa fuses the physical with the spiritual into a living embodiment of Buddhist teachings. This beautifully illustrated, full-colour hardback explores the spread of stupa building across India and Asia, encapsulating the lasting appeal and allure of stupas to travellers, scholars and those interested in architecture and religion.
"This beautiful book takes us around some of the millions of stupas that transform our planet... It is a pleasure and a privilege to wander through its pages from place to place, experiencing the faith and determination of so many individuals in so many civilisations who have shown their aspiration and vision by creatively and patiently placing one stone upon another in the myriad shapes of perfection."
"'Sermons in stone, brick and mortar', one Buddhist scholar has called them. Diagrams of synchronicity in which 'time is fixed, crystallised, rendered static', claims a Western architect. Attempts to encapsulate the durable appeal of the stupa are confounded by the vast and varied history of stupa-building and stupa-worshipping, all but silencing the would-be inquisitor. Yet the stupa continues to inspire us to take photographs, sketch diagrams and write thousands of words, digging to uncover a key that might unlock its mysteries."
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,683
|
1-5/8" 16-gauge 409 stainless steel mandrel-bent tubing for unrestricted flow.
Optimized tube lengths improve flow and performance.
True 2 into 1 mitered merge collector eliminates dead zone for improved flow.
In testing, resulted in an astounding +21 horsepower and +26 lbs. x ft. of torque when combined as verified by our in house dyno.
Savings when purchased as a package!
A direct bolt-on installation which requires no cutting or drilling to provide a hassle-free installation.
aFe POWER headers and Y-pipe's are engineered and manufactured in our Corona, California manufacturing facility.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,751
|
Il Ganges Chasma è una valle presente sulla superficie di Marte, nella parte orientale della regione delle Valles Marineris, il più vasto sistema di canyon del Pianeta rosso.
Si tratta di una valle laterale dell'Eos Chasma, uno dei canyon principali della regione. Il suo letto è ricoperto per la maggior parte da depositi alluvionali provenienti dalle pareti della valle, che sembrano essere collassate e precipitate sul fondo in numerosi punti, come è stato possibile individuare grazie alle fotografie inviate a Terra dalla sonda Mars Global Surveyor nel 2001.
Altri progetti
Collegamenti esterni
Chasmata di Marte
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8
|
The mission got off to a rocky start; Captain S'Ranna of the USS Atlantis was accused by the Korri of the destruction of one of their ships and was escorted to Deep Space 26 in disgrace, with Commander Brell in temporary command. Meanwhile, the USS Blackwell was on a diplomatic assignment. Rear Admiral Renos was accompanied by Lieutenant Pandorn and Lucia Maria Aquilina to Burellion, for a diplomatic function.
While the Admiral and escorts were wining and dining with diplomats and corporate executives, the Blackwell decided to do a bit of exploring. They set off into an asteroid field that was unnaturally dense and, in true Blackwell fashion, were forced to eject their warp core. The Blackwell, now in critical condition, was towed to safety by the mysterious CSS Tempest. Who was the CSS Tempest and why did they appear just at the opportune moment? Only time would tell.
As if that wasn't enough, at a staff meeting to discuss the state of the USS Blackwell, the crew learned of a virus afflicting the members of the crew with telepathic abilities, including the possibility of Captain Zaekia being affected. Commander R'Ven, Lieutenant Junior Grade G'renn and Councilor Didrik Stennes were tasked with identifying and curing this malady before it drives the telepaths out of their minds. And as a precaution, on the suggestion of Councilor Didrik Stennes, Captain Zaekia transferred command of the USS Blackwell to Commander R'Ven.
Also, there is a possibility of a computer virus being transferred from the USS Atlantis to the USS Blackwell after the incident with the robots on the USS Atlantis which may have caused helm control failure, leading to the almost total destruction of the USS Blackwell. Can they isolate and eradicate the virus before it does anymore damage? Time will be of the essence.
Rear Admiral Renos, not to be dissuaded from investigating The Consortium, will return to the planet with Lieutenant Pandorn, Petty Officer Aquilina and Ensign Sarjak to meet with the mysterious Negrid, who Admiral Renos believes has some information on The Consortium that can shed light on the corporation's activities. What will they uncover?
A device was found near the Blackwell's deflector dish and Dr. Kaylessa Yesna, Lt. Commander Randal Shayne, Lt. Jarred Thoran and civilian engineer Charlotte Farnsworth went on an Extravehicular mission to remove said device. The device was found to be attempting to transmit a virus code to the Blackwell and the quartet prepared to remove the device intact,if possible.
Meanwhile on the Blackwell, the holodeck was being converted to a medical lab to find a cure for the virus afflicting the telepathic members of the crew. CMO Lieutenant Junior Grade G'renn, Councilor Didrik Stennes and Acting CO Commander R'Ven began to screen the crew by species. Will they be successful?
With the repairs to the Blackwell just about complete, the crew prepares to get back underway towards the Valcarian front and their primary mission. Lieutenant JG Pandorn and Petty Officer Aquilina were tasked with pulling whatever useful data they can find from the data Negrid passed on to them. What they will find will take quite a while, as they decided to manually go through the data as to protect the Blackwell from any viruses that may be lurking within. New crewmembers Ensign Tya Dirsye and Ensign Asha Sodhi were assigned to the Blackwell during their time in space dock.
The telepathic virus, which has yet to be named, seems to have been cured by the medical staff, led by CMO Lieutenant Junior Grade G'renn, Councilor Didrik Stennes and Acting CO Commander R'Ven, much to the relief of Captain Zaekia. The cure will be dispersed shipwide through the ventilation system. The consensus is that it should not harm the non-telepaths among the crew.
The device that was planted on the Blackwell's deflector dish, one that was transmitting a virus code, was to be removed via transporter by newly assigned engineer, Dr. Kaylessa Yesna, Lt. Commander Randal Shayne, Lt. Jarred Thoran and civilian engineer Charlotte Farnsworth, but was destroyed during transport by some unknown reason.
The crew prepares from some needed shore leave, the highlight of which is the impending wedding of Lt. Commander Nathaniel Wilmer and Charlotte Farnsworth with Admiral Renos officiating.
Lieutenant JG Pandorn and Petty Officer Aquilina may have found a tenuous connection between Allita Systems and Taventa Robotics in the destruction of the Korri vessel in the minefield. An even more tenuous connection between the Consortium and Taventa Robotics was uncovered as well. Will further investigation bear fruit on this news? Will this lead to the exoneration of two Starfleet officers? What will the two intrepid investigators find?
More crew join the USS Blackwell during this time. Ensigns Zoe Cade, Na'Lae Mandak and Sakur arrived on the Freeworlds civilian shuttle along with Lt. Commander Cayden Aydr ready to join the USS Blackwell on its primary mission.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 3,942
|
11111...11111 (1027-digits)
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 7 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
The smallest titanic hexagonal-congruent prime. [Hartley]
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,864
|
import bpy
import logging
from . import BlendManager
def get_property(obj, name):
return obj.game.properties[name]
def create_property(obj, name, type_="STRING"):
bpy.context.scene.objects.active = obj
bpy.ops.object.game_property_new(type=type_, name=name)
return obj.game.properties[name]
def remove_property(obj, name):
bpy.context.scene.objects.active = obj
prop_index = obj.game.properties.find(name)
if prop_index == -1:
raise KeyError("Couldn't find property called {}".format(name))
bpy.ops.object.game_property_remove(index=prop_index)
def get_or_create_property(obj, name, type_="STRING"):
try:
prop = get_property(obj, name)
except KeyError:
prop = create_property(obj, name, type_)
return prop
def ui_hivemap_set(obj, context):
nodetree_name = obj.hive_nodetree
if not nodetree_name:
try:
remove_property(obj, "hivemap")
finally:
return
blend_manager = BlendManager.blendmanager
try:
nodetree_manager = blend_manager.get_nodetree_manager(nodetree_name)
except KeyError:
logging.warning("Couldn't find nodetree for {} called {}".format(obj.name, nodetree_name))
return
textblock_name = blend_manager.get_textblock_name(nodetree_manager)
hivemap_prop = get_or_create_property(obj, "hivemap")
hivemap_prop.value = textblock_name
hivemap_prop.show_debug = 1
_object = None
@bpy.app.handlers.persistent
def tree_switcher(dummy):
scene = bpy.context.scene
if scene is None:
return
if not scene.switch_to_bound_hive_tree:
return
active_object = bpy.context.scene.objects.active
global _object
if active_object == _object or active_object is None:
return
node_tree_name = active_object.hive_nodetree
try:
node_tree = bpy.data.node_groups[node_tree_name]
except KeyError:
return
blend_manager = BlendManager.blendmanager
try:
nodetree_manager = blend_manager.get_nodetree_manager(node_tree.name)
except KeyError:
return
for space in (sp for s in bpy.data.screens for a in s.areas for sp in a.spaces if sp.type == "NODE_EDITOR"):
if space.tree_type != nodetree_manager.tree_bl_idname:
continue
space.node_tree = node_tree
logging.info("Switched to {} {}".format(node_tree_name, node_tree))
_object = active_object
class HivemapSelectionPanel(bpy.types.Panel):
bl_label = "Hive Logic"
bl_idname = "OBJECT_PT_layout"
bl_space_type = 'PROPERTIES'
bl_region_type = 'WINDOW'
bl_context = "object"
COMPAT_ENGINES = {'BLENDER_GAME'}
@classmethod
def on_registered(cls):
# Template list settings
bpy.types.Object.hive_nodetree = bpy.props.StringProperty("", description="NodeTree to bind to",
update=ui_hivemap_set)
BlendManager.blendmanager.on_renamed.append(cls.on_renamed)
BlendManager.blendmanager.on_removed.append(cls.on_removed)
@classmethod
def on_unregistered(cls):
BlendManager.blendmanager.on_renamed.remove(cls.on_renamed)
BlendManager.blendmanager.on_removed.remove(cls.on_removed)
@classmethod
def poll(cls, context):
return context.object is not None
@classmethod
def on_renamed(cls, old_name, new_name):
for obj in bpy.context.scene.objects:
if obj.hive_nodetree != old_name:
continue
obj.hive_nodetree = new_name
@classmethod
def on_removed(cls, name):
for obj in bpy.context.scene.objects:
if obj.hive_nodetree != name:
continue
obj.hive_nodetree = ""
def draw(self, context):
layout = self.layout
scene = context.scene
ob = context.active_object
game = ob.game
row = layout.row()
row.label("Bound NodeTree")
row = layout.row()
row.prop_search(ob, "hive_nodetree", bpy.data, "node_groups", text="")
row = layout.row()
row.label("Properties")
box = layout.box()
is_font = (ob.type == 'FONT')
if is_font:
prop_index = game.properties.find("Text")
if prop_index != -1:
box.operator("object.game_property_remove", text="Remove Text Game Property", icon='X').index = prop_index
row = box.row()
sub = row.row()
sub.enabled = 0
prop = game.properties[prop_index]
sub.prop(prop, "name", text="")
row.prop(prop, "type", text="")
# get the property from the body, not the game property
# note, don't do this - it's too slow and body can potentially be a really long string.
#~ row.prop(ob.data, "body", text="")
row.label("See Text Object")
else:
props = box.operator("object.game_property_new", text="Add Text Game Property", icon='ZOOMIN')
props.name = 'Text'
props.type = 'STRING'
props = box.operator("object.game_property_new", text="Add Game Property", icon='ZOOMIN')
props.name = ''
for i, prop in enumerate(game.properties):
if is_font and i == prop_index:
continue
prop_box = box.box()
row = prop_box.row()
row.prop(prop, "name", text="")
row.prop(prop, "type", text="")
row.prop(prop, "value", text="")
row.prop(prop, "show_debug", text="", toggle=True, icon='INFO')
row.operator("object.game_property_remove", text="", icon='X', emboss=False).index = i
def register():
bpy.utils.register_class(HivemapSelectionPanel)
HivemapSelectionPanel.on_registered()
bpy.app.handlers.scene_update_pre.append(tree_switcher)
def unregister():
bpy.utils.unregister_class(HivemapSelectionPanel)
HivemapSelectionPanel.on_unregistered()
bpy.app.handlers.scene_update_pre.remove(tree_switcher)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,318
|
Бактеріо́ліз — розчинення оболонки бактеріальної клітини і вихід вмісту цієї клітини у довкілля.
Бактеріоліз може бути спонтанним, або зумовленим неспецифічною дією ферментів, фізичних і хімічних факторів. Специфічний бактеріоліз викликається бактеріофагом або бактеріолізинами (специфічні антитіла, що утворюються в організмі людини і тварин при виробленні імунітету).
Література
Мікробіологія
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,288
|
\section{Introduction}
Recently, Transformer has made significant progress on vision tasks. Attempts to introduce the Transformer architecture into the vision community can be broadly classified into two types: some studies regard the Transformer structure as a powerful complement to CNNs, employing a hybrid architecture which combines the attention mechanisms with convolutional networks, attempting to exploit the advantages of both; the other studies, encouraged by Transformer's remarkable success in NLP tasks, devote to explorer a fully attentional model, believe that Transformer will defeat CNN structures in the near future, and attention mechanisms will be served as the fundamental building blocks of the next generation. Several hybrid architectures, such as \cite{dai2021coatnet}\cite{carion2020enddetr}, have rapidly reached state-of-the-art in a variety of tasks, indicating the great potential of the Transformer. In contrast, the fully attentional model did not go so well at the first time. The Vision Transformer\cite{dosovitskiy2020imagevit} (ViT, the first fully attentional model in vision tasks) and many of its successors\cite{deit} were inferior to convnets in terms of performance, until the appearance of the Swin-Transformer\cite{liu2021Swin}.
\begin{figure}
\centering
\includegraphics[width=\linewidth]{image/fig1}
\caption{Comparison with state-of-the-art trackers on LaSOT~\cite{Fan_2019_CVPR} using success (SUC) score and speed. Our SwinTrack-T with light architecture reaches a state-of-the-art performance with 0.667 SUC score and meanwhile runs the fastest at around 100 {\it fps}. Using larger model, SwinTrack-B-384 set a breakthrough new record with 0.702 SUC score and still runs efficiently at approximately 45 {\it fps}. {\it Best viewed in color and by zooming in.}}
\label{auc_fps_fig}
\end{figure}
Swin-Transformer employs a hierarchical window attention-based architecture to address two major challenges in the Transformer architecture: the variety of visual elements in scale and the high computational complexity on high-resolution images. Unlike the ViT family using a fixed-size feature map, Swin-Transformer builds the feature map by gradually merging neighbor patches from large to small. With hierarchical feature maps, traditional multi-scale prediction techniques can be used to overcome the scaling problem. Besides, Swin-Transformer introduces a non-overlapping window partition operation. Self-attention computing is limited within the window. As a result, the computational complexity is greatly reduced. Furthermore, the partition windows are shifted periodically to bridge the windows in preceding layers.
The advantages of Transformer are widely acknowledged to be due to two factors\cite{vaswani2017attention}: The Transformer is a sequence-to-sequence model, which makes it easier to combine multi-modal data, thus providing more flexibility in network architecture design; The capability of long-range modeling from the attention mechanism unleash the limitation of the traditional CNN-based or RNN-based model.
Visual object tracking is a challenging research topic with a long history. Many issues are still not well addressed, including relocation after occlusion or being out of vision, discrimination between similar objects, etc. \cite{TransT} and \cite{stark} are the most advanced trackers in the visual object tracking task. They both use a hybrid architecture, with ResNet serving as the backbone and Transformer serving as the encoder and decoder networks, as previously summarized. We believe that by fully utilizing the power of fully attentional model and the Swin-Transformer backbone, we can significantly boost up the tracker's performance to a new level.
Through the insight of the nature of the attention mechanism and a bunch of thorough experiments, we designed a powerful yet efficient fully attentional tracker - SwinTrack. SwinTrack suppresses the SOTA\cite{stark}\cite{keeptrack} trackers on the challenging long-term dataset LaSOT by 3.1\%, while still having an FPS at 45. We also provide a lighter version of SwinTrack, which provides a SOTA performance at 97 FPS.
The key designs of SwinTrack includes:
\begin{itemize}
\item Swin-Transformer as the backbone network;
\item Proper choices between various candidate network structures for different part of the tracker;
\item Introduce untied positional encoding to provide an accurate positional encoding for concatenation-based feature fusion;
\item Introduce IoU-Aware Classification Score to the classification prediction branch, to select an more accurate bounding box prediction.
\end{itemize}
We believe that SwinTrack has fully revealed the great potential of the Transformer network. We'd like to propose the SwinTrack model as a new baseline network for future research.
\section{Related Work}
\subsection{Transformer in Vision Tasks}
Transformer was first proposed by \cite{vaswani2017attention}, applied in the task of machine text translation. Due to significantly more parallelization and promising performance, Transformer rapidly replaced the LSTM model and soon achieved complete dominance in NLP tasks.
Starting from 2020, Transformer has been vastly introduced to the vision community. DETR\cite{carion2020enddetr} attracted a lot of attention. By modeling the object detection as a direct set prediction problem, DETR removes most hand-crafted processes and reaches a state-of-the-art comparable performance without domain knowledge. Later, the advancing model of DETR\cite{zhu2020deformabledetr} and many other transformer-based models were proposed to the image and video tasks.
The large-scale pre-trained models in NLP tasks have made a great success, such as the well-known BERT\cite{devlin2018bert} and the GPT family\cite{radford2018improvinggpt}. With the attempts to replicate the success, the Vision Transformer(ViT)\cite{dosovitskiy2020imagevit} was proposed. ViT splits the image into multiple fixed-size patches as the token, with a linear projection and a proper positional encoding. The image tokens are then fed into the standard Transformer encoder. With the success of the first applicable convolution-free network architecture and a vision of a shared pre-trained backbone network for CV and NLP tasks, a family of ViT variants was proposed\cite{touvron2021training}\cite{li2021localvit}\cite{chen2021crossvit}\cite{zhou2021deepvit}.
In standard ViTs, the number of tokens is fixed across the layers. To control the computation complexity and open the access to the multi-scale architecture in various vision tasks, multi-scale Vision Transformers with window-based attention were proposed, like \cite{wang2021pyramid}\cite{chu2021twins}\cite{liu2021Swin}. Swin Transformer\cite{liu2021Swin} may be the most famous one since it reached state-of-the-art in multiple tasks when it was first released.
\subsection{Siamese Tracking}
By offline learning a generic matching function from a large set of sequences, tracking is to search for a region that is the most similar to the target template. The Siamese methods formulate object tracking as a matching problem. Especially, the work of \cite{bertinetto2016siamfc} introduces a fully convolutional Siamese network for tracking and shows a good balance off between accuracy and speed. In order to improve \cite{bertinetto2016siamfc} in dealing with scale variation, the method of \cite{li2018highsiamrpn} incorporates the region proposal network into Siamese network and proposes the anchor-based tracker, achieving higher accuracy with faster speed. Later, numerous extensions have been presented to improve \cite{li2018highsiamrpn}, including deeper backbone network \cite{li2019evolutionsiamrpnp}, multi-stage architecture\cite{Fan_2019_CVPR,fan2020cract}, anchor-free Siamese trackers \cite{Ocean_2020_ECCV}.
\subsection{Transformer in Visual Tracking}
Several Transformer based trackers have been proposed. \cite{TransT} \cite{wang2021transformer} \cite{fu2021stmtrack} are the very first works that introduce the Transformer architecture to the visual object tracking. \cite{TransT} propose the ECA and CFA modules. The modules replace the traditional correlation operation with cross attention. \cite{wang2021transformer} improves the Siamese matching and DiMP based tracking frameworks by Transformer enhanced template features and search features. \cite{stark} explores the Spatio-temporal Transformer by integrating the model updating operations into a Transformer module.
\section{Swin Transformer Tracking}
\subsection{Overview}
\begin{figure*}
\centering
\includegraphics[width=\linewidth]{image/fig2}
\caption{Architecture of SwinTrack, which consists of three parts including Swin Transformer-based feature extraction, Transformer-based feature fusion and prediction head. Our SwinTrack is a simple and neat tracking framework without complex designs such as multi-scale feature and temporal update, yet demonstrating state-of-the-art performance.}
\label{overview}
\end{figure*}
Our tracker is based on the Siamese network architecture\cite{bromley1993signature}, as shown in \ref{overview}. Four main components comprise our fully attentional tracker: the Swin-Transformer backbone, the attentional encoder-decoder network, positional encoding, and the head network. During tracking, the backbone network extracts the features of the template image patch and the search region image patch separately with shared weights (for simplification, we call them the template image and the search image for convenience, respectively), the encoder network fuse the feature tokens from the \emph{template image} and the \emph{search image} by concatenation, and enhances the concatenated tokens layer-by-layer by attention mechanism, positional encoding helps the model to distinguish the tokens from the different source and the different position, the decoder network generates the final feature map of the search image and feeds it to the head network to obtain the IoU-Aware classification response map and bounding box estimation map. We will discuss the details of each component in the following sections.
\subsection{Transformer-based Feature Extraction}
The deep convolutional neural network has significantly improved the performance of trackers. Along with the advancement of trackers, the backbone network has evolved twice: AlexNet\cite{krizhevsky2012imagenet} and ResNet\cite{he2016deep}. Swin-Transformer\cite{liu2021Swin}, in comparison to ResNet, can give a more compact feature representation and richer semantic information to assist succeeding networks in better localizing the target objects, which we will demonstrate in the ablation study experimentally.
Our tracker follows the scheme of classic Siamese tracker\cite{bertinetto2016siamfc}, which requires a pair of image patches as the input, one is the template image patch $z\in\mathbb{R}^{H_z\times W_z\times 3}$, the other one is the search region image patch $x\in\mathbb{R}^{H_x\times W_x\times 3}$ (for simplification, we call them the \emph{template image} and the \emph{search image} for convenience respectively).
We denote the feature tokens from the template image as $\mathtt{z}\in\mathbb{R}^{\frac{H_z}{s}\frac{W_z}{s}\times C}$, the feature tokens from the search image as $\mathtt{x}\in\mathbb{R}^{\frac{H_x}{s}\frac{W_x}{s}\times C}$, $s$ is the stride of the backbone network. Since there is no dimension projection in our model, $C$ is also the hidden dimension of the whole model.
\subsection{Transformer-based Feature Fusion}
\vspace{0.2em}
\noindent
{\bf Encoder.} The encoder is composed of a sequence of blocks where each block contains a multi-head self-attention (MSA) module and a feed forward network (FFN). FFN contains a two-layers multi-layer perceptron (MLP), GELU activation layer is inserted after the first layer's output. Layer normalization (LN) is always performed before every module (MSA and FFN). Residual connection is applied on MSA and FFN modules.
Before the feature tokens are fed into the encoder, the tokens from the template image and the search image are concatenated along spatial dimensions to generate a union representation $U$. For each block, the MSA module computes self-attention over the union representation, FFN refines the feature tokens generated by MSA. When the tokens are getting out of the encoder, a de-concatenation operation is performed to recover the template image feature tokens and the search image feature tokens.
The full process can be expressed as:
\begin{equation}
\begin{split}
U^1 & = {\rm Concat}(\mathtt{z}^1,\mathtt{x}^1) \\
&\dots \\
U^{l'} & = U^l+{\rm MSA}({\rm LN}(U^l)) \\
U^{l+1} & = U^{l'}+{\rm FFN}({\rm LN}(U^{l'})) \\
&\dots \\
\mathtt{z}^{L},\mathtt{x}^{L} & = {\rm DeConcat}(U^L),
\end{split}
\end{equation}
where $l$ denotes the $l$-th layer and $L$ the number of blocks.
\vspace{0.2em}
\noindent
{\bf Why concatenated attention?} To simplify the description, we call the method described above \emph{concatenation-based fusion}. To fuse and process features from multiple branches, it is intuitive to perform self-attention on the feature tokens in each branch separately to complete the feature extraction step and then compute cross-attention across feature tokens from different branches to complete the feature fusion step. We call this method \emph{cross-attention-based fusion}. Considering that the Transformer is a sequence-to-sequence model, the Transformer can naturally accept multi-modal data as input. In comparison to \emph{cross-attention-based fusion}, \emph{concatenation-based fusion} can save computation operations through operation combination and reduce model parameters through weight sharing. From this perspective, \emph{concatenation-based fusion} implicitly implements the {\bf Siamese network architecture}. To ensure that the attention mechanism is aware of which branch the token currently being processed belongs to and its location within the branch, we must carefully design the model's positional encoding solution.
\vspace{0.2em}
\noindent
{\bf Why not window-based self/cross-attention?} Since we select stage 3 of the Swin-Transformer as the output, the number of tokens is small enough that the FLOPs between window-based attention and full attention are pretty similar. Furthermore, some tokens may need to go through multiple-layer before computing a correlation, which is too expensive for our tracker.
\vspace{0.2em}
\noindent
{\bf Decoder.} The decoder consists of a multi-head cross-attention($MCA$) module and a feed forward network(FFN). The decoder takes the outputs from the encoder as input, generating the final feature map $\mathbf{x}\in\mathbb{R}^{\frac{H_x}{s}\times \frac{W_x}{s}\times C}$ of the search image by computing cross-attention over $\mathtt{x}_{L}$ and $Concat(\mathtt{z}_L, \mathtt{x}_L)$. The decoder is very similar to a layer in the encoder, except that the correlation from the template image tokens to the search image tokens is dropped, since we do not need to update the features from the template image in the last layer.
We can formulate the process in the decoder by:
\begin{equation}
\begin{split}
U^D & = {\rm Concat}(\mathtt{z}^L,\mathtt{x}^L) \\
\mathtt{x}^{L'} & = \mathtt{x}^L+{\rm MCA}({\rm LN}(\mathtt{x}^L), {\rm LN}(U^D)) \\
\mathbf{x} & = \mathtt{x}^{L'}+{\rm FFN}({\rm LN}(\mathtt{x}^{L'})).
\end{split}
\end{equation}
\vspace{0.2em}
\noindent
{\bf Why not an end-to-end architecture?} Many Transformer-based models have an end-to-end architecture, which means that the model predicts the task's objective directly, without any post-processing steps. However, in our tests, an end-to-end model is still not applicable for our task. In our experiment, when applying a transformer-style decoder, like the one in \cite{carion2020enddetr} to directly predict the bounding box of the target object, the model takes a much longer time to converge and has an inferior tracking performance. The decoder we've chosen can help to improve the performance in three folds: By predicting a response map, we can offload the candidate selection task to the manually designed post-processing step. By dense prediction, we can feed a richer supervision signal to the model, which can fasten the training process. And also, we can use more domain knowledge to help improve the tracking performance, like applying a Hanning penalty window on the response map to introduce the smooth movement assumption.
\vspace{0.2em}
\noindent
{\bf Why not a target query-based decoder?} We also find that the traditional transformer decoder is hard to recover 2D positional information in our experiment.
\subsection{Positional Encoding}
Transformer requires a positional encoding to identify the position of the current processing token\cite{vaswani2017attention}. Through a series of comparison experiments, we choose \emph{untied positional encoding}, which is proposed in TUPE\cite{ke2021rethinking}, as the positional encoding solution of our tracker. In addition, we generalize the \emph{untied positional encoding} to arbitrary dimensions to fit with other components in our tracker.
The original transformer\cite{vaswani2017attention} proposes a absolute positional encoding method to represent the position: a fixed or learnable vector $p_i$ is assigned to each position $i$. Starting from the basic attention module, we have:
\begin{equation}
{\rm Atten}(Q,K,V)={\rm softmax} \Big(\frac{QK^T}{\sqrt{d_k}}V \Big),
\label{attn}
\end{equation}
where $Q$,$K$,$V$ are the $query$ vector, $key$ vector and $value$ vector, which are the parameters of the attention function, $d_k$ is the dimension of $key$. Introducing the linear projection matrix and multi-head attention to the attention module (\ref{attn}), we get the multi-head variant defined in \cite{vaswani2017attention}:
\begin{equation}
{\rm MultiHead}(Q,K,V) ={\rm Concat}({\rm head_1},...,{\rm head_h})W_O,
\end{equation}
where ${\rm head_i} = {\rm Atten}(QW^Q_i,KW^K_i,VW^V_i)$, $W^Q_i\in\mathbb{R}^{d_{{\rm model}}\times d_k}$, $W^K_i\in\mathbb{R}^{d_{{\rm model}}\times d_k}$, $W^V_i\in\mathbb{R}^{d_{{\rm model}}\times d_v}$, $W^O_i\in\mathbb{R}^{hd_v\times d_{\rm model}}$ and $h$ is the number of heads.
For simplicity, as in \cite{ke2021rethinking}, we assume that $d_k=d_v=d_{\rm model}$, and use the single-head version of self-attention module. Denoting the input sequence as $x={x_1, x_2, \dots, x_n}$, where $n$ is the length of sequence, $x_i$ is the $i$-th token in the input data. Denoting the output sequence as $z=(z_1, z_2, \dots, z_n)$. Self-attention module can be rewritten as
\begin{align}
z_i&=\sum_{j=1}^n\frac{{\rm exp}(\alpha_{ij})}{\sum_{j'=1}^n {\rm exp}(\alpha_{ij'})}(x_jW^V), \\
{\rm where}\ \alpha_{ij}&=\frac{1}{\sqrt{d}}(x_iW^Q)(x_jW^K)^T.\label{attn_mat_elem}
\end{align}
Obviously, the self-attention module is permutation-invariance. Thus it can not ``understand" the order of input tokens.
\vspace{0.2em}
\noindent
{\bf Untied absolute positional encoding.} By adding a learnable positional encoding\cite{vaswani2017attention} to the single-head self-attention module, we can obtain the following equation:
\begin{equation}
\begin{split}
\alpha_{ij}^{Abs}&=\frac{((w_i+p_i)W^Q)((w_j+p_j)W^K)^T}{\sqrt{d}} \\
&=\frac{(w_iW^Q)(w_jW^K)^T}{\sqrt{d}} + \frac{(w_iW^Q)(p_jW^K)^T}{\sqrt{d}} \\
&+\frac{(p_iW^Q)(w_jW^K)^T}{\sqrt{d}} + \frac{(p_iW^Q)(p_jW^K)^T}{\sqrt{d}}.\label{attn_mat_elem_with_pos}
\end{split}
\end{equation}
The equation (\ref{attn_mat_elem_with_pos}) is expanded into four terms: token-to-token, token-to-position, position-to-token, position-to-position. \cite{ke2021rethinking} discuss the problems exists in the equation and proposes the \emph{untied absolute positional encoding}, which unties the correlation between tokens and positions by removing the token-position correlation terms in equation (\ref{attn_mat_elem_with_pos}), and using an isolated pair of projection matrices $U^Q$ and $U^K$ to perform linear transformation upon positional embedding vector. The following is the new formula for obtaining $\alpha_{ij}$ using the \emph{untied absolute positional encoding} in the $l$-th layer:
\begin{equation}
\begin{split}
\alpha_{ij}&=\frac{1}{\sqrt{2d}}(x_i^lW^{Q,l})(x_j^lW^{K,l})^T \\
&+\frac{1}{\sqrt{2d}}(p_iU^Q)(p_jU^K)^T.\label{untied_abs_pos_enc}
\end{split}
\end{equation}
where $p_i$ and $p_j$ is the positional embedding at position $i$ and $j$ respectively, $U^Q\in \mathbb{R}^{d\times d}$ and $U^K\in \mathbb{R}^{d\times d}$ are learnable projection matrices for the positional embedding vector. When extending to the multi-head version, the positional embedding $p_i$ is shared across different heads, while $U^Q$ and $U^K$ are different for each head.
\vspace{0.2em}
\noindent
{\bf Relative positional bias.} According to \cite{shaw2018relpos}, relative positional encoding is a necessary supplement to absolute positional encoding. In~\cite{ke2021rethinking}, a relative positional encoding is applied by adding a relative positional bias to equation (\ref{untied_abs_pos_enc}):
\begin{equation}
\begin{split}
\alpha_{ij}&=\frac{1}{\sqrt{2d}}(x_i^lW^{Q,l})(x_j^lW^{K,l})^T \\
&+\frac{1}{\sqrt{2d}}(p_iU^Q)(p_jU^K)^T+b_{j-i},\label{untied_abs_pos_enc_with_rel}
\end{split}
\end{equation}
where for each $j-i$, $b_{j-i}$ is a learnable scalar. The \emph{relative positional bias} is also shared across layers. When extending to the multi-head version, $b_{j-i}$ is different for each head.
\vspace{0.2em}
\noindent
{\bf Generalize to multiple dimensions.} Before working with our tracker's encoder and decoder network, we need to extend the \emph{untied positional encoding} to a multidimensional version. One straightforward method is allocating a positional embedding matrix for every dimension and summing up all embedding vectors from different dimensions at the corresponding index to represent the final embedding vector. Together with \emph{relative positional bias}, for an $\mathtt{n}$-dimensional case, we have:
\begin{equation}
\begin{split}
\alpha_{\underbrace{ij\dots}_\mathtt{n}, \underbrace{\vphantom{j}mn\dots}_\mathtt{n}}&=\frac{1}{\sqrt{2d}}(x_{\underbrace{ij\dots}_\mathtt{n}}W^Q)(x_{\underbrace{\vphantom{j}mn\dots}_\mathtt{n}}W^K)^T \\
&\hspace{-13mm} +\frac{1}{\sqrt{2d}}[\underbrace{(p_i^1+p_j^2+\dots)}_\mathtt{n}U^Q]
[\underbrace{(p_m^1+p_n^2+\dots)}_\mathtt{n}U^K]^T \\
&\hspace{-13mm} +b_{\underbrace{m-i,n-j, \dots}_\mathtt{n}} ~.
\end{split}
\end{equation}
\vspace{0.2em}
\noindent
{\bf Generalize to concatenation-based fusion.} In order to work with \emph{concatenation-based fusion}, the \emph{untied absolute positional encoding} is also concatenated to match the real position, the indexing tuple of \emph{relative positional bias} now appends with a pair of indices to reflect the origination of $query$ and $key$ involved currently.
Taking $l$-th layer in the encoder as the example:
\begin{equation}
\begin{split}
\alpha_{ij,mn,g,h}&=\frac{1}{\sqrt{2d}}(x_{ij,g}^lW^{Q,l})(x_{mn,h}^lW^{K,l})^T \\
& \hspace{-12mm} +\frac{1}{\sqrt{2d}}[(p_{i,g}^1+p_{j,g}^2)U^Q_g]
[(p_{m,h}^1+p_{n,h}^2)U^K_h]^T \\
& \hspace{-12mm} +b_{m-i,n-j,g,h}~,
\end{split}
\end{equation}
where $g$ and $h$ are the index of the origination of $query$ and $key$ respectively, for instance, $1$ for the tokens from the template image, $2$ for the tokens from the search image. The form in the decoder is similar, except that $g$ is fixed. In our implementation, the parameters of \emph{untied positional encoding} are shared inside the encoder and the decoder, respectively.
\subsection{Head and Training Loss}
\noindent
{\bf Head.} The head network is split into two branches: classification branch and bounding box regression branch. Each branch is a three-layer perceptron. One is in charge of foreground-background classification. The other one is in charge of bounding box regression. They are both receiving the feature map $x\in \mathbb{R}^{(H_x\times W_x)\times C}$ from the decoder, predict the classification response map $r_{cls}\in\mathbb{R}^{(H_x\times W_x)\times 1}$ and bounding box regression map $r_{reg}\in\mathbb{R}^{(H_x\times W_x)\times 4}$, respectively.
\vspace{0.2em}
\noindent
{\bf Classification loss.} In classification branch, we employ the \emph{IoU-aware classification score} as the training target and the \emph{varifocal loss}\cite{zhang2020varifocalnet} as the training loss function.
IoU-aware design has been very popular recently, but most works task IoU prediction branch as an auxiliary branch to assist classification branch or bounding box regression branch. To remove the gap between different prediction branches, \cite{zhang2020varifocalnet} and \cite{li2020generalizedfocal} replace the classification target from ground-truth value, i.e., 1 for positive samples, 0 for negative samples, to the IoU between the predicted bounding box and the ground-truth one, which is named the \emph{IoU-aware classification score} (IACS). IACS can help the model select a more accurate bounding box from the candidate pool. Along with the IACS, the varifocal loss was proposed in \cite{zhang2020varifocalnet} to help the IACS approach outperform other IoU-aware designs. The \emph{varifocal loss} has the following form:
\begin{equation}
\rm{VFL}(p,q) \!=\!\left\{\!\!\!
\begin{array}{ll}
-q(qlog(p)+(1-q)log(1-p)) &q>0 \\
-\alpha p^\gamma log(1-p) &q=0,
\end{array}\right.
\end{equation}
where $p$ is the predicted IACS and $q$ is the target score. For positive samples, i.e., the foreground points, $q$ is the IoU between the predicted bounding box and the ground-truth bounding box. For negative samples, $q$ is $0$.
Then the classification loss can be formulated as:
\begin{equation}
\mathbb{L}_{cls}=\rm{VFL}(p, IoU(b, \hat{b})),
\end{equation}
where $b$ denotes the predicted bounding box, $\hat{b}$ denotes the ground-truth bounding box.
\vspace{0.5em}
\noindent
{\bf Regression loss.} For bounding box regression, we employ the generalized IoU loss\cite{GIoU}. The regression loss function can be formulated as:
\begin{equation}
\mathbb{L}_{reg}=\sum_j \mathbbm{1}_{\{q>0\}}[p\mathbb{L}_{\rm{GIoU}}(b_j,\hat{b})].
\end{equation}
The GIoU loss is weighted by $p$ to emphasize the high classification score samples. The training signals from the negative samples are ignored.
\section{Experiments}
\subsection{Implementation}
\noindent
{\bf Model.} We show three variants of SwinTrack with different configurations as follows:
\begin{itemize}
\setlength{\itemsep}{1pt}
\setlength{\parskip}{0pt}
\setlength{\parsep}{0pt}
\item {\bf SwinTrack-T}. \\
Backbone: Swin Transformer-Tiny~\cite{liu2021Swin}; \\
Template size: $[112\times112]$; Search region size: $[224\times224]$;
$C=384$; $N=4$;
\item {\bf SwinTrack-B}. \\
Backbone: Swin Transformer-Base~\cite{liu2021Swin};\\
Template size: $[112\times112]$;
Search region size: $[224\times224]$;
$C=512$; $N=8$;
\item {\bf SwinTrack-B-384}. \\
Backbone: Swin Transformer-Base~\cite{liu2021Swin}; \\
Template size: $[192\times192]$;
Search region size: $[384\times384]$;
$C=512$; $N=8$;
\end{itemize}
where $C$ and $N$ represent the channel number of the hidden layers in the
first stage of Swin Transformer and the number of encoder blocks in feature fusion, respectively. In all variants, we use the output after the third stage of Swin Transformer for feature extraction. Thus, the backbone stride $s$ is set to 16.
\vspace{0.2em}
\noindent
{\bf Training.} We train SwinTrack using training splits of LaSOT~\cite{fan2019lasot}, TrackingNet~\cite{muller2018trackingnet}, GOT-10k~\cite{Huang2021got10k} (1,000 videos are removed for fair comparisons with other trackers~\cite{VOT_TPAMI,stark}) and COCO 2017\cite{lin2014microsoftcoco}. Besides, we also report the performance of SwinTrack-T and SwinTrack-B with GOT-10k training split only to follow the protocol described in \cite{Huang2021got10k}.
The model is optimized with AdamW~\cite{loshchilov2017decoupledadamw}. The learning rate of the backbone is set to 5e-5, and the weight decay is 1e-4. We adopt gradient clipping to prevent very large gradients from misleading the optimization process. We train the network on 8 NVIDIA V100 GPUs for 300 epochs with 131,072 samples per epoch. The learning rate is dropped by a factor of 10 after 210 epochs. To stabilize training process, a warmup strategy is utilized. DropPath~\cite{larsson2016fractalnet} is applied in the latter half of the optimization process.
\vspace{0.2em}
\noindent
{\bf Inference.} We follow the common procedures for Siamese network-based tracking~\cite{bertinetto2016siamfc}. The template image is cropped from the first frame of the video sequence. The target object is in the center of the image with a background area factor of 2. The search region is cropped from the current tracking frame, and the image center is the target center position predicted in previous frame. The background area factor for the search region is 4.
Our SwinTrack takes the template image and search region as inputs and output classification map $r_{cls}$ and regression maps $r_{reg}$. To utilize positional prior in tracking, we apply hanning window penalty on $r_{cls}$, and the final classification map $r_{cls}'$ is obtained via $r_{cls}'=(1-\gamma)\times r_{cls} + \gamma\times h$, where $\gamma$ is the weight parameter and $h$ is a Hanning window with the same size as $r_{cls}$. The target position is determined by the largest value in $r_{cls}'$ and scale is estimated based on the corresponding regression results in $r_{reg}$.
\begin{table*}
\caption{Ablation studies on SwinTrack-T.}
\begin{adjustbox}{width=\textwidth}
\begin{tabular}{c|ccc|ccc|ccc|ccc|c|c|c}
\hline
\multirow{2}{*}{Modification} & \multicolumn{3}{c|}{LaSOT} & \multicolumn{3}{c|}{LaSOT$\rm{_{ext}}$} & \multicolumn{3}{c|}{TrackingNet} & \multicolumn{3}{c|}{GOT-10k} & \multirow{2}{*}{\begin{tabular}[c]{@{}c@{}}Speed\\ (fps)\end{tabular}} & \multirow{2}{*}{\begin{tabular}[c]{@{}c@{}}MACs \\ (G)\end{tabular}} & \multirow{2}{*}{\begin{tabular}[c]{@{}c@{}}Params\\ (M)\end{tabular}} \\ \cline{2-13}
& \begin{tabular}[c]{@{}c@{}}SUC\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}PRE\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}NPRE\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}SUC\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}PRE\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}NPRE\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}SUC\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}PRE\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}NPRE\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}mAO\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}mSR$_{50}$\\ (\%)\end{tabular} & \begin{tabular}[c]{@{}c@{}}mSR$_{75}$\\ (\%)\end{tabular} & & & \\ \hline
(SwinTrack-T) & 66.7 & 70.6 & 75.8 & 46.9 & 52.9 & 57.6 & 80.8 & 77.9 & 85.5 & 70.9 & 81.2 & 64.9 & 98 & 6.4 & 22.7 \\
ResNet-50 & 64.2 & 67.4 & 72.9 & 41.8 & 46.3 & 51.3 & 79.5 & 77.1 & 84.2 & 68.2 & 77.7 & 61.2 & 121 & 21.4 & 20.0 \\
Cross Fusion & 66.6 & 69.9 & 75.6 & 45.4 & 50.8 & 55.8 & 80.2 & 77.7 & 85.3 & 69.3 & 79.4 & 64.2 & 72 & 7.0 & 34.6 \\
Target query & 66.6 & 70.1 & 75.5 & 43.2 & 46.4 & 52.5 & 79.6 & 76.9 & 84.6 & 69.0 & 78.9 & 64.3 & 91 & 5.9 & 25.3 \\
Sine enc. & 65.7 & 68.8 & 74.4 & 45.0 & 50.0 & 55.4 & 80.0 & 77.3 & 85.2 & 70.0 & 80.0 & 64.4 & 103 & 6.2 & 21.6 \\
BCE loss & 66.2 & 69.5 & 75.8 & 46.7 & 52.5 & 57.1 & 79.4 & 76.9 & 84.8 & 68.2 & 78.3 & 63.1 & 98 & 6.4 & 22.7 \\
Weak aug. & 61.6 & 63.4 & 68.4 & 38.7 & 40.5 & 45.8 & 78.6 & 75.7 & 83.2 & 67.9 & 76.8 & 62.5 & 98 & 6.4 & 22.7 \\
No hann. & 65.7 & 69.4 & 74.6 & 46.0 & 51.5 & 56.6 & 80.0 & 77.3 & 85.0 & 69.6 & 79.3 & 65.1 & 98 & 6.4 & 22.7 \\ \hline
\end{tabular}
\end{adjustbox}
\label{tiny_ablation}
\end{table*}
\subsection{Ablations Study and Analysis}
We conduct ablations to study different factors in SwinTrack. To save training time, we perform ablation studies on SwinTrack-T.
\noindent
{\bf Comparison with ResNet backbone.} We compare our Transformer-based backbone with commonly adopted ResNet~\cite{he2016deep} for tracking. As shown in Table \ref{tiny_ablation}, using ResNet-50 lead to a huge drop in each dataset.
\vspace{0.2em}
\noindent
{\bf Feature fusion.} According to Table \ref{tiny_ablation}, compared with the \emph{concatenation-based fusion}, the \emph{cross attention-based fusion} not only perform worse than the \emph{concatenation-based fusion}, but also has a larger number of parameters.
\vspace{0.2em}
\noindent
{\bf Decoder.} We employ a transformer-style decoder, which is introduced in DETR, to our SwinTrack. By computing cross attention with the pre-trained target query tokens, the model can find the potential target objects in the feature. Ideally, it can generate the bounding box of the target object directly without any post-processing steps. However, our empirical results in Table \ref{tiny_ablation} show the tracker with a transformer-style decoder has poor performance in most datasets.
\vspace{0.2em}
\noindent
{\bf Position encoding.} We compare the adopted united positional encoding and the original since encoding in Transformer. As shown in Table \ref{tiny_ablation}, SwinTrack-T with united positional encoding achieves better accuracy with around 1\% improvements over SwinTrack-T with sine encoding on different datasets, while still runs fast in around 98 {\it fps}.
\vspace{0.2em}
\noindent
{\bf Loss function.} From Table \ref{tiny_ablation}, we observe that SwinTrack-T with varifocal loss significantly outperforms the one with binary entropy loss (BCS) without loss of efficiency.
\vspace{0.5em}
\noindent
{\bf Positional Augmentations.}
Inspired by \cite{li2019evolutionsiamrpnp}, we set up an experiment to discover the impact of positional augmentation during image pre-processing.
The "Weak aug." row in Table \ref{tiny_ablation} shows the dataset evaluation results of deducing random scale and random translation during the search image generation in the training phase. The success score evaluated in LaSOT drops by 5.1\%, in LaSOT$_\mathrm{ext}$ even 8.2\%, compared with our fine-tuned hyper-parameters.
\vspace{0.5em}
\noindent
{\bf Post processing.} By removing the hanning penalty window in the post-processing, as shown in Table \ref{tiny_ablation}, the performance is significantly dropped. This suggests that even with a strong backbone network, hanning penalty window is still functional.
\subsection{State-of-the-art Comparison}
We compare our SwinTrack with state-of-the-art trackers on four benchmarks including LaSOT~\cite{fan2019lasot}, LaSOT$_{\mathrm{ext}}$\cite{fan2021lasot}, TrackingNet\cite{muller2018trackingnet} and GOT-10k\cite{Huang2021got10k}.
\vspace{0.2em}
\noindent
{\bf LaSOT.} LaSOT~\cite{fan2019lasot} is a large-scale benchmark containing 280 test sequences. Table \ref{lasot_test_comparison} shows the results of SwinTrack and comparisons with state-of-the-art trackers.
From Table \ref{lasot_test_comparison}, we can see that our SwinTrack-T with light architecture reaches a SOTA performance with 0.667 SUC, 0.706 PRE, and 0.758 NPRE scores, which is competitive compared with other Transformer-based trackers, including STARK-ST101 (0.671 SUC score) and TransT (0.649 SUC and 0.60 PRE scores), and other trackers which utilize complicated designs for tracking, like KeepTrack (0.671 SUC and 0.702 PRE scores) and SiamR-CNN (0.648 SUC score). When using larger backbone and input size, our strongest variant SwinTrack-B-384 gives a breakthrough new record with 0.702 and 0.753 of the SUC score and the PRE score, respectively.
\vspace{0.2em}
\noindent
{\bf LaSOT$_{\mathrm{ext}}$.}
The recently proposed LaSOT$_{\mathrm{ext}}$\cite{fan2021lasot} is an extension of LaSOT by adding 150 extra videos from 15 new categories. These new sequences are challenging because there are many similar distractors that cause difficulties for tracking. KeepTrack designs a complex association technique to deal with the distractors and achieves a promising 0.482 SUC score. Compared with complicated KeepTrack, SwinTrack-T is simple and neat, yet shows comparable performance with 0.469 SUC score. In addition, due to complicated design, KeepTrack runs at less than 20 {\it fps}, while SwinTrack-T runs in 98 {\it fps}, 5$\times$ faster than KeepTrack. When using a larger model, SwinTrack-B shows the best performance with 0.476 SUC score and 0.582 NPRE score in the variants. The SUC score of our SwinTrack-B is still lower than KeepTrack, mainly due to the lack of the utilization of temporal information, while the NPRE score is better than KeepTrack.
\begin{table}\small
\centering
\renewcommand\arraystretch{1}
\caption{Performance comparison on LaSOT~\cite{fan2019lasot}.}
\begin{tabular}{rccc}
\hline
\multicolumn{1}{c}{Tracker} & {\begin{tabular}[c]{@{}c@{}}SUC\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}PRE\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}NPRE\\ (\%)\end{tabular}} \\
\hline \hline
SiamPRN++\cite{li2019evolutionsiamrpnp} & 49.6 & - & 56.9 \\
DiMP\cite{bhat2019learning} & 56.9 & 53.4 & 65.0 \\
Ocean\cite{Ocean_2020_ECCV} & 56.0 & 56.6 & 65.1 \\
SiamR-CNN\cite{voigtlaender2020siam} & 64.8 & - & 72.2 \\
TrSiam\cite{wang2021transformer} & 62.4 & 60.0 & - \\
TrDiMP\cite{wang2021transformer} & 63.9 & 61.4 & - \\
STMTrack\cite{fu2021stmtrack} & 60.6 & 63.3 & 69.3 \\
TransT\cite{TransT} & 64.9 & 69.0 & 73.8 \\
STARK-ST50\cite{stark} & 66.4 & - & - \\
STARK-ST101\cite{stark} & 67.1 & - & 77.0 \\
KeepTrack\cite{keeptrack} & 67.1 & 70.2 & 77.2 \\
\hline
SwinTrack-T & 66.7 & 70.6 & 75.8 \\
SwinTrack-B & 69.6 & 74.1 & \cellcolor{mygray}{\bf 78.6} \\
SwinTrack-B-384 & \cellcolor{mygray}{\bf 70.2} & \cellcolor{mygray}{\bf 75.3} & 78.4 \\
\hline
\end{tabular}
\label{lasot_test_comparison}
\end{table}
\begin{table}\small
\centering
\renewcommand\arraystretch{1}
\caption{Performance comparison on LaSOT$_{\mathrm{ext}}$\cite{fan2021lasot}.}
\begin{tabular}{rccc}
\hline
\multicolumn{1}{c}{Tracker} & {\begin{tabular}[c]{@{}c@{}}SUC\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}PRE\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}NPRE\\ (\%)\end{tabular}}\\
\hline\hline
C-RPN~\cite{Fan_2019_CVPR} & 27.5 & 32.0 & 34.4 \\
DiMP~\cite{bhat2019learning} & 39.2 & 45.1 & 47.5 \\
LTMU~\cite{dai2020high} & 41.4 & 47.3 & 49.9 \\
SuperDiMP~\cite{danelljan2020probabilistic} & 43.7 & - & 52.7 \\
KeepTrack\cite{keeptrack} & \cellcolor{mygray}{\bf 48.2} & - & 58.0 \\
\hline
SwinTrack-T & 46.9 & 52.9 & 57.6 \\
SwinTrack-B & 47.6 & \cellcolor{mygray}{\bf 54.1} & \cellcolor{mygray}{\bf 58.2} \\
SwinTrack-B-384 & 47.5 & 53.3 & 57.7 \\
\hline
\end{tabular}
\label{lasot_ext_results}
\end{table}
\begin{table}\small
\centering
\renewcommand\arraystretch{1}
\caption{Performance comparison on TrackingNet~\cite{muller2018trackingnet}.}
\begin{tabular}{rccc}
\hline
\multicolumn{1}{c}{Tracker} & {\begin{tabular}[c]{@{}c@{}}SUC\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}PRE\\ (\%)\end{tabular}}& {\begin{tabular}[c]{@{}c@{}}NPRE\\ (\%)\end{tabular}} \\
\hline\hline
PrDiMP\cite{danelljan2020probabilistic} & 75.8 & 70.4 & 81.6 \\
SiamFC++\cite{xu2020siamfc++} & 75.4 & 70.5 & 80.0 \\
KYS\cite{bhat2020know} & 74.0 & 68.8 & 80.0 \\
SiamR-CNN\cite{voigtlaender2020siam} & 81.2 & 80.0 & 85.4 \\
TrSiam\cite{wang2021transformer} & 78.1 & 72.7 & 82.9 \\
TrDiMP\cite{wang2021transformer} & 78.4 & 73.1 & 83.3 \\
STMTrack\cite{fu2021stmtrack} & 80.3 & 76.7 & 85.1 \\
TransT\cite{TransT} & 81.4 & 80.3 & 86.7 \\
STARK-ST50\cite{stark} & 81.3 & - & 86.1 \\
STARK-ST101\cite{stark} & 82.0 & - & 86.9 \\
\hline
SwinTrack-T & 80.8 & 77.9 & 85.5 \\
SwinTrack-B & 82.5 & 80.4 & 87.0 \\
SwinTrack-B-384 & \cellcolor{mygray}{\bf 84.0} & \cellcolor{mygray}{\bf 83.2} & \cellcolor{mygray}{\bf 88.2} \\
\hline
\end{tabular}
\label{trackingnettest}
\end{table}
\vspace{0.2em}
\noindent
{\bf TrackingNet.} We evaluate our trackers on the test set of TrackingNet\cite{muller2018trackingnet}. The results are shown in Table \ref{trackingnettest}. From Table \ref{trackingnettest}, we observe that our SwinTrack-T achieves comparable result of 0.808 SUC score. When using larger model and input size, our SwinTrack-B-384 obtains the best performance with 0.840 SUC score, better than STARK-ST101 with 0.820 SUC score and TransT with 0.813 SUC score.
\vspace{0.2em}
\noindent
{\bf GOT-10k.} GOT-10k~\cite{Huang2021got10k} provides 180 sequences for testing and it requires trackers to be trained using GOT-10k train split only. The results and comparisons are displayed in Table \ref{trackingnettest}. From Table \ref{trackingnettest}, we see that SwinTrack-B achieves the best mAO of 0.694, outperforming other Transformer-based counterparts including START-ST101 (0.688 mAO), TransT (0.671 mAO), TrDiMP (0.671 mAP) and TrSiam (0.660 mAO).
\begin{table}\small
\centering
\renewcommand\arraystretch{1}
\caption{Performance comparison on GOT-10k~\cite{Huang2021got10k}.}
\begin{tabular}{rccc}
\hline
\multicolumn{1}{c}{Tracker} & {\begin{tabular}[c]{@{}c@{}}mAO\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}mSR$_{50}$\\ (\%)\end{tabular}} & {\begin{tabular}[c]{@{}c@{}}mSR$_{75}$\\ (\%)\end{tabular}} \\
\hline\hline
SiamPRN++\cite{li2019evolutionsiamrpnp} & 51.7 & 61.6 & 32.5 \\
DiMP\cite{bhat2019learning} & 61.1 & 71.7 & 49.2 \\
Ocean\cite{Ocean_2020_ECCV} & 61.1 & 72.1 & 47.3 \\
SiamR-CNN\cite{voigtlaender2020siam} & 64.9 & 72.8 & 59.7 \\
TrSiam\cite{wang2021transformer} & 66.0 & 76.6 & 57.1 \\
TrDiMP\cite{wang2021transformer} & 67.1 & 77.7 & 58.3 \\
STMTrack\cite{fu2021stmtrack} & 64.2 & 73.7 & 57.5 \\
TransT\cite{TransT} & 67.1 & 76.8 & 60.9 \\
STARK-ST50\cite{stark} & 68.0 & 77.7 & 62.3 \\
STARK-ST101\cite{stark} & 68.8 & \cellcolor{mygray}{\bf 78.1} & 64.1 \\
\hline
SwinTrack-T & 69.0 & \cellcolor{mygray}{\bf 78.1} & 62.1 \\
SwinTrack-B & \cellcolor{mygray}{\bf 69.4} & 78.0 & \cellcolor{mygray}{\bf 64.3} \\
\hline
\end{tabular}
\label{got10k_test_table}
\end{table}
\begin{table}\small
\centering
\renewcommand\arraystretch{1}
\caption{Comparison on average running speed and $\#$ parameters.}
\begin{tabular}{rccc}
\hline
Tracker & Speed ({\it fps}) & Params (M) \\
\hline \hline
SiamRPN++\cite{li2019evolutionsiamrpnp} & 35 & 54 \\
TrSiam~\cite{wang2021transformer} & 35 & - \\
TrDiMP~\cite{wang2021transformer} & 26 & - \\
TransT~\cite{TransT} & 50 & - \\
STARK-ST50\cite{stark} & 42 & 24 \\
STARK-ST101\cite{stark} & 32 & 42 \\
\hline
SwinTrack-T & 98 & 23 \\
SwinTrack-B & 52 & 91 \\
SwinTrack-B-384 & 45 & 91 \\
\hline
\end{tabular}
\label{efficiency_table}
\end{table}
\vspace{0.2em}
\noindent
{\bf Efficiency comparison.} In addition to accuracy comparison, we also report the comparisons of SwinTrack with others trackers on efficiency and complexity. As shown in Table~\ref{efficiency_table}, our SwinTrack-T with a small model runs the fastest with a speed of {\it 98 fps}. Especially, compared with existing state-of-the-art STARK-ST101 and STARK-ST50 with 32 {\it fps} and 42 {\it fps}, SwinTrack-T is 3$\times$ and 2$\times$ faster. Despite using a larger model, our SwinTrack-B-384 is still faster than STARK-ST101 and STARK-ST50.
\section{Conclusion}
In this paper, we propose a strong baseline SwinTrack for Transformer tracker. SwinTrack consists of a Swin-Transformer-based backbone network, a concatenation-based fusion encoder, a general positional encoding solution for any combination of attention operation, and incorporating with some popular training tricks. By achieving state-of-the-art results on multiple challenging benchmarks, we expect SwinTrack can serve as a solid baseline for further research on Transformer-based tracking.
{\small
\bibliographystyle{ieee_fullname}
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"redpajama_set_name": "RedPajamaArXiv"
}
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The King Sabata Dalindyebo Local Municipality council consists of seventy-three members elected by mixed-member proportional representation. Thirty-seven councillors are elected by first-past-the-post voting in thirty-seven wards, while the remaining thirty-six are chosen from party lists so that the total number of party representatives is proportional to the number of votes received. In the election of 1 November 2021 the African National Congress (ANC) won a majority of forty-eight seats.
Results
The following table shows the composition of the council after past elections.
December 2000 election
The following table shows the results of the 2000 election.
March 2006 election
The following table shows the results of the 2006 election.
May 2011 election
The following table shows the results of the 2011 election.
August 2016 election
The following table shows the results of the 2016 election.
November 2021 election
The following table shows the results of the 2021 election.
By-elections from November 2021
The following by-elections were held to fill vacant ward seats in the period from November 2021. In ward seven, a by-election was held in October 2022 after the death of the previous councillor, with the African National Congress retaining its seat. The Economic Freedom Fighters, with 19%, took second spot from the United Democratic Movement, which dropped from 18% to 3%. Turnout was an exceptionally low 17%, with only 3% of registered voters in the Mthatha City district voting.
References
King Sabata Dalindyebo
Elections in the Eastern Cape
elections
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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