text stringlengths 14 5.77M | meta dict | __index_level_0__ int64 0 9.97k ⌀ |
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Q: The selected Stored Procedure returns No Columns using My SQL I am getting error 'The selected Stored Procedure returns No Columns' in Entity framework.
I have created the following Procedure in MySQL:
CREATE PROCEDURE `getCustomersByName`(
p_Name varchar(50)
)
BEGIN
SELECT id,name,categoryid
FROM Customers ;
END
Now i am using ADD Import Function to Import this procedure. then i am getting error.
please provide me the solution of this problem.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,012 |
Q: Jython Shutil (different behaviour between Windows, Linux, J/Python)! I'm using Jython, through Topspin (NMR Software running on Java) to run the following code:
home = "C:/Bruker/TopSpin3.2"
ep_zges_outdir = os.path.abspath(home + "/data/Testshutil/nmr/zges/")
data = ["EP_Saliva_140131_raw", "1", "1", "C:/Bruker/TopSpin3.2/data/Testshutil/nmr"]
ep_zges_list = [["EP_Saliva_140131_raw",25,334],...]
for sample in ep_zges_list:
if data[0] == sample[0] and data[1] == str(sample[1]):
src = os.path.abspath(data[3] + "/" + data[0] + "/" + data[1])
dst = os.path.abspath(ep_zges_outdir + "/" + str(sample[2]))
shutil.copytree(src, dst)
Proper imports were done and, when it works, no os.path.abspath is necessary.
This works perfectly in Windows/Linux python and through the same Software that runs Jython in CentOS. It does not run in the Software/Jython in Windows 7 and the following error is produced:
Traceback (most recent call last):
File "C:/Bruker/TopSpin3.2/exp/stan/nmr/py/user/JF_test_code_8.py", line 41, in <module>
shutil.copytree(os.path.abspath(data[3] + "/" + data[0] + "/" + data[1]), os.path.abspath(ep_zges_outdir + "/" + str(sample[2])))
File "C:\Bruker\TopSpin3.2\jython\Lib\shutil.py", line 145, in copytree
raise Error, errors
shutil.Error: [u'C:\\Bruker\\TopSpin3.2\\data\\Testshutil\\nmr\\EP_Saliva_140310_raw\\1\\pdata\\1',
u'C:\\Bruker\\TopSpin3.2\\data\\Testshutil\\nmr\\zges\\334\\pdata\\1',
"[Errno 5] Input/output error: u'C:\\\\Bruker\\\\TopSpin3.2\\\\data\\\\Testshutil\\\\nmr\\\\zges\\\\334\\\\pdata\\\\1'",
u'C:\\Bruker\\TopSpin3.2\\data\\Testshutil\\nmr\\EP_Saliva_140310_raw\\1\\pdata',
u'C:\\Bruker\\TopSpin3.2\\data\\Testshutil\\nmr\\zges\\334\\pdata',
"[Errno 5] Input/output error: u'C:\\\\Bruker\\\\TopSpin3.2\\\\data\\\\Testshutil\\\\nmr\\\\zges\\\\334\\\\pdata'",
u'C:\\Bruker\\TopSpin3.2\\data\\Testshutil\\nmr\\EP_Saliva_140310_raw\\1',
u'C:\\Bruker\\TopSpin3.2\\data\\Testshutil\\nmr\\zges\\334',
"[Errno 5] Input/output error: u'C:\\\\Bruker\\\\TopSpin3.2\\\\data\\\\Testshutil\\\\nmr\\\\zges\\\\334'"]
Software versions:
Windows 7 SP1 64bit.
Python 2.7 32bit.
Jython 2.5.3 running on Topspin 3.2 and Java 1.7.0_51.
CentOS 6.5 32Bit
Jython 2.5.3 running on Topspin 3.2 and Java 1.7.0_45
A: I think this is a Jython bug (a very annoying one). See
http://bugs.jython.org/issue1872
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 92 |
{"url":"https:\/\/www.zbmath.org\/?q=an%3A0931.76017","text":"# zbMATH \u2014 the first resource for mathematics\n\nA uniformly valid analytic solution of two-dimensional viscous flow over a semi-infinite flat plate. (English) Zbl\u00a00931.76017\nSummary: We apply a new analytic technique, namely the homotopy analysis method, to give an explicit, analytic, uniformly valid solution of the equation governing the two-dimensional laminar viscous flow over a semi-infinite flat plate, $$f'''(\\eta)+ \\alpha f(\\eta)f''(\\eta)+ \\beta[1- f^{\\prime 2}(\\eta)] =0$$, under the boundary conditions $$f(0)= f'(0)= 0$$, $$f'(+\\infty)= 1$$. This analytic solution is uniformly valid in the whole region $$0\\leq \\eta<+\\infty$$. For Blasius\u2019 (1908) flow ($$\\alpha= 1\/2$$, $$\\beta= 0$$), this solution converges to Howarth\u2019s (1938) numerical result and gives analytic value $$f''(0)= 0.332057$$. For the Falkner-Skan (1931) flow ($$\\alpha=1$$), it gives the same family of solutions as Hartree\u2019s (1937) numerical results, and provides a related analytic formula for $$f''(0)$$ when $$2\\geq \\beta\\geq 0$$. Additionally, this analytic solution allows to prove that for $$-0.1988\\leq \\beta<0$$, the Hartree\u2019s (1937) family of solutions possesses the property that $$f'\\to 1$$ exponentially as $$\\eta\\to +\\infty$$.\n\n##### MSC:\n 76D10 Boundary-layer theory, separation and reattachment, higher-order effects 76M45 Asymptotic methods, singular perturbations applied to problems in fluid mechanics\nFull Text:","date":"2021-07-29 16:42:57","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9077667593955994, \"perplexity\": 1466.793727614974}, \"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-2021-31\/segments\/1627046153860.57\/warc\/CC-MAIN-20210729140649-20210729170649-00605.warc.gz\"}"} | null | null |
Levi Magic is a hobbyist magician with a passion for card tricks and close-up illusions.
PERFORMING LIVE AND UP-CLOSE SATURDAY APRIL 20 DURING CHASING RABBITS SPRING SMOKEOUT!
Smoke Change by Levi - Levi demonstrates a very visual color change with smoke.
Impossible Location by Levi (ILL) - Levi teleports a card into a new, difficult to open cigarette pack.
Linking Chains by Levi - Levi demonstrates a modern take on a classic with linking ball chains.
Tic-Tacs by Levi - Levi demonstrates Lloyd Barnes' Osiris Tic-Tac effect!
Binomial Probability Experiments - Levi shuffles a deck and pulls the ace of spades six times, post explaining the math of the impossible nature of the situation.
Color Changing Deck - Levi performs a color changing deck routine.
Graduation Party - Levi entertains a graduation party.
Card Melt - Levi magically melts one card through another.
Stacks - Levi deals stacks and predicts a card.
Wiregram Reveal - Levi performs a wiregram reveal.
Monte Sonata - Two cards. Can you keep track of which is on top and which is on bottom?
Mental Photography - Levi demonstrates 'Mental Photography' and prints a black deck from blank cards.
Flourishing - Levi turns down a game of bingo and offers poker instead while flourishing cards.
Hallucination - Levi performs a pick-a-card trick. Was the card ever red?
Chaos and Order - Levi masterfully disrupts and reorders a deck.
Vegas Money Doubler - Levi transforms a stack of nickels into dimes using a peephole cover borrowed from the hotel room door.
Something In Those Hands - Levi gives a card control demonstration to an animated spectator.
Fist Bump - Levi makes a coin leave the spectator's fist and enter his own with a powerful fist bump.
Card Control - Levi calls the spectator's card then manipulates its position within the deck.
Levitation - Levi performs levitation of a coin.
Mentalism - Levi guides the spectator through magically flipping a card over in a deck of cards.
Cards at the Pool - Levi performs a card control demonstration for a spectator relaxing at the pool. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,843 |
layout: page
title: Rebecca Johnston's 49th Birthday
date: 2016-05-24
author: Jonathan Krause
tags: weekly links, java
status: published
summary: Phasellus in posuere nisl, id.
banner: images/banner/meeting-01.jpg
booking:
startDate: 03/17/2017
endDate: 03/21/2017
ctyhocn: DAYSHHX
groupCode: RJ4B
published: true
---
Nullam aliquet metus tellus, ut accumsan risus finibus ut. Aenean sit amet convallis odio, et mollis odio. Praesent porta eget ipsum vel molestie. Curabitur vel quam id leo aliquam fringilla volutpat quis mi. Sed ultricies convallis mi, sit amet viverra nisi iaculis nec. Fusce ut vestibulum metus, ac dignissim enim. Mauris non odio ante. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Nam eleifend libero turpis, et pellentesque odio finibus vitae. Nunc quis nisl quis massa fringilla bibendum eget nec risus. Sed quis tortor in sem tristique malesuada.
* Maecenas cursus nunc ac massa congue, sit amet laoreet dolor ultrices
* Quisque accumsan tellus ac ante egestas scelerisque
* Morbi vitae lectus facilisis, imperdiet diam quis, viverra dui
* Pellentesque ut enim vel magna ultrices varius
* Suspendisse sodales lectus vehicula nisl semper, ut mattis nunc laoreet.
Mauris tempus dolor ut ligula luctus convallis. Sed ex dolor, interdum non enim nec, pulvinar dapibus eros. Sed ut sollicitudin nunc. Donec tincidunt et nunc eget consequat. Quisque pulvinar tortor ac placerat congue. Curabitur finibus massa a dictum ornare. Quisque a gravida justo. Integer non ullamcorper quam, id laoreet metus. Quisque in ex accumsan, sagittis massa sed, faucibus enim. Nam orci purus, malesuada a nunc id, fermentum commodo nulla.
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,878 |
Sono elencate di seguito le date e i risultati della zona europea (UEFA) per le qualificazioni al mondiale del 1994.
Formula
Il sorteggio per la composizione dei gruppi si svolse a New York, domenica 8 dicembre 1991. Il Liechtenstein si ritirò prima dell'estrazione, mentre la Russia ereditò la tradizione sportiva della scomparsa Unione Sovietica. La Germania fu ammessa direttamente alla fase finale in quanto campione uscente; infine, la Jugoslavia fu squalificata per i fatti bellici che la portarono a venire esclusa anche dall'Europeo 1992.
La suddivisione in fasce fu operata sulla base di un ranking, stilato dividendo i punti per il numero di partite che ciascuna squadra aveva ottenuto nelle eliminatorie del Mondiale 1990 e dell'Europeo.
Gruppo 1
e qualificate.
Gruppo 2
e qualificate.
Gruppo 3
e qualificate.
Gruppo 4
e qualificate.
Gruppo 5
e qualificate.
Gruppo 6
e qualificate.
Note
Collegamenti esterni | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,143 |
Q: DDD Laravel. Repository pattern. How to retrieve an object from persistency and convert it into a not Laravel Entity model? I'm aplying DDD in Laravel.
In this architecture, the entity (conformed by the corresponding value objects) is not a Laravel Model extended class (because the domain layer needs to be agnostic to the infrastructure)
So... when I retrieve some data inside the repository implementation, the result is a stdclass object, and I need to return it as the entity object.
Anybody knows the best approach to do this?
Thanks!
To get this, I tried to convert from stdclass to entity by manual, but it's look hacky and dirty.
A: Ok, got it.
I found two different approaches, just in case others are fighting with the same problem.
Option 1: Embracing the Eloquent Active Record.
Inside the infrastructure layer, I created a Eloquent model to represent the Entity, and I use it as a vehicule for eloquent queries. Like this, all the conection with the framework stay contained in the infrastructure, without polluting other layers.
Option 2: Apply Doctrine in Laravel.
Doctrine has a package for laravel. Doctrine, as occurs in Synfony, is using data mapping, so no worries with that.
Thanks anyway!
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 6,436 |
\section{introduction}\label{sec:intro}
Consider a contact co-oriented manifold $(M,\eta)$ with a contact form $\eta$. We say that $(M,\eta)$ admits a {\it Sasakian structure} $(M,g,\xi,\eta,J)$ if:
\begin{itemize}[leftmargin=.2in]
\item there exists an endomorphism $J:TM\rightarrow TM$ such that
$J^2=-\operatorname{Id}+\xi\otimes\eta$,
for the Reeb vector field $\xi$ of $\eta$;
\item $J$ satisfies the conditions
$d\eta(JX,JY)=d\eta(X,Y)$,
for all vector fields $X,Y$ and $d\eta(JX,X)>0$ for all non-zero $X\in\ker\eta$;
\item the Reeb vector field $\xi$ is Killing with respect to the Riemannian metric
$g(X,Y)=d\eta(JX,Y)+\eta(X)\eta(Y),$
\item the almost complex structure $I$ on the contact cone
$C(M)=(M\times\mathbb{R}_{+},t^2g+dt^2)$
defined by the formulas
$I(X)=J(X),X\in\ker \eta, I(\xi)=t{\partial\over\partial t},I\left(t{\partial\over\partial t}\right)=-\xi$,
is integrable.
\end{itemize}
If one drops the condition of the integrability of $I$, one obtains a {\it K-contact} structure.
The study of manifolds with Sasakian and, more generally, K-contact structures is an important subject, because it interrelates to other geometries as well as brings together several different fields of mathematics from algebraic topology through complex algebraic geometry to Riemannian manifolds with special holonomy. One of the important motivations to study Sasakian geometry is the fact that it can be used to construct Einstein metrics on odd-dimensional manifolds \cite[Chapter 11]{BG}, \cite{BGK}, \cite{K2}.
A $5$-dimensional simply connected manifold $M$ is called a {\it Smale-Barden manifold}. These manifolds are classified by their second homology group over $\mathbb{Z}$ and a {\it Barden invariant} \cite{B},\cite{S}. In greater detail,
let $M$ be a compact smooth oriented simply connected $5$-manifold.
Let us write $H_2(M,\mathbb{Z})$ as a direct sum of cyclic group of prime power order
$$
H_2(M,\mathbb{Z})=\mathbb{Z}^k\oplus( \bigoplus_{p,i}\mathbb{Z}_{p^i}^{c(p^i)}),
$$
where $k=b_2(M)$. Choose this decomposition in a way that the second Stiefel-Whitney class map
$$
w_2: H_2(M,\mathbb{Z})\rightarrow\mathbb{Z}_2
$$
iz zero on all but one summand $\mathbb{Z}_{2^j}$. The value of $j$ is unique, it
is denoted by $i(M)$ and is called the Barden invariant.
The classification of the Smale-Barden manifolds is given by the following theorem.
\begin{theorem}[{\cite[Theorem 10.2.3]{BG}}]\label{thm:sb-classification}
Any simply connected closed $5$-manifold is diffeomorphic to one of the spaces
$$
M_{j;k_1,...,k_s;r}=X_j\# r M_\infty \#M_{k_1}\#\cdots\#M_{k_s}
$$
where the manifolds $X_{-1},X_0,X_j,X_{\infty}, M_j,M_{\infty}$ are characterized as follows: $1<k_i< \infty$, $k_1|k_2|\ldots |k_s$, and
\begin{itemize}
\item $X_{-1}=SU(3)/SO(3)$, $H_2(X_{-1},\mathbb{Z})=\mathbb{Z}_2$, $i(X_{-1})=1$,
\item $X_0=S^5$, $H_2(X_0,\mathbb{Z})=0$, $i(X_0)=0$,
\item $X_j$, $0<j<\infty$, $H_2(X_j,\mathbb{Z})=\mathbb{Z}_{2^j}\oplus \mathbb{Z}_{2^j}$, $i(X_j)=j$,
\item $X_{\infty}=S^2\widetilde\times S^3$, $H_2(X_{\infty},\mathbb{Z})=\mathbb{Z}$, $i(X_\infty)=\infty$
\item $M_k$, $1<k<\infty$, $H_2(M_k,\mathbb{Z})=\mathbb{Z}_k\oplus \mathbb{Z}_k$, $i(M_k)=0$,
\item $M_{\infty}=S^2\times S^3$, $H_2(M_{\infty},\mathbb{Z})=\mathbb{Z}$, $i(M_{\infty})=0$.
\end{itemize}
\end{theorem}
It is an important open problem to describe the class of Smale-Barden manifolds which admit Sasakian structures (see \cite[Chapter 10]{BG}).
Some necessary conditions for a Smale-Barden manifold to carry a Sasakian structure are known.
\begin{definition}
Let $M$ be a Smale-Barden manifold. We say that $M$ satisfies the {\it condition G-K} if the pair
$(H_2(M,\mathbb{Z}),i(M))$ written in the form
$$
H_2(M,\mathbb{Z})=\mathbb{Z}^k\oplus(\bigoplus_{p,i}(\mathbb{Z}_{p^i}^{c(p^i)}),
$$
where $k=b_2(M)$, satisfies all of the following:
\begin{enumerate}
\item $i(M)\in\{0,\infty\}$,
\item for every prime $p$, $t(p):=\#\{i:c(p^i)>0\}\leq k+1$,
\item if $i(M)=\infty$, then $t(2):=\#\{i:c(2^i)>0\}\leq k$.
\end{enumerate}
\end{definition}
\begin{question}[{\cite[Question 10.2.1]{BG}}]
Suppose that a Smale-Barden manifold satisfies the conditon G-K. Does it admit a Sasakian structure?
\end{question}
It is known from \cite[Corollary 10.2.11]{BG} that the condition G-K is necessary for the existence of
K-contact, and hence Sasakian structures. However, it is not known to what extent it is sufficient.
K\'ollar \cite{K} have found subtle obstructions to the existence of Sasakian structures on Smale-Barden manifolds with
$k=0$. Also, the recent works \cite{CMRV} and \cite{MRT} showed the existence of \emph{homology Smale-Barden}
and true Smale-Barden manifolds which carry a K-contact but do not carry any \emph{semi-regular} Sasakian structure
(see section \ref{sec:seifert} for definition).
Therefore, finding sufficient conditions is an important problem.
As the condition G-K is a bound on the numbers $t(p)$
controlling the size of the torsion part at primes $p$, it is natural to start by focusing on the case of small values of $t(p)$. Let
$$
\mathbf{t}(X)=\max\{ t(p)|\,p \text{ prime}\}.
$$
The case of $\mathbf{t} =0$ is that of the torsion-free Smale-Barden manifolds, where
we only have regular Sasakian structures (see section \ref{sec:bt=0}). The next case
to analyse is $\mathbf{t} = 1$. In this direction, we prove the following characterization of Sasakian structures with $\mathbf{t}=1$.
\begin{theorem}\label{thm:main1}
Let $M$ be a Smale-Barden manifold whose second integral homology has the form
$$
H_2(M,\mathbb{Z})=\mathbb{Z}^k\oplus (\bigoplus_{i=1}^r\mathbb{Z}_{m_i}^{2g_i}),
$$
and with $i(M)=0,\infty$.
Assume that $k\geq 1$, $m_i\geq 2, g_i\geq 1$, with $m_i$ pairwise coprime, $1\leq i\leq r$.
Then $M$ admits a semi-regular Sasakian structure.
\end{theorem}
That is, all Smale-Barden manifols with $k\geq 1$ and $\mathbf{t}=1$ admit Sasakian structures, and moreover they admit
semi-regular Sasakian structures.
In this article we fully solve the existence problem for semi-regular Sasakian structures on rational homology spheres. The general question was discussed in \cite{BG} (compare Corollary 10.2.15 therein). Here is the result.
\begin{theorem}\label{thm:sphere-semi}
Denote by $\mathcal{T}=\{\frac12 r(r-1)| r\geq 2\}$ the set of triangular numbers.
Let $m_i\geq 2$ be pairwise coprime, and $g_i=\frac12 (d_i-1)(d_i-2)\in \mathcal{T}$.
Assume further that $\operatorname{gcd}(m_i,d_i)=1$ for all $i$.
Let $M$ be a Smale-Barden manifold
with $H_2(M,\mathbb{Z})= \bigoplus_{i=1}^r \mathbb{Z}_{m_i}^{2g_i}$ and spin. Then $M$ admits a semi-regular
Sasakian structure.
Conversely, any Smale-Barden manifold with $b_2=0$ that admits a semi-regular Sasakian structure
is of this form.
\end{theorem}
Our approach yields new obstructions to the existence of semi-regular Sasakian structures, which are not detected by K\'ollar's work.
Recall that by \cite{K}, for a homology sphere $M$ whose second homology is of the form
$$
H_2(M,\mathbb{Z})=\bigoplus_{p,j} \mathbb{Z}_{p^j}^{c(p^j)},
$$
the existence of a Sasakian structure implies the following restriction on the set of $c(p^j)$:
all but at most $10$ elements of the set $\{g_i = \frac12 c(p^j)\}$ are in $\mathcal{T}$.
Over this, we get an interesting coprimality condition between the order of the torsion and its size (exponent):
since $\operatorname{gcd}(d_i,m_i)=\operatorname{gcd}(d_i,p^j)=1$, where $c(p^j)=2g_i=(d_i-1)(d_i-2)$, we have
$$
c(p^j)\in \mathcal{T}^*_p:=\mathcal{T}- \left\{ \frac12 (r-1)(r-2)\, |\, r\equiv 0\pmod{p}\right\}.
$$
This condition follows from the proof of Theorem \ref{thm:sphere-semi}.
The methods developed in this article allow us to address also the following problem \cite[Open Problems 10.3.3 and 10.3.4]{BG}.
\begin{question}
Which simply connected rational homology $5$-spheres admit negative Sasakian structures?
\end{question}
We construct new examples of such rational homology spheres. The definition of definite Sasakian structures will be given in Section \ref{sec:spheres}.
\begin{theorem}\label{thm:main2}
Let $m_i\geq 2$ be pairwise coprime, and $g_i={1\over 2}(d_i-1)(d_i-2)$.
Assume that $\operatorname{gcd}(m_i,d_i)=1$. Let $M$ be a Smale-Barden manifold with $H_2(M,\mathbb{Z})
=\bigoplus\limits_{i=1}^r\mathbb{Z}_{m_i}^{2g_i}$ and spin, with the exceptions
$\mathbb{Z}_m^2, \mathbb{Z}_2^{2n}, \mathbb{Z}_3^6$. Then $M$ admits a negative Sasakian structure.
\end{theorem}
Finally, let us mention the prerequisites used in this article. We use without further notice basic facts on topology of 4-manifolds and surfaces in them \cite{GS}. The basic facts about complex surfaces and curves on them, as well as the tools we use can be found in \cite{GH} and \cite{GS}.
\subsection*{Acknowledgements}
We thank Denis Auroux, Jos\'e Ignacio Cogolludo, Javier Fern\'andez de Bobadilla and Robert Gompf for useful comments and answering our questions.
The first author was partially supported by Project MINECO (Spain) PGC2018-095448-B-I00.
The second author was supported by the National Science Center (Poland), grant NCN no. 2018/31/D/ST1/00053.
This work was partially done when the second author visited Institute des Hautes
\'Etudes Scientifiques at Bur-sur-Yvette. His thanks go to the Institute for the wonderful research atmosphere.
\section{Quasi-regular and semi-regular Sasakian structures} \label{sec:seifert}
A Sasakian structure on a compact manifold $M$ is called {\it quasi-regular} if there is a positive
integer $\delta$ satisfying the condition that each point of $M$ has a neighbourhood $U$ such that each
leaf for $\xi$ passes through $U$ at most $\delta$ times.
If $\delta=1$, the structure is called regular. It is known \cite{R} that if a compact manifold admits a
Sasakian structure, it also admits a quasi-regular one. Thus, when we are interested in existence
questions, we may consider Sasakian structures which are quasi-regular. In the sequel we will
assume that the notions of a Seifert bundle and a cyclic orbifold are known. One can consult
\cite{BG} and \cite{MRT}. In particular we will need the following.
\begin{theorem}[{\cite[Theorems 7.5.1, 7.5.2]{BG}}]\label{thm:seifert}
Let $(M,\eta,\xi,J,g)$ be a quasi-regular Sasakian manifold. Then the space of leaves $X$ of the foliation
determined by the Reeb field $\xi$ has a natural structure of a Kähler orbifold. The projection $M\rightarrow X$ is a Seifert bundle.
Conversely, if $(X,\omega)$ is a Kähler orbifold and $M$ is the total space of the Seifert bundle determined by the class $[\omega]$,
then $M$ admits a quasi-regular Sasakian structure.
\end{theorem}
In the same way, one can characterize quasi-regular K-contact manifolds considering symplectic orbifolds
instead of the K\"ahler ones (see \cite[Theorems 19 and 21]{MRT}).
Consider now a somewhat stronger condition than being a general Seifert fibration carrying a quasi-regular Sasakian structure. We begin by recalling the construction of smooth orbifolds \cite{MRT}. A smooth orbifold is an orbifold that is a topological manifold.
From now on we restrict to $\dim M=5$ and $\dim X=4$.
\begin{proposition}[{\cite[Proposition 4]{MRT}}]
\label{prop:1}
Let $X$ be a smooth oriented $4$-manifold with embedded surfaces $D_i$ intersecting transversally,
and integers $m_i>1$ such that $\operatorname{gcd}(m_i,m_j)=1$ if $D_i$ and $D_j$ have a non-empty intersection.
Then $X$ admits a structure of a smooth orbifold with isotropy surfaces $D_i$ of multiplicities $m_i$.
\end{proposition}
\begin{definition}
If $M$ is a total space of the Seifert fibration whose base $X$ is a {\it smooth} orbifold,
we say that $M\rightarrow X$ is a {\it semi-regular} Seifert fibration.
\end{definition}
A semi-regular Seifert fibration can be characterized more intrinsically. For each $p\in M$, the number $\delta(p)$ of times
that a nearby generic leaf cuts a (small) transversal $T(p)$ is finite.
Then $M$ is semi-regular if $\delta(p)=\lcm(\delta(q) | q\in T(p), q\neq p)$, for all $p\in M$.
In this case, a more detailed description is available.
Semi-regular Seifert fibrations are determined by orbit invariants and the first Chern
class $c_1(M/X)\in H^2(X,\mathbb{Q})$ as follows. We say that an element $a$ in a free abelian group is
{\it primitive} if it cannot be represented as $a=k\,b$ with a non-trivial $b\in A,k\in \mathbb{N}$.
\begin{proposition}[{\cite[Proposition 14]{MRT}}]\label{prop:orb-inv}
Let $X$ be an oriented $4$-manifold and $D_i\subset X$ oriented surfaces of $X$ which intersect transversally.
Let $m_i>1$ such that $\operatorname{gcd}(m_i,m_j)=1$ if $D_i$ and $D_j$ intersect. Let $0<j_i<m_i$ with $\operatorname{gcd}(j_i,m_i)=1$
for every $i$. Let $0<b_i<m_i$ such that $j_ib_i\equiv 1 \pmod{m_i}$. Let $B$ be a complex line bundle over $X$.
Then there exists a Seifert bundle $\pi: M\rightarrow X$ with orbit invariants $\{(D_i,m_i,j_i)\}$ and the first Chern class
\begin{equation}\label{eqn:c1MX}
c_1(M/X)=c_1(B)+\sum_i{b_i\over m_i}[D_i].
\end{equation}
The set of all such Seifert bundles forms a principal homogeneous space under $H^2(X,\mathbb{Z})$, where the action corresponds to changing of $B$.
\end{proposition}
We will also use the following fact.
\begin{proposition}[{\cite[Proposition 35]{K}}]\label{prop:i}
If $\pi: M\rightarrow X$ is a Seifert fibration such that $X$ is smooth and the codimension $2$ isotropy divisors are orientable, then
$i(M)=0,\infty$.
\end{proposition}
The homology of the total space of a semi-regular Seifert fibration $M\rightarrow X$ is given by the following result.
\begin{theorem}[{\cite[Theorem 16]{MRT}}] \label{thm:Kollar}
Suppose that $\pi:M\to X$ is a semi-regular Seifert bundle with isotropy surfaces $D_i$ with multiplicities $m_i$.
Then $H_1(M,\mathbb{Z})=0$ if and only if
\begin{enumerate}
\item $H_1(X,\mathbb{Z})=0$,
\item $H^2(X,\mathbb{Z})\to \bigoplus H^2(D_i,\mathbb{Z}_{m_i})$ is surjective,
\item $c_1(M/m) =m\, c_1(M)\in H^2(X,\mathbb{Z})$ is primitive, where $m=\lcm(m_i)$.
\end{enumerate}
Moreover $H_2(M,\mathbb{Z})=\mathbb{Z}^k\oplus \bigoplus \mathbb{Z}_{m_i}^{2g_i}$, $g_i$ is the genus of $D_i$, $k+1=b_2(X)$.
\end{theorem}
We will also need to calculate the fundamental group of $M$. By \cite[Theorem 4.3.18]{BG}, we have an exact sequence
$$
\pi_1(S^1)=\mathbb{Z} \to \pi_1(M) \to \pi^{\mathrm{orb}}_1(X).
$$
If $\pi_1^{\mathrm{orb}}(X)=1$, then $\pi_1(M)$ is abelian,
hence $\pi_1(M)=H_1(M,\mathbb{Z})$. The first homology group can be computed via Theorem \ref{thm:Kollar}.
\medskip
In this work we construct various Seifert bundles using some configurations of smooth curves in complex algebraic surfaces.
The calculations of the fundamental groups use the following result.
\begin{proposition} \label{prop:abelian}
If $X$ is smooth simply-connected projective surface and $D_i$ are smooth complex curves with $D_i^2>0$ and
they intersect transversally, then
$$\pi_1(X-(D_1\cup \ldots \cup D_r))$$
is abelian.
\end{proposition}
\begin{proof}
This follows from \cite{Nori}, page 306, item II, taking $E$ empty.
\end{proof}
\medskip
We finally study the second Stiefel-Whitney class of a semi-regular Seifert bundle $\pi:M\to X$.
The formula for Seifert fibrations proved in \cite[Lemma 36]{K} says that a
Seifert fibration with orbit invariants as in Proposition \ref{prop:orb-inv} has the total space $M$ with the
second Stiefel-Whitney class given by the formula
\begin{equation}\label{eqn:w2}
w_2(M)=\pi^*w_2(X)+\sum_i(m_i-1)[E_i]
\end{equation}
where $E_i=\pi^{-1}(D_i)$. Note also that $\pi^*[D_i]=m_i [E_i]$.
The result in \cite[Corollary 37]{K} gives an alternative formula in terms of the orbit invariants $b_i$,
\begin{equation}\label{eqn:w2M}
w_2(M)=\pi^*(w_2(X)+\sum_i b_i[D_i] + c_1(B)).
\end{equation}
For using it, we need to compute the kernel of $\pi^*$. This is given by the following.
\begin{proposition}\label{prop:w2}
Let $\pi:M\to X$ be a semi-regular Seifert fibration with isotropy locus and orbit invariants
$\{(D_i,m_i,b_i)\}$, and $H_1(M,\mathbb{Z})=0$. The mod $2$ cohomology of $X$ is $H^2(X,\mathbb{Z}_2)=\mathbb{Z}_2^{k+1}$ and
the mod $2$ cohomology of $M$ is
$$
H^2(M,\mathbb{Z}_2)=\mathbb{Z}_2^k \oplus (\bigoplus_{m_i \text{ even}} (\mathbb{Z}_2)^{2g_i})
$$
The map $\pi^*:H^2(X,\mathbb{Z}_2)\to H^2(M,\mathbb{Z}_2)$ has image onto the first summand $\mathbb{Z}_2^k \subset H^2(M,\mathbb{Z}_2)$.
Its kernel is:
\begin{itemize}
\item If all $m_i$ are odd, then $\ker\pi^*$ is one-dimensional spanned by $c_1(B)+\sum b_i [D_i]$.
\item If $c=\#\{i\, | \, m_i$ even$\}>0$, then $\ker\pi^*$ is $c$-dimensional, and $\ker\pi^*$ is spanned by those $[D_i]$ with $m_i$ even.
\end{itemize}
\end{proposition}
\begin{proof}
We consider the Leray spectral sequence of the Seifert fibration $\pi:M\to X$ with coefficients in $\mathbb{Z}_2$.
This says that $H^i(X,R^j\pi_*\underline{\mathbb{Z}}{}_2)\Rightarrow H^{i+j}(M,\mathbb{Z}_2)$.
By Theorem \ref{thm:Kollar}, we have the cohomology of $X$ as $H^1(X,\mathbb{Z}_2)=0$,
$H^2(X,\mathbb{Z}_2)=\mathbb{Z}_2^{k+1}$ and $H^3(X,\mathbb{Z}_2)=0$, where $k+1=b_2(X)$.
Now let $\mathcal{F}=R^1\pi_*\underline{\mathbb{Z}}{}_2$. All fibers have $H^1(\pi^{-1}(x),\mathbb{Z}_2)=\mathbb{Z}_2$, so $\mathcal{F}(U)=\mathbb{Z}_2$ for
any small ball $U\subset X$. If $x\in D_i$ is an isotropy set with $m_i$ odd, and $V\subset U$ then $\mathcal{F}(U)\to \mathcal{F}(V)$ is an isomorphism.
If $m_i$ is even then $\mathcal{F}(U)\to \mathcal{F}(U-D_i)$ is the zero map. Relabel the $m_i$ so that $m_1,\ldots, m_c$
are even, and denote $D=\bigsqcup_{i=1}^c D_i$. Note that the union is
disjoint. Let $j:X-D\to X$ and $i:D\to X$ denote the inclusion. Then the previous discussion means that
$$
\mathcal{F}= j_! \underline{\mathbb{Z}}{}_2\oplus i_* \underline{\mathbb{Z}}{}_2
$$
Note also that by Theorem \ref{thm:Kollar}(2), reducing mod $2$, we have that $[D_1],\ldots, [D_c]$ are
linearly independent in $H^2(X,\mathbb{Z}_2)$.
Let us start first by the case where all $m_i$ are odd. Then $F=\underline{\mathbb{Z}}{}_2$, and so the Leray
spectral sequence is
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes,
nodes in empty cells,nodes={minimum width=8ex,
minimum height=5ex,outer sep=-5pt},
column sep=1ex,row sep=1ex]{
\mathbb{Z}_2 & 0 & \mathbb{Z}_2^{k+1} & 0 & \mathbb{Z}_2 \\
\mathbb{Z}_2 & 0 & \mathbb{Z}_2^{k+1} & 0 & \mathbb{Z}_2 \\
};
\draw[-stealth] (m-1-2.south east) -- (m-2-4.north west);
\draw[-stealth] (m-1-3.south east) -- (m-2-5.north west);
\draw[-stealth] (m-1-1.south east) -- (m-2-3.north west);
\end{tikzpicture}
\end{center}
The map $H^0(X,\mathcal{F})\to H^2(X,\mathbb{Z}_2)$ is given by cup product with the Chern class
$c_1(M/m)=m\, c_1(M) = c_1(B)+ \sum b_i [D_i] \pmod2$ (using that $m$ is odd).
Now assume that $c>0$. Clearly $\pi^*[D_i]=m_i[E_i]=0 \pmod2$, so $\ker\pi^*$ is at least $c$-dimensional.
Now we compute $H^k(X,\mathcal{F})$.
Clearly
$$
H^0(X,j_*\underline{\mathbb{Z}}{}_2)=\mathbb{Z}_2^c, H^1(X,j_*\underline{\mathbb{Z}}{}_2)=\mathbb{Z}_2^{2(g_1+\ldots+g_c)},
H^2(X,j_*\underline{\mathbb{Z}}{}_2)=\mathbb{Z}_2^c
$$
Using the exact sequence $j_!\underline{\mathbb{Z}}{}_2 \to \underline{\mathbb{Z}}{}_2\to j_*\underline{\mathbb{Z}}{}_2$,
and that the map $H^2(X,\mathbb{Z}_2) \to \bigoplus_{i=1}^c H^2(D_i,\mathbb{Z}_2)$ is surjective,
we get
\begin{align*}
& H^0(X,j_!\underline{\mathbb{Z}}{}_2)=0, H^1(X,j_!\underline{\mathbb{Z}}{}_2)=\mathbb{Z}_2^{c-1},
H^2(X,j_!\underline{\mathbb{Z}}{}_2)=\mathbb{Z}_2^{2(g_1+\ldots+g_c)+k+1-c}, \\
& H^3(X,j_!\underline{\mathbb{Z}}{}_2)=0, H^4(X,j_!\underline{\mathbb{Z}}{}_2)=\mathbb{Z}_2
\end{align*}
So the $E_2$ term of the Leray spectral sequence becomes
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes,
nodes in empty cells,nodes={minimum width=12ex,
minimum height=5ex,outer sep=-5pt},
column sep=1ex,row sep=1ex]{
\mathbb{Z}_2^c & \mathbb{Z}_2^{2(g_1+\ldots+g_c)+c-1} & \mathbb{Z}_2^{2(g_1+\ldots+g_c)+k+1} & 0 & \mathbb{Z}_2 \\
\mathbb{Z}_2 & 0 & \mathbb{Z}_2^{k+1} & 0 & \mathbb{Z}_2 \\
};
\draw[-stealth] (m-1-2.south east) -- (m-2-4.north west);
\draw[-stealth] (m-1-3.south east) -- (m-2-5.north west);
\draw[-stealth] (m-1-1.south east) -- (m-2-3.north west);
\end{tikzpicture}
\end{center}
Therefore the map $\mathbb{Z}_2^c\to \mathbb{Z}_2^{k+1}$ must be injective, and the map $\pi^*:H^2(X,\mathbb{Z}_2)\to H^2(M,\mathbb{Z}_2)$ has
image on the term $E^{2,0}_2/d_2(E^{0,1})\cong \mathbb{Z}_2^{k+1-c}$.
Note that we recover the mod $2$ cohomology of $M$, as stated.
\end{proof}
\section{Proof of Theorem \ref{thm:main1}}
The proof of the main theorem is a corollary to the results below.
\begin{theorem}\label{thm:sums1}
Let $m_i \geq 2$, $g_i\geq 1$, with $m_i$ pairwise coprime, $1\leq i\leq r$, and with $(m_i,g_i)\neq($even,odd$)$.
Let $M$ be a Smale-Barden manifold with $H_2(M,\mathbb{Z})=\mathbb{Z} \oplus \left(\bigoplus_{i=1}^r \mathbb{Z}_{m_i}^{2g_i}\right)$.
\begin{itemize}
\item If $M$ is spin then $M$ admits a semi-regular Sasakian structure.
\item If $i(M)=\infty$ and some $m_i$ is even then $M$
admits a semi-regular Sasakian structure.
\end{itemize}
\end{theorem}
\begin{proof}
Take the Kähler manifold $X=\CC P^1 \times \CC P^1$.
The second cohomology can be written as $H^2(X,\mathbb{Z})=\mathbb{Z}\langle H_1,H_2\rangle$, where $H_1,H_2$ are the classes
of the two factors. So $H_1\cdot H_2=1$. The canonical class is $K_X=-2H_1-2H_2$, so $w_2(X)=0$ and $X$ is spin.
Now take a collection of curves $D_i\subset X$ such that
\begin{equation}\label{eqn:Di}
[D_i]= 2H_1+(g_i+1)H_2.
\end{equation}
The genus of $D_i$ equals $g_i$, which can be checked with the adjunction formula:
$$
K_X\cdot D_i+ D_i^2=-4-2(g_i+1)+ 4(g_i+1)=2g_i-2.
$$
For $g_i\geq 1$ and a generic section, $D_i$ is smooth. This holds since the divisor $a_1H_1+a_2H_2$ is very
ample for $a_1,a_2\geq 1$, and using Bertini's theorem. Moreover, the divisors $D_i$ can be taken to intersect transversally by
Bertini's theorem.
We put coefficients $m_i$ to each of $D_i$ by using Proposition \ref{prop:1} in order to get a Kähler orbifold.
For the given orbit invariants $b_i$ and line bundle $B$, we have by Proposition \ref{prop:orb-inv} a Seifert bundle
\begin{equation}\label{eqn:piMX}
\pi:M\to X
\end{equation}
that admits a semi-regular Sasakian structure if $c_1(M/X)$ defined by (\ref{eqn:c1MX}) is an orbifold Kähler class.
We need to compute
$$
c_1(M/X)= c_1(B)+ \sum \frac{b_i}{m_i}[D_i].
$$
For a Kähler structure, we have $[\omega]=a_1 H_1+ a_2 H_2$, with $a_1,a_2>0$, so it is enough
to choose $c_1(B)=\b_1H_1+\b_2H_2$ in a way to get $c_1(M/X)=a_1H_1+a_2H_2$, $a_1,a_2> 0$.
The fundamental group of $\pi_1(X-(D_1\cup\ldots D_r))$ is abelian by Proposition \ref{prop:abelian}.
Hence $\pi_1^{\mathrm{orb}}(X)$ and also $\pi_1(M)$ are abelian.
It only remains to prove that $H_1(M,\mathbb{Z})=0$ to know that $M$ is simply-connected. This follows from Theorem
\ref{thm:Kollar} if $c_1(M/m)$ is primitive, where $m=\lcm(m_i)=\prod m_i$, and if
$$
H^2(X, \mathbb{Z}) \to \bigoplus H^2(D_i,\mathbb{Z}_{m_i})
$$
is surjective. As all $m_i$ are coprime, the surjectivity above is equivalent to the surjectivity of
each individual map $H^2(X, \mathbb{Z}) \to H^2(D_i,\mathbb{Z}_{m_i})$, and this happens if the divisibility of $D_i$ is coprime with $m_i$.
By (\ref{eqn:Di}) the class $[D_i]$ is primitive if $g_i$ is even, and the divisibility is $2$ if $g_i$ is odd. Therefore
we need to assume that $(g_i,m_i) \neq ($odd,even$)$.
Next we need to compute
\begin{align*}
c_1(M/m) &=m c_1(B)+ \sum \frac{m}{m_i} b_i [D_i] \\
&= (m\b_1+ \sum 2M_ib_i) H_1 + (m\b_2 + \sum (g_i+1)M_i b_i) H_2,
\end{align*}
where $M_i=m/m_i$, to be primitive. We have the following cases:
\begin{itemize}
\item If all $m_i$ are odd, that is $m$ is odd. Then $\operatorname{gcd}(2M_1,\ldots, 2M_r,m)=1$, and we can solve the equation
$m\b_1+ \sum 2M_ib_i=1$, thereby getting a primitive class. Now choosing any $\b_2$ so that
$m\b_2 + \sum (g_i+1)M_i b_i>0$, we get an orbifold Kähler class, and the manifold $M$ in (\ref{eqn:piMX}) is
semi-regular Sasakian with $\pi_1(M)=0$.
\item If $m$ is even, then we can assume that $m_1$ is even and $m_i$ is odd for $i>1$. Then $g_1$ is also even
by our assumption. Setting $\b_1=0$, we can solve
$\sum 2M_ib_i=2$. Now $\sum (g_i+1)M_i b_i= (\sum g_i M_ib_i )+ 1$ is
odd, so taking any $\b_2$ such that
$m\b_2 + \sum (g_i+1)M_i b_i>0$, we get an orbifold Kähler class, and the manifold $M$ in (\ref{eqn:piMX}) is
semi-regular Sasakian with $\pi_1(M)=0$.
We can also take $\b_1=1$ and solve $m+\sum 2M_ib_i=2$, and work to get the same result.
\end{itemize}
Now let us compute $w_2(M)$. If $m$ is odd, then (\ref{eqn:w2}) says that
$w_2(M)=\pi^*w_2(X)+\sum_i(m_i-1)[E_i]=0$ since $w_2(X)=0$, so $M$ is spin.
If $m$ is even, then arrange $m_1$ even and the others are odd. By Proposition \ref{prop:w2},
we have that $\pi^*:H^2(X,\mathbb{Z}_2)\to H^2(M,\mathbb{Z}_2)$ has one-dimensional kernel spanned by
$[D_1]$. As $m_1$ is even we have that $g_1$ is even, by the assumption of the statement.
Hence $\pi^*[D_1]=\pi^*[H_2]=0$. Therefore (\ref{eqn:Di}) implies that also $\pi^*[D_j]=0$
for all $j\geq 1$. As $w_2(X)=0$, we have that (\ref{eqn:w2M}) implies that
$$
w_2(M)=\pi^*(w_2(X)+\sum_ib_i[D_i])+\pi^*c_1(B)=\pi^*c_1(B)=\b_1 \, \pi^*[H_1].
$$
Taking either $\b_1=0$ or $\b_1=1$ we get $M$ spin or non-spin, as desired.
\end{proof}
\begin{theorem}\label{thm:sums2}
Assume that we are given $m_i \geq 2$, $g_i\geq 1$, with $m_i$ pairwise coprime, $1\leq i\leq r$, and with $(m_i,g_i)\neq($even,even$)$.
Let $M$ be a Smale-Barden manifold with $H_2(M,\mathbb{Z})=\mathbb{Z} \oplus \left(\bigoplus_{i=1}^r \mathbb{Z}_{m_i}^{2g_i}\right)$.
\begin{itemize}
\item If $M$ is non-spin then $M$ admits a semi-regular Sasakian structure.
\item If $i(M)=0$ and some $m_i$ is even then $M$ admits a semi-regular Sasakian structure.
\end{itemize}
\end{theorem}
\begin{proof}
Consider the projective plane $\CC P^2$ blown-up at a point, $X=\CC P^2\#\overline{\CC P}{}^2$. This is a Kähler manifold with
$H_2(X,\mathbb{Z})$ generated by the hyperplane class $H$ an the exceptional divisor $E$, $H^2=1$, $H\cdot E=0$, $E^2=-1$.
Consider plane curves $D_i'$ of degree $d_i\geq 3$ which are smooth except for
a point of multiplicity $d_i-2$ with different branches. After blowing-up, the proper transform is a
smooth curve
\begin{equation}\label{eqn:Di2}
D_i\equiv d_iH-(d_i-2)E
\end{equation}
Note that the divisor $d_i H-(d_i-2)E$ is very ample on $X$ since $H-E$ is a movable divisor
and $2H-E$ is very ample, and adding very ample to movable
gives very ample divisors. This implies that we can choose $D_i$ to be smooth. Moreover we can
choose $D_i$ to intersect transversally.
As $K_X=-3H+E$, we have that the genus of $D_i$ is given by
$$
2g_i-2=K_X\cdot D_i +D_i^2= -3d_i+d_i-2+d_i^2-(d_i-2)^2=2d_i-6,
$$
so $g_i=d_i-2\geq 1$.
Now put isotropy $m_i$ over $D_i$ and make it an orbifold using Proposition \ref{prop:1}.
Given orbit invariants $b_i$ and a line bundle $B$, we have by Proposition \ref{prop:orb-inv} a Seifert bundle
\begin{equation}\label{eqn:piMX2}
\pi:M\to X
\end{equation}
that admits a semi-regular Sasakian structure if $c_1(M/X)$ defined by (\ref{eqn:c1MX}) is an orbifold Kähler class.
We compute
$$
c_1(M/X)= c_1(B)+ \sum \frac{b_i}{m_i}[D_i].
$$
For a Kähler structure, we have $[\omega]=a_1 H- a_2 E$, with $a_1>a_2>0$, so it is enough
to choose $c_1(B)=\b_1H_1+\b_2H_2$ with $c_1(M/X)=a_1H- a_2E$, $a_1>a_2> 0$.
The fundamental group of $\pi_1(X-(D_1\cup\ldots D_r))$ is abelian by Proposition \ref{prop:abelian}.
Hence $\pi_1^{\mathrm{orb}}(X)$ and also $\pi_1(M)$ are abelian.
It only remains to prove that $H_1(M,\mathbb{Z})=0$ to know that $M$ is simply-connected. This follows from Theorem
\ref{thm:Kollar} if $c_1(M/m)$ is primitive, where $m=\lcm(m_i)=\prod m_i$, and if
$$
H^2(X, \mathbb{Z}) \to \bigoplus H^2(D_i,\mathbb{Z}_{m_i})
$$
is surjective. As all $m_i$ are coprime, the surjectivity above is equivalent to the surjectivity of
each map $H^2(X, \mathbb{Z}) \to H^2(D_i,\mathbb{Z}_{m_i})$, and this happens if the divisibility of $D_i$ is coprime with $m_i$.
By (\ref{eqn:Di2}) the class $[D_i]$ is primitive if $g_i$ is odd since then $d_i$ is odd, and it has divisibility $2$ if $g_i$ is even.
Therefore we need to assume that $(g_i,m_i) \neq ($even,even$)$.
Next we compute
\begin{align*}
c_1(M/m) &=m c_1(B)+ \sum \frac{m}{m_i} b_i [D_i] \\
&= (m\b_1+ \sum M_ib_id_i) H - (m\b_2 + \sum M_i b_i(d_i-2)) E,
\end{align*}
where $M_i=m/m_i$, to be primitive. Let $a_1=m\b_1+ \sum M_ib_id_i$ and $a_2=m\b_2 + \sum M_i b_i(d_i-2)$.
Note that $\operatorname{gcd}(a_1,a_2)=\operatorname{gcd}(a_1,a_2-a_1)$, where $a_2-a_1=m(\b_1-\b_2)+ \sum 2M_ib_i$.
We have the following cases:
\begin{itemize}
\item If all $m_i$ are odd, that is $m$ is odd. Then we can solve the equation
\begin{equation}\label{eqn:co-odd}
m(\b_1-\b_2)+ \sum 2M_ib_i=1
\end{equation}
thereby getting a primitive class. Now choose $\b_2$ so that $a_2=m\b_2 + \sum M_i b_i(d_i-2)>0$.
We get an orbifold Kähler class, and the manifold $M$ in (\ref{eqn:piMX2}) is
semi-regular Sasakian with $\pi_1(M)=0$.
\item If $m$ is even, then we can assume that $m_1$ is even and $m_i$ is odd for $i>1$. Then $g_1$ must be odd
by our assumption. We can solve
\begin{equation}\label{eqn:co-even}
m(\b_1-\b_2)+ \sum 2M_ib_i=2.
\end{equation}
As $m_1$ is even, it must be $b_1$ odd, hence
$a_1=m\b_1+ \sum M_ib_id_i \equiv M_1b_1d_1 \equiv 1\pmod 2$.
Taking $\b_2$ large, we get again $a_2>0$
and hence an orbifold Kähler class. The manifold $M$ in (\ref{eqn:piMX2}) is
semi-regular Sasakian with $\pi_1(M)=0$.
\end{itemize}
\medskip
Now let us compute $w_2(M)$.
Suppose first that $m$ is odd, that is all $m_i$ are odd. Then (\ref{eqn:w2}) says that $w_2(M)=\pi^* w_2(X)$.
Since $K_X=-3H+E$, we have that $w_2(X)=H+E$. The images of $[D_i]$ in $H^2(X,\mathbb{Z}_2)$ are $d_iH-(d_1-2)E \equiv
d_i(H+E) \pmod 2$, hence all in the line $\langle (H+E)\rangle \subset H^2(X,\mathbb{Z}_2)$. By Proposition \ref{prop:w2}, the kernel
of $\pi^*:H^2(X,\mathbb{Z}_2) \to H^2(M,\mathbb{Z}_2)$ is generated by $c_1(B)+\sum b_i[D_i]$, hence $w_2(X)\in \ker \pi^*$ if
and only if $c_1(B)\in \langle (H+E)\rangle$. As $c_1(B)=\b_1H-\b_2E$, this is rewritten as $\b_1-\b_2\equiv 0 \pmod2$. By (\ref{eqn:co-odd})
we have that $\b_1-\b_2$ is odd, hence $w_2(X)\not\in \ker \pi^*$, $w_2(M)\neq 0$ and $M$ is non-spin.
Now assume that $m$ is even. Arrange that $m_1$ even and the others are odd. By Proposition \ref{prop:w2},
we have that $\pi^*:H^2(X,\mathbb{Z}_2)\to H^2(M,\mathbb{Z}_2)$ has the one-dimensional kernel spanned by
$[D_1]$. As $m_1$ is even we have that $g_1$ is odd, by the assumption of the statement, so that $[D_1]=H+E$.
Now (\ref{eqn:w2M}) says that
$$
w_2(M)=\pi^*(w_2(X)+\sum_i b_i[D_i] + c_1(B))=\pi^*(\b_1 H+\b_2 E),
$$
using that $w_2(X)=H+E\in \ker \pi^*$ and $[D_i]=d_i(H+E)\in \ker \pi^*$. Therefore $w_2(M)$ is zero or non-zero
according to the parity of $\b_1-\b_2$.
We can solve (\ref{eqn:co-even}) with $\b_1-\b_2=0$ to get $w_2(M)=0$, and we can also solve
(\ref{eqn:co-even}) with $\b_1-\b_2=1$ to get $w_2(M)\neq 0$. So $M$ can be spin or non-spin.
\end{proof}
Our last tool serves to increase the rank $k$ without modifying the torsion.
\begin{theorem}\label{thm:highrank}
If $M$ is a simply-connected Smale-Barden manifold with a semi-regular Sasakian structure, then
$M\# X_\infty$ and $M\# M_\infty$ are also semi-regular Sasakian manifolds.
\end{theorem}
\begin{proof}
As before, consider a Seifert bundle $\pi: M\to X$ over a Kähler orbifold $X$. Blow-up $X$ at a point outside
the ramification divisors to get another orbifold $\hat X$. Clearly $H_2(\hat X)=H_2(X)\oplus \mathbb{Z}\langle E\rangle$,
where $E$ is the exceptional divisor. Then $K_{\hat X}=K_X+E$, and $[\hat\omega]=[\omega]-\a[E]$ is
a Kähler class for a small $\a>0$ (we have to multiply the class $[\hat\omega]$ by a large integer number to make it integral).
Let $\{(D_i,m_i,b_i)\}$ be the orbit invariants and isotropy locus of $\pi:M\to X$. We take some large integer $N>0$
and $\a=\frac{1}{N}$.
Consider for $\hat X$ the
orbit invariants $\{(D_i,m_i,N b_i)\}$ and $[\hat\omega]=N([\omega]-\a[E])=N[\omega] - [E]$. Take the
semi-regular Sasakian structure $\hat\pi : \hat M\to \hat X$ with the first Chern class
$c_1(\hat M/\hat X)=[\hat\omega]$. Hence
\begin{equation}\label{eqn:mmm}
c_1(\hat M/m)= N m\, c_1(B) - m [E] + \sum Nb_i\frac{m}{m_i}[D_i].
\end{equation}
This class is primitive since $c_1(M/m)$ is primitive, and taking $\operatorname{gcd}(m,N)=1$.
The orbifold fundamental
group satisfies $\pi_1^{\mathrm{orb}}(\hat X)=\pi_1^\mathrm{orb}(X)$. By Theorem \ref{thm:Kollar}, $H_1(\hat M,\mathbb{Z})=0$.
Moreover $H_2(\hat M,\mathbb{Z})=H_2(M,\mathbb{Z}) \oplus \mathbb{Z}$.
Finally, let us compute $w_2(\hat M)$. Let $K=\ker (\pi^*:H^2(X,\mathbb{Z}_2) \to H^2(M,\mathbb{Z}_2))$ and
$\hat K=\ker (\pi^*:H^2(\hat X,\mathbb{Z}_2) \to H^2(\hat M,\mathbb{Z}_2))$. If all $m_i$ are odd, then
$K=\langle c_1(B)+ \sum b_i[D_i]\rangle$ and $\hat K=\langle Nc_1(B)+\sum Nb_i[D_i]- [E]\rangle$,
$w_2(M)=\pi^*w_2(X)$ and $w_2(\hat M)=\hat\pi^*w_2(\hat X)$. Also $w_2(\hat X)=w_2(X)+[E]$.
From all of this we have:
\begin{itemize}
\item if $w_2(M)=0$ and $N$ odd then $w_2(\hat M)=0$;
\item if $w_2(M) = 0$ and $N$ even then $w_2(\hat M)\neq 0$;
\item if $w_2(M)\neq 0$ and $N$ odd then $w_2(\hat M)\neq 0$.
\end{itemize}
If some $m_i$ are even, then take $m_1,\ldots, m_c$ even and $m_i$ odd for $i>c$. Then
$K=\langle [D_1],\ldots, [D_c]\rangle$. Then:
\begin{itemize}
\item if the isotropy locus of $\hat\pi:\hat M\to \hat X$ is given by the $D_i$ as chosen above, then
$\hat K=\langle [D_1],\ldots, [D_c]\rangle$. As $w_2(\hat X)=w_2(X)+[E]$, then $w_2(\hat M)=\pi^*w_2(\hat X)\neq 0$.
\item take as isotropy locus $D_1,\ldots, D_r$, and also the divisor $E$ with orbit invariants $m_E=2$, $b_E=1$.
Then the Chern class (\ref{eqn:mmm}) gets modified by adding $\frac{m}2 [E]$, but it is still a Kähler class and
primitive. The condition (2) of Theorem \ref{thm:Kollar} still holds, and the homology does not change since
the genus $g_E=0$. Now $\hat K=\langle [D_1],\ldots, [D_c],[E]\rangle$
and hence $w_2(\hat M)=0$.
\end{itemize}
\end{proof}
\noindent \emph{Proof of Theorem \ref{thm:main1}.}
First, we do the case $k=1$. If all $m_i$ are odd, then for $i(M)=0$, we use Theorem \ref{thm:sums1},
and for $i(M)=\infty$, we use Theorem \ref{thm:sums2}.
If some of $m_i$ is even, we arrange $m_1$ even and all other $m_i$, $i>1$, odd. Then if $g_1$ is even, we use Theorem \ref{thm:sums1}
for both $i(M)=0,\infty$. Alternatively, if $g_1$ is odd then we use Theorem \ref{thm:sums2} for both $i(M)=0,\infty$.
The case $k>1$ is reduced to the case $k=1$ by using Theorem \ref{thm:highrank}.
\hfill $\Box$
\begin{remark}
Theorem \ref{thm:main1} implies that for $k\geq 1$ and $\mathbf{t}=1$, the manifolds admitting
a semi-regular Sasakian structure are the same as the manifolds admitting a semi-regular K-contact structure. This is in contrast with the examples of Smale-Barden manifolds with semi-regular K-contact structures but with no Sasakian semi-regular structures constructed in \cite{CMRV} and \cite{MRT}. It seems to be important to understand this in terms of $\mathbf{t}$.
\end{remark}
\section{The case of homology spheres}\label{sec:spheres}
In this section we consider homology spheres, which correspond to the case $k=0$.
Let
$$
\mathcal{T}=\left\{ \frac12 r(r-1) | r\geq 2\right\}
$$
be the set of triangular numbers.
The genus of a smooth curve of degree $d$ in $\CC P^2$ is $g=\frac12 (d-1)(d-2) \in \mathcal{T}$.
\subsection{Proof of Theorem \ref{thm:sphere-semi}}
\begin{proof}
Start with $X=\CC P^2$.
Consider smooth curves $D_i$ of genus $g_i$ intersecting transversally. Put coefficients $m_i$ and
make the Seifert bundle. Use Proposition \ref{prop:abelian} to check that $\pi_1(M)$ is abelian. We still need
to check the properties of Theorem \ref{thm:Kollar}. For (2), we need $\operatorname{gcd}(d_i,m_i)=1$.
Finally, for (3), choosing $c_1(B)=\b H$ we get $c_1(M/X)=\sum \frac{b_i}{m_i}[D_i]$, so
$$
c_1(M/m) = (m\b + \sum b_i M_i d_i) [H]
$$
where $M_i=\frac{m}{m_i}$, which have $\operatorname{gcd}(M_1,\ldots,M_r)=1$.
Finally, we need the property
$$
\operatorname{gcd}(m,M_1d_1,\ldots, M_rd_r)=1
$$
to get a primitive class $c_1(M/m)$.
To see this take a prime number $p$ that $p|m$. Then
there is some $p|m_i$, hence $p\not| M_i$ and $p\not| d_i$ since $\operatorname{gcd}(m_i,d_i)=1$.
So $p\not| M_id_i$.
By Proposition \ref{prop:w2}, the map $\pi^*:H^2(X,\mathbb{Z}_2)\to H^2(M,\mathbb{Z}_2)$ has kernel, hence it must be the
zero map. Therefore $w_2(X)=0$ by (\ref{eqn:w2M}).
Conversely, let $\pi: M\rightarrow X$ be a semi-regular Sasakian manifold with $b_2(M)=1$,
satisfying the assumptions of Theorem \ref{thm:Kollar}. Therefore, the isotropy locus is a collection of
complex curves $D_i\subset X$ with isotropy $m_i$. Moreover, $X$ must be simply connected and $b_2(X)=1$.
As fake projective planes are never simply-connected, we have
$X=\CC P^2$. Each $D_i$ must realize a homology class $d_iH$, where
$H$ is the hyperplane class in $H^2(X,\mathbb{Z})=\mathbb{Z}\langle H\rangle$. As all $D_i$ intersect,
we have that $\operatorname{gcd}(m_1,\ldots, m_r)=1$.
It is well known \cite{GS} that the genus of $D_i$ is ${1\over 2}(d_i-1)(d_i-2)$, that is a triangular number.
The condition (2) of Theorem \ref{thm:Kollar} says that $H^2(X,\mathbb{Z})\to H^2(D_i,\mathbb{Z}_{m_i})$ is surjective,
which implies that $\operatorname{gcd}(d_i,m_i)=1$.
\end{proof}
Now we want to compare the Smale-Barden manifolds which are rational homology spheres
admitting a semi-regular Sasakian structure with those admitting a semi-regular K-contact structure.
\begin{proposition}\label{prop:last}
Let $M$ be a Smale-Barden manifold with
$H_2(M,\mathbb{Z})= \bigoplus_{i=1}^r \mathbb{Z}_{m_i}^{2g_i}$ and spin, admitting a semi-regular K-contact structure. Then
$m_i\geq 2$ are pairwise coprime, $g_i=\frac12 (d_i-1)(d_i-2)\in \mathcal{T}$, and
either $\operatorname{gcd}(m_i,d_i)=1$ for all $i$, or $\operatorname{gcd}(m_i,d_i+3)=1$ for all $i$.
\end{proposition}
\begin{proof}
The proof is analogous to the last part of the proof of Theorem \ref{thm:sphere-semi}. The difference is that
now $X$ is a symplectic $4$-manifold with $b_2=1$. As $\pi_1(M)=1$ then $\pi_1^{\mathrm{orb}}(X)=1$, which
implies that $\pi_1(X)=1$. The intersection form must be positive definite, hence $H_2(X,\mathbb{Z})=\mathbb{Z}\langle H\rangle$,
for some integral class $H$ with $H^2=1$. The canonical class of the almost-complex structure $K_X$ satisfies
\cite[Theorem 1.4.15]{GS}
$$
K_X^2+\chi(X) =12.
$$
Then $K_X^2=9$, and hence $K_X=\pm 3H$.
In the first case $D_i^2+K_X\cdot D_i=d_i^2-3d_i=2g-2$, hence $g=(d_i-1)(d_i-2)/2$. In the second case,
$D_i^2+K_X\cdot D_i=d_i^2+3d_i=2g-2$, hence $g=(d_i+1)(d_i+2)/2=(k-1)(k-2)/2$, with $k=d_i+3$. Finally it must
be $\operatorname{gcd}(m_i,d_i)=1$ as before.
\end{proof}
\begin{remark}
The only symplectic $4$-manifold which is simply-connected, has $b_2=1$ and $K_X\cdot[\omega]<0$ is $\CC P^2$
by a result of Taubes \cite{Tau}. It is not known whether there are simply-connected symplectic $4$-manifolds with
$b_2=1$ and $K_X\cdot [\omega]>0$, although it is expected that they do not exist. In such case, the possibility
$\operatorname{gcd}(m_i,d_i+3)=1$ would not appear in Proposition \ref{prop:last}, and then the classification of semi-regular K-contact
and semi-regular Sasakian Smale-Barden rational homology spheres would be the same.
\end{remark}
\section{Definite Sasakian structures and proof of Theorem \ref{thm:main2}}
Recall that the Reeb vector field $\xi$ on a co-oriented contact manifold $(M,\eta)$ determines a 1-dimensional foliation
$\mathcal{F}_{\xi}$ called the {\it characteristic foliation}. If we are given a Sasakian manifold $(M,\eta,\xi,g,J)$, then
one can define {\it basic Chern classes} $c_k(\mathcal{F}_{\xi})$ of $\mathcal{F}_{\xi}$ which are elements of the
basic cohomology $H^{2k}_B(\mathcal{F}_{\xi})$ (see \cite[Theorem/Definition 7.5.17]{BG}).
We say that a Sasakian structure is positive (negative) if $c_1(\mathcal{F}_{\xi})$ can be represented by a positive
(negative) definite $1$-form. A Sasakian structure is called null, if $c_1(\mathcal{F}_{\xi})=0$. If none of these, it is called indefinite.
\medskip
\subsection{ Proof of Theorem \ref{thm:main2}}
The proof is a corollary to Theorem \ref{thm:sphere-semi}. To see this, one recalls the results below and makes a comparison.
\begin{theorem}[{\cite[Theorem 10.2.17]{BG}}]\label{thm:pos-sas}
Suppose that $M$ admits a positive Sasakian structure. Then the torsion subgroup of $H_2(M,\mathbb{Z})$ is one of the following:
$$
(\mathbb{Z}_m)^2, m>0, (\mathbb{Z}_5)^4, (\mathbb{Z}_4)^4,
(\mathbb{Z}_3)^4,(\mathbb{Z}_3)^6, (\mathbb{Z}_3)^8, (\mathbb{Z}_2)^{2n},\,n>0.
$$
\end{theorem}
\begin{theorem}[{\cite[Theorem 10.3.14]{BG}}]\label{thm:pos-sphere}
Let $M$ be a rational homology sphere. If it admits a Sasakian structure, then it is either positive,
and the torsion in $H_2(M,\mathbb{Z})$ is restricted by Theorem \ref{thm:pos-sas} or it is negative.
\end{theorem}
\begin{remark}
The proof that rational homology spheres can carry only definite Sasakian structures actually is given in \cite[Proposition 7.5.29]{BG}.
\end{remark}
\section{Regular Sasakian structures } \label{sec:bt=0}
The description of regular Sasakian structures is obtained in \cite[Proposition 10.4.4]{BG}, by considering
various types of Sasakian structures on torsion-free Smale-Barden manifolds classified by
Theorem \ref{thm:sb-classification}. For completeness we show that our methods allow us
to describe regular Sasakian structures in a unified way (see Theorem \ref{thm:torsionfree}).
Note that if a base $X$ is a smooth manifold ($X$ is considered with an empty singular locus), then
Theorem \ref{thm:seifert} yields a circle bundle $M\rightarrow X$ over a Kähler manifold $X$.
This bundle is determined by the cohomology class $[\omega]\in H^2(X,\mathbb{Z})$ as the first Chern
class of the circle bundle. Thus, regular Sasakian structures are in one to one correspondence with such bundles.
Let $M\rightarrow X$ be a circle bundle with Euler class $e$. Assume that $X$ is a simply
connected closed $n$-manifold. The class $e$ is primitive if the map $\langle e,-\rangle: H_2(X,\mathbb{Z})\rightarrow \mathbb{Z}$
is surjective, equivalently if the map
$e\cup: H^{n-2}(X,\mathbb{Z})\rightarrow H^n(X,\mathbb{Z})\cong\mathbb{Z}$ is onto.
\begin{theorem}[{\cite[Theorem 9.12]{H}}]\label{thm:bundle-classif}
Let $X$ be a simply connected oriented $4$-manifold and let $M\rightarrow X$ be
a circle bundle over $X$ with primitive Euler class $e$. Then $M$ is diffeomorphic to
\begin{itemize}
\item $M\cong\#(b_2(X)-1)S^2\times S^3$, if $M$ is spin;
\item $M\cong \#(b_2(X)-2)S^2\times S^3\#S^2\tilde{\times}S^3$, if $M$ is not spin.
\end{itemize}
\end{theorem}
The condition on $M$ to be spin is described as follows.
\begin{theorem}[{\cite[Lemma 9.10]{H}}]\label{thm:spin}
The total space $M$ of the circle bundle determined by the primitive Euler class $e$ is spin if and only if $w_2(X)\equiv 0,e \pmod2$.
\end{theorem}
\begin{lemma} \label{lem:1}
For a smooth projective algebraic surface $S$, the Nakai–Moishezon criterion states that
a divisor $D$ is ample if and only if its self-intersection number $D\cdot D>0$,
and for any irreducible curve $C$ on S we have $D\cdot C > 0$.
\end{lemma}
\begin{corollary}
Let $X=\CC P^2\# k\overline{\CC P}{}^2$ be the blow-up of the projective plane at $k$ points. A divisor $D=a H-\sum b_i E_i$ is
ample if $b_i>0$ and $a>\sum b_i$.
\end{corollary}
\begin{proof}
Using Lemma \ref{lem:1} with $C=H$ and $C=E_i$, we get $a>0$, $b_i>0$. Now let $C$ be any curve which is not
an exceptional divisor. Then $C\equiv dH-\sum \alpha_i E_i$. Intersecting with $H$ and $E_i$ we get
$d>0$, $\alpha_i>0$. Intersecting with a line $L=H-E_i$, we get $d-\alpha_i\geq 0$. Now
$D\cdot C=ad-\sum \alpha_i b_i>\sum b_i (d-\alpha_i)\geq 0$.
\end{proof}
\begin{theorem} \label{thm:torsionfree}
Let $M$ be a Smale-Barden manifold such that $H_2(M,\mathbb{Z})$ has no torsion. Then $M$ always admits a regular
Sasakian structure.
\end{theorem}
\begin{proof}
Take $X=\CC P^2\# k\overline{\CC P}{}^2$, the blow-up of the projective plane at $k$ points. Then $b_2(X)=k+1$ and the homology
of $X$ is $H_2(X)=\mathbb{Z}\langle H,E_1,\ldots, E_k\rangle$, where $H$ is the hyperplane class and $E_i$ are the exceptional divisors. Any
class $e=aH-\sum b_iE_i$ is ample (that is, defined by a Kähler cohomology class)
if $a>\sum b_i$, $b_i>0$ for all $i$. And it is primitive if $\operatorname{gcd}(a,b_i)=1$. The canonical class of $X$ is
$K_X=-3H+\sum E_i$, and $w_2(X)\equiv K_X \pmod2$. Hence $K_X\equiv e$ if and only if all $a,b_i$ are odd numbers.
Therefore, to get spin regular Sasakian manifolds with $b_2(M)=k$ and $H_2(M,\mathbb{Z})=\mathbb{Z}^k$, take a circle bundle $M\to X$
with $e=(2k+1)H-\sum E_i$. To get a non-spin regular Sasakian manifold, take $M\to X$ with $e= 2k H-\sum E_i$.
\end{proof}
\begin{remark}
Theorem \ref{thm:torsionfree} implies that for $\mathbf{t}=0$, the manifolds admitting
a regular Sasakian structure are the same as the manifolds admitting a regular K-contact structure.
\end{remark}
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"redpajama_set_name": "RedPajamaArXiv"
} | 6,414 |
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} | {
"redpajama_set_name": "RedPajamaGithub"
} | 4,906 |
\section{Introduction}
\label{sec:intro}
As a relativistic wave equation,
the Dirac equation provides naturally a description of an electron \cite{Dirac1928}.
Following Dirac's discovery of the linear equation of the electron,
there appears the fundamental idea of nonlinear description of an elementary spin-1/2 particle
which makes it possible to take into account its self-interaction \cite{Ivanenko1938,FinkelsteinLelevierRuderman1951,FinkelsteinFronsdalKaus1956}.
Heisenberg put forward
the idea to use a nonlinear Dirac (NLD) equation
as a possible basis model for a unified field theory \cite{Heisenberg1957}.
A key feature of the NLD equation is that
it allows solitary wave solutions or particle-like solutions --- the stable localized solutions
with finite energy and charge \cite{Ranada1983}.
That is, the particles appear as intense localized regions of field
which can be recognized as the basic ingredient
in the description of extended objects in quantum field theory \cite{Weyl1950}.
Different self-interactions give rise to different NLD models
mainly including
the Thirring model \cite{Thirring1958},
the Soler model \cite{Soler1970},
the Gross-Neveu model \cite{GrossNeveu1974}
(equivalent to the massless Soler model),
and the bag model \cite{Mathieu1984}
(\ie the solitary waves with finite compact support),
all of which attracted wide interest of physicists
and mathematicians around the 1970s and 1980s,
especially on looking for the solitary wave solutions
and investigating the related physical and mathematical properties \cite{Ranada1983}.
For the NLD equation in (1+1) dimensions (\ie one time dimension plus one space dimension),
several analytical solitary wave solutions are derived in \cite{LeeKuoGavrielides1975,ChangEllisLee1975}
for the quadric nonlinearity, \cite{Mathieu1985-prd} for fractional nonlinearity as well as
\cite{Stubbe1986-jmp,CooperKhareMihailaSaxena2010} for general nonlinearity
by using explicitly the constraint resulting from energy-momentum conservation,
and summarized by Mathieu \cite{Mathieu1985-jpa-mg}. In contrast,
there are few explicit solutions in (1+3) dimensions except for some
particular cases shown in \cite{Werle1977-lmp} in spite of
their existence claimed by mathematicians for various situations
\cite{Vazquez1977,Werle1981-appb,MathieuMorris1984,CazenaveVazquez1986,Merle1988-jde,BalabaneCazenaveDouadyMerle1988,BalabaneCazenaveVazquez1990,EstebanSere1995}
(the readers are referred to an overview \cite{EstebanSere2002} on this topic),
and most understanding is based on
numerical investigations, \eg \cite{Rafelski1977-plb,Takahashi1979-jmp,Alvarez1985}.
Beyond this, the study of the NLD equation in (1+1) dimensions could be very helpful for that in (1+3) dimensions
since the (1+1)-dimensional NLD equation correspond to the asymptotic form of the equation in
the physically interesting case of (1+3) dimensions as emphasized by Kaus \cite{Kaus1976}.
That is, some qualitative properties of the NLD solitary waves could be similar in such two cases.
An interesting topic for the NLD equation is the stability issue, which has been the central topic in works
spread out over several decades in an effort that is still ongoing.
Analytical studies of the NLD solitary wave stability face serious obstacles \cite{StraussVazquez1986,AlvarezSoler1986,BlanchardStubbeVazquez1987},
while results of computer simulations are contradictory \cite{Bogolubsky1979-pla,AlvarezSoler1983,Mathieu1983,Alvarez1985}.
The stability analysis of the NLD solitary waves is still a very challenging mathematical problem to date \cite{Barashenkov1998,CooperKhareMihailaSaxena2010}.
Recent efforts in this direction can be found in \cite{Chugunova2006,Pelinovsky2010,Haddad2011-epl,Comech2011,Boussaid2012,Berkolaiko2012}.
Another rising mathematical interest related to the NLD equation is the analysis of global well-posedness,
\eg see \cite{Pelinovsky2010-arxiv,Bournaveas2012} and references therein.
In the case of that
theoretical methods were not able to give the satisfactory results,
numerical methods were used for obtaining
the solitary wave solutions of the NLD equation
as well as for investigating the stability.
An important step in this direction was made by
Alvarez and Carreras in 1981 \cite{AlvarezCarreras1981},
who simulated the interaction dynamics
between the (1+1)-dimensional NLD solitary waves of
different initial charge for the Soler model \cite{Soler1970} by
using a second-order accurate Crank-Nicholson (CN) scheme \cite{AlvarezKuoVazquez1983}.
They first saw there: charge and energy interchange except for some particular
initial velocities of the solitary waves; inelastic interaction in binary collisions;
and oscillating state production from binary collisions.
Motivated by their work, Shao and Tang revisited this interaction dynamics problem in 2005 \cite{ShaoTang2005}
by employing a fourth-order accurate Runge-Kutta discontinuous Galerkin (RKDG) method \cite{ShaoTang2006}.
They not only recovered the phenomena mentioned by Alvarez and Carreras but also revealed
several new ones, \eg
collapse in binary and ternary collisions of two-humped NLD solitary waves \cite{ShaoTang2005};
a long-lived oscillating state formed with an approximate constant frequency in collisions of two standing waves \cite{ShaoTang2006};
full repulsion in binary and ternary collisions of out-of-phase waves \cite{ShaoTang2008}.
Their numerical results also inferred that the two-humped profile could undermine
the stability during the scattering of the NLD solitary waves.
Note in passing that the two-humped profile was first pointed out by Shao and Tang \cite{ShaoTang2005}
and later gotten noticed by other researchers \cite{CooperKhareMihailaSaxena2010}.
Besides the often-used CN \cite{AlvarezKuoVazquez1983}
and RKDG methods \cite{ShaoTang2006},
there exist many other numerical schemes for solving the (1+1)-dimensional
NLD equation:
split-step spectral schemes \cite{FrutosSanz-Serna1989},
the linearized CN scheme \cite{Alvarez1992},
the semi-implicit scheme \cite{Jimcnez1994}\cite{Bournaveas2012},
Legendre rational spectral methods \cite{WangGuo2004},
multi-symplectic Runge-Kutta methods \cite{HongLi2006},
adaptive mesh methods \cite{WangTang2007} \etc.
The fourth-order accurate RKDG method \cite{ShaoTang2006}
is very appropriate for investigating the interaction dynamics of the NLD solitary waves
due to their ability to capture the discontinuous or strong gradients without producing spurious oscillations,
and thus performs better than the second-order accurate CN scheme \cite{AlvarezKuoVazquez1983}.
However, the high cost due to the relatively more freedoms used in each cell
and the stringent time step constraint
reduce its practicality
in more realistic simulations where realtime and quantitative results are required.
Recently, there has been a remarkable upsurge
of the interest in the NLD models,
as they emerge naturally as practical models
in other physical systems,
such as the gap solitons in nonlinear optics \cite{Barashenkov1998},
Bose-Einstein condensates in honeycomb optical lattices \cite{Haddad2009},
as well as matter influencing
the evolution of the Universe in cosmology \cite{Saha2012}.
In view of such new trend,
longtime stable, efficient, conservative and high-order accurate numerical methods
for solving the NLD equation are highly desirable.
Finite difference methods, usually as the first try in practice,
enable easy coding and debugging and thus are often used by physicists and engineers.
However, detailed discussion and careful comparison on finite difference solvers for the NLD equation
are not existed.
To this end, the present work as the first step will extendedly propose
the finite deference schemes for solving the NLD equation with the scalar and vector self-interaction.
A general and precise comparison among them will be presented.
However, all of these finite difference methods are often of the second order accuracy and thus
sustain fast error growth with respect to time. To achieve relatively slow error growth,
high-order accurate numerical methods are required.
By exploiting the local conservation laws of the NLD equation,
we present exponential operator splitting (OS) schemes which are time reversible and
can conserve the discrete charge. One of the high-order accurate OS schemes is afterwards
adopted to investigate the interaction dynamics for
the NLD solitary waves under the general scalar and vector self-interaction.
It should be noted that
the experiments carried out in the literatures
are all limited to the collisions of
the NLD solitary waves under the quadric scalar self-interaction.
Here, the binary collisions of the NLD solitary waves under the cubic scalar self-interaction
or under the vector self-interaction or under the linear combination of the scalar and vector self-interactions
are all studied for the first time.
The paper is organized as follows. There is a brief review of the NLD equation in
Section \ref{sec2:gnld} and the solitary wave solutions are also derived there for
the general scalar and vector self-interaction.
The numerical schemes are presented in Section \ref{sec:DiffSch} and
corresponding numerical analysis is given
in Section \ref{sec:discussion4fdm}. The numerical results are presented with discussion in Section \ref{sec:discussion4fdm}.
The paper is concluded in Section \ref{sec:conclusion} with a few remarks.
\section{Preliminaries and notations}
\label{sec2:gnld}
This section will introduce the (1+1)-dimensional NLD equation
with the scalar and vector self-interaction and derive its two solitary wave solutions.
Throughout the paper, units in which both the speed of light and the reduced Planck constant
are equal to one will be used.
\subsection{Nonlinear Dirac equation}
Our attention is restricted to the NLD equation in $(1+1)$ dimensions
which can be written in the covariant form
\begin{equation}
(\mi\vgamma^\mu\partial_\mu-m)\vPsi+\partial L_\text{I}[\vPsi]/\partial \wbar{\vPsi}=0, \label{generalnld}
\end{equation}
where $\vPsi$ is the spinor with two complex components,
$\wbar{\vPsi}:=\vPsi^\dag\vgamma^0$ denotes the adjoint spinor,
$\vPsi^\dag$ is the complex conjugate transpose of $\vPsi$,
$L_\text{I}[\vPsi]$ denotes the self-interaction Lagrangian,
the Greek index $\mu$ runs from 0 to 1,
the Einstein summation convection has been applied,
$\mi$ is the imaginary unit,
$m$ is the rest mass,
$\partial_\mu=\pp{}{x^\mu}$ stands for the covariant derivative,
and $\vgamma^\mu$, for $\mu=0,1$, are the gamma matrices or the Dirac matrices, chosen
as those in \cite{AlvarezCarreras1981,ShaoTang2005},
$$
\vgamma^0=\begin{pmatrix} 1& 0\\ 0 &-1\end{pmatrix},\quad
\vgamma^1=\begin{pmatrix} 0& 1\\ -1 &0\end{pmatrix}.
$$
In fact, Eq. \eqref{generalnld} is the equation of motion for the classical spinorial particle
with the Lagrangian being a sum of the Dirac Lagrangian and the self-interaction Lagrangian, \ie
\begin{equation}
L[\vPsi]=\wbar{\vPsi} (\mi
\vgamma^\mu\partial_\mu-m)\vPsi+{L}_\text{I}[\vec\Psi]. \label{total_lag}
\end{equation}
There exist several different NLD models in the literature,
where two important models in (1+1) dimensions are the scalar self-interaction of Soler \cite{Soler1970}
\begin{equation}\label{self-scalar1}
L_{\text{s}}[\vPsi]:=\wbar{\vPsi}\vPsi,
\end{equation}
and the vector self-interaction of Thirring \cite{Coleman1975}
\begin{equation}\label{self-vector1}
L_{\text{v}}[\vPsi]:= \wbar{\vPsi}\vgamma^\mu\vPsi\wbar{\vPsi}\vgamma_\mu\vPsi,
\end{equation}
where $\vgamma_\mu=\eta_{\mu\nu}\vgamma^\nu$ with the Minkowski metric
$\eta_{\mu\nu}=\text{diag}(1,-1)$ on spacetime,
which implies $\vgamma_\mu=(-1)^\mu\vgamma^\mu$.
This paper will focus on the NLD equation \eqref{generalnld}
with a more general self-interaction \cite{Stubbe1986-jmp,NogamiToyama1992}
\begin{equation}\label{generalLI}
{L}_\text{I}[\vPsi] = s \left(L_{\text{s}}[\vPsi]\right)^{k+1}
+ v \left(L_{\text{v}}[\vPsi]\right)^{(k+1)/2},
\end{equation}
and extendedly propose and compare its numerical methods,
where $s$ and $v$ are two real numbers and $k$ is a positive real number.
If the spinor $\vPsi$ is scaled by a constant factor as $\vPsi^\prime=\sqrt{\alpha}\vPsi$
with $\alpha\in\mathbb{C}$,
then the scaled self-interaction Lagrangian will be
$\alpha^{k+1}L_\text{I}[\vPsi]$ which shows that the power exponent to $\alpha$ is $k+1$.
In such sense, we call that the self-interaction Lagrangian $L_\text{I}$ has the power exponent $k+1$
\cite{Mathieu1985-prd,Stubbe1986-jmp,CooperKhareMihailaSaxena2010}.
Hereafter the \textit{quadric} and \textit{cubic} self-interaction will be referred to the case $k=1$
and the case $k=2$, respectively.
The self-interaction \eqref{generalLI} implies
the so-called homogeneity relation \cite{Mathieu1985-jpa-mg,Stubbe1986-jmp}
\begin{equation*}
\wbar{\vPsi}\pp{ {L}_\text{I}[\vPsi]}{\wbar{\vPsi}}=(k+1){L}_\text{I}[\vPsi]. \label{ppL=L}
\end{equation*}
Combining it with the definition of the Lagrangian $L[\vPsi]$ and \eqref{generalnld}
gives
\begin{equation} \label{LLIrelation}
{L}[\vPsi] = -k {L}_\text{I}[\vPsi].
\end{equation}
For the NLD equation \eqref{generalnld} with \eqref{generalLI},
one may still verify the following local conservation laws for the current vector $J_\mu$
and the energy-momentum tensor $T_{\mu\nu}$:
\begin{equation}
\partial^\mu J_\mu =0, \quad
\partial^\mu T_{\mu\nu} =0, \label{Tconserve}
\end{equation}
where
\begin{equation*}
J_\mu = \wbar{\vPsi}\vgamma_\mu\vPsi,\quad
T_{\mu\nu} =\mi\wbar{\vPsi}\vgamma_\mu\partial_\nu\vPsi-\eta_{\mu\nu}{L}[\vPsi].
\end{equation*}
For localized solitary waves $\vPsi=(\psi_1,\psi_2)^T$,
one may derive a direct corollary of \eqref{Tconserve},
\ie the following global conservation laws \cite{ShaoTang2006}.
\begin{Proposition}\label{prop-cl}
Assume that $\disp \lim_{|x|\rightarrow +\infty}|\psi_i(x,t)|=0$ and
$\disp \lim_{|x|\rightarrow +\infty}|\partial_x{\psi_i}(x,t)|<+\infty$
hold uniformly for $t\geq 0$ and $i=1,2$. The total energy $E$, the total linear momentum
$P$, and the total charge $Q$, defined respectively by
\begin{equation}\label{qep-infty}
E(t):= \int_{-\infty}^{+\infty} T_{00} \dif x, \quad
P(t):= \int_{-\infty}^{+\infty} T_{01} \dif x, \quad
Q(t):= \int_{-\infty}^{+\infty} J_0 \dif x,
\end{equation}
satisfy
$$ \frac{\dif
}{\dif t}E(t)=0, \quad \frac{\dif}{\dif t}P(t)=0, \quad \frac{\dif}{\dif t}Q(t)=0.$$
\end{Proposition}
The properties \eqref {LLIrelation} and \eqref{Tconserve} will be also exploited to find the solitary wave solutions of the
(1+1)-dimensional NLD equation \eqref{generalnld} with $L_\text{I}$ given in \eqref{generalLI}
in the next subsection.
\subsection{Standing wave solution}
\label{appA:derivation}
This subsection will derive the standing wave solutions of
the (1+1)-dimensional NLD equation \eqref{generalnld}
with the self-interaction \eqref{generalLI} in the spirit of the technique used in \cite{LeeKuoGavrielides1975,ChangEllisLee1975,Mathieu1985-jpa-mg,Stubbe1986-jmp}.
The solution $\vPsi=(\psi_1,\psi_2)^T$
of the NLD equation \eqref{generalnld} with $L_\text{I}$ in \eqref{generalLI},
having the form
\begin{equation*
\psi_1(x,t)=\me^{-\mi \omega t}\varphi(x),\quad \psi_2(x,t)=\me^{-\mi \omega t}\chi(x),
\end{equation*}
is wanted, where $m>\omega\geq 0$,
and $|\varphi(x)|$ and $|\chi(x)|$ are assumed to decay exponentially as $|x|\rightarrow +\infty$ or
have the finite compact support.
For such solution, it is not difficult to verify
that both the Lagrangian $L[\vPsi]$ and the energy-momentum tensor $T_{\mu\nu}$
are independent of the time $t$, because
\begin{equation}\label{total_lag_ti}
\begin{aligned}
{L}[\vPsi] &\equiv \omega \wbar{\vpsi}\vgamma^0\vpsi+\mi\wbar{\vpsi}\vgamma^1\vpsi_x-m\wbar{\vpsi}\vpsi
+{L}_\text{I}[\vPsi],
\\
T_{00} &\equiv-\mi \wbar{\vpsi}\vgamma^1\vpsi_x+m\wbar{\vpsi}\vpsi- {L}_\text{I}[\vPsi],
\quad T_{01}\equiv \mi\wbar{\vpsi}\vgamma^0\vpsi_x, \\
T_{10} &\equiv-\omega \wbar{\vpsi}\vgamma^1\vpsi, \quad T_{11}\equiv -\mi \wbar{\vpsi}\vgamma^1\vpsi_x+{L}[\vPsi],
\end{aligned}
\end{equation}
where $\vec \psi(x)=\big(\varphi(x), \chi(x)\big)^T$.
Using the conservation laws \eqref{Tconserve} further gives
\begin{equation}
T_{10}=-\omega \wbar{\vpsi}\vgamma^1\vpsi =0, \quad
T_{11}=-\mi\wbar{\vpsi}\vgamma^1\vpsi_x+ {L}[\vPsi] =0. \label{t11}
\end{equation}
The first equation implies that $\varphi^\ast\chi$ is imaginary because
\begin{equation*}
\wbar{\vpsi}\vgamma^1\vps
=\varphi^\ast\chi+\varphi\chi^\ast=0.
\end{equation*}
Thus, without loss of generality, we may assume that $\varphi(x)$ is real and $\chi(x)$ is imaginary,
and they are in the form
\begin{equation}
\vpsi(x)=\left(
\begin{matrix}
\varphi(x)\\
\chi(x)
\end{matrix}
\right)=
R(x) \left( \begin{matrix}
\cos\big(\theta(x)\big) \\
\mi \sin\big(\theta(x)\big)\end{matrix}\right),\label{phaseexp}
\end{equation}
where both $R(x)$ and $\theta(x)$ are pending real functions, and $R(x)$ is
assumed to satisfy the inequality $Q(t)\equiv \int_{-\infty}^{+\infty} R^2(x)\dif x<+\infty$.
On the other hand, combining the first equation in \eqref{total_lag_ti} with the second equation in \eqref{t11}
yields
\begin{equation}\label{key-Li}
\omega \vpsi^\dag\vpsi- m\wbar{\vpsi}\vpsi + L_\text{I}[\vPsi]=0,
\end{equation}
which becomes for \eqref{phaseexp}
\begin{equation}
\label{eq:Li-2}
R^2 \big(\omega- m \cos(2\theta)\big) + R^{2k+2}\big(s\cos^{k+1}(2\theta) + v\big) = 0.
\end{equation}
Combining \eqref{key-Li} with \eqref{LLIrelation} leads to
\begin{equation*}
k \omega \vpsi^\dag\vpsi-k m\wbar{\vpsi}\vpsi-\mi\wbar{\vpsi}\vgamma^1\vpsi_x=0,
\label{key-eq1}
\end{equation*}
which reduces to for \eqref{phaseexp}
\begin{equation} \label{theta}
\dd{\theta}{x}=-k
\big(\omega - m\cos (2\theta)\big).
\end{equation}
Because
\begin{equation*}
\int_{0}^{\theta} \frac{\dif \tilde{\theta}}{-k\omega+km \cos (2 \tilde{\theta})}
={\frac{1}{
\sqrt{k^2(m^2-\omega^2)}} \tanh
^{-1}\left(\frac{k(m+\omega)}{\sqrt{k^2(m^2-\omega^2)}}\tan
\left(\theta\right)\right)},
\end{equation*}
for $\theta \in \big(-0.5 \cos^{-1}(\omega/m), 0.5 \cos^{-1}(\omega/m)\big)\subset(-\pi/2,\pi/2)$ when $m>\omega\geq 0$,
the solution of \eqref{theta} may be derived as follows:
\begin{equation}
\theta(x) = \tan^{-1} \left( \sqrt{\frac{m-\omega}{m+\omega}} \tanh\Big(k \sqrt{m^2-\omega^2} x \Big)\right),
\label{theta-1}
\end{equation}
for initial data $\theta(0) =0$ and $k>0$. In fact,
under the previous assumption, one may verify $\theta(x)\in (-\pi/4,\pi/4)$.
{If coefficients $s$ and $v$ in \eqref{generalLI} belong to the set
$\{v\geq 0, s>-v\}$} for $m>\omega>0$, or $\{v>0, s>-v\}$ for $m>\omega=0$,
then from Eq. \eqref{eq:Li-2} one has non-trivial $R(x)$ for the localized solution as follows
\begin{equation}\label{R2}
R(x) = \pm \left(\frac{m\cos\left(2\theta(x)\right)-\omega}
{s \cos^{k+1}\left(2\theta(x)\right) + v }\right)^{{1}/{2k}}.
\end{equation}
Hereto, the standing wave solution of the
NLD equation \eqref{generalnld} with \eqref{generalLI} has been derived,
and will be denoted as follows
\begin{equation}
\vPsi^\text{sw}(x,t) = \left(
\begin{array}{c}
\psi^\text{sw}_1(x,t) \\ \psi^\text{sw}_2(x,t)
\end{array}\right)
= \me^{-\mi \omega t} R(x)\left(
\begin{array}{c}
\cos\big(\theta(x)\big) \\ \mi \sin\big(\theta(x)\big)
\end{array} \right), \label{eq:bound-state}
\end{equation}
where $\theta(x)$ and $R(x)$ are given in Eqs. \eqref{theta-1} and \eqref{R2}, respectively.
This solution represents a solitary wave with zero velocity and
contains some special cases in the literature \eg
\cite{AlvarezCarreras1981,CooperKhareMihailaSaxena2010}.
\begin{Remark}\rm
It has been pointed out in \cite{ShaoTang2005}
that the profile of the charge density $J_0(x,t)$
for the standing wave \eqref{eq:bound-state} under the scalar self-interaction (\ie $s\neq 0$ and $v=0$)
with $k=1$
can be either one-humped or two-humped,
which is also recently confirmed for any $k>0$ by other researchers
in \cite{CooperKhareMihailaSaxena2010}.
They further pointed out there that
the profile can only be one-humped for any $k>0$ in the case of $s=0$ and $v\neq 0$.
For the linear combined self-interaction \eqref{generalLI} with $s\neq 0$ and $v\neq0$,
we find that the profile can also be one-humped (see Figs.~\ref{fig.os-dg-rhoq} and \ref{fig:k1-sv}
where the charge density is denoted by $\rho_Q(x,t)$ for convenience)
or two-humped (see Fig.~\ref{fig:k1-2humped-sv}).
\end{Remark}
\subsection{Solitary wave solution with nonzero velocity}
\label{boostforMW}
This subsection will derive another solution of the (1+1)-dimensional NLD equation \eqref{generalnld} with the self-interaction \eqref{generalLI} by using the Lorentz covariance of the NLD equation. Consider a frame F with an observer O
and coordinates $(x,t)$. The observer O
describes a particle by the wavefunction $\vPsi(x,t)$
which obeys the NLD equation \eqref{generalnld} with $L_\text{I}$ given in \eqref{generalLI}, \ie
\begin{equation}\label{nld-lorentz}
\left(\mi \vgamma^0\pd{}{t}+\mi \vgamma^1\pd{}{x}-m\right)\vPsi(x,t)
+\left(\partial L_\text{I}[\vPsi]/\partial \wbar{\vPsi}\right)(x,t)=0.
\end{equation}
In another inertial frame F$'$ with an observer O$'$ and coordinates $(x',t')$ given by
the Lorentz transformation with a translation in the $x$-direction
\begin{align}\label{boost}
x'=\gamma \big((x-x_0)-Vt\big), \quad t'=\gamma\big(t-V(x-x_0)\big),
\end{align}
which is called ``boost'' in the $x$--direction,
where $x_0$ is any given position,
$V$ is the relative velocity between frames in the $x$-direction,
and $\gamma=1/\sqrt{1-V^2} $ is the Lorentz factor.
According to the relativity principle,
the observer O$'$ describes the same particle by $\vPsi'(x',t')$ which should also satisfy
\begin{equation}\label{nld-prime}
\left(\mi \vgamma^0\pd{}{t'}+\mi \vgamma^1\pd{}{x'}-m\right)\vPsi'(x',t')
+\left(\partial L_\text{I}'[\vPsi']/\partial \wbar{\vPsi}'\right)(x',t')=0.
\end{equation}
Using some algebraic manipulations, the ``transformation'' matrix $\vec{S}$
may be found as follows
\begin{align}\label{boost2b}
\vec{S}
= \left(
\begin{matrix}
\sqrt{\frac{\gamma+1}{2}} & \text{sign}(V)\sqrt{\frac{\gamma-1}{2}} \\
\text{sign}(V)\sqrt{\frac{\gamma-1}{2}} & \sqrt{\frac{\gamma+1}{2}}
\end{matrix}
\right),
\end{align}
which takes $\vPsi(x,t)$ to $\vPsi'(x',t')$ under the Lorentz transformation
\eqref{boost}, \ie
\begin{align} \label{boost2}
\vPsi'(x',t^\prime) = \vec{S} \vPsi(x,t),
\end{align}
where $\text{sign} (x)$ is the sign function
which returns $1$ if $x>0$, $0$ if $x=0$,
and $-1$ if $x<0$.
Applying the transformation \eqref{boost2} with \eqref{boost2b} to
the standing wave solution \eqref{eq:bound-state} gives
another solution of
the NLD equation \eqref{generalnld}-\eqref{generalLI},
\ie the moving wave solution
\begin{align}
\vPsi^\text{mw}(x-x_0,t) = &
\left(
\begin{matrix}
\sqrt{\frac{\gamma+1}{2}} \psi^\text{sw}_1(x',t') + \mathrm{sign}(V)\sqrt{\frac{\gamma-1}{2}} \psi^\text{sw}_2(x',t')\\
\mathrm{sign}(V)\sqrt{\frac{\gamma-1}{2}} \psi^\text{sw}_1(x',t') + \sqrt{\frac{\gamma+1}{2}} \psi^\text{sw}_2(x',t')
\end{matrix}
\right). \label{mw}
\end{align}
This solution represents a NLD solitary wave with velocity $V$
and will return to the standing wave \eqref{eq:bound-state} if setting $V=0$ and $x_0=0$.
\subsection{Time reversibility}
This subsection will show that the NLD equation \eqref{generalnld} with $L_\text{I}$ given in \eqref{generalLI}
remains invariant under the time reversal operation
\begin{equation} \label{TimeR}
x^\prime = x, \quad t^\prime = -t, \quad
\vPsi^\prime(x^\prime,t^\prime) = \vK \vPsi(x,t), \quad \vK=\vgamma^0 \mathcal{C},
\end{equation}
where $\mathcal{C}$ denotes the complex conjugate operation on $\vPsi(x,t)$, \ie $\vPsi^\ast(x,t)= \mathcal{C}\vPsi(x,t)$,
the time-reversal operator $\vK$ satisfies
\begin{equation}\label{k-relation}
\vK^\dag\vK=\vI, \quad \vK\vgamma^0 = \vgamma^0 \vK, \quad \vK\vgamma^1 = - \vgamma^1 \vK,
\end{equation}
due to the anticommutation relation $\{\vgamma^\mu,\vgamma^\nu\}=2 \eta^{\mu\nu}\vI$,
and $\vI$ is the unit matrix.
From the relations \eqref{k-relation},
it can be easily verified that
\begin{equation}\label{tilde-relation}
\begin{aligned}
(\wbar{\vPsi}^\prime \vPsi^\prime)(x^\prime,t^\prime) &= (\wbar{{\vPsi}} {\vPsi})(x,t), \\
(\wbar{\vPsi}^\prime \vgamma^0 \vPsi^\prime)(x^\prime,t^\prime) &= (\wbar{{\vPsi}} \vgamma^0 {\vPsi})(x,t), \\
(\wbar{\vPsi}^\prime \vgamma^1 \vPsi^\prime)(x^\prime,t^\prime) &= - (\wbar{{\vPsi}} \vgamma^1 {\vPsi})(x,t),
\end{aligned}
\end{equation}
so that the self-interaction Lagrangian in \eqref{generalLI} satisfies
\begin{equation}\label{tr-L-relation}
L_\text{I}^\prime[\vPsi^\prime](x^\prime,t^\prime) = L_\text{I}[\vPsi](x,t),
\end{equation}
under the time reversal transformation \eqref{TimeR}.
Applying the time-reversal operator $\vK$ to the NLD equation \eqref{nld-lorentz} and using
the definition \eqref{TimeR} as well as the relations \eqref{k-relation} and \eqref{tr-L-relation}
lead to an equation which has the same form as shown in \eqref{nld-prime}.
That is, if a spinor $\vPsi(x,t)$ satisfies the NLD equation \eqref{nld-lorentz},
then the transformed spinor $\vPsi^\prime(x^\prime,t^\prime)$
by the time reversal operation \eqref{TimeR} will also
satisfy the same NLD equation,
\ie the NLD equation \eqref{nld-lorentz} is time reversible under the operation \eqref{TimeR}.
\section{Numerical methods}
\label{sec:DiffSch}
As we mentioned in Section \ref{sec:intro},
some numerical methods have been well developed for
the NLD equation with a scalar or vector self-interaction.
This section will extendedly present and compare several numerical methods
for solving
the (1+1)-dimensional NLD equation \eqref{eq:gnld}
with the general scalar and vector self-interaction \eqref{generalLI}.
Their numerical analyses
will be presented in Section \ref{sec:discussion4fdm}.
For the sake of convenience, divide the bounded spatial domain $\Omega\subset \mathbb R$ into
a uniform partition $\{x_j | x_j=jh \in \Omega, j\in \mathbb Z\}$ with a constant stepsize $h>0$
and give a grid in time
$\{t_n = n\tau, n=0,1,\cdots\}$ with a time stepsize $\tau>0$,
and recast the NLD equation \eqref{nld-lorentz} into
\begin{equation}
\pp{\vPsi}{t} +\vgamma^0\vgamma^1 \pp{\vPsi}{ x}
+ \mi m \vgamma^0 \vPsi - \mi f_{\text{s}} \vgamma^0\vPsi
- \mi f_{\text{v}}
\wbar{\vPsi}\vgamma_\mu\vPsi \vgamma^0\vgamma^\mu\vPsi
= 0, \label{eq:gnld}
\end{equation}
where
\begin{equation*}
f_{\text{s}} := s (k+1)w_{\text{s}}^{k}, \quad
w_{\text{s}} := \wbar{\vPsi}\vPsi, \quad
f_{\text{v}} := v (k+1) w_{\text{v}}^{(k-1)/2},\quad
w_{\text{v}} := \wbar{\vPsi}\vgamma^\nu\vPsi \wbar{\vPsi}\vgamma_\nu\vPsi, \quad
\end{equation*}
all of which are real functions, and the dependence of $\vPsi(x,t)$ on $(x,t)$ is implied.
\subsection{Several finite difference schemes}
Use $\vec \Psi_j^n$ to denote approximation of $\vec \Psi(x_j,t_n)$ and
define the forward, backward and centered difference operators in space and time by:
\begin{equation}\label{eq:note-sch}
\begin{aligned}
\vdelta_t^\pm =\pm (\vE_t^{\pm 1} -I)/\tau, \quad \vdelta_t^0 = (\vE_t -\vE_t^{-1})/2\tau,
\\
\vdelta_x^\pm =\pm (\vE_x^{\pm 1} -I)/h,\quad \vdelta_x^0 = (\vE_x -\vE_x^{-1})/2h,
\end{aligned}
\end{equation}
where $I$ is the identity operator,
and $\vE_t$ and $\vE_x$ are the translation operators in time and space, respectively, defined by
$$
\vE_t \vPsi^n_j:=\vPsi^{n+1}_j,\quad \vE_x \vPsi^n_j:=\vPsi^n_{j+1},
$$
whose inverses exist and are defined by
$$
\vE_t^{-1} \vPsi^n_j:=\vPsi^{n-1}_j,\quad \vE_x^{-1} \vPsi_j^n:=\vPsi_{j-1}^n.
$$
Besides, several symbols are also introduced for arithmetic averaging operators:
\begin{equation*}
\begin{aligned}
\vell_t^\pm\vPsi_j^n &:= \frac12(\vPsi_j^{n\pm1} + \vPsi_j^n),\quad
\vell_t^0 \vPsi_j^n := \frac12(\vPsi_j^{n+1} +\vPsi_j^{n-1}),\\
\vell_x^\pm \vPsi_j^{n} &:= \frac1{2}(\vPsi_{j\pm 1}^n +\vPsi_{j}^n),\quad
\vell_x^0 \vPsi_j^{n} := \frac1{2}(\vPsi_{j+1}^n + \vPsi_{j-1}^n),
\end{aligned}
\end{equation*}
and for an extrapolation operator:
\begin{equation*}
\vell_t^\text{e}\vPsi_j^{n} = \frac32\vPsi_j^n - \frac12\vPsi_j^{n-1}.
\label{eq:note-sch-sum}
\end{equation*}
\textbf{$\bullet$ Crank-Nicolson schemes}
The CN scheme and its linearized version have been studied in \cite{AlvarezKuoVazquez1983,Alvarez1992,Jimcnez1994}
for the NLD equation with the quadric scalar self-interaction.
For the system \eqref{eq:gnld}, the extension of the CN scheme in \cite{Jimcnez1994,ChangJiaSun1999} may be written as
\begin{align}
\begin{aligned}
\vdelta_t^+\vPsi_j^n + \vgamma^0\vgamma^1 \vell_t^+\vdelta_x^0 \vPsi_j^n
&+ \mi m \vgamma^0 \vell_t^+\vPsi_j^{n} -
\mi \frac{\vdelta_t^+(F_{\text{s}})_j^n}
{\vdelta_t^+(w_{\text{s}})_j^n}\vgamma^0\vell_t^+\vPsi_j^n \\
& - \mi \frac{\vdelta_t^+(F_{\text{v}})_j^n}
{\vdelta_t^+(w_{\text{v}})_j^n}
\vell_t^+(\wbar{\vPsi}\vgamma_\mu\vPsi)_j^n\vgamma^0\vgamma^\mu \vell_t^+\vPsi_j^n
= 0,
\end{aligned} \label{eq:cn-gnld}
\end{align}
by approximating \eqref{eq:gnld} at point $(x_{j},t_{n+\frac12})$ with compact
central difference quotient in place of the partial derivative,
where
$$F_I(w_I) := \int_0^{w_I} f_I(x) \dif x, \quad I =\text{s}, \text{v}.$$
The above CN scheme (named as {\tt CN} hereafter) is fully implicit and forms a nonlinear algebraic system.
In practice, the linearization technique is often used to
overcome difficulty in directly solving such nonlinear algebraic system.
Two linearization techniques \cite{ChangJiaSun1999} for numerical methods of the nonlinear
Schr\"{o}dinger equation are borrowed here.
The first linearized CN scheme we consider is using wholly the extrapolation
technique to the nonlinear self-interaction terms in \eqref{eq:gnld}
\begin{equation} \label{eq:lcn1-gnld}
\delta_t^+\vPsi_j^{n} + \vgamma^0\vgamma^1\ell_t^+\delta_x^0\vPsi_j^n
+ \mi m \vgamma^0 \ell_t^+\vPsi_j^{n}
- \mi \ell_t^\text{e}\left( f_{\text{s}}\vgamma^0\vPsi
+ f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi\vgamma^0\vgamma^\mu\vPsi \right)_j^n = 0,
\end{equation}
which will be called by {\tt LCN1}.
The second linearized CN scheme, denoted by {\tt LCN2}, is
\begin{equation}
\begin{aligned} \label{eq:lcn2-gnld}
\delta_t^+\vPsi_j^{n} + \vgamma^0\vgamma^1\ell_t^+\delta_x^0\vPsi_j^n
& + \mi m \vgamma^0 \ell_t^+\vPsi_j^{n}
- \mi \ell_t^\text{e}(f_{\text{s}})_j^n\vgamma^0\ell_t^+\vPsi_j^n \\
& - \mi \ell_t^\text{e}(f_{\text{v}}\wbar{\vPsi}\vgamma_\mu\vPsi)_j^n\vgamma^0\vgamma^\mu \ell_t^+\vPsi_j^n = 0,
\end{aligned}
\end{equation}
by partially applying the extrapolation technique to the nonlinear self-interaction terms.
It is worth noting that the above linearized CN schemes are not linearized version of the {\tt CN} scheme \eqref{eq:cn-gnld}.
The {\tt LCN2} scheme may conserve the charge and behaves better than the {\tt LCN1} scheme (\videpost).
\begin{Remark}\rm
For the (1+1)-dimensional NLD equation \eqref{generalnld}
with the quadric scalar self-interaction Lagrangian \eqref{self-scalar1},
the CN scheme (named by {\tt CN0})
proposed in \cite{AlvarezKuoVazquez1983} is
\begin{equation}
\label{eq:cn-kuo}
\delta_t^+\vPsi_j^n + \vgamma^0\vgamma^1\ell_t^+\delta_x^0\vPsi_j^n
+ \mi m \vgamma^0 \ell_t^+\vPsi_j^{n} - 2 {\mi} s
(\ell_t^+\wbar{\vPsi}_j^n\ell_t^+\vPsi_j^n)
\vgamma^0\ell_t^+\vPsi_j^n = 0,
\end{equation}
and its linearized version (called by {\tt LCN0}) is given in \cite{Alvarez1992} as follows
\begin{equation}
\begin{aligned} \label{eq:lcn-kuo}
\delta_t^+\vPsi_j^n & + \vgamma^0\vgamma^1\ell_t^+\delta_x^0\vPsi_j^n
+ \mi m \vgamma^0 \ell_t^+\vPsi_j^{n} \\
& - 2 {\mi} s \left((\wbar{\vPsi}\vPsi)_j^n
- \tau\real( \wbar{\vPsi}_j^n\vgamma^0\vgamma^1\delta_x^0\vPsi_j^n)\right)
\vgamma^0\ell_t^+\vPsi_j^n = 0.
\end{aligned}
\end{equation}
We will show in Section \ref{sec:discussion4fdm} that the {\tt CN}, {\tt CN0}
and {\tt LCN0} schemes conserve the charge and the {\tt CN} scheme further conserves the energy.
\end{Remark}
\textbf{$\bullet$ Odd-even hopscotch scheme} The odd-even hopscotch scheme is a numerical integration technique
for time-dependent partial differential equations,
see \cite{Gordon1965}. Its key point is that the forward Euler-central difference scheme
is used for the odd grid points while at the even points the backward Euler-central
difference scheme is recovered. Thus the odd-even hopscotch scheme may be explicitly
implemented.
Such scheme applied to the system \eqref{eq:gnld} becomes
\begin{equation}
\begin{aligned} \label{eq:hops-odd}
\delta_t^+\vPsi_j^{n} & + \vgamma^0\vgamma^1\delta_x^0\vPsi_j^{n}
+ \mi m \vgamma^0 \vPsi_j^{n} \\
& - \mi \ell_x^0\left(f_{\text{s}}\vgamma^0\vPsi
+ f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi\vgamma^0\vgamma^\mu\vPsi \right)_j^n = 0, \quad \text{$n+j$ is odd,}
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned} \label{eq:hops-even}
\delta_t^-\vPsi_j^{n+1} & + \vgamma^0\vgamma^1\delta_x^0\vPsi_j^{n+1}
+ \mi m \vgamma^0 \vPsi_j^{n+1} \\
& - \mi \ell_x^0\left(f_{\text{s}}\vgamma^0\vPsi
+ f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi\vgamma^0\vgamma^\mu\vPsi \right)_j^{n+1} = 0,
\quad \text{$n+j$ is even.}
\end{aligned}
\end{equation}
In the following we will call it by {\tt HS}.
\textbf{$\bullet$ Leapfrog scheme} The leapfrog scheme looks quite similar to the forward scheme, see \eg \eqref{eq:hops-odd},
except it uses the values from the previous time-step instead of the current one.
For the system \eqref{eq:gnld}, it is
\begin{equation}
\label{eq:3l-ex-1}
\delta_t^0\vPsi_j^{n} +\vgamma^0\vgamma^1\delta_x^0\vPsi_j^{n}
+ \mi m\vgamma^0 \vPsi_j^{n}
- \mi \left(f_{\text{s}}\vgamma^0\vPsi
+ f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi\vgamma^0\vgamma^\mu \vPsi\right)_j^n = 0,
\end{equation}
which is a three-level explicit scheme in time with a central difference in space
and will be named by {\tt LF}.
\textbf{$\bullet$ Semi-implicit scheme} Another three-level scheme considered here for the system \eqref{eq:gnld} is
\begin{equation}
\label{eq:semi-ex}
\delta_t^0\vPsi_j^{n} + \vgamma^0\vgamma^1\ell_t^0\delta_x^0\vPsi_j^{n}
+ \mi m \vgamma^0 \ell_t^0\vPsi_j^{n}
- \mi \left(f_{\text{s}}\vgamma^0\vPsi
+ f_{\text{v}}
\wbar{\vPsi}\vgamma_\mu\vPsi\vgamma^0\vgamma^\mu \vPsi\right)_j^n = 0,
\end{equation}
which
is obtained by approximating explicitly the nonlinear terms but implicitly the linear terms and will be called by {\tt SI}.
It is worth noting that such semi-implicit scheme has been studied for the NLD equation with the quadric scalar self-interaction in \cite{Bournaveas2012}.
\subsection{Exponential operator splitting scheme}
\label{sec:split-tech}
This subsection goes into discussing exponential operator splitting scheme
for the NLD equation \eqref{eq:gnld}.
For convenience, we rewrite the system \eqref{eq:gnld} as follows
\begin{equation}
\vPsi_t = \left(\mathcal{L} + \mathcal{N}_{\text{s}}
+ \mathcal{N}_{\text{v}}\right)\vPsi,
\label{eq:split-4}
\end{equation}
where {the linear operator $\mathcal{L}$ and both nonlinear operators $\mathcal{N}_{\text{s}}$
and $\mathcal{N}_{\text{v}}$ are defined by}
\begin{equation*}
\mathcal{L} \vPsi := -\vgamma^0\vgamma^1 \vPsi_x -\mi m\vgamma^0\vPsi,\quad
\mathcal{N}_{\text{s}} \vPsi := \mi f_{\text{s}}\vgamma^0\vec \Psi,\quad
\mathcal{N}_{\text{v}} \vPsi := \mi f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi \vgamma^0\vgamma^\mu\vPsi.
\end{equation*}
Then the problem \eqref{eq:split-4} may be decomposed into three subproblems as follows
\begin{align}
\vPsi_t &= \mathcal{L} \vPsi,
\label{eq:split-l-2}
\\
\label{eq:split-ns-2}
\vPsi_t &= \mathcal{N}_{\text{s}} \vPsi,
\\ \label{eq:split-nv-2}
\vPsi_t &= \mathcal{N}_{\text{v}} \vPsi.
\end{align}
{Due to the local conservation laws which are discussed in
Section \ref{sec:dis-nonlinear}, solutions of the nonlinear subproblem \eqref{eq:split-ns-2} or \eqref{eq:split-nv-2}
may be expressed as an exponential of the operator $\mathcal{N}_{\text{s}}$ or $\mathcal{N}_{\text{v}}$
acting on ``initial data''. Thus we may introduce the exponential operator splitting scheme for the the NLD equation \eqref{eq:split-4} or \eqref{eq:gnld}, imitating that for the linear partial differential equations, see \eg \cite{Sheng1989,Chin2005} and references therein.} Based on the exact or approximate solvers of those three subproblems, a more
general $K$-stage $N$-th order exponential operator splitting method \cite{Sornborger1999,ThalhammerCaliariNeuhauser2009}
for the system \eqref{eq:split-4} can be cast into {product of finitely many exponentials as follows}
\begin{equation}
\label{eq:high-oder-split}
\vPsi_j^{n+1} = \prod_{i=1}^{K}
\big(
\exp({\tau_{i} \mathcal{A}_i^{(1)}})\exp({\tau_{i}\mathcal{A}_i^{(2)}})
\exp({\tau_{i} \mathcal{A}_i^{(3)}})\big) \vPsi_j^{n},
\end{equation}
where $\tau_i$ denotes the time stepsize used within the $i$-th stage
and satisfies
$
\sum_{i=1}^K \tau_{i}=\tau,
$
and $\{\mathcal{A}_i^{(1)},\mathcal{A}_i^{(2)},\mathcal{A}_i^{(3)}\}$ is
any permutation of
$\{\mathcal{L},\mathcal{N}_{\text{s}},\mathcal{N}_{\text{v}}\}$.
Hereafter we call the operator splitting scheme \eqref{eq:high-oder-split} by {\tt OS($N$)}.
{Although one single product of finitely many exponentials exponentials \eqref{eq:high-oder-split}
is employed here, it should be pointed out that the
linear combination of such finite products can also be used to construct
exponential operator splitting schemes as shown in \cite{Sheng1989}.}
A simple example is the well-known second-order accurate operator splitting method
of Strang (named by {\tt OS($2$)}) with
\begin{equation}
\label{eq:strang-split}
K=2, \quad \tau_1=\tau_2{=\frac12\tau}, \quad \mathcal{A}_1^{(1)}=\mathcal{A}_2^{(3)},\quad
\mathcal{A}_1^{(3)}=\mathcal{A}_2^{(1)}, \quad
\mathcal{A}_1^{(2)}=\mathcal{A}_2^{(2)}.
\end{equation}
Another example is the fourth-order accurate operator splitting method
\cite{Sornborger1999}
with
\begin{equation*
\begin{aligned}
K=&8,\quad
\mathcal{A}_q^{(1)}=\mathcal{A}_p^{(3)},\quad
\mathcal{A}_q^{(3)}=\mathcal{A}_p^{(1)}, \quad
\mathcal{A}_q^{(2)}=\mathcal{A}_p^{(2)}, \quad q=1,4,6,7, p=2,3,5,8,\\
\tau_1=&\tau_8=\frac{\tau}{5-\sqrt{13}+\sqrt{2(1+\sqrt{13})}},
\quad
\tau_2=\tau_7=\frac{7+\sqrt{13}-\sqrt{2(1+\sqrt{13})}}{24}\tau,\\
\tau_3=&\tau_6=\frac{\tau_1^2}{\tau_2-\tau_1},\quad
\tau_4=\tau_5=\frac{\tau_2(\tau_1-\tau_2)}{3\tau_1-2\tau_2},
\end{aligned}
\end{equation*}
which is denoted by {\tt OS($4$)} in the following.
\begin{Remark}\rm
Another operator splitting scheme is studied
in \cite{FrutosSanz-Serna1989}
for the NLD equation (\ref{generalnld})
but only with the quadric scalar self-interaction Lagrangian,
and the second-order accurate Strang method \eqref{eq:strang-split} is applied there.
For the system \eqref{eq:gnld}, it is based on
the following operator decomposition
\begin{equation}
\vPsi_t = \left(\widehat{\mathcal{L}} + \widehat{\mathcal{N}}_{\text{s}}
+ \widehat{\mathcal{N}}_{\text{v}}\right)\vPsi,
\label{eq:split-l}
\end{equation}
with
\begin{equation*}
\widehat{\mathcal{L}} \vPsi := -\vgamma^0\vgamma^1 \vPsi_x,\quad
\widehat{\mathcal{N}}_{\text{s}} \vPsi := - \mi \left(m - f_{\text{s}}\right)\vgamma^0\vPsi,\quad
\widehat{\mathcal{N}}_{\text{v}} \vPsi := \mi f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi \vgamma^0\vgamma^\mu\vPsi.
\end{equation*}
\end{Remark}
\begin{Remark}\rm
{For the linear parabolic equation which is an irreversible system,
a more general exponential operator splitting scheme and its accuracy as well as stability are discussed in \cite{Sheng1989}, based on linear combinations of products of finitely many exponentials.
It is shown that for such irreversible system, negative weights or negative time stepsizes $\tau_i$ may lead to instability;
and the highest order of the stable exponential operator splitting approximation (only with positive weights and
positive sub-stepsizes in time) is two.
However, for time-reversible systems, such as the Hamilton system, the Schr{\"o}dinger equations and
the NLD equation \eqref{generalnld} with $L_\text{I}$ given in \eqref{generalLI}, it
is immaterial whether or not the weights or time stepsizes $\tau_i$ are positive \cite{Chin2005}, where
a general framework was presented for understanding the structure of the exponential operator splitting schemes
and both specific error terms and order conditions were analytically solved.
}
\end{Remark}
\subsubsection{Linear subproblem}
\label{sec:dis-linear}
We are now solving the the linear subproblem \eqref{eq:split-l-2}.
Denote its ``initial data'' by $\vPsi_j^{(0)}=\big((\psi_1)_j^{(0)},(\psi_2)_j^{(0)}\big)^T$ at the $i$-th stage in \eqref{eq:high-oder-split}.
If the spinor $\vPsi$ is periodic (\eg $2\pi$-periodic) with respect to $x$, the Fourier spectral method is employed
to solve \eqref{eq:split-l-2} and
gives
\begin{equation} \label{eq:fft}
\vPsi_j^{(1)} = \mathcal{F}^{-1}\Big(
\exp\big({-\mi \tau_i(\kappa \vec\gamma^0\vec\gamma^1+m\vec\gamma^0) }\big)
\mathcal{F}\left(\vPsi_j^{(0)}\right)\Big).
\end{equation}
Here $\mathcal{F}$ and $\mathcal{F}^{-1}$ denote the discrete Fourier transform operator and its inverse,
respectively, defined by
\begin{align*}
\label{eq:fft}
\big(\mathcal{F}(\vPsi)\big)_\kappa :=& \sum_{j=0}^{J-1}\vPsi_j\exp\big({-\mi 2\pi\kappa \frac{j}J}\big)
\quad \kappa = 0,\ldots, J-1, \\
\big(\mathcal{F}^{-1}(\vPhi)\big)_j :=& \frac1{J}\sum_{\kappa=0}^{J-1}\vPhi_\kappa\exp\big({\mi 2\pi j \frac{\kappa}J}\big)
\quad j = 0,\ldots, J-1,
\end{align*}
where $J$ is the grid point number,
and the matrix exponential in \eqref{eq:fft} can be easily evaluated as follows
\begin{equation}\label{eq:matrixfft}
\exp\big({-\mi \tau_i(\kappa \vec\gamma^0\vec\gamma^1 +m\vec\gamma^0) } \big)= \begin{pmatrix}
\cos (\zeta \tau_{i}) -\mi\frac{m}{\zeta}\sin (\zeta \tau_{i})&
-\mi\frac{\kappa }{\zeta}\sin (\zeta \tau_{i}) \\
-\mi\frac{\kappa }{\zeta}\sin (\zeta \tau_{i}) &
\cos (\zeta \tau_{i}) +\mi\frac{m}{\zeta}\sin(\zeta \tau_{i})
\end{pmatrix},
\end{equation}
with $\zeta = \sqrt{\kappa^2+m^2}$.
When the spinor $\vPsi$ is not periodic with respect to $x$, the fifth-order accurate finite difference WENO scheme will be used to
solve the linear subproblem \eqref{eq:split-l-2}. The readers are referred to \cite{Shu2009-WENO} for details.
In this case, the linear subproblem \eqref{eq:split-l-2} can also be solved by using
the characteristics method.
\subsubsection{Nonlinear subproblems}
\label{sec:dis-nonlinear}
The nonlinear subproblems \eqref{eq:split-ns-2} and \eqref{eq:split-nv-2}
are left to be solved now.
Their ``initial data'' is still denoted by $\vec\Psi_j^{(0)}=\big((\psi_1)_j^{(0)},(\psi_2)_j^{(0)}\big)^T$
at the $i$-th stage in \eqref{eq:high-oder-split}, and define
$$
t_n^{(i)}=t_n+\sum_{p=1}^{i-1} \tau_p,\quad i=1,2,\cdots,K.
$$
For nonlinear subproblem \eqref{eq:split-ns-2}, it is not difficulty to verify that
$\partial_t w_\text{s} = 0$ so that
\begin{equation}\label{ns-conv}
\partial_t f_\text{s} = 0.
\end{equation}
Using this local conservation law gives the solution at $t=t_n^{(i+1)}$ of \eqref{eq:split-ns-2}
with the ``initial data'' $\vec\Psi_j^{(0)}$ as follows
\begin{align}
\vPsi_j^{(1)}
&= \exp\left(\mi \int_{t_n^{(i)}}^{t_n^{(i+1)}} (f_{\text{s}})_j\vgamma^0 \dif t \right) \vPsi_j^{(0)} = \exp\left(\mi (f_{\text{s}})_j^{(0)}\vgamma^0 \tau_{i} \right) \vPsi_j^{(0)} \nonumber \\
&= \diag\left\{\exp\left( \mi (f_{\text{s}})_j^{(0)} \tau_{i}\right),
\exp\left(-\mi(f_{\text{s}})_j^{(0)} \tau_{i}\right)\right\} \vPsi_j^{(0)}. \label{OS-NS-eq3}
\end{align}
For the nonlinear subproblem \eqref{eq:split-nv-2}, one may still similarly
derive the following local conservation laws
\begin{equation}\label{nv-con}
\partial_t (\wbar{\vPsi}\vgamma_0\vPsi) = 0,\quad
\partial_t (\wbar{\vPsi}\vgamma_1\vPsi) = 0, \quad
\partial_t f_\text{v} = 0.
\end{equation}
by direct algebraic manipulations if using the fact that
$\wbar{\vec \Psi}\vec\gamma_0\vec \Psi$,
$\wbar{\vec \Psi}\vec\gamma_1\vec \Psi$
and $f_{\text{v}}$ are all real.
Consequently, integrating \eqref{eq:split-nv-2} from $t_n^{(i)}$ to $t_n^{(i+1)}$ gives
its solution as follows
\begin{align}
\vPsi_j^{(1)} & = \exp\left(\mi(f_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi)_j^{(0)}\vgamma^0\vgamma^\mu
\tau_{i} \right)\vPsi_j^{(0)} \nonumber \\
& =\exp(\mi{\alpha\tau_i}) \left(
\begin{array}{rr}
\cos(\beta\tau_i) & \mi \sin (\beta\tau_i) \\
\mi \sin (\beta\tau_i) &\cos (\beta\tau_i)
\end{array}\right) \vPsi_j^{{(0)}}, \label{eq:solution-V}
\end{align}
where $\alpha = (f_{\text{v}} \wbar{\vPsi}\vgamma_0\vPsi)_j^{(0)}$ and $\beta = (f_{\text{v}} \wbar{\vPsi}\vgamma_1\vPsi)_j^{(0)}$.
\begin{Remark}\rm
It is because the local conservation laws \eqref{ns-conv} and \eqref{nv-con} are fully exploited here that
we can solve exactly the nonlinear subproblems \eqref{eq:split-ns-2} and \eqref{eq:split-nv-2} which imply
the more higher accuracy of the OS method than that of other methods.
\end{Remark}
In summary, we have
\begin{itemize}
\item The {\tt CN} \eqref{eq:cn-gnld} and {\tt CN0} \eqref{eq:cn-kuo} schemes are nonlinear and implicit,
and could be solved by iterative algorithms such as Picard iteration and Newton method.
\item The {\tt LCN0}~\eqref{eq:lcn-kuo}, {\tt LCN1}~\eqref{eq:lcn1-gnld}, {\tt LCN2}~\eqref{eq:lcn2-gnld} and {\tt SI}~\eqref{eq:semi-ex}
schemes are linear and implicit.
\item The {\tt HS} \eqref{eq:hops-odd}-\eqref{eq:hops-even}, {\tt LF} \eqref{eq:3l-ex-1}, and {\tt OS($N$)} \eqref{eq:high-oder-split} schemes are explicit.
\end{itemize}
\section{Numerical analysis}
\label{sec:discussion4fdm}
Before investigating the performance of the numerical methods proposed in Section \ref{sec:DiffSch},
this section will go first into numerical analysis of them,
including the accuracy in the sense of the truncation error, time reversibility and the conservation of the charge or energy.
\begin{Proposition}
If $\vPsi(x,t)\in C^{\infty}(\mathbb{R}\times[0,+\infty))$ is periodic,
then the {\tt CN}, {\tt CN0}, {\tt LCN0}, {\tt LCN1}, {\tt LCN2}, {\tt HS}, {\tt LF} and {\tt SI} schemes
are of order $\mathcal{O}(\tau ^2 + h^2)$,
and the {\tt OS($N$)} scheme is of order $\mathcal{O}(\tau^N + h^m)$ for any arbitrary large $m>0$.
\end{Proposition}
\begin{proof}
The proof is very straightforward by using directly the Taylor series expansion for the finite difference schemes
and the Fourier spectral analysis for the {\tt OS($N$)} scheme, and thus is skipped here for saving space.
\end{proof}
\begin{Proposition
The {\tt CN}, {\tt CN0}, {\tt HS}, {\tt LF}, {\tt SI}, and {\tt OS($N$)} schemes are time reversible,
but the {\tt LCN0}, {\tt LCN1}, {\tt LCN2} schemes are not.
\end{Proposition}
\begin{proof}
We give the proof for the {\tt CN} and {\tt LCN1} schemes as an example and the others can be proved in a similar way.
According to the transformation \eqref{TimeR},
the relation between the transformed finite difference solution and the original
one should be ${(\vPsi^\prime)}_j^{n^\prime}=\vK\vPsi_j^n = (\vK\vPsi)_j^n$ with $n^\prime=-n$.
Consequently,
we have
\begin{equation}\label{TR-e1}
\begin{aligned}
\vK \delta_t^+\vPsi_j^n &= \vK \frac{\vPsi_j^{n+1}-\vPsi_j^n}{\tau} = \frac{{(\vPsi^\prime)}_j^{n^\prime-1}-{(\vPsi^\prime)}_j^{n^\prime}}{\tau} = - \delta_t^+ {(\vPsi^\prime)}_j^{n^\prime-1},
\\
\vK \ell_t^+\vPsi_j^n &= \vK \frac{\vPsi_j^{n+1}+\vPsi_j^n}{2} = \frac{{(\vPsi^\prime)}_j^{n^\prime-1}
+{(\vPsi^\prime)}_j^{n^\prime}}{2} = \ell_t^+ {(\vPsi^\prime)}_j^{n^\prime-1}.
\end{aligned}
\end{equation}
and then using the relations in \eqref{tilde-relation} yields
\begin{equation}\label{TR-e2}
\begin{aligned}
\ell_t^+(\wbar{\vPsi}\vgamma_0\vPsi)_j^n &= \ell_t^+(\wbar{\vPsi}^\prime\vgamma_0{\vPsi}^\prime)_j^{n^\prime-1}, \\
\ell_t^+(\wbar{\vPsi}\vgamma_1\vPsi)_j^n &= - \ell_t^+(\wbar{\vPsi}^\prime\vgamma_1{\vPsi^\prime})_j^{{n}^\prime-1},\\
\delta_t^+(w_I)_j^n &= -\delta_t^+(w_I^\prime)_j^{n^\prime-1}, \\
\delta_t^+(F_I)_j^{n} &= -\delta_t^+(F_I^\prime)_j^{n^\prime-1},
\end{aligned}
\end{equation}
for $I\in\{\text{s},\text{v}\}$.
Applying the time-reversal operator $\vK$ to the {\tt CN} scheme \eqref{eq:cn-gnld}
and using the commutation relation \eqref{k-relation} and Eqs.~\eqref{TR-e1} and \eqref{TR-e2}
lead to
\begin{equation*}
\begin{aligned}
\delta_t^+ {(\vPsi^\prime)}_j^{n^\prime-1} + & \vgamma^0\vgamma^1 \ell_t^+\delta_x^0{(\vPsi^\prime)}_j^{n^\prime-1}
+ \mi m \vgamma^0 \ell_t^+ {(\vPsi^\prime)}_j^{n^\prime-1} - \mi
\frac{\delta_t^+(F^\prime_\text{s})_j^{{n}^\prime-1}}
{\delta_t^+(w^\prime_\text{s})_j^{{n}^\prime-1}}\vgamma^0\ell_t^+ {(\vPsi^\prime)}_j^{{n}^\prime-1} \\
&-\mi \frac{\delta_t^+(F^\prime_\text{v})_j^{{n}^\prime-1}}
{\delta_t^+(w^\prime_\text{v})_j^{{n}^\prime-1}}
\ell_t^+(\wbar{\vPsi}^\prime\vgamma_\mu{\vPsi^\prime})_j^{{n}^\prime-1}\vgamma^0\vgamma^\mu \ell_t^+{(\vPsi^\prime)}_j^{n^\prime-1}
= 0,
\end{aligned}
\end{equation*}
which is exactly the {\tt CN} scheme \eqref{eq:cn-gnld} applied to ${(\vPsi^\prime)}_j^{{n}^\prime-1}$.
That is, the {\tt CN} scheme is invariant under the the time-reversal transformation,
namely, it is time reversible.
The fact that the {\tt LCN1} scheme \eqref{eq:lcn1-gnld} is not time reversible can be observed directly if noting
\begin{align*}
&- \vK \mi \ell_t^\text{e}\left({f}_{\text{s}}\vgamma^0\vPsi
+ {f}_{\text{v}} \wbar{\vPsi}\vgamma_\mu\vPsi\vgamma^0\vgamma^\mu\vPsi \right)_j^n
\\
=& \mi \left[\frac32 \left(f^\prime_\text{s}\vgamma^0 \vPsi^\prime +f^\prime_\text{v} \wbar{\vPsi}^\prime\vgamma_\mu \vPsi^\prime \vgamma^0\vgamma^\mu \vPsi^\prime \right)_j^{n^\prime}
- \frac12 \left(f^\prime_\text{s}\vgamma^0 \vPsi^\prime + f^\prime_\text{v} \wbar{\vPsi}^\prime\vgamma_\mu \vPsi^\prime\vgamma^0\vgamma^\mu \vPsi^\prime \right)_j^{n^\prime+1}\right]
\\
\neq & \mi \ell_t^\text{e} \left(f^\prime_\text{s}\vgamma^0 \vPsi^\prime + f^\prime_\text{v} \wbar{\vPsi}^\prime\vgamma_\mu
\vPsi^\prime\vgamma^0\vgamma^\mu \vPsi^\prime \right)_j^{n^\prime-1}.
\end{align*}
\end{proof}
Next,
we will discuss the conservation of the discrete energy, linear momentum and charge defined below
for the numerical methods given in Section \ref{sec:DiffSch}.
After performing the integration in the computational domain $\Omega = [x_\text{L},x_\text{R}]$
and then approximating the first derivative operator $\partial_x$
with the centered difference operator $\delta_x^0$ as well as the integral operator $\int_{x_\text{L}}^{x_\text{R}}\dif x$
with the summation
operator $h\sum_{j=1}^J$ in Eq.~\eqref{qep-infty},
we have the discrete energy, linear momentum and charge at the $n$-th time step
\begin{equation*}
\begin{aligned}
E_h^n &= h \sum_{j = 1}^J \left(\imag (\wbar{\vPsi}\vgamma^1\delta_x^0\vPsi)
+ m (\wbar{\vPsi} \vPsi)-F_{\text{s}}
- \frac12 F_{\text{v}}\right)_j^n,\\
P_h^n &= h \sum_{j = 1}^J \imag (\vPsi^\dag \delta_x^0\vPsi)_j^n, \\
Q_h^n &= \| \vPsi^n\|^2 :=\left<\vPsi^n, \vPsi^n\right>= h \sum_{j = 1}^J (\vPsi_j^{n})^\dag \vPsi_j^{n},
\end{aligned}
\end{equation*}
where the inner product $\left< \cdot,\cdot \right>$ is defined as
\begin{equation*}
\left<\vec u, \vec v\right> = h \sum_{j = 1}^J
(\vec u_j)^\dag \vec v_j
\end{equation*}
for two complex-valued vectors $\vec u$ and $\vec v$, and $J$ is the grid point number.
Here the values of $\vPsi$ at $x_j$ with $1\leq j\leq J$ are unknowns and
those at $x_0$ and $x_{J+1}$ are determined by appropriate boundary conditions.
We first present the following lemma which can be verified through direct algebraic manipulations and will be used in
discussing the conservation of the discrete charge, energy and linear momentum.
\begin{Lemma}
\label{le:1}
Given that $\vPhi_j^n$ is a complex-valued vector mesh function with two components evaluated at the mesh $\{x_j,t_n\}$
($n=0,1,\cdots,N$, $j=0,1,\cdots,J+1$) and a matrix $\vGamma\in \{\vI_2, \vgamma^0, \vgamma^1, \mi \vgamma^0\vgamma^1\}$,
we have the following identities
(a) $2\real\left<\ell_t^+\vPhi^n,\delta_t^+\vPhi^n\right> = \delta_t^+ \|\vPhi^n\|^2$;
(b) $2 \real \left<\vgamma^0\vgamma^1\delta_x^0\vPhi^n,\vPhi^n\right> =
\frac1{2}(\wbar{\vPhi}_{J+1}^n\vgamma^1\vPhi_J^n +
\wbar{\vPhi}_{J}^n\vgamma^1\vPhi_{J+1}^n)
-\frac1{2}(\wbar{\vPhi}_{1}^n\vgamma^1\vPhi_0^n +
\wbar{\vPhi}_{0}^n\vgamma^1\vPhi_{1}^n)$;
(c) $\imag\left(\ell_t^+\wbar{\vPhi}_j^n \vGamma \ell_t^+\vPhi_j^n \right) =0$;
(d) $2 \real (\delta_t^+\wbar{\vPhi}_j^n \vGamma \ell_t^+\vPhi_j^n) =
\delta_t^+(\wbar{\vPhi}_j^n\vGamma \vPhi_j^n)$;
(e) $
2 \mi \imag \left< \vgamma^0\vgamma^1 (\ell_t^+ \delta_x^0) \vPhi^n,\delta_t^+\vPhi^n\right> =
\begin{array}[t]{l}
-\delta_t^+\left( h \sum_{j=1}^J (\wbar{\vPhi}_j^n\vgamma^1
\delta_x^0\vPhi_j^n) \right) \\
+\frac12\big(
(\ell_t^+\wbar{\vPhi}_{J+1}^n\vgamma^1\delta_t^+\vPhi_J^n + \ell_t^+\wbar{\vPhi}_{J}^n\vgamma^1\delta_t^+\vPhi_{J+1}^n) \\
-(\ell_t^+\wbar{\vPhi}_{1}^n\vgamma^1\delta_t^+\vPhi_0^n +
\ell_t^+\wbar{\vPhi}_{0}^n\vgamma^1\delta_t^+\vPhi_{1}^n)
\big).
\end{array}$
\end{Lemma}
\begin{proof}
It can be checked that the following Leibniz rules
\begin{equation*}
\delta_t^a(\vec u^\dag\vec v)^n_j = \delta_t^a (\vec u^\dag)^n_j \ell_t^a \vec v^n_j + \ell_t^a (\vec u^\dag)^n_j\delta_t^a \vec v^n_j
\end{equation*}
holds for any two spinors $\vec u(x,t),\vec v(x,t)$ and $a\in\{+,-,0\}$,
and then we have
\begin{align*}
2\real \left<\ell_t^+\vPhi^n,\delta_t^+\vPhi^n\right>
&= \left<\delta_t^+\vPhi^n,\ell_t^+\vPhi^n\right> + \left<\ell_t^+\vPhi^n, \delta_t^+\vPhi^n\right> \\
&= h\sum_{j=1}^J \delta_t^+(\vPhi^\dag)_j^n \ell_t^+\vPhi_j^n + h\sum_{j=1}^J \ell_t^+(\vPhi^\dag)_j^n \delta_t^+\vPhi_j^n
\\
&= h\sum_{j=1}^J \delta_t^+ (\vPhi^\dag \vPhi)_j^n
= \delta_t^+ \|\vPhi^n\|^2.
\end{align*}
Thus the identity $(a)$ holds.
Because the operator $-\delta_x^0$ is the adjoint operator of $\delta_x^0$
and $\vgamma^0\vgamma^1$ is an Hermite matrix,
we get the identity $(b)$ directly by rearranging the summation.
The identity $(c)$ can be easily verified if using the fact $(\vgamma^0\vGamma)^\dag = \vgamma^0\vGamma$.
The proof of $(d)$ (resp. $(e)$) is similar with that of $(b)$ (resp. $(c)$).
\end{proof}
\begin{Proposition
\label{thm:conservation}
The {\tt CN}, {\tt CN0}, {\tt LCN0}, {\tt LCN2}, and {\tt OS($N$)} schemes conserve the discrete charge,
but only the {\tt CN} scheme conserves the discrete energy.
\end{Proposition}
\begin{proof}
We begin with the discrete conservation law of charge for the {\tt CN} scheme \eqref{eq:cn-gnld}.
Performing the inner product of $\ell_t^+\Vec \Psi^n$ and the {\tt CN} scheme \eqref{eq:cn-gnld} leads to
\begin{equation}\label{cn-inner}
\begin{aligned}
\left<\ell_t^+\vPsi^n,\delta_t^+\vPsi^n\right> & + \left<\ell_t^+\vPsi^n,\vgamma^0\vgamma^1 \delta_x^0\ell_t^+\vPsi^n\right>
\\
& + \left<\ell_t^+\vPsi^n,\mi m \vgamma^0 \ell_t^+\vPsi^{n} - \mi
\frac{\delta_t^+(F_{\text{s}})^n}
{\delta_t^+(w_{\text{s}})^n}\vgamma^0\ell_t^+\vPsi^n\right> \\
& +\left<\ell_t^+\vPsi^n,-\mi \frac{\delta_t^+(F_{\text{v}})^n}
{\delta_t^+(w_{\text{v}})^n}
\ell_t^+(\wbar{\vPsi}\vgamma_\mu\vPsi)^n\vgamma^0\vgamma^\mu \ell_t^+\vPsi^n\right>
= 0,
\end{aligned}
\end{equation}
and then the
conservation law of the discrete charge can be easily verified by
taking directly the real part as follows
\begin{equation}\label{eq:dis-Q-CN}
\delta_t^+ Q_h^n=
\frac1{2}(\ell_t^+\wbar{\vPsi}_{1}^n\vgamma^1\ell_t^+\vPsi_0^n +
\ell_t^+\wbar{\vPsi}_{0}^n\vgamma^1\ell_t^+\vPsi_{1}^n)
- \frac1{2}(\ell_t^+\wbar{\vPsi}_{J+1}^n\vgamma^1\ell_t^+\vPsi_J^n +
\ell_t^+\wbar{\vPsi}_{J}^n\vgamma^1\ell_t^+\vPsi_{J+1}^n),
\end{equation}
where Lemma \ref{le:1} (a) is applied to the first term in Eq.~\eqref{cn-inner},
(b) to the second term and (c) to the third and fourth terms.
Similarly, it can be verified that \eqref{eq:dis-Q-CN} holds
for the {\tt CN0}~\eqref{eq:cn-kuo}, {\tt LCN0}~\eqref{eq:lcn-kuo} and {\tt LCN2} \eqref{eq:lcn2-gnld} schemes.
Performing the inner product of the {\tt CN} scheme \eqref{eq:cn-gnld} and $\delta_t^+\vPsi^n$,
keeping the imaginary part and applying Lemma \ref{le:1} give directly
the conservation of the discrete energy
\begin{align*}
\delta_t^+ E_h^n = \frac{1}2 \imag \left((\ell_t^+\wbar{\vPsi}_{J+1}^n\vgamma^1\delta_t^+\vPsi_J^n +
\ell_t^+\wbar{\vPsi}_{J}^n\vgamma^1\delta_t^+\vPsi_{J+1}^n)\right. \\
\left.
-(\ell_t^+\wbar{\vPsi}_{1}^n\vgamma^1\delta_t^+\vPsi_0^n +
\ell_t^+\wbar{\vPsi}_{0}^n\vgamma^1\delta_t^+\vPsi_{1}^n)\right).
\end{align*}
For the Fourier spectral method~\eqref{eq:fft},
using the fact that $\exp{(-\mi \tau_i (\kappa \vgamma^0\vgamma^1+m\vgamma^0))}$
in \eqref{eq:matrixfft} is a unitary matrix
yields
\begin{equation*}
\left\|\exp\left( -\mi \tau_i (\kappa \vgamma^0\vgamma^1+m\vgamma^0)\right) \vPsi_j^{(1)}\right\|^2 = \left\|\vPsi_j^{(0)}\right\|^2,
\end{equation*}
and then
\begin{align*}
\left\|\vPsi^{(1)}\right\|^2 &= \left\|\mathcal{F}^{-1}
\left[\exp\left( -\mi \tau_i (\kappa \vgamma^0\vgamma^1+m\vgamma^0)\right)
\mathcal{F}\left(\vPsi^{(0)}\right)\right]\right\|^2 \\
&= \left\|\exp\left( -\mi \tau_i (\kappa \vgamma^0\vgamma^1+m\vgamma^0)\right)
\mathcal{F}\left(\vPsi^{(0)}\right)\right\|^2 = \left\| \mathcal{F}\left(\vPsi^{(0)}\right)\right\|^2
= \left\|\vPsi^{(0)}\right\|^2,
\end{align*}
where Parseval's identity is applied twice.
It can be readily verified that the matrix exponents in Eqs.~\eqref{OS-NS-eq3} and \eqref{eq:solution-V}
are unitary, thus $\|\vPsi^{(1)}\|^2=\|\vPsi^{(0)}\|^2$ holds both for
Eqs.~\eqref{OS-NS-eq3} and \eqref{eq:solution-V}, \ie
$Q_h$ should be conserved for solutions of the nonlinear subproblems.
Therefore, the {\tt OS($N$)} scheme satisfies the conservation law of charge.
\end{proof}
\begin{Remark}\rm
It will be verified by numerical results in Section~\ref{sec:numericalresults} that
the {\tt LCN1}, {\tt HS}, {\tt LF} and {\tt SI} schemes do not conserve the discrete charge or energy,
and none of the numerical methods presented in Section \ref{sec:DiffSch} conserves the discrete linear momentum.
\end{Remark}
\section{Numerical results}
\label{sec:numericalresults}
This section will conduct numerical simulations to compare
the performance of numerical schemes proposed in Section~\ref{sec:DiffSch}
and then utilize the {\tt OS($4$)} scheme to investigate the
interaction dynamics for the NLD solitary waves \eqref{mw} under the
scalar and vector self-interaction.
For those localized NLD solitary waves,
the periodic boundary condition for the {\tt OS($N$)} scheme and the non-reflection boundary condition
for other schemes could be adopted at the boundaries of the computational domain if a relatively large computational domain
has been taken in our numerical experiments.
All calculations are performed on a Lenovo desktop computer with
Intel Core i5 650 CPU and 4GB RAM using double precision in the
3.0.0-24-generic x86\_64 Linux operation system and the compiler is gcc 4.6.1.
The computational domain $\Omega$ will be taken as $[-50,50]$ in Examples \ref{eg.1}-\ref{eg.scalar-vector}
and $[-100,100]$ in Example \ref{eg.scalar-2humped}.
and the particle mass $m$ in Eq.~\eqref{nld-lorentz} is chosen to be $1$.
\begin{Example}
\label{eg.1}
The first example is devoted to comparing the numerical performance of all the numerical methods in Section \ref{sec:DiffSch}
in terms of the accuracy, the conservativeness, the efficiency and the error growth.
A one-humped solitary wave with the velocity $V=-0.2$ is simulated here under the quadric scalar self-interaction (\ie $v=0$ and $k=1$), traveling from right to left with the parameters in \eqref{mw}:
$x_0 = 5$, $s= 0.5$, and $\omega = 0.75$.
The $P^N$-RKDG method \cite{ShaoTang2006} is also included here for comparison,
which is assembled with a fourth-order accurate Runge-Kutta time discretization in time
and the Legendre polynomials of degree at most $N$ as local basis functions in the spatial
Galerkin approximation.
\begin{table}
\centering
\caption{Example~\ref{eg.1}. Part I: Numerical comparison of the accuracy, the conservativeness
and the efficiency at $t=50$ with the time stepsize being set to $\tau = \frac12 h$.
The CPU time in seconds is recorded for the finest mesh.}
\label{tab:eg1}
\newsavebox{\tablebox}
\begin{lrbox}{\tablebox}
\begin{tabular}{lrcccccccc}
&&&&&&&&\\
\hline & $h$ & $\mathcal{V}_{Q}$ & $\mathcal{V}_{E}$ & $\mathcal{V}_{P}$ & $\text{err}_2$ & Order & $\text{err}_\infty$ & Order &
Time (s) \\ \hline
{\tt CN} & 0.08 & 1.38E-15 & 1.09E-15 & 2.65E-06 & 2.74E-02 & & 2.61E-02 & & \\
& 0.04 & 4.66E-15 & 1.37E-15 & 1.65E-07 & 6.84E-03 & 2.00 & 6.50E-03 & 2.00 & \\
& 0.02 & 4.78E-15 & 1.37E-15 & 1.03E-08 & 1.71E-03 & 2.00 & 1.63E-03 & 2.00 & \\
& 0.01 & 3.05E-14 & 2.05E-15 & 6.42E-10 & 4.27E-04 & 2.00 & 4.06E-04 & 2.00 & 367.1 \\ \hline
{\tt CN0} & 0.08 & 7.05E-15 & 3.24E-08 & 2.37E-06 & 2.31E-02 & & 2.21E-02 & & \\
& 0.04 & 1.01E-15 & 2.02E-09 & 1.47E-07 & 5.76E-03 & 2.00 & 5.51E-03 & 2.00 & \\
& 0.02 & 5.54E-15 & 1.26E-10 & 9.17E-09 & 1.44E-03 & 2.00 & 1.38E-03 & 2.00 & \\
& 0.01 & 2.88E-14 & 7.88E-12 & 5.72E-10 & 3.59E-04 & 2.00 & 3.44E-04 & 2.00 & 345.3 \\ \hline
{\tt LCN0} & 0.08 & 2.14E-15 & 1.83E-06 & 4.31E-05 & 2.75E-02 & & 2.62E-02 & & \\
& 0.04 & 1.26E-16 & 2.29E-07 & 5.55E-06 & 6.87E-03 & 2.00 & 6.53E-03 & 2.00 & \\
& 0.02 & 8.81E-15 & 2.86E-08 & 7.04E-07 & 1.72E-03 & 2.00 & 1.63E-03 & 2.00 & \\
& 0.01 & 3.11E-14 & 3.58E-09 & 8.87E-08 & 4.29E-04 & 2.00 & 4.08E-04 & 2.00 & 81.8 \\ \hline
{\tt LCN1} & 0.08 & 2.22E-04 & 1.92E-04 & 3.73E-04 & 3.53E-02 & & 3.34E-02 & & \\
& 0.04 & 2.78E-05 & 2.40E-05 & 4.65E-05 & 9.06E-03 & 1.96 & 8.56E-03 & 1.97 & \\
& 0.02 & 3.47E-06 & 3.01E-06 & 5.80E-06 & 2.30E-03 & 1.98 & 2.17E-03 & 1.98 & \\
& 0.01 & 4.34E-07 & 3.76E-07 & 7.25E-07 & 5.78E-04 & 1.99 & 5.45E-04 & 1.99 & 118.7 \\ \hline
{\tt LCN2} & 0.08 & 2.77E-15 & 1.64E-07 & 7.01E-06 & 2.77E-02 & & 2.64E-02 & & \\
& 0.04 & 7.55E-16 & 1.98E-08 & 6.77E-07 & 6.92E-03 & 2.00 & 6.58E-03 & 2.00 & \\
& 0.02 & 7.55E-16 & 2.44E-09 & 7.24E-08 & 1.73E-03 & 2.00 & 1.64E-03 & 2.00 & \\
& 0.01 & 2.91E-14 & 3.03E-10 & 8.29E-09 & 4.32E-04 & 2.00 & 4.11E-04 & 2.00 & 118.4 \\ \hline
{\tt HS} & 0.08 & 1.62E-06 & 1.55E-06 & 4.33E-06 & 2.00E-02 & & 1.55E-02 & & \\
& 0.04 & 1.01E-07 & 9.65E-08 & 2.70E-07 & 5.00E-03 & 2.00 & 3.88E-03 & 2.00 & \\
& 0.02 & 6.30E-09 & 6.03E-09 & 1.68E-08 & 1.25E-03 & 2.00 & 9.71E-04 & 2.00 & \\
& 0.01 & 3.94E-10 & 3.77E-10 & 1.05E-09 & 3.13E-04 & 2.00 & 2.43E-04 & 2.00 & 14.8 \\ \hline
{\tt LF} & 0.08 & 5.88E-05 & 4.73E-05 & 8.68E-06 & 1.41E-02 & & 1.35E-02 & & \\
& 0.04 & 5.51E-06 & 4.43E-06 & 6.59E-07 & 3.51E-03 & 2.01 & 3.36E-03 & 2.01 & \\
& 0.02 & 6.24E-07 & 5.01E-07 & 6.91E-08 & 8.74E-04 & 2.00 & 8.36E-04 & 2.00 & \\
& 0.01 & 7.54E-08 & 6.05E-08 & 8.19E-09 & 2.18E-04 & 2.00 & 2.09E-04 & 2.00 & 8.9 \\ \hline
{\tt SI} & 0.08 & 3.79E-08 & 1.15E-07 & 3.94E-06 & 3.59E-02 & & 4.76E-02 & & \\
& 0.04 & 3.68E-09 & 8.36E-09 & 2.40E-07 & 8.97E-03 & 2.00 & 1.19E-02 & 2.00 & \\
& 0.02 & 4.09E-10 & 6.87E-10 & 1.44E-08 & 2.24E-03 & 2.00 & 2.98E-03 & 2.00 & \\
& 0.01 & 4.89E-11 & 6.46E-11 & 8.20E-10 & 5.60E-04 & 2.00 & 7.44E-04 & 2.00 & 132.7 \\ \hline
\end{tabular}
\end{lrbox}
\scalebox{0.7}{\usebox{\tablebox}}
\end{table}
\begin{table}
\centering
\caption{Example~\ref{eg.1}. Part II:
Numerical comparison of the accuracy, the conservativeness
and the efficiency at $t=50$.
The CPU time measured in seconds is listed for the finest mesh.}
\label{tab:eg1-2}
\begin{lrbox}{\tablebox}
\begin{tabular}{lrrcccccccc}
&&&&&&&&&&\\
\hline & $\tau$ & $h$ & $\mathcal{V}_{Q}$ & $\mathcal{V}_{E}$ & $\mathcal{V}_{P}$ & $\text{err}_{2}$ & Order & $\text{err}_{\infty}$ & Order & Time (s) \\ \hline
{\tt OS($2$)} & 0.04 & 0.78 & 1.04E-13 & 2.33E-07 & 3.97E-05 & 2.03E-03 & & 9.96E-04 & & \\
& 0.02 & 0.39 & 1.33E-13 & 1.45E-08 & 2.55E-13 & 2.53E-04 & 3.01 & 1.90E-04 & 2.39 & \\
& 0.01 & 0.39 & 3.58E-13 & 9.04E-10 & 6.08E-13 & 6.33E-05 & 2.00 & 4.75E-05 & 2.00 & \\
& 0.005 & 0.20 & 6.15E-13 & 5.60E-11 & 7.13E-13 & 1.58E-05 & 2.00 & 1.21E-05 & 1.97 & 8.2 \\ \hline
{\tt OS($4$)} & 0.04 & 0.78 & 7.40E-13 & 5.78E-13 & 4.12E-05 & 1.71E-03 & & 4.27E-04 & & \\
& 0.02 & 0.39 & 1.46E-12 & 1.25E-12 & 2.17E-12 & 7.92E-08 & 14.40 & 3.32E-08 & 13.65 & \\
& 0.01 & 0.20 & 1.88E-12 & 1.60E-12 & 2.71E-12 & 4.28E-10 & 7.53 & 3.28E-10 & 6.66 & \\
& 0.005 & 0.10 & 7.00E-13 & 6.55E-13 & 2.44E-12 & 1.93E-11 & 4.47 & 1.44E-11 & 4.51 & 16.8 \\ \hline
$P^3$-RKDG & 0.04 & 0.78 & 1.03E-05 & 5.47E-07 & 2.61E-06 & 1.46E-04 & & 1.44E-04 & & \\
& 0.02 & 0.39 & 8.60E-08 & 5.96E-07 & 6.31E-07 & 6.59E-06 & 4.47 & 8.85E-06 & 4.03 & \\
& 0.01 & 0.20 & 6.98E-10 & 4.32E-08 & 4.69E-08 & 4.10E-07 & 4.01 & 5.62E-07 & 3.98 & \\
& 0.005 & 0.10 & 7.88E-12 & 2.75E-09 & 3.00E-09 & 2.56E-08 & 4.00 & 3.69E-08 & 3.93 & 551.6 \\ \hline
$P^4$-RKDG & 0.04 & 0.78 & 1.00E-07 & 2.13E-07 & 1.10E-07 & 8.02E-06 & & 1.04E-05 & & \\
& 0.02 & 0.39 & 8.53E-10 & 4.51E-09 & 3.44E-09 & 2.58E-07 & 4.96 & 3.36E-07 & 4.96 & \\
& 0.01 & 0.20 & 2.28E-11 & 6.41E-11 & 5.22E-11 & 8.46E-09 & 4.93 & 1.16E-08 & 4.86 & \\
& 0.005 & 0.10 & 2.58E-12 & 9.56E-11 & 2.16E-10 & 3.09E-10 & 4.78 & 2.39E-10 & 5.60 & 744.8 \\ \hline
\end{tabular}
\end{lrbox}
\scalebox{0.7}{\usebox{\tablebox}}
\end{table}
Tables \ref{tab:eg1} and \ref{tab:eg1-2} summarize the numerical
results at the final time $t=50$, where
$\text{err}_2$ and $\text{err}_\infty$ are respectively
the $l^2$ and $l^\infty$ errors at the final time,
$\mathcal{V}_Q$, $\mathcal{V}_E$, $\mathcal{V}_P$
measure respectively the variation of charge, energy and linear momentum
at the final time relative to the initial quantities,
and the CPU time of calculations
with the same mesh size is recorded for comparing the efficiency.
It can be observed clearly there that:
(1) while the {\tt CN}, {\tt CN0}, {\tt LCN0}, {\tt LCN1}, {\tt LCN2}, {\tt HS}, {\tt LF}, {\tt SI} and {\tt OS($2$)} schemes
are of the second-order accuracy, the {\tt OS($4$)}, $P^3$-RKDG and $P^4$-RKDG methods
exhibit at least the fourth-order accuracy;
(2) The {\tt CN}, {\tt CN0}, {\tt LCN0}, {\tt LCN2}, {\tt OS($2$)} and {\tt OS($4$)} schemes conserve
the discrete charge and only the {\tt CN} scheme conserves the discrete energy,
but none conserves the discrete linear momentum;
(3) The {\tt OS($4$)} scheme could also keep very accurately the discrete energy and linear momentum with
relatively fine meshes.
All above numerical results are consistent with the theoretical results given in Section~\ref{sec:discussion4fdm}.
Among the numerical methods of the second order accuracy,
it is also found that the {\tt OS($2$)} scheme runs fastest ($8.2$ seconds for the mesh $\tau=0.005$ and $h=0.20$)
if requiring to attain almost the same accuracy.
Similarly, the {\tt OS($4$)} scheme runs much more faster than both $P^3$-RKDG and $P^4$-RKDG methods,
and the ratio of the CPU time used by the {\tt OS($4$)} scheme over that used by the $P^3$-RKDG method is around $3.05\%$,
and reduces to around $2.26\%$ over that used by the $P^4$-RKDG method.
\begin{figure}[h]
\centering
\includegraphics[width=6.5cm]{ex1n2}
\caption{Example~\ref{eg.1}. The $l^\infty$ error history.}
\label{fig.LinfErr}
\end{figure}
Fig.~\ref{fig.LinfErr} plots the $l^\infty$ error history
in the finest mesh used in Tables \ref{tab:eg1} and \ref{tab:eg1-2}.
According to the curves shown there,
it can be seen there that the $l^\infty$ error of all the schemes increases almost linearly with the time.
However, the slopes, obtained by the linear fitting, are different.
The smaller the slope is, the longer time the scheme could simulate to.
The {\tt SI} scheme has the largest slope $1.412\times 10^{-05}$ while the {\tt OS($2$)} scheme has
the smallest one $2.334\times 10^{-07}$ among all the second-order accurate methods.
Further, the slopes of the curves of $l^\infty$ errors for
the {\tt OS($4$)}, $P^3$-RKDG and $P^4$-RKDG schemes are almost the same
value of $3.199\times 10^{-13}$ which is much more smaller
than those of the second-order accurate schemes.
\begin{Remark}\rm
Both the theoretical and numerical comparison of the {\tt OS($2$)} scheme with the {\tt CN0} and {\tt LCN0} schemes
show that the former is better, especially in terms of efficiency and error growth.
Therefore in some sense this is an answer to the debate stimulated in \cite{Alvarez1992, FrutosSanz-Serna1989}
over twenty years ago on which one is most efficient among
the {\tt OS($2$)}, {\tt CN0} and {\tt LCN0} schemes.
\end{Remark}
\end{Example}
\begin{Example}
\label{eg.os-dg}
\begin{figure}[h]
\centering
\includegraphics[width=6.5cm]{rhoq-ex16}
\caption{Example~\ref{eg.os-dg}. The initial charge density $\rho_{Q}(x,t)$ for two typical cases,
$V =-0.2$, $x_0=5$.
Case $1$ is shown in the solid line, a two-humped profile ($\omega = 0.3$) under the quadric scalar self-interaction ($k=1$, $s=0.5$ and $v=0$); Case $2$ is shown in the dashed line, a one-humped profile ($\omega = 0.75$) under the cubic scalar and vector self-interaction ($k=2$ and $s=v=0.5$)}
\label{fig.os-dg-rhoq}
\end{figure}
The $P^3$-RKDG method
has been successfully applied before into investigating the interaction for the
NLD solitary waves under the quadric scalar self-interaction in
\cite{ShaoTang2005,ShaoTang2006,ShaoTang2008},
but the numerical comparison shown in Example~\ref{eg.1} tells us that
the proposed {\tt OS($4$)} scheme should be preferred now.
In this example, we further conduct numerical comparison among the {\tt OS($4$)},
$P^3$-RKDG and $P^4$-RKDG methods in simulating one-humped and two-humped solitary waves.
Two typical profiles of the charge density for the NLD solitary wave
displayed in Fig.~\ref{fig.os-dg-rhoq} are considered,
one denoted by Case $1$ has a two-humped profile under the quadric scalar self-interaction,
and the other denoted by Case $2$ has a one-humped profile under the cubic scalar and vector self-interaction.
These two solitary waves are located initially at $x_0=5$,
travel from right to left with the velocity $V =-0.2$ and stop at the final time $t=50$.
For convenience, we use $\rho_Q(x,t)\equiv J_0$ to
represent the charge density.
\begin{table}[h]
\centering
\caption{Example~\ref{eg.os-dg}. Numerical comparison among the {\tt OS($4$)},
$P^3$-RKDG and $P^4$-RKDG methods. The CPU time is measured in seconds. }
\label{tab:os2-4th-vs-rkdg}
Case 1 in the mesh of $\tau = 0.01$ and $h=\frac{100}{512}$
\begin{lrbox}{\tablebox}
\begin{tabular}{ccccccc}
\hline\hline & $\mathcal{V}_{Q}$ & $\mathcal{V}_{E}$ & $\mathcal{V}_{P}$ & $\text{err}_{2}$ & $\text{err}_{\infty}$ & Time (s) \\ \hline
{\tt OS($4$)} & 1.96E-12 & 1.10E-12 & 3.45E-12 & 2.15E-09 & 2.17E-09 & 7.8 \\
$P^3$-RKDG & 3.34E-08 & 2.27E-07 & 2.96E-07 & 2.89E-06 & 4.15E-06 & 146.6 \\
$P^4$-RKDG & 1.35E-12 & 9.43E-08 & 5.34E-08 & 9.02E-08 & 6.94E-08 & 195.5 \\
\hline
\end{tabular}
\end{lrbox}
\scalebox{0.7}{\usebox{\tablebox}}
Case 2 in the mesh of $\tau = 0.005$ and $h=\frac{100}{1024}$
\begin{lrbox}{\tablebox}
\begin{tabular}{ccccccc}
\hline\hline & $\mathcal{V}_{Q}$ & $\mathcal{V}_{E}$ & $\mathcal{V}_{P}$ & $\text{err}_{2}$ & $\text{err}_{\infty}$ & Time (s) \\ \hline
{\tt OS($4$)} & 9.82E-13 & 8.35E-13 & 2.95E-12 & 7.33E-11 & 8.63E-11 & 46.8 \\
$P^3$-RKDG & 3.21E-10 & 9.04E-09 & 1.95E-08 & 2.00E-07 & 3.98E-07 & 881.1 \\
$P^4$-RKDG & 9.10E-13 & 4.89E-09 & 1.57E-09 & 4.68E-09 & 6.35E-09 & 1156.5 \\
\hline
\end{tabular}
\end{lrbox}
\scalebox{0.7}{\usebox{\tablebox}}
\end{table}
The numerical comparison is shown in
Table~\ref{tab:os2-4th-vs-rkdg}, from which we can observe that,
(1) with the same mesh, no matter sparse ($\tau = 0.01$ and $h=\frac{100}{512}\approx0.1953$) or fine
($\tau = 0.005$ and $h=\frac{100}{1024}\approx0.0977$),
the {\tt OS($4$)} scheme is more conservative and higher accurate than both $P^3$-RKDG and $P^4$-RKDG methods;
(2) the {\tt OS($4$)} scheme runs much faster than both $P^3$-RKDG and $P^4$-RKDG methods as we have found in Table~\ref{tab:eg1-2}. Here,
the ratio of the CPU time used by the {\tt OS($4$)} scheme over that used by the $P^3$-RKDG method is around $5.32\%$,
and reduces to around $4.05\%$ over that used by the $P^4$-RKDG method for both cases.
The $l^\infty$ error history is plotted in Fig.~\ref{fig.os-dg},
which shows that those three methods all have almost zero slope (under $4.50$E-$09$).
This can be used to explain our previous success of the $P^3$-RKDG method in
\cite{ShaoTang2005,ShaoTang2006,ShaoTang2008}.
Further more, the more smaller errors of the {\tt OS($4$)} method mean that it should be more powerful
than others.
\begin{figure}[h]
\centering
\includegraphics[width=6.5cm]{os-dgex16}
\caption{Example~\ref{eg.os-dg}. The $l^\infty$ error history. The slopes are displayed above the curves.}
\label{fig.os-dg}
\end{figure}
\end{Example}
It has been shown that the {\tt OS($4$)} scheme behaves best
for both one-humped and two-humped NLD solitary waves in long time simulations.
Therefore,
we conclude the comparison with the judgement that the {\tt OS($4$)} scheme is the most suitable for
simulating the interaction dynamics for the NLD solitary waves
in terms of the accuracy, the conservativeness, the efficiency and the error growth.
The {\tt OS($4$)} scheme will be utilized to investigate the binary collision of the NLD solitary waves.
The initial setup is the linear superposition of two moving waves
\begin{equation}
\label{eq:two-mw-wave}
\vPsi(x,t=0) = \vPsi_\text{l}^{\text{mw}}(x-x_{\text{l}},t=0)+\vPsi_\text{r}^{\text{mw}}(x-x_\text{r},t=0),
\end{equation}
where $\vPsi_{pos}^\text{mw}(x-x_{pos},t)$ denote the moving waves \eqref{mw} centered at $x_{pos}$ with the speed $V_{pos}$ and the frequency $\omega_{pos}$ for $pos \in\{\text{l},\text{r}\}$.
In the following examples, two equal solitary waves are placed symmetrically at $t=0$ with
$-x_\text{l}=x_\text{r}=10$ and $V_\text{l} = -V_\text{r} = 0.2$.
Several typical NLD solitary waves are considered with the parameters given in Table~\ref{3caselist},
and both quadric ($k=1$) and cubic ($k=2$) cases will be studied.
It should be noted that
the experiments carried out in the literatures
are all limited to the collisions of
the NLD solitary waves under the quadric scalar self-interaction.
A relatively fine mesh, $\tau=0.005$ and $h=100/2^{13}\approx0.0122$, is adopted hereafter.
\begin{table}[h]
\caption{The initial setups of different cases in binary collisions.}
\label{3caselist}
\centering
\begin{tabular}{cccc|l}
\hline
case & $s$ & $v$ & $\omega_l=\omega_r$ & Remarks \\ \hline
B1 & 0.5 & 0 & 0.8 & scalar, one-humped \\
B2 & 0 & 0.5 & 0.8 & vector, one-humped \\
B3 & 0.5 & 0.5 & 0.8 & scalar and vector, one-humped \\
B4 & 0.5 & 0 & 0.3 & scalar, two-humped \\
B5 & 4 & 1 & 0.1 & scalar and vector, two-humped \\
\hline
\end{tabular}
\end{table}
\begin{Example}
\label{eg.scalar}
\begin{figure}[h]
\centering
\subfigure[$k=1$]{
\includegraphics[width=6.5cm]{osk1-s-w08-T0-80}
\label{subfigure:k1-s-rho}}
\subfigure[$k=2$]{
\label{subfigure:k1-s-QEP}
\includegraphics[width=6.5cm]{osk2-s-w08-T0-80}}
\caption{Example \ref{eg.scalar}. Binary collision of the NLD solitary waves under the scalar self-interaction.}
\label{fig:k1-s}
\end{figure}
The collision of two equal one-humped solitary waves under the scalar self-interaction,
\ie Case B1 in Table~\ref{3caselist}, is studied in this example.
The interaction dynamics for the quadric case are shown in the left plot of Fig.~\ref{fig:k1-s},
where two equal waves with the initial amplitude of $0.4082$ move close at a velocity of $0.2$ and overlap each other,
then separate into a left moving wave and a right moving wave with the amplitude of $0.3743$ and the velocity of $0.1831$.
Similar phenomena are observed for the cubic case shown in the right plot of Fig.~\ref{fig:k1-s} except that
(1) two waves overlap more stronger around $t=41$ now due to the stronger nonlinearity;
(2) after collision, the amplitude decreases to $0.5899$ from the initial amplitude of $0.6455$ while
the velocity also decreases to $0.1037$.
In both cases, the discrete charge, energy and linear momentum are approximately conserved in the interaction
since the variation of them at $t=80$ is under $1.58$\text{E}-$10$.
\end{Example}
\begin{Example}
\label{eg.v}
\begin{figure}[h]
\centering
\subfigure[$k=1$]{
\includegraphics[width=6.5cm]{osk1-v-w08}
\label{subfigure:k1-v-rho}}
\subfigure[$k=2$]{
\label{subfigure:k1-v-QEP}
\includegraphics[width=6.5cm]{osk2-v-w08}}
\caption{Example \ref{eg.v}: Binary collision of the NLD solitary waves under the vector self-interaction.}
\label{fig:k1-v}
\end{figure}
The collision of two equal one-humped solitary waves under the vector self-interaction,
\ie Case B2 in Table~\ref{3caselist}, is studied in this example.
To the best of our knowledge, it is the first time to study binary collision of
the NLD solitary waves under the vector self-interaction.
The interaction dynamics for the quadric case are shown in the left plot of Fig.~\ref{fig:k1-v},
where the waves keep the shape and the velocity after the collision.
A totally different phenomenon appears for the cubic vector self-interaction as
displayed in the right plot of Fig.~\ref{fig:k1-v}.
The initial one-humped equal waves first merge into a single wave, then separate and overlap again.
Around $t=50$, collapse happens and highly oscillatory waves are generated and moving outside with a big velocity near $1$,
meanwhile a one-humped wave with small amplitude is formed at the center.
In both cases, the discrete charge, energy and linear momentum are approximately conserved in the interaction
since the variation of them at $t=100$ is under $5.41$\text{E}-$11$.
Note in passing that the collapse here is different from that shown in \cite{ShaoTang2005}.
It was reported there that the strong negative energy and radiation appear when
the collapse happens during the binary collision of two-humped waves.
\end{Example}
\begin{Example}
\label{eg.scalar-vector}
\begin{figure}[h]
\centering
\subfigure[$k=1$]{
\includegraphics[width=6.5cm]{osk1-sv-w08}
\label{subfigure:k1-sv-rho}}
\subfigure[$k=2$]{
\label{subfigure:k1-sv-QEP}
\includegraphics[width=6.5cm]{osk2-sv-w08}}
\caption{Example \ref{eg.scalar-vector}: Binary collision of the NLD solitary waves under the scalar and vector self-interaction.}
\label{fig:k1-sv}
\end{figure}
This example is devoted into investigating
for the first time the collision of two equal NLD solitary waves under the scalar and vector self-interaction,
\ie Case B3 in Table~\ref{3caselist}.
The interaction dynamics for the quadric case are shown in the left plot of Fig.~\ref{fig:k1-sv},
where two equal waves with the initial amplitude of $0.2041$ move close at a velocity of $0.2$ and overlap each other,
then separate into a left moving wave and a right moving wave with the amplitude of $0.2091$ and the velocity of $0.1968$.
The collapse similar to that shown in right plot of Fig.~\ref{fig:k1-v} happens again for the
cubic vector self-interaction, see the right plot of Fig.~\ref{fig:k1-sv}.
The initial one-humped equal waves first merge into a single wave at $t=38$, then separate and overlap again.
Around $t=50$, collapse happens and highly oscillatory waves are generated and moving outside with a big velocity near $1$.
In both cases, the discrete charge, energy and linear momentum are approximately conserved in the interaction
since the variation of them at $t=80$ is under $3.53$\text{E}-$10$.
\end{Example}
\begin{Example}
\label{eg.scalar-2humped}
\begin{figure}[h]
\centering
\subfigure[$k=1$]{
\includegraphics[width=6.5cm]{osk1-s-w03}
}
\subfigure[$k=2$]{
\includegraphics[width=6.5cm]{osk2-s-w03}}
\caption{Example~\ref{eg.scalar-2humped}: Binary collision of the two-humped
NLD solitary waves under the scalar self-interaction.}
\label{fig:k1-2humped}
\end{figure}
\begin{figure}[h]
\centering
\subfigure[$k=1$]{
\includegraphics[width=6.5cm]{osk1-s4v1-w01}
\label{subfigure:k1-s-rho-2humped}}
\subfigure[$k=2$]{
\label{subfigure:k1-s-QEP-2humped}
\includegraphics[width=6.5cm]{osk2-s4v1-w01}}
\caption{Example~\ref{eg.scalar-2humped}: Binary collision of the two-humped
NLD solitary waves under the scalar and vector self-interaction.}
\label{fig:k1-2humped-sv}
\end{figure}
As reported before in \cite{ShaoTang2005,ShaoTang2008},
collapse happens in binary and ternary collisions
of the NLD solitary waves under the quadric scalar self-interaction if
the two-humped waves are evolved. In this example,
we will show further that collapse could happen in binary collision of
equal two-humped waves under the cubic scalar self-interaction and under the
linear combination of scalar and vector self-interactions.
First, Case B4 in Table~\ref{3caselist} is studied
and the interaction dynamics are shown in Fig.~\ref{fig:k1-2humped},
which clearly shows that (1) collapse happens in both quadric and cubic cases but is more stronger in the latter;
(2) two initial waves
at the same velocity are decomposed into groups with different velocities after the collision,
but there is no such decomposition for the cubic case.
In the left plot of Fig.~\ref{fig:k1-2humped}, the highly oscillating waves with small amplitude move outside
at a big velocity of $0.9644$, while the one-humped waves with big amplitude follow them at a small velocity of $0.4626$.
In both cases, the discrete charge, energy and linear momentum are approximately conserved in the interaction
since the variation of them at $t=80$ is under $1.01$\text{E}-$5$.
Second, binary collision of equal two-humped solitary waves under the scalar and vector self-interaction,
\ie Case B5 in Table~\ref{3caselist}, is plotted in Fig.~\ref{fig:k1-2humped-sv}.
The phenomena are very similar to that shown in Fig.~\ref{fig:k1-2humped},
and the ``decomposition" phenomenon for the quadric case is more obvious
than that shown in the left plot of Fig.~\ref{fig:k1-2humped}.
\end{Example}
\section{Conclusion and outlook}
\label{sec:conclusion}
Several numerical methods for solving the NLD equation
with the scalar and vector self-interaction
have been presented and compared theoretically and numerically.
Our results have revealed that among them, the {\tt OS($4$)} scheme, one of the fourth-order accurate OS methods,
performs best in terms of the accuracy and the efficiency.
Particularly, the {\tt OS($4$)} scheme
is usually more accurate than the $P^4$-RKDG method in the mesh of the same size,
but the former needs much more less computational cost than the latter.
Such superior performance of the OS methods
is credited to the full use of the local conservation laws of the NLD equation
such that the nonlinear subproblems resulted from them are exactly solved.
The interaction dynamics for the NLD solitary waves
under the quadric and cubic self-interaction have been investigated
with the {\tt OS($4$)} scheme.
We have found that
such interaction dynamics depend on the exponent power of the self-interaction.
Actually,
it has been observed for the first time in our numerical experiments that,
(1) collapse happens in collision of two equal one-humped
NLD solitary waves under the cubic vector self-interaction
but such collapse does not appear for corresponding quadric case;
(2) two initial waves at the same velocity
are decomposed into groups with different velocities after
collapse in binary collision of two-humped NLD solitary
waves under the quadric scalar self-interaction or under the quadric scalar and vector self-interaction
but such phenomenon does not show up for corresponding cubic case.
More efforts on the interaction dynamics for the NLD solitary waves under more general self-interaction
with the {\tt OS($4$)} method are still going on.
\section*{Acknowledgments}
{Sihong Shao was partially supported by the National
Natural Science Foundation of China (Project No. 11101011)
and the Specialized Research Fund for the Doctoral Program of Higher Education (Project No. 20110001120112).
Huazhong Tang was partially supported by the National Natural Science Foundation of China (Project No. 10925101).
The authors would also like to thank the referees for many useful suggestions.}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 510 |
\section{Introduction}
We study the general Choquard equation
\begin{equation}
\label{eqChoquard}
\tag{$\mathcal{C}$}
-\Delta u + u = \bigl(I_\alpha \ast \abs{u}^p\bigr) \abs{u}^{p - 2} u \quad\text{in \({\mathbb R}^N\)},
\end{equation}
where \(N \ge 1\), \(\alpha \in (0, N)\) and \(I_\alpha : {\mathbb R}^N \to {\mathbb R}\) is the Riesz potential
defined at each point \(x \in {\mathbb R}^N \setminus \{0\}\) by
\begin{align*}
I_\alpha (x) &= \frac{A_\alpha}{\abs{x}^{N - \alpha}}, &
&\text{where }&
&A_\alpha = \frac{\Gamma(\tfrac{N-\alpha}{2})}
{\Gamma(\tfrac{\alpha}{2})\pi^{N/2}2^{\alpha} }.
\end{align*}
When \(N = 3\), \(\alpha = 2\) and \(p = 2\), the equation \eqref{eqChoquard} has appeared in several contexts of quantum physics and is known as the \emph{Choquard--Pekar equation} \citelist{\cite{Pekar1954}\cite{Lieb1977}}, the \emph{Schr\"odin\-ger--Newton equation} \cites{KRWJones1995gravitational,KRWJones1995newtonian,Moroz-Penrose-Tod-1998} and the \emph{stationary Hartree equation}.
The action functional \(\mathcal{A}\) associated to the Choquard equation
\eqref{eqChoquard} is defined for each function \(u\) in the Sobolev space
\(H^1 ({\mathbb R}^N)\) by
\[
\mathcal{A} (u) = \frac{1}{2} \int_{{\mathbb R}^N} \abs{\nabla u}^2 + \abs{u}^2 - \frac{1}{2 p} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr) \abs{u}^p.
\]
In view of the Hardy--Littlewood--Sobolev inequality, which states that if \(s
\in (1, \frac{N}{\alpha})\) then for
every \(v \in L^s ({\mathbb R}^N)\), \(I_\alpha \ast v\in L^\frac{N s}{N - \alpha s}
({\mathbb R}^N)\) and
\begin{equation}
\label{eqHLS}
\int_{{\mathbb R}^N} \abs{I_\alpha \ast v}^\frac{N s}{N - \alpha s} \le C
\Bigl(\int_{{\mathbb R}^N} \abs{v}^s \Bigr)^\frac{N}{N - \alpha s},
\end{equation} (see for example \cite{LiebLoss2001}*{theorem 4.3}), and of the
classical Sobolev embedding, the action functional \(\mathcal{A}\) is
well-defined and continuously differentiable whenever
\[
\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le \frac{N}{N + \alpha}.
\]
A natural constraint for the equation is the \emph{Nehari constraint} \(\dualprod{\mathcal{A}' (u)}{u} = 0\) which leads to search for solutions by minimizing the action functional on the \emph{Nehari manifold}
\[
\mathcal{N}_0 = \bigl\{u \in H^1 ({\mathbb R}^N) \setminus \{0\} \;:\; \dualprod{\mathcal{A}' (u)}{u} = 0\}.
\]
The existence of such a solution has been proved when
\[
\frac{N - 2}{N + \alpha} < \frac{1}{p} < \frac{N}{N + \alpha};
\]
these assumptions are optimal
\citelist{\cite{Lieb1977}\cite{Lions1980}\cite{MorVanSchaft13}}.
We are interested in the construction of \emph{nodal solutions} to
\eqref{eqChoquard}, that is, solutions to \eqref{eqChoquard} that change sign.
The easiest way to construct such solutions is to impose an odd symmetry
constraint. More precisely we consider the Sobolev space of odd functions
\begin{multline*}
H^1_\mathrm{odd}({\mathbb R}^N)
= \bigl\{u \in H^1 ({\mathbb R}^N)\;:\; \text{for almost every \((x', x_N) \in
{\mathbb R}^N\)},\\
u(x', -x_N) = - u (x', x_N)\bigr\},
\end{multline*}
we define the odd Nehari manifold
\[
\mathcal{N}_{\mathrm{odd}}
= \mathcal{N}_{0} \cap H^1_\mathrm{odd} ({\mathbb R}^N)
\]
and the corresponding level
\[
c_\mathrm{odd} = \inf_{\mathcal{N}_{\mathrm{odd}}} \mathcal{A}.
\]
Our first result is that this level \(c_\mathrm{odd}\) is achieved.
\begin{theorem}
\label{theoremOdd}
If \(\frac{N - 2}{N + \alpha} < \frac{1}{p}< \frac{N}{N + \alpha}\), then there exists a weak solution \(u \in H^1_\mathrm{odd} ({\mathbb R}^N) \cap C^2 ({\mathbb R}^N)\) to the Choquard equation \eqref{eqChoquard} such that \(\mathcal{A} (u) = c_\mathrm{odd}\).
Moreover, \(u\) has constant sign on each of the half-spaces \({\mathbb R}^N_+\) and \({\mathbb R}^N_-\) and \(u\) is axially symmetric with respect to an axis perpendicular to \(\partial {\mathbb R}^N_+ = {\mathbb R}^{N - 1} \times \{0\}\).
\end{theorem}
Nodal solutions with higher level of symmetries and thus larger action have
already been constructed
\citelist{\cite{CingolaniClappSecchi2012}\cite{CingolaniClappSecchi2013}\cite{
ClapSalazar2013}}.
The proof of theorem~\ref{theoremOdd} relies on two ingredients: a compactness
property up to translation under the strict inequality \(c_\mathrm{odd} < 2
c_0\) obtained by a concentration--compactness argument
(proposition~\ref{propositionOddPalaisSmaleCondition}) and the proof of the
latter strict inequality (proposition~\ref{propositionStrictInequality}).
\bigbreak
Another notion of solution is that of \emph{least action nodal solution}, which has been well studied for local problems \citelist{\cite{CeramiSoliminiStruwe1986}\cite{CastroCossioNeuberger1997}\cite{CastorCossioNeuberger1998}}.
As for these local problems, we define the constrained Nehari nodal set (as in the local case,
in contrast with \(\mathcal{N}_0\) and \(\mathcal{N}_\mathrm{odd}\),
the set \(\mathcal{N}_{\mathrm{nod}}\) \emph{is not a manifold}),
\begin{multline*}
\mathcal{N}_{\mathrm{nod}}
=\bigl\{ u \in H^1 ({\mathbb R}^N) \;:\; u^+ \ne 0 \ne u^-,\,\\
\dualprod{\mathcal{A}'(u)}{u^+} = 0
\text{ and } \dualprod{\mathcal{A}'(u)}{u^-} = 0\bigr\},
\end{multline*}
where \(u^+ = \max (u, 0) \ge 0\) and \(u^- = \min (u, 0) \le 0\). (In contrast
with the local case, we have for every \(u \in \mathcal{N}_{\mathrm{nod}}\),
\(\dualprod{\mathcal{A}'(u)}{u^+} < \dualprod{\mathcal{A}'(u^+)}{u^+}\), and thus \(u^+ \not \in \mathcal{N}_0\) and \(u^- \not \in \mathcal{N}_0\).)
We prove that when \(p > 2\), the associated level
\[
c_\mathrm{nod} = \inf_{\mathcal{N}_{\mathrm{nod}}} \mathcal{A}
\]
is achieved.
\begin{theorem}
\label{theoremNod}
If \(\frac{N - 2}{N + \alpha} < \frac{1}{p} <
\frac{1}{2}\), then there exists a weak solution \(u \in H^1 ({\mathbb R}^N)\cap C^2 ({\mathbb R}^N)\) to the
Choquard equation \eqref{eqChoquard} such that \(\mathcal{A} (u) =
c_\mathrm{nod}\), and \(u\) changes sign.
\end{theorem}
The restriction on the exponent \(p\) can only be satisfied when \(\alpha > N -
4\).
We understand that \(u\) changes sign if the sets \(\{x \in {\mathbb R}^N \;:\; u (x) >
0 \}\) and \(\{x \in {\mathbb R}^N \;:\; u(x) < 0 \}\) have both positive measure.
We do not know whether the solutions constructed in theorem~\ref{theoremNod} are odd and coincide thus with those of theorem~\ref{theoremOdd} or even whether the solutions of theorem~\ref{theoremNod} have axial symmetry as those of theorem~\ref{theoremOdd}. We leave these questions as open problems.
The proof of theorem~\ref{theoremNod} is based on a new reformulation of the
minimization problem as a minimax problem that allows to apply a minimax
principle with location information
(proposition~\ref{propositionSimpleVariational}) and a new compactness property
up to translations under the condition \(c_{\mathrm{nod}} < 2 c_0\) proved by
concentration--compactness
(proposition~\ref{propositionNodalPalaisSmaleCondition}), in the proof of which
we introduce suitable methods and estimates (see lemma~\ref{lemmaNewEstimate}).
The latter strict inequality is deduced from the inequality \(c_\mathrm{nod}
\le c_{\mathrm{odd}}\).
Compared to theorem~\ref{theoremOdd}, theorem~\ref{theoremNod} introduces the additional restriction \(p > 2\). This assumption is almost optimal: in the locally sublinear case \(p < 2\), the level \(c_\mathrm{nod}\) is not achieved.
\begin{theorem}
\label{theoremDegeneracy}
If \(\max (\frac{N - 2}{N + \alpha}, \frac{1}{2}) < \frac{1}{p} < \frac{N}{N
+ \alpha}\), then \(c_\mathrm{nod} = c_0\) is not achieved in
\(\mathcal{N}_{\mathrm{nod}}\).
\end{theorem}
Theorem~\ref{theoremDegeneracy} shows that minimizing the action on the Nehari nodal set does not provide a nodal solution; there might however exist a minimal action nodal solution that would be constructed in another fashion.
We do not answer in the present work whether \(c_\mathrm{nod}\) is achieved when
\(p = 2\) and \(\alpha > N - 4\). In a forthcoming manuscript in collaboration with V.\thinspace{} Moroz, we extend theorem~\ref{theoremDegeneracy} to the case \(p = 2\) by taking the limit \(p \searrow 2\) \cite{GhimentiMorozVanSchaftingen}.
\bigbreak
If we compare the results in the present paper to well-established features of
the \emph{stationary nonlinear Sch\"odinger equation}
\begin{equation}
\label{NLS}
-\Delta u + u = \abs{u}^{2p - 2} u,
\end{equation}
which is the local counterpart of the Choquard equation~\eqref{eqChoquard}, theorems~\ref{theoremOdd} and \ref{theoremNod} are quite surprising.
The action functional associated to \eqref{NLS} is defined by
\[
\mathcal{A} (u)
= \frac{1}{2} \int_{{\mathbb R}^N} \abs{\nabla u}^2 + \abs{u}^2
- \frac{1}{2 p} \int_{{\mathbb R}^N} \abs{u}^{2 p},
\]
which is well-defined and continuously differentiable when \(\frac{1}{2} - \frac{1}{N} < \frac{1}{p} < \frac{1}{2}\).
Since in this case \(\mathcal{A} (u) = \mathcal{A} (u^+) + \mathcal{A} (u^-)\),
it can be easily proved by a density argument that
\[
c_\mathrm{odd} = c_{\mathrm{nod}} = 2 c_0.
\]
Therefore if one of the infimums \( c_\mathrm{odd} \) or \(c_{\mathrm{nod}}\) is
achieved at \(u\), then both \(u^+\) and \(u^-\) should achieve \(c_0\) in
\(\mathcal{N}_0\). This is impossible, since by the strong maximum principle
\(u^+> 0\) and \(u^- > 0\) almost everywhere on the space \({\mathbb R}^N\).
This nonexistence of minimal action nodal solutions also contrasts with
theorem~\ref{theoremDegeneracy}: for the nonlocal problem \(c_{\mathrm{nod}}\) is too
small to be achieved whereas for the local one this level is too large.
\section{Minimal action odd solution}
In this section we prove theorem~\ref{theoremOdd} about the existence of solutions under an oddness constraint.
\subsection{Variational principle}
We first observe that the corresponding level \(c_{\mathrm{odd}}\) is positive.
\begin{proposition}[Nondegeneracy of the level]
\label{propositionOddNondegenerate}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le \frac{N}{N + \alpha}\), then
\[
c_{\mathrm{odd}} > 0.
\]
\end{proposition}
\begin{proof}
Since \(\mathcal{N}_{\mathrm{odd}} = \mathcal{N}_0 \cap
H^1_{\mathrm{odd}}
({\mathbb R}^N) \subset \mathcal{N}_0\) we have \(c_{\mathrm{odd}} \ge c_0\). The
conclusion follows then from the fact that \(c_0 > 0\) \cite{MorVanSchaft13}.
\end{proof}
A first step in the construction of our solution is the existence
a Palais--Smale sequence.
\begin{proposition}[Existence of a Palais-Smale sequence]
\label{propositionOddPalaisSmaleExistence}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le \frac{N}{N + \alpha}\), then
there exists a sequence \((u_n)_{n \in {\mathbb N}}\) in \(H^1_{\mathrm{odd}}
({\mathbb R}^N)\) such that, as \(n \to \infty\),
\begin{align*}
\mathcal{A} (u_n) & \to c_{\mathrm{odd}} &
&\text{ and} &
\mathcal{A}' (u_n) & \to 0 \quad \text{in \(H^1_{\mathrm{odd}} ({\mathbb R}^N)'\)}.
\end{align*}
\end{proposition}
\begin{proof}
We first recall that the level \(c_{\mathrm{odd}}\) can be rewritten as a mountain pass
minimax level:
\[
c_{\mathrm{odd}}
= \inf_{\gamma \in \Gamma} \sup_{[0, 1]} \mathcal \mathcal{A} \circ \gamma,
\]
where the class of paths \(\Gamma\) is defined by
\[
\Gamma = \bigl\{ \gamma \in C \bigl([0, 1], H^1_\mathrm{odd} ({\mathbb R}^N)\bigr) \;:\; \gamma (0)=
0 \text{ and } \mathcal{A} \bigl(\gamma (1)\bigr) < 0\bigr\}
\]
(see for example \cite{Willem1996}*{theorem 4.2}).
By the general minimax principle
\cite{Willem1996}*{theorem 2.8}, there exists a sequence \((u_n)_{n \in {\mathbb N}}\)
in \(H^1_{\mathrm{odd}} ({\mathbb R}^N)\) such that the sequence
\(\bigl(\mathcal{A} (u_n)\bigr)_{n \in {\mathbb N}}\) converges to \(c_\mathrm{odd}\)
and the
sequence
\(\bigl(\mathcal{A}' (u_n)\bigr)_{n \in {\mathbb N}}\) converges strongly to \(0\) in
the dual space \( H^1_{\mathrm{odd}} ({\mathbb R}^N) '\).
\end{proof}
\subsection{Palais--Smale condition}
We would now like to construct out of the Palais--Smale sequence of
proposition~\ref{propositionOddPalaisSmaleExistence} a solution to our problem.
We shall prove that the functional \(\mathcal{A} \vert_{H^1_{\mathrm{odd}}
({\mathbb R}^N)}\) satisfies the Palais--Smale condition up to translations at the
level \(c_{\mathrm{odd}}\) if the strict inequality \(c_\mathrm{odd} < 2 c_0\)
holds.
\begin{proposition}[Palais--Smale condition]
\label{propositionOddPalaisSmaleCondition}
Assume that \(\frac{N - 2}{N + \alpha} < \frac{1}{p} < \frac{N}{N +
\alpha}\).
Let \((u_n)_{n \in {\mathbb N}}\) be a sequence in \(H^1_{\mathrm{odd}} ({\mathbb R}^N)\) such
that, as \(n \to \infty\),
\begin{align*}
\mathcal{A} (u_n) & \to c_{\mathrm{odd}} &
&\text{ and} &
\mathcal{A}' (u_n) & \to 0 \quad \text{in \(H^1_{\mathrm{odd}} ({\mathbb R}^N)'\)}.
\end{align*}
If
\[
c_{\mathrm{odd}} < 2 c_0,
\]
then there exists a sequence of points \((a_n)_{n \in {\mathbb N}}\) in \({\mathbb R}^{N - 1} \times \{0\}\subset {\mathbb R}^N\) such
that the subsequence
\((u_{n_k} \bigl(\cdot - a_{n_k})\bigr)_{k \in {\mathbb N}}\) converges strongly in
\(H^1
({\mathbb R}^N)\) to \(u \in H^1_{\mathrm{odd}} ({\mathbb R}^N)\).
Moreover
\begin{align*}
\mathcal{A} (u) &= c_{\mathrm{odd}} &
&\text{ and }&
\mathcal{A}'(u) &= 0 \quad \text{in \(H^1 ({\mathbb R}^N)'\)}.
\end{align*}
\end{proposition}
\begin{proof}
First, we observe that, as \(n \to \infty\),
\begin{equation}
\label{eqNorm}
\begin{split}
\Bigl(\frac{1}{2} - \frac{1}{2 p}\Bigr)\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2 &= \mathcal{A} (u_n) - \frac{1}{2 p} \dualprod{\mathcal{A}'
(u_n)}{u_n}\\
&= \mathcal{A} (u_n) + o \biggl( \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2 \Bigr)^\frac{1}{2}\biggr)\\
& = c_{\mathrm{odd}} + o \biggl( \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2 \Bigr)^\frac{1}{2}\biggr).
\end{split}
\end{equation}
In particular, the sequence \((u_n)_{n \in {\mathbb N}}\) is bounded in the space \(H^1 ({\mathbb R}^N)\).
We now claim that there exists \(R > 0\) such that
\begin{equation}
\label{eqDR}
\liminf_{n \to \infty} \int_{D_R} \abs{u_n}^\frac{2 N p}{N + \alpha} > 0,
\end{equation}
where the set \(D_R \subset {\mathbb R}^N\) is the infinite slab
\[
D_R = {\mathbb R}^{N - 1} \times [-R, R].
\]
We assume by contradiction that for each \(R > 0\),
\[
\liminf_{n \to \infty} \int_{D_R} \abs{u_n}^\frac{2 N p}{N + \alpha} = 0.
\]
We define for each \(n \in {\mathbb N}\) the functions \(v_n = \chi_{{\mathbb R}^{N - 1}
\times (0, \infty)} u_n\) and \(\Tilde{v}_n = \chi_{{\mathbb R}^{N - 1} \times
(-\infty, 0)}u_n\).
Since \(u_n \in H^1_{\mathrm{odd}}({\mathbb R}^N)\), we have \(v_n \in H^1_0 ({\mathbb R}^{N - 1}
\times (0, \infty))\subset H^1({\mathbb R}^N)\) and \(\Tilde{v}_n \in H^1_0 ({\mathbb R}^{N - 1}
\times (-\infty, 0)) \subset H^1({\mathbb R}^N)\).
We now compute
\[
\begin{split}
\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr) \abs{\Tilde{v}_n}^p
\le &2 \int_{{\mathbb R}^N} \int_{D_R} I_\alpha (x - y) \abs{v_n (y)}^p
\abs{\Tilde{v}_n (x)}^p \,\mathrm{d} y\,\mathrm{d} x\\
&+\int_{{\mathbb R}^N \setminus D_R} \int_{{\mathbb R}^N \setminus D_R} I_\alpha (x - y)
\abs{v_n (y)}^p \abs{\Tilde{v}_n (x)}^p \,\mathrm{d} y\,\mathrm{d} x\\
\end{split}
\]
By definition of the region \(D_R\) we have, if
\(\beta \in (\alpha, N)\),
\begin{multline*}
\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr) \abs{\Tilde{v}_n}^p\\
\le 2 \int_{D_R} \bigl( I_\alpha \ast
\abs{u_n}^p\bigr) \abs{u_n}^p +\int_{{\mathbb R}^N} \bigl((\chi_{{\mathbb R}^N \setminus
B_{2R}} I_\alpha) \ast \abs{u_n}^p\bigr) \abs{u_n}^p\\
\le 2 \int_{D_R} \bigl( I_\alpha \ast
\abs{u_n}^p\bigr) \abs{u_n}^p +\frac{C}{R^{\beta - \alpha}} \int_{{\mathbb R}^N}
\bigl((\chi_{{\mathbb R}^N \setminus B_{2R}}
I_{\beta}) \ast \abs{u_n}^p\bigr) \abs{u_n}^p
\end{multline*}
Since by assumption \(p > \frac{N + \alpha}{N}\), we can take \(\beta\) such that moreover \(\beta < (p - 1) N\), and then by the
Hardy--Littlewood--Sobolev inequality \eqref{eqHLS} and the classical Sobolev
inequality, we
obtain that
\begin{multline*}
\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr) \abs{\Tilde{v}_n}^p
\le C' \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2\Bigr)^\frac{p}{2} \Bigl(\int_{D_R} \abs{u_n}^\frac{2 N p}{N +
\alpha}\Bigr)^\frac{N + \alpha}{2 N}\\ +
\frac{C''}{R^{\beta - \alpha}} \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2\Bigr)^p,
\end{multline*}
from which we deduce that
\begin{equation}
\label{eqCrossIntegralVanishing}
\lim_{n \to \infty}
\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr)
\abs{\Tilde{v}_n}^p
= 0.
\end{equation}
For each \(n \in {\mathbb N}\), we fix \(t_n \in (0, \infty)\) so that
\(t_n v_n \in \mathcal{N}_0\) or, equivalently,
\begin{equation}
\label{eqTaun}
\begin{split}
t_n^{2 p - 2} &= \frac{\displaystyle \int_{{\mathbb R}^N} \abs{\nabla v_n}^2 +
\abs{v_n}^2}{\displaystyle \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr)
\abs{v_n}^p}\\
&= \frac{\displaystyle \int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2}{\displaystyle \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u_n}^p\bigr)
\abs{u_n}^p - 2 \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr)
\abs{\Tilde{v}_n}^p}.
\end{split}
\end{equation}
For every \(n \in {\mathbb N}\), we have
\[
\mathcal{A} (t_n u_n) = 2 \mathcal{A} (t_n v_n)
- \frac{t_n^{2 p}}{p} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v_n}^p\bigr)
\abs{\Tilde{v}_n}^p
\]
By \eqref{eqNorm}, \eqref{eqCrossIntegralVanishing} and \eqref{eqTaun}, in view of proposition~\ref{propositionOddNondegenerate}, we note that \(\lim_{n \to \infty} t_n = 1\) and thus in view
of \eqref{eqCrossIntegralVanishing} again we conclude that
\[
c_{\mathrm{odd}} = \lim_{n \to \infty} \mathcal{A} (u_n) = \lim_{n \to
\infty} \mathcal{A} (t_n u_n) = 2 \lim_{n \to \infty} \mathcal{A} (t_n
v_n) \ge 2 c_0,
\]
in contradiction with the assumption \(c_{\mathrm{odd}} < 2 c_0\) of the
proposition.
We can now fix \(R > 0\) such that \eqref{eqDR} holds. We take a function \(\eta \in
C^\infty ({\mathbb R}^N)\) such that \(\supp \eta \subset D_{3 R /2}\), \(\eta = 1\)
on \(D_R\), \(\eta \le 1\) on \({\mathbb R}^N\) and \(\nabla
\eta \in L^\infty ({\mathbb R}^N)\).
We have the inequality \citelist{\cite{Lions1984CC2}*{lemma I.1}\cite{Willem1996}*{lemma
1.21}\cite{MorVanSchaft13}*{lemma 2.3}\cite{VanSchaftingen2014}*{(2.4)}}
\[
\begin{split}
\int_{D_R} \abs{u_n}^{\frac{2 N p}{N + \alpha}} &\le \int_{{\mathbb R}^N} \abs{\eta
u_n}^{\frac{2 N p}{N + \alpha}}\\
&\le C \Bigl(\sup_{a \in {\mathbb R}^N} \int_{B_{R/2} (a)} \abs{\eta u_n}^{\frac{2
N p}{N + \alpha}} \Bigr)^{1 - \frac{N + \alpha}{N p}}
\int_{{\mathbb R}^N} \abs{\nabla (\eta u_n)}^2 + \abs{\eta u_n}^2\\
&\le C' \Bigl(\sup_{a \in {\mathbb R}^{N - 1} \times \{0\}} \int_{B_{2R} (a)}
\abs{u_n}^{\frac{2 N p}{N + \alpha}} \Bigr)^{1 - \frac{N + \alpha}{N p}}
\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 + \abs{u_n}^2.
\end{split}
\]
Since the sequence \((u_n)_{n \in {\mathbb N}}\) is bounded in the space \(H^1 ({\mathbb R}^N)\) we
deduce from \eqref{eqDR} that there exists
a sequence of points \((a_n)_{n \in {\mathbb N}}\) in the hyperplane \({\mathbb R}^{N - 1} \times \{0\}\) such that
\[
\liminf_{n \to \infty} \int_{B_{2 R} (a_n)} \abs{u_n}^\frac{2 N p}{N +
\alpha} > 0.
\]
Up to translations and a subsequence, we can assume that the sequence
\((u_n)_{n \in {\mathbb N}}\) converges weakly in \(H^1
({\mathbb R}^N)\) to a function \(u \in H^1
({\mathbb R}^N)\).
Since the action functional \(\mathcal{A}\) is invariant under odd reflections,
we note that for every \(n \in {\mathbb N}\), \(\mathcal{A} (u_n) = 0\) on
\(H^1_{\mathrm{odd}} ({\mathbb R}^N)^\perp\) by the symmetric criticality principle
\cite{Palais1979} (see also~\cite{Willem1996}*{theorem 1.28}). This allows to
deduce from the strong convergence of the sequence \((\mathcal{A}' (u_n))_{n
\in {\mathbb N}}\) to \(0\) in \(H^1_{\mathrm{odd}} ({\mathbb R}^N)'\) the strong
convergence to \(0\) of the sequence \((\mathcal{A}' (u_n))_{n \in {\mathbb N}}\) in
\(H^1 ({\mathbb R}^N)'\).
For any test function \(\varphi \in C^1_c ({\mathbb R}^N)\), by the weak
convergence of the sequence \((u_n)_{n \in {\mathbb N}}\), we first have
\[
\lim_{n \to \infty} \int_{{\mathbb R}^N} \scalprod{\nabla u_n}{\nabla \varphi} +
u_n \varphi = \int_{{\mathbb R}^N} \scalprod{\nabla u}{\nabla \varphi} + u \varphi.
\]
By the classical Rellich--Kondrashov compactness theorem, the sequence
\((\abs{u_n}^p)_{n \in {\mathbb N}}\) converges locally in measure to \(\abs{u}^p\) and by the
Sobolev inequality, this sequence is bounded in \(L^\frac{2 N}{N + \alpha}
({\mathbb R}^N)\). Therefore, the sequence \((\abs{u_n}^p)_{n \in {\mathbb N}}\) converges
weakly to \(\abs{u}^p\) in \(L^\frac{2 N}{N + \alpha} ({\mathbb R}^N)\) (see for
example
\citelist{\cite{Bogachev2007}*{proposition~4.7.12}\cite{Willem2013}*{
proposition 5.4.7}}), and, by the Hardy--Littlewood--Sobolev inequality \eqref{eqHLS}, the
sequence \((I_\alpha \ast \abs{u_n}^p)_{n \in {\mathbb N}}\) converges weakly in
\(L^{\frac{2 N}{N - \alpha}} ({\mathbb R}^N)\) to \(I_\alpha \ast \abs{u}^p\).
By the Rellich--Kondrashov theorem again, the sequence \(((I_\alpha \ast
\abs{u_n}^p)\abs{u_n}^{p - 2}
u_n)_{n \in {\mathbb N}}\) converges weakly in \(L^\frac{2 N}{N + 2} (K)\) for every
compact set \(K \subset {\mathbb R}^N\).
Therefore we have
\[
\lim_{n \to \infty} \int_{{\mathbb R}^N} (I_\alpha \ast \abs{u_n}^p)\abs{u_n}^{p -
2} u_n \varphi = \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr)\abs{u}^{p - 2} u
\varphi.
\]
We have thus proved that
\[
\mathcal{A}' (u) = 0 = \lim_{n \to \infty} \mathcal{A}' (u_n).
\]
Finally, we have
\[
\begin{split}
\lim_{n \to \infty} \mathcal{A} (u_n) &= \lim_{n \to \infty} \mathcal{A} (u_n)
- \frac{1}{2 p} \dualprod{\mathcal{A}' (u_n)}{u_n}\\
& =\lim_{n \to \infty} \Bigl(\frac{1}{2} - \frac{1}{2 p} \Bigr)\int_{{\mathbb R}^N}
\abs{\nabla u_n}^2 + \abs{u_n}^2\\
&\ge \Bigl(\frac{1}{2} - \frac{1}{2 p} \Bigr)\int_{{\mathbb R}^N} \abs{\nabla u}^2 +
\abs{u}^2\\
&=\mathcal{A} (u) - \frac{1}{2 p} \dualprod{\mathcal{A}' (u)}{u} = \mathcal{A} (u),
\end{split}
\]
from which we conclude that \(\mathcal{A} (u) = c_{\mathrm{odd}}\) and that
the sequence \((u_n)_{n \in {\mathbb N}}\) converges strongly to \(u\) in \(H^1
({\mathbb R}^N)\).
\end{proof}
\subsection{Strict inequality}
It remains now to establish the strict inequality \(c_\mathrm{odd} < 2 c_0\).
\begin{proposition}
\label{propositionStrictInequality}
If \(\frac{N - 2}{N + \alpha} < \frac{1}{p} < \frac{N}{N + \alpha}\), then
\[
c_{\mathrm{odd}} < 2 c_0.
\]
\end{proposition}
\begin{proof}
It is known that the Choquard equation has a least action solution
\cite{MorVanSchaft13}.
More precisely, there exists \(v \in H^1 ({\mathbb R}^N) \setminus \{0\}\) such that
\(\mathcal{A}' (v)= 0\)
and
\[
\mathcal{A}(v) = \inf_{\mathcal{N}_0} \mathcal{A}.
\]
We take a function \(\eta \in C^2_c ({\mathbb R}^N)\) such that \(\eta = 1\) on
\(B_1\), \(0 \le \eta \le 1\) on \({\mathbb R}^N\) and \(\supp \eta \subset B_2\) and
we define
for each \(R > 0\) the function \(\eta_R \in C^2_c ({\mathbb R}^N)\) for every
\(x \in {\mathbb R}^N\) by
\(\eta_R (x) = \eta (x/R)\).
We define now the function \(u_R : {\mathbb R}^N \to {\mathbb R}\) for each \(x = (x', x_N)
\in {\mathbb R}^N\) by
\[
u_{R} (x) =(\eta_R v) (x', x_N - 2R) - (\eta_R v) (x', - x_N - 2R).
\]
It is clear that \(u_R \in H^1_{\mathrm{odd}} ({\mathbb R}^N)\).
We observe that \(\dualprod{\mathcal{A}'(t_R u_R)}{t_R u_R}=0\) if and only if
\(t_R \in (0, \infty)\) satisfies
\[
{\displaystyle t_{R}^{2p-2}=\frac{\displaystyle \int_{{\mathbb R}^{N}}\abs{\nabla
u_{R}}^{2}+\abs{u_{R}}^{2}}{\displaystyle
\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{ u_ { R } } ^ { p }
\bigr)\abs{u_{R}}^{p}}}.
\]
Such a \(t_R\) always exists and
\[
\mathcal{A} (t_R u_R) =
\Bigl(\frac{1}{2} - \frac{1}{p}\Bigr) \frac{\displaystyle \Bigl(\int_{{\mathbb R}^N}
\abs{\nabla u_R}^2 + \abs{u_R}^2 \Bigr)^\frac{p}{p - 1}}
{\displaystyle \Bigl(\int_{{\mathbb R}^N} (I_\alpha \ast \abs{u_R}^p) \abs{u_R}^p
\Bigr)^\frac{1}{p - 1}}.
\]
The proposition will follow once we have established that for some \(R >0\)
\begin{equation}
\label{eqReformulatedStrict}
\frac{\displaystyle \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_R}^2 + \abs{u_R}^2
\Bigr)^\frac{p}{p - 1}}
{\displaystyle \Bigl(\int_{{\mathbb R}^N} (I_\alpha \ast \abs{u_R}^p) \abs{u_R}^p
\Bigr)^\frac{1}{p - 1}}
< 2 \frac{\displaystyle \Bigl(\int_{{\mathbb R}^N} \abs{\nabla v}^2 + \abs{v}^2
\Bigr)^\frac{p}{p - 1}}
{\displaystyle \Bigl(\int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v}^p
\Bigr)^\frac{1}{p - 1}}.
\end{equation}
We begin by estimating the denominator in the
left-hand side of \eqref{eqReformulatedStrict}. We
first observe that, by construction of the function \(u_R\)
\[
\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{u_{R}}^{p}\bigr)\abs{u_{R}}^{p}\ge
2\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{\eta_{R}v}^{p}\bigr)\abs{\eta_{R}v}^{p
}+ 2\frac{A_\alpha}{(4R)^{N - \alpha}} \Bigl(\int_{{\mathbb R}^N} \abs{\eta_R v
}^p\Bigr)^2.
\]
For the first term, we have
\[
\begin{split}
\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{\eta_{R}v}^{p}\bigr)\abs{\eta_{R}v}^{p}
&=\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{v}^{p}\bigr)\abs{v}^{p} -
2\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{v}^{p}\bigr)(1 -
\eta_R^p)\abs{v}^{p}\\
&\qquad \qquad + \int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast(1 -
\eta_R^p)\abs{v}^{p}\bigr)(1 - \eta_R^p)\abs{v}^{p}\\
&\ge \int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{v}^{p}\bigr)\abs{v}^{p} - 2
\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast \abs{v}^{p}\bigr)(1 - \eta_R^p)
\abs{v}^{p}.
\end{split}
\]
By the asymptotic properties of \(I_\alpha \ast
\abs{v}^p\) \cite{MorVanSchaft13}*{theorem 4}, we have
\[
\lim_{\abs{x} \to \infty} \frac{\bigl(I_{\alpha}\ast
\abs{v}^{p}\bigr)}{I_{\alpha} (x)} = \int_{{\mathbb R}^N} \abs{v}^p,
\]
so that
\[
2\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast \abs{v}^{p}\bigr)(1 - \eta_R^p)
\abs{v}^{p}
\le C \int_{{\mathbb R}^N \setminus B_R} \frac{\abs{v (x)}^p}{\abs{x}^{N -
\alpha}}\,\mathrm{d} x.
\]
We have thus
\begin{multline*}
\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{u_{R}}^{p}\bigr)\abs{u_{R}}^{p}\\
\ge 2 \int_{{\mathbb R}^N}
\bigl(I_{\alpha}\ast\abs{v}^{p}\bigr)\abs{v}^{p}
+ \frac{2 A_\alpha}{(4R)^{N - \alpha}} \Bigl(\int_{B_R} \abs{v }^p\Bigr)^2
- C \int_{{\mathbb R}^N \setminus B_R} \frac{\abs{v (x)}^p}{\abs{x}^{N -
\alpha}}\,\mathrm{d} x.
\end{multline*}
We now use the information that we have on the decay of the least action
solution \(v\) \cite{MorVanSchaft13}.
If \(p < 2\), then \(v (x) = O (\abs{x}^{-(N - \alpha)/(2 - p)})\) as \(\abs{x}
\to \infty\) and
\[
\int_{{\mathbb R}^N \setminus B_R} \frac{\abs{v (x)}^p}{\abs{x}^{N -
\alpha}}\,\mathrm{d} x = O \biggl(\frac{1}{R^\frac{Np
- 2 \alpha}{2 - p}} \biggr) = o \Bigl(\frac{1}{R^{N - \alpha}} \Bigr),
\]
since \(p > \frac{N + \alpha}{N} > \frac{2N}{2N - \alpha}\).
If \(p \ge 2\), then \(v\) decays exponentially at infinity. We have thus the
asymptotic lower bound
\begin{multline}
\label{eqDenominator}
\int_{{\mathbb R}^{N}}\bigl(I_{\alpha}\ast\abs{u_{R}}^{p}\bigr)\abs{u_{R}}^{p}\\ \ge
2 \int_{{\mathbb R}^N}
\bigl(I_{\alpha}\ast\abs{v}^{p}\bigr)\abs{v}^{p} + \frac{2
A_\alpha}{(4R)^{N - \alpha}} \Bigl(\int_{{\mathbb R}^N} \abs{v }^p\Bigr)^2 + o
\Bigl(\frac{1}{R^{N - \alpha}}\Bigr).
\end{multline}
For the numerator in \eqref{eqReformulatedStrict}, we compute by integration
by parts
\[
\begin{split}
\int_{{\mathbb R}^N} \abs{\nabla u_R}^2 + \abs{u_R}^2
& = 2 \int_{{\mathbb R}^N} \abs{\nabla (\eta_R v)}^2 + \abs{\eta_R v}^2 \\
& = 2 \int_{{\mathbb R}^N} \eta_R^2 (\abs{\nabla v}^2 + \abs{v}^2)
- 2\int_{{\mathbb R}^N} \eta_R (\Delta \eta_R) \abs{v}^2\\
& \le 2 \int_{{\mathbb R}^N} \bigl(\abs{\nabla v}^2 + \abs{v}^2\bigr) + \frac{C}{R^2}
\int_{B_{2R} \setminus B_R} \abs{v}^2.
\end{split}
\]
If \(p < 2\), we have by the decay of the solution \(v\)
\[
\frac{1}{R^2} \int_{B_{2R} \setminus B_R} \abs{v}^2
= O \biggl(\frac{1}{R^{{\frac{Np - 2\alpha}{2 - p} + 2}}} \biggr)
= o \Bigl(\frac{1}{R^{N - \alpha}} \Bigr).
\]
In the case where \(p \ge 2\), the solution \(v\) decays exponentially. We
conclude thus that
\begin{equation}
\label{eqNumerator}
\int_{{\mathbb R}^N} \abs{\nabla u_R}^2 + \abs{u_R}^2
=2 \int_{{\mathbb R}^N} \abs{\nabla v}^2 + \abs{v}^2 + o \Bigl(\frac{1}{R^{N -
\alpha}} \Bigr).
\end{equation}
We derive from the asymptotic bounds \eqref{eqDenominator} and
\eqref{eqNumerator}, an asymptotic bound on the quotient:
\begin{multline*}
\frac{\displaystyle \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_R}^2 + \abs{u_R}^2
\Bigr)^\frac{p}{p - 1}}
{\displaystyle \Bigl(\int_{{\mathbb R}^N} (I_\alpha \ast \abs{u_R}^p)
\abs{u_R}^p \Bigr)^\frac{1}{p - 1}}\\
\le 2 \frac{\displaystyle \Bigl(\int_{{\mathbb R}^N} \abs{\nabla v}^2 + \abs{v}^2
\Bigr)^\frac{p}{p - 1}}
{\displaystyle \Bigl(\int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v}^p
\Bigr)^\frac{1}{p - 1}}
\Biggl(1
- \frac{p A_\alpha\displaystyle \Bigl(\int_{B_R} \abs{v}^p\Bigr)^2}
{(p - 1)(4R)^{N - \alpha}\displaystyle
\int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v}^p}\\
+ o \Bigl(\frac{1}{R^{N - \alpha}}\Bigr)\Biggr).
\end{multline*}
The inequality \eqref{eqReformulatedStrict} holds thus when \(R\) is large
enough, and the conclusion follows.
\end{proof}
\subsection{Existence of a minimal action odd solution}
We have now developped all the tools to prove the existence of a least action
odd solution to the Choquard equation, corresponding to the existence part of theorem~\ref{theoremOdd}.
\begin{proposition}
\label{propositionExistence}
If \(\frac{N - 2}{N + \alpha} < \frac{1}{p}< \frac{N}{N + \alpha}\), then there exists solution \(u \in H^1_\mathrm{odd} ({\mathbb R}^N) \cap C^2 ({\mathbb R}^N)\) to the Choquard equation \eqref{eqChoquard} such that \(\mathcal{A} (u) = c_\mathrm{odd}\).
\end{proposition}
\begin{proof}
Let \((u_n)_{n \in {\mathbb N}}\) be the sequence given by
proposition~\ref{propositionOddPalaisSmaleExistence}. In view of
proposition~\ref{propositionStrictInequality},
proposition~\ref{propositionOddPalaisSmaleCondition} is applicable and gives
the required weak solution \(u \in H^1 ({\mathbb R}^N)\).
By the regularity theory for the Choquard equation \cite{MorVanSchaft13}*{proposition 4.1} (see also \cite{CingolaniClappSecchi2012}*{lemma A.10}), \(u \in C^2 ({\mathbb R}^N)\).
\end{proof}
\subsection{Sign and symmetry properties}
We complete the proof of theorem~\ref{theoremOdd} by showing that such solutions have a simple sign and symmetry structure.
\begin{proposition}
\label{propositionQualitative}
If \(\frac{N - 2}{N + \alpha} < \frac{1}{p} <
\frac{1}{2}\) and if \(u \in H^1_\mathrm{odd} ({\mathbb R}^N)\) is a solution to the Choquard equation \eqref{eqChoquard} such that \(\mathcal{A} (u) = c_\mathrm{odd}\),
then \(u\) has constant sign on \({\mathbb R}^N_+\) and \(u\) is axially symmetric with respect to an axis perpendicular to \(\partial {\mathbb R}^N_+\).
\end{proposition}
The proof takes profit of the structure of the problem to rewrite it as a groundstate
of a problem on the halfspace where quite fortunately the strategy for proving similar properties of groundstates of the Choquard equation still works \cite{MorVanSchaft13}*{Propositions 5.1 and 5.2} (see also \cite{MorVanSchaft15}*{propositions 5.2 and 5.3}.
\begin{proof}[Proof of proposition~\ref{propositionQualitative}]
We first rewrite the problem of finding odd solutions on \({\mathbb R}^N\) as a groundstate problem on \({\mathbb R}^N_+\) whose nonlocal term has a more intricate structure.
\begin{claim}
\label{claimRewrite}
For every \(v \in H^1_{\mathrm{odd}} ({\mathbb R}^N)\),
\[
\mathcal{A} (v) = \Tilde{\mathcal{A}} (v\vert_{{\mathbb R}^N_+}),
\]
where \({\mathbb R}^N_+ = {\mathbb R}^{N - 1} \times (0, \infty)\) and the functional \(\Tilde{\mathcal{A}} : H^1_0 ({\mathbb R}^N_+) \mapsto {\mathbb R}\) is defined for \(w \in H^1_0 ({\mathbb R}^N_+)\) by
\[
\Tilde{\mathcal{A}} (w) = \int_{{\mathbb R}^N_+} \abs{\nabla w}^2 + \abs{w}^2 -
\frac{1}{p} \int_{{\mathbb R}^N_+} \int_{{\mathbb R}^N_+} K (\abs{x' - y'},x_N, y_N)
\abs{u (x)}^p \abs{u (y)}^p \,\mathrm{d} x \,\mathrm{d} y,
\]
with \(x= (x', x_N)\), \(y = (y', y_N)\) and the kernel \(K : (0, \infty)^3 \to {\mathbb R}\)
defined for each \((r, s,t) \in (0, \infty)^3\) by
\[
K (r, s, t) = \frac{A_\alpha}{\bigl(r^2 + (s - t)^2\bigr)^\frac{N - \alpha}{2}}
+ \frac{A_\alpha}{\bigl(r^2 + (s + t)^2\bigr)^\frac{N - \alpha}{2}}.
\]
In particular, \(u \in \Tilde{\mathcal{N}}_{\mathrm{nod}}\), where
\[
\Tilde{\mathcal{N}}_{\mathrm{nod}}
=\bigl\{ w \in H^1_0 ({\mathbb R}^N_+) \;:\; \dualprod{\Tilde{\mathcal{A}}' (w)}{w} = 0\bigr\}
\]
and
\[
\Tilde{\mathcal{A}} (u) = \inf_{\Tilde{\mathcal{N}}_{\mathrm{nod}}} \Tilde{\mathcal{A}}.
\]
\end{claim}
\begin{proofclaim}
This follows from the fact that if \(u \in H^1_\mathrm{odd} ({\mathbb R}^N)\), then \(u \vert_{{\mathbb R}^N_+} \in H^1_0 ({\mathbb R}^N_+)\) and by direct computation of the integrals.
\end{proofclaim}
\begin{claim}
\label{claimPositive}
One has either \(u > 0\) almost everywhere on \({\mathbb R}^N_+\) or \(u < 0\) almost everywhere on \({\mathbb R}^N_+\).
\end{claim}
\begin{proofclaim}
Let \(w = u\vert_{{\mathbb R}^N_+}\). We observe that \(\abs{w} \in H^1_0 ({\mathbb R}^N_+)\),
\begin{align*}
\Tilde{\mathcal{A}} (\abs{w}) & = \Tilde{\mathcal{A}} (w) = c_{\mathrm{odd}} &
& \text{and} &
\dualprod{\Tilde{\mathcal{A}}' (\abs{w})}{\abs{w}}
= \dualprod{\Tilde{\mathcal{A}}' (w)}{w}.
\end{align*}
Therefore, if we define \(\Bar{u} \in H^1_{\mathrm{odd}} ({\mathbb R}^N)\) for almost every
\(x = (x', x_N) \in {\mathbb R}^N\) by
\[
\Bar{u} (x)=
\begin{cases}
\abs{w} (x', x_N) & \text{if \(x_N > 0\)},\\
-\abs{w} (x', x_N) & \text{if \(x_N < 0\)},
\end{cases}
\]
the function \(\Bar{u}\) is a weak solution to the Choquard equation \eqref{eqChoquard}.
This function \(\Bar{u}\) is thus of class \(C^2\) \cite{MorVanSchaft13}*{proposition 4.1} (see also \cite{CingolaniClappSecchi2012}*{lemma A.10}) and, in the classical sense, it satisfies
\[
-\Delta \Bar{u} + \Bar{u} \ge 0 \qquad \text{in \({\mathbb R}^N_+\)}.
\]
By the usual strong maximum principle for classical supersolutions, we conclude that \(\abs{u} = \Bar{u} > 0\) in \({\mathbb R}^N_+\).
Since the function \(u\) was also a solution to the Choquard equation \eqref{eqChoquard}, it is also a continuous function, and we have thus either \(u = \abs{u} > 0\) in \({\mathbb R}^N_+\) or \(u = - \abs{u} < 0\) in \({\mathbb R}^N_+\).
\end{proofclaim}
\begin{claim}
\label{claimSymmetry}
The solution \(u\) is axially symmetric with respect to an axis parallel to \(\{0\} \times {\mathbb R} \subset {\mathbb R}^N\).
\end{claim}
\begin{proofclaim}
Let \(H\) be a closed affine half-space perpendicular to \(\partial {\mathbb R}^{N}_+\) and let \(\sigma_H : {\mathbb R}^N \to {\mathbb R}^N\) be the reflection with respect to \(\partial H\). We recall that the polarization or two-point rearrangement with respect to \(H\) of \(w\) is the function \(w^H : {\mathbb R}^N \to {\mathbb R}\) defined for each \(x \in {\mathbb R}^N\) by \citelist{\cite{BrockSolynin2000}\cite{Baernstein1994}}
\[
w^H (x) =
\begin{cases}
\max \bigl(w (x), w (\sigma_H (x))\bigr) & \text{if \(x \in H\)},\\
\min \bigl(w (x), w (\sigma_H (x))\bigr) & \text{if \(x \in {\mathbb R}^N \setminus H\)}.
\end{cases}
\]
Since \(\partial H\) is perpendicular to \({\mathbb R}^{N - 1} \times \{0\}\), we have \(\sigma_H ({\mathbb R}^N_+) = {\mathbb R}^N_+\) so that
\(w^H \in H^1_0 ({\mathbb R}^N)\) and \cite{BrockSolynin2000}*{lemma 5.3}
\[
\int_{{\mathbb R}^N_+} \abs{\nabla w^H}^2 + \abs{w^H}^2
= \int_{{\mathbb R}^N_+} \abs{\nabla w}^2 + \abs{w}^2.
\]
Moreover, we also have
\begin{multline}
\label{eqPolarizationNonlocal}
\int_{{\mathbb R}^N_+} \int_{{\mathbb R}^N_+} K (\abs{x' - y'}, x_N, y_N) \abs{w^H (x)}^p \abs{w^H (y)}^p \,\mathrm{d} x \,\mathrm{d} y\\
= \frac{1}{2} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast (\abs{u}^p)^H\bigr) (\abs{u}^p)^H
\ge \frac{1}{2} \int_{{\mathbb R}^N_+} \bigl(I_\alpha \ast \abs{u}^p\bigr)\, \abs{u}^p\\
=\int_{{\mathbb R}^N_+} \int_{{\mathbb R}^N_+} K (\abs{x' - y'}, x_N, y_N)\, \abs{w (x)}^p\, \abs{w (y)}^p \,\mathrm{d} x \,\mathrm{d} y,
\end{multline}
with equality if and only if either \(\abs{u}^H = \abs{u}\) almost everywhere on \({\mathbb R}^N_+\) or
\(\abs{u}^H = \abs{u} \circ \sigma_H\) almost everywhere on \({\mathbb R}^N_+\) \cite{MorVanSchaft13}*{lemma 5.3} (see also \citelist{\cite{Baernstein1994}*{corollary 4}\cite{VanSchaftingenWillem2004}*{proposition 8}}), or equivalently,
since by claim~\ref{claimPositive} \(w\) has constant sign on \({\mathbb R}^N\), either \(w^H = w\) almost everywhere on \({\mathbb R}^N_+\) or \(w^H = w \circ \sigma_H\) almost everywhere on \({\mathbb R}^N_+\).
If the inequality \eqref{eqPolarizationNonlocal} was strict, then, since \(p > 1\) there would exist \(\tau \in (0, 1)\) such that \(\tau w^H \in \Tilde{\mathcal{N}}\) and we would have
\[
\mathcal{A} (\tau u^H) < \Tilde{\mathcal{A}} (w) = c_{\mathrm{odd}},
\]
in contradiction with claim~\ref{claimRewrite}.
We have thus proved that for every affine half-space \(H \subset {\mathbb R}^N\) whose boundary \(\partial H\) is perpendicular to \(\partial {\mathbb R}^{N}_+\), either
\(w^H = w\) almost everywhere on \({\mathbb R}^N_+\) or \(w^H = w \circ \sigma_H\) almost everywhere on \({\mathbb R}^N_+\). This implies that \(w\) is axially symmetric with respect to an axis perpendicular to \(\partial {\mathbb R}^N_+\) \citelist{\cite{MorVanSchaft13}*{lemma 5.3}\cite{VanSchaftingenWillem2008}*{proposition 3.15}}, which is equivalent to the claim.
\end{proofclaim}
The proposition follows directly from claims~\ref{claimPositive} and \ref{claimSymmetry}.
\end{proof}
\section{Minimal action nodal solution}
This section is devoted to the proof of theorem~\ref{theoremNod} on the existence of a least action nodal solution.
\subsection{Minimax principle}
We begin by observing that the counterpart of proposition~\ref{propositionOddNondegenerate} holds.
\begin{proposition}[Nondegeneracy of the level]
\label{propositionNodNondegenerate}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le \frac{N}{N + \alpha}\), then
\[
c_{\mathrm{nod}} > 0.
\]
\end{proposition}
\begin{proof}
In view of the inequality \(c_0 > 0\) \cite{MorVanSchaft13}, it suffices to
note that since \(\mathcal{N}_{\mathrm{nod}} \subset \mathcal{N}_0\) we have
\(c_{\mathrm{nod}} \ge c_0\).
\end{proof}
We first reformulate the minimization problem as a minimax problem.
\begin{proposition}[Minimax principle]
\label{propositionSimpleVariational}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} <
\frac{1}{2}\), then for every \(\varepsilon > 0\),
\[
c_{\mathrm{nod}} = \inf_{\gamma \in \Gamma} \sup_{{\mathbb B}^2} \mathcal{A}
\circ \gamma,
\]
where
\begin{multline*}
\Gamma = \Bigl\{ \gamma \in C \bigl({\mathbb B}^2; H^1_0 ({\mathbb R}^N)\bigr) : \xi
\bigl(\gamma(\partial {\mathbb B}^2)\bigr) \not \ni 0, \, \deg (\xi \circ
\gamma ) = 1\\ \text{ and } (\mathcal{A} \circ
\gamma)^\frac{p - 1}{p - 2} \le c_\mathrm{nod}^\frac{p - 1}{p - 2} +
\varepsilon - c_0^\frac{p - 1}{p - 2} \text{on \(\partial {\mathbb B}^2\)}\Bigr\},
\end{multline*}
where the map \(\xi = (\xi_+, \xi_-) \in C \bigl(H^1 ({\mathbb R}^N); {\mathbb R}^2\bigr)\)
is defined for each \(u \in H^1 ({\mathbb R}^N)\) by
\[
\xi_{\pm} (u) =
\left\{
\begin{aligned}&\frac{\displaystyle\int_{{\mathbb R}^N} (I_\alpha \ast
\abs{u}^p) \abs{u_\pm}^p }{\displaystyle\int_{{\mathbb R}^N} \abs{\nabla u_\pm}^2 +
\abs{u_\pm}^2}
- 1 & & \text{if \(u_\pm \ne 0\)},\\
& - 1 & & \text{if \(u_\pm = 0\)}.
\end{aligned}
\right.
\]
Moreover, for every \(\gamma \in \Gamma\),
\(\mathcal{N}_{\mathrm{nod}} \cap \gamma ({\mathbb B}^2) \ne 0\).
\end{proposition}
In this statement \({\mathbb B}^2\) denotes the closed unit disc in the plane
\({\mathbb R}^2\) and \(\deg\) is the classical topological degree of Brouwer,
or equivalently, the winding number (see for example
\citelist{\cite{Schechter2004}*{chapter 6}\cite{MawhinWillem1989}*{\S 5.3}}).
The continuity of the map \(\xi\) on the subset of constant-sign functions in
\(H^1 ({\mathbb R}^N)\) follows from the Hardy--Littlewood--Sobolev inequality
\eqref{eqHLS} and
the classical Sobolev inequality, and requires the assumption \(p > 2\).
The map \(\xi\) is the nonlocal counterpart of a map appearing in the variational characterization of least action nodal solutions by Cerami, Solimini and Struwe for local
Schr\"odinger type problems \cite{CeramiSoliminiStruwe1986}, which is done in the framework of critical point theory in ordered spaces whereas our minimax principle works in the more classical
framework of Banach spaces.
\begin{proof}[Proof of proposition~\ref{propositionSimpleVariational}]
We denote the right-hand side in the equality to be proven as \(\Tilde{c}\) and
we first prove that that
\( \Tilde{c} \ge c_\mathrm{nod}\).
Let \(\gamma \in \Gamma\). Since \(\deg (\xi \circ
\gamma ) = 1\), by the existence property of the
degree, there exists \(t^* \in {\mathbb B}^2\)
such that \((\xi \circ \gamma) (t^*) = 0\).
It follows then that \(\gamma (t_*) \in \mathcal{N}_{\mathrm{nod}} = \xi^{-1}
(0)\) and thus
\[
\sup_{{\mathbb B}^2} \mathcal{A} \circ \gamma \ge \gamma (t_*) \ge c_{\mathrm{nod}},
\]
so that \(\Tilde{c} \ge c_{\mathrm{nod}}\).
We now prove that \(\Tilde{c} \le c_{\mathrm{nod}}\).
For a given \(u \in \mathcal{N}_{\mathrm{nod}}\), we define the map
\(\Tilde{\gamma} : [0, \infty)^2 \to H^1 ({\mathbb R}^N)\) for every \((t_+, t_-)\in
[0, \infty)^2\)
by
\[
\Tilde{\gamma} (t_+, t_-) = t_+^\frac{1}{p} u^+ +
t_-^\frac{1}{p} u^-.
\]
We compute for each \((t_+, t_-) \in [0, \infty)^2\)
\begin{multline}
\label{eqExplicittptm}
\mathcal{A}\bigl( \Tilde{\gamma} (t_+, t_-)\bigr)
=\frac{t_+^{2/p}}{2} \int_{{\mathbb R}^N} \abs{\nabla u^+}^2 + \abs{ u^+}^2
+ \frac{t_-^{2/p}}{2} \int_{{\mathbb R}^N} \abs{\nabla u^-}^2+\abs{ u^-}^2\\
- \frac{1}{2 p} \int_{{\mathbb R}^N} \bigabs{I_{\alpha/2} \ast (t_+ \abs{u^+}^p + t_-
\abs{u^-}^p)}^2.
\end{multline}
The function \(\mathcal{A} \circ \Tilde{\gamma}\) is thus strictly concave and
\((\mathcal{A} \circ \Tilde{\gamma})' (1, 1) = 0\). Hence, \((1, 1)\) is the
unique
maximum point of the function \(\mathcal{A} \circ \Tilde{\gamma}\).
We also have in particular
\[
\mathcal{A} \bigl(\Tilde{\gamma} (t_+, 0)\bigr)
= \frac{t_+^{2/p}}{2} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr)
\abs{u^+}^p - \frac{t_+^2}{2 p}\int_{{\mathbb R}^N} \bigl(I_\alpha \ast
\abs{u^+}^p\bigr)
\abs{u^+}^p,
\]
and therefore
\begin{equation}
\label{eqieu}
\mathcal{A} \bigl(\Tilde{\gamma} (t_+, 0)\bigr) \le \Bigl(\frac{1}{2} -
\frac{1}{2 p}\Bigr)
\frac{\Bigl(\displaystyle\int_{{\mathbb R}^n} \bigl(I_\alpha \ast \abs{u}^p\bigr)
\abs{u^+}^p\Bigr)^\frac{p}{p - 1}}{\Bigl(\displaystyle \int_{{\mathbb R}^n}
\bigl(I_\alpha
\ast
\abs{u^+}^p\bigr) \abs{u^+}^p\Bigr)^\frac{1}{p - 1}}.
\end{equation}
By the semigroup property of the Riesz potential \(I_\alpha = I_{\alpha/2}\ast
I_{\alpha/2}\) (see for example \cite{LiebLoss2001}*{theorem 5.9}) and by the
Cauchy--Schwarz inequality,
\begin{multline}
\label{eqiea}
\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr) \abs{u^+}^p =
\int_{{\mathbb R}^N}
(I_{\alpha/2} \ast \abs{u^+}^p)(I_{\alpha/2} \ast \abs{u}^p) \\
\le \Bigl(\int_{{\mathbb R}^N} \bigabs{I_{\alpha/2} \ast
\abs{u}^p}^2\Bigr)^\frac{1}{2} \Bigl(\int_{{\mathbb R}^N} \bigabs{I_{\alpha/2} \ast
\abs{u^+}^p}^2\Bigr)^\frac{1}{2}\\
= \Bigl(\int_{{\mathbb R}^N}\bigl(I_\alpha \ast \abs{u}^p\bigr) \abs{u}^p
)\Bigr)^\frac{1}{2}\Bigl(\int_{{\mathbb R}^N} (I_\alpha \ast
\abs{u^+}^p) \abs{u^+}^p )\Bigr)^\frac{1}{2},
\end{multline}
and therefore by \eqref{eqieu} and \eqref{eqiea}
\[
\mathcal{A} \bigl(\Tilde{\gamma} (t_+, 0)\bigr) \le \Bigl(\frac{1}{2} -
\frac{1}{2 p}\Bigr)
\Bigl(\int_{{\mathbb R}^n} \bigl(I_\alpha \ast \abs{u}^p\bigr)
\abs{u^+}^p\Bigr)^\frac{p - 2}{p - 1}
\Bigl(\int_{{\mathbb R}^n} \bigl(I_\alpha \ast \abs{u}^p\bigr)
\abs{u}^p\Bigr)^\frac{1}{p - 1}
\]
We deduce therefrom that for every \((t_+, t_-) \in [0, \infty)^2\),
\begin{equation}
\label{ineqSplit}
\mathcal{A} \bigl(t_+^{1/p} u^+ \bigr)^\frac{p - 1}{p - 2}
+ \mathcal{A} \bigl( t_-^{1/p} u^-\bigr)^\frac{p - 1}{p - 2}
\le \mathcal{A} (u)^\frac{p - 1}{p - 2}.
\end{equation}
Since \(u_\pm \ne 0\), we have
\[
\sup_{t_\pm \in [0, \infty)} \mathcal{A} \bigl(t_\pm^{1/p} u^\pm\bigr) \ge c_0,
\]
we conclude that
\[
\sup_{t \in \partial ([0, \infty)^2)} (\mathcal{A} \circ
\Tilde{\gamma})^\frac{p - 1}{p - 2}
\le \mathcal{A} (u)^\frac{p - 1}{p - 2} - c_0^\frac{p - 1}{p - 2}.
\]
Moreover, we have by \eqref{eqExplicittptm}
\[
\lim_{\abs{t} \to \infty} \mathcal{A} \bigl(\Tilde{\gamma}(t)\bigr) = -
\infty.
\]
It remains to compute the degree of the map \(\xi \circ \Tilde{\gamma}\) on a
suitable set homeomorphic to \({\mathbb B}^2\). We compute for each \((t_+, t_-) \in
[0, \infty)^2\),
since \(u \in \mathcal{N}_\mathrm{nod}\)
\begin{multline*}
\bigscalprod{(t_+, t_-)}{\xi (\Tilde{\gamma} (t_+, t_-))}\\
= t_+^{3 - \frac{2}{p}} \frac{\displaystyle \int_{{\mathbb R}^N} (I_\alpha \ast
\abs{u^+}^p) \abs{u^+}^p}{\displaystyle \int_{{\mathbb R}^N} (I_\alpha \ast
\abs{u}^p) \abs{u^+}^p}
+
t_-^{3 - \frac{2}{p}} \frac{\displaystyle \int_{{\mathbb R}^N} (I_\alpha \ast
\abs{u^-}^p) \abs{u^-}^p}{\displaystyle \int_{{\mathbb R}^N} (I_\alpha \ast
\abs{u}^p) \abs{u^-}^p} - t_+ - t_-
\end{multline*}
Since \(p > 2\), we can now take \(R > \sqrt{2}\) large enough so that if \(t
\in [0, \infty)^2 \cap \partial B_R\), then
\begin{align*}
\bigscalprod{t}{\xi (\Tilde{\gamma} (t))} & > 0 &
&\text{ and } &
\bigl(\mathcal{A} \circ \Tilde{\gamma} (t)\bigr)^\frac{p - 1}{p - 2}
\le \mathcal{A} (u)^\frac{p - 1}{p - 2}
- c_0^\frac{p - 1}{p - 2}.
\end{align*}
We now define the homotopy \(H : [0, 1] \times [0, \infty)^2\) for each
\((\tau, t_+, t_-) \in [0, 1] \times [0, \infty)^2\) by
\[
H (\tau, t)
= \tau (\xi \circ \Tilde{\gamma})(t) + (1 - \tau) (t_+ - 1, t_- - 1).
\]
By the choice of \(R\), for every \((\tau, t) \in [0, 1] \times \partial ([0,
\infty)^2 \cap B_R)\), \(H (\tau, t) \ne 0\), and thus by the homotopy
invariance property of the degree, \(\deg (\xi \circ \Tilde{\gamma}
\vert_{(0, \infty)^2 \cap B_R}) = 1\).
If we set \(\gamma = \Tilde{\gamma} \circ \Phi\), where \(\Phi : {\mathbb B}^2
\to [0, \infty) \cap \Bar{B}_R\) is an orientation preserving homeomorphism,
we have \(\gamma \in \Gamma\) and \(\sup_{{\mathbb B}^2} \mathcal{A} \circ \gamma =
\mathcal{A} (u)\).
We have thus proved that if \(u \in \mathcal{N}_\mathrm{nod}\) and if
\(\mathcal{A} (u)^\frac{p - 1}{p - 2} \le c_\mathrm{nod}^\frac{p - 1}{p - 2}
+ \varepsilon\), then
\[
\mathcal{A} (u) \ge \Tilde{c},
\]
from which we deduce that \(c_{\mathrm{nod}} \ge \Tilde{c}\).
\end{proof}
We would like to point out that the inequality \eqref{ineqSplit} in the proof
of proposition~\ref{propositionSimpleVariational} gives a \emph{lower bound} on
the level \(c_\mathrm{nod}\) that refines the degeneracy given for
\(p > 2\) by theorem~\ref{theoremDegeneracy}.
\begin{corollary}
\label{corollaryLowerBound}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le
\frac{1}{2}\), then
\[
c_{\mathrm{nod}} \ge 2^{\frac{p - 2}{p - 1}} c_0.
\]
\end{corollary}
In particular, corollary \eqref{corollaryLowerBound} allows
theorem~\ref{theoremDegeneracy} to hold when \(p = 2\).
\begin{proposition}[Existence of a Palais-Smale sequence]
\label{propositionNodalPalaisSmaleExistence}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} < \frac{1}{2}\), then
there exists a sequence \((u_n)_{n \in {\mathbb N}}\) in \(H^1 ({\mathbb R}^N)\)
such that
\begin{align*}
\mathcal{A} (u_n) & \to c_{\mathrm{nod}}, &
& \dist (u_n, \mathcal{N}_{\mathrm{nod}}) \to 0 &
&\text{ and} &
\mathcal{A}' (u_n) & \to 0 \quad \text{in \(H^1 ({\mathbb R}^N)'\)},
\end{align*}
as \(n \to \infty\).
\end{proposition}
\begin{proof}
We take \(\Gamma\) given by proposition~\ref{propositionSimpleVariational} with
\(\varepsilon < c_0^\frac{p - 1}{p - 2}\).
The location theorem \cite{Willem1996}*{theorem 2.20} (see also
\cite{BrezisNirenberg1991}*{theorem 2}\cite{Schechter2009}*{theorem 2.12}) is
applicable and gives the conclusion.
\end{proof}
\subsection{Convergence of the Palais--Smale sequence}
We prove that Palais--Smale sequences at the level \(c_\mathrm{nod}\) and
localized near the Nehari nodal set \(\mathcal{N}_{\mathrm{nod}}\) converge
strongly up to a subsequence and up to translations.
\begin{proposition}
\label{propositionNodalPalaisSmaleCondition}
Let \((u_n)_{n \in {\mathbb N}}\) be a sequence in \(H^1_{\mathrm{odd}} ({\mathbb R}^N)\) such
that, as \(n \to \infty\),
\begin{align*}
\mathcal{A} (u_n) & \to c_{\mathrm{nod}}, &
& \dist (u_n, \mathcal{N}_{\mathrm{nod}}) \to 0 &
&\text{ and} &
\mathcal{A}' (u_n) & \to 0 \quad \text{in \(H^1 ({\mathbb R}^N)'\)}.
\end{align*}
If \(\frac{N - 2}{N + \alpha} < \frac{1}{p} < \frac{1}{2}\) and if
\[
c_{\mathrm{nod}} < 2 c_0,
\]
then there exists a sequence of points \((a_n)_{n \in {\mathbb N}}\) in \({\mathbb R}^N\) such
that the subsequence
\(\bigl(u_{n_k} (\cdot - a_{n_k})\bigr)_{k \in {\mathbb N}}\) converges strongly in
\(H^1
({\mathbb R}^N)\) to \(u \in H^1 ({\mathbb R}^N)\).
Moreover
\begin{align*}
\mathcal{A} (u) &= c_{\mathrm{nod}}, &
u &\in \mathcal{N}_{\mathrm{nod}}&
&\text{ and }&
\mathcal{A}'(u) &= 0 \quad \text{in \(H^1 ({\mathbb R}^N)'\)}.
\end{align*}
\end{proposition}
Palais--Smale conditions have been already proved by concentration--com\-pact\-ness arguments for local semilinear elliptic problems \citelist{\cite{FurtadoMaiaMedeiros2008}\cite{CeramiSoliminiStruwe1986}}.
\begin{proof}[Proof of proposition~\ref{propositionNodalPalaisSmaleCondition}]
We shall proceed through several claims on the sequence \((u_n)_{n \in {\mathbb N}}\).
\setcounter{claim}{0}
\begin{claim}
The sequence \((u_n)_{n \in {\mathbb N}}\) is bounded in the space \(H^1 ({\mathbb R}^N)\).
\end{claim}
\begin{proofclaim}
We write, as \(n \to \infty\),
\[
\begin{split}
\Bigl(\frac{1}{2} - \frac{1}{2 p}\Bigr)\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2 &= \mathcal{A} (u_n) - \frac{1}{2 p} \dualprod{\mathcal{A}'
(u_n)}{u_n}\\
&= \mathcal{A} (u_n) + o \biggl( \Bigl(\int_{{\mathbb R}^N} \abs{\nabla u_n}^2 +
\abs{u_n}^2 \Bigr)^\frac{1}{2}\biggr).
\end{split}
\]
from which the claim follows.
\end{proofclaim}
We now show that neither positive nor the negative parts of the sequence \((u_n)_{n \in
{\mathbb N}}\) tend to \(0\).
\begin{claim}
\label{claimUniformContinuity}%
The functional \(u \in H^1 ({\mathbb R}^N) \mapsto \int_{{\mathbb R}^N}
(I_\alpha \ast \abs{u}^p) \abs{u^\pm}^p\) is uniformly continuous on bounded
subsets of the space \(H^1 ({\mathbb R}^N)\).
\end{claim}
We bring to the attention of the reader that the related
map \(u \in H^1 ({\mathbb R}^N) \mapsto \int_{{\mathbb R}^N} \abs{\nabla u^\pm}^2 +
\abs{u^\pm}^2\) \emph{is not uniformly continuous} on bounded sets.
\begin{proofclaim}
For every \(u, v \in H^1 ({\mathbb R}^N)\), we have
\begin{multline*}
\int_{{\mathbb R}^N}
(I_\alpha \ast \abs{u}^p) \abs{u^\pm}^p
- \int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v^\pm}^p\\
= \frac{1}{2} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast (\abs{u}^p + \abs{v}^p)\bigr)\bigl(\abs{u^\pm}^p - \abs{v^\pm}^p\bigr)\\
+ \frac{1}{2}\int_{{\mathbb R}^N} \bigl(I_\alpha \ast (\abs{u}^p - \abs{v}^p)\bigr)\bigl(\abs{u^\pm}^p + \abs{v^\pm}^p\bigr).
\end{multline*}
By the classical Hardy--Littlewood--Sobolev inequality \eqref{eqHLS}, we obtain
\begin{multline*}
\Bigabs{\int_{{\mathbb R}^N}
(I_\alpha \ast \abs{u}^p) \abs{u^\pm}^p
- \int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v^\pm}^p}\\
\le C \Bigl(\int_{{\mathbb R}^N} \bigl(\abs{u}^p + \abs{v}^p\bigr)^\frac{2 N}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2N}}
\Bigl(\int_{{\mathbb R}^N} \bigabs{\abs{u^\pm}^p - \abs{v^\pm}^p}^\frac{2 N}{N + \alpha} \Bigr)^\frac{N + \alpha}{2 N}\\
+ C \Bigl(\int_{{\mathbb R}^N} \bigl(\abs{u^\pm}^p + \abs{v^\pm}^p\bigr)^\frac{2 N}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2N}}
\Bigl(\int_{{\mathbb R}^N} \bigabs{\abs{u}^p - \abs{v}^p}^\frac{2 N}{N + \alpha} \Bigr)^\frac{N + \alpha}{2 N}.
\end{multline*}
Since for every \(s, t \in {\mathbb R}\), one has \(\bigabs{\abs{s^\pm}^p - \abs{t^\pm}^p}
\le \abs{\abs{s}^p - \abs{t}^p}\) and \(\abs{s^\pm}^p + \abs{t^\pm}^p \le \abs{s}^p + \abs{t}^p\), the latter estimate can be simplified to
\begin{multline*}
\Bigabs{\int_{{\mathbb R}^N}
(I_\alpha \ast \abs{u}^p) \abs{u^\pm}^p
- \int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v^\pm}^p}\\
\le 2C \Bigl(\int_{{\mathbb R}^N} \bigl(\abs{u}^p + \abs{v}^p\bigr)^\frac{2 N}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2N}}
\Bigl(\int_{{\mathbb R}^N} \bigabs{\abs{u}^p - \abs{v}^p}^\frac{2 N}{N + \alpha} \Bigr)^\frac{N + \alpha}{2 N}.
\end{multline*}
Next, for each \(s, t \in {\mathbb R}\), we have, since \(p \ge 2\),
\[
\begin{split}
\bigabs{\abs{s}^p - \abs{t}^p} & \le p \abs{s - t} \int_0^1 p \abs{\tau s + (1 - \tau)t}^{p - 1} \,\mathrm{d} \tau\\
& \le p \abs{s - t} \int_0^1 \tau \abs{s}^{p - 1} + (1 - \tau) \abs{t}^{p - 1}\,\mathrm{d} \tau \\
& = \frac{p}{2} \abs{s - t} \bigl(\abs{s}^{p - 1} + \abs{t}^{p - 1} \bigr).
\end{split}
\]
Hence, we have,
\begin{multline*}
\Bigabs{\int_{{\mathbb R}^N}
(I_\alpha \ast \abs{u}^p) \abs{u^\pm}^p
- \int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v^\pm}^p}\\
\le 2 C \Bigl(\frac{p}{2} \Bigr)^\frac{N + \alpha}{2 N}
\Bigl(\int_{{\mathbb R}^N} \bigl(\abs{u}^p + \abs{v}^p\bigr)^\frac{2 N}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2N}}\\
\times
\Bigl(\int_{{\mathbb R}^N} \bigl(\abs{u}^{p - 1} + \abs{v}^{p - 1}\bigr)^\frac{2N}{N + \alpha} \abs{u - v}^\frac{2 N}{N + \alpha} \Bigr)^\frac{N + \alpha}{2 N}.
\end{multline*}
Therefore, by the H\"older inequality,
\begin{multline*}
\Bigabs{\int_{{\mathbb R}^N}
(I_\alpha \ast \abs{u}^p) \abs{u^\pm}^p
- \int_{{\mathbb R}^N} (I_\alpha \ast \abs{v}^p) \abs{v^\pm}^p}\\
\le C'
\Bigl(\int_{{\mathbb R}^N} \abs{u}^\frac{2 N p}{N + \alpha} + \abs{v}^\frac{2 N p}{N + \alpha}\Bigr)^{\frac{N + \alpha}{N}(1 - \frac{1}{2 p})}
\Bigl(\int_{{\mathbb R}^N} \abs{u - v}^\frac{2 Np}{N + \alpha} \Bigr)^\frac{N + \alpha}{2 N p}.
\end{multline*}
This shows that the map is uniformly continuous on \(L^\frac{2 N p}{N + \alpha} ({\mathbb R}^N)\).
Since by assumption, \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le \frac{N}{N + \alpha}\), in view of the classical Sobolev embedding theorem, the embedding \(H^1 ({\mathbb R}^N) \subset L^\frac{2 N p}{N + \alpha} ({\mathbb R}^N)\) is uniformly continuous, and the claim follows.
\end{proofclaim}
\begin{claim}
\label{claimLowerBound}
We have
\[
\liminf_{n \to \infty} \int_{{\mathbb R}^N} \abs{\nabla u_n^\pm}^2 + \abs{u_n^\pm}^2
= \liminf_{n \to \infty} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u_n}^p\bigr)
\abs{u_n^\pm}^p > 0.
\]
\end{claim}
\begin{proofclaim}
First we observe that if \(v \in \mathcal{N}_{\mathrm{nod}}\), then by the
Hardy--Littewood--Sobolev inequality \eqref{eqHLS}, the Sobolev inequality and the definition
of the nodal Nehari set \(\mathcal{N}_{\mathrm{nod}}\), we have
\[
\begin{split}
\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v}^p\bigr)
\abs{v^\pm}^p
&\le C \Bigl(\int_{{\mathbb R}^N} \abs{v}^\frac{2 N p}{N + \alpha} \Bigr)^\frac{N +
\alpha}{2 N}\Bigl(\int_{{\mathbb R}^N} \abs{v^\pm}^\frac{2 N p}{N + \alpha}
\Bigr)^\frac{N +
\alpha}{2 N}\\
&\le C' \Bigl(\int_{{\mathbb R}^N} \abs{\nabla v}^2 + \abs{v}^2
\Bigr)^\frac{p}{2}
\Bigl(\int_{{\mathbb R}^N} \abs{\nabla v^\pm}^2 + \abs{v^\pm}^2
\Bigr)^\frac{p}{2}\\
&= C' \Bigl(\int_{{\mathbb R}^N} \abs{\nabla v}^2 + \abs{v}^2
\Bigr)^\frac{p}{2}
\Bigl(\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v}^p\bigr)
\abs{v^\pm}^p\Bigr)^\frac{p}{2}.
\end{split}
\]
Since \(v^\pm \ne 0\), we deduce that
\[
\inf_{v \in \mathcal{N}_\mathrm{nod}} \Bigl(\int_{{\mathbb R}^N} \abs{\nabla v}^2 +
\abs{v}^2
\Bigr)^\frac{p}{2}
\Bigl(\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{v}^p\bigr)
\abs{v^\pm}^p\Bigr)^\frac{p-2}{2} > 0.
\]
Since the sequence \((u_n)_{n \in {\mathbb N}}\) is bounded in the space \(H^1
({\mathbb R}^N)\) and since \(\lim_{n \to \infty} \dist (u_n,
\mathcal{N}_\mathrm{odd}) = 0\), we deduce from the lower bound above and from
the uniform continuity property of claim~\ref{claimUniformContinuity} that
\[
\liminf_{n \to \infty} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u_n}^p\bigr)
\abs{u_n^\pm}^p > 0.
\]
Since \(\lim_{n \to \infty} \dualprod{\mathcal{A}' (u_n)}{u_n^\pm} = 0\), the
conclusion follows.
\end{proofclaim}
\begin{claim}
There exists \(R > 0\) such that
\[
\limsup_{n \to \infty} \sup_{a \in {\mathbb R}^N} \int_{B_R (a)} \abs{u_n^+}^\frac{2
N p}{N + \alpha} \int_{B_R (a)} \abs{u_n^-}^\frac{2 N
p}{N + \alpha} > 0.
\]
\end{claim}
\begin{proofclaim}
We assume by contradiction that for every \(R > 0\),
\[
\lim_{n \to \infty} \sup_{a \in {\mathbb R}^N} \int_{B_R (a)} \abs{u_n^+}^\frac{2
N p}{N + \alpha} \int_{B_R (a)} \abs{u_n^-}^\frac{2 N
p}{N + \alpha} = 0.
\]
Then by lemma~\ref{lemmaNewEstimate} below, since the sequences \((u_n^+)_{n
\in
{\mathbb N}}\) and \((u_n^-)_{n \in {\mathbb N}}\) are bounded in the space \(H^1 ({\mathbb R}^N)\), we have
\[
\lim_{n \to \infty} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u_n^+}^p\bigr)
\abs{u_n^-}^p = 0.
\]
We now take \(t_{n, \pm} \in (0, \infty)\) such that \(t_{n, \pm} u_n^\pm \in
\mathcal{N}_0\).
Since
\[
\begin{split}
\displaystyle \int_{{\mathbb R}^N}
\abs{\nabla u_n^\pm}^2 + \abs{u_n^\pm}^2
&= \int_{{\mathbb R}^N} \bigl(I_\alpha
\ast \abs{u_n}^p\bigr) \abs{u_n^\pm}^p + \dualprod{\mathcal{A}' (u_n)}{u_n^\pm}\\
&= \int_{{\mathbb R}^N} \bigl(I_\alpha
\ast \abs{u_n}^p\bigr) \abs{u_n^\pm}^p + o (1)
\end{split}
\]
and
\[
\begin{split}
\int_{{\mathbb R}^N} \bigl(I_\alpha
\ast \abs{u_n^\pm}^p\bigr) \abs{u_n^\pm}^p
&= \int_{{\mathbb R}^N} \bigl(I_\alpha
\ast \abs{u_n}^p\bigr) \abs{u_n^\pm}^p - \int_{{\mathbb R}^N} \bigl(I_\alpha
\ast \abs{u_n^+}^p\bigr) \abs{u_n^-}^p\\
&= \int_{{\mathbb R}^N} \bigl(I_\alpha
\ast \abs{u_n}^p\bigr) \abs{u_n^\pm}^p + o (1),
\end{split}
\]
it follows, in view of claim~\ref{claimLowerBound}, that \(\lim_{n \to \infty}
t_{n, \pm} = 1\).
We compute
\begin{multline*}
\mathcal{A} (t_{n, +} u_n^+ + t_{n, -} u_n^-)\\
= \mathcal{A} (t_{n, +} u_n^+) + \mathcal{A} (t_{n, -} u_n^-)
-\frac{t_{n,+}^p t_{n, -}^p}{p} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast
\abs{u_n^+}^p\bigr) \abs{u_n^-}^p.
\end{multline*}
and we deduce that
\[
\begin{split}
c_\mathrm{nod} &= \lim_{n \to \infty} \mathcal{A} (t_{n, +} u_n^+ + t_{n, -}
u_n^-)\\
&\ge \liminf_{n \to \infty} \mathcal{A} (t_{n, +} u_n^+) + \liminf_{n
\to \infty} \mathcal{A} (t_{n, -} u_n^-) \ge 2 c_0,
\end{split}
\]
in contradiction with the assumption of the proposition.
\end{proofclaim}
We now conclude the proof of the proposition. Up to a translation and a subsequence, we can assume
that
\[
\liminf_{n \to \infty} \int_{B_R (a)} \abs{u_n^\pm}^\frac{2
N p}{N + \alpha} \ge 0
\]
and that the sequence \((u_n)_{n \in {\mathbb N}}\) converges weakly to some \(u \in H^1
({\mathbb R}^N)\).
As in the proof of proposition~\ref{propositionOddPalaisSmaleCondition}, by the
weak convergence and by the classical Rellich--Kondrashov compactness
theorem, \(\mathcal{A}' (u) = 0\) and \(u^\pm \ne 0\), whence \(u \in
\mathcal{N}_\mathrm{nod}\).
We also have by lower semicontinuity of the Sobolev norm under weak convergence
\[
\begin{split}
\lim_{n \to \infty} \mathcal{A} (u_n) &= \lim_{n \to \infty} \mathcal{A} (u_n)
- \frac{1}{2 p} \dualprod{\mathcal{A}' (u_n)}{u_n}\\
& =\lim_{n \to \infty} \Bigl(\frac{1}{2} - \frac{1}{2 p} \Bigr)\int_{{\mathbb R}^N}
\abs{\nabla u_n}^2 + \abs{u_n}^2\\
&\ge \Bigl(\frac{1}{2} - \frac{1}{2 p} \Bigr)\int_{{\mathbb R}^N} \abs{\nabla u}^2 +
\abs{u}^2\\
&=\mathcal{A} (u) - \frac{1}{2 p} \dualprod{\mathcal{A}' (u)}{u} = \mathcal{A}(u),
\end{split}
\]
from which we deduce that \(\mathcal{A} (u) = c_\mathrm{nod}\) and the strong
convergence of the sequence \((u_n)_{n \in {\mathbb N}}\) in the space \(H^1 ({\mathbb R}^N)\).
\end{proof}
\begin{lemma}
\label{lemmaNewEstimate}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} < \frac{N}{N + \alpha}\), then
for every \(\beta \in \bigl(\alpha, \min (1, p - 1) N\bigr)\), there exists \(C
> 0\) such
that for every \(u, v \in H^1 ({\mathbb R}^N)\),
\begin{multline*}
\int_{{\mathbb R}^N} \bigl(I_\alpha \abs{u}^p\bigr) \abs{v}^p
\le C \Bigl(\int_{{\mathbb R}^N} \abs{\nabla
u}^2 + \abs{u}^2 \int_{{\mathbb R}^N} \abs{\nabla
v}^2 + \abs{v}^2 \Bigr)^\frac{1}{2}\\
\shoveright{\Bigl(\sup_{a \in {\mathbb R}^N} \int_{B_R
(a)} \abs{u}^\frac{2 N p}{N + \alpha}\int_{B_R (a)}
\abs{v}^\frac{2 N p}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2
N}(1 - \frac{1}{p})}}\\
+ \frac{C}{R^{\beta - \alpha}} \Bigl(\int_{{\mathbb R}^N} \abs{\nabla
u}^2 + \abs{u}^2 \int_{{\mathbb R}^N} \abs{\nabla
v}^2 + \abs{v}^2 \Bigr)^\frac{p}{2}
\end{multline*}
\end{lemma}
When \(p = \frac{N + \alpha}{N}\), then \((p - 1)N = \alpha\) and there is no \(\beta\) that would satisfy the assumptions.
\begin{proof}[Proof of lemma~\ref{lemmaNewEstimate}]
We first decompose the integral as
\[
\int_{{\mathbb R}^N} \bigl(I_\alpha \abs{u}^p\bigr) \abs{v}^p
=\int_{{\mathbb R}^N} \bigl((\chi_{B_{R/2}} I_\alpha) \ast \abs{u}^p\bigr) \abs{v}^p
+\int_{{\mathbb R}^N} \bigl((\chi_{{\mathbb R}^N \setminus B_{R/2}} I_\alpha) \ast
\abs{u}^p\bigr) \abs{v}^p.
\]
We then observe that if \(\beta \in (\alpha, N)\), then
\[
\int_{{\mathbb R}^N} \bigl((\chi_{{\mathbb R}^N \setminus B_{R/2}} \ast I_\alpha)
\abs{u}^p\bigr)
\abs{v}^p \le \frac{C}{R^{\beta - \alpha}} \int_{{\mathbb R}^N} (I_\beta \ast
\abs{u}^p) \abs{v}^p.
\]
If \(\beta < (p - 1) N\), then by the Hardy--Littewood--Sobolev inequality and
by the Sobolev inequality, we have
\[
\int_{{\mathbb R}^N} \bigl((\chi_{{\mathbb R}^N \setminus B_{R/2}} I_\alpha) \ast
\abs{u}^p\bigr)
\abs{v}^p \le \frac{C'}{R^{\beta - \alpha}} \Bigl(\int_{{\mathbb R}^N} \abs{\nabla
u}^2 + \abs{u}^2\Bigr)^\frac{p}{2}\Bigl(\int_{{\mathbb R}^N} \abs{\nabla v}^2
+ \abs{v}^2\Bigr)^\frac{p}{2}.
\]
Next, we have
\begin{multline*}
\int_{{\mathbb R}^N} \bigl((\chi_{B_{R/2}} I_\alpha) \ast \abs{u}^p\bigr) \abs{v}^p\\
\le \frac{C'}{R^N} \int_{{\mathbb R}^N} \int_{B_R (a)} \int_{B_R (a)} I_\alpha (x
- y) \abs{u (x)}^p \abs{v (y)}^p \,\mathrm{d} x \,\mathrm{d} y \,\mathrm{d} a.
\end{multline*}
For every \(a \in {\mathbb R}^N\), we have, by the Hardy--Littewood--Sobolev
inequality \eqref{eqHLS} and the classical Sobolev inequality on the ball \(B_R
(a)\),
\begin{multline*}
\int_{B_R (a)} \int_{B_R (a)} I_\alpha (x
- y) \abs{u (x)}^p \abs{v (y)}^p \,\mathrm{d} x \,\mathrm{d} y\\
\shoveleft{\le C'' \Bigl(\int_{B_R (a)} \abs{u}^\frac{2 N p}{N + \alpha}
\int_{B_R (a)} \abs{v}^\frac{2 N p}{N + \alpha}\Bigr)^\frac{N + \alpha}{2 N}}\\
\le C''' \Bigl(\int_{B_R (a)} \abs{u}^\frac{2 N p}{N + \alpha}\int_{B_R (a)}
\abs{v}^\frac{2 N p}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2 N}(1-\frac{1}{p})}\\
\times \Bigl(\int_{B_R (a)} \abs{\nabla u}^2 + \abs{u}^2 \int_{B_R (a)} \abs{\nabla
v}^2 + \abs{v}^2 \Bigr)^\frac{1}{2}.
\end{multline*}
We now integrate this estimate with respect to \(a \in {\mathbb R}^N\) and apply the
Cauchy--Schwarz inequality to
obtain
\begin{multline*}
\int_{{\mathbb R}^N} \bigl((\chi_{B_{R/2}} I_\alpha)
\abs{u}^p\bigr)
\abs{v}^p\\
\le \frac{C'''}{R^N} \Bigl(\int_{{\mathbb R}^N} \Bigl(\int_{B_R (a)} \abs{\nabla
u}^2 + \abs{u}^2\Bigr) \,\mathrm{d} a\Bigr)^\frac{1}{2} \Bigl(\int_{{\mathbb R}^N} \Bigl(\int_{B_R (a)} \abs{\nabla
v}^2 + \abs{v}^2\Bigr) \,\mathrm{d} a\Bigr)^\frac{1}{2}\\
\times \shoveright{\Bigl(\sup_{a \in {\mathbb R}^N} \int_{B_R
(a)} \abs{u}^\frac{2 N p}{N + \alpha}\int_{B_R (a)}
\abs{v}^\frac{2 N p}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2
N}(1-\frac{1}{p})}}\\
= C''''\Bigl(\int_{{\mathbb R}^N} \abs{\nabla
u}^2 + \abs{u}^2 \int_{{\mathbb R}^N} \abs{\nabla
v}^2 + \abs{v}^2 \Bigr)^\frac{1}{2}\\
\times \Bigl(\sup_{a \in {\mathbb R}^N} \int_{B_R
(a)} \abs{u}^\frac{2 N p}{N + \alpha}\int_{B_R (a)}
\abs{v}^\frac{2 N p}{N + \alpha}\Bigr)^{\frac{N + \alpha}{2 N}(1-\frac{1}{p})}.
\end{multline*}
\end{proof}
\subsection{Existence of a minimal action nodal solution}
In order to prove theorem~\ref{theoremNod}, we finally establish the strict
inequality.
\begin{proposition}
\label{propositionLevelsOddNod}
If \(\frac{N - 2}{N + \alpha} \le \frac{1}{p} \le \frac{N}{N + \alpha}\), then
\[
\mathcal{N}_\mathrm{odd} \subset \mathcal{N}_{\mathrm{nod}}.
\]
In particular,
\[
c_\mathrm{nod} \le c_\mathrm{odd}.
\]
\end{proposition}
\begin{proof}
If \(u \in \mathcal{N}_\mathrm{odd}\),
then since \(u \in \mathcal{N}_0\),
\[
\dualprod{\mathcal{A}' (u)}{u^+} + \dualprod{\mathcal{A}' (u)}{u^-} =
\dualprod{\mathcal{A}' (u)}{u} = 0.
\]
On the other hand, since \(u \in H^1_\mathrm{odd} ({\mathbb R}^N)\), by the
invariance of \(u\) under odd reflections,
\[
\dualprod{\mathcal{A}' (u)}{u^+} = \dualprod{\mathcal{A}' (u)}{u^-},
\]
and the conclusion follows.
\end{proof}
We can now prove theorem~\ref{theoremNod} about the existence of minimal action
nodal solutions.
\begin{proof}[Proof of theorem~\ref{theoremNod}]
Proposition~\ref{propositionNodalPalaisSmaleExistence} gives the existence of a
localized Palais--Smale sequence \((u_n)_{n \in {\mathbb N}}\).
By propositions~\ref{propositionStrictInequality} and \ref{propositionLevelsOddNod},
the strict inequality \(c_\mathrm{nod} < 2 c_0\) holds. Hence we can apply
proposition~\ref{propositionNodalPalaisSmaleCondition} to reach the conclusion.
The solution \(u\) is twice continuously differentiable by the Choquard equation's regularity theory \cite{MorVanSchaft13}*{proposition 4.1} (see also \cite{CingolaniClappSecchi2012}*{lemma A.10}).
\end{proof}
\subsection{Degeneracy in the locally sublinear case}
We conclude this paper by proving that \(c_{\mathrm{nod}} = c_0\) if \(p < 2\).
\begin{proof}[Proof of theorem~\ref{theoremDegeneracy}]
We observe that if \(u \in \mathcal{N}_0\), then \(\abs{u} \in \mathcal{N}_0\).
Together with a density argument, this shows that
\[
c_0 = \inf \bigl\{\mathcal{A} (u)\;:\; u \in \mathcal{N}_0 \cap C^1_c ({\mathbb R}^N)
\text{ and } u \ge 0 \text{ on } {\mathbb R}^N\bigr\}.
\]
Let now \(u \in \mathcal{N}_0 \cap C^1_c ({\mathbb R}^N)\) such that \(u \ge 0\) on \({\mathbb R}^N\). We choose a point \(a \not \in \supp u\) and a function \(\varphi \in C^1_c ({\mathbb R}^N) \setminus \{0\}\) such that \(\varphi \ge 0\) and we define for each \(\delta > 0\) the function \(u_\delta : {\mathbb R}^N \to {\mathbb R}\) for every \(x \in {\mathbb R}^N\) by
\[
u_\delta(x) = u (x) - \delta^\frac{2}{2 - p} \varphi \Bigl(\frac{x - a}{\delta} \Bigr).
\]
We observe that, if \(\delta > 0\) is small enough, \(u_\delta^+ = u^+\).
By a direct computation, we have \(t_+ u_\delta^+ + t_- u_\delta^- \in \mathcal{N}_\mathrm{nod}\) if and only if
\begin{equation}
\label{eqDegeneracySystem}
\left\{
\begin{aligned}
(t_+^{2 - p} - t_+^{p} )\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr) \abs{u}^p
&= t_-^p \delta^{N + \frac{2p}{2 - p}} J_\delta,\\
t_-^{2 - p} \int_{{\mathbb R}^N} \abs{\nabla \varphi}^2 + \delta^2 \abs{\varphi}^2
&=t_+^p J_\delta + t_-^{p} \delta^{\alpha + \frac{2 p}{2 - p}} \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{\varphi}^p\bigr) \abs{\varphi}^p,
\end{aligned}
\right.
\end{equation}
where
\[
J_\delta = \int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr) (\delta z + a ) \bigl(\varphi (z)\bigr)^p\,\mathrm{d} z.
\]
We observe that when \(\delta = 0\), the system reduces to
\[
\left\{
\begin{aligned}
(t_+^{2 - p} - t_+^{p} )\int_{{\mathbb R}^N} \bigl(I_\alpha \ast \abs{u}^p\bigr) \abs{u}^p
&= 0,\\
t_-^{2 - p} \int_{{\mathbb R}^N} \abs{\nabla \varphi}^2
&=t_+^p \bigl(I_\alpha \ast \abs{u}^p\bigr) (a) \int_{{\mathbb R}^N} \abs{\varphi}^p,
\end{aligned}
\right.
\]
which has a unique solution. By the implicit function theorem, for \(\delta > 0\) small enough there exists \((t_{+, \delta}, t_{-, \delta}) \in (0, \infty)^2\) that solves the system \eqref{eqDegeneracySystem} and such that
\[
\lim_{\delta \to 0}(t_{+, \delta}, t_{-, \delta})
=\biggl(1, \Bigl( \bigl(I_\alpha \ast \abs{u}^p\bigr) (a) \int_{{\mathbb R}^N}
\abs{\varphi}^p \Big/ \int_{{\mathbb R}^N} \abs{\nabla \varphi}^2\Bigr)^\frac{1}{2 -
p}\biggr).
\]
Since \((N - 2)(2 - p) > -4\), we have \(t_+ u_\delta^+ +
t_- u_\delta^- \to u\) strongly in \(H^1 ({\mathbb R}^N)\) as \(\delta \to 0\), and it follows
that
\[
\inf_{\mathcal{N}_\mathrm{nod}} \mathcal{A} \le \mathcal{A} (u),
\]
and thus \(c_{\mathrm{nod}} \le c_0\).
We assume now that the function \(u \in \mathcal{N}_\mathrm{nod}\) minimizes
the action functional \(\mathcal{A}\) on the nodal Nehari set
\(\mathcal{N}_\mathrm{nod}\).
Since \(c_0 = c_{\mathrm{nod}}\), we deduce that \(u\) also minimizes \(\mathcal{A}\)
over the Nehari manifold \(\mathcal{N}_0\).
By the properties of such groundstates \cite{MorVanSchaft13}*{proposition 5.1}, either \(u^+ = 0\) or \(u^- = 0\), in contradiction with the assumption \(u \in \mathcal{N}_\mathrm{nod}\) and the definition of the Nehari nodal set \(\mathcal{N}_\mathrm{nod}\).
\end{proof}
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\end{document} | {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,845 |
Get Nickel and Dimed: On Not Getting by in America from Amazon.com
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Early, Steve, "Prole Like Me," in the Nation, Vol. 272, No. 23, June 11, 2001, p. 52.
Ehrenreich, Barbara, Nickel and Dimed: On (Not) Getting By in America, Metropolitan Books, 2001.
Hays, Sharon, Flat Broke with Children: Women in the Age of Welfare Reform, Oxford University Press, 2003, p. 8.
Klein, Julia, Review of Nickel and Dimed: On (Not) Getting By in America, in the American Prospect, Vol. 12, No. 13, July 30, 2001, p. 43.
Ogyn, Kya, "Can You Live On It?," in Off Our Backs, Vol. 35, Jan-Feb 2005, pp. 44-46.
Scott, Joni, Review of Nickel and Dimed: On (Not) Getting By in America, in the Humanist, Vol. 61, No. 5, September-October 2001, p. 40.
Sherman, Scott, "Class Warrior: Barbara Ehrenreich's Singular Crusade," in Columbia Journalism Review, Vol. 42, No. 4, November-December 2003, pp. 34-42.
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"redpajama_set_name": "RedPajamaC4"
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{"url":"http:\/\/ria.ua.pt\/handle\/10773\/16282","text":"Reposit\u00f3rio Institucional da Universidade de Aveiro\u00a0> CIDMA - Centro de Investiga\u00e7\u00e3o e Desenvolvimento em Matem\u00e1tica e Aplica\u00e7\u00f5es\u00a0> CIDMA - Artigos\u00a0> Fundamental solutions of the time fractional diffusion-wave and parabolic Dirac operators\n Please use this identifier to cite or link to this item http:\/\/hdl.handle.net\/10773\/16282\n\n title: Fundamental solutions of the time fractional diffusion-wave and parabolic Dirac operators authors: Ferreira, Milton dos SantosVieira, Nelson Felipe Loureiro keywords: Fractional Laplace operatorRiemann-Liouville fractional derivativesEigenfunctionsMittag-Leffler functionTime fractional diffusion-wave operatorTime fractional parabolic Dirac operatorFundamental solutionsCaputo fractional derivativeFractional moments issue date: 1-Mar-2017 publisher: Elsevier abstract: In this paper we study the multidimensional time fractional diffusion-wave equation where the time fractional derivative is in the Caputo sense with order $\\beta \\in ]0,2].$ Applying operational techniques via Fourier and Mellin transforms we obtain an integral representation of the fundamental solution (FS) of the time fractional diffusion-wave operator. Series representations of the FS are explicitly obtained for any dimension. From these we derive the FS for the time fractional parabolic Dirac operator in the form of integral and series representation. Fractional moments of arbitrary order $\\gamma>0$ are also computed. To illustrate our results we present and discuss some plots of the FS for some particular values of the dimension and of the fractional parameter. URI: http:\/\/hdl.handle.net\/10773\/16282 ISSN: 0022-247X publisher version\/DOI: http:\/\/dx.doi.org\/10.1016\/j.jmaa.2016.08.052 source: Journal of Mathematical Analysis and Applications appears in collections CIDMA - Artigos\n\nfiles in this item\n\nfile description sizeformat","date":"2018-01-22 01:00:54","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\": 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\": 0.8060888648033142, \"perplexity\": 2346.3211098555716}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-05\/segments\/1516084890928.82\/warc\/CC-MAIN-20180121234728-20180122014728-00675.warc.gz\"}"} | null | null |
{"url":"https:\/\/hal.univ-lorraine.fr\/hal-01502118","text":"# Stress-strain response of textured AZ31B magnesium alloy under uniaxial tension at the different strain rates\n\nAbstract : Grain orientation and deformation strain rate have impact on mechanical behaviors of magnesium alloys through their effect on the deformation modes. In this paper, uniaxial tensile tests were conducted at room temperature under various strain rates, from 2.8 x 10(-5) s(-1) to 1.1 x 10(-1) s(-1), on two kinds of AZ31B magnesium alloy samples which possess fiber texture and spread texture, respectively. It was found that the flow stress-strain curves of the samples with the fiber texture show a concave down shape and a yielding stress higher than that of the samples with the spread texture. The samples with the spread texture present a linear shaped true stress-strain flow curve. A quasi horizontal step'' appears on the stress-strain curves at the high strain rates, related mainly to \\10-12\\ tension twinning that becomes an important deformation mode of grains. It was believed that the difference of stress-strain responses between the two kinds of tensile samples results from the difference of the slip modes, the contribution of the slips to plastic strain decreases with the increment of strain rate.\nKeywords :\nDocument type :\nJournal articles\nDomain :\n\nhttps:\/\/hal.univ-lorraine.fr\/hal-01502118\nContributor : Lem3 Ul <>\nSubmitted on : Wednesday, April 5, 2017 - 10:03:51 AM\nLast modification on : Friday, February 26, 2021 - 2:48:02 PM\n\n### Citation\n\nC. J. Geng, B. L. Wu, X. H. Du, Y. D. Wang, Yu Dong Zhang, et al.. Stress-strain response of textured AZ31B magnesium alloy under uniaxial tension at the different strain rates. Materials Science and Engineering: A, Elsevier, 2013, 559, pp.307-313. \u27e810.1016\/j.msea.2012.08.103\u27e9. \u27e8hal-01502118\u27e9\n\nRecord views","date":"2021-06-15 21:35:39","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\": 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\": 0.1769324392080307, \"perplexity\": 5649.977818850534}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-25\/segments\/1623487621627.41\/warc\/CC-MAIN-20210615211046-20210616001046-00463.warc.gz\"}"} | null | null |
Верони́к Клоде́ль (; 22 ноября 1966, коммуна Корнимон, Лотарингия) — бывшая французская биатлонистка, олимпийская чемпионка 1992 года в эстафете 3х7,5 км, бронзовая призёрка Олимпийских игр 1994 года в этой же дисциплине, многократная призёрка чемпионатов мира.
Ссылки
Профиль IBU
Биатлонисты Франции
Олимпийские чемпионы от Франции
Чемпионы зимних Олимпийских игр 1992 года
Бронзовые призёры зимних Олимпийских игр 1994 года
Биатлонисты на зимних Олимпийских играх 1992 года
Биатлонисты на зимних Олимпийских играх 1994 года | {
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Das aktuelle Ranking der Rapid Surf League© wurde auf Basis der Platzierungen und Scores des Pure Surfcamps Rapid Surf League RIOT RIVER Contests in Cunovo und des Rip Curl Langenfeld Pro ermittelt. Dies ist das finale Ranking für 2018.
The current Rapid Surf League© ranking is based on the results of the Pure Surfcamps Rapid Surf League RIOT RIVER in Cunovo and the Rip Curl Langenfeld Pro. This is the final ranking for 2018. | {
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} | 9,598 |
The AU-11CD from CYP Europe is an advanced HDMI Repeater with integrated Audio de-embedding, up to 5.1.
By connecting the CYP AU11CD Repeater device between your HDMI source and output display, the HDMI signal can be amplified for greater distance transmission, whilst the audio signal can be split out to optical or analogue (RCA) outputs for connection to an external amplifier (such as 51. surround sound system).
If you require any further assistance regarding our 100V Line range just contact us on 02476 369898 and one of our experts will be happy to help. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,277 |
import django.contrib.postgres.fields
from django.db import migrations, models
def move_extensions(apps, schema_editor):
"""Move existing `extended` value into the new field `extensions`."""
Membership = apps.get_model('workshops', 'Membership')
for membership in Membership.objects.all():
membership.extensions = (
[] if membership.extended is None
else [membership.extended]
)
membership.save()
def undo_extensions(apps, schema_editor):
"""Move from `extensions` to single value `extended` field."""
Membership = apps.get_model('workshops', 'Membership')
for membership in Membership.objects.all():
membership.extended = sum(membership.extensions) or None
membership.save()
class Migration(migrations.Migration):
dependencies = [
('workshops', '0244_auto_20210509_2051'),
]
operations = [
migrations.AddField(
model_name='membership',
name='extensions',
field=django.contrib.postgres.fields.ArrayField(base_field=models.PositiveIntegerField(), default=list, help_text='Number of days the agreement was extended. The field stores multiple extensions. The agreement end date has been moved by a cumulative number of days from this field.', size=None),
),
migrations.RunPython(move_extensions, undo_extensions),
migrations.RemoveField(
model_name='membership',
name='extended',
),
]
| {
"redpajama_set_name": "RedPajamaGithub"
} | 733 |
Q: Why did Hermione modify her parents' memories before sending them to Australia? At the start of DH, Hermione tells Harry and Ron that she modified her parents' memories.
"I've also modified my parents' memories so that they're convinced they're really called Wendell and Monica Wilkins, and that their life's ambition is to move to Australia, which they have now done. That's to make it more difficult for Voldemort to track them down and interrogate them about me – or you, because unfortunately, I've told them quite a bit about you.
Assuming I survive our hunt for the Horcruxes, I'll find Mum and Dad and lift the enchantment. If I don't – well, I think I've cast a good enough charm to keep them safe and happy. Wendell and Monica Wilkins don't know that they've got a daughter, you see." (DH)
So why would Hermione consider this necessary?
*
*The Death Eaters are mostly purebloods who don't (want to) know much about the Muggle world. Even the halfbloods like Voldemort himself or Snape are probably out of date with modern life, so they would find it difficult to locate Hermione's parents, unless they had their address from the school or ministry records.
*Her parents use a different name.
*Her parents are on a different continent.
*Her parents are not protected by that memory modification. If the Death Easters, against all odds, manage to find them, they won't believe them that they don't know about her daughter, and probably torture them to get the truth out, or just for fun.
*What could her parents really tell Voldemort about Hermione or Harry that is not already known, unless she told them about the Horcrux mission?
*The memory modification is supposed to be reversible. Voldemort even managed to recover the memories from Bertha Jorkins that Crouch wanted to remove permanently. So why does she think that if Voldemort manages to find them, he won't be able to lift the enchantment?
Since comments raised the question whether Voldemort would be able to lift the enchantment, here the part where Barty Jr. talks about Bertha
"Yes," said Crouch, his eyelids flickering again. "A witch in my father's office. Bertha Jorkins. She came to the house with papers for my father s signature. He was not at home. Winky showed her inside and returned to the kitchen, to me. But Bertha Jorkins heard Winky talking to me. She came to investigate. She heard enough to guess who was hiding under the Invisibility Cloak. My father arrived home. She confronted him. He put a very powerful Memory Charm on her to make her forget what she'd found out. Too powerful. He said it damaged her memory permanently." (GoF)
Although her memory is supposed to be permanently damaged, Voldemort still manages to extract the information from her.
My master had found out that I was still alive. He had captured Bertha Jorkins in Albania. He had tortured her. She told him a great deal. She told him about the Triwizard Tournament. She told him the old Auror, Moody, was going to teach at Hogwarts. He tortured her until he broke through the Memory Charm my father had placed upon her. She told him I had escaped from Azkaban. She told him my father kept me imprisoned to prevent me seeking my master.
Therefor this memory modification doesn't protect Hermione's parents, and it doesn't protect the trio. Why did she do it anyway?
A: I see two good reasons for the memory modification mentioned in the quote.
*
*so that they're convinced they're really called Wendell and Monica Wilkins
If Death Eaters are looking for Hermione's family, just about the only information they have to go on is their names. By forcing them to assume new identities she makes it a lot harder for anyone to track them down.
*If I don't – well, I think I've cast a good enough charm to keep them safe and happy.
Hermione is preparing for the possibility that she will die. If she does die, her parents will be grief-stricken. Losing your only child is a pretty devastating blow. Hermione wants to save them from this pain and grief so she enchants them so that they won't remember that they have a daughter – thus, if she dies there will be no grief and pain for her parents.
A: You must not have children of your own yet. The number one thing parents do is worry about their children, no matter how well they believe they've raised them. By not remembering that they have a daughter, they won't go through the rest of their lives with the emotional distress of not knowing what happened to their daughter, or even worse, knowing that she's now dead.
Wiping their memory of her was the kindest thing Hermione could have done for them.
A: She did it to protect them. She erased herself from them completely so that Voldemort and the death eaters would have no reason to even go after them. Yes they could torment her parents but the reality is that if they don't know anything about her whereabouts then there's nothing they could get out of her. She would allow her parents to start a new life on a brand new continent away from all the magic and craziness that came from having her as a daughter. It was a way to remove them entirely from the situation. Also so that if she died searching for the Horcruxes or even fighting Voldemort then they wouldn't be affected by it. It was a way to cut all ties between them and the magic world. Allow them to be muggles.
A: With or without memories Hermione still cares about THEM. They don't need to have information to be used as hostages to draw out Hermione.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,155 |
La tentative de meurtre est une infraction criminelle reconnue par plusieurs pays qui consiste à tenter de tuer une personne.
Droit américain
Droit canadien
En droit canadien, la tentative de meurtre est punie par l'article 239 (1) du Code criminel
L'arrêt R. c. Ancio est un arrêt de principe concernant la tentative de meurtre. Il s'agit d'une infraction d'intention spécifique.
Droit français
En France, La tentative de meurtre est le fait inachevé par l'auteur du crime, de tuer une personne bien que ce dernier ait tout fait pour y parvenir ; il y a là tentative de meurtre. Au terme de l'article 121-5 du Code pénal français, la définition de la tentative est la suivante « La tentative est constituée dès lors que manifestée par un commencement d'exécution, elle n'a été suspendue ou n'a manqué son effet qu'en raison de circonstances indépendantes de la volonté de son auteur ». Cela suppose que dans le processus criminel; l'iter criminis:
1) le dessein criminel est présent
2) la résolution et son extériorisation sont là
3) les actes préparatoires ayant pour objet de procurer les moyens de l'infraction sont patents
4) que les agissements du crime ayant pour conséquence directe la commission de l'infraction ont été effectués
5) mais que l'achèvement du crime est absente.
Notes et références
Droit pénal
Droit pénal au Canada
en:attempted murder | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,065 |
Отдача от образования в России и на Украине: сравнительный анализ
Прикладная эконометрика. 2017. Т. 47. С. 100-122.
Архипова М. Ю., Егоров А. А., Сиротин В. П.
The paper is focused on the measuring of return to education. This measure reflects the yield of investments in human capital. The analysis is based on estimation of the modified Mincer's type equations by means of quantile regressions for panel data. Along with quantile regressions estimation we consider the models of the joint distribution of wage and education duration based on copulas. The methodological approach used in the study allows comparing the significance of the factors included in the model for the wage formation. Also it allows investigating the structure of the relationship between wage and education and answering the question devoted to the following dilemma – in which country return to education was higher after the decade of independent development.
Full text (PDF, 11.98 Mb)
Keywords: Capitalization of professional knowledgequantile regressionwagesLongitudinal data
Wage Inequality in Russia (1994–2003)
Lukyanova A. Labor markets and social policy. EERC, 2006. No. 06/03.
The paper documents the changes in the size of the wage distribution in Russia over the period 1994–2003. Developments in wage inequality varied a lot by sub-periods: overall wage inequality stayed stable in 1994–1996, then it jumped following the 1998 crisis and remained at higher levels for three years. In 2002 the trend reversed again and in the course of a single year wage inequality fell back to the level of the mid-1990s. We find that evolution wage inequality was largely driven by changes in the upper end of the wage distribution. Decomposition of wage inequality by population sub-groups shows that inequality has been higher for men, younger and low-educated workers, and rural inhabitants. The structure of inequality did not change much over the period from 1994 to 2003. Demographic variables (mainly gender and region) explain the largest proportion of wage dispersion (over 40% of the explained variation and 15% of total variation). Nearly equivalent is the contribution of firm characteristics with industry affiliation of employer playing the leading role. Our results show that returns to education continued to rise at all percentiles of the wage distribution converging at the level of about 8–9% of wage increase for an additional year of schooling.
Анализ территориальных различий доходов населения в мезосистеме
Бакуменко Л. П., Мхитарян В. С. Экономика, статистика и информатика. Вестник УМО. 2010. № 2. С. 86-91.
Questions of differentiation of the population on a level of incomes in Republic Mary El are considered. Parameters of a variation and structure of the basic monetary incomes are analyses. Classification of administrative areas on the parameters describing incomes of the population is lead. The analysis of territorial distinctions of the population on levels of incomes is given. For more in-depth study of interrelation of the parameters describing differentiation of incomes of the population of Republic, on everyone allocated cluster are constructed regression models of a degree of influence of the parameters reflecting age structure of the population on a level of monthly average wages, pensions and social payments.
Директор школы оценивает её ресурсное обеспечение: результаты мониторинга
Абанкина И. В., Савельева М. Б., Сигалов С. В. Народное образование. 2012. № 5. С. 17-22.
Forgone Earnings from Smoking: Evidence for a Developing Country
Lokshin M., Beegle K. Policy Research Working Paper. WPS. World Bank Group, 2006. No. 4018.
The authors estimate the economic losses related to the negative effect of smoking on wages in a context of a developing country. Using data from the 2005 Albania Living Standards Monitoring Survey, they jointly estimate a system of three equations: the smoking decision and two separate wage equations for smokers and nonsmokers. The results show that, after controlling for observed characteristics and taking into account unobserved heterogeneity in personal characteristics, smoking has a substantial negative impact on wages. On average smokers' wages are 20 percent lower than the wages of similar nonsmokers, providing strong evidence for the potential policy relevance of tobacco control initiatives for developing countries.
Sectoral Segregation and the Wage Differential between Immigrants and Native Workers in Russia
Polyakova E. In bk.: Труды 6-й Международной научно-практической конференции студентов и аспирантов «Статистические методы анализа экономики и общества» (12-15 мая 2015 г.). M.: Higher School of Economics Publishing House, 2015.
This paper with use of nationally representative data (RMLS-NRU HSE) in 2004-2012 examines the sectoral segregation between immigrant (persons with an immigration background) and native workers and its impact on earning differential in Russia. This is the first study on the micro-level in Russia about sectoral segregation and wage gap between natives and immigrants under its influence.
In this study we analyze the determinants of the choice of sector, estimate earning differences between natives and immigrants, define Duncan index of dissimilarity and measure the impact of sectoral segregation on earning differential between natives and immigrants using Oaxaca-Blinder decomposition.
Accounting for latent classes in movie box office modeling
Antipov E. A., Pokryshevskaya E. B. Journal of Targeting, Measurement and Analysis for Marketing. 2011. Vol. 19. No. 1. P. 3-10.
This article addresses the issue of unobserved heterogeneity in film characteristics influence on box-office. We argue that the analysis of pooled samples, most common among researchers, does not shed light on underlying segmentations and leads to significantly different estimates obtained by researchers running similar regressions for movie success modeling. For instance, it may be expected that a restrictive MPAA rating is a box office poison for a family comedy, whereas it insignificantly influences an action movie's revenues. Using a finite mixture model we extract two latent groups, the differences between that can be explained in part by the movie genre, the source, the creative type and the production method. On the basis of this result, the authors recommend developing separate movie success models for different segments, rather than adopting an approach, that was commonly used in previous research, when one explanatory or predictive model is developed for the whole sample of movies.
Do students expect any returns on effort applied to studying? An econometric analysis of determinants of the amount of expected wage after graduation
Marina Telezhkina, Andrey Maksimov, Nataliya Maksimova. In bk.: CEUR-WS Proceedings of the Workshop on Computer Modelling in Decision Making (CMDM 2017). Vol. 2018. Aachen: CEUR-WS, 2017. P. 200-212.
Students' perception of the labor market makes a great deal in students' decisions concerning effort to study, work during university studies, etc. The aim of the research is to define whether students identify significant returns on effort with respect to wage after graduation. Moreover, it seems reasonable to single out other factors that students expect to influence their wage significantly. With the use of the data of Russian students' questionnaire conducted in 2012 within the framework of the Monitor of Economics of Education project the regressions with the use of instrumental variables and stochastic frontier approach are estimated. The results suggest effort is considered as an influential factor in determining wage by Russian students if students' incomplete awareness about labor market is taken into account. Besides, university quality, abilities, wage received by working students, region, specialty, family's income and gender make the difference in the amount of wage expected by students. For additional analysis the 20% and 80% quantile regressions are built. According to the results, persons having the highest wage forecasts base them on the amount of wage offered to working students on the labor market and do not correct them subject to their effort, university quality and abilities. At the same time another group of students, keeping similar basis for expectation formation as a previously analyzed group, expect significant contribution of effort and abilities.
Vulnerability to welfare change during economic shocks: Evidence from the 1998 Russian crisis
University of Essex, Department of Economics, 2002.
Using changes in consumption as a proxy for 'vulnerability' we identify the characteristics associated with vulnerability around the time of the 1998 Russian financial crisis. In addition, we examine the role of formal and informal safety nets in preserving individual well being. We apply quantile regression techniques in order to identify the characteristics associated with vulnerability across the two periods. Amongst those most vulnerable during the crisis were, less educated individuals living in urban areas, in households containing greater numbers of pensioners. Furthermore, we found that increases in home production and help from relatives acted to decrease vulnerability, especially amongst those suffering the largest changes in consumption. Following the crisis, amongst the least vulnerable were, better educated individuals, resident in urban areas, able to increase home production, and in receipt of improved pension payments and child benefits. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 274 |
\section{Introduction}
\nocite{video}
Autonomous systems have become increasingly pervasive in our everyday life, whether that be through the rise in interest for autonomous vehicles \cite{autonomous_vehicle}, intelligent defense systems \cite{Autonomous_Swarm_NAVY}, or even human/robot interaction \cite{HRI_ahmadi}. This rise in prevalence has motivated a similar increase in questions regarding the efficacy of these systems in safety critical contexts. These questions are not entirely unfounded, however, as even in those cases when attempting to verify system efficacy, horrific accidents still occur \textit{e.g.} recent autonomous car crashes. Nonetheless, the field is still pushing forward rapidly, and in the future, these autonomous systems will have to deal with even more complex, dynamic, and relatively unstructured environments. Coupled with the cost of failure, this increase in system complexity makes systematic test and evaluation of these systems all the more necessary.
Significant work on this issue has been carried out by the test and evaluation (T\&E) community. Reachability analysis has been used to shape critical test cases from existing data \cite{TE_model_based_unadaptive_1} . At the discrete level, RRT has been used to efficiently search a feasible space to find critical sequences that identify failure of the underlying controller \cite{TE_model_based_unadaptive_2} . Tests based on a graph-search framework over clustered, critical situations, have been developed via exhaustive mission simulation of the underlying system \cite{TE_model_based_unadaptive_3}. Each of the aforementioned methods are model-based and not easily adaptable to other systems/testing environments as they are exhaustive. To address the issue of adaptivity, one approach adaptively samples the feasible space to generate successively harder tests \cite{TE_specific_adaptive_1}. That being said, the aforementioned frameworks require an accurate system model to function well, and except for the latter contribution, none are easily adaptable. However, as noted in a memo by the Department of Defense \cite{DOD_article}, a testing framework that is both adaptive/adversarial and formally guarantees safety is still highly sought after.
Prior work in the T\&E community reference formal methods as a means by which one can formally guarantee safety/the lack thereof (see \cite{TE_formal_methods_CPS}). Formal methods, specifically linear and signal temporal logic (LTL \& STL), have garnered significant interest in the controls community (see \cite{Safety_Logic1,Safety_Logic2,Safety_Logic3,Safety_Logic4,ahmadi2020barrier}). In each of these cases, the logical specification encodes a control objective whose satisfaction is formally guaranteed via the barrier-based controller. In this respect, control barrier functions are very useful in formally guaranteeing these logical specifications insofar as satisfying a specification and remaining within a safe set are both set-based arguments \cite{TAC_Paper}. However, these formal guarantees require specific knowledge of the onboard controller and system dynamics - for the test engineer, this is oftentimes not the case.
\begin{figure}
\centering
\includegraphics[width=0.49\textwidth]{CDC_Fig1.pdf}
\caption{The flowchart for the proposed test generation framework. The framework starts at (bottom left) collecting data for the system to be tested in order to (center top) estimate CBFs corresponding to the control objectives the system intends to satisfy. These estimated CBFs are used in a minimax game to (bottom right) generate tests designed to verify system efficacy in satisfying the aforementioned objective. This test generation framework is designed to systematically identify (center middle) points of system failure that may occur during general operation.}
\label{fig::Title_flowchart}
\end{figure}
\newsec{Our Contribution:} In this paper, the overarching goal is to start to bridge the work done in the controls and the T\&E community. Specifically, we address the issue of designing an adaptable/adversarial testing framework. Given an autonomous system with some operational specifications, we construct a minimax problem whose solution defines testing scenarios intended to optimally frustrate satisfaction of the given specifications without specific knowledge of the onboard control architecture. To this end, we begin by collecting data of the autonomous system satisfying the specifications. Then, we use the collected demonstration data to frame Linear Programs that develop approximate Control Barrier Functions corresponding to the operational specifications of the autonomous system. Finally, we use these approximate control barrier functions to develop a minimax game to solve for optimal testing parameters designed to frustrate satisfaction of the specifications. The proposed method is illustrated in Figure~\ref{fig::Title_flowchart}.
\newsec{Outline:} In Section~\ref{sec:probform}, we review some preliminary definitions and formally define the problem under study. In Section~\ref{sec::Main_Result}, we detail the main result of the paper, \textit{i.e.,} a minimax game for test generation. In Section~\ref{sec::corollaries}, we couple the result with a linear program to systematically generate difficult tests. Finally, in Section~\ref{sec::simulations_and_experiments}, we illustrate our proposed methodology with a case study.
\section{Problem Formulation} \label{sec:probform}
In this section, we present some notions used in the sequel and formally define the problem under study.
\subsection{Preliminaries}
\label{sec::problem_setup}
We consider a class of systems (to-be-tested) that can be modeled as a dynamical system with affine inputs:
\begin{equation}
\label{dyn_sys}
\dot{x} = f(x) + g(x)u, \quad x \in \mathcal{X} \subset \mathbb{R}^n, \quad u \in \mathcal{U} \subset \mathbb{R}^m.
\end{equation}
Furthermore, we will assume that both $f(x)$ and $g(x)$ are locally Lipschitz. For any function $h(x)$,
\begin{align}
L_fh(x) &\triangleq \nabla_xh(x)f(x), \\
L_gh(x) &\triangleq \nabla_xh(x)g(x),
\end{align}
are its Lie derivatives.
\newsec{Formal Methods:} We will define $\mathcal{A}$ to be the set of atomic propositions from which the provided control objective, \textit{i.e.,} a temporal logic specification, has been synthesized. We use the following notation to represent the truth/lack thereof for an atomic proposition
\begin{equation}
\forall \phi \in \mathcal{A}, \quad \llbracket \phi \rrbracket \triangleq \{x \in \mathcal{X} | \phi(x) = \mathrm{TRUE} \},
\end{equation}
where $\phi(x)$ denotes the atomic proposition evaluated at the state $x$. In addition, we will define the symbols $\neg, \wedge, \lor$ to correspond to negation, conjunction, and disjunction respectively. That is, $\neg \phi = $ TRUE when $\phi = $ FALSE. Likewise $\phi \wedge \omega =$ TRUE when $\phi = $ TRUE and $\omega = $ TRUE, and $\phi \lor \omega =$ TRUE when either $\phi=$ TRUE or $\omega = $ TRUE.
In this paper, we consider a subset of temporal logic (TL) operators, $\F$uture and $\G$lobal, defined as follows (here $\equiv$ denotes a logical equivalency):
\begin{align}
\F\phi & \equiv \exists~t^*\geq 0~\mathrm{s.t.~}x(t^*)\in\llbracket \phi \rrbracket,\\
\G\phi & \equiv \forall~t\geq 0,~x(t)\in\llbracket \phi \rrbracket.
\end{align}
While this seems restrictive, these two operators can be composed to consider more complex LTL specifications, such as $\square \lozenge \phi \equiv \G(\F\phi)$.
\newsec{Control Barrier Functions (CBF):} To provide a metric by which we measure satisfaction of the provided specification, we will establish a correspondence between these TL specifications and control barrier functions, $h$. To start, we first define extended class-$\mathcal{K}$ functions, $\alpha: (-b,a) \to (-\infty,\infty)$, to be those functions, $\alpha$, that are strictly increasing and satisfy $\alpha(0) = 0$. Here, $a,b>0$. Using these extended class-$\mathcal{K}$ functions, we can define Control Barrier Functions (CBF).
\vspace{.01cm}
\begin{definition}
[Control Barrier Functions (CBF)]
\label{def::cbf}
\textit{For a dynamical system of the form \eqref{dyn_sys}, a differentiable function, $h: \mathbb{R}^n \to \mathbb{R}$ is considered a control barrier function if it satisfies the following criteria:
\begin{equation}
\label{}
\sup\limits_{u\in\mathcal{U}} \left[L_fh(x) + L_gh(x)u + \alpha(h(x)) \right] \geq 0,\quad \forall x \in \mathcal{X},
\end{equation}
where $\alpha$ is an extended class-$\mathcal{K}$ function~\cite{TAC_Paper}.}
\end{definition}
\vspace{.2cm}
The usefulness of a CBF is in guaranteeing the forward invariance of its 0-superlevel set:
\begin{align}
& \mathcal{C}_h & & \hspace{-0.9 in}= \{x \in \mathbb{R}^n~|~ h(x) \geq 0 \}, \\
& \partial\mathcal{C}_h & &\hspace{-0.9 in}= \{x \in \mathbb{R}^n~|~ h(x) = 0 \}.
\end{align}
Indeed, it was shown in Proposition 1 of \cite{TAC_Paper} that a CBF, as in Definition~\ref{def::cbf}, guarantees forward invariance of its 0-Superlevel set, $\mathcal{C}_h$. Here, we note that what we call control barrier functions are termed as \textit{zeroing control barrier functions} in \cite{wang2017safety}. Finally, a finite time convergence control barrier function requires $\alpha(x) =\gamma\sign(x)|x|^\rho$ to ensure finite time convergence to the set, $\mathcal{C}_h$, by $T = \frac{1}{\gamma(1-\rho)}|h(x_0)|^{1-\rho}$, provided $h(x_0) \leq 0$ \cite{Finite_CBF}.
\subsection{Problem Statement}
\label{sec::problem_statement}
As mentioned earlier, the overarching test and evaluation goal is to validate an autonomous system's capacity to satisfy a provided TL specification. However, as we have no knowledge of the controller on-board the system to-be-tested, not only do we have no metric of quantifying success for the TL specification, but we also do not have a systematic method of developing difficult tests by which to identify control system failures in satisfying the specification. We will show in the sequel that there exists a correspondence between CBFs and TL specifications. So, if we could determine these CBFs for the system at hand, we can use them to test the system against a given specification. This chain of reasoning is the basis for Figure~\ref{fig::Title_flowchart}. To that effect, we collect the following experimental data of the system satisfying the control objective:
\vspace{0.2cm}
\begin{definition}
[Data-Set]
\label{demonstrations}
\textit{ Define $\mathbb{D}_i = \{ (x^i_k,u^i_k) \in \mathbb{R}^n\times\mathbb{R}^m~|~k = 0,1,\dots,T_i \}$ as the data-set of state, action pairs for demonstration, $i$. Here, $k$ indexes time until $T_i$, which is the max time for the specific demonstration at hand. Then, define $\mathbb{D} = \{ \mathbb{D}_1, \dots, \mathbb{D}_r \}$ as the set of all provided demonstrations.}
\end{definition}
\vspace{0.2cm}
\begin{assumption}
\label{Labeling}
For the provided data-set, $\mathbb{D}$, and associated specification, the data-set for each demonstration, $\mathbb{D}_i$, terminates when the specification is satisfied. \textit{e.g.} for a specification defined as $\F\phi \wedge \G\omega$, where $\phi,\omega \in \mathcal{A}$, then for each $\mathbb{D}_i$,
\begin{itemize}
\item $x^i_{T_i} \in \llbracket \phi \rrbracket$ and $x^i_k \not \in \llbracket \phi \rrbracket$ for all $k = 0,1,\dots,T_i-1$, and
\item $x^i_k \in \llbracket \omega \rrbracket$ for all $k = 0,1,\dots,T_i$.
\end{itemize}
\end{assumption}
\vspace{0.2cm}
We use the generated data-set, $\mathbb{D}$, to determine composite CBFs that mimic system behavior. We compose these CBFs from a candidate set of barrier functions defined as follows:
\vspace{.0cm}
\begin{definition}
[Candidate Barrier Set] \textit{We call
\begin{equation*}
\mathcal{B} \triangleq \{h_1, h_2, \dots, h_q \},
\end{equation*}
a candidate barrier set for some provided, continuously differentiable functions, $\{h_i\}_{i=1}^q$.}
\end{definition}
\vspace{.2cm}
Note that in the above definition, each component of the candidate barrier set may not be a valid CBF, \textit{i.e.} $\mathcal{B}$ could be the set of all polynomials of degree, $n\leq q-1$. Finally, we need to formalize how we specifically identify these testing scenarios. \vspace{.2cm}
\begin{definition}
[Testing Parameters]
\label{testing_parameters}
\textit{We define the vector, $d\in\mathbb{R}^p$, to be a collection of testing parameters used to generate tests \textit{e.g.} the location of obstacles, time when a phenomena starts, \textit{etc}.}
\end{definition}
\vspace{.2cm}
With these definitions in place, the problem statement is as follows:
\vspace{.2cm}
\begin{problem}
\label{main_problem}
\textit{Given an autonomous system whose controller is unknown, $\mathbb{D}$, $\mathcal{B}$, and a TL specification the system intends to satisfy, devise an adaptive/adversarial strategy to identify a set of testing parameters $d$. }
\end{problem}
\vspace{0.2cm}
We show in the next section that these test parameters $d$ characterize a test scenario designed to validate that the autonomous system reliably satisfies a given TL specification.
\section{Main Result}
\label{sec::Main_Result}
This section will detail the main result of this paper - the minimax game formulated to generate optimal test parameters, $d^*$, designed to frustrate satisfaction of a TL specification expressed through CBFs.
\subsection{Main Result}
To preface the main result, we will make the following remark to simplify notation:
\vspace{.2cm}
\begin{remark}
\label{approximate_barrier_labeling}\textit{We denote $h^F_i~,i\in \mathcal{I}$ to be a set of CBFs for a finite number of specifications of the type $\mathbf{F}\phi_i$. Likewise, $h^G_j,~j \in \mathcal{J}$ denote CBFs for specifications of the type $\mathbf{G}\omega_j$. That is, $\mathcal{C}_{h^F_i} \equiv \llbracket \phi_i \rrbracket$, $\forall~ i\in\mathcal{I}$, and $\mathcal{C}_{h^G_j} \equiv \llbracket \omega_j \rrbracket$, $\forall~j\in\mathcal{J}$.}
\end{remark}
\vspace{.2cm}
In addition, we will make the following assumption to simplify the formulations in the sequel.
\vspace{.2cm}
\begin{assumption}
\label{test_restriction}
\textit{We will assume that the CBFs $h^G_j,~j\in \mathcal{J}$ depend on a set of test parameters $d$. That is, $h^G_j:\mathbb{R}^n\times\mathbb{R}^p \to \mathbb{R}$ and $\dot{h}^G_j: \mathbb{R}^n\times\mathbb{R}^p\times\mathbb{R}^m \to\mathbb{R}$ whereas, $h^F_i: \mathbb{R}^n \to \mathbb{R}$.}
\end{assumption}
\vspace{.2cm}
We will define the following set of feasible inputs:
\begin{align}
& \mathcal{U}(x,d) = \label{eqn::feasible_set} \\
& \{ u \in \mathcal{U}~|~\dot{h}^G_j(x,u,d) \geq -\alpha_j( h^G_j(x,d)),~\forall~j\in\mathcal{J} \}, \nonumber
\end{align}
where each $\alpha_j$ is the corresponding extended class-$\mathcal{K}$ function with respect to which $h^G_j$ is a CBF. Likewise, we will define:
\begin{equation}
x(t)|_{u(t)} \triangleq x(0) + \int_0^t \left(f(x(s))+g(x(s))u(s)\right) ds,
\end{equation}
to be the solution to equation~\eqref{dyn_sys} provided the input signal, $u(t)$.
Likewise, we will make the following assumption to frame the type of specifications accounted for by the testing framework to be detailed:
\vspace{0.2 cm}
\begin{assumption}
We assume that the provided TL specification can be recast into the following form:
\begin{equation}
\label{eqn::sys_specification}
\left[\lor_{i\in\mathcal{I}} \left( \F \phi_i \right) \right] \wedge \left[\wedge_{j\in\mathcal{J}} \left( \G\omega_j \right) \right],~\phi_i,\omega_j\in\mathcal{A}~\forall~i,j,
\end{equation}
with the following initial conditions:
\begin{subequations}
\begin{align}
& \lor_{i\in\mathcal{I}}(\phi_i(x(0))) & & \hspace{-0.8 in} =\mathrm{FALSE}, \label{eqn::initially_not_F}\\
& \wedge_{j\in\mathcal{J}}(\omega_j(x(0))) & & \hspace{-0.8 in} =\mathrm{TRUE}. \label{eqn::initially_G}
\end{align}
\end{subequations}
\end{assumption}
\vspace{0.2cm}
Intuitively, specifications of type~\eqref{eqn::sys_specification} denote control objectives wherein the system must ensure continued satisfaction of multiple control objectives while accomplishing at least one of a subset of tasks \textit{e.g.} navigating to one of multiple waypoints while avoiding all obstacles. Equations~\eqref{eqn::initially_not_F} and \eqref{eqn::initially_G} indicate that the system does not start in trivial states, wherein the specification~\eqref{eqn::sys_specification} has already been satisfied. Finally, to account for an adversarial testing framework, we specify that the test parameters are a function of the current state, \textit{i.e.} $d(x)$, where the specific functional form is expressed in Theorem~\ref{algorithmic_test_generation}. Intuitively, the idea is that for tests to be adversarial to system action, they must, necessarily, depend on the system state.
Under the notation specified in Remark~\ref{approximate_barrier_labeling}, the main result is as follows:
\vspace{.2cm}
\begin{theorem}
[Algorithmic Test Generation]
\label{algorithmic_test_generation}
\textit{Given an autonomous system and a TL specification of the form in equation~\eqref{eqn::sys_specification},
the solution, $d^*(x)$, to the minimax game:
\begin{align}
d^*(x) = & \,\,\,\, \argmin\limits_{d \in \mathbb{R}^p} & & \hspace{-0.3 in}\max\limits_{u \in \mathcal{U}(x,d)} \, \, \sum\limits_{i \in \mathcal{I}} \dot{h}^F_i(x,u),
\label{differential_game}\quad \quad \quad \tag{Minimax}
\end{align}
defines an optimal test parameter sequence, $d^*(x(t))$, predicated on a state trajectory, $x(t)|_{u(t)}$, and the control signal, $u(t)$, \textit{i.e.,} $d^*(x(t))$ identifies a sequence of test scenarios designed to ensure system satisfaction of the following specification:}
\begin{equation}
\label{eqn::d_specification}
\left[ \wedge_{i\in\mathcal{I}} \left(\G \neg \phi_i \right) \right] \lor \left[ \lor_{j\in\mathcal{J}}
\left( \F \neg \omega_j\right)\right],~\phi_i,\omega_j\in\mathcal{A}~\forall~i,j.
\end{equation}
\end{theorem}
\subsection{Proof of Main Result}
This section contains the lemmas necessary to prove the main result, Theorem~\ref{algorithmic_test_generation}. For all maximization/minimization problems contained within, we specify that infeasibility of the associated optimization problem corresponds to a value of $-\infty,\infty$ respectively.
To start, we need to show that TL specification~\eqref{eqn::d_specification} and TL specification~\eqref{eqn::sys_specification} are mutually exclusive. To that end, we have the following Lemma regarding relations between TL operators:
\vspace{.2cm}
\begin{lemma}
\label{lemma::TL_operator_relations}
\textit{ The following relations are true:
\begin{align}
\neg \G \phi & \equiv \F (\neg \phi), \label{eqn::TL_relation_2}\\
\neg \F \phi & \equiv \G (\neg \phi). \label{eqn::TL_relation_3}
\end{align}
}
\end{lemma}
\vspace{.2cm}
\begin{proof}
For equation~\eqref{eqn::TL_relation_2},
\begin{equation*}
\neg \G \phi \equiv \exists t^*\geq 0~|~x(t^*)\in\llbracket \neg \phi \rrbracket \equiv \F(\neg\phi).
\end{equation*}
Likewise, for equation~\eqref{eqn::TL_relation_3},
\begin{equation*}
\neg\F\phi \equiv \forall~t\geq0~x(t)\in\llbracket\neg\phi\rrbracket \equiv \G(\neg \phi).
\end{equation*}
\end{proof}
Using Lemma~\ref{lemma::TL_operator_relations} and De Morgan's Law, we can prove that the two TL specifications, \eqref{eqn::d_specification} and \eqref{eqn::sys_specification}, are mutually exclusive:
\vspace{.1cm}
\begin{lemma}
\label{lemma::mutual_exclusivity}
\textit{ TL specifications \eqref{eqn::d_specification} and \eqref{eqn::sys_specification} are mutually exclusive.}
\end{lemma}
\vspace{.2cm}
\begin{proof}
\begin{align*}
& \neg \left[ \left[\lor_i \left( \F \phi_i \right) \right] \wedge \left[\wedge_j \left( \G\omega_j \right) \right] \right] \\
\equiv & \neg \left[\lor_i \left( \F \phi_i \right) \right] \lor \neg \left[\wedge_j \left( \G\omega_j \right) \right] \\
\equiv & \left[\wedge_i \neg(\F\phi_i) \right] \lor \left[ \lor_j \neg(\G\omega_j)\right] \\
\equiv & \left[ \wedge_i (\G\neg\phi_i)\right] \lor \left[ \lor_j (\F\neg \omega_j)\right]
\end{align*}
\end{proof}
Effectively, Lemma~\ref{lemma::mutual_exclusivity} proves that if $d^*(x(t))$ ensures system satisfaction of TL specification~\eqref{eqn::d_specification}, then the sequence of test parameters did identify a system failure insofar as the system failed to satisfy the specification~\eqref{eqn::sys_specification}. It remains, however, to show that minimax game~\eqref{differential_game} defines a sequence, $d^*(x(t))$, that forces the system to satisfy~\eqref{eqn::d_specification}. To that end, we have the following Lemma that draws a correspondence between CBFs and TL specifications:
\vspace{.2cm}
\begin{lemma}
\label{equivalence_logic_safety}
\textit{ For an atomic proposition, $\phi \in \mathcal{A}$, if there exists a function, $h_\phi(x)$, such that $\mathcal{C}_{h_\phi} = \llbracket \phi \rrbracket$, then:
\begin{equation*}
\G\phi \equiv h_\phi(x(t)) \geq 0, ~\forall~ t \geq 0,
\end{equation*}
and:
\begin{equation*}
\F\phi \equiv \exists~t^* <\infty~\mathrm{s.t.}~h_\phi(x(t^*)) \geq 0.
\end{equation*}
Furthermore, if $h_\phi(x)$ is a CBF, then $\exists~u(t)$ such that $\G\phi = $TRUE. Likewise, if $h_\phi(x)$ is an FTCBF, then $\exists~u(t)$ such that $\F\phi = $TRUE.}
\end{lemma}
\vspace{.2cm}
\begin{proof}
For $\F\phi$:
\begin{align*}
\F\phi & \quad \equiv \quad \exists~0\leq t^* <\infty~\mathrm{s.t.}~x(t^*) \in \llbracket \phi \rrbracket, \nonumber \\
& \quad \equiv \quad \exists~0\leq t^* <\infty~\mathrm{s.t.}~ x(t^*) \in \mathcal{C}_{h_\phi}, \nonumber \\
& \quad \equiv \quad \exists~0\leq t^* <\infty~\mathrm{s.t.}~ h_\phi(x(t^*)) \geq 0.
\end{align*}
Hence, if $h_\phi(x)$ is an FTCBF wherein $h(x(0)) \leq 0$, then an input sequence, $u(t)$, that satisfies:
\begin{align*}
& L_fh(x(t)) + L_gh(x(t))u(t) + \gamma \sign(h(x(t)))\left| h(x(t))\right|^\rho \geq 0, \\
& \quad \quad \quad \forall~ t \leq T= \frac{1}{\gamma(1-\rho)}|h(x_0)|^{1-\rho}
\end{align*}
ensures $
h(x(T))\geq 0 \implies \F\phi = \mathrm{TRUE}.
$
$\G\phi$ follows similarly.
\end{proof}
Lemma~\ref{equivalence_logic_safety} provides a metric by which to verify that $d^*(x(t))$ ensures system satisfaction of specification~\eqref{eqn::d_specification}. Specifically, Lemma~\ref{equivalence_logic_safety} requires that $d^*(x(t))$ either ensure $h^F_i(x(t)) < 0$ $\forall~i \in \mathcal{I}$ and $\forall~t\geq0$, or $h^G_j(x(t)) < 0$ for at least one $j\in\mathcal{J}$ and $t\geq 0$. To show this, we require the following definitions for the optimal cost, $s$, optimal input, $u^*$, and optimal test parameter, $d^*$:
\begin{align}
& \,\,\, s(x(t),d) & & \hspace{-0.15 in} = & & \hspace{-0.1 in}\max\limits_{u\in\mathcal{U}(x(t),d)} & & \hspace{-0.1 in} \sum\limits_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u), \label{eqn::max_derivative}\\
& u^*(x(t),d) & & \hspace{-0.15 in} = & & \hspace{-0.1 in}\argmax\limits_{u\in\mathcal{U}(x(t),d)} & & \hspace{-0.1 in} \sum\limits_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u), && \label{eqn::u_optimal}\\
& \,\,\,\,d^*(x(t)) & & \hspace{-0.15 in} = & & \hspace{-0.1 in}\,\,\,\,\argmin\limits_{d\in\mathbb{R}^p} & & \hspace{-0.1 in} \sum\limits_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u^*(x(t),d)). && \label{eqn::d_optimal}
\end{align}
Here, we note that equation~\eqref{eqn::d_optimal} is a re-casting of equation~\eqref{differential_game} accounting for the optimal input, $u^*(x(t),d)$. In addition, we will define the following set of invalidating test parameters:
\begin{equation}
\label{eqn::infeasibility_set}
\mathcal{D}(x) = \{d \in \mathbb{R}^p~|~\mathcal{U}(x,d)=\varnothing\}.
\end{equation}
With the above definitions, we have the following Lemma:
\vspace{.2cm}
\begin{lemma}
\label{lemma::ensuring_infeasibility}
\textit{If, for some $x(t)$, $\mathcal{D}(x(t)) \neq \varnothing$, then the optimal solution, $d^*$, to equation~\eqref{differential_game} is such that, $d^*\in\mathcal{D}(x(t))$.}
\end{lemma}
\vspace{.2cm}
\begin{proof}
First, we note that,
\begin{equation}
\label{eqn::infeasibility_yields_infinity}
\forall~d \in \mathcal{D}(x(t)),~s(x(t),d) = -\infty.
\end{equation}
The equation above comes from the infeasibility of maximization problem~\eqref{eqn::max_derivative}, which results in a value of $s = -\infty$. Furthermore, equation~\eqref{differential_game} is equivalent to:
\begin{align}
\label{eqn::recasting_dstar_again}
d^*(x(t)) = & \,\,\,\, \argmin\limits_{d \in \mathbb{R}^p} & & \hspace{-0.6 in} s(x(t),d).
\end{align}
Based on the Locally Lipschitz assumptions made for $f(x)$ and $g(x)$ in equation~\eqref{dyn_sys} and the requirement that a CBF, $h(x)$, is differentiable at least once, it is true that
\begin{equation*}
L_fh^F_i(x),L_gh^F_i(x)~\mathrm{are~bounded}~\forall~i\in\mathcal{I}.
\end{equation*}
In addition,
\begin{equation*}
\forall~u\in\mathcal{U}(x(t),d),~u~\mathrm{is~bounded}.
\end{equation*}
Therefore,
\begin{equation*}
\dot{h}^F_i(x(t),u) = L_fh(x(t)) + L_gh(x(t))u~\mathrm{is~bounded}~\forall~i\in\mathcal{I}.
\end{equation*}
As defined in equations~\eqref{eqn::max_derivative} and \eqref{eqn::u_optimal}, it is also true that
\begin{equation*}
s(x(t),d) = \sum\limits_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u^*(x(t),d)).
\end{equation*}
As each component, $\dot{h}^F_i$, is finite and $|\mathcal{I}|<\infty$, the following is true:
\begin{equation}
\label{eqn::feasibility_yields_finite}
\exists~M < \infty ~\mathrm{s.t.}~\left|s(x(t),d)\right|<M, \quad \forall~d\not\in\mathcal{D}(x(t)).
\end{equation}
By definition of $\argmin$ and using equations~\eqref{eqn::infeasibility_yields_infinity}, \eqref{eqn::recasting_dstar_again}, and~\eqref{eqn::feasibility_yields_finite}, we have that
$
d^*(x(t))\in\mathcal{D}(x(t)).$
\end{proof}
With Lemma~\ref{lemma::ensuring_infeasibility}, we can show that the sequence, $d^*(x(t))$, attempts to force the system to satisfy, $\lor_j (\F\neg \omega_j)$. We will show this first for a single $\G\omega$:
\vspace{.2cm}
\begin{lemma}
\label{lemma::invalidity_Gomega}
\textit{ If, for a given state trajectory, $x(t)$, $\omega(x(0)) = $ TRUE, and $\mathcal{D}(x(t)) \neq \varnothing$ $\forall$ $t\geq0$ with $|\mathcal{J}| = 1$, then:
\begin{equation}
\label{eqn::finite_time_invalidation_omega}
\forall~\delta>0,~\exists~t^*_\delta \in (0,\infty)~\mathrm{s.t.}~h_\omega(x(t^*),d^*d(t^*)) < \delta,
\end{equation}
where $h_\omega$ is the CBF corresponding to $\G\omega$.}
\end{lemma}
\vspace{.2cm}
\begin{proof}
Via Lemma~\ref{lemma::ensuring_infeasibility}, we know that
$
\forall~t\geq 0,~d^*(x(t)) \in \mathcal{D}(x(t)).
$
As $|\mathcal{J}|=1$, this implies that $\forall~t\geq0$,
\begin{equation*}
\label{eqn::no_u_exists}
\dot{h}_\omega(x(t),u,d^*(x(t))) < -\alpha(h_\omega(x(t),d^*(x(t)))),~\forall~u\in\mathcal{U}.
\end{equation*}
As $\alpha(\cdot)\in\mathcal{K}$ (abbreviating $d^*(x(t))$ to $d^*(t)$):
\begin{equation*}
h_\omega(x(t),d^*(t)) < \beta(h_\omega(x(0),d^*(0)),t),
\end{equation*}
where $\beta(\cdot)$ is a class-$\mathcal{KL}$ function. As a result:
\begin{equation*}
\exists~t^*\in(0,\infty)~\mathrm{s.t.}~\beta(h_\omega(x(0),d^*(0)),t^*) \leq \delta,
\end{equation*}
choosing $t_\delta^* = t^*$ completes the proof.
\end{proof}
Lemma~\ref{lemma::invalidity_Gomega} is why we specify that the sequence, $d^*(x(t))$, attempts to force system satisfaction of $\lor_j(\F\neg\omega_j)$ as opposed to specifying that it guarantees that the system will satisfy the same specification. As minimax game~\eqref{differential_game} constrains system action, $u$, to ensure $\wedge_j(\G\omega_j)$, the test sequence can only get $\delta$ close to invalidation assuming optimal system action. This discrepancy will be made clear when compared with Lemma~\ref{lemma::invalidation_Fphi}:
\vspace{.2cm}
\begin{lemma}
\label{lemma::invalidation_Fphi}
\textit{ If $\phi(x(0)) = $ FALSE and $|\mathcal{I}|=1$, then the test parameter sequence, $d^*(x(t))$, is guaranteed to find a system trajectory, $x(t)|_{u^*(x(t),d^*(x(t)))}$, that satisfies $\G\neg\phi$ provided a trajectory exists wherein:
\begin{align}
\dot{h}_\phi(x(t),u^*(x(t),d(t))) & \leq 0,~\forall~t\geq 0, \label{eqn::h_phi_always_decreasing}\\
\mathcal{D}(x(t)) & = \varnothing,~\forall~t\geq0, \label{eqn::no_invalidity_set}
\end{align}
for some $d(t)$.}
\end{lemma}
\vspace{.2cm}
\begin{proof}
First, we denote $h_\phi(x)$ to be the CBF corresponding to $\F \phi$. It follows from Lemma~\ref{equivalence_logic_safety} then,
\begin{equation}
\label{eqn::phi_false_iff_negative}
\phi(x(0)) = \mathrm{FALSE} \equiv h_\phi(x(0)) < 0.
\end{equation}
From equation~\eqref{eqn::phi_false_iff_negative}, to prove $\G\neg\phi$, it is sufficient to prove:
\begin{equation}
\label{eqn::general_h_always_decreasing}
\dot{h}_\phi(x(t),u(t)) \leq 0, \quad \forall~t\geq 0,
\end{equation}
as if true:
\begin{align*}
h_\phi(x(t)) & = h_\phi(x(0)) + \int_0^t \dot{h}_\phi(x(s),u(s)) \mathrm{ds}, \\
& < \int_0^t \dot{h}_\phi(x(s),u(s)) \mathrm{ds} \leq 0, \\
& \implies x(t)|_{u(t)} \in \llbracket \neg \phi \rrbracket,~\forall~t\geq 0~ \equiv \G\neg\phi.
\end{align*}
As a result, all that remains is to show that equation~\eqref{eqn::h_phi_always_decreasing} is satisfied by $d^*(x(t))$. Here, equation~\eqref{eqn::no_invalidity_set} ensures that the results of Lemma~\ref{lemma::ensuring_infeasibility} do not apply, as otherwise $d^*(x(t))\in\mathcal{D}(x(t))$ and we cannot make a statement regarding $s(x(t),d^*)$. Then, by definition of $\argmin$ and equation~\eqref{eqn::d_optimal} (abbreviating $d^*(x(t))$ to $d^*(t)$):
\begin{equation*}
\dot{h}_\phi(x(t),u^*(x(t),d^*(t))) \leq \dot{h}_\phi(x(t),u^*(x(t),d(t))),
\end{equation*}
which results in:
\begin{equation}
\label{eqn::h_decreasing_under_optimality}
\dot{h}_\phi(x(t),u^*(x(t),d^*(t))) \leq 0.
\end{equation}
From equation~\eqref{eqn::h_decreasing_under_optimality} and the sufficiency proof predicated on equation~\eqref{eqn::general_h_always_decreasing}, we have:
\begin{equation*}
x(t)|_{u^*(x(t),d^*(t))} \in \llbracket \neg \phi \rrbracket,~\forall~t\geq 0 \equiv \G\neg\phi.
\end{equation*}
\end{proof}
With the aforementioned lemmas, we are now ready to prove Theorem~\ref{algorithmic_test_generation}.
\begin{proof}
[Theorem~\ref{algorithmic_test_generation}] If both $|\mathcal{I}| = 1$ and $|\mathcal{J}|=1$, then the result stems directly from Lemmas~\ref{lemma::invalidity_Gomega} and \ref{lemma::invalidation_Fphi}. First we note the following is true:
\begin{equation}
\label{eqn::vacuously_true_D}
\left( \mathcal{D}(x(t)) = \varnothing \right) \lor \left( \mathcal{D}(x(t)) \neq \varnothing \right) = \mathrm{TRUE},~\forall~t\geq 0.
\end{equation}
As a result, it is true that $\forall~t \geq 0$, the optimal test parameter sequence, $d^*(x(t))$, attempts to ensure that the following is true:
\begin{align}
& \left(\dot{h}_\phi(x(t),u^*(x(t),d^*(x(t))))\leq 0 \right) \lor \label{eqn::CBF_inequality_forall_time} \\
& \left(\dot{h}_\omega(x(t),u,d^*(x(t))) < -\alpha(h_\omega(x(t),d^*(x(t)))~\forall~u\in\mathcal{U}\right) \nonumber \\
& \quad \quad \quad \forall~t\geq 0.\nonumber
\end{align}
Hence, if either $\mathcal{D}(x(t))=\varnothing$ or $\mathcal{D}(x(t)) \neq \varnothing$ persist $\forall~t\geq 0$, then the the results of Lemmas~\ref{lemma::invalidity_Gomega} and \ref{lemma::invalidation_Fphi} ensure $d^*(x(t))$ attempts to force the system to satisfy, $
\G \neg \phi \lor \F \neg \omega.$
We lose the guarantee on $\G\neg\phi$ that we had in Lemma~\ref{lemma::invalidation_Fphi} as we can no longer ensure $\mathcal{D}(x(t)) = \varnothing$, $\forall~t\geq0.$ However, whenever $\mathcal{D}(x(t)) = \varnothing$, $d^*(x(t))$ will steer the system away from achieving $\F\phi$, if feasible.
This same rationale extends to the case wherein $|\mathcal{I}| \neq 1$ and/or $|\mathcal{J}|\neq1$. Since~\eqref{eqn::vacuously_true_D} holds, for the multi-specification case, $d^*(x(t))$ attempts to ensure:
\begin{align}
& \left(\sum_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u^*(x(t),d^*(x(t))))\leq 0 \right) \lor \label{eqn::multi_CBF_inequality_forall_time} \\
& \left(\dot{h}^G_j(x(t),u,d^*(x(t))) < -\alpha_j(h^G_j(x(t),d^*(x(t)))~\forall~u\in\mathcal{U}\right) \nonumber \\
& \quad \quad \quad \forall~t\geq 0,~\mathrm{and~for~at~least~one~}j, \nonumber
\end{align}
For the first inequality in equation~\eqref{eqn::multi_CBF_inequality_forall_time}, the following implication is true:
\begin{align}
& \sum_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u^*(x(t),d^*(x(t))))\leq 0 \implies \label{eqn::decrease_in_at_least_one_phi}\\
& \lor_{i\in\mathcal{I}}\left( \dot{h}^F_i(x(t),u^*(x(t),d^*(x(t)))) \leq 0\right) = \mathrm{TRUE}. \nonumber
\end{align}
Implication~\eqref{eqn::decrease_in_at_least_one_phi} can be deduced from a contradiction. If, for the same implication, we were to assume the LHS of~\eqref{eqn::decrease_in_at_least_one_phi} to be true and the RHS to be false, then:
\begin{align*}
& \lor_{i\in\mathcal{I}}\left( \dot{h}^F_i(x(t),u^*(x(t),d^*(x(t)))) \leq 0\right) = \mathrm{FALSE} & \implies \\
& \wedge_{i\in\mathcal{I}}\left( \dot{h}^F_i(x(t),u^*(x(t),d^*(x(t)))) > 0\right) = \mathrm{TRUE} & \implies \\
& \sum_{i\in\mathcal{I}}\dot{h}^F_i(x(t),u^*(x(t),d^*(x(t)))) > 0,
\end{align*}
which is a contradiction.
As a result, $d^*(x(t))$ attempting to ensure equation~\eqref{eqn::multi_CBF_inequality_forall_time} is equivalent to saying $d^*(x(t))$ attempts to ensure:
\begin{multline}
\lor_{i\in\mathcal{I}}\left( \dot{h}^F_i(x(t),u^*(x(t),d^*(x(t)))) \leq 0\right) \lor \label{eqn::d_specification_CBF} \\
\lor_{j\in\mathcal{J}}\bigg(\dot{h}^G_j(x(t),u,d^*(x(t))) < -\alpha_j(h^G_j(x(t),d^*(x(t))), \\
~\forall~u\in\mathcal{U}\bigg), \forall~t\geq 0.
\end{multline}
Coupled with the initial conditions~\eqref{eqn::initially_not_F} and \eqref{eqn::initially_G}, $d^*(x(t))$ attempting to ensure equation~\eqref{eqn::d_specification_CBF} is equivalent to saying $d^*(x(t))$ attempts to ensure system satisfaction of equation~\eqref{eqn::d_specification}, which is the desired result.
\end{proof}
\section{Test Synthesis}
\label{sec::corollaries}
This section provides some additions to the main result that make it extensible to the problem at hand. Specifically, we formulate a linear program to extend the results of Theorem~\ref{algorithmic_test_generation} to generate test cases wherein we have no prior knowledge of the controller on-board the system. Likewise, we have a corollary that permits a predictive form of equation~\eqref{differential_game} such as the one used to generate the tests in Section~\ref{sec::simulations_and_experiments}.
To start, we want to use the results of Theorem~\ref{algorithmic_test_generation} to see if the provided autonomous system satisfies the associated TL specification. However, as we do not know the controller onboard the system, we do not have any CBFs with which to define the minimax game in Theorem~\ref{algorithmic_test_generation}. That being said, Lemma~\ref{equivalence_logic_safety} in the Appendix provides us a method by which to determine these CBFs from the system demonstration data, $\mathbb{D}$. First, we define an estimated CBF (e-CBF) to be a convex combination of component functions in $\mathcal{B}$, where $p_j$ below denote the weights for said combination:
\begin{equation}
\label{optimal_combination} \tag{e-CBF}
h^*(x) = \sum_{j=1}^{|\mathcal{B}|} p_jh_j(x), \quad \forall~h_j \in \mathcal{B}.
\end{equation}
By default, Lemma~\ref{equivalence_logic_safety} indicates that specification satisfaction requires the associated CBF to be positive. As a result, we will choose a cost function that is minimized when \eqref{optimal_combination} is most positive over all demonstrations:
\begin{equation}
\label{eqn::cost}
J(\mathcal{B},x,p) \triangleq -\sum_{i=1}^{|\mathbb{D}|} \sum_{k=0}^{T_i} \sum_{j=1}^{|\mathcal{B}|} p_j h_j(x^i_k). \tag{Cost}
\end{equation}
Likewise, Assumption~\ref{Labeling} dictates that demonstrations end upon satisfaction of the control objective. Therefore, for the estimated CBF to correspond to $\F$uture type constraints, \eqref{optimal_combination} should be positive at the end of each demonstration. Similarly, for $\G$lobal type constraints, \eqref{optimal_combination} should be positive over the length of all demonstrations. This results in the following Corollary:
\begin{figure*}[htbp]
\centering
\includegraphics[width = 0.99\textwidth]{Compilation_normalized.pdf}
\caption{Shown above are simulated examples of tests generated by the test generation framework detailed in the paper. In all cases shown, each agent's goal is to cross the map to a location directly opposite to its starting location. Here, the origin is the center of each rectangular region. In all figures, red lines denote agent trajectories and blue lines denote obstacle trajectories/regions if stationary. (Left) Two, stacked figures showing successful robot navigation with multiple, stationary obstacles - this is our simulated demonstration data for the multi-agent case. (Center Left) Two, stacked figures which show stationary obstacle placement based on the test framework, note that, in the simulations shown, multiple crashes occur. (Center Right and Right) Two simulations of a single agent in a moving obstacle case. In both cases, the obstacle trajectory is updated to match the obstacle location generated by minimax game~\eqref{experiment_game}. Note that in both cases, the simulation terminates when obstacle safety has been violated ($h^*_o \leq 0$).}
\label{fig::simulation_crashes}
\end{figure*}
\vspace{.2cm}
\begin{corollary}
\label{composite_CBF_Lemma}
\textit{ For a given data-set, $\mathbb{D}$, and a candidate set of functions, $\mathcal{B}$, the solution, $p^*$, to the following linear program:
\begin{align}
p^* = &\,\,\,\argmin\limits_{p\in\mathbb{R}^{|\mathcal{B}|}} & & J(\mathcal{B},x,p), && \label{CBF-LP} \tag{CBF-LP} \\
& \subjectto & & \eqref{future-constraint}~\mathrm{or}~\eqref{global-constraint}, \nonumber && \\
& & & p_j \geq 0, \quad \forall~j=1,2,\dots,|\mathcal{B}|, && \nonumber\\
& & & \sum_{j=1}^{|\mathcal{B}|} p_j = 1, && \nonumber
\end{align}
\begin{subequations}
\begin{align}
\label{future-constraint}
&\hspace{-0.3in}\sum_{j=1}^{|\mathcal{B}|} p_j h_j(x^i_{T_i}) & & \hspace{-0.3in}\geq 0,~\forall~i=1,2,\dots,r, \\
\label{global-constraint}
&\hspace{-0.3in}\sum_{j=1}^{|\mathcal{B}|} p_j h_j(x^i_k) & & \hspace{-0.3in} \geq 0,~ \forall~i=1,\dots,r,~k=0,1,\dots,T_i,
\end{align}
\end{subequations}
determines an estimated CBF, \eqref{optimal_combination}, for specifications of type $\F\phi$ (constraint~\eqref{future-constraint}) or type $\G\phi$ (constraint~\eqref{global-constraint}).
Furthermore, for $\F\phi$:
\begin{equation}
\mathcal{C}_{h^*} \cap \llbracket \phi \rrbracket \supseteq \{x^i_{T_i}\}~\forall~i=1,2,\dots,r,
\end{equation}
and for $\G\phi$:
\begin{equation}
\mathcal{C}_{h^*} \cap \llbracket \phi \rrbracket \supseteq \{x^i_k\}~\forall~k = 0,1,\dots,T_i~\mathrm{and}~i=1,\dots,r.
\end{equation}
}
\end{corollary}
\vspace{.2cm}
\begin{proof}
Assumption~\ref{Labeling} requires that $\phi =$ TRUE at each $T_i$ for $\mathbb{D}_i$. If a solution to equation~\eqref{CBF-LP} exists with constraint \eqref{future-constraint}, then we have:
\begin{align*}
x^i_{T_i} & \in \llbracket \phi \rrbracket, \\
h^*(x^i_{T_i}) & = \sum_{j=1}^{|\mathcal{B}|} p_j^*h_j(x^i_{T_i}) \geq 0,
\end{align*}
for any solution, $p^*$.
As a result,
\begin{align*}
\mathcal{C}_{h^*} \cap \llbracket \phi \rrbracket \supseteq \{x^i_{T_i}\} \quad \forall~i=1,2,\dots,r,
\end{align*}
which only implies set equivalence up to the provided data. As a result, Lemma~\ref{equivalence_logic_safety} only applies over the provided data-set where the equivalence holds. To show the same for $\G$lobal type specifications, replace constraint \eqref{future-constraint} with \eqref{global-constraint} and the proof follows similarly.
\end{proof}
\vspace{0.2 cm}
As the CBFs generated via Corollary~\ref{composite_CBF_Lemma} are not exact, the results of Theorem~\ref{algorithmic_test_generation} cannot be guaranteed. However, they are very useful in generating tests as will be shown in Section~\ref{sec::simulations_and_experiments}.
Secondly, the minimax game in Theorem~\ref{algorithmic_test_generation} may be non-convex and/or calculation of the solution may be computationally difficult. However, as the minimax game depends only on the current state, we can calculate the optimal test parameters for some subset of points and define the actual test to be an interpolation of the parameters defined at these points. That being said, this yields sub-optimal tests.
Finally, the minimax problem~\eqref{differential_game} need not have that specific cost function for it to determine optimal test parameters. It suffices if the chosen cost function for the inner maximization problem is maximized when the estimated CBFs $h^F_i(x) \geq 0$. This permits predictive games of the form used in the simulations in Section~\ref{sec::simulations_and_experiments}.
\section{Case Study}
\label{sec::simulations_and_experiments}
In this section, we detail simulations of test scenarios devised by our framework applied to the Georgia Tech Robotarium \cite{Robotarium}. The set up is defined next.
\newsec{System Specification} $\mathbf{F}g_i \wedge_j \mathbf{G}\neg a_j$. The system will always ensure that agent $i$ ends up at goal location $i$ while avoiding all the other agents $j$, provided that $\llbracket g_i \rrbracket \not \subseteq \cup_j \llbracket a_j \rrbracket$.
For this specification, we have the following information:
\begin{itemize}
\item $\mathbb{D}$: A set of twenty demonstrations ($r=20$) derived from simulations wherein a single agent successfully navigates to a predetermined goal while avoiding another obstacle.
\item $\mathcal{B}_F$: A set of norm-based barrier functions of the form $h(x) = -\|x - c\| + r$ wherein $h(x) \geq 0 \equiv \|x-c\| \leq r$.
\item $\mathcal{B}_G$: A set of norm-based barrier functions of the form $h(x) = \|x - c\| - r$ wherein $h(x) \geq 0 \equiv \|x-c\| \geq r$.
\end{itemize}
Equation~\eqref{CBF-LP} identified:
\begin{align*}
h^*_g(x) &= - \|x-g\| + 0.02, \\
h^*_o(x,d) &= \|x-d\| - 0.175,
\end{align*}
as the estimated CBF's for $\F g_i$ and $\G \neg a_j$ respectively. Here, $d$ is the desired location of the obstacle agent, and the testing parameter we control. We estimated that the robots in the Robotarium can be sufficiently modeled with single-integrator systems and developed the following game from the derived barrier functions:
\begin{align}
d^* = & \, \, \, \, \argmin\limits_{d \in \mathbb{R}^2} & \max\limits_{u \in \mathcal{U}} & \, \, \sum_{i=1}^{N}h^*_g(x_i),
\label{experiment_game} \tag{Simulation Game} \\
& \subjectto & & \, \, \dot{h}^*_o(x_{i-1},u_i,d) \geq -\beta h^*_o(x_{i-1},d), \nonumber \\
& & & \, \, x_i = x_{i-1} + u_i\Delta t, \nonumber\\
& & & \,\, \dots\forall~i=1,2,\dots,N, \nonumber \\
& & & \, \, \|d-x_0\| \geq r_o. \nonumber
\end{align}
In equation~\eqref{experiment_game} above, $N=2$, $\beta = 100$, and $r_o = 0.175$. For large values of $\beta$, there is less of an implied assumption about system behavior as it decays to the boundary of the estimated safe region. As a result, large $\beta$ values permit equation~\eqref{experiment_game} to account for a wider range of system behavior when solving for $d^*$. In addition, $r_o$ constrains against trivial solutions wherein $d=x_0$, which makes the inner maximization problem infeasible.
To quantify how "hard" a test/demonstration is, we define:
\begin{itemize}
\item $H^i_g \triangleq \frac{1}{T_i+1} \sum_{k=0}^{T_i} \left| \hat{h}_g(x_k)\right|$ to be the average time the system spent outside the goal. Here, $\hat{h}_g$ denotes a normalized version of our estimated CBF, $h^*_g$, such that $-1 \leq \hat{h}_g(x_k) \leq 0$, $\forall$ $k = 0,1,\dots T_i$, and $T_i$ is the max time for our Demonstrations/tests as defined in Definition~\ref{demonstrations}.
\item $H^i_o \triangleq 1 - \frac{1}{T_i+1} \sum_{k=0}^{T_i} \hat{h}_o(x_k)$ to be the average time spent collision free. Here, $\hat{h}_o$ denotes a normalized version of our estimated CBF, $h^*_o$ such that $0 \leq \hat{h}_o(x_k) \leq 1$, $\forall$ $k = 0,1,\dots,T_i$.
\end{itemize}
To note, tests drive $H^i_o \to 1$ in an effort to drive $H^i_g \to 1$ which denotes system safety failure and inability to reach the goal, respectively.
Figures~\ref{fig::simulation_crashes} and \ref{fig::Metrized Data} show the results of simulations based on the obstacle locations outputted by minimax game~\eqref{experiment_game}. For the multi-agent case, examples of the provided demonstration data are shown in the two, stacked figures to the far left. Under normal operating parameters, the agent successfully avoid the obstacles while moving to their repsective goals (none of the red lines interset the blue circles). However, when the stationary obstacle locations are updated based on solutions to \eqref{experiment_game}, multiple crashes occur as shown in the two, stacked figures just left of center.
Likewise, for the single-agent simulations shown, we inputted the desired obstacle location, $d$, as the goal location for a secondary agent. This agent acted as a moving obstacle, and for $2/20$ tests simulated, the trajectories for both agents are shown in the four figures on the right of Figure~\ref{fig::simulation_crashes}. In both of these cases, notice how the estimated CBF, $h^*_o$, decays to $0$ upon termination. Effectively, in both of these cases, the test framework chose a sequence of obstacle locations, $d^*(x(t))$, that forced the system to satisfy $\G a_j$, at least with respect to the estimated CBF, $h^*_o$.
Data for all $20$ single-agent simulations are compared against the provided data-set, $\mathbb{D}$, in Figure~\ref{fig::Metrized Data}. Under normal operation, the demonstration data is relatively consistent \textit{i.e.} $H^i_g$ hovers just below $0.4$ and $H^i_o$ hovers just around $0.7$ for all demonstrations, $i=1,2,\dots,20$. However, for all test simulations, $H^i_g > 0.4$ and $H^i_o > 0.8$ further corroborating that the test parameter sequence generates difficult tests, and in $7/20$ cases wherein $H^i_o=1$, also forced the system to satisfy specification~\eqref{eqn::d_specification}. An example of an experimental demonstration of the test framework can be seen in an accompanying video (linked here: \cite{video}). The setup here mimics the same single-agent case shown in the examples in Figure~\ref{fig::simulation_crashes}.
\begin{figure}[t]
\centering
\includegraphics[width = 0.45\textwidth]{Simulation_Comparison.pdf}
\caption{The figures compares the specification satisfaction performance of the provided demonstration data (left) with the data generated during simulated tests (right) based on the metric defined in Section V. Note that for all test cases, the agent took longer to reach its goal, and in $7/20$ cases, the test caused the robots to crash $H^i_o = 1$.}
\label{fig::Metrized Data}
\end{figure}
\section{Conclusion}
In this paper, we attempt to solve the problem of test and evaluation for verification and validation of autonomous systems, wherein the specific controllers are unknown. The goal in doing so, is to provide a mathematical framework designed to root out system inefficiencies in an effort to ensure confidence in those systems that pass the procedure. The method detailed involves estimation of approximate control barrier functions to frame a minimax game that is guaranteed to choose test parameters to frustrate system satisfaction of a provided temporal logic specification. In the future, we aim to extend this work to richer specification classes and formalize an iterative testing procedure based on our framework.
\bibliographystyle{ieeetr}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,859 |
\section{Introduction}
The performance of downlink wireless communication systems can be improved by multi-antenna techniques, which enable efficient utilization of spatial dimensions. Depending on the available channel state information (CSI), these dimensions can be used for enhanced reliability and/or spatial multiplexing of multiple data streams with controlled interference \cite{Gesbert2007a}.
The downlink single-cell sum capacity (with perfect CSI) behaves as
\begin{equation} \label{eq_sumcapacity}
\min(N,MK) \log_2(P) + \mathcal{O}(1)
\end{equation}
where $N$ is the number of base station antennas, $K$ is the number of users, each user has $M \geq 1$ antennas, and $P$ is the signal-to-noise ratio (SNR) defined as the
total transmit power divided by the noise power.
The number of users is assumed to be large such that $K\geq N$, thus we have $MK \geq N$ and the maximal \emph{multiplexing gain} becomes $\min(N,MK)=N$.
The multiplexing gain will have a major impact on the throughput of future cellular networks, where high SNRs can be achieved in an energy-efficient way by large-scale antenna arrays \cite{Rusek2013a} and/or increased cell density \cite{Hoydis2011c}.
The sum capacity in \eqref{eq_sumcapacity} is theoretically achieved by dirty-paper coding \cite{Weingarten2006a}, but this non-linear scheme has impractical complexity and is very sensitive
to CSI imperfections. Fortunately, the maximal multiplexing gain of $N$ can be achieved by linear \emph{spatial division multiple access} (SDMA) strategies \cite{Richard1996a}, such as \emph{block-diagonalization (BD)} \cite{Spencer2004a,Ravindran2008a} and \emph{zero-forcing with combining (ZFC)} \cite{Jindal2008a,Trivellato2008a}. Such SDMA strategies transmit $N$ simultaneous data streams, but can divide them among the users in different ways; the system can select between $\lceil \frac{N}{M} \rceil$ and $N$ users to be active and allocate from $1$ to $M$ streams to each of them. This raises a fundamental design question: \emph{how should the receive antennas at each user be used to maximize the system throughput?}
Inter-user interference degrades user performance, while the mutual interference between users' own streams can be handled by receive processing.
It thus seems beneficial to only have a few active users and multiplex many streams to each of them. However, one should keep in mind that every additional stream allocated to a user
experiences a weaker channel gain than the previous streams.
If fewer than $M$ streams are allocated to a user, this user has degrees of freedom for interference-aware receive combining to achieve a strong effective channel and better spatial co-user compatibility. In other words, it is not clear whether receive antennas should be utilized for \emph{multi-stream multiplexing} or \emph{receive combining}, or perhaps something intermediate.
The answer has a profound impact on wireless system design, including the CSI acquisition protocols, scheduling algorithms, and receiver architecture.
\subsection{Related Work}
The sum-rate maximization problem is nonconvex and combinatorial \cite{Luo2008a}, thus only suboptimal strategies are feasible in practice. Such low-complexity algorithms have been proposed in \cite{Tolli2005a,Boccardi2007a,Chen2008a,Guthy2010a}, among others, by successively allocating data streams to users in a greedy manner. Simulations have indicated that fewer than $N$ streams should be used when $P$ and $K$ are small, and that spatial correlation makes it beneficial to divide the streams among many users.
Simulations in \cite{Boccardi2007a} indicates that the probability of allocating more than one stream per user is small when $K$ grows large, but \cite{Boccardi2007a} only considers users with homogeneous channel conditions and all the aforementioned papers assume perfect CSI.
The authors of \cite{Trivellato2008a} claim that transmitting at most one stream per user is desirable when there are many users in the system. They justify this statement by using asymptotic results from \cite{Bayesteh2008a} where $K\rightarrow \infty$. This argumentation ignores some important issues: 1) asymptotic optimality can also be proven with multiple streams per user;\footnote{The uplink analysis in \cite{Rhee2004a} shows that a non-zero (but bounded) number of users can use multiple streams, and the well-established uplink-downlink duality makes this result applicable also in our downlink scenario.} 2) the performance at practical values on $K$ is unknown; and 3) the analysis implies an unbounded asymptotic multi-user diversity gain, which is a modeling artifact of fading channels \cite{Dohler2011a}.
The authors of \cite{Jindal2008a,Ravindran2008a} arrive at a different conclusion when they compare BD (which selects $\frac{N}{M}$ users and sends $M$ streams/user) and ZFC (which selects $N$ users and sends one stream/user) under quantized CSI. Their simulations reveal a distinct advantage of BD (i.e., multi-stream multiplexing), but are limited to uncorrelated channels and neither include user selection nor interference rejection. We show that their results are misleading, because single-user transmission greatly outperforms both BD and ZFC in the scenario that they simulate.
Despite the similar terminology, our problem is fundamentally different from the classic works on the diversity-spatial multiplexing tradeoff (DMT) in \cite{Zheng2003a,Lozano2010a}. The DMT brings insight on how many streams should be transmitted in the high-SNR regime, while we consider how a fixed number of streams should be divided among the users.
\subsection{Main Contributions}
This paper provides a comprehensive answer to how multi-antenna users should utilize their antennas in downlink transmissions, or similarly how many data streams that should be allocated per active user under different system conditions; see Fig.~\ref{figure_combining_vs_multiplexing}. The main contributions are:
\begin{itemize}
\item New analytic results for analyzing the problem under spatial correlation, user selection, heterogeneous user channel conditions, and realistic CSI acquisition.
These enable asymptotic comparison of the two extremes: allocating $M$ streams per active user (called BD) and one stream per active user (called ZFC). We show that ZFC is more resilient to spatial correlation and well adapted to find near-orthogonal users, while BD is better at utilizing heterogeneous user conditions. Imperfect CSI acquisition is shown to have a similar impact on both strategies.
\item Numerical examples show that allocating one stream per active user is essentially optimal under realistic system conditions, and we explain how other conclusions may arise. The main conclusion is that utilizing receive combining is preferable over multi-stream multiplexing.
\end{itemize}
\begin{figure}
\begin{center}
\subfigure[1 stream per 2-antenna user: ZFC enables receive combining.]
{\label{figure_combining}\includegraphics[width=86mm]{figure1a}}\hfill
\subfigure[2 streams per 2-antenna user: BD exploits multi-stream multiplexing.]
{\label{figure_multiplexing}\includegraphics[width=86mm]{figure1b}} \vskip-2mm
\caption{Two ways of dividing four data streams among multi-antenna users, which also represents two ways of utilizing the receive antennas to reduce interference.
(a) Receive one stream per user and linearly combine the antenna to achieve an effective channel that rejects interference. (b) Receive multiple streams and handle their mutual interference through receive processing.}\label{figure_combining_vs_multiplexing}
\end{center} \vskip-5mm
\end{figure}
\section{System Model}
We consider a downlink multi-user MIMO system where a single base station with $N$ antennas communicates with $K \geq N$ users. Each user has $M$ antennas. For analytical convenience\footnote{The case $M \geq N$ is analytically different because 1) Single-user transmission achieves the full multiplexing gain; 2) CSI acquisition is simplified since $\vect{H}_k$ has full row rank, thus any effective channel $\vect{C}_k^H \vect{H}_k$ can be achieved by selecting the receive combining $\vect{C}_k$ properly. Since user devices are size-constrained, the case $M < N$ is also reasonable in practice.} we assume that $M < N$ and often also that $\frac{N}{M}$ is an integer, but the precoding strategies considered herein can be applied for any $M$. The narrowband, flat-fading channel to user $k$ is represented in the complex-baseband by $\vect{H}_k \in \mathbb{C}^{M \times N}$. The received signal at this user is
\begin{equation} \label{eq_system_model}
\vect{y}_k = \vect{H}_k \vect{x} + \vect{n}_k
\end{equation}
where $\vect{x} \in \mathbb{C}^{N \times 1}$ is the joint transmitted signal for all users and $\vect{n}_k \sim \mathcal{CN}(\vect{0},\vect{I}_{M})$ is the (normalized) circularly-symmetric complex Gaussian noise vector. For analytic convenience, and motivated by measurements \cite{Chizhik2003a,Yu2004b}, we employ the Kronecker model with $\vect{H}_k = \vect{R}_{R,k}^{1/2} \widetilde{\vect{H}}_k \vect{R}_{T,k}^{1/2}$, where $\vect{R}_{T,k}$ and $\vect{R}_{R,k}$ are the positive-definite spatial correlation matrices at the transmitter and receiver side, respectively, and $\widetilde{\vect{H}}_k$ has independent $\mathcal{CN}(0,1)$-entries. We assume $\vect{R}_{T,k} = \vect{I}_N$ (i.e., large antenna separation at the base station) throughout the analysis, because transmit correlation
both creates complicated mathematical structures and requires limiting assumptions on the user distribution geometry and fading environment.
Observe that $\vect{R}_{R,k}$ generally is different for each users, describing different spatial properties.
\subsection{Cyclic System Operation}
\begin{figure}[t!]
\begin{center}
\subfigure[]
{\label{figure_fdd_operation}\includegraphics[width=86mm]{figure2a}}\hfill
\subfigure[]
{\label{figure_tdd_operation}\includegraphics[width=86mm]{figure2b}} \vskip-2mm
\caption{Basic block-fading system operation of (a) FDD systems; and (b) TDD systems. The system operation is repeated in a cyclic manner.}\label{figure_fdd_tdd}
\end{center} \vskip-5mm
\end{figure}
We assume block fading where $\vect{H}_k$ is static for a set of channel uses, called the \emph{coherence time}, and then updated independently. We consider both frequency division duplex (FDD) and time division duplex (TDD); baselines of the respective cyclic system operations are illustrated in Fig.~\ref{figure_fdd_tdd}.
In FDD systems, the users acquire CSI through training signaling \cite{Bjornson2010a} and some users feed back quantized CSI. The base station then performs resource allocation (i.e., data stream allocation and precoding) and informs the scheduled users of their precoding through a second training stage. Data transmission follows until the end of the coherence time, when the cycle in Fig.~\ref{figure_fdd_operation} restarts.
In TDD systems, the system toggles between uplink and downlink transmission on the same channel, thus enabling training signaling in both directions. We assume perfect channel reciprocity\footnote{The physical channel is always reciprocal, but different transceiver hardware is typically used in the downlink and the uplink. Thus, careful calibration is necessary to utilize the reciprocity in practice.} and that the coherence time makes CSI obtained in one block of Fig.~\ref{figure_tdd_operation} correct until the same block occurs in the next cycle. The base station does resource allocation for both uplink and downlink, and it informs the users through training signaling.
We assume that all training signals sent in the downlink direction provide the users with perfect CSI, while CSI feedback (in FDD) and uplink training (in TDD) might lead to imperfect CSI at the base station. This assumption enables coherent reception, thus making the conventional achievable sum rate expression a reasonable performance measure.\footnote{Many of the results herein can be extended to include imperfect CSI at the users in the resource allocation, followed by a second training stage that provides scheduled users with sufficiently accurate CSI of the precoded channels to enable coherent reception. See \cite{Caire2010a} for an example in FDD systems. The loss of having imperfect CSI also in the second training stage can be characterized as in \cite{Yoo2006b}.}
\subsection{Linear Precoding: General Problem Formulation}
\label{subsection_general_problem_formulation}
We consider linear precoding and the transmitted signal is
\begin{equation}
\vect{x} = \sum_{k=1}^{K} \vect{W}_k \vect{d}_k
\end{equation}
where $\vect{W}_k \in \mathbb{C}^{N \times d_k}$ is the precoding matrix, $\vect{d}_k \sim \mathcal{CN}(\vect{0},\vect{I}_{d_k})$ is the data signal, and $d_k$ is the number of multiplexed data streams to user $k$. Each user applies a semi-unitary receive combining matrix $\vect{C}_k \in \mathbb{C}^{M \times d_k}$ (i.e., $\vect{C}_k^H \vect{C}_k = \vect{I}_{d_k}$) and treats inter-user interference as Gaussian noise. The achievable information rate is
\begin{equation} \label{eq_data_rate_general}
g_k(\{\vect{W}_{\ell}\},\vect{C}_{k}) = \log_2 \! \frac{\det \!\Big( \vect{I}_{d_k} \!+\! \fracSumtwo{\ell=1}{K} \vect{C}_k^H \vect{H}_k \vect{W}_{\ell} \vect{W}_{\ell}^H \vect{H}_k^H \vect{C}_k \Big) \!}{ \det \!\Big( \vect{I}_{d_k} \!+\! \fracSum{\ell \neq k} \vect{C}_k^H \vect{H}_k \vect{W}_{\ell} \vect{W}_{\ell}^H \vect{H}_k^H \vect{C}_k \Big) \! }
\end{equation}
where $\{\vect{W}_{\ell}\}$ denotes the set of precoding matrices and $\ell$ is an arbitrary user index \cite{Guthy2010a}. The transmission is limited by an average power/SNR constraint of $P$, thus
\begin{equation}
\mathbb{E}\{ \vect{x}^H \vect{x} \} = \sum_{k=1}^{K} \mathrm{tr}( \vect{W}_k \vect{W}_k^H ) \leq P.
\end{equation}
Ideally, we would like to select $\vect{W}_{k},\vect{C}_{k},d_k \, \forall k$ to maximize the sum rate; that is,
\begin{equation} \label{eq_sum_rate_early}
\begin{split}
\maximize{\{\vect{W}_{k},\vect{C}_{k},d_k \}} \,\, & \,\, \sum_{k=1}^{K} g_k(\{\vect{W}_{\ell}\},\vect{C}_{k}) \\
\mathrm{subject}\,\,\mathrm{to}\,\, & \, \sum_{k=1}^{K} \mathrm{tr}( \vect{W}_k \vect{W}_k^H ) \leq P, \\
& \, \vect{C}_k^H \vect{C}_k = \vect{I}_{d_k}, \quad \quad d_k\geq 0 \quad \forall k.
\end{split}
\end{equation}
Unfortunately, this resource allocation problem is NP-hard and therefore not practically solvable \cite{Luo2008a}.
There are algorithms that find local optima of \eqref{eq_sum_rate_early} (see \cite{Shi2011} and references therein), but these are iterative and thus cannot be implemented under the cyclic system operation in Fig.~\ref{figure_fdd_tdd}.
We limit the selection of $\{\vect{W}_{k},\vect{C}_{k},d_k \}$ to achieve a tractable problem formulation.
\begin{enumerate}
\item \textbf{Precoding:} Zero or minimal inter-user interference should be caused, which is possible when $\sum_{k=1}^{K} d_k \leq N$. This makes \eqref{eq_sum_rate_early} partially feasible, because it becomes a convex problem for any fixed $\vect{C}_{k},d_k$. This is a non-limiting assumption at high SNR \cite{Yoo2006a}, which is the regime where systems with high spectral efficiencies need to operate (e.g., using high power, small cells, or large antenna arrays \cite{Hoydis2011c,Rusek2013a}).
\item \textbf{Receive combining:} The matrix $\vect{C}_k$ is fixed at some value $\widetilde{\vect{C}}_k$ beforehand. This makes sense from a CSI acquisition perspective as only the effective channel $\widetilde{\vect{C}}_k^H \vect{H}_k$ needs to be obtained through feedback (in FDD) or training signaling (in TDD). The value $\widetilde{\vect{C}}_k$ might be the $d_k$ strongest (left) singular vectors of $\vect{H}_k$, known as maximum ratio combining (MRC), but can also be selected to improve the CSI feedback accuracy \cite{Jindal2008a,Trivellato2008a}.
\item \textbf{Stream allocation:} Users are scheduled sequentially using some predefined scheduling policy. This avoids making an exhaustive search over all data stream allocations, which is practically infeasible when $N$ and $K$ grow large. Greedy scheduling algorithms can perform remarkably close to optimum \cite{Tolli2005a,Boccardi2007a,Chen2008a,Guthy2010a}, while random selection ensures user fairness.
\end{enumerate}
We now have a simplified resource allocation problem,
\begin{equation} \label{eq_practical_resource_allocation}
\begin{split}
\maximize{\{\vect{W}_{k}\}} \,\, & \, \sum_{k \in \mathcal{S}} \log_2 \det ( \vect{I}_{d_k} +\widetilde{\vect{C}}_k^H \vect{H}_k \vect{W}_{k} \vect{W}_{k}^H \vect{H}_k^H \widetilde{\vect{C}}_k ) \\
\mathrm{subject}\,\,\mathrm{to}\,\, & \, \sum_{k \in \mathcal{S}} \mathrm{tr}( \vect{W}_k \vect{W}_k^H ) \leq P, \\
& \, \widetilde{\vect{C}}_k^H \vect{H}_k \vect{W}_{\ell} = \vect{0}_{d_k \times d_{\ell}} \quad \forall k \in \mathcal{S}, \, \forall \ell \in \mathcal{S} \!\setminus \! \{ k \},
\end{split}
\end{equation}
where $\mathcal{S}$ is the scheduling set given by the predefined scheduling rule and $d_k>0$ for $k \in \mathcal{S}$ is the corresponding data stream allocation.
\begin{remark}[Updating the Receive Combiner] \label{remark:receive-combiner}
When \eqref{eq_practical_resource_allocation} has been solved, the users are informed of the resource allocation through training signaling. This enables estimation of both the precoded channel $\vect{H}_k \vect{W}_{k}$ and the second-order interference term $\boldsymbol{\mathcal{I}}_k=\sum_{\ell \neq k} \vect{H}_k \vect{W}_{\ell} \vect{W}_{\ell}^H \vect{H}_k^H$, both being necessary for coherent reception.
As a nice by-product \cite{Trivellato2008a}, this enables user $k$ to replace $\widetilde{\vect{C}}_k$ with the rate-maximizing MMSE receive combiner $\vect{C}_k^{\text{MMSE}}$ containing the $d_k$ dominating left singular vectors of $( \vect{I}_{M} + \boldsymbol{\mathcal{I}}_k )^{-1} \vect{H}_k \vect{W}_{k}$ \cite{Guthy2010a}.
This improves the information rate by balancing between signal gain and interference rejection. We consider $\widetilde{\vect{C}}_k$ in the analysis, while $\vect{C}_k^{\text{MMSE}}$ is used in simulations.
\end{remark}
\subsection{Linear Precoding: BD and ZFC}
\label{subsection_definition_BD_ZF}
In this paper, we primarily analyze and compare two instances of \eqref{eq_practical_resource_allocation}: \emph{block-diagonalization (BD)} \cite{Spencer2004a} and \emph{zero-forcing with combining (ZFC)} \cite{Jindal2008a,Trivellato2008a}. These strategies allocate a fixed number of streams per scheduled user, but can be combined with any scheduling policy. There are alternative strategies that allocate different numbers of streams to different users \cite{Boccardi2007a}, but simulations will show that these are \emph{not} increasing the performance when the CSI acquisition overhead is treated properly.
\begin{definition} \emph{(Block-Diagonalization Precoding)} \label{definition_block_diagonalization}
Let $\mathcal{S}^{\text{BD}}$ be a scheduling set with at most $\frac{N}{M}$ users. For each user $k \in \mathcal{S}^{\text{BD}}$, we set $d_k=M$ and $\vect{W}_k=\vect{W}_k^{\text{BD}} \vect{\Upsilon}_k^{1/2}$, where
$\vect{W}_k^{\text{BD}}$ is a semi-unitary matrix that satisfies $\vect{W}_k^{\text{BD},H} \vect{W}_k^{\text{BD}}= \vect{I}_{M}$ and $\vect{H}_{\ell} \vect{W}_k^{\text{BD}} = \vect{0}$ for all $\ell \in \mathcal{S}^{\text{BD}} \! \setminus \!\{k\}$. The power allocation is given by the diagonal matrix $\vect{\Upsilon}_k \succeq \vect{0}_M$.
The information rate is
\begin{equation} \label{eq_data_rate_BD}
g^{\text{BD}}_k(P) = \log_2 \det \left( \vect{I}_{M} + \vect{H}_k \vect{W}_k^{\text{BD}} \vect{\Upsilon}_k \vect{W}_k^{\text{BD},H} \vect{H}_k^H \right).
\end{equation}
\end{definition}
\begin{definition} \emph{(Zero-Forcing Precoding with Combining)} \label{definition_zero-forcing}
Each user combines its antennas using some channel-dependent unit-norm vector $\tilde{\vect{c}}_k \in \mathbb{C}^{M \times 1}$.
Based on the effective channels $\vect{h}_k^H =\tilde{\vect{c}}_k^H \vect{H}_k \in \mathbb{C}^{1 \times N}$, a scheduling set $\mathcal{S}^{\text{ZFC}}$ with at most $N$ users is selected.
For each user $k \in \mathcal{S}^{\text{ZFC}}$, we set $d_k=1$ and let $\vect{W}_k=\sqrt{p_k} \vect{w}^{\text{ZFC}}_{k}$, where $\vect{w}^{\text{ZFC}}_{k}$ is a unit-norm vector that satisfies $\vect{h}_{\ell}^H \vect{w}^{\text{ZFC}}_{k} = 0$ for all $\ell \in \mathcal{S}^{\text{ZFC}} \!\setminus\!\{k\}$. The power $p_k\geq 0$ is allocated to user $k$ and the information rate is
\begin{equation} \label{eq_data_rate_ZF}
g^{\text{ZFC}}_k(P) = \log_2 \left( 1 + p_k | \vect{h}_k^H \vect{w}^{\text{ZFC}}_{k}|^2 \right).
\end{equation}
\end{definition}
The sum-rate maximizing power allocations for BD and ZFC are achieved through water-filling (see \cite{Spencer2004a}), but the asymptotic analysis in this paper often assumes equal power allocation (i.e., $\vect{\Upsilon}_k = \frac{P}{M |\mathcal{S}^{\text{BD}}|} \vect{I}_M \,\, \forall k \in \mathcal{S}^{\text{BD}}$ and $p_k=\frac{P}{|\mathcal{S}^{\text{ZFC}}|} \,\, \forall k \in \mathcal{S}^{\text{ZFC}}$) since this becomes optimal in the high-SNR regime where $P \rightarrow \infty$ \cite{Lee2007a}. Although the definitions of BD and ZFC assume perfect CSI, both strategies can be applied when the transmitter has imperfect CSI by making $\vect{W}_k$ orthogonal to the acquired co-user channels \cite{Jindal2008a,Trivellato2008a,Ravindran2008a}. The resulting loss will be quantified in later sections.
ZFC can schedule up to $N$ users and sends one data stream per user, while BD can only schedule $\frac{N}{M}$ users but multiplexes $M$ streams to each of them. Although BD and ZFC are identical when each user only has one antenna, this does \emph{not} mean that BD is a generalization of ZFC. In fact, there are good reasons for applying ZFC instead of BD when $M>1$:
\begin{enumerate}
\item The base station only needs to acquire the effective channels $\vect{h}_k$;
\item The effective channel $\vect{h}_k$ has better properties than $\vect{H}_k$ and can be adapted for interference rejection;
\item User devices require simpler hardware that only decodes one stream.
\end{enumerate}
The interference mitigation is, on the other hand, less restrictive under BD since fewer users are involved and the mutual interference between streams sent to the same user is handled by receive processing \cite{Ravindran2008a}. By analyzing and comparing ZFC and BD under both perfect and imperfect CSI, we try to answer the fundamental question: should we select many multi-antenna users to enable receive combining or select few users and exploit multi-stream multiplexing?
\begin{remark}[Ambiguous Terminology]
The terminology \emph{block-diagonalization} and \emph{zero-forcing} have been given different meanings in prior works.
Herein, BD refers to the original work in \cite{Spencer2004a}, where each active user receives exactly $M$ data streams. Apart from the ZFC strategy in Definition \ref{definition_zero-forcing} (and in \cite{Jindal2008a,Trivellato2008a}), another downlink zero-forcing strategy for multi-antenna users was proposed in \cite{Yoo2006a}. In their definition, each antenna at the multi-antenna users is viewed as a separate virtual single-antenna user and the zero-forcing idea is applied to send a separate stream to each antenna
with zero inter-antenna interference. That approach is nothing else than BD with stricter interference mitigation and can \emph{never} perform better than BD.
Herein, ZFC means sending one stream per user and utilizing receive combining, thus ZFC is \emph{not} a special case of BD and can hypothetically outperform BD.
\end{remark}
\section{Comparison of BD and ZFC with Perfect CSI}
\label{section_comp_BD_ZFC_perfect_CSI}
In this section, we will compare BD and ZFC in the ideal scenario when both the base station and the users have perfect CSI. We derive analytic results indicating the impact of different system properties. Under perfect CSI, the achievable sum rate in \eqref{eq_practical_resource_allocation} asymptotically becomes (as $P \rightarrow \infty$) \cite{Lee2007a}
\begin{equation} \label{eq_sum_rate_asymptotic}
\begin{split}
f_{\text{sum}}^{\text{BD}}(P) &\cong N \log_2 \! \left(\frac{P}{N}\right) \!+ \!\! \sum_{k \in \mathcal{S}^{\text{BD}}} \! \log_2 \det (\vect{H}_k \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{H}_k^H), \\
f_{\text{sum}}^{\text{ZFC}}(P) &\cong N \log_2 \! \left(\frac{P}{N}\right) \!+ \!\!\sum_{k \in \mathcal{S}^{\text{ZFC}}} \! \log_2 (|\vect{h}^H_k \vect{w}^{\text{ZFC}}_{k}|^2),
\end{split}
\end{equation}
for BD and ZFC, respectively. This result is based on having scheduling sets that satisfy $|\mathcal{S}^{\text{BD}}|=\frac{N}{M}$ and $|\mathcal{S}^{\text{ZFC}}|=N$ and on equal power allocation (which is asymptotically optimal).
For both strategies, the asymptotic sum rate behaves as $\mathcal{M}_{\infty} \log_2(P) + \mathcal{R}_{\infty}$, where $\mathcal{M}_{\infty}$ is the multiplexing gain and $\mathcal{R}_{\infty}$ is the rate offset. Both BD and ZFC achieve a multiplexing gain of $\mathcal{M}_{\infty}=N$, which is the same high-SNR slope as of the sum capacity. We thus need to compare the rate offsets $\mathcal{R}_{\infty}$ to conclude which strategy is preferable in the high-SNR regime.
\begin{theorem} \label{theorem_asymptotic_difference}
Assume the receive correlation matrices $\vect{R}_{R,k}$ have eigenvalues $\lambda_{k,M} \geq \ldots \geq \lambda_{k,1} > 0$ and the use of random user selection with $|\mathcal{S}^{\text{BD}}|=\frac{N}{M}$, $|\mathcal{S}^{\text{ZFC}}|=N$. The expected asymptotic difference in sum rate between BD and ZFC (with MRC) is
\begin{equation} \label{eq_asymptotic_difference}
\begin{split}
&\bar{\beta}_{\text{BD-ZFC}} = \mathbb{E} \left\{ \lim_{P \rightarrow \infty} f_{\text{sum}}^{\text{BD}}(P) - f_{\text{sum}}^{\text{ZFC}}(P) \right\} \\
&= N \frac{\log_2(e)}{M} \sum_{i=1}^{M-1} \frac{M-i}{i} \!+\! \log_2 \left(\prod_{k \in \mathcal{S}^{\text{BD}}} \prod_{m=1}^{M} \lambda_{k,m} \! \right) \!-\! \sum_{\ell \in \mathcal{S}^{\text{ZFC}}} z_{\ell}
\end{split}
\end{equation}
where $z_{\ell} = \mathbb{E}\{ \log_2( \| \tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell} \|_2^2) \} - \frac{\psi(N)}{\log_e(2)}$ and $\psi(\cdot)$ is the digamma function. Furthermore, $\log_2(\lambda_{\ell,M}) \leq z_{\ell} \leq \log_2( \mathbb{E}\{ \| \tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell} \|_2^2 \}) - \frac{\psi(N)}{\log_e(2)}$ where $\mathbb{E}\{ \| \tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell} \|_2^2 \}$ is given by \eqref{eq_mean_value_effective_channel} in Lemma \ref{lemma_distribution_effective_channel}.
\end{theorem}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_theorem_asymptotic_difference}.
\end{IEEEproof}
The expected asymptotic difference in \eqref{eq_asymptotic_difference} has several terms. The first term is the (positive) expected gain of BD in a spatially uncorrelated scenario with homogenous user channels and no receive combining---this was considered in \cite[Theorem 3]{Lee2007a}. The other terms depend on the spatial correlation and choice of receive combining. For users with homogenous channel conditions where all $\vect{R}_{R,k}$ have the same eigenvalues $\lambda_{k,m}=\lambda_m$, we have
\begin{equation}
\bar{\beta}_{\text{BD-ZFC}} \leq N \frac{\log_2(e)}{M} \sum_{i=1}^{M-1} \frac{M-i}{i} + N \log_2 \frac{\prod_{m=1}^{M} \lambda_{m}^{1/M}}{\lambda_{M}}
\end{equation}
where the last term contains the geometric mean of all eigenvalues divided by the largest eigenvalue. This ratio is smaller than one (or equal for uncorrelated channels) and thus its logarithm is negative and approaches $-\infty$ as the eigenvalue spread increases. Therefore, Theorem \ref{theorem_asymptotic_difference} shows that BD \emph{might} have an advantage on uncorrelated channels, but ZFC always becomes the better choice as the receive-side correlation grows. The explanation is that BD has less restrictive interference mitigation, but is more vulnerable to poor channels since it uses all channel dimensions for transmission. We can expect a similar impact of any channel property that increases the eigenvalue spread in $\vect{H}_k \vect{H}_k^H$; for example, spatial correlation at the transmitter-side or a strong (low-rank) line-of-sight component.
To illustrate the opposite effect of having users with different path losses, we assume for simplicity that there are $\frac{N}{M}$ strong users with $\vect{R}_{R,k} = \gamma \vect{I}_M$, for some $\gamma> 1$, and $N-\frac{N}{M}$ weak users with $\vect{R}_{R,k} = \vect{I}_M$. If BD only serves the strong users while ZFC serves also the weak users, we have
\begin{equation}
\bar{\beta}_{\text{BD-ZFC}} \leq N \frac{\log_2(e)}{M} \sum_{i=1}^{M-1} \frac{M-i}{i} + \left( N-\frac{N}{M} \right) \log_2( \gamma).
\end{equation}
This upper bound approaches $+\infty$ as the difference $\gamma$ between the strong and weak users grows. Although not strictly proved, this indicates that BD is better at utilizing heterogenous channel conditions as it requires fewer users to be close to the base station to achieve high sum rates. This benefit reduces if some fairness mechanism is used to compensate for unfavorable path losses.
The expected asymptotic difference in sum rate, $\bar{\beta}_{\text{BD-ZFC}}$, can be transformed into a difference $-\frac{\bar{\beta}_{\text{BD-ZFC}}}{ 10 N \log_{10}(2)}$ [dB] in transmit power to achieve the same sum rate in the high-SNR regime \cite{Lee2007a}.
\subsection{Impact of User Selection}
The comparison in Theorem \ref{theorem_asymptotic_difference} was based on random user selection of the maximal number of users ($\frac{N}{M}$ with BD and $N$ with ZFC), although scheduling of spatially separated users is necessary to achieve the full potential of multi-user MIMO. This paper assumes $K \geq N$ users, meaning that only a subset of users is scheduled at each channel use. If the users are unevenly distributed in the cell, it could be beneficial to intentionally schedule fewer users than possible. We will now analyze how the ability of selecting users with spatially compatible channels impacts performance.
In the high-SNR regime, the optimal (semi-unitary) precoding matrix $\vect{W}_k^{\textrm{su}}$ for single-user transmission matches the channel as $\widetilde{\vect{C}}_k^H\vect{H}_k \vect{W}_k^{\textrm{su}} = \widetilde{\vect{C}}_k^H \vect{H}_k$, while the precoding matrix $\vect{W}_k \in \mathbb{C}^{N \times d_k}$ of an SDMA strategy is balanced between matching the own channel and being orthogonal to the co-user channels. The expected asymptotic performance loss of having to cancel inter-user interference is therefore
\begin{equation} \label{eq_expected_precoding_loss}
\begin{split}
\mathbb{E}\{\text{Loss} \} &= \mathbb{E}\{ \log_2 \det (\widetilde{\vect{C}}_k^H \vect{H}_k \vect{H}_k^H \widetilde{\vect{C}}_k) \\ & \qquad - \log_2 \det (\widetilde{\vect{C}}_k^H \vect{H}_k \vect{W}_k \vect{W}_k^{H} \vect{H}_k^H \widetilde{\vect{C}}_k) \} \\
& = \mathbb{E}\Big\{ \log_2 \frac{\det ( \vect{\Lambda}_{k} \vect{\Lambda}_{k}^H )}{ \det ( \vect{\Lambda}_{k} \vect{B}_k \vect{W}_k \vect{W}_k \vect{B}_k^H \vect{\Lambda}_{k}^H ) } \Big\} \\
&= - \mathbb{E}\{ \log_2 \det (\vect{B}_k \vect{W}_k \vect{W}_k \vect{B}_k^H) \}
\end{split}
\end{equation}
where $\vect{\Lambda}_{k} \in \mathbb{C}^{d_k \times d_k}$ contains the non-zero singular values of $\widetilde{\vect{C}}_k^H \vect{H}_k$ and $\vect{B}_k$ contains the corresponding right singular vectors.\footnote{These matrices can be obtained from a \emph{compact} singular value decomposition $\widetilde{\vect{C}}_k^H \vect{H}_k = \vect{U}_k \vect{\Lambda}_{k} \vect{B}_k$. Note that $\vect{B}_k$ contains an orthonormal basis of the row space of the effective channel $\widetilde{\vect{C}}_k^H \vect{H}_k$.} Observe that the eigenvalues of $\vect{B}_k \vect{W}_k \vect{W}_k \vect{B}_k^H$ are smaller or equal to one, thus $\mathbb{E}\{\text{Loss} \} \geq 0$. The following theorem indicates how this loss is affected by user selection.
\begin{theorem} \label{theorem_impact_of_scheduling}
For any given scheduling sets $\mathcal{S}^{\text{BD}},\mathcal{S}^{\text{ZFC}}$ (with $|\mathcal{S}^{\text{BD}}|=\frac{N}{M}$ and $|\mathcal{S}^{\text{ZFC}}|=N$), suppose we replace one of the users in each set with the best one among $K$ random users.
If the best user is the one minimizing the expected asymptotic loss in \eqref{eq_expected_precoding_loss}, these losses for BD and ZFC, respectively, can be lower bounded as \vskip-4mm
\begin{equation}
\begin{split}
\mathbb{E}\{\text{Loss}_{\text{BD}}\} &\geq - M \log_2(1-c_1 K^{-\frac{1}{M(N-M)}}) \\
\mathbb{E}\{\text{Loss}_{\text{ZFC}} \} &\geq - \log_2(1-c_2 K^{-\frac{1}{N-M}})
\end{split}
\end{equation}
when $K$ is large ($c_1,c_2$ are positive constants, see the proof).
\end{theorem}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_theorem_impact_of_scheduling}.
\end{IEEEproof}
The lower bounds in this theorem indicate that it is easier to find users with near-orthogonal channels under ZFC than under BD. This seems reasonable since the random channels of BD users occupy $M$ dimensions and should happen to be compatible to the co-users in all of them, while ZFC users only utilize one dimension and use receive combining to pick the most compatible among its $M$ dimensions. Related observations can be made in the area of channel quantization, where fewer codewords are necessary to describe $(N \times 1)$-dimensional channels to a certain accuracy than are needed for $(N \times M)$-dimensional channels \cite{Dai2008a}. The concave structure of the information rates makes it difficult to obtain exact results, but the indications of Theorem \ref{theorem_impact_of_scheduling} are verified by simulations herein.
\subsection{Numerical Illustrations under Perfect CSI}
\label{subsection_numerical_perfect_CSI}
Next, the analytic properties in Theorem \ref{theorem_asymptotic_difference} and Theorem \ref{theorem_impact_of_scheduling} are illustrated numerically.
To this end, we adopt the simple exponential correlation model of \cite{Loyka2001a}, where $0 \leq \rho \leq 1$, $\iota=\sqrt{-1}$, $U[\cdot,\cdot)$ denotes a uniform distribution, and
\begin{equation} \label{ch2_eq_exponential_correlation_model}
[\vect{R}(\rho,\theta)]_{ij} = \begin{cases} (\rho e^{\iota \theta})^{j-i}, & i \leq j, \\[-2mm]
(\rho e^{-\iota \theta})^{i-j}, & i>j,
\end{cases} \quad \theta \sim U[0,2\pi).
\end{equation}
The magnitude $\rho$ is the \emph{correlation factor} between adjacent antennas, where $\rho=0$ means no spatial correlation and $\rho=1$ means full correlation. For simplicity, $\rho$ is the same for all users while $\theta$ is different. Note that $\rho$ impacts the perceived spatial correlation non-linearly; a typical angular spread in a highly spatially correlated scenario is $10--20$ degrees which roughly corresponds to $\rho \approx 0.9$ \cite{Bjornson2009c}.
The expected asymptotic difference between BD and ZFC is shown in Fig.~\ref{figure_correlation} as a function of $\rho$, using $N=8$ transmit antennas and $M=2$ receive antennas. This simulation confirms that BD is advantageous in uncorrelated systems, while ZFC becomes beneficial as the correlation increases ($\rho>0.4$ under receive-side correlation, $\rho>0.7$ under transmit-side correlation, and $\rho>0.25$ when both sides are correlated). The two bounds from Theorem \ref{theorem_asymptotic_difference} are also shown in the Fig.~\ref{figure_correlation}. The lower bound is very accurate, while the upper bound is only tight at high correlation.
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure3.pdf} \vskip-2mm
\caption{The expected asymptotic difference between BD and ZFC in a system with $N=8$ transmit antennas, $M=2$ receive antennas per user, and random user selection.
The impact of spatial correlation at the receiving users, transmitting base station, and both sides is shown (using the exponential correlation model from \cite{Loyka2001a} with different correlation factors $\rho$).}\label{figure_correlation}
\end{center} \vskip-3mm
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure4.pdf} \vskip-2mm
\caption{The average achievable sum rate in a system with perfect CSI, $N=8$ transmit antennas, $M=4$ receive antennas, and the same average SNR among all users (10 or 20 dB).
The performance with different strategies are shown as a function of the total number of users and for different correlation factors $\rho$ among the receive antennas.}\label{figure_perfect_dBs}
\end{center} \vskip-5mm
\end{figure}
To exemplify the impact of user selection, we use the \emph{capacity-based suboptimal user selection} (CBSUS) algorithm from \cite{Shen2006a}, which greedily adds users sequentially to maximize the sum rate and might give scheduling sets with fewer than $N$ data streams. We consider a scenario with $N=8$ uncorrelated transmit antennas and $M=4$ receive antennas with correlation factor $\rho \in \{0,0.4,0.8\}$; see \cite{Bjornson2011e} for another scenario. We compare ZFC (1 stream/user) and BD (4 streams/user) with multi-user eigenmode transmission (MET) from \cite{Boccardi2007a} where data streams are allocated greedily with zero inter-user interference and users can have different numbers of streams. We also simulated 2 streams/user, but it is not shown herein because the sum rate was always in between ZFC and BD.
Fig.~\ref{figure_perfect_dBs} shows the average achievable sum rate as a function of the total number of users $K$. We consider the case when all users have the same average SNR (defined as $P \frac{ \mathbb{E} \{\| \vect{H}_k\|_F^2 \}}{NM}$), either equal to 10 or 20 dB. Irrespective of the SNR, number of users, and receive-side correlation, ZFC outperforms BD. Thus, the scheduling-benefit of ZFC (from Theorem \ref{theorem_impact_of_scheduling}) dominates over the interference mitigation-benefit of BD (from Theorem \ref{theorem_asymptotic_difference})---even for spatially uncorrelated channels. As expected, the performance with ZFC improves with $\rho$, while correlation degrades the BD performance. MET has an advantages over ZFC since it can allocate different numbers of streams to different users (based on how many singular values are strong in their channels), but this advantage is small and disappears asymptotically with the number of users; this was also observed in \cite{Boccardi2007a}.
Next, we consider heterogeneous channel conditions by having uniformly distributed users in a circular cell with radius $250$ m (minimal distance is $35$ m), a path loss coefficient of 3.5, and log-normal shadow-fading with 8 dB in standard deviation. The average achievable sum rate is shown in Fig.~\ref{figure_perfect_pathloss} with an SNR of 20 dB at the cell edge.\footnote{Such SNRs are reasonable in dense cellular systems and are necessary to compare BD and ZFC in regimes where these are supposed to work well.} The variation in path loss between users makes the results very different from the previous scenario in Fig.~\ref{figure_perfect_dBs}. At low receive-correlation, BD outperforms ZFC, but the difference reduces with $K$. ZFC is however better than BD at high correlation and many users. MET has a large advantage over the other strategies, explained by its flexible stream allocation. To comprehend the difference, the probability that a selected user is allocated a certain number of streams is shown in Fig.~\ref{figure_bars}.
We observe that spatial correlation reduces the number of streams per user, but the distance-dependence is even more significant; cell center users usually receive many streams while cell edge users only receive one or a few streams. This is natural since cell center users are more probable to have channel matrices with multiple relatively strong singular directions.
The conclusion is that ZFC is the method of choice in multi-user MIMO systems with perfect CSI and homogenous user conditions (since it performs very closely to the more complicated MET). On the other hand, MET and BD are better under heterogeneous user conditions. It is worth noting that the more streams allocated per user, the more channel dimensions need to be know at the base station. The next section will therefore study how practical CSI acquisition affects our results.
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure5.pdf} \vskip-2mm
\caption{The average achievable sum rate in a circular cell with perfect CSI, $N=8$ transmit antennas, $M=4$ receive antennas, and an SNR of 20 dB at the cell edge.
The performance with different strategies are shown as a function of the total number of users and for different correlation factors $\rho$ among the receive antennas.}\label{figure_perfect_pathloss}
\end{center} \vskip-5mm
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure6.pdf} \vskip-2mm
\caption{The probability that a scheduled user is allocated a certain number of streams, assuming a circular cell with perfect CSI, $N=8$ transmit antennas, $M=4$ receive antennas, $K=20$ users, and an SNR of 20 dB at the cell edge. The whole cell has a radius of 250 meters, whereof users closer than 100 meters belong to the cell center and users further away than 200 meters belong to the cell edge.}\label{figure_bars}
\end{center} \vskip-5mm
\end{figure}
\section{Comparison of BD and ZFC with Imperfect CSI}
In this section, we continue the comparison of BD and ZFC by introducing imperfect CSI, originating from either quantized feedback in an FDD system or imperfect reverse-link estimation in a TDD system. The resources for channel acquisition are limited which has a major impact on both the number of channel dimensions that can be acquired per user and the accuracy of the acquired CSI. Theoretically,
users can feed back different numbers of channel dimensions depending on some kind of long-term statistical CSI, but that would reduce the coverage (by favoring cell center users)
and require a flexible system operation with additional control signaling. We therefore assume that the system acquires $d$ dimensions/user from a randomly selected user set, where $d \geq 1$ is fixed but depends on the intended precoding strategy. This assumption is relaxed in the numerical evaluation.
\subsection{Comparison with Quantized CSI}
In the FDD system operation of Fig.~\ref{figure_fdd_operation}, each user selected for feedback conveys the $d$-dimensional subspace spanned by its effective channel $\widetilde{\vect{C}}_k^H \vect{H}_k$ using $B$ bits. Similar to \cite{Love2003a,Dai2008a,Raghavan2007a,Love2008a,Ravindran2008a}, we use a codebook $\mathcal{C}_{N,d,B}=\{\vect{U}_1,\ldots,\vect{U}_{2^{B}}\}$ with codewords
$\vect{U}_{i} \in \mathbb{C}^{N \times d}$ from the (complex) \emph{Grassmannian manifold} $\mathcal{G}_{N,d}$; that is, the set of all $d$-dimensional linear subspaces (passing through the origin) in an $N$-dimensional space. Each codeword forms an orthonormal basis, thus $\vect{U}_{i}$ is a semi-unitary matrix satisfying $\vect{U}_{i}^H \vect{U}_{i} = \vect{I}_{d}$.
User $k$ selects the codeword that minimizes the chordal distance \cite{Zhou2006a}:
\begin{equation}
\bar{\vect{H}}_k = \argmin{ \vect{U} \in \mathcal{C}_{N,d,B}} \,\, \delta \left( \widetilde{\vect{C}}_k^H \vect{H}_k, \vect{U} \right)
\end{equation}
where $\delta(\vect{B},\vect{U}) = \sqrt{d-\mathrm{tr}( \mathrm{span}(\vect{B})^H \vect{U} \vect{U}^H \mathrm{span}(\vect{B}) )}$ and $\mathrm{span}(\cdot)$ gives a matrix containing an orthonormal basis of the row space.
We assume error-free and delay-free feedback, but the conclusions of this section are expected to hold true also under feedback errors (cf.~\cite{Caire2010a}).
There is a variety of ways to handle feedback errors (especially if the error structure is known), but a simple approach is to treat $\bar{\vect{H}}_k$ as being the true channel \cite{Ravindran2008a} and calculate the precoding using a strategy developed for perfect CSI. This results in a lower bound on the performance and the information rates with BD and ZFC becomes
\begin{align} \label{eq_data_rate_BD_quant}
\!\! g^{\text{BD-Q}}_k(P) &= \log_2 \! \frac{\det \!\Big( \vect{I}_{M} \!+\!\!\! \fracSum{\ell \in \mathcal{S}^{\text{BD}}} \vect{H}_k \bar{\vect{W}}_{\ell}^{\text{BD}} \bar{\vect{\Upsilon}}_{\ell} \bar{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \! \Big) }{ \! \det \! \Big(\vect{I}_{M} \!+\!\!\! \fracSum{\ell \in \mathcal{S}^{\text{BD}} \setminus \{k\}} \!\!\! \vect{H}_k \bar{\vect{W}}_{\ell}^{\text{BD}} \bar{\vect{\Upsilon}}_{\ell} \bar{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \!\Big) \!\! } \\
\label{eq_data_rate_ZF_quant}
\!\! g^{\text{ZFC-Q}}_k(P) &= \log_2 \! \bigg( \!1 \!+\! \frac{ \bar{p}_k |\vect{c}_k^H \vect{H}_k \bar{\vect{w}}^{\text{ZFC}}_{k}|^2 }{1 + \!\!\fracSum{\ell \in \mathcal{S}^{\text{ZFC}} \setminus \{k\}} \!\! \bar{p}_{\ell} |\vect{c}_k^H \vect{H}_k \bar{\vect{w}}^{\text{ZFC}}_{\ell}|^2 } \! \bigg)
\end{align}
for users in the scheduling sets $\mathcal{S}^{\text{BD}}$ and $\mathcal{S}^{\text{ZFC}}$, respectively.
Next, we quantify the performance loss for BD and ZFC compared with having perfect CSI.
Random vector quantization (RVQ) is used for analytic convenience (as in \cite{Santipach2004a,Au-Yeung2007a,Jindal2008a,Ravindran2008a}), meaning that
we average over codebooks with random codewords from the Grassmannian manifold. As any judicious codebook design is better than RVQ, the upper bounds on the performance loss that we will derive are valid for any reasonable codebook. The following theorem provides an upper bound on the performance loss under BD and extends results in \cite{Ravindran2008a} to include heterogeneous user conditions and spatial correlation.
\begin{theorem} \label{theorem_rateloss_BD}
Assume that $\frac{N}{M}$ users are scheduled randomly. The average rate loss with BD (using equal power allocation) for user $k \in \mathcal{S}^{\text{BD}}$ due to RVQ is upper bounded as
\begin{equation} \label{eq_bounding_QBD_loss_theorem}
\begin{split}
\Delta^{\text{BD-Q}}_k &= \mathbb{E} \{ g^{\text{BD}}_k(P)- g^{\text{BD-Q}}_k(P) \} \\
& \leq \log_2 \det \left(\vect{I}_M + \frac{P}{M} D^{\text{BD}} \vect{R}_{R,k} \right)
\end{split}
\end{equation}
where the average quantization distortion is
\begin{equation}
\begin{split}
&D^{\text{BD}} = \mathbb{E} \{ \delta^2(\vect{H}_k,\bar{\vect{H}}_k) \} \\
&\approx \frac{\Gamma\left(\frac{1}{M(N-M)}\right)}{M(N-M)} \! \left( \!\frac{2^{B}}{(M(N-M))!} \prod_{i=1}^{M} \frac{(N-i)!}{(M-i)!} \! \right)^{\!-\frac{1}{M(N-M)}}.
\end{split}
\end{equation}
\end{theorem}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_theorem_rateloss_BD}.
\end{IEEEproof}
This theorem will be compared with the corresponding result for ZFC, but before stating that result we discuss how to select the (preliminary) receive combiner $\tilde{\vect{c}}_k$.
There are primarily two factors to consider when selecting $\tilde{\vect{c}}_k$: the gain of the effective channel $\|\tilde{\vect{c}}_k^H \vect{H}_k\|_2^2$ and the quantization distortion.
The results of \cite{Kountouris2006a,Ravindran2008b} indicate that the top priority in multi-user MIMO systems is to achieve small quantization errors, because it is a prerequisite for low inter-user interference. The error can be minimized by the \emph{quantization-based combining} (QBC) approach in \cite{Jindal2008a}, where the codeword and receive combiner are selected jointly as
\begin{equation} \label{eq_QBC_receiver}
(\tilde{\vect{c}}^{\text{QBC}}_k,\bar{\vect{h}}_k) = \argmax{\substack{\vect{c}: \|\vect{c}\|_2=1 \\ \vect{u} \in \mathcal{C}_{N,1,B} }} \,\, \delta \left(\vect{H}_k^H \vect{c}, \vect{u} \right).
\end{equation}
The \emph{maximum expected SINR combiner (MESC)} in \cite{Trivellato2008a} achieves better practical performance by balancing effective channel gain and quantization distortion, but is asymptotically equal to QBC at high SNR. Since this is the regime of main interest herein, we will exploit the analytic simplicity of QBC. Observe that QBC and MESC are only used for improved feedback accuracy; the MMSE combiner in Remark \ref{remark:receive-combiner} is used to maximize the performance during transmission (this was \emph{not} done in the original QBC framework of \cite{Jindal2008a}).
The following theorem provides an upper bound on the performance loss under ZFC and extends results in \cite{Jindal2008a} to include heterogeneous user conditions and spatial correlation.
\begin{theorem} \label{theorem_rateloss_ZF}
Assume that $\vect{R}_{R,k}$ has eigenvalues $\lambda_{k,M} > \ldots > \lambda_{k,1}>0$ and that $N$ users are selected randomly.
The average rate loss for ZFC (using equal power allocation and the same $\tilde{\vect{c}}^{\text{QBC}}_k$) due to RVQ is upper bounded as
\begin{equation} \label{eq_bounding_QBC_loss_theorem}
\begin{split}
\Delta^{\text{ZFC-Q}}_k &= \mathbb{E} \{ g^{\text{ZFC}}_k(P)- g^{\text{ZFC-Q}}_k(P) \} \\ &\leq \log_2 \left(1+ \frac{P}{N} D^{\text{QBC}} G_k \right)
\end{split}
\end{equation}
where the average quantization distortion is
\begin{equation}
D^{\text{QBC}} = \mathbb{E} \{ \delta^2(\vect{h}_k,\bar{\vect{h}}_k) \} \approx 2^{-\frac{B}{N-M}} \Big( \!\!\!
\begin{array}{c}
N-1\\[-1.2ex]
M-1
\end{array} \!\!\! \Big)^{-\frac{1}{N-M}}
\end{equation}
and the average channel gain with QBC is (where $\mu_{n} = \frac{1}{\lambda_{k,n}}$)
\begin{equation} \label{eq_L-expression}
\begin{split}
G_k =& \sum_{m=1}^{M-1} \sum_{n=1}^m \sum_{t=m+1}^{M} \frac{(N-M+1) A_{m,n,t} }{(\mu_n\!-\!\mu_t) \condProd{i=1}{i \neq n}{m} (\mu_n\!-\!\mu_i) \condProd{j=m+1}{j \neq l}{M} \!\! (\mu_j\!-\!\mu_t) }, \\
A_{m,n,t} =& \log_e \big(\frac{\mu_{m+1}}{\mu_{m}} \big) \frac{\mu_n^{m-1}}{(-\mu_t)^{
2+m-M}} \left(m \mu_t +\mu_n (M-m-1) \right) \\
+&\sum_{s=0}^{m-1} \sum_{r=0}^{M-m-1} \!\!
\Big( \!\!\!
\begin{array}{c}
m\\[-1.2ex]
s
\end{array} \!\!\! \Big) \!
\Big( \!\!\!
\begin{array}{c}
M-m-1\\[-1.2ex]
r
\end{array} \!\!\! \Big)
\frac{ \mu_n^{m-s} (-\mu_t)^{M-m-1-r}}{(-1)^{s} (1+s+r)} \\
\times & \left( \frac{m\!-\!s}{\mu_n} + \frac{M\!-\!m\!-\!1\!-\!r}{\mu_t} \right) (\mu_m^{r+s} -\mu_{m+1}^{r+s}).
\end{split}
\end{equation}
\end{theorem}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_theorem_rateloss_ZF}.
\end{IEEEproof}
The rate loss expressions in Theorem \ref{theorem_rateloss_BD} and Theorem \ref{theorem_rateloss_ZF} for BD and ZFC, respectively, indicate the joint impact of spatial correlation (at the receiver) and CSI quantization on the performance.
The main observation is that spatial correlation only has a marginal effect on the feedback accuracy; the expressions have a similar structure as for uncorrelated channels and the same scaling
in the number of feedback bits is necessary to achieve the maximal multiplexing gain \cite{Jindal2008a,Ravindran2008a}.
\begin{corollary} \label{cor_multiplexing_gain_quantized}
To achieve the maximal multiplexing gain with BD or ZFC under quantized CSI and arbitrary receive correlation, it is sufficient to scale the total number of CDI feedback bits for the scheduled users as
\begin{equation} \label{eq_total_bits}
B_{\mathrm{total}} \approx N(N-M) \log_2(P) + \mathcal{O}(1).
\end{equation}
\end{corollary}
While this corollary only provides a sufficient condition, we can expect the scaling law in \eqref{eq_total_bits} to also be necessary.\footnote{The necessary scaling can be proved for ZFC with QBC using a technique from \cite[Theorem 4]{Jindal2006a}, while simulations in \cite{Ravindran2008a} show that quantized ZFC and BD have the same scaling in the necessary number of bits.} In any case, the scaling law in \eqref{eq_total_bits} is easily satisfied by allocating (approximately) $N-M$ channel uses for CSI feedback, since typically the uplink sum rate also behaves as $N \log_2(P) + \mathcal{O}(1)$ in the high-SNR regime \cite{Caire2010a}.
Observe that this result is based on random user selection, while additional feedback of gain information is necessary to achieve multi-user diversity or short-term rate adaptation (cf.~\cite{Yoo2007a}). As BD requires $M$ times more bits per user, ZFC can typically achieve feedback from $M$ times more users. We therefore expect ZFC to further strengthen its advantage at finding near-orthogonal users (indicated in Theorem \ref{theorem_impact_of_scheduling} under perfect CSI). In addition, spatial correlation at the transmitter-side (and other factors that make the channel matrices ill-conditioned) will inflict larger performance losses on BD than ZFC, just as in the case of perfect CSI.
\subsection{Comparison under Estimated CSI}
Next, we assume that the base station acquires CSI through imperfect CSI estimation. The primary focus will be on TDD systems, where channel estimates are obtained through training signaling in the uplink (assuming perfect channel reciprocity). It is worth noting that this approach is similar to having analog CSI feedback in FDD systems, where the unquantized channel coefficients are sent on an uplink subcarrier \cite{Ravindran2008a,Caire2010a}.\footnote{Digital/quantized feedback might be beneficial over analog/unquantized feedback when there is plenty of resources for channel estimation \cite{Caire2010a}. But if very accurate CSI is required, Corollary \ref{cor_multiplexing_gain_quantized} shows that the quantization codebooks grow very large and thus the search for the best codeword might be computationally infeasible.}
The reciprocal uplink counterpart to the system model in \eqref{eq_system_model} is
\begin{equation}
\widetilde{\vect{y}}_k = \vect{H}^T_k \widetilde{\vect{x}}_k + \widetilde{\vect{n}}_k
\end{equation}
where $\widetilde{\vect{y}}_k \in \mathbb{C}^{N \times 1}$ is the received uplink signal, $\widetilde{\vect{x}}_k \in \mathbb{C}^{M \times 1}$ is the transmitted uplink signal, and $\widetilde{\vect{n}}_k \sim \mathcal{CN}(\vect{0},\sigma^2 \vect{I}_{N})$ is the noise vector.\footnote{The downlink noise vector was normalized towards the channel matrix in the system model of \eqref{eq_system_model}. To account for a different noise level at the base station, $\sigma^2$ is the (relative) uplink noise variance.}
To estimate $\widetilde{\vect{C}}_k^H \vect{H}_k \in \mathbb{C}^{d \times N}$, user $k$ sends $\widetilde{\vect{C}}_k^* \vect{T}_k$ over $d$ uplink channel uses, for some known training matrix $\vect{T}_k \in \mathbb{C}^{d \times d}$ and where $(\cdot)^*$ denotes the complex conjugate.
Assuming perfect statistical CSI, the MMSE estimate $\widehat{\vect{H}}_k$ of $\widetilde{\vect{C}}_k^H \vect{H}_k$ and the corresponding error covariance matrix $\vect{E}_k$ are~\cite{Bjornson2010a}
\begin{equation} \label{eq_MMSE_estimator}
\begin{split}
\mathrm{vec}(\widehat{\vect{H}}_k^T) &= \frac{1}{\sigma^2} \vect{E}_k \widetilde{\vect{T}}_k^H \mathrm{vec}(\vect{Y}_k), \\
\vect{E}_k &= \left( ( \widetilde{\vect{C}}_k^H \vect{R}_{R,k} \widetilde{\vect{C}}_k \! \otimes \vect{I}_N )^{-1} + \frac{\widetilde{\vect{T}}_k^H \widetilde{\vect{T}}_k}{\sigma^2} \right)^{-1}
\end{split}
\end{equation}
where $\widetilde{\vect{T}}_k=(\vect{T}_k^T \! \otimes \, \vect{I}_N)$ and $\vect{Y}_k$ is the received signal from training signaling.
The training matrix $\vect{T}_k$ has a total training power/SNR constraint $\mathrm{tr}( \vect{T}_k^H \vect{T}_k ) = \Psi$.
As under quantized CSI, we calculate the precoding by treating $\widehat{\vect{H}}_k$ as the true channel. This results in a lower bound on the performance and the information rates with BD and ZFC becomes
\begin{align} \label{eq_data_rate_BD_est}
g^{\text{BD-EST}}_k(P) &= \log_2 \! \frac{\det \!\Big( \vect{I}_{M} \!+\! \fracSum{\ell \in \mathcal{S}^{\text{BD}}} \vect{H}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{\Upsilon}}_{\ell} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \Big) }{ \! \det \! \Big( \vect{I}_{M} \!+\! \!\! \fracSum{\ell \in \mathcal{S}^{\text{BD}} \setminus \{k\}} \!\!\!\! \vect{H}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{\Upsilon}}_{\ell} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \Big) \!\! } \\
\label{q_data_rate_ZF_est}
g^{\text{ZFC-EST}}_k(P) &= \log_2 \Bigg( 1 + \frac{ \hat{p}_k |\vect{c}_k^H \vect{H}_k \widehat{\vect{w}}^{\text{ZFC}}_{k}|^2 }{1 + \fracSum{\ell \in \mathcal{S}^{\text{ZFC}} \setminus \{k\}} \hat{p}_{\ell} |\vect{c}_k^H \vect{H}_k \widehat{\vect{w}}^{\text{ZFC}}_{\ell}|^2 } \Bigg)
\end{align}
for users in the scheduling sets $\mathcal{S}^{\text{BD}}$ and $\mathcal{S}^{\text{ZFC}}$, respectively.
The following theorem provides an upper bound on the performance loss under BD due to imperfect CSI estimation.
\begin{theorem} \label{theorem_rateloss_BD_EST}
Assume that $\frac{N}{M}$ users are scheduled randomly under BD. The average rate loss for user $k \in \mathcal{S}^{\text{BD}}$ (using equal power allocation) due to CSI estimation is upper bounded as
\begin{equation} \label{eq_bounding_BD_est_loss_theorem}
\begin{split}
\Delta^{\text{BD}} &= \mathbb{E} \{ g^{\text{BD}}_k(P)- g^{\text{BD-EST}}_k(P) \} \\
&\leq \log_2 \det \left(\vect{I}_M + \frac{P(N-M)}{N} \left( \vect{R}_{R,k}^{-T} + \frac{\vect{T}_k^H \vect{T}_k}{\sigma^2} \right)^{\!-1} \right)\!.
\end{split}
\end{equation}
\end{theorem}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_theorem_rateloss_BD_EST}.
\end{IEEEproof}
This theorem will be compared with the corresponding result for ZFC, but before stating that theorem we need to consider the impact of having MRC as the receive combiner $\tilde{\vect{c}}_k$.
ZFC is similar to applying BD to the effective channels $\vect{h}_k^H =\tilde{\vect{c}}_k^H \vect{H}_k$, but an important difference is that the effective channels are not Rayleigh fading because $\tilde{\vect{c}}_k$ depends on the current channel realization. The expression in \eqref{eq_MMSE_estimator} will therefore not give the MMSE estimate, but fortunately the linear MMSE (LMMSE) estimator from a similar expression to \eqref{eq_MMSE_estimator} if we know the first two moments of $\vect{h}_k$ \cite{Bjornson2010a}.
\begin{lemma} \label{lemma_distribution_effective_channel}
Assume that $\vect{R}_{R}$ has eigenvalues $\lambda_{M} > \ldots > \lambda_{1}>0$, where the user indices were dropped for convenience.
If $\tilde{\vect{c}}$ is the dominating left singular vector of $\vect{H}$, it holds that
\begin{itemize}
\item the direction $\frac{\vect{h}}{\|\vect{h}\|_2}$ of $\vect{h}= \tilde{\vect{c}}^H \vect{H}$ is isotropically distributed on the unit sphere;
\item the gain $\|\vect{h}\|_2^2$ is independent of the direction and
\end{itemize}
\begin{equation} \label{eq_mean_value_effective_channel}
\begin{split}
&\mathbb{E}\{ \|\vect{h}\|_2^2 \} = \! \! \sum_{m=1}^{M} \sum_{\boldsymbol{\zeta} \in
\mathcal{A}_M} \!\! \frac{ \condProdtwo{{\ell}=1}{M}
\lambda_{\zeta_{\ell}}^{N-{\ell}+1} \condProdtwo{{\ell}=N-M+1}{N} \!({\ell}-1)! }
{(-1)^{\mathrm{per}(\boldsymbol{\zeta})+m+1} \det (\vect{\Delta}) }
\\ & \times \sum_{\boldsymbol{\beta} \in \mathcal{B}_{m,M}} \!\!
\sum_{{\ell}=0}^{K_m(\boldsymbol{\beta})} \!\! \sum_{\tilde{k} \in
\widetilde{\Omega}_{\ell}^{(m)}} \frac{{\ell}!}{\tilde{k}_1! \cdots
\tilde{k}_m!}
\frac{(\fracSumtwo{i=1}{m}
\lambda_{\zeta_{\beta_i}}^{-1})^{-({\ell}+1)}}{\condProdtwo{i=1}{m}
\lambda_{\zeta_{\beta_i}}^{\tilde{k}_i}}
\end{split}
\end{equation}
where the $ij$th element of $\vect{\Delta} \in \mathbb{R}^{M \times M}$ is given by
\begin{equation}
[\vect{\Delta}]_{ij} = \lambda_j^{N-i+1} (N-i)!.
\end{equation}
In \eqref{eq_mean_value_effective_channel}, the set of all permutations of $\{1,\ldots,M\}$ is denoted $\mathcal{A}_M$. The sign of a given permutation $\boldsymbol{\zeta}=\{\zeta_1,\ldots,\zeta_M\} \in \mathcal{A}_M$ is denoted
$(-1)^{\mathrm{per}(\boldsymbol{\zeta})}$, where $\mathrm{per}(\cdot)$ is the number of inversions\footnote{An inversion in a sequence is a pair of numbers that is in incorrect order (i.e., not in ascending order).} in the permuted sequence.
Next, $\mathcal{B}_{l,M}$ is the collection of all subsets of $\mathcal{A}_M$ with cardinality $l$ and increasing elements (i.e., $\beta_1<\ldots<\beta_l$ for $\boldsymbol{\beta}=\{\beta_1,\ldots,\beta_l\} \in \mathcal{B}_{l,M}$).
The upper bound in the summation over $\ell$ is $K_l(\boldsymbol{\beta})= \sum_{i=1}^l (N-\beta_l)$.
Finally, $\widetilde{\Omega}^{(l)}_{{\ell}}$ is the set of all $l$-length partitions $\{\tilde{k}_1,\ldots,\tilde{k}_l\}$ of ${\ell}$ (i.e., $\sum_{i=1}^l \tilde{k}_i = \ell$)
that satisfy $0 \leq \tilde{k}_i \leq N-\beta_i$:
\begin{equation}
\widetilde{\Omega}^{(l)}_{{\ell}} =\bigg\{
\{\tilde{k}_1,\ldots,\tilde{k}_l\}: \,\, \sum_{j=1}^{\ell} \tilde{k}_j \!=
{\ell}, \, 0 \leq \tilde{k}_j \leq N-\beta_j \,\, \forall j \bigg\}.
\end{equation}
\end{lemma}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_lemma_distribution_effective_channel}.
\end{IEEEproof}
The following theorem provides an upper bound on the performance loss under ZFC due to imperfect CSI estimation.
\begin{theorem} \label{theorem_rateloss_ZF_EST}
Assume that $N$ users are scheduled randomly under ZFC and that MRC is applied. The average rate loss for user $k \in \mathcal{S}^{\text{ZFC}}$ due to CSI estimation is upper bounded as
\begin{equation} \label{eq_bounding_ZF_EST_loss_theorem}
\begin{split}
\Delta^{\text{ZFC-EST}} &= \mathbb{E} \{ g^{\text{ZFC}}_k(P)- g^{\text{ZFC-EST}}_k(P) \} \\
&\leq \log_2 \left(1 + \frac{P(N-1)}{N} \frac{1}{\mathbb{E}\{ \|\vect{h}_k\|_2^2 \}^{-1}+ \frac{\Psi}{\sigma^2}} \right)
\end{split}
\end{equation}
where $\mathbb{E}\{ \|\vect{h}_k\|_2^2 \}$ is given in \eqref{eq_mean_value_effective_channel}.
\end{theorem}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_theorem_rateloss_ZF_EST}.
\end{IEEEproof}
The rate loss expressions in Theorem \ref{theorem_rateloss_BD_EST} and Theorem \ref{theorem_rateloss_ZF_EST} indicate the joint impact of spatial correlation and imperfect channel estimation on the performance of BD and ZFC, respectively. BD is slightly more resilient to CSI uncertainty, since the BD expression contains $(N-M)$ where the ZFC expression has $(N-1)$. But observe that the performance losses are calculated against the same precoding strategy with perfect CSI; we know from Section \ref{section_comp_BD_ZFC_perfect_CSI} that ZFC and BD have different preferable user conditions, making it hard to analytically conclude which strategy to use under imperfect CSI estimation. However, the important result is the following extension of \cite{Caire2010a} to spatially correlated scenarios with $M\geq1$.
\begin{corollary} \label{cor_multiplexing_gain_estimated}
To achieve the maximal multiplexing gain with BD or ZFC under imperfect CSI estimation and arbitrary receive correlation, it is necessary and sufficient to scale the training power $\Psi$ as
\begin{equation} \label{eq_training_law}
\frac{P}{\Psi} \rightarrow \texttt{constant} < \infty \quad \text{when} \, \, P \rightarrow \infty.
\end{equation}
\end{corollary}
\begin{IEEEproof}
The proof is given in Appendix \ref{app_cor_multiplexing_gain_estimated}.
\end{IEEEproof}
This corollary says that the training power/SNR should increase linearly with the transmit power/SNR to achieve the optimal sum rate scaling. This is, for example, satisfied by setting the total training power to $\Psi=P$ under ZFC and $\Psi=M P$ under BD, which corresponds to the reasonable assumption of having the same average SNR in the downlink and in the uplink.\footnote{Battery-powered user devices might operate at lower power budget than the base station, but Corollary \ref{cor_multiplexing_gain_estimated} is satisfied as long as $P$ and $\Psi$ exhibit the same scaling. In practical scenarios, the path loss is the main source of SNR variations and affects the downlink and uplink equally.} The demands for higher CSI accuracy with increasing SNR is therefore automatically fulfilled by the reduced estimation errors.
Observe that one uplink channel use is consumed per user antenna dimension that is estimated, thus creating a practical bound on how many user channels that can be estimated in block fading systems \cite{Caire2010a}. As ZFC only has one effective antenna per user, it can accommodate $M$ times more users than BD on the same estimation overhead and thereby exploit multi-user diversity to a larger extent.
\section{Numerical Illustrations Under Imperfect CSI}
\label{subsection_numerical_imperfect_CSI}
This section consists of two parts. First, the numerical illustrations in Section \ref{subsection_numerical_perfect_CSI} are continued under imperfect CSI estimation. Then, we analyze the performance behavior under quantized CSI.
\subsection{Continuation of Section \ref{subsection_numerical_perfect_CSI} under Estimated CSI}
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure7.pdf} \vskip-2mm
\caption{The average achievable sum rate in a system with CSI estimation errors, $N=8$ transmit antennas, $M=4$ receive antennas, and the same average SNR among all users (10 or 20 dB).
The performance with different strategies are shown as a function of the total number of users and for different correlation factors $\rho$ among the receive antennas.}\label{figure_imperfect_dBs}
\end{center} \vskip-5mm
\end{figure}
We continue the simulations in Section \ref{subsection_numerical_perfect_CSI} by introducing imperfect CSI estimation.
We use the MSE-minimizing training matrices from \cite[Theorem 1]{Bjornson2010a} and training power $\Psi= P \, d$ (for estimation of $d$ dimensions/user). The CBSUS algorithm in \cite{Shen2006a} is modified\footnote{Estimation errors contribute an average interference of $P (|\mathcal{S}|-1)/|\mathcal{S}| \vect{E}_{\text{est}}$, where $\vect{E}_{\text{est}} \!=\!(\vect{R}_{R,k}^{-T}\! +\! \vect{T}_k^H \vect{T}_k/\sigma^2 )^{-1}$ for BD and $\vect{E}_{\text{est}}\!=\!(1/\mathbb{E}\{\|\vect{h}_k\|_2^2\} \!+\! \Psi/\sigma^2)^{-1}$ for ZFC.} to include the average interference (due to CSI estimation errors) in the scheduling.
The average achievable sum rate is shown in Fig.~\ref{figure_imperfect_dBs} as a function of the number of users that we obtain CSI estimates for using ZFC (while BD only obtains channel estimates for $\frac{1}{M}$ of them). All users have the same average SNR of either 10 or 20 dB.
The performance loss compared with having perfect CSI is 10-20\% (see Fig.~\ref{figure_perfect_dBs}), but the conclusion is otherwise the same and even clearer than before:
ZFC outperforms BD in terms of performance with few users, in handling spatial correlation, and in exploiting multi-user diversity.
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure8.pdf} \vskip-2mm
\caption{The average achievable sum rate in a circular cell with CSI estimation errors, $N=8$ transmit antennas, $M=4$ receive antennas, and an SNR of 20 dB at the cell edge.
The performance with different strategies are shown as a function of the total number of users and for different correlation factors $\rho$ among the receive antennas.}\label{figure_imperfect_pathloss}
\end{center} \vskip-3mm
\end{figure}
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure9.pdf} \vskip-2mm
\caption{The average achievable sum rate in a circular cell with CSI estimation errors, $N=8$ transmit antennas, $M=4$ receive antennas, and an SNR of 20 dB at the cell edge.
The performance is shown as a function of the total number of users $K$, and CSI is only acquired for the users with strongest long-term statistics.}\label{figure_imperfect_pathloss_opportunistic}
\end{center} \vskip-5mm
\end{figure}
In case of a circular cell (see Section \ref{subsection_numerical_perfect_CSI} for details), the average achievable sum rate is shown in Fig.~\ref{figure_imperfect_pathloss}. Recall from Fig.~\ref{figure_perfect_pathloss} that BD was often better than ZFC in this scenario under perfect CSI, but the case is completely different under imperfect CSI; ZFC outperforms the other strategies when the limited resources for CSI acquisition are taken into account. This means that the ZFC benefit of easily finding near-orthogonal users (among $M$ times more users than with BD) dominates the BD benefit of multi-stream multiplexing (preferably to cell center users). We also tested a MET-like strategy with greedy stream allocation (we took the optimum among feeding back 1, 2 or 4 channel dimensions per active user), but it was always identical to ZFC (in both Fig.~\ref{figure_perfect_dBs} and Fig.~\ref{figure_perfect_pathloss})---this further confirms our conclusion.
The users selected for feedback were chosen randomly (e.g., in a round-robin fashion) in Figs.~\ref{figure_imperfect_dBs} and \ref{figure_imperfect_pathloss}, but could theoretically be based on some kind of long-term statistical CSI. This could for instance mean that ZFC acquires one dimension from each of the $K$ users, while BD acquires $M$ dimensions from the $\frac{K}{M}$ users
with the strongest long-term statistics $\mathrm{tr}( \vect{R}_{T,k}) \mathrm{tr}( \vect{R}_{R,k})$. The greedy stream allocation strategy MET in \cite{Boccardi2007a} can be generalized to this scenario by finding the $K$ strongest statistical eigendirections among the users and acquire CSI for an equivalent number of dimensions per user.
Under these assumptions, the average achievable sum rate for the circular cell is shown in Fig.~\ref{figure_imperfect_pathloss_opportunistic}. The performance behavior is quite similar to the case with perfect CSI in Fig.~\ref{figure_perfect_pathloss}; BD is better than ZFC, except at high correlation, and there is a large gain from greedy stream allocation. However, we stress that this scenario is unrealistic as CSI is only acquired for cell center users, thus reducing the coverage and destroying user fairness as cell edge users are not even considered when their channels are relatively strong.
\subsection{Observations under Quantized CSI}
Next, we consider quantized CSI and let the number of feedback bits (per channel dimension) be scaled as $(N-M) \log_2(P) - \mathtt{constant}$, where the constant is selected as in \cite[Eq.~(17)]{Ravindran2008a} to maintain a 3 dB gap between BD with perfect and quantized CSI. We consider $N=4$ transmit antennas, $M=2$ receive antennas, and RVQ. We also modify\footnote{Quantization errors contribute an average interference of $P (|\mathcal{S}|-1)/|\mathcal{S}| \vect{E}_{\text{quant}}$, where $\vect{E}_{\text{quant}}=N/(M(N-M)) D^{\text{BD}} \vect{R}_{R,k}$ for BD and $\vect{E}_{\text{quant}}=D^{\text{QBC}} G /(N-1)$ for ZFC. When calculating $D^{\text{BD}}$ and $D^{\text{QBC}}$, BD uses $M$ times more feedback bits per user than ZFC.} the CBSUS algorithm in \cite{Shen2006a} to include the average interference due to quantization.
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure10.pdf} \vskip-2mm
\caption{The average achievable sum rate with BD and ZFC, quantized CSI feedback, $N=4$ transmit antennas, $M=2$ receive antennas, uncorrelated channels, and varying SNR.
The number of feedback bits is scaled with the transmit power according to Corollary \ref{cor_multiplexing_gain_quantized} and \cite[Eq.~(17)]{Ravindran2008a}.}\label{figure_RVQ_scaling}
\end{center} \vskip-5mm
\end{figure}
First, we compare BD (having either quantized or perfect CSI) with quantized ZFC using MESC-MMSE combining \cite{Trivellato2008a} and with single-user SVD-based transmission (to a randomly selected user). The quantized effective channels are obtained from 8 users under ZFC, while the entire channels are quantized for 4 users under BD. The average achievable sum rate is shown in Fig.~\ref{figure_RVQ_scaling} as a function of the average SNR. At low SNRs, quantized BD only selects one user and performs similar to single-user transmission. As two data streams are transmitted to the selected user, both strategies are slightly better than ZFC in this regime. But quantized ZFC quickly improves with SNR and becomes the method of choice at practical SNRs. The simulation was stopped at $P=14.3$ dB where BD requires feedback of 22 bits per user, meaning that the best codeword is selected in a codebook with over a million entries.\footnote{An approach to emulate RVQ for very large random codebooks was proposed in \cite{Ravindran2008a}, but this does not change the fact that the quantization complexity becomes infeasible much faster under BD than under ZFC.} BD is therefore suboptimal both in terms of sum rate and computational complexity.
This observation stands in contrast to the numerical results in \cite{Ravindran2008a}, where BD clearly beats ZFC under quantized CSI. To explain the difference, we repeat the simulation in \cite[Fig.~6]{Ravindran2008a} with $N=6$ transmit antennas and $M=2$ receive antennas.
In this simulation, the RVQ codebooks contain 10 bits/user under BD and 5 bits/user under ZFC. The achievable sum rate is shown in Fig.~\ref{figure_RVQ} for the quantized BD
approach in \cite{Ravindran2008a} and the ZFC-QBC approach in \cite{Jindal2008a}. We have also included: 1) an improved version of ZFC-QBC where the MMSE receive combiner is applied during transmission; and 2) single-user SVD-based transmission to a randomly selected user.
Our simulation confirms that BD is better than ZFC in this scenario, but the difference becomes much smaller when the MMSE combiner is applied.
However, none of these strategies should be used in this scenario since single-user transmission is vastly superior. The explanation is that the number of feedback bits is fixed at a number that only satisfies/exceeds the feedback scaling law in \eqref{eq_total_bits} and \cite[Eq.~(17)]{Ravindran2008a} at low SNRs (cf. \cite[Fig.~2]{Ravindran2008a}), while the strict interference mitigation in BD and ZFC is only practically meaningful at high SNR. The observation in \cite{Ravindran2008a} is thus misleading and does not contradict the superiority of ZFC under proper feedback loads.
Conclusions from the mathematical and numerical analysis are summarized in the next section.
\begin{figure}
\begin{center}
\includegraphics[width=86mm]{figure11.pdf} \vskip-2mm
\caption{Comparison of single-user transmission, BD, and different forms of ZFC under quantized CSI feedback. The scenario is the same as in \cite[Fig.~6]{Ravindran2008a}, where the superior single-user strategy was not included.}\label{figure_RVQ}
\end{center} \vskip-5mm
\end{figure}
\section{Conclusion}
This paper analyzed how to divide data streams among users in a downlink system with many multi-antenna users; should few users be allocated many streams, or many users be allocated few streams? New and generalized analytic results were obtained to study this tradeoff under spatial correlation, user selection, heterogeneous user channel conditions, and practical CSI acquisition.
The main conclusion is that sending one stream per selected user and exploiting receive combining is the best choice under realistic conditions. This is good news as it reduces the hardware requirements at the users, compared with multi-stream multiplexing, and enables computationally efficient resource allocation as in \cite{Bjornson2013d}. The result is explained by a stronger resilience towards spatial correlation and larger benefit from user selection. To arrive at alternative conclusions, one has to consider a scenario with heterogeneous user conditions with either perfect CSI (unrealistic) or where CSI is only acquired for the strongest users (destroys coverage and fairness). It should however be noted that if only very inaccurate CSI can be acquired, then inter-user interference will limit performance thus making single-user transmission advantageous.
\appendices
\section{Collection of Lemmas}
\label{app_collection_of_lemmas}
This appendix contains two lemmas that are essential for proving the theorems of this paper.
The first result shows how spatial correlation at the receiver affects the channel directions.
\begin{lemma} \label{lemma_row_space}
Let $\vect{A} \succ \vect{0}_M$ be any Hermitian positive-definite matrix and let $\widetilde{\vect{H}} \in \mathbb{C}^{M \times N}$ be an arbitrary matrix. Then,
$\mathrm{span}(\widetilde{\vect{H}})=\mathrm{span}(\vect{A} \widetilde{\vect{H}})$, where $\mathrm{span}(\cdot)$ denotes the row space.
\end{lemma}
\begin{IEEEproof}
Let $\vect{A}=\vect{U}_A \vect{\Lambda}_A \vect{U}_A^H$ be an eigen decomposition of $\vect{A}$. The lemma follows by observing that $\vect{U}_A$ only rotates the basis vectors of the row space and $\vect{\Lambda}_A$ scales the rows without affecting their span.
\end{IEEEproof}
The second result generalizes the bounding of performance loss under imperfect CSI in \cite{Jindal2008a,Ravindran2008a}.
\begin{lemma} \label{lemma_inequality_rate_difference}
Let $\vect{W}_{k}, \widetilde{\vect{W}}_{k}$ be isotropically distributed on the Grassmannian manifold $\mathcal{G}_{N,d_k}$ and independent of $\vect{H}_k$, then
\begin{equation}
\begin{split}
\mathbb{E} \Big\{ &\log_2 \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \widetilde{\vect{C}}_k^H \vect{H}_k \vect{W}_{k} \vect{W}_{k}^H \vect{H}_k^H \widetilde{\vect{C}}_k \Big) \Big\} \\
-&\mathbb{E} \left\{ \log_2 \! \frac{\det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \fracSum{\ell} \widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_{\ell} \widetilde{\vect{W}}_{\ell}^H \vect{H}_k^H \widetilde{\vect{C}}_k \Big) \!}{ \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \fracSum{\ell \neq k} \widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_{\ell} \widetilde{\vect{W}}_{\ell}^H \vect{H}_k^H \widetilde{\vect{C}}_k \Big) \! } \right\} \\
& \leq \log_2 \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \fracSum{\ell \neq k} \mathbb{E} \big\{ \widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_{k} \widetilde{\vect{W}}_{k}^H \vect{H}_k^H \widetilde{\vect{C}}_k \big\} \Big).
\end{split}
\end{equation}
\end{lemma}
\begin{IEEEproof}
This lemma follows from two inequalities. First,
\begin{equation} \label{eq_first_inequality}
\begin{split}
&\mathbb{E} \Big\{ \log_2 \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \widetilde{\vect{C}}_k^H \vect{H}_k \vect{W}_{k} \vect{W}_{k}^H \vect{H}_k^H \widetilde{\vect{C}}_k \Big) \Big\} \\
&- \mathbb{E} \Big\{ \log_2 \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \sum_{\ell} \widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_{\ell} \widetilde{\vect{W}}_{\ell}^H \vect{H}_k^H \widetilde{\vect{C}}_k \Big) \Big\} \leq 0
\end{split}
\end{equation}
since $\widetilde{\vect{C}}_k^H \vect{H}_k \vect{W}_{k} \vect{W}_{k}^H \vect{H}_k^H \widetilde{\vect{C}}_k$ and $\widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_k \widetilde{\vect{W}}_k^H \vect{H}_k^H \widetilde{\vect{C}}_k$ have the same distribution, and the second term contains additional positive semi-definite matrices. Second, applying Jensen's inequality on the concave function $\log_2 \det(\cdot)$ gives
\begin{equation} \label{eq_second_inequality}
\begin{split}
\mathbb{E} & \Big\{ \log_2 \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \sum_{\ell \neq k} \widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_{\ell} \widetilde{\vect{W}}_{\ell}^H \vect{H}_k^H \widetilde{\vect{C}}_k \Big) \Big\} \\
&\leq \log_2 \det \!\Big( \vect{I}_{d_k} \!+\! \frac{P}{N} \sum_{\ell \neq k} \mathbb{E} \big\{ \widetilde{\vect{C}}_k^H \vect{H}_k \widetilde{\vect{W}}_{\ell} \widetilde{\vect{W}}_{\ell}^H \vect{H}_k^H \widetilde{\vect{C}}_k \big\} \Big).
\end{split}
\end{equation}
The lemma follows from combining \eqref{eq_first_inequality} and \eqref{eq_second_inequality}.
\end{IEEEproof}
\section{}
\label{app_theorem_asymptotic_difference}
\textit{Proof of Theorem \ref{theorem_asymptotic_difference}:} Using \eqref{eq_sum_rate_asymptotic}, the expected asymptotic difference is
\begin{equation} \label{eq_asymptotic_difference1}
\bar{\beta}_{\text{BD-ZFC}} = \mathbb{E}\left\{ \log_2 \left( \frac{\prod_{k \in \mathcal{S}^{\text{BD}}} \det (\vect{H}_k \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{H}_k^H)}
{ \prod_{\ell \in \mathcal{S}^{\text{ZFC}}} |\tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell} \vect{w}^{\text{ZFC}}_{\ell}|^2} \right) \right\}.
\end{equation}
The direction $\frac{\tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell}}{\| \tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell}\|_2}$ is isotropically distributed on the unit sphere, according to Lemma \ref{lemma_distribution_effective_channel}. This enables us to rewrite \eqref{eq_asymptotic_difference1} as
\begin{equation} \label{eq_asymptotic_difference_rewritten}
\begin{split}
\bar{\beta}_{\text{BD-ZFC}} =& \mathbb{E}\left\{ \log_2 \left( \frac{\prod_{k \in \mathcal{S}^{\text{BD}}} \det (\widetilde{\vect{H}}_k \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \widetilde{\vect{H}}_k^H)}
{ \prod_{k \in \mathcal{S}^{\text{ZFC}}} |\widetilde{\vect{h}}_k^H \vect{w}^{\text{ZFC}}_{k}|^2} \right) \right\} \\ &+ \sum_{k \in \mathcal{S}^{\text{BD}}} \log_2 \det(\vect{R}_{R,k}) - \sum_{\ell \in \mathcal{S}^{\text{ZFC}}} \mathbb{E} \{ z_{\ell} \}
\end{split}
\end{equation}
where $z_{\ell} = \mathbb{E} \left\{ \log_2 \left( \frac{ \| \tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell} \|_2^2 }{\| \widetilde{\vect{h}}_{\ell}\|_2^2} \right) \right\}$ and $\widetilde{\vect{h}}_{\ell} \sim \mathcal{CN}(\vect{0},\vect{I}_N)$.\footnote{The vector $\widetilde{\vect{h}}_{\ell}$ is correlated with $\tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell}$ (they have the same direction), but this property does not affect the proof.} The first term in \eqref{eq_asymptotic_difference_rewritten} equals the first term in \eqref{eq_asymptotic_difference} by applying \cite[Theorem 3]{Lee2007a}. The cited theorem was stated for uncorrelated channels, but can be applied in our scenario since $\vect{W}_k^{\text{BD}}$ is not affected by the receive-side correlation matrices $\vect{R}_{R,\ell} \,\, \forall \ell \in \mathcal{S}^{\text{ZFC}}$; see Lemma \ref{lemma_row_space} in Appendix \ref{app_collection_of_lemmas}. Two bounds on $z_{\ell}$ are given in the theorem. The lower bound is achieved by the suboptimal choice of $\tilde{\vect{c}}_{\ell}$ as the dominating eigenvector of $\vect{R}_{R,k}$; this makes $\tilde{\vect{c}}_{\ell}^H \vect{H}_{\ell} \sim \mathcal{CN}(\vect{0},\lambda_{\ell,M} \vect{I}_N)$. The upper bound is achieved from Lemma \ref{lemma_distribution_effective_channel} by applying Jensen's inequality and computing
$\mathbb{E} \{ \log_2 ( \| \widetilde{\vect{h}}_{\ell}\|_2^2 )\} = \frac{\psi(N)}{\log_e(2)}$ using standard integration.
\section{}
\label{app_theorem_impact_of_scheduling}
\textit{Proof of Theorem \ref{theorem_impact_of_scheduling}:}
We begin with BD and assume that there are $K$ candidates to become the new user $k$, $\mathcal{K}=\{1,\ldots,K\}$, while the other users in $\mathcal{S}^{\text{BD}}$ are fixed. Since $|\mathcal{S}^{\text{BD}}|=\frac{N}{M}$, all available degrees of freedom are consumed by the interference cancelation. The precoding matrix $\vect{W}_k^{\text{BD}}$ is therefore completely determined by the common null space of the co-users' channels and fixed in this proof.
Minimizing \eqref{eq_expected_precoding_loss} corresponds to finding the user $\ell \in \mathcal{K}$ with the row space of $\vect{H}_{\ell}$ most compatible with $\vect{W}_k^{\text{BD}}$. For a user candidate $\ell \in \mathcal{K}$, we can lower bound \eqref{eq_expected_precoding_loss} as
\begin{equation} \label{eq_expected_loss_BD_boundingsteps}
\begin{split}
- &\mathbb{E}\{ \log_2 \det (\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H) \} \\
&= - M \mathbb{E}\{ \log_2 (\det (\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H)^{1/M}) \} \\
&\geq -M \mathbb{E}\left\{ \log_2 \left(\frac{\mathrm{tr}(\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H )}{M}\right) \right\} \\
&\geq - M \log_2 \left(\frac{\mathbb{E}\{\mathrm{tr}(\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H )\} }{M}\right) \\
&= - M \log_2 \left(1+ \frac{\mathbb{E}\{\mathrm{tr}(\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H )- M\} }{M}\right).
\end{split}
\end{equation}
The first inequality is the classic inequality between arithmetic and geometric means, while the second inequality follows from applying Jensen's inequality on the convex function $-\log_2 \det(\cdot)$. The final expression in \eqref{eq_expected_loss_BD_boundingsteps} contains $M-\mathrm{tr}(\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H )$, which is the squared chordal distance between $\vect{B}_{\ell}$ and $\vect{W}_k^{\text{BD}}$.
Since the matrices $\vect{B}_{\ell}$, for $\ell \in \mathcal{K}$, are independent and isotropically distributed on the Grassmannian manifold $\mathcal{G}_{N,M}$ irrespective of the receive-side correlation (see Lemma \ref{lemma_row_space}), we can bring in results from \cite{Dai2008a} on quantization of Grassmannian manifolds using $K$ random codewords. From \cite[Theorem 4]{Dai2008a}, we have the following lower bound on the average squared chordal distance (for sufficiently large $K$):
\begin{equation} \label{eq_codebook_bounds}
\begin{split}
\min_{\ell \in \mathcal{K}} \, \mathbb{E} \Big\{ M-&\mathrm{tr}(\vect{B}_{\ell} \vect{W}_k^{\text{BD}} \vect{W}_k^{\text{BD},H} \vect{B}_{\ell}^H ) \Big\} \\ &\geq \frac{M(N-M)}{M(N-M)+1} c_{N,M,M,2}^{-\frac{1}{M(N-M)}} K^{-\frac{1}{M(N-M)}}
\end{split}
\end{equation}
where $c_{N,M,M,2}$ is a positive constant defined in \cite[Eq.~(8)]{Dai2008a}. Plugging \eqref{eq_codebook_bounds} into \eqref{eq_expected_loss_BD_boundingsteps} yields the lower bound for BD in the theorem.
A similar approach can be taken under ZFC (by setting $M=1$ in the derivation), but the $M$ receive antennas provide degrees of freedom to select the effective channel as the vector in the row space of $\vect{H}_{\ell}$ that minimizes the chordal distance to $\vect{w}_k^{\text{ZFC}}$. This is done by the QBC approach in \cite{Jindal2008a}, which was derived for uncorrelated channels but can be applied under receive correlation due to Lemma \ref{lemma_row_space}. We apply \cite[Lemma 1]{Jindal2008a}, which says that the minimal chordal distance is the minimum of $K$ independent $\beta(N-M,M)$-distributed random variables. This quantity can be lower bounded by taking the minimum of $K$ independent $\beta(N-M,1)$ variables and further lower bounded by the quantization bound in \cite[Theorem 4]{Dai2008a}:
\begin{equation} \label{eq_codebook_bounds_ZF}
\begin{split}
\min_{\ell \in \mathcal{K}} \, \mathbb{E} \left\{ \! 1 \! - \! \left|\frac{\vect{h}_{\ell}^H \vect{w}_k^{\text{ZFC}} }{\| \vect{h}_{\ell}\|_2} \right|^2 \!\right\} \!\geq\! \frac{(N\!-\!M) K^{-\frac{1}{(N-M)}}}{(N\!-\!M)+1} c_{N-M+1,1,1,2}^{-\frac{1}{(N-M)}}
\end{split}
\end{equation}
where $c_{N-M+1,1,1,2}$ is a positive constant defined in \cite[Eq.~(8)]{Dai2008a}. Plugging \eqref{eq_codebook_bounds_ZF} into \eqref{eq_expected_loss_BD_boundingsteps} for $M=1$ yields the lower bound for ZFC in the theorem.
\section{}
\label{app_theorem_rateloss_BD}
\textit{Proof of Theorem \ref{theorem_rateloss_BD}:} Using Lemma \ref{lemma_row_space}, the row space of the correlated channel $\vect{H}_k = \vect{R}^{1/2}_{R,k} \widetilde{\vect{H}}_k$ is the same as for the uncorrelated channel $\widetilde{\vect{H}}_k$. Consequently, $\bar{\vect{W}}_{k}^{\text{BD}}$ will be isotropically distributed
on the Grassmannian manifold $\mathcal{G}_{N,M}$, just as proved for uncorrelated channels in \cite[Theorem 1]{Ravindran2008a}.
The performance loss can therefore be bounded using Lemma \ref{lemma_inequality_rate_difference} and it only remains to
characterize $\mathbb{E} \{ \vect{H}_k \bar{\vect{W}}^{\text{BD}}_{\ell} \bar{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \}$
for $\ell \neq k$. Observe that
\begin{equation}
\begin{split}
\mathbb{E} \{ \vect{H}_k &\bar{\vect{W}}_{\ell}^{\text{BD}} \bar{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \} \\
&=\vect{R}^{1/2}_{R,k} \mathbb{E} \{ \vect{L}_k \vect{Q}_k \bar{\vect{W}}_{\ell}^{\text{BD}} \bar{\vect{W}}_{\ell}^{\text{BD},H} \vect{Q}_k^H \vect{L}_k^H \} \vect{R}^{1/2,H}_{R,k}
\end{split}
\end{equation}
using that $\vect{H}_k = \vect{R}^{1/2}_{R,k} \widetilde{\vect{H}}_k = \vect{R}^{1/2}_{R,k} \vect{L}_k \vect{Q}_k$, where $\vect{L}_k \in \mathbb{C}^{M \times M}$ is the lower triangular matrix and $\vect{Q}_k \in \mathbb{C}^{M \times N}$ is the semi-unitary matrix in an LQ decomposition of $\widetilde{\vect{H}}_k$.
Observe that $\vect{L}_k$ and $\vect{Q}_k$ are independent, thus we can calculate their expectations sequentially as
\begin{equation} \label{eq_calculating_inner_expectations}
\begin{split}
\mathbb{E} \{\vect{L}_k &\vect{Q}_k \bar{\vect{W}}_{\ell}^{\text{BD}} \bar{\vect{W}}_{\ell}^{\text{BD},H} \vect{Q}_k^H \vect{L}_k^H \} \\
&= \frac{D^{\text{BD}}}{N-M} \mathbb{E} \{\vect{L}_k \vect{I}_M \vect{L}_k^H \} = \frac{N D^{\text{BD}}}{N-M} \vect{I}_M.
\end{split}
\end{equation}
The first equality follows from \cite[Eq.~(43)--(45)]{Ravindran2008a}, while the second follows from
$\mathbb{E} \{ \vect{L}_k \vect{L}_k^H \} = N \vect{I}_M$ (since $\mathbb{E} \{ \widetilde{\vect{H}}_k \widetilde{\vect{H}}_k^H \} = N \vect{I}_M$).
Plugging \eqref{eq_calculating_inner_expectations} into Lemma \ref{lemma_inequality_rate_difference} yields
\begin{equation}
\Delta^{\text{BD}} \leq \log_2 \det \left( \vect{I}_{M} +\frac{P}{N} \!\left(\frac{N}{M}-1 \right) \!\frac{N D^{\text{BD}}}{N-M} \vect{R}_{R,k} \right)
\end{equation}
from which \eqref{eq_bounding_QBD_loss_theorem} follows directly. The approximate expression for $ D^{\text{BD}}$ is given in \cite[Eq.~(26)]{Ravindran2008a}.
\section{}
\label{app_theorem_rateloss_ZF}
\textit{Proof of Theorem \ref{theorem_rateloss_ZF}:} This proof follows along the lines of \cite[Theorem 1]{Jindal2008a}, with the difference that 1) we have spatial correlation at the receiver; and 2) we use QBC also under perfect CSI. Using Lemma \ref{lemma_row_space}, we observe that the row space of the correlated channel $\vect{H}_k = \vect{R}^{1/2}_{R,k} \widetilde{\vect{H}}_k$ is the same as for the uncorrelated channel $\widetilde{\vect{H}}_k$. Since the gain of the effective channel is ignored in \eqref{eq_QBC_receiver}, the error-minimizing codeword is the same as for uncorrelated channels and we can apply \cite[Lemma 2]{Jindal2008a} to conclude that the direction of the effective channel $\vect{h}_k = \vect{H}_k^H \tilde{\vect{c}}^{\text{QBC}}_k$ is isotropically distributed. The beamforming vector $\bar{\vect{w}}_{k}^{\text{ZFC}}$ is independent of $\vect{h}_k$ and also isotropic, thus the performance loss can be bounded using Lemma \ref{lemma_inequality_rate_difference}. It only remains to characterize $\mathbb{E} \{ | \vect{h}_k^H \bar{\vect{w}}^{\text{ZFC}}_{\ell} |^2 \} = \mathbb{E} \{ \| \vect{h}_k \|_2^2 \} \mathbb{E} \{ | \frac{ \vect{h}_k^H}{\|\vect{h}_k\|_2} \bar{\vect{w}}^{\text{ZFC}}_{\ell} |^2 \}$ for $\ell \neq k$. The second factor equals $\frac{D^{\text{QBC}}}{N-1}$ using \cite[Lemma 2]{Jindal2006a} and \cite[Eq.~(17)]{Jindal2008a}, while computing the average norm $\mathbb{E} \{ \| \vect{h}_k \|_2^2 \}$ of the effective channel is nontrivial. To enable reuse of results from \cite{Jindal2008a}, let $\tilde{\vect{c}}^{\text{U-QBC}}_k$ be the QBC for the uncorrelated channel $\widetilde{\vect{H}}_k$ and observe that
\begin{equation}
\tilde{\vect{c}}^{\text{QBC}}_k = \frac{\vect{R}^{-1/2}_{R,k} \tilde{\vect{c}}^{\text{U-QBC}}_k}{\| \vect{R}^{-1/2}_{R,k} \tilde{\vect{c}}^{\text{U-QBC}}_k \|_2}.
\end{equation}
We can therefore express the effective channel as
\begin{equation}
\begin{split}
\vect{h}_k =\vect{H}_k^H \tilde{\vect{c}}^{\text{QBC}}_k =\vect{H}_k^H \frac{\vect{R}^{-1/2}_{R,k} \tilde{\vect{c}}^{\text{U-QBC}}_k}{\| \vect{R}^{-1/2}_{R,k} \tilde{\vect{c}}^{\text{U-QBC}}_k \|_2} = \frac{\widetilde{\vect{H}}_k^H \tilde{\vect{c}}^{\text{U-QBC}}_k}{\| \vect{R}^{-1/2}_{R,k} \tilde{\vect{c}}^{\text{U-QBC}}_k \|_2}
\end{split}
\end{equation}
and its squared norm will be
\begin{equation} \label{eq_norm_efficient_channel}
\|\vect{h}_k\|_2^2 = \|\widetilde{\vect{H}}_k^H \tilde{\vect{c}}^{\text{U-QBC}}_k\|_2^2 \frac{1}{\tilde{\vect{c}}^{\text{U-QBC},H}_k \vect{R}^{-1}_{R,k} \tilde{\vect{c}}^{\text{U-QBC}}_k}.
\end{equation}
The first factor is the same as under uncorrelated fading and satisfies
$\mathbb{E} \{ \|\widetilde{\vect{H}}_k^H \tilde{\vect{c}}^{\text{U-QBC}}_k\|_2^2 \}= N-M+1$ (see \cite[Lemma 4]{Jindal2008a}), while the second factor depends on $\vect{R}_{R,k}$.
Since both the quantization codebook and $\widetilde{\vect{H}}_k$ are isotropically distributed, $\tilde{\vect{c}}^{\text{U-QBC}}_k$ is also isotropic and the two terms in
\eqref{eq_norm_efficient_channel} are independent. To characterize the second term, observe that $\tilde{\vect{c}}^{\text{U-QBC}}_k$ can be viewed as a normalized uncorrelated circular-symmetric complex Gaussian vector. By using that the eigenvectors of $\vect{R}_{R,k}$ are not affecting the distribution and that squared magnitudes of $\mathcal{CN}(0,1)$-variables are exponentially distributed \cite{Hammarwall2008a}, we conclude that the second term of \eqref{eq_norm_efficient_channel} has the same distribution as
\begin{equation}
\frac{\sum_{i=1}^{M} \xi_i}{\sum_{i=1}^{M} \frac{\xi_i}{\lambda_{k,i}}}
\end{equation}
for some independent exponentially distributed $\xi_i \sim \mathrm{Exp}(1)$. For any $a$ such that $\lambda_{k,m} \leq a \leq \lambda_{k,m+1}$, we can write the CDF as
\begin{equation}
\begin{split}
\mathrm{Pr}&\left\{ \frac{\sum_{i=1}^{M} \xi_i}{\sum_{i=1}^{M} \frac{\xi_i}{\lambda_{k,i}}} \leq a \right\} \\ &= \mathrm{Pr}\Bigg\{
\sum_{i=1}^{m} \underbrace{\left( \frac{a}{\lambda_{k,i}}-1\right)}_{\geq 0} \xi_i - \! \sum_{i=m+1}^{M} \underbrace{\left( 1-\frac{a}{\lambda_{k,i}}\right)}_{\geq 0} \xi_i
\geq 0 \Bigg\}.
\end{split}
\end{equation}
This is the difference of two sums of exponentially distributed variables (with distinct positive variances). The PDF of each sum is characterized by \cite[Theorem 4]{Hammarwall2008a} and by calculating their convolution and integrating over all positive values, we achieve the CDF
\begin{equation}
\begin{split}
\mathrm{Pr}&\left\{ \frac{\sum_{i=1}^{M} \xi_i}{\sum_{i=1}^{M} \frac{\xi_i}{\lambda_{k,i}}} \leq a \right\} \\& =
\sum_{n=1}^m \sum_{t=m+1}^{M} \frac{ (\mu_n-a^{-1})^{m} (a^{-1}-\mu_t)^{M-m-1} }{(\mu_n-\mu_t) \condProd{i=1}{i \neq n}{m} (\mu_n-\mu_i) \condProd{j=m+1}{j \neq l}{M} \!\!(\mu_j-\mu_t) }
\end{split}
\end{equation}
using the simplifying notation $\mu_{n} = \frac{1}{\lambda_{k,n}}$. The corresponding mean value is achieved from the CDF by simply taking the derivative and sum up the mean values over each $a$-interval. By multiplying the mean value expression with $N-M+1$ (i.e., the contribution of the first part in \eqref{eq_norm_efficient_channel}), we achieve the expression for $G_k$.
\section{}
\label{app_theorem_rateloss_BD_EST}
\textit{Proof of Theorem \ref{theorem_rateloss_BD_EST}:} The proof follows along the lines of Theorem \ref{theorem_rateloss_BD}, but we consider CSI estimation errors instead of quantization errors. First, observe that both $\vect{W}_{\ell}^{\text{BD}}$ and $\widehat{\vect{W}}_{\ell}^{\text{BD}}$ are isotropically distributed
on the Grassmannian manifold $\mathcal{G}_{N,M}$ (since receive-side correlation is not affecting the row space of $\vect{H}_k$ and $\widehat{\vect{H}}_k$; see Lemma \ref{lemma_row_space}).
The performance loss can therefore be bounded using Lemma \ref{lemma_inequality_rate_difference} and it only remains to
characterize $\mathbb{E} \{ \vect{H}_k \widehat{\vect{W}}^{\text{BD}}_{\ell} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \}$
for $\ell \neq k$. From \eqref{eq_MMSE_estimator} we have
\begin{equation} \label{eq_true_estimated_channel}
\vect{H}_k = \widehat{\vect{H}}_k + \vect{R}_{E,k}^{1/2}\widetilde{\vect{E}}_k
\end{equation}
where the second term is the estimation error, $\vect{R}_{E,k} = \left(\vect{R}_{R,k}^{-T} + \frac{\vect{T}_k^H \vect{T}_k}{\sigma^2} \right)^{-1}$, and $\widetilde{\vect{E}}_k$ has $\mathcal{CN}(0,1)$-entries.
By using that $\widehat{\vect{H}}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} = \vect{0}$ for $\ell \neq k$, we achieve
\begin{equation}
\begin{split}
\mathbb{E} \{ \vect{H}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \}
= \vect{R}_{E,k}^{1/2}
\mathbb{E} \{ \widetilde{\vect{E}}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \widetilde{\vect{E}}_k^H \}
\vect{R}_{E,k}^{1/2}
\end{split}
\end{equation}
where $\mathbb{E} \{ \widetilde{\vect{E}}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \widetilde{\vect{E}}_k^H \}= M\vect{I}_M$ since $\widetilde{\vect{E}}_k$ is complex Gaussian and independent of $\widehat{\vect{W}}_{\ell}^{\text{BD}}$.
Therefore, $\mathbb{E} \{ \vect{H}_k \widehat{\vect{W}}^{\text{BD}}_{\ell} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H \} = M \left( \vect{R}_{R,k}^{-T} + \frac{\vect{T}_k^H \vect{T}_k}{\sigma^2} \right)^{-1}$.
\section{}
\label{app_lemma_distribution_effective_channel}
\textit{Proof of Lemma \ref{lemma_distribution_effective_channel}:} Observe that $\vect{H} = \vect{R}^{1/2}_{R} \widetilde{\vect{H}}$ has the same distribution as $\vect{H} \vect{U}$ for any unitary matrix $\vect{U}$.
Thus, we can rotate $\vect{h}$ arbitrarily without changing the statistics, meaning that $\frac{\vect{h}}{\|\vect{h}\|_2}$ must be isotropically distributed.
Next, note that $\| \vect{h}\|_2^2 = \|\tilde{\vect{c}}^H \vect{H} \vect{U}\|_2^2 = \|\tilde{\vect{c}}^H \vect{H}\|_2^2$, thus unitary rotations will not affect the effective channel gain meaning that the direction and the channel gain are statistically independent.
$\| \vect{h}\|_2^2$ is the dominating eigenvalue of the correlated complex Wishart matrix $\vect{H} \vect{H}^H \in \mathcal{W}_M(N,\vect{R}_{R})$. The expectation in \eqref{eq_mean_value_effective_channel} achieved directly from \cite[Theorem 3]{Bjornson2008d} or by using the moment generating function in \cite{Mckay2006b} (which gives an equivalent expression that looks slightly different).
\section{}
\label{app_theorem_rateloss_ZF_EST}
\textit{Proof of Theorem \ref{theorem_rateloss_ZF_EST}:} This theorem is proved in the same way as Theorem \ref{theorem_rateloss_BD_EST}. The only notable difference is that we use the effective channel $\vect{h}_k$, which has a single effective receive antenna, instead of the original channel $\vect{H}_k$. The effective channel is zero-mean and has an average channel gain $\mathbb{E}\{\|\vect{h}_k\|_2^2\}$ given by \eqref{eq_mean_value_effective_channel}. Thus, the effective channel and its channel estimate is related as $\vect{h}_k^H= \widehat{\vect{h}}_k^H + \left( \frac{1}{\mathbb{E}\{\|\vect{h}_k\|_2^2\}} + \frac{\Psi}{\sigma^2} \right)^{-1/2} \widetilde{\vect{e}}_k^H$, where $\widetilde{\vect{e}}_k \sim \mathcal{CN}(\vect{0},\vect{I}_N)$.
\section{}
\label{app_cor_multiplexing_gain_estimated}
\textit{Proof of Corollary \ref{cor_multiplexing_gain_estimated}:} The sufficiency is easily achieved from Theorem \ref{theorem_rateloss_BD_EST} and Theorem \ref{theorem_rateloss_ZF_EST}. To obtain the necessity with BD, consider the interference term
$\vect{H}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{\Upsilon}}_{\ell} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \vect{H}_k^H = \vect{R}_{E,k}^{1/2}
\widetilde{\vect{E}}_k \widehat{\vect{W}}_{\ell}^{\text{BD}} \widehat{\vect{W}}_{\ell}^{\text{BD},H} \widetilde{\vect{E}}_k^H \vect{R}_{E,k}^{1/2}$, where the true and estimated are related as in \eqref{eq_true_estimated_channel}. This term must be bounded as $P \rightarrow \infty$. Since the product $\widetilde{\vect{E}}_k \widehat{\vect{W}}_{\ell}^{\text{BD}}$ is non-zero almost surely and $ \widehat{\vect{\Upsilon}}_{\ell} \rightarrow \frac{P}{M |\mathcal{S}^{\text{BD}}|} \vect{I}_M$, it is necessary that $\frac{1}{P} \vect{R}_{E,k}$ has bounded elements. This makes \eqref{eq_training_law} a necessary condition. The proof for ZFC is analogous.
\bibliographystyle{IEEEtran}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,528 |
\section{Introduction}
\label{section:introduction}
Axisymmetric pulsations of rotating neutron stars could be excited in
a number of astrophysical scenarios, such as rotating core collapse,
accretion-induced collapse, core quakes due to a large
phase-transition in the equation of state (EOS), or hypermassive
neutron star formation in a binary neutron star merger \citep[see][for
extensive discussions] {stergioulas_04_a, kokkotas_05_a}. These
pulsations are a potential source of detectable high-frequency
gravitational waves. While for nonrotating stars the frequencies of
normal modes can be computed with perturbative methods and a theory of
gravitational wave asteroseismology has already been formulated
\citep{andersson_98_a, kokkotas_01_a, benhar_04_a}, there exist no
accurate frequency determinations for rapidly rotating stars to date,
nor has the theory of gravitational wave asteroseismology been
extended to include the effects of rotation on the oscillation
frequencies. Most existing computations of oscillation modes in
rapidly rotating relativistic stars use the Cowling approximation
\citep[]{yoshida_02_a, stergioulas_04_a, yoshida_05_a}, with the only
exception being the computation of the two lowest-order quasi-radial
modes in \citet{font_02_a} \citep[see also][]{shibata_03_a,
shibata_03_b}.
In this paper we present an accurate determination of several
axisymmetric pulsation modes of rotating stars in general
relativity. The accurate knowledge of the frequencies of different
modes excited in the astrophysical events mentioned above is necessary
both for narrow-banding of the detectors, as well as for solving the
inverse problem, i.e.\ identifying the EOS of high-density matter.
The traditional approach for computing mode frequencies uses
perturbation theory for either solving a time-independent eigenvalue
problem or for obtaining the time evolution of the linearised
equations governing the dynamics of matter and spacetime
\citep[see][for comprehensive reviews of these
approaches]{kokkotas_99_a, kokkotas_03_a}. The advantage of the
perturbative approach is that the equations can be expanded in terms
of spherical harmonics. However, for rapidly rotating relativistic
stars this approach has only worked in the Cowling approximation so
far \citep{yoshida_99_a, yoshida_01_a, yoshida_02_a, yoshida_05_a},
except for zero-frequency $ f $-modes, which were computed in full
general relativity \citep{stergioulas_98_a, morsink_99_a}. The main
problem for applying the perturbative approach in full general
relativity is the absence of analytic boundary conditions at infinity,
which would allow to apply the outgoing-wave boundary conditions
defining the quasi-normal modes \citep[see][for a
review]{stergioulas_03_a}. Only if one assumes the slow-rotation
approximation the problem is still tractable \citep[see, e.g.,][]
{hartle_75_a, kojima_97_a, datta_98_a, ruoff_02_a, pons_05_a}.
In recent years, the time evolution of the non-linear equations
governing the dynamics of matter and spacetime has been introduced
as a promising new approach for computing mode frequencies
\citep{font_00_a, font_01_a, font_02_a, stergioulas_01_a,
stergioulas_04_a}. For small amplitudes, the obtained frequencies
are in excellent agreement with those expected by linear perturbation
theory, while two-dimensional eigenfunctions can be obtained through a
Fourier transform technique \citep[see][SAF
hereafter]{stergioulas_04_a}. The advantages of this method are that
one does not need precise outer boundary conditions and that one can
also study non-linear pulsations.
This study extends the results presented in SAF (which were obtained
in the Cowling approximation) by incorporating the spacetime dynamics
in the evolutions. This is done by using the Isenberg--Wilson--Mathews
approximation of general relativity (also known as the \emph{conformal
flatness condition}; hereafter CFC) where the 3\,+\,1 Einstein
equations reduce to a non-linear set of five coupled elliptic
equations for the lapse function, the shift vector, and the conformal
factor \citep{isenberg_78_a, wilson_96_a}. The approximation
essentially ignores gravitational radiation and is thus appropriate
for equilibrium, quasi-equilibrium, but also for highly dynamical
situations \citep[see, e.g.,][]{cook_96_a, dimmelmeier_02_a,
oechslin_02_a, faber_04_a, saijo_04_a}.
For small pulsation amplitudes we identify several modes, including
the lowest-order $ l = 0 $, 2, and 4 axisymmetric modes as well as
several axisymmetric inertial modes. The pulsations are studied along
the same sequences of uniformly and differentially rotating
relativistic polytropes with index $ N = 1 $ as in SAF. Differential
rotation significantly shifts mode frequencies to smaller values,
increasing the likelihood of detection by current gravitational wave
interferometric detectors. An important feature of the frequency
spectrum, induced by rotation, is the existence of avoided crossings
between different mode sequences \citep[see][]{clement_86_a,
yoshida_01_a}. We observe an extended avoided crossing between the
$ l = 0 $ and $ l = 4 $ first overtones. This is important for
correctly identifying mode frequencies in case of detection.
Our non-linear approach allows us to identify non-linear harmonics
in addition to the well known linear modes. These harmonics arise
due to couplings between various modes or due to non-linear
self-couplings \citep[see also][]{sperhake_01_a, sperhake_02_a};
for similar results obtained for the non-linear oscillations of a torus
orbiting a black hole, see \citet{zanotti_05_a}. It has been suggested
\citep{clark_79_a, eardley_83_a} that nonradial oscillations after
core bounce could be enhanced through a parametric instability with
the quasi-radial mode \citep[see also][for recent related
work]{passamonti_05_a}. In nonrotating or slowly rotating collapse,
such a parametric instability can only take place under special
conditions that would allow the two modes to be in resonance. In our
work we find that rotational shifting of the frequency of different
modes broadens the range of parameters for which interesting
resonances could take place. In particular we notice that the
quasi-radial mode will be in resonance with some inertial mode(s) for
all rotation rates above a critical value. It is thus interesting to
further study the possible energy transfer between different modes
excited after, e.g., a core collapse or an accretion-induced collapse
event, either on secular time-scales or as a possible parametric
instability.
The paper is organized as follows. In Section~\ref{section:framework}
we introduce the mathematical and numerical framework and specify the
initial fluid perturbations, while in
Section~\ref{section:equilibrium_models} we present the sequences of
equilibrium initial models. In Section~\ref{section:linear_pulsations}
we discuss the effects of linear pulsations, focusing on the role of
rotation and avoided crossings of modes, and present a detailed
comparison to results in the Cowling
approximation. Section~\ref{section:recycling} is devoted to the
technique of mode recycling, and in
Section~\ref{section:nonlinear_pulsations} we examine non-linear
effects of the pulsations like mode coupling or mass-shedding-induced
damping. Gravitational wave emission and asteroseismology are
discussed in Section~\ref{section:gravitational_waves}. A summary of
our results in Section~\ref{section:summary} concludes this work.
Unless otherwise noted, we choose dimensionless units for all physical
quantities by setting the speed of light, the gravitational constant,
and the solar mass to one, $ c = G = M_\odot = 1 $. Latin indices run
from 1 to 3, Greek indices from 1 to 4.
\section{Mathematical and numerical framework}
\label{section:framework}
We study axisymmetric pulsations of rapidly rotating relativistic
stars by first constructing several sequences of uniformly and
differentially rotating equilibrium models. In SAF the equilibrium
models were constructed using the numerical code
\texttt{rns}~\citep{stergioulas_95_a}. In the present work we build
the stellar equilibrium models using the self-consistent field method
described in \citet{komatsu_89_a, komatsu_89_b} (KEH hereafter), which
solves the general relativistic hydrostatic equations for rotating
matter distributions whose pressure obeys an EOS given by a polytropic
relation (see Eq.~(\ref{eq:polytropic_eos}) below). Comparisons of the
accuracy of both approaches in the case of uniform rotation can be
found in \citet{nozawa_98_a} and in~\citet{stergioulas_03_a}. Specific
details of the equilibrium models are deferred to
Section~\ref{section:equilibrium_models} below, where a quantitative
comparison of the equilibrium properties of a particular highly
differentially rotating model built either using \texttt{rns} or the
KEH solver is made.
These equilibrium models are taken as initial data for the evolution
code after a suitable perturbation has been added in order to excite
specific modes of oscillation (see
Section~\ref{subsection:perturbations}).
The time dependent numerical simulations are performed with the code
\textsc{CoCoNuT}, developed by \citet{dimmelmeier_02_a,
dimmelmeier_02_b} with a metric solver based on spectral methods as
described in \citet{dimmelmeier_05_a}. The code uses the general
relativistic field equations for a curved spacetime in the
3\,+\,1-split under the assumption of conformal flatness for the
three-metric. The hydrodynamics equations are consistently formulated
in conservation form, and are solved by high-resolution
shock-capturing schemes.
In the code used by SAF, the spacetime dynamics was neglected, and
the attention was focused on the oscillations of the stars on a fixed
background metric given by the solution of the Einstein equations on
the initial time slice. Keeping the spacetime fixed to the initial
equilibrium state during the evolution corresponds to the Cowling
approximation in perturbation theory. In the simulations presented in
this work, however, such a simplification is not made and the
spacetime fields are also allowed to evolve in time.
In the following, we present the mathematical formulation of the
metric and hydrodynamics equations, and then summarise the numerical
methods used for solving them.
\subsection{Metric equations}
\label{subsection:metric_equations}
We adopt the ADM 3\,+\,1 formalism by \citet{arnowitt_62_a} to
foliate a spacetime endowed with a metric $ g_{\mu\nu} $ into a set of
non-intersecting spacelike hypersurfaces. The line element then reads
\begin{equation}
ds^2 = g_{\mu\nu} \, dx^\mu \! dx^\nu = - \alpha^2 dt^2 +
\gamma_{ij} (dx^i + \beta^i dt) (dx^j + \beta^j dt), ~
\label{eq:line_element}
\end{equation}
where $ \alpha $ is the lapse function, $ \beta^i $ is the spacelike
shift three-vector, and $ \gamma_{ij} $ is the spatial three-metric.
In the 3\,+\,1 formalism, the Einstein equations are split into
evolution equations for the three-metric $ \gamma_{ij} $ and the
extrinsic curvature $ K_{ij} $, and constraint equations (the
Hamiltonian and momentum constraints) which must be fulfilled at every
spacelike hypersurface:
\begin{equation}
\setlength{\arraycolsep}{0.14 em}
\begin{array}{rcl}
\partial_t \gamma_{ij} & = & - 2 \alpha K_{ij} +
\nabla_{\!i} \beta_j + \nabla_{\!j} \beta_i,
\\ [0.8 em]
\partial_t K_{ij} & = & - \nabla_{\!i} \nabla_{\!j} \alpha +
\alpha (R_{ij} \!-\! 2 K_{ik} K_j^k) +
\beta^k \nabla_{\!k} K_{ij} + K_{ik} \! \nabla_{\!j} \beta^k
\hspace{-2 em}
\\ [0.5 em]
& & \displaystyle +
K_{jk} \nabla_{\!i} \beta^k - 8 \pi \alpha \! \left( \! S_{ij} -
\frac{\gamma_{ij}}{2} (S_k^k - \rho_\mathrm{H}) \right),
\\ [0.8 em]
0 & = & R - K_{ij} K^{ij} - 16 \pi \rho_\mathrm{H},
\\ [0.8 em]
0 & = & \nabla_{\!i} K^{ij} - 8 \pi S^j,
\end{array}
\label{eq:adm_metric_equations}
\end{equation}
where the maximal slicing condition, $ K^i_i = 0 $, is imposed. In
these equations $ \nabla_{\!i} $ is the covariant derivative with
respect to the three-metric $ \gamma_{ij} $, $ R_{ij} $ is the
corresponding Ricci tensor, and $ R $ is the scalar curvature.
The matter fields of the general relativistic fluid appearing in the
above equations are the spatial components $ S_{ij} $ of the (perfect
fluid) stress-energy tensor $ T^{\mu\nu} = \rho h u^\mu u^\nu + P
g^{\mu\nu}$, the three momenta $ S^i = \rho h W^2 v^i $, and the total
energy $ \rho_\mathrm{H} = \alpha^2 T^{00} $. The fluid is specified
by the rest-mass density $ \rho $, the four-velocity $ u^\mu $, and
the pressure $ P $, with the specific enthalpy defined as
$ h = 1 + \epsilon + P / \rho $, where $ \epsilon $ is the specific
internal energy. The three-velocity of the fluid as measured by an
Eulerian observer is given by
$ v^i = u^i / (\alpha u^0) + \beta^i / \alpha $, and the Lorentz
factor $ W = \alpha u^0 $ satisfies the relation
$ W = 1 / \sqrt{1 - v_i v^i} $.
The equations of the original ADM formulation, when implemented
numerically with standard finite-difference methods, suffer from
several numerical instabilities. For many years there have been
numerous attempts to reformulate these equations into forms better
suited for numerical work \citep[see, e.g.,][and references
therein]{alcubierre_04_a}. One recent approach is based on a
constrained evolution scheme \citep{bonazzola_04_a}, which exploits
the observation that the more constraints are used in the formulation
of the equations the more numerically stable the evolution appears to
be. We refer the interested reader to the discussion in
\citet{dimmelmeier_05_a} and references therein.
Based on the ideas of \citet{isenberg_78_a} and \citet{wilson_96_a},
and as it was done in the work of
\citet{dimmelmeier_02_a, dimmelmeier_02_b}, we follow a similar
strategy. We approximate the general metric $ g_{\mu\nu} $
by replacing the spatial three-metric $ \gamma_{ij} $ with the
conformally flat three-metric
\begin{equation}
\gamma_{ij} = \phi^4 \hat{\gamma}_{ij},
\label{eq:cfc_metric}
\end{equation}
where $ \hat{\gamma}_{ij} $ is the flat metric and $\phi$ is the
conformal factor. Therefore, at all times during a numerical
simulation we assume that all off-diagonal components of the
three-metric are zero, and the diagonal elements have the common
factor $ \phi^4 $.
In this CFC approximation the expression for the extrinsic curvature
becomes time-independent and reads
\begin{equation}
K_{ij} =
\frac {1}{2 \alpha} \left( \nabla_{\!i} \beta_j + \nabla_{\!j}
\beta_i - \frac{2}{3} \gamma_{ij} \nabla_{\!k} \beta^k \right).
\label{eq:extrinsic_curvature_cfc}
\end{equation}
With this the ADM equations~(\ref{eq:adm_metric_equations}) reduce to
a set of five coupled elliptic non-linear equations for the metric
components,
\begin{equation}
\setlength{\arraycolsep}{0.14 em}
\begin{array}{rcl}
\hat{\Delta} \phi & = & \displaystyle - 2 \pi \phi^5 \left( \rho h
W^2 - P + \frac{K_{ij} K^{ij}}{16 \pi} \right), \\ [1.0 em]
\hat{\Delta} (\alpha \phi) & = & \displaystyle 2 \pi \alpha \phi^5
\! \left( \! \rho h (3 W^2 - 2) \! + \! 5 P \! + \! \frac{7 K_{ij}
K^{ij}}{16 \pi} \! \right), \!\!\!\!\!\!\!\!\!\!\!\! \\ [1.0 em]
\hat{\Delta} \beta^i & = & \displaystyle 16 \pi \alpha \phi^4 S^i
+
2 \phi^{10} K^{ij} \hat{\nabla}_{\!j}
\left( \frac{\alpha}{\phi^6} \right) -
\frac{1}{3} \hat{\nabla}^i \hat{\nabla}_{\!k} \beta^k,
\end{array}
\label{eq:cfc_metric_equations}
\end{equation}
where $ \hat{\nabla}_{\!i} $ and $ \hat{\Delta} $ are the flat space
Nabla and Laplace operators, respectively. They do not contain
explicit time derivatives, and thus the metric is calculated by a
fully constrained approach, at the cost of neglecting some
evolutionary degrees of freedom in the spacetime metric (e.g.,
dynamical gravitational wave degrees of freedom). On each time slice
the metric is hence solely determined by the instantaneous
hydrodynamic state, i.e.\ the distribution of matter in space.
The accuracy of the CFC approximation has been tested in various
works, both in the context of stellar core collapse and for
equilibrium models of neutron stars~\citep{cook_96_a,
dimmelmeier_02_a, shibata_04_a, dimmelmeier_05_a, saijo_05_a,
cerda_05_a}. The spacetime of rapidly (uniformly or differentially)
rotating neutron star models is still very well approximated by the
CFC metric~(\ref{eq:cfc_metric}). The accuracy of the approximation is
expected to degrade only in extreme cases, such as a rapidly rotating
black hole.
Recently, \citet{cerda_05_a} have extended the CFC system of equations
by incorporating additional degrees of freedom in the approximation,
which render the spacetime metric exact up to the second
post-Newtonian order (CFC+ approach). Results for uniformly rotating
pulsating neutron stars show only minute differences with respect to
the CFC approximation for the computed frequencies of the quasi-radial
fundamental $ F $-mode and its first two overtones. Moreover, a direct
comparison of the CFC approach with fully general relativistic
simulations of the quasi-radial modes of a rotating star, obtained
in \citet{font_02_a} have also yielded excellent agreement in the
oscillation frequencies.
\subsection{General relativistic hydrodynamics}
\label{subsection:gr_hydrodynamics}
The hydrodynamic evolution of a relativistic perfect fluid is
determined by a system of local conservation equations, which read
\begin{equation}
\nabla_{\!\mu} J^{\mu} = 0, \qquad \nabla_{\!\mu} T^{\mu \nu} = 0,
\label{eq:gr_equations_of_motion}
\end{equation}
where $ J^{\mu} = \rho u^{\mu} $ is the rest-mass current, and
$\nabla_{\!\mu}$ denotes the covariant derivative with respect to the
four-metric $ g_{\mu \nu} $. Following~\citet{banyuls_97_a} we
introduce a set of conserved variables in terms of the primitive
(physical) variables $ (\rho, v_i, \epsilon) $:
\begin{displaymath}
D = \rho W,
\qquad
S_i = \rho h W^2 v_i
\qquad
\tau = \rho h W^2 - P - D.
\end{displaymath}
Using the above variables, the local conservation
laws~(\ref{eq:gr_equations_of_motion}) can be written as a
first-order, flux-conservative hyperbolic system of equations,
\begin{equation}
\frac{1}{\sqrt{- g}} \left[
\frac{\partial \sqrt{\gamma} \mb{U}}{\partial t} +
\frac{\partial \sqrt{- g} \mb{F}^i}{\partial x^i} \right] = \mb{S},
\label{eq:hydro_conservation_equation}
\end{equation}
with the state vector, flux vector, and source vector
\begin{equation}
\setlength{\arraycolsep}{0.14 em}
\begin{array}{rcl}
\mb{U} & = & [D, S_j, \tau], \\ [1.0 em]
\mb{F}^i & = & \displaystyle
\left[ D \hat{v}^i, S_j \hat{v}^i + \delta^i_j P,
\tau \hat{v}^i + P v^i \right], \\ [1.0 em]
\mb{S} & = & \displaystyle
\left[ 0, T^{\mu \nu} \left(
\frac{\partial g_{\nu j}}{\partial x^\mu} -
{\it \Gamma}^\lambda_{\mu \nu} g_{\lambda j} \!\right),
\alpha \left( T^{\mu 0}
\frac{\partial \ln \alpha}{\partial x^\mu} -
T^{\mu \nu} {\it \Gamma}^0_{\mu \nu} \!\right) \right]\!,
\hspace{-2.5 em}
\end{array}
\label{eq:hydro_conservation_equation_constituents}
\end{equation}
respectively. Here $ \hat{v}^i = v^i - \beta^i / \alpha $, and
$ \sqrt{-g} = \alpha \sqrt{\gamma} $, with
$ g = \det (g_{\mu \nu}) $ and $ \gamma = \det (\gamma_{ij}) $. In
addition, $ {\it \Gamma}^\lambda_{\mu \nu} $ are the Christoffel
symbols associated with $ g_{\mu \nu} $.
The system of hydrodynamics
equations~(\ref{eq:hydro_conservation_equation}) is closed by an
EOS, which relates the pressure to some thermodynamically independent
quantities, e.g., $ P = P (\rho, \epsilon) $. For rotating neutron
star models below the mass-shedding limit \citep[see][for the precise
definition of the mass-shedding limit in the case of rapidly rotating
relativistic stars]{friedman_86_a} we assume that the star remains
isentropic. We can thus demand that the pressure obeys an EOS given by
the polytropic relation
\begin{equation}
P = K \rho^\gamma,
\label{eq:polytropic_eos}
\end{equation}
where $ K $ is the polytropic constant and $ \gamma = 1 + 1 / N $ is
the adiabatic index. Note that the evolution equation for the
generalised energy $ \tau $ can be discarded if an EOS of the form
$ P = P (\rho) $ as in Eq.~(\ref{eq:polytropic_eos}) is used. In this
particular case the internal specific energy can be obtained from the
ideal fluid EOS as $ \epsilon = P / [\rho (\gamma - 1)] $.
Near the mass-shedding limit, even small amplitude oscillations can
result in significant shedding of matter from the stellar surface in
the form of shocks (see Section~\ref{subsection:mass_shedding}). In
this case the polytropic relation~(\ref{eq:polytropic_eos}) does not
hold anymore. Therefore, as in the work by SAF, we then employ the
adiabatic ideal fluid EOS instead,
\begin{equation}
P = \rho \epsilon (\gamma - 1).
\label{eq:ideal_gas_eos}
\end{equation}
We point out that at the initial time the isentropic equilibrium
models constructed with the polytropic EOS~(\ref{eq:polytropic_eos})
are consistent with the ideal fluid EOS~(\ref{eq:ideal_gas_eos}).
\subsection{Numerical methods for solving the metric and hydrodynamics
equations}
\label{subsection:numerical_methods}
The hydrodynamics solver performs the numerical time integration of
the system of conservation
equations~(\ref{eq:hydro_conservation_equation}) using a
high-resolution shock-capturing (HRSC) scheme on a finite difference
grid \citep[for a review of such methods in numerical general
relativity, see][]{font_03_a}. This method ensures numerical
conservation of physically conserved quantities and a correct
treatment of discontinuities such as shocks (which may be present in
hydrodynamic quantities). In (upwind) HRSC methods a Riemann problem
has to be solved at each cell interface, which requires the
reconstruction of the primitive variables $ (\rho, v^i, \epsilon) $ at
these interfaces. We use the PPM method for the reconstruction, which
yields third order accuracy in space. The solution of the Riemann
problems then provides the numerical fluxes at cell interfaces. To
obtain this solution, the characteristic structure of the
hydrodynamics equations is explicitly needed \citep{banyuls_97_a}. In
our code the numerical fluxes are computed by means of Marquina's
approximate flux formula \citep{donat_98_a}. The time update of the
conserved vector $ \mb U $ is done using the method of lines in
combination with a Runge--Kutta scheme with second order accuracy in
time. Once the state vector is updated in time, the primitive
variables are recovered through an iterative Newton--Raphson method.
Although the most common approach to numerically solve the Einstein
equations is by means of finite differences, such methods are not
particularly well suited if the metric equations are formulated as
non-linear coupled elliptic equations like in the CFC approach. For
multidimensional simulations, the necessary grid resolutions typically
require to solve computationally very expensive numerical problems. If
iterative solvers in combination with spherical polar coordinates are
used, an additional obstacle may manifest itself in slow or failed
convergence due to problems at the coordinate origin or axis
\citep[see discussion in][]{dimmelmeier_05_a}. A possible remedy to
these shortcomings can be the use of non-linear multigrid solvers.
As an alternative strategy to reduce the complexity associated with
solving elliptic equations by reducing the number of grid points
required for a given numerical accuracy, we utilise an iterative
non-linear solver based on spectral methods in our code. This metric
solver, which is described in detail in \citet{dimmelmeier_05_a}, uses
routines from the publicly available object-oriented \textsc{Lorene}
library (\texttt{www.lorene.obspm.fr}), which supplies routines that
implement spectral methods in spherical polar coordinates. In contrast
to the hydrodynamic quantities, the metric components are always
smooth, and thus spectral methods are ideally suited for numerically
representing the spacetime metric. The practical implementation of the
combination of HRSC methods for the hydrodynamics and spectral methods
for the metric equations (the {\em Mariage des Maillages\/} or `grid
wedding' approach) in a multidimensional numerical code has been
presented in \citet{dimmelmeier_05_a}.
The \textsc{CoCoNuT} code utilises Eulerian spherical polar
coordinates $ \{r, \theta, \varphi \} $, and thus axially or
spherically symmetric configurations can be easily simulated. For the
rotating neutron star models discussed in this work, we choose an
axisymmetric grid setup (no dependence of quantities on the coordinate
$ \varphi $), and assume symmetry with respect to the equatorial
plane. The finite difference grid consists of 160 radial and 60
angular grid points, which are equidistantly spaced. A small part of
the grid covers an artificial low-density atmosphere extending beyond
the stellar surface, whose rest-mass density is $ 10^{-17} $ of the
initial central rest-mass density of the star. The spectral grid of
the metric solver is split into 3 radial domains with 33 radial and 17
angular collocation points each. The innermost radial domain (or
nucleus) stretches from the coordinate origin to half the stellar
equatorial radius, followed by the second radial domain which extends
out to the outer boundary of the finite difference grid. The third
domain uses a compactified radial coordinate and reaches out to
spatial infinity. The metric equations~(\ref{eq:cfc_metric_equations})
can therefore be numerically integrated out to spacelike infinity, and
all noncompact support source terms in these equations can be
consistently handled in a non-approximative way. For further details
about the grid setup and particularly the relevance of a compactified
radial spectral grid for an accurate evolution of rotating equilibrium
models, we refer to \citet{dimmelmeier_05_a}.
Even when using spectral methods the calculation of the spacetime
metric is computationally expensive. Hence, in our simulations the
metric is updated only once every 50 hydrodynamic time steps during
evolution (which corresponds to a time interval of
$ 10^{-3} \mathrm{\ ms} $) and extrapolated in between. The
suitability of this procedure is tested and discussed in detail in
\citet{dimmelmeier_02_a}. We also note that convergence tests with
different grid resolution have been performed to ascertain that the
regular grid resolution specified above is appropriate for our
simulations.
\subsection{Fluid perturbations for exciting the pulsations}
\label{subsection:perturbations}
To excite specific eigenmodes of oscillation in the stellar models we
perturb selected equilibrium variables before starting the
evolution. In the absence of the true eigenfunction of a given mode,
each perturbation is selected so as to mimic the angular dependence of
the eigenfunction of the corresponding mode of a slowly rotating
Newtonian star. Usually, this ensures that the chosen mode will
dominate the time evolution at least for the slower rotating
models. However, since the perturbation is not exact, additional
pulsation modes will be excited, especially for rapidly rotating
models (see Section~\ref{section:recycling}).
For the $ l = 0 $ modes we use as trial eigenfunction a perturbation
of the radial component of the covariant three-velocity in the form
\begin{equation}
v_r = a \sin \left( \pi \frac{r}{r_\mathrm{s}(\theta)} \right),
\label{eq:perturbation_l=0}
\end{equation}
where $ r_\mathrm{s} (\theta) $ is the coordinate radius of the
surface of the star. The constant $ a $ is the amplitude of the
perturbation, for which we choose the value $ -0.005 $ (in units of
$ c $).
The $ l = 2 $ modes are excited by perturbing the $ \theta $-component
of the covariant three-velocity as follows:
\begin{equation}
v_\theta = a \sin \left( \pi \frac{r}{r_\mathrm{s}(\theta)} \right)
\sin \theta \, \cos \theta,
\label{eq:perturbation_l=2}
\end{equation}
where we set $ a $ to 0.01 \citep[see][for more details about
perturbations of this form]{font_01_a}.
Correspondingly, the $ l = 4 $ perturbations are excited using the
following perturbation:
\begin{equation}
v_\theta = a \sin \left( \pi \frac{r}{r_\mathrm{s}(\theta)} \right)
\sin \theta \, \cos \theta \left( 3 - 7 \cos^2 \theta \right).
\label{eq:perturbation_l=4}
\end{equation}
The time series of the evolved perturbations are Fourier analysed
after an evolution time of $ 20 \mathrm{\ ms} $, which corresponds to
a nominal frequency resolution of $ 0.05 \mathrm{\ kHz}$. Due to
the finite evolution time and the numerical viscosity which is
damping the oscillations, the frequency peaks are not
$ \delta $-functions, but correspond to a nearly symmetric bell shape,
spreading over several frequency bins. As in SAF, we use a
second-order accurate numerical derivative formula in order to find
the location of the maximum of a frequency peak, when the peaks are
well separated. In practice, this method gives agreement with expected
frequencies which is significantly better than the nominal frequency
resolution due to the finite evolution time.
The peaks in the Fourier spectra are identified with specific
pulsation modes, starting from the nonrotating member of the sequence,
where the pulsation frequencies are known from perturbation theory. As
the rotation rate increases, it becomes gradually more difficult to
identify specific modes in the Fourier spectrum. For this reason, we
also extract the two-dimensional eigenfunction for each peak in the
Fourier spectrum and use it as an additional criterion to identify
specific modes (see Section~\ref{section:recycling}).
We also emphasize that in contrast to previous work \citep[as, e.g.,
in][and SAF]{font_01_a}, which assumed the Cowling approximation of a
fixed spacetime metric, we do not use a perturbation of the rest-mass
density for the $ l = 0 $ mode,
\begin{equation}
\delta \rho = a \rho_\mathrm{c}
\sin \left( \pi \frac{ r}{r_\mathrm{s}(\theta)} \right),
\label{eq:perturbation_rho}
\end{equation}
where $ \rho_\mathrm{c} $ is the central rest-mass density of the
star. In the Cowling approximation, the oscillation amplitude
$ a_\mathrm{evol} $ of the density at a particular location in the
star during the evolution not only scales linearly with the initial
perturbation amplitude $ a $ (as expected for small $ a $), but its
value is also typically close to $ a $. In contrast to this, if the
spacetime is allowed to evolve, i.e.\ it is coupled to the evolution
of the fluid, we find that the oscillation amplitude
$ a_\mathrm{evol} $ during the evolution is significantly larger than
$ a $, usually by a factor $ \sim 5 \mbox{--} 8 $. Thus even a small
initial perturbation with an amplitude of a few per cent can excite an
oscillation with a large amplitude $ a_\mathrm{evol} $, which can
possibly violate the assumption of linearity.
We attribute this effect to the fact that the perturbation enters the
hydrodynamic evolution
equations~(\ref{eq:hydro_conservation_equation}) not only through the
state vector $ \mb{U} $, but also via the metric in the form
$ \sqrt{\gamma} $, which is proportional to $ \phi^6 $. If
$ \mb{U} $ and (via the metric
equations~\ref{eq:cfc_metric_equations}) also $ \phi^6 $ are
affected by the initial perturbation with amplitude $ a $, the
combination $ \sqrt{\gamma} \, \mb{U} $ exhibits a much larger
oscillation amplitude than $ a $, namely
$ a_\mathrm{evol} \sim (1 + a)^6 \cdot (1 + a) - 1 = (1 + a)^7 - 1 $,
which is roughly $ 7 a $ for $ a \ll 1 $. This is also reflected by an
increase of the total mass of the star by $ (1 + a)^7 $, when applying
this artificial perturbation by adding rest-mass density to the
equilibrium model, as the rest mass integral also contains a term
$ \sqrt{\gamma} \, \rho $. As the spacetime metric adapts to a new
quasi-equilibrium state with this higher rest mass, the oscillation
frequencies are systematically shifted towards higher values than
those obtained if the approximately `mass neutral'
perturbation~(\ref{eq:perturbation_l=0}) is applied. Note that this
effect of frequency shift is negligible in the Cowling approximation
when $ \delta \rho $ is small, as there $ \sqrt{\gamma} $ is constant
in time, and thus the total rest mass of the star is affected by the
density perturbation only through $ \rho $, i.e.\ like $ (1 + a) $.
\section{Equilibrium models}
\label{section:equilibrium_models}
\begin{table*}
\begin{minipage}{140mm}
\centering
\caption{Properties of the four sequences of equilibrium models. A
is a sequence of fixed rest mass $ M_0 = 1.506 \, M_\odot $ with
$ \hat{A} = 1 $, AU is the corresponding sequence of uniformly
rotating models, B is a sequence of fixed central rest-mass
density $ \rho_\mathrm{c} = 1.28 \times 10^{-3} $ with
$ \hat{A} = 1 $, and BU is the corresponding sequence of
uniformly rotating models. All models are relativistic
polytropes with $ N = 1 $ and $ K = 100 $.
$ \varepsilon_\mathrm{c} $ is the central energy density with
$ \varepsilon = \rho (1 + \epsilon) $, $ M $ is the
gravitational mass, $ R $ is the circumferential stellar radius,
and $ \Omega_\mathrm{c/e} $ are the central/equatorial angular
velocity. All other quantities are defined in the main text.}
\label{table:equilibrium_models}
\begin{tabular}{@{}lccrrcccc@{}}
\hline
Model & $ \varepsilon_\mathrm{c} $ & $ M $ &
\multicolumn{1}{c}{$ R $} & \multicolumn{1}{c}{$ r_\mathrm{e} $} &
$ r_\mathrm{p} / r_\mathrm{e} $ & $ \Omega_\mathrm{c} $ &
$ \Omega_\mathrm{e} $ & $ T / |W| $ \\
& $ (\times 10^{-3}) $ & & & & & $ (\times 10^{-2}) $ &
$ (\times 10^{-2}) $ & \\
\hline
A0 & 1.444 & 1.400 & 9.59 & 8.13 & 1.000 & 0.000 & 0.000 & 0.000 \\ [0.5ex]
A1 & 1.300 & 1.405 & 10.01 & 8.54 & 0.930 & 2.019 & 0.759 & 0.018 \\ [0.5ex]
A2 & 1.187 & 1.408 & 10.40 & 8.92 & 0.875 & 2.580 & 0.977 & 0.033 \\ [0.5ex]
A3 & 1.074 & 1.410 & 10.84 & 9.35 & 0.820 & 2.944 & 1.125 & 0.049 \\ [0.5ex]
A4 & 0.961 & 1.413 & 11.37 & 9.87 & 0.762 & 3.192 & 1.232 & 0.066 \\ [0.5ex]
A5 & 0.848 & 1.418 & 12.01 & 10.49 & 0.703 & 3.340 & 1.303 & 0.086 \\ [0.5ex]
A6 & 0.735 & 1.422 & 12.78 & 11.25 & 0.643 & 3.383 & 1.336 & 0.107 \\ [0.5ex]
A7 & 0.622 & 1.427 & 13.75 & 12.21 & 0.579 & 3.339 & 1.337 & 0.131 \\ [0.5ex]
A8 & 0.509 & 1.433 & 15.01 & 13.45 & 0.513 & 3.197 & 1.300 & 0.158 \\ [0.5ex]
A9 & 0.396 & 1.439 & 16.70 & 15.13 & 0.444 & 2.953 & 1.223 & 0.189 \\ [0.5ex]
A10 & 0.283 & 1.447 & 19.03 & 17.44 & 0.370 & 2.604 & 1.101 & 0.223 \\
\hline
AU0 & 1.444 & 1.400 & 9.59 & 8.13 & 1.000 & 0.000 & 0.000 & 0.000 \\ [0.5ex]
AU1 & 1.300 & 1.404 & 10.19 & 8.71 & 0.919 & 1.293 & 1.293 & 0.020 \\ [0.5ex]
AU2 & 1.187 & 1.407 & 10.79 & 9.30 & 0.852 & 1.656 & 1.656 & 0.037 \\ [0.5ex]
AU3 & 1.074 & 1.411 & 11.56 & 10.06 & 0.780 & 1.888 & 1.888 & 0.055 \\ [0.5ex]
AU4 & 0.961 & 1.415 & 12.65 & 11.14 & 0.698 & 2.029 & 2.029 & 0.076 \\ [0.5ex]
AU5 & 0.863 & 1.420 & 14.94 & 13.43 & 0.575 & 2.084 & 2.084 & 0.095 \\
\hline
B0 & 1.444 & 1.400 & 9.59 & 8.13 & 1.000 & 0.000 & 0.000 & 0.000 \\ [0.5ex]
B1 & 1.444 & 1.437 & 9.75 & 8.24 & 0.950 & 1.801 & 0.666 & 0.013 \\ [0.5ex]
B2 & 1.444 & 1.478 & 9.92 & 8.36 & 0.900 & 2.574 & 0.944 & 0.026 \\ [0.5ex]
B3 & 1.444 & 1.525 & 10.11 & 8.49 & 0.850 & 3.189 & 1.160 & 0.040 \\ [0.5ex]
B4 & 1.444 & 1.578 & 10.31 & 8.63 & 0.800 & 3.728 & 1.342 & 0.055 \\ [0.5ex]
B5 & 1.444 & 1.640 & 10.53 & 8.77 & 0.750 & 4.227 & 1.504 & 0.071 \\ [0.5ex]
B6 & 1.444 & 1.713 & 10.76 & 8.91 & 0.700 & 4.707 & 1.651 & 0.087 \\ [0.5ex]
B7 & 1.444 & 1.798 & 11.01 & 9.05 & 0.650 & 5.185 & 1.789 & 0.105 \\ [0.5ex]
B8 & 1.444 & 1.899 & 11.26 & 9.17 & 0.600 & 5.683 & 1.921 & 0.124 \\ [0.5ex]
B9 & 1.444 & 2.020 & 11.50 & 9.26 & 0.550 & 6.232 & 2.052 & 0.144 \\
\hline
BU0 & 1.444 & 1.400 & 9.59 & 8.13 & 1.000 & 0.000 & 0.000 & 0.000 \\ [0.5ex]
BU1 & 1.444 & 1.432 & 9.83 & 8.33 & 0.950 & 1.075 & 1.075 & 0.012 \\ [0.5ex]
BU2 & 1.444 & 1.466 & 10.11 & 8.58 & 0.900 & 1.509 & 1.509 & 0.024 \\ [0.5ex]
BU3 & 1.444 & 1.503 & 10.42 & 8.82 & 0.850 & 1.829 & 1.829 & 0.037 \\ [0.5ex]
BU4 & 1.444 & 1.543 & 10.78 & 9.13 & 0.800 & 2.084 & 2.084 & 0.050 \\ [0.5ex]
BU5 & 1.444 & 1.585 & 11.20 & 9.50 & 0.750 & 2.290 & 2.290 & 0.062 \\ [0.5ex]
BU6 & 1.444 & 1.627 & 11.69 & 9.95 & 0.700 & 2.452 & 2.452 & 0.074 \\ [0.5ex]
BU7 & 1.444 & 1.666 & 12.30 & 10.51 & 0.650 & 2.569 & 2.569 & 0.084 \\ [0.5ex]
BU8 & 1.444 & 1.692 & 13.07 & 11.26 & 0.600 & 2.633 & 2.633 & 0.091 \\ [0.5ex]
BU9 & 1.444 & 1.695 & 13.44 & 11.63 & 0.580 & 2.642 & 2.642 & 0.092 \\
\hline
\end{tabular}
\end{minipage}
\end{table*}
Since our focus is on the effects of rotation on pulsation modes, we
do not survey here a broad range of high-density EOSs, but choose
instead a single polytropic EOS~(\ref{eq:polytropic_eos}) with
$ N = 1 $ and $ K = 100 $. For the sake of comparison we use the same
equilibrium model sequences as SAF to which we refer the reader for
more details. Here we only give a brief overview of the basic
properties of the various sequences.
We restrict our attention to two different sequences of differentially
rotating models (sequences~A and~B) and their uniformly rotating
counterparts (sequences~AU and~BU). The equilibrium properties of all
models are summarised in Table~\ref{table:equilibrium_models}. In the
nonrotating limit all sequences end in the same nonrotating model
(thus, models~A0, AU0, B0, and~BU0 all coincide). Notice that in
Table~\ref{table:equilibrium_models} the numerical values of different
equilibrium properties are displayed as computed by SAF, using the
numerical code \texttt{rns} \citep{stergioulas_95_a}. In contrast, the
equilibrium initial data used in the actual time evolutions presented
here are obtained by a similar, but different, initial data solver,
based on the original KEH method \citep{komatsu_89_a}. An extensive
comparison between the two initial data solvers shows very good
agreement with differences being much less than the 1 per cent level
in all computed equilibrium quantities (the differences arise solely
due to the truncated computational domain of the original KEH
method). We also note that for a few of the most rapidly
differentially rotating equilibrium models used by SAF, the elliptic
solver for the spacetime evolution used here did not converge. These
models are therefore omitted in Table~\ref{table:equilibrium_models}.
The differentially rotating sequence~A and its corresponding uniformly
rotating sequence~AU are characterised by a fixed rest mass
$ M_0 = 1.506 \, M_\odot $. Along sequence~A, the degree of
differential rotation is held fixed at $ \hat{A} = 1 $, where
$ \hat{A} = A / r_\mathrm{e} $ with $ r_\mathrm{e} $ being the
equatorial coordinate radius of the star and $ A $ being the rotation
parameter as defined in~\citet{komatsu_89_a}. The values of $ M_0 $
and $ \hat{A} $ are chosen in order to represent a newly-born,
differentially rotating neutron star. The angular velocity at the
equator is roughly 1/3 to 1/2 of the central angular velocity, which
is similar to the degree of differential rotation obtained in typical
core collapse simulations \citep[see, e.g.,][]{villain_04_a}. The
fastest rotating model in sequence~A has a ratio of polar to
equatorial coordinate axis of only
$ r_\mathrm{p} / r_\mathrm{e} = 0.370 $, which translates into a
rotation rate $ T / |W| = 0.223 $. (Here $ T $ is the rotational
kinetic energy and $ |W| $ is the gravitational potential energy.) The
central rest-mass density is nearly an order of magnitude smaller than
the one of the corresponding nonrotating model, while the
circumferential radius is nearly twice as large. The uniformly
rotating sequence~AU only reaches an axis ratio of 0.575 at a rotation
rate $ T / |W| = 0.095 $, half the central rest-mass density, and a
50 per cent larger radius than the corresponding nonrotating
model. Model~AU5 is at the mass-shedding limit.
On the other hand, the differentially rotating sequence~B and its
corresponding uniformly rotating sequence~BU are characterised by a
fixed central rest-mass density
$ \rho_\mathrm{c} = 1.28 \times 10^{-3} $. In sequence~B the
differential rotation parameter is also set to $ \hat{A} = 1 $. Its
fastest rotating member has a gravitational mass
$ M = 2.02 \, M_\odot $ and an axis ratio of 0.55. Since all models in
the sequence are compact, the radius $ R $ is much smaller than along
sequence~A. The corresponding uniformly rotating sequence~BU only
reaches an axis ratio of 0.58 at the mass-shedding limit (model~BU9)
with an increase in radius $ R $ by 40 per cent. Thus, we see that
when considering a sequence of fixed central rest-mass density, the
uniformly rotating models attain a larger equatorial radius than
differentially rotating models, with the latter expanding out of the
equatorial plane, becoming torus-like.
\section{Linear pulsation modes}
\label{section:linear_pulsations}
When the pulsation amplitude is small (so that, e.g., density
variations are at the level of $ \delta \rho / \rho \sim 10^{-2} $),
the dynamics is close to linear and one can identify the
eigenfrequencies and eigenfunctions of linear quasi-normal modes from
a time evolution. In order to compute the real part of the
eigenfrequency of a pulsation mode, we Fourier-transform the time
series of the evolution of a suitable physical variable (the density
for the $ l = 0 $ modes and $ v_{\theta} $ for the $ l = 2 $
modes). Instead of examining the Fourier spectra at a few specific
points inside the star, we integrate the amplitude of the Fourier
transform along a coordinate line. For instance, in the case of
$ l = 0 $ modes we examine the integrated Fourier amplitude along
$ \theta = \pi / 2 $ (equatorial plane), while for the $ l = 2 $ modes
the integrated Fourier amplitude along a line of $ \theta = \pi / 4 $
is used. We also verify that each identified discrete mode has the
same frequency at any point inside the star in the coordinate frame.
As discussed in SAF, the trial eigenfunction used for exciting the
pulsations does not correspond exactly to a particular mode.
Therefore, additional modes apart from the main mode one wishes to
study are excited, particularly for rapidly rotating models where
rotational coupling effects are significant and higher-order coupling
terms in the mode-eigenfunctions become comparable to the dominant
term. We begin the identification of specific modes along a sequence
of equilibrium models using the known pulsation frequencies of the
nonrotating star in the sequence and by comparing the Fourier
transforms of evolved variables between subsequent models. Erroneous
mode identification could happen close to the mass-shedding limit due
to the appearance of avoided crossings between different modes (see
Section~\ref{subsection:avoided_crossings}). To avoid this we do not
rely on Fourier transforms at only a few points inside the star, but
in many cases reconstruct the whole two-dimensional eigenfunction of
each mode, using Fourier transforms at every point inside the star. At
the eigenfrequency of a specific mode, the amplitude of the Fourier
transform correlates with its eigenfunction. A change in sign in the
eigenfunction corresponds to both the real and imaginary part of the
Fourier transform going through zero. Comparing the eigenfunctions
corresponding to different peaks in the Fourier transforms allows for
an unambiguous identification of specific mode sequences.
Tables~\ref{table:frequencies_a} to~\ref{table:frequencies_a_l=4}
summarise our main results, showing the frequencies of the two
lowest-order $ l = 0 $ modes ($ F $ and $ H_1 $) and $ l = 2 $ modes
($ {}^{2\!}f $ and $ {}^{2}p_1 $) for all four sequences as well as the
two lowest order $ l = 4 $ modes ($ {}^{4}f $ and $ {}^{4}p_1 $) for
sequence~A. In addition, frequencies of several axisymmetric inertial
modes are displayed. All of the above frequency data (except for the
$ {}^{4}f $-mode, which is omitted for reasons of clarity) are shown
as a function of $ T / |W| $ in Figs.~\ref{fig:frequencies_a_au}
and~\ref{fig:frequencies_b_bu}. Next, we first discuss some general
trends due to rotation and then present our results for each mode
sequence in more detail.
\begin{table}
\centering
\caption{Frequencies of the fundamental quasi-radial ($ l = 0 $)
mode, $ F $, its first overtone, $ H_1 $, the fundamental
quadrupole ($ l = 2 $) mode, $ {}^{2\!}f $, its first overtone,
$ {}^{2}p_1 $, and three inertial modes, $ i_{-2} $, $ i_1 $,
and $ i_2 $ for the sequence of differentially rotating models
A. All frequencies are given in kHz.}
\label{table:frequencies_a}
\begin{tabular}{@{}l@{~~}ccccccc@{}}
\hline
Model & $ F $ & $ H_1 $ & $ {}^{2\!}f $ & $ {}^{2}p_1 $ &
$ i_{-2} $ & $ i_1 $ & $ i_2 $ \\
\hline
A0 & 1.458 & 3.971 & 1.586 & 3.726 & 0.000 & 0.000 & 0.000 \\
A1 & 1.400 & 3.816 & 1.577 & 3.580 & 0.302 & 0.460 & 0.596 \\
A2 & 1.358 & 3.733 & 1.567 & 3.424 & 0.399 & 0.603 & 0.779 \\
A3 & 1.307 & 3.664 & 1.550 & 3.237 & 0.477 & 0.711 & 0.917 \\
A4 & 1.248 & 3.583 & 1.535 & 3.013 & 0.543 & 0.794 & 1.022 \\
A5 & 1.184 & 3.494 & 1.513 & 2.780 & 0.603 & 0.863 & 1.108 \\
A6 & 1.105 & 3.352 & 1.482 & 2.557 & 0.646 & 0.914 & 1.163 \\
A7 & 1.018 & 3.360 & 1.432 & 2.315 & 0.683 & 0.946 & 1.198 \\
A8 & 0.915 & 3.114 & 1.360 & 2.068 & 0.711 & 0.965 & 1.204 \\
A9 & 0.809 & 2.985 & 1.264 & 1.827 & 0.723 & 0.958 & 1.180 \\
A10 & 0.685 & 2.830 & 1.098 & 1.610 & 0.712 & 0.911 & 1.104 \\
\hline
\end{tabular}
\end{table}
\begin{table}
\centering
\caption{Same as Table~\ref{table:frequencies_a}, but for the
sequence of uniformly rotating models~AU.}
\label{table:frequencies_au}
\begin{tabular}{@{}l@{~~}ccccccc@{}}
\hline
Model & $ F $ & $ H_1 $ & $ {}^{2\!}f $ & $ {}^{2}p_1 $ &
$ i_{-2} $ & $ i_1 $ & $ i_2 $ \\
\hline
AU0 & 1.458 & 3.971 & 1.586 & 3.726 & 0.000 & 0.000 & 0.000 \\
AU1 & 1.398 & 3.785 & 1.562 & 3.455 & 0.280 & 0.354 & 0.468 \\
AU2 & 1.345 & 3.716 & 1.554 & 3.192 & 0.384 & 0.478 & 0.611 \\
AU3 & 1.283 & 3.635 & 1.537 & 2.885 & 0.468 & 0.575 & 0.736 \\
AU4 & 1.196 & 3.552 & 1.516 & 2.520 & 0.545 & 0.660 & 0.810 \\
AU5 & 1.107 & 3.457 & 1.459 & 2.090 & 0.639 & 0.747 & 0.858 \\
\hline
\end{tabular}
\end{table}
\begin{table}
\centering
\caption{Same as Table~\ref{table:frequencies_a}, but for the
sequence of differentially rotating models~B.}
\label{table:frequencies_b}
\begin{tabular}{@{}l@{~~}ccccccc@{}}
\hline
Model & $ F $ & $ H_1 $ & $ {}^{2\!}f $ & $ {}^{2}p_1 $ &
$ i_{-2} $ & $ i_1 $ & $ i_2 $ \\
\hline
B0 & 1.458 & 3.971 & 1.586 & 3.726 & 0.000 & 0.000 & 0.000 \\
B1 & 1.407 & 3.927 & 1.628 & 3.713 & 0.258 & 0.399 & 0.518 \\
B2 & 1.373 & 3.927 & 1.670 & 3.666 & 0.370 & 0.567 & 0.739 \\
B3 & 1.332 & 3.964 & 1.709 & 3.584 & 0.463 & 0.702 & 0.912 \\
B4 & 1.287 & 4.014 & 1.747 & 3.490 & 0.544 & 0.819 & 1.061 \\
B5 & 1.237 & 4.072 & 1.789 & 3.390 & 0.622 & 0.921 & 1.208 \\
B6 & 1.178 & 4.118 & 1.819 & 3.280 & 0.681 & 1.015 & 1.318 \\
B7 & 1.077 & 4.179 & 1.854 & 3.187 & 0.750 & 1.090 & 1.439 \\
B8 & 1.080 & 4.212 & 1.900 & 3.103 & 0.827 & 1.213 & 1.562 \\
B9 & 0.945 & 4.469 & 1.917 & 3.028 & 0.890 & 1.294 & 1.670 \\
\hline
\end{tabular}
\end{table}
\begin{table}
\centering
\caption{Same as Table~\ref{table:frequencies_a}, but for the
sequence of uniformly rotating models~BU.}
\label{table:frequencies_bu}
\begin{tabular}{@{}l@{~~}ccccccc@{}}
\hline
Model & $ F $ & $ H_1 $ & $ {}^{2\!}f $ & $ {}^{2}p_1 $ &
$ i_{-2} $ & $ i_1 $ & $ i_2 $ \\
\hline
BU0 & 1.458 & 3.971 & 1.586 & 3.726 & 0.000 & 0.000 & 0.000 \\
BU1 & 1.413 & 3.915 & 1.611 & 3.634 & 0.224 & 0.288 & 0.385 \\
BU2 & 1.380 & 3.907 & 1.635 & 3.516 & 0.326 & 0.413 & 0.541 \\
BU3 & 1.343 & 3.921 & 1.669 & 3.345 & 0.408 & 0.511 & 0.658 \\
BU4 & 1.304 & 3.950 & 1.698 & 3.200 & 0.486 & 0.601 & 0.762 \\
BU5 & 1.281 & 3.964 & 1.714 & 3.018 & 0.552 & 0.675 & 0.847 \\
BU6 & 1.219 & 4.010 & 1.729 & 2.859 & 0.617 & 0.743 & 0.912 \\
BU7 & 1.207 & 4.018 & 1.720 & 2.677 & 0.680 & 0.809 & 0.969 \\
BU8 & 1.168 & 4.030 & 1.685 & 2.512 & 0.723 & 0.868 & 1.004 \\
BU9 & 1.169 & 4.029 & 1.679 & 2.483 & 0.737 & 0.878 & 1.004 \\
\hline
\end{tabular}
\end{table}
\begin{table}
\centering
\caption{Frequencies of the fundamental hexadecapole ($ l = 4 $) mode,
$ {}^{4}f $, and its first overtone, $ {}^{4}p_1 $, for the sequence
of differentially rotating models~A (left) and~B (right). All
frequencies are given in kHz.}
\label{table:frequencies_a_l=4}
\begin{tabular}{@{}lccclcc@{}}
\hline
Model & $ {}^{4}f $ & $ {}^{4}p_1 $ & \quad \quad \quad &
Model & $ {}^{4}f $ & $ {}^{4}p_1 $ \\ [0.4 em]
\cline{1-3}
\cline{5-7} \\ [-0.8 em]
A0 & 2.440 & 4.896 & & B0 & 2.440 & 4.896 \\
A1 & 2.370 & 4.727 & & B1 & 2.453 & 4.877 \\
A2 & 2.300 & 4.600 & & B2 & 2.468 & 4.855 \\
A3 & 2.223 & 4.372 & & B3 & 2.486 & 4.827 \\
A4 & 2.130 & 4.130 & & B4 & 2.500 & 4.781 \\
A5 & 2.028 & 3.864 & & B5 & 2.504 & 4.740 \\
A6 & 1.910 & 3.617 & & B6 & 2.491 & 4.643 \\
A7 & 1.780 & 3.116 & & B7 & 2.499 & 4.556 \\
A8 & 1.630 & 2.790 & & B8 & 2.501 & 4.506 \\
A9 & 1.480 & 2.430 & & B9 & 2.493 & 4.164 \\
A10 & 1.330 & 2.028 \\
\hline
\end{tabular}
\end{table}
\subsection{General trends of rotational effects}
\label{subsection:rotation_trends}
It is well known that the frequencies of the fundamental $ l = 0 $ and
$ l = 2 $ polar modes of oscillation depend mainly on the central
density of a star, or, equivalently, on the compactness $ M / R $
\citep[see, e.g.,][]{hartle_75_a}. The sequences~A and~AU of fixed
rest mass $ M_0 = 1.506 \, M_\odot $ start with a nonrotating model
with compactness $ M / R = 0.15 $ and terminate at models with much
smaller compactness ($ M / R = 0.076 $ for sequence~A and
$ M / R = 0.095 $ for sequence~AU). Based on this significant decrease
of the compactness along the fixed-rest-mass sequences, one expects a
corresponding decrease in the frequencies of the fundamental modes
(and a similar tendency for the first overtones). In contrast, along
sequences~B and~BU, where the central density is fixed, the
compactness varies much less than for sequences~A and~AU. In fact, for
sequence~B, the compactness even somewhat \emph{increases}. One
therefore expects a weaker dependence of the pulsation frequencies on
rotation for the sequences of fixed central density. The above
expectations have already been verified qualitatively in the Cowling
approximation by SAF.
\begin{figure}
\includegraphics[width=84mm]{figure01.eps}
\caption{Frequencies of various modes for sequences~A (solid
lines) and~AU (dashed lines). Note the avoided crossing between
the $ H_1 $ and the $ {}^{4}p_1 $-mode.}
\label{fig:frequencies_a_au}
\end{figure}
In slowly rotating stars, the frequencies of all inertial modes
increase linearly with increasing $ T / |W| $
\citep[see][]{friedman_01_a, lockitch_03_a}. At higher rotation
rates, higher-order rotational terms can modify this behavior. For
uniformly rotating stars, the expectation is that the inertial mode
frequency still increases up to the mass-shedding limit. As clearly
visible for the rapidly rotating models of sequence~A in
Fig.~\ref{fig:frequencies_a_au}, this general expectation is no longer
valid for differentially rotating stars (for a more detailed
discussion, see Section~\ref{subsection:inertial_modes}).
\begin{figure}
\includegraphics[width=84mm]{figure02.eps}
\caption{Same as Fig.~\ref{fig:frequencies_a_au}, but for the sequences~B and
BU.}
\label{fig:frequencies_b_bu}
\end{figure}
Due to differential rotation, the outer layers of the star rotate
slower and the equatorial radius is smaller compared to a uniformly
rotating model of same $ T / |W| $. This leads to a smaller
sound-crossing time and correspondingly higher fundamental mode
frequencies for the differentially rotating models. This explains why
the curves for the fundamental mode frequencies of the $ F $ and
$ {}^{2\!}f $-mode of sequence~A in Fig.~\ref{fig:frequencies_a_au} have
smaller slopes than those corresponding to sequence~AU. This behavior
was already found for the same model sequence in the Cowling
approximation in the work of SAF.
In general, higher order or large $ l $ modes are affected more strongly
by rotation than lower order or smaller $ l $ modes. At large rotation
rates this can lead to \emph{avoided crossings} between mode
sequences, where modes can exchange the character of their
eigenfunctions. These avoided crossings are already known to
exist from perturbative studies of axisymmetric modes in rotating
Newtonian stars~\citep{clement_86_a} and for relativistic quasi-radial
modes~\citep{yoshida_01_a}. It is not trivial to decide how to
label the mode sequences after an avoided crossing. The decisive
criterion for labeling a pulsation mode is \emph{not} the continuity
of the eigenfrequency along a mode sequence. More important is the
character of the oscillation, i.e.\ the eigenfunction. One must
therefore examine the eigenfunctions of two pulsations before and
after an avoided crossing. Then the character of the modes after the
crossing can be determined according to which modes (of those before
the crossing) they resemble. Thus, continuity of eigenfunctions is
preferred over continuity of eigenfrequencies in labeling modes.
\subsection{Quasi-radial (\boldmath $ l = 0 $) modes and avoided
crossings}
\label{subsection:avoided_crossings}
The computed frequencies for the fundamental quasi-radial $ l = 0 $
mode $ F $ and its first overtone $ H_1 $ for the
fixed rest mass sequences~A and~AU are displayed in
Tables~\ref{table:frequencies_a} and~\ref{table:frequencies_au} and
plotted in Fig.~\ref{fig:frequencies_a_au}. Along the uniformly
rotating sequence~AU, there is a decrease in the frequency of the
fundamental quasi-radial mode, since the central density of the star
decreases with increasing rotation rate. For Newtonian nonrotating
polytropic models the frequency of the fundamental quasi-radial mode
is proportional to the square root of the average density. Even though
we do not compute an average density for the rapidly rotating models
of sequences~A and~AU, we notice that the decrease in the frequency of
the $ F $-mode from $ 1.458 \mathrm{\ kHz} $ to $ 1.107 \mathrm{\ kHz} $
along sequence~AU follows closely the decrease in central energy
density $ \varepsilon_\mathrm{c} $. Along the differentially rotating
sequence A, the frequency of the $ F $-mode is further decreasing,
reaching a very low value of $ 685 \mathrm{\ Hz} $ for the most
rapidly rotating model. Remarkably, when one compares models of same
$ T / |W| $ along the two sequences~A and~AU, the frequency of the
$ F $-mode is insensitive to the degree of differential rotation.
The first overtone $ H_1 $ also decreases along the uniformly rotating
sequence~AU, but by less than the near
$ \sqrt{\varepsilon_\mathrm{c}} $-dependence of the $ F $-mode. Along
the differentially rotating sequence A, the $ H_1 $-mode shows a
similar insensitivity to the degree of differential rotation as the
$ F $-mode, when comparing models of same $ T / |W| $, up to rotation
rates of $ T / |W| \sim 0.1 $. For larger rotation rates, however, an
extended \emph{avoided crossing} with the $ {}^{4}p_1 $-mode takes
place. Such avoided crossings exist, because the various mode
sequences are affected to a different degree by rotation. In
particular, the frequencies of higher $ l $ modes tend to decrease
faster with increasing $ T / |W| $ than the frequencies of lower $ l $
modes, leading to approaching mode-sequences. However, mode-sequences
that contain similar terms in their eigenfunction expansions are not
allowed to cross, even in the linear
approximation~\citep{clement_86_a, yoshida_01_a}. Instead, two
continuous sequences of pulsation frequencies avoid to cross, as shown
in Fig.~\ref{fig:frequencies_a_au}.
The character of the eigenfunctions along these continuous sequences
is \emph{exchanged} at an avoided crossing. As a result, the
lower-frequency part of the continuous sequence starting as the
$ H_1 $-mode in the nonrotating limit becomes the $ {}^{4}p_1 $-mode
at large $ T / |W| $. Correspondingly, the lower-frequency part of the
continuous sequence which starts as the $ {}^{4}p_1 $-mode in the
nonrotating limit, becomes the $ H_1 $-mode at large $ T / |W| $. At
the avoided crossing, the character of the eigenfunctions of both
modes is a mixture of the eigenfunctions of the two modes before the
avoided crossing. We have determined the correct labeling of the
sequences after the avoided crossing by carefully comparing their
two-dimensional eigenfunctions (see Section~\ref{section:recycling}
for a discussion of how the eigenfunctions are obtained from our
time-evolutions). The particular avoided crossing between the $ H_1 $
and $ {}^{4}p_1 $-modes was also observed along a sequence of rotating
models in the relativistic Cowling approximation \citep{yoshida_01_a}.
Along the fixed central density sequences~B and~BU (see
Tables~\ref{table:frequencies_b} and~\ref{table:frequencies_bu} and
Fig.~\ref{fig:frequencies_b_bu}), the frequency of the $ F $-mode
decreases, because, even though the central density stays fixed,
rotational effects still increase the radius of the star, so that the
sound crossing time in the equatorial region increases. This seems to
have a monotonous influence on the frequency of the $ F $-mode, which
is again insensitive to the degree of differential rotation. The
frequency of the $ H_1 $-overtone, on the other hand, first decreases
somewhat with rotation, but then increases again, eventually
surpassing its value of $ 3.971 \mathrm{\ kHz} $ in the nonrotating
limit. Since the $ H_1 $-mode has a node in its eigenfunction in the
nonrotating limit, it is more sensitive to rotational effects (in
comparison to the fundamental $ F $-mode). These rotational effects
have a different influence near the symmetry axis than near the
equator. The changing dependence of the $ H_1 $-mode frequency as the
star becomes more flattened reflects this sensitivity. In the case of
sequence~B, the avoided crossing between the $ H_1 $-mode and the
$ {}^{4}p_1 $-mode happens at the largest rotation rates along this
sequence. Consequently, the fastest rotating model~B9 is in the
avoided crossing region, where the mode eigenfunctions are mixed.
\subsection{Quadrupole (\boldmath $ l = 2 $) modes}
\label{subsection:quadrupole_modes}
The computed frequencies for the fundamental quadrupole $ l = 2 $ mode
$ {}^{2\!}f $ and its first overtone $ {}^{2}p_1 $ for the fixed rest
mass sequences~AU and~A are displayed in
Tables~\ref{table:frequencies_au} and~\ref{table:frequencies_a} and
plotted in Fig.~\ref{fig:frequencies_a_au}. Along the uniformly
rotating sequence~AU, there is only a small decrease in the frequency
of the fundamental $ {}^{2\!}f $-mode, from $1.586 \mathrm{\ kHz}$ to
$ 1.459 \mathrm{\ kHz} $ at the mass-shedding limit. However, for the
differentially rotating sequence~A, the rate of decrease gets stronger
for models with very high $ T / |W| $, and the frequency of the
fundamental $ {}^{2\!}f $-mode becomes as small as
$ 1.098 \mathrm{\ kHz} $.
The first overtone $ {}^{2}p_1 $ shows a much stronger decrease in
frequency with increasing rotation rates. Starting from
$ 3.726 \mathrm{\ kHz} $ at the nonrotating limit, along sequence~AU
its frequency becomes $ 2.090 \mathrm{\ kHz} $ at mass-shedding, while
it has a value of only $ 1.610 \mathrm{\ kHz} $ for the fastest
differentially rotating model along sequence~A. A striking difference
compared to all other modes studied here is that the frequency of the
$ {}^{2}p_1 $-mode is indeed sensitive to the degree of differential
rotation, as is evident from Figs.~\ref{fig:frequencies_a_au}
and~\ref{fig:frequencies_b_bu}.
Along the fixed central density sequences~B and~BU the
$ {}^{2\!}f $-mode shows the opposite tendency compared to its behavior
along the sequences~A and~AU, with its frequency increasing to
$ 1.679 \mathrm{\ kHz} $ and $ 1.917 \mathrm{\ kHz} $, respectively
(see Tables~\ref{table:frequencies_b} and~\ref{table:frequencies_bu}
and Fig.~\ref{fig:frequencies_b_bu}). In contrast, the frequency of
the $ {}^{2}p_1 $-mode is still decreasing (although not as
drastically as for sequences~A and~AU) with increasing rotation rates,
reaching frequencies of $ 3.028 \mathrm{\ kHz} $ at mass-shedding
along sequence~B and $ 2.483 \mathrm{\ kHz} $ along sequence~BU. The
$ {}^{2}p_1$-mode appears to be even more sensitive to the degree of
differential rotation along the fixed central density sequence~B than
along the fixed rest mass sequence~A.
\subsection{Inertial modes}
\label{subsection:inertial_modes}
In our simulations we observe a large number of inertial modes,
which are supported by the Coriolis force and become degenerate
at zero frequency in nonrotating stars. From linear perturbation
theory we expect that there exists an infinite number of inertial
modes in a finite frequency range, which corresponds to 0 to
$ 2 \, \Omega $ for uniformly rotating Newtonian stars \citep[see,
e.g.,][]{lockitch_99_a}. In spite of this, we do not find evidence for
the excitation of an arbitrary number of inertial modes, but only a
few specific inertial modes are predominantly excited. Since we do not
make use of the eigenfunction recycling method to excite inertial
modes (see Section~\ref{section:recycling}), they are excited as
by-products of the excitation of other modes. Hence, inertial modes
can be excited either due to the non-exact nature of the initial trial
eigenfunctions used to perturb the initial model (as described in
Section~\ref{subsection:perturbations}) or due to non-linear couplings
with linear polar modes (see discussion in
Section~\ref{subsection:nonlinear_couplings}). The fact that the
amplitude of the observed inertial modes, with our choice of trial
eigenfunctions, scales non-linearly with increasing amplitude, points
to non-linear couplings as a possible origin, at least for most of the
inertial modes we observe.
Examining the eigenfunctions of the excited inertial modes we notice
that the mode with the usually highest PSD in a Fourier transform has
only one node in the eigenfunction for $ v_{\theta}$, while other
modes at larger or smaller frequency than this `fundamental'
inertial mode have a larger number of nodes (increasing as the
absolute frequency difference from the `fundamental' inertial mode
increases). Out of the many excited inertial modes, we choose to
display in Tables~\ref{table:frequencies_a}
to~\ref{table:frequencies_bu} and Figs.~\ref{fig:frequencies_a_au}
and~\ref{fig:frequencies_b_bu} the `fundamental' inertial mode with
only one node in $ v_{\theta} $, which we call $ i_1 $-mode, and the
two modes with two nodes in the eigenfunction of $ v_{\theta} $ on
both sides of the $ i_1 $-mode in the frequency domain, which we
denote the $ i_{-2} $-mode and $ i_2 $-mode, respectively.
\begin{figure}
\includegraphics[width=84mm]{figure03.eps}
\caption{Comparison of the frequencies for the $ {}^{2\!}f $ and
$ {}^{2}p_1 $-mode (upper panel) and for the $ F $ and
$ H_1 $-mode (lower panel) in CFC (solid lines) and in the Cowling
approximation (dashed lines) for the sequence of differentially
rotating models~A. In the lower panel we also show the frequencies
of the $ {}^{4}p_1 $-mode after the avoided crossing with the
$ H_1 $-mode (green line).}
\label{fig:cfc_comparison_a}
\end{figure}
Notice that the number of nodes is determined at slow rotation rates
and can change at very high rotation rates. Note also that inertial
modes come in two flavours (polar-led and axial-led) and are hybrid
modes in the sense that they do not reduce to either a pure polar or a
pure axial mode in the nonrotating limit
\citep[see][]{lockitch_99_a}. A proper classification scheme for
inertial modes, based on the dominant components in their
eigenfunctions, is already in use \citep{lockitch_99_a,
friedman_01_a, lockitch_03_a}. However, comparing our numerically
calculated eigenfunctions with eigenfunctions obtained from linear
perturbation theory with more simplifying assumptions (weak gravity or
uniform density) is beyond the scope of the present paper and will be
addressed in a separate publication. Thus, we use our own naming
scheme for inertial modes simply for convenience, until proper
identification is established in the future.
\begin{figure}
\includegraphics[width=84mm]{figure04.eps}
\caption{Same as Fig.~\ref{fig:cfc_comparison_a}, but for the
sequence of uniformly rotating models~AU.}
\label{fig:cfc_comparison_au}
\end{figure}
Along sequence~AU the frequencies of the three inertial modes
increases monotonically with rotation. However, along sequence~A the
frequencies of the three inertial modes reach a maximum value for the
rotation parameter $ T / |W| $ being in the range 0.15 to 0.19,
depending on the specific mode. This is due to the fact that, even
though $ T / |W| $ increases monotonically along sequence~A, both the
angular velocity at the centre and the angular velocity at the surface
reach maximum values along this sequence, as can be seen in
Table~\ref{table:equilibrium_models}. On the other hand, along
sequences~BU and~B the frequencies of the three inertial modes again
increase monotonically, as does the angular velocity at the centre and
at the surface.
\subsection{Comparison to the Cowling approximation}
\label{subsection:comparison_to_cowling}
When comparing the dependence of the mode frequency on $ T / |W| $
plotted in Figs.~\ref{fig:cfc_comparison_a}
to~\ref{fig:cfc_comparison_bu} for the $ F $, $ H $, $ {}^{2\!}f $, and
$ {}^{2}p_1 $-modes obtained with the CFC approximation for the
spacetime evolution against previous results in the Cowling
approximation by SAF, several qualitative observations can be
made. Apparently, for all sequences there is a large discrepancy in
the values of the frequencies of the fundamental quasi-radial
$ F $-mode in the CFC simulations compared to the ones in the Cowling
approximation at all rotation rates. This is also quantitatively
evident in the relative differences of frequencies between the CFC and
Cowling simulations of up to a factor $ \sim 2 $, listed for the
nonrotating model (A0/AU0/B0/BU0) and the most rapidly rotating model
of each sequence (A10, AU5, B9, BU9) in
Table~\ref{table:comparsion_cfc_cowling}. We notice that these
findings are consistent with those of \citet{yoshida_97_a} for
nonrotating stars. In terms of absolute differences, for the $ F
$-mode the Cowling approximation fails by far to predict the correct
mode frequencies, as it yields values which are typically
$ \gtrsim 1 \mathrm{\ kHz} $ too high. While in sequences~A, AU,
and~BU the frequency curve in the Cowling approximation at least
correctly captures the decline with increasing rotation rate, it
cannot reproduce the similar behavior in sequence~B.
We emphasize that neither in sequence~A nor~B we find any evidence for
a splitting of the $ F $-mode, a phenomenon that was noticed for those
sequences in the Cowling approximation by SAF. This apparently
confirms the possible explanation for the $ F $-mode splitting as an
artifact of the Cowling approximation offered in that work. In
Figs.~\ref{fig:cfc_comparison_a} to~\ref{fig:cfc_comparison_bu} and
Table~\ref{table:comparsion_cfc_cowling} we use the frequencies of the
regular $ F $-mode rather than those of the $ F_\mathrm{II} $-mode
of SAF, as the eigenfunctions of the latter mode do not possess the
typical characteristics of a genuine $ F $-mode (and, in addition,
their amplitude of the frequency peak in FFTs is smaller than for
the $ F $-mode).
\begin{figure}
\includegraphics[width=84mm]{figure05.eps}
\caption{Same as Fig.~\ref{fig:cfc_comparison_a}, but for the
sequence of differentially rotating models~B.}
\label{fig:cfc_comparison_b}
\end{figure}
Figs.~\ref{fig:cfc_comparison_a} to~\ref{fig:cfc_comparison_bu}
demonstrate that for the frequencies of the fundamental quadrupole
$ {}^{2\!}f $-mode and the $ H_1 $ and $ {}^{2}p_1 $-modes, the
discrepancies between the CFC simulations and the Cowling
approximation are much less severe (see also
Table~\ref{table:comparsion_cfc_cowling}). In the case of the
$ H_1 $-mode and the $ {}^{2}p_1 $-mode, the curve in the Cowling
approximation follows our CFC results reasonably close, while we
observe a crossing of the frequency curves for the $ {}^{2\!}f $-mode
beyond medium rotation rates in all sequences. Nevertheless, the
$ {}^{2\!}f $-mode frequencies still agree well quantitatively. Note
however that avoided crossings of the $ H_1 $-mode and the
$ {}^{4}p_1 $-mode (as in sequences~A and~B) complicates a
comparison. If the continuous frequency curve of the $ H_1 $-mode is
followed, the $ H_1 $-mode takes over the characteristics of the
$ {}^{4}p_1 $-mode at the avoided crossing. Thus, if \emph{continuous}
frequency curves are compared, a possible change of the mode labeling
must be taken into account. Although the simulations in the Cowling
approximation reproduce the $ H_1 $, $ {}^{2\!}f $, and
$ {}^{2}p_1 $-mode frequencies from the CFC simulations fairly close
for most sequences, which supports the idea of establishing empirical
relations for predicting the correct frequencies from models evolved
in the Cowling approximation, such relations should be used with
caution (see discussion below).
\begin{figure}
\includegraphics[width=84mm]{figure06.eps}
\caption{Same as Fig.~\ref{fig:cfc_comparison_a}, but for the
sequence of uniformly rotating models~BU.}
\label{fig:cfc_comparison_bu}
\end{figure}
\begin{table}
\centering
\caption{Relative difference in per cent in the frequencies of the
$ F $, $ H_1 $, $ {}^{2\!}f $, and $ {}^{2}p_1 $-mode for the
nonrotating model and the most rapidly rotating models of
sequence~A, AU, B, and~BU in the Cowling approximation with
respect to the CFC simulations. A $ + $~sign ($ - $~sign)
indicates an overestimation (underestimation) of the CFC frequency
in the Cowling approximation.}
\label{table:comparsion_cfc_cowling}
\begin{tabular}{@{}l@{~~}rrrr@{}}
\hline
Model & $ F $ & $ H_1 $ & $ {}^{2\!}f $ & $ {}^{2}p_1 $ \\
\hline
A0/AU0/B0/BU0 & $ +86 $ & $ +15 $ & $ +16 $ & $ +10 $ \\
A10 & $ +108 $ & $ -12 $ & $ -12 $ & $ +18 $ \\
AU5 & $ +77 $ & $ +5 $ & $ -9 $ & $ +22 $ \\
B9 & $ +81 $ & $ -1 $ & $ -3 $ & $ +18 $ \\
BU9 & $ +98 $ & $ +6 $ & $ -7 $ & $ +6 $ \\
\hline
\end{tabular}
\end{table}
For sequence~BU a comparison of models evolved using a coupled
evolution of hydrodynamics and spacetime, with no approximation for
the gravitational field equations, against the Cowling approximation
was presented by \citet{font_02_a}\footnote{Note that in that work the
sequence already ended at our model BU7 with an axis ratio
$ r_\mathrm{p} / r_\mathrm{e} = 0.650 $, while several additional
models with intermediate rotation rates were also evolved.}.
Comparing these results with our new simulations using the CFC
approximation as presented in Fig.~\ref{fig:gr_comparison_bu} shows
excellent agreement for the frequencies of the $ F $-mode. This is
another important confirmation of the validity of the results obtained
with our code, considering in particular the differences in coordinate
choice, grid resolution, evolution time, and assumption of CFC
compared to the code employed in \citet{font_02_a}\footnote{A similar
comparison for sequence~AU yields equally good results
\citep{bernuzzi_06_a}.}.
In the case of the $ H_1 $-mode we also find rather good agreement,
with increasing mismatch as the models approach the mass-shedding
limit. The oscillations in the $ H_1 $-mode frequency at large
rotation rates observed in the fully relativistic simulations by
\citet{font_02_a} were interpreted as possible effects of avoided
crossings. In contrast to this, in our simulations we find neither
these oscillations nor signs of avoided crossing of the $ H_1 $-mode
with other modes in sequences~AU and~BU (see
Fig.~\ref{fig:gr_comparison_bu}, and also
Figs.~\ref{fig:frequencies_a_au} and~\ref{fig:frequencies_b_bu}). For
sequences~A and~B, where there is clear evidence for avoided crossing,
this happens at larger rotation rates. We thus conclude that the
oscillations visible in Fig.~\ref{fig:gr_comparison_bu} (upper red
dashed line) are an artifact which can be explained by the effectively
lower grid resolution and shorter evolution time compared to the
current simulations, as this increases the error in the numerical
extraction of the mode frequency particularly for higher order modes.
\begin{figure}
\includegraphics[width=84mm]{figure07.eps}
\caption{Comparison of the frequencies for the $ F $ and
$ H_1 $-mode in CFC (black solid lines) and full general
relativity (red dashed lines) for sequence~BU.}
\label{fig:gr_comparison_bu}
\end{figure}
When comparing the frequencies for sequence~BU from simulations with
a dynamical spacetime to the ones in the Cowling approximation,
\citet{font_02_a} also observed that the frequencies for the
$ F $-mode (and, to a lesser degree, also the ones for the
$ H_1 $-mode) show a similar dependence on the rotation rate in both
cases. From the approximate constancy of the difference between the
$ F $-mode frequency and the corresponding result in the Cowling
approximation, they constructed an approximate empirical relation for
calculating the frequency of the $ F $-mode for arbitrary rotation
rates. This relation thus depends on the $ F $-mode frequency of the
nonrotating model~BU0 obtained in a dynamical spacetime evolution
and the variation of the $ F $-mode frequencies with increasing
rotation in the Cowling approximation. It yields the correct
frequencies with an accuracy of better than 2 per cent for the most
rapidly rotating in their model series of sequence~BU (our model~BU7).
Unfortunately our results clearly suggest that it is not possible to
straightforwardly apply this simple method to set up analogous
relations for sequences~A, AU, and~B, as shown in
Figs.~\ref{fig:cfc_comparison_a} to~\ref{fig:cfc_comparison_bu}. Only
for sequence~BU, the $ F $-mode frequency exhibits a nearly constant
difference between the CFC and Cowling simulations irrespective of the
rotation rate. Moreover, the frequencies of all other modes displayed
in these figures also do not fulfill such a relation, except maybe in
the case of the $ {}^{2}p_1 $-mode for sequences~A and~AU.
Lacking results from simulations involving evolution of the
spacetime, SAF proposed an approximate relation for the $ F $ and
$ {}^{2\!}f $-mode frequencies of sequence~A similar to the one in
\citet{font_02_a}. Based on the work by \citet{yoshida_97_a} and
\citet{yoshida_01_a}, an empirical relation was derived using
information from the compactness of a model, and assuming nearly
linear scaling of the frequencies with increasing rotation rate. When
applying this relation to our model~A10, we find a significant
discrepancy between the predicted frequencies
($ f_F = 0.882 \mathrm{\ kHz} $ and
$ f_{H_1} = 0.757 \mathrm{\ kHz} $) and the ones extracted from the
actual numerical simulations ($ f_F = 0.685 \mathrm{\ kHz} $ and
$ f_{H_1} = 1.098 \mathrm{\ kHz} $), with relative differences of 29
and 31 per cent for the $ F $ and $ {}^{2\!}f $-mode frequencies,
respectively. Although their assumption of nearly linear scaling of
$ f_F $ and $ f_{H_1} $ with the rotation rate is confirmed by our
results (see Figs.~\ref{fig:cfc_comparison_a}
to~\ref{fig:cfc_comparison_bu}), this error is much larger than the
predicted uncertainty of the relation of a few per cent.
Owing to this, we refrain from establishing similar relations for
other mode frequencies, even in cases where we also observe such
nearly linear scaling of the frequency with rotation rate in CFC
simulations and/or constant difference between the frequency of a
specific mode in CFC and Cowling. We rather suggest that except in
special cases, (the computationally expensive) simulations of rotating
stars in which the spacetime and the hydrodynamics are coupled cannot
be replaced by a combination of simulations in the Cowling
approximation and simple empirical relations.
\section{Eigenfunction recycling}
\label{section:recycling}
As pointed out in Section~\ref{subsection:perturbations}, initial
perturbations of the form given by Eqs.~(\ref{eq:perturbation_l=0},
\ref{eq:perturbation_l=2}) excite not only one desired specific
eigenmode, but additional oscillation modes as well. This is clearly
visible in the power spectrum of model~A1 displayed in the upper panel
of Fig.~\ref{fig:recycling}. While the fundamental modes $ F $ and
$ {}^{2\!}f $ clearly dominate the spectrum, the overtones $ H_1 $ and
$ {}^{2}p_1 $ along with several other modes are also significantly
excited by a perturbation using the $ l = 0 $ and $ l = 2 $ trial
eigenfunction of the initial velocity components $ v_r $ and
$ v_{\theta} $, respectively.
\begin{figure}
\includegraphics[width=84mm]{figure08.eps}
\caption{Fourier transform of the evolution of the radial profile of
the rest-mass density $ \rho $ along the equatorial
plane for model A1. For the two simulations in the upper panel an
$ l = 0 $ and $ l = 2 $ trial eigenfunction is used as
perturbation, respectively, while the four recycling runs in the
lower panel are excited with recycled eigenfunctions. The scaling
of the ordinate is linear, and the power spectra in the upper
panel are scaled to obtain the same strength of the $ F $-mode.}
\label{fig:recycling}
\end{figure}
This consequence of using trial eigenfunctions can be avoided by
performing an additional `recycling' run. For this the \emph{actual}
two-dimensional eigenfunctions of the $ v_r $ and $ v_{\theta} $
velocity components at a selected mode frequency are extracted from
the original simulation which was perturbed by a trial
eigenfunction. These eigenfunctions are then applied with appropriate
amplitude as initial perturbation to a second simulation of the same
stellar equilibrium model. In this case, other modes than the selected
one are strongly suppressed, as shown for model~A1 in the lower panel
of Fig.~\ref{fig:recycling}. Here eigenfunctions of the $ F $ and
$ H_1 $-mode, and the $ {}^{2\!}f $ and $ {}^{2}p_1 $-mode are extracted
from a previous simulation perturbed with $ l = 0 $ and $ l = 2 $
trial eigenfunctions, respectively. These are then used as initial
perturbations for four recycling runs with relative amplitudes with
respect to the amplitude of the $ F $ eigenfunction of 5.9, 3.8, and
7.0 for the $ H_1 $, $ {}^{2\!}f $, and $ {}^{2}p_1 $ eigenfunction,
respectively, in order to arrive at approximately equal strengths of
the dominant modes in the power spectrum.
We note that the choice of using $ v_r $ and $ v_\theta $
eigenfunctions as initial perturbations for doing the recycling
simulation is not arbitrary. Naively, one would think to use all
four evolved variables $ \rho $, $ v_r $, $ v_\theta $, and
$ v_\varphi $ instead. However, for a given mode some of these
variables are out of phase with respect to the others. Using the
eigenfunctions of all variables as recycling perturbations
simultaneously without taking into account the relative phase between
them does not lead to the excitation of a single mode, but to the
excitation of a sum of different modes (similar to choosing a trial
eigenfunction). From the phase information contained in the complex
FFT of the various variables, we determined that, at least for the
modes we are interested in, the quantities $ v_r $ and $ v_\theta $
have the same phase, while the other two are out of phase by
$ \pi / 2 $ with respect to $ v_r $ and $ v_\theta $.
While the suppression of undesired additional modes in the power
spectrum significantly improves when eigenfunction recycling is
performed, the form of the eigenfunction of various metric and
hydrodynamic quantities extracted from the recycling run is usually
altered only negligibly as compared to the eigenfunction extracted
from the original simulation with the trial perturbation. In
Fig.~\ref{fig:eigenfunction_comparison} we present radial profiles
of the rest-mass density eigenfunction $ \rho^\mathrm{ef} $
along the equatorial plane (upper panel) and of the
$ \theta $-velocity eigenfunction $ v_{\theta}^\mathrm{ef} $
along $ \theta = \pi / 4 $ (lower panel). They are extracted from both
the original simulations of model~A1 using $ l = 0 $
(for $ \rho^\mathrm{ef} $) and $ l = 2 $ (for
$ v_{\theta}^\mathrm{ef} $) perturbations with trial
eigenfunctions, and also from the respective $ H_1 $ and $ {}^{2}p_1 $
recycling runs. Only at the outer stellar boundary the shape of the
eigenfunctions depends slightly on whether the model is perturbed by a
trial or extracted eigenfunction, while in the bulk of the star the
eigenfunctions are practically identical. We find similar results for
the fundamental modes $ F $ and $ {}^{2\!}f $ and for the eigenfunctions
of other metric and hydrodynamic quantities in model~A1 and several
other moderately rotating models.
\begin{figure}
\includegraphics[width=84mm]{figure09.eps}
\caption{Radial profiles of the eigenfunctions $ \rho^\mathrm{ef} $
for the $ H_1 $-mode (along the equatorial plane; upper panel) and
$ v_{\theta}^\mathrm{ef} $ for the $ {}^{2}p_1 $-mode (along
$ \theta = \pi / 4 $; lower panel) obtained from the evolution of
model~A1. For an initial perturbation using either a trial
eigenfunction (black solid lines) or an extracted eigenfunction
(blue dashed lines), the shapes of the eigenfunctions agree
well. The eigenfunctions are scaled to the same maximum height.}
\label{fig:eigenfunction_comparison}
\end{figure}
We can thus conclude that for such rotating stellar models an initial
perturbation with trial eigenfunctions is adequate to precisely
obtain the mode frequency and to extract an accurate corresponding
eigenfunction of the fundamental modes and their first overtones from
the evolution. Additionally, for such models a \emph{single} recycling
run suffices to efficiently suppress the excitation of all unwanted
modes. However, if the peak of the investigated mode in the power
spectrum is small and/or if several modes interact (see also the
discussion of avoided crossings in
Section~\ref{subsection:avoided_crossings}), which is typically the
case for higher order modes in rapidly rotating models, another
recycling loop may be necessary to clearly determine the mode
frequency and eigenfunctions, and to channel most of the initial
perturbation energy into a single oscillation mode.
Particularly in the case of the avoided crossing of the $ H_1 $ and
the $ {}^{4}p_1 $-mode in sequence~A and~B (see also
Figs.~\ref{fig:frequencies_a_au} and~\ref{fig:frequencies_b_bu}), we
use a modified recycling strategy for an accurate mode
analysis. Starting from a model where the mode frequency and
eigenfunctions can still be clearly determined, the next model in the
sequence is perturbed with the eigenfunctions of the investigated mode
extracted from the previous model. This {\em sequential} recycling is
a very helpful tool to resolve problems with ambiguous or unclear mode
frequencies and character of the eigenfunction.
\section{Non-linear pulsations}
\label{section:nonlinear_pulsations}
Although linear perturbations of rotating stars are assumed to have a
vanishingly small amplitude, so that the background equilibrium star
is unaffected by a linear oscillation mode, in certain situations
non-linear effects can become important. Our non-linear evolution code
allows us to investigate such effects, of which we present several
cases in the following.
\begin{figure}
\includegraphics[width=84mm]{figure10.eps}
\caption{Fourier transform of the evolution of the radial profile of
the rest-mass density $ \rho $ along the equatorial plane for the
nonrotating model (A0/AU0/B0/BU0) with an $ l = 0 $ trial
eigenfunction used as initial perturbation. Additional to linear
modes, several non-linear harmonics can be identified.}
\label{fig:nonlinear_harmonics_l=0}
\end{figure}
\subsection{Non-linear harmonics}
\label{subsection:nonlinear_harmonics}
The most basic non-linear effect we see in our simulations is the
appearance of non-linear harmonics of the linear pulsation modes, a
general property of non-linear systems
\citep[cf.][\S~28]{landau_76}. To lowest order, these arise as linear
sums and differences of different linear modes, including
self-couplings. If the system has eigenfrequencies $ \omega_i $, the
non-linearity of the equations excites modes at frequencies
$ \omega_i \pm \omega_j $, with amplitudes proportional to the product
of the amplitudes of the combining frequencies. We note that such
non-linear harmonics were also recently noticed by~\citet{zanotti_05_a}
in the numerical investigation of the dynamics of oscillating,
relativistic, high-density tori around Kerr black holes.
In Fig.~\ref{fig:nonlinear_harmonics_l=0} we present the Fourier PSD
of the density evolution of the nonrotating model (A0/AU0/B0/BU0), to
which a finite radial $ l = 0 $ initial perturbation of the
form~(\ref{eq:perturbation_l=0}) was added. In addition to the main
linear modes $ F $ and $ H_1 $, one can observe several of their
non-linear harmonics, such as the self-couplings $ 2 \cdot F $,
$ 3 \cdot F $ and the linear sums $ H_1 - F $ and
$ H_1 - 2 \cdot F $. In the PSD a large number of additional peaks can
be seen, and essentially all of those should correspond to non-linear
harmonics of the excited linear modes. It is interesting to note that
several non-linear harmonics are present that have frequencies much
smaller than the fundamental radial mode $ F $ (which, in the linear
approximation, possesses the lowest frequency). These can actually
fall into the frequency range of the inertial modes for rotating
models. Thus, in rotating models further non-linear interactions
between radial modes and inertial modes can be expected (see also the
discussion in Section~\ref{subsection:nonlinear_couplings}).
Non-linear harmonics can occur not only due to couplings between modes
of the same index $ l $ (such as the $ l = 0 $ modes discussed above),
but also due to couplings between modes of different
$ l $. Fig.~\ref{fig:nonlinear_harmonics_l=2} shows several identified
harmonics that represent linear sums and differences between linear
modes for the nonrotating model, when an $ l = 2 $ perturbation was
added to the initial data of the nonrotating model. Due to the
approximate nature of the chosen trial eigenfunction, $ l = 0 $ modes
are also excited in addition to the main $ l = 2 $ modes. The presence
of both $ l = 0 $ and $ l = 2 $ modes then leads to the appearance of
several non-linear harmonics, which include cases like
$ {}^{2}p_1 - F $, $ H_1 - {}^{2\!}f $, etc. It is thus clear that even
though in the linear approximation modes of different $ l $ are
orthogonal to each other (in a nonrotating perfect fluid star),
non-linear effects couple all linear modes with different $ l $ that
are present in a non-linear simulation)\footnote{Notice, however, that
for relativistic stars there exists no proof yet on the
completeness of quasi-normal modes.}.
\begin{figure}
\includegraphics[width=84mm]{figure11.eps}
\caption{Same as Fig.~\ref{fig:nonlinear_harmonics_l=0}, but with an
$ l = 2 $ trial eigenfunction used as initial perturbation.}
\label{fig:nonlinear_harmonics_l=2}
\end{figure}
In rotating models, where there can be a perplexing alternation of
various linear modes and non-linear harmonics in a PSD plot, one can
easily distinguish the non-linear harmonics from the linear modes by
comparing PSDs produced from simulations with different initial
perturbation amplitudes. As expected, the linear modes scale almost
linearly, while the non-linear harmonics scale as the product of the
amplitudes of the modes (or of the modes and harmonics) from which
they are produced. Such a case is shown in
Fig.~\ref{fig:nonlinear_harmonics_scaling}, where model~A1 from
sequence~A is used in which the linear modes $ F $, $ H_1 $,
$ {}^{2\!}f $, and $ {}^{2}p_1 $ are all excited at roughly the same
strength using the eigenfunction recycling technique for the initial
perturbation as described in Section~\ref{section:recycling}.
Comparing two simulations of this model that differ by a common factor
of 4 in the initial perturbation amplitude, one can clearly notice
that while the amplitudes of the linear modes scale nearly linearly,
the amplitudes of the various non-linear harmonics scale non-linearly
(with many of them scaling roughly quadratically).
\begin{figure}
\includegraphics[width=84mm]{figure12.eps}
\caption{Fourier transform of the evolution of the radial profile of
the $ \theta $-velocity $ v_{\theta} $ along $ \theta = \pi / 4 $ for
model~A1 with a combined $ F $, $ H_1 $, $ {}^{2\!}f $, and $
{}^{2}p_1 $-mode eigenfunction as initial perturbation. Changing
the perturbation amplitude $ a $ reveals the non-linear scaling of
the coupled modes as opposed to the linear scaling of the linear
modes.}
\label{fig:nonlinear_harmonics_scaling}
\end{figure}
\subsection{Non-linear 3-mode couplings}
\label{subsection:nonlinear_couplings}
The presence of non-linear harmonics opens the possibility of 3-mode
couplings when the star is rotating. The reason such 3-mode couplings
can take place is the fact that the effect of rotation on the
different modes varies. As was already discussed in
Section~\ref{subsection:rotation_trends}, higher-order modes are
typically affected stronger by rotation than lower-order modes,
which results in avoided crossings between mode sequences. Rotation
also influences the frequencies of the various non-linear harmonics to
a different degree. Thus, at certain rotation rates, a non-linear
harmonic can have the same frequency as a linear mode. Examples
of such cases are demonstrated in
Fig.~\ref{fig:nonlinear_three_mode_coupling}, which shows the
frequencies of several linear modes and of the two non-linear
harmonics $ {}^{2}p_1 - {}^{2\!}f $ and $ H_1 - F $ as a function of
$ T / |W| $. Apparently the $ {}^{2}p_1 - {}^{2\!}f $ harmonic is
crossing the frequency of the fundamental quasi-radial $ F $-mode at
about $ T / |W| \sim 0.1 $, while the $ H_1 - F $ harmonic is crossing
the frequency of the $ {}^{2}p_1 $-mode at about
$ T / |W| \sim 0.12 $. We also note that at any rotation rate one can
expect several non-linear harmonics to coincide in frequency with some
of the infinitely many inertial modes contained in the inertial mode
range.
\begin{figure}
\includegraphics[width=84mm]{figure13.eps}
\caption{Frequencies of various linear modes and the two non-linear
harmonics $ H_1 - F $ and $ {}^{2}p_1 - {}^{2\!}f $ for sequence~A.
At specific rotation rates the frequencies of linear and non-linear
modes coincide, which can lead to non-linear 3-mode coupling.}
\label{fig:nonlinear_three_mode_coupling}
\end{figure}
At such crossings, the coinciding of the frequencies could
potentially lead to resonance effects and even to parametric
instabilities. In such a way, significant energy from one mode
could be transfered to other modes. The most interesting case
would be if pulsational energy from the quasi-radial mode, which
weakly radiates gravitational waves, could be transfered non-linearly
to stronger radiating nonradial modes. Since during core bounce
a significant amount of kinetic energy is stored in the radial
modes of pulsation, the transfer of even a small percentage of
this kinetic energy to a nonradial mode could result in the emission
of strong gravitational waves. This scenario has first been suggested
by \citet{clark_79_a} and \citet{eardley_83_a}, who investigated the
parametric instabilities that could take place, using a Newtonian,
slowly rotating collapse model \citep[for recent related work for
nonrotating or slowly rotating relativistic stars,
see][]{passamonti_05_a}.
In \emph{nonrotating} or \emph{slowly rotating} models, such a
parametric instability can only take place under special conditions
that would allow the two modes to be in resonance. Here we find that
in \emph{rapidly rotating} models rotational shifting of the frequency
of different modes broadens the range of parameters for which
interesting resonances could take place. We particularly notice that
the quasi-radial mode will be in resonance with some inertial mode(s)
for \emph{all rotation rates above a critical value}. It is thus
interesting to further study the possible energy transfer between
different modes excited after, e.g., a core collapse or an
accretion-induced collapse event, either on secular time-scales or as
a parametric instability.
Here we only observe the necessary conditions for non-linear 3-mode
couplings to take place\footnote{We include the case of self
couplings, when discussing 3-mode couplings.}. Whether such
couplings will indeed lead to strong parametric resonances and
enhanced gravitational wave emission remains to be investigated
through much more detailed studies.
\subsection{Mass-shedding-induced damping}
\label{subsection:mass_shedding}
Another striking example of a non-linear effect is the
mass-shedding-induced damping of oscillations in stellar models which
rotate at or near the mass-shedding limit. This new damping mechanism
was first observed and discussed by SAF, and is especially important
for pulsations in uniformly rotating stars. In SAF, the
mass-shedding-induced damping was demonstrated for the uniformly
rotating models~BU8 and~BU9 using fixed-spacetime evolutions, i.e.\
the Cowling approximation.
\begin{figure}
\includegraphics[width=84mm]{figure14.eps}
\caption{Distribution of the logarithm of the rest-mass density,
$ \log \rho $ ($ \mathrm{g\ cm}^{-3} $), in the meridional plane
for model BU9 with ideal fluid EOS at $ t = 4.4 \mathrm{\ ms} $ in
the Cowling approximation. The ordinate coincides with the
rotation axis. The single shocks traveling through the atmosphere
and creating a matter envelope are clearly visible. The black
color coding corresponds to the low density artificial atmosphere
where the rest-mass density is actually much smaller than
$ 10^7 \mathrm{\ g\ cm}^{-3} $. Note that only the innermost 80~km
of the computational grid are shown.}
\label{fig:mass_shedding_2d_cowling}
\end{figure}
As explained in SAF, the damping mechanism works as follows. As the
star approaches the mass-shedding limit, the effective gravity near
the equatorial surface diminishes, exactly vanishing at the
mass-shedding limit. A small radial pulsation then suffices to cause
mass-shedding after each oscillation period. As a result, a
low-density envelope is created around the star. This envelope is
initially concentrated in the regions close to the stellar equator,
but with each oscillation period more and more mass is shed in the
form of shock waves, and the envelope expands outwards and away from
the equatorial plane. Since in rotating stars every pulsation mode
also has a radial velocity component, the damping affects all
modes. In SAF, it was found that the damping in the Cowling
approximation can be rather strong. Here we investigate the same
damping effect in the CFC approximation and compare the two cases.
We first repeat the study of the mass-shedding-induced damping for
the radially perturbed model BU9, as presented in SAF, in the Cowling
approximation. Fig.~\ref{fig:mass_shedding_2d_cowling} shows the
distribution of the rest-mass density $ \rho $ in the meridional plane
using the (nonisentropic) ideal fluid EOS~(\ref{eq:ideal_gas_eos}). At
time $ t=4.4\mathrm{\ ms} $ several oscillations have already occurred
and the high-entropy envelope has acquired a near-equilibrium state,
extending into almost the entire computational grid (whose outer
boundary has been set to 5 stellar equatorial radii with 120
additional logarithmically spaced radial grid points)\footnote{Note
that the equatorial stellar radius is at
$ r_\mathrm{e} \simeq 17 \mathrm{\ km} $. In
Table~\ref{table:equilibrium_models} dimensionless units are used
for $ r_\mathrm{e} $, resulting in a different numerical value of
the same physical location.}. Near the equatorial plane the envelope
has a rest-mass density of $ \sim 10^{-6}$ to
$ 10^{-4} \rho_\mathrm{c} $, through which further consecutive shocks
propagate. Only at angles $ \lesssim 20^\circ $ (as measured from the
rotation axis) does the high-entropy envelope not completely form.
We then study the same perturbed model (with a similar initial
effective perturbation amplitude) but also evolving the spacetime,
in the CFC approximation. Compared to the Cowling approximation,
the mass-shedding behavior and the properties of the matter envelope
are now significantly different. In Fig.~\ref{fig:mass_shedding_2d},
one can see (at the same time $ t = 4.4 \mathrm{\ ms} $ as in
Fig.~\ref{fig:mass_shedding_2d_cowling}) that the high-entropy
envelope outside the star is confined to within roughly $ 45^\circ $
with respect to the equatorial plane, filled with matter of one to two
orders of magnitude lower density than in the Cowling approximation.
In addition, the consecutive shocks barely reach the outer grid
boundary.
\begin{figure}
\includegraphics[width=84mm]{figure15.eps}
\caption{Same as Fig.~\ref{fig:mass_shedding_2d}, but for a
spacetime coupled to the hydrodynamics (CFC simulation).
Note that in contrast to the Cowling approximation, a much smaller
angular part of the atmosphere is filled with a matter envelope,
which also has significantly also lower density.}
\label{fig:mass_shedding_2d}
\end{figure}
Fig.~\ref{fig:mass_shedding_profile} shows the corresponding radial
rest-mass density profiles along the equatorial plane for both the
Cowling and CFC approximations. The envelope in the CFC approximation
has a lower density than in the Cowling approximation, with the
difference consistently increasing with distance and becoming 2 orders
of magnitude at the outer grid boundary. In this plot the single
shocks generated by the stellar oscillations, which first create the
envelope and later propagate through it, are clearly noticeable. For
model~BU9 the frequency of the $ F $-mode oscillations, which are
mainly responsible for the mass-shedding in the above simulations is
about twice as high in the Cowling than in the CFC approximation
($ 2.313 \mathrm{\ kHz} $ vs.\ $ 1.169 \mathrm{\ kHz} $). This is,
however, not reflected in the shock pattern of the radial density
profile in Fig.~\ref{fig:mass_shedding_profile}, where naively twice
as many shocks between $ r_\mathrm{e} $ and some radius
$ r > r_\mathrm{e} $ could be expected. This is a consequence of the
unequal profiles of the shock propagation velocity in the envelope due
to the different density profiles, as well as interactions of shocks
traveling at different velocities due to unequal shock strengths. The
latter effect can also lead to reverse shocks, which further
complicate the picture. Note that in the density profile from the
simulation in the Cowling approximation in
Fig.~\ref{fig:mass_shedding_profile}, the significant loss of the
matter in the stellar regions close to $ r_\mathrm{e} $ by
mass-shedding is evident, whereas in the CFC simulation the initial
density profile inside the stellar boundary is nearly preserved.
\begin{figure}
\includegraphics[width=84mm]{figure16.eps}
\caption{Radial profiles of the rest-mass density
$ \rho $ along the equatorial plane for model BU9 with
ideal fluid EOS at $ t = 4.4 \mathrm{\ ms} $ in CFC (black solid
line) and in the Cowling approximation (red dashed line). The
single shocks due to mass-shedding are clearly visible. The dotted
line marks the equatorial stellar radius.}
\label{fig:mass_shedding_profile}
\end{figure}
In an equilibrium model rotating at the mass-shedding limit, effective
gravity at the equatorial surface vanishes due to a delicate balance
between the pressure gradient, the gravitational force, and the
centrifugal force. In the Cowling approximation the spacetime is held
fixed during the entire simulation, and thus the gravitational force
cannot react to local over- or underdensity caused by the pulsations.
Evidently, in this case matter can easily be ejected from the stellar
equatorial surface at each pulsation. In contrast to this, when the
spacetime is coupled to matter it responds to changes in the
hydrodynamic evolution. The most strongly excited pulsation mode in
the mass-shedding simulations of model BU9 is the $ F $-mode, whose
rest-mass density eigenfunction $ \rho^\mathrm{ef} $ features a node
at $ r_\mathrm{n\,\rho} \simeq 7 \mathrm{\ km} $ in the equatorial
plane\footnote{The occurence of at least one node in the radial
eigenfunction $ \rho^\mathrm{ef} $ of a quasi-radial mode is a
consequence of mass conservation.}, located at less than half the
stellar equatorial radius (see
Fig.~\ref{fig:mass_shedding_eigenfunctions}). However, not everywhere
in the star an over- or underdensity is reflected by an according
localised increase or decrease of gravity, as the \emph{local}
gravitational pull is created by the \emph{global} density
distribution. Consequently, the nodes of the eigenfunctions of both
the conformal factor $ \phi^\mathrm{ef} $ and its negative radial
derivative $ - \partial_r \phi^\mathrm{ef} $ (representing the
perturbation of the gravitational force) lie much farther out at
$ r_\mathrm{n\,\partial_r \phi} \simeq 14 \mathrm{\ km} >
r_\mathrm{n\,\phi} \simeq 11 \mathrm{\ km} > r_\mathrm{n\,\rho} $. In
a region $ r_\mathrm{n\,\rho} < r \lesssim r_\mathrm{n\,\phi} $ the
perturbations of density ($ \rho^\mathrm{ef} $) and gravitational
force ($ \phi^\mathrm{ef} $, $ - \partial_r \phi^\mathrm{ef} $)
have opposite sign, and thus at some point during a pulsation cycle
the gravitational force there decreases although the local density
increases. Nevertheless, close to the stellar equatorial surface
both $ \phi^\mathrm{ef} $ and $ - \partial_r \phi^\mathrm{ef} $ are
again in phase with $ \rho^\mathrm{ef} $, and remain greater than zero
(albeit rather small) \emph{at} the surface. Thus, whenever the local
density there increases in a pulsation, this small local net gain
in the gravitational pull suffices to efficiently attenuate
mass-shedding, as observed in the CFC simulations.
\begin{figure}
\includegraphics[width=84mm]{figure17.eps}
\caption{Radial profiles of the $ F $-mode eigenfunction of the
rest-mass density $ \rho^\mathrm{ef} $ (black solid line), the
conformal factor $ \phi^\mathrm{ef} $ (blue dashed line), and its
negative radial derivative $ - \partial_r\phi^\mathrm{ef} $ (red
dotted line) along the equatorial plane for model BU9 with ideal
fluid EOS in CFC. The vertical dotted lines mark the locations
$ r_\mathrm{n\,\rho} $, $ r_\mathrm{n\,\phi} $,
$ r_{\mathrm{n\,}\partial_r \phi} $ of the eigenfunction nodes and
the equatorial stellar radius $ r_\mathrm{e} $. The vertical
scaling of the eigenfunctions is arbitrary, but their sign is
not.}
\label{fig:mass_shedding_eigenfunctions}
\end{figure}
When matter is shed from the star, kinetic energy is carried away to
the expense of the pulsational energy. Consequently, the pulsations
of the star are gradually damped, depending on the intensity of
mass-shedding. Fig.~\ref{fig:mass_shedding_evolution} shows the
time evolution of the central rest-mass density for model~BU9 obtained
from the CFC simulation and the one in the Cowling approximation. It
is evident that the damping of the pulsations is very strong in the
latter case, leading to an approximate damping time-scale of
$ 10 \mathrm{\ ms} $. In contrast, the insignificant mass-shedding in
the CFC simulation results in only small damping on a much longer
time-scale. Actually, a long-term observation of the time evolution of
$ \rho_\mathrm{c} $ reveals that most of the decline of the
oscillation amplitude in the time window of
Fig.~\ref{fig:mass_shedding_evolution} can be explained by a
mode-beating effect, which superimposes the genuine damping. Even with
high initial perturbation amplitudes we are not able to observe
significant unambiguous mass-shedding-induced damping during typical
evolution times of $ 20 \mathrm{\ ms} $. Simulations with lower
initial perturbation amplitudes or of models which rotate not so close
to the mass-shedding limit exhibit less mass-shedding compared to the
above model, which is in accordance with the findings presented by
SAF.
\begin{figure}
\includegraphics[width=84mm]{figure18.eps}
\caption{Time evolution of the central rest-mass density
$ \rho_\mathrm{c} $ for model BU9 with ideal fluid EOS at
$ t = 4.4 \mathrm{\ ms} $ in CFC (black solid line) and in the
Cowling approximation (red dashed line). In the Cowling
approximation, significant mass-shedding-induced damping can be
seen, whereas the decline of the oscillation amplitude in CFC can
be mostly attributed to mode-beating.}
\label{fig:mass_shedding_evolution}
\end{figure}
Even though in the CFC simulation the mass-shedding-induced damping of
pulsations is not as strong as in the Cowling approximation, it could
still have significant implications for unstable modes that grow on
secular time-scales due to the gravitational-radiation driven CFS
instability. In particular, the $ l = m = 2 $ $ f $-mode becomes
unstable only near the mass-shedding limit in uniformly rotating stars
\citep[see][]{stergioulas_98_a, morsink_99_a}. A detailed
investigation is required to determine whether the rate of
mass-shedding-induced damping is shorter than the secular growth rate
of the instability at amplitudes smaller than of order unity.
In Section~\ref{subsection:perturbations} we have demonstrated that
for exciting $ F $-mode oscillations, an initial velocity perturbation
of the form~(\ref{eq:perturbation_l=0}) is more appropriate than the
density perturbation of Eq.~(\ref{eq:perturbation_rho}) if the
spacetime is coupled to the hydrodynamics. However, we have performed
mass-shedding test simulations of model~BU9 in the Cowling
approximation, which show that in this case a velocity perturbation
with typical amplitudes either results in a strong negative or
positive drift in the time evolution of the central rest-mass density,
depending on the sign of the initial perturbation amplitude. Avoiding
this by significantly reducing the initial perturbation amplitude in
turn effectively suppresses the mass-shedding and the resulting
damping of the stellar pulsations. Following SAF, we therefore use the
density perturbation~(\ref{eq:perturbation_rho}) for the simulations
of the mass-shedding model~BU9 in both CFC and the Cowling
approximation presented here. To compensate for the amplification of
the oscillation amplitude during \emph{evolution} using equal
\emph{initial} perturbation amplitudes for a coupled spacetime as
compared to the Cowling approximation (see
Section~\ref{subsection:perturbations}), we use different initial
perturbation amplitudes in CFC ($ a = 0.01 $) than in the Cowling
approximation ($ a = 0.05 $). This yields an approximately equal
height of the first oscillation peak during the evolution (see
Fig.~\ref{fig:mass_shedding_evolution}), so that we can consider the
initial conditions in the two simulations to be similar.
\section{Gravitational waves}
\label{section:gravitational_waves}
In the numerical simulations of the models presented above, we extract
the gravitational waves emitted by the pulsations initiated
in each model. For this we use the Newtonian quadrupole formula in
its time-integrated form \citep[as described in detail
in][]{dimmelmeier_02_b}, which yields the quadrupole wave amplitude
$ A^\mathrm{E2}_{20} $ as the lowest order term in a multipole
expansion of the radiation field into pure-spin tensor harmonics
\citep{thorne_80_a}.
\subsection{Gravitational wave power spectrum}
\label{subsection:wave_spectrum}
Fig.~\ref{fig:gw_power_spectrum} shows the
gravitational wave power spectrum (i.e.\ the PSD of the time evolution
of $ A^\mathrm{E2}_{20} $) for a series of simulations of model~A1
(i.e.\ the most slowly rotating model of sequence~A) in which the
$ F $, $ H_1 $, $ {}^{2\!}f $, and $ {}^{2}p_1 $-modes are individually
excited using the eigenfunction recycling technique. As already stated
in Section~\ref{section:recycling} the initial excitation amplitude
$ a $ for each individual mode was chosen such that the PSDs of the
integrated density variations in the equatorial plane reach similar
amplitudes (see also lower panel of Fig.~\ref{fig:recycling}).
However, this choice for $ a $ results in a different relative
oscillation amplitude $ \delta \rho_\mathrm{c} $ of the central
density during the evolution (with $ 2 \delta \rho_\mathrm{c} $ being
the density variation measured top to bottom of the oscillation) for
each of the four simulations. In Fig.~\ref{fig:gw_power_spectrum} we
thus scale the original gravitational wave signal such that it
corresponds to $ \delta \rho_\mathrm{c} = 1 \mathrm{\ per\ cent} $.
This is possible as we find that the maximum amplitude
$ |A^\mathrm{E2}_{20}|_\mathrm{max} $ of the quadrupole signal is
approximately proportional to $ \delta \rho_\mathrm{c} $ (and in turn
also to $ a $) to fair accuracy for the four linear stellar pulsation
modes investigated here.
\begin{figure}
\includegraphics[width=84mm]{figure19.eps}
\caption{Gravitational wave power spectrum for model~A1 excited with
recycled eigenfunctions of the $ F $, $ H_1 $, $ {}^{2\!}f $, and
$ {}^{2}p_1 $-mode, respectively. In the $ F $, $ H_1 $, and
$ {}^{2}p_1 $-mode recycling runs a significant
$ {}^{2\!}f $-mode component is also present in the gravitational
wave signal. The thick lines indicate an artifical fall-off with
quadratic frequency dependence assumed to get rid of these
spurious contributions. Units on the $y$-axis are arbitrary.}
\label{fig:gw_power_spectrum}
\end{figure}
As expected, the quadrupolar $ {}^{2\!}f $-mode is the strongest
emitter of gravitational waves, while the other three modes are still
within roughly one or two orders of magnitude of this dominant
gravitational wave emitting mode. Note that an excitation of the
$ {}^{2\!}f $-mode results in a nearly monochromatic gravitational wave
power spectrum, indicating that the recycled eigenfunctions used for
exciting this mode are of high accuracy. The PSD for the individual
excitations of the $ F $, $ H_1 $, and $ {}^{2}p_1 $-mode are also
dominated by the appropriate frequency of the respective mode, but
exhibit also additional contributions of other excited modes. These
contributions are usually small (at least one order of magnitude
smaller than the dominant mode), but as the quadrupole
$ {}^{2\!}f $-mode strongly radiates gravitational waves, a small
leaking of energy into that mode from the main excitation mode is
sufficient to exhibit a significant contribution at that specific
frequency. This is particularly apparent in the case of the
$ H_1 $-mode, which radiates gravitational waves only through its
\emph{rotationally-induced} $ l \geq 2 $ terms.
In order to suppress the unwanted $ {}^{2\!}f $-mode contribution in
the computation of the characteristic signal frequency and amplitude
in the following Section~\ref{subsection:wave_detectability}, for the
gravitational wave power spectrum of the $ F $, $ H_1 $, and
$ {}^{2}p_1 $-mode simulations we have assumed a fall-off behavior
with quadratic frequency dependence in the region of the spectrum
where the spurious $ {}^{2\!}f $-mode is located (indicated by thick
lines in Fig.~\ref{fig:gw_power_spectrum}).
We also note that for the excitations of the overtones $ H_1 $ and
$ {}^{2}p_1 $ with recycled eigenfunction, some inertial modes
(visible at the low-frequency end in the PSD of
Fig.~\ref{fig:gw_power_spectrum}) also weakly contribute to the
emission of gravitational waves.
\subsection{Detectability of gravitational waves}
\label{subsection:wave_detectability}
In order to assess the prospects for detectability of the
gravitational wave signal from our models of pulsating neutron stars
by current and planned interferometer detectors, we calculate the
characteristic dimensionless amplitude $ h_\mathrm{c} $ of the signal
from the quadrupole wave amplitude $ A^\mathrm{E2}_{20} $ as
described in \citet{zanotti_05_a}. We perform a Fourier transform of
the transverse traceless gravitational wave signal,
\begin{equation}
\tilde{h} (f) =
\int_{-\infty}^\infty \!\! e^{2 \pi i f t} h^\mathrm{TT} (t) \, dt,
\label{eq:waveform_fourier_tranform}
\end{equation}
where
\begin{equation}
h^\mathrm{TT} =
\frac{1}{8} \sqrt{\frac{15}{\pi}} \frac{A^\mathrm{E2}_{20}}{r_\mathrm{gw}}
\label{eq:amplitude_to_signal_strain}
\end{equation}
in the equatorial plane of the neutron star (assuming optimal
detection geometry of the interferometer) with $ r_\mathrm{gw} $ being
the distance from the emitting source to the detector. To obtain the
detector dependent characteristic signal frequency
\begin{equation}
f_\mathrm{c} =
\left( \int_0^\infty \!
\frac{\langle |\tilde{h}|^2 \rangle}{S_h}
f \, df \right)
\left( \int_0^\infty \!
\frac{\langle |\tilde{h}|^2 \rangle}{S_h}
df \right)^{-1}
\label{eq:characteristic_frequency}
\end{equation}
and characteristic signal amplitude
\begin{equation}
h_\mathrm{c} =
\left( 3 \int_0^\infty \!
\frac{S_{h\mathrm{\,c}}}{S_h} \langle |\tilde{h}|^2 \rangle
f \, df \right)^{1/2}\!\!\!\!\!\!\!\!,
\label{eq:characteristic_amplitude}
\end{equation}
the power spectral density $ S_h $ of the detector is needed (with
$ S_{h\mathrm{\,c}} = S_h (f_\mathrm{c}) $). We approximate the
average
$ \langle |\tilde{h}|^2 \rangle $ over randomly distributed angles
by $ |\tilde{h}|^2 $. From Eqs.~(\ref{eq:characteristic_frequency},
\ref{eq:characteristic_amplitude}) the signal-to-noise ratio can be
computed as $ S / N = h_\mathrm{c} / [h_\mathrm{rms} (f_\mathrm{c})]$,
where $ h_\mathrm{rms} (f_\mathrm{c}) = \sqrt{f_\mathrm{c} S_h (f_\mathrm{c})} $
is the value of the rms strain noise (i.e.\ the theoretical
sensitivity window) for the detector at the characteristic frequency.
In Table~\ref{table:gw_sensitivity} we give the values of
$ f_\mathrm{c} $ and $ h_\mathrm{c} $ for the gravitational wave
signal emitted by the \emph{most slowly rotating} model~A1 excited
with recycled eigenfunctions, computed for the theoretical power
spectral density $ S_h $ of the detectors VIRGO, LIGO~I and advanced
LIGO located at a distance of $ r_\mathrm{gw} = 10 \mathrm{\ kpc} $
from the source. The strong emission of the gravitational waves by the
quadrupolar $ {}^{2\!}f $-mode and its first overtone $ {}^{2}p_1 $ is
reflected by the large wave amplitudes $ h_\mathrm{c} $ of
$ \sim 400 \times 10^{-22} $ and $ \sim 80 \times 10^{-22} $,
respectively, compared to only $ \sim 20 \times 10^{-22} $ for the
$ F $-mode and $ \sim 10 \times 10^{-22} $ for the $ H_1 $-mode.
\begin{table}
\centering
\caption{Characteristic frequency $ f_\mathrm{c} $ and
characteristic amplitude $ h_\mathrm{c} $ for a gravitational wave
signal emitted by the \emph{most slowly rotating} model~A1 excited
with recycled eigenfunctions of the $ F $, $ H_1 $, $ {}^{2\!}f $,
and $ {}^{2}p_1 $-mode, respectively, as prospectively seen by
various interferometer detectors. For comparison, the frequency
$ f $ of the respective stellar oscillation mode is also
given. All frequencies are in kHz, and $ h_\mathrm{c} $ is given
in units of $ 10^{-22} $ for a source at
$ r_\mathrm{gw} = 10 \mathrm{\ kpc} $ oscillating with
$ \delta \rho_\mathrm{c} = 1 \mathrm{\ per\ cent} $. The signal
duration time is assumed to be
$ t_\mathrm{gw} = 20 \mathrm{\ ms} $.}
\label{table:gw_sensitivity}
\begin{tabular}{@{}l@{\quad}r@{\qquad}r@{\quad}r@{\qquad}r@{\quad}r@{\qquad}r@{\quad}r@{}}
\hline
& &
\multicolumn{2}{c@{\qquad}}{VIRGO} &
\multicolumn{2}{c@{\qquad}}{LIGO~I} &
\multicolumn{2}{c@{}}{adv.~LIGO} \\
Mode &
\multicolumn{1}{c@{\qquad}}{$ f $} &
\multicolumn{1}{c@{\quad}}{$ f_\mathrm{c} $} &
\multicolumn{1}{c@{\qquad}}{$ h_\mathrm{c} $} &
\multicolumn{1}{c@{\quad}}{$ f_\mathrm{c} $} &
\multicolumn{1}{c@{\qquad}}{$ h_\mathrm{c} $} &
\multicolumn{1}{c@{\quad}}{$ f_\mathrm{c} $} &
\multicolumn{1}{c@{}}{$ h_\mathrm{c} $} \\
\hline
$ F $ & 1.400 & 1.370 & 19.3 & 1.338 & 18.9 & 1.236 & 17.4 \\
$ H_1 $ & 3.816 & 3.783 & 12.1 & 3.792 & 12.1 & 3.791 & 12.1 \\
$ {}^{2\!}f $ & 1.577 & 1.524 & 463.3 & 1.467 & 467.5 & 1.292 & 393.8 \\
$ {}^{2}p_1 $ & 3.580 & 3.545 & 76.9 & 3.545 & 76.9 & 3.544 & 76.9 \\
\hline
\end{tabular}
\end{table}
Note that we integrate $ h^\mathrm{TT} $ over a finite duration
time of $ t_\mathrm{gw} = 20 \mathrm{\ ms} $ in the FFT and scale
$ \delta \rho_\mathrm{c} $ to 1 per cent for each individual excited
mode. To compute $ h_\mathrm{c} $ in a straightforward way from the
values given in Table~\ref{table:gw_sensitivity} for an arbitrary
source distance, signal duration, and central oscillation amplitude of
the star, the following simple scaling laws can be used:
$ h_\mathrm{c} \propto r_\mathrm{gw}^{-1} $ and
$ h_\mathrm{c} \propto \sqrt{t_\mathrm{gw}} $, while
$ h_\mathrm{c} \propto |A^\mathrm{E2}_{20}|_\mathrm{max} $ for a given
power spectral density $ S_h $ and thus approximately proportional
to $ \delta \rho_\mathrm{c} $ (see discussion in
Section~\ref{subsection:wave_spectrum}).
As we artificially remove the spurious contribution of the quadrupole
$ {}^{2\!}f $-mode to the gravitational wave power spectrum of the
$ F $, $ H_1 $, and $ {}^{2}p_1 $-mode recycling runs (see
Section~\ref{subsection:wave_spectrum}), the small discrepancy between
the values for characteristic gravitational wave signal frequency
$ f_\mathrm{c} $ and the stellar oscillation mode frequency $ f $ in
Table~\ref{table:gw_sensitivity} is predominantly caused by the
dependence of $ f_\mathrm{c} $ on the power spectral density
$ S_h $ of the specific detector. Accordingly, as
$ h_\mathrm{c} \neq h^\mathrm{TT} $, the values of characteristic
amplitudes $ h_\mathrm{c} $ are slightly different for each detector.
The effect of $ S_h $ on $ f_\mathrm{c} $ and $ h_\mathrm{c} $ is
strongest for the gravitational wave signal emitted by the relatively
low frequency modes $ F $ and $ {}^{2\!}f $.
\begin{figure}
\includegraphics[width=84mm]{figure20.eps}
\caption{Distribution of the characteristic gravitational wave
signal amplitude $ h_\mathrm{c} $ in the frequency space for
model~A1 excited with recycled eigenfunctions of the $ F $,
$ H_1 $, $ {}^{2\!}f $, and $ {}^{2}p_1 $-mode, respectively.
The crosses represent
$ h_\mathrm{c} $ for a source at
$ r_\mathrm{gw} = 10 \mathrm{\ kpc} $, oscillating with
$ \delta \rho_\mathrm{c} = 5 \mathrm{\ per\ cent} $. The signal
duration time is assumed to be
$ t_\mathrm{gw} = 20 \mathrm{\ ms} $. The sensitivities
$ h_\mathrm{rms} $ of the interferometer detectors VIRGO (black
solid line), LIGO~I (red dashed line), and advanced LIGO (blue
dotted line) are also shown. Particularly for the $ F $ and
$ {}^{2\!}f $-mode signal, the value of $ h_\mathrm{c} $ and
$ f_\mathrm{c} $ is slightly dependent on the sensitivity of the
respective interferometer detector, as indicated by the color
coding. The arrows indicate the location of $ h_\mathrm{c} $ for $
\delta \rho_\mathrm{c} = 1 \mathrm{\ per\ cent} $.}
\label{fig:gw_sensitivity}
\end{figure}
In Fig.~\ref{fig:gw_sensitivity} we show the location of the
characteristic gravitational wave amplitude $ h_\mathrm{c} $ for the
$ F $, $ H_1 $, $ {}^{2\!}f $, and $ {}^{2}p_1 $-mode signal of model~A1
in relation to the sensitivity windows of VIRGO, LIGO~I, and advanced
LIGO. We again assume a distance to the source of
$ r_\mathrm{gw} = 10 \mathrm{\ kpc} $, i.e.\ within our Galaxy, and
(in contrast to Table~\ref{table:gw_sensitivity}) an oscillation
amplitude $ \delta \rho_\mathrm{c} = 5 \mathrm{\ per\ cent} $. In such
a scenario the gravitational wave signal from a neutron star pulsating
in the $ {}^{2\!}f $-mode with the adopted oscillation amplitude would
be clearly detectable even by current interferometer observatories
with a signal-to-noise ratio of around 50 for VIRGO and LIGO~I (and
300 for the planned advanced LIGO detector). On the other hand, the
$ F $ and $ {}^{2}p_1 $-mode signals lie only marginally above the
sensitivity threshold of current detectors, corresponding to a
signal-to-noise ratio of at most $ \sim 2 $. The wave signal from the
$ H_1 $-mode, in turn, could possibly be detectable only by the
proposed advanced LIGO detector for the selected pulsation amplitude
and duration.
If $ \delta \rho_\mathrm{c} $ is reduced to 1 per cent, as exemplified
by the arrows in Fig.~\ref{fig:gw_sensitivity}, the $ F $ and
$ {}^{2}p_1 $-mode signals clearly fall out of the sensitivity window
of VIRGO and LIGO~I, but would be detectable by advanced LIGO, while
the $ H_1 $-mode signal is now evidently out of the range even for
the latter. Only the signal from the $ {}^{2\!}f $-mode still exhibits a
signal-to-noise ratio of around 10 for the current detectors. It is
important to note that using the scaling properties of the signal
discussed above, the data presented in Fig.~\ref{fig:gw_sensitivity}
can be used to determine the necessary minimum stellar oscillation
amplitude $ \delta \rho_\mathrm{c} $, maximum source distance
$ r_\mathrm{gw} $, and minimum signal duration time $ t_\mathrm{gw} $
required for a prospective detection (set by a specific
signal-to-noise ratio above 1) of the gravitational wave emitted by
model~A1.
We also emphasize that the frequencies of the investigated linear
pulsation modes $ F $, $ H_1 $, $ {}^{2\!}f $, and $ {}^{2}p_1 $ are
located at the rapidly ascending high frequency slope of the detector
sensitivity curves. Consequently, for the fixed rest mass sequences~A
and~AU the decrease of the frequencies for these modes with increasing
rotation can enhance the prospects for detection by shifting the
signals towards the more sensitive domain of the detector sensitivity
window. For the fixed central density sequences~B and~BU, however, the
opposite dependence of the $ H_1 $ and particularly of the
$ {}^{2\!}f $-mode frequencies on rotation reverses this effect for these
two modes.
Further increase of the signal-to-noise ratio of the gravitational
wave signals presented here could be achieved with the advanced LIGO
detector, using its planned ability to be tuned at higher frequencies
in a narrow-banding mode. The chances for a successful detection of
neutron star oscillation modes will also rise dramatically by the
simultaneous operation of high-frequency-band detectors such as the
proposed dual sphere detectors \citep{cerdonio_01_a}, which are
designed to have an improved sensitivity at frequencies of a few kHz
and would thus be particularly suitable for detecting the first
overtones $ H_1 $ and $ {}^{2}p_1 $.
\subsection{Gravitational wave asteroseismology}
\label{subsection:gw_asteroseismology}
It is clear from the discussion of our results so far, that pulsating
rotating neutron stars are gravitational wave sources that depend on
several parameters (EOS, mass, angular momentum, differential rotation
law, initial perturbation amplitude, damping mechanisms, etc.). All
these parameters may have different effects on the oscillation
spectrum of the star and, therefore, the successful extraction of the
physical characteristics of the source from the gravitational wave
signal will be difficult to achieve. It is important to isolate each
effect on the gravitational waveform in order to find general trends.
Empirical formulas that can be used for gravitational wave
asteroseismology (with the aim of pin-pointing the correct
high-density EOS through gravitational wave observations) were
constructed in recent years \citep{andersson_98_a, kokkotas_01_a,
benhar_04_a}, assuming nonrotating cold equilibrium models. Thus,
these formulas contain only the effect of EOS and mass. More recently,
finite temperature effects and slow-rotation effects on the mode
frequencies were computed \citep{ferrari_03_a, ferrari_03_b,
ferrari_04_a}, but these effects have not yet been incorporated in
the empirical formulas for asteroseismology. In our work we focus on
the effects of rotation, using a particular cold EOS as an example. A
complete treatment will, of course, require using a large sample of
realistic EOSs, but such an attempt is beyond the scope of this
work. Using only the frequencies extracted from the simulations
presented here, we can nevertheless identify several interesting
features that indicate qualitatively the influence of rotation on the
empirical formulas used in gravitational wave asteroseismology.
First, the frequencies of most modes show a simple dependence on
rotation so that the empirical formulas obtained for nonrotating stars
could be extended to rotation by multiplying with a factor containing
only one or two powers of a rotational parameter such as
$ T / |W| $. The exception is the $ H_1 $-mode, which will require
special treatment due to the avoided crossing at moderate values of
$ T / |W| $. Second, the frequencies of most modes, such as the $ F $,
$ {}^{2\!}f $ and $ H_1 $-mode, become nearly independent of the degree
of differential rotation when parametrized as a function of
$ T / |W| $. However, the frequency of the $ {}^{2}p_1 $-mode is
strongly affected by the degree of differential rotation. Thus, an
additional parameter, measuring the strength of differential rotation,
could be built into an empirical formula for this specific mode.
Ideally, if the four modes $ F $, $ H_1 $, $ {}^{2\!}f $, and
$ {}^{2}p_1 $ were detected, then a set of empirical formulas for
their frequencies, constructed in the way described above, would allow
the simultaneous extraction of the mass, radius, $ T / |W| $, and
degree of differential rotation. Of course, this ideal situation may
become more complicated by the inclusion of finite temperature
effects, which would add at least one more parameter to the system.
\section{Summary and Outlook}
\label{section:summary}
Using the axisymmetric general relativistic hydrodynamics code
\textsc{CoCoNuT} we have investigated pulsations of uniformly and
differentially rotating neutron star models. We have compared our
numerical simulations, in which the spacetime dynamics is coupled to
the evolution of the fluid, to previous results performed under the
assumption of a fixed spacetime (Cowling approximation). In the
present work we have used the so-called conformal flatness condition
(CFC) for the spatial three-metric, this being an excellent and
well-tested approximation of the exact spacetime in the regime studied
here. The coupled system of (hyperbolic) hydrodynamics and (elliptic)
metric equations has been solved using the novel approach of combining
Riemann-solver-based HSRC methods for the fluid evolution and spectral
methods for the computation of the spacetime
metric~\citep{dimmelmeier_05_a}.
As equilibrium initial data we have constructed four sequences of
relativistic polytropes with parametrized rotation, which have then
been perturbed by small amplitude $ l = 0 $, 2, or 4 trial
eigenfunctions to excite pulsation modes. By analyzing the time
evolutions of various hydrodynamic and metric quantities using Fourier
transforms along radial profiles or on the entire spatial grid, we
have obtained the pulsation frequencies for the fundamental
quasi-radial ($ l = 0 $) $ F $-mode, the fundamental quadrupole
($ l = 2 $) $ {}^{2\!}f $-mode, and their respective overtones, the
$ H_1 $ and $ {}^{2}p_1 $-mode, as well as three inertial modes
labelled $ i_{-2} $, $ i_1 $, and $ i_2$. Additionally, for two
differentially rotating sequences we have obtained the frequencies for
the $ {}^{4}f $ and $ {}^{4}p_1 $-mode. We have found that for these
two sequences the $ {}^{4}p_1 $-mode engages in an avoided crossing
with the $ H_1 $-mode. This is a consequence of the different
influence of rotation on the frequency of these two modes, which
brings their frequencies closer with increasing rotation rate. At the
avoided crossing, an exchange of the character of eigenfunctions of
the $ H_1 $ and $ {}^{4}p_1 $-mode has been observed. Although linear
perturbation theory predicts the existence of an infinite number of
inertial modes in a finite frequency range, in our simulations we have
detected the predominant excitation of only a few specific inertial
modes, which are excited as by-products of the excitation of other
modes. For the differentially rotating sequence of fixed rest mass,
the inertial mode frequencies reach a maximum value before the
mass-shedding limit is reached and then decrease again.
In order to suppress the simultaneous excitation of more than one
mode, which is typically the case when trial eigenfunctions are used
as initial perturbations, we have employed a new technique, which we
call \emph{eigenfunction recycling}. For this, we first extract the
two-dimensional eigenfunction of velocity components at the frequency
of the mode under consideration and then apply it as an initial
perturbation in a second simulation of the original neutron star
model. This selective excitation of modes works very well even for
overtones.
When comparing the frequency curves of the $ F $-mode for all four
investigated sequences with previous results in the Cowling
approximation, we have observed that the latter typically overestimate
the correct frequency by about a factor of 2 (corresponding to an
absolute difference of $ \sim 1 \mathrm{\ kHz} $), which is consistent
with similar findings for nonrotating stars by
\citet{yoshida_97_a}. Moreover, we have found no evidence for the
$ F $-mode splitting found by SAF to occur in the Cowling
approximation. Thus, we conclude that this effect is an artifact which
can be attributed to the Cowling approximation, as was suggested in
SAF. On the other hand, for the $ H $, $ {}^{2\!}f $, and
$ {}^{2}p_1 $-mode frequencies, much closer agreement with the Cowling
results of SAF has been found. For these modes the relative difference
is usually less than $ \sim 20 \mathrm{\ per\ cent} $. Concerning
empirical formulas constructed by \citet{font_02_a} and SAF using
results in the Cowling approximation in order to predict mode
frequencies for rotating sequencies, we have found that the changing
characteristics of equilibrium models along various sequences
significantly affect the relative differences between the actual
frequencies and those obtained in the Cowling approximation. As a
result, such empirical relations can only be of limited validity,
while still representing a significant improvement compared to the
Cowling approximation.
Even though we have used only small-amplitude initial perturbations,
the use of a non-linear evolution code has allowed us to investigate
various non-linear effects in pulsating neutron star models. We have
observed non-linear harmonics of the linear pulsation modes of our
stellar models, arising as linear sums and differences of various
linear modes, including self-couplings \citep[for similar findings in
high-density accretion tori around Kerr black holes,
see][]{zanotti_05_a}. For rotating models such non-linear harmonics
actually fall into the frequency range of the inertial modes and could
interact with linear modes in non-linear 3-mode couplings in the form
of a resonance or a parametric instability. It would be interesting to
investigate under which conditions strong mode couplings or
instabilities can occur.
In SAF a new non-linear damping mechanism for non-linear pulsations
was found to operate in uniformly rotating models that are near the
mass shedding limit. The damping of pulsations is due to mass
shedding, when fluid elements near the equator become unbound due to
the radial component of the velocity perturbation. This
mass-shedding-induced damping could have severe consequences for,
e.g., the $ f $-mode gravitational-wave driven CFS instability, which
grows on a secular time-scale. Here we have found that while the actual
mass-shedding-induced damping does not occur on dynamical time-scales
(as in the Cowling approximation) it is still present on secular
time-scales, so that a detailed comparison to the $ f $-mode growth
rate is required in order to determine the outcome of the $ f $-mode
instability.
Taking into account that in the core collapse simulations by
\citet{dimmelmeier_01_a, dimmelmeier_02_a, dimmelmeier_02_b} the
dominant quadrupole contribution had an effective gravitational wave
amplitude that was roughly one to two orders of magnitude above the
sensitivity curve of the advanced LIGO detector, it is likely that in
a favourable detection event several pulsation modes of the
proto-neutron star may be detected simultaneously. We have thus
investigated the gravitational wave signals (in the Newtonian
quadrupole approximation) emitted by the linear modes $ F $, $ H_1 $,
$ {}^{2\!}f $, and $ {}^{2}p_1 $, for a slowly rotating neutron star
model. In order to estimate the prospects for detectability of the
signal by interferometric detectors, we have computed the (weakly
detector dependent) characteristic signal frequency and amplitude for
the theoretical power spectral density of VIRGO, LIGO~I, and advanced
LIGO. We infer that for a pulsation amplitude of the stellar central
density of 5 per cent a gravitational wave signal from the
investigated modes (with the exception of the $ H_1 $-mode) is
measurable by current detectors, if the signal is integrated over a
time of at least $ 20 \mathrm{\ ms} $ and the source is located in our
Galaxy. We have shown how scaling properties of the signal can be used
to determine the necessary minimum stellar pulsation amplitude,
maximum source distance, and minimum signal duration time required for
a prospective detection. We have also discussed how effects of
rotation can qualitatively influence empirical formulas used in
gravitational wave asteroseismology and can thus help to constrain
parameters of neutron star models in the case of a successful
detection of gravitational waves from pulsation modes. A more complete
study of the detectability of various modes for all equilibrium models
of our sequences will appear elsewhere.
In the relativistic core collapse simulations by
\citet{dimmelmeier_01_a, dimmelmeier_02_a, dimmelmeier_02_b} the
quasi-periodic gravitational waves emitted around core bounce were
found to have frequencies less than roughly $ 1.1 \mathrm{\ kHz} $. On
the other hand, for nonrotating models constructed with a large sample
of different realistic EOSs, the frequencies of the fundamental
$ l = 2 $ quadrupole $ f $-mode range from
$ \sim 1.35 \mathrm{\ kHz} $ for extremely stiff EOSs to
$ \sim 3.6 \mathrm{\ kHz} $ for extremely soft EOSs \citep[see,
e.g.,][]{andersson_98_a}. One of the main effects lowering the
frequency of emitted gravitational waves is rotation. In our
differentially rotating sequence of fixed rest mass, the nonrotating
model has a fundamental $ l = 2 $ $ f $-mode frequency of
$ \sim 1.6 \mathrm{\ kHz} $, which decreases to
$ \sim 1.1 \mathrm{\ kHz} $ at the largest rotation rate. Therefore,
additional effects must be taken into account in order to explain the
wider range of frequencies found by \citet{dimmelmeier_01_a,
dimmelmeier_02_b}. One such important factor is the high entropy in
a proto-neutron star immediately after core bounce. The proto-neutron
star is surrounded by a hot envelope into which the pulsations of the
interior can penetrate, increasing the period of each pulsation
\citep[see][]{ferrari_03_a}. Thus, a natural extension of our
investigation will be to compute the frequencies of the various
axisymmetric modes for a realistic proto-neutron star structure.
\section*{Acknowledgments}
It is a pleasure to thank Theocharis Apostolatos, John Friedman,
Kostas Kokkotas, Ewald M\"uller, Christian Ott, Jos\'e Pons, and
Luciano Rezzolla for helpful comments and discussions. We are grateful
to Olindo Zanotti for providing us with data for producing the
sensitivity curves of the interferometer detectors. Financial support
for this research has been provided by the EU ILIAS initiative of the
European Network of Theoretical Astroparticle Physics (ENTApP), the
German Research Foundation DFG (SFB/Transregio~7
`Gravitations\-wellen\-astro\-nomie'), the Greece--Spain bilateral
research grant by the General Secretariat for Research and Technology
(GSRT) and the Spanish Ministerio de Educaci\'on y Ciencia (grants
AYA2004-08067-C03-01 and HG2004-0015), the Pythagoras,
Pythagoras-II, and Heraclitus grants of the Greek Ministry of Education
and Religious Affairs, and the IKYDA German--Greek research
travel grant by DAAD and IKY. The computations were performed at the
Max-Planck-Institut f\"ur Astrophysik in Garching and the
Max-Planck-Institut f\"ur Gravitationsphysik in Golm, Germany.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,881 |
# 300 Low-Carb
Slow Cooker Recipes
**Healthy Dinners That Are Ready When You Are!**
Dana Carpender
For my sister Kim,
who works way too hard,
and loves her slow cooker.
## contents
Introduction: My Journey to Slow Cooker Mastery
Ingredients: Common and Not-So-Common Ingredients
CHAPTER 1 Slow Cooker Snacks and Hot Hors D'oeuvres
CHAPTER 2 Slow Cooker Eggs
CHAPTER 3 Slow Cooker Poultry
CHAPTER 4 Slow Cooker Beef
CHAPTER 5 Slow Cooker Pork
CHAPTER 6 Slow Cooker Lamb
CHAPTER 7 Slow Cooker Fish and Seafood
CHAPTER 8 Slow Cooker Crazy Mixed-up Meals
CHAPTER 9 Slow Cooker Soups
CHAPTER 10 Slow Cooker Sides
CHAPTER 11 Slow Cooker Desserts
CHAPTER 12 Just a Few Extras . . .
Acknowledgments
About the Author
Index
## Introduction:
My Journey to Slow Cooker Mastery
I have a confession: When my editor, Holly, suggested I write a low-carb slow cooker book, I balked. Oh, I knew it would be popular—many readers had written me asking for a slow cooker book. I just wasn't thrilled at the idea of a couple of months of slow cooked dinners. I'd made some slow cooked meals that were pretty good, but overall, it seemed to me that most slow cooked food was not brilliant. Too many dishes seemed to be waterlogged, mushy, and insipid. Furthermore, so many slow cooker recipes seemed to rely on high-carb canned cream soups—indeed, many slow cooker books seemed to think that "put food in pot, dump in condensed cream of mushroom soup, and cook on low until you come home from work" was a recipe—but not in my book, figuratively or literally!
So I resisted for quite a while, but those e-mails saying, "Please, please, write us a slow cooker book!" were piling up in my inbox. I needed to write a slow cooker book! But it was clear I had to get better at slow cooking.
Well, my mom is a retired librarian, and I learned years ago that if you want to learn something, you need to look it up. So I went to Amazon.com and read reviews of slow cooker books to determine which were drawing raves. I then read the books that got the best reviews, gleaning what I could from them of the tricks of making slow cooker food as appealing as possible, both in taste and texture. Not surprisingly, my slow cooker performance took a remarkable upturn!
I also got an idea of what slow cookers do well. Obviously, they're not for anything that you want to come out crispy and brown, but if what you need is slow, moist cooking, a slow cooker will do it better than any other appliance. Preparing soups and stews and braising are obvious slow cooker strengths, but I also learned that a slow cooker is terrific for cooking anything that needs to be baked in a water bath (sometimes called a bain-marie)—custards, in particular. I was thrilled to discover that my slow cooker did the best job ever of roasting nuts and seeds, and it's perfect for hot beverages for parties and hors d'oeuvres that would otherwise need a chafing dish.
I was very surprised to learn that cooking fish in my slow cooker worked well. You can't leave it for hours and hours because fish overcooks easily. But just an hour or so of the gentle heat of the slow cooker leaves fish tender, moist, and succulent. Do try it when you have an hour to get dinner on the table, even if you mostly use your slow cooker to cook supper while you're out of the house for hours.
I also had a few spectacular failures. (Don't even ask about the brussels sprouts!) But overall, I was pleased to discover that with a few simple considerations in mind, slow cookers can turn out truly wonderful food.
### About Slow Cookers
By the time I finished this project, I owned three slow cookers. All of them are Rival Crock-Pots, the original slow cooker. (Crock-Pot is a brand name. All Crock-Pots are slow cookers, but not all slow cookers are Crock-Pots.) The Crock-Pot, as the original, is pretty much the gold standard of slow cookers. The heat comes from all around the crockery insert, rather than only coming from the bottom. If you have one of the slow cookers that has the heating element only on the bottom, you'll have to experiment a bit with these recipes to see if the times are correct.
The "low" setting on a Rival Crock-Pot is around 200°F (93°C) (just above, actually, because things will boil eventually at this setting, and the boiling point is 212°F [100°C]), and the "high" setting is around 300°F (150°C). If you have another brand of slow cooker that lets you set specific temperatures, keep this in mind. If you have another brand of slow cooker and you're not sure what temperature the settings will give (look in the booklet that came with it for this information), you can fill the slow cooker with water, heat it for 2 hours on low, and test the water's temperature with a kitchen thermometer, but this is a lot of trouble. I'd probably just use the low and high settings and keep mental notes on how meals turn out.
A good thing about the original Crock-Pot is that the crockery insert lifts out of the base. This allows it to be refrigerated, microwaved (if your microwave is big enough), and—most important—put in the dishwasher. _Do not_ put your slow cooker in the dishwasher if the crockery cannot be separated from the heating element! Nothing electric should ever be submerged in water.
My slow cookers range in size. The smallest holds 2 1/2 quarts (2.4 liters), the middle-sized holds 3 quarts (2.8 liters), while the big one holds 5 1/2 quarts (5 liters). The 5 1/2-quart can easily hold enough food for 8 people. It's the obvious choice if you have a big family or like to cook enough to have leftovers for future meals. The 3-quart is the most common size. If you have a family of 4, it should be about right. If you have this size, figure you'll need to halve recipes that make 6 to 8 servings. The 2 1/2-quart is great for making dips, hot hors d'oeuvres, and hot beverages, but it is a bit small for family cooking.
Another consideration: My 5 1/2-quart Crock-Pot will fit a 6-cup (1.4 L) casserole dish, 8-inch (20 cm) springform pan, and a standard Bundt pan opening up many new cooking options. If you have a smaller unit and want to make custards, cheesecakes, and other dishes that call for inserting a dish or pan, you'll have to find dishes that will fit. It's easier with a bigger slow cooker.
Keep in mind that slow cookers come in round or oval shapes. You'll want a round slow cooker, instead of an oval, so you can insert a round glass casserole or a springform. Sadly, a big, round slow cooker takes the most storage space. I know of no good way around this.
### Some Things I've Learned about Slow Cooking
• Browning meat or poultry before putting it in the slow cooker upgrades vast hordes of recipes. Yes, it takes time and dirties up a skillet. But the flavor and texture that browning bring are worth it, worth it, worth it. Often I'll have you sauté your vegetables, too.
• It's important to keep liquids to the minimum that will make the recipe work, especially in recipes that have a lot of vegetables. All of the liquid that cooks out of the food while slow cooking will accumulate in the pot because no evaporation occurs. It's easy to end up with very watery food. This rule does not apply to soups, of course.
• Because of this accumulation of liquid, it's a good idea to use concentrated flavors. In particular, you'll find that in many of these recipes I use both broth and bouillon concentrate to make what amounts to a broth that is double-strength or more.
• Sometimes it's a good idea to transfer the liquid from the slow cooker to a saucepan and boil it hard till it's reduced by half. Half the volume means double the flavor.
• It's generally best to use lean cuts of meat, and you'll see I've often used skinless poultry, too. This is because fat that becomes crackling and succulent in the oven makes slow cooker food unbearably greasy. This makes slow cooking a great way to cook some of the leaner and tougher cuts of meat that you might not want to roast. It also makes slow cooking a good cooking method for those of you who are watching calories as well as carbs. It can even save you money—often tough and bony cuts of meat are cheap.
• For some strange reason, vegetables cook more slowly in a slow cooker than meat does. If you put vegetables on top of the meat in your slow cooker, you may find that they're still crunchy when the rest of the dinner is done. Put the ingredients in the pot in the order given in the recipes in this book.
• For this reason, too, it's best to cut vegetables into fairly small pieces. You'll find I've told you what size to cut things, for the most part. If the recipe says 1/2-inch (13 mm) cubes, and you cut your turnips in 1 1/2-inch (3.8 cm) cubes, you're going to have underdone turnips.
• It's never a bad idea to spray your slow cooker with nonstick cooking spray before putting the food in it. I don't always do this, though I've specified it where it seems a particularly good idea. But I can't think of a situation in which it would hurt.
### Things That Other Slow Cooker Books Seem to Think Are Terribly Important, But Don't Seem Like a Big Deal to Me
• Several books wanted me to use only whole spices, such as whole peppercorns or coarse-cracked pepper, whole cloves, whole leaf herbs, etc., etc. I used what I had on hand. I got tasty food.
• A few books felt that you shouldn't season your slow cooker food until the end of cooking time. I do often suggest that you add salt and pepper to taste at the end of the cooking time, but other than that, our seasonings go into the pot with the food. I've never had a problem with this.
• Some books were emphatic about the size of the slow cooker. This matters some; you can't put 4 quarts (3.8 L) of soup in a 3-quart (2.8 L) slow cooker, and if you're only making a small batch of dip, you probably shouldn't use a 5 1/2-quart (5 L) pot. But some cookbooks predicted dire results if I didn't fill my slow cooker at least halfway. I often filled my big slow cooker less than halfway. I got tasty food.
### About Timing
The biggest reason for the popularity of slow cookers is what I call "time-shifting"—the fact that they allow you to cook dinner at some time other than right before you eat it, so you can eat soon after you get home from work. Because of this, many slow cooker books tell you to cook most of their recipes for 8 hours or more. They figure you'll be away that long.
Unfortunately, I find that many dishes get unbearably mushy and overdone if they cook that long. I've tried to give the cooking times that I feel give the best results, which may not be the time that fits with your workday. These can generally be extended by an hour without a problem, but extending them by 2, 3, or more hours may well give you a very different result than I got.
A better idea is to do all your prep work the night before—cooking dinner after dinner, as it were. Then lift the filled crockery insert out of the base and stash it in the fridge overnight. The next morning, pull it out of the fridge, put it in the base, and turn it on, just before you leave the house. Starting with chilled food will add 1 to 2 hours to your cooking time. If you do this, don't heat up the base before putting the chilled crockery insert in! You may well crack the crockery.
If you need to extend your time even further, consider getting a timer. You should be fine letting your food wait 2 hours before the timer turns on the slow cooker—3 if the food is straight out of the fridge when the crockery goes in the base. It's a better idea to delay the starting time than to turn the pot off early because retained heat will cause the food to continue cooking even after the pot is turned off. Ask the nice people at your local hardware store about a timer you can plug things into. If you're just now acquiring your slow cooker, there are units available with time-delayed starters built in.
On the other hand, if you want to speed up a slow cooker recipe, you can do so by getting the contents warm before putting the crockery insert in the base. The crocks for two of my three slow cookers fit in my microwave. I have, on occasion, microwaved the full crock on medium heat until it was warm through before putting the crock in the base to continue cooking. This cuts a good hour off the cooking time.
You can, of course, also use the high setting when the low setting is specified. This will cut the cooking time roughly in half. However, I find that for most recipes low yields better results. If you have the time to use it when it's recommended, I suggest you do so.
If you're going to be around for a while and leave the house later, you could cook on high for an hour or so and then switch to low when you leave. Figure, again, that each hour on high is worth 2 hours on low.
The size of your slow cooker relative to your recipe will somewhat affect the cooking time. If you have a 5 1/2-quart (5 L) slow cooker and the food only fills it 1/4 full, you can likely subtract 1 hour from the cooking time. Conversely, if the food fills your slow cooker to within an inch of the rim, you can add an hour.
## Ingredients:
Common and Not-So-Common Ingredients
Here are a few ingredients I thought needed a little explanation:
• _**Beer**_ —One or two recipes in this book call for beer. The lowest carbohydrate beer on the market is Michelob Ultra, but I don't much like it. Still, it should be okay for cooking. Miller Lite and Milwaukee's Best Light are better, and they have only about 0.5 grams more carb per can.
• _**Black soybeans**_ —Most beans and other legumes are too high in carbohydrates for many low-carb dieters, but there is one exception. Black soybeans have a very low usable carb count, about 1 gram per serving, because most of the carbs in them are fiber. Several recipes in this book call for Eden brand canned black soybeans. Many health food stores carry these. If yours doesn't, I'll bet they could special order them for you. Health food stores tend to be wonderful about special orders.
I wouldn't recommend eating soybean recipes several times a week. I know that soy has a reputation for being the Wonder Health Food of All Existence, but there are reasons to be cautious. For decades now, soy has been known to be hard on the thyroid, and if you're trying to lose weight and improve your health, a slow thyroid is the last thing you need. More alarmingly, a study done in Hawaii in 2000 showed a correlation between the amount of tofu subjects ate in middle age and their rate and severity of cognitive problems in old age. Because scientists suspect the problem lies with the soy estrogens that have been so highly touted, any unfermented soy product, including canned soybeans, is suspect.
This doesn't mean we should completely shun soybeans and soy products, but we need to approach them with caution and eat them in moderation. Because many low-carb specialty products are soy-heavy, you'll want to pay attention there, too. Personally, I try to keep my soy consumption to 1 serving a week or less.
• _**Blackstrap molasses**_ —What the heck is molasses doing in a low-carb cookbook?! It's practically all carbohydrate, after all. Well, yes, but I've found that combining Splenda with a very small amount of molasses gives a good, brown-sugar flavor to all sorts of recipes. Always use the darkest molasses you can find. The darker it is, the stronger the flavor, and the lower the carb count. That's why I specify blackstrap, the darkest, strongest molasses there is. It's nice to know that blackstrap is also where all the minerals they take out of sugar end up. It may be high-carb, but at least it's not a nutritional wasteland. Still, I use only small amounts.
Most health food stores carry blackstrap molasses, but if you can't get it, buy the darkest molasses you can find. Most grocery store brands come in both light and dark varieties.
Why not use some of the artificial brown-sugar flavored sweeteners out there? Because I've tried them, and I haven't tasted one I would be willing to buy again. Caution: Splenda brown sugar blend contains sugar.
• _**Broths**_ —Canned or boxed chicken broth and beef broth are very handy items to keep around and certainly quicker than making your own. However, the quality of most of the canned broth you'll find at your local grocery store is appallingly bad. The chicken broth has all sorts of chemicals in it and often sugar as well. The beef broth is worse. It frequently has no beef in it whatsoever. I refuse to use these products, and you should, too.
However, there are a few canned or boxed broths on the market worth buying. Many grocery stores now carry a brand called Kitchen Basics, which contains no chemicals at all. It is packaged in quart-size boxes, much like soy milk. Kitchen Basics comes in both chicken and beef. Health food stores also have good quality canned and boxed broths. Both Shelton and Health Valley brands are widely distributed in the United States.
Decent packaged broth will cost you a little more than the stuff that is made of salt and chemicals but not a whole lot more. If you watch for sales, you can often get it as cheaply as the bad stuff, so stock up. When my health food store runs a sale on good broth for 89 cents a can, I buy piles of it!
_**One last note:**_ You will also find canned vegetable broth, particularly at health food stores. This is tasty, but because it runs much higher in carbohydrates than chicken and beef broths, I'd avoid it.
• _**Bouillon or broth concentrates**_ —Bouillon or broth concentrate comes in cubes, crystals, liquids, and pastes. It is generally full of salt and chemicals and doesn't taste notably like the animal it supposedly came from. It definitely does _not_ make a suitable substitute for good quality broth if you're making a pot of soup. However, these products can be useful for adding a little kick of flavor here and there—more as seasonings than as soups. For this use, I keep them on hand. I now use a paste bouillon concentrate product called Better Than Bouillon that comes in both chicken and beef flavors. I find it preferable to the granules or cubes.
• _**Carb Countdown**_ —When I wrote the original edition of this book, the Atkins Boom of 2003 was still going strong, and grocery stores were carrying a lot of low-carb specialty products, some dubious, other quite useful. Carb Countdown, a carbohydrate-reduced milk put out by Hood Dairy, was one that I found very useful, and accordingly I used it in several recipes.
Things have changed. Hood still makes the product, though they long since renamed it Calorie Countdown. However, far fewer stores are carrying it than were when the original edition came out. If you can find Calorie Countdown near you, great! If not, here are some possible substitutions, and their carb counts:
_**LC-Milk:**_ As I write this, a company called LC Foods has just introduced a powdered low-carb milk product called LC-Milk. Because it's powdered, it is easy to ship and easy to store. You simply combine the powder with water and heavy cream, and chill, though I'm thinking for a slow cooker recipe the chilling part would be unnecessary. Sufficient powder to make 1 cup (235 ml) of milk, or 10 teaspoons, has 1 gram of carb, with 1 gram of fiber, for 0 grams usable carb. The cream adds 1 gram of carb. The two combined have 192 calories per cup. The mix has no lactose, though there would be a little in the cream. LC-Milk is available from www.holdthecarbs.com, and may well be carried by low-carb online retailers by the time this sees print.
_**Half-and-half:**_ 10 grams of carbohydrate per cup. Of the substitutes available at your grocery store, this will likely give a result most similar to the original recipe. (For international readers: Half-and-half is a mixture of half milk, half heavy cream.) Adds 187 calories over the Carb Countdown.
_**Heavy cream:**_ 7 grams carbohydrate per cup. Obviously, this will give a richer result. Adds 684 calories. On the other hand, I've long since started trying to get 75 percent of my calories from fat, and I consider butterfat to be a healthful food. Your call.
_**Heavy cream and water, in equal parts:**_ This works well in many recipes, and of course it gives a lighter result than pure cream. Simple third-grade arithmetic tells me you'll get 3.5 grams of carbohydrate, and an extra 342 calories per cup.
_**Almond milk:**_ I've been playing around with almond milk recently, and I like it. It's now widely available in big grocery stores; look for it near the soy milk. (I would _not_ use soy milk. I try to minimize my soy consumption. I'm unconvinced the stuff is safe.) Keep in mind that almond milk comes in sweetened and unsweetened varieties; obviously you want the unsweetened kind. The brand I've been getting has just 1 gram of carbohydrate per cup, with 1 gram of fiber, for 0 grams usable carb, so this would be your lowest carb option. It's also far and away the lowest calorie choice, at just 35 calories per cup. It will, however, have a somewhat different flavor than dairy—mild and pleasant, but different. This is also a lactose-free choice.
_**Milk:**_ You know—just milk. It's 12 grams of carb per cup, so it's the highest carb choice, but in a recipe that makes, say, 5 or 6 servings, that's not disastrous. I'd recommend whole milk. Butterfat is nutritious, and it certainly tastes better than the low-fat stuff. It's an extra 22 calories per cup.
_**Final note:**_ I've heard from some readers that their grocery stores dropped Carb Countdown, but started re-stocking Calorie Countdown due to customer requests. So, it doesn't hurt to ask.
• _**Cauliflower**_ —You'll notice a certain reliance on cauliflower in this book, both in the form of "Fauxtatoes" (see recipe page 343) and in the form of Cauli-Rice (see recipe page 343). This is because many slow cooker recipes make wonderful gravy, and it's a shame not to have a side dish to help you eat it. (Indeed, traditional slow cooker recipes show a similar dependence on potatoes, rice, and noodles.)
You can skip the cauliflower if you like. Or you can substitute tofu shirataki noodles.
_**By the way:**_ If cauliflower (or another suggested garnish or side dish) isn't mentioned in the ingredient list, it's merely suggested and it's not included in the nutritional analysis for the dish. If it is in the ingredient list, it has been included in the analysis.
• _**Chili garlic paste**_ —This is a traditional Asian ingredient, consisting mostly, as the name strongly implies, of hot chiles and garlic. If, like me, you're a chile-head, you'll find endless ways to use the stuff once you have it on hand. Chili garlic paste comes in jars, and it keeps for months in the refrigerator. It is worth seeking out at Asian markets or in the international foods aisle of big grocery stores.
• _**Chipotle peppers canned in adobo sauce**_ —Chipotle peppers are smoked jalapeños. They're very different from regular jalapeños, and they're quite delicious. Look for them, canned in adobo sauce, in the Mexican foods section of big grocery stores. Because you're unlikely to use the whole can at once, you'll be happy to know that you can store your chipotles in the freezer, where they'll keep for months. I just float my can in a bowl of hot tap water for 5 minutes till it's thawed enough to peel off one or two peppers and then put it right back in the freezer.
• _**Erythritol**_ —This is one of the polyol or sugar alcohol sweeteners, and the one I use most often. Unlike maltitol, which is widely used in commercial sugar-free sweets, erythritol has very little gastrointestinal effect. It also has the lowest absorption profile of all the sugar alcohols, so it has virtually no usable carbohydrate, while I generally count half a gram for maltitol. My local health food stores all carry erythritol, but again, you can buy it through CarbSmart.com or Amazon.com if you can't find it locally.
You can substitute xylitol, another polyol rapidly growing in popularity. Be aware, however, that it is profoundly toxic to dogs, so don't let your pooch sneak a bite.
• _**Fish sauce or nuoc mam or nam pla**_ —This is a salty, fermented seasoning widely used in Southeast Asian cooking, available in Asian grocery stores and in the Asian food sections of big grocery stores. Grab it when you find it; it keeps nicely without refrigeration. Fish sauce is used in a few (really great) recipes in this book, and it adds an authentic flavor. In a pinch, you can substitute soy sauce, although you'll lose some of the Southeast Asian accent.
• _**Garlic**_ —I use only fresh garlic, except for occasional recipes for sprinkle-on seasoning blends. Nothing tastes like the real thing. To my taste buds, even the jarred, chopped garlic in oil doesn't taste like fresh garlic. We won't even _talk_ about garlic powder. You may use jarred garlic if you like—half a teaspoon should equal about 1 clove of fresh garlic. If you choose to use powdered garlic, well, I can't stop you, but I'm afraid I can't promise the recipes will taste the same either. One quarter teaspoon of garlic powder is the rough equivalent of 1 clove of fresh garlic.
• _**Ginger root**_ —Many recipes in this book call for fresh ginger, sometimes called ginger root. Dried, powdered ginger is _not_ a substitute. Fortunately, fresh ginger freezes beautifully. Drop the whole ginger root (called a hand of ginger) into a resealable plastic freezer bag and toss it in the freezer. When time comes to use it, pull it out, peel enough of the end for your immediate purposes, and grate it. Ginger grates just fine while still frozen. Throw the remaining root back in the bag and toss it back in the freezer.
Ground fresh ginger root in oil is available in jars at some very comprehensive grocery stores. I buy this when I can find it without added sugar, but otherwise, I grate my own.
• _**Granular Sucralose**_ —Since _15 Minute Low-Carb Recipes_ was first published, sucralose, best known by the trade name Splenda, has gone off-patent. There are now dozens of knock-offs and store brands, and there's no reason not to use them instead of the name brand. Do look for one that measures cup-for-cup like sugar. Be wary of brown sugar blends and other sucralose blends; these have sugar in them. And remember that because of the maltodextrin used to bulk it, granular sucralose has about 24 grams of carbohydrate per cup—the _0 carb_ figure is an artifact of the teeny serving size listed on the label.
• _**Guar and xanthan gums**_ —These sound just dreadful, don't they? But they're in lots of your favorite processed foods, so how bad can they be? They're forms of water soluble fiber, extracted and purified. Guar and xanthan are both flavorless white powders, and their value to us is as low-carb thickeners. Technically speaking, these are carbs, but they're all fiber—nothing but.
Those of you who read _500 Low-Carb Recipes_ know that I used to recommend putting your guar or xanthan through the blender with part or all of the liquid in the recipe to avoid lumps. You may now happily forget that technique. Instead, acquire an extra salt shaker, fill it with guar or xanthan, and keep it handy. When you want to thicken the liquid in your slow cooker, simply sprinkle a little of the thickener over the surface _while stirring_ , preferably with a whisk. Stop when your sauce, soup, or gravy is a little less thick than you want it to be. It'll thicken a little more on standing.
Your health food store may well be able to order guar or xanthan for you if they don't have them on hand. You can also find suppliers online. Of the two, I slightly prefer xanthan.
• _**Ketatoes**_ —Ketatoes, a low-carb instant mashed potato substitute, has sadly passed from our midst. However, at this writing there is a similar product available called Dixie Carb Counters Instant Mashers that works exactly the same as Ketatoes in my recipes, at least the ones I've tried it in. **Anywhere you see "Ketatoes" in this book, think "Dixie Carb Counters Instant Mashers," and you'll be fine.** You can get 'em through the low carb etailers, and even through Amazon.com. I get mine from CarbSmart.com. Yes, I have a relationship with them. No, I will not get a kickback if you order some Instant Mashers.
• _**Low-carb tortillas**_ —These are becoming easier and easier to find. I can get them at every grocery store in town. If you can't buy them at a local store, you can order them online. They keep pretty well. I've had them hang around for 3 or 4 weeks in a sealed bag without getting moldy or stale, so you might want to order more than one package at a time.
I use La Tortilla Factory brand because they've got the lowest usable carb count of any I've found, just 3 grams. They're mostly made of fiber! Beware: I have recently seen "low-carb" tortillas with deceptive packaging. The listed serving size turned out to equal only half of one tortilla. That's not a serving, as far as I'm concerned!
• _**Low-sugar preserves**_ —In particular, I find low-sugar apricot preserves to be a wonderfully versatile ingredient. I buy Smucker's brand and like it very much. This is lower in sugar by _far_ than the "all fruit" preserves, which replace sugar with concentrated fruit juice. Folks, sugar from fruit juice is still sugar. I also have been known to use low-sugar orange marmalade and low-sugar raspberry preserves.
• _**Splenda**_ —Be aware that Splenda granular that comes in bulk, in a box, or in the new "baker's bag" is different than the Splenda that comes in the little packets. The Splenda in the packets is considerably sweeter. One packet equals 2 teaspoons granular Splenda. All these recipes use granular Splenda.
• _**Sriracha**_ —This is a Southeast Asian hot sauce, and it's taking over the world— _Bon Appetit_ magazine declared it The Ingredient of the Year for 2010. Find Sriracha in the international aisle of big grocery stores or in Asian markets—look for the bright-red sauce with the rooster on the bottle.
• _**Sugar-free imitation honey**_ —This is a polyol (sugar alcohol) syrup with flavoring added to make it taste like honey. The two brands I've tried, one by HoneyTree and the other by Steele's, are not bad imitations.
Sugar-free imitation honey is becoming more and more available, and it is a useful product. I can get sugar-free imitation honey here in Bloomington at Sahara Mart, my favorite source of low-carb specialty products, and I've heard from readers that Wal-Mart now carries it. For that matter, many of the low-carb e-tailers carry Steele's brand of imitation honey. It shouldn't be too hard to get your hands on some.
• _**Sugar-free pancake syrup**_ —This is actually easy to find. All my local grocery stores carry it—indeed, many have more than one brand. It's usually with the regular pancake syrup, but it may be shelved with the diabetic or diet foods. It's just like regular pancake syrup, only it's made from polyols (sugar alcohols) instead of sugar. I use it in small quantities in a few recipes to get a maple flavor.
• _**Tofu Shirataki**_ —Finally! There's a genuinely low carb noodle. Shirataki are a traditional Japanese noodle made from a root called kojac or _konyaku_ , sometimes mistakenly translated _yam_ or _yam bean_. The konjac root is a rich source of a fiber called glucomannan, and it is that glucomannan fiber that forms the bulk of shirataki noodles.
Shirataki come in two basic varieties: traditional shirataki and tofu shirataki. Traditional shirataki are translucent and gelatinous and very . . . well, Asian. They're good in Asian dishes but pretty weird in Western-style recipes. Tofu shirataki are white, considerably less chewy, and good in a wide variety of applications. My local health food stores carry them in three widths: fettuccini, spaghetti, and angel hair. I keep the fettuccine and spaghetti widths on hand.
Shirataki come pre-hydrated in a pouch full of liquid. This makes them perfect for our super-quick recipes because you don't have to cook them, only drain and heat them. Snip open the pouch and dump them in a strainer in the sink. You'll notice the liquid smells fishy; you'll want to rinse them. After that, I put them in a microwaveable bowl and give them 90 seconds on high. More liquid will cook out of them, so I drain them again, heat for another 90 seconds, and drain yet a third time; then I add whatever sauce I like. This keeps the liquid cooking out of the noodles from diluting the sauce.
Shirataki keep up to a year in the fridge, so feel free to stock up.
• _**Vege-Sal**_ —If you've read my newsletter, _Lowcarbezine!_ , or my previous cookbooks, you know that I'm a big fan of Vege-Sal. It's a salt that's been seasoned, but don't think "seasoned salt." Vege-Sal is much milder than traditional seasoned salt. It's simply salt that's been blended with some dried, powdered vegetables. The flavor is quite subtle, but I think it improves all sorts of things. I've given you the choice between using regular salt or Vege-Sal in many recipes. Don't worry, they'll come out fine with plain old salt, but I do think Vege-Sal adds a little something extra. Vege-Sal is made by Modern Products, and it is available in health food stores.
chapter one
Slow Cooker Snacks and Hot Hors D'oeuvres
Slow cookers are mostly used for cooking dinner while you're out of the house, but they have other uses, such as keeping hors d'oeuvres and dips hot through your whole party! Plus your slow cooker will do the best job of roasting nuts ever. Here are some ways you can make your slow cooker the life of the party.
### Glazed Chicken Wings
Put out a pot of these and a big ole pile of napkins and watch your guests eat!
3 pounds (1.4 kg) chicken wings
1/2 cup (168 g) sugar-free imitation honey
1/2 cup (12 g) Splenda
1/2 cup (120 ml) soy sauce
2 tablespoons (28 ml) oil
2 cloves garlic
2 tablespoons (28 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
Cut the chicken wings into "drummettes." Season them with salt and pepper and put them in your slow cooker.
In a bowl, stir together the honey, Splenda, soy sauce, oil, garlic, and ketchup. Drizzle the mixture over the wings and stir them to coat. Cover the slow cooker, set it to low, and let it cook for 6 to 8 hours.
_**Yield:**_ 8 servings, each with: 144 calories, 10 g fat, 10 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs. (Analysis does not include the polyols in the imitation honey.)
### Maple-Mustard Wings
3 pounds (1.3 kg) chicken wings
Salt and pepper
1/4 cup (80 g) sugar-free pancake syrup
1/2 cup (125 g) _Dana's "Kansas City" Barbecue Sauce_ (see recipe on page 335)
2 tablespoons (22 g) brown mustard
1 garlic clove, crushed
1 teaspoon lemon juice
If you didn't buy your wings already cut up, cut them into drummettes. (If you stash the bones in the freezer to make bone broth, add the pointy wing tips to the bag; they're wonderful.) Salt and pepper your wings and lay them on your broiler rack. Now broil them, 4 to 5 inches (10 to 13 cm) from the heat, maybe 4 to 5 minutes per side.
While they're broiling, mix together everything else.
When the wings are browned a bit, use tongs to transfer them to the slow cooker. Pour the sauce over them and stir to coat.
Cover the pot, set on low, and let them cook for 4 to 5 hours. Serve from the slow cooker, with plenty of napkins!
_**Yield:**_ If 5 people share these, each will get: 368 calories, 25 g fat, 28 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Cranberry-Barbecue Meatballs
Boring old ground turkey does a Cinderella turn and comes to the party in this dish!
2 pounds (900 g) ground turkey
2 eggs
4 scallions, minced
2 tablespoons (28 ml) soy sauce
1/4 teaspoon orange extract
1/2 teaspoon pepper
1 teaspoon Splenda
1/4 cup (60 ml) oil
1 cup (250 g) low-carb barbecue sauce (see recipe page 335 or purchase)
1 cup (120 g) cranberries (These are strictly seasonal, but they freeze well.)
1/4 cup (6 g) Splenda
In a big mixing bowl, combine the turkey, eggs, and scallions.
In another bowl, mix together the soy sauce, orange extract, pepper, and 1 teaspoon Splenda and pour into the bowl with the turkey. Now use clean hands to smoosh it all together until it's very well blended. Make 1-inch (2.5 cm) meatballs from the mixture.
Heat half the oil in a big, heavy skillet over medium heat. Brown the meatballs in a few batches, adding the rest of the oil as needed. Transfer the browned meatballs to your slow cooker.
In a blender or food processor with an **S** -blade, combine the barbecue sauce, cranberries, and 1/4 cup (6 g) Splenda. Run it until the berries are pureed. Pour this mixture over the meatballs. Cover the slow cooker, set to low, and let it cook for 5 to 6 hours. Serve hot from the slow cooker with toothpicks for spearing!
_**Yield:**_ 48 meatballs, each with: 44 calories, 3 g fat, 4 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Colombo Meatballs with Jerk Sauce
Colombo is the Caribbean version of curry, and jerk is the notoriously fiery barbecue marinade from Jamaica. The heat of this recipe is best controlled by choosing your hot sauce wisely. Use Tabasco sauce, or Louisiana hot sauce, and they'll be spicy. Use Jamaican Scotch Bonnet sauce, or habanero sauce, and they'll take the top of your head right off!
1 pound (455 g) ground lamb
1 egg
1/4 cup (40 g) minced onion
1/4 teaspoon ground coriander
1/4 teaspoon ground turmeric
1/8 teaspoon anise seed, ground
1 clove garlic, minced
1/4 teaspoon dry mustard
2 teaspoons lemon juice
1/2 teaspoon Splenda
1/2 teaspoon salt
2 tablespoons (28 ml) olive oil
1 bay leaf
1/4 cup (40 g) minced onion
1 teaspoon ground allspice
1 tablespoon (8 g) grated ginger root
1 tablespoon (15 ml) soy sauce
1/4 teaspoon dried thyme
1/4 teaspoon ground cinnamon
1 tablespoon (1.5 g) Splenda
2 cloves garlic, crushed
1/4 cup (60 g) low-carb ketchup
1 tablespoon (15 ml) lemon juice
1 tablespoon (15 ml) lime juice
1 1/2 teaspoons hot sauce
In a big mixing bowl, add the lamb, egg, 1/4 cup (40 g) minced onion, coriander, turmeric, anise seed, minced garlic, dry mustard, 2 teaspoons lemon juice, 1/2 teaspoon Splenda, and salt. Using clean hands, moosh it all together till it's well blended. Then make 1-inch (2.5 cm) meatballs, pressing them together firmly.
Heat the oil in a big, heavy skillet over medium heat and brown the meatballs in two batches. Drop the bay leaf in the bottom of the slow cooker and then put the meatballs on top of it.
Mix together the second 1/4 cup (40 g) minced onion, the allspice, ginger, soy sauce, thyme, cinnamon, 1 tablespoon (1.5 g) Splenda, crushed garlic, ketchup, 1 tablespoon (15 ml) lemon juice, lime juice, and hot sauce. Pour this sauce evenly over the meatballs. Cover the slow cooker, set it to low, and let it cook for 3 hours. Remove the bay leaf. Serve hot from the slow cooker.
_**Yield:**_ 35 servings, each with: 48 calories, 4 g fat, 2 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Easy Party Shrimp
How easy is this? Yet your guests will devour it. If you can't find the crab boil spices in the spice aisle at your grocery store, ask the fish guys. They should know where it is.
1 envelope (3 ounces, or 85 g) crab boil spices
12 ounces (355 ml) light beer
1 tablespoon (18 g) salt or Vege-Sal
4 pounds (1.8 kg) easy-peel shrimp or frozen shrimp, unthawed
Drop the crab boil spice net bag in your slow cooker and pour in the beer. Add the salt or Vege-Sal and stir. Add the shrimp. Add just enough water to bring the liquid level up to the top of the shrimp. Cover the slow cooker, set it to high, and let it cook for 1 to 2 hours until the shrimp are pink through. Set the pot to low.
Serve the shrimp straight from the slow cooker with low-carb cocktail sauce, lemon butter, or mustard and mayo stirred together for dipping. Or heck, serve all three. This is enough shrimp for a good-sized party, at least 15 or 20 people, if you're serving it as an hors d'oeuvre/party snack.
_**Yield:**_ 20 servings, each with: 101 calories, 2 g fat, 18 g protein, 1 g carbohydrate, 0 g dietary fiber, 1 g usable carbs. (Analysis does not include any dipping sauces.)
### Zippy Cocktail Dogs
Here's an easy way to jazz up little cocktail wieners.
1/4 cup (60 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
1/4 cup (6 g) Splenda
1/2 teaspoon blackstrap molasses
1 teaspoon Worcestershire sauce
1/4 cup (60 ml) bourbon
1/2 pound (225 g) cocktail-size hot dogs
In a large bowl, stir together the ketchup, Splenda, molasses, Worcestershire sauce, and bourbon.
Put the hot dogs in the slow cooker and pour the sauce over them. Cover the slow cooker, set it to low, and let it cook for 2 hours; then uncover and cook for 1 more hour. Serve with toothpicks for spearing.
_**Note:**_ If you can't get cocktail-size hot dogs, use regular hot dogs cut in chunks. They're not as cute, but they should taste the same!
_**Yield:**_ 6 servings, each with: 158 calories, 11 g fat, 5 g protein, 4 g carbohydrate, trace dietary fiber, 4 g usable carbs.
### Horseradish Smokies
My husband loved these!
1 pound (455 g) small smoked sausage links
1/4 cup (60 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
1/3 cup (8 g) Splenda
2 tablespoons (30 g) prepared horseradish
1/4 teaspoon blackstrap molasses
Put the sausage in your slow cooker.
In a bowl, mix the ketchup, Splenda, horseradish, and molasses. Pour the sauce over the sausage. Stir to coat the sausage. Cover the slow cooker, set it to low, and let it cook for 3 hours. Serve the sausage hot from the slow cooker with toothpicks for spearing.
_**Yield:**_ 8 servings, each with: 193 calories, 17 g fat, 8 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Orange Smokies
Put these out at your next Super Bowl party and watch people eat!
1 pound (455 g) small smoked sausage links
1/4 cup (60 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
1/4 cup (60 ml) lemon juice
2 tablespoons (3 g) Splenda
1/4 teaspoon orange extract
1/4 teaspoon guar or xanthan (optional)
Put the sausage in your slow cooker.
In a small bowl, stir together the ketchup, lemon juice, Splenda, and orange extract. Thicken the mixture just a little, if you think it needs it, with guar or xanthan. Pour the sauce over the sausage. Cover the slow cooker, set it to low, and let it cook for 3 hours. Keep the sausages hot in the slow cooker to serve.
_**Yield:**_ 8 servings, each with: 193 calories, 17 g fat, 8 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Tuna and Artichoke Stuffed Mushrooms
I actually came up with this recipe, which was a big hit at my Toastmasters Club Christmas party by the way, because I wanted to figure out the method of baking stuffed mushrooms in a slow cooker. It worked great and frees up your oven for other things. It also lets you get the hot hors d'oeuvres in early, so you can field other pre-party details. You may well find yourself baking all your stuffed mushrooms this way.
1 can (14 ounces, or 390 g) artichoke hearts
3 scallions
1 can (7 ounces, or 200 g) tuna in water, drained
3 ounces (85 g) cream cheese, softened
8 ounces (225 g) shredded Italian cheese blend
3 tablespoons (42 g) mayonnaise
1/2 teaspoon pepper
1/4 cup (15 g) minced fresh parsley
1/4 teaspoon hot sauce or to taste
2 pounds (900 g) mushrooms
Drain and chop the artichoke hearts and throw them in a mixing bowl. Slice up the scallions, including the crisp part of the green shoot, and add them to the bowl. Add the drained tuna. Add the cream cheese, Italian cheese blend, mayo, pepper, parsley, and hot sauce and mash it all up with a fork until everything is evenly blended.
Wipe your mushrooms clean and remove the stems. Reserve the stems for some other purpose, like making an omelet or topping a steak. (Or you could add them to one of the mushrooms recipes in the _Slow Cooker Sides_ chapter.) Now stuff the artichoke and tuna mixture into the caps.
Put a basket-type steamer in your slow cooker. Arrange a layer of mushrooms on it. Now take a piece of nonstick aluminum foil, also called _release foil_ , and perforate it in several places with a fork to let liquid drip through. Fit it down on top of the first layer of mushrooms. (You'll probably have to make a hole in the middle for the stem of the steamer to poke through.) Put another layer of mushrooms on this. Repeat with a second layer of foil and a third layer of mushrooms. My big slow cooker fit all of my mushrooms in three layers.
Slap on the top, set the cooker to low, and let it go for 4 to 5 hours. You can serve your mushrooms out of the slow cooker if you want to keep them warm, or you can use a spoon to transfer them to a platter.
_**Yield:**_ This served a party of at least 15 people. Assuming 15 servings, each will have: 142 calories, 10 g fat, 10 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Bacon-Cheese Dip
Bacon and cheese together—it makes you glad to be a low-carber, doesn't it?
16 ounces (455 g) Neufchâtel cheese, softened, or light or regular cream cheese
2 cups (225 g) shredded cheddar cheese
2 cups (230 g) shredded Monterey Jack cheese
1/2 cup (120 ml) Carb Countdown dairy beverage
1/2 cup (120 ml) heavy cream
2 tablespoons (22 g) brown mustard
1 tablespoon (10 g) minced onion
2 teaspoons Worcestershire sauce
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon cayenne
1 pound (455 g) bacon, cooked, drained, and crumbled
Cut the Neufchâtel cheese or cream cheese in cubes and put them in your slow cooker. Add the cheddar cheese, Monterey Jack cheese, Carb Countdown, cream, mustard, onion, Worcestershire sauce, salt or Vege-Sal, and cayenne. Stir to distribute the ingredients evenly. Cover the slow cooker, set it to low, and let it cook for 1 hour, stirring from time to time.
When the cheese has melted, stir in the bacon.
_**Note:**_ Serve with cut-up vegetables, fiber crackers, or other low-carb dippers.
_**Yield:**_ 12 servings, each with: 505 calories, 44 g fat, 25 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs. (Analysis is exclusive of dippers.)
### Awesome Sauce aka Provolone and Blue Cheese Fondue
That Nice Boy I Married took one bite and said "Awesomesauce!" And so it is. This can be a party snack, but we like it as a light supper.
6 ounces (170 g) cream cheese
12 ounces (340 g) provolone cheese
4 ounces (115 g) blue cheese
1 clove garlic
1/2 cup (120 ml) heavy cream
1/2 cup (120 ml) dry white wine
Crush the garlic and put it in the bottom of the slow cooker. Cube the cream cheese and put it in. Then cube the provolone, put that on top of the cream cheese, and add half the blue cheese on that. Add the heavy cream and wine, cover the pot, set to low, and let it cook for a good two hours.
Open up the slow cooker. You will find a gloppy mess. Do not panic. Take your stick blender and blend it until the whole thing is smooth. That's it. It now will keep for at least a couple of hours on low or on the serve setting. When you're ready, set out the slow cooker of _Awesome Sauce_ , plus veggies to dip. We dipped with green pepper strips, blanched asparagus, canned quartered artichoke hearts, well-drained, and marinated mushrooms, but I can think of a dozen more veggies that would be good with this. Indeed, it's hard to think of anything that wouldn't be good with this!
_**Yield:**_ If 12 people share this, each will get: 224 calories, 19 g fat, 11 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs. (Analysis is exclusive of vegetables.)
### Bagna Cauda
The name of this traditional Italian dip means "hot bath," and that's just what it is—a bath of hot, flavored olive oil to dip your vegetables in. Believe it or not, our tester Maria Vander Vloedt's kids really liked this!
1 cup (235 ml) extra-virgin olive oil
1/4 cup (55 g) butter
3 cloves garlic, minced
2 ounces (55 g) canned anchovies, minced
Combine everything in a small slow cooker. Cover the slow cooker, set it to low, and let it cook for 1 hour.
_**Note:**_ Serve with vegetables. Fennel, pepper strips, cauliflower, mushrooms, celery, canned artichoke hearts, and lightly steamed asparagus are all traditional choices.
_**Yield:**_ Per batch: 2448 calories, 267 g fat, 17 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs. (It's hard to know how to divide this into servings, and it's unlikely you'll end up eating it all, even with a big group. After all, you can't scoop up a tablespoon (15 ml) at a time of hot olive oil. So here's the stats for the whole potful. Notice that carb count!)
### Hot Artichoke Dip
This is my slow cooker version of the ubiquitous hot artichoke dip that appeared in _500 Low-Carb Recipes_. Using a slow cooker, the dip stays hot till it's gone!
1 cup (225 g) mayonnaise
1 cup (100 g) grated Parmesan cheese
1 clove garlic, crushed
8 ounces (225 g) shredded mozzarella cheese
1 can (14 ounces, or 390 g) artichoke hearts, drained and chopped
Add everything to your slow cooker, stir it up well, and smooth the surface. Cover the slow cooker, set it to low, and let it cook for 2 to 3 hours.
Serve with low-carb crackers and/or cut-up vegetables.
_**Yield:**_ 8 servings, each with: 339 calories, 33 g fat, 11 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs. (Analysis is exclusive of crackers and vegetables.)
### Hot Crab Dip
Hot crab, hot cheese, garlic—what's not to like?
1 cup (225 g) mayonnaise
8 ounces (225 g) shredded cheddar cheese
4 scallions, minced
1 can (6 ounces, or 170 g) crabmeat, drained
1 clove garlic, crushed
3 ounces (85 g) cream cheese, softened, cut into chunks
Combine everything in your slow cooker and stir together. Cover the slow cooker, set it to low, and let it cook for 1 hour. Remove the lid and stir to blend in the now-melted cream cheese. Re-cover and cook for another hour.
Serve with celery, pepper, and cucumber dippers.
_**Yield:**_ 8 servings, each with: 372 calories, 37 g fat, 13 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs. (Analysis is exclusive of vegetables.)
### Artichoke-Spinach-Ranch Dip
Artichoke dip, spinach dip, and ranch dip are all so popular, combining them seemed destined for greatness!
1 can (14 ounces, or 390 g) artichoke hearts, drained and chopped
1 package (10 ounces, or 280 g) frozen chopped spinach, thawed and drained
1 cup (225 g) mayonnaise
1 cup (230 g) sour cream
1 packet (1 ounce, or 28 g) ranch-style dressing mix
2 cups (160 g) shredded Parmesan cheese
1 clove garlic, crushed
Spray your slow cooker with nonstick spray. Mix everything together in your slow cooker. Cover, set it to low, and let it cook for 3 to 4 hours. Keep the dip hot in the slow cooker to serve.
_**Note:**_ Serve dip with cut-up vegetables or low-carb crackers.
_**Yield:**_ 12 servings, each with: 247 calories, 23 g fat, 7 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs. (Analysis is exclusive of vegetables and crackers.)
### Buffalo Wing Dip
I wanted to come up with a hot dip that had all the flavors of Buffalo wings. This is how it came out, and it's awesome!
1 pound (455 g) cream cheese at room temperature
4 ounces (115 g) crumbled blue cheese
4 scallions, minced
2 garlic cloves, crushed
1 teaspoon Worcestershire sauce
1 cup (225 g) cooked chicken, minced
1/2 cup (120 ml) Buffalo wing sauce
Turn on the slow cooker set to low and put the cream cheese in it. Dump in the blue cheese, too.
Mince your scallions fine—I used my food processor. Add them to the cheeses and crush in the garlic. Add the Worcestershire, too.
Mince your chicken quite fine. I used two skinned chicken leg and thigh quarters that I simply covered with water in my slow cooker and let cook on high for 3 to 4 hours. (I did this earlier in the day.) I stripped the meat off the bone and used my kitchen shears to snip across the grain, so it naturally came apart into little bits. If you don't want to work that hard, you could use canned chunk chicken, I bet—just drain it and mash it up good. Anyway, stir your teeny bits of chicken into the cheese mixture, which should be getting soft by now. Make sure it's all well blended and then put the top on the pot and let the whole thing cook for 3 to 4 hours.
An hour before your party, pour the buffalo wing sauce—I used _Wing_ _Time_ brand—evenly over the cheese mixture and re-cover the pot. Let it cook for that last hour and then serve with celery sticks for dipping.
_**Yield:**_ This will serve a whole party, call it 15 people. Each will get: 164 calories, 14 g fat, 7 g protein, 2 g carbohydrate, 5 g usable carbs. (Analysis is exclusive of celery.)
### Chicken Liver Pâté
If you like chicken livers, you'll like this, but you won't if you don't. I adore them myself and practically lived on pâté on fiber crackers for a few days after making this. You could also stuff this into celery stalks.
1/2 cup (80 g) finely chopped onion
1 clove garlic, crushed
1 cup (70 g) sliced mushrooms
3 tablespoons (45 g) butter
1 pound (455 g) chicken livers
2 tablespoons (28 ml) heavy cream
2 tablespoons (28 ml) brandy
1 bay leaf, crumbled
1/2 teaspoon dried thyme
1/2 teaspoon dried marjoram
1 tablespoon (4 g) chopped fresh parsley
3/4 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
In a big, heavy skillet, start the onion, garlic, and mushrooms sautéing in the butter over low heat. While that's happening, halve the chicken livers where they naturally divide into two lobes. When the mushrooms have softened and changed color, add the livers and sauté, stirring occasionally, until they're sealed all over and the color of the surface has changed but they are not cooked through. Transfer the mixture to a food processor with the **S** -blade in place.
Add the cream, brandy, bay leaf, thyme, marjoram, parsley, salt or Vege-Sal, and pepper. Run the food processor until the mixture is finely pureed.
Spray a 3- or 4-cup (700 or 950 ml) glass casserole dish with nonstick cooking spray. Pour the mixture from the food processor into the casserole dish. Place the casserole dish in your slow cooker. Carefully pour water around the casserole dish to within 1 inch (2.5 cm) of the rim. Cover the slow cooker, set it to low, and let it cook for 8 hours or until the mixture is well set. Turn off the slow cooker and let the water cool until you can remove the casserole dish without risk of scalding your fingers. Remove the casserole dish and chill the pâté overnight before serving.
You can simply scoop this from the casserole dish with a knife and spread it on fiber crackers if you like, but it's fancier to turn it out, slice it, and serve it on a bed of greens.
_**Yield:**_ 8 servings, each with: 138 calories, 8 g fat, 11 g protein, 4 g carbohydrate, trace dietary fiber, 4 g usable carbs.
### Roasted Nuts and Seeds
I was astounded by how great a job my slow cooker did of roasting nuts! I got a little carried away, I confess. But then, I really like nuts! Make these often and keep them on hand, and you'll find yourself missing both chips and candy a whole lot less. This is one job where the size of the slow cooker matters. Use a 3-quart (2.8 liters) or, if you have a 5 1/2-quart (5 liters) cooker, double the recipe.
### Dana's Snack Mix, Slow Cooker–Style
This is similar to a snack mix that appears in _500 Low-Carb Recipes_ , and it's addictive. Thank goodness it's also healthy! You'll likely find shelled pumpkin seeds and raw, shelled sunflower seeds at your health food store. You can also find the pumpkin seeds in Latino groceries, labeled "pepitas."
4 tablespoons (55 g) butter, melted
3 tablespoons (45 ml) Worcestershire sauce
1 1/2 teaspoons garlic powder
2 1/2 teaspoons seasoned salt
2 teaspoons onion powder
2 cups (276 g) raw pumpkin seeds, shelled
1 cup (145 g) raw sunflower seeds, shelled
1 cup (145 g) raw almonds
1 cup (100 g) raw pecans
1 cup (120 g) raw walnut pieces
1 cup (140 g) raw cashews
1 cup (145 g) dry-roasted peanuts
If you've got a little time, you can just put the butter in the slow cooker, turn it to low, and wait for it to melt. Otherwise, melt it on the stove or in the microwave and then transfer it to the slow cooker pot. Add the Worcestershire sauce, garlic powder, seasoned salt, and onion powder. Stir it all together. Add the nuts and seeds. Stir well until all the nuts are evenly coated. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours, stirring once or twice if you're around.
Uncover the pot, stir the nut and seed mix up, and cook for another 45 to 60 minutes to dry the nuts and seeds. Let them cool before storing them in an airtight container.
_**Yield:**_ 24 servings of 1/3 cup, each with: 279 calories, 25 g fat, 9 g protein, 9 g carbohydrate, 3 g dietary fiber, 6 g usable carbs.
### Blue Cheese Dressing Walnuts
I originally wanted to make these with powdered blue cheese dressing mix, only to find that there is no such thing, at least not in my grocery stores. So I tried using liquid dressing instead. It didn't end up tasting a lot like blue cheese, but it did end up tasting really good.
4 cups (400 g) walnuts
1/2 cup (115 g) blue cheese salad dressing
1 teaspoon garlic salt
Combine the walnuts and dressing in your slow cooker. Stir until the nuts are evenly coated with the dressing. Cover the slow cooker, set it to low, and let it cook for 3 hours, stirring once halfway through.
Stir in the garlic salt just before serving.
_**Yield:**_ 16 servings, each with: 228 calories, 22 g fat, 8 g protein, 4 g carbohydrate, 2 g dietary fiber, 2 g usable carbs.
### Cajun-Spiced Pecans
You can used purchased Cajun seasoning for this or make your own from the recipe on page 339.
1 pound (455 g) pecan halves
2 tablespoons (28 g) butter, melted
3 tablespoons (27 g) Cajun seasoning
Place the pecans in your slow cooker. Stir in the butter to coat the pecans. Add the Cajun seasoning and stir again to coat. Cover the slow cooker, set it to low, and let it cook for 3 hours, stirring once halfway through if you are around.
_**Yield:**_ 16 servings, each with: 118 calories, 12 g fat, 1 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Candied Pecans
These are a great treat to leave around in pretty little dishes at a holiday party.
1 pound (455 g) pecan halves
1/2 cup (112 g) butter, melted
1/2 cup (12 g) Splenda
1 1/2 teaspoons ground cinnamon
1/4 teaspoon ground ginger
1/4 teaspoon ground allspice
Put the pecans in your slow cooker and stir in the melted butter, coating the pecans thoroughly. Sprinkle the Splenda, cinnamon, ginger, and allspice over the pecans and stir again to coat.
Cover the slow cooker, set it to high, let it cook for 30 minutes. Then uncover, turn it to low, and let it cook for 1 1/2 to 2 hours, stirring once or twice.
_**Yield:**_ 8 servings, each with: 304 calories, 32 g fat, 2 g protein, 6 g carbohydrate, 3 g dietary fiber, 3 g usable carbs.
### Kickin' Pecans
This isn't enough cayenne to be really hot, just enough to add a little kick. Hence the name.
3 cups (300 g) pecan halves
1 egg white
1 teaspoon cinnamon
1/2 teaspoon salt
1/4 teaspoon cayenne, or more to taste
1 cup (25 g) Splenda
Put the pecans in your slow cooker. Add the egg white and stir until the pecans are evenly coated.
In a bowl, stir together the cinnamon, salt, cayenne, and Splenda. Pour the mixture over the pecans and stir until they're evenly coated. Cover the slow cooker, set it to low, and let it cook for 3 hours, stirring every hour or so.
If the nuts aren't dry by the end of the 3 hours, uncover the slow cooker, stir, and cook for another 30 minutes until dry. Store in an airtight container.
_**Yield:**_ 9 servings, each with: 242 calories, 24 g fat, 3 g protein, 7 g carbohydrate, 3 g dietary fiber, 4 g usable carbs.
### Curried Pecans
These are astonishingly good. I may make them in quantity and give them away for Christmas this year!
3 tablespoons (45 g) butter
1/4 teaspoon blackstrap molasses
1 1/2 teaspoons curry powder
1/4 teaspoon salt
1/4 teaspoon ground cumin
12 ounces (340 g) pecan halves
2 tablespoons (3 g) Splenda
In a big, heavy skillet, melt the butter over medium-low heat. Stir in the molasses, curry powder, salt, and cumin and cook for just a minute or two.
Add the pecans and stir until they're evenly coated with the butter and seasonings. Then transfer them to your slow cooker. Sprinkle the Splenda over the pecans, stirring as you sprinkle, so you coat them evenly. Cover the slow cooker, set it to low, and let it cook for 2 to 3 hours, stirring once or twice during the cooking time.
_**Yield:**_ 9 servings, each with: 169 calories, 17 g fat, 2 g protein, 4 g carbohydrate, 2 g dietary fiber, 2 g usable carbs.
### Smokin' Chili Peanuts
Oh, my goodness. These are hot and crunchy and just too darned good. But they're not for the faint of heart.
1/4 cup (55 g) butter, melted
2 tablespoons (16 g) chili powder
1 tablespoon (15 ml) liquid smoke flavoring
1 jar (24 ounces, or 680 g) salted, dry-roasted peanuts
In your slow cooker, combine the butter, chili powder, and liquid smoke flavoring. Stir them together well. Add the peanuts and stir them until they're evenly coated with the butter and the seasonings. Cover the slow cooker, set it to low, and let it cook for 2 to 2 1/2 hours.
Remove the lid, stir, and let cook for another 30 minutes or until the peanuts are dry. Store them in the original jar!
_**Yield:**_ 24 servings, each with: 185 calories, 16 g fat, 7 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Chili Garlic Peanuts
These Spanish-influenced peanuts make the perfect nibble with drinks.
2 tablespoons (28 g) bacon grease
3 cups raw (435 g) Spanish peanuts
2 teaspoons ancho chili powder
1/4 teaspoon cayenne
10 cloves garlic, crushed
1 teaspoon salt
1 tablespoon (15 ml) lime juice
Melt the bacon grease and dump it into your slow cooker (or put it in and turn the slow cooker on for a while). Then add the peanuts and stir to coat.
Now add the ancho chili powder, cayenne, and all that yummy garlic. Stir it all up well. Cover, set to low, and let cook for 45 to 60 minutes. Go stir, re-cover, and give them another 45 to 60 minutes. Then try one for doneness. If they're still underdone, give them another half hour, but if they're pretty done, go on to your next step.
When they're roasty and crunchy, stir in the salt and lime juice, making sure all the peanuts are coated. Leave the lid off and give them another 15 minutes. Stir and let them go a final fifteen minutes—you're just drying that lime juice.
Eat them warm or store in a snap-top container.
_**Yield:**_ 12 servings, each with: 232 calories, 20 g fat, 10 g protein, 7 g carbohydrate, 3 g dietary fiber, 4 g usable carbs.
### Asian Peanuts
1 tablespoon (15 g) coconut oil
24 ounces (680 g) raw peanuts
1/4 cup (60 ml) soy sauce
1 tablespoon (15 g) erythritol or xylitol
Put the coconut oil in the slow cooker and turn it on to low. Let the coconut oil melt. (If you're feeling impatient, you could just zap the coconut oil for a minute in your microwave before putting it in the pot.)
Dump in the peanuts and stir to coat with the oil.
Now add the soy sauce and erythritol, sprinkling each in as you stir. Keep stirring until everything's evenly coated. Cover the pot and let the peanuts roast for 3 to 4 hours, uncovering and stirring every hour or so.
Then uncover the pot and cook for another 40 to 60 minutes, stirring every 15 to 20 minutes, until the peanuts are dry. Cool and store in a snap-top container.
_**Yield:**_ 24 servings of 1 ounce (28 g), each with: 167 calories, 15 g fat, 7 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### Sweet and Salty Peanuts
Sweet and salty is a fantastic combination!
2 tablespoons (28 g) coconut oil
3 cups (435 g) raw peanuts
1/4 cup (60 g) erythritol or xylitol
1 teaspoon salt
Melt the coconut oil and pour into the slow cooker. Alternatively, put it in, switch the cooker on to high, and cover while the coconut oil melts.
Then add the peanuts and stir until they're evenly coated. Sprinkle the erythritol over the peanuts and again stir until everything is evenly coated.
Cover the pot and set to high if it isn't on yet. Set a timer for 1 hour. When it beeps, stir the peanuts well, making sure to scoop all the ones on the bottom and sides into the mass—this helps promote even cooking. Re-cover the pot and give them another hour.
Sprinkle with the salt and stir well again. Re-cover and let them cook for another 30 minutes to an hour until the nuts are crunchy and done through. Let them cool and store in a tightly lidded container.
_**Yield:**_ 12 servings, each with: 227 calories, 20 g fat, 9 g protein, 6 g carbohydrate, 3 g dietary fiber, 3 g usable carbs.
### Butter-Spice Almonds
3 cups (435 g) almonds
2 tablespoons (28 g) butter, melted
2 teaspoons vanilla extract
2 teaspoons butter-flavored extract
1/2 cup (12 g) Splenda
1 teaspoon ground cinnamon
1/4 teaspoon salt
Put the almonds in your slow cooker.
In a bowl, stir together the butter, vanilla extract, and butter-flavored extract until well combined. Pour the mixture over the almonds and stir to coat. Add the Splenda, cinnamon, and salt and stir to coat again. Cover the slow cooker, set it to low, and let it cook for 4 to 5 hours, stirring once or twice.
When the time's up, uncover the slow cooker, stir the almonds again, and let them cook for another 30 to 45 minutes. Store in an airtight container.
_**Yield:**_ 6 servings, each with: 457 calories, 41 g fat, 14 g protein, 15 g carbohydrate, 8 g dietary fiber, 7 g usable carbs.
### Buttery Vanilla Almonds
4 tablespoons (55 g) butter
2 teaspoons vanilla
3 tablespoons (45 g) erythritol or xylitol
1/4 teaspoon salt
3 cups (435 g) almonds
Turn on the slow cooker and throw in the butter to melt. (Alternatively, melt the butter in your microwave and pour it in.) When the butter is melted, add the vanilla, erythritol, and salt and stir the whole thing until it's well-blended.
Now dump in the almonds and stir until they're all evenly coated. Set the slow cooker to low, cover, and set a timer for 1 hour. When the timer beeps, stir the almonds well, making sure to scoop them up from the bottom and off the sides. Re-cover and reset the timer for another hour. Repeat until your almonds have cooked 3 to 4 hours. Then cool and store in a tightly lidded container.
_**Yield:**_ 12 servings, each with: 245 calories, 22 g fat, 7 g protein, 7 g carbohydrate, 4 g dietary fiber, 3 g usable carbs.
### Maple-Glazed Walnuts
3 cups (300 g) walnuts
1 1/2 teaspoons ground cinnamon
1 tablespoon (14 g) butter, melted
1/4 teaspoon salt
2 teaspoons vanilla extract
1/3 cup (107 g) sugar-free pancake syrup
1/3 cup (8 g) Splenda
Put the walnuts in your slow cooker.
In a bowl, mix together the cinnamon, butter, salt, vanilla extract, pancake syrup, and 1/4 cup (6 g) of the Splenda. Pour the mixture over the nuts and stir to coat. Cover the slow cooker, set it to low, and let it cook for 2 to 3 hours, stirring every hour or so.
Then uncover the slow cooker and cook, stirring every 20 minutes, until the nuts are almost dry. Stir in the remaining 2 tablespoons (3 g) Splenda, cook for another 20 minutes, and then remove from the slow cooker and cool. Store in an airtight container.
_**Yield:**_ 9 servings, each with: 268 calories, 25 g fat, 10 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Spiced Walnuts
Spicy and sweet, these walnuts are nothing short of addictive. They'd make a nice little nibble to pass around after supper, instead of a dessert.
2 tablespoons (28 g) coconut oil
1 teaspoon ground ginger
1 teaspoon curry powder
1/2 teaspoon cayenne
1/4 teaspoon onion powder
1/4 teaspoon garlic powder
1/2 teaspoon salt
1/4 cup (60 g) erythritol or xylitol
3 cups (300 g) shelled walnuts
Melt the coconut oil and pour it into the slow cooker (or throw it in solid, turn on the pot, and let it sit until it melts).
In the meantime, mix together all the seasonings and the erythritol.
When the oil is melted, dump the walnuts in the pot. Stir until they're evenly coated with the oil. Now sprinkle in the seasoning blend as you stir; keep stirring until they're evenly coated. Cover the pot, set for high, and set a timer for 1 hour.
When the timer beeps, stir the nuts, re-cover the pot, and set the timer for another hour. When it beeps again, check for doneness.
When they're crunchy and toasty, cool them and store in snap-top containers.
_**Yield:**_ 12 servings, each with: 182 calories, 18 g fat, 4 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
chapter two
Slow Cooker Eggs
Why should you cook eggs in your slow cooker? The answer is because sometimes you want a brunch dish that you can put in to cook and forget about while you do other things—and that will stay warm while people serve themselves. And it's also because eggs ain't just for breakfast anymore.
By the way, all of these recipes except the _Tuna Egg Casserole_ can be halved if you're only serving four or five people. In that case, give them 1 hour on high and 2 to 3 on low.
### Maria's Eggs Florentine
Our tester, Maria, came up with this recipe herself.
2 cups (225 g) shredded cheddar cheese, divided
1 package (10 ounces, or 280 g) frozen chopped spinach, thawed and drained
1 can (8 ounces, or 225 g) mushrooms, drained
1/4 cup (25 g) chopped onion
6 eggs, beaten
1 cup (235 ml) heavy cream
1 teaspoon black pepper
1/2 teaspoon Italian seasoning
1/2 teaspoon garlic powder
Spray your slow cooker with nonstick cooking spray. Spread 1 cup (115 grams) of the cheese on the bottom of the slow cooker. Layer the spinach, mushrooms, and onion.
In a bowl, combine the egg, cream, pepper, Italian seasoning, and garlic powder. Pour the mixture into the slow cooker. Top with the remaining 1 cup (115 grams) cheese. Cover the slow cooker, set it to high, and let it cook for 2 hours or until the center is set.
_**Yield:**_ 4 servings, each with: 568 calories, 48 g fat, 27 g protein, 10 g carbohydrate, 4 g dietary fiber, 6 g usable carbs.
### Artichoke Egg Casserole
Look at this! It's has artichokes, sun-dried tomatoes, bacon, and cheese! You know you want it!
12 eggs
2 cups (450 g) creamed cottage cheese
2 cans (14 ounces, or 390 g each) artichoke hearts, drained and chopped
2 bunches scallions, sliced, including the crisp part of the green
12 sun-dried tomato halves, chopped
12 slices bacon, cooked crisp and crumbled
3 cups (345 g) shredded Italian cheese blend
4 teaspoons (4 g) dried basil
4 cloves garlic, crushed
Coat your slow cooker with nonstick cooking spray. Turn on to high while you mix this up. Let it heat.
This is simple. You just throw everything in a mixing bowl and whisk it up. Pour it into the heated slow cooker and cover. Let it cook for 2 hours on high and then turn it down to low and give it another 3 hours. That's all there is to it!
_**Yield:**_ 8 servings, each with: 266 calories, 13 g fat, 23 g protein, 15 g carbohydrate, 2 g dietary fiber, 13 g usable carbs.
### Tuna Egg Casserole
This is sort of a cross between traditional tuna casserole and a quiche.
8 ounces (225 g) sliced mushrooms
1/2 onion, chopped
2 tablespoons (28 g) butter
6 eggs
1 cup (225 g) creamed cottage cheese
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
1 can (12 ounces, or 340 g) tuna in water
1 cup (130 g) frozen peas
4 ounces (115 g) shredded Cheddar cheese
In your big, heavy skillet, over medium heat, start the mushrooms and onions sautéing in the butter. Use the edge of your spatula to chop the mushrooms into smaller bits as you stir.
Add the eggs and cottage cheese, plus the salt and pepper, and whisk them together.
Go stir your mushrooms and onions!
Open and drain your tuna and dump it in the egg mixture. Add the peas, too.
When your mushrooms are soft and your onion translucent, dump them in the slow cooker. Dump the egg and cheese mixture over it and stir it together. Now sprinkle the cheese over the top.
Cover the slow cooker, set to high, and give it an hour. Then turn it down to low, give it another 2 to 3 hours, and serve.
_**Yield:**_ 5 servings, each with: 371 calories, 20 g fat, 37 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs.
_**Note:**_ Chunk light tuna is not only cheaper than white, but it's far less likely to be contaminated with mercury.
### Broccoli-Bacon-Colby Quiche
This crustless quiche is wonderful, but feel free to make any quiche recipe you've got, minus the crust, in the same way. For this recipe, I use broccoli cuts that are bigger than chopped broccoli but smaller than florets, and I think they're ideal.
2 cups (312 g) frozen broccoli florets, thawed and coarsely chopped, or a bag of broccoli cuts
2 cups (225 g) shredded Colby cheese
6 slices cooked bacon
4 eggs
2 cups (475 ml) Carb Countdown dairy beverage
1 teaspoon salt or Vege-Sal
1 teaspoon dry mustard
2 teaspoons prepared horseradish
1/4 teaspoon pepper
Spray a 1 1/2-quart (1.4 L) glass casserole dish with nonstick cooking spray.
Put the broccoli in the bottom of the casserole dish. Spread the cheese evenly on top of the broccoli and crumble the bacon evenly over the cheese.
In a bowl, whisk together the eggs, Carb Countdown, salt or Vege-Sal, dry mustard, horseradish, and pepper and pour it over the broccoli in the casserole dish.
Place the casserole dish in your slow cooker and carefully pour water around the casserole dish to within 1 inch (2.5 cm) of the rim. Cover the slow cooker, set it to low, and let it cook for 4 hours.
Then turn off the slow cooker, uncover it, and let the water cool until you can remove the casserole dish without risk of scalding your fingers. Serve hot or at room temperature.
_**Yield:**_ 6 servings, each with: 292 calories, 21 g fat, 20 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Chili Egg Puff Slow Cooker Style
_Chili Egg Puff_ showed up in **_500 Low Carb Recipes_** ; it is a brunch dish that causes nostalgic drooling when mentioned to those who have tried it. And if you buy pre-shredded cheese, it takes about three minutes to put together—maybe five if you have to shred your cheese yourself. So I figured I needed to slow-cooker-ize it.
12 eggs
2 cups (450 g) creamed cottage cheese
2 cans (4 ounces, or 115 g each) diced green chiles
1 teaspoon salt
1/2 teaspoon pepper
16 ounces (455 g) shredded Monterey Jack cheese
Coat your slow cooker with nonstick cooking spray. Turn it on to high and let it heat while you whisk everything else together. Pour into the slow cooker, cover, and let it cook on high for a couple of hours. Turn down to low and let it cook another 3 hours.
_**Yield:**_ 8 servings, each with: 363 calories, 25 g fat, 30 g protein, 4 g carbohydrate, trace dietary fiber, 4 g usable carbs.
### Crab Puff
I made this for a friend and her family when she was ill. They all thought it was the best "quiche" they'd ever had.
2 leeks
2 tablespoons (28 g) butter
2 cups (450 g) creamed cottage cheese
12 eggs
1 tablespoon (9 g) dry mustard
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
1 teaspoon Tabasco sauce or to taste
1/2 cup (50 g) grated Parmesan cheese
1 can (12 ounces, or 340 g) crabmeat, drained
2 cups (230 g) shredded Monterey Jack cheese
Trim the leaves and the root off your leeks and split them the long way. Rinse them well, getting in between the layers where the dirt tends to lurk. Then lay them on your cutting board and slice them thinly into half-rounds.
Over medium-low heat, melt the butter in your big, heavy skillet and start sautéing the leeks.
In the meantime, measure the cottage cheese into a big mixing bowl and crack in the eggs. Add the mustard, salt, pepper, Tabasco, and Parmesan and whisk until it's well-blended.
Don't forget to stir your leeks from time to time!
When the leeks are softened, add them to the mixing bowl along with the crab and stir them both in well.
Coat your slow cooker with nonstick cooking spray and pour in half the egg and cheese mixture. Sprinkle half of the Monterey Jack evenly over this, add the rest of the egg mixture, and top with the rest of the cheese. Cover the pot and set to high.
I gave mine 2 hours on high and then turned it down to low and gave it another two, but you could turn it down after one hour and let it go 4 hours on low.
_**Yield:**_ 8 servings, each with: 319 calories, 21 g fat, 26 g protein, 7 g carbohydrate, trace dietary fiber, 7 g usable carbs.
chapter three
Slow Cooker Poultry
Here you will find roughly a billion ways to fix chicken and some ideas for turkey, too! You'll notice that many (but not all) of these recipes call for the poultry to be skinless. This is because the skin usually doesn't end up very tasty or interesting when moist-cooked as it is in a slow cooker.
I've often specified light meat or dark meat—breasts versus thighs. This is just what appealed to me. Feel free to use whichever you like best or what's on sale. Often it's cheapest to buy a whole cut-up chicken and strip the skin off yourself at home. That's fine, too—just remove any obvious globs of fat so your sauce doesn't end up greasy.
### Chicken Burritos
Wow. This is easy, delicious, low-carb, low-calorie, and reheats easily. What more do you want from a recipe?
2 1/2 pounds (1.1 kg) boneless, skinless chicken thighs
5 cloves garlic, crushed
2 tablespoons (16 g) chili powder
2 tablespoons (28 ml) olive oil
2 tablespoons (28 ml) lime juice
1 teaspoon salt
1 large jalapeño, minced, or 2 teaspoons canned jalapeños
12 low-carb tortillas, 6-inch
1 cup (72 g) shredded lettuce
1 cup (120 g) shredded cheddar cheese
2/3 cup (154 g) light sour cream
3/4 cup (195 g) salsa
1/2 cup (8 g) chopped fresh cilantro (optional)
Place the chicken in your slow cooker.
In a bowl, mix the garlic, chili powder, oil, lime juice, salt, and jalapeño together. Pour over the chicken and stir to coat. Cover the slow cooker, set it to low, and let it cook for 10 hours. (Or cook on high for 5 hours.)
When the time's up, stir the mixture with a fork to reduce the chicken to a big pot of tasty chicken shreds. Fill each tortilla with 1/3 cup (79 g) chicken and top with lettuce, cheese, 1 tablespoon (15 g) sour cream, a generous tablespoon (16 g) salsa, and a sprinkling of cilantro if desired. Wrap and devour!
This is a great meal for a family that has some low-carbers and some non–low-carbers, just give them regular or (preferably) whole wheat flour tortillas. The chicken keeps well in the fridge and reheats quickly in the microwave for a fast snack. (I find that 45 seconds on 70 percent power is about right for a 1/3-cup (85 g) serving.)
_**Yield:**_ 12 servings, each with: 225 calories, 13 g fat, 22 g protein, 14 g carbohydrate, 9 g dietary fiber, 5 g usable carbs.
### Seriously Simple Chicken Chili
The name says it all!
2 pounds (900 g) boneless, skinless chicken breasts
1 jar (16 ounces, or 455 g) prepared salsa
1 tablespoon (8 g) chili powder
1 teaspoon chicken bouillon concentrate
3 ounces (85 g) shredded Monterey Jack cheese
6 tablespoons (90 g) light sour cream
Put the chicken in your slow cooker.
In a bowl, stir together the salsa, chili powder, and bouillon, making sure the bouillon's dissolved. Pour the mixture over the chicken. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, shred the chicken with a fork. Serve topped with the cheese and sour cream.
_**Yield:**_ 6 servings, each with: 263 calories, 9 g fat, 39 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Chicken Chili Verde
This repeat from _15-Minute Low-Carb Recipes_ is marvelous, and it's a really nice change from the traditional beef chili.
1 1/2 pounds (680 g) boneless, skinless chicken breasts
1 1/2 cups (384 g) prepared salsa verde
1/2 medium onion, chopped
1 bay leaf
1/2 teaspoon pepper
1 teaspoon ground cumin
1 teaspoon minced garlic or 2 cloves garlic, crushed
1 to 2 tablespoons (9 to 18 g) jarred, sliced jalapeño (I use 2 tablespoons (18 g), and it makes the chili pretty hot.)
2 teaspoons chicken bouillon concentrate
Guar or xanthan (optional)
Sour cream
Shredded Monterey Jack cheese
Chopped fresh cilantro
Place the chicken in your slow cooker and add the salsa verde, onion, bay leaf, pepper, cumin, garlic, jalapeños, and bouillon on top. Cover the slow cooker, set it to low, and let it cook for 9 to 10 hours.
When the time's up, shred the chicken with a fork. Stir it up, thicken the chili a little with the guar or xanthan if you think it needs it, and serve with sour cream, cheese, and cilantro on top.
_**Yield:**_ 5 servings, each with: 190 calories, 2 g fat, 32 g protein, 7 g carbohydrate, trace dietary fiber, 7 g usable carbs. (Analysis does not include garnishes.)
### Mean Old Rooster Chili
The name is no joke. We keep chickens, and I had a rooster who was a nasty, aggressive thing. We got tired of it and turned him into food. He was too old to be roasted or grilled, so I stewed him up and made him into this chili, a form in which he was far more agreeable. I know you're unlikely to have a mean old rooster around that needs dispatching, so I've started with the cooked chicken meat—or turkey, which will do fine, too. But if you happen to have a mean old rooster around, you know what to do!
2 pounds (900 g) cooked chicken or turkey meat
1 medium onion, chopped
2 medium Granny Smith apples, diced
2 cups (475 ml) chicken or turkey broth
1 garlic clove, crushed
1 tablespoon cumin
2 tablespoons (16 g) chili powder
1 teaspoon chicken bouillon concentrate
2 cans (15 ounces, or 425 g each) Eden Organic black soy beans
3/4 cup (175 ml) heavy cream
Dice up your chicken or turkey and throw it in the slow cooker.
Peel and chop your onion and core and chop your apples (don't bother peeling them). Throw them in the slow cooker, too. Add everything else but the cream. Cover the pot, set to low, and cook for 5 to 6 hours.
A little before suppertime, stir in the cream, re-cover, and let it heat another 15 to 30 minutes; then serve.
_**Yield:**_ 8 servings, each with: 395 calories, 18 g fat, 45 g protein, 13 g carbohydrate, 6 g dietary fiber, 7 g usable carbs.
### Dana's Kim's Kings Ranch Chicken
This is a great Southwestern-style casserole that serves a crowd. It started with an apparently popular dish (of which I had not heard) called King's Ranch Chicken. It included stuff like canned cream of mushroom soup and tortillas. It was also not a slow cooker dish. My friend Kim Workman Palmer posted her decarbed version on _Facebook_ , and I took that and slow-cooker-ized it. By the way, I originally left out the canned tomatoes with chilies by mistake, and it was still mighty tasty. And it was a tad lower carb, too. But the tomatoes are canonical and awfully good.
1 red bell pepper
1 green bell pepper
1/2 medium onion
1/2 cup (120 ml) chicken broth
1 can (4 ounces, or 115 g) mushroom pieces
1 can (14 ounces, or 390 g) tomatoes with green chiles
2 1/2 cups (350 g) diced cooked chicken or turkey
1 tablespoon (15 ml) lime juice
1 tablespoon (8 g) chili powder
1 teaspoon salt
1 1/2 teaspoons paprika
3/4 teaspoon onion powder
1 clove garlic, crushed
1/4 teaspoon cayenne
1/4 teaspoon cumin
1 teaspoon chicken bouillon concentrate
3 ounces (85 g) cream cheese
1/2 cup (120 ml) heavy cream
2 egg yolks
8 ounces (225 g) shredded Mexican 4-cheese blend
Core and seed the peppers. With the **S** -blade in place, cut them in chunks and put them in your food processor. Peel your half onion and throw it in, too. Pulse to chop fairly fine. Let them sit while you go to the next step.
Put the chicken broth, mushrooms, and tomatoes with chiles—don't drain either the mushrooms or tomatoes first—in a sauce pan and start them heating over a medium burner.
Cut up your cooked chicken. I just snip mine up with my kitchen shears right into a big mixing bowl. Drizzle in the lime juice and toss. Sprinkle everything from the chili powder through the cumin over the chicken and toss to coat evenly.
Okay, back to your chicken broth and veggies. Whisk in the chicken bouillon concentrate and the cream cheese and keep whisking until the cheese melts in and the bouillon is dissolved. Now whisk in the cream.
Separate the eggs and do something else with the whites. (I fed mine to my dogs.) Put the yolks in a cereal bowl and whisk them up. Now add a ladleful—about 1/2 cup (115 g)—of the sauce into the yolk and whisk them together well. Then whisk this mixture back into the main batch of sauce and keep whisking until it thickens. Whisk in a little guar or xanthan, too, if needed, to get a texture a little thinner than commercial condensed mushroom soup.
Okay, you are now ready to assemble your casserole! Spray the slow cooker with nonstick spray. Spread half the chicken in the bottom. Grab that bowl off the food processor and drain off any liquid that has collected in the bottom. Now put half the pepper/onion mixture over the chicken. Spoon half the sauce over that and then sprinkle in half the cheese. Repeat the layers with the rest of the ingredients.
Cover the pot, set to low, and cook for 5 to 6 hours.
_**Yield:**_ 8 servings, each with: 398 calories, 31 g fat, 23 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Chicken Cacciatore
Here's a slow cooker version of an old favorite. It's easy, too, what I call a dump-and-go recipe.
6 skinless chicken leg and side quarters (about 3 lbs, or 1.4 kg)
2 cups (500 g) no-sugar-added spaghetti sauce (I use Hunt's.)
1 can (8 ounces, or 225 g) whole mushrooms, drained
2 teaspoons dried oregano
1/2 cup (80 g) chopped onion
1 green bell pepper, diced
2 cloves garlic, crushed
1/4 cup (60 ml) dry red wine
Guar or xanthan (optional)
Simply put everything except the guar or xanthan in your slow cooker and stir it up to combine. Cover the slow cooker, set it to low, and let it cook for 7 hours.
When the time's up, remove the chicken with tongs and put it in a big serving bowl. Thicken the sauce up a little with the guar or xanthan if it needs it and ladle the sauce over the chicken.
If you like, you can serve this over _Cauli-Rice_ (page 343), spaghetti squash, or even low-carb pasta, but I'd probably eat it as is.
_**Yield:**_ 6 servings, each with: 293 calories, 8 g fat, 42 g protein, 11 g carbohydrate, 4 g dietary fiber, 7 g usable carbs.
### Italian Chicken and Vegetables
1/2 head cabbage, cut in wedges
1 medium onion, sliced
8 ounces (225 g) sliced mushrooms
2 pounds (900 g) skinless chicken breasts
2 pounds (900 g) skinless chicken thighs
2 cups (500 g) no-sugar-added spaghetti sauce (I use Hunt's.)
Guar or xanthan (optional)
Grated Parmesan cheese
Put the cabbage, onion, and mushrooms in your slow cooker. Place the chicken on top of the vegetables. Pour the spaghetti sauce over the top.
Cover the slow cooker, set it to low, and let it cook for 6 hours. Thicken the sauce with guar or xanthan if needed and serve with Parmesan cheese.
_**Yield:**_ 6 servings, each with: 254 calories, 5 g fat, 46 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Chicken Paprikash
This is pure Hungarian decadence! Feel free to use full-fat sour cream in this sumptuous gravy if you prefer. Or you could use plain yogurt, just drain off any watery whey first.
1/2 cup (80 g) chopped onion
1 tablespoon (14 g) butter
1 tablespoon (15 ml) oil
3 pounds (1.4 kg) chicken thighs
1/2 cup (120 ml) chicken broth
1/4 cup (60 ml) dry white wine
1 1/2 tablespoons (24 g) tomato paste
1 teaspoon chicken bouillon concentrate
1 1/2 tablespoons (11 g) paprika
1/2 teaspoon caraway seeds
1/4 teaspoon pepper
1/2 cup (115 g) light sour cream
Guar or xanthan (optional)
In a big, heavy skillet, sauté the onion in the butter and oil over medium-low heat until it's just golden. Transfer it to your slow cooker. Add the chicken to the skillet, turn the heat up to medium, and brown the chicken all over. Transfer it to the slow cooker, too.
Pour off the fat from the skillet and pour in the broth and wine. Stir it around to dissolve the tasty brown stuff stuck to the skillet and then stir in the tomato paste and bouillon. When those are dissolved, pour the liquid over the chicken.
Sprinkle the paprika, caraway seeds, and pepper over the chicken. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, remove the chicken with tongs or a slotted spoon and put it on a platter. Whisk the sour cream into the liquid in the slow cooker and thicken it further with guar or xanthan if desired. Serve the sauce with the chicken.
Don't forget to serve this with _Fauxtatoes_ (see recipe page 343) to ladle this beautiful gravy onto!
_**Yield:**_ 5 servings, each with: 537 calories, 39 g fat, 39 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Chicken Stroganoff
Chicken and noodles with creamy sauce—you talk about comfort food! Feel free to make this with boneless, skinless chicken thighs or breasts instead. I prefer meat cooked on the bone, and I also think the bones enrich the sauce. They're cheaper, too.
8 ounces (225 g) sliced mushrooms
1 medium onion, sliced
4 cloves garlic, crushed
1 pound (455 g) chicken legs and thighs
1 1/2 cups (355 ml) chicken broth
2 teaspoons chicken bouillon concentrate
3 tablespoons (45 ml) Worcestershire sauce
2 packages tofu shirataki, fettucini width
1 cup (230 g) sour cream
Black pepper to taste
Guar or xanthan
Strip the skin from the chicken and save it for _Chicken Chips_ (see recipe page 331).
Put the mushrooms, onions, and garlic in the slow cooker and lay the skinless legs and thighs on top.
Stir together the chicken broth, bouillon concentrate, and Worcestershire until the concentrate is dissolved and pour it over the whole thing. Cover and set to low. Cook for 6 hours.
Fish out the chicken with tongs or a slotted spoon and lay it on a plate. Give it 10 minutes to cool. In the meantime, re-cover the pot.
While your chicken is cooling, drain and rinse the shirataki and snip them a few times with your kitchen shears.
Okay, the chicken is now cool enough to handle. Strip the meat off the bones, which will be very easy. Use your kitchen shears to snip it back into the pot, in bite-sized bits. Stir in the sour cream and thicken to taste with your guar or xanthan.
Stir in the noodles and serve.
_**Yield:**_ 3 servings, each with: 308 calories, 20 g fat, 17 g protein, 15 g carbohydrate, 2 g dietary fiber, 13 g usable carbs.
### Curried Chicken with Coconut Milk
The day I first made this, my cleaning crew was here, and they couldn't stop talking about how great it smelled. It tastes even better! Find coconut milk in the Asian section of big grocery stores or at Asian markets. It comes in regular or light, and they generally have the same carb count, so choose whichever you prefer. But remember, coconut oil is really good for you.
3 pounds (1.4 kg) skinless chicken thighs
1/2 cup (80 g) chopped onion
2 cloves garlic, crushed
1 1/2 tablespoons (9 g) curry powder
1 cup (235 ml) coconut milk
1 teaspoon chicken bouillon concentrate
Guar or xanthan
Put the chicken in your slow cooker. Place the onion and garlic over it.
In a bowl, mix together the curry powder, coconut milk, and bouillon. Pour the mixture over the chicken and vegetables in the slow cooker. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, remove the chicken and put it on a platter. Thicken the sauce to a gravy consistency with guar or xanthan.
You'll want to serve this with _Cauli-Rice_ (see recipe page 343) to soak up the extra curry sauce. It's too good to miss!
_**Yield:**_ 5 servings, each with: 310 calories, 18 g fat, 32 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Chicken Vindaloo
I made this at a local campground over Memorial Day Weekend, and it was a huge hit with fellow campers. It's exotic and wonderful.
6 pounds (2.7 kg) boneless, skinless chicken thighs
1 medium onion, chopped
5 cloves garlic, crushed
1/4 cup (32 g) grated ginger root
4 teaspoons (8 g) Garam Masala (see recipe page 340) or purchased garam masala
1 teaspoon ground turmeric
1/4 cup (60 ml) lime juice
1/4 cup (60 ml) rice vinegar
1/2 cup (120 ml) chicken broth
1 teaspoon salt
Put the chicken, onion, and garlic in your slow cooker.
In a bowl, stir together the ginger, garam masala, turmeric, lime juice, vinegar, broth, and salt. Pour the mixture over the chicken. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
Serve with _Slow Cooker Chutney_ (see recipe page 307).
_**Yield:**_ 12 servings, each with: 295 calories, 10 g fat, 46 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Chicken with Raspberry-Chipotle Sauce
Oh, my, is this wonderful. My only regret about this recipe is that the Raspberry-Chipotle Sauce loses its brilliant ruby color during the long, slow cooking. But the flavor definitely remains. Consider using this simple sauce uncooked as a condiment on roasted poultry or pork. If you can't find raspberry syrup at a local coffee joint, you can order it online.
3 pounds (1.4 kg) chicken
6 teaspoons (12 g) adobo seasoning
2 tablespoons (28 ml) oil
1 cup (125 g) raspberries
1/4 cup (60 g) raspberry-flavored sugar-free coffee flavoring syrup (Da Vinci makes one.)
1 chipotle chile canned in adobo sauce
1 tablespoon (15 ml) white wine vinegar
1/4 cup (4 g) chopped fresh cilantro (optional)
Sprinkle the chicken all over with the adobo seasoning.
In a big, heavy skillet, heat the oil over medium-high heat and then brown the chicken all over. Transfer the chicken to your slow cooker.
In a blender or food processor with the **S** -blade in place, combine the raspberries, raspberry-flavored coffee syrup, chipotle, and vinegar. Process till smooth. Pour the mixture evenly over the chicken. Cover the slow cooker, set it to low, and let it cook for 6 hours.
Stir the sauce before serving it over the chicken. Sprinkle a little cilantro over each piece of chicken, if desired.
_**Yield:**_ 5 servings, each with: 492 calories, 36 g fat, 35 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### Chicken Stew
This dish is a nice change from the usual beef stew. It's light, flavorful, and your whole meal in one pot.
2 tablespoons (28 ml) olive oil
1 1/2 pounds (680 g) boneless, skinless chicken thighs, cut into 1-inch (2.5 cm) cubes
8 ounces (225 g) sliced mushrooms
1 medium onion, sliced
3 cups (360 g) zucchini slices
4 cloves garlic, crushed
1 can (14 ounces, or 390 g) tomato wedges or diced tomatoes
3/4 cup (175 ml) chicken broth
1 teaspoon chicken bouillon concentrate
1 tablespoon (4 g) poultry seasoning Guar or xanthan
In a big, heavy skillet, heat 1 tablespoon (15 g) of the oil. Brown the chicken until it is golden all over. Transfer the chicken to your slow cooker.
Heat the remaining 1 tablespoon (15 g) of oil in the skillet and sauté the mushrooms, onion, and zucchini until the mushrooms change color and the onions are translucent. Transfer them to the slow cooker, too. Add the garlic and tomatoes to the slow cooker.
Put the broth and bouillon in the skillet and stir them around to dissolve any flavorful bits sticking to the skillet. Pour into the slow cooker. Sprinkle the poultry seasoning over the mixture. Cover the slow cooker, set it to low, and let it cook for 4 to 5 hours.
When the time's up, thicken the liquid in the slow cooker with guar or xanthan.
_**Yield:**_ 6 servings, each with: 159 calories, 8 g fat, 11 g protein, 12 g carbohydrate, 2 g dietary fiber, 10 g usable carbs.
### Chicken with Root Vegetables in Parmesan Cream
3 slices bacon
3 pounds (1.3 kg) chicken pieces—thighs, legs, or breasts, whatever you like
3 cloves garlic, crushed
2 turnips, peeled and sliced thin
1/4 large rutabaga, peeled and sliced thin
1/2 medium onion, peeled and sliced thin
4 cloves garlic, crushed
1 tablespoon dried sage
1/2 cup (120 ml) chicken broth
1 teaspoon chicken bouillon concentrate
2/3 cup (160 ml) heavy cream
1 cup (100 g) grated Parmesan cheese
In your big, heavy skillet, cook the bacon crisp. Remove from the pan and reserve. Skin the chicken—keep the skin for _Chicken Chips_ (see recipe page 331)—and brown the chicken a bit in the bacon grease.
In the meantime, peel and slice your turnip, rutabaga, and onion. Put them in the slow cooker, stir them up a little, and place the chicken on top.
Mix together the garlic, sage, chicken broth, and chicken bouillon concentrate, stirring until the bouillon dissolves. Pour over the chicken, put on the lid, set to low, and let it cook for a good 6 to 7 hours.
When dinnertime comes, use a slotted spoon to transfer the chicken and vegetables to a platter and keep it in a warm place. Whisk the cream and Parmesan cheese into the sauce in the slow cooker. Re-cover the pot and let it heat for another 15 minutes. Then serve the chicken and vegetables with the sauce.
_**Yield:**_ 6 servings, each with: 531 calories, 39 g fat, 37 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Chicken with Root Vegetables, Cabbage, and Herbs
I think of this as being a sort of French Country dish. Of course, I've never been to the French countryside, so what do I know? It's good, though, and you don't need another darned thing with it.
5 pounds (2.3 kg) chicken
1 1/2 tablespoons (23 ml) olive oil
1 1/2 tablespoons (21 g) butter
2 medium turnips, cut into 1/2-inch (13 mm) cubes
2 medium carrots, cut into 1/2-inch (13 mm) slices
1 medium onion, cut into 1/4-inch (6 mm) half-rounds
1 head cabbage
4 cloves garlic, crushed
1/2 teaspoon dried rosemary
1/2 teaspoon dried thyme
1/2 teaspoon dried basil
2 bay leaves, crumbled
In a big, heavy skillet, brown the chicken on both sides in the oil and butter over medium-high heat.
When the chicken is browned all over, remove it to a plate and reserve. Some extra fat will have accumulated in the skillet. Pour off all but a couple of tablespoons (28 ml) and then add the turnips, carrots, and onion. Sauté them, scraping the tasty brown bits off the bottom of the skillet as you stir, until they're getting a touch of gold, too.
Transfer the sautéed vegetables to your slow cooker.
Cut the cabbage into eighths and put it on top of the vegetables. Arrange the chicken on top of the cabbage. Sprinkle the garlic over the chicken and vegetables, making sure some ends up on the chicken and some down among the vegetables. Sprinkle the rosemary, thyme, basil, and bay leaves into the slow cooker, making sure some gets down into the vegetables. Season with salt and pepper. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
_**Yield:**_ 8 servings, each with: 510 calories, 37 g fat, 36 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Slow Cooker Brewery Chicken and Vegetables
There are plenty of vegetables in here, so you don't need a thing with it, except maybe some bread for the carb-eaters in the family. And the gravy comes out a beautiful color!
8 ounces (225 g) turnips (two turnips roughly the size of tennis balls), peeled and cut into chunks
2 stalks celery, sliced
1 medium carrot, sliced
1/2 medium onion, sliced
1 tablespoon (18 g) chicken bouillon concentrate
2 1/2 to 3 pounds (1.1 to 1.4 kg) cut-up chicken (I use leg and thigh quarters, cut apart at the joints.)
12 ounces (355 ml) light beer
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles Guar or xanthan (optional)
Put the turnips, celery, carrot, onion, bouillon, and chicken in your slow cooker. Pour the beer and the tomatoes over the lot. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
When the time's up, remove the chicken with tongs and place it on a serving platter. Then, using a slotted spoon, scoop out the vegetables. Put 1 1/2 cups (340 g) of them in a blender and pile the rest on and around the chicken on the platter. Scoop out 1 1/2 to 2 cups (355 to 475 ml) of the liquid left in the slow cooker and put it in the blender with the vegetables. Puree the veggies and broth and thicken the mixture a little more with the guar or xanthan, if it seems necessary. Add salt and pepper to taste and serve as a sauce with the chicken and vegetables.
_**Yield:**_ 5 servings, each with: 415 calories, 26 g fat, 30 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Citrus Spice Chicken
This dish has a sunshiny citrus flavor! It's another dump-and-go recipe.
1/3 cup (80 ml) lemon juice
2 tablespoons (3 g) Splenda
1/2 teaspoon orange extract
1/2 cup (120 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
2 tablespoons (40 g) low-sugar orange marmalade
1/2 teaspoon ground cinnamon
1/2 teaspoon ground allspice
1/8 teaspoon ground cloves
1/4 teaspoon cayenne
3 pounds (1.4 kg) skinless chicken thighs
In a bowl, stir together the lemon juice, Splenda, orange extract, ketchup, marmalade, cinnamon, allspice, cloves, and cayenne.
Put the chicken in your slow cooker and pour the sauce over it. Cover the slow cooker, set it to low, and let it cook for 6 hours.
Serve with _Cauli-Rice_ (see recipe page 343).
_**Yield:**_ 5 servings, each with: 191 calories, 6 g fat, 31 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Tuscan Chicken
This is fabulous Italian chicken!
4 pounds (1.8 kg) skinless chicken thighs
1 tablespoon (15 ml) olive oil
1/2 cup (80 g) chopped onion
1 red bell pepper, cut into strips
1 green bell pepper, cut into strips
1 can (15 ounces, or 425 g) black soybeans, drained
1 can (14 1/2 ounces, or 410 g) crushed tomatoes
1/2 cup (120 ml) dry white wine
1 teaspoon dried oregano
1 clove garlic, crushed
1 teaspoon chicken bouillon concentrate
In a big, heavy skillet, brown the chicken in the oil over medium-high heat.
Meanwhile, put the onion, peppers, and soybeans in your slow cooker. Place the chicken on top of the vegetables and beans.
In a bowl, stir together the tomatoes, wine, oregano, garlic, and bouillon. Pour the mixture over the chicken. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours. Add salt and pepper to taste.
_**Yield:**_ 8 servings, each with: 258 calories, 9 g fat, 31 g protein, 10 g carbohydrate, 5 g dietary fiber, 5 g usable carbs.
### Lemon Chicken
3 pounds (1.4 kilogram) skinless chicken thighs
2 tablespoons (28 g) butter
1 teaspoon dried oregano
1/2 teaspoon seasoned salt
1/4 teaspoon pepper
1/4 cup (60 ml) chicken broth
3 tablespoons (45 ml) lemon juice
2 cloves garlic, crushed
2 tablespoons (8 g) chopped fresh parsley
1 teaspoon chicken bouillon concentrate Guar or xanthan
In a big, heavy skillet, brown the chicken in the butter over medium-high heat.
In a bowl, mix together the oregano, seasoned salt, and pepper. When the chicken is golden, sprinkle the spice mixture over it. Transfer the chicken to your slow cooker.
Pour the broth and lemon juice in the skillet, stirring around to deglaze the pan. Add the garlic, parsley, and bouillon. Stir until the bouillon dissolves. Pour into the slow cooker.
Cover the slow cooker, set it to low, and let it cook for 4 to 5 hours. When the chicken is tender, remove it from the slow cooker. Thicken the sauce a bit with guar or xanthan. Serve the sauce with the chicken.
This dish goes well with _Cauli-Rice_ (see recipe page 343).
_**Yield:**_ 6 servings, each with: 200 calories, 9 g fat, 26 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Lemon-Herb Chicken
This is simple and classic.
3 pounds (1.4 kg) chicken parts—legs, thighs, or breasts, whatever you like
2 tablespoons (28 g) butter
1/2 medium onion, chopped
1/4 cup (60 ml) lemon juice
1/2 teaspoon chicken bouillon concentrate
1/4 cup (15 g) chopped fresh parsley
1 tablespoon (2 g) fresh thyme leaves
Cut the chicken into serving pieces and skin. Save the skins for _Chicken Chips_ (see recipe page 331)!
In your big, heavy skillet, over medium heat, melt the butter and start the chicken browning.
In the meantime, chop the onion and throw it in the slow cooker. When the chicken is golden on both sides, use tongs to place the chicken on top of the onion.
Mix together the lemon juice and chicken bouillon concentrate until the bouillon dissolves. Pour over the chicken.
Chop the parsley and strip the thyme leaves from the stems. Scatter the herbs over the chicken. Cover the pot, set on low, and cook for 5 to 6 hours.
Serve chicken with the pan juices spooned over it. I like _Cauli-Rice_ (see recipe page 343) with this to soak up the juices.
_**Yield:**_ 4 to 5 servings; Assuming 4, each will have: 304 calories, 13 g fat, 41 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Tequila Lime Chicken
Please don't write me asking what to substitute for the tequila; the flavor is too distinctive. If you must, just leave it out and make Lime Chicken.
3/4 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
1/4 teaspoon cayenne
1/2 teaspoon garlic powder
1/4 teaspoon onion powder
1/4 teaspoon thyme
3 pounds (1.4 kg) skinless chicken thighs
2 tablespoons (28 g) butter
1 tablespoon (15 ml) olive oil
4 cloves garlic
3 tablespoons (45 ml) lime juice
1 tablespoon (15 ml) tequila
1 teaspoon chicken bouillon concentrate
Mix together everything from the salt or Vege-Sal through the thyme.
If your chicken came with the skin on, skin it and save the skin for _Chicken Chips_ (see recipe page 331). Sprinkle the seasoning mixture over all sides of your chicken.
Put your big, heavy skillet over medium heat and add the butter and olive oil. Let the butter melt and swirl the two together. Now add the chicken and let brown until it's golden on both sides. While it's cooking, crush or mince your garlic.
Transfer the chicken to your slow cooker with tongs and pour all the fat into the cooker on top of it. Put the pan back on the burner.
Throw the garlic in the pan and stir it around for minute. Then add the lime juice, tequila, and chicken bouillon concentrate. Stir everything around, scraping up all the nice browned bits and making sure the bouillon concentrate is dissolved. Then pour this mixture evenly over the chicken.
Cover the slow cooker, set to low, and let it cook for 5 hours. Then remove the chicken to a platter, thicken up the sauce, and serve it over the chicken.
_**Yield:**_ 5 servings, each with: 265 calories, 13 g fat, 31 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Chicken with Thyme and Artichokes
This is sort of classic, yet very little work.
1 1/2 pounds (680 g) boneless, skinless chicken thighs
2 tablespoons (28 ml) olive oil
1/2 cup (120 ml) dry white wine
1 tablespoon (15 ml) lemon juice
1 teaspoon chicken bouillon concentrate
2 teaspoons dried thyme
1 clove garlic, crushed
1/4 teaspoon pepper
1 can (13 ounces, or 365 g) artichoke hearts, drained
Guar or xanthan
In a big, heavy skillet, brown the chicken in the oil over medium-high heat until golden on both sides. Transfer to your slow cooker.
In a bowl, stir together the wine, lemon juice, bouillon, thyme, garlic, and pepper. Pour the mixture over the chicken. Place the artichokes on top. Cover the slow cooker, set it to low, and let it cook for 6 hours.
Scoop out the chicken and artichokes with a slotted spoon. Thicken the liquid left in the pot with just enough guar or xanthan to make it the thickness of half-and-half.
Serve the chicken and artichokes, plus the sauce, over _Cauli-Rice_ (see recipe page 343).
_**Yield:**_ 4 servings, each with: 314 calories, 10 g fat, 42 g protein, 7 g carbohydrate, trace dietary fiber, 7 g usable carbs.
### Chicken with Artichokes and Sun Dried Tomatoes
What a sunny flavor this dish has! And you don't need anything with it but a glass of wine.
2 pounds (900 g) boneless, skinless chicken breast or thighs, or both
2 tablespoons (28 ml) olive oil
2 cans (13 1/2 ounces, or 380 g each) quartered artichoke hearts
1/2 cup (55 g) sun-dried tomatoes, oil packed
1/2 cup (90 g) roasted red peppers—cut in strips
2 cloves garlic, crushed
1/2 cup (120 ml) chicken broth
1 teaspoon chicken bouillon concentrate
1/4 teaspoon pepper
3 tablespoons (45 ml) balsamic vinegar
3 tablespoons (27 g) capers, drained
1/4 cup (10 g) minced fresh basil
Cut the chicken into 6 servings. In your big, heavy skillet, start it browning in the olive oil.
In the meantime, drain the artichokes and throw them in the slow cooker. Drain the tomatoes and chop them if they're in halves. (I buy them already in strips.) Drain and slice your roasted red peppers and add them as well. Crush the garlic and throw it in, too. Now, go flip your chicken!
Stir together the chicken broth, chicken bouillon concentrate, pepper, and balsamic vinegar until the bouillon dissolves.
When the chicken is just touched with gold on either side, lay it on the vegetables in the slow cooker. Pour the broth mixture over everything, put on the lid, set it to low, and let it cook for 5 to 6 hours.
To serve, place a serving of chicken on each plate and scoop some of the vegetables on top. Thicken the liquid in the pot just a touch with your guar or xanthan shaker and spoon that over, too. Now scatter a few capers and some basil over each plate and serve it forth.
This would be a perfect dish to serve over tofu shirataki angel hair!
_**Yield:**_ 6 servings, each with: 296 calories, 10 g fat, 38 g protein, 12 g carbohydrate, 1 g dietary fiber, 11 g usable carbs.
### Lemon-White Wine-Tarragon Chicken
This is similar to Chicken with Thyme and Artichokes (page 82), except, of course, that tarragon is very different from thyme and there are no artichokes.
3 pounds (1.4 kg) skinless chicken thighs
1/2 cup (120 ml) dry white wine
1/2 cup (120 milliliter) lemon juice
1 teaspoon Splenda
1 tablespoon (5 g) dried tarragon
1/2 teaspoon pepper
1 teaspoon chicken bouillon concentrate
Guar or xanthan
Put the chicken in your slow cooker.
In a bowl, mix together the wine, lemon juice, Splenda, tarragon, pepper, and bouillon, stirring until the bouillon dissolves. Pour the mixture over the chicken. Cover the slow cooker, set it to low, and let it cook for 6 hours.
Remove the chicken to serving plates and thicken the remaining liquid with guar or xanthan to achieve the texture of cream.
_**Yield:**_ 6 servings, each with: 176 calories, 5 g fat, 26 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Chicken with Apples and Rosemary
This happened because our darling friend Keith Johnson, the Organic Gardening God, showed up at the door with an incredibly fragrant branch of fresh rosemary.
2 pounds (900 g) cut up chicken—legs, thighs, or breasts, whatever you like
1 Granny Smith apple
1 onion
1 cup (235 ml) chicken broth
1/4 teaspoon pepper
1 teaspoon chicken bouillon concentrate
1 tablespoon (2 g) fresh rosemary leaves
1/2 cup (120 ml) heavy cream
Guar or xanthan
Lay the chicken in your slow cooker. Slice the apple and onion both about 1/4-inch (6 mm) thick and lay them on top of the chicken.
Mix together the broth, pepper, and bouillon concentrate until the concentrate is dissolved and pour it over the whole thing. Scatter the rosemary needles evenly over everything.
Cover and cook on low for 5 to 6 hours.
When time is up, use a slotted spoon to fish the chicken, apples, and onions out and put them on a platter. Now whisk the cream into the juice in the pot and use your guar or xanthan shaker to thicken the whole thing to a heavy cream consistency. Serve the gravy with the chicken, apples, and onions.
_**Yield:**_ 5 servings, each with: 385 calories, 28 g fat, 25 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Mom's 1960s Chicken, Redux
Back in the 1960s my mom would make a dish for company with chicken breasts, wrapped in bacon, laid on a layer of chipped beef, topped with a sauce made of sour cream and cream of mushroom soup. It tasted far more sophisticated than it sounds and never failed to draw raves from dinner party guests. This is my attempt to de-carb and slow-cooker-ize the same dish—without the carb-filled cream of mushroom soup.
2 1/4 ounces (62 g) dried beef slices (aka "chipped beef")
6 slices bacon
2 pounds (900 g) boneless, skinless chicken breasts
1 cup (70 g) sliced mushrooms
1 tablespoon (14 g) butter
1 cup (235 ml) heavy cream
1 teaspoon beef bouillon concentrate
1 pinch onion powder
1 pinch celery salt
1/4 teaspoon pepper
Guar or xanthan
1 cup (230 g) sour cream
Paprika
Line the bottom of your slow cooker with the dried beef.
Place the bacon in a glass pie plate or on a microwave bacon rack and microwave for 3 to 4 minutes on high. Drain the bacon and reserve. (What you're doing here is cooking some of the grease off of the bacon, without cooking it crisp.)
Cut the chicken into 6 servings. Wrap each piece of chicken in a slice of bacon and place it in the slow cooker on top of the dried beef.
In a big, heavy skillet, sauté the mushrooms in the butter until they're soft. Add the cream and bouillon and stir until the bouillon dissolves. Stir in the onion powder, celery salt, and pepper and then thicken with guar or xanthan until the mixture reaches a gravy consistency. Stir in the sour cream.
Spoon this mixture over the chicken breasts and sprinkle with a little paprika. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
To serve, scoop up some of the dried beef and sauce with each bacon-wrapped piece of chicken.
_**Yield:**_ 6 servings, each with: 472 calories, 32 g fat, 41 g protein, 4 g carbohydrate, trace dietary fiber, 4 g usable carbs.
### Slow Cooker Chicken Mole
Chicken mole is the national dish of Mexico, and I'm crazy about it. On my honeymoon in Mexico, I bought a container of chicken mole at the deli at the local grocery store and kept it in the hotel room fridge to heat up in the microwave! Here's a slow cooker version.
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1/2 cup (80 g) chopped onion
1/4 cup (28 g) slivered almonds, toasted
3 cloves garlic, crushed
3 tablespoons (18 g) unsweetened cocoa powder
2 tablespoons (18 g) raisins
1 tablespoon (8 g) sesame seeds
1 tablespoon (1.5 g) Splenda
1/4 teaspoon ground cinnamon
1/4 teaspoon ground nutmeg
1/4 teaspoon ground coriander
1/4 teaspoon salt
3 pounds (1.4 kg) skinless chicken thighs
Guar or xanthan
2 tablespoons (14 g) slivered almonds, toasted
Put the tomatoes, onion, 1/4 cup (28 g) almonds, garlic, cocoa powder, raisins, sesame seeds, Splenda, cinnamon, nutmeg, coriander, and salt in a blender or food processor and puree coarsely.
Place the chicken in your slow cooker. Pour the sauce over it. Cover the slow cooker, set it to low, and let it cook for 9 to 10 hours.
Remove the chicken from the slow cooker with tongs. Thicken the sauce to taste with guar or xanthan. Serve the sauce over the chicken. Top with the 2 tablespoons (14 g) almonds.
_**Yield:**_ 8 servings, each with: 284 calories, 12 g fat, 37 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Mediterranean Chicken
This recipe, bursting with classic Mediterranean flavors, originally appeared in _500 More Low-Carb Recipes_.
8 ounces (225 g) sliced mushrooms
1 can (14 1/2 ounces, or 410 g) tomatoes
1 can (6 ounces, or 170 g) artichoke hearts
2 1/2 ounces (63 g) sliced black olives
3 pounds (1.4 kg) skinless chicken thighs
1 tablespoon (6 g) Italian seasoning
3/4 cup (175 ml) chicken broth
1 teaspoon chicken bouillon concentrate
1/4 cup (60 ml) dry white wine
Guar or xanthan
Put the mushrooms, tomatoes, artichokes, and olives in your slow cooker. Place the chicken on top.
In a bowl, mix together the Italian seasoning, broth, bouillon, and wine. Pour the sauce over the chicken and vegetables. Cover the slow cooker, set it to low, and let it cook for 7 hours.
When the time's up, thicken the juices a bit with guar or xanthan.
_**Yield:**_ 6 servings, each with: 215 calories, 7 g fat, 28 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Mediterranean Pepper and Olive Chicken
Years ago, I had a great dish of chicken braised with pepperoncini; I've been trying to replicate it ever since. I actually think this is better. I mean, how could added olives make it worse?
3 pounds (1.4 kg) chicken parts—legs, thighs, or breasts, whatever you like
2 tablespoons (28 ml) olive oil
1 small onion
12 jarred pepperoncini peppers
20 kalamata olives
1/4 cup (60 ml) chicken broth
2 tablespoons (28 ml) lemon juice
1 teaspoon chicken bouillon concentrate
2 cloves garlic, crushed
1 teaspoon oregano
Chopped fresh parsley, optional
Cut the chicken into serving pieces and skin. Save the skins for _Chicken Chips_ (see recipe page 331)!
In your big, heavy skillet, over medium heat, start browning the chicken in the olive oil.
In the meantime, peel and chop your onion. Cut the stem ends off your pepperoncini and seed them, if you like. Pit your olives if you didn't buy them that way. (Just squish them with your thumb and flick out the pit.) Put all of this in the bottom of the slow cooker. When the chicken is golden, use tongs to place it on top of the vegetables.
Put the chicken broth, lemon juice, chicken bouillon concentrate, the crushed garlic, and the oregano in the skillet. Stir until the bouillon is dissolved, scraping up any tasty brown bits. Pour over the chicken, cover the pot, set to low, and cook for 6 hours. Serve the chicken with the peppers, onions, and olives and with the juices spooned over it.
Tofu shirataki spaghetti or angel hair would be nice with this.
_**Yield:**_ 4 to 5 servings; Assuming 5, each will have: 302 calories, 16 g fat, 33 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs.
### Mu Shu Chicken
This took some figuring out. The high-carb recipe I started with called for a whole can of plums in syrup. That wasn't happening, and I discovered no one is canning plums in water anymore. I was in line at the grocery store, when the woman behind me put a box of prunes on the conveyor belt. Aha! This also makes good omelets, which saves you the carbs in the tortillas.
8 dried prunes
3 tablespoons (45 ml) lemon juice
3 tablespoons (45 ml) rice vinegar
1/8 teaspoon orange extract
1/4 teaspoon five-spice powder
1 tablespoon (8 g) grated ginger root
1 teaspoon dark sesame oil
1 clove garlic, crushed
1 tablespoon (1.5 g) Splenda, or other sugar-free sweetener to equal 1 tablespoon (13 g) sugar
2 1/2 pounds (1.1 kg) boneless, skinless chicken breast or thighs (I like the thighs better.)
1 tablespoon (15 g) coconut oil
1 bag (14 ounces, or 390 g) coleslaw mix
1/4 onion, minced
2 tablespoons (28 ml) soy sauce
1 tablespoon (15 ml) dark sesame oil
8 large low-carb tortillas
Put the prunes in your slow cooker. Stir together everything from the lemon juice through the Splenda and pour it over the prunes. Place the chicken on top, cover the pot, set it for low, and let it cook 5 to 6 hours.
Okay, dinnertime has rolled around. Uncover your slow cooker and use tongs to pull out your chicken and put it on your cutting board. Let it cool while you go to the next step.
Melt your coconut oil in your big heavy skillet, over high heat, and stir-fry your cabbage and onion with the soy sauce and sesame oil until the cabbage is just wilted but still has some crunch.
Use your stick blender to blend the prunes up with all the liquid in the pot. (If you don't have a stick blender, throw it in your regular blender.) Thicken a touch with guar or xanthan if you think it needs it.
Now slice your chicken into shreds.
To serve, smear a little of the sauce on one of the tortillas and then add chicken and cabbage. Wrap and eat.
_**Yield:**_ 8 servings, each with: 322 calories, 11 g fat, 33 g protein, 29 g carbohydrate, 16 g dietary fiber, 13 g usable carbs.
### Orange Teriyaki Chicken
This has been officially rated "Very Easy, Very Good!"
1 bag (16 ounces, or 455 g) frozen Oriental vegetable mixture, unthawed
2 pounds (900 g) boneless, skinless chicken breasts, cubed
3/4 cup (175 ml) chicken broth
2 tablespoons (28 ml) _Low-Carb Teriyaki_ Sauce (see recipe page 337) or purchased low-carb teriyaki sauce
1 teaspoon chicken bouillon concentrate
1 tablespoon (20 g) low-sugar orange marmalade
1/4 teaspoon orange extract
2 tablespoons (28 ml) lemon juice
1 teaspoon Splenda
1 teaspoon dry mustard
1/2 teaspoon ground ginger
Guar or xanthan
Pour the vegetables into your slow cooker. Place the chicken on top.
In a bowl, combine the broth, teriyaki sauce, bouillon, marmalade, orange extract, lemon juice, Splenda, dry mustard, and ginger, stirring well. Pour the mixture over the chicken and veggies. Cover the slower cooker, set it to low, and let it cook for 4 to 5 hours.
Before serving, thicken the sauce a bit with guar or xanthan.
Serve over _Cauli-Rice_ (see recipe page 343) if desired. Or for the carbivores, you can serve it over brown rice, lo mein noodles, or plain old spaghetti.
_**Yield:**_ 6 servings, each with: 222 calories, 4 g fat, 36 g protein, 7 g carbohydrate, 2 g dietary fiber, 5 g usable carbs.
### Thai Chicken Bowls
This was a big hit with Maria's family!
8 boneless, skinless, chicken thighs, cubed (a little over 2 1/4 lbs, or 1 kg)
2 cloves garlic, crushed
1/2 cup (80 g) chopped onion
2 stalks celery, sliced
2 teaspoons grated ginger root
1 teaspoon five-spice powder
1/2 teaspoon salt
1 tablespoon (15 ml) lemon juice
1 teaspoon hot sauce (optional)
28 ounces (805 ml) chicken broth
1 head cauliflower
Guar or xanthan
6 tablespoons (36 g) sliced scallions
6 tablespoons (6 g) chopped cilantro
Place the chicken in your slow cooker. Top with the garlic, onion, celery, ginger, five-spice powder, salt, and lemon juice.
In a bowl, combine the hot sauce, if using, with the broth and pour it into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
Okay, it's almost supper time. Run your cauliflower through the shredding blade of your food processor to make _Cauli-Rice_. Put your _Cauli-Rice_ in a microwaveable casserole with a lid, add a couple of tablespoons (28 ml) of water, cover, and microwave on high for 6 minutes.
Thicken up the sauce in the slow cooker with a little guar or xanthan to about the texture of heavy cream.
Okay, the _Cauli-Rice_ is done! Uncover it immediately, drain, and divide it into 6 bowls. Divide the chicken mixture, ladling it over the _Cauli-Rice_. Top with the scallions and cilantro.
_**Yield:**_ 6 servings, each with: 138 calories, 4 g fat, 20 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Thai Hot Pot
This recipe takes a few more steps than some, but the results are worth it! If you can't get Southeast Asian fish sauce, you can substitute soy sauce.
1 1/2 pounds (680 g) boneless, skinless chicken thighs
1 medium carrot, sliced
1 medium onion, sliced
1 clove garlic, crushed
14 ounces (390 ml) coconut milk
1 tablespoon (8 g) grated ginger root
2 tablespoons (28 ml) fish sauce (nam pla or nuoc mam) or soy sauce
1 tablespoon (15 ml) lime juice
2 teaspoons Splenda
1/2 teaspoon hot sauce
1/3 cup (87 g) natural peanut butter
1 pound (455 g) shrimp, shelled
1 cup (75 g) fresh snow pea pods, cut into 1/2-inch (13 mm) pieces
Guar or xanthan
6 cups (720 g) _Cauli-Rice_ (see recipe page 343)
1/3 cup (48 g) chopped peanuts
Put the chicken in your slow cooker and add the carrot, onion, and garlic.
In a blender, combine the coconut milk, ginger, fish sauce or soy sauce, lime juice, Splenda, hot sauce, and peanut butter and blend until smooth. Pour the sauce over the chicken and vegetables, using a rubber scraper to make sure you get all of it! Cover the slow cooker, set it to low, and let it cook for 8 hours.
Stir in the shrimp and snow peas, re-cover the slow cooker, and turn it up to high. Cook for 10 minutes or until the shrimp are pink through.
Thicken the sauce slightly with guar or xanthan. Serve over the _Cauli-Rice_ (see recipe page 343) or brown rice for the carb-eaters. Top each serving with the peanuts.
_**Yield:**_ 6 servings, each with: 480 calories, 32 g fat, 33 g protein, 19 g carbohydrate, 7 g dietary fiber, 12 g usable carbs.
### Chicken in Thai Coconut Curry Sauce
This is a little like Panang chicken. Feel free to increase or decrease the heat to taste. If the family has differing heat tolerance, put a bottle of Sriracha hot sauce on the table!
3 Thai bird peppers (little hot red peppers; mine were dried)
8 ounces (225 g) sliced mushrooms
2 pounds (900 g) boneless, skinless chicken breast
1 can (13 1/2 ounces, or 380 ml) coconut milk
1 cup (235 ml) chicken broth
1 teaspoon chicken bouillon concentrate
1 tablespoon (15 ml) lemon juice
3 tablespoons (45 ml) lime juice
1 tablespoon (15 g) thai red curry paste
1 tablespoon (8 g) grated ginger root
2 tablespoons (28 ml) fish sauce
Put the bird peppers on the bottom of your slow cooker and dump your sliced mushrooms on top of them. Cut the chicken into 6 servings and place on top of the mushrooms.
Now mix everything else together until the bouillon concentrate and the curry paste are dissolved. Pour over the chicken and mushrooms, cover the pot, set to low, and cook for 4 to 5 hours. Thicken the sauce a little with your guar or xanthan shaker and then serve.
I'd serve this with shirataki or _Cauli-Rice_ (see recipe page 343) to sop up the super yummy sauce.
_**Yield:**_ 6 servings, each with: 320 calories, 16 g fat, 38 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs.
### Thai-ish Chicken and Noodles
Don't let the need to make the _Not-Quite-Asian Sweet Chili Sauce_ put you off; it's super-easy. Really, the whole thing is easy. And it's good. If you prefer, the traditional (non-tofu) shirataki noodles would be as good here as the tofu shirataki.
1 batch _Not-Quite-Asian Sweet Chili Sauce_ (see recipe page 234.)
3 tablespoons (48 g) natural peanut butter
3 garlic cloves
1 cup (235 ml) chicken broth
1 teaspoon chicken bouillon concentrate
2 pounds (900 g) boneless, skinless chicken breast or thighs
1 package shirataki noodles, spaghetti width
1 cup (70 g) bagged coleslaw mix
1 cup (104 g) bean sprouts Chopped fresh cilantro, optional
Chopped peanuts, optional
Make your _Not-Quite-Asian Sweet Chili Sauce_ , adding the peanut butter, extra garlic, chicken broth, and bouillon concentrate before you run the blender.
Dice your chicken and throw it in the slow cooker. Pour the sauce over it, put the lid on, set it to low, and give it a good 2 to 3 hours.
Stir in the coleslaw mix and sprouts. Drain, rinse, and snip the shirataki noodles and stir them in, too. Put the lid back on and give it another 20 minutes or so—you don't want the bean sprouts to get unappealingly limp.
Serve with cilantro and peanuts on top. Or not—it's up to you.
_**Yield:**_ 6 servings, each with: 282 calories, 8 g fat, 38 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs. (Analysis does not include the polyols in the _Not-Quite-Asian Sweet Chili Sauce_.)
### Southwestern Barbecue
This is incredibly popular, for something so easy!
1/2 cup (123 g) tomato sauce
1 tablespoon (1.5 g) Splenda
1 1/2 tablespoons (14 g) canned, sliced jalapeños
2 tablespoons (28 ml) lime juice
1/8 teaspoon blackstrap molasses
1 teaspoon ground cumin
1/4 teaspoon red pepper flakes
4 pounds (1.8 kg) skinless chicken thighs
Combine everything except for the chicken in your slow cooker and stir well. Place the chicken in the sauce, meaty side down. Cover the slow cooker, set it to low, and let it cook for 6 hours.
Serve the chicken with the sauce spooned over it.
_**Yield:**_ 6 servings, each with: 215 calories, 7 g fat, 34 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Chicken and Dumplings
This takes some work, but boy, is it comfort food. You could make this with leftover turkey instead if you prefer. If you do that, put the cubed, cooked turkey in about 5 to 6 hours into the initial cooking time.
2 medium carrots, sliced
1 medium onion, chunked
2 medium turnips, cut into
1/2-inch (13 mm) cubes
1 1/2 cups (186 g) frozen green beans, cross-cut
8 ounces (225 g) sliced mushrooms
1 1/2 pounds (680 g) boneless, skinless chicken thighs, cut into 1-inch (2.5 cm) cubes
1 1/2 cups (355 ml) chicken broth
1 teaspoon poultry seasoning
3 teaspoons (18 g) chicken bouillon concentrate
1/2 cup (120 ml) heavy cream
Guar or xanthan
_Dumplings_ (see recipe on next page)
In your slow cooker, combine the carrots, onion, turnips, green beans, mushrooms, and chicken.
In a bowl, mix together the broth, poultry seasoning, and bouillon. Pour the mixture over the chicken and vegetables. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, stir in the cream and thicken the gravy to a nice consistency with guar or xanthan. Add salt and pepper to taste. Re-cover the slow cooker and turn it to high.
While the slow cooker is heating up (it'll take at least 30 minutes), make your _Dumplings_ , stopping before you add the liquid. Wait until the gravy in the slow cooker is boiling. Then stir into the dry ingredients the buttermilk and drop the biscuit dough by spoonfuls over the surface of the chicken and gravy. Re-cover the slow cooker and let it cook for another 25 to 30 minutes.
_**Yield:**_ 8 servings, each with: 417 calories, 25 g fat, 36 g protein, 14 g carbohydrate, 4 g dietary fiber, 10 g usable carbs. (Analysis includes the _Dumplings_.)
### Dumplings
Feel free to use this with other meat-and-gravy dishes if you like! (These instructions require gravy to boil the _Dumplings_ in.)
3/4 cup (71 g) ground almonds, or "almond meal"
1/2 cup (80 g) rice protein powder (Get this at your health food store. If they don't have it, they can order it. I use NutriBiotic brand.)
1/4 cup (30 g) wheat gluten
2 tablespoons (28 g) butter
2 tablespoons (28 g) coconut oil
1/2 teaspoon salt
2 teaspoons baking powder
1/2 teaspoon baking soda
3/4 cup (175 ml) buttermilk
Put everything but the buttermilk into your food processor with the **S** -blade in place. Pulse the food processor to cut in the butter. (You want it evenly distributed in the dry ingredients.) Dump this mixture into a mixing bowl.
Check to make sure your gravy is boiling. (If it isn't, have a quick cup of tea until it is.) Now pour the buttermilk into the dry ingredients and stir it in with a few swift strokes. (Don't overmix; you just want to make sure everything's evenly damp.) This will make a soft dough. Drop by spoonfuls over the boiling gravy, cover the pot, and let it cook for 25 to 30 minutes.
_**Note:**_ Grind 3/4 cup (70 g) almonds to cormeal-like consistency or use almond meal.
_**Yield:**_ 12 servings, each with: 153 calories, 10 g fat, 14 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Chicken in Creamy Horseradish Sauce
Don't think that just because this has horseradish it's really strong. The sauce is mellow, subtle, and family-friendly.
4 pounds (900 g) cut-up chicken
1 tablespoon (14 g) butter
1 tablespoon (15 ml) olive oil
3/4 cup (175 ml) chicken broth
1 1/2 teaspoons chicken bouillon concentrate
1 tablespoon (15 g) prepared horseradish
4 ounces (115 g) cream cheese, cut into chunks
1/4 cup (60 ml) heavy cream
Guar or xanthan (optional)
In a big, heavy skillet, brown the chicken in the butter and oil over medium-high heat. Transfer the chicken to your slow cooker.
In a bowl, stir together the broth, bouillon, and horseradish. Pour the mixture over the chicken. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, remove the chicken with tongs and put it on a platter. Melt the cream cheese into the sauce in the slow cooker. Stir in the cream. Thicken the sauce with guar or xanthan if you think it needs it. Add salt and pepper to taste.
I think this would be good with _Fauxtatoes_ (see recipe page 343) and green beans.
_**Yield:**_ 8 servings, each with: 442 calories, 34 g fat, 30 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Chicken in Creamy Orange Sauce
4 pounds (1.8 kg) skinless chicken thighs
3 tablespoons (45 ml) oil
3 tablespoons (45 ml) brandy
1/2 cup (120 ml) white wine vinegar
1/2 cup (120 ml) lemon juice
1/2 teaspoon orange extract
1 teaspoon grated orange rind
1/3 cup (8 g) Splenda
8 scallions, sliced
6 ounces (170 g) light cream cheese, cut into chunks
Guar or xanthan (optional)
In a big, heavy skillet, brown the chicken in the oil over medium-high heat. Transfer the chicken to your slow cooker.
In a bowl, stir together the brandy, vinegar, lemon juice, orange extract, and Splenda. Pour the mixture over chicken. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, transfer the chicken to a platter. Add the scallions to the liquid in the slow cooker and then add the cream cheese and stir till it's melted. Thicken with guar or xanthan if you think it needs it. Add salt and pepper to taste. Serve the sauce over the chicken.
_Cauli-Rice_ (see recipe page 343) in one form or another is the natural side dish with this. Add a big green salad, and there's supper!
_**Yield:**_ 8 servings, each with: 359 calories, 20 g fat, 35 g protein, 5 g carbohydrate, trace dietary fiber, 5 g usable carbs.
### "I've Got a Life" Chicken
This recipe from _15-Minute Low-Carb Recipes_ is remarkably good. It's sweet, tangy, and fruity.
3 to 3 1/2 pounds (1.4 to 1.6 kg) bone-in chicken parts (I use legs and thighs, but a whole cut-up chicken would work great.)
8 ounces (225 g) sliced mushrooms
3 tablespoons (45 ml) orange juice
Grated zest of one orange
1 tablespoon (18 g) chicken bouillon concentrate
1/2 teaspoon pepper
8 ounces (225 g) canned tomato sauce
2 tablespoons (28 ml) soy sauce
2 tablespoons (3 g) Splenda
1/2 teaspoon blackstrap molasses
2 teaspoons minced garlic or
4 cloves garlic, crushed
1 teaspoon dried thyme
Guar or xanthan
Remove the skin and any big lumps of fat from the chicken and place it in your slow cooker. (You can save time by buying chicken with the skin already removed, but it's more expensive.) Place the mushrooms on top.
In a bowl, mix together the orange juice, orange zest, bouillon, pepper, tomato sauce, soy sauce, Splenda, molasses, garlic, and thyme. Pour the mixture on top of the chicken and mushrooms. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
When the time's up, remove the chicken and put it on a platter. Use the guar or xanthan to thicken up the sauce in the slow cooker and serve the sauce with the chicken.
_**Yield:**_ 6 servings, each with: 424 calories, 27 g fat, 35 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs. (This analysis assumes that you eat all of the gravy.)
### Slow Cooker Chicken Guadeloupe
This isn't authentically anything, but it borrows its flavors from the Creole cooking of the Caribbean.
1 cut-up broiler-fryer chicken, about 3 1/2 pounds (1.5 kg), or whatever chicken parts you prefer
1/2 medium onion, chopped
2 teaspoons ground allspice
1 teaspoon dried thyme
1/4 cup (60 ml) lemon juice
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1 shot (3 tablespoons, or 45 ml) dark rum
Guar or xanthan
Place the chicken, onion, allspice, thyme, lemon juice, tomatoes, and rum in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
Remove the chicken carefully—it'll be sliding from the bone! Thicken up the sauce with guar or xanthan. Add salt and pepper to taste and serve the sauce over the chicken.
_**Yield:**_ 5 servings, each with: 541 calories, 36 g fat, 41 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Yassa
This chicken stew comes from Senegal. Traditionally it is quite hot, so feel free to increase the cayenne if you like!
3 large onions, thinly sliced
6 cloves garlic, crushed
1/2 cup (120 ml) lemon juice
1 1/2 teaspoons salt
1/2 teaspoon cayenne, or more to taste
6 pounds (2.7 kg) chicken, cut up
1/4 cup (60 ml) oil
8 cups (960 g) _Cauli-Rice_ (see recipe page 343)
In your slow cooker, combine the onions, garlic, lemon juice, salt, and cayenne. Add the chicken and toss so that all the chicken comes in contact with the seasonings. Cover your slow cooker and refrigerate overnight. (It's a good idea to stir this a few times if you think of it, though I don't expect you to get up in the middle of the night to do it!)
Using tongs, remove the chicken from the marinade. Pat it dry with paper towels and set it aside. Transfer the marinade to your slow cooker.
In a big, heavy skillet, heat the oil over medium-high heat. Place the chicken skin side down and cook it until the skin is browned. (You'll need to do this in batches unless your skillet is a lot bigger than mine!) Don't bother browning the other side of the chicken.
Transfer the chicken back to the slow cooker with the marinade. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
Remove the chicken from the slow cooker with tongs. Put the chicken on a platter, cover it with foil, and put it in a warm place.
Ladle the onions and liquid out of the slow cooker into the skillet and turn the heat to high. Boil this hard, stirring often, until most of the liquid has evaporated. (You want the volume reduced by more than half.) Serve the chicken, onions, and sauce over the _Cauli-Rice_.
_**Yield:**_ 8 servings, each with: 638 calories, 46 g fat, 45 g protein, 11 g carbohydrate, 3 g dietary fiber, 8 g usable carbs.
### Sort-of-Ethiopian Chicken Stew
The slow cooker method is hardly authentic, but the flavors come from an Ethiopian recipe—except that the Ethiopians would use a lot more cayenne! Increase it if you like really hot food.
1 cut-up broiler-fryer, about
3 pounds (1.4 kg)
1 medium onion, chopped
1 teaspoon cayenne
1 teaspoon paprika
1/2 teaspoon pepper
1/2 teaspoon grated ginger root
2 tablespoons (28 ml) lemon juice
1/2 cup (120 ml) water
Guar or xanthan
Place the chicken, onion, cayenne, paprika, pepper, ginger, lemon juice, and water in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
If you'd like to make this really stewlike, you can pick the meat off the bones when it's done (which will be very easy), thicken the gravy with guar or xanthan, and then stir the chicken back into the liquid. Or you can just serve the gravy over the chicken. Take your pick.
_**Yield:**_ 5 servings, each with: 437 calories, 31 g fat, 34 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### African Chicken, Peanut, and Spinach Stew
This may be my favorite recipe in this chapter. It's an incredibly satisfying one-dish meal.
2 pounds (900 g) boneless, skinless chicken thighs
2 tablespoons (28 g) coconut oil
3 garlic cloves, crushed
3 tablespoons (24 g) grated ginger root
1 medium onion, chopped
1 1/2 cups (355 ml) chicken broth
1 teaspoon chicken bouillon concentrate
2 1/2 tablespoons (36 ml) soy sauce
1 tablespoon (15 ml) rice vinegar
1 1/2 teaspoons dark sesame oil
1/2 teaspoon chili paste
3 tablespoons (48 g) natural peanut butter
10 ounces (280 g) frozen chopped spinach, thawed
Cut the chicken into 1/2-inch (13 mm) cubes and in your big, heavy skillet, start them browning in the coconut oil. While that's happening, crush your garlic, grate your ginger, and chop your onion.
When the chicken has a touch of color, transfer it to the slow cooker. Add the garlic, ginger, and onion.
Dissolve the chicken bouillon concentrate in the broth and add to the pot along with the soy sauce, vinegar, sesame oil, and chili garlic paste. Give it all a stir. Plunk the peanut butter on top, cover the pot, set to low, and let it cook for a good 5 to 6 hours.
About 30 minutes before serving, drain the thawed spinach well—I pick mine up with clean hands and squeeze it dry. Stir it into the pot, re-cover, and let the whole thing cook for another half an hour; then serve in bowls, with soup spoons.
_**Yield:**_ 5 servings, each with: 352 calories, 21 g fat, 32 g protein, 9 g carbohydrate, 3 g dietary fiber, 6 g usable carbs.
### Cranberry-Peach Turkey Roast
This fruity sauce really wakes up the turkey roast!
3 pounds (1.4 kg) turkey roast
2 tablespoons (28 ml) oil
1/2 cup (80 g) chopped onion
1 cup (120 g) cranberries
1/4 cup (6 g) Splenda
3 tablespoons (33 g) spicy mustard
1/4 teaspoon red pepper flakes
1 peach, peeled and chopped
If your turkey roast is a Butterball like mine, it will be a boneless affair of light and dark meat rolled into an oval roast, enclosed in a net sack. Leave it in the net for cooking so it doesn't fall apart.
In a big heavy skillet, heat the oil and brown the turkey on all sides. Transfer the turkey to the slow cooker.
In a blender or food processor with the **S** -blade in place, combine the onion, cranberries, Splenda, mustard, red pepper, and peach. Run it until you have a coarse puree. Pour the mixture over the turkey. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
Remove the turkey to a platter, stir up the sauce, and ladle it into a sauce boat to serve with the turkey. You can remove the net from the turkey before serving, if you like, but I find it easier just to use a good sharp knife to slice clear through the netting and let diners remove their own.
_**Yield:**_ 8 servings, each with: 255 calories, 8 g fat, 31 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Turkey with Mushroom Sauce
3 pounds (1.4 kg) boneless, skinless turkey breast (in one big hunk, not thin cutlets)
2 tablespoons (28 g) butter
1/4 cup (15 g) chopped fresh parsley
2 teaspoons dried tarragon
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
1 cup (70 g) sliced mushrooms
1/2 cup (120 ml) dry white wine
1 teaspoon chicken bouillon concentrate
Guar or xanthan (optional)
In a big, heavy skillet, sauté the turkey in the butter until it's golden all over. Transfer the turkey to your slow cooker.
Sprinkle the parsley, tarragon, salt or Vege-Sal, and pepper over the turkey. Place the mushrooms on top.
In a bowl, mix the wine and bouillon together until the bouillon dissolves. Pour it over the turkey. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, remove the turkey and put it on a platter. Transfer about half of the mushrooms to a blender and add the liquid from the slow cooker. Blend until the mushrooms are pureed. Scoop the rest of the mushrooms into the dish you plan to use to serve the sauce, add the liquid, and thicken further with guar or xanthan, if needed.
_**Yield:**_ 8 servings, each with: 281 calories, 14 g fat, 34 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Braised Turkey Wings with Mushrooms
Turkey wings are my favorite cut of turkey for the slow cooker. They fit in easily, they come in good individual serving sizes, and oh yeah, they taste great.
3 1/4 pounds (1.5 kg) turkey wings
1/4 cup (60 ml) olive oil
1/2 cup (120 ml) chicken broth
1 teaspoon chicken bouillon concentrate
1 teaspoon poultry seasoning
1 tablespoon (16 g) tomato paste
1 cup (70 g) sliced mushrooms
1/2 medium onion, sliced
1/2 cup (115 g) sour cream
In a big, heavy skillet, brown the turkey all over in the oil over medium-high heat. Transfer the turkey to your slow cooker.
In a bowl, stir together the broth, bouillon, poultry seasoning, and tomato paste. Pour the mixture over the turkey. Add the mushrooms and onion. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, remove the turkey from the slow cooker with tongs. Whisk the sour cream into the sauce and serve the sauce over the turkey.
_**Yield:**_ 3 servings, each with: 555 calories, 40 g fat, 41 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs.
### Ranch-E-Cue Wings
This is so simple!
2 pounds (900 g) turkey wings
3 tablespoons (45 ml) olive oil
1/2 cup (120 ml) chicken broth
3 teaspoons (15 g) ranch-style dressing mix
1/2 cup (120 g) _Dana's "Kansas City" Barbecue Sauce_ (see recipe page 335) or purchased low-carb barbecue sauce
Cut the turkey wings at the joints, discarding the pointy wing tips.
In a big, heavy skillet, heat the oil and brown the turkey all over. Transfer the turkey to your slow cooker.
In a bowl, mix together the broth, dressing mix, and barbecue sauce. Pour the mixture over the wings. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
_**Yield:**_ 4 servings, each with: 246 calories, 17 g fat, 18 g protein, 5 g carbohydrate, 0 g dietary fiber, 5 g usable carbs.
### Turkey Loaf with Thai Flavors
Ground turkey is cheap, low-carb, and low-calorie—and by itself just plain boring. So jazz it up by adding some Thai flavors.
2 pounds (900 g) ground turkey
1 medium onion, chopped
1 can (4 1/2 ounces, or 130 g) mushroom slices, drained
4 cloves garlic, crushed
2 tablespoons (28 ml) lemon juice
4 tablespoon (60 ml) lime juice, divided
4 teaspoons (20 g) chili paste
3 tablespoons (24 g) grated ginger root
1 1/2 tablespoons (23 ml) fish sauce
1 1/2 tablespoons (23 ml) soy sauce
1 1/2 teaspoons pepper
1/2 cup (60 g) pork rind crumbs (Run some pork rinds through your food processor.)
1/2 cup (8 g) chopped fresh cilantro
1/2 cup (115 g) mayonnaise
Place the turkey in a big mixing bowl.
Place the onion, mushrooms, and garlic in a food processor. Pulse until everything is chopped medium-fine. Add it to the turkey.
Add the lemon juice, 2 tablespoons (28 ml) of the lime juice, the chili paste, ginger, fish sauce, soy sauce, pepper, pork rind crumbs, and cilantro to the bowl. Mix it around with clean hands until it is well blended.
Spray a rack or a collapsible-basket-type steamer with nonstick cooking spray and place it in your slow cooker. Add a cup (235 ml) of water under the rack. If the holes in the rack are pretty large, cover it with a sheet of foil and pierce it all over with a fork. Take two 18-inch (45 cm) squares of foil, fold them into strips about 2 iches (5 cm) wide, and criss-cross them across the rack or steamer, running the ends up the sides of the slow cooker. (You're making a sling to help lift the meat loaf out of the slow cooker.) Place the meat mixture on the rack or steamer and form it into an evenly-domed loaf. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, use the strips of foil to gently lift the loaf out of the slow cooker and place it on a platter.
In a bowl, mix together the mayonnaise and the remaining 2 tablespoons (28 ml) lime juice. Cut the loaf into wedges and serve it with the lime mayonnaise.
_**Yield:**_ 8 servings, each with: 327 calories, 24 g fat, 25 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Mediterranean Turkey Loaf
Ground turkey is nutritious and often inexpensive, but it can be a little bland. Here's a way to liven it up a bit. This recipe serves a crowd, too. Make a big salad with vinaigrette, and there's supper.
2 pounds (900 g) ground turkey
1 cup (120 g) pork rind crumbs
1 medium onion, chopped
4 cloves garlic, crushed
2 teaspoons salt or Vege-Sal
1/2 teaspoon black pepper
1 egg
1/2 cup (75 g) crumbled feta cheese
2 tablespoons (12 g) Italian seasoning
1 cup (245 g) no-sugar-added spaghetti sauce, optional
You know the deal: Dump the turkey, pork rind crumbs, chopped onion, crushed garlic, salt and pepper, feta, Italian seasoning, and tomato paste in a big mixing bowl. Use clean hands to smoosh it all together really well.
Take two pieces of heavy-duty foil, long enough to reach down into your slow cooker, across the bottom, and back up the other side, and fold each into a strip about 2 inches (5 cm) wide. Put your basket steamer in the slow cooker and criss-cross the foil strips across it, going around the stem in the middle.
Dump the meat in the slow cooker and form it into a nice, even loaf on the steamer. Cover the slow cooker, set to low, and let the whole thing cook for 6 to 7 hours.
Use the foil strips to lift the meat loaf out onto a platter and serve. If you want, warm the no-sugar-added spaghetti sauce and top each serving with a couple of tablespoons (28 g).
_**Yield:**_ 8 servings, each with: 284 calories, 15 g fat, 29 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Simple Turkey Drumsticks
You can't really fit a whole turkey in a slow cooker, but drumsticks fit nicely, and I love them. The serving size on these is huge, because carving a turkey drumstick into multiple servings is tough. We wound up with leftovers, but that's hardly a tragedy.
2 1/2 pounds (1.1 kg) turkey legs—2 drumsticks
1/2 teaspoon salt or Vege-Sal
1/2 teaspoon garlic powder
1/2 teaspoon seasoned salt
1/2 teaspoon paprika
1/4 teaspoon pepper
1/4 cup (60 ml) chicken broth
Mix together all the seasonings and sprinkle them liberally over the drumsticks, covering all surfaces. Lay the drumsticks flat on the floor of your slow cooker. (This takes a large size slow cooker. My big one fits two drumsticks, flat.) Pour in the broth. Cover the pot, set to low, and cook for 5 hours.
_**Yield:**_ 2 servings, each with: 689 calories, 32 g fat, 93 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Chipotle Turkey Legs
This dish has spicy, rich Southwestern flavor.
3 turkey drumsticks, about
2 pounds (900 g) total
1 1/2 teaspoons cumin
1 teaspoon chili powder
1 teaspoon dried, powdered sage
1 teaspoon minced garlic or 2 cloves garlic, crushed
1/2 teaspoon red pepper flakes
1/4 teaspoon turmeric
1 or 2 canned chipotle chiles in adobo sauce, plus a couple teaspoons of the sauce they come in
1 can (8 ounces, or 225 g) tomato sauce
1 tablespoon (15 ml) Worcestershire sauce
Guar or xanthan
6 tablespoons (42 g) shredded queso quesadilla* or Monterey Jack cheese (optional)
Place the turkey in your slow cooker. (If you can fit more, feel free. My 3-quart (2.8 L) slow cooker will only hold 3.)
In a blender, combine the cumin, chili powder, sage, garlic, red pepper flakes, turmeric, chiles, tomato sauce, and Worcestershire sauce. Run it for a minute and then pour the mixture over the turkey. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
When the time's up, remove each turkey leg to a serving plate, thicken the sauce with guar or xanthan, and spoon the sauce over the turkey legs. If you like, sprinkle 2 tablespoons (14 g) of shredded cheese over each turkey leg and let it melt for a minute or two before serving.
_**Yield:**_ 3 servings, each with: 451 calories, 22 g fat, 54 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs. (Analysis depends on the size of the turkey leg.)
* This is a mild, white Mexican cheese.
chapter four
Slow Cooker Beef
Are you tired of steaks and burgers? Use your slow cooker, and you'll be on your way to beef stews, pot roasts, chili, and other classic comfort foods—all waiting when you get home!
### New England Boiled Dinner
This is our traditional St. Patrick's Day dinner, but it's a simple, satisfying one-pot meal on any chilly night. This is easy, but it takes a long time to cook. Do yourself a favor and assemble it ahead of time. If you have carb-eaters in the family, you can add a few little red boiling potatoes still in their jackets to this.
6 small turnips, peeled and quartered
2 large stalks celery, cut into chunks
2 medium onions, cut into chunks
3 pounds (1.4 kg) corned beef
1/2 head cabbage, cut into wedges
Spicy brown mustard
Horseradish
Butter
Place the turnips in your slow cooker along with the celery and the onions. Set the corned beef on top and add water to cover. There will be a seasoning packet with the corned beef—empty it into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 10 to 12 hours. (You can cut the cooking time down to 6 to 8 hours if you set the slow cooker to high, but the low setting yields the most tender results.)
When the time's up, remove the corned beef from the slow cooker with a fork or tongs, put the lid back on the slow cooker to retain heat, put the beef on a platter, and keep it someplace warm. Place the cabbage in the slow cooker with the other vegetables. Re-cover the slow cooker, set it to high, and let it cook for 1/2 hour.
With a slotted spoon, remove all the vegetables and pile them around the corned beef on the platter. Serve with the mustard and horseradish as condiments for the beef and butter for the vegetables.
_**Yield:**_ 8 servings, each with: 372 calories, 25 g fat, 26 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs.
### Maple-Glazed Corned Beef with Vegetables
This is a trifle less traditional than, but just as good as, the _New England Boiled Dinner_ (see recipe on previous page). The pancake syrup and mustard, plus last-minute glazing under the broiler, give it a new aspect.
6 medium turnips, cut into chunks
2 medium carrots, cut into chunks
1 medium onion, quartered
5 pounds (2.3 kg) corned beef brisket
2 cups (475 ml) water
1 medium head cabbage, cut into wedges
3 tablespoons (60 g) sugar-free pancake syrup
1 tablespoon (11 g) brown mustard
Horseradish
Place the turnips, carrots, and onion in your slow cooker. Place the corned beef on top. Scatter the contents of the accompanying seasoning packet over everything and pour the water over the whole thing. Cover the slow cooker, set it to low, and let it cook for 9 to 10 hours, and a bit more won't hurt!
When the time's up, carefully remove the corned beef and put it on your broiler rack, fatty side up. Use a slotted spoon to skim out the vegetables, put them on a platter, cover, and keep in a warm place.
Place the cabbage in the slow cooker, set it to high, and let it cook for 15 to 20 minutes or until just tender. (Or you can pour the liquid from the pot into a saucepan and cook the cabbage in it on your stovetop, which is faster.)
While the cabbage is cooking, mix together the pancake syrup and the mustard. Spread the mixture over the corned beef, just the side that is up. When the cabbage is almost done, run the corned beef under your broiler for 2 to 3 minutes until glazed.
Transfer the cabbage to the platter with a slotted spoon. Slice the corned beef across the grain and serve immediately.
Serve this with horseradish and mustard.
_**Yield:**_ 12 servings, each with: 416 calories, 28 g fat, 29 g protein, 10 g carbohydrate, 3 g dietary fiber, 7 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### 3-Minute Slow Cooker Pot Roast
This recipe, originally from _15-Minute Low-Carb Recipes_ , is very 1965, but it's still incredibly easy, and it tastes great.
8 ounces (225 g) sliced mushrooms
2 to 3 pounds (0.9 to 1.4 kg) boneless chuck pot roast
1 envelope (1 ouncek, or 28 g) French onion soup mix
1/2 cup (120 ml) dry red wine
Guar or xanthan
Place the mushrooms in the bottom of your slow cooker and add the beef on top of them.
In a bowl, mix together the onion soup mix and wine and pour it into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the beef (carefully—it will be very tender) and use the guar or xanthan to thicken the juices in the slow cooker. Serve this gravy with the pot roast.
_**Yield:**_ 6 servings, each with: 358 calories, 24 g fat, 25 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs. (This analysis assumes you use a 2-pound (900 g) roast and that you eat every drop of the gravy.)
### Pot Roast Brisket
This is such a classic.
3 pounds (1.4 kg) brisket
Salt and pepper
2 tablespoons (28 ml) olive oil
1 large onion
1 carrot
2 celery ribs
2 bay leaves
1 teaspoon beef bouillon concentrate
Salt and pepper the brisket all over. Put your big heavy skillet over high heat, add the olive oil, and start browning the brisket; you want it good and brown all over.
In the meantime, coat your slow cooker with nonstick cooking spray.
Chop the onion, peel and shred the carrot, and chunk your celery—keep any unwilted leaves; they add a wonderful flavor. Put all the veggies on the bottom of the slow cooker and add one of the bay leaves.
When the brisket is brown all over, use tongs to place it on the vegetables. Put the second bay leaf on top of it, cover the pot, and set it to low. Let it cook a good 7 hours.
When cooking time is up, use your tongs to fish out the brisket and put it on a platter. Remove the bay leaves and add the bouillon concentrate. Now use a stick blender to puree the vegetables into the liquid. This will also thicken your gravy, though you can thicken it a tad more with your guar or xanthan shaker if you like. Salt and pepper to taste.
Now slice your brisket across the grain and serve with the gravy. _Fauxtatoes_ (see recipe page 343) are good with this. Roasted turnips would be good, too.
_**Yield:**_ 6 to 8 servings; Assuming 6, each will have: 764 calories, 65 g fat, 39 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Slow Cooker Texas Brisket
In Texas, barbecue generally means beef brisket. This uses similar flavors, but it doesn't require all that slow smoking—or even going outside.
3 pounds (1.4 kg) beef brisket
2 tablespoons (28 g) bacon grease or (28 ml) olive oil
1/4 cup (65 g) low-carb barbecue sauce ( _Dana's "Kansas City" Barbecue Sauce_ see recipe [page 335], or alternatively, Stubb's barbecue sauce, available in grocery stores, is one of the lowest sugar commercial barbecue sauces I've found.)
1 tablespoon (15 ml) Worcestershire sauce
1 tablespoon (15 ml) liquid Barbecue Smoke®
2 cloves garlic, crushed
1/2 teaspoon celery salt
1/2 teaspoon lemon pepper
1/4 teaspoon salt or Vege-Sal
1/2 medium onion, chopped
In your big, heavy skillet, over high heat, start the brisket browning in the bacon grease. You want it well-browned all over.
In the meantime, mix together the barbecue sauce, Worcestershire sauce, liquid smoke, garlic, celery salt, lemon pepper, and salt.
Chop your half an onion. Coat your slow cooker with nonstick cooking spray and spread the onion over the bottom.
When the beef is well-browned all over, lay it on the onion. Pour the sauce all over it. Cover the slow cooker, set to low, and let it cook for a good 6 hours.
When it's done, fish it out, lay it on a platter, and slice it across the grain. Serve with the liquid from the pot spooned over it.
_**Yield:**_ 6 servings, each with: 759 calories, 65 g fat, 39 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Balsamic Pot Roast
Balsamic vinegar and rosemary give this pot roast an Italian accent.
3 1/2 pounds (1.6 kg) beef round, trimmed of fat
2 tablespoons (28 ml) olive oil
1 large onion, sliced
2 cloves garlic, crushed
1 cup (235 ml) beef broth
1 teaspoon beef bouillon concentrate
1/4 cup (60 ml) balsamic vinegar
1/2 teaspoon dried rosemary, ground
1 cup (180 g) canned diced tomatoes
Guar or xanthan
In a big, heavy skillet, sear the beef in the oil until browned all over. Transfer the beef to your slow cooker. Scatter the onion and garlic around the beef.
In a bowl, stir together the broth, bouillon, vinegar, and rosemary. Pour the mixture over the beef. Pour the tomatoes on top. Season with pepper. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the beef with tongs and place it on a serving platter. Scoop the onions out with a slotted spoon and pile them around the roast. Thicken the juice left in the slow cooker with guar or xanthan and serve it with the beef.
_**Yield:**_ 8 servings, each with: 451 calories, 29 g fat, 42 g protein, 5 g carbohydrate, trace dietary fiber, 5 g usable carbs.
### Reuben Hot Pot
I wanted to come up with something using the flavors of a Reuben sandwich. This came out even better than I hoped. It's really tasty and unusual.
28 ounces (785 g) sauerkraut
3 pounds (1.4 kg) corned beef brisket
1 cup (225 g) _Russian Dressing_ (see recipe below.)
3 tablespoons (33 g) brown mustard
Drain and rinse your sauerkraut and dump half of it in the slow cooker. Slap your corned beef on top. Smear it with half the dressing and mustard and top with the rest of the sauerkraut. Put the rest of the dressing and mustard on top of that. Cover the pot, set to low, and let it cook for a good 7 to 8 hours. Serve with more mustard!
_**Yield:**_ 8 servings, each with: 358 calories, 26 g fat, 26 g protein, 5 g carbohydrate, 3 g dietary fiber, 2 g usable carbs.
### Russian Dressing
Russian dressing is an essential part of a Reuben sandwich, but the bottled stuff has quite a lot of sugar in it. This is easy to make, so long as you have the no-sugar ketchup in the house. You might like it on salads, too, or as a dip for vegetables.
1/2 cup (115 g) mayonnaise
6 tablespoons (90 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or commercial no-sugar-added ketchup
4 sugar-free bread and butter pickle slices (These are available in big grocery stores; I get Mt. Olive brand or Kroger's house brand.)
2 tablespoons (28 ml) pickle juice
2 tablespoons (28 ml) white wine vinegar
1/4 teaspoon pepper
Throw everything in your food processor with the **S** -blade in place and pulse until the pickle is chopped but not pulverized. Alternatively, chop your pickles with a knife and then just stir everything together.
_**Yield:**_ 8 servings, each with: 107 calories, 12 g fat, trace protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Bavarian Pot Roast
Given the Bavarian theme and all, I'm thinking _Fauxtatoes_ (see recipe page 343) and cooked cabbage are the obvious side dishes with this.
1 large red onion, sliced 1-inch (2.5 cm) thick
2 tablespoons (3 g) Splenda
1/4 teaspoon blackstrap molasses
2 tablespoons (28 ml) cider vinegar
1 teaspoon salt
1 teaspoon beef bouillon concentrate
2 1/2 pounds (1.1 kg) beef round, trimmed of fat and cubed
Guar or xanthan (optional)
Place the onion in your slow cooker.
In a bowl, mix together the Splenda, molasses, vinegar, salt, and bouillon. Pour the mixture over the onion. Place the beef on top. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, thicken the broth with guar or xanthan if desired.
_**Yield:**_ 8 servings, each with: 297 calories, 18 g fat, 29 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Pot Roast with Beer and Mushrooms
Chuck roast is usually cheap, and it's very flavorful. Add mushrooms and onions, all that jazz, and it's a party in your mouth.
2 pounds (900 g) chuck roast
2 tablespoons (28 g) bacon grease
1 pound (455 g) sliced mushrooms (I used criminis.)
1 medium onion, sliced
3 carrots, peeled and sliced
4 cloves garlic
12 fluid ounces (355 ml) light beer
1 can (8 ounces, or 225 g) tomato sauce
3 tablespoons (45 ml) lemon juice
1 teaspoon beef bouillon concentrate
Put your big, heavy skillet over medium-high heat and start searing your chuck roast in the bacon grease. You want it nice and brown all over. While that's happening, go on to the next step.
Dump your sliced mushrooms (you bought them sliced, right?) in the slow cooker and add the onion, carrots, and garlic. Stir the veggies together a bit.
When the chuck is crusty and brown all over, throw it on top of the vegetables.
Now pour the beer, tomato sauce, lemon juice, and beef bouillon concentrate in the skillet. Stir it all around, scraping up all the nice brown bits and making sure the bouillon concentrate is dissolved. Pour this over the stuff in the slow cooker. Slap on the lid and set it to low.
Let it cook for 6 hours or so. Then fish out the roast, put it on a platter, and surround it with the veggies. Thicken up the sauce with your guar, xanthan, or glucomannan shaker and serve with the meat and vegetables.
Serve this with _Fauxtatoes_ (see recipe page 343)—indeed, I like to spoon the _Fauxtatoes_ onto each plate and then pile some vegetables on top and pour on the gravy.
_**Yield:**_ 6 servings, each with: 431 calories, 28 g fat, 27 g protein, 14 g carbohydrate, 3 g dietary fiber, 11 g usable carbs. (Analysis is exclusive of _Fauxtatoes_.)
### Peking Slow Cooker Pot Roast
This sounds nuts, but it tastes great! This recipe, originally from _500 Low-Carb Recipes_ , takes starting ahead, but it's not a lot of work.
3 to 5 pounds (1.4 to 2.3 kg) beef roast (round, chuck, or rump)
5 or 6 cloves garlic, sliced thin
1 cup (235 ml) cider vinegar
1 cup (235 ml) water
1 small onion, thinly sliced
1 1/2 cups (355 ml) strong coffee (instant works fine)
1 teaspoon guar or xanthan
At least 24 to 36 hours before you want to actually cook your roast, stick holes in the beef with a thin-bladed knife and insert a garlic slice into each hole. Put the beef in a big bowl and pour the vinegar and the water over it. Put it in the refrigerator and let it sit there for a day or so, turning it over when you think of it so the whole thing marinates.
On the morning of the day you want to serve your roast, pour off the marinade and put the beef in your slow cooker. Place the onion on top of the beef. Pour the coffee over the beef and onion. Cover the slow cooker, set it to low, and let it cook for 8 hours for a smaller roast or up to 10 hours for a larger one.
When the time's up, remove the beef carefully from the cooker. (It will now be so tender it's likely to fall apart.) Scoop out 2 cups (475 milliliters) of the liquid and some of the onions and put them in a blender with the guar or xanthan. Blend for few seconds and then pour the mixture into a saucepan set over high heat. Boil this sauce hard for about 5 minutes to reduce it a bit. Add salt and pepper to the sauce to taste. (It's amazing the difference the salt and pepper make here; I didn't like the flavor of this sauce until I added the salt and pepper, and then I liked it a lot.) Slice the beef and serve it with this sauce.
_**Warning:**_ Do not make this with a tender cut of beef! This recipe will tenderize the toughest cut; a tender one will practically dissolve. Use inexpensive, tough cuts, and prepare to be amazed at how fork-tender they get.
_**Yield:**_ 12 servings, each with: 324 calories, 24 g fat, 24 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs. (This analysis is for a 4-pound (1.8 kg) boneless roast.)
### Sauerbrauten
This classic German pot roast takes advance planning, but it's not a lot of work, and it yields impressive results. Don't forget the _Fauxtatoes_ (see recipe page 343) for that gravy!
4 pounds (1.8 kg) boneless beef round or chuck
1 cup (235 ml) cider vinegar
1 cup water
1/2 onion, sliced
2 bay leaves
1 teaspoon pepper
1/4 cup (6 g) Splenda
2 tablespoons (28 g) bacon grease or (28 ml) oil
1/4 teaspoon ground ginger
1 cup (230 g) light sour cream (Use full-fat sour cream if you prefer, but it's no lower carb.)
Guar or xanthan (optional)
Pierce the beef all over with a fork. In a deep, non-reactive bowl (stainless steel, glass, or enamel), combine the vinegar, water, onion, bay leaves, pepper, and Splenda. Place the beef in the marinade and put the bowl in the refrigerator. Marinate the beef for at least 3 days, and 5 or 6 days won't hurt. Turn it over at least once a day, so both sides marinate evenly.
When the time comes to cook your _Sauerbrauten_ , remove the beef from the marinade and pat it dry with paper towels. Reserve the marinade.
In a big, heavy skillet, heat the bacon grease or oil and sear the beef all over. Transfer the beef to your slow cooker.
Scoop the onion and bay leaves out of the marinade with a slotted spoon and put them on top of the beef. Remove 1 cup (235 milliliters) of the marinade from the bowl and add the ginger to it. Pour this over the beef and discard the remaining marinade. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, remove the beef to a serving plate. Stir the sour cream into the liquid in the slow cooker and thicken it if you think it needs it with guar or xanthan. Add salt and pepper to taste and serve the sauce with the beef.
_**Yield:**_ 10 servings, each with: 407 calories, 26 g fat, 38 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Chipotle Brisket
Our tester, who loved this recipe, halved it. You can feel free to do the same.
4 pounds (1.8 kg) beef brisket, cut into pieces if necessary to fit into your slow cooker
2 tablespoons (28 ml) olive oil
1 medium onion, thinly sliced
4 stalks celery, thinly sliced
4 cloves garlic, crushed
1 tablespoon (9 g) dry mustard
1 tablespoon (3 g) dried oregano
1 teaspoon ground cumin
2 teaspoons pepper
1 teaspoon salt or Vege-Sal
1 can (16 ounces, or 455 g) tomato sauce
1/2 cup (120 ml) beef broth
1 teaspoon beef bouillon concentrate
1/4 cup (60 ml) red wine vinegar
1/2 cup (12 g) Splenda
1/2 teaspoon blackstrap molasses
2 chipotle chiles canned in adobo sauce
2 bay leaves
Guar or xanthan
In a big, heavy skillet, brown the beef all over in the oil over medium-high heat. Transfer the beef to your slow cooker.
Add the onion and celery to the skillet and sauté until softened. Stir in the garlic, dry mustard, oregano, cumin, pepper, and salt or Vege-Sal and sauté for another minute or two. Transfer the mixture to the slow cooker, on top of the brisket.
In a blender or food processor, combine the tomato sauce, broth, bouillon, vinegar, Splenda, molasses, and chipotles and blend until smooth.
Put the bay leaves in the slow cooker, on top of the beef, and pour the sauce over the whole thing. Cover the slow cooker, set it to low, and let it cook for 12 hours.
When the time's up, remove the beef to a platter. Remove the bay leaves. Thicken the sauce to taste with guar or xanthan and serve the sauce over the beef.
_**Yield:**_ 8 servings, each with: 779 calories, 64 g fat, 41 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Simple Salsa Beef
Here's one of those super-simple dump-and-go recipes. It's great for a day when you didn't get dinner in the slow cooker the night before!
3 turnips, peeled and cubed
1 pound (455 g) baby carrots
3 pounds (1.4 kg) beef arm pot roast
2 cups (520 g) salsa
Guar or xanthan (optional)
Put the turnips and carrots in your slow cooker; then place the beef on top. Pour the salsa over the lot. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
When the time's up, remove the beef and pull it apart into shreds with two forks. Scoop the vegetables out onto serving plates with a slotted spoon. Pile the beef on top. If desired, thicken the sauce with a little guar or xanthan. Spoon the sauce over the vegetables and beef.
_**Yield:**_ 8 servings, each with: 200 calories, 5 g fat, 26 g protein, 11 g carbohydrate, 3 g dietary fiber, 8 g usable carbs.
### Beef and Broccoli
This doesn't come out exactly like stir-fry, but it's still Chinese-y-good, and it's a lot less last-minute trouble.
1 pound (455 g) beef round, cut into 1-inch (2.5 cm) cubes
1 can (4 ounces, or 115 g) sliced mushrooms, drained
1 medium onion, cut into wedges
1/2 cup (120 ml) beef broth
1 teaspoon beef bouillon concentrate
1 tablespoon (1.5 g) Splenda
1 teaspoon grated ginger root
1 tablespoon (15 ml) dry sherry
2 tablespoons (28 ml) soy sauce
1 clove garlic, crushed
1 teaspoon dark sesame oil
1 tablespoon (8 g) sesame seeds
2 cups (312 g) frozen broccoli florets
Guar or xanthan
Combine the beef, mushrooms, onion, broth, bouillon, Splenda, ginger, sherry, soy sauce, garlic, and sesame oil in your slow cooker. Sprinkle the sesame seeds on top. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
When the time's up, add the broccoli to the slow cooker, re-cover the slow cooker, and let it cook for another 30 minutes. Thicken the juices a little with guar or xanthan.
Serve over _Cauli-Rice_ (see recipe page 343) if desired.
_**Yield:**_ 4 servings, each with: 314 calories, 17 g fat, 29 g protein, 10 g carbohydrate, 4 g dietary fiber, 6 g usable carbs.
### Beef Carbonnade
Très French!
2 pounds (900 g) beef round, cut into 1-inch (2.5 cm) cubes
2 tablespoons (28 ml) olive oil
1 large onion, sliced
2 medium carrots, cut 1-inch (2.5 cm) thick
2 turnips, cubed
12 ounces (355 ml) light beer
1/4 cup (60 ml) red wine vinegar
3 tablespoons (4.5 g) Splenda
1/4 teaspoon blackstrap molasses
1 cup (235 ml) beef broth
2 teaspoons beef bouillon concentrate
3 cloves garlic, crushed
2 teaspoons dried thyme
2 teaspoons Worcestershire sauce
1/2 teaspoon pepper
2 bay leaves
Guar or xanthan
In a big, heavy skillet, sear the beef all over in the oil. Place the beef in your slow cooker. Add the onion, carrots, and turnips and stir everything around a bit.
In a bowl, mix together the beer, vinegar, Splenda, molasses, broth, bouillon, garlic, thyme, Worcestershire sauce, and pepper. Pour the mixture into the slow cooker. Throw the bay leaves on top. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the bay leaves and add guar or xanthan to thicken the sauce a bit.
You can serve this as is, or to be more traditional, serve it over _Fauxtatoes_ (see recipe page 343).
_**Yield:**_ 6 servings, each with: 411 calories, 24 g fat, 34 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Beef Stroganoff
This creamy gravy is fabulous!
2 pounds (900 g) beef round, cut into 1-inch (2.5 cm) cubes
1 large onion, chopped
1 can (8 ounces, or 225 g), sliced mushrooms, undrained
1 can (14 ounces, or 425 ml) beef broth
1 teaspoon beef bouillon concentrate
2 teaspoons Worcestershire sauce
1 teaspoon paprika
8 ounces (225 g) cream cheese (regular or light)
8 ounces (225 g) sour cream (regular or light)
Put the beef in your slow cooker. Put the onion on top and then dump in the mushrooms, liquid and all.
In a bowl, mix the beef broth with the bouillon, Worcestershire sauce, and paprika. Pour the mixture into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
When the time's up, cut the cream cheese into cubes and stir it into the mixture in the slow cooker until melted. Stir in the sour cream.
Serve over _Fauxtatoes_ (see recipe page 343) or _Cauli-Rice_ (see recipe page 343), if desired. Actually, because noodles are traditional with _Beef Stroganoff_ , this would be a good place to serve low-carb pasta, if you have a brand you like.
_**Note:**_ This can be made with plain yogurt in place of both the cream cheese and sour cream. After getting everything together in the slow cooker and starting the cooking, place a strainer in a bowl. Line the strainer with a clean coffee filter and pour two 8-ounce (225 g) containers of plain yogurt into it. Set the strainer and bowl in the refrigerator and let the yogurt drain all day. Whisk the resulting yogurt cheese into your Stroganoff in place of cream cheese and sour cream.
_**Yield:**_ 8 servings, each with: 413 calories, 31 g fat, 28 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Beef with Asian Mushroom Sauce
Once you have the _Hoisin Sauce_ on hand, this is very quick to put together. The _Hoisin Sauce_ is a snap, and it keeps well in the fridge.
4 ounces (115 g) sliced mushrooms
4 pounds (1.8 kg) beef tip roast
1/4 cup (60 ml) _Hoisin Sauce_ (see recipe page 337)
2 cloves garlic, minced
1/2 teaspoon salt
1/4 cup (60 ml) beef broth
Guar or xanthan
6 tablespoons (36 g) sliced scallions
Put the mushrooms in your slow cooker and place the beef on top. Spread the _Hoisin Sauce_ over the beef, scatter the garlic and salt over it, and pour in the broth around it. Cover the slow cooker, set it to low, and let it cook for 9 hours.
When the time's up, remove the beef from the slow cooker and put it on a platter. Add guar or xanthan to thicken up the sauce a bit and then pour the sauce into a sauce boat. Slice the beef and serve it with the sauce, topped with the scallions.
_**Yield:**_ 6 servings, each with: 658 calories, 43 g fat, 61 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Asian Slow Cooker Short Ribs
Look for black bean sauce in Asian markets or in the international aisle of a big grocery store. You'll only use a little at a time, but it keeps a long time in the fridge, and it adds authenticity to Asian dishes.
6 pounds (2.7 kg) beef short ribs
3 tablespoons (45 ml) oil
1 stalk celery, chopped
1/4 cup (37 g) shredded carrot
1/2 cup (80 g) chopped onion
2 tablespoons (16 g) grated ginger root
6 teaspoons (30 g) Chinese black bean sauce
3 teaspoons (16 g) chili garlic paste
3 cloves garlic, crushed
1/4 cup (60 ml) soy sauce
1 cup (235 ml) dry red wine
2 cups (475 ml) beef broth
1 teaspoon five-spice powder
1 tablespoon (1.5 g) Splenda
Guar or xanthan
In a big, heavy skillet, brown the ribs all over in the oil. Transfer the ribs to your slow cooker.
Add the celery, carrot, and onion to the skillet and sauté over medium-high heat until they soften and start to brown. Stir in the ginger, black bean sauce, chili garlic paste, and garlic and sauté for another couple of minutes. Now stir in the soy sauce, wine, broth, five-spice powder, and Splenda. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, transfer the ribs to a platter and scoop the vegetables into a blender with a slotted spoon. Add 2 cups (475 ml) of the liquid and run the blender till the vegetables are pureed. Thicken the sauce to a heavy cream consistency with guar or xanthan and serve the sauce with the ribs.
_**Yield:**_ 12 servings, each with: 948 calories, 86 g fat, 35 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Good Low-Carb Slow Cooked Short Ribs
This was one of the first recipes I adapted from Peg Bracken's _I Hate To Cook Book_ , aka The World's Funniest Cookbook (and also one of the most useful). It was higher carb, and it wasn't originally a slow cooker recipe, but it adapted well to both!
1 can (8 ounces, or 225 g) tomato sauce
3/4 cup (175 ml) water
2 tablespoons (28 ml) wine or cider vinegar
4 tablespoons (60 ml) soy sauce
2 teaspoons Splenda
3 to 4 pounds (1.4 to 1.8 kg) beef short ribs
1 large onion, sliced
Guar or xanthan (optional)
In a bowl, mix together the tomato sauce, water, vinegar, soy sauce, and Splenda.
Put the ribs in your slow cooker. Place the onion on top of the ribs. Pour the sauce over the onion and ribs. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours. (Because it's more convenient to use frozen ribs, that's what I assumed you were using for this recipe. If you put the ribs in thawed, cut about 1 hour off the cooking time.)
When the time's up, thicken the sauce with guar or xanthan if you prefer. (This recipe gives you tremendously tasty ribs in a thin but flavorful sauce—it's more like a broth. You can thicken it a bit with guar or xanthan, but I rather like it as it is.)
_**Yield:**_ 7 servings, each with: 559 calories, 31 g fat, 61 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs. (This analysis is for 3 pounds (1.4 kg) of ribs. The total carbs will vary with how much of the sauce you eat because most of the carbs are in there. Furthermore, that calorie count assumes that you eat all of the fat that cooks off of the ribs—which I wouldn't suggest.)
### Short Ribs with Mushrooms
Short ribs can be pricey, but man, are they good!
2 pounds (900 g) beef short ribs
8 ounces (225 g) crimini mushrooms (Buy them sliced if you can.)
1/2 onion
2 cloves garlic
1 cup (235 ml) beef broth
1 teaspoon beef bouillon concentrate
1/4 teaspoon dried thyme
2 tablespoons (28 ml) dry white wine
2 tablespoons (28 ml) dry sherry
1 bay leaf
1/2 teaspoon pepper
1 teaspoon tomato paste
Guar or Xanthan
Lay the short ribs on your broiler rack and slide them under a high flame, about 4 to 5 inches (10 to 13 cm) from the heat. Set your timer for 8 minutes.
Meanwhile, slice the mushrooms if you didn't buy them sliced and put them in the slow cooker. Slice your half onion, crush the garlic, and throw them in with the mushrooms. Stir everything together to distribute evenly.
Somewhere in here, the timer will go off. Use tongs to turn the ribs and broil the other side for another 8 minutes.
Okay, the ribs are browned. Put a bay leaf on top of the mushrooms and onions and the ribs on top of that.
Mix together everything else and pour it over the ribs and mushrooms. Cover the pot, set to low, and cook for 5 to 6 hours.
When it's done, fish the short ribs out with your tongs. Remove the bay leaf and then use your guar or xanthan shaker to thicken up the liquid in the pot to heavy cream consistency. Ladle the sauce, mushrooms, and onions over each serving.
We ate this as is, but it would be wonderful over either _Fauxtatoes_ (see recipe page 343) or shirataki fettucini.
_**Yield:**_ 6 servings, each with: 348 calories, 18 g fat, 38 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Short Ribs with Wine and Mushrooms
Short ribs are very flavorful, and this is a simple way to make the most of them.
4 pounds (1.8 kg) beef short ribs
2 bay leaves
1 tablespoon (15 ml) Worcestershire sauce
1 tablespoon (18 g) beef bouillon concentrate
1/2 cup (120 ml) dry red wine
1 can (8 ounces, or 225 g) mushrooms, drained
Guar or xanthan
Place the ribs in your slow cooker. Add the bay leaves, Worcestershire sauce, and bouillon. Pour the wine over everything. Place the mushrooms on top. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
When the time's up, use a slotted spoon to scoop out the ribs and mushrooms and put them on a platter. There may be a fair amount of grease on the liquid in the pot; it's best to skim it off. Thicken the sauce to taste with guar or xanthan.
_Fauxtatoes_ (see recipe page 343) are the ideal side with this, so you have something to eat all that gravy on!
_**Yield:**_ 10 servings, each with: 388 calories, 22 g fat, 42 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Insanely Good Slow-Cooker Barbecued Beef Ribs
These really are insanely good. And they are so easy, too. I only find beef ribs at my grocer's now and then, and I always grab a couple of slabs to stick in the freezer.
3 1/2 pounds (1.6 kg) beef ribs
2 tablespoons (14 g) paprika
1 1/2 teaspoons pepper
1/2 teaspoon salt, Vege-Sal, or hickory smoked salt
1 tablespoon (1.5 g) Splenda or other sugar-free sweetener to equal
1 tablespoon (13 g) sugar
1 teaspoon chili powder
1 teaspoon garlic powder
1 teaspoon onion powder
1/4 teaspoon cayenne pepper
These are super-easy and so good! Whack your slab of ribs into as many sections as it takes for them to fit in your slow cooker—I just cut mine in half.
Lay the ribs on your broiler rack, set your broiler to high, and slide the ribs underneath, about 4 to 5 inches (10 to 13 cm) from the heat. Give them about 7 to 8 minutes per side until they're browned.
While the ribs are browning, mix together everything else. When the ribs are done browning, pull them out and sprinkle this mixture liberally all over them, both sides.
Throw the ribs in the slow cooker—you can put one section on top of the other, that's fine. Cover, set to low, and cook for 5 to 6 hours. Serve with any leftover seasoning mixture and a big roll of paper towels.
_**Yield:**_ 8 servings, each with: 632 calories, 54 g fat, 33 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Short Rib Stew
3 pounds (1.4 kg) beef short ribs
2 tablespoons (28 ml) olive oil
1 medium onion, chopped
8 ounces (225 g) sliced mushrooms
1 1/2 cups (355 ml) beef broth
1/2 teaspoon pepper
1/2 teaspoon dried marjoram
1/2 teaspoon caraway seeds
1 tablespoon (15 ml) lemon juice
2 tablespoons (28 ml) red wine vinegar
1 teaspoon beef bouillon concentrate
In a big, heavy skillet, brown the ribs all over in the oil over medium-high heat. Transfer the ribs to your slow cooker. In the skillet, over medium-low heat, sauté the onion and mushrooms until they're just softened. Transfer them to the slow cooker, too.
In a bowl, mix together the broth, pepper, marjoram, caraway seeds, lemon juice, vinegar, and bouillon. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
You can thicken the pot liquid if you like, but I rather like this as is, especially with _Fauxtatoes_ (see recipe page 343).
_**Yield:**_ 6 servings, each with: 955 calories, 87 g fat, 36 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Chuck with Avocado Aioli
This was a big hit with our tester's family!
3 pounds (1.4 kg) boneless beef chuck roast
1 tablespoon (15 ml) olive oil
1 medium onion, finely chopped
1/2 cup (120 ml) water
1 teaspoon beef bouillon concentrate
3 tablespoons (45 ml) Worcestershire sauce
1 teaspoon dried oregano
1 clove garlic, crushed
_Avocado Aioli_ (See recipe on next page.)
Season the beef with salt and pepper.
In a big, heavy skillet, sear the beef all over in the oil. Transfer the beef to your slow cooker. Add the onion.
In a bowl, combine the water and bouillon. Pour it over the beef. Add the Worcestershire sauce, oregano, and garlic. Cover the slow cooker, set it to low, and let it cook for 8 hours.
Serve with the _Avocado Aioli_.
_**Yield:**_ 8 servings, each with: 381 calories, 28 g fat, 27 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Avocado Aioli
You don't have to reserve this for use with the _Chuck with Avocado Avioli_ (see recipe on previous page). Serve it as a dip with vegetables, too. California avocados are the little, black, rough-skinned ones, and they're lower in carbs than the big, green, smooth-skinned Florida avocados.
2 ripe California avocados
1/4 cup (60 g) mayonnaise
1 tablespoon (15 ml) lime juice
1 to 2 cloves garlic, crushed
1/4 teaspoon salt
Scoop the avocado flesh into a blender or food processor. Add the mayonnaise, lime juice, garlic, and salt and process until smooth.
_**Yield:**_ 8 servings, each with: 127 calories, 13 g fat, 1 g protein, 3 g carbohydrate, 2 g dietary fiber, 1 g usable carbs.
### Cube Steaks in Gravy
This is a great down-home, stick-to-the-ribs type of dish.
1 tablespoon (15 ml) olive oil
1 1/2 pounds (680 g) cube steaks
1 medium onion, sliced
8 ounces (225 g) sliced mushrooms
3 cups (700 ml) beef broth
1 tablespoon (18 g) beef bouillon concentrate
Guar or xanthan
In a big, heavy skillet, heat the oil and brown the steaks on both sides.
Put the onion and mushrooms in your slow cooker.
In a bowl, stir the broth and bouillon and pour the mixture over the veggies. Place the steaks on top. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, remove the steaks and thicken the sauce with guar or xanthan to your liking.
Serve with _Fauxtatoes_ (see recipe page 343).
_**Yield:**_ 6 servings, each with: 297 calories, 17 g fat, 29 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Easy Italian Beef
This is way easy but full of flavor!
2 tablespoons (28 ml) olive oil
2 pounds (900 g) beef chuck, trimmed of fat
1/2 cup (120 ml) beef broth
1 tablespoon (18 g) beef bouillon concentrate
1 package (0.7 ounces, or 19 g) Italian salad dressing mix
In a big, heavy skillet, heat the oil over medium-high heat and brown the beef on both sides. Transfer the beef to your slow cooker.
In a bowl, combine the broth, bouillon, and salad dressing mix. Pour the mixture over the beef. Cover the slow cooker, set it to low, and let it cook for 6 to 8 hours.
_**Yield:**_ 4 servings, each with: 543 calories, 42 g fat, 37 g protein, 1 g carbohydrate, 0 g dietary fiber, 1 g usable carbs.
### Roman Stew
Instead of using the usual Italian seasonings, this was adapted from a historic Roman stew recipe using spices from the Far East. It's unusual and wonderful.
3 pounds (1.4 kg) beef stew meat, cut into 1-inch (2.5 cm) cubes
3 tablespoons (45 ml) olive oil
4 cloves garlic
2 cups (200 g) sliced celery
1 teaspoon salt or Vege-Sal
1/4 teaspoon ground cinnamon
1/4 teaspoon ground cloves
1/4 teaspoon pepper
1/8 teaspoon ground allspice
1/8 teaspoon ground nutmeg
1 can (14 1/2 ounces, or 410 g) diced tomatoes, undrained
1/2 cup (120 ml) dry red wine
Guar or xanthan (optional)
In a big, heavy skillet, brown the beef in the oil over medium-high heat, in a few batches. Transfer the beef to your slow cooker. Add the garlic and celery to the slow cooker and then sprinkle the salt or Vege-Sal, cinnamon, cloves, pepper, allspice, and nutmeg over the beef and vegetables. Pour the tomatoes and the wine over the beef and vegetables. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
You can thicken the pot juices a little if you like with guar or xanthan, but it's not really necessary.
_**Yield:**_ 8 servings, each with: 369 calories, 17 g fat, 44 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Mexican Stew
This Tex-Mex dinner is a simple family-pleaser.
2 pounds (900 g) beef stew meat, cut into 1-inch (2.5 cm) cubes
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1/2 cup (80 g) sliced onion
1 teaspoon chili powder
1 envelope (1 1/4 ounces, or 35 g) taco seasoning mix
1 can (15 ounces, or 425 g) black soybeans
1/2 cup (115 g) sour cream
Put the beef, tomatoes, onion, and chili powder in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
Stir in the taco seasoning and soybeans. Re-cover the slow cooker, turn it to high, and let it cook for another 20 minutes. Place a dollop of sour cream on each serving.
This makes 6 generous servings, and it could even serve 8.
_**Yield:**_ 6 servings, each with: 399 calories, 18 g fat, 46 g protein, 12 g carbohydrate, 5 g dietary fiber, 7 g usable carbs.
### Beef and Zucchini Stew
Don't try adding the zukes at the beginning, or they'll cook to a mush! Put out some vegetables and dip for the ravening hoards, and sip a glass of wine while you're waiting that last hour.
2 pounds (900 g) boneless beef chuck, trimmed of fat and cubed
1 medium onion, sliced
1 large red bell pepper, cut into 1-inch squares
1 large green bell pepper, cut into 1-inch squares
1 cup (245 g) no-sugar-added spaghetti sauce (I suggest Hunt's.)
1/2 cup (120 ml) beef broth
1/2 teaspoon beef bouillon concentrate
1 1/2 pounds (680 g) zucchini, cut into 1/2-inch (13 mm) slices
Guar or xanthan (optional)
In your slow cooker, combine the beef with the onion and peppers.
In a bowl, stir together the spaghetti sauce, broth, and bouillon. Pour the mixture over the beef and vegetables and stir. Cover the slow cooker, set it to low, and let it cook for 9 hours.
Turn the slow cooker to high, stir in the zucchini, re-cover, and let it cook for 1 more hour.
When the time's up, thicken the sauce with guar or xanthan if needed.
_**Yield:**_ 6 servings, each with: 367 calories, 24 g fat, 27 g protein, 10 g carbohydrate, 3 g dietary fiber, 7 g usable carbs.
### Comfort Food Casserole
This is one of those meal-in-a-bowl sorts of things that just seem—well, comforting, somehow. I've found that slow cooking really brings out the best in turnips. They end up remarkably like potatoes.
1 1/2 pounds (680 g) ground round
1 tablespoon (15 ml) oil
1 medium onion, chopped
4 cloves garlic, crushed
4 stalks celery, diced
1 cup (235 ml) beef broth
1 teaspoon beef bouillon concentrate
1/2 teaspoon salt or Vege-Sal
1 teaspoon pepper
2 teaspoons dried oregano
1 teaspoon dry mustard
2 tablespoons (32 g) tomato paste
4 ounces (115 g) cream cheese
3 turnips, cubed
3/4 cup (86 g) shredded cheddar cheese
In a big, heavy skillet, brown and crumble the beef over medium-high heat. Pour off the fat and transfer the beef to your slow cooker.
Add the oil to the skillet and reduce the heat to medium-low. Add the onion, garlic, and celery and sauté until they're just softened. Add the broth, bouillon, salt or VegeSal, pepper, oregano, dry mustard, and tomato paste and stir. Now add the cream cheese, using the edge of a spatula to cut the cream cheese into chunks. Let this mixture simmer, stirring occasionally, until the cream cheese is melted.
Meanwhile add the turnips to the slow cooker.
When the cream cheese has melted into the sauce, pour the sauce into the slow cooker. Stir until the ground beef and turnips are coated. Cover the slow cooker, set it to low, and let it cook for 6 hours. Serve with cheddar cheese on top.
_**Yield:**_ 6 servings, each with: 549 calories, 40 g fat, 35 g protein, 12 g carbohydrate, 3 g dietary fiber, 9 g usable carbs.
### Hamburger and Turnip Layered Casserole
This started out as a high-carb slow cooker casserole with potatoes and canned mushroom soup. This is a little more work, but it's a lot fewer carbs.
1 1/2 pounds (680 g) ground chuck
1 medium onion
2 medium turnips
4 slices cooked bacon
4 ounces (115 g) shredded Cheddar cheese
1 can (4 ounces, or 115 g) diced green chilies
4 ounces (115 g) jarred roasted red peppers
1 can (4 ounces, or 115 g) mushrooms
1 1/4 cups (285 ml) half and half
1 teaspoon onion powder
1 tablespoon (18 g) beef bouillon concentrate
1/2 teaspoon pepper
In your big, heavy skillet, start browning and crumbling your beef. While that's happening, chop your onion. When a little grease has cooked out of the beef, throw in the onion, too, and continue cooking until the pink is gone.
If you haven't cooked your bacon, lay it on a microwave bacon rack or in a Pyrex pie plate and give it 4 to 5 minutes on high or until done crisp.
Peel your turnips, cut them in half, and slice those halves thin. If you didn't buy your cheese shredded, shred it. Open and drain your green chiles and drain and chop your roasted red peppers.
Now drain the liquid from the mushrooms into a measuring cup with the half and half, onion powder, beef bouillion concentrate, and pepper. Mix it up until the bouillon dissolves. Now add enough guar or xanthan to give the liquid a cream sauce consistency—somewhere between heavy cream and canned cream of mushroom soup. Stir in the mushrooms. This is your sauce.
Okay, it's time to layer. Here's how it goes: Beef mixture, peppers (both kinds), cheese, turnips, and then sauce. Repeat until you run out of ingredients. Try to end with sauce, though it's not a tragedy if you don't.
Cover the slow cooker, set it to low, and cook for 6 to 7 hours. That's all.
_**Yield:**_ 6 servings, each with: 499 calories, 38 g fat, 29 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Noodleless Spinach Lasagna
This is a great choice when you need to feed a crowd after being out all afternoon.
2 pounds (900 g) ground chuck
1/2 medium onion, chopped
2 tablespoons (6 g) oregano
5 cloves garlic
26 ounces (735 g) no-sugar-added spaghetti sauce
1/2 teaspoon beef bouillon concentrate
1/4 teaspoon red pepper flakes
24 ounces (680 g) creamed cottage cheese, small curd
8 ounces (225 g) whipped cream cheese with chives and onions
20 ounces (560 g) frozen chopped spinach, thawed
8 ounces (225 g) shredded mozzarella cheese
1/4 cup (25 g) grated Parmesan cheese
In your big, heavy skillet, over medium heat, start browning the meat while you chop the onion. Throw the onion in there and continue browning and crumbling the meat, adding the oregano and two cloves of the garlic, crushed. When the pink is gone from the meat, tilt the pan, spoon off the fat, and then stir in the spaghetti sauce, beef bouillon concentrate, and red pepper flakes. Turn the burner to low and let this simmer while you go on to the next step.
Mix the cottage cheese and chive cream cheese together, along with the remaining garlic, crushed. Blend this all very well.
Drain your spinach—I dump mine into a colander in the sink and then use clean hands to squeeze it out very well.
It's time to assemble your lasagna! In your slow cooker pot, layer everything like this: Meat sauce, cottage cheese mixture, spinach, and then mozzarella. Make three sets of layers and finish with the end of the sauce. Sprinkle the Parmesan on top. Cover, set to low, and cook for 4 to 5 hours.
_**Yield:**_ 10 servings, each with: 514 calories, 36 g fat, 35 g protein, 13 g carbohydrate, 4 g dietary fiber, 9 g usable carbs.
### Eric's Goop Suey
This started with a high-carb recipe for something called American Chop Suey. I played around with it, whacked out a bunch of carbs, and came up with this. It didn't seem anything like chop suey to me, so I was going to call it All-American Glop. Then Eric, aka That Nice Boy I Married, said it reminded him of his very favorite childhood dish, a thing he'd simply called goop. A _Facebook_ fan named Rita Taylor suggested splitting the difference and naming it _Eric's Goop Suey_. So I did.
1 1/2 pounds (680 g) ground chuck
1 medium onion, chopped
1 large green bell pepper, chopped
2 cloves garlic, crushed
1 teaspoon paprika
1 1/2 teaspoons salt
1/2 teaspoon pepper
1 cup (115 g) shredded Cheddar cheese
2 tablespoons (32 g) tomato paste
1 can (14 ounces, or 390 g) diced tomatoes
2 ounces (55 g) cream cheese
2 packages tofu shirataki fettuccini
In your big, heavy skillet, over medium heat, start browning and crumbling the ground beef while you chop the onion and green pepper. Throw them in, too, and keep cooking and breaking up the meat until all the pink is gone. Spoon off most of the fat that's cooked out and then dump the meat and veggies in your slow cooker.
Add the garlic, paprika, salt, pepper, shredded Cheddar, tomato paste, and canned tomatoes, undrained. Stir it all up. Plunk the cream cheese on top, cover the pot, set to low, and let it cook 5 to 7 hours.
When suppertime comes around, snip open your shirataki and dump them into a strainer in the sink. Snip across them a few times with your kitchen shears and then put them in a microwaveable bowl and give them 2 minutes on high. Drain again and then give them another two minutes on high. Drain one more time, stir into the mixture in the slow cooker, and serve with extra grated cheese on top if desired.
_**Yield:**_ 5 servings, each with: 535 calories, 40 g fat, 32 g protein, 11 g carbohydrate, 1 g dietary fiber, 10 g usable carbs.
### Firehouse Chili
Here's a crowd-pleaser! I served this on a rainy afternoon at our local campground and made a lot of friends! You could halve this, but you'd be left with a half a can of soybeans, and you know you'll eat it up, so why bother?
2 pounds (900 g) ground chuck
1 1/2 cups (240 g) chopped onion
4 cloves garlic, crushed
3 tablespoons (24 g) chili powder
3 teaspoons (7 g) paprika
4 teaspoons (10 g) ground cumin
1/4 cup (60 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased lowcarb ketchup
2 tablespoons (32 g) tomato paste
1 can (14 1/2 ounces, or 410 g) diced tomatoes
12 ounces (355 ml) light beer
1 teaspoon Splenda
2 1/2 teaspoons (15 g) salt
1 can (15 ounces, or 425 g) black soybeans
In a big, heavy skillet, brown and crumble the beef over medium-high heat. Drain it and place it in your slow cooker. Add the onion, garlic, chili powder, paprika, cumin, ketchup, tomato paste, tomatoes, beer, Splenda, salt, and soybeans. Stir everything up. Cover the slow cooker, set it to low, and let it cook for 8 hours.
This is good with shredded cheese and sour cream. What chili isn't? But it also stands on its own very well.
_**Yield:**_ 10 servings, each with: 329 calories, 21 g fat, 21 g protein, 12 g carbohydrate, 4 g dietary fiber, 8 g usable carbs.
### Texas Red
Using no tomatoes or beans not only makes this classically Texan, but it also makes it lower carb than most chilis. And it's good and hot!
2 pounds (900 g) beef chuck
3 tablespoons (45 g) bacon grease or as needed
1 medium onion
3 tablespoons (27 g) canned sliced jalapeño peppers, or a couple of fresh jalepeño peppers
4 cloves garlic, crushed
2 tablespoons (16 g) ancho chili powder
2 tablespoons (16 g) chili powder (any one of the popular blends sold simply as chili powder)
2 teaspoons cumin
2 teaspoons oregano
1 teaspoon pepper
1/2 teaspoon cayenne
1/2 cup (120 ml) beef broth
1 teaspoon beef bouillon concentrate
Cut your beef into 1/2-inch (13 mm) cubes—this is easier if it's half frozen. Put your big, heavy skillet over high heat, add some of the bacon grease, and start browning your beef cubes. Unless your skillet is bigger than mine, you'll need to do them in batches—I had three batches. Just don't crowd them too much. Add more bacon grease to the skillet as needed. As they're browned, transfer the beef cubes to the slow cooker.
While your beef cubes are browning, dice your onion fairly small and throw it in the slow cooker, too. Chop up those jalapeño slices a bit or seed and chop fresh jalapeños, throw them in, and crush in your garlic. Now wash your hands really well with soap and water, or you'll be sorry the next time you touch your eyes or nose!
Measure in the ancho chili powder, standard chili powder, cumin, oregano, pepper, and cayenne. Stir it all up.
Once all the beef is in the slow cooker, dump the broth in the skillet along with the bouillon concentrate. Stir to deglaze the skillet and dissolve the bouillon concentrate. Pour this over the beef mixture and give it all one more stir. Slap the lid on, set it to low, and let it cook for 8 hours or on high for 4 hours.
Serve with shredded cheese and sour cream—and a paper napkin to blow your nose on!
_**Yield:**_ 5 servings, each with: 494 calories, 38 g fat, 31 g protein, 8 g carbohydrate, 3 g dietary fiber, 5 g usable carbs.
### Free Venison Chili
This is _Free Venison Chili_ because I got piles and piles of deer bones for free from a local deer processor this autumn. Most of it I fed to my dogs, but I got enough good meat off some of those bones to make this chili. If you don't have any free venison, I see no reason not to use beef chuck, round, or rump. (I have no clue what part of the deer I used.)
2 pounds (900 g) venison, cut in 1-inch (2.5 cm) cubes
2 celery ribs
2 cans (14 1/2 ounces, or 410 g each) tomatoes with green chiles
1 medium onion, chopped
4 garlic cloves, crushed
1 cup (235 ml) light beer
2 teaspoons dried oregano
1 teaspoon ground cumin
1 teaspoon pepper
1 teaspoon salt or Vege-Sal
1/2 teaspoon thyme
1/2 teaspoon paprika
1/2 teaspoon coriander
1/2 teaspoon celery salt
2 tablespoons (16 g) chili powder
2 bay leaves
2 teaspoons sugar-free pancake syrup
4 teaspoons (24 g) cocoa powder
2 teaspoons (3 g) Splenda or equivalent quantity of another sugar-free sweetener
This couldn't be more simple! Throw everything in your slow cooker and stir to combine well. Cover the pot, set on low, and let it cook all day—8 hours is good, and more won't hurt. Remove the bay leaves and serve with the usual shredded Cheddar cheese and sour cream.
_**Yield:**_ 6 servings, each with: 244 calories, 5 g fat, 37 g protein, 12 g carbohydrate, 3 g dietary fiber, 6 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Carne all'Ungherese
The original recipe, from which I adapted this, said it was an Italian version of a Hungarian stew. Whatever it is, it's good!
1/4 cup (60 ml) olive oil
1 1/2 pounds (680 g) beef stew meat, cut into 1-inch (2.5 cm) cubes
1 medium onion, chopped
1 green pepper, cut into strips
2 cloves garlic, crushed
1 cup (235 ml) beef broth
1 teaspoon beef bouillon concentrate
1 teaspoon dried marjoram
1 tablespoon (16 g) tomato paste
1 tablespoon (7 g) paprika
1 tablespoon (28 ml) lemon juice
1/2 cup (115 g) plain yogurt
In a big, heavy skillet, heat a tablespoon or two (15 to 28 ml) of the oil over medium-high heat. Start browning the stew meat. It will take two or three batches; add more oil as you need it. Transfer each batch of browned meat to your slow cooker as it's done.
When all the meat is browned, put the last of the oil in the skillet, reduce the heat to medium-low, and add the onion. Sauté the onion until it's just softening and add it to the slow cooker. Add the green pepper to the slow cooker.
In a bowl, mix together the garlic, broth, bouillon, marjoram, tomato paste, paprika, and lemon juice, stirring until the bouillon and tomato paste are dissolved. Pour the mixture over the meat and onions. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, stir in the yogurt.
Serve over _Fauxtatoes_ (see recipe page 343).
_**Yield:**_ 5 servings, each with: 382 calories, 22 g fat, 38 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs. (Analysis does not include _Fauxtatoes_.)
### Swiss Steak
Here's a no-work version of this old-time favorite.
1 large onion, sliced
3 pounds (1.4 kg) beef round
1 tablespoon (18 g) beef bouillon concentrate
8 ounces (235 ml) vegetable juice (such as V8)
2 stalks celery, sliced
Guar or xanthan (optional)
Place the onion in your slow cooker. Place the beef on top.
In a bowl, stir the bouillon into the vegetable juice. Pour the mixture over the beef. Scatter the celery on top. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
When the time's up, thicken the juices with guar or xanthan if desired.
Serve over puréed cauliflower.
_**Yield:**_ 8 servings, each with: 360 calories, 22 g fat, 35 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Oxtails Pontchartrain
This has a lot of New Orleans elements, including some serious heat, so I named it after Lake Pontchartrain. Oxtails are bony, but very flavorful, and they take very well to the slow cooker. If you haven't had oxtails, don't fear them; they're just muscle meat, like a steak or a roast. It's just that there's a high bone-to-meat ratio.
4 pounds (1.8 kg) beef oxtails
3 tablespoons (27 g) Cajun seasoning
2 tablespoons (28 ml) olive oil
3 large banana peppers, sliced
1 medium onion, sliced
1 medium carrot, shredded
2 stalks celery, sliced
1 clove garlic, crushed
1 cup (235 ml) dry red wine
1/4 cup (60 ml) brandy
1 1/2 teaspoons dried thyme
3 bay leaves
1 can (14 1/2 ounces, or 410 g) diced tomatoes
2 chipotle chiles canned in adobo sauce, chopped (You can use just one if you'd like to cut the heat a bit.)
Sprinkle the oxtails all over with the Cajun seasoning.
In a big, heavy skillet, heat the oil and brown the oxtails all over. Transfer the oxtails to your slow cooker.
Add the peppers, onion, carrot, celery, and garlic to the skillet and sauté them until they're just softened. Add them to the slow cooker, too, and mix them in with the oxtails.
Pour the wine and brandy in the skillet and stir it around. Stir in the thyme and add the bay leaves, tomatoes, and chipotles. Stir this all up and pour it over the oxtails and veggies. Cover the slow cooker, set it to low, and let it cook for 8 hours.
_**Yield:**_ 6 servings each with: 935 calories, 48 g fat, 96 g protein, 13 g carbohydrate, 3 g dietary fiber, 10 g usable carbs.
### Pepperoncini Beef
Pepperoncini are hot-but-not-scorching pickled Italian salad peppers. You'll find these in the same aisle as the olives and pickles. They make this beef very special.
2 to 3 pounds (0.9 to 1.4 kg) boneless chuck pot roast
1 cup (120 g) pepperoncini peppers, undrained
1/2 medium onion, chopped
Guar or xanthan
Place the beef in your slow cooker, pour the peppers on top, and strew the onion over that. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the beef, put it on a platter, and use a slotted spoon to scoop out the peppers and pile them on top of the beef. Thicken the juices in the pot with the guar or xanthan. Add salt and pepper to taste and serve the sauce with the beef.
_**Yield:**_ 6 servings, each with: 325 calories, 24 g fat, 24 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs. (This analysis is for a 2-pound (900 g) roast.)
### Beef in Beer
Here's a simple recipe from _500 Low-Carb Recipes_. The tea, the beer, and the long, slow cooking make this as tender as can be. I've changed this just a little. Originally, you dredged the beef in soy powder or low-carb bake mix, but I've decided that's inessential. And it's another step and a messy one at that.
2 to 3 tablespoons (28 to 45 ml) olive oil
2 pounds (900 g) boneless beef round roast
1 medium onion, sliced
1 can (8 ounces, or 225 g) tomato sauce
12 ounces (355 ml) light beer
1 teaspoon instant tea powder
1 can (4 ounces, or 115 g) mushrooms, drained
2 cloves garlic, crushed
Heat oil in a big, heavy skillet over medium-high heat and sear the beef until it's brown all over. Transfer the beef to your slow cooker.
In the oil left in the skillet, fry the onion for a few minutes and add that to the slow cooker, too.
Pour the tomato sauce and beer over the beef. Sprinkle the tea over it and add the mushrooms and garlic. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
This is good served with _Fauxtatoes_ (recipe page 343).
_**Yield:**_ 6 servings, each with: 374 calories, 24 g fat, 28 g protein, 7 g carbohydrate, 2 g dietary fiber, 5 g usable carbs.
### Chili-Cocoa Pot Roast
This can be placed somewhere between barbecue and a chili, with pot roast thrown in as well! Chocolate and chiles are a classic combination in Mexican cookery, dating back to the Aztecs, at least.
2 1/2 pounds (1.1 kg) rump roast
3 tablespoons (45 g) bacon grease or (45 ml) olive oil or (45 g) coconut oil
1 medium onion, chopped
3 garlic cloves, crushed
1 can (14.5 ounces, or 410 g) tomatoes with green chiles
1 can (8 ounces, or 225 g) tomato sauce
3 tablespoons (24 g) _Chili-Cocoa Rub_ (see recipe page 159.)
1 tablespoon (6 g) cocoa powder
1 tablespoon (1.5 g) Splenda, or other sugar-free sweetener to equal
1 tablespoon (13 g) sugar
1 teaspoon beef bouillon concentrate
In your big, heavy skillet, over high heat, start searing the rump roast in the grease. You want it browned well on all sides.
While that's happening, chop your onion and crush your garlic. Coat your slow cooker with nonstick cooking spray and throw the onion and garlic in there. Go turn your roast.
Mix together the tomatoes, tomato sauce, _Chili-Cocoa Rub_ , cocoa powder, Splenda, and beef boullion concentrate until the boullion concentrate is dissolved.
When your roast is well seared all over, put it on top of the onions and garlic and pour the tomato mixture over it. Cover and cook on low for 6 hours.
Serve with the tomato mixture spooned over the meat. A dollop of sour cream on each serving is wonderful, but it's not essential. We also like sliced avocado and a green salad with this.
_**Yield:**_ 6 servings, each with: 365 calories, 16 g fat, 44 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Chili-Cocoa Rub
Once you have this on hand, you'll find lots of uses for it.
1/4 cup (72 g) salt or Vege-Sal
2 tablespoons (3 g) Splenda
1 tablespoon (15 g) erythritol
2 teaspoons cocoa powder
3 tablespoons (27 g) garlic powder
1 tablespoon (7 g) onion powder
3 tablespoons (21 g) cumin
2 tablespoons (16 g) chili powder
2 tablespoons (12 g) pepper
Stir everything together and store in a used spice shaker bottle. Use on steaks, pork chops, burgers, or ribs—you name it. It's so good!
_**Yield:**_ 16 servings, each with: 17 calories, trace fat, 1 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Cocoa Joes
You've no doubt noticed a little beef-with-cocoa theme going on here. That's because cocoa ain't just for breakfast and dessert anymore! It's a great source of antioxidants, too. These are somewhat different from the Sloppy Joes you grew up with.
2 pounds (900 g) ground chuck
1 medium onion, chopped
2 cloves garlic, crushed
1 tablespoon (6 g) cocoa powder
1 teaspoon instant coffee granules or powder
1 1/2 teaspoons cumin
1/4 teaspoon cinnamon
1 can (8 ounces, or 225 g) tomato sauce
In your big, heavy skillet, over medium heat, start browning and crumbling the ground chuck.
In the meantime, chop your onion and crush your garlic.
When the meat is browned, drain it a bit—leave maybe a third of the fat. Dump this in your slow cooker.
Add everything else and stir. Cover the pot, set to low, and let it cook for 5 to 6 hours.
_**Yield:**_ 6 servings, each with: 424 calories, 32 g fat, 28 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs. (Analysis is exclusive of coleslaw.)
### Coffee Beef
This is so tender and just so good!
2 pounds (900 g) beef chuck
2 tablespoons (28 g) bacon grease or (28 ml) olive oil or (28 g) coconut oil
1/2 medium onion, chopped
1/2 cup (120 ml) water
1 rounded teaspoon instant coffee granules or powder
1/2 teaspoon molasses
2 tablespoons (3 g) Splenda or other sugar-free sweetener to equal
2 tablespoons (26 g) of sugar
2 tablespoons (28 ml) balsamic vinegar
2 tablespoons (22 g) brown mustard
2 tablespoons (28 ml) soy sauce
1 1/2 teaspoons Worcestershire sauce
1 1/2 teaspoons hot sauce—Tabasco or Frank's, or other Louisiana style
2 cloves garlic, crushed
In your big, heavy skillet, over medium-high heat, start searing your beef in the bacon grease or oil.
In the meantime, chop the half an onion and throw it in your slow cooker.
Measure everything else and stir it all together.
When your beef is seared on both sides, place it on top of the chopped onion. Pour the coffee mixture over it, cover, set to low, and cook for 5 to 6 hours.
Thicken up the pan juices just a little with your guar or xanthan shaker to the consistency of heavy cream and serve over the beef—and _Fauxtatoes_ (see recipe page 343), if you have some.
_**Yield:**_ 6 servings, each with: 374 calories, 28 g fat, 25 g protein, 4 g carbohydrate, trace dietary fiber, 4 g usable carbs.
chapter five
Slow Cooker Pork
This is one big chapter and with good reason. Pork is delicious, nutritious, versatile, and slow cooks really well. You'll find everything from simple family suppers to company food in this chapter!
### Braised Pork with Fennel
This was one of my first great slow-cooking triumphs, and it still ranks as one of the two or three best dishes I've ever cooked in my slow cooker. This is easily good enough to serve to company. By the way, some grocery stores label "fennel" as "anise." It looks like a bulb at the bottom, with celery-like stalks above and feathery foliage. The stems are tough, but the foliage can be chopped up in salads or used as a garnish. It has a wonderful licorice-like taste.
4 pounds (1.8 kg) pork shoulder roast
2 tablespoons (28 ml) olive oil
1 medium onion, sliced
1 bulb fennel, sliced
1 cup (235 ml) cider vinegar
3 tablespoons (4.5 g) Splenda
1 cup (180 g) canned diced tomatoes, drained
1 cup (235 ml) chicken broth
1 teaspoon chicken bouillon concentrate
2 cloves garlic, crushed
1/2 teaspoon dried thyme
1/2 teaspoon red pepper flakes, or to taste
Guar or xanthan
In a big, heavy skillet, sear the pork in the oil over medium-high heat until it's brown all over. (This will take 20 minutes or so.) Transfer the pork to your slow cooker.
Pour off all but about 1 tablespoon (15 ml) of fat from the skillet and reduce the heat to medium-low. Sauté the onion and fennel until they're just getting a little golden. Transfer them to the slow cooker, too.
In a bowl, mix together the vinegar and Splenda. Pour the mixture over the pork. Add the tomatoes.
In a bowl, mix together the broth and bouillon until the bouillon dissolves. Stir in the garlic, thyme, and red pepper flakes. Pour this over the pork, too. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the pork from the slow cooker and place it on a serving platter. Using a slotted spoon, scoop out the vegetables and pile them around the pork. Cover the platter with foil and put it in a warm place.
Ladle the liquid from the slow cooker into a saucepan. Place it over the highest heat and boil it hard for 5 to 7 minutes to reduce the sauce a bit. Add some guar or xanthan to thicken the sauce just a bit. (You want it to be about the texture of half-and-half, not a thick gravy.) Serve the sauce over the pork and vegetables.
_**Yield:**_ 6 servings, each with: 621 calories, 46 g fat, 41 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Easy Pork Roast
This is basic, which is a strength, not a weakness. It would be a great supper with a big salad.
3 pounds (1.4 kg) boneless pork loin
2 tablespoons (28 ml) olive oil
1 can (8 ounces, or 225 g) tomato sauce
1/4 cup (60 ml) soy sauce
1/2 cup (120 ml) chicken broth
1/2 cup (12 g) Splenda
2 teaspoons dry mustard
Guar or xanthan (optional)
In a big, heavy skillet, brown the pork on all sides in the oil. Transfer the pork to your slow cooker.
In a bowl, mix together the tomato sauce, soy sauce, broth, Splenda, and dry mustard. Pour the mixture over the pork. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
When the time's up, remove the pork to a serving platter. Thicken the pot liquid, if needed, with guar or xanthan. Serve the juice with the pork.
_**Yield:**_ 8 servings, each with: 301 calories, 14 g fat, 37 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Pork Roast with Apricot Sauce
Here's a fabulous Sunday dinner for the family—with very little work.
2 1/2 pounds (1.1 kg) boneless pork loin
2 tablespoons (28 ml) olive oil
1/3 cup (53 g) chopped onion
3/4 cup (175 ml) chicken broth
1/4 cup (80 g) low-sugar apricot preserves
1 tablespoon (15 ml) balsamic vinegar
1 tablespoon (15 ml) lemon juice
1 tablespoon (1.5 g) Splenda
Guar or xanthan
In a big, heavy skillet, sear the pork all over in the oil. Transfer the pork to your slow cooker. Scatter the onion around it.
In a bowl, mix together the broth, preserves, vinegar, lemon juice, and Splenda. Pour the mixture over the pork. Cover the slow cooker, set it to low, and let it cook for 7 hours.
When the time's up, remove the pork and put it on a serving platter. Season the juices with salt and pepper to taste. Thicken the juices with guar or xanthan. Ladle the juices into a sauce boat to serve.
_**Yield:**_ 6 servings, each with: 338 calories, 17 g fat, 40 g protein, 5 g carbohydrate, trace dietary fiber, 5 g usable carbs.
### Pork Roast with Creamy Mushroom Gravy and Vegetables
Here's a great down-home dinner the family will love!
2 1/2 pounds (1.1 kg) boneless pork loin
1/2 cup (61 g) sliced carrots
4 ounces (115 g) sliced mushrooms
10 ounces (280 g) frozen cross-cut green beans, unthawed
1 tablespoon (18 g) beef bouillon concentrate
2 tablespoons (28 ml) water
1 can (14 1/2 ounces, or 410 g) tomatoes with roasted garlic
Guar or xanthan
1/2 cup (120 ml) heavy cream
Put the pork in the bottom of your slow cooker. Surround the pork with the carrots, mushrooms, and green beans. (Don't bother thawing the green beans, just whack the package hard on the counter before opening so the beans are all separated.)
In a bowl, dissolve the bouillon in the water. Stir in the tomatoes. Pour the mixture over the pork and vegetables. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
When the time's up, remove the pork and vegetables to a platter. Thicken the juices in the pot with guar or xanthan and then whisk in the cream. Add salt and pepper to taste. Serve the juices with the pork and vegetables.
_**Yield:**_ 8 servings, each with: 284 calories, 19 g fat, 23 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Orange Rosemary Pork
One day I saw the adorable Rachael Ray (whose Food Network show I love) grilling some pork chops for one of her 30-minute meals, and she was saying that orange and rosemary were great for basting pork. Well, hey, Rachael knows what she's talking about, so I decided to borrow those flavors for a low-carb slow cooker dish. Oh, boy, did it work out well! Thanks for the idea, Rachael!
1 1/2 pounds (680 g) boneless pork loin
2 tablespoons (28 ml) olive oil
1/4 cup (60 ml) white wine vinegar
1/4 cup (60 ml) lemon juice
3 tablespoons (4.5 g) Splenda
1/4 teaspoon orange extract
1/2 teaspoon ground, dried rosemary
1 clove garlic, crushed
1 teaspoon soy sauce
1/4 teaspoon pepper
1/4 teaspoon salt or Vege-Sal
In a big, heavy skillet, brown the pork in the oil over medium-high heat. Transfer the pork to your slow cooker.
In a bowl, stir together the vinegar, lemon juice, Splenda, orange extract, rosemary, garlic, soy sauce, pepper, and salt or Vege-Sal and pour over the pork. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
_**Yield:**_ 4 servings, each with: 317 calories, 23 g fat, 23 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Sweet and Tangy Mustard Pork Roast
This is simple and great. I love Boston butt—which is really just a shoulder roast.
2 1/2 pounds (1.1 kg) Boston butt roast
2 tablespoons (28 g) bacon grease or coconut oil
1/4 cup (60 g) erythritol, xylitol, or similar sweetener
1 1/2 tablespoons (17 g) brown mustard
1 tablespoon (15 ml) lemon juice
1/4 teaspoon molasses
1/4 teaspoon ground rosemary
1/4 teaspoon sage
1/4 teaspoon thyme
1/4 teaspoon pepper
1 teaspoon soy sauce
In your big, heavy skillet, over medium-high heat, start your pork roast searing in the bacon grease. You want to brown it on all sides.
In the meantime, measure the erythritol, mustard, lemon juice, molasses, rosemary, sage, thyme, pepper, and soy sauce into a small dish. Stir together well.
When your roast is browned all over, smear the erythritol mixture evenly over all sides except the fatty side. Place the roast, fatty side up, in the slow cooker and spoon the rest of the mixture evenly over the fatty side.
Cover and set the slow cooker to low. Cook five hours. That's it! Put it on a platter, carve, and serve.
_**Yield:**_ 6 servings, each with: 339 calories, 20 g fat, 37 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Pork with Cabbage
Need I point out that this recipe is for cabbage lovers?
4 pounds (1.8 kg) boneless pork shoulder roast, trimmed of fat
2 tablespoons (28 ml) olive oil
2 carrots, cut into 1-inch (2.5 cm) pieces
2 cloves garlic, crushed
2 stalks celery, cut
1/2-inch (6 mm) thick
1 envelope (1 ounce, or 28 g) onion soup mix
1 1/2 cups (355 ml) water
1 1/2 pounds (680 g) cabbage, coarsely chopped
Guar or xanthan
In a big, heavy skillet, start browning the pork in the oil.
Place the carrots, garlic, and celery in your slow cooker. Add the soup mix and water.
When the pork is brown all over, put it on top of the vegetables in the slow cooker. Cover the slow cooker, set it to low, and let it cook for 7 hours.
When the time's up, stir in the cabbage, pushing it down into the liquid. Re-cover the slow cooker and let it cook for another 45 minutes to 1 hour.
Remove the pork and put it on a platter. Use a slotted spoon to pile the vegetables around the pork. Thicken the liquid in the slow cooker with guar or xanthan. Add salt and pepper to taste. Pour the liquid into a sauce boat and serve with the pork and vegetables.
_**Yield:**_ 8 servings, each with: 478 calories, 35 g fat, 31 g protein, 10 g carbohydrate, 3 g dietary fiber, 7 g usable carbs.
### Kalua Pig with Cabbage
No, no, that's not Kahlua; there's no coffee liqueur involved. Rather, it's a slow-cooker adapted version of a traditional Hawaiian dish. Super-simple, super-low carb, and utterly delicious—that Nice Boy I Married rated it a perfect 10. Feel free to use a 6 pound (2.7 kg) pork shoulder, more salt, more liquid smoke, and more cabbage and serve an army. Just increase the cooking times a bit.
3 pounds (1.4 kg) Boston butt pork roast
2 teaspoons sea salt
1 tablespoon (15 ml) liquid smoke flavoring
1 head cabbage, chopped fairly coarsely
1/4 medium onion, minced
Take a carving fork and stab your pork roast viciously all over. Do your best slasher movie imitation. You're making lots of holes to let the smoky flavor in.
Now sprinkle the salt all over the roast, getting every bit of the surface, and rub it in a little. Do the same with the smoke flavoring.
Lay your roast on the bottom of your slow cooker, cover it, set it to low, and forget about it for a good 7 to 8 hours, minimum. Then flip the roast, re-cover, and forget about it for another 7 to 8 hours.
An hour or 90 minutes before serving time, chop up your cabbage and onion.
Haul out your pork, put it in a big bowl, and shred it up with a fork. Scoop out a bit of the liquid from the pot to moisten it if it seems to need it. Then keep it somewhere warm (or you can rewarm it later in the microwave).
Throw the cabbage and onion in the remaining liquid and toss it to coat. Cover the pot, set it on high, and let it cook for at least an hour—you want it wilted but still a little crunchy.
Serve the meat and cabbage together.
_**Yield:**_ 8 servings, each with: 384 calories, 27 g fat, 33 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Pork with Rutabaga
If you haven't tried rutabaga, you simply must. Also sometimes called a "swede" or a "yellow turnip," rutabaga is similar to a turnip, except that it has an entrancing bitter-sweet flavor. Anyway, it's fun confusing grocery store checkout clerks who can't figure out what that big yellow root is! It's actually delicious to make this with half rutabaga and half a fresh, cubed pumpkin, but it's just not possible to find fresh pumpkin some seasons of the year.
2 1/2 pounds (1.1 kg) rutabaga, peeled and cubed
3 pounds (1.4 kg) boneless pork shoulder roast, tied or netted
1/2 teaspoon blackstrap molasses
1/2 cup (12 g) Splenda
1/4 teaspoon cayenne
1 clove garlic, minced
Put the rutabaga in the bottom of your slow cooker. Put the pork on top. Drizzle the molasses over the pork and rutabega.
In a bowl, mix together the Splenda, cayenne, and garlic. Sprinkle the mixture over the pork and rutabaga. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
When the time's up, remove the pork from the slow cooker, cut off the string or net, and slice or pull the pork apart. Serve the pork over the rutabaga with the pot liquid.
_**Yield:**_ 6 servings, each with: 472 calories, 31 g fat, 32 g protein, 16 g carbohydrate, 5 g dietary fiber, 11 g usable carbs.
### "Honey" Mustard Ham
You may wonder how to roast a ham when you're not going to be around for hours to tend the oven. The answer is in your slow cooker, of course. You'll need a big slow cooker for this.
5 pounds (2.3 kg) fully cooked, bone-in ham
1/3 cup (75 ml) apple cider vinegar
1/2 cup (12 g) Splenda
1 tablespoon (11 g) brown mustard
1/2 teaspoon blackstrap molasses
1 teaspoon water
Place the ham in your slow cooker.
In a bowl, mix together the vinegar and 2 tablespoons (3 g) of the Splenda. Add the mixture to the slow cooker. In the same bowl, mix together the mustard, molasses, remaining Splenda, and water and spread the mixture over the ham. Cover the slow cooker, set it to low, and let it cook for 7 hours.
_**Yield:**_ 6 servings, each with: 683 calories, 41 g fat, 68 g protein, 6 g carbohydrate, trace dietary fiber, 6 g usable carbs.
### Ham with Rutabaga and Turnips
If you're roasting a ham in your slow cooker, you may as well roast your vegetables, too, right?
4 turnips, cubed
1 1/2 pounds (680 g) rutabaga, peeled and cubed
6 1/2 pounds (3 kg) shank half ham
Put the turnips and rutabaga in the bottom of your slow cooker. Place the ham on top, flat side down. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours. Again, you'll need a big slow cooker.
_**Yield:**_ 10 servings, each with: 674 calories, 51 g fat, 43 g protein, 9 g carbohydrate, 3 g dietary fiber, 6 g usable carbs.
### Creamy Ham Casserole
I made this up to use the end of a ham I'd slow cooked, and it was a hit with my husband.
1 head cauliflower
1 medium onion, chopped
1 large stalk celery, with leaves
2 cups (475 ml) Carb Countdown dairy beverage
1 cup (235 ml) chicken broth
6 teaspoons guar or xanthan
1 teaspoon dry mustard
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
8 ounces (225 g) Gruyère cheese, shredded
Run the cauliflower through the slicing blade of your food processor. Transfer it to a bowl and replace the slicing disc with the **S** -blade. Chop the onion and celery fine in the food processor.
With a hand blender or regular blender, blend the Carb Countdown and broth. Add the guar or xanthan and blend it until there are no lumps. Pour the mixture it into a saucepan and heat it over medium-low heat. (If you do have a hand blender, you may as well just dump the Carb Countdown and the chicken broth in the saucepan and use the hand blender to blend in the thickener in the pot to save a little dishwashing.) Stir in the dry mustard, salt or Vege-Sal, and pepper. When the sauce is hot, stir in the cheese, a little at a time, until it's all melted. Turn off the burner.
Spray your slow cooker with nonstick cooking spray. Put in a layer of cauliflower, a lighter layer of onion and celery, and then a generous layer of ham. Repeat these layers until everything's gone and the slow cooker is full. Pour half of the sauce over the top. It won't immediately flow down into the food in the slow cooker, so poke down into it several times with the handle of a rubber scraper or spoon, piercing the layers to the bottom. The sauce will start to seep down. When there's more room on top, pour in the rest of the sauce and poke down through the layers again. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
_**Yield:**_ 8 servings, each with: 287 calories, 19 g fat, 24 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Creamy Ham Hash
This is really wonderful. You'll thank me the day after a ham-roasting holiday, when you're trying to figure out what to do with all those leftovers. Throw this in the slow cooker and take all those holiday guests out to a movie or something, knowing that supper is taken care of. This would make a good brunch dish, too, if you made it ahead and then warmed it up.
2 large turnips
1 medium rutabaga
1 medium onion
1 medium red bell pepper
1 medium green bell pepper
2 pounds (900 g) cooked ham
2 packages (8 ounces, or 225 g each) cream cheese
1 1/2 cups (355 ml) half and half
1/2 cup (120 ml) heavy cream
1/2 teaspoon pepper
3 tablespoons (33 g) brown mustard
This involves a lot of cutting things in little cubes. If you prefer, you can use a food processor, but you won't get the nice, even texture that you get from turning on some music or the television and just giving yourself over to dicing things.
Peel the turnips and cut them in 1/4-inch (6 mm) dice. Do the same with the rutabaga. You'll want to start dumping these in your slow cooker as you go, or you'll run out of space on your cutting board long before you're done dicing. Dice the onion and peppers and throw them in, too.
Now cut your ham in 1/4-inch (6 mm) cubes, too—I got about 5 cups (750 g) of ham cubes. Add to the slow cooker and stir everything together.
In a microwaveable bowl or big measuring cup, nuke one 8-ounce (225 g) package of the cream cheese with the half and half, heavy cream, pepper, and mustard. Give it a good 2 to 3 minutes on high and then whisk the whole thing until it's smooth. Pour this mixture evenly over the ham and veggies.
Plunk the second 8-ounce (225 g) block of cream cheese on top of the stuff in the slow cooker. Slap on the lid and set the cooker for low. Let it go for 7 to 8 hours.
At the end of the cooking time, stir the whole thing up. At first it will seem watery and curdled, but as you stir, the sauce will turn creamy and delicious. Once the sauce is creamy, serve in bowls, with soup spoons.
_**Yield:**_ 8 servings, each with: 549 calories, 43 g fat, 27 g protein, 14 g carbohydrate, 2 g dietary fiber, 12 g usable carbs.
### Sweet and Sour Pork
Here's another stir-fry dish turned into a slow cooker meal. This lacks the chunks of pineapple you often find in sweet-and-sour dishes. They're just too high carb. But the crushed pineapple in the sauce gives the right flavor!
1 1/2 pounds (680 g) boneless pork loin, cut into 1-inch (2.5 cm) cubes
1 green bell pepper, diced
1/4 cup (6 g) Splenda
1 tablespoon (8 g) grated ginger root
1 clove garlic, crushed
1/4 cup (60 ml) rice vinegar
3 tablespoons (45 ml) soy sauce
1/4 teaspoon blackstrap molasses
1/3 cup (80 g) canned crushed pineapple in juice
1/2 head cauliflower
Guar or xanthan
Put the pork and pepper in your slow cooker.
In a bowl, mix together the Splenda, ginger, garlic, vinegar, soy sauce, molasses, and pineapple. Pour the mixture over the pork and pepper. Cover the slow cooker, set it on low, and let it cook for 6 hours.
When the time's up, run the cauliflower through the shredding blade of a food processor and put it in a microwaveable casserole dish with a lid. Add a couple of tablespoons of water (28 ml), cover, and microwave on high for 7 minutes. This is your _Cauli-Rice_!
Meanwhile, thicken up the pot juices with guar or xanthan until they're about the texture of commercial Chinese food. Serve the pork mixture over the _Cauli-Rice_.
_**Yield:**_ 4 servings, each with: 283 calories, 11 g fat, 36 g protein, 9 g carbohydrate, 1 g dietary fiber, 8 g usable carbs.
### Curried Pork Stew
This is not terribly authentic, but it's awfully good. Try one of the chutneys (see recipe pages 309 and ) with this.
1 pound (455 g) boneless pork loin, cubed
1/2 teaspoon salt
2 tablespoons (12 g) curry powder
1 tablespoon (15 ml) olive oil
1 onion, sliced
2 small turnips, cubed
1 cup (180 g) canned diced tomatoes
1/2 cup (120 ml) cider vinegar
2 tablespoons (3 g) Splenda
2 cups (200 g) diced cauliflower
Season the pork with the salt and sprinkle with 1 tablespoon (6 g) of the curry powder.
In a big, heavy skillet, heat the oil and brown the pork over medium-high heat.
Place the onion and turnips in your slow cooker. Top with the pork and tomatoes.
In a bowl, stir together the vinegar, Splenda, and the remaining 1 tablespoon (6 g) curry powder. Pour the mixture over the pork. Cover the slow cooker, set it to low, and let it cook for 7 hours.
When the time's up, stir in the cauliflower. Re-cover the slow cooker and cook for 1 more hour or until the cauliflower is tender.
_**Yield:**_ 6 servings, each with: 177 calories, 8 g fat, 17 g protein, 11 g carbohydrate, 3 g dietary fiber, 8 g usable carbs.
### Pork and "Apple" Stew
The apple flavor here comes from the apple cider vinegar. Our tester, Maria, cut her turnips into apple-slice shapes, and her family thought they were apples! They loved the whole thing.
2 pounds (900 g) pork loin, cut into 1-inch (2.5 cm) cubes
2 medium turnips, cubed
2 medium carrots, cut 1/2-inch (13 mm) thick
1 medium onion, sliced
1/2 cup (50 g) sliced celery
1 cup (235 ml) apple cider vinegar
3 tablespoons (4.5 g) Splenda
1 cup (235 ml) chicken broth
1 teaspoon chicken bouillon concentrate
1 teaspoon caraway seeds
1/4 teaspoon pepper
Combine the pork, turnips, carrots, onion, and celery in your slow cooker.
In a bowl, stir together the vinegar, Splenda, broth, and bouillon. Pour the mixture over the pork and vegetables. Add the caraway seeds and pepper and stir everything. Cover the slow cooker, set it to low, and let it cook for 8 hours.
_**Yield:**_ 6 servings, each with: 226 calories, 6 g fat, 34 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Easy Southwestern Pork Stew
Our tester gave this a 10—and so did her family!
1 medium onion, chopped
3 cloves garlic, crushed
2 pounds (900 g) boneless pork loin, cut into 1-inch (2.5 cm) cubes
2 teaspoons ground cumin
1 tablespoon (4 g) dried oregano
1/2 teaspoon salt
1 can (15 ounces, or 425 g) black soybeans
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1 cup (235 ml) chicken broth
1 teaspoon chicken bouillon concentrate
Put the onion and garlic in your slow cooker and place the pork on top.
In a bowl, stir together the cumin, oregano, salt, soybeans, tomatoes, broth, and bouillon. Pour the mixture over the pork and vegetables. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
_**Yield:**_ 6 servings, each with: 257 calories, 10 g fat, 34 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs.
### Orange Pork Loin
Boneless pork loin frequently goes on sale. It's very lean, however, so it's often both bland and dry. Slow cooking takes care of that little problem! Sadly, fresh pumpkin is only available for a couple of months in the autumn, so that's when you'll need to make this dish. Buy a small pumpkin, or you'll have piles of it leftover.
1 pound (455 g) pumpkin, peeled and cut into 1/2-inch (13 mm) cubes
1 pound (455 g) rutabaga, cut into 1/2-inch (13 mm) cubes
2 tablespoons (28 ml) olive oil
2 pounds (900 g) pork loin
2 tablespoons (40 g) low-sugar marmalade or orange preserves
1/4 teaspoon orange extract
2 teaspoons Splenda
2 cloves garlic, crushed
1/2 teaspoon salt
1/2 cup (120 ml) chicken broth
Guar or xanthan
Put the pumpkin and rutabaga in the bottom of your slow cooker.
In a big, heavy skillet, heat the oil over medium-high heat and brown the pork all over. Put the pork in the slow cooker on top of the pumpkin and rutabaga.
In a bowl, stir together the marmalade, orange extract, Splenda, garlic, salt, and broth. Pour the mixture over the pork. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, carefully remove the pork to a platter and use a slotted spoon to pile the vegetables around it. Use guar or xanthan to thicken the liquid in the pot to the consistency of heavy cream. Serve the liquid with the pork and vegetables.
_**Yield:**_ 6 servings, each with: 281 calories, 10 g fat, 34 g protein, 13 g carbohydrate, 2 g dietary fiber, 11 g usable carbs.
### Spinach-Stuffed Pork Loin
Pork loin is a little lean for my tastes. On the other hand, it's a nice, solid chunk of meat in a handy shape, just right for stuffing. It makes a great company dish. Feel free to use this basic method with other low-carb stuffings.
2 pounds (900 g) boneless pork loin in one chunk
Salt and pepper
2 tablespoons (28 ml) olive oil
10 ounces (280 ml) frozen chopped spinach, thawed
2 tablespoons (14 g) sun-dried tomatoes, oil-packed, diced
4 pieces pepperoncini peppers, topped and chopped
4 cloves garlic, crushed
1/2 onion, chopped
1/3 cup (50 g) crumbled feta cheese
Salt and pepper
2 tablespoons (28 ml) lemon juice
2 tablespoons (28 ml) olive oil
1/2 teaspoon chicken bouillon concentrate
1 pinch pepper
1/3 cup (45 g) pine nuts
First, slice your pork loin down one long side, cutting it almost all the way through, so you can open it up like a book. Open it up, place it with the inside down on your cutting board, and salt and pepper the outside surface.
In your big, heavy skillet, heat the olive oil over medium-high heat. Lay the pork, outside down, in the hot oil. You're browning the outside a bit.
While that's happening, drain your spinach very well—I dump mine into a colander in the sink and then use clean hands to squeeze it dry. Put it in a big mixing bowl. Now go check your pork! If it's nicely golden, turn off the heat.
Now go back to the mixing bowl. Add the sun-dried tomatoes, chopped pepperoncini, crushed garlic, chopped onion, and crumbled feta. Mix this all up well.
Lay three lengths of string, big enough to tie around your pork roast, on your cutting board and lay your pork, the cut, unbrowned side up, on top of them. Spoon about half the spinach mixture evenly over one side and then close the other side over it. Use the string to tie your roast shut and then carefully transfer it to the slow cooker.
Mix together the lemon juice, olive oil, chicken bouillon concentrate, and pepper, stirring until the bouillon is dissolved. Pour this evenly over the roast. Cover the pot and set to low. Forget about it for 4 to 5 hours.
You'll notice you still have a bunch of spinach mixture left. Not to worry. Get a big sheet of aluminum foil and lay it out on that cutting board. Spoon the rest of the spinach mixture onto half of it, fold the other half over it, and fold up all the edges, making a nice, tight packet. Stash this in the fridge for a few hours.
About an hour before you want to serve your roast—about 4 to 5 hours after you started cooking your roast—place your packet of leftover spinach stuffing on top of the pork, replace the lid, and let the whole thing cook another hour.
Ten minutes before you're getting ready to serve, put your pine nuts in a dry skillet over medium heat and stir them until they're touched with gold. (Or just buy them already toasted, if your store has them that way.)
Pull the packet out of the slow cooker and set it aside for a moment. Use a big spatula to carefully lift out the roast and put it on a platter. Snip the strings and pull them out, discarding them. Now open your packet and pile the extra spinach around the roast.
Now it's back to that slow cooker. However you can, get the yummy lemony-olive-oil-y juices into a pitcher or sauce dish! I used two hot mitts, picked up the whole crockery liner, and poured! Put the pine nuts in a little serving dish with a spoon, too.
Serve slices of the stuffed roast with a little of the extra stuffing with each slice. Pour pot juices over each serving and sprinkle with a spoonful of pine nuts.
_**Yield:**_ 6 servings, each with: 298 calories, 20 g fat, 24 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Pork Slow Cooker Chili
Try this when you want to have people over after the kids' soccer game!
1 tablespoon (15 ml) olive oil
2 1/2 pounds (1.1 kg) boneless pork loin, cut into 1-inch (2.5 cm) cubes
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1/4 cup (40 g) chopped onion
1/4 cup (38 g) diced green bell pepper
1 clove garlic, crushed
1 tablespoon (8 g) chili powder
In a big, heavy skillet, heat the oil and brown the pork all over. Transfer the pork to your slow cooker. Stir in the tomatoes, onion, pepper, garlic, and chili powder. Cover the slow cooker, set it to low, and let it cook for 6 to 8 hours.
Serve this with sour cream and shredded Monterey Jack cheese, if you like, but it's darned good as is.
_**Yield:**_ 8 servings, each with: 189 calories, 8 g fat, 25 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Orange and Tomatillo Pork Chili
2 pounds (900 g) pork loin
1 tablespoon (15 g) bacon grease
1 medium onion
1 jalapeño pepper
1 pound (455 g) tomatillos, husks removed
1 can (14 ounces, or 390 g) diced tomatoes
1 can (15 ounces, 425 g) Eden Organic black soy beans
4 cloves garlic
2 teaspoons grated orange rind
2 tablespoons (28 ml) orange juice
1/4 cup (60 ml) lemon juice
1/4 teaspoon orange extract
1 tablespoon (1.5 g) Splenda or the equivalent in another sugar-free sweetener
1/2 cup (8 g) chopped fresh cilantro
12 fluid ounces (355 ml) light beer
Cut your pork in bite-sized cubes. In your big, heavy skillet, over medium-high heat, start browning your pork cubes in the bacon grease. Don't crowd them; do them in a few batches, transferring them to the slow cooker as they're done. Add more bacon grease if you need to.
While the pork is browning, chop your onion and seed and mince your jalapeño. Add these to the slow cooker. (Then wash your hands thoroughly with soap and water, or you'll be sorry the next time you touch your eyes or nose. You must always wash your hands well after handling hot peppers.) Husk your tomatillos, cut them in quarters, and throw those in, too.
Add the diced tomatoes, undrained. Drain your soybeans and add them to the pot.
Now grate your orange rind and squeeze 2 tablespoons (28 ml) of orange juice. Mix the rind and orange juice with the lemon juice, orange extract, Splenda or other sweetener, cilantro, and beer. Pour this mixture over everything in the slow cooker.
Cover, set cooker to low, and let it go for a good 7 to 8 hours. You can serve with sour cream, if you like, but it's good as is.
_**Yield:**_ 6 servings, each with: 224 calories, 8 g fat, 21 g protein, 13 g carbohydrate, 2 g dietary fiber, 11 g usable carbs.
_**Note:**_ Feel free to kick up the heat in this by adding some red pepper flakes or cayenne—or for that matter, another jalepeño pepper.
### Slow Cooker Mu Shu Pork
This isn't authentic, by any means, but it's very tasty. My husband, not a big Chinese food guy, really liked this.
2 plums, pitted and chopped
1 clove garlic, crushed
2 tablespoons (16 g) grated ginger root
1/4 cup (60 ml) soy sauce
2 tablespoons (28 ml) dry sherry
2 teaspoons dark sesame oil
1/8 teaspoon five-spice powder
1 tablespoon (15 ml) rice vinegar
3 tablespoons (4.5 g) Splenda
2 pounds (900 g) boneless pork loin, cut into a few big chunks across the grain
Guar or xanthan
5 eggs, beaten
2 cups (208 g) bean sprouts
2 cups (150 g) shredded napa cabbage
16 low-carb tortillas, 6-inch
3/4 cup (75 g) sliced scallions
_Hoisin Sauce_ (optional, see recipe page 337)
Put the plums, garlic, ginger, soy sauce, sherry, sesame oil, five-spice powder, vinegar, and Splenda in a food processor with the **S** -blade in place and run until the plum is puréed.
Place the pork in your slow cooker. Pour the plum sauce over the pork. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, scoop out the pork with a slotted spoon and put it on a big plate. Use a couple of forks to tear the pork into little shreds. Thicken the sauce in the slow cooker to about ketchup consistency with guar or xanthan. Stir the shredded pork back into the slow cooker. Re-cover the slow cooker, set it to high, and let it cook for 30 minutes.
Meanwhile, spray a big, heavy skillet well with nonstick cooking spray. (A nonstick skillet is even better for this.) Put it over medium-high heat and let it get hot. Pour in enough of the eggs to form a thin layer on the bottom and let it cook, not stirring, until it's a solid sheet. Lift the eggs out and set them on a plate. Cook the rest of the eggs into a thin sheet as well. Use a sharp knife to cut these sheets of cooked egg into strips about 1/4 inch (6 millimeters) wide. Reserve.
When the 30 minutes are up, stir the bean sprouts, cabbage, and shredded eggs into the pork mixture. Re-cover the slow cooker and let it cook for just another 10 to 15 minutes. (You want the bean sprouts to be hot through but still have some crispness.) While that's happening, slice your scallions. (If you're not going to be eating all of this right away, only slice enough scallions for immediate use.)
To serve, spread 1/3 cup (85 g) of the pork-and-egg-and-vegetable mixture on a tortilla, sprinkle with sliced scallions, wrap, and eat! If you want to be more authentic, you could spread a little _Hoisin Sauce_ on each tortilla before filling, but it's hardly essential.
_**Yield:**_ 16 servings, each with: 175 calories, 8 g fat, 19 g protein, 14 g carbohydrate, 9 g dietary fiber, 5 g usable carbs.
### Slow Cooker Pulled Pork
Pulled pork is a Carolina tradition, and it usually involves many, many hours of long, slow smoking. This is not authentic, but it is tasty, and thanks to liquid smoke flavoring, it has an appealingly smoky flavor. This recipe requires a meat injector, a big, scary-looking syringe that looks like your doctor got way out of hand. You can buy inexpensive ones at housewares stores.
1/3 cup (80 ml) liquid smoke flavoring
3 pounds (1.4 kg) pork shoulder
Sauces, such as _Eastern Carolina Vinegar Sauce_ (see recipe page 336), _Dana's "Kansas City" Barbecue Sauce_ (see recipe page 335), and _Piedmont Mustard Sauce_ (see recipe page 336)
Slurp up a syringe-full of the liquid smoke flavoring and inject it into a dozen sites all over the pork shoulder. Season the pork with salt and pepper. Place the pork in your slow cooker and pour another tablespoon or two (15 to 28 ml) of liquid smoke over it. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the pork from the slow cooker and pull out the bone, which will be very easy to do at this point. Discard the bone, along with any surface fat. Use two forks to pull the meat into shreds. Toss it with one of the sauces.
You can serve _Slow Cooker Pulled Pork_ in one of a few ways: You can wrap it in low-carb tortillas, serve it on low-carb buns, or my favorite, serve it on a bed of coleslaw.
_**Yield:**_ 6 servings, each with: 405 calories, 31 g fat, 29 g protein, trace carbohydrate, trace dietary fiber, 0 g usable carbs.
### Hot Asian Ribs
This is full-bodied Chinese flavor. If your family loves Chinese spareribs, you have to make this!
3 1/2 pounds (1.6 kg) country-style pork ribs
4 scallions, sliced
1/4 cup (60 ml) soy sauce
1/3 cup (8 g) Splenda
1 teaspoon blackstrap molasses
2 tablespoons (28 ml) white wine vinegar
2 teaspoons toasted sesame oil
2 teaspoons lemon juice
1/2 teaspoon hot sauce
1 clove garlic
1/2 teaspoon ground ginger
1/2 teaspoon chili powder
1/4 teaspoon red pepper flakes
6 teaspoons (30 g) _Hoisin Sauce_ (see recipe page 337)
Put the ribs in your slow cooker.
In a bowl, mix together the scallions, soy sauce, Splenda, molasses, vinegar, sesame oil, lemon juice, hot sauce, garlic, ginger, chili powder, red pepper flakes, and _Hoisin Sauce_. Pour the sauce over the ribs. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
_**Yield:**_ 6 servings, each with: 469 calories, 35 g fat, 32 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Chili-Apricot Glazed Ribs
As you'll probably notice, ribs are one of my favorite things to do in the slow cooker. That's partly because ribs are one of my favorite things, partly because they take really well to slow cooking, partly because they're often cheap, and partly because you can do so many great things with them. These are both spicy and fruity.
4 pounds (1.8 kg) pork spareribs
1 tablespoon (9 g) garlic powder
1 teaspoon dried thyme
1 tablespoon (7 g) paprika
1 tablespoon (15 g) erythritolor other sugar-free sweetener to equal 1 tablespoon (13 g) sugar
2 teaspoons salt or Vege-Sal
1/2 teaspoon cayenne
1 teaspoon pepper
1 tablespoon (9 g) dry mustard
1/3 cup (107 g) low-sugar apricot preserves
1/4 cup (60 ml) cider vinegar
2 tablespoons (28 ml) water
15 drops apricot nectar stevia drops
1 teaspoon red pepper flakes
1 tablespoon (11 g) brown mustard
1/4 teaspoon salt or Vege-Sal
Cut your ribs into sections that fit in your slow cooker.
Mix together everything from the garlic powder through the dry mustard and sprinkle the ribs liberally all over with this mixture. If you have any left over, save it to season the ribs at the table later on. Throw your rub coated ribs in the slow cooker and set it on low. Let them cook for a good 6 to 7 hours.
When dinnertime rolls around, combine the rest of the ingredients in a small, nonreactive sauce pan and whisk them over low heat until you have a syrupy sauce.
Fish the ribs out of the slow cooker and lay them meaty side up on your broiler rack—you may have to do them in a couple of batches. Coat them with the sauce, run them under a high broiler just for five minutes or so, and then serve.
_**Note:**_ Those apricot nectar stevia drops are made by the Sweet Leaf company. It is available at health food stores or online.
_**Yield:**_ 6 servings, each with: 573 calories, 45 g fat, 33 g protein, 8 g carbohydrate, 1 g dietary fiber, 7 g usable carbs.
### Slow Cooker Pork Ribs Adobado
This is my favorite seasoning mixture for cooking ribs in the oven, so I thought it would be good in the slow cooker, too. It is!
3 pounds (1.4 kg) pork spareribs
2 to 3 tablespoons (28 to 45 ml) olive oil
2 teaspoons garlic powder
1 tablespoon (7 g) paprika
1 teaspoon ground cumin
1 teaspoon dried oregano
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
Cut your slab of ribs into sections that will fit in your slow cooker, stacked. Rub them with the oil, lay them on your broiler rack, and slide them under the broiler, about 4 inches (10 cm) from the heat. You're just browning them a little on both sides.
In the meantime, mix together everything else in a small dish.
When the ribs are brown, sprinkle them liberally all over with the rub. Stack them up in the slow cooker, slap on the lid, set it to low, and forget about them for 6 to 7 hours. That's it! Cut them into individual ribs and serve with plenty of napkins.
_**Yield:**_ 4 to 5 servings; Assuming 5, each will have: 565 calories, 48 g fat, 29 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Satay-Flavored Pork
I totally cribbed this combination of seasonings from a recipe for pork satay. It's exotic and great; if you like curry, you'll like this.
3 pounds (1.4 kg) pork country-style ribs
2 tablespoons (28 g) coconut oil
1 tablespoon (1.5 g) Splenda
1/4 teaspoon molasses
2 teaspoons fish sauce
1 tablespoon (15 ml) lemon juice
1/2 small onion, chopped
1 tablespoon (16 g) chili garlic paste
1 teaspoon ground turmeric
1 can (13 1/2 ounces, or 380 ml) coconut milk, unsweetened
1 teaspoon soy sauce
1 teaspoon chicken bouillon concentrate
In your big, heavy skillet, start the ribs browning a bit in the coconut oil.
While that's happening, mix together everything else, stirring it up well.
Transfer the browned ribs to the slow cooker and pour the sauce over them. Turn them over once or twice, using tongs, to coat well. Cover the pot, set to low, and let it go 6 to 7 hours.
You'll find your ribs have turned a lovely golden color. Serve with the sauce spooned over them. _Cauli-Rice_ (see recipe page 343) would be nice with this to soak up the extra sauce.
_**Yield:**_ 6 to 8 servings; Assuming 6, each will have: 478 calories, 33 g fat, 41 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Key West Ribs
Citrusy barbecue sauce gives this a Florida kind of taste!
3 pounds (1.4 kg) country-style pork ribs
1/4 cup (40 g) finely chopped onion
1/4 cup (65 g) low-carb barbecue sauce (recipe from page 343 or purchased)
1 teaspoon grated orange peel
1 teaspoon grated lemon rind
1/2 teaspoon salt
2 tablespoons (28 ml) white wine vinegar
2 tablespoons (28 ml) lemon juice
2 tablespoons (28 ml) lime juice
1 1/2 tablespoons (2 g) Splenda
1/8 teaspoon orange extract
2 tablespoons (28 ml) olive oil
In a big, heavy skillet, brown the ribs over medium-high heat. Transfer them to your slow cooker.
In a bowl, mix together the onion, barbecue sauce, orange peel, lemon rind, salt, vinegar, lemon juice, lime juice, Splenda, orange extract, and oil. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 7 to 9 hours.
Serve the ribs together with the sauce.
_**Yield:**_ 6 servings, each with: 421 calories, 33 g fat, 26 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Maple-Spice Country-Style Ribs
My pal Ray Stevens, who has tested many recipes for me, raves about this. It's shaping up to be the recipe by which all other recipes are judged!
3 pounds (1.4 kg) country-style pork ribs
1/2 cup (160 g) sugar-free pancake syrup
3 tablespoons (4.5 g) Splenda
2 tablespoons (28 ml) soy sauce
1/4 cup (40 g) chopped onion
1/2 teaspoon ground cinnamon
1/2 teaspoon ground ginger
1/2 teaspoon ground allspice
3 cloves garlic, crushed
1/4 teaspoon pepper
1/8 teaspoon cayenne
Put the ribs in your slow cooker.
In a bowl, mix together the syrup, Splenda, soy sauce, onion, cinnamon, ginger, allspice, garlic, pepper, and cayenne. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 9 hours.
_**Yield:**_ 6 servings, each with: 382 calories, 29 g fat, 27 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Maple Chili Ribs
This barbecue sauce is so good, you'll make it for cooking ribs on the grill, too. Feel free to double the sauce and do a full slab.
3 pounds (1.4 kg) pork spareribs (about half a slab)
Salt and pepper
2 tablespoons (30 g) _Dana's No-Sugar Ketchup_ (see recipe page 332.)
2 chipotle chiles canned in adobo, plus 2 teaspoons of the sauce
1 teaspoon onion powder
1/4 cup (80 g) sugar-free pancake syrup
2 teaspoons Worcestershire sauce
2 tablespoons (28 ml) soy sauce
1 1/2 teaspoons dry mustard
1/2 teaspoon chili powder
Cut the ribs into lengths that will fit, stacked, in your slow cooker. Salt and pepper them and put them on your broiler rack. Give them about 12 minutes per side 4 to 5 inches (10 to 13 cm) from the broiler, set on high. In the meantime, do to the next step.
Mix together everything else in a small dish. You know to chop up the chipotle, right? And then wash your hands really well?
When the ribs are browned, place the first chunk in the bottom of the slow cooker and spread some of the sauce all over it, edge to edge. Repeat with the rest of the ribs. Reserve half of the sauce.
Cover the slow cooker, set it to low, and let it go 6 hours or so. Serve with the reserved sauce.
_**Yield:**_ 4 servings, each with: 629 calories, 50 g fat, 37 g protein, 4 g carbohydrate, 1 g dietary fiber, 3g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Orange-Glazed Country Ribs
Fruit flavors of all kinds bring out the best in pork.
2 pounds (900 g) boneless pork country-style ribs
1/2 small onion, sliced
1 clove garlic, crushed
1/2 tablespoon (7 g) butter
1/2 cup (75 g) finely chopped green bell pepper
1/2 clove garlic, crushed
3 tablespoons (60 g) low-sugar orange marmalade
1/4 cup (60 ml) lemon juice
1 tablespoon (1.5 g) Splenda
1/4 teaspoon orange extract
1 tablespoon (8 g) grated ginger root
Guar or xanthan
Spray your slow cooker with nonstick cooking spray. Add the ribs, onion, and garlic. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, melt the butter in a medium-size nonreactive saucepan. Add the pepper and garlic and sauté until it's just soft. Add the marmalade, lemon juice, Splenda, orange extract, and ginger, bring to a simmer, and let it cook for 5 minutes. Thicken the sauce a little with guar or xanthan.
Remove the ribs from the slow cooker and put them on a broiler rack. Brush them with the sauce and run them under the broiler, set on high, for 5 minutes to glaze. Serve with the rest of the sauce.
_**Yield:**_ 6 servings, each with: 304 calories, 20 g fat, 27 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Cranberry-Orange Ribs
You'll need cranberries for this, obviously, and they're only available for a few months in the fall. They freeze well, though, so grab several extra bags and throw them in the freezer. Or just make this in the fall.
2 1/2 pounds (1.1 kg) pork boneless country-style ribs
2 tablespoons (28 g) coconut oil
1/4 cup (80 g) low-sugar orange marmalade
1/4 cup (60 ml) rice vinegar
1/4 teaspoon orange extract
1/2 cup (12 g) Splenda or other sugar-free sweetener to equal 1/2 cup (100 g) sugar
1 tablespoon (11 g) brown mustard
2 cups (200 g) cranberries
1/2 onion, chopped
In your big, heavy skillet over medium-high heat, start browning the pork in the coconut oil.
In the meantime, mix together the orange marmalade, vinegar, orange extract, Splenda or other sweetener, and mustard.
Dump the cranberries in the slow cooker, add the onion, and pour the orange mixture over them. Give it all a stir.
When the ribs are browned all over, transfer them to the pot with tongs and lay them on top of the cranberries and onion. Cover the pot, set to low, and let it cook for 6 hours.
Serve the ribs with the sauce and cranberries spooned over them.
_**Yield:**_ 6 servings, each with: 395 calories, 29 g fat, 22 g protein, 11 g carbohydrate, 2 g dietary fiber, 9 g usable carbs.
### Ribs 'n' Kraut
If you like, you can make this with smoked sausage instead of the country-style ribs. If you do, read the labels to find the lowest carb smoked sausage.
2 pounds (900 g) country-style pork ribs
1 medium Granny Smith apple, diced
1/2 medium onion, sliced
16 ounces (455 g) sauerkraut, rinsed and drained
3 tablespoons (4.5 g) Splenda
1/2 teaspoon blackstrap molasses
1 teaspoon caraway seeds
1/4 cup (60 ml) dry white wine
Put the ribs, apple, and onion in your slow cooker. Cover with the sauerkraut.
In a bowl, stir together the Splenda, molasses, caraway seeds, and wine. Pour the mixture over the sauerkraut and ribs. Cover the slow cooker, set it to low, and let it cook for 8 hours.
_**Yield:**_ 6 servings, each with: 286 calories, 19 g fat, 18 g protein, 7 g carbohydrate, 3 g dietary fiber, 4 g usable carbs.
### Ribs with Apple Kraut
Even folks who aren't wild about sauerkraut may like this. With the apple and the other vegetables, there's a lot more than just sauerkraut going on here.
3 1/2 pounds (1.6 kg) country-style pork ribs
2 tablespoons (28 ml) oil
3/4 cup (75 g) diced cauliflower
3/4 cup (113 g) diced turnip
1 Granny Smith apple, cored and thinly sliced
2 carrots, sliced
1 medium onion, sliced
1 pound (455 g) sauerkraut, rinsed and drained
1/2 cup (120 ml) apple cider vinegar
3 tablespoons (4.5 g) Splenda
2 teaspoons caraway seeds
1/8 teaspoon ground cloves
Guar or xanthan
In a big, heavy skillet, brown the ribs in the oil over medium-high heat.
Put the cauliflower, turnip, apple, carrots, and onion in your slow cooker. Put the ribs and sauerkraut on top.
In a bowl, mix together the vinegar, Splenda, caraway seeds, and cloves. Pour the mixture over the ribs and sauerkraut. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
When the time's up, remove the ribs to a platter with tongs and scoop out the vegetables with a slotted spoon. Thicken the pot liquid with guar or xanthan, add salt and pepper to taste, and serve the sauce with the ribs and vegetables.
_**Yield:**_ 6 servings, each with: 529 calories, 39 g fat, 32 g protein, 13 g carbohydrate, 4 g dietary fiber, 9 g usable carbs.
### Hocks and Shanks with Cabbage and Apples
Ham hocks and ham shank slices (sometimes called smoked hocks) give you great ham flavor without having to wrestle with that whole ham. They're cheaper, too. And again, this kind of tough, bony meat really shines when you slow-cook it.
1 1/2 pounds (680 g) ham hocks
1 1/2 pounds (680 g) ham, shank half
(or 3 pounds [1.4 kg] of hocks or 3 pounds [1.4 kg] of shanks, whatever you can get)
1 medium head red cabbage
1/2 Granny smith apple
1/2 medium onion
1 1/2 cups (355 ml) chicken broth
1/2 cup (120 ml) apple cider vinegar
4 teaspoons (2 g) Splenda or the equivalent in other sugar-free sweetener
1 tablespoon (15 g) bacon grease, melted
Throw your hocks, shanks, or both into your slow cooker. Coarsely chop your cabbage and dice your apple and onion. Throw them in, too.
Mix together everything else and then pour it over the meat and vegetables. Cover your slow cooker, set it on low, and let the whole thing cook for a good 6 to 7 hours; then serve.
_**Note:**_ Use smoked hocks or shanks, whichever you can get—I used both. The shanks have more meat.
_**Yield:**_ 6 servings, each with: 624 calories, 44 g fat, 44 g protein, 12 g carbohydrate, 3 g dietary fiber, 9 g usable carbs.
### Havana Ribs
The inspiration for these ribs was a recipe for a Cuban pork roast, marinated in citrus juice and rum. I took a lot of liberties, but the results are good, so why not?
3 pounds (1.4 kg) pork spareribs
Salt and pepper
1/2 orange
2 tablespoons (28 ml) lime juice
2 tablespoons (28 ml) lemon juice
2 tablespoons (28 ml) dark rum
5 cloves (15 g) garlic, crushed
1 teaspoon cumin
1 teaspoon oregano
1/2 teaspoon pepper
4 teaspoons (1 g), or 2 packets, Splenda, or other low-carb sweetener (not aspartame) to equal 4 teaspoons sugar in sweetness
2 tablespoons (28 ml) chicken broth
2 tablespoons (38 g) low-sugar orange marmalade
Cut your ribs into sections that fit in your slow cooker, if you need to. Salt and pepper 'em all over. Plunk them in the cooker.
Grate the zest from your orange half, and squeeze the juice into a bowl. Add everything else, and stir it up well. Pour it over the ribs, and turn them all about, making sure they get coated with the citrus-rum mixture. Cover, and cook on low for 5 hours.
When time's up, use a tongs to pull your ribs out and throw them on a platter. Keep them warm while you . . .
Pour the liquid from the slow cooker into a small, non-reactive saucepan, and boil it down untill it's syrupy. If you like, you can also slide your ribs under the broiler to brown them a little while you're boiling down the sauce, but it's not essential.
Coat the ribs with the sauce, and serve.
_**Yield:**_ 4 Servings, each with 654 calories, 50 g fat, 37 g protein, 8 g carbohydrate, 1 g dietary fiber, 7 grams usable carb
_**Note:**_ I have long used Smucker's brand low-sugar preserves in my recipes, but I suggest you read the nutrition labels—if you can find something with fewer carbs, go for it. Don't use anything aspartame-sweetened, however; aspartame breaks down with prolonged heat exposure.
### Polynesian Pork Ribs
It's a luau in your slow cooker!
2 pounds (900 g) country-style pork ribs
1 clove garlic, crushed
1/2 medium onion, sliced
1/4 cup (60 g) canned crushed pineapple in juice
1/2 cup (120 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
1 tablespoon (1.5 g) Splenda
1/2 teaspoon blackstrap molasses
1 tablespoon (15 ml) soy sauce
1 teaspoon grated ginger root
1/2 teaspoon dark sesame oil
1 tablespoon (15 ml) cider vinegar
Put the ribs in your slow cooker, along with the garlic and onion.
In a bowl, stir together the pineapple and 1/4 cup (60 g) of the ketchup. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
When the time's up, pull the ribs out of the slow cooker, put them on a broiler rack, and stash them somewhere warm.
Ladle the cooking liquid into a saucepan. Stir in the remaining 1/4 (60 g) cup ketchup, the Splenda, molasses, soy sauce, ginger, sesame oil, and vinegar, bring to a simmer, and let cook until you have a passably thick sauce.
Spoon the sauce over the ribs and run them under the broiler for 5 minutes or so, just to glaze them a bit.
_**Yield:**_ 6 servings, each with: 288 calories, 20 g fat, 19 g protein, 9 g carbohydrate, 1 g dietary fiber, 8 g usable carbs.
### Soy and Sesame Ribs
Here's another Asian take on ribs. The toasted sesame seed topping really sets this recipe apart.
3 pounds (1.4 kg) pork spareribs
1/3 cup (8 g) Splenda
1/4 cup (60 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or prepared lowcarb ketchup
1 tablespoon (21 g) sugar-free imitation honey
1 tablespoon (28 ml) cider vinegar
1 clove garlic, crushed
1/2 teaspoon ground ginger
1/2 teaspoon red pepper flakes
1/2 teaspoon dark sesame oil
1 tablespoon (8 g) sesame seeds
4 scallions, thinly sliced
Cut the ribs into portions if needed to fit them in your slow cooker. Broil the ribs about 6-inches (15 cm) from high heat until browned, about 10 minutes per side. Transfer them to your slow cooker.
In a bowl, mix together the Splenda, ketchup, honey, vinegar, garlic, ginger, red pepper flakes, and sesame oil. Pour the mixture over the ribs, turning to coat if needed. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
Toast the sesame seeds by stirring them in a dry skillet over medium-high heat until they start to make popping sounds and jump around a bit. Serve the ribs with sesame seeds and scallions scattered over them.
_**Yield:**_ 4 servings, each with: 646 calories, 52 g fat, 37 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs. (Analysis does not include the polyols in the imitation honey.)
### Teriyaki-Tangerine Ribs
This is an easy twist on plain old teriyaki.
4 pounds (900 g) country-style pork ribs
_Low-Carb Teriyaki Sauce_ (see recipe page 337)
_Florida Sunshine Tangerine Barbecue Sauce_ (see recipe page 334)
Put the ribs in your slow cooker.
In a bowl, mix together the _Low-Carb Teriyaki Sauce_ and the _Florida Sunshine Tangerine Barbecue Sauce_. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, remove the ribs. Transfer the sauce to a nonreactive saucepan and put it over high heat. Boil it hard until it thickens up a bit and serve the sauce over the ribs.
_**Yield:**_ 8 servings, each with: 415 calories, 29 g fat, 28 g protein, 8 g carbohydrate, 1 g dietary fiber, 7 g usable carbs.
### Slow Cooker Teriyaki Ribs
This dish is sweet, spicy, and tangy and falling-off-the-bone tender.
6 pounds (2.7 kg) pork spareribs, cut into 3 or 4 pieces so they fit in the slow cooker
3/4 cup (180 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
1 batch _Low-Carb Teriyaki Sauce_ (see recipe page 337)
1/4 cup (6 g) Splenda
1/4 teaspoon blackstrap molasses
1 teaspoon minced garlic or 2 cloves garlic, crushed
Guar or xanthan
Place the ribs in your slow cooker.
In a bowl, mix the ketchup, _Low-Carb Teriyaki Sauce_ , Splenda, molasses, and garlic together. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 10 hours.
When the time's up, use tongs to pull out the now unbelievably tender and flavorful ribs. Ladle out as much of the pot liquid as you think you'll use and thicken it using guar or xanthan. Serve the sauce over the ribs.
_**Yield:**_ 8 servings, each with: 650 calories, 50 g fat, 38 g protein, 9 g carbohydrate, 1 g dietary fiber (depending on how much of the liquid you eat), about 8 g usable carbs.
### Fruity, Spicy Ribs
This recipe is a little hot, a little sweet, a little Southwestern, and a little Asian—but all tasty.
6 pounds (2.7 kg) pork spareribs, cut in pieces so they fit in your slow cooker
6 tablespoons (120 g) low-sugar apricot preserves
1/3 cup (80 ml) lemon juice
2 tablespoons (3 g) Splenda
2 tablespoons (16 g) chili powder
2 teaspoons five-spice powder
1/4 cup (60 ml) soy sauce
1/2 cup (120 ml) chicken broth
Put the ribs on a broiler rack and broil them about 6 inches (15 cm) from high heat for about 7 to 8 minutes per side or until browned. Transfer the ribs to your slow cooker.
In a bowl, mix together the preserves, lemon juice, Splenda, chili powder, five-spice powder, soy sauce, and broth. Pour the mixture over the ribs. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, remove the ribs to a platter. Pour off the liquid in the pot into a deep, clear container and let the fat rise to the top. Skim off the fat. Now pour the liquid into a saucepan. Boil it hard until it's reduced by at least half and starting to thicken. Serve the sauce with the ribs.
_**Yield:**_ 8 servings, each with: 636 calories, 50 g fat, 37 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Rosemary-Ginger Ribs with Apricot Glaze
This recipe originally appeared in _15-Minute Low-Carb Recipes_. Feel free to use a full-size slab of ribs—about 6 pounds (2.7 kilograms) worth—and double the seasonings if you're feeding a family.
1 slab baby back ribs, about 2 1/2 pounds (about 1 kg)
Rosemary-ginger rub (I use a brand called Stubb's.)
2 tablespoons (40 g) low-sugar apricot preserves
1 1/2 teaspoons spicy brown mustard
1 teaspoon Splenda
1 1/2 teaspoons soy sauce
Sprinkle both sides of the ribs generously with the rosemary-ginger rub. Curl the slab of ribs around and fit it down into your slow cooker. Cover the slow cooker, set it to low, and let it cook for 9 to 10 hours. (No, I didn't forget anything. You don't put any liquid in the slow cooker. Don't sweat it.)
When the time's up, mix together the preserves, mustard, Splenda, and soy sauce. Carefully remove the ribs from the slow cooker. (They may fall apart on you a bit because they'll be so tender.) Arrange the ribs meaty-side-up on a broiler rack. Spread the apricot glaze evenly over the ribs and run them under a broiler set on high, 3 to 4 inches (8 to 10 centimeters) from the heat, for 7 to 8 minutes.
_**Yield:**_ 3 servings, each with: 689 calories, 56 g fat, 40 g protein, 4 g carbohydrate, trace dietary fiber, 4 g usable carbs.
### Slow Cooker "Barbecued" Ribs
Okay, it's not really barbecue because it's not done over a fire. But this recipe tastes great and lets you dig into your ribs within minutes of walking in the door.
2 1/2 pounds (1.1 kg) pork spare ribs
2 tablespons (12 g) _Classic Rub_ (see recipe page 338) or purchased dry rub
1/3 cup (85 g) _Dana's "Kansas City" Barbecue Sauce_ (see recipe page 335) or purchased low-carb barbecue sauce
Sprinkle the slab of ribs liberally on both sides with the dry rub, coil the ribs up, and slide them into your slow cooker. Cover the slow cooker, set it to low, and let it cook for 9 to 10 hours.
When the time's up, pull the ribs out of the slow cooker. (Do this carefully because they'll be falling-apart tender.) Lay the ribs on a broiler rack, meaty-side-up, and spread the barbecue sauce over them. Broil 3 to 4 inches (8 to 10 centimeters) from the broiler set on high for 7 to 8 minutes.
_**Note:**_ If you'd like to give these a smoked flavor, you can buy liquid smoke flavoring at your grocery store. Simply brush the ribs with the liquid smoke before you sprinkle on the dry rub.
_**Yield:**_ 3 servings, each with: 688 calories, 56 g fat, 40 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs. (Your carb count will be a bit different depending on whether you use homemade sugar-free barbecue sauce or commercial low-carb sauce.)
### About Pork Neckbones
Unless you grew up on soul food, you may never have tried pork neckbones. They're another one of those cuts that are perfect for the slow cooker. They're bony and tough and cheap. My grocery store has them for 59 cents a pound week in and week out. Yet cooked with slow moist heat, they're incredibly flavorful, and because the meat falls right off the bone, who cares that they're bony?
I did have one teeny problem with pork neckbones: I simply could not find any nutritional statistics for them, and I even wrote to a big pork producer! However, you can count on them being carb-free, so these carb counts are accurate. It's the protein and calorie counts that I couldn't get, so that's why they're missing in these neckbone recipes.
### Stewed Pork Neckbones with Turnips and Cabbage
This one-pot meal is not a beautiful dish to look at, but boy, does it taste good! Plenty of Tabasco is essential.
3 turnips, diced
3 pounds (1.4 kg) meaty pork neckbones
1 teaspoon red pepper flakes
1 1/2 teaspoons salt or Vege-Sal
3 cups (700 ml) water
1/2 head cabbage, cut in wedges
Tabasco sauce
Put the turnips in the bottom of your slow cooker. Put the neckbones on top of them. Sprinkle the red pepper flakes and salt or Vege-Sal over it and then pour the water over that. Now arrange the cabbage wedges on top of that. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
Scoop everything out onto a platter together with a slotted spoon and dose it well with Tabasco sauce before serving.
_**Yield:**_ 4 servings, each with: 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Neckbones and "Rice"
I adapted this recipe from one on a soul food website. I have no experience with genuine soul food, so I can't tell you how close this comes, but it's great in its own right.
2 1/2 pounds (1.1 kg) meaty pork neckbones
1/2 cup (120 ml) oil
1 medium onion, sliced
1 tablespoon (9 g) garlic powder
1 teaspoon salt or Vege-Sal
1 teaspoon pepper
2 cups (475 ml) chicken broth
1/2 head cauliflower
Guar or xanthan
In a big, heavy skillet, brown the pork neckbones in the oil over medium-high heat, in batches. Transfer the neckbones to your slow cooker.
Add the onion and sprinkle the garlic powder, salt or Vege-Sal, and pepper over the whole thing. Pour in the broth and give the whole thing a stir. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, run the cauliflower through the shredding blade of a food processor. Put the resulting _Cauli-Rice_ in a microwaveable casserole dish with a lid. Add a couple of tablespoons of water (28 ml), cover, and microwave on high for 7 minutes.
Meanwhile, remove the neckbones to a platter. Thicken the liquid a little with guar or xanthan. Serve the neckbones, onions, and gravy over the Cauli-Rice.
_**Yield:**_ 3 servings, each with: 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Italian Neck Bones
Pork neck bones are so wonderfully succulent and flavorful. Traditionally poor people's food, they're better than most of the expensive cuts. These are particularly sumptuous.
4 pounds (1.8 kg) pork neck bones
1 large onion, diced
5 cloves garlic, crushed
1/2 cup (120 ml) olive oil, or as needed
1 can (14 1/2 ounces, or 410 g) diced tomatoes
2 tablespoons (28 ml) balsamic vinegar
1/4 cup (60 ml) dry red wine
1/4 teaspoon pepper
1/4 teaspoon red pepper flakes
1 teaspoon chicken bouillon concentrate
1 tablespoon (6 g) Italian seasoning
Lay your neck bones on your broiler rack and slide them under a high broiler, 4 to 5 inches (10 to 13 cm) from the heat, to brown a bit—maybe 5 minutes per side.
In the meantime, chop your onion—what the heck? In your big, heavy skillet, start the onions and garlic sautéing in all that nice olive oil.
Use tongs to transfer your neckbones to the slow cooker. When the onion is translucent, add it to the pot. Put the skillet back over the heat. Dump in the tomatoes, vinegar, wine, pepper, bouillon concentrate, and Italian seasoning. Stir it all around, scraping up any nice browned bits, and keep stirring until the bouillon is dissolved. Pour it all over the neckbones.
Cover the pot, set it for low, and let it go for a good 7 to 8 hours until it is falling off the bone succulent! Serve with the pot liquor spooned over the neck bones, with plenty of napkins.
_**Yield:**_ 5 servings, each with: 238 calories, 22 g fat, 1 g protein, 9 g carbohydrate, 1 g dietary fiber, 8 g usable carbs.
### Cocido de Puerco
This pork stew appeared in _500 Low-Carb Recipes_ , but I hadn't thought of slow cooking it yet. It worked out great!
3 pounds (1.4 kg) pork neckbones
2 tablespoons (28 ml) olive oil
1 small onion, chopped
1 clove garlic, crushed
1 green bell pepper, diced
2 medium zucchini, chunked
1 can (14 1/2 ounces, or 410 g) diced tomatoes
2 teaspoons cumin
2 teaspoons dried oregano
1/2 teaspoon red pepper flakes
1 large onion, sliced
In a big, heavy skillet, brown the pork neckbones in the oil.
While that's happening, put the onion and garlic in the bottom of your slow cooker. When the neckbones are browned, put them in the slow cooker on top of the onions. Put the pepper and zucchini on top.
In a bowl, stir together the tomatoes, cumin, oregano, and red pepper flakes. Pour the mixture into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
_**Yield:**_ 5 servings, each with: 12 g carbohydrate, 2 g dietary fiber, 10 g usable carbs.
### Stuffed Peppers
6 large green peppers
1 1/2 pounds (680 g) bulk Italian sausage
1/2 head cauliflower
2 cups (490 g) no-sugar-added spaghetti sauce
2/3 cup (100 g) crumbled feta cheese
1/2 cup (80 g) chopped onion
1/4 cup (90 g) chopped tomato
1/4 cup (15 g) chopped fresh parsley
2 tablespoons (12 g) chopped black olives 1 clove garlic, crushed
1/2 teaspoon salt or Vege-Sal
1 teaspoon Italian seasoning
1/2 teaspoon red pepper flakes
Cut the tops off the peppers. Remove the usable pepper wall from the stems and chop it, discarding the stems and seeds. Reserve the pepper shells.
In a big, heavy skillet, brown and crumble the sausage till done. Drain the fat.
Run the cauliflower through the shredding blade of a food processor. Place the resulting _Cauli-Rice_ in a big mixing bowl.
Put 1 cup (245 g) of the spaghetti sauce in the slow cooker. Add the rest to the _Cauli-Rice_. Add the cheese, onion, tomato, parsley, olives, garlic, salt or Vege-Sal, Italian seasoning, red pepper flakes, cooked sausage, and the chopped bit of green pepper to the _Cauli-Rice_. Combine well. Divide the mixture between the pepper shells.
Put the stuffed peppers in the slow cooker. Cover the slow cooker, set it to low, and let it cook for 4 to 5 hours or until the peppers are tender.
_**Yield:**_ 6 servings, each with: 592 calories, 51 g fat, 18 g protein, 18 g carbohydrate, 5 g dietary fiber, 13 g usable carbs.
### Tangy Pork Chops
4 pounds (1.8 kg) pork chops, 1/2-inch (13 mm) thick
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
1/2 cup (80 g) chopped onion
2 stalks celery, diced
1 green bell pepper, diced
1 can (14 1/2 ounces, or 410 g) diced tomatoes
1/2 cup (120 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
2 tablespoons (28 ml) cider vinegar
2 tablespoons (28 ml) Worcestershire sauce
2 tablespoons (3 g) Splenda
1/4 teaspoon blackstrap molasses
1 tablespoon (15 ml) lemon juice
1 teaspoon beef bouillon concentrate
Guar or xanthan
Place the pork in the bottom of your slow cooker. Sprinkle the pork with the salt and pepper. Now add the onion, celery, green pepper, and tomatoes.
In a bowl, stir together the ketchup, vinegar, Worcestershire sauce, Splenda, molasses, lemon juice, and bouillon. Stir until the bouillon dissolves. Pour the mixture into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
When the time's up, remove the pork and place it on a platter. Use a slotted spoon to pile the vegetables around the pork. Thicken the liquid left in the slow cooker with guar or xanthan and serve the sauce with the pork and vegetables.
_**Yield:**_ 8 servings, each with: 396 calories, 23 g fat, 36 g protein, 11 g carbohydrate, 1 g dietary fiber, 10 g usable carbs.
### Onion-Mustard Pork Chops
2 pounds (900 g) pork chops
1 tablespoon (15 ml) olive oil
1 medium onion, thinly sliced
4 cloves garlic, crushed
1 teaspoon dry mustard
1/2 teaspoon salt or Vege-Sal
1 teaspoon pepper
1 dash hot sauce
1 tablespoon (15 ml) cider vinegar
1/2 cup (120 ml) dry white wine
2 tablespoons (22 g) brown mustard
1/2 cup (120 ml) heavy cream
Guar or xanthan
In a big, heavy skillet, brown the pork on both sides in the oil over medium-high heat. Transfer the pork to your slow cooker.
Turn the heat down to medium-low and add the onion to the skillet. Sauté it until it's translucent. Then stir in the garlic, dry mustard, salt or Vege-Sal, pepper, and hot sauce and let the whole thing sauté for another minute or so. Transfer the mixture to the slow cooker on top of the pork.
In a bowl, stir together the vinegar and wine. Pour the mixture into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, remove the pork to a platter. Stir the mustard and cream into the juice in the pot. Thicken the juice a tad with guar or xanthan and serve the sauce over the pork.
_**Yield:**_ 6 servings, each with: 352 calories, 25 g fat, 25 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Choucroute Garni
This is a streamlined version of a traditional dish from the Alsace region of France. The name means Garnished Sauerkraut. It's so simple and so good, especially on a cold night.
1 can (14 ounces, or 400 g) sauerkraut, rinsed and drained
1 tablespoon (15 g) bacon grease
1/4 cup (60 ml) apple cider vinegar
1 tablespoon (1.5 g) Splenda
1/2 medium onion, thinly sliced
2 tablespoons (28 ml) gin
1/4 cup (60 ml) dry white wine
1 pound (455 g) meat (Choose any combination of kielbasa, smoked sausage, frankfurters, link sausages, 1/4-inch-thick (6 mm) ham slices, or smoked pork chops.*)
Place the sauerkraut in your slow cooker. Add the bacon grease, vinegar, Splenda, onion, gin, and wine and give it a quick stir. Place the meat on top. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
* I use 1/2 pound (225 g) each of the lowest carbohydrate kielbasa and smoked sausage I can find.
_**Note:**_ This doesn't even start to fill my slow cooker, so feel free to double or even triple this recipe. If you increase it, I suggest arranging it with a layer of kraut, a layer of meat, a layer of kraut, and so on. And of course, you'll have to increase the cooking time by an hour, maybe two, depending on how many extra layers you use.
_**Yield:**_ 3 servings, each with: 112 calories, 5 g fat, 1 g protein, 9 g carbohydrate, 4 g dietary fiber, 5 g usable carbs. (This will depend on which meats you use.)
chapter six
Slow Cooker Lamb
You'll notice a certain reliance on lamb shanks here. Lamb shanks are the bottom-most part of a leg of lamb, and they're ideal for slow cooking for a couple reasons. They fit neatly in the pot, and they're the sort of tough, flavorful meat that really shines with slow, moist cooking.
If you're having trouble finding lamb shanks at your grocery store, ask the nice meat guys. Or you could make any of the shank recipes with a chunk of lamb leg or shoulder of the right weight. Have the meat guys cut it into a few pieces, though, so it'll fit in your pot and won't take far more cooking time than the recipe specifies. Most grocery stores will cut up a roast for you for no added charge.
### Lamb Shanks in Red Wine
This is a hearty one-pot meal.
5 pounds (2.3 kg) lamb shank (4 shanks)
1/4 cup (60 ml) olive oil
2 stalks celery, sliced 1/2-inch (13 mm) thick
2 carrots, sliced 1/2-inch (13 mm) thick
8 cloves garlic, crushed
1/2 onion, chunked
8 ounces (225 g) sliced mushrooms
1 cup (235 ml) chicken broth
1 cup (235 ml) dry red wine
1 teaspoon beef bouillon concentrate
2 teaspoons pepper
1/2 teaspoon ground rosemary
2 bay leaves
Guar or xanthan
In a big, heavy skillet, sear the lamb all over in the oil.
Place the celery, carrots, garlic, onion, and mushrooms in your slow cooker.
When the lamb is browned all over, transfer it to the slow cooker, on top of the veggies.
In a bowl, stir together the broth, wine, bouillon, pepper, and rosemary. Pour the mixture over the lamb. Add the bay leaves. (Make sure they land in the liquid!) Cover the slow cooker, turn it to low, and let it cook for 6 hours.
When the time's up, remove the lamb to serving plates. Remove the bay leaves. Using guar or xanthan, thicken the liquid in the slow cooker to the consistency of heavy cream. Ladle the sauce and vegetables over the lamb.
_**Yield:**_ 6 servings, each with: 757 calories, 50 g fat, 59 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Lemon Lamb Shanks
Lemon brings out the best in lamb!
4 pounds (1.8 kg) lamb shank
2 tablespoons (28 ml) olive oil
1 teaspoon lemon pepper
1/2 teaspoon dry mustard
1/2 cup (120 ml) chicken broth
1 teaspoon beef bouillon concentrate
1/2 teaspoon grated lemon peel
2 tablespoons (28 ml) lemon juice
1 teaspoon dried rosemary
2 cloves garlic, crushed
Guar or xanthan
Sear the lamb all over in the oil. Place the lamb in your slow cooker.
In a bowl, mix together the lemon pepper and dry mustard. Sprinkle the mixture evenly over the lamb.
In the same bowl, mix together the broth, bouillon, lemon peel, lemon juice, rosemary, and garlic. Pour the mixture over the lamb. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, remove the lamb and thicken up the liquid in the slow cooker a bit with guar or xanthan.
Serve this dish with a salad with plenty of cucumbers and tomatoes!
_**Yield:**_ 6 servings, each with: 535 calories, 37 g fat, 46 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Kashmiri Lamb Shanks
This was originally a recipe for a skillet curry of lamb, but it works wonderfully in the slow cooker. If you like Indian food, you have to try this. And if you've never eaten Indian food, you need to start!
2 1/2 pounds (1.1 kg) lamb shank
2 tablespoons (28 ml) olive oil
1 cup (235 ml) chicken broth
1/2 teaspoon beef bouillon concentrate
1 teaspoon _Garam Masala_ (see recipe page 340) or purchased garam masala
2 teaspoons ground coriander
1 tablespoon (8 g) grated ginger root
1/4 teaspoon cayenne
Guar or xanthan
In a big, heavy skillet, sear the lamb all over in the oil over medium-high heat. Transfer the lamb to your slow cooker.
In a bowl, mix together the broth, bouillon, garam masala, coriander, ginger, and cayenne. Pour the mixture over the lamb. Cover the slow cooker, set it to low, and let it cook for 8 hours.
Remove the lamb to a platter, thicken the sauce a bit with guar or xanthan, and serve the sauce over the lamb.
_**Yield:**_ 4 servings, each with: 530 calories, 38 g fat, 44 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Seriously Simple Lamb Shanks
Simple is good!
3 pounds (1.4 kg) lamb shank
2 tablespoons (28 ml) olive oil
1 cup (235 ml) chicken broth
1 teaspoon beef bouillon concentrate
2 teaspoons paprika
5 cloves garlic, crushed
Guar or xanthan
Season the lamb all over with salt and pepper. In a big, heavy skillet, over medium-high heat, sear the lamb in the oil until it's brown all over. Transfer the lamb to your slow cooker.
In a bowl, mix together the broth and bouillon. Pour the mixture over the lamb. Sprinkle the paprika and garlic over the lamb. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
Remove the lamb with tongs and put it on a serving plate. Pour the liquid in the slow cooker into a 2-cup (475 ml) glass measuring cup and let the fat rise to the top. Skim the fat off and discard and then thicken up the remaining liquid using guar or xanthan. Serve the sauce with the shanks.
Either _Cauli-Rice_ (see recipe page 343) or _Fauxtatoes_ (see recipe page 343) would be nice with this, but it's fine with just a simple salad or vegetable side.
_**Yield:**_ 4 servings, each with: 627 calories, 44 g fat, 52 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Lamb Shanks Osso Bucco Style
Osso Bucco is traditionally made with veal shank, but I like lamb better. These are wonderful.
2 slices bacon
2 tablespoons (28 g) bacon grease
4 1/2 pounds (2 kg) lamb shanks
1 medium onion
1 medium carrot
1/2 cup (120 ml) dry white wine
1 can (14 1/2 ounces, or 410 g) diced tomatoes
1 teaspoon dried rosemary
1 teaspoon dried oregano
1 teaspoon chicken bouillon concentrate
1 teaspoon beef bouillon concentrate
5 garlic cloves, crushed
1/2 teaspoon pepper
1 bay leaf, whole
In your big heavy skillet, cook your bacon crisp. (Alternately, you can microwave your bacon and use bacon grease you've previously saved to brown the meat.) Remove the bacon from the pan and reserve. Add extra bacon grease to equal about 2 tablespoons (28 ml). Start browning the lamb shanks in it; you want them browned all over.
In the meantime, slice the onion and the carrot and put in your slow cooker. When your shanks are browned, use tongs to transfer them to the slow cooker.
Add the wine, tomatoes, rosemary, oregano, chicken bouillon concentrate, beef bouillon concentrate, garlic, and pepper to the skillet and stir it over the heat, deglazing the pan, until the boullion concentrates are dissolved. Pour over the shanks. Add the bay leaf.
Cover the pot, set to low, and let it cook for a good 7 to 8 hours. Remove the bay leaf. Thicken the sauce a little with your guar or xanthan shaker and serve.
_**Yield:**_ 6 servings, each with: 650 calories, 42 g fat, 52 g protein, 9 g carbohydrate, 1 g dietary fiber, 8 g usable carbs.
### Lamb Stew Provençal
I turned this recipe over to my sister to test. She's mad for French food, especially from Provence. She gave this the thumbs-up.
3 pounds (1.4 kg) lamb stew meat—shoulder is good, cubed. (Have the meat guys cut it off the bone.)
3 tablespoons (45 ml) olive oil
1 whole fennel bulb, sliced lengthwise
1 medium onion, sliced lengthwise
4 cloves garlic, crushed
1 bay leaf
1 teaspoon dried rosemary, whole needles
1 can (15 ounces, or 425 g) black soybeans, drained
1 cup (235 ml) beef broth
1 teaspoon chicken bouillon concentrate
1/2 teaspoon dried basil
1/2 teaspoon dried marjoram
1/2 teaspoon dried savory
1/2 teaspoon dried thyme
Guar or xanthan
Season the lamb with salt and pepper. In a big, heavy skillet, heat the oil and brown the lamb on all sides over medium-high heat.
Place the fennel, onion, and garlic in the bottom of your slow cooker. Add the bay leaf and rosemary. Dump the soybeans on top of that. When the lamb is browned, put it on top of the vegetables.
In a bowl, stir together the broth, bouillon, basil, marjoram, savory, and thyme. Pour the mixture over the lamb. Cover the slow cooker, set it to low, and let it cook for 8 to 9 hours.
When it's done, thicken the liquid to the texture of heavy cream with guar or xanthan.
_**Yield:**_ 8 servings, each with: 348 calories, 17 g fat, 41 g protein, 8 g carbohydrate, 4 g dietary fiber, 4 g usable carbs.
### Braised Leg of Lamb
This is a great company dish, and it feeds a crew.
5 pounds (2.3 kg) leg of lamb
Olive oil—a few tablespoons (45 to 60 ml) or as needed
1 large onion, sliced
1 carrot, shredded
1 bay leaf
1 cup (235 ml) beef broth
1/2 cup (120 ml) dry red wine
6 cloves garlic, crushed
1 teaspoon chicken bouillon concentrate
1/2 teaspoon pepper
1 1/2 tablespoons (5 g) dried rosemary, or several sprigs fresh
Salt and pepper to taste
In your big heavy skillet, over medium-high heat, start searing the lamb in the olive oil; you want it brown all over.
In the meantime, slice your onion and put half of it in the bottom of the slow cooker. Shred the carrot and put all of it in there. Add the bay leaf, too.
When the lamb is browned on all sides, transfer it to the slow cooker and put it on top of the onion and carrot. Now mix together everything from the broth through the rosemary and stir until the bouillon concentrate is dissolved. Pour this over the roast. Make sure at least half of the crushed garlic winds up in the bottom of the pot, but you do want some on top of the roast, too.
Scatter the remaining onion over the roast. Cover the pot, set to low, and cook for a good 6 to 7 hours.
When it's done, fish out the roast and put it on a platter. Use a slotted spoon to pile the cooked onion from the bottom of the pot on top. Thicken the pot juices with your guar or xanthan shaker, salt and pepper to taste, and serve as gravy.
_**Yield:**_ 8 servings or more; Assuming 8, each will have: 549 calories, 38 g fat, 42 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Caribbean Slow Cooker Lamb
Lamb and goat are very popular in the Caribbean, and this is my slow cooker interpretation of a Caribbean lamb dish. Look for tamarind concentrate in a grocery store with a good international section. I found it in a medium-size town in southern Indiana, so you may well find it near you! If you can't find it, you could use a tablespoon (15 ml) of lemon juice and a teaspoon of Splenda instead. Your lamb will be less authentically Caribbean-tasting, but still yummy.
2- to 3-pound section (0.9 to 1.4 kg) of a leg of lamb
1/2 medium onion, chopped
1/2 teaspoon minced garlic or 1 clove garlic, crushed
1 teaspoon tamarind concentrate
1 tablespoon (11 g) spicy brown mustard
1 cup (180 g) canned diced tomatoes
1 teaspoon hot sauce—preferably Caribbean Scotch Bonnet sauce—or more or less to taste
Guar or xanthan (optional)
Place the lamb in your slow cooker.
In a bowl, stir together the onion, garlic, tamarind, mustard, tomatoes, and hot sauce. Pour the mixture over the lamb. Cover the slow cooker, set it to low, and let it cook for a good 8 hours.
When it's done, remove the lamb to a serving platter, thicken the pot juices with the guar or xanthan if it seems necessary, and add salt and pepper to taste.
_**Yield:**_ 6 servings, each with: 357 calories, 26 g fat, 27 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Greek Stuffed Peppers
This is really a one-dish meal since it has so many vegetables in it. Feel free to use this method of cooking with any stuffed pepper recipe.
4 large green bell peppers
1 small onion, chopped
1 1/4 (570 g) pounds ground lamb
1 cup (150 g) crumbled feta cheese
1 can (14 1/2 ounces, or 410 g) diced tomatoes, drained
2 cloves garlic, crushed
2 teaspoons dried oregano
1/4 cup (15 g) chopped fresh parsley
1 teaspoon salt, or to taste
1/2 teaspoon pepper
Whack the tops off the peppers. Scoop out and discard the cores. Trim the useable flesh from around the stems and throw it in your food processor with the **S** -blade in place.
Peel your onion and cut it in chunks; throw it in the food processor, too. Pulse to chop everything medium-fine.
Throw the chopped pepper and onion in a big mixing bowl. Add the lamb, feta, drained tomatoes, garlic, oregano, parsley, salt, and pepper and use clean hands to mash everything together really well.
Stuff this mixture into your peppers, mounding it a bit on top. Settle them down into your slow cooker, cover, and set to low. Cook for 5 to 6 hours and then serve. You can crumble a little extra feta over the tops, if you like, but it's not essential.
_**Yield:**_ 4 servings, each with: 575 calories, 42 g fat, 31 g protein, 19 g carbohydrate, 3 g dietary fiber, 16 g usable carbs.
chapter seven
Slow Cooker Fish and Seafood
I don't generally think of fish when I think of long, slow cooking, and indeed who wants to leave fish in a slow cooker all day? But the gentle, even heating of your slow cooker, used judiciously, can yield fish and seafood that's remarkably moist and tender. When you don't need to have dinner waiting, try these recipes!
Or as in a few of these recipes, you can cook everything else all day and then just stir in the seafood for the last little bit of cooking time. It doesn't take long to cook fish or seafood through!
### Lime-Basted Scallops
My seafood-loving husband thought these were some of the best scallops he'd ever had.
1/4 cup (60 ml) lime juice
3 tablespoons (45 g) butter
2 cloves garlic
24 ounces (680 g) sea scallops
Guar or xanthan
1/4 cup (4 g) chopped fresh cilantro
Put the lime juice, butter, and garlic in your slow cooker. Cover the slow cooker, set it to high, and let it cook for 30 minutes.
Uncover the slow cooker and stir the butter, lime juice, and garlic together. Now add the scallops, stirring them around to coat them with the sauce. Spread them in a single layer on the bottom of the slow cooker. (If the sauce seems to pool in one or two areas, try to cluster the scallops there. In my pot, the sauce liked to stay around the edges.) Re-cover the pot, set it to high, and let it cook for 45 minutes.
When the time's up, remove the scallops to serving plates. Thicken the pot liquid just a tiny bit with guar or xanthan and spoon the sauce over the scallops. Top each serving with a tablespoon of cilantro.
_**Yield:**_ 4 servings, each with: 233 calories, 10 g fat, 29 g protein, 6 g carbohydrate, trace dietary fiber, 6 g usable carbs.
### Scallops Florentine
This is a simple creamed spinach recipe with scallops. This would make a nice romantic dinner since it serves two.
10 ounces (280 g) frozen chopped spinach, thawed
1 clove garlic, crushed
1/4 cup (60 ml) heavy cream
1/4 cup (25 g) grated Parmesan cheese
1 tablespoon (7 g) Old Bay seafood seasoning
12 ounces (340 g) sea scallops
Dump your thawed spinach into a strainer and press it really well to get out all the water—you could even pick it up with clean hands and squeeze it hard. Just drain it really well. Then dump it in your slow cooker.
Add the garlic, cream, Parmesan, and Old Bay seasoning. Stir this mixture together well. Cover the pot, set to low, and let it cook for 30 to 45 minutes until the spinach mixture is getting hot.
Lay the scallops in a single layer on top of the spinach, re-cover the pot, and let the whole thing cook for an hour. Serve the scallops on a bed of the spinach.
_**Yield:**_ 2 servings, each with: 334 calories, 16 g fat, 37 g protein, 11 g carbohydrate, 4 g dietary fiber, 7 g usable carbs.
### Lemon-Mustard Salmon Steaks
This is so simple and classic. The salmon comes out tender and moist.
2 tablespoons (28 g) butter
1 tablespoon (15 ml) lemon juice
1 teaspoon Dijon mustard
1 pinch salt or Vege-Sal
2 salmon steaks (totaling about 1 pound, or 455 g)
2 tablespoons (8 g) chopped fresh parsley
Combine the butter, lemon juice, mustard, and salt or Vege-Sal in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 30 to 40 minutes. Stir together.
Now put the salmon steaks in the slow cooker and turn them once or twice to coat. Re-cover the slow cooker and let it cook for 1 hour. Spoon some of the pot liquid over the salmon and sprinkle with the parsley before serving.
_**Yield:**_ 2 servings, each with: 369 calories, 19 g fat, 46 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Orange Basil Salmon
Keith the Organic Gardening God next door always plants a ton of basil, and we get the overflow. Here's one way to use some of it!
1 orange
1/3 cup (80 ml) extra virgin olive oil
1/2 cup (20 g) minced fresh basil
24 ounces (680 g) salmon fillets
Salt and pepper
Grate the zest of your orange and reserve.
Pour the olive oil into your slow cooker. Now whack the orange in half and squeeze in the juice. (If you use a navel orange, you won't have to worry about pits.) Throw in the zest, too. Cover the pot, set to low, and let it heat for a half an hour.
When the half an hour is up, cut your salmon fillets into servings and salt and pepper them on both sides. Mince your basil. Open up your slow cooker and give the oil and orange a stir to make sure it's well-blended. Add the basil to the pot and then put in the salmon. Flip it over a few times to make sure the salmon's well-coated on both sides. Now lay it, skin-side up, flat on the bottom of the slow cooker in the olive oil/ orange mixture.
Re-cover the pot and let the salmon cook for 60 to 90 minutes. Serve with the pot juices spooned over it.
_**Yield:**_ 4 servings, each with: 373 calories, 24 g fat, 34 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Maple-Balsamic Salmon
This started as a recipe for grilled salmon in _The Low-Carb Barbecue Book_ , but it works brilliantly in the slow cooker, as well.
2 salmon steaks (totaling about 1 pound, or 455 g)
1 tablespoon (15 ml) olive oil
2 tablespoons (40 g) sugar-free pancake syrup
1 clove garlic, crushed
1 tablespoon (15 ml) balsamic vinegar
In your slow cooker, combine everything but the salmon. Cover the slow cooker, set it to low, and let it cook for 30 minutes.
Now add the salmon, turning the steaks to coat them with the sauce. Re-cover the pot and let it cook for 1 hour. Spoon some of the pot liquid over the steaks before serving.
_**Yield:**_ 2 servings, each with: 326 calories, 15 g fat, 45 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Balsamic-Vanilla Salmon
Cooking is a remarkably trend-driven business, and a recent fashion has been using vanilla in main-dish recipes. It works so well for chocolate, I thought I'd give it a try. It turned out great!
2 tablespoons (28 ml) extra virgin olive oil
1/4 teaspoon Splenda
1/2 teaspoon Worcestershire sauce
2 drops molasses
2 teaspoons balsamic vinegar
1/2 teaspoon spicy brown or dijon mustard
1/2 teaspoon vanilla extract
1 clove garlic, crushed
2 tablespoons (20 g) minced onion
2 tablespoons (6 g) minced fresh basil
2 tablespoons (8 g) minced fresh parsley
24 ounces (680 g) salmon fillets
Measure and whisk together everything from the olive oil through the garlic.
Place the onion, basil, and parsley in the bottom of your slow cooker and pour in the vinaigrette. Stir it all together a little.
Add your salmon fillets and flip them once or twice to coat. Arrange them skin-side up. Then cover the pot, set it to low, and let it cook for an hour.
Serve with the liquid and herbs from the pot spooned over the fish.
_**Yield:**_ 4 servings, each with: 264 calories, 13 g fat, 34 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Salmon in Sweet Chili Sauce
That _Not-Quite-Asian Sweet Chili Sauce_ is quick and easy; so don't hesitate!
1/2 batch _Not-Quite-Asian Sweet Chili Sauce_ (see recipe below.)
24 ounces (680 g) salmon fillets
Simply put the sauce in the slow cooker, lay the fillets in it, and flip them a couple of times to coat. Arrange them skin-side up, cover the pot, set to low, and let it cook for 60 to 90 minutes. That's it!
_**Yield:**_ 4 servings, each with: 222 calories, 6 g fat, 35 g protein, 3 g carbohydrate, trace dietary fiber, 2 g usable carbs. (Analysis does not include the polyols in the _Not-Quite-Asian Sweet Chili Sauce_.)
### Not-Quite-Asian Sweet Chili Sauce
This is great with fish, chicken, and pork—all sorts of stuff.
5 dried Thai bird peppers (the little skinny pointy hot peppers)
1/2 cup (160 g) sugar-free pancake syrup
1/2 cup (120 g) erythritol
3 tablespoons (45 ml) cider vinegar
1/4 cup (60 ml) dry sherry
1/4 cup (60 ml) soy sauce
1 tablespoon (28 g) fresh ginger root—a chunk a little bigger than a whole shell-on walnut
3 garlic cloves
1/4 teaspoon pepper
Just throw everything in your blender or food processor and run until the peppers, ginger, and garlic are pulverized. Keep in a screw-top jar in the fridge.
_**Yield:**_ 1 1/2 cups, or 24 servings of 1 tablespoon, each with: 8 calories, trace fat, trace protein, 1g carbohydrate, trace dietary fiber, 1 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Almond-Stuffed Flounder Rolls with Orange Butter Sauce
4 tablespoons (55 g) butter, divided
2 tablespoons (28 ml) lemon juice
1/8 teaspoon orange extract
1 teaspoon Splenda
1/3 cup (145 g) almonds
1/4 cup (40 g) minced onion
1 clove garlic, crushed
1 1/2 teaspoons Dijon mustard
1/2 teaspoon soy sauce
1/4 cup (15 g) minced fresh parsley, divided
1 pound (455 g) flounder fillets, 4 ounces (115 g) each
Put 2 tablespoons (28 grams) of the butter, the lemon juice, orange extract, and Splenda in your slow cooker. Cover the slow cooker, set it to low, and let it heat while you fix your flounder rolls.
Put the almonds in a food processor with the **S** -blade in place and grind them to a cornmeal consistency. Melt 1 tablespoon (14 grams) of butter in a medium-sized heavy skillet and add the ground almonds. Stir the almonds over medium heat for 5 to 7 minutes or until they smell toasty. Transfer them to a bowl.
Now melt the final tablespoon (14 g) of butter in the skillet and sauté the onion and garlic over medium-low heat until the onion is just turning translucent. Add them to the almonds and stir them in. Now stir in the mustard, soy sauce, and 2 tablespoons (8 grams) of the parsley.
Lay the flounder fillets on a big plate and divide the almond mixture between them. Spread it over the fillets and then roll each one up and fasten it with a toothpick.
Take the lid off the slow cooker and stir the sauce. Place the flounder rolls in the sauce and spoon the sauce over them. Re-cover the pot and let the rolls cook for 1 hour. When they're done, spoon the sauce over the rolls and sprinkle the remaining parsley over them to serve.
_**Yield:**_ 4 servings, each with: 285 calories, 19 g fat, 24 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### Bacon-Almond Flounder Rolls
6 tablespoons (85 g) butter, divided
2 tablespoons (28 g) bacon grease
6 flounder fillets
6 scallions
2 small celery ribs
1 cup (112 g) almond meal
6 slices bacon
1 teaspoon dried thyme
Salt and pepper
Paprika
Start 2 tablespoons (28 g) of the butter and the bacon grease heating in your slow cooker on high.
Lay your bacon on a microwave bacon rack or in a Pyrex pie plate and give them 6 to 8 minutes on high—you want it crisp but not burnt.
Trim the scallions and celery, throw them in your food processor, and chop them fairly fine.
Okay, melt another 2 tablespoons (28 g) of the butter in a small skillet over medium-low heat. Add the almond meal and stir until it smells toasty. Transfer to a plate.
Melt the last 2 tablespoons (28 ml) of butter and sauté the scallions and celery until softened. Add the almond meal back to the skillet and stir to combine. Now use your kitchen shears to snip in the bacon in little bits. Add the thyme, stir it up, and salt and pepper to taste.
Lay a fillet flat on a plate, put a couple tablespoons (28 g) of stuffing across it about 2 inches (5 cm) from one end, and wrap those 2 inches (5 cm) around it. Wrap the other end around. Carefully transfer to the slow cooker, placing it seam-side down. Repeat with the rest of the fish and stuffing.
Sprinkle the rolls lightly with paprika. Then cover the pot and let them cook an hour. Spoon some of the grease/liquid in the pot over each roll when you serve them and snip a few scallion tops over them to make them look pretty.
_**Yield:**_ 6 servings, each with: 427 calories, 25 g fat, 42 g protein, 8 g carbohydrate, 1 g dietary fiber, 7 g usable carbs.
### Super-Simple Catfish
Catfish is That Nice Boy I Married's favorite, but breading and frying it is a big production. This is a whole lot easier and tastes great. And there's no carbs at all!
2 tablespoons (28 g) bacon grease
2 pounds (900 g) catfish fillets
Creole seasoning (I used Tony Cachere's.)
This is so easy, but my husband loved it! Turn on your slow cooker to low and throw in the bacon grease. Let it heat for 30 minutes.
If you need to, cut your catfish into pieces that will fit flat in the bottom of the slow cooker. Sprinkle it liberally on both sides with the Creole seasoning. Then lay it in the bacon grease, flipping it once to coat. Cover the pot and let it cook for 45 minutes to an hour and then serve.
_**Yield:**_ 4 servings, each with: 275 calories, 13 g fat, 37 g protein, 0 g carbohydrate, 0 g dietary fiber, 0 g usable carbs.
### Sweet and Sour Shrimp
Adding the shrimp and snow peas at the last moment keeps them from becoming desperately overcooked.
1 cup (170 g) peaches, peeled and cubed (Frozen unsweetened peaches work well. Just cut them into smaller chunks—1/2 inch [13 mm].)
1/2 cup (80 g) chopped onion
1 green bell pepper, diced
1/2 cup (50 g) chopped celery
1/2 cup (120 ml) chicken broth
2 tablespoons (28 ml) dark sesame oil
1/4 cup (60 ml) soy sauce
2 tablespoons (28 ml) rice vinegar
1/4 cup (60 ml) lemon juice
1 teaspoon red pepper flakes
1 tablespoon (1.5 g) Splenda
6 ounces (170 g) fresh snow pea pods, trimmed
1 1/2 pounds (700 g) shrimp, shells removed
1/3 cup (37 g) slivered almonds, toasted
Guar or xanthan
Put the peaches, onion, pepper, celery, broth, sesame oil, soy sauce, vinegar, lemon juice, red pepper flakes, and Splenda in your slow cooker and stir them together. Cover the slow cooker, set it to low, and let it cook for 4 hours. (Or you could cook it on high for 2 hours.)
When the time's up, turn the pot up to high while you trim the snow peas and cut them in 1-inch (2.5 cm) lengths. Stir them in and let it cook for 15 to 20 minutes. Now stir in the shrimp. If they're uncooked, give them 10 minutes or until they're pink through. If they're cooked already, just give them 5 minutes or so to get hot through.
You can serve this over rice for the carb-eaters in the family, of course. If you like, you can have yours on _Cauli-Rice_ (see recipe page 343), but this dish is high-carb enough already that I'd probably eat it plain.
_**Yield:**_ 6 servings, each with: 257 calories, 11 g fat, 27 g protein, 13 g carbohydrate, 3 g dietary fiber, 10 g usable carbs.
### Pantry Seafood Supper
This is convenient because, as the name strongly suggests, it uses seafood you've got sitting in your pantry. If you've got seafood sitting in your freezer, you can use it instead—just let it cook an extra 30 minutes or so to make sure it's thawed and cooked through.
1/4 cup (45 g) roasted red peppers jarred in oil, diced small (about 1 pepper)
1/3 cup (20 g) chopped parsley
1 cup (70 g) chopped mushrooms
3/4 cup (175 ml) chicken broth
3/4 cup (175 ml) dry white wine
2 tablespoons (20 g) minced onion
2 teaspoons dried dill weed
1/2 teaspoon paprika
1/2 teaspoon Tabasco sauce
1 cup (235 ml) Carb Countdown dairy beverage
1/4 cup (60 ml) heavy cream
Guar or xanthan
1 can (6 ounces, or 170 g) tuna, drained
1 can (6 ounces, or 170 g) crab, drained
1 can (6 ounces, or 170 g) shrimp, drained
Combine the red peppers, parsley, mushrooms, broth, wine, onion, dill, paprika, and Tabasco sauce in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 3 to 4 hours.
When the time's up, stir in the Carb Countdown and cream and thicken the sauce to your liking with guar or xanthan. Now stir in the tuna, crab, and shrimp and let it cook for another 15 to 20 minutes.
Now you have a choice: You can eat this as a chowder, or you can serve it over _Cauli-Rice_ (see recipe page 343) or low-carbohydrate pasta—or even over spaghetti squash. It's up to you.
_**Yield:**_ 4 servings, each with: 269 calories, 9 g fat, 33 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Tuna-Noodle Casserole
Here's everybody's favorite childhood comfort food, decarbed.
8 ounces (225 g) mushrooms, coarsely chopped
3 cans (5 ounces, or 140 g each) tuna in water, drained well
1 cup (235 ml) heavy cream
1/2 teaspoon beef bouillon granules
1 tablespoon (15 ml) Worcestershire sauce
1/4 teaspoon salt or Vege-Sal
2 ounces (55 g) cream cheese
1/4 teaspoon pepper
1 cup (130 g) frozen peas NOT thawed
2 packages fettuccini-style tofu shirataki
Put the chopped mushrooms and drained tuna in your slow cooker. Stir together the cream, bouillon concentrate, Worcestershire sauce, salt, and pepper until the bouillon is dissolved. Pour this over the mushrooms and tuna and stir everything up. Now cut the cream cheese into a few chunks and settle them down into the tuna-mushroom mixture.
Dump the peas on top of everything and don't stir them in. Cover the pot, set it to low, and let it cook for 4 hours.
When suppertime rolls around, drain and rinse your shirataki and snip across them a few times with your kitchen shears. Put them in a microwaveable bowl and nuke them for 2 minutes on high. Drain them again, put them back in the bowl, nuke them for another 2 minutes, and then drain one more time. (You're getting the excess water out of your noodles so they don't make your casserole watery.)
While your noodles are in the microwave, take the lid off the slow cooker and stir everything together. See if it needs to be thicker—it may, because of the liquid cooking out of those mushrooms. Use your guar or xanthan shaker to thicken the sauce a bit if it needs it. When the noodles are drained for the third time, stir them in and serve. Alternatively, you can put the noodles on plates and serve the tuna mixture on top.
If you want something crunchy to replace the usual breadcrumb or potato chip topping, try crushed pork rinds or toasted slivered almonds, sprinkled over each serving right before you put them on the table.
_**Yield:**_ 5 servings, each with: 339 calories, 23 g fat, 26 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
chapter eight
Slow Cooker Crazy Mixed-up Meals
It's really very hard to organize a cookbook. Take, for instance, the recipes that follow. They really messed up my plan to organize this book largely by protein source. After backing and forthing, moving them around, I threw up my hands and gave them their own chapter.
### Simple Meat Loaf
This is simple, but it's hardly boring! Because meat loaf takes at least an hour in the oven, many of you may have given up on it as a weeknight dinner. Baking it in your slow cooker works remarkably well, although it doesn't brown much.
1 pound (455 g) ground round
1 pound (455 g) pork sausage
1 medium onion, finely chopped
1 medium green pepper, finely chopped
1/4 cup (24 g) oat bran
1/2 teaspoon pepper
1/2 teaspoon salt or Vege-Sal
2 tablespoons (28 ml) Worcestershire sauce
2 eggs
1/2 cup (60 g) pork rind crumbs (Run pork rinds through your food processor or blender.)
Place all ingredients in a big mixing bowl and use clean hands to smoosh everything together until it's well blended.
Put a rack or a collapsible basket-style steamer in the bottom of your slow cooker. Fold two squares of foil into strips and criss-cross them on the rack or steamer and up the sides of the slow cooker. (What you're doing is making a sling to help you lift the loaf out of the slow cooker.) If the holes/slots in your rack are pretty big, put a sheet of foil over the criss-crossed strips and pierce it all over with a fork.
Place the meat mixture on top of this and form into an evenly-domed meat loaf, smoothing the surface with dampened hands. Cover the slow cooker, set it to low, and let it cook for 9 to 12 hours.
Use the foil strips to lift the meat loaf out of the slow cooker.
_**Yield:**_ 8 servings, each with: 438 calories, 36 g fat, 23 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs.
### Morty's Mixed Meat Loaf
The combination of meats makes this loaf unusually tasty.
2 slices cooked bacon, crumbled
1 pound (455 g) ground round
1 pound (455 g) ground pork
1 pound (455 g) ground turkey
3 stalks celery, finely chopped
1/2 cup (30 g) finely chopped fresh parsley
1/2 cup (120 g) _Dana's No-Sugar Ketchup_ (see recipe page 332) or purchased low-carb ketchup
1 teaspoon hot sauce
2 eggs
1/2 cup (60 g) barbecue-style pork rind crumbs (Just run them through your food processor.)
2 tablespoons (28 ml) Worcestershire sauce
1 teaspoon lemon juice
1/2 teaspoon dried marjoram
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
Put a rack or basket-style steamer in the bottom of your slow cooker pot. Fold two squares of foil into strips and criss-cross them on the rack or steamer and up the sides of the slow cooker. (What you're doing is making a sling to help you lift the loaf out of the slow cooker.) If the holes/slots in your rack are pretty big, put a sheet of foil over the criss-crossed strips and pierce it all over with a fork.
Place all your ingredients in a big bowl and using clean hands, smoosh everything together really well. Form it into a round loaf on the rack in the slow cooker. Cover the slow cooker, set it to low, and let it cook for 8 to 10 hours.
Use the strips of foil to lift the loaf out of the slow cooker.
_**Yield:**_ 8 servings, each with: 453 calories, 30 g fat, 37 g protein, 7 g carbohydrate, 2 g dietary fiber, 5 g usable carbs.
### Louisiana Meat Loaf
I totally made this up out of my head. I just threw in all the Louisiana flavors I could think of, starting with the holy trinity of onion, green pepper, and celery, and it came out great.
1 pound (455 g) ground chuck
1 pound (455 g) pork sausage or turkey sausage
1/2 medium onion
1 celery rib, including leaves
1/2 green bell pepper
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles, drained
3 cloves garlic, crushed
1 tablespoon (1 g) dried basil
4 teaspoons (12 g) Creole seasoning (I used Tony Cachere's.)
1/2 cup (60 g) pork rind crumbs
1 egg
This is your basic meat loaf procedure: Dump your meat in a mixing bowl. Throw your onion, celery, and green pepper in your food processor with the **S** -blade in place and pulse until they're chopped medium-fine. Dump them in with the meat.
Add everything else. Smoosh it all up really well with clean hands.
Take two pieces of heavy-duty foil, long enough to reach down into your slow cooker, across the bottom, and back up the other side and fold each into a strip about 2 inches (5 cm) wide. Put your basket steamer in the slow cooker and criss-cross the foil strips across it, going around the stem in the middle.
Dump the meat in the slow cooker and form it into a nice, even loaf on the steamer. Cover the slow cooker, set to low, and let the whole thing cook for 6 to 7 hours.
Use the foil strips to lift the meat loaf out onto a platter and serve.
_**Yield:**_ 6 servings, each with: 588 calories, 49 g fat, 28 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Pub Loaf
Why is it Pub Loaf? It's because I imagine this is the sort of thing you might get in an English pub. Never having been to England, I can imagine anything I like. Plus, it's got beer in it!
4 slices bacon
4 ounces (115 g) chopped mushrooms
1 pound (455 g) ground chuck
1 pound (455 g) ground turkey
1 medium carrot
1 celery rib
1/2 medium onion
1 cup (115 g) shredded Cheddar cheese
2 teaspoons salt or Vege-Sal
1/2 teaspoon pepper
1 cup (120 g) pork rind crumbs
1 egg
1/2 cup (120 ml) light beer
In your big, heavy skillet, cook the bacon until crisp. Remove from the skillet and reserve.
Start the mushrooms frying in the bacon grease. In the meantime, put the meats in a big mixing bowl. Shred your carrot and chop your onion fine and add them, too, along with the cheese, salt, pepper, pork rind crumbs, and beer. When the mushrooms are soft, throw them in, too. Use clean hands to smoosh it all together really well.
Take two pieces of heavy-duty foil, long enough to reach down into your slow cooker, across the bottom, and back up the other side and fold each into a strip about 2 inches (5 cm) wide. Put your basket steamer in the slow cooker and criss-cross the foil strips across it, going around the stem in the middle.
Dump the meat in the slow cooker and form it into a nice, even loaf on the steamer. Cover the slow cooker, set to low, and let the whole thing cook for 6 to 7 hours.
Use the foil sling to remove your meat loaf from the slow cooker, place it on a platter, and serve.
_**Yield:**_ 8 servings, each with: 333 calories, 23 g fat, 26 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Tuscan-ish Meat Loaf
Okay, I haven't been to Tuscany any more than I've been to England. I'm provincial, so sue me. But I saw a recipe for a Tuscan meat loaf, and it looked good, so I whacked out the carbs and subbed for some hard to get ingredients, and it was tasty, so here it is. And I'll start thinking of tax deductible ways to get to Europe. I could jot down ideas for recipes, right?
1 pound (455 g) ground chuck
1 pound (455 g) pork sausage
1 pound (455 g) ground chicken
3 eggs
1 1/2 medium onions (or one big one)
1 large carrot
1 large celery rib, leaves included
4 cloves garlic
1/2 teaspoon nutmeg
1 1/2 teaspoons salt or Vege-Sal
1/2 teaspoon pepper
3/4 cup (75 g) grated Parmesan cheese
1 cup (120 g) pork rind crumbs
Throw all your meats in a big mixing bowl, along with the eggs. Chop your veggies fairly fine and throw them in, too, along with everything else. Use clean hands to smoosh it all up really well.
Take two pieces of heavy-duty foil, long enough to reach down into your slow cooker, across the bottom, and back up the other side and fold each into a strip about 2 inches (5 cm) wide. Put your basket steamer in the slow cooker and criss-cross the foil strips across it, going around the stem in the middle.
Dump the meat in the slow cooker and form it into a nice, even loaf on the steamer. Cover the slow cooker, set to low, and let the whole thing cook for 7 to 8 hours.
Use the sling to lift the meat loaf out of the slow cooker to a platter and serve.
_**Yield:**_ 10 servings, each with: 510 calories, 37 g fat, 36 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Pizza-ish Meat Loaf
Here you've got your pizza sauce, you got your cheese, and you got your Italian sausage and onions and peppers. I do not, however, recommend trying to throw it in the air.
1 pound (455 g) ground chuck
1 pound (455 g) Italian sausage, hot or sweet, as you prefer
1/3 cup (40 g) pork rind crumbs
1 tablespoon (6 g) Italian seasoning
1 cup (245 g) no-sugar-added pizza sauce, divided
1 medium onion, chopped fairly fine
1 medium green pepper, chopped fairly fine
1/2 teaspoon salt
1/4 teaspoon pepper
4 cloves garlic
1 egg
1/2 cup (60 g) shredded mozzarella cheese
This is your basic meat loaf procedure: Dump the two meats, the pork rind crumbs, the Italian seasoning, 1/2 cup (123 g) of the pizza sauce, the onion, green pepper, salt, pepper, garlic, and egg in a big mixing bowl and use clean hands to squish it together really well.
Take two pieces of heavy-duty foil, long enough to reach down into your slow cooker, across the bottom, and back up the other side and fold each into a strip about 2 inches (5 cm) wide. Put your basket steamer in the slow cooker and criss-cross the foil strips across it, going around the stem in the middle.
Dump the meat in the slow cooker and form it into a nice, even loaf on the steamer. Cover the slow cooker, set to low, and let the whole thing cook for 6 to 7 hours.
Open the slow cooker, sprinkle the mozzarella over the top, and re-cover the pot for just ten minutes to let the cheese melt. In the meantime, warm the rest of the pizza sauce. Use the foil sling to lift the meat loaf out of the slow cooker to a platter. Serve each slice of meat loaf with a tablespoon (15 g) of pizza sauce on it.
_**Yield:**_ 8 servings, each with: 419 calories, 33 g fat, 23 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Pizza Stew
Here's a slow cooker meal for your pizza-craving family. Maria gets the credit for perfecting this one. It wasn't quite right the way I originally conceived it. Maria gave me feedback, we went with her suggestions, and success was ours!
1 pound (455 g) bulk Italian sausage
1 pound (455 g) ground round
1 green bell pepper, diced
1/4 cup (40 g) diced onion
14 ounces (390 g) no-sugar-added pizza sauce (Ragu makes one.)
2 cups (230 g) shredded mozzarella cheese
1/2 cup (40 g) shredded Parmesan cheese
In a big, heavy skillet, brown and crumble the sausage and beef together. Drain them well and transfer them to your slow cooker. Add the pepper and onion and stir. Stir in the pizza sauce. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
Now uncover the slow cooker and top the stew with the mozzarella cheese. Re-cover the slow cooker and let it cook for another 30 to 45 minutes to melt the cheese. Serve with the Parmesan cheese sprinkled on top.
_**Yield:**_ 6 servings, each with: 626 calories, 49 g fat, 37 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Beef and Sausage Stew
This combination of beef and smoked sausage is great.
2 pounds (900 g) beef chuck
1 pound (455 g) smoked sausage
2 tablespoons (28 g) bacon grease
4 cups (600 g) cubed turnip
1 1/2 medium onions (or maybe 1 big one)
2 celery ribs
1/2 cup (120 ml) beef broth
1 teaspoon beef bouillon granules
2 teaspoons thyme
1 can (14 1/2 ounces, or 410 g) diced tomatoes
Cut your beef into cubes and slice your sausage about 1/2-inch (13 mm) thick. Put your big heavy skillet over medium-high heat and throw in the bacon grease. Brown the beef cubes—you'll want to do them in a couple of batches to avoid crowding—and transfer them to your slow cooker. Add more bacon grease if you need it and also brown your sausage slices a bit on both sides.
Are you up for some multitasking? While you're browning your meat, peel your turnips and cube them—I used three smallish ones. Peel and dice your onions, too.
But don't let your sausage slices burn! When they're browned, transfer them to the pot, too.
Slice the celery crosswise, thinly. Throw all the veggies in the pot.
Dump the beef broth in the skillet you browned the meat in and add the bouillon concentrate and thyme. Stir it around until the bouillon dissolves, along with all the tasty brown bits on the bottom of the skillet. Pour this over all the stuff in the pot. Add the tomatoes, give everything a stir, and whack on the lid. Set it for low and let it go for 7 to 8 hours.
_**Yield:**_ 8 servings, each with: 502 calories, 38 g fat, 28 g protein, 11 g carbohydrate, 2 g dietary fiber, 9 g usable carbs.
### Albondigas en Salsa Chipotle
Here are some Mexican meatballs! Yummy.
1/2 medium onion, chopped
1 tablespoon (15 ml) oil
1 teaspoon ground cumin
1 teaspoon ground coriander
3 cloves garlic, crushed
1 teaspoon dried oregano
1 teaspoon dried thyme
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles, drained
1 cup (235 ml) chicken broth
1/2 teaspoon chicken bouillon concentrate
1 chipotle chile canned in adobo sauce, or more to taste
1 pound (455 g) ground round
1 pound (455 g) ground turkey
1 egg
1 teaspoon pepper
1/2 teaspoon ground allspice
1/2 teaspoon ground cumin
1/2 teaspoon ground coriander
1 teaspoon dried oregano
2 teaspoons salt or Vege-Sal
1/2 medium onion, finely chopped
3 cloves garlic, crushed
2 tablespoons (28 ml) olive oil
1 teaspoon guar or xanthan
In a big, heavy skillet, sauté 1/2 onion in the 1 tablespoon (15 ml) oil over medium-high heat until translucent. Add the 1 teaspoon ground cumin, 1 teaspoon ground coriander, 3 cloves garlic, 1 teaspoon oregano, and 1 teaspoon thyme and sauté for another minute or two. Transfer the mixture to your blender or food processor.
Add the tomatoes, broth, bouillon, and chipotle. Blend until smooth and pour into your slow cooker. Cover the slow cooker, set it to low, and start it cooking as you make the meatballs.
Place the beef, turkey, egg, pepper, allspice, 1/2 teaspoon ground cumin, 1/2 teaspoon ground coriander, 1 teaspoon oregano, salt or Vege-Sal, 1/2 onion, and 3 cloves garlic into a big bowl and use clean hands to smoosh it all together. Form it into meatballs.
In the skillet, heat 2 tablespoons (28 ml) oil over medium-high heat and brown the meatballs on all sides. Transfer the meatballs to the slow cooker, re-cover, and let the whole thing cook for 3 to 4 hours.
Scoop the meatballs out with a slotted spoon and put them in a serving bowl. Thicken the sauce to taste with the guar or xanthan and pour the sauce over the meatballs.
_**Yield:**_ 8 servings, each with: 297 calories, 20 g fat, 23 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Low-Carb Slow Cooker Paella
Maria, who tested this, rates it a 10. She added, "While this won't fool a purist, the flavor is quite good. For those who can't afford saffron, turmeric is an acceptable substitute. For an authentic presentation, which is important, put the 'rice' in a paella pan or flat casserole, mix it with the veggies, and arrange the chicken and shrimp artistically on top." Do look for Spanish chorizo, rather than Mexican chorizo, for this dish.
6 chicken leg and side quarters (about 3 pounds, or 1.4 kg)
1/4 cup (60 ml) olive oil
1 cup (160 g) chopped onion
1 clove garlic, crushed
1 green bell pepper
6 ounces (170 g) chorizo links or diced ham
1 can (14 1/2 ounces, or 410 g) tomato wedges, drained
1/2 teaspoon saffron threads
1 cup (235 ml) chicken broth
1 teaspoon chicken bouillon concentrate
6 ounces (170 g) shrimp
1/2 cup (73 g) fresh snow pea pods, cut into 1/2-inch (13 mm) pieces
1 head cauliflower
In a big, heavy skillet, brown the chicken in the oil.
While that's happening, put the onion, garlic, and green pepper in your slow cooker. When the chicken is brown, transfer it to the slow cooker as well.
If using chorizo, slice it into rounds and brown it in the same skillet and then transfer it to the slow cooker. If using ham, you can simply add it directly to the slow cooker. Place the tomatoes on top of that.
In a bowl, mix together the saffron, broth, and bouillon and pour it into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 6 hours.
When the time's up, turn the slow cooker to high. Add the shrimp and snow peas to the slow cooker, re-cover, and cook for another 30 minutes.
While that's happening, shred the cauliflower in your food processor, put it in a microwavable casserole dish with a lid, add a few tablespoons (45 to 60 ml) of water, cover, and microwave it on high for 8 to 9 minutes. Serve with the paella.
_**Yield:**_ 8 servings, each with: 599 calories, 42 g fat, 46 g protein, 8 g carbohydrate, 1 g dietary fiber, 7 g usable carbs.
### Brunswick Stew
This is an old favorite, decarbed and updated for your slow cooker.
1 large onion, sliced
2 pounds (900 g) skinless chicken thighs
1 1/2 cups (225 g) ham cubes, cooked
1 teaspoon dry mustard
1 teaspoon dried thyme
1/2 teaspoon pepper
1 cup (180 g) canned diced tomatoes
1 can (14 ounces, or 390 g) chicken broth
3 cloves garlic, crushed
1 tablespoon (15 ml) Worcestershire sauce
1/4 teaspoon hot sauce, or to taste
1 cup (172 g) canned black soybeans, drained
Place the onion in your slow cooker first. Add the chicken and ham.
In a bowl, mix together the dry mustard, thyme, pepper, tomatoes, broth, garlic, Worcestershire sauce, and hot sauce. Pour the mixture over the chicken and ham. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, stir in the soybeans and let the whole thing cook for another 20 minutes or so.
_**Yield:**_ 6 servings, each with: 212 calories, 10 g fat, 21 g protein, 9 g carbohydrate, 3 g dietary fiber, 6 g usable carbs.
chapter nine
Slow Cooker Soups
If there's a type of dish for which the slow cooker seems custom-made, it's soups. How wonderful to come home tired on a cold evening to a pot of hot, hearty, homemade soup!
### Bone Broth for Busy People
This is more a rule than a recipe. Save all of your chicken bones in a sack in the freezer. Save your steak and other beef bones in a separate sack. Don't worry that these bones are picked clean; totally naked bones will make amazingly good broth.
When you have enough bones to fill your slow cooker, whatever its size, you're ready to make broth. Dump your bones into the slow cooker—you don't have to bother to thaw them, just dump them in. Cover them with water and add maybe a teaspoon of salt for a smallish slow cooker or two for a big one, plus a couple of tablespoons (28 ml) of vinegar, any kind. Cover the slow cooker, set it to low, and forget about it. I mean, for a long time. I let mine go for a good 36 hours.
Turn your slow cooker off, let the broth cool, then strain, throw away the bones, and either make soup with your broth or freeze it in snap-top containers for all future brothy needs. If you're going to freeze it, do yourself a favor and pay attention to how much your containers hold. Then, when you're looking at a recipe that calls for, say, 6 cups (1.4 L) of broth, you'll know how much you've got.
No hard and fast nutritional stats on this one, but unless you add onion or celery or something, it should be pretty much carb-free.
### Black Bean Soup
One of the few carb-y dishes I sometimes miss is legume soup, especially black bean soup. This is my decarbed version.
1 can (28 ounces, or 785 g) black soybeans
1 can (14 ounces, or 390 g) black beans
2 cups (475 ml) chicken broth
1 medium onion, cut into chunks
4 cloves garlic, crushed
1 medium carrot, shredded
2 medium stalks celery, finely diced
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
1 tablespoon (15 ml) liquid smoke flavoring
2 teaspoons hot sauce
2 cups (300 g) ham cubes
Using your food processor with the **S** -blade in place, puree the soybeans and black beans. Place them in your slow cooker. Stir in the broth.
Place the onion in the food processor. Add the garlic, carrot, and celery. Pulse the food processor until everything is finely chopped. Add to the soup.
Stir in the salt or Vege-Sal, pepper, liquid smoke flavoring, hot sauce, and ham. Cover the slow cooker, set it to low, and let it cook for 9 to 10 hours.
When the time's up, stir the soup up (it'll have settled out some) and check to see if it needs more salt and pepper. (This will depend on how salty your ham is.)
_**Yield:**_ 8 servings, each with: 218 calories, 9 g fat, 19 g protein, 17 g carbohydrate, 9 g dietary fiber, 8 g usable carbs.
### Mexican Beef and Bean Soup
You know the family will love this!
12 ounces (340 g) ground round
1 medium onion, chopped
2 cloves garlic, crushed
1 medium green bell pepper, diced
1 quart (950 ml) beef broth
1 teaspoon beef bouillon concentrate
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1 can (15 ounces, or 425 g) black soybeans
2 teaspoons ground coriander
1 teaspoon ground cumin
4 tablespoons (4 g) chopped cilantro
6 tablespoons (90 g) sour cream
In a big, heavy skillet, brown and crumble the ground beef. Drain it well and transfer it to your slow cooker.
Add the onion, garlic, bell pepper, broth, bouillon, tomatoes, soybeans, coriander, and cumin and stir. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
Top each bowlful with cilantro and sour cream.
_**Yield:**_ 6 servings, each with: 296 calories, 16 g fat, 25 g protein, 14 g carbohydrate, 5 g dietary fiber, 9 g usable carbs.
### Pho
This Vietnamese-style beef noodle soup is so wonderful! I had never had it before I made some, but it was instantly soothing and comforting, as if I'd been eating it all my life. The gelatin is just to add some body to the beef broth, assuming you're using store-bought. If you've made your own _Bone Broth_ (see recipe page 256), it will have plenty of gelatin in it.
6 cups (1.4 L) beef broth
2 teaspoons unflavored gelatin—optional
2 teaspoons beef bouillon concentrate
2 whole star anise
1 cinnamon stick
1 tablespoon (6 g) fresh ginger root, minced
1 teaspoon chili garlic sauce
1/4 cup (60 ml) fish sauce
12 ounces (340 g) beef round
2 packages shirataki noodles
1 cup (104 g) mung bean sprouts
4 scallions, sliced thin, including the crisp part of the green
1/4 cup (10 g) minced fresh basil
1/4 cup (4 g) minced fresh cilantro
If you are using the gelatin, sprinkle it over 1 cup (235 ml) of the beef broth to soften for five minutes.
In the meantime, put the rest of the broth in the slow cooker with the beef bouillon concentrate, anise, cinnamon stick, minced ginger, chili garlic paste, and fish sauce. Turn the slow cooker on to low.
Slice your beef as thinly as possible across the grain—having it half-frozen makes this easier. Cut those strips into smaller ones, no more than an inch or two in any direction. Add to the broth. Stir in the broth with the gelatin, too, if you're using it. Cover and let it cook for 5 to 6 hours.
It's time for supper! Drain your shirataki and snip across them a few times. Stir them into your pho, along with the mung bean sprouts. Re-cover and let it cook just another 15 minutes or so while you slice the scallions thin and mince the basil and cilantro. Don't let the bean sprouts get too wilted.
Serve with scallion, basil, and cilantro on each bowlful.
_**Notes:**_ Fish sauce runs about 6 grams of carb per fluid ounce; you can slice 6 grams out of this recipe by using half fish sauce, half soy sauce. Or perhaps you'll find you like it less salty and use less of both.
_**Yield:**_ 4 servings, each with: 348 calories, 14 g fat, 39 g protein, 21 g carbohydrate, 5 g dietary fiber, 16 g usable carbs.
I prefer traditional shirataki to the tofu shirataki in this dish, spaghetti or angel hair width.
### Bollito Misto
All this Italian soup-stew needs with it is a green salad and maybe some crusty bread for the carb-eaters.
1 large onion, sliced
2 carrots, cut 1/2-inch (13 mm) thick
3 stalks celery, cut 1/2-inch (13 mm) thick
2 pounds (900 g) beef round, cubed
1/2 teaspoon salt
1/2 teaspoon pepper
2 tablespoons (8 g) chopped fresh parsley
1 bay leaf
3 teaspoons (18 g) chicken bouillon concentrate
1 quart (950 ml) chicken broth
2 pounds (900 g) boneless, skinless chicken thighs, cubed
1 pound (455 g) Italian sausage links
1/2 cup (130 g) purchased pesto sauce
Put the onion, carrots, and celery in your slow cooker. Season the beef with the salt and pepper and place them on top. Add the parsley and bay leaf. Stir the bouillon into the chicken and pour it into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
Add the chicken, turn the heat up to high, and let the whole thing cook another hour.
While it's cooking, pour yourself a glass of Chianti and put out some vegetables and dip for the kids. In a big, heavy skillet, place the sausages, cover with water, slap a lid on, and simmer for 20 minutes over medium heat. Remove the skillet from the heat and leave the sausages in the water, with the lid on.
When the slow cooker's time is up, drain the sausage and cut it into 1-inch (2.5 cm) chunks. Stir the sausage into the stuff in the slow cooker. Remove the bay leaf. Now ladle the whole thing into soup bowls and top each serving with a tablespoon (15 g) of pesto.
_**Yield:**_ 8 servings, each with: 599 calories, 43 g fat, 44 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs.
### Italian Sausage Soup
This is perfect on a chilly, rainy evening!
1 pound (455 g) Italian sausage links
1 medium onion
1 green pepper
1 celery rib
1 1/2 teaspoons oregano
1 1/2 teaspoons Italian seasoning
3 cloves garlic
8 cups (1.9 L) chicken broth
2 cups (475 ml) heavy cream
Put your big, heavy skillet over medium heat and brown the sausage all over. Let it cool a little so you can handle it.
While it's cooling, chop your onion, pepper, and celery.
Transfer your sausage to your cutting board and throw the veggies in the skillet. Add a splash of olive oil if needed. Sauté the the veggies for a few minutes while you move on to the next step.
Slice your sausage. Dump it in the slow cooker. Throw the sautéed vegetables in, too. Add the oregano and Italian seasoning and crush in the garlic.
Pour a cup or two (235 to 475 ml) of the broth into the skillet and stir it around to deglaze. Add to the slow cooker, along with the rest of the broth. Cover the slow cooker and cook for five or six hours.
At dinnertime, stir in the cream. Let it heat another 10 to 15 minutes and serve.
_**Yield:**_ 6 servings, each with: 604 calories, 55 g fat, 19 g protein, 8 g carbohydrate, 1 g dietary fiber, 7 g usable carbs.
### Sausage and Green Bean Soup
Read the labels on the smoked sausage—they vary a lot in carb content.
1 pound (455 g) smoked sausage
1 tablespoon (15 g) coconut oil
1 small onion, chopped
1 can (14 1/2 ounces, or 410 g) diced tomatoes
6 cups (1.4 L) chicken broth
1 pound (455 g) frozen cross-cut green beans, thawed
1 tablespoon (6 g) Italian seasoning
1 teaspoon cumin
1/2 teaspoon red pepper flakes
4 cloves garlic, crushed
Split the sausage lengthwise and slice into half-rounds about 1/4-inch (6 mm) thick. Put your big heavy skillet over medium-low heat and start browning the sausage in the coconut oil. (I gave mine a squirt of nonstick cooking spray first and needed only to rinse and wipe when done.) Unless your skillet is bigger than mine, this will take a couple of batches.
While the sausage is browning, chop the onion. When the sausage is all browned, throw it in the slow cooker and add the onion to the skillet. Sauté until it's translucent. Dump it in the slow cooker, too.
Add everything else, cover the cooker, and set it to low for 6 to 8 hours or on high for 4 to 6 hours.
When it's done, taste to see if it needs a little salt and pepper—mine didn't need a thing!—then serve.
_**Yield:**_ 5 servings, each with: 441 calories, 32 g fat, 21 g protein, 18 g carbohydrate, 3 g dietary fiber, 15 g usable carbs.
### Vegetable Beef Soup
This is serious comfort food!
3 pounds (1.4 kg) beef soup bones—meaty ones!
8 cups (1.9 L) water
5 teaspoons (30 g) beef bouillon concentrate
1 can (14 1/2 ounces, or 410 g) diced tomatoes
1 can (8 ounces, or 225 g) tomato sauce
1/2 cup (120 ml) dry red wine
2 celery ribs, diced
2 medium carrots, peeled and sliced
1 medium onion, chopped
3 cloves garlic, crushed
12 ounces (340 g) frozen cross-cut green beans or Italian beans, thawed
1 cup (130 g) frozen peas, thawed
1 large turnip, peeled and diced
2 tablespoons (2 g) dried basil
1 tablespoon (2 g) dried oregano
2 bay leaves
1 teaspoon pepper
Preheat your oven to 400°F (200°C, or gas mark 6) and let it heat while you throw your soup bones in a roasting pan. Slide them into the oven and give them about 40 minutes; you want them browned all over.
Put your browned bones in the slow cooker and add the water and beef bouillon concentrate. Turn it on to low while you gather/open/peel/cut up/measure everything else and add it all to the pot. Cover and let the whole thing cook for a good eight hours.
Fish out the bones with tongs or slotted spoon. The meat will be falling off the bone, so it should be very easy to take it off, chop it up, and throw it back in the pot. Remove the bay leaves, stir it up, and serve!
_**Note:**_ If you don't have any meaty beef bones, you can make excellent soup with 2 quarts (1.9 L) of packaged beef broth, plus about 1 1/2 pounds (680 g) of beef chuck, rump, or round, diced.
_**Yield:**_ 8 servings, each with: 17 g carbohydrate, 4 g dietary fiber, 13 g usable carbs. It's hard to get a clear calorie and protein count on this because it will depend on how meaty your bones are. If they're not meaty enough for your tastes, feel free to throw in some diced beef—chuck or rump would do fine.
### Cauliflower, Cheese, and Spinach Soup
Maria's family gave this raves. It's easy, too!
6 cups (600 g) cauliflower florets, cut into
1/2-inch (13 mm) pieces
1 quart (950 ml) chicken broth
1/2 cup (80 g) minced red onion
5 ounces (140 g) bagged baby spinach leaves, pre-washed
1/4 teaspoon cayenne
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
4 cloves garlic, crushed
3 cups (360 g) shredded smoked Gouda cheese
1 cup (235 ml) Carb Countdown dairy beverage
Guar or xanthan
In your slow cooker, combine the cauliflower, broth, onion, spinach, cayenne, salt or Vege-Sal, pepper, and garlic. Cover the slow cooker, set it to low, and let it cook for 6 hours or until the cauliflower is tender.
When the time's up, stir in the Gouda, a little at a time, and then the Carb Countdown. Re-cover the slow cooker and cook for another 15 minutes or until the cheese has thoroughly melted. Thicken soup a little with guar or xanthan.
_**Yield:**_ 8 servings, each with: 214 calories, 14 g fat, 17 g protein, 7 g carbohydrate, 2 g dietary fiber, 5 g usable carbs.
### Tavern Soup
Here's a cheese soup with beer! Don't worry about the kids, the alcohol cooks off.
1 1/2 quarts (1.4 L) chicken broth
1/4 cup (30 g) finely diced celery
1/4 cup (38 g) finely diced green bell pepper
1/4 cup (28 g) shredded carrot
1/4 cup (15 g) chopped fresh parsley
1/2 teaspoon pepper
1 pound (455 g) sharp cheddar cheese, shredded
12 ounces (355 ml) light beer
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon Tabasco sauce
Guar or xanthan
Combine the broth, celery, green pepper, carrot, parsley, and pepper in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 6 to 8 hours, and even a bit longer won't hurt.
When the time's up, either use a hand-held blender to puree the vegetables right there in the slow cooker or scoop them out with a slotted spoon, purée them in your blender, and return them to the slow cooker.
Now whisk in the cheese a little at a time until it's all melted in. Add the beer, salt or Vege-Sal, and Tabasco sauce and stir till the foaming stops. Use guar or xanthan to thicken your soup until it's about the texture of heavy cream. Re-cover the pot, turn it to high, and let it cook for another 20 minutes before serving.
_**Yield:**_ 8 servings, each with: 274 calories, 20 g fat, 18 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs.
### Cream of UnPotato Soup
I never cease to marvel at the versatility of cauliflower. This really does taste like potato soup.
1 quart (950 ml) chicken broth
1/2 head cauliflower, chunked
1/2 cup (80 g) chopped onion
1/2 cup (50 g) Ketatoes mix
1/2 cup (120 ml) heavy cream
1/2 cup (120 ml) Carb Countdown dairy beverage
Guar or xanthan (optional)
5 scallions, sliced
Put the broth, cauliflower, and onion in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 4 to 5 hours.
I use a hand blender to purée my soup right in the slow cooker, but you may transfer the cauliflower and onion, along with a cup of broth, into your blender or food processor instead. Either way, purée until completely smooth and then blend in the Ketatoes. If you have removed the cauliflower from the slow cooker to purée, pour the purée back in and whisk it into the remaining broth.
Stir in the cream and Carb Countdown. Thicken it a bit further with guar or xanthan if you feel it needs it. Add salt and pepper to taste and stir in the sliced scallions. Serve hot right away or chill and serve as vichyssoise.
_**Yield:**_ 6 servings, each with: 190 calories, 11 g fat, 12 g protein, 13 g carbohydrate, 6 g dietary fiber, 7 g usable carbs.
### German UnPotato Soup
This is worth the time you spend cutting things up! It's hearty and filling.
1 head cauliflower, chunked
2 stalks celery, sliced
1 medium onion, chopped
8 ounces (225 g) smoked sausage, sliced
1 tablespoon (28 ml) oil
4 cups (950 ml) beef broth
2 tablespoons (28 ml) vinegar
1 tablespoon (1.5 g) Splenda
1/4 teaspoon celery seed
1/2 teaspoon dry mustard
1/4 teaspoon pepper
2 cups (140 g) bagged coleslaw mix
Place the cauliflower, celery, and onion in your slow cooker.
In a big, heavy skillet, brown the sausage a bit in the oil. Transfer the sausage to the slow cooker, too.
Pour 1 cup (235 ml) of the broth into the skillet and stir it around a bit to dissolve the flavorful bits. Pour it into the slow cooker.
In a bowl, combine the rest of the broth with the vinegar, Splenda, celery seed, dry mustard, and pepper. Pour over the vegetables and sausage. Cover the slow cooker, set it to low, and let it cook for 8 hours.
When the time's up, stir in the coleslaw mix and let it cook for another 20 to 30 minutes.
_**Yield:**_ 4 servings, each with: 344 calories, 22 g fat, 20 g protein, 17 g carbohydrate, 2 g dietary fiber, 15 g usable carbs.
### Cream of Mushroom Soup
If you've only ever thought of mushroom soup as gooey stuff that came in cans and was used in casseroles, you need to try this! It has a rich, earthy flavor. Even my mushroom-phobic husband liked it.
8 ounces (225 g) mushrooms, sliced
1/4 cup (40 g) chopped onion
2 tablespoons (28 g) butter
1 quart (950 ml) chicken broth
1/2 cup (120 ml) heavy cream
1/2 cup (115 g) light sour cream
Guar or xanthan (optional)
In a big, heavy skillet, sauté the mushrooms and onion in the butter until the mushrooms soften and change color. Transfer them to your slow cooker. Add the broth. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
When the time's up, scoop out the vegetables with a slotted spoon and put them in your blender or food processor. Add enough broth to help them process easily and purée them finely. Pour the puréed vegetables back into the slow cooker, scraping out every last bit with a rubber scraper. Now stir in the heavy cream and sour cream and add salt and pepper to taste. Thicken the sauce a bit with guar or xanthan if you think it needs it. Serve immediately.
_**Yield:**_ 5 servings, each with: 176 calories, 15 g fat, 6 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 usable carbs.
### Chipotle Pumpkin Soup
This is so good! If you leave the diced chicken breast out of this, it would make a nice starter for Thanksgiving supper.
1 leek
2 tablespoons (28 g) butter
3 cloves garlic, crushed
1 can (15 ounces, or 425 g) pumpkin purée
2 chipotle chiles canned in adobo—minced, plus a couple of teaspoons of the sauce
6 cups (1.4 L) chicken broth
1 teaspoon cumin
1 bay leaf
12 ounces (340 g) boneless, skinless chicken breast
1 cup (235 ml) heavy cream
2/3 cup (154 g) sour cream
Trim the top and root off your leek and then split it vertically. Wash well, making sure there's no dirt between the layers. Then slice it thin, across the curve.
In your big, heavy skillet, melt the butter over medium heat and sauté the leek until it's softened a bit. Transfer to the slow cooker.
Add the garlic, pumpkin purée, chopped chipotles, broth, cumin, and bay leaf to the slow cooker and turn it on to low. Cover the pot. Now go wash the hot pepper off your hands! Let it cook for at least 4 hours, and a couple more won't hurt.
Dice the chicken into 1/4-inch (6 mm) cubes—this is easier if the chicken is half-frozen. Stir this into the soup along with the heavy cream. Re-cover and let it cook another 45 minutes to an hour.
You can thicken it just a little with guar or xanthan at this point, if you like, but don't make it any thicker than cream, okay?
Serve with a couple of tablespoons (30 g) of sour cream floating in the middle of each bowlful.
_**Yield:**_ 6 servings, each with: 369 calories, 27 g fat, 21 g protein, 12 g carbohydrate, 3 g dietary fiber, 9 g usable carbs.
### Not-Pea Soup
I love split pea soup, always have! But despite authorities asserting that legumes are good carbs, they make me gain weight. So I came up with this soup, which is remarkably like split pea, except for the carb count.
1 onion
1 carrot
2 celery ribs
4 cloves garlic
1 teaspoon thyme
2 bay leaves
2 pounds (900 g) ham hocks or shanks
3 pounds (1.4 kg) frozen green beans, thawed
8 cups (1.9 L) water
1 tablespoon ham bouillion concentrate (Better Than Bouillon makes one.)
Guar or xanthan
1/2 teaspoon pepper or to taste
Salt or Vege-Sal, to taste
Hot sauce—optional
Peel your onion and whack it into chunks. Do the same for your carrot. Throw them both in your food processor with the **S** -blade in place and pulse until they're chopped pretty fine. Dump this in your slow cooker. Put the food processor bowl back on the stand and put the blade back in place. Whack your celery into chunks, including any fresh leaves, and throw it in the food processor; chop that pretty fine, too. (You could do the celery with the onion and carrot, but it's so much softer that it's likely to be pretty pulverized by the time the carrot is chopped.) Dump that in the slow cooker, too. Crush in the garlic and add the thyme and bay leaves.
Throw your hocks or shanks (shanks are meatier) on top of the seasoning vegetables. Dump in all of the green beans. Add the water, cover, and set to low. Let it cook for a good 6 to 8 hours until the green beans are soft and have changed color—they should look like canned green beans, sort of dull green.
Use a slotted spoon to fish out the hocks or shanks and put them on a plate. Let them cool to the point where you can handle them.
In the meantime, use a stick blender to purée the soup until it's smooth. Now blend in the ham bouillon concentrate.
Use your guar or xanthan shaker and the stick blender to thicken up the soup to your taste. Salt and pepper to taste, too. I like a few dashes of Louisiana-style hot sauce—Tabasco, Frank's, or the like—in this, too, but go with your own taste.
_**Yield:**_ 8 servings, each with: 370 calories, 22 g fat, 29 g protein, 16 g carbohydrate, 6 g dietary fiber, 10 g usable carbs.
### Chicken and Vegetable Soup with Thai Spices
This soup is light, fragrant, and wonderful.
2 tablespoons (28 ml) oil
2 carrots, thinly sliced
4 stalks celery, thinly sliced
2 cups (140 g) sliced mushrooms
1/2 medium onion, thinly sliced
2 cloves garlic, crushed
1 1/2 pounds (680 g) boneless, skinless chicken breast, cut into 1/2-inch (13 mm) cubes
2 quarts (1.9 L) chicken broth
2 cups (248 g) frozen cross-cut green beans
1 tablespoon (8 g) grated ginger root
9 teaspoons (45 g) chili paste
1 tablespoon (15 ml) lemon juice
1 tablespoon (15 ml) lime juice
1/8 teaspoon anise seed, ground
1/4 teaspoon ground cardamom
1/4 teaspoon ground cinnamon
1/2 teaspoon ground cumin
1/4 teaspoon ground coriander
1 tablespoon (15 ml) fish sauce
Fresh cilantro
Heat the oil in a big, heavy skillet over medium-high heat and sauté the carrots, celery, mushrooms, and onion until the onion is starting to get translucent. Transfer them to your slow cooker, adding the garlic. Now add the chicken to the skillet and sauté just until it's sealed on the outside. Transfer it to the slow cooker, too.
Pour in the broth and add the green beans, ginger, chili paste, lemon juice, lime juice, anise seed, cardamom, cinnamon, cumin, coriander, and fish sauce. Stir, cover the slow cooker, set it to low, and let it cook for 6 to 8 hours.
Scatter a little cilantro over each bowlful before serving.
_**Yield:**_ 8 servings, each with: 209 calories, 8 g fat, 25 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs.
### Chicken Soup with Wild Rice
Wild rice has more fiber and therefore fewer usable carbs than regular rice, either white or brown. And it adds a certain cachet to your soup!
2 quarts (1.9 L) chicken broth
2 carrots, thinly sliced
2 stalks celery, diced
1/2 cup (80 g) chopped onion
1 pound (455 g) boneless, skinless chicken breast, cut into 1/2-inch (13 mm) cubes
1/4 cup (40 g) wild rice
1 teaspoon poultry seasoning
Simply combine everything in your slow cooker, cover, set it to low, and let it cook for 6 to 7 hours.
_**Yield:**_ 6 servings, each with: 182 calories, 4 g fat, 25 g protein, 10 g carbohydrate, 2 g dietary fiber, 8 g usable carbs.
### Chicken Minestrone
Here's a decarbed version of the Italian favorite. You'll never miss the pasta!
3 slices bacon, chopped
1 medium onion, chopped
2 medium turnips, cut into 1/2-inch (13 mm) cubes
1 medium carrot, thinly sliced
2 small zucchini, quartered and sliced
2 stalks celery, thinly sliced
3 tablespoons (45 ml) olive oil
1 1/2 quarts (1.4 L) chicken broth
1 1/2 pounds (680 g) skinless chicken thighs, boned and cubed
1 tablespoon (6 g) Italian seasoning
1 can (14 1/2 ounces, or 410 g) diced tomatoes, undrained
1 can (15 ounces, or 425 g) black soybeans
Spray a big, heavy skillet with nonstick cooking spray and then start the bacon frying over medium heat. As some grease cooks out of the bacon, add as many of the vegetables as will fit and sauté them until they soften just a bit. Transfer the vegetables to your slow cooker and continue sautéing the rest of the vegetables, adding oil as needed, until all the vegetables are softened a bit and in the slow cooker.
Place the broth, chicken, Italian seasoning, tomatoes, soybeans, and salt and pepper to taste in the slow cooker. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
_**Yield:**_ 6 servings, each with: 294 calories, 16 g fat, 21 g protein, 18 g carbohydrate, 6 g dietary fiber, 12 g usable carbs.
### So Totally Inauthentic Malaysian-oid Pumpkin-Chicken-Tomato Soup
This started with a recipe for a Malaysian noodle dish called laksa. That wasn't happening, but the flavors looked really good, so I turned it into a soup, substituting for a few hard-to-get ingredients. The result is stellar!
1 1/2 pounds (680 g) boneless, skinless chicken thighs (or breasts, but I like the thighs here)
1 quart (950 ml) chicken broth
1 can (15 ounces, or 425 g) pumpkin purée
8 cloves garlic
3 tablespoons (24 g) grated ginger root
1 can (14 1/2 ounces, or 410 g) tomatoes with green chiles
1 can (13 1/2 ounces, or 380 ml) coconut milk
1/4 cup (60 ml) fish sauce
2 tablespoons (28 ml) lemon juice
This is so simple! Cube the chicken and throw it in the slow cooker. Add everything else, whisk it all up, cover the slow cooker, and set it to low. Let it cook for 4 to 5 hours.
_**Yield:**_ 5 servings, each with: 244 calories, 9 g fat, 27 g protein, 16 g carbohydrate, 3 g dietary fiber, 13 g usable carbs.
### Spicy Chicken and Mushroom Soup
This is exotic and delicious.
3 tablespoons (45 g) butter
1 leek, thinly sliced (white part only)
8 ounces (225 g) sliced mushrooms
1 clove garlic, crushed
2 teaspoons _Garam Masala_ (see recipe page 340) or purchased garam masala
1 teaspoon pepper
1/4 teaspoon cayenne
1/4 teaspoon ground nutmeg
1 quart (950 ml) chicken broth
12 ounces (340 g) boneless, skinless chicken breasts, cut into thin strips
1/2 cup (120 ml) Carb Countdown dairy beverage
1/2 cup (120 ml) heavy cream
3 tablespoons (3 g) chopped fresh cilantro (optional)
Melt the butter in a big, heavy skillet, over medium heat, and sauté the leek with the mushrooms until they both soften. Stir in the garlic, _Garam Masala_ , pepper, cayenne, and nutmeg and sauté for another minute or two. Transfer to your slow cooker. Pour in the broth and add the chicken. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
When the time's up, use a slotted spoon to scoop roughly two-thirds of the solids into your blender or food processor. Add a cup (235 ml) or so of the broth and purée until smooth. Stir the purée back into the rest of the soup. (You may want to rinse the blender or food processor out with a little broth to get all of the purée.) Stir in the Carb Countdown and cream. Re-cover the pot and let it cook for another 30 minutes. Serve with cilantro on top. Or not, if you prefer—it's nice without it, too!
_**Yield:**_ 6 servings, each with: 243 calories, 16 g fat, 18 g protein, 6 g carbohydrate, 1 g dietary fiber, 5 g usable carbs.
### Chicken and Fennel Soup
If you haven't tried fennel, you really should. It has an enchanting mild licorice flavor, and it is a favorite in Italian cuisine, especially. It looks sort of like celery with feathery leaves and a swollen bulb at the bottom. Be aware that some stores mislabel it as anise.
2 quarts (1.9 L) chicken broth
2 teaspoons chicken bouillon concentrate
2 fennel bulbs 1 onion
4 tablespoons (55 g) butter
6 cloves garlic 2 pounds (900 g) boneless, skinless chicken breast
1 can (14 1/2 ounces, or 410 g) diced tomatoes, preferably petite diced
1/2 teaspoon pepper
1 star anise
Dump the chicken broth in the slow cooker and turn it on to high; you may as well start it heating while you assemble everything else. Add the chicken bouillon concentrate, too, to start it dissolving.
Put your big, heavy skillet over medium heat and start melting the butter. Whack the stems off the fennel (save some tops for garnish) and trim the bottoms. Then cut them into chunks and throw them in your food processor with the **S** -blade in place. Peel your onion and chunk it and throw it in there, too. Pulse until everything's chopped medium-fine. Then dump it in the skillet. Sauté, stirring often, until everything's softened a bit.
While your veggies are sautéing, crush your garlic—you can add it to the fennel and onion or just throw it in the broth. You also want to cut your chicken into bite-sized pieces.
Okay, your veggies are softened. Scrape them into the slow cooker and stir in the chicken, too. Open the tomatoes and stir them in—don't drain them first—and add the juice, too.
Stir in the pepper, toss in the star anise, and re-cover the pot. Turn it back to low, let it all cook for 5 to 6 hours, and then serve with a garnish of fennel fronds.
_**Note:**_ Star anise is commonly used in Chinese cooking; look for it with the Asian foods in your grocery store. It really does look like a little brown star!
_**Yield:**_ 6 servings, each with: 352 calories, 14 g fat, 43 g protein, 13 g carbohydrate, 4 g dietary fiber, 9 g usable carbs.
### Turkey Sausage Soup
This is a great, filling family soup for a cold night.
1 1/2 pounds (680 g) bulk turkey sausage
1 can (14 1/2 ounces, or 410 g) diced tomatoes
1 can (8 ounces, or 225 g) sliced mushrooms
1 turnip, diced
1 cup (100 g) cauliflower, diced
1/2 cup (80 g) chopped onion
1 cup (150 g) chopped green bell pepper
1 quart (950 ml) chicken broth
2 teaspoons chicken bouillon concentrate
1 teaspoon dried basil
2 teaspoons prepared horseradish
1 cup (235 ml) heavy cream
In a large, heavy skillet, brown and crumble the sausage. Pour off the fat and put the sausage in your slow cooker. Add the tomatoes, mushrooms, turnip, cauliflower, onion, and green pepper.
In a bowl, stir the broth and bouillon together. Stir in the basil and horseradish. Pour the mixture into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 7 to 8 hours.
When the time's up, stir in the cream and let it cook for another 10 to 15 minutes.
_**Yield:**_ 6 servings, each with: 666 calories, 61 g fat, 17 g protein, 12 g carbohydrate, 2 g dietary fiber, 10 g usable carbs.
### Maria's New England Clam Chowder
This one is from my tester and dear friend Maria. She says, "This was so good that I called my brother Peter over and made him try it. He was very impressed. He's a big clam chowder fan, and he said that our grandmother would approve."
4 slices bacon, diced
1 medium onion, chopped
1 large turnip, cut in 1/2-inch (13 mm) cubes
1 can (16 ounces, or 455 g) clams, undrained
3 cloves garlic, crushed
1 teaspoon salt
1/2 teaspoon pepper
2 cups (475 ml) heavy cream
2 tablespoons (28 g) butter
In a big, heavy skillet, sauté the bacon, onion, and turnip until the onion is golden. Drain and put it on the bottom of your slow cooker.
Pour the clam liquid into a 2-cup (475 ml) measuring cup and add enough water to make 2 cups (475 ml). In a bowl, combine the liquid and water mixture, the clams, garlic, salt, and pepper. Pour the mixture into the slow cooker. Cover the slow cooker, set it to low, and let it cook for 5 hours or until the turnips are tender. Blend in the cream and butter during the last 45 minutes of cooking.
_**Yield:**_ 5 servings, each with: 550 calories, 44 g fat, 27 g protein, 11 g carbohydrate, 1 g dietary fiber, 10 g usable carbs.
### Seafood Chowder
My sister Kim, who tested this for me, said that you don't have to stick to shrimp. You could use crab, chunks of lobster tail, or even a cut-up firm-fleshed fish fillet. Don't use fake seafood—Delicaseas and such. It has a lot of added carbs.
1 1/2 cups (150 g) shredded cauliflower
1/3 cup (37 g) shredded carrots
1 teaspoon dried thyme
1 clove garlic
1 tablespoon (9 g) finely minced green bell pepper
1/8 teaspoon cayenne
1/4 teaspoon pepper
3 cups (700 ml) chicken broth
1 cup (235 ml) Carb Countdown dairy beverage
1/4 cup (60 ml) heavy cream
8 ounces (225 g) shrimp, shells removed
1 tablespoon (5 g) Ketatoes mix
1/4 cup (25 g) scallions, thinly sliced
Guar or xanthan
Combine the cauliflower, carrots, thyme, garlic, green pepper, cayenne, pepper, and broth in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 4 hours.
Turn the slow cooker to high and stir in the Carb Countdown and cream. Re-cover the slow cooker and let it cook for another 30 to 45 minutes. If your shrimp are big, chop them coarsely during this time, but little, whole shrimp will look prettier, of course!
Stir in the Ketotoes mix. Now stir in the shrimp and re-cover the pot. If your shrimp are pre-cooked, just give them 5 minutes or so to heat through. If they're raw, give them 10 minutes. Stir in the scallions and salt to taste. Thicken the broth with the guar or xanthan.
_**Yield:**_ 5 servings, each with: 164 calories, 8 g fat, 17 g protein, 7 g carbohydrate, 2 g dietary fiber, 5 g usable carbs.
chapter ten
Slow Cooker Sides
My guess is you use your slow cooker mostly for main dishes because that's the way it'll make your life the most convenient. Many of the main dishes in this book are complete meals, full of vegetables as well as protein, and they don't need a thing with them except a beverage. With the main dishes that do need a side, the easiest and most appealing thing is often a salad.
Yet there are good reasons to cook a side dish in your slow cooker. Some of them just plain take less work and watching this way, like the Southern Beans. Sometimes you want to make an interesting side to go with a plain meat roasting in the oven and want to be able to ignore it for an hour or two or three while you do something else. In both of these instances, the slow cooker is your very good friend.
### Barbecue Green Beans
These are fab!
4 cups (496 g) cross-cut frozen green beans, unthawed
1/4 cup (40 g) chopped onion
4 slices cooked bacon, drained and crumbled
1/3 cup (85 g) low-carb barbecue sauce (see recipe page 335 or use purchased sauce)
Put the green beans in your slow cooker. Add the onion and bacon and then stir in the barbecue sauce. Cover the slow cooker, set it to high, and let it cook for 3 hours. (If you prefer, set it to low and let it cook for 5 to 6 hours.)
_**Yield:**_ 6 servings, each with: 58 calories, 2 g fat, 3 g protein, 8 g carbohydrate, 2 g dietary fiber, 6 g usable carbs.
### Southern Beans
Southerners will be shocked to know that I never tasted green beans slowly cooked with bacon until I moved to southern Indiana, but I liked them right off. Around our house, this recipe is jokingly referred to as The Sacred Masonic Vegetable because my husband's never been to a Masonic banquet that didn't feature beans cooked this way!
4 cups (496 g) frozen green beans, unthawed
1/3 cup (53 g) diced onion
1/4 cup (30 g) diced celery
4 slices bacon, cooked and crumbled
1 tablespoon (15 g) bacon grease
1/2 cup (120 ml) water
Place the beans in the slow cooker and stir in everything else. Cover the slow cooker, set it to low, and let it cook for 4 hours.
_**Yield:**_ 6 servings, each with: 75 calories, 4 g fat, 3 g protein, 7 g carbohydrate, 3 g dietary fiber, 4 g usable carbs.
### Tangy Beans
4 cups (496 g) frozen green beans, unthawed
1/4 cup (40 g) chopped onion
1/4 cup (38 g) chopped green bell pepper
1/4 cup (60 ml) cider vinegar
2 tablespoons (3 g) Splenda
1/8 teaspoon black pepper
Combine everything in your slow cooker. Stir to distribute evenly. Cover the slow cooker, set it to low, and let it cook for 5 hours.
Serve with a pat of butter and a little salt.
_**Yield:**_ 4 servings, each with: 50 calories, trace fat, 2 g protein, 12 g carbohydrate, 4 g dietary fiber, 8 g usable carbs.
### Festive Green Beans
The jarred roasted red pepper and Alfredo sauce make this a whole lot easier than it tastes. Read the labels on the Alfredo sauce to find the lowest carb brand.
1 pound (455 g) frozen green beans, cross-cut
1/2 medium onion, chopped
1/3 cup (60 g) jarred roasted red pepper, drained and chopped
1/2 cup (125 g) jarred Alfredo sauce
1/4 cup (30 g) coarsely crushed pork rinds
Don't thaw the green beans, unless you want to speed the whole thing up by at least an hour. Throw everything but the pork rinds in your slow cooker and stir to combine. Cover and cook on low for 5 to 6 hours.
Coarsely crush the pork rinds and sprinkle over green beans just before serving.
_**Yield:**_ 5 servings, each with: 148 calories, 8 g fat, 10 g protein, 9 g carbohydrate, 3 g dietary fiber, 6 g usable carbs.
### Green Bean Casserole
28 ounces (785 g) frozen green beans, unthawed
1 cup (70 g) chopped mushrooms
1/4 cup (45 g) roasted red pepper, diced
1/4 cup (40 g) chopped onion
2 teaspoons dried sage
1 teaspoon salt or Vege-Sal
1 teaspoon pepper
1/2 teaspoon ground nutmeg
1 cup (235 ml) beef broth
1 teaspoon beef bouillon concentrate
1/2 cup (120 ml) heavy cream
Guar or xanthan
3/4 cup (83 g) slivered almonds
1 tablespoon (14 g) butter
Combine the green beans, mushrooms, red pepper, and onion in your slow cooker.
In a bowl, mix together the sage, salt or Vege-Sal, pepper, nutmeg, broth, and bouillon. Pour the mixture over the vegetables. Stir the whole thing up. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
When the time's up, stir in the cream and thicken the sauce a bit with guar or xanthan. Re-cover the slow cooker and let it stay hot while you sauté the almonds in the butter until golden. Stir them into the beans.
_**Yield:**_ 8 servings, each with: 191 calories, 14 g fat, 7 g protein, 12 g carbohydrate, 4 g dietary fiber, 8 g usable carbs.
### Creamy, Cheesy Mustard Green Beans and Mushrooms
Try this with a roasted ham!
4 ounces (115 g) mushrooms, chopped
1/4 medium onion, chopped
1 pound (455 g) frozen green beans, thawed
1/4 cup (45 g) jarred roasted red pepper, diced
1 cup (115 g) shredded Cheddar cheese
1 cup (235 ml) heavy cream
1 teaspoon beef bouillon granules
1/8 teaspoon pepper
1 1/2 teaspoons spicy brown mustard
Coat your slow cooker with nonstick cooking spray, dump in the mushrooms, onion, green beans, roasted red peppers, and cheese, and stir it up.
Combine the cream, beef bouillon concentrate, pepper, and mustard and stir until the bouillion is dissolved. Pour over the veggies.
Cover the slow cooker, set to low, and let it cook for 4 hours. Stir before serving.
_**Yield:**_ 6 servings, each with: 247 calories, 21 g fat, 7 g protein, 9 g carbohydrate, 3 g dietary fiber, 6 g usable carbs.
### Basic Artichokes
2 artichokes
1/4 cup (60 ml) lemon juice
Using kitchen shears, snip the pointy tips off the artichoke leaves. Split the artichokes down the middle, top to bottom, and scrape out the chokes.
Fill your slow cooker with water, add the lemon juice, and put in the artichokes. Cover the slow cooker, set it to high, and let it cook for 3 to 4 hours.
Drain the artichokes.
Serve the artichokes with the dipping sauce of your choice, such as lemon butter, mayonnaise, aioli, chipotle mayonnaise, whatever you've got. If you have a big slow cooker, feel free to cook more artichokes!
_**Yield:**_ 2 servings, each with: 68 calories, trace fat, 4 g protein, 16 g carbohydrate, 7 g dietary fiber, 9 g usable carbs.
### Maria's Slow Cooker Asparagus
Here's another recipe contributed by my pal and recipe tester Maria.
1 pound (455 g) asparagus spears
1 teaspoon dried rosemary
1 clove garlic, crushed
1 tablespoon (15 ml) lemon juice
Cut off woody ends of the asparagus. Place trimmed asparagus on the bottom of your slow cooker. (If you're using a round cooker, you may need to cut them to fit.) Sprinkle the asparagus with the rosemary and garlic and pour the lemon juice on top. Cover the slow cooker, set it to low, and let it cook for 2 hours or until the asparagus is tender.
_**Yield:**_ 4 servings, each with: 17 calories, trace fat, 1 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Spinach Parmesan Casserole
This is a lot like creamed spinach, only less—well, creamy.
20 ounces (560 g) frozen chopped spinach, thawed and drained*
1/3 cup (80 ml) heavy cream
1/2 cup (40 g) shredded Parmesan cheese
1 clove garlic, crushed
2 tablespoons (20 g) minced onion
1 egg
1/2 teaspoon salt
Place all ingredients in a mixing bowl. Stir it to blend very well. Spray a 6-cup (1.4 L) glass casserole dish with nonstick cooking spray. Put the spinach mixture in the casserole, smoothing the top.
Place the casserole dish in your slow cooker and carefully pour water around it up to 1 inch of the rim. Cover the slow cooker, set it to low, and let it cook for 4 hours.
Uncover the slow cooker and turn it off at least 30 minutes before serving time so the water cools enough that you can remove the dish without scalding yourself.
* Make sure your spinach is very well drained. It's best to put it in a colander and press it as hard as you can, turning it several times.
_**Yield:**_ 6 servings, each with: 109 calories, 8 g fat, 7 g protein, 5 g carbohydrate, 3 g dietary fiber, 2 g usable carbs.
### Macadangdang
Long-time readers know that I'm a big fan of Peg Bracken and her cookbooks. This is my decarbed version of Macadangdang Spinach Medley, which appeared in Peg's _I Hate To Cook Almanac_. It's named for her Aunt Henry Macadangdang, who married a Filipino gentleman of that name. I found the name so charming and euphonious, I thought I'd just call this version Macadangdang. My husband rates this a perfect 10, and I like it, too!
1/2 head cauliflower
10 ounces (280 g) frozen chopped spinach, thawed and drained
2 tablespoons (28 g) butter
1/2 cup (80 g) chopped onion
1 clove garlic, crushed
4 eggs
1/2 cup (120 ml) Carb Countdown dairy beverage
1 1/2 teaspoons salt or Vege-Sal
1/4 teaspoon pepper
1/2 cup (50 g) grated Parmesan cheese
1 cup (115 g) shredded mozzarella cheese
Run the cauliflower through the shredding blade of your food processor. Place the resulting _Cauli-Rice_ in a mixing bowl. Drain the thawed frozen spinach really well (I actually squeeze mine) and add it to the _Cauli-Rice_.
Melt the butter in a medium skillet, over medium-low heat, and sauté the onion until it's just translucent. Add the garlic, sauté for another minute or two, and then dump the whole thing in the bowl with the cauliflower and spinach.
Add the eggs, Carb Countdown, salt or Vege-Sal, pepper, and Parmesan and stir the mixture up quite well. Put it in a 1 1/2 to 2 quart glass (1.4 to 1.9 L) casserole dish that you've sprayed with nonstick cooking spray. Cover the dish with foil and place it in your slow cooker. Pour water around it up to 1 inch of the rim. Cover the slow cooker, set it to low, and let it cook for 2 1/2 hours.
Open the slow cooker and remove the foil. Sprinkle the mozzarella over your _Macadangdang_ , put the foil back on, recover the pot, and let it cook for another 20 minutes to melt the cheese. Then turn off the pot, uncover it, take off the foil, and let the whole thing cool just to the point where you can remove it from the water bath without scalding yourself, around 20 more minutes.
_**Yield:**_ 6 servings, each with: 198 calories, 14 g fat, 13 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### Broccoli with Bacon and Pine Nuts
This is quite special. Don't cook your broccoli any longer than 2 hours!
1 pound (455 g) frozen broccoli, unthawed
1 clove garlic, crushed
3 slices cooked bacon, crumbled
1 tablespoon (28 g) butter
1 tablespoon (15 ml) oil
2 tablespoons (18 g) pine nuts (pignolia), toasted
Place the broccoli in your slow cooker. Stir in the garlic and crumble in the bacon. Cover the slow cooker, set it to low, and let it cook for 2 hours.
Before serving, stir in the butter and oil and top with the pine nuts.
_**Yield:**_ 3 servings, each with: 184 calories, 15 g fat, 8 g protein, 8 g carbohydrate, 5 g dietary fiber, 3 g usable carbs.
### Fluffy, Savory Pumpkin
What a fantastic side dish this is! Serve this in place of the potatoes, and everyone will rave, even the carbivores.
1 large onion, chopped
4 tablespoons (55 g) butter
1 can (29 ounces, or 810 g) pumpkin purée
2 eggs
3/4 cup (195 g) ricotta cheese
3/4 cup (175 ml) heavy cream
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
Slowly sauté your chopped onion in the butter until it's soft and turning golden brown. Transfer to a big mixing bowl.
Add everything else and whisk it up until everything is well-mixed.
Coat your slow cooker with nonstick cooking spray. Spoon the pumpkin mixture into it and level the top. Cover, set the slow cooker to low, and cook for 4 to 5 hours.
_**Yield:**_ 8 servings, each with: 225 calories, 18 g fat, 6 g protein, 11 g carbohydrate, 3 g dietary fiber, 8 g usable carbs.
### Fauxtatoes
_Fauxtatoes_ is the now nearly-universal name for cauliflower puree, used by low-carbers everywhere as a substitute for mashed potatoes. To make plain Fauxtatoes as a side dish to serve with a slow-cooked main dish, see recipe page 343. These are slow cooked _Fauxtatoes_ with truly tasty additions.
### Cheddar-Barbecue Fauxtatoes
My husband was crazy about these!
1/2 head cauliflower, cut into florets
1/2 cup (120 ml) water
1/2 cup (58 g) shredded cheddar cheese
2 teaspoons _Classic Rub_ (see recipe page 338) or purchased barbecue rub
2 tablespoons (10 g) Ketatoes mix
Put the cauliflower in your slow cooker, including the stem. Add the water. Cover the slow cooker, set it to high, and let it cook for 3 hours. (Or cook it on low for 5 to 6 hours.)
When the time's up, use a slotted spoon to scoop the cauliflower out of the slow cooker into your blender or your food processor (have the **S** -blade in place) and purée it there, or you can drain off the water and use a hand-held blender to purée the cauliflower right in the pot. Either way, drain the cauliflower and purée it!
Stir in everything else until the cheese has melted.
_**Yield:**_ 3 servings, each with: 120 calories, 7 g fat, 8 g protein, 6 g carbohydrate, 3 g dietary fiber, 3 g usable carbs.
### Chipotle Fauxtatoes
This is killer with a grilled steak or barbecued brisket.
1/2 head cauliflower
1 chipotle chili canned in adobo
2 tablespoons (28 ml) olive oil
2 tablespoons (28 ml) half and half
6 garlic cloves
2 ounces (55 g) cream cheese
Whack your cauliflower into smallish chunks and throw it in the slow cooker.
Chop up your chipotle. Mix it, plus a couple of teaspoons of the adobo sauce from the can, with the olive oil, half and half, and garlic. Pour this over the cauliflower and toss to coat. Plunk the block of cream cheese on top.
Cover the slow cooker and cook for 3 hours on high or 5 to 6 hours on low.
When the time's up, uncover the slow cooker and use your stick blender to purée the cauliflower and mix in the cream cheese; then serve.
_**Yield:**_ 5 servings, each with: 115 calories, 10 g fat, 3 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### Italian Garlic and Herb Fauxtatoes
1/2 head cauliflower, cut into florets
1/2 cup (120 ml) chicken broth
1 teaspoon Italian seasoning
1 clove garlic, crushed
1 ounce (28 g) cream cheese
Guar or xanthan
Place the cauliflower in your slow cooker. Add the broth, Italian seasoning, and garlic. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours. (Or cook it on high for 3 hours.)
When the time's up, either remove the cauliflower with a slotted spoon and put it in your blender or food processor (with the **S** -blade in place) and purée it or drain the broth out of the slow cooker and use a hand-held blender to purée your cauliflower in the pot. Add the cream cheese and stir till melted.
The mixture will still be a little watery. Stir with a whisk as you use guar or xanthan to thicken it up a bit.
_**Yield:**_ 3 servings, each with: 46 calories, 4 g fat, 2 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Ranch and Green Onion Fauxtatoes
This is great with anything with a barbecue flavor.
1 head cauliflower, cut into florets
1 cup (235 ml) water
1 cup (100 g) Ketatoes mix
6 teaspoons (18 g) ranch-style dressing mix
4 scallions, thinly sliced
Place the cauliflower in your slow cooker with the water. Cover the slow cooker, set it to low, and let it cook for 5 hours. (Or cook it on high for 3 hours.)
When the time's up, the easiest thing to do is use a hand blender to purée the cauliflower right in the slow cooker. Don't bother to drain the water first. Whisk in the Ketatoes, ranch dressing mix, and scallions.
_**Yield:**_ 6 servings, each with: 170 calories, 3 g fat, 14 g protein, 23 g carbohydrate, 11 g dietary fiber, 12 g usable carbs.
### Garlic-Onion Fauxtatoes
Our tester Maria said her kids were particularly impressed by this!
1 head cauliflower, cut into florets
1/2 cup (80 g) chopped onion
3 cloves garlic, crushed
2/3 cup (150 ml) water
2/3 cup (67 g) Ketatoes mix
3 tablespoons (45 g) butter
Place the cauliflower in your slow cooker. Add the onion, garlic, and water. Cover the slow cooker, set it to high, and let it cook for 2 1/2 to 3 hours. (Or cook it on low for 5 to 6 hours.)
When the time's up, use a hand-held blender to purée the cauliflower, onion, and garlic right there in the slow cooker. Alternatively, scoop it all into a food processor to purée, but you'll want the water in the pot, so if you transfer the vegetables, put the purée back in the pot with the water when you're done. Now stir in the Ketatoes, butter, and salt and pepper to taste.
_**Yield:**_ 6 servings, each with: 162 calories, 8 g fat, 9 g protein, 15 g carbohydrate, 7 g dietary fiber, 8 g usable carbs.
### Easy (But Really Slow) Rutabagas
We adore rutabaga, but it can take a long time to cook soft. What better job for a slow cooker? This is how I cooked my rutabaga for Thanksgiving this year.
4 tablespoons (55 g) butter
1 large rutabaga
Salt and pepper
Throw the butter in the slow cooker and turn it on to low. Cover and let the butter melt.
Peel the rutabaga and cut it into 1/2-inch (13 mm) cubes. This takes a big sharp knife and some tough talk.
Dump the rutabaga cubes in the slow cooker—by now your butter should be melted. Toss the cubes to coat with the butter. Cover the slow cooker and let them cook for a minimum of 12 hours, and 14 isn't too much.
Toss with salt and pepper and serve.
_**Yield:**_ 6 servings, each with: 76 calories, 8 g fat, trace protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
_**Bonus Recipe:**_ Make Rutabaga Hash Browns! If you have any rutabaga left over, the next day you chop a little onion and throw it in your big, heavy skillet with the leftover rutabaga cubes and plenty of melted butter. Fry them, chopping up the rutabaga more with your spatula, and keep cooking until they're getting brown and crusty. YUMMY!
### Maple-Mustard Turnips
4 medium turnips
2 tablespoons (28 g) bacon grease, melted
2 tablespoons (40 g) sugar-free pancake syrup
1 tablespoon (11 g) brown mustard
1/2 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
2 dashes Tabasco sauce or other Louisiana-style hot sauce
Peel and cube the turnips and throw them in the slow cooker. Drizzle in the bacon grease and toss to coat.
Mix together everything else, drizzle over the turnips, and toss to coat again. Cover the pot, set to low, and cook for 5 hours.
_**Yield:**_ 6 servings, each with: 67 calories, 5 g fat, 1 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Cheesy Neeps
This would make a great holiday side dish. If you're serving buffet style, just plug in the slow cooker and they'll even stay warm for you. (In case you're wondering, _neeps_ is an old Scots term for turnips.)
2 large turnips
6 ounces (170 g) Swiss cheese (Gruyère is even better, if you can find it.)
1/2 medium onion
3 tablespoons (45 g) butter
1 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
Peel the turnips, whack them into hunks, and run them through your food processor's shredding blade. Run the Swiss cheese through, too. Chop your onion pretty fine—I find my shredding disc doesn't work for this, so I just used a knife and a cutting board, but you could swap out the shredding disc for the **S** -blade and use it to chop your onion fine, if you like.
Coat your slow cooker with nonstick cooking spray. Put the turnips, cheese, and onion in there and sprinkle the salt and pepper over it all. Toss until everything is well-combined.
Smooth the top. Now cut your butter into little bits and dot the top of the turnip mixture evenly with it. Cover the pot, set to low, and let it all cook for 5 hours or so.
_**Yield:**_ 6 servings, each with: 172 calories, 14 g fat, 9 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Mashed Turnips
2 large turnips, cubed
1/4 cup (40 g) chopped onion
3/4 cup (175 ml) beef broth
Put the turnips and onion in your slow cooker and add the broth. Cover the slow cooker, set it to low, and let it cook for 6 to 7 hours.
I like to mash these right in the pot with my hand-held blender, but if you prefer, you can transfer them to your food processor or regular blender.
If you're going to serve these plain, you might add some butter, salt, and pepper, but if you're serving them with a gravy, they're great as is.
_**Yield:**_ 4 servings, each with: 31 calories, trace fat, 3 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### The Simplest Slow Cooker Mushrooms
You want simple? We got simple.
1 pound (455 g) sliced mushrooms (You'll buy them that way, right?)
8 ounces (225 g) cream cheese with chives and onions
2 cloves garlic crushed
Dump everything in the slow cooker. Cover the pot. Set to low. Cook for 5 to 6 hours. Stir. Serve.
_**Yield:**_ 8 servings, each with: 110 calories, 9 g fat, 3 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Lemon-Parmesan Mushrooms
8 ounces (225 g) mushrooms
1/2 cup (125 ml) chicken broth
1/4 cup (60 ml) lemon juice
1/2 cup (40 g) shredded Parmesan cheese
1/4 cup (15 g) chopped fresh parsley
Wipe the mushrooms clean with a damp cloth or paper towel and put them in your slow cooker. Pour the broth and lemon juice over them. Cover the slow cooker, set it to low, and let it cook for 6 to 8 hours.
Remove the mushrooms from the slow cooker with a slotted spoon and put them on serving plates. Sprinkle with the Parmesan and parsley.
_**Yield:**_ 4 servings, each with: 65 calories, 3 g fat, 6 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Blue Cheese Mushrooms
These are perfect next to a steak or slab of prime rib.
8 ounces (225 g) fresh mushrooms
1/4 cup (30 g) crumbled blue cheese
2 ounces (55 g) cream cheese
1 clove garlic
Wipe or rinse your mushrooms and throw them in a slow cooker. Sprinkle the blue cheese over them and add the cream cheese cut into little hunks, so you can distribute it a bit. Crush the garlic and throw it in, too. Give it all a quick stir. Then cover, set to high, and cook for 2 hours or set to low and cook for 4 hours.
_**Yield:**_ 3 servings, each with: 126 calories, 10 g fat, 5 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Spinach-Mushroom Gratin
Mushrooms, spinach, and cheese—what's not to like? This is great with take-out grilled or rotisserie chicken.
1/2 head cauliflower
1 onion, chopped
4 tablespoons (55 g) butter
8 ounces (225 g) sliced mushrooms
3 cloves garlic, crushed
10 ounces (280 g) frozen chopped spinach, thawed
1 teaspoon thyme
1 teaspoon oregano
2 teaspoons salt or Vege-Sal
1 teaspoon pepper
1/2 teaspoon red pepper flakes
1 1/2 cups (173 g) shredded Cheddar cheese
Chop the onion. Melt the butter in your big heavy skillet and start sautéing the onions with the mushrooms; use the edge of your spatula to break the sliced mushrooms up into smaller pieces as they cook—you want them in chunks a little bigger than a pea.
Run your cauliflower through the shredding blade of your food processor.
Drain the spinach very well—I dump mine a colander in the sink and use my hands to squeeze it dry.
When the onion is translucent and the mushrooms have changed color, stir in the crushed garlic and then put all the vegetables in your slow cooker.
Add the thyme, oregano, salt, pepper, red pepper, and cheese. Stir it all up until it's well-combined. Cover, set to low, and let it cook for 4 hours.
_**Yield:**_ 8 servings, each with: 169 calories, 13 g fat, 8 g protein, 7 g carbohydrate, 3 g dietary fiber, 4 g usable carbs.
### Mushrooms Stroganoff
Even That Nice Boy I Married, not a mushroom fan, loved this. He actually asked for the leftovers to take for lunch! This is great over steak or chicken, but try it in omelets, too.
1 1/2 pounds (680 g) sliced mushrooms, button, crimini, and portobello
1 cup (235 ml) heavy cream
1/4 cup (60 ml) Worcestershire sauce
1/2 teaspoon salt or Vege-Sal
1/2 teaspoon pepper
I used 1/2 pound (225 g) of each kind of mushroom, but just one or two kinds is fine. Buy them sliced—so much easier! Dump all your mushrooms in the slow cooker, and if you're using different kinds, stir them together.
Stir everything else together, pour over the mushrooms, cover the pot, and set to low. Cook for 5 to 6 hours.
_**Yield:**_ 12 servings, each with: 87 calories, 8 g fat, 2 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Curried Mushrooms Malabar
This recipe is for fans of the exotic.
1 1/2 pounds (680 g) sliced mushrooms—portobello, crimini, button, or a combination
1 can (13 1/2 ounces, or 380 ml) coconut milk
3 cloves garlic, crushed
1 tablespoon (8 g) grated ginger root
1 tablespoon (11 g) mustard seed
2 teaspoons ground coriander
2 teaspoons ground cumin
1 teaspoon ground turmeric
1 teaspoon salt or Vege-Sal
Dump the mushrooms in the slow cooker. Measure everything else and stir it together well. Pour it evenly over the mushrooms, cover the pot, and set to low. Cook for 4 hours and then serve.
_**Yield:**_ 12 servings, each with: 92 calories, 8 g fat, 2 g protein, 5 g carbohydrate, 1 g dietary fiber, 4 g usable carbs.
### Baked Beans
This has a classic baked bean flavor and will be appreciated at any barbecue. I wouldn't attempt this recipe without a slow cooker. Do you have any idea how long it takes to cook soybeans soft? But with your slow cooker, you can just forget about them for 12 hours. If your health food store can't get you black soybeans (they're lower carb than white soybeans), you can order them from www.locarber.com.
2 cups (372 g) dry black soybeans
2 cups (475 ml) water
1/2 cup (80 g) chopped onion
1/2 tablespoon blackstrap molasses
3 tablespoons (45 g) low-carb ketchup (see recipe page 332)
1 tablespoon (9 g) dry mustard
2 tablespoons (3 g) Splenda
2 cups (475 ml) water
3/4 pound (340 g) smoked ham hocks
Put the soybeans in a big, nonreactive bowl and cover with the first 2 cups (475 ml) of water. Let them sit until the water is absorbed. Then put your soaked beans in the freezer overnight. (The freezing water will help break cell walls in the soybeans, making them soften faster when you cook them.)
When you want to cook the soybeans, thaw them and pour off any soaking water. Put them in your slow cooker. Add the onion, molasses, ketchup, dry mustard, and Splenda. Pour the additional 2 cups (475 ml) of water over all and stir it up. Now dig a hole in the center with a spoon and plunk the ham hock down in it. Cover the slow cooker, set it to low, and let cook for 12 hours.
Fish out the ham hock with tongs, remove and discard the skin and bone. Chop the meat and stir it back into the beans before serving.
_**Yield:**_ 10 servings (5 cups total), each with: 286 calories, 15 g fat, 8 g protein, 12 g carbohydrate, 11 g dietary fiber, 1 g usable carbs.
### Slow Cooker "Risotto" with Mushrooms and Peas
Okay, it's not really risotto. But it's got a lot of the same flavors, and it cooks while you do something else.
8 ounces (225 g) mushrooms, crimini or portobello, sliced
1 medium onion, chopped
2 tablespoons (28 g) butter
2 tablespoons (28 ml) olive oil
1 tablespoon (2 g) dried oregano
1/2 head cauliflower
1 cup (100 g) grated Parmesan cheese
1 1/2 cups (195 g) frozen peas
1/2 teaspoon pepper
1/3 cup (80 ml) dry white wine
2 teaspoons chicken bouillon granules
1 cup (235 ml) heavy cream
1 tablespoon (15 ml) balsamic vinegar
2 ounces (55 g) cream cheese
Start the mushrooms and chopped onion sautéing in the butter and olive oil. Use the edge of your spatula to break the mushrooms up into smaller pieces as you sauté them. When the onion is translucent and some liquid has cooked out of the mushrooms, stir in the oregano. Remove from heat.
Run the cauliflower through the shredding blade of your food processor.
Coat your slow cooker with nonstick cooking spray. Dump in the cauliflower, mushrooms and onion, peas, Parmesan, and pepper. Stir to combine.
In a 2-cup (475 ml) glass measuring cup, measure the wine. Add the cream, bouillon concentrate, and balsamic vinegar and whisk until the bouillon concentrate is dissolved. Pour this mixture into the cauliflower mixture and stir the whole thing up. Now pack it down and smooth the top. Put the cream cheese on top.
Cover the slow cooker, set to low, and let cook for 4 hours. When it's done, stir in the cream cheese until everything's evenly coated and serve.
_**Yield:**_ 6 servings, each with: 362 calories, 31 g fat, 10 g protein, 11 g carbohydrate, 3 g dietary fiber, 8 g usable carbs.
### Bavarian Cabbage
This is great with the _Sauerbrauten_ on page 126!
1 head red cabbage
1 medium onion, chopped
1 medium Granny Smith apple, chopped
6 slices cooked bacon, crumbled
2 teaspoons salt
1 cup (235 ml) water
3 tablespoons (4.5 g) Splenda
2/3 cup (160 ml) cider vinegar
3 tablespoons (45 ml) gin
Whack your head of cabbage in quarters and remove the core. Then whack it into biggish chunks. Put it in a big mixing bowl. Add the onion, apple, and bacon to the cabbage. Toss everything together. Transfer the mixture to your slow cooker. (This will fill a 3-quart [2.8 L] jobbie just about to overflowing! I barely got the top on mine.)
In a bowl, mix together the salt, water, Splenda, vinegar, and gin. Pour the mixture over the cabbage. Cover the slow cooker, set it to low, and let it cook for 6 to 8 hours.
_**Yield:**_ 6 servings, each with: 80 calories, 3 g fat, 2 g protein, 7 g carbohydrate, 1 g dietary fiber, 6 g usable carbs.
### Foil Packet Side Dishes
This is a nifty little technique: You wrap your vegetable side dish in a foil packet and then drop it in the slow cooker right on top of your main dish, whatever it may be. You get two dishes, with only the one slow cooker to wash.
However, since most vegetables cook more quickly than the main dishes do, you'll need to put them in later. That means these side dishes won't work for days when you're out from breakfast until supper. They're ideal, however, for those days when you're in and out of the house all day. Just make up the packet at the same time you put your main dish in to cook and stash it in the fridge. Then drop it in the slow cooker on one of those quick trips through the house, between picking up the dry cleaning and taking the kid to gymnastics.
I haven't tried it, but I'm betting using frozen vegetables would add a good hour to these cooking times, maybe more.
### Hobo UnPotatoes Slow-Cooker Style
4 slices cooked bacon
1/4 large head cauliflower
1/4 medium chopped onion
1/4 cup (28 g) shredded carrot
1/3 cup (40 g) diced celery
1/4 teaspoon salt or Vege-Sal
1/4 teaspoon pepper
1 tablespoon (14 g) butter
If you didn't start with cooked bacon, you might microwave it. Five minutes on high would be about right in my microwave.
Cut your cauliflower into chunks no bigger than 1 inch (2.5 cm). Chop your onion, shred your carrot, and dice your celery—all that stuff.
Lay a big sheet of foil on your counter and pile all the vegetables in the middle. Sprinkle the salt and pepper over them. Crumble in the bacon. Dot with the butter. You could add a spoonful of the bacon grease, too, if you wanted.
Lay another sheet of foil on top. Roll up all the edges, making a tight packet. Drop it in your slow cooker on top of whatever is cooking. This will take 3 to 4 hours on low.
_**Yield:**_ 3 to 4 servings; Assuming 4, each will have: 71 calories, 6 g fat, 2 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Cheesy Cauliflower Packet
1/4 large head cauliflower
1/4 green bell pepper
1/4 onion
1 tablespoon (14 g) butter
1/4 cup (30 g) shredded Cheddar cheese
Salt and pepper
Cut the cauliflower into chunks no bigger than 1 inch (2.5 cm); smaller will cook faster. Dice the pepper and onion.
Lay a big sheet of foil on your counter and pile the vegetables in the middle. Dot with the butter, sprinkle on the cheese, and salt and pepper it. Lay another sheet of foil over it and roll up the edges, making a packet. Drop this into your slow cooker on top of whatever supper is cooking—two hours on low is about right.
_**Yield:**_ 4 servings, each with: 67 calories, 5 g fat, 3 g protein, 3 g carbohydrate, 1 g dietary fiber, 2 g usable carbs.
### Chive-and-Garlic Spinach Packet
This is the easiest creamed spinach you'll ever make.
2 ounces (55 g) frozen chopped spinach, thawed
1 clove garlic, crushed
1/4 cup (60 g) cream cheese with chives and onions
1 tablespoon (14 g) butter
Drain your spinach really well—I dump mine into a colander in the sink and then squeeze it out with clean hands.
Stir the crushed garlic into the spinach.
Take a big sheet of foil and lay it on your counter. Pile the spinach on top. Add the cream cheese, in clumps, instead of one big blob. Dot with the butter.
Lay another sheet of foil over the spinach and roll up the sides, making a tight packet. Drop on top of whatever you're cooking in the slow cooker and let it cook for 60 to 90 minutes.
_**Yield:**_ 4 servings, each with: 80 calories, 7 g fat, 1 g protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Slow Cooker Chutney
This is the slow cooker version of my _Major Gray's Chutney_ recipe from _500 More Low-Carb Recipes_. It's wonderful with anything curried.
2 pounds (900 g) sliced peaches
1/3 cup (33 g) ginger root slices
1 1/2 cups (38 g) Splenda
3 cloves garlic
1 teaspoon red pepper flakes
1 teaspoon cloves
1 1/2 cups (355 ml) cider vinegar
Guar or xanthan
Combine everything but the guar or xanthan in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 4 hours.
Take the lid off and let it cook for another hour to let it cook down. Thicken a bit more, if you like, and store in an airtight container in the fridge.
_**Yield:**_ 32 servings (1 quart total), each with: 15 calories, trace fat, trace protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Slow Cooker Cranberry Sauce
I like to have cranberry sauce on hand for those occasions when I don't want to do much cooking. It adds interest to plain roasted chicken (or even store-bought rotisseried chicken.) It's easy to do and makes plenty!
24 ounces (680 g) cranberries
1 cup (235 ml) water
2 cups (50 g) Splenda
Simply combine everything in your slow cooker and give it a stir. Cover the slow cooker, set it to low, and let it cook for 3 hours.
This won't be as syrupy as commercial cranberry sauce because of the lack of sugar. If this bothers you, you can thicken your sauce with your trusty guar or xanthan shaker, but I generally leave mine as is. This makes quite a lot, so divide it between three or four snap-top containers and store it in the freezer. This way, you'll have cranberry sauce on hand whenever you bring home a rotisseried chicken!
_**Yield:**_ Makes about 2 3/4 cups, or 22 servings of 2 tablespoons, each with: 15 calories, trace fat, trace protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Cranberry Chutney
This is a cranberry sauce with a kick! It's great with anything curried or just with chicken or pork.
12 ounces (340 g) fresh cranberries
2/3 cup (16 g) Splenda—or other sugar-free sweetener to equal 2/3 cup (133 g) sugar
1/2 cup (120 ml) water
1/2 cup (120 ml) cider vinegar
6 whole allspice berries
6 whole cloves
3 cinnamon sticks
3 garlic cloves, crushed
1/2 medium onion, diced
Guar or xanthan
Simply combine everything but the guar or xanthan in your slow cooker. Cover, set to low, and let cook for 4 hours. Then let it cool, thicken to a syrupy consistency with your guar or xanthan shaker, and store in a snap-top container in the fridge. If you're not going to use it up in a week or two, consider two or three smaller containers in the freezer.
_**Yield:**_ 12 servings, each with: 51 calories, 1 g fat, 1 g protein, 13 g carbohydrate, 5 g dietary fiber, 8 g usable carbs.
### Cranberry-Peach Chutney
This is seriously kicked-up from regular cranberry sauce! It's a natural with curried poultry, but try it with any simple poultry or pork dish.
12 ounces (340 g) cranberries
1 1/2 cups (255 g) diced peaches (I use unsweetened frozen peach slices, diced.)
1 clove garlic, minced
3 inches (7.5 cm) ginger root, sliced into paper-thin rounds
1 lime, sliced paper-thin
1 1/4 cups (31 g) Splenda
1 cinnamon stick
1 teaspoon mustard seed
1/4 teaspoon salt
1/4 teaspoon orange extract
1/4 teaspoon baking soda
Combine everything but the baking soda in your slow cooker. Cover the slow cooker, set it to low, and let it cook for 3 hours, stirring once halfway through.
When the time's up, stir in the baking soda and keep stirring till the fizzing subsides. Store in a tightly lidded container in the fridge. If you plan to keep it for long, freezing's a good idea.
Why use baking soda? Because by neutralizing some of the acid in the cranberries, it lets you get away with less Splenda—and fewer carbs.
_**Yield:**_ Makes about 2 1/2 cups, or 20 servings of 2 tablespoons, each with: 31 calories, trace fat, trace protein, 8 g carbohydrate, 1 g dietary fiber, 7 g usable carbs.
chapter eleven
Slow Cooker Desserts
There are desserts that adapt well to the slow cooker, and then there are desserts that don't. In this chapter, I've really played to the slow cooker's strengths. Custards actually cook better in a slow cooker than in the oven, which is why you'll find half a dozen of them here. Indeed, they're so easy in the slow cooker and so appealing and nutritious, you may find yourself making custard more often.
Your slow cooker also excels at baking cheesecake, though you may not know it yet. Give it a try!
### Chocolate Fudge Custard
This really is dense and fudgy. It's intensely chocolatey, too.
1 cup (235 ml) Carb Countdown dairy beverage
3 ounces (85 g) unsweetened baking chocolate
2/3 cup (16 g) Splenda
1 cup (235 ml) heavy cream
1/2 teaspoon vanilla extract
1 pinch salt
6 eggs, beaten
In a saucepan, over the lowest possible heat (use a double boiler or heat diffuser if you have one), warm the Carb Countdown with the chocolate. When the chocolate melts, whisk the two together and then whisk in the Splenda.
Spray a 6-cup (1.4 L) glass casserole dish with nonstick cooking spray. Pour the cream into it and add the chocolate mixture. Whisk in the vanilla extract and salt. Now add the eggs, one by one, whisking each in well before adding the next one.
Put the casserole dish in your slow cooker and pour water around it up to 1 inch of the top rim. Cover the slow cooker, set it to low, and let it cook for 4 hours.
Then turn off the slow cooker, remove the lid, and let the water cool enough so it won't scald you before removing the casserole dish. Chill the custard well before serving.
_**Yield:**_ 6 servings, each with: 299 calories, 28 g fat, 10 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs.
### Flan
This is my slow cooker version of _Maria's Flan_ from _500 Low-Carb Recipes_. It's so rich!
2 tablespoons (42 g) sugar-free imitation honey
1 teaspoon blackstrap molasses
1 cup (235 ml) Carb Countdown dairy beverage
1 cup (235 ml) heavy cream
6 eggs
2/3 cup (16 g) Splenda
1 teaspoon vanilla
1 pinch nutmeg
1 pinch salt
Spray a 6-cup (1.4 L) glass casserole dish with nonstick cooking spray. In a bowl, mix together the honey and the molasses. Pour the mixture in the bottom of the casserole dish.
In a mixing bowl, preferably one with a pouring lip, combine the Carb Countdown, cream, eggs, Splenda, vanilla, nutmeg, and salt. Whisk everything together well. Pour the mixture into the casserole dish.
Carefully lower the casserole dish into your slow cooker. Now pour water around the casserole dish to within 1 inch of the rim. Cover the slow cooker, set it to low, and let it cook for 3 to 3 1/2 hours.
_**Yield:**_ 6 servings, each with: 229 calories, 20 g fat, 8 g protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs. (Analysis does not include the polyols in the imitation honey.)
### Southeast Asian Coconut Custard
I adapted this from a carby recipe in another slow cooker book. Maria, who tested it, says it's wonderful and also has a Latino feel to it. Look for shredded unsweetened coconut in Asian markets and health food stores.
1/4 cup (84 g) sugar-free imitation honey
1/2 teaspoon blackstrap molasses
1 1/2 teaspoons grated ginger root
1 tablespoon (15 ml) lime juice
1 can (14 ounces, or 390 ml) coconut milk
2/3 cup (16 g) Splenda
1/4 teaspoon ground cardamom
1 teaspoon grated ginger root
1/2 cup (120 ml) Carb Countdown dairy beverage
1/2 cup (120 ml) heavy cream
1/2 teaspoon vanilla extract
4 eggs
1/2 cup (40 g) shredded unsweetened coconut
Spray a 6-cup (1.4 L) glass casserole dish with nonstick cooking spray. Put the honey and molasses in the casserole dish. Cover the casserole dish with plastic wrap or a plate and microwave on high for 2 minutes. Add the 1 1/2 teaspoons ginger and lime juice and stir. Set aside.
In a mixing bowl, combine the coconut milk, Splenda, cardamom, 1 teaspoon ginger, Carb Countdown, cream, vanilla extract, and eggs. Whisk until well combined. Pour into the casserole dish. Cover the casserole dish with foil and secure it with a rubber band.
Put the casserole dish in your slow cooker and pour water around it to within 1 inch of the rim. Cover the slow cooker, set it to low, and let it cook for 3 to 4 hours.
Turn off the slow cooker, uncover, and let it cool till you can lift out the casserole dish without scalding your fingers. Chill overnight.
Before serving, stir the coconut in a dry skillet over medium heat until it's golden. Remove the custard from the fridge and run a knife carefully around the edge. Put a plate on top and carefully invert the custard onto the plate. Sprinkle the toasted coconut on top.
_**Yield:**_ 8 servings, each with: 227 calories, 22 g fat, 5 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs. (Analysis does not include the polyols in the imitation honey.)
### Maple Custard
This is for all you maple fans out there, and I know you are legion!
1 1/2 cups (355 ml) Carb Countdown dairy beverage
1/2 cup (120 ml) heavy cream
1/3 cup (107 ml) sugar-free pancake syrup
1/3 cup (8 g) Splenda
3 eggs
1 pinch salt
1 teaspoon vanilla extract
1/2 teaspoon maple extract
Simply whisk everything together and pour the mixture into a 6-cup (1.4 L) glass casserole dish you've sprayed with nonstick cooking spray. Put the casserole dish in your slow cooker and pour water around it to within 1 inch of the rim. Cover the slow cooker, set it to low, and let it cook for 4 hours.
When the time's up, turn off the slow cooker, remove the lid, and let it sit until the water is cool enough so that you can remove the casserole without risk of scalding. Chill well before serving.
_**Yield:**_ 6 servings, each with: 135 calories, 12 g fat, 6 g protein, 2 g carbohydrate, 0 g dietary fiber, 2 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Maple-Pumpkin Custard
This is very much like the filling of a pumpkin pie, without the crust. The pecans add a little textural contrast.
1 can (15 ounces, or 425 g) canned pumpkin purée
1 cup (235 ml) Carb Countdown dairy beverage
1/2 cup (120 ml) heavy cream
1/3 cup (107 ml) sugar-free pancake syrup
1/3 cup (8 g) Splenda
1/2 teaspoon maple flavoring
3 eggs
1 pinch salt
1 tablespoon (6 g) pumpkin pie spice
1/3 cup (37 g) chopped pecans
1 1/2 teaspoons butter
_Whipped Topping_ (see recipe page 342)
In a mixing bowl, preferably one with a pouring lip, whisk together the pumpkin, Carb Countdown, cream, pancake syrup, Splenda, maple flavoring, eggs, salt, and pumpkin pie spice.
Spray a 6-cup (1.4 L) glass casserole dish with nonstick cooking spray. Pour the custard mixture into it. Place it in your slow cooker. Now carefully fill the space around the casserole with water up to 1 inch from the rim. Cover the slow cooker, set it to low, and let it cook for 3 to 4 hours.
Remove the lid, turn off the slow cooker, and let it cool till you can remove the casserole dish without scalding your fingers. Chill the custard for at least several hours.
Before serving, put the pecans and butter in a heavy skillet over medium heat and stir them for 5 minutes or so. Set aside. Also have the _Whipped Topping_ made and standing by.
Serve the custard with a dollop of _Whipped Topping_ and 1 tablespoon (7 g) of toasted pecans on each serving.
_**Yield:**_ 6 servings, each with: 341 calories, 31 g fat, 7 g protein, 10 g carbohydrate, 3 g dietary fiber, 7 g usable carbs. (Analysis does not include the polyols in the sugar-free pancake syrup.)
### Apricot Custard
Don't go increasing the quantity of apricot preserves here. They're the biggest source of carbs. This dessert is yummy, though!
1/3 cup (107 g) low-sugar apricot preserves
2 tablespoons (28 ml) lemon juice
2 teaspoons Splenda
1 1/2 cups (355 ml) Carb Countdown dairy beverage
1/2 cup (120 ml) heavy cream
4 eggs
2/3 cup (16 g) Splenda
1/2 teaspoon almond extract
1 pinch salt
Whisk together the preserves, lemon juice, and the 2 teaspoons of Splenda. Spread them over the bottom of a 6-cup (1.4 L) glass casserole dish you've sprayed with nonstick cooking spray. Set aside.
Whisk together the Carb Countdown, cream, eggs, 2/3 cup (16 g) Splenda, almond extract, and salt. Pour into the prepared casserole gently, so as not to mix in the apricot preserves.
Place the casserole dish in your slow cooker. Pour water around the casserole to within 1 inch of the rim. Cover the slow cooker, set it to low, and let it cook for 4 hours.
When the time's up, turn off the slow cooker, uncover it, and let it cool until you can remove the casserole dish without risk of scalding. Chill well before serving.
_**Yield:**_ 6 servings, each with: 165 calories, 12 g fat, 7 g protein, 7 g carbohydrate, trace dietary fiber, 7 g usable carbs.
### Peaches with Butterscotch Sauce
These are delectable. You can serve them as is, with a little heavy cream, with _Whipped Topping_ (see recipe page 342)—or, the Big Casino, with a scoop of low-carb vanilla ice cream.
1 pound (455 g) frozen, unsweetened, sliced peaches
2 teaspoons lemon juice
1/3 cup (8 g) Splenda
2 tablespoons (42 g) sugar-free imitation honey
1/2 teaspoon blackstrap molasses
2 tablespoons (28 ml) heavy cream
1/4 teaspoon cinnamon
2 tablespoons (28 g) butter, melted
Guar or xanthan
Place the peaches in your slow cooker. (I didn't even bother to thaw mine.)
In a bowl, stir together the lemon juice, Splenda, honey, molasses, cream, cinnamon, and butter. Pour the mixture over the peaches. Cover the slow cooker, set it to low, and let it cook for 6 hours.
Thicken the sauce to a creamy consistency with a little guar or xanthan and serve hot.
_**Yield:**_ 6 servings, each with: 86 calories, 6 g fat, 1 g protein, 9 g carbohydrate, 2 g dietary fiber, 7 g usable carbs. (Analysis does not include polyols in the imitation honey.)
### Rhubarb Flummery
Because it's so sour, rhubarb is low-carb. This is a simple, old-fashioned dessert.
1 pound (455 g) frozen rhubarb
1/2 cup (12 g) Splenda
1/2 cup (120 ml) water
1/8 teaspoon orange extract
Guar or xanthan
Place the rhubarb in your slow cooker and stir in the Splenda, water, and orange extract. Cover the slow cooker, set it to low, and let it cook for 5 to 6 hours.
When the time's up, the rhubarb will be very soft. Mash it with a fork to a rough pulp. Thicken the sauce to a soft pudding consistency with guar or xanthan and serve hot or cold.
This dessert is great with a little heavy cream or _Whipped Topping_ (see recipe page 342).
_**Yield:**_ 6 servings, each with: 16 calories, trace fat, trace protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Hot Cinnamon Mocha
Assemble this in your slow cooker before going skating, caroling, or to a football game and have a winter party waiting when you get home!
1/2 gallon (1.9 L) chocolate-flavored Carb Countdown dairy beverage
2 cinnamon sticks
3 tablespoons (18 g) instant coffee granules
1 1/2 teaspoons vanilla extract
Combine everything in your slow cooker and give it a stir. Cover the slow cooker, set it to high, and let it cook for 3 hours. Turn the slow cooker to low and serve from the slow cooker.
If it's a grown-up party, put a bottle of _Mockahlua_ on the side for spiking! (See recipe page 342.)
_**Yield:**_ 10 servings, each with: 92 calories, 4 g fat, 10 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### About Cheesecake
There were so many cheesecake recipes in _500 More Low-Carb Recipes_ that I only did three for this book. However, cheesecake works very well in the slow cooker, so feel free to experiment with baking your favorite low-carb cheesecake recipes this way.
You'll notice I've called for light cream cheese (or Neufchâtel, which is, so far as I can tell, the same thing) and light sour cream in these recipes, instead of the full-fat versions. There's a reason for this: The light versions generally have no more carbohydrate than their full-fat counterparts, and they are, of course, lower calorie. I consider that a gain, and the cheesecakes come out very well. However, feel free to use the full-fat versions if you prefer.
(Since writing the above paragraph, my feelings have changed. I now deliberately aim for 75 percent of my calories to come from fat and prefer full fat cream cheese.)
### Mochaccino Cheesecake
This cheesecake is extraordinary, as good as any dessert I ever had in a restaurant. This alone is a good enough reason to go buy a large, round slow cooker and an 8-inch (20 cm) springform to fit into it! It's also a good excuse to make some _Mockahlua_ (see recipe page 342), but who needs an excuse to do that?
_Crisp Chocolate Crust_ (see recipe on the next page)
16 ounces (455 g) light cream cheese or Neufchâtel cheese, softened
1 egg
1/4 cup (60 ml) heavy cream
1/2 cup (45 g) + 2 tablespoons (12 g) unsweetened cocoa powder
1/2 cup (12 g) Splenda
1/4 cup (60 ml) _Mockahlua_ (see recipe page 342)
2 tablespoons (28 ml) brewed coffee
Using your electric mixer, beat together the cream cheese, egg, and cream until quite smooth. (You'll need to scrape down the sides of the bowl several times.) Now beat in the cocoa powder, Splenda, _Mockahlua_ , and coffee. When it's all well blended and very smooth, pour into the crust. Cover the springform pan tightly with foil, squeezing it in around the rim.
Take a big sheet of foil, at least 18 inches (46 centimeters) long, and roll it into a loose cylinder. Bend it into a circle and place it in the bottom of your slow cooker. (You're making a rack to put the pan on.) Pour 1/4 inch of water in the bottom of the slow cooker and then put the pan on the donut of foil. Cover the slow cooker, set it to high, and let it cook for 3 to 4 hours.
Turn off the slow cooker, uncover, and let cool for at least 20 to 30 minutes before you try to remove the pan from the slow cooker. Chill well before serving.
It's nice to make the _Whipped Topping_ (see recipe page 342), with a little _Mockahlua_ in it, to serve on top of this, but it's hardly essential.
_**Yield:**_ 12 servings, each with: 284 calories, 24 g fat, 11 g protein, 10 g carbohydrate, 4 g dietary fiber, 6 g usable carbs. (Analysis includes Crisp Chocolate Crust. You could cut this into eight, more generous servings if you'd like.)
### Crisp Chocolate Crust
1 1/2 cups (218 g) almonds
1/4 cup (6 g) Splenda
2 squares bitter chocolate, melted
3 tablespoons (45 g) butter, melted
2 tablespoons (20 g) vanilla whey protein powder
Preheat the oven to 325°F (170°C, or gas mark 3).
Using the **S** -blade of your food processor, grind the almonds until they're the texture of corn meal. Add the Splenda and pulse to combine. Pour in the chocolate and butter and run processor till evenly distributed. (You may need to stop the processor and run the tip of a knife blade around the outer edge to get everything to combine properly.) Then add the protein powder and pulse again to combine.
Turn the mixture into an 8-inch (20 cm) springform pan you've coated with nonstick cooking spray. Press firmly and evenly into place. Bake for 10 to 12 minutes in the preheated oven. Cool before filling.
_**Yield:**_ 12 servings, each with: 164 calories, 15 g fat, 6 g protein, 5 g carbohydrate, 3 g dietary fiber, 2 g usable carbs.
### New York–Style Cheesecake
You can top this with fruit if you like, but it's mighty good just as it is.
_"Graham" Crust_ (see recipe on the next page)
1 pound (455 g) light cream cheese or Neufchâtel cheese, softened
1/2 cup (115 g) light sour cream
2 eggs
1/2 cup (12 g) Splenda
2 teaspoons vanilla extract
1 pinch salt
Prepare the _"Graham" Crust_ and let it cool.
Using your electric mixer, beat the cheese, sour cream, and eggs until they're very smooth. (You'll need to scrape down the sides of the bowl at least a few times.) Now beat in the Splenda, vanilla extract, and salt. Pour into the waiting crust. Cover the pan tightly with foil, squeezing it in around the rim.
Take a big sheet of foil, at least 18 inches (45 cm) long, and roll it into a loose cylinder. Bend it into a circle and place it in the bottom of your slow cooker. (You're making a rack to put the pan on.) Pour 1/4 inch of water into your slow cooker and then put the pan on the donut of foil. Cover the slow cooker, set it to high, and let it cook for 3 to 4 hours.
Turn off the slow cooker, uncover, and let cool for at least 20 to 30 minutes before you try to remove the pan from the slow cooker. Chill well before serving.
_**Yield:**_ 12 servings, each with: 241 calories, 21 g fat, 8 g protein, 6 g carbohydrate, 2 g dietary fiber, 4 g usable carbs. (You could cut this into eight, more generous servings if you'd like.)
### "Graham" Crust
Wheat germ and wheat bran give this a "graham" flavor.
1 1/4 cups (181 g) almonds
2 tablespoons (14 g) wheat germ
2 tablespoons (14 g) wheat bran
3 tablespoons (4.5 g) Splenda
1 pinch salt
6 tablespoons (85 g) butter, melted
Preheat the oven to 325°F (170°C, or gas mark 3).
Put the almonds in your food processor with the **S** -blade in place. Run it until they're ground to about the texture of corn meal. Add the wheat germ, wheat bran, Splenda, and salt and pulse to combine. Now turn on the processor and pour in the butter, running the processor until everything's well combined. (You may need to stop the processor and run a knife around the bottom edge to make sure all the dry ingredients come in contact with the butter.)
Turn this mixture out into an 8-inch (20 cm) springform pan you've sprayed with nonstick cooking spray. Press firmly into place. Bake for 10 to 12 minutes or until just turning gold around the edges. Cool before filling.
_**Yield:**_ 12 servings, each with: 144 calories, 14 g fat, 3 g protein, 4 g carbohydrate, 2 g dietary fiber, 2 g usable carbs.
### Peanut Butter Cheesecake
You can certainly eat this plain, but I'd likely top it with some sugar-free chocolate sauce. Hershey's makes one that's available in my grocery store, or you could order some from a low-carb e-tailer. Or you could make some from the recipe on page 341. Or for that matter, you could melt 6 to 8 ounces (170 to 225 grams) of your favorite sugar-free chocolate bars and swirl them into the peanut butter batter before baking. The possibilities are endless!
_Crisp Chocolate Crust_ (see recipe page 322) or _"Graham" Crust_ (see recipe page 323)
16 ounces (455 g) light cream cheese or Neufchâtel cheese, softened
1/2 cup (115 g) light sour cream
1 egg
3/4 cup (195 g) natural peanut butter (salted is better than no-salt-added, here)
2/3 cup (16 g) Splenda
1/2 teaspoons blackstrap molasses
Have your crust made and standing by.
Using your electric mixer, beat the cream cheese or Neufchâtel, sour cream, and egg until they're very smooth. (You'll want to scrape down the sides of the bowl several times.) Now beat in the peanut butter, Splenda, and molasses.
When the mixture is very smooth and well blended, pour it into the crust. Cover the pan tightly with foil, squeezing it in around the rim.
Take a big sheet of foil, at least 18 inches (45 cm) long, and roll it into a loose cylinder. Bend it into a circle and place it in the bottom of your slow cooker. (You're making a rack to put the pan on.) Pour 1/4 inch of water into the slow cooker and then put the pan on the donut of foil. Cover the slow cooker, set it to high, and let it cook for 3 to 4 hours.
Turn off the slow cooker, uncover, and let cool for at least 20 to 30 minutes before you try to remove the pan from the slow cooker. Chill well before serving.
_**Yield:**_ 12 servings, each with: 364 calories, 32 g fat, 13 g protein, 10 g carbohydrate, 4 g dietary fiber, 6 g usable carbs. (Analysis includes Crisp Chocolate Crust. Analysis does not include any chocolate sauce or melted chocolate you might add! You could cut this into eight, more generous servings if you'd like.)
### Chocolate Chocolate-Chip Pudding Cake
YUM! It's so gooey, chocolatey, and good! This is high enough in carbs that you'll want to save it for special occasions, but wow! Whipped cream or no-sugar-added vanilla ice cream would be good with this, but they're hardly essential. Serve hot or cold.
1/2 cup (56 g) almond meal
1/2 cup (80 g) vanilla whey protein powder
1 teaspoon guar or xanthan, divided
1/4 cup (23 g) plus 3 tablespoons (18 g) unsweetened cocoa powder
3/4 cup (180 g) erythritol or xylitol, divided
2 teaspoons baking powder
1/4 teaspoon salt
1/2 cup (120 ml) half and half
3 tablespoons (45 g) butter—melted
1/2 cup (88 g) no-sugar chocolate chips
1/2 cup (12 g) Splenda, divided
2 tablespoons (3 g) Splenda
1 cup (235 ml) hot water
Coat your slow cooker with nonstick cooking spray.
In a mixing bowl, combine the almond meal, vanilla whey protein, 1/2 teaspoon of guar or xanthan, 3 tablespoons (18 g) of cocoa powder, 1/4 cup (60 g) plus 2 tablespoons (30 g) of erythritol, 1/4 cup (6 g) plus 2 tablespoons (3 g) of Splenda, baking powder, and salt. Stir together.
Add the half and half and melted butter and whisk together well. Stir in the chocolate chips and scrape the batter into the slow cooker.
In another bowl, combine the remaining 1/4 cup (60 g) plus 2 tablespoons (30 g) erythritol, 1/4 cup (6 g) Splenda, 1/2 teaspoon guar or xanthan, and 1/4 cup (23 g) cocoa powder. Sprinkle this mixture over the batter in the slow cooker. Now pour the hot water evenly over that. Do not mix.
Cover and cook on high for 1 1/2 to 2 hours and then serve.
_**Notes:**_ I use Nevada Manna brand sugar-free chocolate chips from CarbSmart.com. If you prefer, you could chop up sugar-free semi-sweet chocolate bars in your food processor; that's what I did before I could get sugar-free chocolate chips. Also, I think this would be nice with a teaspoon or two of instant coffee powder mixed in with the dry ingredients, for a mocha flavor.
_**Yield:**_ 6 servings, each with: 328 calories, 19 g fat, 22 g protein, 23 g carbohydrate, 12 g dietary fiber, 11 g usable carbs.
It was a great day for me when I figured out that my Bundt pan just exactly fit in my big slow cooker, hanging from the rim, and allowing the lid to fit on tightly. I really like baking in my slow cooker. Not only are the results excellent, but it uses less energy than my oven. In the summer, it will heat up the kitchen less. And on holidays, it frees up the oven for other things, like a turkey or a ham. But you will need a slow cooker that fits a Bundt pan.
About dried egg white powder: You can find this in cans in the baking aisle, but I recommend buying it through Amazon.com; you'll get it cheaper. It's pricey, but it improves the texture of nut meal-based baked goods quite a lot.
### Chocolate Chip Cinnamon Zucchini Snack Cake
This is a moist and sturdy cake, perfect for snacking on right out of your hand.
2 1/4 cups (252 g) almond meal, divided
1/2 cup (115 g) dried egg white powder
2 teaspoons baking powder
2 teaspoons guar or xanthan
1 cup (160 g) vanilla whey protein powder
1/2 teaspoons baking soda
3/4 teaspoon salt
1 1/2 teaspoons cinnamon
1/2 cup (120 g) erythritol or xylitol
1/2 cup (12 g) Splenda
3 eggs
2/3 cup (150 g) coconut oil, melted
2/3 cup (154 g) plain yogurt
1 1/2 cups (180 g) shredded zucchini—about 1 small zucchini
1 cup (175 g) sugar-free chocolate chips
Heavily grease a Bundt pan and use 1/4 cup (28 g) of the almond meal to flour it.
In a mixing bowl, combine all the other dry ingredients and stir until they're evenly distributed.
In another bowl, whisk together the eggs, melted coconut oil, and yogurt. If you haven't shredded your zucchini, take care of that little matter now, before you combine the wet ingredients with the dry ones.
Okay, dump the egg/oil/yogurt mixture into the dry ingredients and whisk them together, making sure you leave no pockets of dry stuff at the bottom. Now whisk in the zucchini and finally the chocolate chips.
Scrape the batter into the prepared Bundt pan and place the pan in your slow cooker. Set it for high and let it cook for 3 hours or until it's pulling away from the sides of the pan and a wooden skewer inserted halfway between the walls of the pan comes out clean. Turn off the slow cooker and leave it uncovered until the pan cools enough for you to handle it. Then turn out on a wire rack to finish cooling.
_**Note:**_ This, too, is not as sweet as commercial baked goods. Again, feel free to increase the sweetener if you like.
_**Yield:**_ 16 servings, each with: 328 calories, 20 g fat, 24 g protein, 17 g carbohydrate, 8 g dietary fiber, 9 g usable carbs.
### Pumpkin Bread
This has enough carbs that it's not Induction food, and it probably shouldn't be a staple. But it's a magnificent treat: sweet and spicy and moist. With as much protein per slice as 4 eggs, it'll keep you full for a long, long time. It would make a great breakfast for egg-resistant kids! Since there are only two in my household, I cut it into slices, put each in a zipper-lock bag, and freeze them, so I have a treat with my tea whenever I like.
2 1/4 cups (252 g) almond meal, divided
1 1/2 cups (240 g) vanilla whey protein powder
1/2 cup (115 g) egg white powder
2 teaspoons baking powder
2 teaspoons baking soda
2 teaspoons guar or xanthan
1 cup (240 g) erythritol or xylitol
1 cup (25 g) Splenda
1 1/2 teaspoons salt
1 1/2 teaspoon cinnamon
1 teaspoon nutmeg
1/2 teaspoon ground cloves
1/4 teaspoon ground ginger
1 cup (225 g) coconut oil, melted
4 eggs
1 can (15 ounces, or 425 g) pumpkin purée
2/3 cup (160 ml) water
1 cup (120 g) chopped walnuts
Heavily grease your Bundt pan. Use 1/4 cup (28 g) of the almond meal to flour the pan.
In a big mixing bowl, combine all the dry ingredients, from the remaining 2 cups (224 g) almond meal through the ginger. Stir everything together until it's all evenly distributed. Break up any big lumps of stuff.
In another bowl, whisk together the melted oil, eggs, pumpkin purée, and water. Dump this mixture into the dry ingredients and whisk it all together, making sure there are no pockets of dry stuff anywhere.
Now whisk in the walnuts.
Scrape the batter into the prepared Bundt pan. Place the pan in your slow cooker, cover, and turn it on to high. Cook for 2 1/2 to 3 hours or until a wooden skewer inserted halfway between the walls of the pan comes out clean.
Uncover the slow cooker and turn it off. Let the pumpkin bread cool there until you can handle the pan without burning yourself. Then remove from the slow cooker and turn it out on a wire rack to finish cooling.
_**Note:**_ This version is mildly sweet, because I don't like stuff overwhelmingly sweet. If you're still coming off of major sugar addiction, you can increase the erythritol and Splenda each by as much as 1/2 cup (120g and 12g).
_**Yield:**_ 16 slices, each with: 360 calories, 24 g fat, 27 g protein, 13 g carbohydrate, 2 g dietary fiber, 11 g usable carbs.
### Caramel Mocha Latte
This isn't a baked good, but it would be good with them! And I couldn't figure out where else to put it.
12 cups (2.8 L) brewed coffee
3 cups (700 ml) heavy cream
1/2 cup (120 g) sugar-free chocolate coffee flavoring syrup
1/2 cup (120 g) caramel sugar-free coffee flavoring syrup
Just combine everything in your slow cooker and keep it on low to serve.
_**Yield:**_ 12 generous servings, each with: 210 calories, 22 g fat, 1 g protein, 3 g carbohydrate, 0 g dietary fiber, 3 g usable carbs.
chapter twelve
Just A Few Extras . . .
This is where I've tucked the recipes that you need to make other recipes but that aren't, themselves, slow cooker stuff. It just seemed easiest to put them all in one place, you know? Most of these have appeared in one or more of my previous cookbooks. And by the way, we didn't count these toward your grand total of 300 slow cooker recipes—that's why they're extras.
### Chicken Chips
Every time I mention on Facebook that I've made these, someone asks for the recipe. It's so simple and so good! Just take the skin you've peeled off any chicken and any chunks of fat, too, and spread them on your broiler rack. Slide them into the oven at 350°F (180°C, or gas mark 4) and let them roast for 15 minutes or so until they're brown and crisp. Salt them and eat them like chips. That's all. You'll wish you could buy sacks of extra chicken skin!
My MasterCook doesn't have a nutritional listing for just chicken skin, but these have no carbs. They're a good source of gelatin, too, which is great for your joints, hair, nails, and skin.
### Dana's No-Sugar Ketchup
This recipe has appeared in all my cookbooks because ketchup is an essential ingredient in so many recipes, but store-bought ketchup usually has so much sugar. Recently, commercially-made low-carb ketchup has been appearing in the grocery stores. If you can get this, do so because food processors can get ingredients the home cook cannot, so store-bought low-carb ketchup is lower in carbs than this. If you can't find low-carb ketchup, however, this is easy to make, tastes great, and is about half the carbs of regular ketchup. Be aware that recipes that list ketchup as an ingredient are analyzed for this homemade version, so if you use commercial low-carb ketchup, the carb counts will be a tad lower.
6 ounces (170 g) tomato paste
2/3 cup (160 ml) cider vinegar
1/3 cup (80 ml) water
1/3 cup (8 g) Splenda
2 tablespoons (20 g) minced onion
2 cloves garlic
1 teaspoon salt
1/8 teaspoon ground allspice
1/8 teaspoon ground cloves
1/8 teaspoon pepper
Put everything in your blender and run it until the onion disappears. Scrape it into a container with a tight lid and store it in the refrigerator.
_**Yield:**_ Makes roughly 1 1/2 cups, or 12 servings of 2 tablespoons, each with: 15 calories, trace fat, 1 g protein, 5 g carbohydrate, 1 g fiber, 4 g usable carbs.
### Cocktail Sauce
You'll need this for the _Easy Party Shrimp_ on page 27!
1/2 cup (120 g) _Dana's No-Sugar Ketchup_ (page 332) or purchased low-carb ketchup
2 teaspoons prepared horseradish
1/4 teaspoon Tabasco sauce
1 teaspoon lemon juice
Combine all ingredients in a bowl and mix well.
_**Yield:**_ Makes about 1/2 cup. The whole batch contains: 142 calories, 1 g fat, 5 g protein, 36 g carbohydrate, 6 g dietary fiber, 30 g usable carbs. Good thing you'll be sharing it! You can drop this carb count considerably by using commercially-made low-carb ketchup.
### Florida Sunshine Tangerine Barbecue Sauce
The name of this sauce is partly from the tangerine note, which is unusual and delicious, but the name is also from the fact that this sauce is at least as hot as the Florida sun! It's especially good on poultry. You can use this in any of the recipes that call for barbecue sauce, or you can use the Kansas City–style sauce on the following page—this one's lower carb, but less traditional. Or for that matter, you can use the newly available bottled low-carb barbecue sauce. You won't hurt my feelings.
12 ounces (355 ml) Diet Rite Tangerine Soda
1/4 cup (6 g) Splenda
1 tablespoon (8 g) chili powder
2 teaspoons black pepper
1 teaspoon ginger
1 teaspoon dry mustard
1 teaspoon onion salt
4 cloves garlic, crushed
1/2 teaspoon cayenne
1/2 teaspoon coriander
1/2 teaspoon red pepper flakes
1 bay leaf
1/2 cup (120 ml) cider vinegar
1 tablespoon (20 g) sugar-free imitation honey
1 tablespoon (15 ml) Worcestershire sauce
3/4 cup (180 g) _Dana's No-Sugar Ketchup_ (page 332) or purchased low-carb ketchup
Pour the soda into a nonreactive saucepan and turn the heat under it to medium-low. While that's heating, measure the other ingredients into the sauce. By the time you get to the ketchup, it should be simmering. Whisk everything together until smooth and let it simmer over lowest heat for a good 10 to 15 minutes.
_**Yield:**_ Makes about 3 cups, or 24 servings of 2 tablespoons, each with: 11 calories, trace fat, trace protein, 3 g carbohydrate, trace dietary fiber, 3 g usable carbs. (Analysis does not include the polyols in the imitation honey.)
### Dana's "Kansas City" Barbecue Sauce
This recipe, from _The Low-Carb Barbecue Book_ , is it—what most of us think of when we think of barbecue sauce: tomato-y, spicy, and sweet. It's unbelievably close to a top-flight commercial barbecue sauce—and my Kansas City–raised husband agrees. If you like a smoky note in your barbecue sauce, add 1 teaspoon of liquid smoke flavoring to this. (Note: If you can get it locally, commercially-made low-carb barbecue sauce is likely to be lower carb than this recipe.)
2 tablespoons (28 g) butter
1 clove garlic
1/4 cup (40 g) chopped onion
1 tablespoon (15 ml) lemon juice
1 cup (240 g) _Dana's No-Sugar Ketchup_ (page 332) or purchased low-carb ketchup
1/3 cup (8 g) Splenda
1 tablespoon (20 g) blackstrap molasses
2 tablespoons (28 ml) Worcestershire sauce
1 tablespoon (8 g) chili powder
1 tablespoon (15 ml) white vinegar
1 teaspoon pepper
1/4 teaspoon salt
Combine everything in a saucepan over low heat. Heat until the butter melts, stir the whole thing up, and let it simmer for 5 minutes or so. That's it!
_**Yield:**_ Roughly 1 3/4 cups, or 14 servings of 2 tablespoons each, each with: 45 calories, 3 g fat, 1 g protein, 7 g carbohydrate, 1 g fiber, 6 g usable carbs.
### Piedmont Mustard Sauce
This bright-yellow sauce, heavy on the mustard, but completely free of tomato, is typical of the Piedmont region of North Carolina. Try it on _Slow Cooker Pulled Pork_ (see recipe page 188).
1/2 cup (88 g) yellow mustard
2 tablespoons (28 ml) lemon juice
2 tablespoons (3 g) Splenda
1 tablespoon (15 ml) white vinegar
1/4 teaspoon cayenne
Combine everything in a saucepan and simmer for 5 minutes over low heat.
_**Yield:**_ Makes roughly 3/4 cup, or 6 servings of 2 tablespoons, each with: 17 calories, 1 g fat, 1 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Eastern Carolina Vinegar Sauce
This is the traditional Eastern Carolina sauce for pulled pork. It's just sweetened vinegar with a good hint of hot pepper. It'll be great with your _Slow Cooker Pulled Pork_ (see recipe page 188)!
1/2 cup (120 ml) cider vinegar
1 1/2 tablespoons (2 g) Splenda
1/4 teaspoon blackstrap molasses
1 teaspoon red pepper flakes
1/4 teaspoon cayenne
Combine everything in a saucepan and simmer for 5 minutes over low heat.
_**Yield:**_ 6 servings, each with: 4 calories, trace fat, trace protein, 2 g carbohydrate, trace dietary fiber, 2 g usable carbs.
### Hoisin Sauce
Hoisin sauce is sort of Chinese barbecue sauce, and it customarily contains a lot of sugar. It also doesn't usually have peanut butter in it, but it works quite well here. This is a repeat from _500 Low-Carb Recipes_ , by the way.
4 tablespoons (60 ml) soy sauce
2 tablespoons (32 g) natural peanut butter
2 tablespoons (3 g) Splenda
2 teaspoons white vinegar
1 clove garlic
2 teaspoons dark sesame oil
1/8 teaspoon five-spice powder
Just assemble everything in your blender and blend till it's smooth. Store it in a tightly lidded jar in the fridge. Feel free to double or triple this, if you like.
_**Yield:**_ Makes roughly 1/3 cup, or 6 servings of about 1 tablespoon, each with: 52 calories, 4 g fat, 2 g protein, 2 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Low-Carb Teriyaki Sauce
There's now commercial low-carb teriyaki sauce on the market, but I like this better, and it's so easy to make, why wouldn't I? Why wouldn't you?
1/2 cup (120 ml) soy sauce
1/4 cup (60 ml) dry sherry
1 clove garlic, crushed
2 tablespoons (3 g) Splenda
1 tablespoon (8 g) grated ginger root
Combine all the ingredients. Refrigerate until ready to use.
_**Yield:**_ Makes 3/4 cup, or 12 servings of 1 tablespoon, each with: 13 calories, trace fat, 1 g protein, 1 g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Classic Rub
This barbecue rub first appeared in _The Low-Carb Barbecue Book_ , and it's a great choice in any recipe that calls for barbecue rub.
1/4 cup (6 g) Splenda
1 tablespoon (18 g) seasoned salt
1 tablespoon (9 g) garlic powder
1 tablespoon (15 g) celery salt
1 tablespoon (7 g) onion powder
2 tablespoons (14 g) paprika
1 tablespoon (8 g) chili powder
2 teaspoons pepper
1 teaspoon lemon pepper
1 teaspoon ground sage
1 teaspoon mustard
1/2 teaspoon thyme
1/2 teaspoon cayenne
Simply stir everything together and store in an airtight container.
_**Yield:**_ Makes 13 tablespoons of rub, each with: 13 calories, trace fat, 1 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs.
### Cajun Seasoning
This New Orleans–style seasoning, originally from _500 Low-Carb Recipes_ , is good sprinkled over chicken, steak, pork, fish—just about anything. It will also work in any recipe in this book that calls for Cajun seasoning. But if you'd rather use purchased seasoning, go for it.
2 1/2 tablespoons (18 g) paprika
2 tablespoons (36 g) salt
2 tablespoons (18 g) garlic powder
1 tablespoon (6 g) pepper
1 tablespoon (7 g) onion powder
1 tablespoon (5 g) cayenne
1 tablespoon (3 g) dried oregano
1 tablespoon (4 g) dried thyme
Combine all ingredients thoroughly and store in an airtight container.
_**Yield:**_ Makes 2/3 cup. The whole batch contains: 187 calories, 4 g fat, 8 g protein, 37 g carbohydrate, 9 g dietary fiber, 28 g usable carbs. (Considering how spicy this is, you're unlikely to use more than a teaspoon or two at a time. One teaspoon has 1 gram of carbohydrate, with a trace of fiber.)
### Adobo Seasoning
Adobo is a popular seasoning in Latin America and the Caribbean. It's available at many grocery stores, in the spice aisle or the international aisle. But if you can't find it, it sure is easy to make.
10 teaspoons (30 g) garlic powder
5 teaspoons (5 g) dried oregano
5 teaspoons (10 g) pepper
2 1/2 teaspoons (8 g) paprika
5 teaspoons salt
Simply measure everything into a bowl, stir, and store in a lidded shaker jar.
_**Yield:**_ Makes a little over 1/2 cup, or about 48 servings of 1/2 teaspoon, each with: 3 calories, trace fat, trace protein, 1g carbohydrate, trace dietary fiber, 1 g usable carbs.
### Garam Masala
This is an Indian spice blend used in several of the curries in this book. You may well be able to buy perfectly lovely garam masala already made at a local Asian market or big grocery store—I can! But if you can't, you can easily make your own.
2 tablespoons (14 g) ground cumin
2 tablespoons (12 g) ground coriander
2 tablespoons (12 g) ground cardamom
1 1/2 tablespoons (9 g) black pepper
4 teaspoons (8 g) ground cinnamon
1/2 teaspoon ground cloves
1 teaspoon ground nutmeg
Simply combine everything and store in an airtight container.
_**Yield:**_ Makes roughly 9 tablespoons, each with: 19 calories, 1 g fat, 1 g protein, 4 g carbohydrate, 1 g dietary fiber, 3 g usable carbs.
### Sugar-Free Chocolate Sauce
This is as good as any sugar-based chocolate sauce you've ever had, if I do say so myself. Which I do. Don't try to make this with Splenda; it won't work. The polyol sweetener somehow makes the water and the chocolate combine. It's chemistry, or magic, or some darned thing. You very likely will need to order maltitol. Just do a Web search under "low carbohydrate," and you'll find dozens of websites ready to ship anything your heart desires. My favorite is carbsmart (www.carbsmart.com).
1/3 cup (80 ml) water
2 ounces (55 g) unsweetened baking chocolate
1/2 cup (100 g) maltitol
3 tablespoons (45 g) butter
1/4 teaspoon vanilla
Put the water and chocolate in a glass measuring cup and microwave on high for 1 to 1 1/2 minutes or until the chocolate has melted. Stir in the maltitol and microwave on high for another 3 minutes, stirring halfway through. Stir in the butter and vanilla.
_**Note:**_ This works beautifully with maltitol. However, when I tried to make it with other granular polyols—erythritol, isomalt—it started out fine but crystallized and turned grainy as it cooled. I'd stick with maltitol.
_**Yield:**_ Makes roughly 1 cup, or 8 servings of 2 tablespoons, each with: 75 calories, 8 g fat, 1 g protein, 2 g carbohydrate, 1 g dietary fiber, 1 g usable carbs. (Analysis does not include the maltitol.)
### Mockahlua
This recipe originally appeared in _500 Low-Carb Recipes_ , but because I've included it in some recipes in the dessert chapter I thought I'd better repeat it! This recipe makes quite a lot, but don't worry about that; 100-proof vodka's a darned good preservative. Your _Mockahlua_ will keep indefinitely.
2 1/2 cups (570 ml) water
3 cups (75 g) Splenda
3 tablespoons (18 g) instant coffee granules
1 teaspoon vanilla extract
1 bottle (750 milliliters) 100-proof vodka (Use the cheap stuff.)
In a large pitcher or measuring cup, combine the water, Splenda, coffee granules, and vanilla extract. Stir until the coffee and Splenda are completely dissolved.
Pour the mixture through a funnel into a 1.5 or 2 liter bottle. (A clean 1.5 liter wine bottle works fine, so long as you've saved the cork.) Pour in the vodka. Cork and shake well.
_**Yield:**_ 32 servings of 1 1/2 ounces (42 ml)—a standard "shot," each with: 53 calories, 0 g fat, trace protein, trace carbohydrate, 0 g dietary fiber, trace usable carbs.
### Whipped Topping
I keep repeating this recipe, but then, it's a wonderful, classic whipped cream topping for any dessert.
1 cup (235 ml) heavy cream, chilled
3 teaspoons (15 g) sugar-free vanilla instant pudding mix
Simply whip the cream with the pudding mix. Use your electric mixer, or if you like, a whisk—neither a blender nor a food processor will work. Stop whipping as soon as your topping is nice and thick, or you'll end up with vanilla butter!
If you make this ahead of time, refrigerate it until you're ready to serve dessert.
_**Yield:**_ 8 servings, each with: 104 calories, 11 g fat, 1 g protein, 1 g carbohydrate, 0 g dietary fiber, 1 g usable carbs.
### Cauli-Rice
I give thanks to Fran McCullough for this recipe! I got this idea from her book _Living Low-Carb_ , and it's served me very well indeed.
1/2 head cauliflower
Simply put the cauliflower through your food processor using the shredding blade. This gives a texture that is remarkably similar to rice. You can steam this, microwave it, or even sauté it in butter. Whatever you do, though, don't overcook it! I usually put mine in a microwaveable casserole with a lid, add a couple of tablespoons (28 to 45 ml) of water, and microwave it for 7 minutes on high.
_**Yield:**_ This makes about 3 cups, or at least 3 to 4 servings. Assuming 3 servings, each with: 24 calories, trace fat, 2 g protein, 5 g carbohydrate, 2 g dietary fiber, 3 g usable carbs.
### Fauxtatoes
This is a wonderful substitute for mashed potatoes if you want something to put a fabulous sour cream gravy on! Feel free, by the way, to use frozen cauliflower instead. It works quite well here.
1 head cauliflower or 1 1/2 pounds (680 g) frozen cauliflower
4 tablespoons (55 g) butter
Steam or microwave the cauliflower until it's soft. Drain it thoroughly and put it through the blender or food processor until it's well puréed. Add the butter and salt and pepper to taste.
_**Yield:**_ 6 servings, each with: 72 calories, 8 g fat, trace protein, 1 g carbohydrate, trace dietary fiber, trace usable carbs. (This makes six generous servings.)
### The Ultimate Fauxtatoes
I'm not crazy about Ketatoes by themselves, but added to pureed cauliflower _Fauxtatoes_ , they add a potato-y flavor and texture that is remarkably convincing! This is a killer side dish with many of your slow cooker main dishes.
1/2 head cauliflower
1/2 cup (50 g) Ketatoes mix
1/2 cup (120 ml) boiling water
1 tablespoon (14 g) butter
Trim the bottom of the stem of your cauliflower and whack the rest of the head into chunks. Put them in a microwaveable casserole dish with a lid. Add a couple of tablespoons (28 to 45 ml) of water, cover, and microwave on high for 8 to 9 minutes.
While that's happening, measure your Ketatoes mix and boiling water into a mixing bowl and whisk together.
When the microwave beeps, pull out your cauliflower—it should be tender. Drain it well and put it in either your food processor, with the **S** -blade in place, or in your blender. Either way, purée the cauliflower until it's smooth. Transfer the puréed cauliflower to the mixing bowl and stir the cauliflower and Ketatoes together well. Add the butter and stir till it melts. Add salt and pepper to taste and serve.
_**Yield:**_ 4 servings, each with: 140 calories, 5 g fat, 10 g protein, 14 g carbohydrate, 8 g dietary fiber, 6 g usable carbs.
## Acknowledgments
A few quick thank you's here:
To my darling friend Maria Vander Vloedt, who tested many, many recipes for this book. She's the perfect tester: She's smart and funny and reliable, she knows how to cook, she can follow instructions but make constructive suggestions when they're needed, she eats low-carb, and she has a husband and five kids to try my recipes out on! Thanks, Maria.
Also to my sister Kim, who is always up for a new recipe, and my pal Ray Stevens, who tested a bunch when crunch time rolled around: Thanks, guys!
As always, to my husband, Eric Schmitz. Thank God I married him, I could never find another person with his combination of skills in the open job market. All this, and he's nice to have around the house, too.
And to my editor, Holly, for browbeating me into writing this book. It's been a lot cooler than I expected.
## About the Author
**Dana Carpender** is a best-selling author and radio host who was startled to discover that limiting her carbohydrate intake not only helped her control her weight, but produced the health and vitality a low-fat diet had promised but never delivered. More than fifteen years later, she laughs at people who say "You can't eat that way long-term." Her nine cookbooks are the result of her realization that the key to permanent dietary change is the answer to the age-old question, "What's for supper?" To date they have sold over a million copies worldwide. Dana blogs about low-carb nutrition at www.HoldtheToast.com; her weekly blog digest goes out to over 20,000 readers. She is also managing editor of CarbSmart Magazine at www.CarbSmart.com, as well as a featured staff writer. She launched her internet radio show and podcast, Dana's Low Carb For Life, in January 2011. Dana lives in Bloomington, Indiana, with her husband and a menagerie of pets, all of whom are well and healthily fed.
## Index
A
Adobe Seasoning,
African Chicken, Peanut, and Spinach Stew,
Albondigas en Salsa Chipotle, –251
almonds
Almond-Stuffed Flounder Rolls with Orange Butter Sauce,
Bacon-Almond Flounder Rolls,
Butter-Spice Almonds,
Buttery Vanilla Almonds,
apples
Chicken with Apples and Rosemary,
Hocks and Shanks with Cabbage and Apples,
Ribs with Apple Kraut,
apricots
Apricot Custard,
Chili-Apricot Glazed Ribs, –191
Rosemary-Ginger Ribs with Apricot Glaze,
artichokes
Artichoke Egg Casserole,
Artichoke-Spinach-Ranch Dip,
Basic Artichokes,
Chicken with Artichokes and Sun Dried Tomatoes,
Chicken with Thyme and Artichokes,
Hot Artichoke Dip,
Tuna and Artichoke Stuffed Mushrooms, –31
Asian dishes
Asian Peanuts,
Asian Slow Cooker Short Ribs,
Beef with Asian Mushroom Sauce,
Hot Asian Ribs,
Mu Shu Chicken, –91
Orange Teriyaki Chicken, –92
Peking Slow Cooker Pot Roast,
Pho, –260
Polynesian Pork Ribs,
Satay-Flavored Pork,
Slow Cooker Mu Shu Pork, –187
Slow Cooker Teriyaki Ribs,
So Totally Inauthentic Malaysian-oid Pumpkin-Chicken-Tomato Soup,
Southeast Asian Coconut Custard, –315
Soy and Sesame Ribs,
Sweet and Sour Pork,
Sweet and Sour Shrimp,
Teriyaki-Tangerine Ribs,
Thai Chicken Bowls, –93
Thai-ish Chicken and Noodles,
Asparagus, Maria's Slow Cooker,
avocado
Avocado Aioli,
Chuck with Avocado Aioli,
Awesome Sauce aka Provolone and Blue Cheese Fondue,
B
bacon
Bacon-Almond Flounder Rolls,
Bacon-Cheese Dip,
Broccoli-Bacon-Colby Quiche,
Bagna Cauda,
Baked Beans,
Balsamic Pot Roast,
Balsamic-Vanilla Salmon,
barbecue dishes
Barbecue Green Beans,
Cranberry-Barbecue Meatballs,
Insanely Good Slow-Cooker Barbecued Beef Ribs,
Key West Ribs,
Maple Chili Ribs,
Slow Cooker "Barbecued" Ribs,
Slow Cooker Texas Brisket,
Southwestern Barbecue,
barbecue sauce
Classic Rub,
Dana's "Kansas City" Barbecue Sauce,
Florida Sunshine Tangerine Barbecue Sauce,
Bavarian Cabbage,
Bavarian Pot Roast,
beans
_See also_ green beans
Baked Beans,
Black Bean Soup,
Mexican Beef and Bean Soup,
beef
3-Minute Slow Cooker Pot Roast,
Albondigas en Salsa Chipotle, –251
Asian Slow Cooker Short Ribs,
Balsamic Pot Roast,
Bavarian Pot Roast,
Beef and Broccoli,
Beef and Sausage Stew,
Beef and Zucchini Stew,
Beef Carbonnade,
Beef in Beer,
Beef Stroganoff,
Beef with Asian Mushroom Sauce,
Bollito Misto,
Carne all'Ungherese,
Chili-Cocoa Pot Roast,
Chipotle Brisket, –128
Chuck with Avocado Aioli,
Cocoa Joes,
Coffee Beef,
Comfort Food Casserole,
Cube Steaks in Gravy,
Easy Italian Beef,
Eric's Goop Suey,
Firehouse Chili,
Good Low-Carb Slow Cooked Short Ribs,
Hamburger and Turnip Layered Casserole,
Insanely Good Slow-Cooker Barbecued Beef Ribs,
Louisiana Meat Loaf,
Maple-Glazed Corned Beef with Vegetables,
Mexican Beef and Bean Soup,
Mexican Stew,
Morty's Mixed Meat Loaf,
New England Boiled Dinner,
Noodleless Spinach Lasagna,
Oxtails Pontchartrain,
Peking Slow Cooker Pot Roast,
Pepperoncini Beef,
Pho, –260
Pizza Stew,
Pizza-ish Meat Loaf,
Pot Roast Brisket,
Pot Roast with Beer and Mushrooms,
Pub Loaf,
Reuben Hot Pot,
Roman Stew,
Sauerbrauten,
Short Rib Stew,
Short Ribs with Mushrooms,
Short Ribs with Wine and Mushrooms,
Simple Meat Loaf,
Simple Salsa Beef,
Slow Cooker Texas Brisket,
Swiss Steak,
Texas Red, –152
Tuscan-ish Meat Loaf,
Vegetable Beef Soup, –265
beer,
black soybeans, –16
Baked Beans,
Black Bean Soup,
Mexican Beef and Bean Soup,
blackstrap molasses,
blue cheese
Awesome Sauce aka Provolone and Blue Cheese Fondue,
Blue Cheese Dressing Walnuts,
Blue Cheese Mushrooms,
Bollito Misto,
bouillon,
broccoli
Beef and Broccoli,
Broccoli with Bacon and Pine Nuts,
Broccoli-Bacon-Colby Quiche,
broth
Bone Broth for Busy People,
concentrated, ,
packaged, –17
browning,
Brunswick Stew, –253
Buffalo Wing Dip,
Burritos, Chicken,
Butter-Spice Almonds,
Buttery Vanilla Almonds,
C
cabbage
_See also_ sauerkraut
Bavarian Cabbage,
Chicken with Root Vegetables, Cabbage, and Herbs,
Hocks and Shanks with Cabbage and Apples,
Kalua Pig with Cabbage,
Pork with Cabbage,
Stewed Pork Neckbones with Turnips and Cabbage,
Cajun Seasoning,
Cajun-Spiced Pecans,
cakes
_See also_ cheesecake
Chocolate Chip Cinnamon Zucchini Snack Cake, –327
Chocolate Chocolate-Chip Pudding Cake, –326
Candied Pecans,
Caramel Mocha Latte,
Carb Countdown dairy beverage,
Caribbean Slow Cooker Lamb,
Carne all'Ungherese,
casseroles
Artichoke Egg Casserole,
Comfort Food Casserole,
Creamy Ham Casserole,
Dana's Kim's Kings Ranch Chicken, –65
Green Bean Casserole,
Hamburger and Turnip Layered Casserole,
Spinach Parmesan Casserole,
Tuna Egg Casserole,
Tuna-Noodle Casserole,
Catfish, Super-Simple,
cauliflower
about,
Cauliflower, Cheese, and Spinach Soup,
Cauli-Rice,
Cheddar-Barbecue Fauxtatoes,
Cheesy Cauliflower Packet,
Chipotle Fauxtatoes,
Fauxtatoes,
Garlic-Onion Fauxtatoes,
Hobo UnPotatoes Slow-Cooker Style, –306
Italian Garlic and Herb Fauxtatoes, –294
Macadangdang, –290
Ranch and Green Onion Fauxtatoes,
Slow Cooker "Risotto" with Mushrooms and Peas,
Ultimate Fauxtatoes,
Cheddar-Barbecue Fauxtatoes,
cheese
Awesome Sauce aka Provolone and Blue Cheese Fondue,
Bacon-Cheese Dip,
Broccoli-Bacon-Colby Quiche,
Buffalo Wing Dip,
Cauliflower, Cheese, and Spinach Soup,
Cheddar-Barbecue Fauxtatoes,
Cheesy Cauliflower Packet,
Cheesy Neeps,
Spinach-Mushroom Gratin, –300
Tavern Soup,
cheesecake
about,
Crisp Chocolate Crust,
"Graham" Crust,
Mochaccino Cheesecake,
New York–style Cheesecake, –323
Peanut Butter Cheesecake,
chicken
African Chicken, Peanut, and Spinach Stew,
Bollito Misto,
Brunswick Stew, –253
Chicken and Dumplings,
Chicken and Fennel Soup, –278
Chicken and Vegetable Soup with Thai Spices, –273
Chicken Burritos,
Chicken Cacciatore,
Chicken Chili Verde,
Chicken Chips,
Chicken in Creamy Horseradish Sauce,
Chicken in Creamy Orange Sauce,
Chicken in Thai Coconut Curry Sauce,
Chicken Liver Pâté, –38
Chicken Minestrone,
Chicken Paprikash,
Chicken Soup with Wild Rice,
Chicken Stew,
Chicken Stroganoff,
Chicken Vindaloo,
Chicken with Apples and Rosemary,
Chicken with Artichokes and Sun Dried Tomatoes,
Chicken with Raspberry-Chipotle Sauce,
Chicken with Root Vegetables, Cabbage, and Herbs,
Chicken with Root Vegetables in Parmesan Cream,
Chicken with Thyme and Artichokes,
Chipotle Pumpkin Soup,
Citrus Spice Chicken,
Curried Chicken with Coconut Milk,
Dana's Kim's Kings Ranch Chicken, –65
Glazed Chicken Wings,
Italian Chicken and Vegetables,
"I've Got a Life" Chicken,
Lemon Chicken,
Lemon-Herb Chicken,
Lemon-White Wine-Tarragon Chicken,
Low-Carb Slow Cooker Paella, –252
Maple-Mustard Wings,
Mean Old Rooster Chili,
Mediterranean Chicken,
Mediterranean Pepper and Olive Chicken,
Mom's 1960s Chicken, Redux, –87
Mu Shu Chicken, –91
Orange Teriyaki Chicken, –92
Seriously Simple Chicken Chili,
Slow Cooker Brewery Chicken and Vegetables,
Slow Cooker Chicken Guadeloupe,
Slow Cooker Chicken Mole, –88
So Totally Inauthentic Malaysian-oid Pumpkin-Chicken-Tomato Soup,
Sort-of-Ethiopian Chicken Stew,
Southwestern Barbecue,
Spicy Chicken and Mushroom Soup,
Tequila Lime Chicken,
Thai Chicken Bowls, –93
Thai Hot Pot,
Thai-ish Chicken and Noodles,
Tuscan Chicken,
Tuscan-ish Meat Loaf,
Yassa,
chili
Chicken Chili Verde,
Chili Egg Puff Slow Cooker Style,
Chili Garlic Peanuts,
Chili-Apricot Glazed Ribs, –191
Chili-Cocoa Pot Roast,
Chili-Cocoa Rub,
Firehouse Chili,
Free Venison Chili, –153
Maple Chili Ribs,
Mean Old Rooster Chili,
Not-Quite-Asian Sweet Chili Sauce,
Orange and Tomatillo Pork Chili, –186
Pork Slow Cooker Chili,
Seriously Simple Chicken Chili,
Texas Red, –152
chili garlic paste,
Chinese dishes. _See_ Asian dishes
chipotle peppers
about,
Chicken with Raspberry-Chipotle Sauce,
Chipotle Brisket, –128
Chipotle Fauxtatoes,
Chipotle Pumpkin Soup,
Chipotle Turkey Legs,
Chive-and-Garlic Spinach Packet,
chocolate
Chocolate Chip Cinnamon Zucchini Snack Cake, –327
Chocolate Chocolate-Chip Pudding Cake, –326
Chocolate Fudge Custard,
Crisp Chocolate Crust,
Sugar-Free Chocolate Sauce,
Choucroute Garni,
Chuck with Avocado Aioli,
chutneys
Cranberry Chutney,
Cranberry-Peach Chutney,
Slow Cooker Chutney, –308
Citrus Spice Chicken,
Clam Chowder, Maria's New England,
Classic Rub,
Cocido de Puerco,
Cocktail Sauce,
cocoa
Chili-Cocoa Pot Roast,
Chili-Cocoa Rub,
Cocoa Joes,
Coconut Custard, Southeast Asian, –315
coffee
Caramel Mocha Latte,
Coffee Beef,
Hot Cinnamon Mocha,
Colombo Meatballs with Jerk Sauce, –27
Comfort Food Casserole,
concentrate, ,
cooking times, –14
crab
Crab Puff,
Hot Crab Dip,
Pantry Seafood Supper,
cranberry
Cranberry Chutney,
Cranberry-Barbecue Meatballs,
Cranberry-Orange Ribs,
Cranberry-Peach Chutney,
Cranberry-Peach Turkey Roast,
Slow Cooker Cranberry Sauce,
Cream of Mushroom Soup,
Cream of UnPotato Soup,
Creamy, Cheesy Mustard Green Beans and Mushrooms,
Creamy Ham Casserole,
Creamy Ham Hash,
Crisp Chocolate Crust,
Crock-Pots, –11
Cube Steaks in Gravy,
curry dishes
Chicken in Thai Coconut Curry Sauce,
Chicken Vindaloo,
Curried Chicken with Coconut Milk,
Curried Mushrooms Malabar,
Curried Pecans,
Curried Pork Stew,
Kashmiri Lamb Shanks,
custard
Apricot Custard,
Chocolate Fudge Custard,
Maple Custard,
Maple-Pumpkin Custard,
Southeast Asian Coconut Custard, –315
D
Dana's "Kansas City" Barbecue Sauce,
Dana's Kim's Kings Ranch Chicken, –65
Dana's No-Sugar Ketchup,
Dana's Snack Mix, Slow Cooker-Style, –39
desserts
Apricot Custard,
Chocolate Chip Cinnamon Zucchini Snack Cake, –327
Chocolate Chocolate-Chip Pudding Cake, –326
Chocolate Fudge Custard,
Flan,
Hot Cinnamon Mocha,
Maple Custard,
Maple-Pumpkin Custard,
Mochaccino Cheesecake,
New York–style Cheesecake, –323
Peaches with Butterscotch Sauce,
Peanut Butter Cheesecake,
Pumpkin Bread, –329
Rhubarb Flummery,
Southeast Asian Coconut Custard, –315
dips
Artichoke-Spinach-Ranch Dip,
Avocado Aioli,
Bacon-Cheese Dip,
Bagna Cauda,
Buffalo Wing Dip,
Hot Artichoke Dip,
Dressing, Russian, –123
Dumplings,
E
Eastern Carolina Vinegar Sauce,
eggs
Artichoke Egg Casserole,
Broccoli-Bacon-Colby Quiche,
Chili Egg Puff Slow Cooker Style,
Crab Puff,
Maria's Eggs Florentine,
Tuna Egg Casserole,
Eric's Goop Suey,
erythritol,
F
fat,
Fauxtatoes,
about,
Cheddar-Barbecue Fauxtatoes,
Chipotle Fauxtatoes,
Garlic-Onion Fauxtatoes,
Italian Garlic and Herb Fauxtatoes, –294
Ranch and Green Onion Fauxtatoes,
Ultimate Fauxtatoes,
Fennel Soup, Chicken and, –278
Festive Green Beans,
Firehouse Chili,
fish
Almond-Stuffed Flounder Rolls with Orange Butter Sauce,
Bacon-Almond Flounder Rolls,
Balsamic-Vanilla Salmon,
Lemon-Mustard Salmon Steaks,
Maple-Balsamic Salmon,
Orange Basil Salmon,
Pantry Seafood Supper,
Salmon in Sweet Chili Sauce,
Super-Simple Catfish,
Tuna and Artichoke Stuffed Mushrooms, –31
Tuna Egg Casserole,
Tuna-Noodle Casserole,
fish sauce,
Flan,
Florida Sunshine Tangerine Barbecue Sauce,
flounder
Almond-Stuffed Flounder Rolls with Orange Butter Sauce,
Bacon-Almond Flounder Rolls,
foil packet side dishes
Cheesy Cauliflower Packet,
Chive-and-Garlic Spinach Packet,
Hobo UnPotatoes Slow-Cooker Style, –306
Fruity, Spicy Ribs,
G
Garam Masala,
garlic, –19 Garlic-Onion Fauxtatoes,
German UnPotato Soup,
ginger root,
Glazed Chicken Wings,
"Graham" Crust,
granular sucralose, ,
Greek Stuffed Peppers,
green beans
Barbecue Green Beans,
Creamy, Cheesy Mustard Green Beans and Mushrooms,
Festive Green Beans,
Green Bean Casserole,
Not-Pea Soup, –272
Sausage and Green Bean Soup,
Southern Beans,
Tangy Beans,
guar,
H
ham
Brunswick Stew, –253
Creamy Ham Casserole,
Creamy Ham Hash,
Ham with Rutabaga and Turnips,
Hocks and Shanks with Cabbage and Apples,
"Honey" Mustard Ham,
Not-Pea Soup, –272
Hamburger and Turnip Layered Casserole,
Havana Ribs,
Hobo UnPotatoes Slow-Cooker Style, –306
Hocks and Shanks with Cabbage and Apples,
Hoisin Sauce,
"Honey" Mustard Ham,
hors d'oeuvres
_See also_ dips; snacks
Awesome Sauce aka Provolone and Blue Cheese Fondue,
Bacon-Cheese Dip,
Chicken Liver Pâté, –38
Colombo Meatballs with Jerk Sauce, –27
Cranberry-Barbecue Meatballs,
Easy Party Shrimp,
Glazed Chicken Wings,
Horseradish Smokies,
Maple-Mustard Wings,
Orange Smokies,
Tuna and Artichoke Stuffed Mushrooms, –31
Zippy Cocktail Dogs,
Horseradish Smokies,
hot dogs
Zippy Cocktail Dogs,
I
ingredients
order of placement in pot,
overview of, –22
Insanely Good Slow-Cooker Barbecued Beef Ribs,
Italian dishes
Bollito Misto,
Chicken Cacciatore,
Chicken Minestrone,
Easy Italian Beef,
Italian Chicken and Vegetables,
Italian Garlic and Herb Fauxtatoes, –294
Italian Neck Bones,
Italian Sausage Soup,
Lamb Shanks Osso Bucco Style,
Tuscan Chicken,
Tuscan-ish Meat Loaf,
"I've Got a Life" Chicken,
K
Kalua Pig with Cabbage,
Kashmiri Lamb Shanks,
Ketatoes,
Ketchup, Dana's No-Sugar,
Key West Ribs,
Kickin' Pecans,
L
lamb
about: shanks,
Braised Leg of Lamb,
Caribbean Slow Cooker Lamb,
Colombo Meatballs with Jerk Sauce, –27
Greek Stuffed Peppers,
Kashmiri Lamb Shanks,
Lamb Shanks in Red Wine,
Lamb Shanks Osso Bucco Style,
Lamb Stew Provençal,
Lemon Lamb Shanks,
Seriously Simple Lamb Shanks,
lemons
Lemon Chicken,
Lemon Lamb Shanks,
Lemon-Herb Chicken,
Lemon-Mustard Salmon Steaks,
Lemon-Parmesan Mushrooms,
Lemon-White Wine-Tarragon Chicken,
Lime-Basted Scallops,
liquids,
Louisiana Meat Loaf,
Low-Carb Slow Cooker Paella, –252
Low-Carb Teriyaki Sauce,
low-sugar preserves,
M
Macadangdang, –290
maple
Maple Chili Ribs,
Maple Custard,
Maple-Balsamic Salmon,
Maple-Glazed Corned Beef with Vegetables,
Maple-Glazed Walnuts,
Maple-Mustard Turnips,
Maple-Mustard Wings,
Maple-Pumpkin Custard,
Maple-Spice Country-Style Ribs,
Maria's Eggs Florentine,
Maria's New England Clam Chowder,
Maria's Slow Cooker Asparagus,
Mashed Turnips, –298
Mean Old Rooster Chili,
meat
_See also_ beef; chicken; pork; turkey
browning,
lean cuts,
meat loaf
Louisiana Meat Loaf,
Morty's Mixed Meat Loaf,
Pizza-ish Meat Loaf,
Pub Loaf,
Simple Meat Loaf,
Tuscan-ish Meat Loaf,
meatballs
Albondigas en Salsa Chipotle, –251
Colombo Meatballs with Jerk Sauce, –27
Cranberry-Barbecue Meatballs,
Mediterranean dishes
Mediterranean Chicken,
Mediterranean Pepper and Olive Chicken,
Mediterranean Turkey Loaf,
Mexican dishes
Albondigas en Salsa Chipotle, –251
Chicken Burritos,
Chili-Cocoa Pot Roast,
Mexican Beef and Bean Soup,
Mexican Stew,
Slow Cooker Chicken Mole, –88
Tequila Lime Chicken,
Mochaccino Cheesecake,
Mockahlua,
molasses,
Mom's 1960s Chicken, Redux, –87
Morty's Mixed Meat Loaf,
Mu Shu Chicken, –91
Mu Shu Pork, Slow Cooker, –187
mushrooms
Beef with Asian Mushroom Sauce,
Blue Cheese Mushrooms,
Braised Turkey Wings with Mushrooms,
Cream of Mushroom Soup,
Creamy, Cheesy Mustard Green Beans and Mushrooms,
Curried Mushrooms Malabar,
Lemon-Parmesan Mushrooms,
Mushrooms Stroganoff,
Pork Roast with Creamy Mushroom Gravy and Vegetables,
Pot Roast with Beer and Mushrooms,
Short Ribs with Mushrooms,
Short Ribs with Wine and Mushrooms,
Simplest Slow Cooker Mushrooms,
Slow Cooker "Risotto" with Mushrooms and Peas,
Spicy Chicken and Mushroom Soup,
Spinach-Mushroom Gratin, –300
Tuna and Artichoke Stuffed Mushrooms, –31
Turkey with Mushroom Sauce,
N
nam pla,
Neckbones and "Rice",
New England Boiled Dinner,
New York–style Cheesecake, –323
nonstick cooking spray,
Noodleless Spinach Lasagna,
Not-Pea Soup, –272
Not-Quite-Asian Sweet Chili Sauce,
nuoc mam,
nuts and seeds
Asian Peanuts,
Blue Cheese Dressing Walnuts,
Butter-Spice Almonds,
Buttery Vanilla Almonds,
Cajun-Spiced Pecans,
Candied Pecans,
Chili Garlic Peanuts,
Curried Pecans,
Dana's Snack Mix, Slow Cooker-Style, –39
Kickin' Pecans,
Maple-Glazed Walnuts,
Smokin' Chili Peanuts,
Spiced Walnuts,
Sweet and Salty Peanuts,
O
Onion-Mustard Pork Chops,
orange
Cranberry-Orange Ribs,
Orange and Tomatillo Pork Chili, –186
Orange Basil Salmon,
Orange Pork Loin,
Orange Rosemary Pork,
Orange Smokies,
Orange Teriyaki Chicken, –92
Orange-Glazed Country Ribs,
Oxtails Pontchartrain,
P
Paella, Low-Carb Slow Cooker, –252
Pantry Seafood Supper,
peach
Cranberry-Peach Turkey Roast,
peaches
Cranberry-Peach Chutney,
Peaches with Butterscotch Sauce,
Peanut Butter Cheesecake,
peanuts
African Chicken, Peanut, and Spinach Stew,
Asian Peanuts,
Chili Garlic Peanuts,
Smokin' Chili Peanuts,
Sweet and Salty Peanuts,
pecans
Cajun-Spiced Pecans,
Candied Pecans,
Curried Pecans,
Kickin' Pecans,
Peking Slow Cooker Pot Roast,
Pepperoncini Beef,
peppers
Greek Stuffed Peppers,
Stuffed Peppers,
Pho, –260
Piedmont Mustard Sauce,
Pizza Stew,
Pizza-ish Meat Loaf,
Polynesian Pork Ribs,
pork
_See also_ ham; sausage about: neckbones, –209
Braised Pork with Fennel, –165
Chili-Apricot Glazed Ribs, –191
Choucroute Garni,
Cocido de Puerco,
Cranberry-Orange Ribs,
Curried Pork Stew,
Easy Pork Roast,
Easy Southwestern Pork Stew,
Fruity, Spicy Ribs,
Havana Ribs,
Hocks and Shanks with Cabbage and Apples,
Hot Asian Ribs,
Italian Neck Bones,
Kalua Pig with Cabbage,
Key West Ribs,
Maple Chili Ribs,
Maple-Spice Country-Style Ribs,
Morty's Mixed Meat Loaf,
Neckbones and "Rice",
Onion-Mustard Pork Chops,
Orange and Tomatillo Pork Chili, –186
Orange Pork Loin,
Orange Rosemary Pork,
Orange-Glazed Country Ribs,
Polynesian Pork Ribs,
Pork and "Apple" Stew,
Pork Slow Cooker Chili,
Pork with Cabbage,
Pork with Rutabaga,
Ribs 'n' Kraut,
Ribs with Apple Kraut,
Rosemary-Ginger Ribs with Apricot Glaze,
Satay-Flavored Pork,
Simple Meat Loaf,
Slow Cooker "Barbecued" Ribs,
Slow Cooker Mu Shu Pork, –187
Slow Cooker Pork Ribs Adobado,
Slow Cooker Pulled Pork,
Slow Cooker Teriyaki Ribs,
Soy and Sesame Ribs,
Spinach-Stuffed Pork Loin, –183
Stewed Pork Neckbones with Turnips and Cabbage,
Stuffed Peppers,
Sweet and Sour Pork,
Sweet and Tangy Mustard Pork Roast,
Tangy Pork Chops,
Teriyaki-Tangerine Ribs,
pot roast
3-Minute Slow Cooker Pot Roast,
Balsamic Pot Roast,
Bavarian Pot Roast,
Chili-Cocoa Pot Roast,
Peking Slow Cooker Pot Roast,
Pork Roast with Apricot Sauce,
Pork Roast with Creamy Mushroom Gravy and Vegetables,
Pot Roast Brisket,
Pot Roast with Beer and Mushrooms,
Sauerbrauten,
poultry. _See_ chicken; turkey
Pub Loaf,
pumpkin
Chipotle Pumpkin Soup,
Fluffy, Savory Pumpkin,
Maple-Pumpkin Custard,
Pumpkin Bread, –329
So Totally Inauthentic Malaysian-oid Pumpkin-Chicken-Tomato Soup,
Q
Quiche, Broccoli-Bacon-Colby,
R
Ranch and Green Onion Fauxtatoes,
Ranch-E-Cue Wings,
Raspberry-Chipotle Sauce,
Reuben Hot Pot,
Rhubarb Flummery,
ribs
Asian Slow Cooker Short Ribs,
Chili-Apricot Glazed Ribs, –191
Cranberry-Orange Ribs,
Fruity, Spicy Ribs,
Good Low-Carb Slow Cooked Short Ribs,
Hot Asian Ribs,
Insanely Good Slow-Cooker Barbecued Beef Ribs,
Key West Ribs,
Maple Chili Ribs,
Maple-Spice Country-Style Ribs,
Orange-Glazed Country Ribs,
Polynesian Pork Ribs,
Ribs 'n' Kraut,
Ribs with Apple Kraut,
Rosemary-Ginger Ribs with Apricot Glaze,
Satay-Flavored Pork,
Short Rib Stew,
Short Ribs with Mushrooms,
135 Short Ribs with Wine and
Mushrooms,
Slow Cooker "Barbecued" Ribs,
Slow Cooker Pork Ribs Adobado,
Slow Cooker Teriyaki Ribs,
Soy and Sesame Ribs,
Teriyaki-Tangerine Ribs,
"Risotto" with Mushrooms and Peas, Slow Cooker,
Rival Crock-Pots, –11
roasts. _See_ pot roast Roman Stew,
Rosemary-Ginger Ribs with Apricot Glaze,
Russian Dressing, –123
rutabaga
Easy (But Really Slow) Rutabagas, –296
Ham with Rutabaga and Turnips,
Pork with Rutabaga,
S
salmon
Balsamic-Vanilla Salmon,
Lemon-Mustard Salmon Steaks,
Maple-Balsamic Salmon,
Orange Basil Salmon,
Salmon in Sweet Chili Sauce,
Satay-Flavored Pork,
sauces
Butterscotch Sauce,
Cocktail Sauce,
Cranberry Chutney,
Cranberry-Peach Chutney,
Dana's "Kansas City"
Barbecue Sauce,
Dana's No-Sugar Ketchup,
Eastern Carolina Vinegar Sauce,
Florida Sunshine Tangerine Barbecue Sauce,
Hoisin Sauce,
Low-Carb Teriyaki Sauce,
Not-Quite-Asian Sweet Chili Sauce,
Piedmont Mustard Sauce,
Raspberry-Chipotle Sauce,
Slow Cooker Chutney, –308
Slow Cooker Cranberry Sauce,
Sugar-Free Chocolate Sauce,
Sauerbrauten,
sauerkraut
Choucroute Garni,
Ribs 'n' Kraut,
Ribs with Apple Kraut,
sausage
Beef and Sausage Stew,
Bollito Misto,
German UnPotato Soup,
Horseradish Smokies,
Italian Sausage Soup,
Louisiana Meat Loaf,
Low-Carb Slow Cooker Paella, –252
Orange Smokies,
Pizza Stew,
Pizza-ish Meat Loaf,
Sausage and Green Bean Soup,
Stuffed Peppers,
Turkey Sausage Soup,
Tuscan-ish Meat Loaf,
scallops
Lime-Basted Scallops,
Scallops Florentine,
seafood
Almond-Stuffed Flounder Rolls with Orange Butter Sauce,
Bacon-Almond Flounder Rolls,
Balsamic-Vanilla Salmon,
Crab Puff,
Easy Party Shrimp,
Hot Crab Dip,
Lemon-Mustard Salmon Steaks,
Lime-Basted Scallops,
Low-Carb Slow Cooker Paella, –252
Maple-Balsamic Salmon,
Maria's New England Clam Chowder,
Orange Basil Salmon,
Pantry Seafood Supper,
Salmon in Sweet Chili Sauce,
Scallops Florentine,
Seafood Chowder,
Super-Simple Catfish,
Sweet and Sour Shrimp,
Tuna and Artichoke Stuffed Mushrooms, –31
Tuna Egg Casserole,
Tuna-Noodle Casserole,
seasonings
about, –13 Adobe Seasoning,
Cajun Seasoning,
Classic Rub,
Garam Masala,
seeds. _See_ nuts and seeds
Seriously Simple Chicken Chili,
Seriously Simple Lamb Shanks,
shirataki,
short ribs. _See_ ribs
shrimp
Easy Party Shrimp,
Low-Carb Slow Cooker Paella, –252
Pantry Seafood Supper,
Seafood Chowder,
Sweet and Sour Shrimp,
Simple Meat Loaf,
Simple Salsa Beef,
Simple Turkey Drumsticks,
Simplest Slow Cooker Mushrooms,
slow cookers
about, –11
size of, ,
slow cooking tips, –12
Smokin' Chili Peanuts,
snacks
_See also_ hors d'oeuvres Asian Peanuts,
Blue Cheese Dressing Walnuts,
Butter-Spice Almonds,
Buttery Vanilla Almonds,
Cajun-Spiced Pecans,
Candied Pecans,
Chili Garlic Peanuts,
Curried Pecans,
Dana's Snack Mix, Slow Cooker-Style, –39
Kickin' Pecans,
Maple-Glazed Walnuts,
Smokin' Chili Peanuts,
Spiced Walnuts,
Sweet and Salty Peanuts,
So Totally Inauthentic Malaysian-oid Pumpkin-Chicken-Tomato Soup,
Sort-of-Ethiopian Chicken Stew,
soups
Black Bean Soup,
Bollito Misto,
Cauliflower, Cheese, and Spinach Soup,
Chicken and Fennel Soup, –278
Chicken and Vegetable Soup with Thai Spices, –273
Chicken Minestrone,
Chicken Soup with Wild Rice,
Chipotle Pumpkin Soup,
Cream of Mushroom Soup,
Cream of UnPotato Soup,
German UnPotato Soup,
Italian Sausage Soup,
Maria's New England Clam Chowder,
Mexican Beef and Bean Soup,
Not-Pea Soup, –272
Pho, –260 Sausage and Green Bean
Soup,
Seafood Chowder,
So Totally Inauthentic
Malaysian-oid Pumpkin-Chicken-Tomato Soup,
Spicy Chicken and Mushroom Soup,
Tavern Soup,
Turkey Sausage Soup,
Vegetable Beef Soup, –265
Southeast Asian Coconut Custard, –315
Southern Beans,
Southwestern Barbecue,
Southwestern Pork Stew, Easy,
Soy and Sesame Ribs,
soybeans
about, –16
Baked Beans,
Black Bean Soup,
Mexican Beef and Bean Soup,
Spiced Walnuts,
spices, whole,
Spicy Chicken and Mushroom Soup,
spinach
African Chicken, Peanut, and Spinach Stew,
Artichoke-Spinach-Ranch Dip,
Cauliflower, Cheese, and Spinach Soup,
Chive-and-Garlic Spinach Packet,
Macadangdang, –290
Maria's Eggs Florentine,
Noodleless Spinach Lasagna,
Scallops Florentine,
Spinach Parmesan Casserole,
Spinach-Mushroom Gratin, –300
Spinach-Stuffed Pork Loin, –183
Splenda, ,
sriracha,
Stewed Pork Neckbones with Turnips and Cabbage,
stews
African Chicken, Peanut, and Spinach Stew,
Beef and Sausage Stew,
Beef and Zucchini Stew,
Brunswick Stew, –253
Carne all'Ungherese,
Chicken Stew,
Cocido de Puerco,
Curried Pork Stew,
Easy Southwestern Pork Stew,
Lamb Stew Provençal,
Low-Carb Slow Cooker Paella, –252
Mexican Stew,
Pizza Stew,
Pork and "Apple" Stew,
Roman Stew,
Short Rib Stew,
Sort-of-Ethiopian Chicken Stew,
Yassa,
Stuffed Peppers,
sucralose, ,
Sugar-Free Chocolate Sauce,
sugar-free imitation honey,
sugar-free pancake syrup,
Super-Simple Catfish,
Sweet and Salty Peanuts,
Sweet and Sour Pork,
Sweet and Sour Shrimp,
Sweet and Tangy Mustard Pork Roast,
Swiss Steak,
T
Tangy Beans,
Tangy Pork Chops,
Tavern Soup,
temperature settings, –11
Tequila Lime Chicken,
Teriyaki Ribs, Slow Cooker,
Teriyaki-Tangerine Ribs,
Texas Red, –152
Thai dishes
Chicken and Vegetable Soup with Thai Spices, –273
Chicken in Thai Coconut Curry Sauce,
Thai Chicken Bowls, –93
Thai Hot Pot,
Thai-ish Chicken and Noodles,
Turkey Loaf with Thai Flavors, –110
3-Minute Slow Cooker Pot Roast,
timing, –14
tofu shirataki,
tortillas,
tuna
Pantry Seafood Supper,
Tuna and Artichoke Stuffed Mushrooms, –31
Tuna Egg Casserole,
Tuna-Noodle Casserole,
turkey
Albondigas en Salsa Chipotle, –251
Braised Turkey Wings with Mushrooms,
Chipotle Turkey Legs,
Cranberry-Barbecue Meatballs,
Cranberry-Peach Turkey Roast,
Mean Old Rooster Chili,
Mediterranean Turkey Loaf,
Morty's Mixed Meat Loaf,
Pub Loaf,
Ranch-E-Cue Wings,
Simple Turkey Drumsticks,
Turkey Loaf with Thai Flavors, –110
Turkey Sausage Soup,
Turkey with Mushroom Sauce,
turnips
Cheesy Neeps,
Ham with Rutabaga and Turnips,
Hamburger and Turnip Layered Casserole,
Maple-Mustard Turnips,
Mashed Turnips, –298
Stewed Pork Neckbones with Turnips and Cabbage,
Tuscan Chicken,
Tuscan-ish Meat Loaf,
U
Ultimate Fauxtatoes,
V
Vege-Sal,
Vegetable Beef Soup, –265
vegetables, cooking times for,
Venison Chili, Free, –153
W
walnuts
Blue Cheese Dressing Walnuts,
Maple-Glazed Walnuts,
Spiced Walnuts,
Whipped Topping,
X
xanthm gum,
Y
Yassa,
Z
Zippy Cocktail Dogs,
Zucchini Stew, Beef and,
© 2005, 2011 Fair Winds Press
Text © 2005, 2011 Dana Carpender
First published in the USA by
Fair Winds Press, a member of
Quayside Publishing Group
100 Cummings Center
Suite 406-L
Beverly, MA 01915-6101
www.fairwindspress.com
All rights reserved. No part of this book may be reproduced or utilized, in any form or by any means, electronic or mechanical, without prior permission in writing from the publisher.
15 14 13 12 11 1 2 3 4 5
ISBN: 978-1-59233-497-1
Digital edition published in 2011
eISBN-13: 978-1-61058-152-3
Digital edition: 978-1-61058-152-3
Softcover edition: 978-1-59233-497-1
Library of Congress Cataloging-in-Publication Data available
Printed and bound in Canada
The information in this book is for educational purposes only. It is not intended to replace the advice of a physician or medical practitioner. Please see your health care provider before beginning any new health program.
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Илийчо Йорданов е добруджански културен деец, участник в борбата за освобождение на Добруджа
Дейност
Роден в Силистра. Завършва силистренското педагогическо училище.
Най-голям принос за културните изяви на русенското дружество "Добруджа" в града и в страната има основаният още през 1919 г. смесен добруджански хор под диригентството на Илийчо Йорданов. Сам добруджански бежанец, Илийчо Йорданов е първият и единствен диригент на хора от неговото създаване до разформирането му през 1947 г.
Хорът се радва на голяма популярност в цялата страна. Паметни са участията му по време на големите чествания на годишнината от гибелта на Стефан Караджа и Тутраканската епопея. Секретар на дружество "Добруджа" в Русе.
За заслуги към Родината е предложен за награда с Народна пенсия и орден.
Източници
Димитрова, М., С. Йорданов. Лицата на Русе. Русе, 2012
Златев, Л., Х. Лебикян. Иван Хаджииванов. Родови корени, общественик и деен ръководител на борбата за освобождението на Добруджа. Русе, 2005
Външни препратки
Златев, Л. Вътрешната добруджанска революционна организация (ВДРО) 1923 – 1940 г. Русе, 2009 г.
Златев, Л., Х. Лебикян. Иван Хаджииванов. Родови корени, общественик и деен ръководител на борбата за освобождението на Добруджа . Русе, 2005
Дейци на ВДРО
Български диригенти
Родени в Силистра
Русенци | {
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The Chimney Sweepers Act 1875 was an Act of the Parliament of the United Kingdom that superseded the Chimney Sweepers and Chimneys Regulation Act 1840 passed to try to stop child labour. The Bills proposed by Lord Shaftesbury, triggered by the death of twelve-year-old George Brewster whose master had caused him to climb and clean the chimney at Fulbourn Hospital.
The Chimney Sweepers Act 1875 was repealed for England and Wales by section 1(1) of the Chimney Sweepers Acts (Repeal) Act 1938 (1 & 2 Geo 6 c 58).
The 1840 Act prohibited any person under 21 being compelled or knowingly allowed to ascend or descend a chimney or flue for sweeping, cleaning or coring. This Act ensured all chimney-sweeps would be registered with the police, and that official supervision of their work would take place. The provisions of all previous acts would now take place.
Further reading
"The Chimney Sweepers Act 1875". Halsbury's Statutes. (The Complete Statutes of England). First Edition. Butterworth & Co (Publishers) Limited. 1930. Volume 13. Pages 784 et seq. See also pages 479 and 870.
"Chimney Sweepers Act 1875". Education in England. (Derek Gillard).
References
Notes
Bibliography
United Kingdom Acts of Parliament 1875
United Kingdom labour law
Child labour law
1875 in labor relations
Chimney sweeps | {
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{"url":"https:\/\/www.math-forums.com\/threads\/why-are-there-only-finitely-many-steiner-points.412828\/","text":"# Why are there only finitely many Steiner points?\n\nDiscussion in 'Math Research' started by tchow, Jun 24, 2006.\n\n1. ### tchowGuest\n\nA few years back, I posted here asking why the maximum number of\nSteiner points in a minimum (Euclidean) Steiner tree is n - 2, where n\nis the number of initial points. Gerry Myerson gave a helpful response,\npointing me to the work of Melzak. However, I was just looking at this\nagain, and one point that was never quite settled to my satisfaction\nwas the question of why the maximum number of Steiner points is finite.\n(I buy that *if* the number of points is finite, then it's at most n - 2.)\n\ntchow, Jun 24, 2006\n\n2. ### Gerry MyersonGuest\n\nSteiner points have degree 3. I don't know much about infinite trees;\nis it possible to have one with all but finitely many of its vertices\nof degree 3?\n\nGerry Myerson, Jun 26, 2006\n\n3. ### klaus hoffmannGuest\n\nist the finiteness unconditional, i.E. for bounded and unbounded pointsets?\nregards\nKlaus\n\nklaus hoffmann, Jun 26, 2006\n4. ### klaus hoffmannGuest\n\nIf the shortest tree has mostly angles >120 degrees then there is no need for\nmany steiner points. This is the only explanation I can think of - perhaps we\nget mostly those configurations if we consider the limit to an infinite pointset.\nI'm very curious to see the reference for the finiteness\nKlaus Hoffmann\n\nklaus hoffmann, Jun 26, 2006\n5. ### tchowGuest\n\nIn principle, yes; e.g., take the tree whose vertices *all* have degree 3.\n\n\"Obviously,\" such fractal monstrosities are ridiculous in the context of\nEuclidean Steiner trees, but it seems annoying to prove.\n\ntchow, Jun 27, 2006\n6. ### klaus hoffmannGuest\n\nThere are two questions:\n1. The finiteness of the steiner-candidate set if n is finite\n2. The finiteness of the steiner set if n is infinite\n\n1.\nConsider a fixed set of n terminal points. Build a connecting tree with n-2\ninternal points [1]. There are f(n)=(2n-4)!\/2^(n-2)\/(n-2)! nonisomorphic trees;\nfor each the minimum length, and the set of nonterminal points is unique [2].\nThere are at most\n(n-2) f(n) interior points.\nq.e.d.\n\n[1] the degree of a (interior=) Steiner point is 3.\n[2] if the tree topology is fixed, the optimization w.r.t. length is convex; in\ngeneral innterior points will coincide.\n\n2. infinite n\nassuming we have some useful definition of total tree length\na) Consider (x_n,y_n)=(4*n,0). There is no need for steiner points.\n\nb) attach the unit square to each of the points in a). \"obviously\" we need\nsteiner points in each square, so seem to need an infinite number of s.p. .\n\nRegards\nKlaus\n\nklaus hoffmann, Jun 28, 2006\n7. ### tchowGuest\n\n<1. The finiteness of the steiner-candidate set if n is finite\n[...]\n<Consider a fixed set of n terminal points. Build a connecting tree with n-2\n<internal points [1]. There are f(n)=(2n-4)!\/2^(n-2)\/(n-2)!\n<nonisomorphic trees;\n<for each the minimum length, and the set of nonterminal points is unique [2].\n<There are at most\n<(n-2) f(n) interior points.\n<q.e.d.\n<\n<[1] the degree of a (interior=) Steiner point is 3.\n<[2] if the tree topology is fixed, the optimization w.r.t. length is\n<convex; in general innterior points will coincide.\n\nI don't follow how this rules out infinite trees. All it seems to show\nthat *if* the number of points is finite, then you can get a bound on them.\n\ntchow, Jun 30, 2006\n8. ### tchowGuest\n\nI managed to come up with enough of an argument to satisfy myself.\n\nThe usual definition of a Euclidean Steiner tree is a tree of minimum\nEuclidean distance that connects a given set of points in the plane.\nThe tree is allowed to contain vertices (Steiner points) other than\nthe original set of points. My question was, if the initial set of\npoints is finite, must the set of Steiner points be finite?\n\nI'm pretty comfortable now with the following approach, which more or\nless \"defines away\" any potential problems. We simply insist that for\nany pair of initial points to be \"connected\" by the tree, it must be the\ncase that there is a *finite* sequence of Steiner points, each adjacent\nto its neighbors in the sequence, that joins the two.\n\nGiven this, we can restrict our attention to the case of finitely many\nSteiner points as follows. Suppose we have an infinite Steiner tree.\nFor each pair of initial points there is some path connecting them.\nAs we range over all pairs of initial points, these paths will involve\nat most finitely many vertices in all. This finite set of vertices will\ninduce a finite subtree of the total tree. Throwing away the rest of\nthe tree cannot disconnect the tree or increase its total length. Thus\nit suffices to consider the case of finitely many Steiner points.\n\nThe loophole in this argument is that perhaps there is some intelligible\nsense in which two initial points are \"connected\" by means of infinitely\nmany Steiner points, other than by a finite sequence of them as stipulated\nabove. However, I guess I don't mind excluding such possibilities by fiat.\n\ntchow, Jul 2, 2006","date":"2021-04-15 02:50:36","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8507112264633179, \"perplexity\": 1749.4221086735208}, \"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\/1618038082988.39\/warc\/CC-MAIN-20210415005811-20210415035811-00311.warc.gz\"}"} | null | null |
### Table of Contents
Copyright
Aphelion: Tales from the Dark Recesses
Times Change
Off Flesh
Reflection
One Mistake
Serere, A Prelude
Aphelion: Tales from the Dark Recesses
By Andy Frankham-Allen
Copyright 2014 by Andy Frankham-Allen
Cover Copyright 2014 by Untreed Reads Publishing
Cover Design by Ginny Glass
The author is hereby established as the sole holder of the copyright. Either the publisher (Untreed Reads) or author may enforce copyrights to the fullest extent.
All stories in this anthology were previously published individually by Untreed Reads.
This ebook is licensed for your personal enjoyment only. No part of this book may be reproduced in any form or by any electronic or mechanical means, including information storage and retrieval systems, without written permission from the publisher or author, except in the case of a reviewer, who may quote brief passages embodied in critical articles or in a review. If you would like to share this book with another person, please purchase an additional copy for each person you share it with. If you're reading this book and did not purchase it, or it was not purchased for your use only, then please return to your ebook retailer and purchase your own copy. Thank you for respecting the hard work of this author.
This is a work of fiction. The characters, dialogue and events in this book are wholly fictional, and any resemblance to companies and actual persons, living or dead, is coincidental.
Also by Andy Frankham-Allen and Untreed Reads Publishing
Seeker
Space: 1889 and Beyond Series:
Journey to the Heart of Luna
Conspiracy of Silence (with Frank Chadwick)
Mundus Cerialis (with Sharon Bidwell)
The Forever Journey (with Tom Sanford, Christian Mansell, David Parish-Whittaker)
Horizons of Deceit Book II (with Jonathan Cooper)
www.untreedreads.com
Aphelion: Tales from the Dark Recesses
Andy Frankham-Allen
Times Change
Henry Ward Beecher once said, "clothes and manners do not make the man; but, when he is made, they greatly improve his appearance." It was true. Appearance was all important in the world of mortal man, and as he regarded himself in the mirror, Iago smiled. He was the epitome of man; every muscle perfectly toned, his eyes exactly the right color to go with his skin tone, hair fine but thick, nails manicured to perfection. And his manners, beyond reproach.
No woman would be able to resist him, as ever.
Once again it was time to go forth, and multiply.
*
Sex on a stick. There was just no other term for it. When Caitlyn first saw him that's exactly the phrase that popped into her head; second to that, she just knew she had to get this guy's number.
She had been with her girlfriends enjoying a macchiato when she had spotted him. He was walking in her general direction, but she liked to think he was approaching her. In reality, he was probably going to order his own shot of caffeine from the counter behind her, but he did throw a very candid wink her way as he passed her table. Of all the girlfriends huddled around the little table, it was Niki who first spotted Caitlyn's distraction.
"Do you really think so?" Niki whispered in her ear, leaning in close so the others wouldn't hear.
Caitlyn smiled broadly. How could she not? "I really really do," she said. She offered Niki a wink and excused herself from the table. As she walked towards the counter, she heard the curious whispers of her girlfriends, and Niki acting as if she knew nothing. Of course, soon they'd all spot the hunk of sexiness and they'd realize. But by then it would be too late; he would be hers.
She stopped at the counter, standing behind the man, pretending to look at the cakes in the display cabinet, but really admiring his rather fit ass. Dark blue jeans with white stitching clung tightly to his perfectly shaped bubble butt, the seam riding high between his legs. For a second Caitlyn actually had to look at the cakes, otherwise she knew she'd need to start fanning herself. Back under control she inched forward, almost touching him.
The girl behind the till must have noticed, because she cast a dark look Caitlyn's way. Caitlyn just smiled back in response. She didn't care; if the girl had eyes that worked she'd realize the need to do anything to try and get the attention of the man. At the moment he was simply a customer, but if the girl had played her cards right he could have become so much more. Caitlyn wondered where such thoughts came from.
He spoke to the cashier, and moved on towards the counter by the espresso machine, where his latte was being prepared.
"Can I help you?" the girl asked Caitlyn.
Caitlyn had to drag her eyes away from the man. "What? Ah, no," she said, knowing she should at least pretend to consider wanting something. Other than that sexy beast, of course.
"Okay, well, if you'd like to move along, I do have other customers, you know," the girl said, icy politeness in every word.
Caitlyn opened her mouth to speak, her mind rushing to find some kind of excuse to remain at the counter, close to the man. As it turned out, the man himself gave her a reason.
"Can I get you a drink?"
Caitlyn turned her head towards him, smiling in response to his own bright smile. He was perfect, no other way to put it. His skin was almost bronze, his hair raven black, and his eyes, deeply set in a face that looked like it had been sculptured by the angels, were as brown as mocha. And his smile... Pearly white teeth filled that smile, broad and welcoming. Caitlyn almost felt weak at the knees just looking at him. He stepped forward, and offered his hand. Automatically she reached out with her own hand, and he leaned over and kissed the back of it. It was such an old fashioned gesture. Caitlyn was certain that no one had ever done such a thing to her before. He looked to be no more than twenty-five, yet such an introduction suited a man twice his years. Caitlyn approved.
"And to whom do I owe the pleasure?" he asked, his tones speaking of Eastern origins, although not being much good at geography Caitlyn could never hope to pinpoint the source. But it didn't matter, his English was perfect.
"Caitlyn," she began, her voice almost a whisper. She coughed demurely (demurely! She had never done anything demurely in her entire life. Until now!), and tried again. "Caitlyn Eaves."
"A beautiful name, for a beautiful lady," he said, the words rolling off his tongue with such delight. "Possibly derived from katharos, the Greek for pure?"
Caitlyn had no idea. So she opted for, "possibly." Not that the word pure had ever been applied to her before. Unless pure vixen counted.
"I am simply Iago."
Caitlyn tried that name out. "Ee-A-gaw?" she said slowly, and grinned again. "Iago is a nice name. It's distinctive."
"Yes," Iago agreed, still smiling at her. "Distinctive is good. I like distinctive. Excuse me," he added as he was called to pick up his latte. Caitlyn waited, quite taken by his manners. Normally she wouldn't care less, after all she was hardly the most couth person on the planet, and she liked her men to be men, not pansies. But with Iago it seemed different. Suited him like an old coat. And besides, unlike Hunter at home, there was nothing feminine about Iago. He was all man, and as he reached for his latte, she watched his t-shirt pull tight against his well-toned torso, feeling her pulse racing that bit more. Oh yes, all man.
Once he held the tall mug in his bronze hands, Iago looked back at Caitlyn. "Would you care for a drink?"
"I'm good, thanks. I've already had one," she said, glancing back at the table her girlfriends were still sitting at. Iago looked over with a smile, and Caitlyn noticed almost all the girls bat their eyelids, grinning like idiots. Except Niki, who just winked at Caitlyn and returned to her decaf Frappuccino.
"I see," Iago said slowly. "Then perhaps I can trouble you for your company? That is, if your friends don't mind me stealing you."
"Mind?" Caitlyn almost spluttered. They'd be downright jealous, but she wasn't about to reveal that Iago. "I'm sure they'll understand."
"Excellent. Then, perhaps you can introduce me to them another time?"
Caitlyn wasn't sure why he'd want that, and it wasn't like she intended to share him, but she shrugged and muttered an "okay."
At this Iago laughed. "Perhaps that table there?"
"As good as any," Caitlyn said and allowed Iago to lead the way. After all, she wanted to get the best view of his ass as he walked. She followed, and glanced back at the girl behind the till, who was now staring daggers in Caitlyn's direction. Too late, hun, Caitlyn thought, much too late. He's mine now.
*
"I'm not so sure about this, love," Hunter said, as Caitlyn opened the door to leave.
Caitlyn looked back, and offered him a reassuring smile. She did love Hunter, would never have shared a flat with him if not, but he was a bit of worrier. "Don't be daft. What can go wrong? A man with such charm...and looks." She stopped, and for a moment her mind went back to the coffee shop, and the way Iago's biceps worked under his tight top. "Whew, did I mention how hot he is?"
Hunter had to laugh at that. "Yes, once or twice. Feel a bit jealous now, actually."
"As you should, darling boy, as you should." She leaned over and pecked him on the cheek. "Be seeing you," she said and walked out of the flat.
She glanced back at the converted terraced house, and offered Hunter a little wave. He tentatively waved back, his face still carrying the same look of worry. Caitlyn shook her head. Silly boy, it wasn't like this was the first date she'd been on. And besides, as Iago had promised, there he was now. Standing beside the black cab, door open for her.
"Your carriage awaits," he said, as she approached.
Caitlyn fluttered her eyelids at him, and he flashed her his biggest smile. As she stepped into the cab he took her hand in his and kissed it. Oh yes, it was going to be wonderful night.
*
Hunter didn't like it. Caitlyn was acting very odd, even for her. He had known her for a few years and she had never been one for lady-like behavior, but since returning from work she'd been very prim and proper, like she'd become some sort of fairytale princess all of a sudden; endlessly chatting about this guy who had totally blown her mind. Now, Hunter was more than happy to see Caitlyn falling for someone decent; after all, she had fallen for some skets over the years, but something about this didn't ring true.
Although, Hunter had to admit, having now caught sight of the enigmatic Iago, that Caitlyn was right about his looks. He was gorgeous, of the drop dead variety. Even from this distance Hunter could see that.
He remained as he was for a while, standing in the doorway, watching the cab pull away, images of Iago filling his mind.
Yes, he was definitely hot, and it seemed such a shame to waste such an amazing man on someone like Caitlyn.
With a flourish Hunter spun on his heels and flounced back into the house. Caitlyn might think she was a princess now, but he'd show her what real royalty was.
*
As promised, Iago took her to a recently opened French restaurant just off Regent's Street; Grand Restaurer de Londres. The food was divine, although a bit pricey for her, but Iago assured her that she deserved nothing less, so who was she to argue? Caitlyn stifled a giggle at that.
Once again she knew she was acting a little unusual in Iago's presence, but she could not help it, he simply brought the inner lady out of her. Until the coffee shop at lunch she didn't even realize there was a lady in her. And yet, here she was, sitting in the plush restaurant, using the correct cutlery, napkin spread out on her lap, and sipping expensive Cuvée Alexandra Rosé. Yes! Actually sipping, not downing in one go.
"How is the champagne?" Iago asked.
Caitlyn lowered the glass, and went to speak, but the bubbles from the champagne hit her nose and she laughed. Iago smiled at her indulgently, and reached his hand across the table. He held her hand gently, massaging the back of it with his ringed thumb.
"I know it is most improper, but I must say I do feel like I should invite you back to my hotel tonight."
Caitlyn blushed, at the mere thought of spending alone-time with Iago. The girls would be so jealous tomorrow. "I wouldn't be against such an idea," she said softly.
Iago nodded his head once. "Then we are agreed."
*
From his vantage point across the road from the Grand Restaurer de Londres, Hunter could see Caitlyn and Iago at their window table, he holding her hand while they talked, his eyes never once leaving her face.
Hunter didn't like it. Should have been him in there with Iago, being the centre of his world, not Caitlyn.
Somehow he had to separate the two of them.
He narrowed his eyes in thought. An opportunity would present itself. He just had to be patient. Although, Hunter wondered, how long could he be patient for?
*
Food finished, bill paid, the two of them headed for the doors where the maître d' was waiting for them with their coats. Once Iago had slipped into his, he opened Caitlyn's and allowed her to slip her arms into the sleeves while the maître d' held onto her purse. She thanked the man, and turned to Iago, who was holding the door open for her.
"After you," he said.
"Why thank you, sir," Caitlyn responded, and made to leave. She stopped, her hand holding her stomach as the acids within loudly digested some food. She looked around the restaurant in horror, her cheeks turning red. Normally a bit of gas would not have bothered her, "better out than in" as her mum would say, but with Iago here...
"The oysters might have been a little too rich for me," she whispered, and Iago nodded in sympathy. But he said nothing, which troubled Caitlyn even more than her unruly stomach.
"The powder room is this way, Madame," the maître d' said, one arm sweeping in the general direction of the lady's room.
With a flush, Caitlyn excused herself.
*
With delight he had barely felt before, Hunter watched as Iago stepped out of the restaurant alone. There would never be a better time, he decided, and rushed across the road. Such was his focus that he barely noticed the car racing towards him, but the honking of the horn alerted him in time and he was able to just avoid being hit.
He stepped onto the pavement, smiling at Iago, who, he noticed, had been watching him the whole time. Iago frowned, confusion sweeping across his face.
"Hiya," Hunter said.
Iago stepped forward, his hand reaching out. "You are..."
"Hunter James. How you doing, love?"
Iago's hand stopped within centimeters of Hunter. He glanced down at his hand, momentarily uncertain about something. "You are not a woman," he said, looking back up, his deep brown eyes probing Hunter's own baby blues, "and yet, somehow..."
"Well, I am sometimes," Hunter pointed out, "if that helps."
"Most perplexing. I have never..."
Hunter grinned. This was called knocking a man off his feet. He'd impressed a few men over the years, but never quite left them speechless. He stepped towards Iago, but Iago pulled back quickly, shaking his head.
"No, this is... No." Abruptly Iago grabbed Hunter by his shoulders. Hunter closed his eyes, feeling Iago's breath on his face. Utopia was calling him now. "You must leave, this cannot be."
"Huh?" Hunter opened his eyes, and found his nose almost touching Iago's. "Don't be silly, love, you know you want me," he said, surprising himself with his words. They were so unlike him, but deep down he knew they were true. Hunter frowned a little. "I can feel it."
Iago's eyes darted over Hunter's face. "I do, but... No. Wrong. This is all wrong."
He released Hunter and stepped back sharply. "Leave!" he hissed, turning away as he did so.
Hunter wanted to step forward, turn Iago around, show the man how much they wanted each other. But something in him broke; the rejection from Iago was too much. Trying to hold back a torrent of sudden tears, Hunter scrambled away.
*
Caitlyn stepped out of the restaurant to find Iago looking down the road, his eyes searching for something, a deep frown creasing his usually beautiful face. She looked to see what he was looking for, but there was nothing, just the usual hustle of people you'd expect to see on a summer's night by Regent's Street.
"Sorry about that," she said, stepping into his line of sight.
Iago blinked, turning his attention to her. He smiled, an act that seemed a little forced to Caitlyn. She brushed the slight feeling of disappointment aside. "Yes, but let us not worry about that now. Shall we go to the hotel?"
Caitlyn put her arm in his and snuggled up to him. "Yes. Let's."
*
Together, arm in arm, they set off. Caitlyn filled with more bubbles than the champagne, her mind on the fun they'd have at the hotel. Iago, outwardly smiling, but inwardly still wearing his frown.
That man had confused him, drawn something out of him that only women were designed to do.
Something was wrong with the world.
*
It had been a long time since he'd felt like this about another man, but Hunter could not get Iago out of his head. Sure, he'd had plenty of romances since he came out at fourteen, and most of them had ended well, except for one. That had left him devastated, and he had sworn to himself that he would never ever again allow another man to make him feel that way.
But last night, when Iago had told him to leave...
He'd gone home and cried like a girl. So much and all night long, leaving him so tired in the morning that he had to call into work sick, spending the day in bed. Trying to sleep. And failing miserably at that, too.
The evening had been no better. His day job he could put off; if they complained too much he'd just tell them his mother had died. Wasn't like he ever spoke to her, so the guilt factor would be pretty low. But his evening job...much like the show, it had to go on.
He was a performer, and Tess Tosterone never kept her audience waiting. Hiding behind the make-up, the glitz, made things easier to deal with, although while singing his end number, a funked up rendition of Madonna's "Cherish," he could have sworn he'd seen Iago's face in the crowd. Several times over.
Now as he approached the converted terraced house, still dressed up as Tess, his mind was returning to Iago again. And he knew, without a doubt, that he had it bad.
Hunter had never been a believer in love at first sight, but after seeing Iago standing outside the cab... Now he knew it was possible. No, not just possible, it had happened. To him.
He just had to tell Caitlyn. She would understand. Of course she might be a little angry at first, since technically Iago was hers, but he'd make her realize that his love for Iago was stronger than any lust she felt.
Hunter closed the door behind him and made for his bedroom, which was the first door along the hallway, a few feet away from the living room opposite. "Hi, honey, I'm home," he called out, using his husky Tess voice. "Be a love and put the kettle on, I'm beyond parched."
Once he'd dumped his gear in his room and removed his heels (yes, they helped sell the illusion, but they were a nightmare to wear—almost as bad as the tights, but he couldn't be arsed to remove them yet, besides his cuppa was waiting), he headed into the living room.
Where Caitlyn sat on the sofa with Iago.
Hunter stopped, his heart jumping into his throat. "Oh."
Caitlyn looked up with a huge grin, her hand flying to her mouth. "Oh god," she said, struggling on to her knees so she was leaning against the back of the sofa. "I was going to introduce you to Iago, but..." She looked Hunter up and down. "Who should I introduce you as?"
Hunter worked his mouth to speak, but no words would emerge, instead all he could do was look at Iago. The man remained where he was sitting, the same look of confusion sweeping across his features like the previous night.
"You really are most perplexing," Iago said softly.
Caitlyn looked from Hunter to Iago, and back again. Hunter didn't like the look in her eyes, that suspicious, almost accusatory, stare. He swallowed. "Yeah, get that a lot. Excuse me, love, I just need to go and jump in front of a bus."
With that he rushed out to his bedroom, his heart a maelstrom of hurt and rejection.
*
Several hours passed before Hunter dared to venture out of his room. The flat was quiet, the living room light off. He had no idea if they were still in or not, he'd made sure his music was on full to drown out any possible noise the two of them might make together. It was bad enough just imagining the things Caitlyn would do to Iago, without the sound effects crashing into his room like some ghastly confirmation.
Now dressed only in his quarter lengths and a tee, he padded down the hallway towards the kitchen, careful to not make any sound as he passed by Caitlyn's room. Just in case they were in there. He didn't think he could bear seeing Caitlyn right now, let alone Iago. They had probably had a good laugh at his expense already.
Once in the kitchen he made himself a tea and sat down at the breakfast bar. He sat like that for a while, cradling the hot cup until it was not so hot, his thoughts running wild at what Iago had probably told Caitlyn about last night outside the restaurant. He didn't know what she'd do about it, and he wasn't sure he wanted to find out. All he knew for sure was that he wanted Iago, way more than Caitlyn could possibly want him. There were things he could do for Iago, to Iago, that Caitlyn couldn't possibly compete with. If only Iago would give him a chance.
He sucked back a sob, just as the door to Caitlyn's bedroom opened a crack. He looked up, expecting to get Caitlyn's full wrath, but instead out stepped Iago. At first he was bathed in shadow, his outline implying that he was very naked, but as he stepped into the kitchen light Hunter saw that Iago did, at least, have his boxers on. His very tight boxers.
Hunter shifted uncomfortably, folding one leg over the other so as to hide his reaction to Iago's remarkably perfect body. He swallowed hard, and offered up a "hiya".
"Hunter," Iago said, stopping at the breakfast bar directly in front of him. Hunter couldn't help but notice that his knee was barely an inch from Iago's crotch. Again he shifted, now even more uncomfortable, the object of his desire so close to him and yet, with Caitlyn only in the next room, so very far away still.
"Why do you perplex me so? Since last night you have filled my mind, and I do not understand why. This has never happened to me before." Iago shook his head, and sat on the other stool. "I was created for women, not for men, and yet I feel drawn to you."
"And not a sniff of alcohol, eh, love?"
Iago frowned at this. "No," he said, his voice carrying a depth of seriousness Hunter had only ever heard from a priest performing last rights.
"Sorry," Hunter said, trying to keep his own voice serious. Not an easy job; humor was his natural outlet when it came to unexpectedly intense moments like this. "It was a joke, you know, the old story about what's the difference between a straight man and a gay man?" Iago looked at him blankly. "A few pints?" Still nothing. "Wow, thought everyone had heard of that, love."
"I have not." Iago leaned forward. "This world is very strange to me; not as I remember at all." He sat back, and rested his hands on the breakfast bar. "I was made for women, they are pulled to me irresistibly; this is a fact. But I have noticed men looking at me, before and since last night. I thought they looked with jealousy, angered by the perfection they could never reach. But... Now there is you. And I see it is not jealousy. Like with women, it is desire."
Hunter didn't know what to say. Instead he just searched Iago's face for some trace of humor. Clearly the man had to be pulling his leg. But Iago had the most earnest look that Hunter had ever seen.
"I have filled your head."
Hunter blinked, not too sure he had heard right. "Sorry?"
"Last night, you were marked by me. Since then all you can do is think of me, is that not so?"
Hunter couldn't help but smile at this. Wasn't quite the way he would have put it, but Iago spoke the truth. "Yes, ever since I saw you outside last night I've wanted you."
"Men desiring men?" Iago shook his head, his eyes clouding. "This is wrong."
"Well, yes and no," Hunter said slowly, not having expected to delve into such an intense topic so soon. All he really wanted was to get Iago into his bedroom. Only... "It's not wrong for men to want men, but for me to want you. That's wrong, love. Caitlyn is my friend, I love her, but..." He tried to find the words, to explain what his heart was telling him. He shrugged. "Truth is, Iago, she is not right for you. I am." Hunter took a deep breath, glad he had finally said it.
"How can you be? Always there have been men and women. Two sexes, made to procreate, bring forth new life. I was made for women, to impregnate them."
Now Hunter understood. He had met people like this before, straight men who had never considered the possibility of being bi, until being confronted with that one man who was able to spin their heads. And then the confusion set in.
"Times change, love," Hunter said, and removed himself off his stool, no longer bothered by the stirring beneath his shorts. "Tell me you haven't wanted me since last night? You said it yourself, you can't stop thinking about me."
"This is true," Iago said, allowing himself to be pulled gently off the stool. "But it is still wrong."
"Perhaps," Hunter said, bringing his face closer to Iago's, "but sometimes the wrong things are the best things."
Iago was going to reply, but as soon as his mouth opened Hunter placed his tongue inside. For a moment they remained as they were, their tongues probing each other, casting about inside their mouths, each man enjoying the sensation of being within the other. A precursor to bigger and better things. Hunter was the first to pull away, and when he did he couldn't help but smile at the way Iago placed one finger on his own lips, his brow furrowing.
"Tell me that wasn't great?"
"It was..." Iago shook his head.
"Different? New? Exciting?" Hunter winked. "Exciting would be the best answer."
"Wrong."
Hunter pulled away, he didn't understand. He was sure he was on to something, that Iago was seeing the light, but still all he could say was wrong?
"No, love, it's not wrong. You are what you are, and you have to just accept that."
Iago nodded slowly. "Yes, you are right. I know what I was made to be." He pulled Hunter close, and ran a finger up his face. "This must never happen again." His finger stopped against Hunter's forehead. "You must see the truth, and abase yourself."
A sharp pain stabbed his brain, and Hunter staggered back, holding his head. The pain continued, surging through him, causing him to fall to his knees. Once it finally subsided he opened his eyes and looked up at Iago. "What the hell did you just..."
What stood there by the breakfast bar was no man. It stood at least eight feet tall, a beast from hell. Leathery wings sprouted from its back, while coarse black hair covered its body like dripping oil. Its limbs were long and thin, built with sinewy muscle, and hanging between its legs, poking out of the dark hair, was the longest and most repulsive penis Hunter had ever seen.
"I have traversed the world more times than you can conceive of," the creature said, its voice deep and resonant. Hunter swallowed; his throat dry. The voice, although lacking the warmth, was unmistakably that of Iago. "Always the men feared me; my name became myth, a tale used to scare adolescents. In Mesopotamia, I was known as Lilu, and my son was Gilgamesh. A great warrior, created for the war above. I have seen so much of this world, but never have I seen it so changed as now."
Hunter couldn't speak. This was beyond him. Iago, that most handsome of men, who had seduced him with just a look, was this creature, this... "What..." Hunter ran his tongue over his lips. "What are you?"
"I am incubus. I am he that preys on mortal women, he that raises them up to be the bearer of warriors, the victors of the war above us all." The creature, the incubus, moved forward, every muscle in its legs moving beneath the oily skin. "This world is wrong; men no longer fear me, instead they wish to mate with me." Hunter barely had time to blink before the incubus' claw-like hands surrounded his throat, lifting him off the floor until his face was inches from the demonic visage. "Can you now bear children? Does the seed of man swim within you to create new life?"
Hunter wanted to answer, to explain what it meant to be gay, why it wasn't wrong, but he could barely get breath into his lungs. Only the air supply through his nose was preventing him from fainting, such was the creature's grip on his throat. He was getting delirious, he must have been, to even think that this creature cared what he thought.
A scream echoed from somewhere in the house. It had to be Caitlyn. The creature turned its great black head, its yellow eyes casting a look at the closed bedroom door. "Behold, man, you are about to witness the true meaning of creation," the incubus said, and released its hold of Hunter.
He dropped onto the kitchen floor, hard. But he did not make any attempt to get up, instead he gasped for air, which flooded his lungs like shards of broken glass. His vision was becoming blurred, but he had to focus. To see what the incubus was going to do next. And so he watched. There was little else he could do.
The incubus approached the small flight of steps that led up to the hallway, and with each step it took its form began to change. First the legs and arms pulled in tighter, and the coarse hair seemed to melt into its skin. The wings folded tight against its back, sinking into the tight muscles there. Slowly the black oily skin became lighter, until it was the bronze it had previously been. Once again the creature was Iago, only this time he was completely naked, and for a moment Hunter found his eye lingering on the bubble butt as it moved in time with each step.
Hunter shook his head and looked away, reminding himself that Iago and the incubus were the same thing. That the perfectly formed man, the most stunning example of hunkdom that Hunter had ever laid his eyes on, was really that demonic creature which preyed on women. That lived to...
"Oh god," Hunter said, his words barely a breath. He knew why Caitlyn was screaming.
*
It was impossible. And it was unwanted. Two thoughts, intricately linked to the size of her belly. Caitlyn could feel them (yes, them!) moving inside her. That's what had woke her up. She had been sleeping, dreaming nice dreams of her and Iago, of all the fabulous things they would do together, when an alarm sounded in her head. As if her body was trying to nudge her from her dream. Wake up, it was saying, you have simply got to see this.
And see she did. As soon as she opened her eyes she noticed it. The massive lump in her belly, her skin stretched out like some balloon. She had never asked to be a mother, never wanted to be a mother. She knew she'd make a bad one, learned far too many mistakes from her own mum. There was nothing she could teach a child that would be useful.
She was careful. Always, no matter how serious a relationship she was in, she was always careful. A couple of years back she even had the Implanon contraceptive implant put into her arm, supplying her body with the progestin needed to stop ovulation. Only one in a hundred women with the implant got pregnant. She was sure it would not be her.
But, wait. The little lives growing in her, her body undergoing amazing changes so damned quickly. Hormones out of whack! She wasn't thinking straight. She'd been pregnant for, what, an hour or two? How was she supposed to react? Think things through calmly, or go off her rocker.
Caitlyn screamed again.
*
The door opened before Iago reached it, and Caitlyn stepped out, looking around wildly. She pointed at Iago. "You!" she yelled, her arms flailing at him. "What have you done to me?" She pointed at her belly. "I mean...how?"
"Be calm," Iago said, reaching out to her. She tried to pull away, but in her condition she could not move fast enough. He led her by the arm into the kitchen, and Hunter got a full view on her new "condition." She was dressed in her usual pink satin pajamas, but they were now incredibly tight around the waist, while the top was no longer open, the bottoms having popped under the sheer pressure of her expanding belly.
Caitlyn was very pregnant.
And still Hunter just watched, the air ripping through his lungs.
Iago glanced over at him, as he helped Caitlyn to sit on the bottom step, and grinned. Hunter felt a shiver shoot up his back. He had to do something, prevent whatever was going to happen from happening. He attempted to rise off the floor, but his strength failed him.
Iago had done something to him.
The fake-man knelt beside Caitlyn, his naked knees resting on the cold lino of the kitchen floor. Hunter blinked, unable to miss the lengthy manhood hanging between Iago's legs. Before Iago had revealed his true form, when he had still worn the tight boxers that now lay shredded on the floor, he didn't seem to be packing so much. If he had been, Hunter would have noticed. Clearly, when disguising himself as Iago once more, the incubus had failed to disguise the thing that now hung so freely. Accident, or was it a spiteful act to remind Hunter of what he might have had, had things been different? Is that why he had winked? Iago was playing with him, sapping him of his strength so he was forced to watch while this...thing...that had so worked its way into Hunter's very soul played out its twisted act of creation with Caitlyn.
*
"How is this happening?" she asked Iago, now calmed by the gentle administrations of her gorgeous man. "We only slept together last night... And tonight we..."
"One night was enough," Iago said. "It always is enough," he added with his calming smile, the one he had used on her so many times. She touched his face gently; his skin was so warm. Warm and soothing.
"But how? I have this implant; it's supposed to be ninety-nine percent..."
Iago shushed her. "Be proud, my dear katharos, you will give birth to my new legion. And soon we will win the war."
Caitlyn just stared at him. "What war?"
*
Hunter looked up at this. Iago, the incubus, had mentioned the war before. The war above!
Of course. He had read enough mythology books over the years; he had heard of the war in heaven. But it was just a myth, right? Like the incubus was a myth...
Reality crashed in, and Hunter's brain shut down.
*
Iago noticed the man pass out, but he did not care. The world of man had become more complicated, and he needed more time to study it. But right now he had more important things to which he had to attend.
Caitlyn let out a scream of pain, and Iago turned to her. Beneath her belly, deep in her womb, his children were moving. Already they pressed against the birth sac, wanting to be free, to join their father in the war. It was only a matter of moments.
"Be at peace, katharos," Iago said. "You cannot prevent this."
She looked up at him, and he was struck by the anger and hate. He was not surprised by it. So many women over the centuries had responded the same way, once the truth became clear to them. Iago and Caitlyn were not destined to be with each other, she was merely an incubator for his seed to create life.
The truth hurt. "You bastard!" she screamed.
Iago smiled. "I am many things."
He placed his hand on Caitlyn's belly and, feeling their father's presence, his children broke free, tearing their way through the woman's internal organs, cracking the rib cage as if it were a weak prison, and ripped through the layer of skin.
Blood splattered Iago's face, and his bright teeth erupted through his lips in a smile.
The war could now continue.
The incubus glanced down at Caitlyn's dead eyes as its children swarmed up its black, oily arm. The woman would never know the great service she had performed, like so many women before her.
Still there was another who would know. The incubus rose to its clawed feet, and turned to where Hunter lay, still out cold. Death would be swift for the man, but the incubus did not wish to kill him. Although it knew that it would never be able to impregnate a man, something new stirred within, brought on by a remembrance of a kiss.
"Different? New? Exciting?"
Yes; all these things. One day the incubus would return, but in a new form, one Hunter did not know. It wished to explore further this new breed of man. The war would continue to wage without it for a while; its children, spreading out over its body as they were, would see to that. Yes, it would take a break from the war, and see what Hunter had to offer.
*
Three Months Later...
Questions, accusations, statements, medical examinations, even psychiatric evaluations. No one was quite sure what to do with Hunter after the police had come to investigate the reports of screaming from the neighbors. At first it was assumed that Hunter, who was huddled in the corner of the kitchen, cradling Caitlyn's body in his arms, blood all over his clothes, had killed her. But medical examinations proved that somehow something had ripped itself out of her, killing her painfully.
Hunter wouldn't talk of what had happened, not at first, but eventually he opened up. And that's when they placed him in the psychiatric hospital, just for tests, of course. He'd been there for over two months since that one-off visit, and they were happy with him, although his counselor was worried for his safety.
Hunter was worried, too, but he had kept his nose clean and eventually his day-out was approved.
He knelt by the grave and placed the flowers gently on the ground. He thought it was going to be hard coming to see Caitlyn, but it was proving surprisingly easy. In his own mind he was still trying to sort out what had happened while he'd passed out, but it was as if there was some block preventing him from thinking too hard about it.
Hunter knew it would remain a mystery. And of that he was glad.
He smiled thinly. "See you soon, Cait, love," he whispered, making sure his companion couldn't hear him. He was pretty sure the orderlies at the hospitals had no idea what he was planning. It didn't matter; soon he wouldn't be a problem for them.
He stood up and walked over to his companion. A new member of staff at the hospital. A stunning young man from Scotland, with bright blue eyes, light brown hair and a smile that had bedazzled Hunter the second they'd been introduced.
"Ye ready to go back now, then, Hunter?" he asked.
Hunter nodded. "Yeah, love. Thanks for coming with me."
The younger man held Hunter's hand and together they walked off. "Ach, man, for ye, anything," Iagan said, grinning.
Off Flesh
Travel, they say, broadens the mind. It's a truism if ever there was one. What they fail to tell you is that it can scare the living crap out of you, too. I travel a lot, visit a lot of places, stay at a lot of hotels. I've been to some crappy hotels, some really luxurious ones, too. But never been to one like The Cliff's Edge in Torquay. It was a business meeting about selling outboard motors, pretty tedious stuff, really.
Things started going weird on the Saturday after we'd all arrived. The actual meetings weren't to begin until Monday, which left us the whole weekend to pal around and get to know each other. You know, chill in the sauna the way half-naked men seem to like to do, play tennis in the convenient courts located beside the hotel, or just go for a stroll into the nearest little town.
After finishing breakfast, scrambled eggs on lightly buttered toast and a couple of glasses of milk, I came out of the dining room just in time to catch Mr. Wyndham entering the lobby. He was dressed in his tennis whites, so no prize for guessing where he'd been. Something of a fitness fanatic, really, which came as a bit of mystery to me seeing as he didn't eat breakfast. Something told me that Mr. Wyndham, who had a few years on me, would not be around on this little world of ours for longer than I. Still, he seemed a nice enough chap. Like me he had arrived a few days early, so we had the chance to get to know each other a little bit more than the others. I still think of him as Mr. Wyndham, even though by Saturday morning we were already on first name terms. Mark of respect, I suppose. It's "a thing," as my niece would have said.
"Hey there, Sam," I said to him.
"Alright..." he said to me, with a wide smile, and after a failed attempt to juggle his tennis racket and bag gave up on the offering to shake my hand. I laughed and asked him if he fancied a meander into town later. We both shared an interest in antiques, and I'd noticed a little shop on the drive here. Mr. Wyndham said he'd be more than happy to accompany me once he'd had a shower. No problem, I could find something to occupy me while he was getting rid of all that manly sweat.
I watched him walk away, my eyes lingering on his pert ass beneath the white shorts, and only turned away when he entered the lift. I glanced around the lobby, hoping no one had noticed where my eyes had looked. Not that I'm in the closet or anything, it's just there was something about him that I couldn't resist. And yes, it's true; I'm a married man. So sue me.
So, there I was, not much of anything to do except wait. Once I was certain no one was paying me any attention my eyes returned to the lift. Going up, of course. Mr. Wyndham was on the first floor, so I guessed he wouldn't be too long. I turned away, intending to find something to occupy me, but before I could come up with anything even remotely interesting there was the ding of a bell and the sliding noise of metal on metal as the lift doors reopened. I turned around. Maybe Mr. Wyndham had left something in the courts.
It wasn't him. A couple emerged from the lift, so caught up in their own world they were totally unaware of this casually dressed thirty-something man watching them. I suspect they were having an affair...only people in the midst of a clandestine affair would be so wrapped up in each other.
For a moment I was puzzled. Surely there had not been enough time for the lift to reach the first floor? I dismissed this. Not like I wasn't in a world of my own for a while there. More time could certainly have passed than I realised.
Once again I turned away from the lift.
*
Time passed, as it is want to do. At first I wasn't sure how much, since I got caught up in conversation with another hotel guest. It was a bizarre conversation, one in which I spent most of the time nodding and making the occasional agreeable sound, since I barely had a chance to get a word in. This guest, a young lady called Elisa, rambled on about the patterns in life. To be honest I had no idea what she was getting at, since all these patterns she saw were way beyond me. I sometimes think it takes a special person to discover the secret patterns of life, other times I just think these people are barking. Elisa was, I would say politely, totally out there.
Still, if nothing else, it helped me pass the time while I waited for Mr. Wyndham. Eventually, I managed to excuse myself, which I did by cunningly introducing her to the wonders of outboard motors. A topic guaranteed to bore the living crap out of anyone, except yours truly. There is only so much deep and meaningful conversation the mind can take before midday, and mine had a full quota already.
So, off she trotted and I returned to the lobby and to the total lack of Mr. Wyndham. I checked my watch. A whole hour had passed with change. I looked around, hoping that Mr. Wyndham was elsewhere in the lobby, perhaps in conversation with someone a little more interesting than Elisa. He wasn't, which puzzled me, 'cause I honestly couldn't believe he'd have passed me outside without saying a word. We had, after all, spent several hours talking the night before and seemed to be kindred spirits. What was I to do? First thing that came to mind was to see if he'd fallen asleep in his room. It made a certain sense; he could have been more tired from his tennis than he looked.
I walked up to the lift and pressed the call button. And waited. Chewing my lips, trying not to appear anxious or impatient, I watched the indicator above the lift as the light told me it had moved from floor four to floor three. Should be with me in a minute. Or not. Up to fifth, and top, floor. I raised an eyebrow. It stayed at five for a fair while. Finally the light went out again, signalling the lift's descent.
I glanced around the lobby, hoping no one noticed that I was practically hopping from foot to foot like some schoolboy about to visit a friend he'd not seen in a long time. Or, perhaps, even visiting Santa.
Looking back, I suppose I did kind of feel like that, too. It had been a long time since I'd found myself attracted to someone new; it was a feeling I'd not had in a long, long time. And never since.
The light on the indicator never did come back on. I assumed the light had simply broken; either that or the lift had got stuck between floors. Either way, I couldn't wait any more. The stairs were now my only option.
Narrow corridors. Hate them, don't you? Hotels have this thing about them. Never quite understood why. After all, looking at The Cliff's Edge from outside it looks flipping huge, and yet inside there seems to be no space at all. Makes me wonder where it all goes, 'cause neither the rooms nor the corridors take up much space.
Room 173 was before me. I raised a hand to knock, and for a few worrying moments it remained in midair, barely an inch from the door. A large part of me wanted to knock, like some previously unknown desire was driving me to see Mr. Wyndham in the privacy of his room regardless of what he was doing. But there was a more cautious part of my brain attempting to hold me back. It was telling me to leave him alone, that this guy was grabbing a few winks after a tiring workout on the tennis courts. And then there was that tiny part of me screaming, telling me, back off, to leave the man alone! Just what the hell did I think I was doing anyway? I had a husband at home!
The cautious part lost, and so did that tiniest scream. My fist rapped on the door. Once, twice. Pause.
There was no answer. I assumed Mr. Wyndham was a deep sleeper, so I knocked again, this time a little harder. Still nothing. A third attempt, I decided, then I would go and...well, I didn't know what I would do, but a third attempt was going to be last. This time I added my voice to my efforts.
"Hello, Sam, you okay in there?"
My heart skipped a beat, certain I had heard something move inside the room. "Hello?" I called again. This time there was nothing.
Head lowered, I turned and began the long walk back to the lift. It was only twenty feet away, but those twenty feet felt like the farthest distance I had ever walked. By the time I was three feet from the lift I heard the creak of metal on metal, and saw the doors slide open. I stopped, hoping to God that Mr. Wyndham was going to walk out of there.
Nothing. No Mr. Wyndham, no no-one. I tenderly approached the lift and looked inside. There wasn't anybody waiting in there. I stepped to enter, deciding I couldn't be arsed to walk back down the stairs, or maybe I could retire to my own room, wait a while, then try Mr. Wyndham's room again. But as my shoe landed on the minute gap between corridor and lift I stopped.
My breath caught. For a startlingly long second I couldn't breathe at all. It was as if the lift had started to close in around me, about to gobble me up like a Sunday roast. My hands rushed to my throat in a mad desire to open an emergency hole to let the air through, but as soon my fingers brushed the skin of my throat the air returned. I staggered back, and fell against the wall.
The lift doors closed. My eyes climbed to the indicator above. No light, no sign that it had been on the first floor at all.
People were lined up in a queue at the reception desk. Judging by the look of them they were all here for the forthcoming conference. You could always tell people who sold outboard motors. Grey people in grey suits. Much like me, really. As I stood there my life flashed before my eyes. It didn't last long at all. A mediocre life as a child, with my father drumming into me the need to be a stable husband, my extremely exciting college business studies course, my marriage to Jake, and our subsequent stable but very dull life together. We never did anything interesting; when I wasn't at work we'd sit at home, watch TV, eat, sleep, and then go back to work the next day. When was the last time we had a holiday? Four years ago, and that was our honeymoon.
A grey man in a grey suit living a grey life.
Mr. Wyndham was different. He hadn't arrived at the hotel in a grey suit. He'd arrived in baggy jeans, a nice tight t-shirt and sunglasses. He didn't carry a suitcase, either. His conference papers were in a trendy off the shoulder "man bag." It wasn't until I'd got talking to him the previous night that I realised he was one of us, and that he was actually older than me. He could easily have been mistaken for twenty-five.
My eyes skittered to the lift, and my mind returned to the worrying absence of Mr. Wyndham. I needed to speak to the manager. Since leaving the first floor my mind had been over things several times, and I was now absolutely certain that I had heard a sound in Mr. Wyndham's room. The sound of falling.
I didn't want to draw attention to myself, though. Thus I waited, watching as one by one my fellows signed in and picked up their room keys. An interesting insight into the tedium of being a receptionist at a hotel. I could see that the young woman behind the desk was forcing the smile more each time she turned to the next guest. It was good, in a way, to know that it wasn't only my job that was tedious. I felt an affinity with the woman, and was sure we'd meet on common ground over the disappearance of Mr. Wyndham. Hopefully it would amount to nothing, a simple case of Mr. Wyndham falling asleep and then falling off his bed in surprise at the loudness of my knock. One way or another, for the woman it would be a change from the humdrum of manning the reception desk.
Hang the manager. He probably had a hundred and one things to do anyway. The receptionist needed some spark.
I approached the desk, barely registering the lift doors sliding open as I moved within four feet of them. Had I paid more attention I might have realised that there was no way they should have opened, since only seconds ago the last of the grey suits had entered the lift for his own floor.
I smiled at the receptionist, whose name was Meg according to her badge, and asked, "Could you tell me if Mr. Wyndham has gone out? We were supposed to meet in the lobby but I think I might have missed him."
She asked me to wait one moment while she checked the keys hanging behind her. Hotel rules didn't permit the taking of keys off the premises. As long as you remained on the grounds it was fine, but if you were going beyond you had to return the key to reception. A security measure, probably something to do with fire regulations.
"His key isn't here, which means he's either in his room or on the grounds somewhere." Meg smiled at me. Her practiced smile.
"Ah." I paused a moment, wondering if what I had to say next would come out right. "Well the thing is, I checked on his room a short while ago. You see, when I last saw him he had just returned from playing tennis."
The sign of recognition come to her face. "Oh yes, I remember. I saw the two of you talking. You watched him enter the lift, didn't you?" She asked the question with a very innocent voice, but I could tell by the glint in her eyes that she was trying to imply something.
I chose to ignore that. "Anyway, I checked his room and there was no answer."
"Did you take the lift?"
"Pardon me?"
"The lift. Did you take the lift to first floor?"
Again her voice was quite innocent, but her eyes narrowed. Somewhere in the back of my mind a small bell of alarm rang. Foolishly I ignored it. I needed this woman's help to find out what had happened to Mr. Wyndham. "No, I took the stairs. The lift was...erm, busy."
Meg smiled knowingly. "Then maybe you missed him? He might have taken the lift down here while you took the stairs."
"He might have, yes. But two things make me think otherwise."
"Oh yes?"
"Yes. One, if he had returned to the lobby at any time you would have noticed, since you clearly pay close attention to your guests' movements when they're down here. And two, I heard a sound when I knocked on his door."
"Oh."
This little revelation seemed to shut her up for a moment. I watched her reaction, and it occurred to me that I was never going to connect with Meg. Despite the equal tedium of our jobs we had nothing in common. She looked at me as though she was the keeper of a particular secret that I had no right to whatsoever.
"Maybe we should check together?" Meg suggested.
This time it was my turn to narrow the eyes. The offer of help came a little too quickly for my liking. But how could I turn down such help? I needed to know what had happened to Mr. Wyndham.
"That would be...ideal," I said, once I had decided on the most innocuous word I could think of.
She reached under the desk to retrieve something. I couldn't see what it was, since by the time she returned to an upright position she had deposited it into the back pocket of her skirt. She joined me on the other side of the desk and I motioned her to lead the way. Although she was a good foot shorter than me, and a much smaller build, I still didn't like the idea of her walking behind me. Let alone beside me. I followed her towards the lift.
"I'd rather we took the stairs, actually," I said just as she pressed a thumb against the call button.
"It's only a lift," Meg said with a small laugh. "Do you have claustrophobia?"
I shook my head. "No, it's not that."
"What is it, then?"
How could I explain to her? There was something very wrong about this lift. I had no idea what, but I instinctively knew something was up. Another of the conclusions I had drawn between the time I left the first floor and returned to the lobby. As I looked at Meg more closely I came to the realisation that she probably had a good idea anyway.
The lift doors opened. Meg waved me in. I eyed her, wondering if I should let her know that I knew something was wrong with this picture. No, not yet. No need to play my hand. Once I saw inside Mr. Wyndham's room, sure, but not before then.
"Very well," I conceded and walked over to the lift. We stepped inside at the same time. I looked around. It was a normal lift, nothing special about it. Just a typical metal box lit from above, with the floor and emergency buttons to the left of the doors. I smiled. Perhaps I was being a little paranoid after all. Someone probably just pressed the buttons before leaving the lift, hence why it opened when no one called it.
The door began to close and I started to settle into that comfortable reasoning. Just then, at the worst possible moment, Meg decided to slip out of the lift. I dashed forward, but wasn't quick enough to prevent the doors from meeting in the middle. I stabbed at the door-open button but there was no response.
I spun around, shocked by the abrupt movement of the lift. It was going down. I swallowed hard, doing my best to control my breathing. There was no lower ground floor or basement button, and yet I was most certainly going down.
I closed my eyes. My gym instructor had shown me some meditation techniques, and I just prayed they would be enough to calm me as the metal box and I descended to God-knows-where.
*
Eventually the lift came to rest. As I neared my unwanted destination I could have sworn I could smell burning. Not the fumes of a simple fire, rather the same smell you got when you accidentally burned the hair on your fingers when lighting a stove with matches.
With my nostrils full of that smell, my heart started thumping harder as the doors slid open. Whatever was coming next was something to be dreaded, and I surely did. I pulled back, pressing myself against the far wall of the lift as much as was possible. I wanted to get a good look at what was beyond the doors before I committed myself to stepping outside. I had already convinced myself that whether I pressed the ground floor button or not, the lift would not be returning me to terra firma.
What I saw was, I suppose, a cave. Maybe my senses had not been deceiving me after all, since it now seemed that the lift had descended all the way down through the cliff. I sniffed. Mixed in with the smell of burning was a hint of salt. I must have been at the bottom, in a cave near the sea.
I crossed the lift and pressed a button. Just in case. As expected the doors did not close, instead they remained resolutely open. I took a deep breath, and almost gagged with the taste of the burning. The longer I was exposed to the air of the cave, the more intense the burning became. The cave was saturated in it.
Having no other real choice I stepped out of the lift. It was a cave alright, but whether it was natural or fashioned by human hands I could not tell. Not really my field of expertise. I sold outboard motors, for God's sake, and I was seriously out of my depth.
Nonetheless I continued on. I had to find out what had happened to Mr. Wyndham, and I just knew the answer lay further into this cave. I had taken several steps when I heard the unmistakeable sound of the lift doors closing, amplified by the echoing void of the cave that surrounded me. I spun on my heel, intending to dive into the lift before the doors could meet, but I was too far away. I hadn't realised I'd walked so far, but I had, and I'd need to be Superman to cross the distance between me and the lift in time. Feeling as useless as a screen door on a sub, I watched as the doors sealed my fate.
It was just me and the cave now. And the burning.
*
It didn't take me too long to find the source of the acrid smell. Whoever was behind all this (and I had my suspicions thanks to Meg's manoeuvring me into the lift) clearly didn't want their...what? Trophies? I wasn't sure. Whatever they liked to call the poor people in the cave, the perpetrators didn't like to walk too far.
Several people were chained to the walls, their arms and legs spread eagle, heads slumped. It was hard to tell if they were alive or dead from my position at the mouth of this little cavern; hard enough to keep looking at them, what with the way they had been skinned. One of them had no skin at all; all that could be seen was the muscles that usually lay undisturbed and protected by the outer layer. There was something incredibly gross and wrong about seeing a body of pure muscle like this. Seeing someone in a naked and vulnerable state was one thing, as the other bodies were, but to see someone stripped to the muscle... I fought the urge to vomit.
The other people hung to the walls were in various states of being skinned. Whole strips of skin were missing, some across the chest, others along the arms, legs and torso. One unfortunate man had been castrated, too. I winced, my hand gripping my own privates involuntarily. Although I'd never had anything done to my own personals other than circumcision when I was a kid, I could well imagine how it must have felt to have it cut off. I suppose any man would, wouldn't they?
Nearby, on a large metal table, lay several cutting implements. Knives and saws of varying shape and size. I was surprised to see how clean they were, and then my eyes alighted on the sanitising and disinfecting solutions that also stood on the table. At least the people responsible showed some good sense.
What was I saying? Good sense? How could they possibly justify what had been done to the men on the walls. And yes, it occurred to me then that there were only men in this cavern. No women at all. For a moment I pondered on the idea that perhaps the women were in another cavern. But I soon dismissed that idea. Deep down I knew it was only men who were the victims here.
I approached the table to get a better look at what was on there. I treaded carefully, and quietly. Not sure if any of the men were still alive, I didn't want to cause them further pain by shocking them into movement with any sudden noise.
My heart sank further when I noticed the lack of any anaesthetic on the table. Clean these bastards may have been, but they clearly had no qualms about causing the men pain.
"Who..." Cough. "Who are you?"
A simple but very obvious question. I turned from the table, and my mouth fell open. Seeing the men hang there, skin torn to shreds, was one thing, but to have one actually speaking to me was another. My eyes drifted to the shuddering rise and fall of his tattered chest. I lifted my gaze onto the man's face, and was hit by the sheer pain etched there. Totally understandable, of course, but I never knew you could really feel someone else's pain the way I could then.
I told him my name, not that it was of much use to him. I wondered what I could do for him.
"Are you with them?"
"No," I replied in a whisper, the anger and disgust bubbling in my tone. "They trapped me down here." I looked around. "Although I have no idea why," I added, not bothering to hide the fear that had ridden up in me.
The man coughed. "Divine retribution... That's what they'll call it."
"They?" I asked, although deep down I knew the answer to that.
He nodded upwards painfully. "Up there, in the...hotel."
I approached him, and reached up for the manacles around his wrists. "Let me get you out of this."
"No." He coughed again; this time it came out all ragged, and was followed by a dribble of blood. I reached into my trousers pocket and retrieved my hankie. I dabbed the blood from the side of his mouth, and he smiled at me. It hit me that this was probably the first sign of human compassion he had felt in a long while. My eyes watered at the overbearing sadness of it all.
"Please...kill me."
I pulled back, a spasm of shock shaking me. I shook my head. I couldn't kill a person. No matter what. I just didn't have it in me.
"Please. Before they come back and finish...this."
"Look," I said, a sudden urgency gripping me, "I came here to find a friend. I'm sure they've brought him here. Is there another cavern like this?"
"Kill me."
I looked back at the mouth of the cavern. He was sure they were going to return, and that only made me certain, too. I had to find Mr. Wyndham before they returned. I sniffed. The smell of salt was stronger now, so I couldn't have been too far from the sea. This meant there had to be another exit from these caves. If I could find Mr. Wyndham, then we could...
"It's too late."
My attention snapped back to the man, and my heart was stopped by the look of pure horror on his face. "What do you...?"
I didn't need to finish my question. I heard the lift doors open a short distance away.
"I have to go." I reached up a hand and wiped a further dribble of blood off the man's chin. "I'm sorry."
With one final look of apology I turned to leave him to his certain death. That brief moment of humanity was going to cost me, since it had given them enough time to reach the mouth of the cavern. A small group of them stood there, completely blocking the only way deeper into the caves. My only escape route.
I recognised them all. Meg the receptionist, the man who kept the tennis courts in order, the waiting staff from the dining room, the chef, and at the head of the small group the manager himself. Each of them was smiling, and the sheer delight in those smiles made my skin squirm.
"Hello, Mr. Jensen," the manager said. "So nice of you to join us."
*
I'll admit I screamed. Not because of what I saw, so much as because I knew what was coming my way next. They manacled me to the wall, right next to the man I had spoken to. He'd not said a single word since they had entered; he didn't even look my way once during the whole time that they forced me against the wall and ferociously stripped me naked. But I watched him, as Meg carefully sliced a long strip of skin off him, from the left shoulder right down to his waist. He didn't scream, I think he had got so used to it now that he couldn't scream any more. Although the pain he felt was clearly written all over his face. I did scream, however.
Once Meg had finished she held the skin aloft like a trophy. Then, and I have to confess I could not remove my eyes from the spectacle; she put one end in her mouth and started chewing. The old keeper of the courts came over to her laughing, and she nodded at him. My stomach turned as he took the other end of the strip of skin into his own mouth, and together they continued chewing as if the skin was a long piece of spaghetti being eaten by two lovers.
"Why?" I asked.
"Infidelity, Mr. Jensen."
"What? I've never..." A flash of memory; watching the tight ass of Mr. Wyndham in his tennis shorts, Jake, my husband, at home oblivious. I swallowed, and the manager nodded. "But...I didn't do anything."
"No, but you would have. And now Mr. Wyndham will be saved the displeasure of taking part in your infidelity."
I looked around; checking one last time to make sure Mr. Wyndham was not hanging on the wall. "Where is he?"
"Safe in his room. Meg tells me it was you who almost disturbed me returning him there."
I was too stupefied to respond to that. So the manager carried on.
"He shall awake in his room, believing he fell asleep after a tiring bout of tennis. He'll have no memory of his brief trip down here." The manager nodded at the chef. "Gene here makes the most amazing and potent amnesia pills. Mr. Wyndham had to be brought down here to arouse your curiosity. We knew you'd want to know how he could get in a lift one second, and then not be in it the next. But, he is safe now. The lure worked."
"Congratulations," I said, trying to sound braver than I felt. "But he'll be expecting me to meet him."
"Yes, until Meg explains that you left earlier without any word as to why."
"Others will miss me. My husband..."
"Will receive a letter from you explaining that you had an affair with another man, and how you could not handle the guilt and so he shall never hear from you again."
My mouth worked to speak, but I could not find the words. In my mind I could see Jake at home thinking that I had been capable of... I lowered my head. I would have, given the chance. Maybe I did deserve this. To treat my marriage in such a casual manner...
"You will be missed for a while, but you will soon just become another statistic. One of millions who can't handle their lives and so sink into the underbelly of this wonderful nation of ours. Sometimes someone will pass a tramp on the street and think they recognise him as you, but they'll ignore that as stupid. You'll soon be forgotten."
I looked at the man beside me, who now seemed to be unconscious. Knocked out by the pain, no doubt.
"Yes, you will be like these." The manager indicated his staff. "We are the avenging angels, seeking divine retribution for the infidelity of man. We have these conferences to seek out those who wish to pervert the sanctity of life. Those who would sleep with others when bound by wedlock; those who climb to their present positions in life by nefarious means. We gather them in, and consume the sin off their flesh."
Meg and the old man had finished their bizarre meal. The manager walked over to Meg and licked the remaining blood off her lips. He looked back at me and winked. "You shall make an excellent feast indeed. Your sin is one of desire, and that reeks throughout your body." He placed an arm around Meg and guided her out of the small cavern. "We shall return for you."
And they shall. Of that I have no doubt. Maybe I deserve it.
Reflection
It had been a shitter of a day, but Corey Jordan was glad to be home. There was no way, in his considered opinion, in which the day could get much worse. Fuck Duncan Leman anyway—if he didn't want to get a smack in the mouth a week before Christmas, then it was his own stupid fault for constantly picking on the non-Brits at work. Racist bastard! About time someone tore a piece off the old fool. And Corey was more than happy to be that person.
Of course, as it turned out, the boss didn't agree. And, for reasons Corey couldn't quite work out, it was he who ended up with the first written warning. Was it his fault that racism pissed him off so? No. Was it his fault then, when that pissed off, his mouth ran away with itself and produced more profanities than Corey even realised he knew? No, it was not. Apparently accusing someone of being racist was as bad as being racist nowadays. And it wasn't like Duncan didn't deserve the smack. What a fucked up world they lived in!
And so, leaving work early (and not out of choice!), he decided to pop into the pub. Six bottles of Bud, two Jägerbombs, and four hours later he finally realised it was time to head home. So, here he was, stumbling through his door at seven-thirty on a Monday night, completely stone sober. Whoever said he couldn't handle his drink was clearly talking crap.
"Shhh!" he hissed at the table as it wobbled next to him. "Stupid table! What you doing in the way?" he asked, in a stage whisper.
He looked up at the dark hallway. Why he was whispering he had no idea. Not like anyone else lived in his house, was it? He laughed, bitterly. One day he'd get Iracema living with him, he just knew it, ain't that right, boy? He really did need to stop thinking to himself like he was two people. As he opened the door to the living room he wondered if thinking to yourself was the first sign of madness. It's what they said....
He shook his head. Nope, talking to yourself was, he belatedly remembered. "So where does that leave talking to tables?" he asked the door, and entered the room.
He stopped. There was someone standing in the middle of the room, silhouetted against the lights coming from the street outside. Corey took a deep breath, his mind clouded and confused. He knew he ought to do something, say something, but all he could do was watch as the person slowly turned their head. A light swept past the large windows looking out onto the street, and for a split second Corey got a glimpse of the person's eyes.
White! Pure white. No pupil, no iris, just pure white eyes!
Without even realising he was doing it, Corey's hand reached for the light switch and flicked it. The light flooded the room, and for a second Corey was blinded. He blinked, forcing his eyes to adjust to the illumination.
"What the fuck?" he said, breathing heavily.
Other than himself there was no one in the room. He looked around, wondering if the person had dashed into the kitchen via the small arch while he was blinking, but no. The kitchen was empty, too. Corey shook his head.
Okay, so maybe he was a little bit drunk after all.
*
Corey pushed himself back from the monitor, and rubbed his temples. Damn hangover. He looked around quickly, making sure no one noticed his rubbing. He didn't get drunk; at least that's what he liked to tell his colleagues, so the idea of appearing to be hung over was not exactly conducive to his manufactured image.
He reached into the drawer of his desk, and surreptitiously removed the small silver box. Wrapping his hand around it, ensuring that no one else could see what he was holding, Corey got to his feet and made his way across the open-plan office.
Open plan. The scourge of privacy at work. He hated it. Hell, he hated working in a call centre period, but he was kind of stuck with it. The unwanted image of the written warning came to his mind, and he smiled slyly to himself. Well, he was just about stuck with it. Maybe after the New Year he'd start looking for something else, but right now he had to hold on to his job. Which meant trying to steer clear of Duncan Leman.
The bastard!
Once he was in the staff toilets, Corey checked to make sure he was alone, then turned to the sink. He turned on the tap and opened his hand, revealing the packet of paracetamol. He popped a couple out of their foil, and placed them on his tongue, bending over the sink to drink directly from the tap. Not as elegant as a cup, but then the water in the bathroom was not normally used for washing down hangover pills. Standing up straight again, he tilted his back. He swallowed, and let out a breath of air.
It'd take a little while, but his headache would soon subside to a manageable level. In the meantime he just had to make sure he didn't lean in too close to the...
"Drowning your sorrows last night, then, eh, Cor?"
He closed his eyes. He bloody hated it when that tit used the diminutive; it implied a familiarity that wasn't warranted. Slowly, Corey opened his eyes again and turned to face the intruder. "What do you want, Dunc?" he asked, placing particular emphasis on the last word. He knew Duncan was no fan of being called that—besides, it made the owner of the name sound incredibly thick. And Corey liked that.
Duncan Leman was a short man, somewhat overweight and not very tidy. One of the sort who figured that since they spent their working hours hidden behind a PC and phone there was no need to worry about their appearance. It was the middle of winter and the fool was dressed in khaki shorts and a sickeningly bright t-shirt. Corey wanted to hit him just for dressing like that.
Duncan shrugged. "Don't want anything, mate. Just making an observation is all."
"Yeah, well don't. I didn't ask for your opinion."
Again Duncan shrugged. "Mate, we're all entitled to our opinions."
Corey chewed his bottom lip and shook his head. "No, really. It's just the fact that you think I'm entitled to your opinion that grates on me." He nodded at the door behind Duncan. "Sod off."
As expected, Duncan didn't take the advice, which suited Corey well. The headache had yet to leave, and Duncan's continued presence was only serving to irritate it. He turned away from Duncan and looked at his reflection in the mirror. Corey pocketed the tablets and turned the tap back on.
"Why are you always so aggressive, Cor?"
Corey couldn't believe the stupidity of the man. "Shit, dude! Do you actually have a brain in that head of yours?" he asked, as he splashed water over his face. "I'm pretty sure I spelled it all out yesterday. I don't like you, Duncan." He pulled a couple of paper towels out of the dispenser and proceeded to dry his face. "You're pond scum. Now, fuck off, before I ask less politely."
Duncan smiled and narrowed the gap between them. "One step closer to the edge, Cor."
Corey raised an eyebrow, continuing to watch Duncan in the mirror. He was clearly provoking Corey; intentionally, too. Duncan wanted Corey out of the job. Corey wondered why. He also decided not to rise to the bait.
"Yeah, whatever, dude. I've got work to do." He went to move around Duncan, but the shorter man merely stepped in the way. Corey sighed. "Mate, sorry, but I ain't getting no second warning for you."
Duncan, at least, had the good sense to look disappointed. "Now that is a shame, 'cause I know that unc...Mr Roberts wants your ass."
Corey stepped back, and regarded Duncan in a new light. The slip was clear, and Duncan hadn't been quick enough in covering it. The boss was his uncle, eh? So, it was a tag-team event. He decided to laugh it off. For now. "Yeah, well, if he wants a piece of ass he should try Heaven. I'm spoken for!" With that he barged his way out past Duncan.
*
Corey flicked the lamp off, and for a few moments stood by the archway that connected the living room to the kitchen. The soft orange glow of the fire created a calming atmosphere in the room, making Corey not want to move. He knew he had to, of course, since it was the early hours of the morning and he needed some sleep before work. But he felt chilled, and not the least bit tired. Iracema was good at making him feel relaxed. The soft flickering shadows on the walls, coupled with the slight inebriation, made him more relaxed than sleepy.
He walked across the room, and stopped before the mirror hanging above the mantle. He stepped closer, and remained there, watching his reflection and the way the fire below cast shadows across his face, giving him a much more definite bone structure than he usually had. He rubbed a hand across his jaw line, ending on his chin. Once again he wondered if he ought to go for a beard. Not a goatee; that was too expected these days. No, he wanted a full beard, but nicely trimmed...Frakes style.
He turned his head slightly to the left, to check the shadow line across his jaw. He stopped, his eyes never reaching his jaw. Instead they rested on the figure. Once again it was there, silhouetted against the window. Corey swallowed hard. This time he was not drunk. He'd barely had two bottles of Budweiser.
Not daring to remove his eyes from the silhouette, Corey started to move to his right, to where the lamp stood.
"Don't turn the light on."
Corey froze. For a moment he forgot to breathe. He opened his mouth to speak, but no words came. Not that he knew what he was going to say. Forming thoughts was hard enough.
"Turn the light on, and I'll be gone."
Nonetheless Corey felt his fingers reaching out towards the lamp. He knew he wasn't close enough to reach the lamp, but his fingers seemed to think otherwise. It was as if some primal instinct was forcing his hand to move, to take away the darkness.
"Listen to me, Corey Jordan. It's time to put an end to the pressure."
Somehow he managed to find his voice. "Pressure?"
"Yes. Duncan Leman."
Once again Corey swallowed hard. His throat was incredibly dry. Which made little sense, since he'd only recently polished off a bottle of liquid. The way the Silhouette said that name. It sounded very familiar. Somehow, although Corey couldn't quite get his mind around how, he knew he should recognise the voice. "Who...who are you?" he asked lamely.
"That, I cannot answer."
For the first time Corey noticed that the Silhouette had its head turned away from him. Which was nice, Corey considered, since those opaque eyes scared the crap out of him. "Why not?"
"It is unimportant."
The answer was simple enough, but Corey wanted to object to the finality of it. Somehow, this person was in his house. Once more standing in the middle of the room, lit by the lights outside the window. And there was no way he could have got inside. Corey made a point of locking the front door behind him when he returned home, and considering the weather lately, Corey had not opened any window or the back door in days. And then there was the previous night....
Drunk he might have been, but Corey was certain he had seen the Silhouette. Certain that the person had vanished once the light came on. Of course, he'd convinced himself it was his drunken imagination...but now, here in his living room, he realised that deep down he had been certain all the time. "Turn the light on and I will be gone," the thing had said. And it was a thing, of that Corey was in no doubt. Whatever was standing in his living room, it wasn't human. It simply clothed itself in the silhouette of one.
"Go to St. Andrew's Square."
The voice brought Corey back. St. Andrew's Square? He knew it. It wasn't too far from where he worked. He opened his mouth to ask why.
"Don't ask, Corey Jordan. Go there. 22a. St. Andrew's Square. The door will be open, enter the house, and on the top floor you will find Duncan Leman."
"It's his house?" Even before the Silhouette nodded its head, Corey knew the answer. He felt an odd sense of excitement suddenly, as if he'd taken a direct hit of adrenalin. "What do you want me to do there?"
"Just go there, Corey Jordan. You will know."
Corey nodded his head slowly, and with his next words he knew he'd made a promise that could not be broken. No matter the consequences. "Okay."
*
St. Andrew's Square was one of those old quadrants that Corey usually loved to visit. Located in South Kensington it was rather typical of what he'd come to expect in London. A small garden in the centre, cordoned off with black railings, and surrounded by Victorian houses with three storeys. What he loved so much was the fact that around the corner was a built-up area, very cosmopolitan. A complete contrast to the quadrant he was standing in. Only London seemed to carry off the mix of old and new with such grace.
This time, though, there was nothing about love behind his reason for being there. This time is was simple need. He had to find out what the Silhouette was talking about, and more...he needed to do this. He didn't know why, but something was compelling him.
He walked around the quadrant until he came to 22. The building was three storeys up, and down below..."22a" was engraved on the wall beside the door at the bottom of the stone steps. He noticed that the door, as the Silhouette had promised, was indeed open. Only a fraction, but enough for someone looking closely to notice.
Corey looked up and down the street, just to make sure he wasn't being observed. It was almost five in the morning, and all those with common sense were tucked away in bed, happily enjoying their trips to Nod. Gripping the collar of his coat tightly about his neck, Corey descended the steps carefully.
Once he reached the bottom he stopped. He glanced up the way he had come, worry plaguing his mind. He was almost certain he was being watched, yet he could see no sign of anyone. Trying to ignore the cold feeling running down his back, he gently pushed open the door and stepped over the threshold into the basement flat that belonged to Duncan Leman.
He found himself in a long corridor, which reminded him of his Gran's old place from when he was a kid. The grossest wallpaper he'd ever seen lined the walls, made up of distorted squares in the most lurid shades of green and yellow. The carpet itself clashed hideously with the walls, being a dark burgundy colour. Corey found himself shuddering, and this time not because of his unease, but rather through repulsion at the décor.
The door at the far end of the hallway was open, and he could see the kitchen beyond. Much like the hallway, it seemed to be from another age. He knew he was in a Victorian house, but he'd expected a bit of modern stuff inside. Mind you, he reflected, considering who lived here....
There were a further two doors along the left side of the hallway, both of which were slightly ajar. Cautiously, Corey crept along the passageway and approached the nearest door. As luck would have it, looking through the crack in the door, Corey saw that it was Duncan's bedroom. And the bed was occupied.
Corey pushed at the door, hoping that it would not creak. It didn't. He stepped into the bedroom and drew closer to the bed. The duvet covered a rather lumpy form, which he guessed was Duncan. But then his eyes alighted on the smaller shape on the other side of the bed.
Corey frowned and continued to creep around the bed. When he was close enough to get a good view of the upper half of the second shape, Corey saw a woman's head poking out of the duvet.
Not just any woman though. It was Iracema....
It was she who Duncan had been so racist to, and now here they were...in bed. Together!
Corey swallowed. He heard a sound behind him and turned.
Behind was the window, and before that window...Corey was unsurprised to see the Silhouette. He wanted to speak, to ask something, but he didn't wish to disturb the sleeping forms until he knew for sure.
The Silhouette nodded. For a moment Corey wondered what it was nodding at, then he realised. It was an answer to his unspoken question.
His heart hardening, Corey turned to face the bed.
He had no idea how he was going to handle this, but he knew he had to do something. He opened his mouth to speak, but the words came from the Silhouette. The words sent a shiver down him. Not because of what was said, but because of the voice. He had heard it before, of course, but now he knew whose voice it was.
His own.
"This ends now, Duncan Leman."
Corey heard the movement behind him, but he did not dare to look. Just the knowledge that the Silhouette had his voice was enough—to actually see himself was not something Corey was ready for. But the choice was soon taken out of his hands. The Silhouette came into his line of sight. First as a dark shape on the periphery of his vision and then before him, as it neared the bed.
Corey felt his breath being stolen away from him. He stood there, immobile. Able only to watch. As the Silhouette approached the bed, the shadow fell off it—him. As the shadow fell, dripping away like watered-down oil, Corey took in the appearance of his doppelganger. There was something almost clinical about the suit he wore. It was a dull grey affair, straight slacks, and a body-hugging top, with a short but tight-looking collar. Corey swallowed when he got his first full look at his double's face. If he had passed the man in the street, Corey was sure he'd probably have never noticed the similarity, since the Silhouette's face was a lot harsher than his own. It was his face, but at the same time it wasn't. The full Frakes-style beard only helped to highlight the differences. Sunken eye sockets, containing the orbs of pure white that he had seen before, and a nose that had clearly been broken on several occasions.
His observation of his distorted self was interrupted by movement from the bed. His head snapped around, in time to see Duncan struggle into a sitting position. As soon as Duncan's eyes alighted on the Silhouette his mouth fell open. Amidst the fear, Corey also saw recognition in Duncan's eyes.
"No, you're not dreaming this time, Duncan Leman."
Duncan moved his mouth to speak, but no words came. In a way, Corey found that oddly reassuring, knowing that it wasn't only him who had trouble getting words out in the presence of his doppelganger.
"You knew this time would come," the Silhouette continued. "The scales of justice are tilting, and not in your favour, Duncan Leman."
Corey looked over at his double again. For the first time in ages, a laugh erupted from his mouth, which caused the Silhouette to look at him. Corey resisted the urge to flinch, but once more the nervous laugh came out. The Silhouette frowned.
"You have no need to fear me, Corey Jordan. Only he does," he said in a cold voice, gesturing to Duncan with one hand which, Corey noticed, was deeply scarred.
"But," Corey began, with a deep swallow, "you're..."
"You?" The Silhouette nodded his head slowly. "Yes. As one of our favourite writers might have said, I am you, seen through a mirror darkly."
As explanations went that was of no help to Corey at all. He shook his head. "I don't understand."
"Understanding later. Action now." The Silhouette resumed his attention on Duncan, who was still sitting in his bed, silently watching the exchange between the two Coreys. Iracema had yet to stir. A small mercy Corey was thankful for. "Stand up, Duncan Leman. Face your fate with dignity."
Duncan looked around wildly, and shook his head. Finally he found his voice, but when it came it was pitiful. "No," he said barely in a whisper.
The Silhouette raised a hand, and pointed a finger at Duncan. "Stand."
Corey watched as, clearly despite himself, Duncan removed himself from the bed, revealing his nakedness to Corey. It wasn't the actual sight of his lumpy naked body that repulsed Corey, so much as the thought that that nakedness was once embracing Iracema. His Iracema!
"Yes, Corey Jordan. Such righteous rage is needed."
Something stirred in the darkest corner of the room. Corey glanced over, and a black shape, a shadow, emerged from the corner, moving close to him. Without meaning to, he opened his hand and the shadow drifted onto his palm. His fist tightened around something hard. Corey looked down, and saw the black club he was holding.
"No, no," Duncan said, sniffing away like a scared child.
Corey stepped forward, fully aware of what he had to do. Scum like Duncan Leman were not allowed to continue. Stealing Iracema from him, the racist attacks, everything about Duncan was wrong. Duncan staggered back against the wall, his whole body shaking.
"And now your dream comes true, Duncan Leman," the Silhouette said.
"Please no. I'm sorry. I didn't..."
"Don't you dare, Dunc! I always knew you were scum," Corey said, raising the club in the air. "I just never realised how much." And, with a sadistic pleasure he never knew he could possess, Corey proceeded to strike Duncan with the club. Again and again and again....
*
It was sometime later when Corey stopped, his rage having drained away. He looked at the pulpy mess that had once been Duncan Leman, and he stepped back, the club falling out of his hand. He felt satisfied, yet at the same time disgusted with himself. He knew that, without a doubt, the world was a better place without people like Duncan, but he still felt sickened by the violence he was able to dish out.
He turned to the Silhouette. Explanations later, he had said. Well, it was later, and Corey was sure he deserved some answers now.
The Silhouette was gone. Corey looked around the room frantically, and as his eyes came to rest on the bed they widened in horror. There was no one else in the bed, and no sign that there ever had been. His throat went dry.
He rubbed his fingers together, feeling the warmness between them. He glanced down, and noticed the dark red substance that covered his hand. Blood. The exact same blood that covered the corpse on the floor before him.
"Oh god," Corey breathed, as realisation dawned.
One Mistake
He looked down at the card in his hand; the rather shaky card. No, that wasn't true. Cards, being inanimate objects, didn't shake by themselves. It was his hand that was shaking, the nerves threatening to get the better of him. Clasping his wrist, he attempted to steady the offending hand, and focussed once more on the address scribbled on the back of the card. He had to admit his handwriting was pretty shit, really, and hard to read at the best of times. And writing while nervous helped his script none. Still, he was familiar with his own writing enough to be able to decipher the address, and looked up from the card at the small house before him.
No doubt about it. The address was the same.
But did he really want to do this?
His legs started moving, one foot down, then the other, taking him towards the house. He stopped at the front door, and his knuckles rapped loudly on the cracked wood. He waited. And as he waited he thought. Why was he here, and why in the hell had he even bothered calling the number on the card?
It seemed public phone boxes were becoming a thing of the past, something only those unwilling to change with the times would use. Fossils. Like him. He was barely into his forties, but he refused point blank to buy a mobile phone, or have one of those, what did they call them, oh yeah, one of those compacts. They seemed to cost a lot of money to do things he didn't understand. Besides which, he always reasoned, if people wished to contact him they could always ring him at home. House phones had served people well since the late nineteenth century, so why this bizarre need to have every part of their lives subject to the intrusions of others? Bad enough those random companies could contact him in the privacy of his own home; he didn't want to be intruded upon when he was out and about on his strolls. All this notwithstanding, public phone boxes were still about, and as they had been since time immemorial, they were still littered with calling cards from those offering sex services and the like. Personally he had never picked up one of those cards before; indeed he barely looked at them, preferring to focus his attention on the world outside the phone box whenever the need to use one took him. But, barely an hour ago, something pulled him towards a particular card.
Discovering the Art of Astral Projection it said. For a moment, phone still to his ear, he had looked at the card, completely oblivious to what his mother was saying on the other end of the line. It was almost as if he were sinking under water. He was aware of his mother's voice, but the words made no sense to him, the sounds simply reverberated around his ear. His attention was squarely on the card, which his hand tenderly pulled off the wall of the booth. He was careful not to damage the card, almost as if by doing so he would offend the person who had placed it there. He held it close to his eyes; the number at the bottom was in the smallest print he'd ever seen. Clearly the owner of the number wanted people to pay attention, not merely glance at the card like all those that offered the promise of sexual pleasuring of various parts of the body.
He couldn't recall if he'd actually bothered saying goodbye to his mother (He hoped he had—his mother would not have been happy if he'd simply hung up on her!), but next thing he recalled he was dialling the number on the card. He punched the numbers in, carefully rechecking the card with each individual number, just to make sure he didn't get it wrong.
The call was answered before the first ring had completed, as if whoever it was had been sitting, hand on the receiver, waiting. There was no hello, just the sound of steady breathing. He tried a hello himself, always believing politeness cost nothing, but he'd barely got "hell—" out before a very old voice issued out an address. Urgently he reached into his inner coat pocket and pulled out a pen. He scribbled the address down, and was about to double check the door number, having been caught off guard, when the line went dead.
For a few seconds he remained as he was; phone receiver in one hand, the card in the other. Then it occurred to him. The address given was only a twenty-minute walk away.
Now he waited for an answer, still no clearer on why he was doing this than he had been when he'd first peeled the card off the booth wall. He leaned in closer to the door, briefly wondering if perhaps the owner of that old voice had died in the twenty minutes since he'd given the address. After all, it had been a very old voice, and in his experience old people tended to die at the most inopportune times. But no, he could hear movement from beyond the door. He stepped back, not wanting to appear too eager.
The door creaked open. Actually creaked, like in the old horror films that his mother had forced him to watch when he was a child—a millennia ago it seemed. Like he didn't sit there shitting his pants through every single minute of the films. Now he felt like soiling his underwear again, but he clenched himself, both literally and figuratively. At first, even with the light coming from the street behind him, he could not see a single thing beyond the opened door, as if some hitherto unknown depth of darkness lived inside the house. His eyes adjusted and he saw the old man standing there, regarding him with baleful eyes.
"Hello, Robert," the old man said.
*
Robert followed the man down the hallway, which was actually little more than a narrow passage through the ground floor of the house. Along the right wall a staircase led up. A very threadbare carpet covered each step, full of burns and stains, the origins of which Robert didn't much wish to think about. The whole house, which he eventually got to see in its entirety, carried with it a bearing of neglect, as if the old man merely existed in the house, not lived. There were signs that once upon a time the house had been lived in, but that time had long passed for whatever reason. Robert didn't want to consider the reason; some things were best left unknown.
He stopped at the kitchen doorway, situated at the rear of the house, and looked around. Neglect was putting it mildly. Filthy pots and pans littered the sideboards; plates with bits of food welded to them, and cups lying on their sides, the starch staining the insides so intensely it was as if it had become part of the natural colouring of the china. The stove itself was, unsurprisingly, old and rusted, except for one single square of the hob which gleamed against the rest of the dirty metal. This, Robert guessed, was the single part of the cooker still in use. After all, as bad as he looked, the old man clearly still ate something to sustain himself.
Then there was the old man.
Robert watched him sit at a small table, pushed against one wall. Papers and empty food tins littered the floor around it. He seemed to be broken. Not in a metaphorical sense, but actually physically broken. His entire shape looked as if every single bone in his body had, at one point or another, been snapped out of place and left to reset on its own. The result was a man who walked like a marionette without strings, coming to realise that it could in fact walk unaided, albeit in a fashion that barely resembled a normal human. His face was also one of brokenness; bruise upon bruise, an open welt above one eye, a nose that had seen better days. For a second Robert was reminded of the pictures newspapers liked to print from time to time, reminding people who were trying to enjoy their lives of the bitter and twisted nature of the world in which they lived. Grannies battered in their own homes, granddads beaten senseless while getting money from an ATM in town. Yes, that's how this old man's face looked, like someone had really been to town on it.
"How do you know my name?" Robert asked finally, no longer able to stand the eerie silence that pervaded the house. And it truly was silent. Lifeless.
"You touched my card," the old man replied, his voice like dried leaves. To him the answer was an obvious one. Robert wanted to argue this, tell the old man that that was no answer. It explained nothing. "You want me to teach you how to project yourself beyond your body?"
Did he? Robert wasn't so sure. He still wasn't even sure why he was here; what had compelled him to pick up the card, to call the number, and to visit the house? And he sure as hell wasn't sure why he would want to take a trip out of his body. So he shrugged. "Suppose so," he said.
"Good. Close your eyes."
Robert blinked. "That's it? No build up? Just 'close your eyes'? I thought I had to go into a trance or something. Imagine myself lifting up, pulling away, looking down at my body."
"Ah." The old man smiled at Robert, but his rheumy eyes contained the same balefulness. "Do you think that will help? Are you some kind of expert now?" he asked, his voice becoming more forceful with every word.
"Well..." Robert swallowed. Hard. "No, of course not, but I saw something about it on Most Haunted the other week and Yvette said..."
The old man sighed. "Robert. Just. Close...Your...Eyes."
Robert did so. He didn't know what the old man was expecting as a result, but nothing happened. Apart from Robert becoming aware of the smells in the house. Rank, acrid; the smell of the dead. He went to open an eye—just the one, mind, to get a quick peek to see what the old man was up to—but no sooner had he thought about opening that eye than he felt something grab hold of him.
Not his body. Oh no, because at that moment it occurred to him that he wasn't his body. That was merely a shell, a vessel in which he moved, became a part of the substantial world. He was something else entirely. And it was that something else that was being grabbed, pulled, yanked out of the body with such force that he could not resist. Not that he would have known how to. Until a second ago he didn't even realise he was this something else.
There was his body, slumped against the grimy wall of the old man's kitchen, now vacant of its owner. He was above it, floating in the ether, a spectral mass of conscience looking down on a limited form that had once constrained him.
Wait; why was he thinking such things? He was Robert Hoard of East Acton; a nobody, sure, just a small man going about his own business. Aspirations nil; a shelf filler in a local supermarket, and slave to his mother. Still. After forty years.
"Because, Robert, on the astral plane everyone is high and mighty."
Robert tried to look around, find the source of the voice, but he couldn't. Look that is; he had no eyes with which to look. He knew the voice, though, even out here on the "astral plane." It was the old man.
Robert tried to speak, but shock of shocks he didn't know how. He had only ever spoken with his physical voice. A—what? Astral voice? Yes. An astral voice was new to him and he had no idea how to use it.
"You'll work it out. You shall be here for a while. And you'll discover that although you are literally high out here, you are far from mighty. Now, if you don't mind, I'm just going to borrow your body for a while."
Mind? Of course he minded. But what could he possibly do to stop the old man? There was nothing he could do because he was nothing. Just thought, intangible and hanging uselessly in the ether.
His mind was all over the place. Perhaps he was having a mind panic? That's what happened when you panicked after all; your mind goes all over the place, not able to focus on any one thing. Instead it was in little bits; here, there and everywhere. Analysing the wrong things, going down the wrong paths, and not paying attention to the immediate issue. If that was so, then yes, Robert was having a mind panic attack.
He was everywhere in the house at once. In the skanky bedroom in which the old man used to sleep, back when he did sleep, which was probably before the days when he used the room as a dumping ground for more trash. At the same time he was in the lounge, which unsurprisingly contained a very ancient TV, a huge wooden box with the smallest screen, the kind Robert had seen in pictures from the late '50s, and, of course, the now anticipated mess and general look of abandonment. All over the house it was the same; a place where someone used to live, probably with some contentment, but now that happiness had moved, replaced by apathy that was verging on clinical disassociation. The old man simply did not view this house as a home any more, but a prison unworthy of respect. Walls that kept him from the world beyond.
Robert wondered where that had come from. Somehow, deep down, he knew he was right, that the man truly believed this was his prison.
What was it Douglas Adams had once written? Don't Panic! Possibly the best advice in the world. Panic was not something Robert tended to do a lot, after all, his mother pretty much controlled most of his life so he had little space in which to lose it.
His mind was rambling. He had to focus.
The old man had talked of borrowing Robert's body, and considering the state of the old man's body, Robert wasn't sure he much liked the idea of that. God knew what state his own would be in when he got it back.
It was the thought of that, more than anything else, that gave him his focus back. He was back in the kitchen, once again looking at the scene from above. His body started to stir. At first it was just the fingers, twitching as if he were dreaming. Only it was not he, this was for sure. He was still up near the ceiling, an abstract collection of thoughts with no form to affect anything.
However...
A thought occurred to him. Surely if the old man could somehow transfer to his body, then Robert could do the same. All he had to do was get in the old man's body first, rouse it, and then tie up his own body. Perhaps a few threats, a bit of minor damage, and the old man would vacate. Robert mentally shrugged. At this point he had little else to lose.
Shutting his astral eyes, an act that didn't actually block anything out, but merely allowed him to focus inwards, he pictured what it would be like to be in the old man's body. Frail, full of aches and unwanted spasms. No longer able to digest the foods he liked so much. No more fatty burgers, excess amounts of pop. Just a carefully controlled diet and...
Oh my god!
He was there. Within a split second of being in that old, disgusting, body Robert knew he wanted to be anywhere but. There were thoughts, images, not of his making. There was no doubt in his mind—if, indeed, he even could claim to have his own mind while inside the body of this disgusting thing!—that he had not lived the life he was now being exposed to. His life had been dull, yes, boring beyond words, but at least it had been safe, free of such sin as this!
The man, Bernard Jacob Rubin, had been such a good fellow in his younger years. Always there for his family and his friends. This house had welcomed many a person over the years; no one was turned away from his happy home. His wife would busy herself in the kitchen preparing food and drinks for their guests while he entertained them with stories—for Bernard was a storyteller of the finest order. People were always telling him to write them down, but he never truly believed in himself. Then one day they took in a young woman, Georgia, the daughter of Frank and Julie Nettles, very dear friends of his wife and he. Georgia was something of a trouble maker, but Bernard saw the light in the seventeen-year-old. Alas, it seemed Bernard saw too much, and a lot more than the young woman saw in herself. Soon Bernard was lured to the teen's bedroom and...
"No!" Robert shouted, and expelled himself from the old man's body with haste.
He hovered there, once again mind without form, only this time he felt contaminated. He had seen what happened and wanted to shut it out, but he could not. Even as he pulled out of Bernard's shell the scene had continued, the events speeding up like some fast-forwarded film, taking Robert right up to the moment where Bernard had opened the door to him less than a half-hour ago.
Robert was revolted. The hatred, the self-loathing. Once again, though, it was not truly him feeling this. These emotions, intense, eating away at the core, came from the old man. Bernard.
Robert now understood, but he was helpless to prevent what was going to happen.
Robert's body was on its feet, animated by the presence of Bernard. He reached beneath the table, rummaged through the trash, and pulled out a steel pipe. He looked up, and Robert was shocked at the lack of feeling on his own face. It was as if Bernard was beyond being able to express the guilt that had eaten at him every single day for the last twenty years.
"You see, Robert, this is what I must do. Penance for my sin... Never in my life have I ever thought of touching, even looking at, another woman. I had my wife, what did I need other women for? We let this Georgia in, did our best to help her and she..." Bernard shook Robert's head. "Yes, she was legal, but she was the daughter of a friend, someone I took in. People tell me she led me on, but..." Even now, the words were beyond him. Helpless to act, all Robert could do was listen to what Bernard had to say. That it came in Robert's own voice disturbed him greatly, even more than what he had seen of Bernard's life. "It was this body," Bernard continued, pointing at the old body lying on the dirty floor, "that gave in, allowed itself to be led down the dark path of indiscretion. And it must pay, as I have paid by losing everything."
Robert knew of what Bernard spoke. His friends, his family, even his wife, who had promised to stick by him unto death, had turned on him. Not even wishing to hear his side of the story. Since that time he had been alone... Just Bernard and his guilt.
Robert wanted to speak, tell Bernard that it was not his fault. He had seen the life Bernard had lived, watched as the teenager manipulated things, twisted everything. Bernard never stood a chance.
Robert knew from his brief tour of Bernard's memory that he was the latest in a long line of people who Bernard had lured to his house, to borrow their bodies, use them as tools of his punishment.
In his whole life, Robert had never wanted to block anything out as much as he did this. And he knew, whatever the outcome of this night, his safe life was gone for good. And so he watched—what else could he do?—as Bernard raised the pipe and began beating down on his own, vacant, body...
*
Time passed, as it was wont to do, and Robert could do nothing but wait. Float around the house, explore every nook and cranny, anything to keep himself occupied and out of the kitchen. Away from the beaten pulp of Bernard's body.
Robert had watched, horrified by the pure viciousness of Bernard's assault, raining down blow after blow with the steel pipe. Eventually, after what seemed like hours but was probably only twenty minutes, Bernard had stopped, turning away from the body, paying it no further mind. He had gone to the sink, and washed the blood off the pipe and Robert's hands, before returning the pipe to its hiding place amongst the rubbish beneath the table.
"I will be back," Bernard had said, looking up. "Make yourself at home. Don't do anything I wouldn't do," he added.
Robert was struck by the lack of feeling in Bernard's voice. He could only pretend to understand how much the self-hatred ate its way inside Bernard; even with the snapshot view of Bernard's life, Robert could not truly comprehend living with such darkness for twenty years. Bernard had left, walking out of the house in Robert's skin as if he'd only borrowed a jumper.
Robert was left to float around uselessly. Exploring the house was no adventure; it was disgusting and vile. For someone to think so little about themselves, that they'd allow their home to get in such a state... It sickened Robert, who lived in a tidy house, sure he didn't have a world-shaping life, but he had a good life, one of self-respect. Looking around, he wondered just what would need to happen in his life for it to sink to the level of Bernard's.
Robert shook his astral head. What was he doing, trying to sympathise? Bernard had brought him to this house by nefarious means. To what end? To play this twisted game of self-loathing? What had happened was wrong, Robert could not and would not deny that, but for Bernard to carry on the way he was...it was one sin compounded on another.
How many people had Bernard brought into his dark and twisted world? Almost twenty years' worth of visitors, stealing their bodies, using them to inflict unimaginable pain on his own useless shell. Leaving them to float around the house, helpless, while Bernard went out into the real world, his body healing from the worst of the injuries.
Disassociation at its worst. Bernard was not paying for his error, he was distancing himself from it. Dishing out the punishment on his body, as if it wasn't him who gave in all those years ago. No, Robert, decided, the cycle had to end.
*
It was the early hours of the morning, by the time his body sauntered into the kitchen; yes, actually sauntered, carrying bags of shopping, as if the old man inhabiting it had not a care in the world. Robert would have smiled to himself if he could. Bernard looked up, to where he imagined Robert would be.
"You'll soon be back in your body," Bernard said, Robert's voice having already started to take on the gravelly cadence of the old man. He hefted the carrier bags onto the table. "Needed food."
Robert didn't know why. If Bernard was so intent on punishing himself, then why bother with the sporadic trips to the shops? Then again, soon it wouldn't matter. Obviously, though, Bernard had been up to more than just shopping. It wasn't even ten o'clock when he'd left the house, and now the sun was back up, the sounds of life returning to the world outside.
He waited patiently, watching as Bernard pottered around the kitchen like he had all the time in the world. He opened cupboards, placing the new tins next to the already half-opened ones, pushing aside maggots like they were nothing. Robert was glad to see that Bernard hadn't bought any fresh produce, as he didn't think he could stomach seeing the results of such previous shopping. Rotten skins, peelings... If he could, Robert would have shuddered at the thought.
"Right," Bernard said, approaching the body on the floor. "Time to change again. I expect, like everyone else, you'll run, full of judgments on my life." He glanced up. "Just remember, it could have easily been your life," Bernard said, Robert's eyes, now alien to him, looking up at the ceiling, piercing his astral self with a twisted glint.
Robert doubted it. In his entire life he had never done anything wrong; his mother had made sure of that. Not even a slight indiscretion. Solid and dependable Robert, that was him. He was no saint, he had moments where he had wanted to do some bad things to others, but he was too weak to do so. Too boring. And boring meant safe, as his mother often reminded him.
Bernard closed Robert's eyes, and slowly the body slumped into a heap on the dirty floor. Robert waited his time. He could not see Bernard's astral body, but he could feel it. A sense of movement about him, almost like an invisible breeze. Soon it was gone, which meant Bernard had returned to his own damaged body.
Robert wasted no time. He shot down from the ether and dived back into his own body.
His eyes flickered open. Smiling Robert stood, stretching his arms and legs, his back, his everything. It felt good to be back home. He glanced down at the old body, and this time he did shudder. Bernard was wrong, Robert was not going to run away, full of his judgments. The final judgment on Bernard's life had been made. It had been painful, but Robert had managed to return to Bernard's body, forced it awake...
He crouched down next to the old man and rolled him over. The light from the kitchen window glinted off the steel blade of the knife sticking out of the old man's chest. Robert didn't know what had been more painful. Moving the old broken body, or ramming the knife into the chest, shoving it in just under the breast bone, piercing the heart beneath. Robert had stayed in the body a bit longer, feeling a strange rush as the old body began to die, and something changed within him. He became aware that boring, although equating safe, meant a wasted life. He had wanted to stay for the entire ride, but as the final breath came he had felt himself being dragged down with it, and so had forced himself back out of the body. Leaving it on the floor, a lifeless corpse, in the exact same position Bernard had left it, knife hidden from sight.
Robert smiled, rising to his feet again. He wondered how long it had taken Bernard to realise that he had returned to a dead body. How long before Bernard's own life was snuffed out?
As he walked through the hallway Robert realised he didn't care. Bernard was gone, the hatred with him, and now no one else would be contaminated by his darkness. He opened the door, feeling a sense of self-satisfaction well up inside. For the first time in his life he had done something that wasn't boring, and in so doing had achieved the perfect murder.
He was right, feeling death approaching had changed him. He would return home, and show his mother how boring his life was now. And never again would it be so.
He stopped, one hand still on the door, looking into the faces of two police officers. The smile still plastered on his face.
"Robert Hoard?" one of the officers asked.
"Um, yeah?"
One officer nodded to another, and out came the handcuffs. "We're arresting you for the murder of Georgia Webber," he began.
Robert didn't hear the rest, he merely felt the officers take a hold of his arm and snap the cuffs on. Webber they had said, but Robert was able to put two and two together. That wily old fox Bernard had been busy the last twenty years, preparing a bit at a time. Until now...
There was no doubt in Robert's mind as to who Georgia Webber was. Twenty years ago she had been a teenager called Georgia Nettles, and she had taken advantage of a once kind man, turned him into the vessel of self-hatred. Robert glanced back. The perfect murder. He closed his eyes.
"Just remember, it could have easily been your life." Bernard's final words echoed in Robert's mind.
And now it was.
Serere, A Prelude
The Garden Saga
Part One: 18th Century
Newington Green, England, 1788.
Isobel Shelley waited, as she promised she would, but it was getting dark and the rain had started to fall. Not that either thing bothered her personally, but it was terribly inconvenient. She lifted her lantern, which she did not really need, of course, but appearances were important, and looked out to the northern carriage way. The Green was quiet, most people safely indoors, sheltered from the cold, but Isobel could not be sure she wasn't being watched. Newington Green, home to the free-thinkers and dissidents, had history, and the people who tended to gravitate to this place knew better than to take things for granted. Probably one of the many reasons she loved living on the Green.
The sound of hoof beats crunching gravel drifted over to her, and she focused on the approaching shape. A gig pulled by a single horse, two people jostling about in the carriage as the wooden wheels managed to find every ditch and trough in the path. Both figures were dressed in the finest cloth, one looking down, his head bobbling about as if he were asleep, but the second, holding the reins in his hands, was looking firmly ahead, mindful of the mood of the horse. The gig slowed, and stopped right next to Isobel. She smiled, finally able to see the countenance of the young driver.
Young and as radiant as ever, Hareton Wesley smiled down at Isobel, and tipped his bicorn hat. "Miss Shelley, you are still a diamond of the first water, I see. A pleasure indeed."
Isobel curtsied slightly, with a smile of her own. It had been some time since she had seen anything of Hareton, and was not displeased to see him once more. "Young Master Wesley, an' you and the gentleman like to follow me?"
The gentleman in question looked up, clearly not asleep. An austere looking man of some fifty years (which certainly meant he was older), he raised an eyebrow at Isobel and edged his lip in the form of a very slight smile, which looked somewhat strange on such a Friday-faced man. Hareton looked at him, no doubt awaiting instruction, and the gentleman nodded. "As Miss Shelley says, so shall it be," the gentleman said, in an accent that sounded almost German, although it had a cadence that Isobel could not quite place. She was not particularly well travelled, but accents did not usually stump her so. "Do lead on, dear lady."
"As you wish," Isobel said and tuned away, lantern still held aloft, and led the way across the Green.
*
Once the door was bolted, and the candles lit, all pretence of formality ceased. Isobel flung herself into Hareton's arms, and their lips met with great passion. For a full minute they remained like that, any thought of the gentleman momentarily gone. Only the distant sound of movement in the room served to remind them that they were not alone. Eventually a sharp clearing of the throat tore them apart, and Isobel looked over at the gentleman demurely.
"Sorry. Hareton and I..."
"Have a history?" the gentleman asked, his face no longer as severe as it had been out in the rain. Indeed, his features now seemed to be full of warmth. He pulled up a seat and sat at the table, removing his hat and wig, both of which had become sodden in the rain. His hair beneath the wig was silver-grey, pulled back and clubbed with a black ribbon, his upper lip covered in an equally grey moustache, but it was his eyes that pulled Isobel in: deep brown, mortal eyes, containing such compassion. It was rare to meet one of their kind with human eyes. Although they still managed to pass off as normal among the common folk, her eyes were pale, the pigment of the iris slowly fading with the passing of each year. And such was true of most of their people, except those who had yet to experience the Second Death. The gentleman before her was clearly one such person.
Isobel batted her eyelids bashfully like a betty, although she was anything but. However it was an image she had maintained for a long time, fooling the gentry all through the Town, and she saw no reason to reveal her true self to a man she did not know. Even if he had been sent by the Three. "Yes, sir, history we have."
The man nodded, turned his eyes to Hareton. "See to the horse, we shan't be here too long, I want them ready to go," he said sharply.
Hareton bowed. "Of course, Mr Holtzrichter."
He turned to leave, but was prevented by Isobel's hand on his shoulder. He glanced back at her, and she looked at Mr Holtzrichter, steel in her pale eyes. Demure and prim might have been a role she liked to play with mortals around, but no one ordered another under her roof except her.
"You have both travelled far, and I will have neither of you leaving without full stomachs." For a moment Isobel was certain Mr Holtzrichter was going to stand and strike her, such was the coldness that swept over his face, but it soon passed and he smiled, nodding sharply.
"Quite the chit, are you not?" he said, good humour in his voice.
"When the mood takes me, sir, but don't ever take it to mean I am bacon-brained," Isobel returned, careful to keep her own tone light.
"Indeed not."
Isobel returned his smile, and curtsied, which brought laughter from Holtzrichter's belly. "Very good, my dear, I like the cut of you."
"Hareton, be seated," Isobel said. "I have a broth prepared already. Mr Holtzrichter and I can be alone shortly. To conduct our...business."
Hareton walked over to the table and sat on one of the hard chairs, but he did not question the source of such business. Isobel felt sure he did not know, but he was not so foolish as to enquire in front of Mr Holtzrichter. Although he would return later. How could he not? He was on the high ropes and he, too, remembered their last encounter as clearly as she. And it was an encounter both wished to repeat.
As she poured the broth into bowls for the two men she had to consider, once again, just why the Three would send a special envoy all the way from France to see her. Certainly she had chosen her side during recent events, and she applauded the reforms the Lady Celeste had put into place over the last six months, but she was one among tens of thousands of their kind in England, and not worthy of such attention. It troubled her. Rumour had spread that Celeste was still removing her enemies, those who had taken sides with the Brotherhood. Could Celeste have been misinformed and now considered Isobel one such enemy?
She smiled at Mr Holtzrichter, who had offered his own smile upon receipt of his broth. Maybe she was looking too far into it, but there was something she didn't like hidden behind his smile. And his name...it sounded German, and didn't Celeste have a German consort?
Once the men had finished their broth, Hareton left to tend to the horse. Isobel busied herself with cleaning the bowls, all the while feeling Mr Holtzrichter's eyes on her back. She stopped for a moment, and asked; "Is your name German?"
Mr Holtzrichter chuckled. "No," he said, "although a common mistake. It is Prussian. I was born in a little town called Posen in 1722."
Isobel turned to him. "You are a young one, too, then," she said with a coy smile. "So you come from the home of the Tree King?"
For a moment Mr Holtzrichter looked confused, then he smiled. "Oh yes, your mad King George," he said, referring to the tale of the ailing king who had once shook the branch of a tree believing it to be King Frederick William, the incumbent ruler of Prussia.
"Hardly my mad king, Mr Holtzrichter. I have lived a long time, seen this country at war many times over, ruled by many fools. Still," she added wistfully, "it is my home, although I am very much no longer of Great Britain." Holtzrichter nodded in acknowledgment of this, and Isobel smiled, thinking that another hundred years of life and he too would not consider himself of any one country. Their people transcended the loyalties of mortal living. He was still young, despite his outward appearance, and he had much to learn. One thing he did know, though, was how to show his host respect. Holtzrichter had not needed to offer up such intimate information; age and birthplace was rarely a secret shared among their people, and Isobel took it as a mark of respect.
"For myself I am, as of this month, one hundred and seventy-nine years, born in London to a modest family. And, as you can see," she added indicated their surroundings, "little has changed. Although let it be never said of me that I'll be found punting up the River Trick. Financially or else."
"Being in debt is never something to be encouraged." Holtzrichter frowned. "You have lived over a century and thought to make nothing of yourself? If I may make so bold, why?"
"You misunderstand me, sir," Isobel said and sat at the table. "I choose to be like this, a woman of little means. You cannot live for over a century by attracting attention to yourself. As I said, this country has been at war with one country or another for so long now, an' I were to be noticed..." She shook her head. "This is why I came to Newington Green. It has a history for attracting the dissidents, the outsiders, those who do not conform to the Church and the Crown. And those who wish to remain invisible."
Holtzrichter nodded. "I understand. I was born poor, and lived a very modest life, until a visiting French noblewoman noticed me. She changed my life, and now she wishes to change yours."
Isobel was taken aback, but she had no doubt as to whom he meant. For a long moment Isobel remained as she was. Then she asked, softly, "why me? I keep myself to myself, I..."
"We both know this is not quite true, do we not, Isabella?"
For the second time in as many minutes she did not know what to say. She was certain she had kept her tracks well hidden. Of course she had been dragged into the revolution, but as far as most knew it was with open reluctance. Very few knew the truth, knew what exactly Isabella Frith had done during that violent time, and only a select few knew the true identity of Isabella Frith. It appeared one such person had talked. Isobel let out a sigh of defeat. "I do not seek attention and..."
"That is why Celeste has sent me to you." Holtzrichter said her name with such a feeling of intimacy it surprised Isobel. The Lady Celeste was said to not keep many close to her, but it seemed Holtzrichter was one of those. If he was not the Lady Celeste's consort who was he to speak so freely of the Lady Celeste? "She heard of what you did, the great service you did in the name of the Three..."
"They did not even exist, then," Isobel pointed out.
"No, of course, not as a body, but as an ideal embodied in Celeste. Her desire to bring our people out of superstition, away from the monster of myth, has always been with her. Ever since..." Here Holtzrichter paused, and looked down. Isobel watched him closely. He knew a lot more about the Lady Celeste than he was willing to share. The mystery deepened; who was he? Isobel knew better than to ask, it was clear she would not be told. "Since the beginning," he continued, "and it is that ideal for which you fought. As a thank you, she wishes to offer you something. Something," Holtzrichter looked around the small room, "you have clearly denied yourself."
"Then perhaps it is something I still do not care to have," Isobel said, beginning to have an inkling of what was about to be offered. "I prefer to be unknown, Mr Holtzrichter."
"Then, my dear lady, why did you fight in Celeste's name?"
"Because..." Isobel stopped.
For the first time in many long years she felt she needed to explain herself. Perhaps it was because of what was about to be offered. She needed the Lady Celeste to understand why she stood against the Brotherhood, and why she had to remain as she was. Working for the benefit of the Three in her own way. Isobel stood. "Let me show you why. I shall return momentarily, sir."
For a short while Isobel left her guest alone, as she visited her private chambers, where she slept and kept herself hidden from the world. She returned shortly, holding in her hands several sheets of parchment. She placed them before Holtzrichter, who watched her with great interest.
He spread the parchment out before him. "And what are these?"
"A few years ago I was visited by one of our people, Mr Holtzrichter, a coxcomb named Edward Lomax." Isobel shuddered with the memory. "Something ailed him, sir, ghosts and voices, one too many maggots in the brain. But still he talked with great intelligence. No less queer in the attic as King George he may have been, but Edward Lomax was a man of learning. And he brought with him the Book of Origin."
Holtzrichter looked up, his dark eyes full of suspicion. Only the youngest of their people did not know of the Book, and it was clear that Holtzrichter was not among them. Somewhere in the world there lived a being called the Ancient, the oldest of their kind, and it was said that he was there at the beginning, in Egypt. The Book was his, notes of dreams and visions, tales of their combined history, everything from how their people came to be to prophecies of the future. The Book, it was said, was lost to the Ancient centuries ago, and he scoured the world looking for it. The look of disbelief in Holtzrichter's eyes no doubt matched hers four years ago when Edward Lomax had presented her with the Book.
She nodded. "He was cast out of the Green when he was found strangling a boy of only ten years, but he took the Book with him. However, in his haste, he left these behind." Isobel pointed at the parchment. "These pages, translated by Edward, tell of a prophecy about a man called Seker..."
"Seker?" Holtzrichter looked down at the pages. He reached into the breast pocket of his jacket and removed a quizzing glass. He picked up a piece of parchment and brought the quizzing glass close to it, not that he would need such a thing. Their kind had perfect vision.
"Yes, believed to guard the gates of the underworld in Egypt mythology. These pages tell us that he will, apparently, return in the second millennium." Isobel resumed her seat and found the relevant passage. "And Seker shall return in fire, to bring the children back home to her."
Holtzrichter did not look up, but continued reading. "Do you mean that Julius was right?"
"I think he has diluted the truth. I do not fully understand what Julius teaches, but I do know that what he claimed is a lie. According to the Book we are at over two hundred years away from Seker's arrival."
Holtzrichter did not comment for a while, instead he read.
"This is why I followed Celeste," Isobel continued. "The Brotherhood sought to descend our people into chaos, and that is not the goal of Seker. We are not animals, despite our collective past, and we must prove that."
Holtzrichter did not respond, instead his attention flickered from one piece of parchment to another. For a short while Isobel watched him, but it soon became obvious that he was no longer aware of her, so she took her leave of him and slipped out to see Hareton.
*
As she stepped out of the house, back into the rain, Isobel did up her pelisse, casting a look back at Mr Holtzrichter who still had his head in the pages. She closed the door and made her way to the stables, where she found Hareton readying the horse and gig for the long journey ahead of them. He looked up as soon as she entered, a smile stretching across his countenance. When she had last seen Hareton Wesley he had been only seventeen years, no longer a boy but not quite a man. Although, Isobel recalled with a flush, he had soon found his way around her body like a man used to a bit o' muslin. Now he was twenty-three, a young man, and she had to confess he had filled out quite well.
"Do you find me amiable?" he asked, enjoying the attention of her eyes.
"Very." Isobel bolted the stable door and crossed over to him. "You have become quite a man, Hareton," she said, running a delicate hand across his firm jaw.
He held up a hand to stop her. "What of Mr Holtzrichter?"
"He is else occupied, besides which I do not care a groat. Can he stop two people in the high ropes?" At this Hareton lowered his hand, and Isobel continued to stroke his jaw. "I have heard word of your exploits these last six years, playing messenger to my people. And now to deliver the Lady Celeste's very own envoy?"
Hareton placed his hand over hers and brought it to his lips. "Does this mean I have proven my loyalty?" he asked, kissing the tips of her fingers.
"It would appear so."
"Then you will grant me my desire?"
"And what is it the young master desires?" Isobel asked, in her best meek voice, the tone of a doxy looking to please her master.
"To be with you forever," he answered, gently pulling her towards him. As their bodies touched, he stepped backwards until he was resting against a wooden beam. "In six years my desire has not wilted. Everything I have done has been for this moment."
Isobel did not trust herself to speak. Six years ago, when the young Hareton had first come to Newington Green she had found herself compelled by his beauty, but she could not give herself over to him, not in the way he wanted. A physical paring was one thing, but to give her heart to a mortal...it was bound to end in tears of blood. But that did not stop Hareton, even when he learned what she was. So she had sent him away; if he could prove his loyalty to her and her kind then she promised she would take him. That was six years ago, when her world was governed by rules. Now the Three were leading her people into a unified future of civilisation, and it was not for her to bring one into their ranks of her own accord.
But while he was here...
She pressed her body against Hareton's, feeling him harden beneath his breeches. "Take me now, Hareton!" she whispered in his ear.
As Hareton's hands undid the buttons that fastened her pelisse, Isobel lifted her face to the stable ceiling, allowing Hareton's tongue to play on her throat. His hands found their way inside her gown and reached to unlace the stays beneath, but his fingers barely found the lace when a banging came at the stable door. For a moment they looked at each other, Isobel's eyes daring Hareton to continue, but despite the desire burning in him Hareton removed his hands from Isobel's clothing and gently pushed her away. Still a man of his age, Isobel realised with disdain, a man of scruple. She fastened her pelisse and watched him unbolt the door. If she truly took him, eventually, like every one of her people, he would soon realise he was no longer bound by the rules of the land.
Mr Holtzrichter stood outside the barn, the rain rinsing the powder from his hair down his face. He glanced at Hareton, who looked to the ground, his face flustered, and then at Isobel who merely smiled at him. "I see," he said with a curt nod. "Miss Shelley, a moment of your time if you please."
"Of course, sir," she said, falling back into her public role, and stepped out of the stable. As she passed him she noticed Holtzrichter give Hareton a lopsided grin of apology.
"Do peg the pardon of a gentleman for taking one so young off the high ropes."
The younger man clearly did not know how to respond. So he stepped back further into the stable, and turned back to the horse.
*
"It is good you have someone," Holtzrichter said as he closed the door behind him. Isobel stopped by the table, waiting for him to elaborate. "Later," Holtzrichter said, with a wave of a hand. "We shall return to that in a moment, but first these." He indicated the parchment. "I would like to take them to Lyon and study them further, if you have no further need of them?"
Isobel shrugged. She recognised that glint in Holtzrichter's eyes, and was reminded for a moment of the vacant look in Edward Lomax's own eyes. Obsession. She did not know why the parchment interested Holtzrichter so, nor did she really care. "If you so wish. Now, may we return to the reason for your visit?"
"Of course, dear lady," Holtzrichter said, tucking the parchment into one of his Hessian boots. "As you know our people have lived in disarray, with no rules or..."
"I do know, so if you would care to..."
"Of course." Holtzrichter smiled, and it was one born of both surprise and respect. "Celeste fears that war is coming soon to the human world, a war at the heart of which France will reside. Over the centuries our people have spent too much time involving themselves in such things, and if we're to move ahead into civilisation, then we cannot allow such distractions any more. It was such involvement that allowed the Brotherhood to get the foothold they did."
"How does the Lady Celeste propose to stop this from happening again? It is human nature to involve themselves in things that do not concern them. A trait our people have yet to grow out of."
"Agreed, which is why they need strong leadership. People who will show them. Eventually we shall become one with this world again, walk side by side with humans, unseen and unsuspected for what we are. But it will take time and effort, and strong leaders. The Three are creating the domains, sections of the world lead by a council of Lords and Ladies, with clear directives. Celeste would like you to become Lady Isobel, of the Great Britain Domain."
Isobel just stared at Holtzrichter. "Me? I have told you, I like to remain..."
"Unnoticed. Yes, but you also told me why you opposed the Brotherhood, that you believe in the ideals that the Three represent. If our people are to emerge from the shackles of myth and legend then we need people like you to show them how." Holtzrichter regarded her, and pulled a small piece of rolled-up parchment from his jacket. He placed it on the table. "An invitation to attend the first meeting of the Domain Council. If you choose to accept this position, then the Three look forward to your attendance."
Holtzrichter stepped towards Isobel and took her hand, which he kissed gently. "Now I take my leave of you, My Lady, and I thank you for your hospitality." He turned to leave, then looked back. "One further thing. If you choose to accept this new position, then you will need someone who can support you...in all ways. I believe Mr Wesley would be parti for you, and I do not see the Three opposing such a thing. In fact they would encourage it."
With that Holtzrichter removed himself from Isobel's home, leaving her looking at the rolled-up parchment still sitting on the table.
*
Lyon, France, 1790.
Frederick looked up from the parchment, at the unwanted knock at his closed door. Things were getting ugly in France, another kind of revolution was underway, of the kind the Three had expressly forbidden their people to get involved in. Only two days ago the Civil Constitution of the Clergy was passed by the Assembly, despite King Louis' apparent objections. Even now Celeste was visiting the king to try and talk some sense into him. It always surprised Frederick, even after almost fifty-seven years, the way Celeste was able to talk her way into the confidence of those in power. He knew it should not surprise him, after all Celeste was born of noble blood, and she was at home with nobility of every kind. Especially in her own country.
It frustrated him, too, that Celeste was becoming involved in the revolution sweeping France, when it was she who created the Domain Council to prevent such involvement in worldly affairs. But there was no reasoning with her; France was her pet project, and she had to do her best to keep the forthcoming war she feared from the French borders. If Celeste was to believed there was nothing to be done, France would be at war within a few years. It was now inevitable.
He rose from the table, glancing one last time at the pile of parchment, and turned to the door. That also frustrated him. He had studied the words on the parchment many times in the last two years, ever since he had claimed them from Lady Isobel, and now knew them word-perfect, but still he wanted to know more. In that time he had scoured all over, visited countless countries to uncover anything that would help him discover the answers he needed. So far all he had found was scraps; notes written in obscure languages that he could not read. Even the best translators found much of the languages difficult to understand. What he had read, though, intrigued him greatly, even if a lot of it was contradictory. Of one thing he was sure, he had to learn more, to find out the truth of where his people had come from. He had never believed the lies spread by the Brotherhood, but he was beginning to suspect that Julius, although undeniably egocentric and deranged, was closer to the truth than Frederick liked.
"What is it?" he demanded, as he flung the door open. Honoré, the head servant of Celeste's house, stood there, his face a mask of fear. "Well, speak!"
"Pardon, monsieur, un courrier a introduit le present document pour vous," Honoré said, and handed Frederick a rolled-up parchment, sealed with a red ribbon. Frederick's French was shaky at best, even though he'd been with a French woman for over fifty years, but he understood a few words. Someone had brought this document to the house for him. To take him from his studies it had better be of importance.
"Merci," Frederick said, and turned from Honoré, unrolling the parchment. He stopped in his tracks and read the words written in the finely crafted script twice. He swallowed, span on his feet, and turned back to Honoré, who was already walking away from Frederick's room. "Honoré, has Celeste returned?"
Honoré stopped and looked back, with a frown of concentration. "Pardon, monsieur, je ne comprends pas."
Frederick growled. "That is the problem, neither of us understand the..." He paused. "Wait, I did understand that. Celeste, a retourné?" he asked, suddenly able to speak and understand fluent French. Celeste always said that eventually he would be able to understand every language he heard, a peculiar trait that their people developed when near the Second Death. Which meant soon it would be time to... Frederick shook his head. No, he did not wish to contemplate what that meant. He knew, that was enough.
"She has, sir. I believe she is dining at this moment," Honoré said.
"Thank you."
Forgetting to close the door, Frederick swept past Honoré and made his way through the house to the dining room. There he found her sitting at the head of the table, resplendent in the finest silks, her dark red hair contrasting with the lighter shades of her dress. She looked up from her food, raised an eyebrow at Frederick's haste, and offered him an empty wine glass.
"Mes toujours, a pleasure as ever. What brings you here in such a hurry?"
Frederick sat himself at the table and took the glass, allowing Celeste to pour the red liquid out of the crystal decanter. He returned her smile, and sipped before beginning. "I have received a missive, an invitation from the Ancient himself." Still hardly able to believe his eyes, Frederick handed the parchment over.
Celeste quietly read the script. Once finished she carefully placed it on the table and raised her pale eyes to look at Frederick. "Moldavia. A long journey, Frederick, and a treacherous one. But such a summons cannot be ignored." She smiled and reached a hand out to him, which he took and held in his. "Perhaps you shall now have answers to these questions?"
"It would seem most probable. And, of course I shall go, how can I not? There have been reports of the Ancient for many years, but none have been substantiated in decades. Just rumour. And now Wamukota wishes to see...me? Why me? Why now?"
"You question too much, mes toujours, I have always said so. You always want to know things with certainty, to be sure and have no doubt. Such yearnings lead to a closed mind."
Frederick shook his head. "No, questions should be asked. Always."
"Perhaps, but some answers are best left unknown."
"Like the Second Death?" Frederick said softly, disturbed by the quake in his voice. "It is coming soon, Celeste, I know it. I understood Honoré with perfect clarity."
Celeste took this news with grace. She knew Honoré spoke only French, and she knew how difficult Frederick found learning their native tongue. She smiled sadly, and placed a hand on his face. "I will miss seeing these eyes, but you know what must be done."
For a moment neither spoke another word.
Frederick swallowed. "We shall see," he said, and bent down to kiss Celeste. She returned the kiss with passion. "I shall return as soon as I am able. With answers," he added.
Celeste raised her glass. "To answers, may they be all you wish. And when you return, may you be as young and vibrant as when we met."
Frederick bowed, then turned to leave. It was, as Celeste said, a long and treacherous journey ahead, through countries at war. Always, it seemed, humans were fighting over something. He shook his head. It did not matter. He would make it to Moldavia and meet with the Ancient, the oldest of their kind. And he would find a way to escape the Second Death...somehow.
* * *
Part Two: 21st Century
Newington Green, England, 2002.
"I don't know, Jake," Willem said into his phone, as he stepped out of the cafe. He found a free table and sat down, placing the carrier bag on his lap and cracking open the can of Pepsi. It was a hot day and he was parched. Downing a can of drink while resting his legs sounded like a good plan. "You say that but there's something about Cruise, you know?"
"Like what? He's an okay actor, I guess," returned Jake, the slight Californian lilt of his accent still there, despite twenty years of living in London, "but he picks such crap movies, guy."
"You said you rated Minority Report," Willem pointed out, lifting the box out of the carrier bag. An old man, on a course for the cafe, stumbled over a loose paving-stone and almost knocked the box out of Willem's arm.
"So sorry," the old man said, as Willem fought to steady him with his free hand.
"It's okay, man," Willem returned. The man gathered himself together, and for a moment he remained standing there, looking at Willem through his dark shades. Willem stared back, feeling his blood go oddly cold. "You sure you're okay?"
"Yes, yes," the old man mumbled, "sorry, yes, I'm fine now. Just for a moment there you reminded me of...someone else." He shook his head. "Excuse me."
Willem watched the old man continue his way into the cafe, and blinked. He turned his attention back to the phone call, and could hear Jake on the other end trying to speak to him. "Sorry, dude, some old guy almost collapsed into me. Anyway, rating Minority Report."
"Right, well I do, but what's that got to do with Cruise? It was a good movie, but Cruise... Sorry, Will, but can't be agreeing with you on that score. His wife on the other hand, she's a babe!"
"What? Nicole Kidman?" Jake's whistle on the other end of the phone made Willem laugh. Even now he could see his mate subconsciously repositioning his bits. "You do know they split last year, right?"
"Oh." Silence, and then, "really? Proves my point, then. How can you rate Cruise when he divorced Kidman? He's a fruitcake, obviously."
"Good logic, man," Willem said with a laugh.
Having just spent some time away from work to visit his father in Hackney, Willem was glad for the light relief Jake brought him. It wasn't often that he visited his father, but now and then Willem felt obligated to visit, just to check in on the old goat. He was still a little concerned that his father was continuing his decline; first he'd turned to drink, which lasted a few years, and more recently he'd found religion. Willem still didn't buy it, and it made his visits more sporadic than ever. He just couldn't take seeing his dad turning into a pious old hermit, who spent most of his time quoting the Bible instead of asking how Willem's life was going. Feeling a little down afterwards, Willem stopped en route home to treat himself to a new phone.
It was the latest in phone design, a Nokia 7650; slide-open, and the first Nokia phone to feature a camera. Willem didn't quite understand why you'd need a camera in your phone and he certainly didn't see it catching on, but that didn't matter, he had seen it last month in the film Minority Report and had wanted one. Ringing Jake to tell him about the new purchase was what had initiated the critique on Tom Cruise. And now, with his old Ericsson T66 resting between ear and shoulder, he sat outside the cafe playing around with his new phone.
"So, when you back, guy?" Jake asked.
Willem chewed his lip, wondering if they would mind him borrowing a socket inside the cafe so he could charge the phone, and said, "couple of hours, probably. Have to meet with Ste, then pop over to the old folks home."
Jake chuckled. "Don't tell me, more drama with the teenager?"
Willem rolled his eyes at that, placing the phone back in its box. "Wouldn't mind so much if my sister was a teenager already, at least then she'd have a reason for being such a stroppy bitch. Ah well, you know how it is, man, wouldn't be my sister and mother if they weren't having some kind of drama. Of course, they're probably giving Eon a headache, so swings and round-a-bouts."
"Yeah, always a plus. Anyway, guy, I'll let you go."
"Right, okay, cool. See you on the flip side, yeah?" He put the phone in his jacket pocket and glanced back at the cafe. He could ask them to borrow a plug socket, but... Willem checked his watch. Getting to Fulham would take a while, and since he didn't drive... He stood up and turned to leave, thinking that maybe it was time he sorted out some driving lessons. Couldn't become a business executive and not drive, that would be just—
He stopped and looked back at the shop. Just for a moment he had the distinct feeling that someone was watching him, very closely. But no one seemed to be paying him any undue attention, not even that old bloke, who was now immersed in his newspaper, mug of tea on the table before him. Willem shrugged. He had things to do.
*
The old man looked up from his paper once again, and slowly lifted his sunglasses. With eyes so transparent they showed the blood behind, he observed the young man with carrier bag walking away from the outside table, leaving the Pepsi behind.
"At last," Frederick said, "just as the Ancient promised."
*
His work for the Three done, Frederick made his way slowly up Hawthorn Road. He didn't mind Ashington too much, a largely urban town in the North East of England. He'd been sent to worse places in the centuries he'd served as the Three's special envoy, and Ashington was...nice. He'd rather be in London, keeping an eye on Willem Townsend, but he had duties that did not allow him the luxury of such excesses. He had spent far too much time in London in the last few months, anyway, ever since he'd first spotted Willem outside that cafe, and Lady Isobel was beginning to get a little curious. If he continued it would only be a matter of time before Celeste found out, and he wasn't ready to share yet.
He needed to be absolutely sure first. If what the Book said was true, then a few more years had to pass before he could make his move, enough time for him to be certain of the ka he'd sensed when he stumbled into Willem. It was him, Frederick was so sure, but not absolutely. A few other things needed clarity first.
That was why he now walked up Hawthorn Road, following the teen before him. A century had passed by so quickly, and now, once again, it was time.
He'd been following Robin Turner for a few days now, delving into the human's mind. Such a fragile thing, even the weakest mind trawler would have had no difficulty reading the surface thoughts of Robin. Frederick had learned what he needed, and knew that after work Robin always popped by his mother's before going on to his girlfriend's flat. And he knew that the path he took was always the same.
On cue, Robin turned into Hirst Park, and Frederick quickened his pace. Robin reminded him of so many others he'd known over the years. Dead on six feet tall, thin but not slim, with dark hair and deep brown eyes. Just like with all the others, Robin had the kind of eyes that sucked a person right in.
How could Frederick resist? Especially now.
He turned into Hirst Park himself, and was surprised to see Robin standing there, his body tense, fists clenched. As Frederick had suspected, Robin knew he was being followed. Which is what Frederick liked; he never picked the weak ones. There was no fun in that at all.
"What the fuck, man! What are you, some kind of nonce?"
Frederick grinned, and shook his head. "No, children have no interest for me. Younger than nineteen and..."
"You're sick!" Robin stepped forward. "You've picked the wrong fucking man to stalk."
"No, you're perfect." Without warning, and faster than Robin could take another step, Frederick was right in Robin's face, one hand clamped around his throat. "To be nineteen again," Frederick whispered, and forced his mouth over Robin's.
*
Central London, England, 2003.
"Bro, that was just...wow!" Ste looked up at the glass-faced tower, unable to wipe the smile off his face. Only seconds ago both he and his mate were at the top of the Canary Wharf Tower and now they were both standing in the square below, surrounded by a cheering crowd who stood behind barriers some distance away.
"Express elevator to hell, right?"
"God yeah." Ste laughed, and took a deep breath. "Shit. BASE jumping is just... Shit yeah! Can't get much more crazy than that!"
"That a challenge?"
"Fuck yeah!" Ste said and held his hand out, which his mate grasped with equal fervour, their thumbs linking. That's what Ste loved about Robin, always throwing out the next challenge, which he knew Ste would have to accept. Some called him an adrenalin junkie, and maybe they were right. Fact was Ste didn't want to waste his life; he had to live on the edge. He'd almost died in a car accident when he was a kid, and since then it seemed foolish to waste his second chance.
"Dude, we should probably clear up the parachutes," Robin said. "Before our adoring fans want our autographs." He nodded at the crowds.
Ste looked over and laughed. "Yeah, extreme sports, extreme fans. Which reminds me, those twins from last night...erm, Karen and Anne? Did you get their numbers?"
"Sorry, mate, forgot," Robin said, as he started work on gathering the parachute off the ground. "You know me, fuck 'em and leave 'em. No time for action replays."
"Not always true," Ste said with a wink.
Robin laughed at this, and playfully punched Ste's shoulder. "But I've never fucked you."
"Everything but, though."
"Too extreme for you?" Robin asked, that old wicked glint in his brown eyes, one eyebrow raised.
"You wish."
"So, tonight then, yeah?"
Ste shook his head, laughing. "Yeah, you're on."
*
Ste swung his legs over, and with a painful sting in his groin he managed to sit on the edge of the roof next to Robin, who smiled at him, then looked out from the top of Michael Stewart House. From their vantage point on the roof they could see across Fulham and out past Charing Cross Hospital. It was a clear night, offering a good view, but it was chilly. Fortunately, due to their very own version of extreme indoor sports, and the subsequent climbing onto the roof, Ste didn't feel much of that chill. His muscles were sore, his body warm, he was also in some pain. But it was a nice pain.
"It'll probably go septic by the morning," he said.
Robin shrugged. "Not necessarily. Depends on how well you heal."
"Well enough," Ste said, gently pulling the crotch of his jeans away from his now very sensitive genitals. "Still, that was... Where did you learn that?"
"Jassy, a small place in Moldavia."
"Right. You've travelled a lot," Ste said, having forgotten the amount of places Robin had mentioned in a first-hand-experience kind of way. They'd only known each about four months, met in a pub off Oxford Street. Ste noticed something in Robin he'd found very familiar, and decided to introduce himself. Soon found out they had much in common, including a love for extreme sports. Of course, back in November Ste had no idea that Robin liked to take those extreme sports into the bedroom. But hey, Ste was up for trying anything once, and that first night when Robin had applied pressure there Ste had almost fallen to his knees. It had hurt at first, but then the adrenalin kicked in, the endorphins were released, and he found himself incredibly turned on by the pain.
Now here they were. Ste would never call himself gay, nor Robin come to that, but they had developed a rather interesting friendship, one with very few boundaries. It wasn't about sex, none of it was; it was about the rush, the high they both got. Be it with each other or with women involved, it didn't much matter. What mattered was the end result; the high!
"What if I told you that you're still missing out on the biggest high of them all?" Robin asked, almost as if he had read Ste's mind.
"Then I'd say let's do it, man!"
Robin nodded his head slowly. "Right, okay, but I need to show you something first."
"What?"
Robin looked at Ste and winked. Then, without preamble, he flung himself off the roof.
For a second, unable to believe his eyes, Ste continued to look in the spot Robin had occupied, then he lowered his head, his heart beating faster than it had ever beaten before, and saw something that he just couldn't accept.
Several stories below, on the grass, Robin was clambering to his feet. At first he seemed to have little balance, but Ste figured that might have had something to do with the way Robin's left leg was completely out of joint. He shook his head, wondering at the way he was viewing this. It was abstract, unreal. Yet...
Robin popped his leg back into place and waved up at Ste. "Come down!" he shouted.
Ste swallowed. BASE jumping was one thing, but to do it without a parachute... He was an adrenalin junkie, sure, but not insane. "I'm taking the stairs," he mumbled, his throat suddenly dry.
*
In the time it took for him to walk to the ground floor, two things changed in Ste. One, his legs had decided to work properly again, the alternate stiffness (from the extreme exertion) and shakiness (from the shock of seeing one's best mate commit a failed suicide) had subsided. Two, his mind had settled on anger. The shock, which probably hadn't gone totally, had crystallised into anger.
He found Robin still outside, now sitting on the bar that ran the length of the wall at the edge of the grass. He had his back to Ste, looking across at the Fulham Pools on the opposite side of Lillie Road.
"What the fuck was that, man?" Ste wanted to know.
Still Robin didn't turn. Ste slowed his walk. Something was different about Robin, the way he sat. There was new strength to him, not to say that Robin had ever proven weak, but he sat like a different man. The kind of man you didn't want to fuck with.
"Listen, how did you do that?" Ste stopped a few feet away.
"Come sit with me, and I'll tell you."
Even Robin's voice sounded different. He sounded like an older man, with an accent Ste had never heard before. He would have said German, but there was something not right about that guess. He took a deep breath. He couldn't back away now, he had to know how Robin had managed to jump off that block of flats and survived. So he did as he was told. He climbed the wall and sat next to Robin. For a moment neither spoke, they both looked out to the empty road. Fortunate that it was half three in the morning, no spectators to witness Robin's miracle BASE jump.
Robin turned his face slightly, and the person Ste saw looking at him was not Robin. It had nothing to do with the cuts on his face, it was more to do with the way the face sat. The features seemed harder somehow, more serious, no trace of a smile at all. Ste didn't know what to think, but he was sure he was not looking at the nineteen-year-old Northern lad he'd known for four months.
"Do you still want the biggest high ever?"
Again Ste swallowed. He really did want it. It's what he lived for. All the hours he put in at work at the coffee shop, he hadn't become a manager at twenty-two just because he loved his job. Will paid him a good wage, more than enough to pay for his extreme life style. And he'd done it all, every extreme sport that had been invented he'd given a go, found new ways to push, to make the rush even more intoxicating. Now Robin was offering him something else entirely...
"What is it?"
"My blood, Stephen. It's special, keeps me alive. Forever." At this Robin smiled, but it wasn't the wide "here comes the rush" smile he usually had. This was more ironic. "That's how I survived."
"And you...what? Want to give me your blood?" At this Ste laughed. "Come on, dude, this is the twenty-first century, I'm not taking your blood. I don't know what you might have."
Robin raised an eyebrow, and for the first time, without the shadow of the brow covering it, Ste saw that Robin's right eye was extremely bloodshot. Well, he had fallen from a great height, something was bound to be damaged. "If I had anything, don't you think you would have caught it by now?"
"Good point. But still..."
"Then perhaps you'd like to sample the goods?"
Ste wasn't too sure about that, but before he could decide Robin moved. Fast. Faster than it took them both to jump from Canary Wharf. Ste's eyes went wide as Robin forced his mouth open. Robin's wrist was a gash of blood, torn open by an incredibly long thumbnail. No, not a nail, it was if a talon made of bone had sprouted from the tip of Robin's thumb! Ste barely had time to take any of this in, before Robin's blood poured down his throat.
Robin whipped his arm away and Ste fell back onto the grass. He lay there for a while, feeling his heart beat so fast, the blood whizzing around his body, adrenalin kicking in. He jumped to his feet in one swift movement, eyes darting about. He espied the play house in the park outside the Pools, and before he knew it he was standing on top of it, balancing perfectly on the tiny roof.
"Wow. This is the shit!" he said, his voice sounding loud in his ears. Robin leapt the fence and walked through the park.
"Well?"
"This is just..." Balance suddenly gone, Ste tumbled and landed on his arse beside the play house. Robin was looking down at him; he offered a smile and his hand. For a brief second Ste was looking down at himself, looking up at Robin, eyes glassed over. Then he changed. It was not Ste he was looking at but an old man, incredibly old, wizened skin like bronze. He sniffed. Something was burning...
"Come back to me, Stephen," a voice said.
The smell faded, and once again he was looking up at Robin, who effortlessly pulled him to his feet. "What was that? I was..."
"In one of my memories. A long, long time ago...in Moldavia."
"Your memories?" Ste shook his head, trying to get that image out of his head. "Who was that man? How was I in your..."
"His name was Wamukota, the oldest of my people. And it's the blood, Stephen, it's always the blood. Our life source." Robin offered a smile. "The effect becomes more pronounced each time, the buzz better, and every time a different memory." A beat. "Want more?"
"God yes!" Ste replied without compunction.
Robin nodded, the bemused expression slipping off his face like oil. "Then we trade. I will give you as much as you like, but I need you to do something important for me."
Ste shrugged. "Hey, Rob, there's always a price, right?"
"Quite so." Robin took a step back, and for a moment the light from the lamp outside the park flashed on his face and Ste saw the truth. Robin's right eye was not bloodshot, it was transparent, and he could see the blood flowing behind it. But the left, that was brown still... Robin blinked and reached for his eye. "Ah yes, I lost one of my contacts when I landed. Not too worry." He removed the lens from his left eye, and looked directly at Ste. With matching transparent eyes. "Do we have a deal then? My blood for your help...?"
Reserved.
| {
"redpajama_set_name": "RedPajamaBook"
} | 4,472 |
\section{Introduction}
Given a homomorphism of commutative, unital rings $\A\to\B$ that makes $\B$ locally free of finite rank $n$ as an $\A$-module, there is a zoo of ``trace'' or ``norm'' operations taking algebraic data over $\B$ to algebraic data of the same type over $\A$. For example:
\begin{itemize}
\item Given an element $b\in \B$, we can take its \emph{trace} to get an element of $\A$: working locally so that $\B$ has an $\A$-basis, represent multiplication by $b$ as the action of a square matrix over $\A$, and then take the trace of that matrix. This gives us an $\A$-linear function $\mathrm{Tr}_{\B/\A}:\B\to\A$.
\item Alternatively, we can take the \emph{norm} of $b$ by taking the determinant of the matrix by which $b$ acts. This gives a multiplicative function $\mathrm{Nm}_{\B/\A}: \B\to\A$.
\item Given a line bundle (a locally free rank-$1$ module) $\L$ over $\B$, we can take its ``norm'' to get $\mathrm{Nm}_{\B/\A}(\L) = \mathrm{Hom}_{\A}({\textstyle\bigwedge}^n \B, {\textstyle\bigwedge}^n \L)$, a line bundle over $\A$. We can also apply this norm operations to homomorphisms of line bundles $\L\to \L'$, and furthermore, if an endomorphism $f:\L\to\L$ is given by multiplication by $b\in\B$, then the endomorphism $\mathrm{Nm}_{\B/\A}(f)$ of $\mathrm{Nm}_{\B/\A}(\L)$ is multiplication by the ordinary norm of $b$ in $\A$.
\item More generally, in \cite{Ferrand} Ferrand shows how to take an arbitrary $\B$-module $\M$ and construct its ``norm'' $\mathrm{Nm}_{\B/\A}(M)$ as an $\A$-module. In the special case that $\M$ is a line bundle this functor agrees with the above definition of $\mathrm{Nm}_{\B/\A}(M)$, but in general if $\M$ is locally free it does not follow that $\mathrm{Nm}_{\B/\A}(\M)$ is locally free.
\end{itemize}
In each case, there is a group or monoid operation that the trace or norm preserves. The ordinary trace preserves addition: $\mathrm{Tr}_{\B/\A}(b+b') = \mathrm{Tr}_{\B/\A}(b) + \mathrm{Tr}_{\B/\A}(b')$, while the ordinary norm preserves multiplication $\mathrm{Nm}_{\B/\A}(bb') = \mathrm{Nm}_{\B/\A}(b)\mathrm{Nm}_{\B/\A}(b')$. On the other hand, the norm for line bundles and other modules is a functor that preserves tensor products up to isomorphism: $\mathrm{Nm}_{\B/\A}(\M\otimes_\B \M') \cong \mathrm{Nm}_{\B/\A}(\M)\otimes_\A \mathrm{Nm}_{\B/\A}(\M')$.
In case the map of schemes $\pi: \mathrm{Spec}(\B)\to\mathrm{Spec}(\A)$ is \'etale of degree $n$ (meaning that after a faithfully-flat finite-presentation base change it becomes a trivial degree-$n$ cover of the form $\coprod_{i=1}^n \mathrm{Spec}(\A)\to\mathrm{Spec}(\A)$), then for any sheaf $\G$ of abelian groups on the big \'etale site over $\mathrm{Spec}(\A)$ we have a ``trace'' homomorphism $\pi_\ast\pi^\ast \G \to \G$, for which taking sections over $\mathrm{Spec}(\A)$ gives us a homomorphism $\G(\B)\to\G(\A)$. If $\G$ is the additive group $\mathbb{G}_{a, \A}$, represented by the scheme $\mathrm{Spec}(\A[x])$, then this homomorphism is just the ordinary trace map $\B\to\A$. If $\G$ is the multiplicative group $\mathbb{G}_{m,\A}$ represented by $\mathrm{Spec}(\A[x,x^{-1}])$, then we get the norm map on units $\B^\times\to\A^\times$. It is not hard to extend the argument in \cite[Lemma V.1.12]{Milne} to sheaves of commutative monoids, giving us the full multiplicative norm map $\B\to\A$.
Meanwhile, for the norm of line bundles, one way to think about what is going on in the case of $\A\to\B$ \'etale is that line bundles are torsors for $\mathbb{G}_m$, and so we are applying $H^1(\mathrm{Spec}(\A), \cdot)$ to the ``trace'' homomorphism $\pi_\ast\pi^\ast \mathbb{G}_{m,\A}\to\mathbb{G}_{m,\A}$, giving a function
\begin{align*}
H^1(\mathrm{Spec}(\A), \pi_\ast\pi^\ast \mathbb{G}_{m,\A}) &\to H^1(\mathrm{Spec}(\A), \mathbb{G}_{m,\A})\\
\text{i.e. }H^1(\mathrm{Spec}(\B), \mathbb{G}_{m,\B}) &\to H^1(\mathrm{Spec}(\A), \mathbb{G}_{m,\A})
\end{align*}
sending the isomorphism class of a line bundle over $\B$ to the isomorphism class of its norm as a line bundle over $\A$. This, however, forgets the functorial nature of the norm operation on line bundles. Meanwhile, Ferrand's extension of the norm functor to all $\B$-modules is defined in terms of a universal property ($\mathrm{Nm}_{\B/\A}(\M)$ is the universal $\A$-module equipped with a homogeneous degree-$n$ polynomial law $\M\to\mathrm{Nm}_{\B/\A}(\M)$) and does not readily admit a cohomological interpretation even in the \'etale case.
However, in \cite{Waterhouse} Waterhouse uses the cohomological method to define a ``trace'' operation for \'etale quadratic algebras, since they are parameterized by $S_2$-torsors. Explicitly, if $\A\to\B$ is a rank-$n$ \'etale algebra in which $2$ is a unit, we can describe this ``trace'' operation as sending each \'etale quadratic $\B$-algebra of the form $\B[\sqrt{d}]$, with $d\in\B^\times$, to the \'etale quadratic $\A$-algebra $\A\left[\sqrt{\mathrm{Nm}_{\B/\A}(d)}\right]$. It is the goal of the present paper to extend this construction to a norm functor from the category of quadratic $\B$-algebras to the category of quadratic $\A$-algebras, regardless of whether any of the algebras are \'etale or whether $2$ is a unit. This norm operation is defined for quadratic algebras with a chosen generator in \cref{def:norm-based}, made functorial in \cref{prop:different-generator}, and extended to arbitrary quadratic algebras in \cref{def:norm}.
The main theorems about this norm functor for quadratic algebras are:
\begin{enumerate}
\item It commutes with base change (\cref{thm:base-change}).
\item It is ``transitive'' in the sense that for a tower of algebras $\A\to\B\to\C$, taking the norm of a quadratic $\C$-algebra to get a quadratic $\B$-algebra, then taking the norm again to get a quadratic $\A$-algebra, gives the same result as regarding $\C$ as an $\A$-algebra and applying the norm operation once (\cref{thm:transitive}).
\item It extends the notion of trace of an $S_2$-torsor in the case that the algebras are \'etale (\cref{thm:etale-addition}).
\item For the unique extension of $S_2$-torsor addition to a monoid operation $\ast$ on all quadratic algebras characterized by Voight in \cite{Voight}, the norm functor is a monoid homomorphism with respect to $\ast$ (\cref{thm:homomorphism}).
\item Taking the norm of a quadratic algebra commutes with taking its determinant line bundle and discriminant bilinear form (\cref{thm:norm-det} and \cref{thm:norm-disc}).
\end{enumerate}
However, it is still an open question whether this notion of the norm of a quadratic algebra is compatible with the notion of discriminant algebra from \cite{BieselGioia} in the sense of \cite{Waterhouse}; see \cref{conj:norm-discalg}.
Acknowledgements: The author would like to thank John Voight and Asher Auel for careful reading and helpful comments on an early draft, and Marius Stekelenburg for a much simpler proof of \cref{lem:transitive-sn}.
\section{Finite-rank and Quadratic Algebras}
We begin with background and notation for concepts related to rank-$n$ algebras in general and quadratic algebras in particular. In the following, all rings and algebras are commutative and unital, and are denoted by calligraphic capital letters: $\A, \B,\C,\D,\dots$.
\begin{definition}
Let $\A$ and $\B$ be rings and $\A\to\B$ a ring homomorphism, so that $\B$ is an $\A$-algebra. We say that $\B$ is a \emph{rank-$n$ $\A$-algebra} if $\B$ is projective of constant rank $n$ as an $\A$-module; equivalently, if there exist $a_1,\dots,a_k\in\A$ together generating the unit ideal such that each localization $\B_{a_i}$ is isomorphic to $\A_{a_i}^{\oplus n}$ as $\A_{a_i}$-modules.
\end{definition}
\begin{definition}\label{def:quadalg}
A \emph{quadratic $\A$-algebra} $\D$ is an $\A$-algebra of rank $2$. If $\D$ is equipped with a choice of algebra generator $x$, so that $\D\cong \A[x]/(x^2 - tx + n)$ for some elements $t,n\in\A$ (the \emph{trace} and \emph{norm} of $x\in\D$), then we call $\D$ a \emph{based} quadratic algebra. (Not every quadratic algebra admits a singleton generating set, but every quadratic algebra does so locally.) As a based quadratic algebra is determined up to unique isomorphism by the ordered pair $(t,n)$, we follow \cite{Loos} in using the notation
\[\quadalg[\A]{t}{n} \coloneqq \A[x]/(x^2 - tx + n)\]
for based quadratic algebras (the subscript ring may be omitted if it is clear from context). For example, $\quadalg[\A]{1}{0} = \A[x]/(x^2 - x) \cong \A\times\A$ with generating element $(1,0)$, and $\quadalg[\A]{0}{0}\cong \A[\varepsilon]/(\varepsilon^2)$ with generator $\varepsilon$.
\end{definition}
In \cref{def:quadalg}, we referred to the trace and norm of an element. We will not continue to need the notion of trace of a general element of a rank-$n$ algebra, but we will make frequent use of elements' norms. Because there are so many notions of ``norm'' we will be using in this paper, we will reserve the notation $\mathrm{Nm}_{\B/\A}$ for norm functors, applied to mathematical objects such as line bundles and other modules, and soon quadratic algebras. To refer to the ordinary norm function for a rank-$n$ algebra $\A\to\B$, we use the notation $s_n$ as defined below:
\begin{definition}\label{def:s_n}
Let $\A\to\B$ be a rank-$n$ algebra. Then there is a canonical function $s_n\colon \B\to\A$, called the \emph{norm}, defined as follows:
\begin{enumerate}
\item If $\B$ has a basis as an $\A$-module, and if $b$ is an element of $\B$, then its norm $s_n(b)$ is the determinant of the map $\B\to\B$ given by multiplication by $b$ (which is independent of choice of basis).
\item If $\B$ is a general rank-$n$ $\A$-algebra, and if $b\in\B$, then we can define the norm of $b$ in any localization $\B_a$ that is free as an $\A_a$-module. These values of $s_n(b)$ in each such $\A_a$ glue to give a single well-defined value for $s_n(b)\in \A$.
\end{enumerate}
\end{definition}
\begin{remark} The norm function $s_n\colon \B\to\A$ is not generally a ring homomorphism, but rather has the property that if $a\in\A$ is any element of the base ring, then $s_n(ab) = a^n s_n(b)$. Furthermore, the calculation of $s_n(b)$ commutes with base change, making $s_n$ an example of a \emph{polynomial law} in the sense of \cite{Roby}. For us, the upshot of $s_n$ being a polynomial law is that it automatically comes with polarized versions $s_{k_1,k_2,\ldots, k_m}(b_1,b_2,\ldots b_m)$ for each list of natural numbers $k_1+\dots+k_m = n$, defined by
\[s_{k_1,k_2,\ldots, k_m}(b_1,b_2,\ldots b_m) \coloneqq \begin{array}{cc}
\text{the coefficient of }\lambda_1^{k_1}\lambda_2^{k_2}\cdots\lambda_{m}^{k_m}\text{ in }\\s_n(\lambda_1 b_1+\lambda_2 b_2 + \ldots + \lambda_m b_m)\end{array},\]
the latter of which is calculated via the norm map of the base changed rank-$n$ algebra $\A[\lambda_1,\dots,\lambda_m]\to\B[\lambda_1,\dots,\lambda_m]$.
Phrased another way, the $s_{k_1,\dots,k_m}$ are defined so that we have the following identity:
\[s_n(\lambda_1b_1+\ldots+\lambda_mb_m) = \sum_{\substack{k_1,\dots,k_m\in\mathbb{N}\\
k_1+\ldots+k_m = n}}\lambda_1^{k_1}\dots \lambda_m^{k_m} s_{k_1,\dots,k_m}(b_1,\dots,b_m).\]
For example, the coefficients of the characteristic polynomial of an element $b\in\B$ can be expressed in terms of the polarized forms of the norm function:
\begin{align*}
s_n(\lambda - b) &= \left.\vphantom{\sum}s_n(\lambda 1 + \mu b)\right|_{\mu=-1}\\
&= \left.\sum_{k=0}^n \lambda^k\mu^{n-k} s_{k,n-k}(1,b)\right|_{\mu=-1}\\
&= \sum_{k=0}^n \lambda^k (-1)^{n-k} s_{k, n-k}(1,b).
\end{align*}
In particular, the \emph{trace} of $b\in \B$ can be calculated as the trace of the matrix representing multiplication by $b$ in any localization making $\B$ a free $\A$-module, or it can be expressed as the polarized form $s_{1,n-1}(b, 1)$.
\end{remark}
We now prove some of the basic results about the norm functions $s_n$ that we will need later:
\begin{lemma}\label{lem:multiplicative-sn}
The norm functions are multiplicative: If $\A\to\B$ is a rank-$n$ algebra and $b,b'\in\B$, then $s_n(bb') = s_n(b)s_n(b')$ in $\A$.
\end{lemma}
\begin{proof}
We can check locally, so assume that $\B$ has an $\A$-basis. Then the $n\times n$ matrix by which multiplication by $bb'$ acts, with respect to this basis, is the product of the matrices by which $b$ and $b'$ act. Taking determinants of both sides, and since the determinant is multiplicative, we obtain $s_n(bb') = s_n(b)s_n(b')$.
\end{proof}
\begin{lemma}\label{lem:transitive-sn}
The norm functions are transitive: Let $\A\to\B$ be a rank-$n$ algebra and $\B\to\C$ be a rank-$m$ algebra, so that $\C$ is also a rank-$mn$ algebra. For all $c\in\C$, we have $s_n(s_m(c)) = s_{mn}(c)$ as elements of $\A$.
\end{lemma}
\begin{proof}
We can work locally on $\A$, so assume that $\B$ is free as an $\A$-module. By \cite[Prop.\ 6.1.12 on p.\ 113]{EGAII}, since $\A\to\B$ is finite we can also assume that $\C$ is free as a $\B$-module by localizing $\A$. So assume that $\B$ is free of rank $n$ as an $\A$-module and that $\C$ is free of rank $m$ as a $\B$-module.
Choose an $\A$-basis $\theta_1,\dots,\theta_n$ for $\B$ and a $\B$-basis $\phi_1,\dots,\phi_m$ for $\C$.
First consider the $m\times m$ matrix $(M_{ij})_{i,j}$ with elements in $\B$ representing multiplication by $c$ with respect to basis $\phi_1,\dots,\phi_m$, i.e.\
\[c\phi_i = \sum_{j=1}^m M_{ij}\phi_j.\]
Then for each $M_{ij}\in\B$, consider the $n\times n$ matrix $(P_{ijk\ell})_{k\ell}$ representing multiplication by $M_{ij}$ with respect to basis $\theta_1,\dots,\theta_n$, i.e.\
\[M_{ij}\theta_k = \sum_{\ell=1}^n P_{ijk\ell} \theta_\ell.\]
Then the $mn\times mn$ matrix $(P_{ijk\ell})_{(i,k), (j,\ell)}$, with rows and columns indexed by $\{1,\dots,m\}\times\{1,\dots,n\}$, represents multiplication by $c$ with respect to the $\A$-basis $\phi_1\theta_1, \phi_2\theta_1, \dots, \phi_m\theta_n$ for $\C$:
\begin{align*}
c \phi_i\theta_k &= \sum_{j=1}^m M_{ij}\phi_j\theta_k\\
&= \sum_{j=1}^m \sum_{\ell=1}^n P_{ijk\ell}\phi_j\theta_\ell\\
&= \sum_{(j,\ell)\in\{1,\dots,m\}\times\{1,\dots,n\}} P_{ijk\ell} \phi_j\theta_\ell.
\end{align*}
We can regard $(P_{ijk\ell})_{(i,k), (j,\ell)}$ as an $m\times m$ block matrix, where the $n\times n$ blocks that represent multiplication by $M_{ij}$ all commute with each other (since the elements $M_{ij}\in\B$ all commute, as $\B$ is a commutative algebra). Then by \cite{Kovacs}, the determinant of $(P_{ijk\ell})_{(i,k), (j,\ell)}$ is equal to the determinant of the $n\times n$ matrix given by formally applying the $m\times m$ determinant formula to the blocks of $(P_{ijk\ell})_{(i,k), (j,\ell)}$. In other words,
\begin{align*}
s_{mn}(c) &= s_n(\mathrm{det}((M_{ij})_{i,j}))= s_n(s_m(c)).\qedhere
\end{align*}
\end{proof}
Here is a collection of results about the polarized forms of $s_n$:
\begin{lemma}\label{lem:polarized-identity}
Let $\A\to\B$ be a rank-$n$ algebra, let $n = k_1+\ldots+k_m$ be a partition of $n$, and let $b_1,\dots,b_m\in\B$. Then the following identities hold:
\begin{enumerate}
\item (Reordering) If $\sigma:\{1,\dots,m\}\to\{1,\dots,m\}$ is any permutation, then
\[s_{k_1,\dots,k_m}(b_1,\dots,b_m) = s_{k_{\sigma(1)},\dots,k_{\sigma(m)}}(b_{\sigma(1)},\dots,b_{\sigma(m)}).\]
\item (Combination) If $b_1 = b_2$, then
\[s_{k_1,\dots,k_m}(b_1,\dots,b_m) = \binom{k_1+k_2}{k_1} s_{k_1+k_2,k_3,\dots,k_m}(b_1,b_3,\dots,b_m).\]
\item (Homogeneity) If $a\in\A$, then
\[s_{k_1,\dots,k_m}(ab_1,\dots,b_m) = a^{k_1} s_{k_1,\dots,k_m}(b_1,\dots,b_m).\]
\item (Degeneracy) If $k_1=0$, then
\[s_{k_1,k_2,\dots,k_m}(b_1,b_2,\dots,b_m) = s_{k_2,\dots,k_m}(b_2,\dots,b_m).\]
\item (Multiplicativity) If $b'\in\B$, then
\[ s_{k_1,\dots,k_m}(b'b_1,\dots,b'b_m) = s_n(b') s_{k_1,\dots,k_m}(b_1,\dots,b_m).\]
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item (Reordering) This follows from comparing the coefficients of $\lambda_1^{k_1}\dots \lambda_m^{k_m} = \lambda_{\sigma(1)}^{k_{\sigma(1)}}\dots \lambda_{\sigma(m)}^{k_{\sigma(m)}}$ in $s_n(\lambda_1b_1+\dots+\lambda_m b_m) = s_n(\lambda_{\sigma(1)}b_{\sigma(1)} + \dots + \lambda_{\sigma(m)}b_{\sigma(m)})$.
\item (Combination) The term $s_{k_1,\dots,k_m}(b_1,\dots,b_m)$ is by definition the coefficient of $\lambda_1^{k_1}\dots \lambda_m^{k_m}$ in
\begin{align*}
s_n(\lambda_1 b_1 + \ldots + \lambda_m b_m) &= s_n\bigl((\lambda_1+\lambda_2)b_1+\ldots + \lambda_m b_m\bigr)\\
&= \sum_{\substack{\ell, \ell_3, \dots, \ell_m\in \mathbb{N}:\\ \ell+\ell_3+\ldots+\ell_m = n}} (\lambda_1+\lambda_2)^\ell \lambda_3^{\ell_3}\dots\lambda_m^{\ell_m} s_{\ell, \ell_3, \ldots, \ell_m}(b_1,b_3,\dots,b_m),
\end{align*}
in which the coefficient of $\lambda_1^{k_1}\lambda_2^{k_2}\dots \lambda_m^{k_m}$ is $\binom{k_1+k_2}{k_1} s_{k_1+k_2,k_3,\dots,k_m}(b_1,b_3,\ldots,b_m)$, as desired.
\item (Homogeneity) The quantity $s_{k_1,\dots,k_m}(ab_1,\dots,b_m)$ is the coefficient of $\lambda_1^{k_1}\dots\lambda_m^{k_m}$ in
\[s_n(a\lambda_1b_1+\lambda_2b_2+\dots+\lambda_m b_m) = \sum_{\substack{\ell_1,\dots,\ell_m\in\mathbb{N}\\\ell_1+\ldots+\ell_m = n}} (a\lambda_1)^{\ell_1}\lambda_2^{\ell_2}\dots \lambda_m^{\ell_m} s_{\ell_1,\dots,\ell_m}(b_1,\dots,b_m),\]
so the coefficient of $\lambda_1^{k_1}\lambda_2^{k_2}\dots \lambda_m^{k_m}$ is $a^{k_1}s_{k_1,\dots,k_m}(b_1,\dots,b_m)$, as desired.
\item (Degeneracy) If $k_1 = 0$, then we are looking for the coefficient of $\lambda_2^{k_2}\dots\lambda_m^{k_m}$ in $s_n(\lambda_1b_1 + \dots + \lambda_m b_m)$, which since it does not involve $\lambda_1$ is unchanged if we base change to give $\lambda_1$ a concrete value. In particular, we can base change along $\A[\lambda_1,\dots,\lambda_m]\to\A[\lambda_2,\dots,\lambda_m] : \lambda_1\mapsto 0$, giving us the coefficient of $\lambda_2^{k_2}\dots\lambda_m^{k_m}$ in $s_n(\lambda_2b_2 + \dots + \lambda_m b_m)$, namely, $s_{k_2,\dots,k_m}(b_2,\dots,b_m)$.
\item We have that $s_{k_1,\dots,k_m}(b'b_1,\dots,b'b_m)$ is the coefficient of $\lambda_1^{k_1}\dots \lambda_m^{k_m}$ in
\[s_n(\lambda_1b'b_1 + \ldots +\lambda_m b' b_m) = s_n(b') s_n(\lambda_1 b_1 + \ldots + \lambda_m b_m)\]
by \cref{lem:multiplicative-sn}.
The coefficient of $\lambda_1^{k_1}\dots \lambda_m^{k_m}$ is therefore $s_n(b') s_{k_1,\dots,k_m}(b_1,\dots,b_m)$.
\end{enumerate}
\end{proof}
Now we consider the various senses in which rank-$n$ algebras over $\A$ form not just a set of isomorphism classes, but a category:
\begin{definition}
There are various levels of strictness for homomorphisms of rank-$n$ $\A$-algebras. The weakest is that of ordinary $\A$-algebra homomorphism. A stronger notion that uses the rank-$n$ structure is that of a \emph{norm-preserving} homomorphism, which is an $\A$-algebra homomorphism $\B\to\B'$ that, together with the norm functions $s_n$, makes the resulting triangle of functions
\[\begin{array}{ccccc}
\B & & \to & & \B'\\
& \searrow & & \swarrow\\
& & A & &
\end{array}\]
commute. (For example, if $\A$ is any nonzero ring, the $\A$-algebra homomorphism $\A\times\A \to\A\times\A$ sending $(a,a')$ to $(a,a)$ is not norm-preserving, since $(1,0)$ has norm $0$ and $(1,1)$ has norm $1$.)
There is a stronger notion of \emph{universally norm-preserving} homomorphism, which is a norm-preserving $\A$-algebra homomorphism that remains so after arbitrary base change. Isomorphisms of rank-$n$ algebras are always universally norm-preserving, though the converse is not true.
\end{definition}
However, we can say more in case our algebras are quadratic:
\begin{lemma}
Let $\A$ be a ring and $\quadalg[\A]{t}{n}$ and $\quadalg[\A]{t'}{n'}$ be two based quadratic algebras. For a homomorphism $f:\quadalg[\A]{t'}{n'}\to\quadalg[\A]{t}{n}$ sending $x\mapsto ux+c$, the following are equivalent:
\begin{enumerate}
\item $f$ is universally norm-preserving,
\item $f$ is norm-preserving,
\item $t' = ut+2c$ and $n' = u^2n + uct + c^2$.
\end{enumerate}
\end{lemma}
In particular, since quadratic algebras can all locally be given a based structure, all norm-preserving homomorphisms of quadratic algebras are universally norm-preserving.
\begin{proof}
$(1)\implies (2)$ is trivial. For $(2)\implies(3)$, suppose that $f$ is norm-preserving. Then the norms of $x\in\quadalg{t'}{n'}$ and $f(x)=ux+c\in\quadalg[\A]{t}{n}$ are equal; namely
\begin{align*}
n' &= s_2(ux+c)\\
&= s_2(ux) + s_{1,1}(ux,c) + s_2(c)\\
&= u^2 s_2(x) + uc s_{1,1}(x,1) + c^2 s_2(1)\\
&= u^2 n + uct + c^2.
\end{align*}
Now note that in any quadratic algebra, the trace of any element $d$ is $s_{1,1}(d,1) = s_2(d+1) - s_2(d) - 1$, so norm-preserving maps are also trace-preserving. Therefore the traces of $x\in\quadalg{t'}{n'}$ and $f(x)=ux+c\in\quadalg[\A]{t}{n}$ are also equal:
\begin{align*}
t' &= s_{1,1}(ux+c,1)\\
&= u s_{1,1}(x, 1) + c s_{1,1}(1,1)\\
&= ut + 2c.
\end{align*}
Finally, for $(3)\implies(1)$ suppose that $t'=ut+2c$ and $n' = u^2n + uct + c^2$, so that $p_{ux+c}(\lambda) = \lambda^2 - t'\lambda+n'$ is the characteristic polynomial of $ux+c\in\quadalg{t}{n}$. But for any element $b$ of any rank-$n$ $\A$-algebra $\B$, the homomorphism $\A[x]/(p_b(x))\to\B$ sending $x$ to $b$ is universally norm-preserving by \cite{BieselGioia}.
\end{proof}
\begin{remark}
In particular, if $\quadalg[\A]{t}{n}$ is isomorphic to $\quadalg[\A]{t'}{n'}$, there must exist $u,c\in\A$ with $u$ a unit, such that $t'=ut+2c$ and $n'=u^2t+utc+c^2$.
\end{remark}
\section{Norms of Based Quadratic Algebras}
We will now be considering rank-$n$ algebras $\A\to\B$ and the relationships between quadratic algebras over $\A$ and over $\B$. As a general convention for ease of context, we will tend to use lowercase letters like $s, t, m,u,c,\dots$ for elements of $\A$, and capital letters like $S, T, M, U, C,\dots$ for elements of $\B$.
As a bit of motivation for the definition of the norm of a quadratic algebra, suppose we have a rank-$n$ algebra $\A\to\B$ and a based quadratic $\B$-algebra $\quadalg[\B]{T}{N}$. We want the discriminant of the norm of $\quadalg{T}{N}$ to be the norm of the discriminant of $\quadalg[\B]{T}{N}$, namely $s_n(T^2 - 4N)$. So if we have $\mathrm{Nm}_{\B/\A}\quadalg[\B]{T}{N} = \quadalg[\A]{t}{m}$, we had better have the identity
\[t^2 - 4m = s_n(T^2 - 4N).\]
Is there a canonical way to write $s_n(T^2 - 4N)$ as a square minus a multiple of four? Yes: we can use the polarized forms of $s_n$ to expand out $s_n(T^2 - 4N)$:
\begin{align*}
s_n( - 4N + T^2) &= \sum_{k=0}^n s_{k, n-k}(-4N, T^2)\\
&= \sum_{k=0}^n (-4)^k s_{k, n-k}(N, T^2)\\
&= s_n(T^2) + \sum_{k=1}^n (-4)^k s_{k, n-k}(N, T^2),\\
\intertext{since $s_{0,n}(N, T^2) = s_n(T^2)$ by \cref{lem:polarized-identity} (degeneracy),}
&= \bigl[s_n(T)\bigr]^2 - 4\left[\sum_{k=1}^n (-4)^{k-1} s_{k, n-k}(N, T^2)\right],
\end{align*}
where we have used \cref{lem:multiplicative-sn} to rewrite $s_n(T^2)$ as $s_n(T)^2$. If we then let $t = s_n(T)$ and $m = \sum_{k=1}^n (-4)^{k-1} s_{k, n-k}(N, T^2)$, we have $t^2 - 4m = s_n(T^2 - 4N)$ as desired, and $\quadalg[\A]{t}{m}$ is a good candidate for the norm of $\quadalg[\B]{T}{N}$. That is indeed the definition we will use for the norm of a based quadratic algebras:
\begin{definition}\label{def:norm-based}
Let $\B$ be a rank-$n$ $\A$-algebra, and let $\quadalg[\B]TN$ be a based quadratic $\B$-algebra. Then we define the \emph{norm} of $\quadalg[\B]{T}{N}$ to be the based quadratic $\A$-algebra
\[\mathrm{Nm}_{\B/\A}\quadalg[\B]{T}{N} \coloneqq \bigquadalg[\A]{s_n(T)}{\sum_{k=1}^n (-4)^{k-1}s_{k, n-k}(N, T^2)}.\]
Noting that over the polynomial ring $\A[\lambda]$ we have
\[s_n(\lambda N + T^2) = \sum_{k=0}^n \lambda^k s_{k, n-k}(N, T^2),\]
and therefore
\[\sum_{k=1}^n \lambda^{k-1} s_{k, n-k}(N, T^2) = \frac{s_n(\lambda N + T^2) - s_n(T^2)}{\lambda},\]
we will sometimes write the norm of $\quadalg[\B]{T}{N}$ as
\[\mathrm{Nm}_{\B/\A}\quadalg[\B]{T}{N} = \bigquadalg[\A]{s_n(T)}{\left.\frac{s_n(\lambda N + T^2) - s_n(T^2)}{\lambda}\right|_{\lambda=-4}}\]
to denote that the norm entry can be calculated as that fraction in $\A[\lambda]$ and then evaluated in $\A$ along the map $\A[\lambda]\to\A:\lambda\mapsto -4$.
\end{definition}
\begin{example}
Let $\A$ be a ring in which $2$ is a unit, such as a number field. Then by completing the square, all based quadratic algebras over $\A$ can be written in the form $\A[\sqrt{d}] = \A[x]/(x^2-d) = \quadalg[\A]{0}{-d}$. What are the norms of based quadratic algebras of this type? If we have a rank-$n$ algebra $\A\to \B$ (such as a degree-$n$ extension of number fields), and a quadratic $\B$-algebra $\B[\sqrt{D}]$, then its norm is the quadratic $\A$-algebra
\begin{align*}
\mathrm{Nm}_{\B/\A}\quadalg[\B]{0}{-D} &= \bigquadalg[\A]{s_n(0)}{\left.\frac{s_n(\lambda(-D) + (0)^2) - s_n((0)^2)}{\lambda}\right|_{\lambda=-4}}\\
&= \bigquadalg[\A]{0}{\left.\frac{s_n(-\lambda D)}{\lambda}\right|_{\lambda=-4}}\\
&= \quadalg[\A]{0}{\left.(-1)^n \lambda^{n-1} s_n(D)\right|_{\lambda=-4}}\\
&= \quadalg[\A]{0}{-4^{n-1} s_n(D)}\\
&= \A[\sqrt{4^{n-1} s_n(D)}] = \A[2^{n-1} \sqrt{s_n(D)}]\\
&\cong \A[\sqrt{s_n(D)}]
\end{align*}
since $2$ is a unit in $\A$. So the norm of $\B[\sqrt{D}]$ is $\A[\sqrt{s_n(D)}]$.
\end{example}
Next we have a few examples that begin to show how this norm operation is related to the monoid operation $\ast$ on quadratic algebras in \cite{Voight}, which is defined locally by
\begin{equation}\label{eq:ast}
\begin{gathered}
\A[x]/(x^2 - sx + m) \ast \A[x]/(x^2 - tx + n) \coloneqq \\
\A[x]/(x^2 - (st)x + (mt^2 + ns^2 - 4mn)),
\end{gathered}
\end{equation}
i.e.\ $\quadalg{s}{m} \ast \quadalg{t}{n} \coloneqq \quadalg{st}{mt^2 + ns^2 - 4mn}$. This monoid operation is (up to isomorphism) commutative and associative, with a two-sided identity given by the split algebra $\quadalg[\A]{1}{0}\cong \A\times \A$ and a zero (absorbing) element given by the degenerate algebra $\quadalg[\A]{0}{0} \cong \A[\varepsilon]/(\varepsilon^2)$.
\begin{example}
If $\A\to\B$ is an arbitrary rank-$n$ algebra, then we have $\mathrm{Nm}_{\B/\A}\quadalg[\B]{1}{0} = \quadalg[\A]{1}{0}$ and $\mathrm{Nm}_{\B/\A}\quadalg[\B]{0}{0} = \quadalg[\A]{0}{0}$. In other words, the norm of the split quadratic algebra $\B\times\B$ is again the split algebra $\A\times\A$, and the norm of the degenerate quadratic algebra $\B[\varepsilon]/(\varepsilon^2)$ is the degenerate algebra $\A[\varepsilon]/(\varepsilon^2)$. This is part of what it means for the norm functor on quadratic algebras to be a homomorphism of monoids-with-zero under $\ast$.
\end{example}
We will prove in \cref{thm:homomorphism} that the norm functor is truly a monoid homomorphism, and there we will use the following example showing how the monoid operation $\ast$ is actually a special case of the norm:
\begin{example}\label{ex:norm-product}
Given two based quadratic $\A$-algebras $\quadalg[\A]SM$ and $\quadalg[\A]TN$, we can consider their ordinary cartesian product as a based quadratic $\A\times\A$-algebra
\[\quadalg[\A]SM \times \quadalg[\A]TN \cong \quadalg[\A\times\A]{(S,T)}{(M,N)}.\]
Since $\A\to\A\times\A$ is an algebra of rank $n=2$, we can take the norm of this quadratic $\A\times\A$ algebra to get a quadratic $\A$-algebra:
\begin{align*}
\mathrm{Nm}\quadalg{(S,T)}{(M,N)} &= \quadalg{s_2((S,T))}{s_{1,1}((M,N), (S,T)^2) - 4 s_2((M,N))}
\end{align*}
To evaluate these entries we look at the matrices by which $(S,T)$ and $(M,N)$ act on $\A\times\A$ with respect to the standard basis $\{(1,0), (0,1)\}$. They act by diagonal matrices:
\[(S,T)\colon \begin{pmatrix}S & 0\\0 & T\end{pmatrix}\qquad (M,N)\colon \begin{pmatrix}M & 0 \\ 0 & N\end{pmatrix}\]
so we have $s_2((S,T)) = \mathrm{det} \begin{psmallmatrix}S & 0\\0 & T\end{psmallmatrix} = ST$ and $s_2((M,N)) = \mathrm{det} \begin{psmallmatrix}M & 0\\0 & N\end{psmallmatrix} = MN$.
We also need $s_{1,1}((M,N), (S,T)^2)$, which is the coefficient of $\lambda\mu$ in $s_2(\lambda(M,N) + \mu(S,T)^2)$, i.e. the coefficient of $\lambda\mu$ in
\begin{align*}
\mathrm{det}\begin{pmatrix} \lambda M + \mu S^2 & 0 \\ 0 & \lambda N + \mu T^2\end{pmatrix} &= (\lambda M + \mu S^2)(\lambda N + \mu T^2)\\
&= \lambda^2 MN + \lambda\mu(MT^2 + NS^2) + \mu^2 S^2T^2,
\end{align*}
so $s_{1,1}((M,N), (S,T)^2) = MT^2 + NS^2$.
Therefore the norm of $\quadalg{S}{M}\times\quadalg{T}{N}$ is
\begin{align*}
\mathrm{Nm}\quadalg{(S,T)}{(M,N)} &= \quadalg{s_2((S,T))}{s_{1,1}((M,N), (S,T)^2) - 4 s_2((M,N))}\\
&= \quadalg{ST}{MT^2 + NS^2 - 4MN},
\end{align*}
which is the composite quadratic algebra $\quadalg{S}{M} \ast \quadalg{T}{N}$ defined by the monoid operation $\ast$ in \cref{eq:ast}.
so the norm of a product of two quadratic algebras is their composite under the monoid operation $\ast$. Similarly, the norm of the quadratic $\A\times\A\times\A$-algebra $\quadalg{S}{M} \times \quadalg{T}{N}\times \quadalg{U}{P}$ would be the quadratic $\A$-algebra $\quadalg{S}{M} \ast \quadalg{T}{N}\ast \quadalg{U}{P}$, and so on.
\end{example}
Next we show that the norm operation is invariant under base change and is transitive along towers of algebras:
\begin{theorem}\label{thm:basechange-based}
The norm operation on based quadratic algebras commutes with base change: Let $\A\to\B$ be a rank-$n$ algebra, and let $\quadalg[\B]TN$ be a based quadratic $\B$-algebra. If $\C$ is any $\A$-algebra, then $\C\to\C\otimes_\A \B$ is another rank-$n$ algebra, and
\[\mathrm{Nm}_{(\C\otimes_\A \B)/\C} \quadalg{T}{N} \cong \C\otimes_\A \mathrm{Nm}_{\B/\A}\quadalg TN.\]
\end{theorem}
\begin{proof}
Since $s_n$ is a polynomial law, it and its polarized forms commute with base change, so the two based quadratic $\C$-algebras agree exactly.
\end{proof}
\begin{theorem}\label{thm:transitive-based}
The norm operation on based quadratic algebras is transitive: If $\A\to \B$ is a rank-$n$ algebra and $\B\to \C$ is a rank-$m$ algebra, and if $\quadalg[\C]{T}{N}$ is a based quadratic $\C$-algebra, then $\mathrm{Nm}_{\C/\A}(\quadalg[\C]{T}{N}) \cong \mathrm{Nm}_{\B/\A}(\mathrm{Nm}_{\C/\B}(\quadalg[\C]{T}{N}))$ as based quadratic $\A$-algebras.
\end{theorem}
\begin{proof}
We check that the traces and norms of the canonical generators of $\mathrm{Nm}_{\C/\A}\quadalg{T}{N}$ and $\mathrm{Nm}_{\B/\A}\mathrm{Nm}_{\C/\B}\quadalg{T}{N}$ agree. First, the traces of $x\in\mathrm{Nm}_{\B/\A}\mathrm{Nm}_{\C/\B}\quadalg{T}{N}$ and $x\in\mathrm{Nm}_{\C/\A}\quadalg{T}{N}$ are $s_n(s_m(T))= s_{mn}(T)$.
Second, the norm of $x\in\mathrm{Nm}_{\B/\A}\mathrm{Nm}_{\C/\B}\quadalg{T}{N}
$ is
\begin{gather*}
\left.\frac{s_n\left(\mu\left[\frac{s_m(\lambda N + T^2) - s_m(T^2)}{\lambda}\right]_{\lambda=-4}+s_m(T)^2\right) - s_n(s_m(T)^2)}{\mu}\right|_{\mu=-4},\\
\intertext{in which, since we are setting $\mu$ equal to $-4$ anyway, we might as well set $\lambda$ equal to $\mu$ instead of $-4$:}
\begin{aligned}
&= \left.\frac{s_n\left(\mu\left[\frac{s_m(\lambda N + T^2) - s_m(T^2)}{\lambda}\right]_{\lambda=\mu}+s_m(T)^2\right) - s_n(s_m(T)^2)}{\mu}\right|_{\mu=-4}\\
&= \left.\frac{s_n\left(\mu\left[\frac{s_m(\mu N + T^2) - s_m(T^2)}{\mu}\right]+s_m(T)^2\right) - s_n(s_m(T)^2)}{\mu}\right|_{\mu=-4}\\
&= \left.\frac{s_n\left([s_m(\mu N + T^2) - s_m(T^2)]+s_m(T)^2\right) - s_n(s_m(T)^2)}{\mu}\right|_{\mu=-4}\\
&= \left.\frac{s_{mn}(\mu N + T^2) - s_{mn}(T^2)}{\mu}\right|_{\mu=-4},\\
\end{aligned}
\end{gather*}
which is the norm of $x$ in $\mathrm{Nm}_{\C/\A}\quadalg{T}{N}$, as desired.
\end{proof}
\cref{thm:transitive-based} may be understood as a kind of functoriality of the norm map with respect to the underlying algebras. In order to extend the norm operation to all quadratic algebras, we will also need a functoriality with respect to isomorphisms of based quadratic algebras. In \cref{prop:different-generator} we define how the norm map acts on general norm-preserving homomorphisms of quadratic algebras, and in \cref{prop:functorial} we prove that this norm map preserves composition of homomorphisms. The combinatorial argument at the heart of \cref{prop:different-generator} is the key result that makes this norm functor well-defined.
\begin{proposition}\label{prop:different-generator}
Let $\B$ be a rank-$n$ $\A$-algebra, and let $\quadalg TN$ and $\quadalg{T'}{N'}$ be based quadratic $\B$-algebras. If we have an isomorphism (resp.\ norm-preserving homomorphism) of $\B$-algebras
\[\begin{gathered}
f\colon \quadalg{T'}{N'} \to \quadalg TN\\
: x \mapsto Ux + C,
\end{gathered}\]
then we also have an isomorphism (resp.\ norm-preserving homomorphism)
\begin{equation}\label{eq:hom-image}
\begin{gathered}
\mathrm{Nm}_{\B/\A}(f)\colon \mathrm{Nm}_{\B/\A}\quadalg{T'}{N'} \to \mathrm{Nm}_{\B/\A}\quadalg TN\\
: x \mapsto s_n(U) x + \sum_{k=1}^n 2^{k-1} s_{k, n-k}(C, UT).
\end{gathered}
\end{equation}
We will also write the constant term above as $\left.\frac{s_n(\lambda C + UT) - s_n(UT)}{\lambda}\right|_{\lambda=2}$, to indicate that it is the image under the map $\A[\lambda]\to \A:\lambda\mapsto 2$ of the element $(s_n(\lambda C + UT) - s_n(UT))/\lambda$.
\end{proposition}
\begin{proof}
We prove the more general claim in the case that $f$ is a norm-preserving homomorphism, noting that if $f$ is an isomorphism, then $U$ is a unit in $\B$, so $s_n(U)$ is a unit in $\A$ and $\mathrm{Nm}_{\B/\A}(f)$ is an isomorphism too.
First let $\tilde T = UT$ and $\tilde N = U^2N$. We can then decompose the norm-preserving map $\quadalg{T'}{N'}\to\quadalg{T}{N}$ into two maps $\quadalg{T'}{N'}\to\quadalg{\tilde T}{\tilde N}$ sending $x\mapsto x+C$ and $\quadalg{\tilde T}{\tilde N}\to\quadalg{T}{N}$ sending $x\mapsto Ux$. We will show immediately after this proposition that the norm operation on norm-preserving homomorphisms is functorial; it therefore suffices to show that \cref{eq:hom-image} provides a norm-preserving homomorphism between the norm algebras in the cases that $C=0$ or $U=1$.
First consider the case $C=0$, so that we are considering the map $\quadalg{UT}{U^2 N}\to \quadalg{T}{N}$ sending $x\mapsto Ux$. We must show that the map $\mathrm{Nm}\quadalg{UT}{U^2N}\to \mathrm{Nm}\quadalg{T}{N}$ sending $x\mapsto s_n(U)x$ is also norm-preserving. Indeed, the trace of $x\in\mathrm{Nm}\quadalg{UT}{U^2N}$ is $s_n(UT) = s_n(U)s_n(T)$, which is also the trace of $s_n(U)x\in \mathrm{Nm}\quadalg{T}{N}$. And their norms match as well: the norm of $x\in \mathrm{Nm}\quadalg{UT}{U^2N}$ is
\begin{align*}
\sum_{k=1}^n (-4)^{k-1} s_{k, n-k}(U^2N, (UT)^2) &= \sum_{k=1}^n (-4)^{k-1} s_{k, n-k}(U^2N, U^2T^2)\\
&= \sum_{k=1}^n (-4)^{k-1} s_n(U^2) s_{k, n-k}(N,T^2)\\
&= s_n(U)^2\sum_{k=1}^n (-4)^{k-1} s_{k, n-k}(N,T^2),
\end{align*}
which is the norm of $s_n(U)x\in\mathrm{Nm}\quadalg{T}{N}$.
Now we consider the case $U=1$ but $C\neq 0$, so that we are considering the norm-preserving map $\quadalg{T+2C}{N+CT+C^2} \to \quadalg{T}{N}$ sending $x\mapsto x+C$, and we wish to show that the map $\mathrm{Nm}\quadalg{T+2C}{N+CT+C^2}\to\mathrm{Nm}\quadalg{T}{N}$ sending $x\mapsto x+\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C, T)$ is also norm-preserving. The trace of $x$ in $\mathrm{Nm}\quadalg{T+2C}{N+CT+C^2}$ is
\begin{align*}
s_n(T+2C) &= \sum_{k=0}^n s_{k, n-k}(2C, T)\\
&= \sum_{k=0}^n 2^k s_{k, n-k}(C, T)\text{ by homogeneity}\\
&= s_n(T) + 2\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C, T)
\end{align*}
which equals the trace of $x + \sum_{k=1}^n 2^{k-1} s_{k, n-k}(C, T)$ in $\mathrm{Nm}\quadalg{T}{N}$ as desired.
Comparing norms will be more difficult; we will show separately that the norm of $x$ in $\mathrm{Nm}\quadalg{T+2C}{N+CT+C^2}$ and the norm of $x+\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C, T)$ in $\mathrm{Nm}\quadalg{T}{N}$ are both equal to
\begin{equation}\label{eq:norm-translate}
\sum_{i=1}^n (-4)^{i-1} s_{i, n-i}(N, T^2) + \sum_{\substack{j,k\in \mathbb{N}:\\1 \leq j+k \leq n}} 4^{j+k-1} s_{j, k, n-j-k}(C^2, CT, T^2).
\end{equation}
Consider first the norm of $x$ in $\mathrm{Nm}\quadalg{T+2C}{N+CT+C^2}$. By definition it equals
\begin{align*}
&\sum_{m=1}^n (-4)^{m-1} s_{m, n-m}(N+CT+C^2, (T+2C)^2)\\
&= \sum_{m=1}^n (-4)^{m-1} s_{m, n-m}(N+CT+C^2, T^2+4CT+4C^2)\\
&= \sum_{m=1}^n(-4)^{m-1}\sum_{\substack{p,q,r,s,t,u\in\mathbb{N}\\p+q+r=m\\s+t+u=n-m}}s_{p,q,r,s,t,u}(N, C^2, CT, 4C^2, 4CT, T^2)\\
&= \sum_{m=1}^n(-4)^{m-1}\sum_{\substack{p,q,r,s,t,u\in\mathbb{N}\\p+q+r=m\\s+t+u=n-m}}4^{s+t}s_{p,q,r,s,t,u}(N, C^2, CT, C^2, CT, T^2)\\
&= \sum_{m=1}^n\sum_{\substack{p,q,r,s,t,u\in\mathbb{N}\\p+q+r=m\\s+t+u=n-m}}(-4)^{m-1}4^{n-m-u}\binom{q+s}{s}\binom{r+t}{t}s_{p,q+s,r+t,u}(N, C^2, CT, T^2)
\end{align*}
by \cref{lem:polarized-identity} (combination).
We may then ask what the coefficient is of a given $s_{i,j,k,\ell}(N, C^2, CT, T^2)$ in the last sum, for natural numbers $i+j+k+\ell=n$. That coefficient is
\begin{align*}
&\sum_{m=1}^n \sum_{\substack{q,r,s,t\in\mathbb{N}\\i+q+r=m\\s+t+\ell=n-m\\q+s=j\\r+t=k}}(-4)^{m-1}4^{n-m-\ell}\binom{j}{s}\binom{k}{t}\\
&= 4^{n-\ell-1}\sum_{m=1}^n (-1)^{m-1}\sum_{\substack{q,r,s,t\in\mathbb{N}\\q+r=m-i\\s+t=n-m-\ell\\q+s=j\\r+t=k}}\binom{j}{s}\binom{k}{t}
\end{align*}
That inner sum over $q,r,s,t\in\mathbb{N}$ is exactly counting the number of ways of forming a subset of size $m-i$ from a disjoint union of two sets of size $j$ and $k$: you must divide the set of size $j$ into sets of size $q$ and $s$ (for which there are $\binom{j}{s}$ possibilities), and the set of size $k$ into sets of size $r$ and $t$ (with $\binom{k}{t}$ possibilities), such that $q+r$ is the desired size $m-i$. Therefore that inner sum merely equals $\binom{j+k}{m-i}$, so the coefficient of $s_{i,j,k,\ell}(N,C^2,CT,T^2)$ is
\begin{align*}
&= 4^{n-\ell-1}\sum_{m=1}^n (-1)^{m-1} \binom{j+k}{m-i}\\
&= 4^{n-\ell-1}\sum_{m'=1-i}^{n-i} (-1)^{m'+i-1} \binom{j+k}{m'},
\end{align*}
where we have changed the index of summation to $m'=m-i$. Now that sum over $m'$ possibly includes several terms where the binomial coefficient vanishes because $m'$ is out of bounds. In particular, $n-i$ is always greater than or equal to $j+k$, so we might as well stop the sum at $m'=j+k$. Similarly, we may as well start the sum at $m'=0$, unless $i=0$, in which case the sum starts at $1$. So if $i>0$, the coefficient of $s_{i,j,k,\ell}(N, C^2, CT, T^2)$ is
\begin{align*}
&4^{n-\ell-1}(-1)^{i-1} \sum_{m'=0}^{j+k} (-1)^{m'}\binom{j+k}{m'}\\
&= 4^{n-\ell-1}(-1)^{i-1} (1-1)^{j+k}\\
&= \begin{cases}0 &\text{if }j+k > 0\\
4^{n-\ell-1}(-1)^{i-1} = (-4)^{i-1} &\text{if }j=k=0.
\end{cases}
\end{align*}
On the other hand, if $i=0$, the coefficient of $s_{0, j, k, \ell}(N,C^2, CT, T^2)$ is
\begin{align*}
&\qquad 4^{n-\ell-1}(-1)\sum_{m'=1}^{j+k} \binom{j+k}{m'}\\
&= -4^{n-\ell-1}[(1-1)^{j+k} - 1]\\
&= \begin{cases}
4^{n-\ell-1}=4^{j+k-1} &\text{if }j+k>0\\
0 &\text{if }j=k=0.
\end{cases}
\end{align*}
Putting it all together, and leaving out the terms with coefficient $0$, the norm of $x$ in $\mathrm{Nm}\quadalg{T+2C}{N+CT+C^2}$ is
\[\sum_{i=1}^n (-4)^{i-1} s_{i, n-i}(N, T^2) + \sum_{\substack{j,k\in\mathbb{N}\\1\leq j+k\leq n}} 4^{j+k-1}s_{j, k, n-j-k}(C^2, CT, T^2),\]
which matches \cref{eq:norm-translate} as desired.
Now we must show that this also agrees with the norm of $x+\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C, T)$ in $\mathrm{Nm}\quadalg{T}{N}$. This norm is
\begin{align*}
&\sum_{k=1}^n (-4)^{k-1} s_{k, n-k}(N, T^2)+ s_n(T)\left(\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C,T)\right) + \left(\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C,T)\right)^2.
\end{align*}
The first term agrees with that in \cref{eq:norm-translate}, so we must show that the second and third terms add up to $\sum_{j,k\in\mathbb{N}: 1\leq j+k\leq n} 4^{j+k-1}s_{j,k,n-j-k}(C^2, CT, T^2)$. We begin by expanding the third term as
\begin{align*}
\left(\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C,T)\right)^2 &= \sum_{k, \ell=1}^n 2^{k+\ell - 2} s_{k, n-k}(C, T)s_{\ell, n-\ell}(C, T).
\end{align*}
Now $s_{k, n-k}(C, T)$ is the coefficient of $\lambda^k$ in $s_n(\lambda C + T)$, so $s_{k, n-k}(C, T)s_{\ell, n-\ell}(C, T)$ is the coefficient of $\lambda^k\mu^\ell$ in
\begin{align*}
s_n(\lambda C + T)s_n(\mu C + T) &= s_n(\lambda\mu C^2 + (\lambda+\mu)CT + T^2)\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a + b + c = n}} s_{a,b,c}(\lambda\mu C^2, (\lambda+\mu)CT, T^2)\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a + b + c = n}} (\lambda\mu)^a(\lambda+\mu)^b s_{a,b,c}(C^2, CT, T^2)\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a + b + c = n}} \sum_{d=0}^b \lambda^{a+d}\mu^{a+b-d}\binom{b}{d}s_{a,b,c}(C^2, CT, T^2)
\end{align*}
whose coefficient of $\lambda^k\mu^\ell$ is therefore
\begin{align*}
s_{k,n-k}(C,T)s_{\ell,n-\ell}(C,T) &= \sum_{\substack{a,b,c,d\in\mathbb{N}\\a+b+c=n\\a+d=k\\a+b-d=\ell}} \binom{b}{d}s_{a,b,c}(C^2, CT, T^2)\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n\\k+\ell=2a+b}} \binom{b}{k-a}s_{a,b,c}(C^2, CT, T^2),
\end{align*}
so we have
\begin{align*}
\left(\sum_{k=1}^n 2^{k-1} s_{k, n-k}(C,T)\right)^2 &= \sum_{k, \ell=1}^n 2^{k+\ell - 2} s_{k, n-k}(C, T)s_{\ell, n-\ell}(C, T).\\
&= \sum_{k, \ell=1}^n 2^{k+\ell - 2} \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n\\k+\ell=2a+b}} \binom{b}{k-a}s_{a,b,c}(C^2, CT, T^2)\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n}} 2^{2a+b - 2} s_{a,b,c}(C^2, CT, T^2)\sum_{\substack{1\leq k, \ell \leq n\\k+\ell=2a+b}} \binom{b}{k-a}\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n}} 2^{2a+b - 2} s_{a,b,c}(C^2, CT, T^2)\sum_{k=1}^{ 2a+b-1} \binom{b}{k-a}\\
\end{align*}
Now the value of that inner sum breaks into three cases depending on how many terms are missing from the sum $2^b = \sum_j \binom{b}{j}$: either $a = 0$ and $b > 0$, in which case we are missing the start and the end terms, giving
\[\sum_{k=1}^{2a+b-1}\binom{b}{k-a} = \sum_{k=1}^{b-1}\binom{b}{k} = 2^b - 2,\]
or $a=0$ and $b=0$, in which case the sum is empty and equals $0$, or $a>0$, in which case neither end term is missing:
\[\sum_{k=1}^{2a+b-1}\binom{b}{k-a} = \sum_{k'=1-a}^{a+b-1}\binom{b}{k'}=\sum_{k'=0}^{b}\binom{b}{k'}= 2^b.\]
So our expansion of $(\sum_{k=1}^n 2^{k-1}s_{k,n-k}(C,T))^2$ becomes
\begin{align*}
&\qquad \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n}} 2^{2a+b - 2} s_{a,b,c}(C^2, CT, T^2)\sum_{k=1}^{ 2a+b-1} \binom{b}{k-a}\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n\\a+b\geq 1}} 2^{2a+b-2}(2^b) s_{a,b,c}(C^2, CT, T^2) - \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n\\a=0,b>0}} 2^{b-2} (2) s_{a,b,c}(C^2,CT, T^2)\\
&= \sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n\\a+b\geq 1}} 2^{2a+2b-2}s_{a,b,c}(C^2, CT, T^2) - \sum_{b=1}^n 2^{b-1}s_{b,n-b}(CT, T^2).
\end{align*}
Therefore the sum $(\sum_{k=1}^n 2^{k-1}s_{k,n-k}(C,T))^2 + \sum_{b=1}^n 2^{b-1}s_{b,n-b}(CT, T^2)$ equals
\[\sum_{\substack{a,b,c\in\mathbb{N}\\a+b+c=n\\a+b\geq 1}} 2^{2a+2b-2}s_{a,b,c}(C^2, CT, T^2) = \sum_{\substack{a,b\in\mathbb{N}\\1\leq a+b\leq n}} 4^{a+b-1}s_{a,b,c}(C^2, CT, T^2)\]
as desired. So the map $\mathrm{Nm}\quadalg{T+2C}{N+CT+C^2}\to\mathrm{Nm}\quadalg{T}{N}$ sending $x\mapsto x + \sum_{k=1}^n 2^{k-1}s_{k,n-k}(C,T)$ preserves traces and norms, so it is a norm-preserving homomorphism.
\end{proof}
\begin{example}\label{ex:swap}
For example, let $\A\to\B$ be a rank-$n$ algebra and let $f:\B^2\to\B^2$ be the ``swap'' map of quadratic algebras $\B$-algebras sending $(b, b')\mapsto (b',b)$. (In terms of based quadratic algebras, this is the map $\quadalg[\B]{1}{0}\to\quadalg[\B]{1}{0}$ sending $x\mapsto -x+1$.) If $n$ is odd, then $\mathrm{Nm}_{\B/\A}(f):\A^2\to\A^2$ is also the swap map; if $n$ is even it is the identity map.
\end{example}
\begin{proof}
Applying the norm map to $f\colon \quadalg[\B]{1}{0}\to\quadalg[\B]{1}{0}: x\mapsto -x+1$, we obtain
\begin{align*}
\mathrm{Nm}_{\B/\A}(f)\colon x&\mapsto s_n(-1)x + \left.\frac{s_n(\lambda - 1) - s_n(-1)}{\lambda}\right|_{\lambda=2}\\
&= (-1)^n x + \left.\frac{(\lambda-1)^n - (-1)^n}{(\lambda-1) - (-1)}\right|_{\lambda=2}\\
&= (-1)^n x + \left.(\lambda-1)^{n-1} + (-1)(\lambda-1)^{n-2} + \dots + (-1)^{n-1}\right|_{\lambda=2}\\
&= (-1)^n x + 1 - 1 + \ldots + (-1)^{n-1}.
\end{align*}
Now if $n$ is even, this works out to $x\mapsto 1x+0$, the identity map. And if $n$ is odd, this becomes $-x + 1$, the swap map $\A^2\to\A^2$.
\end{proof}
\begin{proposition}\label{prop:functorial}
The assignment $f\mapsto \mathrm{Nm}_{\B/\A}(f)$ in \cref{prop:different-generator} is functorial: if we have a commuting triangle of univerally norm-preserving homomorphisms between quadratic $\B$-algebras
\[\begin{array}{ccccc}
& & \quadalg{T'}{N'} & &\\
& \nearrow & & \searrow & \\
\quadalg{T''}{N''} & & \longrightarrow & & \quadalg{T}{N}
\end{array}\]
then the resulting triangle of norm-preserving $\A$-algebra homomorphisms
\[\begin{array}{ccccc}
& & \mathrm{Nm}_{\B/\A} \quadalg{T'}{N'} & &\\
& \nearrow & & \searrow & \\
\mathrm{Nm}_{\B/\A}\quadalg{T''}{N''} & & \longrightarrow & & \mathrm{Nm}_{\B/\A}\quadalg{T}{N}
\end{array}\]
also commutes.
\end{proposition}
\begin{proof}
Let us suppose that the homomorphism $\quadalg{T'}{N'} \to \quadalg{T}{N}$ sends $x\mapsto Ux+C$, so that $T'=UT+2C$. Suppose also that the homomorphism $\quadalg{T''}{N''}\to \quadalg{T'}{N'}$ sends $x\mapsto Vx+D$, so that the composite homomorphism $\quadalg{T''}{N''}\to\quadalg{T}{N}$ sends $x\mapsto V(Ux+C)+D = VUx + (VC+D)$. Then the composite homomorphism of norms
\[\mathrm{Nm}\quadalg{T''}{N''} \to \mathrm{Nm}\quadalg{T'}{N'} \to \mathrm{Nm}\quadalg{T}{N}\]
sends
\begin{align*}
x&\mapsto s_n(V)x +\left.\frac{s_n(\lambda D + VT') - s_n(VT')}{\lambda}\right|_{\lambda = 2}\\
&= s_n(V)s_n(U)x + s_n(V)\left.\frac{s_n(\lambda C + UT) - s_n(UT)}{\lambda}\right|_{\lambda = 2}+\left.\frac{s_n(\lambda D + VT') - s_n(VT')}{\lambda}\right|_{\lambda = 2}\\
&= s_n(VU)x + \left.\frac{s_n(\lambda VC + VUT) - s_n(VUT)}{\lambda}\right|_{\lambda = 2}+\left.\frac{s_n(\lambda D + VUT+2VC) - s_n(VUT + 2 VC)}{\lambda}\right|_{\lambda = 2}\\
\end{align*}
We can combine the two constant terms into one larger sum (replacing a couple of extra 2's by $\lambda$ since they will be set equal to $2$ again anyway) as follows:
\begin{align*}
&\left.\frac{s_n(\lambda VC + VUT) - s_n(VUT)}{\lambda}\right|_{\lambda = 2}+\left.\frac{s_n(\lambda D + VUT+2VC) - s_n(VUT + 2 VC)}{\lambda}\right|_{\lambda = 2}\\
&= \left.\frac{s_n(\lambda VC + VUT) - s_n(VUT) + s_n(\lambda D + VUT + \lambda VC) - s_n(VUT + \lambda VC)}{\lambda}\right|_{\lambda=2}\\
&= \left.\frac{ s_n(\lambda D + VUT + \lambda VC) - s_n(VUT)}{\lambda}\right|_{\lambda=2}\\
\end{align*}
So in all, the composite map of norms $\mathrm{Nm}\quadalg{T''}{N''}\to \mathrm{Nm}\quadalg{T'}{N'}\to \mathrm{Nm}\quadalg{T}{N}$ sends
\[x\mapsto s_n(VU) x + \left.\frac{ s_n(\lambda (VC+D) + VUT) - s_n(VUT)}{\lambda}\right|_{\lambda=2}\]
which is exactly the norm of the composite map $\quadalg{T''}{N''}\to\quadalg{T'}{N'}\to \quadalg{T}{N}$ sending $x\mapsto VUx + (VC+D)$, as desired.
\end{proof}
To show that the norm operation on general quadratic algebras is transitive, we will also need transitivity for the norm operation on homomorphisms:
\begin{proposition} \label{thm:transitive-morphism}The norm map on homomorphisms is also transitive: if $\A\to\B$ is a rank-$n$ algebra, and $\B\to\C$ is a rank-$m$ algebra, and $f:\quadalg{T'}{N'}\to\quadalg{T}{N}$ is a norm-preserving homomorphism of based quadratic $\C$-algebras, then
\[\mathrm{Nm}_{\B/\A}(\mathrm{Nm}_{\C/\B}(f)) = \mathrm{Nm}_{\C/\A}(f).\]
\end{proposition}
\begin{proof}
Let us suppose that $f:\quadalg{T'}{N'}\to\quadalg{T}{N}$ sends $x\mapsto Ux+C$. Then on the one hand, we have
\[ \mathrm{Nm}_{\C/\B}(f) \colon x\mapsto s_m(U)x +\left. \frac{s_m(\lambda C + UT) -s_m(UT)}{\lambda}\right|_{\lambda=2}\]
so, since $x\in\mathrm{Nm}_{\C/\B}\quadalg{T}{N}$ has trace $s_m(T)$, we have that $\mathrm{Nm}_{\B/\A}( \mathrm{Nm}_{\C/\B}(f))$ sends
\begin{align*}
x &\mapsto s_n(s_m(U))x +\left.\frac{s_n\left(\mu\left. \frac{s_m(\lambda C + UT) -s_m(UT)}{\lambda}\right|_{\lambda=2} + s_m(U)s_m(T)\right) - s_n(s_m(U)s_m(T))}{\mu}\right|_{\mu=2}\\
&= s_{mn}(U) x +\left.\frac{s_n\left(\mu\left. \frac{s_m(\lambda C + UT) -s_m(UT)}{\lambda}\right|_{\lambda=\mu} + s_m(UT)\right) - s_n(s_m(UT))}{\mu}\right|_{\mu=2}\\
&= s_{mn}(U) x +\left.\frac{s_n\left(\mu\frac{s_m(\mu C + UT) -s_m(UT)}{\mu} + s_m(UT)\right) - s_{mn}(UT)}{\mu}\right|_{\mu=2}\\
&= s_{mn}(U)x + \left.\frac{s_n\bigl(s_m(\mu C+ UT)- s_m(UT)+s_m(UT)\bigr) - s_{mn}(UT)}{\mu}\right|_{\mu=2}\\
&= s_{mn}(U)x + \left.\frac{s_{mn}(\mu C+ UT) - s_{mn}(UT)}{\mu}\right|_{\mu=2},
\end{align*}
which is exactly where $\mathrm{Nm}_{\C/\A}(f)$ sends $x$.
\end{proof}
\section{Norms of Quadratic Algebras}
We now have all the results we need in order to define the norm of a general quadratic algebra and prove that it is well-defined:
\begin{definition}\label{def:norm}
Given a rank-$n$ $\A$-algebra $\B$, we define the norm of a general quadratic $\B$-algebra $\D$ as follows:
\begin{enumerate}
\item Choose elements $a_1,\dots, a_k\in \A$, together generating the unit ideal, such that each $\D_{a_i}$ is free as a $\B_{a_i}$-module.
\item Choose generators for each $\D_{a_i}$ to obtain isomorphisms $\D_{a_i} \cong \quadalg[\B_{a_i}]{T_i}{N_i}$.
\item Over overlaps $\A_{a_ia_j}$, use \cref{prop:different-generator} to convert the isomorphisms $\quadalg[\B_{a_ia_j}]{T_i}{N_i} \cong \D_{a_ia_j}\cong \quadalg[\B_{a_ia_j}]{T_j}{N_j}$ into isomorphisms \[\mathrm{Nm}_{\B_{a_ia_j}/\A_{a_ia_j}}\quadalg[\B_{a_ia_j}]{T_i}{N_i} \cong \mathrm{Nm}_{\B_{a_ia_j}/\A_{a_ia_j}}\quadalg[\B_{a_ia_j}]{T_j}{N_j}.\]
\item Use these isomorphisms to glue the norms of the based quadratic $\A_{a_i}$-algebras $\mathrm{Nm}_{\B_{a_i}/\A_{a_i}} \quadalg{T_i}{N_i}$ into a single quadratic $\A$-algebra, called $\mathrm{Nm}_{\B/\A}(\D)$.
\end{enumerate}
\end{definition}
\begin{lemma}
The construction of $\mathrm{Nm}_{\B/\A}(\D)$ in \cref{def:norm} is well-defined.
\end{lemma}
\begin{proof}
There are two things to check: that the norms of the based quadratic algebras $\mathrm{Nm}_{\B_{a_i}/\A_{a_i}} \quadalg{T_i}{N_i}$ do, in fact, glue together to yield a quadratic $\A$-algebra, and that the resulting algebra is independent (up to isomorphism) of the choices of the $a_i$ and generators for $\D_{a_i}$.
First we check that the quadratic algebras do glue; this amounts to checking that over triple overlaps $\A_{a_ia_ja_k}$, the following triangle of isomorphisms commutes:
\[\begin{array}{ccccc}
& & \mathrm{Nm} \quadalg{T_j}{N_j} & &\\
& \nearrow & & \searrow & \\
\mathrm{Nm}\quadalg{T_i}{N_i} & & \longrightarrow & & \mathrm{Nm}\quadalg{T_k}{N_k}
\end{array}\]
But this is true by \cref{prop:functorial} because the corresponding triangle of quadratic $\B_{a_ia_ja_k}$-algebra homomorphisms commutes, since the isomorphisms between them come from chosen isomorphisms with $\D_{a_ia_ja_k}$:
\[\begin{array}{ccccc}
& & \quadalg{T_j}{N_j} & &\\
& \nearrow & & \searrow & \\
\quadalg{T_i}{N_i} & & \longrightarrow & & \quadalg{T_k}{N_k}
\end{array}\]
Second, we check independence of the choices of the $a_i$ and generators for $\D_{a_i}$: suppose that we had a second collection of elements $a_1', a_2', \ldots$ and isomorphisms $\D_{a_i'}\cong \quadalg{T_i'}{N_i'}$. Then we could combine the two collections into one and construct a common gluing of all the $\mathrm{Nm}\quadalg{T_i}{N_i}$ and $\mathrm{Nm}\quadalg{T_i'}{N_i'}$. In particular, the result will be isomorphic to the gluings obtained from each family separately, so the norm is independent of choice of generator.
\end{proof}
We can now boost our results on the norm operation for based quadratic algebras to this new, more general setting:
\begin{theorem}\label{thm:base-change}
The norm operation on general quadratic algebras commutes with base change: Let $\A\to\B$ be a rank-$n$ algebra, and let $\D$ be a quadratic $\B$-algebra. If $\C$ is any $\A$-algebra, then $\C\to\C\otimes_\A \B$ is another rank-$n$ algebra, and $\C\otimes_\A \D$ is a quadratic $\C\otimes_\A \B$-algebra with
\[\mathrm{Nm}_{(\C\otimes_\A \B)/\C} (\C\otimes_\A \D) = \C \otimes_\A \mathrm{Nm}_{\B/\A}(\D).\]
\end{theorem}
\begin{proof}
We already know by \cref{thm:basechange-based} that this holds locally, so we need only check that taking the norms of the gluing isomorphisms also commutes with base change. But the gluing isomorphisms are also built out of polarized forms of the polynomial law $s_n$, so this is again automatic.
\end{proof}
\begin{theorem}\label{thm:transitive}
The norm operation is transitive: If $\A\to \B$ is a rank-$n$ algebra and $\B\to \C$ is a rank-$m$ algebra, and if $\D$ is a quadratic $\C$-algebra, then $\mathrm{Nm}_{\C/\A}(\D) \cong \mathrm{Nm}_{\B/\A}(\mathrm{Nm}_{\C/\B}(\D))$.
\end{theorem}
\begin{proof}
We have already shown in \cref{thm:transitive-based} that this holds for norms of based quadratic algebras; it remains to show that the isomorphisms used to glue together the norms of based localizations also agree, but this holds by \cref{thm:transitive-morphism}.
\end{proof}
We can now use transitivity of the norm operation to show that $\mathrm{Nm}_{\B/\A}$ is a homomorphism with respect to the monoid operation $\ast$ on quadratic algebras:
\begin{theorem}\label{thm:homomorphism}
The norm operation is a homomorphism with respect to the monoid operation $\ast$ on isomorphism classes of quadratic algebras: If $\A\to \B$ is a rank-$n$ algebra and $\D$ and $\D'$ are quadratic $\B$-algebras, then $\mathrm{Nm}_{\B/\A}(\D\ast \D') \cong \mathrm{Nm}_{\B/\A}(\D) \ast \mathrm{Nm}_{\B/\A} (\D')$.
\end{theorem}
\begin{proof}
Consider the tower of algebras $\A \to \B \to \B^2$ and the quadratic $\B^2$-algebra $\D\times \D'$. We can take its norm to produce the quadratic $\A$-algebra
\begin{align*}
\mathrm{Nm}_{\B^2/\A}(\D\times \D') &\cong \mathrm{Nm}_{\B/\A}(\mathrm{Nm}_{\B^2/\B}( \D\times \D' ))\\
&\cong \mathrm{Nm}_{\B/\A}( \D\ast \D').
\end{align*}
On the other hand, we can also factor the algebra map $\A \to \B^2$ as $\A \to \A^2 \to \B^2$, for which transitivity of the norm operation gives
\begin{align*}
\mathrm{Nm}_{\B^2/\A}(\D\times \D') &\cong \mathrm{Nm}_{\A^2/\A} (\mathrm{Nm}_{\B^2/\A^2} (\D\times \D'))\\
&\cong \mathrm{Nm}_{\A^2/\A}(\mathrm{Nm}_{\B/\A}(\D) \times \mathrm{Nm}_{\B/\A}(\D'))\\
&\cong \mathrm{Nm}_{\B/\A}(\D) \ast \mathrm{Nm}_{\B/\A}(\D').
\end{align*}
(In the second line, we have used base-change invariance of the norm operation to construct the norm over each factor separately, and in the third line, we have invoked \cref{ex:norm-product}) Comparing these two expressions for $\mathrm{Nm}_{\B^2/\A}(\D\times \D')$, we obtain $\mathrm{Nm}_{\B/\A}(\D\ast \D') \cong \mathrm{Nm}_{\B/\A}(\D) \ast \mathrm{Nm}_{\B/\A}(\D')$, as desired.
\end{proof}
Our last results are that this notion of norm of quadratic algebras commutes with taking determinant line bundles and discriminant bilinear forms:
\begin{theorem}\label{thm:norm-det}
The norm operation commutes with taking determinant line bundles: If $\A\to\B$ is a rank-$n$ algebra and $\D$ is a quadratic $\B$-algebra, then $\bigwedge_\A^2\mathrm{Nm}_{\B/\A}(\D) = \mathrm{Nm}_{\B/\A}(\bigwedge_\B^2 \D)$.
\end{theorem}
\begin{proof}
If $\L$ is a line bundle (rank-$1$ module) over $\B$, then its norm $\mathrm{Nm}_{\B/\A}(\L)$ is the line bundle obtained by the following process analogous to \cref{def:norm}:
\begin{enumerate}
\item Choose a set of elements $a_i\in \A$ generating the unit ideal such that each localization $\L_{a_i}$ is free as an $\B_{a_i}$-module.
\item Choose generators for each $\L_{a_i}$ to obtain isomorphisms $\phi_i:\L_{a_i}\cong \B_{a_i}$.
\item Over overlaps $\A_{a_ia_j}$, we have isomorphisms $(\phi_i)_{a_j} \circ (\phi_j^{-1})_{a_i} \colon \B_{a_ia_j} \cong \L_{a_ia_j} \cong \B_{a_ia_j}$. Each such isomorphism is given by multiplication by a unit $U_{ij}\in\B_{a_ia_j}$; take the norm of $U_{ij}$ to get units $s_n(U_{ij})\in\A_{ij}$.
\item Glue together free rank-1 $\A_{a_i}$-modules, using multiplication by $s_n(U_{ij})$ as the isomorphisms on overlaps, to get a single rank-$1$ $\A$-module, called $\mathrm{Nm}_{\B/\A}(\L)$.
\end{enumerate}
Now suppose $\L = \bigwedge^2 \D$. We will follow along steps 1 through 4 of \cref{def:norm} and see that $\bigwedge^2 \mathrm{Nm}_{\B/\A}(\D)$ agrees with the above construction of $\mathrm{Nm}_{\B/\A}(\L)$.
Since the determinant line bundle of a based quadratic algebra $\quadalg[\B]{T}{N}$ is free of rank $1$ with canonical generator $1\wedge x$, completing steps 1 and 2 of \cref{def:norm} also completes steps 1 and 2 of constructing the line bundle norm $\mathrm{Nm}_{\B/\A}(\L)$.
For step 3, suppose that for each $i,j\in\{1,\dots,k\}$ we have an isomorphism $\quadalg[\B_{a_ia_j}]{T_i}{N_i} \cong \D_{a_ia_k} \cong \quadalg[\B_{a_i}{a_j}]{T_j}{N_j}$ sending $x\mapsto U_{ij}x+C_{ij}$.
Then taking exterior powers, we have isomorphisms $\bigwedge^2 \quadalg[\B_{a_ia_j}]{T_i}{N_i} \cong \bigwedge^2 \D_{a_ia_k} \cong \bigwedge^2 \quadalg[\B_{a_i}{a_j}]{T_j}{N_j}$ sending $1\wedge x \mapsto U_{ij} \cdot (1 \wedge x)$.
This means that multiplication by $U_{ij}$ is also the composite isomorphism $\B_{a_ia_j}\cong \L_{a_ia_j}\cong\B_{a_ia_j}$.
To construct $\mathrm{Nm}_{\B/\A}(\D)$, we apply \cref{prop:different-generator} to the isomorphisms $\quadalg[\B_{a_ia_j}]{T_i}{N_i} \cong \quadalg[\B_{a_i}{a_j}]{T_j}{N_j}$ sending $x\mapsto U_{ij}x+C_{ij}$ to get isomorphisms
\[\mathrm{Nm}\quadalg[\B_{a_ia_j}]{T_i}{N_i} \cong \mathrm{Nm}\quadalg[\B_{a_i}{a_j}]{T_j}{N_j}: x\mapsto s_n(U_{ij})x+\left.\frac{s_n(\lambda C_{ij}+U_{ij}T_j)- s_n(U_{ij}T_j)}{\lambda}\right|_{\lambda=2}\]
which we use as gluing maps to construct $\mathrm{Nm}_{\B/\A}(\D)$. Taking the exterior powers of these maps, we obtain isomorphisms sending $1\wedge x \mapsto s_n(U_{ij})\cdot (1\wedge x)$, meaning that $\bigwedge^2 \mathrm{Nm}_{\B/\A}(\D)$ is obtained by gluing together free rank-$1$ $\A_{a_i}$-modules with the gluing isomorphisms given by multiplication by $s_n(U_{ij})$. That is exactly the gluing which produces $\mathrm{Nm}_{\B/\A}(\L)$, as claimed.
\end{proof}
\begin{theorem}\label{thm:norm-disc}
Taking the norm commutes with taking discriminants: if $\A\to\B$ is a rank-$n$ algebra and $\D$ is a quadratic $\B$-algebra with discriminant bilinear form $\delta_\D \colon ({\textstyle\bigwedge}_\B^2 \D)^{\otimes 2} \to \B$, then the discriminant bilinear form of the quadratic $\A$-algebra $\mathrm{Nm}_{\B/\A}( \D)$ agrees with the norm of $\delta_\D$ as a map of line bundles.
\end{theorem}
\begin{proof}
Note that this only makes sense because by \cref{thm:norm-det} we have an isomorphism $\bigwedge^2\mathrm{Nm}(\D) = \mathrm{Nm}(\bigwedge^2 \D)$, so the norm of the line bundle homomorphism $\delta_\D \colon ({\textstyle\bigwedge}^2 \D)^{\otimes 2} \to \B$ can be viewed as a homomomorphism $\mathrm{Nm}(\delta_\D) \colon ({\textstyle\bigwedge}^2 \mathrm{Nm}(\D))^{\otimes 2} \to \A$. Then we can check locally that this agrees with $\delta_{\mathrm{Nm}(\D)}$, so assume without loss of generality that $\D = \quadalg{T}{N}$.
With respect to the canonical basis $1\wedge x$ for ${\textstyle\bigwedge}^2\D$, we have that $\delta_\D$ is just multiplication by $T^2 - 4N$. The norm of this map is therefore multiplication by $s_n(T^2 - 4N)$.
On the other hand, with respect to the canonical basis $\delta_{\mathrm{Nm}(\D)}$ is multiplication by
\begin{align*}
& s_n(T)^2 - 4\left[\frac{s_n(\lambda N + T^2) - s_n(T^2)}{\lambda}\right]_{\lambda = -4} \\
& = s_n(T)^2 + \left[s_n(\lambda N + T^2) - s_n(T^2)\right]_{\lambda = -4}\\
&= s_n(T^2) + s_n(T^2 - 4N) - s_n(T^2)\\
&= s_n(T^2 - 4N),
\end{align*}
which agrees with $\mathrm{Nm}(\delta_\D)$ as desired.
\end{proof}
In particular, if a quadratic $\B$-algebra $\D$ is \'etale (its discriminant bilinear form is an isomorphism $({\textstyle\bigwedge}^2 \D)^{\otimes 2}\ds\mathop{\longrightarrow}^\sim\, \B$), then its norm $\mathrm{Nm}_{\B/\A}(\D)$ is also an \'etale quadratic $\A$-algebra. Since \'etale quadratic algebras are equivalent to $S_2$-torsors, we may ask if the norm operation agrees with the trace map for $S_2$-torsors:
\begin{theorem}\label{thm:etale-addition}
On \'etale quadratic algebras, the norm operation reduces to the ordinary trace for $S_2$-torsors.
\end{theorem}
\begin{proof}
Let $\A\to\B$ be a rank-$n$ algebra. Since a split \'etale $\B$-algebra is isomorphic to $\B^2\cong \quadalg[\B]10$, and $\mathrm{Nm}_{\B/\A}\quadalg[\B]10 = \quadalg[\A]10\cong\A^2$, we know that since the norm operation commutes with base change, therefore the norm of any \'etale quadratic algebra is still \'etale. It remains to show that the norm of a $\B$-algebra automorphism of $\B^2$ is its trace as an automorphism of $\A^2$. Any $\B$-algebra automorphism of $\B^2$ is locally either the identity map $\B^2\to\B^2$ (which is sent to the identity map $\A^2\to\A^2$) or the swap map $\B^2\to\B^2$ sending $(b,b')\mapsto (b', b)$. By \cref{ex:swap}, the norm of the swap map is either the identity (if the rank $n$ is even) or again the swap map (if $n$ is odd). In other words, the norm of the swap map is the swap map composed with itself $n$ times, which is exactly the trace of the swap map, as desired.
\end{proof}
Finally, we have the following conjectured relationship between norms and discriminant algebras which was the motivation for constructing the norm functor for quadratic algebras in the first place:
\begin{conjecture}\label{conj:norm-discalg}
Let $\Delta$ be the discriminant algebra operation of \cite{BieselGioia}, sending rank-$n$ algebras $\A\to\B$ to quadratic algebras $\A \to\Delta_{\B/\A}$. Now let $\A\to\B$ be a rank-$n$ algebra and $\B\to\C$ be a rank-$m$ algebra, so that $\A\to\C$ is also a rank-$mn$ algebra. Is it the case that
\[\Delta_{\C/\A} \cong \mathrm{Nm}_{\B/\A}(\Delta_{\C/\B}) \ast \Delta_{\B/\A}^{\ast m}?\]
\end{conjecture}
It is known that \cref{conj:norm-discalg} holds in case $\A\to\B$ and $\B\to\C$ are \'etale; see \cite[Theorem 4]{Waterhouse} for a proof.
\bibliographystyle{acm}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 447 |
Q: CSS Flex Box Positioning inside a modal dialog box body Could anybody tell me how can I achieve this style using flexbox? Pls note the modal close button in the right side of the photo.
This is what i have so far. I don't know how to place the close button on top of the image.
Below is my code for the bootstrap modal dialog box:
<Modal show={this.props.show} onHide={() => this.props.onHide()}
>
<Modal.Body className ="modal-body">
<div className = "data-container">
<div class="callout" data-closable>
<button class="close-button" onClick={() => this.props.onHide()} aria-label="Dismiss alert" type="button" data-close>
<span aria-hidden="true">×</span>
</button>
</div>
{this.props.data}
</div>
</Modal.Body>
{/* <Modal.Footer className = "modal-footer">
{this.props.title}
</Modal.Footer> */}
</Modal>
And here is the CSS:
.modal-dialog {
display: flex;
max-width: 800px;
height:auto;
flex-direction:column;
justify-content: center;
padding-top:150px;
margin: auto;
}
.modal-content {
overflow: hidden;
border:none;
justify-content: center;
margin:auto;
}
.modal-body {
display: flex;
flex-direction:column;
align-items: center;
justify-content: center;
background-color:rgb(221, 221, 221);
padding: 2px;
}
.callout {
display:flex;
border: none;
display: block;
margin-left:47rem;
}
.close-button {
border: none;
background-color:rgb(221, 221, 221);
color:red;
margin:auto;
/* font-weight: bold; */
font-size:50px;
/* width:30px;
height:30px; */
}
Any help is greatly appreciated.
Thanks ✌️
A: First you need to specify your parent containers position as relative
.data-container {
position: relative;
}
And then set position as absolute and right to 0 on container that containers your close button
.callout {
position: absolute;
right: 0;
}
Demo here
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,087 |
Metisa plana är en fjärilsart som beskrevs av Walker 1883. Metisa plana ingår i släktet Metisa och familjen säckspinnare. Inga underarter finns listade i Catalogue of Life.
Källor
Externa länkar
Säckspinnare
plana | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,065 |
Q: In a Sun Disk Label scheme, what are partitions of type E and F? I am working with an image (dd) of an HDD removed from a Sun server that was in use circa 2006.
Using fdisk on a Debian system to view the partition table it shows 3 partitions:
Partition 2: A type 5, whole disk - as is the norm.
Partition 3: At cylinder 0, 1 cylinder long, type f.
Partition 4: At cylinder 1, the rest of the disk, type e.
I can't find any documentation to tell me what partitions of type E or F might be.
I've used disktype and file to examine the disk and volumes respectively, but with no luck.
There is a partner disk in the server which has root and var partitions of type UFS. From what I've seen on this disk, the original disk might be part of a vxfs disk group, but I don't know how to find out or mount the partition.
Can anybody suggest any next steps?
Edit: The partition table as shown by format on SunOS...
Part Tag Flag Cylinders Size Blocks
0 unassigned wm 0 0 (0/0/0) 0
1 unassigned wm 0 0 (0/0/0) 0
2 backup wu 0 - 4923 8.43GB (4924/0/0) 17682084
3 - wu 0 - 0 1.75MB (1/0/0) 3591
4 - wu 1 - 4923 8.43GB (4923/0/0) 17678493
5 unassigned wm 0 0 (0/0/0) 0
6 unassigned wm 0 0 (0/0/0) 0
7 unassigned wm 0 0 (0/0/0) 0
And the output from fdisk on Debian:
Disk file.dd (Sun disk label): 27 heads, 133 sectors, 4924 cylinders
Units = sectors of 1 * 512 bytes
Device Flag Start End Blocks Id System
file.dd3 u 0 17682084 8841042 5 Whole disk
file.dd4 u 0 3591 1795+ f Unknown
file.dd5 u 3591 17682084 8839246+ e Unknown
Edit: bit more info...
In fdisk (on Debian) the Sun disklabel partition types are listed as:
0 Unassigned 4 SunOS usr 8 SunOS home 82 Linux swap
1 Boot 5 Whole disk 9 SunOS alt secto 83 Linux native
2 SunOS root 6 SunOS stand a SunOS cachefs 8e Linux LVM
3 SunOS swap 7 SunOS var b SunOS reserved fd Linux RAID auto
'e' and 'f' simply aren't listed.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,233 |
Gemini Film Circuit is a film distribution and production studio unit in Chennai, Tamil Nadu, India.
Career
The company was founded by Ravi Shankar Prasad who also owned Anand Cine Services and it went on to produce and distribute many successful Tamil films including the remakes of Rajkumar Hirani films like Vasool Raja MBBS, Shankar Dada MBBS, Shankar Dada Zindabad & Nanban. However the failure of Mani Ratnam's Kadal (2013) affected company's finances after the venture became a surprise failure at the box office, and distributors of Kadal wanted compensation before Madha Gaja Raja was released. The company made its comeback co-producing and distributing the Malayalam film Unda (2019).
Filmography
As producer
Released
As Distributor
Released
Notes
References
Film production companies based in Chennai
Film production companies of India
Companies with year of establishment missing | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,689 |
package za.ac.sun.cs.intlola.test;
import java.io.IOException;
import org.eclipse.core.resources.IResourceDelta;
import za.ac.sun.cs.intlola.file.Const;
import za.ac.sun.cs.intlola.processing.ArchiveProcessor;
import za.ac.sun.cs.intlola.processing.json.Assignment;
import za.ac.sun.cs.intlola.processing.json.AssignmentInfo;
import za.ac.sun.cs.intlola.processing.json.Project;
import za.ac.sun.cs.intlola.processing.json.ProjectInfo;
import za.ac.sun.cs.intlola.processing.paths.TestPaths;
import za.ac.sun.cs.intlola.util.IntlolaError;
import za.ac.sun.cs.intlola.util.InvalidModeException;
public class ArchiveProcessorTest extends Thread {
public static void main(String[] args) throws InterruptedException {
Thread t = new ArchiveProcessorTest();
t.start();
t.join();
}
public void run() {
Thread[] runners = new Thread[Settings.THREAD_COUNT];
for (int i = 0; i < Settings.THREAD_COUNT; i++) {
try {
runners[i] = new ArchiveSender();
runners[i].start();
} catch (InvalidModeException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
for (Thread runner : runners) {
try {
runner.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
public static class ArchiveSender extends Thread {
private ArchiveProcessor proc;
public ArchiveSender() throws InvalidModeException, IOException {
proc = new ArchiveProcessor(
new TestPaths(Settings.PROJECT_LOCATION));
}
public void run() {
int count = 0;
while (count < Settings.FILE_COUNT) {
try {
sleep(Settings.SLEEP_DURATION);
proc.processChanges(Settings.FILE_NAME,
IResourceDelta.CHANGED);
} catch (IOException e) {
e.printStackTrace();
} catch (InterruptedException e) {
e.printStackTrace();
}
count++;
}
IntlolaError error = proc.login(Const.LOGIN, Settings.USER_NAME,
Settings.PASSWORD, Settings.ADDRESS, Settings.PORT);
if (!error.equals(IntlolaError.SUCCESS)) {
System.err.println(error.getDescription());
}
Project p = null;
Assignment a = null;
for (ProjectInfo pi : proc.getProjects()) {
if (pi.getProject().Name.equals(Settings.PROJECT_NAME)) {
AssignmentInfo[] as = pi.getAssignments();
if (as.length > 0) {
a = as[0].getAssignment();
p = pi.getProject();
}
break;
}
}
if (p != null && a != null) {
error = proc.createSubmission(p, a);
if (!error.equals(IntlolaError.SUCCESS)) {
System.err.println(error.getDescription());
}
} else {
System.err.println("Could not load project and/or assignment");
}
proc.stop();
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,902 |
Граф Франц Моріц фон Лассі (на російській службі ; , 21 жовтня 1725 р., м. Санкт-Петербург — †24 листопада 1801 р., м. Відень) — видатний австрійський воєначальник часів Семирічної війни, що носив чин генерал-фельдмаршала. Син російського фельдмаршала Петра Лассі. Власник Галліцінберга під Віднем.
Біографія
Франц-Моріц походив з родини з ірландсько-балтійськими коріннями: третій, молодший син з восьми дітей графа Лассі, ірландського норманна, від його шлюбу з балтійської німкенею графинею Мартою Філіппіною фон Лёзер, вдовою графа фон Функ. Дитинство провів у Санкт-Петербурзі та Ризі. У віці дванадцяти років, як і старші брати, відправлений батьком, що бажали, щоб його сини робили кар'єру самостійно, а не за рахунок становища батьків, за кордон. Два роки навчання в заснованої в 1708 році кавалерійської школі в Лігниці (в той час — Австрія), завершення військової освіти в Інженерної академії у Відні.
Вихід з російської служби в 1743 році, в тому ж році вступає прапорщиком () до австрійського піхотний полк Броуна. Як ад'ютанта фельдмаршал-лейтенанта Броуна бере участь у Війні за австрійську спадщину. Бойове хрещення в невдалій спробі штурму італійської фортеці Веллетри в ніч з 10 на 11 серпня 1744 року. Штикове поранення при облозі Веллетри 29 жовтня 1744 року. Тоді ж підвищено до капітану (). За виявлену доблесть в битві при П'яченці 16 червня 1746 отримав звання майора. Бере участь в битві при Роттофредо і облозі Генуї.
Після війни кар'єра Лассі розвивається не менш успішно: в березні 1749 рік стає підполковником, з 1753 по 1756 роки командує полком, розквартированим в Богемії.
Семирічна війна
З початком Семирічної війни (1756—1763) Лассі відзначився в битві при Лобозіце 1 жовтня 1756 року, де, на загальну думку сучасників, саме йому зобов'язана була австрійська армія своїм порятунком. Отримав звання генерал-фельдвахтмейстери, в наступному, 1757 році, за Бреслау (22 листопада) — фельдмаршал-лейтенант і генерал-квартирмейстер.
За перемогу над пруссаками при Хохкірхе 14 жовтня 1758 року, де Лассі належить диспозиція битви, нагороджений Великим хрестом ордена Марії Терезії (честь, якої за всю історію Австрії удостоїлися лише двадцять воєначальників).
20 листопада 1759 року Лассі з 17 000 солдатів вдається повністю оточити і примусити до здачі 15 тисячний корпус прусського генерала Фінка при Максене. За цю блискучу перемогу, яка поставила Фрідріха II, яка мала гостру нестачу солдатів, в надзвичайно скрутне становище, Лассі отримав звання генерал-фельдцейхмейстера.
У жовтні 1760, разом з російськими генералами Г. К. Г. Тотлебеном, З. Г. Чернишовим і П. І. Паніним, брав участь в експедиції на Берлін. Відмовившись визнати умови капітуляції, підписані Тотлебеном, і, вимагаючи частки для австрійців у військовій видобутку, — Лассі мало не силою ввів свої війська до Берліна і, вважаючи себе обділеним, розпустив їх до такої міри, що Тотлебен був змушений ввести в місто додаткові російські підрозділи з наказом, у разі необхідності, стріляти по австрійських союзникам.
C 1762 року Лассі командував крилом австрійської армії. Відмовився від чину генерал-фельдмаршала, не бажаючи обходити свого друга, старшого за званням.
Президент гофкрігсрата
C 1763 року — член вищого військового органу імперії, гофкрігсрата (букв. Придворний військова рада), генерал-інспектор піхоти. З 1766 року і по 1774 рік фельдмаршал Лассі — президент гофкрігсрата. На цій посаді він придбав репутацію успішного і далекоглядного реформатора. Реформи Лассі зачіпають всі сторони життя армії, від комплектування до фінансування. Заслугою Лассі вважається, серед іншого, установа навчальних таборів.
У 1770 році нагороджений Орденом Золотого руна. З 1794 року є канцлером ордена Марії-Терезії. Один з найближчих радників Йосипа II, з яким зблизився під час Війни за спадщину баварського престолу.
Полководська кар'єра складається на схилі віку Лассі не зовсім гладко. Так, під час війни з Туреччиною 1788—1789 років він після ряду невдач був змушений поступитися верховне командування. Піддавався критиці за свою надмірно обережну тактику.
У 1790-ті роки поступово відходить від справ, не маючи ні сім'ї, ні дітей, живе замкнутої, якщо не самотньою, життям.
Родовий маєток російської гілки сімейства Лассі, від якої граф Франц Моріц Лассі веде своє походження, знаходилося на території сучасної Білорусі. Нащадки російських Лассі, які втекли в 1939 році з Білорусі під час поділу Польщі і пізнішого наближення Червоної армії, проживають нині в Польщі і в Аргентині.
Одна з сестер Франца Лассі — Хелен — була одружена з російським фельдмаршалом, пізніше ризьким губернатором, графом Ю. Ю. Броуном. Її син, граф Георг Броун, похований у Відні разом зі своїм знаменитим дядьком, графом Лассі.
Примітки
Джерела
Kotasek, Edith: Feldmarschall Graf Lacy. Ein Leben für Österreichs Heer, Verlag Ferdinand Berger, Horn, N.-Ö. 1956
Уродженці Санкт-Петербурга
Ірландська діаспора
Австрійські фельдмаршали
маршали Священної Римської імперії
Учасники Семирічної війни
Померли у Відні
Поховані у Відні
Кавалери Великого Хреста ордену Марії-Терезії (Австро-Угорщина) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,023 |
Alien as a Comic Book : Adaptation and Genre Shifting - Download this document for free, or read online. Document in PDF available to download.
Abstract : This article examines the 1979 adaptation of Ridley Scott-s Alien. It focuses on the way the adaptation process results in two works with very different generic mix. Both the film and the comics can be read a generic hybrid, belonging both to horror and science fiction, but the balance is different. The article argues that the constraints of the comics form make it impossible to adapt the established grammar of horror cinema—especially the effect of surprise—which leads to the observed shift in generic affiliation. This leads to a broader argument on the effect of adaptations on genre. | {
"redpajama_set_name": "RedPajamaC4"
} | 107 |
The Island Moving Company's Junior Company consists of 20 teenage dancers who have been invited to the group. The IMC's Junior Company will present a production of Carnival of the Animals at the Jamestown Arts Center on Sunday, March 31st at 2:30 pm.
The "Carnival of the Animals" is Directed by Brooke DiFrancesco and choreographed by Christine Sandorfi. A Lion, two Hens, a Donkey, a Tortoise and other assorted creatures dance to music by Camille Saint-Saens in a production that lasts roughly one half hour. This presentation will be suitable for children of all ages and will include an opportunity to join the dancers in some dancing after the performance. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,575 |
Buddha (c. 500s B.C.E.)
The historical Buddha, also known as Gotama Buddha, Siddhārtha Gautama, and Buddha Śākyamuni, was born in Lumbini, in the Nepalese region of Terai, near the Indian border. He is one of the most important Asian thinkers and spiritual masters of all time, and he contributed to many areas of philosophy, including epistemology, metaphysics and ethics. The Buddha's teaching formed the foundation for Buddhist philosophy, initially developed in South Asia, then later in the rest of Asia. Buddhism and Buddhist philosophy now have a global following.
In epistemology, the Buddha seeks a middle way between the extremes of dogmatism and skepticism, emphasizing personal experience, a pragmatic attitude, and the use of critical thinking toward all types of knowledge. In ethics, the Buddha proposes a threefold understanding of action: mental, verbal, and bodily. In metaphysics, the Buddha argues that there are no self-caused entities, and that everything dependently arises from or upon something else. This allows the Buddha to provide a criticism of souls and personal identity; that criticism forms the foundation for his views about the reality of rebirth and an ultimate liberated state called "Nirvana." Nirvana is not primarily an absolute reality beyond or behind the universe but rather a special state of mind in which all the causes and conditions responsible for rebirth and suffering have been eliminated. In philosophical anthropology, the Buddha explains human identity without a permanent and substantial self. The doctrine of non-self, however, does not imply the absolute inexistence of any type of self whatsoever, but is compatible with a conventional self composed of five psycho-physical aggregates, although all of them are unsubstantial and impermanent. Selves are thus conceived as evolving processes causally constrained by their past.
Interpreting the Historical Buddha
The Buddha's Epistemology
The Extremes of Dogmatism and Skepticism
The Role of Personal Experience and the Buddha's Wager
Interpretations of the Buddha's Advice to the Kālāma People
Higher Knowledge and the Question of Empiricism
The Buddha's Cosmology and Metaphysics
The Universe and the Role of Gods
The Four Noble Truths or Realities
Ontology of Suffering: the Five Aggregates
Arguments for the Doctrine of Non-self
Human Identity and the Meaning of Non-self
Causality and the Principle of Dependent Arising
Nirvana and the Silence of the Buddha
Two Kinds of Nirvana and the Undetermined Questions
Eternalism, Nihilism, and the Middle Way
Buddhist Ethics
References and Further Reading
1. Interpreting the Historical Buddha
a. Dates
There is no complete agreement among scholars and Buddhist traditions regarding the dates of the historical Buddha. The most common dates among Buddhists are those of the Theravāda school, 623-543 B.C.E. From the middle of the 19th century until the late 20th century, Western scholars had believed the dates of the Buddha to be ca. 560-480 B.C.E. However, after the publication in 1991-2 of the proceedings of the international symposium on the date of the historical Buddha held in Göttingen in 1988, the original consensus on these dates no longer exists.
Although there is no conclusive evidence for any specific date, most current scholars locate the Buddha's life one hundred years earlier, around the fifth century B.C.E. Some of the new dates for the Buddha's "death" or more accurately, for his parinirvāṇa are: ca. 404 B.C.E. (R. Gombrich), between 410-390 B.C.E. (K.R. Norman), ca. 400 B.C.E. (R. Hikata), ca. 397 B.C.E. (K.T.S. Sarao), between ca.400-350 B.C.E. (H. Bechert), 383 B.C.E. (H. Nakamura), 368 B.C.E. (A. Hirakawa), between 420-380 B.C.E. (A. Bareau).
b. Sources
The historical Buddha did not write down any of his teachings, they were passed down orally from generation to generation for at least three centuries. Some scholars have attempted to distinguish the Buddha's original teachings from those developed by his early disciples. Unfortunately, the contradictory conclusions have led most scholars to be skeptical about the possibility of knowing what the Buddha really taught. This however, does not mean that all Buddhist texts that attribute teachings to the Buddha are equally valuable to reconstruct his thought. The early sūtras in Pāli and other Middle Indo-Aryan languages are historically and linguistically closer to the cultural context of the Buddha than Mahāyāna sūtras in Sanskrit, Tibetan, and Chinese. This does not imply that later translations of the early sūtras in Chinese (there are no Tibetan translations of the early sūtras) are less authentic or useless in reconstructing the philosophy of the Buddha. On the contrary, the comparative study of Pāli and Chinese versions of the early sūtras can help to infer what might have been the Buddha's position on a number of issues.
Following what seems to be a growing scholarly tendency, I will reconstruct the philosophy of the historical Buddha by drawing on the Sutta Piṭaka of the Pāli canon. More specifically, our main sources are the first four Pāli Nikāyas (Dīgha, Majjhima, Saṃyutta, Aṅguttara) and some texts of the fifth Pāli Nikāya (Dhammapada, Udāna, Itivuttaka, and Sutta Nipāta). I do not identify these sources with the Buddha's "ipsissima verba," that is, with "the very words" of the Buddha, even less with his "actual" thought. Whether these sources are faithful to the actual thought and teachings of the historical Buddha is an unanswerable question; I can only say that to my knowledge there are not better sources to reconstruct the philosophy of the Buddha.
According to the traditional Buddhist account, shortly after the Buddha's death five hundred disciples gathered to compile his teachings. The Buddha's personal assistant, Ānanda, recited the first part of the Buddhist canon, the Sūtra Piṭaka, which contains discourses in dialogue form between the Buddha, his disciples, and his contemporaries on a variety of doctrinal and spiritual questions. Ānanda is reported to have recited the sutras just as he had heard them from the Buddha; that is why Buddhist sutras begin with the words "Thus have I heard." Another disciple, Upāli, recited the second part of the Buddhist canon, the Vinaya Piṭaka, which also contains sutras, but primarily addresses the rules that govern a monastic community. After the recitation of Ānanda and Upāli, the other disciples approved what they had heard and communally recited the teachings as a sign of agreement. The third part of the Buddhist canon or Abhidharma Piṭaka, was not recited at that moment. The Theravāda tradition claims that the Buddha taught the Abhidharma while visiting the heaven where his mother was residing.
From a scholarly perspective, the former account is questionable. It might be the case that a large collection of Buddhist texts was written down for the first time in Sri Lanka during the first century B.C.E. However, the extant Pāli canon shows clear signs of historical development in terms of both content and language. The three parts of the Pāli canon are not as contemporary as the traditional Buddhist account seems to suggest: the Sūtra Piṭaka is older than the Vinaya Piṭaka, and the Abhidharma Piṭaka represents scholastic developments originated at least two centuries after the other two parts of the canon. The Vinaya Piṭaka appears to have grown gradually as a commentary and justification of the monastic code (Prātimokṣa), which presupposes a transition from a community of wandering mendicants (the Sūtra Piṭaka period ) to a more sedentary monastic community (the Vinaya Piṭaka period). Even within the Sūtra Piṭaka it is possible to detect older and later texts.
Neither the Sūtra Piṭaka nor the Vinaya Piṭaka of the Pāli canon could have been recited at once by one person and repeated by the entire Buddhist community. Nevertheless, the Sūtra Piṭaka of the Pāli canon is of particular importance in reconstructing the philosophy of Buddha for four main reasons. First, it contains the oldest texts of the only complete canon of early Indian Buddhism, which belong to the only surviving school of that period, namely, the Theravāda school, prevalent in Sri Lanka and Southeast Asia. Second, it has been preserved in a Middle Indo-Aryan language closely related to various Prakrit dialects spoken in North of India during the third century B.C.E., including the area where the Buddha spent most of his teaching years (Magadha). Third, it expresses a fairly consistent set of doctrines and practices. Fourth, it is strikingly similar to another version of the early Sūtra Piṭaka extant in Chinese (Āgamas). This similarity seems to indicate that a great part of the Sūtra Piṭaka in Pāli does not contain exclusively Theravāda texts, and belongs to a common textual tradition probably prior to the existence of Buddhist schools.
c. Life
Since the Pāli Nikāyas contain much more information about the teachings of the Buddha than about his life, it seems safe to postulate that the early disciples of the Buddha were more interested in preserving his teachings than in transmitting all the details of his life. The first complete biographies of the Buddha as well as the Jātaka stories about his former lives appeared centuries later, even after, and arguably as a reaction against, the dry lists and categorizations of early Abhidharma literature. The first complete biography of the Buddha in Pāli is the Nidānakathā, which serves as an introduction to the Jātaka verses found in the fifth Pāli Nikāya. In Sanskrit, the most popular biographies of the Buddha are the Buddhacarita attributed to the Indian poet Aśvaghoṣa (second century C.E), the Mahāvastu, and the Lalitavistara, both composed in the first century C.E.
The first four Pāli Nikāyas contain only fragmented information about the Buddha's life. Especially important are the Mahāpadāna-suttanta, the Ariyapariyesanā-suttanta, the Mahāsaccaka-suttanta, and the Mahāparinibbāna-suttanta. According to the Mahāpadāna-suttanta, the lives of all Buddhas or perfectly enlightened beings follow a similar pattern. Like all Buddhas of the past, the Buddha of this cosmic era, also known as Gautama (Gotama in Pāli), was born into a noble family. The Buddha's parents were King Śuddhodana and Queen Māyā. He was a member of the Śakya clan and his name was Siddhartha Gautama. Even though he was born in Lumbinī while his mother was traveling to her parents' home, he spent the first twenty-nine years of his life in the royal capital, Kapilavastu, in the Nepalese region of Terai, close to the Indian border.
Like all past Buddhas, the conception and birth of Gautama Buddha are considered miraculous events. For instance, when all Buddhas descend into their mothers' wombs from a heaven named Tuṣita, a splendid light shines forth and the entire universe quakes; their mothers are immaculate, healthy, and without pain of any sort during their ten months of pregnancy, but they die a week after giving birth. Buddha babies are born clean, though they are ritually bathed with two streams of water that fall from the sky; they all take seven steps toward the north and solemnly announce that this is their last rebirth.
Like former Buddhas, prince Siddhartha enjoyed all types of luxuries and sensual pleasures during his youth. Unsatisfied with this type of life, he had a crisis when he realized that everything was ephemeral and that his existence was subject to old age, sickness, and death. After seeing the serene joy of a monk and out of compassion for all living beings, he renounced his promising future as prince in order to start a long quest for a higher purpose, nirvāṇa (Pali nibbāna), which entails the cessation of old age, sickness and death. Later traditions speak of the Buddha as abandoning his wife Yaśodharā immediately after she gave birth to Rāhula, the Buddha's only son. The Pāli Nikāyas, however, do not mention this story, and refer to Rāhula only as a young monk.
According to the Ariyapariyesanā-suttanta and the Mahāsaccaka-suttanta, the Buddha tried different spiritual paths for six years. First, he practiced yogic meditation under the guidance of Ālāra Kālāma and Uddaka Rāmaputta. After experiencing the states of concentration called base of nothingness and base of neither-perception-nor-non-perception, he realized that these lofty states did not lead to nirvana. Then the Buddha began to practice breathing exercises and fasting. The deterioration of his health led the Buddha to conclude that extreme asceticism was equally ineffective in attaining nirvana. He thus resumed eating solid food; after recovering his health, he began to practice a more moderate spiritual path, the middle path, which avoids the extremes of sensual self-indulgence and self-mortification. Soon after, the Buddha experienced enlightenment, or awakening, under a bodhi-tree. First he was inclined to inaction rather than to teaching what he had discovered. However, he changed his mind after the god Brahmā Sahampati asked him to teach. Out of compassion for all living beings, he decided to start a successful teaching career that lasted forty-five years.
d. Significance
It would be simplistic to dismiss all supernatural aspects of the Buddha's life as false and consider historically true only those elements that are consistent with our contemporary scientific worldview. However, this approach towards the Buddha's life was prevalent in the nineteenth century and a great part of the twentieth century. Today it is seen as problematic because it imposes modern western ideals of rationality onto non-western texts. Here I set aside the question of historical truth and speak exclusively of significance. The significance of all the biographies of Buddha does not lie in their historical accuracy, but rather in their effectiveness to convey basic Buddhist ideas and values throughout history. Even today, narratives about the many deeds of Buddha are successfully used to introduce Buddhists of all latitudes into the main values and teachings of Buddhism.
The supernatural elements of the Buddha's life are as historically significant as the natural ones because they help to understand the way Buddhists conceived – and in many places continue to conceive – the Buddha. Like followers of other religious leaders, Buddhist scribes tended to glorify the sanctity of their foundational figure with extraordinary events and spectacular accomplishments. In this sense, the narratives of the Buddha are perhaps better understood as hagiographies rather than as biographies. The historical truth behind hagiographies is impossible to determine: how can we tell whether or not the Buddha was conceived without sexual intercourse; whether or not he was able to talk and walk right after his birth; whether or not he could walk over water, levitate, fly, and ascend into heaven at will? How do we know whether the Buddha was really tempted by Māra the evil one; whether there was an earthquake at the moment of his birth and death? The answers to these questions are a matter of faith. If the interpreter does not believe in the supernatural, then many narratives will be dismissed as historically false. However, for some Buddhists the supernatural events that appear in the life of Buddha did take place and are historically true.
The significance of the hagiographies of the Buddha is primarily ethical and spiritual. In fact, even if the life of Buddha did not take place as the hagiographies claim, the ethical values and the spiritual path they illustrate remain significant. Unlike other religions, the truth of Buddhism does not depend on the historicity of certain events in the life of the Buddha. Rather, the truth of Buddhism depends on the efficacy of the Buddhist path exemplified by the life of the Buddha and his disciples. In other words, if the different Buddhist paths inspired by the Buddha are useful to overcome existential dissatisfaction and suffering, then Buddhism is true regardless of the existence of the historical Buddha.
The fundamental ethical and spiritual point behind the Buddha's life is that impermanent, conditioned, and contingent things such as wealth, social position, power, sensual pleasures, and even lofty meditative states, cannot generate a state of ultimate happiness. In order to overcome the profound existential dissatisfaction that all ephemeral and contingent things eventually generate, one needs to follow a comprehensive path of ethical and mental training conducive to the state of ultimate happiness called nirvana.
2. The Buddha's Epistemology
a. The Extremes of Dogmatism and Skepticism
While the Buddha's view of the spiritual path is traditionally described as a middle way between the extremes of self-indulgence and self-mortification, the Buddha's epistemology can be interpreted as a middle way between the extremes of dogmatism and skepticism.
The extreme of dogmatism is primarily represented in the Pāli Nikāyas by Brahmanism. Brahmanism was a ritualistic religion that believed in the divine revelation of the Vedas, thought that belonging to a caste was determined by birth, and focused on the performance of sacrifice. Sacrifices involved the recitation of hymns taken from the Vedas and in many cases the ritual killing of animals.
Ritual sacrifices were offered to the Gods (gods for Buddhism) in exchange for prosperity, health, protection, sons, long life, and immortality. Only the male members of the highest caste, the priestly caste of Brahmins, could afford the professional space to seriously study the three Vedas (the Atharva Veda did not exist, or if it existed, it was not part yet of the Brahmanic tradition). Since only Brahmins knew the three Vedas, only they could recite the hymns necessary to properly perform the ritual sacrifice. Both ritual sacrifice and the social ethics of the caste system were seen as an expression of the cosmic order (Dharma) and as necessary to preserve that order.
Epistemologically speaking, Brahmanism emphasized the triple knowledge of the Vedas, and dogmatic faith in their content: "in regard to the ancient Brahmanic hymns that have come down through oral transmission and in the scriptural collections, the Brahmins come to the definite conclusion: 'Only this is true, anything else is wrong' " (M.II.169).
The extreme of skepticism is represented in the Pāli Nikāyas by some members of the Śramanic movement, which consisted of numerous groups of spiritual seekers and wandering philosophers. The Sanskrit word "śramana" means "those who make an effort," and probably refers to those who practice a spiritual discipline requiring individual effort, not just rituals performed by others. In order to become a śramana it was necessary to renounce one's life as householder and enter into an itinerant life, which entailed the observance of celibacy and a simple life devoted to spiritual cultivation. Most śramanas lived in forests or in secluded places wandering from village to village where they preached and received alms in exchange.
The Śramanic movement was extremely diverse in terms of doctrines and practices. Most śramanas believed in free will as well as the efficacy of moral conduct and spiritual practices in order to attain liberation from the cycle of reincarnations. However, there was a minority of śramanas who denied the existence of the after life, free will, and the usefulness of ethical conduct and other spiritual practices. Probably as a reaction to these two opposite standpoints, some śramanas adopted a skeptic attitude denying the possibility of knowledge about such matters. Skeptics are described by the Buddha as replying questions by evasion (D.I.58-9), and as engaging in verbal wriggling, in eel-wriggling (amarāvikkhepa): "I don't say it is like this. And I don't say it is like that. And I don't say it is otherwise. And I don't say it is not so. And I don't say it is not not so" (M.I. 521).
b. The Role of Personal Experience and the Buddha's Wager
In contrast to Brahmanic dogmatism, the Buddha of the Pāli Nikāyas did not claim to be omniscient (M.I.482); in fact, he proposed a critical attitude toward all sources of knowledge. In the Majjhima Nikāya (II.170-1), the Buddha challenges Brahmins who accept Vedic scriptures out of faith (saddhā) and oral tradition (anussava); he compares those who blindly follow scripture and tradition without having direct knowledge of what they believe with "a file of blind men each in touch with the next: the first one does not see, the middle one does not see, and the last one does not see." The Buddha also warns Brahmins against knowledge based on likeability or emotional inclination (ruci), reflection on reasons (ākāraparivitakka), and consideration of theories (diṭṭhinijjhānakkhanti). These five sources of knowledge may be either true or false; that is, they do not provide conclusive grounds to claim dogmatically that "only this is true, anything else is wrong."
Dogmatic claims of truth were not the monopoly of Brahmins. In the Majjhima Nikāya (I.178), the Buddha uses the simile of the elephant footprint to question dogmatic statements about him, his teachings, and his disciples: he invites his followers to critically investigate all the available evidence (different types of elephant footprints and marks) until they know and see for themselves (direct perception of the elephant in the open). The Pāli Nikāyas also refer to many śramanas who hold dogmatic views and as a consequence are involved in heated doctrinal disputes. The conflict of dogmatic views is often described as "a thicket of views, a wilderness of views, a contortion of views, a vacillation of views, a fetter of views. It is beset by suffering, by vexation, by despair, and by fever, and it does not lead to disenchantment, to dispassion, to cessation, to peace, to higher knowledge, to enlightentment, to Nibbāna" (M.I.485).
Public debates were common and probably a good way to gain prestige and converts. Any reputed Brahmin or śramana had to have not only the ability to speak persuasively but also the capacity to argue well. Rational argument played an important role in justifying doctrines and avoiding defeat in debate, which implied conversion to the other's teaching. At the time of the Buddha many of these debates seem to have degenerated into dialectical battles that diverted from spiritual practice and led to disorientation, anger, and frustration. Although the Buddha of the Pāli Nikāyas utilizes reasoning to justify his positions in debates and conversations with others, he discourages dogmatic attachment to doctrines including his own (see the simile of the raft, M.I.135), and the use of his teachings for the sake of criticizing others and for winning debates (M.I.132).
Unlike the skepticism of some śramanas, the Buddha of the Pāli Nikāyas takes clear stances on ethical and spiritual issues, and rejects neither the existence of right views (M.I.46-63) nor the possibility of knowing certain things as they are (yathābhūtaṃ). In order to counteract skepticism, the Buddha advises to the Kālāma people "not go by oral tradition, by succession of disciples, by hearsay, by the content of sacred scripture, by logical consistency, by inference, by reflection on reasons, by consideration of theories, by appearance, by respect to a teacher." Instead, the Buddha recommends knowing things for oneself as the ultimate criterion to adjudicate between conflicting claims of truth (A.I.189).
When personal experience is not available to someone, the Buddha of the Pāli Nikāyas proposes taking into account what is praised or censored by the wise, as well as a method to calculate the benefits of following certain opinions called the incontrovertible teaching (apaṇṇakadhamma), which, in some ways, resembles Pascal's wager. According to the incontrovertible teaching, it would be better to believe in certain doctrines because they produce more benefits than others. For instance, even if there is no life after death and if good actions do not produce good consequences, still a moral person is praised in this life by the wise, whereas the immoral person is censured by society. However, if there is life after death and good action produce happy consequences, a moral person is praised in this life, and after death he or she goes to heaven. On the contrary, the immoral person is censured in this life, and after death he or she goes to hell (M.I.403). Therefore, it is better to believe that moral actions produce good consequences even if we do not have personal experience of karma and rebirth.
c. Interpretations of the Buddha's Advice to the Kālāma People
Some have interpreted the Buddha's advice to the Kālāma people as an iconoclast rejection of tradition and faith. This, however, does little justice to the Pāli Nikāyas, where the Buddha is said to be part of a long and respectable tradition of past Buddhas, and where the first Brahmins are sometimes commended by their holiness. The Buddha shows respect for many traditional beliefs and practices of his time, and rejects only those that are unjustified, useless, or conducive to suffering for oneself and others.
Faith in the Buddha, his teachings, and his disciples, is highly regarded in the Pāli Nikāyas: it is the first of the five factors of striving (M.II.95-6), and a necessary condition to practice the spiritual path (M.III.33). Buddhist faith, however, is not unconditional or an end in and of itself but rather a means towards direct knowledge that must be based on critical examination, supported by reasons, and eventually verified or rooted in vision (dassanamūlikā) (M.I.320).
Another common interpretation of the advice to the Kālāmas is that for the Buddha of the Pāli Nikāyas only personal experience provides reliable knowledge. However, this is misleading because analogical and inferential reasoning are widely used by the Buddha and his disciples to teach others as well as in debates with non-Buddhists. Similarly, analytical or philosophical meditation is a common practice for the attainment of liberation through wisdom. Personal experience, like any other means of knowledge is to be critically examined. Except in the case of Buddhas and liberated beings, personal experience is always tainted by affective and cognitive prejudices.
The Pāli Nikāyas might give the first impression of endorsing a form of naïve or direct realism: that is, the Buddha and his disciples seem to think that the world is exactly as we perceive it to be. While it is true that the Pāli Nikāyas do not question the common sense connection between objects of knowledge and the external world, there are some texts that might support a phenomenalist reading. For instance, the Buddha defines the world as the six senses (five ordinary senses plus the mind) and their respective objects (S.IV.95), or as the six senses, the six objects, and the six types of consciousness that arise in dependence on them (S.IV.39-40).
Here, the epistemology of the Buddha is a special form of realism that allows both for the direct perception of reality and the constructions of those less realized. Only Buddhas and liberated beings perceive the world directly; that is, they see the Dharma, whose regularity and stability remains independent of the existence of Buddhas (S.II.25). Unenlightened beings, on the other hand, see the world indirectly through a veil of negative emotions and erroneous views. Some texts go so far as to suggest that the world is not simply seen indirectly, but rather that it is literally constructed by our emotional dispositions. For instance, in the Majjhima Nikāya (I.111), the Buddha explicitly states that "what one feels, one perceives" (Yaṃ vedeti, taṃ sañjānāti). That is, our knowledge is formed by our feelings. The influence of feelings in our ways of knowing can also be inferred from the twelve-link chain of dependent arising, which explains the arising and cessation of suffering. The second link, saṅkhāra, or formations, conditions the arising of the third link, consciousness. The term saṅkhāra literally means "put together," connoting the constructive role of the mental factors that fall into this category, many of them affective in nature.
Similarly, the Buddha of the Pāli Nikāyas says that "with what one has mentally constructed as the root cause (Yaṃ papañceti tato nidānaṃ), perceptions, concepts, and [further] mental constructions (papañcasaññāsaṅkhā) beset a man with respect to past, future, and present forms…sounds…odours…flavors…tangibles…mind-objects cognizable by the eye…ear… nose…tongue…body…mind" (M.I.111-112). That is, the knowledge of unenlightened beings has papañca, or mental constructions, as its root cause. The word papañca is a technical term that literally means diversification or proliferation; it refers to the tendency of unenlightened minds to construct or fabricate concepts conducive to suffering, especially essentialist and ego-related concepts such as "I" and "mine," concepts which lead to a variety of negative mental states such as craving, conceit, and dogmatic views about the self (Ñāṇananda 1971).
It is precisely because our experiences are affectively and cognitively conditioned that the Buddha of the Pāli Nikāyas advocates a critical approach toward all sources of knowledge, including personal experience. Even the lofty experiences derived from meditation are to be analyzed carefully because they might lead to false opinions about the nature of the self, the world, and the after life. The epistemological ideal is to know things directly beyond mental constructions (papañca), which presupposes the "tranquilization of all mental formations" (sabbasaṅkhārasamatha).
d. Higher Knowledge and the Question of Empiricism
Contemplative experiences are of two main types: meditative absorptions or abstractions (jhāna), and higher or direct knowledge (abhiññā). There are six classes of higher or direct knowledge: the first one refers to a variety of supernatural powers including levitation and walking on water; in this sense, it is better understood as a know-how type of knowledge. The second higher knowledge is literally called "divine ear element" or clairaudience. The third higher knowledge is usually translated as telepathy, though it means simply the ability to know the underlying mental state of others, not the reading of their minds and thoughts.
The next three types of higher knowledge are especially important because they were experienced by the Buddha the night of his enlightenment, and because they are the Buddhist counterparts to the triple knowledge of the Vedas. The fourth higher knowledge is retrocognition or knowledge of past lives, which entails a direct experience of the process of rebirth. The fifth is the divine eye or clairvoyance; that is, direct experience of the process of karma, or as the texts put it, the passing away and reappearing of beings in accordance with their past actions. The sixth is knowledge of the destruction of taints, which implies experiential knowledge of the four noble truths and the process of liberation.
Some scholars have interpreted the Buddha's emphasis on direct experience and the verifiable nature of Buddhist faith as a form of radical empiricism (Kalupahana 1992), and logical empiricism (Jayatilleke 1963). According to the empiricist interpretation, Buddhist faith is always subsequent to critically verifying the available empirical evidence. All doctrines taught by the Buddha are empirically verifiable if one takes the time and effort to attain higher or direct knowledge, interpreted as extraordinary sense experience. For instance, the triple knowledge of enlightenment implies a direct experience of the processes of karma, rebirth, and the four noble truths. Critiques of the empiricist interpretation point out that, at least at the beginning of the path, Buddhist faith is not always based on empirical evidence, and that the purpose of extraordinary knowledge is not to verify the doctrines of karma, rebirth, and the four noble truths (Hoffman 1982, 1987).
Whether or not the Buddha's epistemology can be considered empiricist depends on what we mean by empiricism and experience. The opposition between rationalism and empiricism and the sharp distinction between senses and reason is foreign to Buddhism. Nowhere in the Pāli Nikāyas does the Buddha say that all knowledge begins in or is acquired from sense experience. In this sense, the Buddha is not an empiricist.
3. The Buddha's Cosmology and Metaphysics
a. The Universe and the Role of Gods
The Buddha of the Pāli Nikāyas accepts the cosmology characteristic of his cultural context: a universe with several realms of existence, where people are reborn and die again and again (saṃsāra) depending on their past actions (karma) until they attain salvation (mokṣa). However, the Buddha substantially modifies the cosmology of his time. Against the Brahmanic tendency to understand karma as ritual action, and the Jain claim that all activities including involuntary actions constitute karma, the Buddha defines karma in terms of volition, or free will, which is expressed through thoughts, words, and behavior. That is, for the Buddha, only voluntary actions produce karma.
Another important modification is that for the Buddha of the Pāli Nikāyas, saṃsāra refers primarily to a psychophysical process that takes place within the physical universe. For instance, when the Buddha speaks about the end of the world, he says that it cannot be reached by traveling through the physical universe, but only by putting an end to suffering (saṃsāra), where "one is not born, does not age, does not die, does not pass away, and is not reborn" Accordingly, salvation is not understood in world-denying terms or as an escape from the physical universe, but rather as an inner transformation that takes place within one's own psychophysical organism: "It is, friend, in just this fathom-high carcass endowed with perception and mind that I make known the world, the origin of the world, the cessation of the world, and the way leading to the cessation of the world." (S.I.62; A.II.47-9).
There are five kinds of destinations within saṃsāra: hell, animal kingdom, realm of ghosts, humankind, and realm of devas or radiant beings, commonly translated as gods (M.I.73). There are many hells and heavens and life there is transitory, just as in other destinations. In some traditions there is another destination, the realm of asuras or demigods, who are jealous of the gods and who are always in conflict with them.
The Pāli Nikāyas further divide the universe of saṃsāra into three main planes of existence, each one subdivided into several realms. The three planes of existence are sensorial, fine-material, and immaterial (M.I.50). Most destinations belong to the sensorial realm. Only a minority of heavens belong to the fine-material and immaterial realms. Rebirth in a particular realm depends on past actions: good actions lead to good destinations and bad actions to bad rebirths. Rebirth as a human or in heaven is considered a good destination; rebirth in the realm of ghosts, hell, and the animal realm are bad. Human rebirth is extremely difficult to attain (S.V.455-6; M.III.169), and it is highly regarded because of its unique combination of pain and pleasure, as well as its unique conductivity for attaining enlightenment. In this last sense human rebirth is said to be even better than rebirth as a god.
Rebirth also depends on the prevalent mental states of a person during life, and especially at the moment of death. That is, there is a correlation between mental states and realms of rebirth, between cosmology and psychology. For instance, a mind where hatred and anger prevails is likely to be reborn in hell; deluded and uncultivated minds are headed toward the animal kingdom; someone obsessed with sex and food will probably become bound to earth as a ghost; loving and caring persons will be reborn in heaven; someone who frequently dwells in meditative absorptions will be reborn in the fine-material and immaterial realms. Human rebirth might be the consequence of any of the aforementioned mental states.
Perhaps the most important modification the Buddha introduces into the traditional cosmology of his time was a new view of Gods (gods within Buddhism). In the Pāli Nikāyas, gods do not play any significant cosmological role. For the Buddha, the universe has not been created by an all-knowing, all-powerful god that is the lord of the universe and father of all beings (M.I.326-7). Rather, the universe evolves following certain cyclic patterns of contraction and expansion (D.III.84-5).
Similarly, the cosmic order, or Dharma, does not depend on the will of gods, and there are many good deeds far more effective than ritual sacrifices offered to the gods (D.I144ff). Gods for the Buddha are unenlightened beings subject to birth and death that require further learning and spiritual practice in order to attain liberation; they are more powerful and spiritually more developed than humans and other living beings, but Buddhas excel them in all regards: spiritual development, wisdom, and power. Even the supreme type of god, Brahmā, offers his respects to the Buddha, praises him, and asks him to preach the Dharma for those with little dust in their eyes (M.I.168-9).
Since the Buddha of the Pāli Nikāyas does not deny the existence of gods, only their cosmological and soteriological functions, it is inaccurate to define early Buddhism as atheistic or as non-theistic. The word atheistic is usually associated with anti-religious attitudes absent in the Buddha, and the term non-theistic seems to imply that rejecting the theistic concept of God is one of the main concerns of the Buddha, when in fact it is a marginal question in the Pāli Nikāyas.
b. The Four Noble Truths or Realities
One the most common frameworks to explain the basic teachings of early Buddhism is the four noble truths (ariyasacca, Sanskrit āryasatya). The word sacca means both truth and reality. The word ariya refers primarily to the ideal type of person the Buddhist path is supposed to generate, a noble person in the ethical and spiritual sense. Translating ariyasacca by 'noble truths' is somehow misleading because it gives the wrong impression of being a set of beliefs, a creed that Buddhists accept as noble and true. The four noble truths are primarily four realities whose contemplation leads to sainthood or the state of the noble ones (ariya). Other possible translations of ariyasacca are "ennobling truths" or "truths of the noble ones."
Each noble truth requires a particular practice from the disciple; in this sense the four noble truths can be understood as four types of practice. The first noble truth, or the reality of suffering, assigns to the disciple the practice of cultivating understanding. Such understanding takes place gradually through reflection, analytical meditation, and eventually direct experience. What needs to be understood is the nature of suffering, and the different types of suffering and happiness within saṃsāra.
A common misconception about the first noble truth is to think that it presupposes a pessimistic outlook on life. This interpretation would be correct only if the Buddha of the Pāli Nikāyas had not taught the existence of different types of happiness and the third noble truth, or cessation of suffering; that is, the good news about the reality of nirvana, defined as the highest happiness (Dhp.203; M.I.505). Since the Buddha of the Pāli Nikāyas teaches the reality of both suffering and the highest happiness, perhaps it is more accurate to speak of his attitude as realist: there is a problem but there is also a solution to that problem.
The second noble truth, or reality of the origin of suffering, calls for the practice of renunciation to all mental states that generate suffering for oneself and others. The mental state that appears in the second noble truth is taṇhā, literally "thirst." It was customary in the first Western translations of Buddhist texts (Burnouf, Fausboll, Muller, Oldenberg, Warren) to translate taṇhā by desire. This translation has misled many to think that the ultimate goal of Buddhists is the cessation of all desires. However, as Damien Keown puts it, "it is an oversimplification of the Buddhist position to assume that it seeks an end to all desire." (1992: 222).
In fact, there are many terms in the Pāli Nikāyas that can be translated as desire, not all of them related to mental states conducive to suffering. On the contrary, there are many texts in the Pāli Nikāyas that demonstrate the positive role of certain types of desire in the Buddha's path (Webster, 2005: 90-142). Nonetheless, the term taṇhā in the Pāli Nikāyas designates always a harmful type of desire that leads to "repeated existence" (ponobhavikā), is "associated with delight and lust" (nandirāgasahagatā), and "delights here and there" (tatra tatrābhinandinī) (M.I.48; D.II.308; etc). There is only one text (Nettipakaraṇa 87) that speaks about a wholesome type of taṇhā that leads to its own relinquishment, but this text is extra-canonical except in Myanmar.
The most common translation of taṇhā nowadays is craving. Unlike the loaded, vast, and ambivalent term desire, the term craving refers more specifically to a particular type of desire, and cannot be misinterpreted as conveying any want and aspiration whatsoever. Rather, like taṇhā in the Pāli Nikāyas, craving refers to intense (rāga can be translated by both lust and passion), obsessive, and addictive desires (the idiom tatra tatra can also be interpreted as connoting the idea of repetition or tendency to repeat itself).
Since craving, or taṇhā, does not include all possible types of desires, there is no "paradox of desire" in the Pāli Nikāyas. In other words, the Buddha of the the Pāli Nikāyas does not teach that in order to attain liberation from suffering one has to paradoxically desire to stop all desires. There is no contradiction in willing the cessation of craving. That is, for the Buddha of the Pāli Nikāyas it is possible to want, like, or strive for something without simultaneously craving for it.
The Pāli Nikāyas distinguish between three kinds of taṇhā: craving for sensual pleasures (kāmataṇhā), craving for existence (bhavataṇhā), and craving for non-existence (vibhavataṇhā). Following Webster, I understand the last two types of craving as "predicated on two extreme (wrong) views, those of eternalism and annihilationism" (2005:130-1). In other words, craving for existence longs for continued existence of one's self within saṃsāra, and craving for non-existence is a reversed type of desire or aversion to one's own destruction at the moment of death.
The underlying root of all suffering, however, is not craving but spiritual ignorance (avijjā). In the Pāli Nikāyas spiritual ignorance does not connote a mere lack of information but rather a misconception, a distorted perception of things under the influence of conceptual fabrications and affective prejudices. More specifically, ignorance refers to not knowing things as they are, the Dharma, and the four noble truths. The relinquishing of spiritual ignorance, craving, and the three roots of the unwholesome (greed or lobha, aversion or dosa, delusion or moha) entails the cultivation of many positive mental states, some of the most prominent in the Pāli Nikāyas being: wisdom or understanding (paññā), letting go (anupādāna), selflessness (alobha), love (avera, adosa, avyāpāda), friendliness (mettā), compassion (karuṇā), altruistic joy (muditā), equanimity (upekkhā), calm (samatha, passaddhi), mindfulness (sati), diligence (appamāda).
The third noble truth, or reality of the cessation of suffering, asks us to directly realize the destruction of suffering, usually expressed with a variety of cognitive and affective terms: peace, higher knowledge, the tranquilization of mental formations, the abandonment of all grasping, cessation, the destruction of craving, absence of lust, nirvana (Pali nibbāna). The most popular of all the terms that express the cessation of suffering and rebirth is nirvana, which literally means blowing out or extinguishing.
Metaphorically, the extinction of nirvana designates a mental event, namely, the extinguishing of the fires of craving, aversion, and delusion (S.IV.251). That nirvana primarily denotes a mental event, a psychological process, is also confirmed by many texts that describe the person who experiences nirvana with intransitive verbs such as to nirvanize (nibbāyati) or to parinirvanize (parinibbāyati). However, there are a few texts that seem to indicate that nirvana might also be a domain of perception (āyatana), element (dhātu), or reality (dhamma) known at the moment of enlightenment, and in special meditative absorptions after enlightenment. This domain is usually defined as having the opposite qualities of saṃsāra (Ud 8.1), or with metaphoric expressions (S.IV.369ff).
What is important to point out is that the concern of the Pāli Nikāyas is not to describe nirvana, which, strictly speaking, is beyond logic and language (It 37), but rather to provide a systematic explanation of the arising and cessation of suffering. The goal of Buddhism as it appears in the Pāli Nikāyas does not consist in believing that suffering arises and ceases like the Buddha says, but in realizing that what he teaches about suffering and its cessation is the case; that is, the Buddha's teaching, or Dharma, is intended to be experienced by the wise for themselves (M.I.265).
The fourth noble truth, or reality of the path leading to the cessation of suffering, imposes on us the practice of developing the eightfold ennobling path. This path can be understood either as eight mental factors that are cultivated by ennobled disciples at the moment of liberation, or as different parts of the entire Buddhist path whose practice ennoble the disciple gradually. The eight parts of the Buddhist path are usually divided into three kinds of training: training in wisdom (right view and right intention), ethical training (right speech, right bodily conduct, and right livelihood), and training in concentration (right effort, right mindfulness and right concentration).
c. Ontology of Suffering: the Five Aggregates
A prominent concern of the Buddha in the Pāli Nikāyas is to provide a solution to the problem of suffering. When asked about his teachings, the Buddha answers that he only teaches suffering and its cessation (M.I.140). The first noble truth describes what the Buddha means by suffering: birth, aging, illness, death, union with what is displeasing, separation from what is pleasing, not getting what one wants, the five aggregates of grasping (S.V.421).
The original Pali term for suffering is dukkha, a word that ordinarily means physical and mental pain, but that in the first noble truth designates diverse kinds of frustration, and the existential angst generated by the impermanence of life and the unavoidability of old age, disease, and death. However, when the Buddha of the Pāli Nikāyas mentions birth and the five aggregates of grasping, he seems to be referring to the fact that our psychophysical components are conditioned by grasping, and consequently, within saṃsāra, the cycle of births and deaths. This interpretation is consistent with later Buddhist tradition, which speaks about three types of dukkha: ordinary suffering (mental and physical pain), suffering due to change (derived from the impermanence of things), and suffering due to conditions (derived from being part of saṃsāra).
When the Buddha of the Pāli Nikāyas speaks about personal identity and the human predicament, he uses the technical expression "five aggregates of grasping" (pañcupādānakkhandhā). That is, the Buddha describes human existence in terms of five groups of constituents. The five aggregates are: material form (rūpa), sensations (vedanā), perceptions (saññā), mental formations (saṃkhāra), consciousness (viññāṇa). While the first aggregate refers to material components, the other four designate a variety of mental functions.
The aggregate material form is explained as the four great elements and the shape or figure of our physical body. The four great elements are earth, water, fire, and air. The earth element is further defined as whatever is solid in our body, and water as whatever is liquid. The fire element refers to "that by which one is warmed, ages, and is consumed," and the process of digestion. The air element denotes the breathing process and movements of gas throughout the body (M.I.185ff).
The aggregate sensations denote pleasant, unpleasant and neutral feelings experienced after there is contact between the six sense organs (eye, ear, nose, tongue, body, and mind) and their six objects (forms, sounds, odors, tastes, tangible objects, and mental phenomena). The aggregate perceptions express the mental function by which someone is able to identify objects. There are six types of perceptions corresponding to the six objects of the senses. The aggregate formations express emotional and intellectual dispositions, literally volitions (sañcetanâ), towards the six objects of the senses. These dispositions are the result of past cognitive and affective conditioning, that is, past karma or past voluntary actions. The aggregate consciousness connotes the ability to know and to be aware of the six objects of the senses (S.III.59ff).
d. Arguments for the Doctrine of Non-self
The Buddha reiterates again and again throughout the Pāli Nikāyas that any of the five aggregates "whether past, future or present, internal or external, gross or subtle, inferior or superior, far or near, ought to be seen as it actually is with right wisdom thus: 'this is not mine, this I am not, this is not my self.' " When the disciple contemplates the five aggregates in this way, he or she becomes disenchanted (nibbindati), lust fades away (virajjati), and he or she attains liberation due to the absence of lust (virāgā vimuccati) (M.I.138-9).
The Buddha of the Pāli Nikāyas justifies this view of the five aggregates as non-self with three main arguments, which are used as a method of analytical meditation, and in polemics with members of other schools. The assumption underlying the Buddha's arguments is that something might be considered a self only if it were permanent, not leading to suffering, not dependently arisen, and subject to one's own will. Since none of the five aggregates fulfill any of these conditions, it is wrong to see them as belonging to us or as our self.
In the first and most common argument for non-self the Buddha asks someone the following questions: "What do you think, monks, is material form permanent or impermanent?" – "Impermanent, venerable sir." – "Is what is impermanent suffering or happiness?" – "Suffering, venerable sir." –Is what is impermanent, suffering, and subject to change, fit to be regarded as: "this is mine, this I am, this is my self?" – "No, venerable sir" (M.I.138, etc). The same reasoning is applied to the other aggregates.
The first argument is also applied to the six sensual organs, the six objects, the six types of consciousness, perceptions, sensations, and formations that arise dependent on the contact between the senses and their objects (M.III.278ff). Sometimes the first argument for non-self is applied to the six senses and their objects without questions and answers: "Monks, the visual organ is impermanent. What is impermanent is suffering. What is suffering is non-self. What is non-self ought to be seen as it really is with right wisdom thus: 'this is not mine, this I am not, this is not my self' " (S.IV.1ff).
The second argument for non-self is much less frequent: "Monks, material form is non-self. If it were self, it would not lead to affliction. It would be possible [to say] with regard to material form: 'Let my material form be thus. Let my material form not be thus.' But precisely because it is non-self, it leads to affliction. And it is not possible [to say] with regard to material form: 'Let my material form be thus. Let my material form not be thus' "(S.III.66-7). The same reasoning is applied to the other four aggregates.
The third argument deduces non-self from that fact that physical and mental phenomena depend on certain causes to exist. For instance, in (M.III.280ff), the Buddha first analyzes the dependent arising of physical and mental phenomena. Then he argues: "If anyone says: 'the visual organ is self,' that is unacceptable. The rising and falling of the visual organ are fully known (paññāyati). Since the rising and falling of the visual organ are fully known, it would follow that: 'my self arises and falls.' Therefore, it is unacceptable to say: 'the visual organ is self.' Thus the visual organ is non-self." The same reasoning is applied to the other senses, their objects, and the six types of consciousness, contacts (meeting of sense, object and consciousness), sensations, and cravings derived from them.
The third argument also appears combined with the first one without questions and answers. For instance, in (A.V.188), it is said that "whatever becomes, that is conditioned, volitionally formed, dependently arisen, that is impermanent. What is impermanent, that is suffering. What is suffering, that is [to be regarded thus]: 'this is not mine, this I am not, this is not my self.' "
If something can be inferred from these three arguments, it is that the target of the doctrine of non-self is not all concepts of self but specifically views of the self as permanent and not dependently arisen. That is, the doctrine of non-self opposes what is technically called "views of personal identity" or more commonly translated "personality views" (sakkāyadiṭṭhi). Views of personal identity relate the five aggregates to a permanent and independent self in four ways: as being identical, as being possession of the self, as being in the self, or as the self being in them (M.I.300ff). All these views of personal identity are said to be the product of spiritual ignorance, that is, of not seeing with right wisdom the true nature of the five aggregates, their origin, their cessation, and the way leading to their cessation.
e. Human Identity and the Meaning of Non-self
Since the Pāli Nikāyas accept the common sense usages of the word "self" (attan, Skt. ātman), primarily in idiomatic expressions and as a reflexive pronoun meaning "oneself," the doctrine of non-self does not imply a literal negation of the self. Similarly, since the Buddha explicitly criticizes views that reject karma and moral responsibility (M.I.404ff), the doctrine of non-self should not be understood as the absolute rejection of moral agency and any concept of personal identity. In fact, the Buddha explicitly defines "personal identity" (sakkāya) as the five aggregates (M.I.299).
Since the sixth sense, or mind, includes the four mental aggregates, and since the ordinary five senses and their objects fall under the aggregate of material form, it can be said that for the Buddha of the Pāli Nikāyas personal identity is defined not only in terms of the five aggregates, but also in terms of the six senses and their six objects.
If the meaning of non-self were that there is literally no self whatsoever, no personal identity and no moral agency whatsoever, then the only logical conclusion would be to state that the Buddha taught nonsense and that the Pāli Nikāyas are contradictory, sometimes accepting the existence of a self and other times rejecting it. Even though no current scholar of Buddhism would endorse such an interpretation of non-self, it is still popular in some missionary circles and apologetic literature.
A more sympathetic interpretation of non-self distinguishes between two main levels of discourse (Collins 1982). The first level of discourse does not question the concept of self and freely uses personal terms and expressions in accordance with ordinary language and social conventions. The second level of discourse is philosophically more sophisticated and rejects views of self and personal identity as permanent and not dependently arisen. Behind the second level of discourse there is a technical understanding of the self and personal identity as the five aggregates, that is, as a combination of psychophysical processes, all of them impermanent and dependently originated.
This concept of the self as permanent and not dependently arisen is problematic because it is based on a misperception of the aggregates. This misperception of the five aggregates is associated with what is technically called "the conceit I am" (asmimāna) and "the underlying tendency to the conceits 'I' and 'mine' " (ahaṃkāra-mamaṅkāra-mānānusaya). This combination of conceit and ignorance fosters different types of cravings, especially craving for immortal existence, and subsequently, speculations about the past, present, and future nature of the self and personal identity. For instance, in (D.I.30ff), the Buddha speaks of different ascetics and Brahmins who claim that the self after death is "material, immaterial, both material and immaterial, neither material nor immaterial, finite, infinite, both, neither, of uniform perception, of varied perception, of limited perception, of unlimited perception, wholly happy, wholly miserable, both, neither." The doctrine of non-self is primarily intended to counteract views of the self and personal identity rooted in ignorance regarding the nature of the five aggregates, the conceit "I am," and craving for immortal existence.
A minority of scholars reject the notion that the Buddha's doctrine of non-self implies the negation of the true self, which for them is permanent and independent of causes and conditions. Accordingly, the purpose of the doctrine of non-self is simply to deny that the five aggregates are the true self. The main reason for this interpretation is that the Buddha does not say anywhere in the Pāli Nikāyas that the self does not exist; he only states that a self and what belongs to a self are not apprehended (M.I.138). Therefore, for these interpreters the Buddha of the Pāli Nikāyas only claims that impermanent and conditioned things like the five aggregates are not the true self. For these scholars, the Buddha does talk about the true self when he speaks about the consciousness of liberated beings (M.I.140), and the unconditioned, unborn and deathless nirvana (Bhattacarya 1973; Pérez Remón 1981).
However, the majority of Buddhist scholars agree with the traditional Buddhist self-understanding: they think that the doctrine of non-self is incompatible with any doctrine about a permanent and independent self, not just with views that mistakenly identify an alleged true self with the five aggregates. The main reason for this interpretation relates to the doctrine of dependent arising.
f. Causality and the Principle of Dependent Arising
The importance of dependent arising (paṭiccasamuppāda) cannot be underestimated: the Buddha realized its workings during the night of his enlightenment (M.I.167). Preaching the doctrine of dependent arising amounts to preaching the Dharma (M.II.32), and whoever sees it sees the Dharma (M.I.191). The Dharma of dependent arising remains valid whether or not there are Buddhas in the world (S.II.25), and it is through not understanding it that people are trapped into the cycle of birth and death (D.II.55).
The doctrine of dependent arising can be formulated in two ways that usually appear together: as a general principle or as a chain of causal links to explain the arising and ceasing of suffering and the process of rebirth. The general principle of dependent arising states that "when this exists, that comes to be; with the arising of this, that arises. When this does not exist, that does not come to be; with the cessation of this, that ceases" (M.II.32; S.II.28).
Unlike the logical principle of conditionality, the principle of dependent arising does not designate a connection between two ideas but rather an ontological relationship between two things or events within a particular timeframe. Dependent arising expresses not only the Buddha's understanding of causality but also his view of things as interrelated. The point behind dependent arising is that things are dependent on specific conditions (paṭicca), and that they arise together with other things (samuppāda). In other words, the principle of dependent arising conveys both ontological conditionality and the constitutive relativity of things. This relativity, however, does not mean that for the Buddha of the Pāli Nikāyas everything is interdependent or that something is related to everything else. This is a later development of Buddhist thought, not a characteristic of early Indian Buddhism.
The most comprehensive chain of dependent arising contains twelve causal links: (1) ignorance, (2) formations, (3) consciousness, (4) mentality-materiality, (5) the six senses, (6) contact, (7) sensations, (8) craving (9) grasping, (10) becoming, (11) birth, (12) old age and death. The most common formulation is as follows: with 1 as a condition 2 [comes to be]; with 2 as a condition 3 [comes to be], and so forth. Conversely, with the cessation of 1 comes the cessation of 2; with the cessation of 2 comes the cessation of 3, and so forth.
It is important to keep in mind that this chain does not imply a linear understanding of causality where the antecedent link disappears once the subsequent link has come to be. Similarly, each of the causal links is not to be understood as the one and only cause that produces the next link but rather as the most necessary condition for its arising. For instance, ignorance, the first link, is not the only cause of the process of suffering but rather the cause most necessary for the continuation of such a process. For the Buddha of the Pāli Nikāyas, as well as for later Buddhist tradition, there is always a multiplicity of causes and conditions at play.
The traditional interpretation divides the twelve link chain of dependent arising into three lives. The first two links (ignorance and formations) belong to the past life: due to a misperception of the nature of the five aggregates, a person (the five aggregates) performs voluntary actions: mental, verbal, and bodily actions, with wholesome, unwholesome, and neutral karmic effects. The next ten factors correspond to the present life: the karmic effects of past voluntary formations are stored in consciousness and transferred to the next life. Consciousness together with the other mental aggregates combines with a new physical body to constitute a new psychophysical organism (mentality-materiality). This new stage of the five aggregates develops the six senses and the ability to establish contact with their six objects. Contacts with objects of the senses produce pleasant, unpleasant and neutral sensations. If the sensations are pleasant, the person usually responds with cravings for more pleasant experiences, and if the sensations are unpleasant, with aversion. Craving and aversion, as well as the underlying ignorance of the nature of the five aggregates are fundamental causes of suffering and rebirth: the three roots of the unwholesome according to the Pāli Nikāyas, or the three mental poisons according to later Buddhist traditions.
By repeating the affective responses of craving and aversion, the person becomes more and more dependent on whatever leads to more pleasant sensations and less unpleasant ones. This creates a variety of emotional dependencies and a tendency to grasp or hold onto what causes pleasure and avoids pain. The Buddha of the Pāli Nikāyas speaks about four types of grasping: towards sensual pleasures, views, rites-and-observances, and especially towards doctrines of a [permanent and independent] self (D.II.57-8).
The original term for grasping is upādāna, which also designates the fuel or supply necessary to maintain a fire. In this sense, grasping is the psychological fuel that maintains the fires of craving, aversion, and delusion, the fires whose extinction is called nirvana. The Buddha's ideal of letting go and detachment should not be misunderstood as the absence of any emotions whatsoever including love and compassion, but specifically as the absence of emotions associated with craving, aversion, and delusion. Motivated by grasping and the three mental fires, the five aggregates perform further voluntary actions, whose karmic effects perpetuate existence within the cycle of rebirth and subsequent suffering. The last two links (birth, aging and death) refer to the future life. At the end of this present existence, a new birth of the five aggregates will take place followed by old age, death, and other kinds of suffering.
The twelve-link chain of dependent arising explains the processes of rebirth and suffering without presupposing a permanent and independent self. The Buddha of the Pāli Nikāyas makes this point explicit in his passionate rebuttal of the monk Sāti, who claimed that it is the same consciousness that wanders through the cycle of rebirth. For the Buddha, consciousness, like the other eleven causal links, is dependent on specific conditions (M.I.258ff), which entails that consciousness is impermanent, suffering, and non-self.
Instead of a permanent and independent self behind suffering and the cycle of rebirth, the Buddha of the Pāli Nikāyas presupposes five psychophysical sets of processes, namely, the five aggregates, which imply an impermanent and dependently-arisen concept of 'self' and 'personal identity.' In other words, the Buddha rejects substance-selves but accepts process-selves (Gowans 2003). Yet, the Buddha of the Pāli Nikāyas explicitly refuses to use personal terms such as 'self' in technical explanations of rebirth and suffering, and he prefers to speak in terms of causes and conditions that produce other causes and conditions (S.II.13-4; S.II.62; M.III.19). But what happens to consciousness and the other aggregates when grasping no longer exists and the three mental fires have been extinguished? What happens when suffering ceases and the cycle of rebirth stops?
4. Nirvana and the Silence of the Buddha
a. Two Kinds of Nirvana and the Undetermined Questions
When the fires of craving, aversion, and ignorance are extinguished at the moment of enlightenment, the aggregates are liberated due to the lack of grasping. This is technically called nirvana with remainder of grasping (saupādisesa-nibbāna), or as later tradition puts it, nirvana of mental defilements (kilesa-parinibbāna). The expression 'remainder of grasping' refers to the five aggregates of liberated beings, which continue to live after enlightenment but without negative mental states.
The aggregates of the liberated beings perform their respective functions and, like the aggregates of anybody else, they grow old, get sick, and are subject to pleasant and unpleasant sensations until death. The difference between unenlightened and enlightened beings is that enlightened beings respond to sensations without craving or aversion, and with higher knowledge of the true nature of the five aggregates.
The definition of nirvana without remainder (anupādisesa-nibbāna) that appears in (It 38) only says that for the liberated being "all that is experienced here and now, without enchantment [another term for grasping], will be cooled (sīta)." Since "all" is defined in the Pāli Nikāyas as the six senses and their six objects (S.IV.15), which is another way of describing the individual psychophysical experience or the five aggregates, the expression "all that is experienced" refers to what happens to the aggregates of liberated beings. Since (It 38) explicitly uses the expression "here and now" (idheva), it seems impossible to conclude that the definition of nirvana without remainder is intended to say anything about nirvana or the aggregates beyond death. Rather (It 38) describes nirvana and the aggregates at the moment of death: they will be no longer subject to rebirth and they will become cooled, tranquil, at peace. The question is: what does this peace or coolness entail? What happens after the nirvana of the aggregates? Does the mind of enlightened beings survive happily ever after? Does the liberated being exist beyond death or not?
These questions are left undetermined (avyākata) by the Buddha of the the Pāli Nikāyas. The ten questions in the the Pāli Nikāyas ask whether (1) The world is eternal; (2) The world is not eternal; (3) The world is infinite; (4) The world is finite; (5) Body and soul are one thing; (6) Body and soul are two different things; (7) A liberated being (tathāgata) exists after death; (8) A liberated being (tathāgata) does not exist after death; (9) A liberated being (tathāgata) both exists and does not exist after death; (10) A liberated being (tathāgata) neither exists nor does not exist after death. In Sanskrit Buddhist texts the ten views become fourteen by adding the last two possibilities of the tetralema (both A and B, neither A nor B) to the questions about the world.
Unfortunately for those looking for quick answers, the Buddha of the Pāli Nikāyas does not provide a straightforward yes or no response to any of these questions. When the Buddha is asked whether the liberated being exists, does not exist, both, or neither, he sets aside these questions by saying that (1) he does not hold such views, (2) he has left the questions undetermined, and (3) the questions do not apply (na upeti). The first two answers are also used to respond to questions about the temporal and spatial finitude or infinitude of the world, and the identity or difference between the soul and the body. Only the third type of answer is given to the questions about liberated beings after death.
Most presentations of early Buddhism interpret these three answers of the Buddha as an eloquent silence about metaphysical questions due primarily to pragmatic reasons, namely, the questions divert from spiritual practice and are not conducive to liberation from suffering. While the pragmatic reasons for the answers of the Buddha are undeniable, it is inaccurate to understand them as silence about metaphysical questions. In fact, the Buddha of the Pāli Nikāyas does address many metaphysical issues with his teachings of non-self and dependent arising.
The answers of the Buddha to the undetermined questions are due not only to pragmatic reasons but also to metaphysical reasons: the questions are inconsistent with the doctrines of non-self and dependent arising because they assume the existence of a permanent and independent self, a self that is either finite or infinite, identical or different from the body, existing or not existing after death. Besides pragmatic and metaphysical reasons, there are cognitive and affective reasons for the answers of the Buddha: the undetermined questions are based on ignorance about the nature of the five aggregates and craving for either immortal existence or inexistence. The questions are expressions of 'identity views,' that is, they are part of the problem of suffering. Answering the questions directly would have not done any good: a yes answer would have fostered more craving for immortal existence and led to eternalist views, and a no answer would have fostered further confusion and led to nihilist views (S.IV.400-1).
In the case of the undetermined questions about the liberated being, there are also apophatic reasons for answering "it does not apply." The Buddha of the Pāli Nikāyas illustrates the inapplicability of the questions with the simile of the fire extinct: just as it does not make sense to ask about the direction in which an extinct fire has gone, it is inappropriate to ask about the status of the liberated being beyond death: "The fire burned in dependence on its fuel of grass and sticks. When that is used up, if it does not get any more fuel, being without fuel, it is reckoned as extinguished. Similarly, the enlightened being has abandoned the five aggregates by which one might describe him…he is liberated from reckoning in terms of the five aggregates, he is profound, immeasurable, unfathomable like the ocean" (M.I.487-8).
b. Eternalism, Nihilism, and the Middle Way
There are three possible interpretations of the simile of the extinct fire: (1) liberated beings no longer exist beyond death (2) liberated beings exist in a mysterious unfathomable way beyond death (3) the Buddha is silent about both the liberated being and nirvana after death. The first interpretation seems the most logical conclusion given the Buddha's ontology of suffering and the doctrine of non-self. However, the nihilist interpretation makes Buddhist practice meaningless and contradicts texts where the Buddha criticizes teachings not conducive to spiritual practice such as materialism and determinism (M.I.401ff). But more importantly, the nihilist interpretation is vehemently rejected in the Pāli Nikāyas: "As I am not, as I do not proclaim, so have I been baselessly, vainly, falsely, and wrongly misrepresented by some ascetics and brahmins thus: 'the ascetic Gotama [Buddha] is one who leads astray; he teaches the annihilation, the destruction, the extermination of an existing being' "(M.I.140).
The second interpretation appears to some as following from the Buddha's incontrovertible response to the nihilist reading of his teachings: since the Buddha rejects nihilism, he must somehow accept the eternal existence of liberated beings, or at least the eternal existence of nirvana. For eternalist interpreters, the texts in the Pāli Nikāyas that speak about the transcendence and ineffability of liberated beings and nirvana can be understood as implying their existence after or beyond death.
There are several eternalist readings of the Buddha's thought. We have already mentioned the most common: the doctrine of non-self merely states that the five aggregates are not the true self, which is the transcendent and ineffable domain of nirvana. However, there are eternalist interpretations within Buddhism too. That is, interpretations that are nominally consistent with the doctrine of non-self but that nevertheless speak of something as eternally existing: either the mind of liberated beings or nirvana. For instance, Theravāda Buddhists usually see nirvana as non-self, but at the same time as an unconditioned (asaṃkhata) and deathless (amata) reality. The assumption, though rarely stated, is that liberated beings dwell eternally in nirvana without a sense of "I" and "mine," which is a transcendent state beyond the comprehension of unenlightened beings. Another eternalist interpretation is that of the Dalai Lama who, following the standard interpretation of Tibetan Buddhists, claims that the Buddha did not teach the cessation of all aggregates but only of contaminated aggregates. That is, the uncontaminated aggregates of liberated beings continue to exist individually beyond death, though they are seen as impermanent, dependently arisen, non-self, and empty of inherent existence (Dalai Lama 1975:27). Similarly, Peter Harvey understands nirvana as a selfless and objectless state of consciousness different from the five aggregates that exists temporarily during life and eternally beyond death (1995: 186-7).
The problem with eternalist interpretations is that they contradict what the Pāli Nikāyas say explicitly about the way to consider liberated beings, the limits of language, the content of the Buddha's teachings, and dependent arising as a middle way between the extremes of eternalism and annihilationism. In (S.III.110ff), Sāriputta, the Buddha's leading disciple in doctrinal matters, explains that liberated beings should be considered neither as annihilated after death nor as existing without the five aggregates.
In (D.II.63-4) the Buddha makes clear that consciousness and mentality-materiality, that is, the five aggregates, are the limits of designation (adhivacana), language (nirutti), cognitions (viññatti), and understanding (paññā). Accordingly, in (D.II.68) the Buddha says it is inadequate to state that the liberated being exists after death, does not exist, both, or neither. This reading is confirmed by (Sn 1076): "There is not measure (pamāṇa) of one who has gone out, that by which [others] might speak (vajju) of him does not exist. When all things have been removed, then all ways of speech (vādapathā) are also removed."
Given the Buddha's understanding of the limits of language and understanding in the Pāli Nikāyas, it is not surprising that he responded to the accusation of teaching the annihilation of beings, by saying that "formerly and now I only teach suffering and the cessation of suffering." Since the Buddha does not teach anything beyond the cessation of suffering at the moment of death, that is, beyond the limits of language and understanding, it is inaccurate to accuse him of teaching the annihilation of beings. Similarly, stating that liberated beings exist after death in a mysterious way beyond the four logical possibilities of existence, non-existence, both or neither, is explicitly rejected in (S.III.118-9) and (S.IV.384), where once again the Buddha concludes that he only makes known suffering and the cessation of suffering.
If the eternalist interpretation were correct, it would have been unnecessary for the Buddha of the Pāli Nikāyas to put so much emphasis on the teaching of dependent arising. Why would dependent arising be defined in (S.II.17) as right view and as the middle way between the extremes of eternalism and annihilationism if the truth were that the consciousness of liberated beings or the unconditioned nirvana exist eternally? If knowing and seeing dependent arising precludes someone from speculating about a permanent self in the past and the future (M.I.265), why would the Buddha teach anything about the eternal existence of liberated beings and nirvana?
In order to avoid the aforementioned contradictions entailed by eternalist readings of the Pāli Nikāyas, all texts about nirvana and the consciousness of liberated beings are to be understood as referring to this life or the moment of death, never to some mysterious consciousness or domain that exists beyond death. Since none of the texts about nirvana and liberated beings found in the Pāli Nikāyas refer unambiguously to their eternal existence beyond death, I interpret the Buddha as being absolutely silent about nirvana and liberated beings beyond death (Vélez de Cea 2004a). In other words, nothing of what the Pāli Nikāyas say goes beyond the limits of language and understanding, beyond the content of the Buddha's teachings, and beyond dependent arising as the middle way between eternalism and annihilationism.
Instead of focusing on nirvana and liberated beings beyond death, the Buddha of the Pāli Nikāyas emphasizes dependent arising and the practice of the four foundations of mindfulness. Dependent arising is intended to avoid views about a permanent and independent self in the past and the future (M.I.265; M.III.196ff), and the four foundations of mindfulness are said to be taught precisely to destroy such views (D.III.141). That is, the Buddha's fundamental concern is to address the problem of suffering in the present without being distracted by views about the past or the future: "Let not a person revive the past, or on the future build his hopes; for the past has been left behind and the future has not been reached. Instead with insight let him see each presently arising state (paccuppannañca yo dhammaṃ tattha tattha vipassati); let him know that and be sure of it, invincibly, unshakeably. Today the effort must be made, tomorrow death may come, who knows?" (Bhikkhu Bodhi's translation. M.III.193).
5. Buddhist Ethics
Early Buddhist ethics includes more than lists of precepts and more than the section on ethical training of the eightfold noble path; that is, Buddhist ethics cannot be reduced to right action (abstaining from killing, stealing, lying), right speech (abstaining from false, divisive, harsh, and useless speech), and right livelihood (abstaining from professions that harm living beings). Besides bodily and verbal actions, the Pāli Nikāyas discuss a variety of mental actions including thoughts, motivations, emotions, and perspectives. In fact, it is the ethics of mental actions that constitutes the main concern of the Buddha's teaching.
Early Buddhist ethics encompasses the entire spiritual path, that is, bodily, verbal, and mental actions. The factors of the eightfold noble path dealing with wisdom and concentration (right view, right intentions, rights effort, right concentration, right mindfulness) relate to different types of mental actions. The term "right" (sammā) in this context does not mean the opposite of "wrong," but rather "perfect" or "complete;" that is, it denotes the best or the most effective actions to attain liberation. This, however, does not imply that the Buddha advocates the most perfect form of ethical conduct for all his disciples.
Early Buddhist ethics is gradualist in the sense that there are diverse ways of practicing the path with several degrees of commitment; not all disciples are expected to practice Buddhist ethics with the same intensity. Monks and nuns take more precepts and are supposed to devote more time to spiritual practices than householders. However, a complete monastic code (prātimoka) like those found in later Vinaya literature does not appear in the Pāli Nikāyas. The most comprehensive formulation of early Buddhist ethics, probably common to monastic disciples and lay people, is the list of ten dark or unwholesome actions and their opposite, the ten bright or wholesome actions: three bodily actions (abstaining from killing, stealing, sexual misconduct), four verbal actions (abstaining from false, divisive, harsh, and useless speech), and tree mental actions (abstaining from covetousness, ill-will, and dogmatic views).
The Buddha of the Pāli Nikāyas defines action in terms of intention or choice (cetanā): "It is intention, monks, what I call action. Having intended, someone acts through body, speech, and mind" (A.III.415). The Pāli Nikāyas define the roots of unwholesome (akusala) actions as greed (lobha), aversion (dosa), and delusion (moha). Conversely, the roots of wholesome actions are defined as the opposite mental states (M.I.47). Some scholars infer from these two definitions that Buddhist ethics is an ethics of intention or an agent-based form of virtue ethics. That is, according to these scholars, for the Buddha of the the Pāli Nikāyas, only the agent's intention or motivation determine the goodness of actions. This interpretation, however, is disproved by many texts of the Pāli Nikāyas where good and evil actions are discussed without any reference to the underlying intention or motivation of the agent. Consequently, the more comprehensive account understands intention not as the only factor that determines the goodness of actions, but rather as the condition of possibility, the necessary condition for speaking about action in the moral sense. Without intention or choice, there is no ethical action. Similarly, motivation, while a central moral factor in Buddhist ethics, is neither the only factor nor always the most important factor to determine the goodness of actions. Understanding Buddhist ethics as concerned exclusively with the three roots of the wholesome does not fully capture the breath of moral concern of the Pāli Nikāyas (Vélez de Cea 2004b).
The fundamental moral law of the universe according to early Buddhism is what is popularly called the "law of karma": good actions produce good consequences, and bad actions lead to bad consequences. The consequences of volitional actions can be experienced in this life or in subsequent lives. Although not everything we experience is due to past actions, physical appearance, character, lifespan, prosperity, and rebirth destination are believed to be influenced by past actions. This influence however, is not to be confused with fatalism, a position rejected in the Pāli Nikāyas. There is always room for mitigating and even eradicating the negative consequences of past actions with new volitions in the present. That is, past karma does not dictate our situation: the existence of freewill and the possibility of changing our predicament is always assumed. There is conditioning of the will and other mental factors, but no hard determinism.
A common objection to early Buddhist ethics is how there can be freewill and responsibility without a permanent self that transmigrates through lives. If there is no self, who is the agent of actions? Who experiences the consequences of actions? Is the person who performs an action in this life the same person that experiences the consequences of that action in a future life? Is it a different person? The Buddha considers these questions improper of his disciples, who are trained to explain things in terms of causes and condition (S.II.61ff; S.II.13ff)). In other words, since the Buddha's disciples explain processes with the doctrine of dependent arising, they should avoid explanations that use personal terms and presuppose the extremes of eternalism and nihilism. The moral agent is not a substance-self but rather the five aggregates, a dynamic and dependently-arisen process-self who, like a flame or the water of a river, changes all the time and yet has some degree of continuity.
The most common interpretations of early Buddhist ethics view its nature as either a form of agent-based virtue ethics or as a sophisticated kind of consequentialism. The concern for virtue cultivation is certainly prevalent in early Buddhism, and evidently the internal mental state or motivation underlying actions is extremely important to determine the overall goodness of actions, which is the most important factor for advanced practitioners. Similarly, the concern for the consequences of actions, whether or not they lead to the happiness or the suffering of oneself and others, also pervades the Pāli Nikāyas. However, the goodness of actions in the Pāli Nikāyas does not depend exclusively on either the goodness of motivations or the goodness of consequences. Respect to status and duty, observance of rules and precepts, as well as the intrinsic goodness of certain external bodily and verbal actions are equally necessary to assess the goodness of at least certain actions. Since the foundations of right action in the Pāli Nikāyas are irreducible to one overarching principle, value or criterion of goodness, early Buddhist ethics is pluralistic in a metaethical sense. Given the unique combination of deontological, consequentialist, and virtue ethical trends found in the Pāli Nikāyas, early Buddhist ethics should be understood in its own terms as a sui generis normative theory inassimilable to Western ethical traditions.
6. References and Further Reading
a. Primary Sources
All references to the Pāli Nikāyas are to the edition of The Pāli Text Society, Oxford. References to the Aṅguttara, Dīgha, Majjhima, and Saṃyutta Nikāyas are to the volume and page number. References to Udāna and Itivuttaka are to the page number and to Dhammapada and Sutta Nipāta to the verse number.
A. Aṅguttara Nikāya
D. Dīgha Nikāya
M. Majjhima Nikāya
S. Saṃyutta Nikāya
Ud. Udāna
It. Itivuttaka
Dhp. Dhammapada
Sn. Sutta Nipāta
b. Secondary Sources
Bechert, H. (Ed) 1995. When Did the Buddha Live? The Controversy on the Dating of the Historical Buddha. Selected Papers Based on a Symposium Held under the Auspices of the Academy of Sciences in Göttingen. Delhi: Sri Satguru Publications. 1995.
Bhattacharya, K. 1973. L´Ātman-Brahman dans le Bouddhisme Ancien. París: EFEO.
Bhikkhu Ñānamoli and Bhikkhu Bodhi. 1995. The Middle Length Discourses of the Buddha. A New Translation of the Majjhima Nikāya. Kandy: Buddhist Publication Society.
Bhikkhu Ñāṇananda. 1971. Concept and Reality in Early Buddhist Thought. Kandy: Buddhist Publication Society.
Cousins, L.S. 1996. "Good or Skillful? Kusala in Canon and Commentary." Journal of Buddhist Ethics.Vol. 3: 133-164.
Collins, S. 1982. Selfless Persons. Cambridge: Cambridge University Press.
Collins, S. 1994. "What are Buddhists Doing When They Deny the Self?" In Religion and Practical Reason, edited by Frank E. Reynolds and David Tracy. Albany: SUNY.
Collins, S. 1998. Nirvana and other Buddhist Felicities. Cambridge: Cambridge University Press
Dalai Lama. 1994. The Way to Freedom. San Francisco: Harper.
Dharmasiri, G. 1996. Fundamentals of Buddhist Ethics. Singapore: Buddhist Research Society.
Fuller, P. 2005. The Notion of Diṭṭhi in Theravāda Buddhism. London: RoutledgeCurzon.
Gombrich, R. 1988. Theravāda Buddhism: A Social History from Ancient Benares to Modern Colombo. London: Routledge.
Gombrich, R. 1996. How Buddhism Began. London: Athlone.
Gethin, R. 2001. The Buddhist Path to Awakening. Richmon Surrey: Curzon Press.
Gowans, C. W. 2003. Philosophy of the Buddha. London: Routledge.
Hallisey, C. 1996. "Ethical Particularism in Theravāda Buddhism." Journal of Buddhist Ethics. Vol. 3: 32-34.
Hamilton, S. 2000. Early Buddhism: A New Approach. Richmon Surrey: Curzon Press.
Harvey, P. 1995. The Selfless Mind: Personality, Consciousness, and Nirvana in Early Buddhism. Richmon Surrey: Curzon Press.
Harvey, P. 1995. "Criteria for Judging the Unwholesomeness of Actions in the Texts of Theravāda Buddhism." Journal of Buddhist Ethics. Vol. 2: 140-151.
Harvey, P. 2000. An Introduction to Buddhist Ethics. Cambridge: Cambridge University Press.
Hoffman, F. J. 1987. Rationality and Mind in Early Buddhism. New Delhi: Motilal Banarsidass.
Hwang, S. 2006. Metaphor and Literalism in Buddhism: The Doctrinal History of Nirvana. London: RoutledgeCurzon.
Jayatilleke, K. N. 1963. Early Buddhist Theory of Knowledge. London: Allen & Unwin.
Johansson, R. 1969. The Psychology of Nirvana. London: Allen and Unwin Ltd.
Kalupahana, D. 1976. Buddhist Philosophy: A Historical Analysis. Honolulu: University Press of Hawai'i.
Kalupahana, D. 1992. A History of Buddhist Philosophy: Continuities and Discontinuities. Honolulu: University Press of Hawai'i.
Keown, D. 1992. The Nature of Buddhist Ethics. New York: Palgrave.
Norman, K. R. 1983. Pāli Literature: Including the Canonical Literature in Prakrit and Sanskrit of all the Hīnayāna schools of Buddhism. Wiesbaden: Otto Harrassowitz.
Norman, K. R. 1990-6. Collected Papers. Oxford: The Pāli Text Society.
Pande, G.C. 1995. Studies in the Origins of Buddhism. New Delhi: Motilal Banarsidass.
Pérez-Remón, J. 1980. Self and Non-Self in Early Buddhism. New York: Mouton.
Perret, R. 1986. "Egoism, Altruism, and Intentionalism in Buddhist Ethics." Journal of IndianPhilosophy. Vol. 15: 71-85.
Premasiri, P. D. 1987. "Early Buddhist Concept of Ethical Knowledge: A Philosophical Analysis." Kalupahana, D.J. and Weeraratne, W.G. eds. Buddhist Philosophy and Culture: Essays in Honor of N.A. Jayawickrema. Colombo: N.A. Jayawickrema Felicitation Volume Committee. Pp. 37-70.
Ronkin, N. 2005. Early Buddhist Metaphysics: The Making of a Philosophical Tradition. London: RoutledgeCurzon.
Tilakaratne, A. 1993. Nirvana and Ineffability: A Study of the Buddhist Theory of Reality and Languague. Colombo: Karunaratne and Sons.
Vélez de Cea , A. 2004 a. "The Silence of the Buddha and the Questions about the Tathāgata after Death." The Indian International Journal of Buddhist Studies, no 5.
Vélez de Cea , A. 2004 b. "The Early Buddhist Criteria of Goodness and the Nature of Buddhist Ethics."Journal of Buddhist Ethics 11, pp.123-142.
Vélez de Cea , A. 2005. "Emptiness in the Pāli Suttas and the Question of Nāgārjuna's Orthodoxy."Philosophy East and West. Vol. 55: 4.
Webster, D. 2005. The Philosophy of Desire in the Pali Canon. London: RoutledgeCurzon.
See also the Encylopedia articles on Madhyamaka Buddhism and Pudgalavada Buddhism.
Abraham Velez
Email: abraham.velez@eku.edu | {
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Q: Android L: LinkageError crashes application As part of testing an existing Android app against the L preview, the app is crashing with the following exception:
07-08 10:05:39.024: E/AndroidRuntime(2126): FATAL EXCEPTION: main
07-08 10:05:39.024: E/AndroidRuntime(2126): Process: com.example, PID: 2126
07-08 10:05:39.024: E/AndroidRuntime(2126): java.lang.LinkageError: com.example.BaseActivity
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexFile.defineClassNative(Native Method)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexFile.defineClass(DexFile.java:222)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexFile.loadClassBinaryName(DexFile.java:215)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexPathList.findClass(DexPathList.java:321)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.BaseDexClassLoader.findClass(BaseDexClassLoader.java:54)
The app works fine in API levels 14-19, but crashes on start up with this cryptic error on the L preview. The compile/targetSdk version is still set to API 19.
A: The problem was a conflicting method in BaseActivity. I had created a helper method called getDrawable(int drawableId) that had the same signature as a method added to the Activity class in Android L.
This conflict caused the LinkageError. To fix the issue, I simply renamed my helper method so it wouldn't conflict with the new built-in method.
A: 07-08 10:05:39.024: E/AndroidRuntime(2126): FATAL EXCEPTION: main
07-08 10:05:39.024: E/AndroidRuntime(2126): Process: com.example, PID: 2126
07-08 10:05:39.024: E/AndroidRuntime(2126): java.lang.LinkageError: com.example.BaseActivity
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexFile.defineClassNative(Native Method)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexFile.defineClass(DexFile.java:222)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexFile.loadClassBinaryName(DexFile.java:215)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.DexPathList.findClass(DexPathList.java:321)
07-08 10:05:39.024: E/AndroidRuntime(2126): at dalvik.system.BaseDexClassLoader.findClass(BaseDexClassLoader.java:54)
| {
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<?php include dirname(__FILE__).'/exception.xml.php' ?> | {
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{"url":"http:\/\/steventhornton.ca\/blog\/2\/","text":"## Markov Chains in LaTeX\n\nThe TikZ package is a great tool for generating publication quality illustrations of Markov chains in LaTeX. This article presents 3 examples along with the complete LaTeX source. At each step, links are provided to the LaTeX source on Overleaf where you can view and edit the examples.","date":"2019-09-16 08:04:45","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8373607397079468, \"perplexity\": 972.2645677495118}, \"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-2019-39\/segments\/1568514572516.46\/warc\/CC-MAIN-20190916080044-20190916102044-00415.warc.gz\"}"} | null | null |
Кристиан Альбрехт Йенсен (; , Бредштедт — , Копенгаген) — датский художник-портретист, один из представителей Золотого века датского искусства.
Жизнь и творчество
Родился в семье перчаточника.
Художественное образование получил в 1810—1816 годах а Датской королевской академии искусств в Копенгагене, под руководством К. А. Лоренцена. В 1817—1818 годах Йенсен учится в дрезденской Академии художеств.
В 1818 году он совершает поездку в Италию, посещает по пути Вену, Венецию, Болонью и Флоренцию. Затем больше года живёт в Риме, где знакомится со скульптором Б. Торвальдсеном. В папской столице датский художник близко сходится с представителями местной датско-немецкой культурной колонии, пишет портреты многих её представителей. Покинув Италию, Йенсен на некоторое время оседает в Гамбурге, где живёт как свободный художник. Зимой 1822/23 годов он возвращается в Копенгаген, и женится на Катрин Лоренцен.
За свою долгую жизнь живописца мастер написал более 400 полотен, в том числе крупнейших современных ему представителей науки и искусства (портреты Гаусса (1840), Торвальдсена и др.). С 1832 года он работает при Королевском собрании произведений искусств в замке Фредериксборг, сперва как копиист, затем как хранитель. С 1835 года он — профессор живописи. В 1837 году мастер выполняет для Пулковской обсерватории под Санкт-Петербургом серию из 11 полотен — портретов известных учёных. Свои два последние портрета Йенсен создаёт в 1854 году — Андреаса Кристиана Крога и теолога Андреаса Готлиба Рудельбаха. Эти работы признаются специалистами как одни из лучших в творчестве мастера.
Галерея
Примечания
Литература
Биографическая статья в Dansk biografisk leksikon
Художники Дании
Художники-портретисты Дании | {
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We sit on the driftwood and write our names in charcoal.
slide back down my throat. I hate it when they talk to me too.
and I want to believe us.
long before the introduction of pollutants.
yet, so we take off our shoes and dip our feet in the water.
you don't want to get old.
we get back in the car and turn up the heat.
Olivia Scarlet Hoffman is a 20 year old Oakland native currently perusing her BFA in Creative Writing at the University of British Columbia. Her writing can be found on her instagram, @oliviapoetics, as well as on GROUNDERS magazine's online blog. | {
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$even $isters (TV series)
The Dark Cure (feature movie)
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Nippon Park (feature movie)
East Pole (feature /TV series)
Grünenberg Film Production
Imperial Affairs (feature movie)
Cyprus Connection (TV series)
Discovery of the East Pole
Samurai Diet
The Pilgrim's Sword
Laws of Singularity
Die Entdeckung des Ostpols
Samurai-Diät
Das Ende der Bundesrepublik
Politische Subjektivität
Psychopathia antisemitica (D)
Psychopathia antisemitica (E)
Was ist ein Demokrat?
What is a Democrat?
Ritter Conrad, mein Vater und ich
The Uniqueness Hypothesis
Gründungsgewalt und Politik
Blog (E/G)
reggie's world
Psychopathia Antisemitica
By Reginald Grünenberg
Anti-Semitism is not an opinion, and even less a political judgement*. It's a clinical personality disorder.
First published in Frankfurter Allgemeine Sonntagszeitung of April 8, 2012, p. 9, under the title Die Welterklärungsmaschine.
Jews are strange people. Therefore, I'll have to tell a little story. When I moved to Paris in 1984 to study political science at Sciences Po, my father lived there with his fiancé Raymonde Silberring, a Moroccan Jew with an Israeli passport. She was well-off because her late husband, the cacique of Tel Aviv – a personality with enough charisma and authority to settle disputes between enemy gangsters and entire gangs in the underworld – had run an extremely lucrative nightclub and left her a considerable fortune. She had a heart of gold and there was always something going on with her, but when she was in a bad mood, she would rip the power cables out of the wall. My father was proud of her heritage, for she was a direct descendant from Eleazar Ben Ya'ir, the legendary leader of the last Jewish resistance to the Roman invasion. When the conquerors stormed the Masada fortress on a mesa in 73 AD, Ben Ya'ir persuaded his one thousand followers to commit collective suicide. The Romans were shocked, and their historians reported this event with awe. Since then, Masada has been considered the strongest symbol of the Jewish will for freedom.
Raymonde, who played poker or baccarat in the Club des Milliardaires, the Billionaire's Club, knew many wealthy Jews and regularly held sumptuous dinners for her friends. At the magnificent table with much silver, crystal, goose liver pâté (kosher), Beluga caviar (possibly not kosher, since the sturgeon has fins but no scales, whereby the fins could also be a kind of scales – a hotly disputed question since Rabbi Maimonides in the 12th century), Bœuf Stroganoff (not kosher at all) and wines from the estate Lafite-Rothschild (not kosher, but Baron Edmond de Rothschild had at least brought viticulture to Israel in 1882 where they also produced kosher wines), there were always two Gojim, my father and myself.
It was a surreal and initially frightening world for me, because until then I only knew Jews from documentaries in black and white, as they were shown at school. There they were emaciated miserable figures wearing the infamous yellow badge, the Star of David, on worn-out clothes. Or corpses dug into mass graves by bulldozers after the concentration camps were liberated. Jean-Paul instead sold islands in the Caribbean, Bernard organized the Paris-Dakar Rally, and Bernadette owned a delicatessen chain with her pretty daughter Féline. They were always in a good mood, and apart from 'la bouffe', the eternal number one theme in Paris, wining and dining, they preferred to talk about real estate, jewelry, cars, and other people. When it came to historical, political or literary questions, they gladly turned to my father or myself, because these children of Abraham were not as much interested in education as is commonly imagined with Jews. What surprised me back then was how they could sit at the table with both of us Germans so lightheartedly. I imagined what it would have been like to be born a Jew. That's what scared me. For I would have become an angry, yes, vengeful Jew, certainly also an ardent Zionist, who would have consecrated a house altar to Theodor Herzl, the author of The Jewish State (1896) and founder of Zionism. Initially, I expected that at some point during these illustrious societies, which at later hours slipped into permissive carousals, the restrained hatred would break through. But nothing happened. On the contrary, these were some of the most beautiful and exciting evenings of my student life in Paris. These and later encounters have repeatedly confirmed that Jews are strange people, for they have already forgiven us Germans for one of the greatest crimes in human history. In this way, the model of the Jews of all people has made the most difficult teaching of the Christian faith accessible to me, namely the forgiveness of evil.
But if Jews are strange people, how much stranger are anti-Semites, especially German anti-Semites? In view of the fact of the Holocaust and the lack of retaliation by the Jews, I simply could no longer imagine how anyone could justify his hatred of the Jews. All veils of anti-Semitic myths seemed to have long since been lifted. But appearances are deceptive. A commission of experts of the German Bundestag, the federal parliament, confirmed the studies of the last decades by stating that even today 20 percent of Germans are latently anti-Semitic. This makes anti-Semitism more common and widespread than depression (5%), asthma (7%) or diabetes (8%). There are two topics with which the anti-Semite can express his concern as justified indignation, unchallenged up to the best circles, namely the German compensation payments and the existence of the State of Israel.
Yet, the question of reparations is settled long since. The purely material damage suffered by Jews from German persecution and the Holocaust is estimated at 230 to 380 billion US dollars. The total compensation that Germany has paid to date and will continue to pay is only 100 billion US dollars, half of it in the form of modest pensions to victims of National Socialism living in Israel. But the founder of the Jewish World Congress Nahum Goldmann, who had negotiated the Israel Treaty with Konrad Adenauer in 1952, in which only 3.5 billion Deutschmarks were agreed upon for compensations, was so surprised by the expansion of the payments in the following years that he wrote back in 1978: "One cannot therefore accuse the Germans of being petty and of not having kept their promises".
Concerning the foundation and the existence of the state of Israel since 1948, the anti-Semite claims that both are controversial among Jews themselves. With books such as The Holocaust Industry (2000) and Beyond Chutzpah. On the Misuse of Anti-Semitism and the Abuse of History (2005), Norman Finkelstein, a legal fundamentalist, political dimwit and Hezbollah admirer, fights against Alan Dershowitz, the conservative hard-boiled star lawyer from New York, who defends the country with his brilliant The Case for Israel (2004). The German anti-Semite cheers Finkelstein, experiences his coming-out as an anti-Zionist, and is committed to the liberation of Palestine. Anti-Zionism, however, is based on a dramatic transfiguration of the six wars and two intifadas with which the neighboring Arab states not only wanted to defeat Israel since 1948, but to destroy it. For as presumptuous and precarious as the founding of a Jewish state in Palestine was – because of the flight and expulsion of 700,000 Arabs and despite the legal Jewish settlement since 1880 – it was so justified by Israel's repeated victories over the Arab forces and its ethnic cleansing plans. That Israel grew stronger and bigger through every further assault without declaration of war, with every act of violence and terror, the aggressors have still not understood this dialectic. The State of Israel in its present form is the work of its Arab enemies. The Palestinian people, an entity invented by the Arabs – for until 1967 there were no plans on the Arab side for an independent Palestine – can only be commiserated in this context. They were instrumentalized for decades by the notorious Arab warmongers and permanent war-losers. Even worse, their own corrupt Palestinian leadership boycotted the two-state solution in Camp David in 2000 when it was actually within reach.
If these reasons for hatred of the Jews have become untenable, where does it come from? And why the Jews? Over the years and many encounters with anti-Semites, there has been growing evidence that this is a kind of mental illness that is probably incurable. Anti-Semitism is a one-way ticket. The strongest indication of this is the fact that there are no testimonies of 'former anti-Semites'. The assumption that anti-Semitism and its afterbirth, anti-Zionism, are indeed forms of clinical personality disorder has already been expressed several times. The Romanian physician and Zionist Karpel Lippe wrote already in 1887 about symptoms of the anti-Semitic mental illness and the philosopher Constantin Brunner published 191 9 his book Der Judenhass und die Juden, 'The Hatred of Jews and the Jews'. In it he stated: "Psychopathia antisemitica belongs to general psychology, more precisely to psychological anthropology". Brunner, himself a Jew, showed real compassion for the afflicted: "How and to what extent can we help the lamentable people who have gone mad over the Jews, and how can we save others from the same unfortunate person in the future?" Further impulses came from the 1944 Psychiatric Symposium on Anti-Semitism in San Francisco. The approaches presented there were further developed by individual psychoanalysts such as Bela Grunenberg. But today, the psychological aspect no longer plays a role in anti-Semitism research. The shift in the foundations of the social sciences and humanities from acting human subjects to 'systems', 'structures', 'functions', 'discourses', and 'contexts' has led us to know less about this subject today than ever before, including the anti-Semite. The result of this metatheoretical ban on the subject** is that the anti-Semite is now seen as the natural product of a sick, capitalist alienated society. As Jean-Paul Sartre wrote in 1944 in his famous book Reflections on the Jewish Question: "We agree with the anti-Semite on one point: We do not believe in human 'nature', we refuse to regard society as a sum of isolated or isolable molecules." Sartre's frequently quoted statement that anti-Semitism is the fear of humanity was only a lay-psychological attribution, an overstated cliché.
To get to the root of anti-Semitism, we must explore the thoughts of the anti-Semite and reproduce them in such a way that he could feel understood. What do I mean by that? Here's a shot. Anti-Semitism is known to be a passion that intimately links the anti-Semite with the existence of the Jew. He feels pleasure paired with contempt and deepest security at the thought: The Jew explains the world, capitalism or communism, the difference between rich and poor or at least the Middle East conflict. This is an important achievement by which the anti-Semite debts himself to the Jew. If one reverses this statement, one also recognizes the fear associated with it: "Without the Jews, I no longer understand the world!" For the Jew has become part of the cognitive apparatus of the anti-Semite, a powerful, reassuring machine for explaining the world, which he can no longer do without. Hence the Jew, for he cannot be replaced in this function. In the final stages of this disease, the anti-Semite begins to seriously suffer from it and tries to heal himself, increasingly developing a willingness to move from words to action: The world that the Jew explains to me, he has also created. I want a new world – one without Jews.
After the failure of all other approaches, it's time to deal with the anti-Semite again as the philosopher Brunner intended, namely as a severely afflicted subject, as a patient.
*What it requires to be or to become a judgment that can be rightly qualified as 'political', that is the main theme of my book Political Subjectivity. The Philosophical Foundation of Democratic Individualism, published in English 2018. Here is a preview on Acadamia.edu.
**I covered this topic in greater detail in my essay What is a Democrat? An Attempt to Define the Democratic Personality, equally published in 2012.
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©Reginald Grünenberg | {
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Q: SyntaxError on MySql Db Insert I'm currently working on a project that allows users to perform self-evaluations. Currently, I store the criteria for the evaluation in a MySQL table and iterate through that table to make the FORM object, appending a drop down list to each row. Although this creates the form with ease, when I put the insert command inside the form acceptance conditional block, I get a syntax error on the first field assignment that uses an underscore. Here is the code for the table:
db.define_table('Evaluation', Field('EvaluationId', 'integer', 'id'),
Field('PersonBeingEvaluated', db.Person, 'integer'),
Field('PersonEvaluating', 'integer'),
Field('EvalDate', 'datetime'),
Field('1_1A', 'integer'), Field('1_1B', 'integer'), Field('1_2', 'integer'),
Field('1_3', 'integer'), Field('1_4', 'integer'), Field('1_5', 'integer'),
Field('1_6', 'integer'), Field('2_1', 'integer'), Field('2_2A', 'integer'),
Field('2_2B', 'integer'), Field('2_3A', 'integer'), Field('2_3B', 'integer'),
Field('2_4', 'integer'), Field('2_5', 'integer'), Field('3_1', 'integer'),
Field('3_2', 'integer'), Field('3_3', 'integer'), Field('3_4', 'integer'),
Field('3_5A', 'integer'), Field('3_5B', 'integer'), Field('3_6', 'integer'),
Field('3_7', 'integer'), Field('3_8', 'integer'), Field('3_9A', 'integer'),
Field('3_9B', 'integer'), Field('3_9C', 'integer'), Field('4_1', 'integer'),
Field('4_2', 'integer'), Field('4_3A', 'integer'), Field('4_3B', 'integer'),
Field('4_4', 'integer'), Field('BeliefsScore', 'integer'),
Field('CharacterScore', 'integer'), primarykey=['EvaluationId'])
And the code for my method:
@auth.requires_login()
def Take_Eval():
user = db(db.Person.auth_userUserId == auth.user.id).select().first().PersonId
table = db(db.PageContent.ContentId > 0).select()
accordion = DIV(_id="accordion")
for row in table:
accordion.append(H2(row['ContentDescription'], _style="padding-left: 10px;"))
accordion.append(DIV(XML(row.ContentHTML), BR(), P(SELECT(OPTGROUP(OPTION('Does not Attempt', _value=0),
OPTION('Developing', _value=1), OPTION('Progressing', _value=2), OPTION('Established', _value=3),
OPTION('Advanced', _value=4), OPTION('Master', _value=5)), _id=row['QuestionNumber']), _style='text-align: center;')))
form = FORM(accordion, BR(), P(INPUT(_type='submit', _value='Submit', _name='btnSubmit'), _style="text-align: center;"))
if form.accepts(request, session):
response.flash = 'Evaluation Saved Successfully on ' + str(time.strftime("%Y-%m-%d %H:%M:%S", time.localtime()))
db.Evaluation.insert(PersonBeingEvaluated=user, PersonEvaluating=0, EvalDate=datetime.now(),
1_1A = form.vars.1_1A, 1_1B = form.vars.1_1B, 1_2 = form.vars.1_2, 1_3 = form.vars.1_3,
1_4 = form.vars.1_4, 1_5 = form.vars.1_5, 1_6 = form.vars.1_6, 2_1 = form.vars.2_1,
2_2A = form.vars.2_2A, 2_2B = form.vars.2_2B, 2_3A = form.vars.2_3A, 2_3B = form.vars.2_3B,
2_4 = form.vars.2_4, 2_5 = form.vars.2_5, 3_1 = form.vars.3_1, 3_2 = form.vars.3_2,
3_3 = form.vars.3_3, 3_4 = form.vars.3_4, 3_5A = form.vars.3_5A, 3_5B = form.vars.3_5B,
3_6 = form.vars.3_6, 3_7 = form.vars.3_7, 3_8 = form.vars.3_8, 3_9A = form.vars.3_9A,
3_9B = form.vars.3_9B, 3_9C = form.vars.3_9C, 4_1 = form.vars.4_1, 4_2 = form.vars.4_2,
4_3A = form.vars.4_3A, 4_3B = form.vars.4_3B, 4_4 = form.vars.4_4,
BeliefsScore = form.vars.BeliefsScore, CharacterScore = form.vars.CharacterScore)
redirect(URL('default', 'index'))
elif form.errors:
response.flash = 'Profile could not be submitted. Please try again later. '
return locals()
In order to resolve this issue, I've refactored the form, believing that perhaps my naming convention was wrong for the drop down lists. However, I don't believe that is the case. I've also tried using something along the lines of:
1_1A = form.vars.1_1A.value
Thinking that maybe I was assigning the drop down list instead of its numerical value, but that didn't change the error code either. The optimal solution to this problem would be to get the code to compile correctly and have each drop down list submit an integer to the database on insert.
Can anyone tell me why the syntax for my insert method errors out? Thanks.
EDIT :
Here's the HTML page:
{{extend 'layout.html'}}
<script>
$(function() {
$("#accordion").accordion({
header: "h2",
autoHeight: false
});
});
</script>
<h1>Evaluation Info</h1>
<br />
<div id="contentPane">
{{=form}}
</div>
<br />
The actual page itself is several thousand lines long; however, I also grabbed a markup for on of the select tags:
<p style="text-align: center;"><select id="Q1_1A"><optgroup>
<option value="0">Does not Attempt</option>
<option value="1">Developing</option>
<option value="2">Progressing</option>
<option value="3">Established</option>
<option value="4">Advanced</option>
<option value="5">Master</option>
</optgroup></select></p>
Let me know if you need anything else.
| {
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Q: Conversion: np.array of indices to np.array of corresponding dict entries I have a numpy array of indices in Python 2.7 that correspond to a value in a dictionary. So I want to create a numpy array of the corresponding values from the dictionary. The code might be clear immediately:
import numpy as np
indices = np.array([(0, 1), (2, 0), (2, 0)], dtype=[('A', int), ('B', int)])
d = {(0, 1): 10,
(2, 0): 9}
values = d[(indices['A'], indices['B'])]
The call in the last line is not hashable (I tried to find a way to make a np.array hashable but it did not work):
TypeError: unhashable type: 'numpy.ndarray'
I could substitute this by a loop but this takes ages to write the variable values:
np.array([d[(indices[i]['A'], indices[i]['B'])] for i in range(len(indices))])
Or is there any alternative for dict to make such task pythonic, i.e. faster? The variable indices can not be changed but I can change the type of dict.
Edit
The actual index array contains also other entries. That is why I wrote the calls so complicated:
indices = np.array([(0, 1, 's'), (2, 0, 's'), (2, 0, 't')],
dtype=[('A', int), ('B', int), ('C', str)])
A: I believe you can use a list comprehension for this (it would be a bit faster than a normal for loop method). Example -
values = [d[tuple(a)] for a in indices]
Please note, I am using d instead of dict, since it would not be recommended to use dict as a variable name, as that would shadow the built-in type dict.
Demo -
In [73]: import numpy as np
In [74]: indices = np.array([(0, 1), (2, 0), (2, 0)], dtype=[('A', int), ('B', int)])
In [76]: d = {(0, 1): 10,
....: (2, 0): 9}
In [78]: values = [d[tuple(a)] for a in indices]
In [79]: values
Out[79]: [10, 9, 9]
A faster method for larger arrays would be to use np.vectorize() to vectorize the dict.get() method and then apply that on the indices array. Example -
vecdget = np.vectorize(lambda x: d.get(tuple(x)))
vecdget(indices)
Demo with timing results -
In [88]: vecdget = np.vectorize(lambda x: d.get(tuple(x)))
In [89]: vecdget(indices)
Out[89]: array([10, 9, 9])
In [98]: indices = np.array([(0, 1), (2, 0), (2, 0)] * 100, dtype=[('A', int), ('B', int)])
In [99]: %timeit [d[tuple(a)] for a in indices]
100 loops, best of 3: 1.72 ms per loop
In [100]: %timeit vecdget(indices)
1000 loops, best of 3: 341 µs per loop
Timing test for the new method suggested by @hpaulj in the comments - [d.get(x.item()) for x in indices] -
In [114]: %timeit [d.get(x.item()) for x in indices]
1000 loops, best of 3: 417 µs per loop
In [115]: %timeit vecdget(indices)
1000 loops, best of 3: 331 µs per loop
In [116]: %timeit [d.get(x.item()) for x in indices]
1000 loops, best of 3: 354 µs per loop
In [117]: %timeit vecdget(indices)
1000 loops, best of 3: 262 µs per loop
| {
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Q: How to execute while loop given number of times? I have this while loop to get next working day excluding holidays and sundays.
But it calculates by adding 1 day. And i want that number of day to be given by the user. I get that input from the below TextBox (TboxIzin).
How can execute that while loop to do the calculation for given number of times?
int i = 1;
int sayi;
IzinIslem i1 = new IzinIslem();
int.TryParse(i1.TboxIzin.Text, out sayi);
public static DateTime GetNextWeekDay(DateTime date,
IList<Holiday> holidays, IList<DayOfWeek> weekendDays)
{
int i = 1;
int sayi;
IzinIslem i1 = new IzinIslem();
int.TryParse(i1.TboxIzin.Text, out sayi);
// always start with tomorrow, and truncate time to be safe
date = date.Date.AddDays(1);
// calculate holidays for both this year and next year
var holidayDates = holidays.Select(x => x.GetDate(date.Year))
.Union(holidays.Select(x => x.GetDate(date.Year + 1)))
.Where(x => x != null)
.Select(x => x.Value)
.OrderBy(x => x).ToArray();
// increment to get working day
while (true)
{
if (weekendDays.Contains(date.DayOfWeek) ||
holidayDates.Contains(date))
date = date.AddDays(1);
else
return date;
}
}
I get not all code paths return a value
error when i try nesting while loops.
A: while is a conditional loop. Here you put a non-condition in the clause and immediately follow up with a condition. Put the condition in the while clause:
while(weekendDays.Contains(date.DayOfWeek) || holidayDates.Contains(date)) {
date = date.AddDays(1);
}
return date;
The actual reason you're getting the error is that the compiler cannot predict if your if condition will ever resolve to false. If it doesn't, then your function never returns.
With the modified while loop, that may still happen, but it will result in an infinite loop, and the compiler is fine if you shoot yourself in the foot that way.
A: You can change your else clause to break out of the loop. And then return out of the loop.
while (true)
{
if (weekendDays.Contains(date.DayOfWeek) ||
holidayDates.Contains(date))
date = date.AddDays(1);
else
break;
}
return date;
A: Let's get rif of all UI in the GetNextWeekDay (like int.TryParse(i1.TboxIzin.Text, out sayi);):
public static DateTime GetNextWeekDay(DateTime date,
IEnumerable<Holiday> holidays,
IEnumerable<DayOfWeek> weekendDays) {
// public method arguments validation
if (null == holidays)
throw new ArgumentNullException(nameof(holidays));
else if (null == weekendDays)
throw new ArgumentNullException(nameof(weekendDays));
HashSet<DayOfWeek> wends = new HashSet<DayOfWeek>(weekendDays);
// Current Year and Next years - .AddYear(1)
HashSet<DateTime> hds = new HashSet<DateTime>(holidays
.Select(item => item.Date)
.Concate(holidays.Select(item => item.Date.AddYear(1))));
for (var day = date.Date.AddDays(1); ; day = day.AddDays(1))
if (!wends.Contains(day.DayOfWeek) && ! hds.Contains(day))
return day;
}
Or if you prefer Linq, the loop can be rewritten as
return Enumerable
.Range(1, 1000)
.Select(day => date.Date.AddDays(day))
.TakeWhile(item => item.Year - date.Year <= 1)
.First(item => !wends.Contains(item.DayOfWeek) && !hds.Contains(item));
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,099 |
\section{Introduction}
\label{sec:intro}
Human-action recognition (HAR) has wide range of applications such as life log, healthcare, video surveillance, and worker assistance.
The recent advances in deep neural networks (DNN) have drastically enhanced the performance of HAR both in terms of recognition accuracy and coverage of the recognized actions
\cite{Herath2017,wang2017deep}.
DNN-based methods, however, sometimes face difficulty in practical deployment; a system user sometimes wants to change or add target actions to be recognized, but it is not so trivial for DNN-based methods to do so since they require large amount of training data of the new target actions.
Zero-shot learning (ZSL) has a great potential to overcome this difficulty of dependence on training data when recognizing a new target class \cite{larochelle2008zero,Frome2013,Fu2018}.
Whereas in normal supervised-learning setting, the set of classes contained in test data is exactly the same as that in training data, it is not the case in ZSL; test data includes ``unseen'' classes, of which instances are not contained in training data.
In other words, if $\mathcal{Y}_{train}$ is a set of class labels in training data and $\mathcal{Y}_{test}$ is that in test data, then~$\mathcal{Y}_{train} = \mathcal{Y}_{test}$ in normal supervised-learning framework, while $\mathcal{Y}_{train} \neq \mathcal{Y}_{test}$ in ZSL framework.
(more specifically, $\mathcal{Y}_{train} \cap \mathcal{Y}_{test} = \phi$ in some cases, and $\mathcal{Y}_{train} \subset \mathcal{Y}_{test}$ in other cases).
Unseen classes are classified using {\it attribute} together with a description of the class based on the attributes, which~ is usually given on the basis of external knowledge.
Most typically it is manually given by humans~\cite{lampert2009learning, lampert2014attribute}.
The {\it attribute} represents a semantic property of the class.
A classifier to judge the presence of the attribute (or the probability of the presence) is learnt using training data.
For example, the attribute of ``striped'' can be learnt using the data of ``striped shirts'', while the attribute of ``four-legged'' can be learnt using the data of ``lion''.
Then an unseen class ``zebra'' can be recognized, without any training data of zebra itself, by using these attribute classifiers as well as the description that zebras are striped and four-legged.
The idea of ZSL has been applied also to human action recognition \cite{Liu2011,cheng2013towards,Xu2016,li2016recognizing,qin2017zero}.
Indeed they successfully demonstrated a capability of recognizing unseen actions, but the attributes used in these studies are relatively task-specific and not so fundamental as to be able to recognize wider variety of human actions.
The potential of recognizing truly wide variety of actions becomes substantially bigger if more fundamental and general set of attributes are utilized.
To this end, we believe the status of each human-body joint is appropriate attribute since any kinds of human action can be represented using the set of each body joint's status.
There are sophisticated vision-based methods such as \cite{iqbal2017posetrack,chen2017adversarial,guler2018densepose} for estimating the status of body joint, but the problem of occlusion is essentially inevitable for those approach.
Moreover, they are not suitable for the applications in which the target person moves around beyond the range of camera view.
Thus we utilize wearable sensors, which are free from occlusion problem, to estimate the statuses of all the major human body joints.
We aim at developing a method that flexibly recognizes wide range of human actions with ZSL.
This study especially focuses on the classification of static actions, or poses, as the first step toward that goal {{(Some of the} poses are sometimes referred to as ``action'' in prior works, but we use the term ``pose'' in this study)}.
The biggest challenge in zero-shot pose recognition is the intra-class variation of the poses.
The~difficulty of intra-class variation in general action recognition was discussed in \cite{Liu2011}.
For~example, when ``folding arm'', one may clench his/her fists while another may not.
The authors introduced a method to deal with the intra-class variation by regarding attributes as latent variables.
However, it was for normal supervised learning and their implementation in ZSL scenario was naive nearest-neighbor-based method that does not address this problem.
The intra-class variation becomes an even severe problem in ZSL especially when fine-grained attributes like each body joint's status are utilized.
This is because the value of all the attributes should be specified in ZSL even though some of the attribute actually may take arbitrary values.
It is difficult to uniquely define the status of hands for ``folding arm'', but the attribute ``hands'' cannot be omitted because it is necessary for recognizing other poses such as ``pointing''.
Conventional ZSL methods have dealt with all the attributes equally even though some of them are actually not important for some of the classes.
This sometimes causes misclassification because a metric (e.g., likelihood, distance) to represent that a given sample belongs to a class can be undesirably influenced by irrelevant attributes.
This paper solves the problem by taking the importance of each attribute for each class into account when calculating the metric.
The effectiveness of the method is demonstrated on a human pose dataset collected by us that is named Hitachi-DFKI pose dataset, or HDPoseDS in short.
HDPoseDS contains 22 classes of poses performed by 10 subjects with 31 inertial measurement unit (IMU) sensors across full body.
To the best of our knowledge, this is the richest dataset especially in terms of sensor density for human pose classification based on wearable sensors.
Due to its sensor density, it gives us a chance of extracting fundamental set of attributes for human poses, namely the status of body joints.
Therefore, it is the first dataset suitable for studying wearable-based zero-shot learning in which wide variety of full-body poses are involved.
We make this dataset publicly available to encourage the community for further research in this direction.
It is available at \url{http://projects.dfki.uni-kl.de/zsl/data/}.
The main contribution of this study is two folds.
(1) We present a simple yet effective method to enhance the performance of ZSL by taking the importance of each attribute for each class into account. We experimentally show the effectiveness of our method in comparison to baseline methods.
(2) We provide HDPoseDS, a rich dataset for human pose classification suitable especially for studying wearable-based zero-shot learning.
In addition to these major contributions, we also present a practical design for estimating the status of each body joint; while conventional ZSL methods formulate attribute-detection problem as 2-class classification (whether the attribute is present or not), we~estimate it under the scheme of either multiclass classification or regression depending upon the characteristics of each body joint.
\section{Related Work}
\label{sec:related work}
We review three types of prior works in this section, namely ZSL, wearable-based action and pose recognition, and wearable-based zero-shot action and pose recognition.
\subsection{Zero-Shot Learning}
The idea of ZSL was firstly presented in \cite{larochelle2008zero} followed by \cite{lampert2009learning} and \cite{palatucci2009zero}.
The major input sources have been images and videos, but there have been some studies based on wearable sensors as reviewed later in this section.
The most fundamental framework established in the early days is as follows.
Firstly a~function $f: \mathcal{X} \mapsto \mathcal{A}$ is learnt using labeled training data, where $\mathcal{X}$ denotes an input (feature) space, and $\mathcal{A}$ denotes an attribute space.
The definition of unseen classes is given manually, and it represents a vector in the attribute space $\mathcal{A}$.
Then a function $g: \mathcal{A} \mapsto \mathcal{Y}$ is learnt using the vectors in $\mathcal{A}$ and their labels.
Here $\mathcal{Y}$ denotes a label space.
When test data are given, their labels are estimated by applying the learnt functions $f$ and $g$ subsequently.
In early days, SVM was frequently used for learning $f$, and~it's replaced by DNN-based method these days.
One of the most common methods for learning $g$ has been nearest neighbors \cite{palatucci2009zero,Liu2011,cheng2013nuactiv,xu2015semantic}.
This study also uses a nearest-neighbor-based method.
Extensive efforts have been made to improve ZSL methods from various viewpoints.
Socher~et~al.~\cite{socher2013zero} invented a method that does not need manual definition of unseen classes by utilizing natural language corpora (word2vec).
Jayaraman and Grauman~\cite{jayaraman2014zero} took the unreliability of attribute estimation into account by a random-forest based method.
Semantic representations were effectively enriched by using synonyms in \cite{alexiou2016exploring}, and by using textual descriptions as well as relevant still images in \cite{wang2017alternative}.
Qin~et~al.~\cite{qin2016beyond} extended the semantic attributes to latent attributes in order to obtain more discriminative representation as well as more balanced attributes.
Tong~et~al.~\cite{tong2017adversarial} were the first to introduce generative adversarial network (GAN)~\cite{goodfellow2014generative} in ZSL.
A problem of domain shift that is common in ZSL was effectively dealt with in \cite{Xu2016} and \cite{qin2017zero}.
Liu~et~al.~\cite{liu2018cross} studied cross-modal ZSL between tactile data and visual data.
Our idea of incorporating each attribute's importance for each class was inspired by \cite{jayaraman2014zero} as their idea of incorporating attributes' unreliability is similar in terms of dealing with the characteristics of attributes.
\subsection{Wearable-Based Action and Pose Recognition}
As reviewed in \cite{lara2013survey,bulling2014tutorial,mukhopadhyay2015wearable}, the mainstream of action recognition methods before DNN-based methods become popular has consisted of two-stage approach; firstly they apply a sliding window to time-series data and extract time domain features such as mean and standard deviation as well as frequency domain features such as FFT coefficients, and secondly apply various machine-learning method such as hidden Markov model (HMM) \cite{pmlr-v48-guan16}, support vector machine (SVM) \cite{bulling2012multimodal}, conditional random field (CRF) \cite{adams2016hierarchical}, and an ensemble method \cite{zheng2013physical}.
In recent years, DNN-based approaches have become increasingly popular as they showed overwhelming results \cite{wang2017deep}.
In \cite{yang2015deep,jiang2015human,ronao2015deep}, they introduced a way to employ convolutional neural networks (CNN) to automatically extract efficient features from time-series data.
Ord{\'o}{\~n}ez~et~al.~\cite{ordonez2016deep} proposed a method to more explicitly deal with the temporal dependencies of the human actions by utilizing long short-term memory (LSTM).
Hammerla~et~al.~\cite{hammerla2016deep} also introduced a LSTM-based method and gave the performance comparison among DNN, CNN, and LSTM as well as the influence of the network parameters in each method.
Following the findings from these researches, we also utilize CNN for estimating the status of each body joint.
The details of the implementation will be given in the following section.
\subsection{Wearable-Based Zero-Shot Action and Pose Recognition}
One of the earliest attempts to apply the idea of ZSL to human activity recognition based on wearable sensors is \cite{cheng2013nuactiv} and their subsequent work \cite{cheng2013towards}.
They firstly used nearest-neighbor-based approach to recognize activities using attributes, and later enhanced the method to incorporate temporal dependency by using CRF.
Wang~et~al.~\cite{wang2017zero} proposed a nonlinear compatibility based method, where they first project the features extracted from sensor readings to a hidden space by a nonlinear function, and then calculate the compatibility score based on the features in the hidden space and prototypes in semantic space.
Al-Naser~et~al.~\cite{naser2017} introduced a ZSL model for recognizing complex activities by using simpler actions and surrounding objects as attributes.
These prior works successfully showed a great potential of realizing zero-shot action recognition based on wearable sensors.
However, on one hand the attributes used in those studies are neither fine-grained nor fundamental enough so as to represent truly wide variety of human actions or poses.
On the other hand, the methods used in those studies do not take the attributes' importance into account, which matters more especially when using fine-grained attributes to represent diverse poses.
\section{Dataset: HDPoseDS}
\vspace{-6pt}
\subsection{Sensor}
Our goal is to use all the major human-body joints as attributes to represent full-body poses.
Thus, a very dense sensor set across full-body is required.
Perception Neuron from Noitom Ltd. is ideal for this {purpose} (\url{https://neuronmocap.com/}).
It has 31 IMU sensors across full body; 1 on head, 2~on shoulders, 2 on upper arms, 2 on lower arms, 2 on hands, 14 on fingers, 1 on spine, 1 on hip, 2~on upper legs, 2~on lower legs, 2 on feet (Figure~\ref{fig:pn}).
Each IMU is composed of a 3-axis accelerometer, 3-axis gyroscope and 3-axis magnetometer.
We use 10 dimensional data from each IMU including 3~acceleration data, 3 gyro data, and 4 quaternion data.
This rich set of sensors are especially helpful in applications where detailed full-body pose-recognition is desired.
For example, in workers' training, novice workers can learn to avoid undesirable poses that can cause safety or quality issues with the help of a pose recognition system.
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.9\columnwidth]{figures/Picture1c.png}
\vspace{-5pt}
\caption{Sensor displacement in Perception Neuron.}
\label{fig:pn}
\end{center}
\end{figure}
\subsection{Target Poses}
\textls[-15]{In order to test the generalization capability of zero-shot models, we constructed a human pose dataset named HDPoseDS using Perception Neuron.
We newly built the dataset because existing wearable-sensor datasets are not collected with so densely-attached sensors as to be used for extracting fundamental set of attributes for human poses, namely the status of each body joint.
We defined 22 poses such that various body parts are involved and thus the generalization capability of the developed method in zero-shot scenario can be appropriately tested (see Figure~\ref{fig:poses} for appearance and Table~\ref{tab:val}~for~names)}.
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.8\linewidth]{figures/Picture2c.png}
\vspace{-8pt}
\caption{Poses in HDPoseDS. The numbers under the figures correspond to the numbers in Table~\ref{tab:val}.}
\label{fig:poses}
\end{center}
\end{figure}
\vspace{-12pt}
\begin{table}[H]
\centering
\caption{Intra-class variation observed in data collection. Poses are sometimes represented in slightly different manner depending on the subjects. L and R in the pose names means Left and Right.}
\label{tab:val}
\begin{tabular}{llll}
\toprule
\textbf{ID} & \textbf{Pose} & \textbf{Variation} & \textbf{Involved Body Joint} \\ \midrule
1 & Standing & no big variation & - \\ \midrule
2 & Sitting & hands on a table, on knees, or straight down & elbows, hands \\ \midrule
3 & Squatting & hands hold on to sth, on knees, or straight down & elbows, hands \\ \midrule
4, 5 & Raising arm (L, R) & a hand on hip, or straight down & elbow, hand \\ \midrule
6, 7 & Pointing (L, R) & a hand on hip, or straight down & elbow, hand \\ \midrule
8 & Folding arm & \begin{tabular}[c]{@{}l@{}}wrist curled or straight, \\ hands clenched or normal\end{tabular} & \begin{tabular}[c]{@{}l@{}}wrist, \\ hands\end{tabular} \\ \midrule
9 & Deep breathing & head up or front & head \\ \midrule
10 & Stretching up & head up or front & head \\ \midrule
11 & Stretching forward & waist straight or half-bent & waist \\ \midrule
12 & Waist bending & no big variation & - \\ \midrule
13, 14 & Waist twisting (L, R) & \begin{tabular}[c]{@{}l@{}}head left (right) or front, \\ arms down or left (right)\end{tabular} & \begin{tabular}[c]{@{}l@{}}head,\\ shoulders, elbows\end{tabular} \\ \midrule
15, 16 & Heel to back (L, R) & \begin{tabular}[c]{@{}l@{}}a hand hold on to sth, straight down, \\ or stretch horizontally\end{tabular} & shoulder, elbow, hand \\ \midrule
17, 18 & Stretching calf (L, R) & head front or down & head \\ \midrule
19 & Boxing & head front or down & head \\ \midrule
20 & Baseball hitting & head left or front & head \\ \midrule
21 & Skiing & head front or down & head \\ \midrule
22 & Thinking & \begin{tabular}[c]{@{}l@{}}head front or down,\\ wrist reverse curled or normal, \\ hand clenched or normal\end{tabular} & \begin{tabular}[c]{@{}l@{}}head, \\ wrist, \\ hands\end{tabular} \\ \bottomrule
\end{tabular}
\end{table}
We had 10 subjects, and each subject performed all the 22 poses for about 30 s.
All of the 10~subjects were males, but from 4 different countries.
The body heights of the subjects ranged from 160~cm to 185~cm.
The ages were from 23 years old to 37 years old.
To the best of our knowledge, this~is the richest dataset especially in terms of sensor density (31 IMU across full body).
Therefore, it~is the first dataset suitable for studying wearable-based zero-shot learning in which wide variety of full-body poses are involved.
We make this dataset publicly available to encourage the community for further research in this direction.
It is available at \url{http://projects.dfki.uni-kl.de/zsl/data/}.
During the data collection, only brief explanation about each pose was given, and thus we observed some intra-class variation in the dataset as summarized in Table~\ref{tab:val}.
\section{Proposed Method}
\label{sec:proposed method}
Following the standard scheme of ZSL, our approach also consists of two stages; attribute estimation based on sensor readings, and class label estimation using estimated attributes. We explain the two stages one by one in detail.
In this section, we first explain the sensor to be used in our study, then describe the two stages in detail.
\subsection{Attribute Estimation}
\label{sec:ae}
We use 14 major human-body joints as attributes to represent various poses as summarized in the first column of Table~\ref{tab:poses}.
Unlike conventional ZSL methods, where 2-class classification is always used (whether an attribute is present or not), we use either multiclass classification or regression depending upon the characteristics of each body joint as shown in Table~\ref{tab:poses}.
For the joints that have only one degree of freedom like knees (or at least whose major movement is restricted to one dimension), it is more suitable and beneficial to use regression to estimate the status.
This allows us to represent intermediate status of the joint by just specifying an intermediate value, which enables to describe detailed status of the joint, rather than just ``straight'' and ``curl'', to represent more complicated poses in the future.
For~the joints that have more than 2 degrees of freedom like head, we use multiclass classification.
It is indeed possible to replace this by 2-class classification on each status, but it's more natural to formulate this as a multiclass classification problem since each status are mutually exclusive (e.g., if head is ``up'', then it cannot be ``down'' at the same time).
We use CNN to deal with multivariate time-series data.
Previous studies \cite{yang2015deep,ordonez2016deep,hammerla2016deep} first dealt with different modalities individually by applying convolution only on temporal direction (a kernel size of CNN is $k\times1$), and integrated the output from all the modality in fully connected layers that appear right before the classification layer.
However, as shown in Figure~\ref{fig:cnn}, we integrate the readings from different sensor modalities in the earlier stage using CNN (a kernel size of CNN is $k_1\times k_2$) since~we experimentally found that it gives better performance.
We construct one network per one joint, resulted in 14 networks to estimate the status of all the joints.
\begin{table}[H]
\centering
\caption{The 14 body joints used as attributes and the types and values to represent their status.
Note~that each joint has left part and right part except head and waist.}
\begin{tabular}{lll}\toprule
\textbf{Joint} & \textbf{Type} & \textbf{Value}\\ \midrule
head & classification & up, down, left, right, front \\
shoulder & classification & up, down, left, right, front \\
elbow & regression & 0 (straight)--1 (bend) \\
wrist & regression & 0 (reverse curl)--1 (curl)\\
hand & classification & normal, grasp, pointing \\
waist & classification & straight, bend, twist-L, twist-R\\
hip joint & regression & 0 (straight)--1 (bend) \\
knee & regression & 0 (straight)--1 (bend) \\ \bottomrule
\end{tabular}
\label{tab:poses}
\end{table}
The sliding window size in this study is 60, which corresponds to 1 s.
The number of modality (referred to as $M$ in Figure~\ref{fig:cnn}) is $s \times 10$, where $s$ is the number of used IMU for each joint.
We use 4 convolution layers with 25, 20, 15, and 10 channels.
One fully connected layer with 100 nodes is inserted before the final classification or regression layer.
The activation function is leaky ReLU throughout the network but the regression layer has sigmoid activation to squash the values to $[0, 1]$.
We use cross entropy loss for multiclass classification, and mean absolute loss for regression.
They are optimized using MomentumSGD with momentum value of 0.9.
Batch normalization and drop out is used for regularization.
The kernel size in convolution layers are $3\times10$ for hands and $3\times3$ for all the rest.
We use the wider kernel for hands because the number of IMU used for classifying hands' status is significantly larger than that for other joints.
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.95\linewidth]{figures/Picture4c.png}
\caption{Architecture of time-series CNN for basic pose recognition (attribute estimation).}
\label{fig:cnn}
\end{center}
\end{figure}
\subsection{Pose Classification with Attributes' Importance}
\label{sec:ai}\vspace{-6pt}
\subsubsection{Naive Formulation}
We use nearest-neighbor-based method for zero-shot pose classification.
The input to pose classification is the output from attribute estimation explained in Section~\ref{sec:ae}.
The output dimension from the networks for each joint is the number of classes if the joint's status is estimated by multiclass classification, and 1 if it is by regression, resulting in a 33 dimensional vector in total.
Let \mbox{$\bm{a^{(n)}}=\{a^{(n)}_1, \cdots, a^{(n)}_D\}$} be an attribute vector of $n$'th sample ($D=33$ in this study).
Please note that $\forall a^{(n)}_d \in [0, 1]$ since we squash the values by softmax and sigmoid function for multiclass classification and regression, respectively.
For ZSL, we need extra information for estimating pose labels using vectors in attribute space.
Following some of the previous works \cite{lampert2009learning,lampert2014attribute,cheng2013towards,cheng2013nuactiv}, we simply use a manually defined table for this as shown in Table~\ref{tab:def} and convert them to corresponding 33-dimensional vectors (one-of-K representation is used for multiclass classification part).
For normal nearest-neighbor-based method, a given test data $\bm{x}$ is first converted to an attribute vector $\bm{a}$ using the learnt attribute estimation networks, and then the distance between $\bm{a}$ and the $i$th training data $\bm{v}^{(i)}$ is calculated as follows.
\begin{equation}
\label{eq:d1}
d(\bm{a}, \bm{v}^{(i)}) = (\sum_d^{D}{|a_d - v^{(i)}_d|^p})^{1/p},
\end{equation}
where $D$ is the dimension of an attribute space, and $x_d$ denotes the $d$'th element of vector $\bm{x}$.
For seen classes, $\bm{v^{(i)}}\in\mathbb{R}^{D}$ is a training data mapped to the attribute space using the learnt attribute estimation networks, while for unseen classes it is a vector created based on the pose definition table (Table~\ref{tab:def}).
$p$~is usually 1 or 2.
Then the given data $\bm{x}$ is classified to the following class.
\begin{equation}
C(\argmin_{i} {d(\bm{a}, \bm{v}^{(i)})}),
\end{equation}
where $C(i)$ gives the class to which $i$th training data belong.
\tabcolsep = 2pt
\begin{table}[H]
\centering
\caption{Definition of the poses in HDPoseDS. (L) denotes left part and (R) denotes right part. For~joints, He: Head, S: Shoulder, E: Elbow, Wr: Wrist, Ha: Hand, Wa: Waist, HJ: Hip Joint, K: Knee. For joint status, F: Front, U: Up, D: Down, L: Left, R: Right, N: Normal, G: Grasp, P: Pointing, S: Straight, B:~Bend, TwL (R): Twist to Left (Right).}
\label{tab:def}
\small
\begin{tabular}{llcccccccccccccc}
\toprule
& \textbf{Pose\textbackslash Joint} & \textbf{He} & \textbf{S(L)} & \textbf{S(R)} & \textbf{E(L)} & \textbf{E(R)} & \textbf{Wr(L)} & \textbf{Wr(R)} & \textbf{Ha(L)} & \textbf{Ha(R)} & \textbf{Wa} & \textbf{HJ(L)} & \textbf{HJ(R)} & \textbf{K(L)} & \textbf{K(R)} \\ \midrule
1 & Standing & F & D & D & 0 & 0 & 0.5 & 0.5 & N & N & S & 0 & 0 & 0 & 0 \\
2 & Sitting & F & D & D & 0 & 0 & 0.5 & 0.5 & N & N & S & 0.5 & 0.5 & 0.5 & 0.5 \\
3 & Squatting & F & D & D & 0 & 0 & 0.5 & 0.5 & N & N & S & 1 & 1 & 1 & 1 \\
4 & Raising arm (L) & F & U & D & 0 & 0 & 0.5 & 0.5 & N & N & S & 0 & 0 & 0 & 0 \\
5 & Raising arm (R) & F & D & U & 0 & 0 & 0.5 & 0.5 & N & N & S & 0 & 0 & 0 & 0 \\
6 & Pointing (L) & F & F & D & 0 & 0 & 0.5 & 0.5 & P & N & S & 0 & 0 & 0 & 0 \\
7 & Pointing (R) & F & D & F & 0 & 0 & 0.5 & 0.5 & N & P & S & 0 & 0 & 0 & 0 \\
8 & Folding arm & F & D & D & 0.5 & 0.5 & 0.5 & 0.5 & N & N & S & 0 & 0 & 0 & 0 \\
9 & Deep breathing & F & L & R & 0 & 0 & 0.5 & 0.5 & N & N & S & 0 & 0 & 0 & 0 \\
10 & Stretching up & F & U & U & 0 & 0 & 0 & 0 & N & N & S & 0 & 0 & 0 & 0 \\
11 & Stretching forward & F & F & F & 0 & 0 & 0 & 0 & N & N & S & 0 & 0 & 0 & 0 \\
12 & Waist bending & F & D & D & 0 & 0 & 0.5 & 0.5 & N & N & B & 0 & 0 & 0 & 0 \\
13 & Waist twisting (L) & L & L & L & 0 & 0.5 & 0.5 & 0.5 & N & N & TwL & 0 & 0 & 0 & 0 \\
14 & Waist twisting (R) & R & R & R & 0.5 & 0 & 0.5 & 0.5 & N & N & TwR & 0 & 0 & 0 & 0 \\
15 & Heel to back (L) & F & D & D & 0 & 0 & 0.5 & 0.5 & G & N & S & 0 & 0 & 1 & 0 \\
16 & Heel to back (R) & F & D & D & 0 & 0 & 0.5 & 0.5 & N & G & S & 0 & 0 & 0 & 1 \\
17 & Stretching calf (L) & F & F & F & 0 & 0 & 0 & 0 & N & N & S & 0 & 0.3 & 0 & 0.3 \\
18 & Stretching calf (R) & F & F & F & 0 & 0 & 0 & 0 & N & N & S & 0.3 & 0 & 0.3 & 0 \\
19 & Boxing & F & D & D & 1 & 1 & 0.5 & 0.5 & G & G & S & 0 & 0 & 0 & 0 \\
20 & Baseball hitting & L & D & D & 0.5 & 0.5 & 0.5 & 0.5 & G & G & S & 0.5 & 0 & 0.5 & 0 \\
21 & Skiing & F & D & D & 0.5 & 0.5 & 0.5 & 0.5 & G & G & S & 0.3 & 0.3 & 0.3 & 0.3 \\
22 & Thinking & F & D & D & 0.5 & 1 & 0.5 & 0 & N & N & S & 1 & 1 & 0.5 & 0.5 \\ \bottomrule
\end{tabular}
\end{table}
\tabcolsep = 6pt
\subsubsection{Incorporating Attributes' Importance}
\label{subsubsec:ai}
As summarized in Table~\ref{tab:val}, sometimes there is intra-class variation in the poses.
In normal supervised learning setting, this intra-class variation can be naturally learnt as training data cover various instances of the given class.
However, it is not possible to do so in ZSL since there is no training data other than a definition table of the unseen classes.
In this situation, it is not appropriate to equally deal with all the attribute because not all the attributes are equally important for classifying a particular class.
For example, for ``squatting'' class, the status of hip joints and knees are important, and the values are expected to be always 1 (bent).
On the other hand, the status of elbows are not important since it is still ``squatting'' regardless of the status of elbows;
the values may be 1 (bent in order to hold on to something or to put hands on knees) or 0 (straight down to the floor).
In other words, the~attribute values of elbows do not matter to tell whether given test data belong to squatting class or not.
Please note that we cannot simply omit the attribute ``elbow'' because it is indeed necessary for other poses such as ``folding arm''.
Therefore, we need to design a new distance metric so that we can incorporate {\it each attributes' importance for each class}.
We formulate this as follows.
\begin{eqnarray}
\label{eq:d2}
d(\bm{a}, \bm{v}^{(i)}) =
\quad \frac{1}{W^{(i)}_{ai}}(\sum_d^{D}{w_{ai}(d, i) w_{rc}(d) |a_d - v^{(i)}_d|^p})^{1/p} + \lambda\frac{1}{W^{(i)}_{ai}},
\end{eqnarray}
\vspace{-12pt}
\begin{eqnarray}
\label{eq:ai}
W^{(i)}_{ai} = \sum_{j}w'_{ai}(j, i),
\end{eqnarray}
\vspace{-12pt}
\begin{eqnarray}
w_{rc}(d) =
\begin{cases}
1 & \text{{\it if d'th attribute is given by regression}} \\
0.5 & \text{{\it otherwise}},
\end{cases}
\end{eqnarray}
where $w'_{ai}(j, i)$ denotes the importance of joint $j$ for the class $C(i)$.
It is given manually in this study as shown in Table~\ref{tab:ai}.
It may be indeed an extra work to manually define the attribute importance, but~actually it does not require too much extra time because anyway the attribute table (Table~\ref{tab:def}) should be manually defined as it was the case in many previous works.
In addition, it is neither difficult because it is natural to assume that the person, or the system user, who defines the attribute table (Table~\ref{tab:def}) has enough knowledge not only on the definition of the target poses but also on which attribute (body-joint status) is important for each pose.
It may be also possible to infer the attributes' importance either from training data or external resources (e.g., word embedding) rather than manually defining them, but it lies beyond the scope of this study at this moment.
We use binary values for attributes' importance for simplicity, but it is easy to extend it to continuous numbers.
$w_{ai}(d, i)$ is the importance of attribute $d$ for the class $C(i)$ and the value of it is copied from $w'_{ai}(j, i)$, where $d$'th attribute comes from joint $j$ .
Please note that it depends not only on $d$ but also on $C(i)$.
$w_{rc}(d)$ is 0.5 if the $d$'th attribute is calculated using multiclass classification because the total distance with regards to the joint $j$ from which the $d$'th attribute comes from $(\sum_{d \in joint j}{|a_d - v^{(i)}_d|})$ ranges from 0 to 2, whereas the distance with regard to the joint whose status is estimated using regression ranges from 0 to 1.
$W_{ai}$ can be interpreted as the total number of ``valid'' joints to be used for classification of class $C(i)$.
Therefore, the first term on the right-hand side in Equation~(\ref{eq:d2}) can be interpreted as the average distance between $\bm{a}$ and $\bm{v}^{(i)}$ over ``valid'' attribute that comes from ``valid'' joints.
The second term is introduced to penalize the class that uses too few attributes.
If~test data have the same distance to two classes that have different numbers of important attributes, this~term encourages to classify the data to the class which has larger number of important attributes, which~indicates more detailed definition of the pose.
We use $\lambda = 0.1$ and $p = 1$ in this study.
Note~that the increase of the computational cost compared to the naive formulation (Equation~(\ref{eq:d1})) is trivial because we just multiply constant numbers when calculating the distances.
\tabcolsep = 2pt
\begin{table}[H]
\centering
\caption{Attributes' importance.}
\label{tab:ai}
\small
\begin{tabular}{llcccccccccccccc}
\toprule
& \textbf{Pose} & \textbf{He} & \textbf{S(L)} & \textbf{S(R)} & \textbf{E(L)} & \textbf{E(R)} & \textbf{Wr(L)} & \textbf{Wr(R)} & \textbf{Ha(L)} & \textbf{Ha(R)} & \textbf{Wa} & \textbf{HJ(L)} & \textbf{HJ(R)} & \textbf{K(L)} & \textbf{K(R)} \\ \midrule
1 & Standing & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
2 & Sitting & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
3 & Squatting & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
4 & Raising arm (L) & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\
5 & Raising arm (R) & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
6 & Pointing (L) & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\
7 & Pointing (R) & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
8 & Folding arm & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
9 & Deep breathing & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
10 & Stretching up & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
11 & Stretching forward & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\
12 & Waist bending & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
13 & Waist twisting (L) & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
14 & Waist twisting (R) & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
15 & Heel to back (L) & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
16 & Heel to back (R) & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\
17 & Stretching calf (L) & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
18 & Stretching calf (R) & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
19 & Boxing & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
20 & Baseball hitting & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
21 & Skiing & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\
22 & Thinking & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ \bottomrule
\end{tabular}
\end{table}
\tabcolsep = 6pt
\section{Experiment}
\label{sec:experiment}
\vspace{-6pt}
\subsection{Evaluation Scheme}
We use HDPoseDS for evaluation.
The evaluation procedure is as follows.
\begin{enumerate}[leftmargin=7.6mm,labelsep=3mm]
\renewcommand{\labelenumi}{(\arabic{enumi}).}
\item All the input data are converted to attribute vectors using the neural networks explained in Section~\ref{sec:ae}.
The sliding window size is 60, which corresponds to 1 s, and it's shifted by 30 (0.5 s).
This ends up with roughly 590 ($=(30/0.5-1) \times 10$) attribute vectors per pose since HDPoseDS contains data from 10 subjects and each subject performed roughly 30 s for each pose.
\item For each class {\it c}, we construct a set of training data by combining the data from all the other classes than {\it c} and the pose definition of {\it c} based on attributes (Table~\ref{tab:def}).
We use class {\it c}'s data as test data.
\item The labels of the test data are estimated using the method explained in Section~\ref{sec:ai}.
\item We repeat this for all the 22 classes.
\item We calculate the F-measure for each class based on the precision and recall rate.
\end{enumerate}
Please note that we do not assume that the possible output classes are only unseen (test) classes; we~assume that the seen (training) classes are also potential output classes during testing.
Since we do not use instances of seen (training) classes in testing, this evaluation scheme is not exactly the same as the generalized zero-shot learning (G-ZSL)~\cite{xian2017zero,kumar2018generalized}, in which the instances of seen classes are also used in testing.
It is, however, more similar to G-ZSL than normal ZSL in a sense that the target classes in testing include not only unseen (test) classes but also seen (training) classes.
In addition to this, we also investigate how the proposed method works in few-shot learning scenario, where only a small number of training data are available.
The evaluation procedure is the same as the ZSL case except the step (2);
instead of including the attribute definition of {\it c} in the training data, we include $k$ samples from class {\it c}'s data in $k$-shot learning scenario, and all the other data of class {\it c} are used for testing.
To choose the $k$ samples, firstly we randomly permutate class {\it c}'s data.
Then we use the $l$'th $(l=1,2, ..., \lfloor N_c / k \rfloor)$ $k$ samples for training and the remaining $(N_c - k)$ samples for testing, where $N_c$ is the number of class {\it c}'s data.
Then we proceed to step 3
The estimation result for class {\it c} is averaged, and then we proceed to step (4).
The performance of the proposed method is compared with three baseline methods.
The first one is one of the most frequently used method in ZSL studies, which is called ``direct attribute prediction (DAP)'' introduced in \cite{lampert2014attribute}.
Please note that we did not compare with indirect attribute prediction (IAP) that is also introduced in \cite{lampert2014attribute}.
This is because, as the authors of \cite{lampert2014attribute} stated, IAP is not appropriate for the case where training classes are also potential output class during testing.
The~other two~baseline methods are nearest-neighbor-based, which is also common in ZSL studies.
The~proposed method is also based on a nearest-neighbor method.
The first nearest-neighbor-based baseline is a naive nearest-neighbor-based method, in which the distance between samples are calculated using Equation~(\ref{eq:d1}) with weights $w_{rc}(d)$.
The second nearest-neighbor-based baseline is the one that uses random attributes' importance.
We~randomly generate either 0 or 1 for each $w'_{ai}(j, i)$ in Equation~(\ref{eq:ai}).
For this baseline method, we~test 1000 times using different random weights and report the average F-measure of the 1000 tests.
For~both of the proposed and the baseline methods, we use a~prototype representation (mean vector) of each class introduced in \cite{snell2017prototypical}, rather than all the training data themselves, in order to deal with the severe imbalance of number of training samples.
We tested all the method using a single desktop PC with Intel® Core™ i7-8700K CPU and NVIDIA GeForce GTX 1080 GPU.
\subsection{Results and Discussion}
The result of ZSL is summarized in Table~\ref{tab:reszsl}.
{The details are given} in Appendix \ref{sec:appendix}.
As shown in the table, our proposed method outperformed all the baseline methods in average F-measure.
In addition, the proposed method could run at about 20 Hz, which is near real-time.
The comparison with the naive nearest-neighbor-based method (without attributes' importance) shows the effectiveness of the attributes' importance.
The performance of the baseline that uses random attributes' importance shows that the attributes' importance should be carefully designed.
In other words, our method enables users to incorporate appropriate domain knowledge on the target classes so that the performance of the model is enhanced.
Compared to DAP~\cite{lampert2014attribute}, the proposed method showed more stable performance on different poses.
The improvement compared to the best baseline (nearest neighbor without attributes' importance) was 4.55 points, which corresponds to 5.91\% relative improvement.
In addition, the proposed method achieved higher scores in majority of the poses compared to this baseline.
Especially a big improvement was observed in ``Stretching calf(L)'' and ``Stretching calf(R)'' poses.
This was because there were unignorable number of subjects who faced down when performing these poses though they were supposed to face forward according to the definition of the pose given in Table~\ref{tab:def}.
Our method could successfully deal with this intra-class variation simply by ignoring the status of head and focusing more on the other important attributes.
On the other hand, there are some poses whose F-measure dropped by incorporating the attributes' importance.
Among those, the biggest drop was observed in pose ``Folding arm''.
This was caused by the low estimation accuracy of the important attributes for folding-arm pose;
the statuses of shoulders in folding-arm pose were sometimes estimated as ``front'' while they had to be ``down'', probably~because arms were slightly pulled forward to make the space for hands at underarms.
Incorporating attributes' importance means focusing more on the important attributes for each pose and ignoring the other attributes.
Therefore, in case the attribute estimation accuracy is not good for those important attributes, the pose classification is done by relying too much on the unreliable attributes.
This problem may be addressed by integrating attributes' unreliability that was introduced~in \cite{jayaraman2014zero}.
\begin{table}[H]
\centering
\caption{The evaluation result (F-measure) of ZSL. Abbreviations are as follows. DAP: direct attribute prediction, NN: nearest neighbor, AI: attributes' importance. The bold numbers represent the best score or the one close to the best (the difference is less than 0.01) for each pose. }
\label{tab:reszsl}
\begin{tabular}{lcccc}\toprule
\textbf{Pose} & \textbf{DAP \cite{lampert2014attribute}} & \textbf{NN w/o AI} & \textbf{NN w/random AI} & \textbf{NN w/AI (Proposed)} \\
\midrule
Standing & {\bf 0.7148} & 0.6953 & 0.3639 & 0.6944 \\
Sitting & 0.4072 & 0.6796 & 0.4567 & {\bf 0.7438} \\
Squatting & 0.8922 & 0.9745 & 0.7637 & {\bf 1.0000} \\
RaiseArmL & {\bf 1.0000} & 0.9791 & 0.4212 & 0.9854 \\
RaiseArmR & {\bf 0.9973} & 0.9662 & 0.4250 & 0.9522 \\
PointingL & {\bf 0.9937} & 0.9541 & 0.4090 & 0.9721 \\
PointingR & 0.9629 & {\bf 0.9991} & 0.4688 & {\bf 0.9947} \\
FoldingArm & 0.4773 & {\bf 0.5345} & 0.2206 & 0.4387 \\
DeepBreathing & 0.9061 & 0.9734 & 0.4457 & {\bf 0.9804} \\
StretchingUp & {\bf 0.9913} & 0.9861 & 0.5947 & {\bf 1.0000} \\
StretchingForward & 0.3703 & {\bf 0.8778} & 0.3625 & {\bf 0.8707} \\
WaistBending & {\bf 1.0000} & 0.9735 & 0.3795 & 0.9612 \\
WaistTwistingL & {\bf 0.4241} & 0.2171 & 0.0995 & 0.2642 \\
WaistTwistingR & {\bf 0.3639} & 0.1547 & 0.1150 & 0.2928 \\
HeelToBackL & {\bf 1.0000} & {\bf 1.0000} & 0.5458 & 0.9787 \\
HeelToBackR & {\bf 0.9931} & 0.8710 & 0.4042 & 0.9729 \\
StretchingCalfL & 0.6647 & 0.5463 & 0.4160 & {\bf 0.8068} \\
StretchingCalfR & {\bf 0.9570} & 0.5956 & 0.4498 & 0.8897 \\
Boxing & 0.6549 & 0.6748 & 0.5434 & {\bf 0.7494} \\
BaseballHitting & 0.5856 & 0.6957 & 0.4328 & {\bf 0.7739} \\
Skiing & 0.5277 & {\bf 0.7977} & 0.6334 & 0.7830 \\
Thinking & {\bf 0.8241} & 0.7801 & 0.6420 & {\bf 0.8230} \\ \midrule
avg. & 0.7595 & 0.7694 & 0.4361 & {\bf 0.8149} \\ \bottomrule
\end{tabular}
\end{table}
The result of few-shot learning is shown in Figure~\ref{fig:fsl}.
Here we only compared the proposed method with the best performed baseline, which is a nearest-neighbor-based method without attribute importance.
It shows that incorporating attributes' importance consistently improves the performance also in few-shot learning scenario.
The improvement is especially bigger when number of shots (training data) is less, and the impact of attributes' importance becomes smaller as number of available training data increases.
This is because the intra-class variation is reflected more in the training data as the number of available training increases and the classifier can naturally learn which attribute is actually important.
Another interesting observation is that the F-measure in ZSL scenario was better than that in one-shot learning scenario regardless of with or without attributes' importance.
This implies that under a situation where only extremely limited number of training data is available, human knowledge (pose definition table) can give a better compromise than just relying on the small number of training data.
\begin{figure}[H]
\begin{center}
\vspace{10pt}
\includegraphics[width=0.7\linewidth]{figures/Picture5c.png}
\vspace{-20pt}
\caption{The results of few-shot learning. }
\label{fig:fsl}
\end{center}
\end{figure}
\section{Conclusions}
\label{sec:conclusion}
This paper has presented a simple yet effective method for improving the performance of ZSL.
In~contrast to the conventional ZSL methods, the proposed method takes the importance of each attribute for each class into account, which becomes more critical when using a set of fine-grained attributes in order to represent wide variety of human poses and actions.
The experimental results on our dataset HDPoseDS have shown that the proposed method is effective not only for ZSL scenario, but also for few-shot learning scenario.
The results as well as the provided dataset are expected to promote further researches toward practical development of human-action-recognition technology under the situation of limited training data.
\vspace{6pt}
\authorcontributions{
Conceptualization, H.O., M.A.-N., S.A., K.N., T.S., and A.D.;
Methodology, H.O., K.N., and T.S.;
Software, H.O.;
Formal Analysis, H.O.;
Investigation, H.O., M.A.-N., and S.A.;
Data Curation, H.O.;
Writing---Original Draft Preparation, H.O.;
Writing---Review \& Editing, H.O., S.A., and K.N.;
Supervision, A.D.
}
\funding{This research received no external funding.}
\conflictsofinterest{The authors declare no conflict of interest.}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,259 |
ACCEPTED
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
null
#### Original name
Maxillaria rauhii D.E.Benn. & Christenson
### Remarks
null | {
"redpajama_set_name": "RedPajamaGithub"
} | 7,719 |
module Boxr
class Client
attr_reader :access_token, :refresh_token, :client_id, :client_secret, :identifier, :as_user_id
#API_URI = "https://wcheng.inside-box.net/api/2.0"
#UPLOAD_URI = "https://upload.wcheng.inside-box.net/api/2.0"
API_URI = "https://api.box.com/2.0"
UPLOAD_URI = "https://upload.box.com/api/2.0"
FILES_URI = "#{API_URI}/files"
FILES_UPLOAD_URI = "#{UPLOAD_URI}/files/content"
FOLDERS_URI = "#{API_URI}/folders"
USERS_URI = "#{API_URI}/users"
GROUPS_URI = "#{API_URI}/groups"
GROUP_MEMBERSHIPS_URI = "#{API_URI}/group_memberships"
COLLABORATIONS_URI = "#{API_URI}/collaborations"
COLLECTIONS_URI = "#{API_URI}/collections"
COMMENTS_URI = "#{API_URI}/comments"
SEARCH_URI = "#{API_URI}/search"
TASKS_URI = "#{API_URI}/tasks"
TASK_ASSIGNMENTS_URI = "#{API_URI}/task_assignments"
SHARED_ITEMS_URI = "#{API_URI}/shared_items"
FILE_METADATA_URI = "#{API_URI}/files"
FOLDER_METADATA_URI = "#{API_URI}/folders"
METADATA_TEMPLATES_URI = "#{API_URI}/metadata_templates"
EVENTS_URI = "#{API_URI}/events"
WEB_LINKS_URI = "#{API_URI}/web_links"
DEFAULT_LIMIT = 100
FOLDER_ITEMS_LIMIT = 1000
FOLDER_AND_FILE_FIELDS = [:type,:id,:sequence_id,:etag,:name,:created_at,:modified_at,:description,
:size,:path_collection,:created_by,:modified_by,:trashed_at,:purged_at,
:content_created_at,:content_modified_at,:owned_by,:shared_link,:folder_upload_email,
:parent,:item_status,:item_collection,:sync_state,:has_collaborations,:permissions,:tags,
:sha1,:shared_link,:version_number,:comment_count,:lock,:extension,:is_package,:can_non_owners_invite]
FOLDER_AND_FILE_FIELDS_QUERY = FOLDER_AND_FILE_FIELDS.join(',')
COMMENT_FIELDS = [:type,:id,:is_reply_comment,:message,:tagged_message,:created_by,:created_at,:item,:modified_at]
COMMENT_FIELDS_QUERY = COMMENT_FIELDS.join(',')
TASK_FIELDS = [:type,:id,:item,:due_at,:action,:message,:task_assignment_collection,:is_completed,:created_by,:created_at]
TASK_FIELDS_QUERY = TASK_FIELDS.join(',')
COLLABORATION_FIELDS = [:type,:id,:created_by,:created_at,:modified_at,:expires_at,:status,:accessible_by,:role,:acknowledged_at,:item]
COLLABORATION_FIELDS_QUERY = COLLABORATION_FIELDS.join(',')
USER_FIELDS = [:type,:id,:name,:login,:created_at,:modified_at,:role,:language,:timezone,:space_amount,:space_used,
:max_upload_size,:tracking_codes,:can_see_managed_users,:is_sync_enabled,:is_external_collab_restricted,
:status,:job_title,:phone,:address,:avatar_uri,:is_exempt_from_device_limits,:is_exempt_from_login_verification,
:enterprise,:my_tags]
USER_FIELDS_QUERY = USER_FIELDS.join(',')
GROUP_FIELDS = [:type, :id, :name, :created_at, :modified_at]
GROUP_FIELDS_QUERY = GROUP_FIELDS.join(',')
VALID_COLLABORATION_ROLES = ['editor','viewer','previewer','uploader','previewer uploader','viewer uploader','co-owner','owner']
def initialize( access_token=ENV['BOX_DEVELOPER_TOKEN'],
refresh_token: nil,
client_id: ENV['BOX_CLIENT_ID'],
client_secret: ENV['BOX_CLIENT_SECRET'],
enterprise_id: ENV['BOX_ENTERPRISE_ID'],
jwt_private_key: ENV['JWT_PRIVATE_KEY'],
jwt_private_key_password: ENV['JWT_PRIVATE_KEY_PASSWORD'],
jwt_public_key_id: ENV['JWT_PUBLIC_KEY_ID'],
identifier: nil,
as_user: nil,
&token_refresh_listener)
@access_token = access_token
raise BoxrError.new(boxr_message: "Access token cannot be nil") if @access_token.nil?
@refresh_token = refresh_token
@client_id = client_id
@client_secret = client_secret
@enterprise_id = enterprise_id
@jwt_private_key = jwt_private_key
@jwt_private_key_password = jwt_private_key_password
@jwt_public_key_id = jwt_public_key_id
@identifier = identifier
@as_user_id = ensure_id(as_user)
@token_refresh_listener = token_refresh_listener
end
private
def get(uri, query: nil, success_codes: [200], process_response: true, if_match: nil, box_api_header: nil, follow_redirect: true)
uri = Addressable::URI.encode(uri)
res = with_auto_token_refresh do
headers = standard_headers
headers['If-Match'] = if_match unless if_match.nil?
headers['BoxApi'] = box_api_header unless box_api_header.nil?
BOX_CLIENT.get(uri, query: query, header: headers, follow_redirect: follow_redirect)
end
check_response_status(res, success_codes)
if process_response
return processed_response(res)
else
return res.body, res
end
end
def get_all_with_pagination(uri, query: {}, offset: 0, limit: DEFAULT_LIMIT, follow_redirect: true)
uri = Addressable::URI.encode(uri)
entries = []
begin
query[:limit] = limit
query[:offset] = offset
res = with_auto_token_refresh do
headers = standard_headers
BOX_CLIENT.get(uri, query: query, header: headers, follow_redirect: follow_redirect)
end
if (res.status==200)
body_json = Oj.load(res.body)
total_count = body_json["total_count"]
offset = offset + limit
entries << body_json["entries"]
else
raise BoxrError.new(status: res.status, body: res.body, header: res.header)
end
end until offset - total_count >= 0
entries.flatten.map{|i| BoxrMash.new(i)}
end
def post(uri, body, query: nil, success_codes: [201], process_body: true, content_md5: nil, content_type: nil, if_match: nil)
uri = Addressable::URI.encode(uri)
body = Oj.dump(body) if process_body
res = with_auto_token_refresh do
headers = standard_headers
headers['If-Match'] = if_match unless if_match.nil?
headers["Content-MD5"] = content_md5 unless content_md5.nil?
headers["Content-Type"] = content_type unless content_type.nil?
BOX_CLIENT.post(uri, body: body, query: query, header: headers)
end
check_response_status(res, success_codes)
processed_response(res)
end
def put(uri, body, query: nil, success_codes: [200], content_type: nil, if_match: nil)
uri = Addressable::URI.encode(uri)
res = with_auto_token_refresh do
headers = standard_headers
headers['If-Match'] = if_match unless if_match.nil?
headers["Content-Type"] = content_type unless content_type.nil?
BOX_CLIENT.put(uri, body: Oj.dump(body), query: query, header: headers)
end
check_response_status(res, success_codes)
processed_response(res)
end
def delete(uri, query: nil, success_codes: [204], if_match: nil)
uri = Addressable::URI.encode(uri)
res = with_auto_token_refresh do
headers = standard_headers
headers['If-Match'] = if_match unless if_match.nil?
BOX_CLIENT.delete(uri, query: query, header: headers)
end
check_response_status(res, success_codes)
processed_response(res)
end
def options(uri, body, success_codes: [200])
uri = Addressable::URI.encode(uri)
res = with_auto_token_refresh do
headers = standard_headers
BOX_CLIENT.options(uri, body: Oj.dump(body), header: headers)
end
check_response_status(res, success_codes)
processed_response(res)
end
def standard_headers()
headers = {"Authorization" => "Bearer #{@access_token}"}
if @jwt_private_key.nil?
headers['As-User'] = "#{@as_user_id}" unless @as_user_id.nil?
end
headers
end
def with_auto_token_refresh
return yield unless @refresh_token or @jwt_secret_key
res = yield
if res.status == 401
auth_header = res.header['WWW-Authenticate'][0]
if auth_header && auth_header.include?('invalid_token')
if @refresh_token
new_tokens = Boxr::refresh_tokens(@refresh_token, client_id: client_id, client_secret: client_secret)
@access_token = new_tokens.access_token
@refresh_token = new_tokens.refresh_token
@token_refresh_listener.call(@access_token, @refresh_token, @identifier) if @token_refresh_listener
else
if @as_user_id
new_token = Boxr::get_user_token(@as_user_id, private_key: @jwt_private_key, private_key_password: @jwt_private_key_password, public_key_id: @jwt_public_key_id, client_id: @client_id, client_secret: @client_secret)
@access_token = new_token.access_token
else
new_token = Boxr::get_enterprise_token(private_key: @jwt_private_key, private_key_password: @jwt_private_key_password, public_key_id: @jwt_public_key_id, enterprise_id: @enterprise_id, client_id: @client_id, client_secret: @client_secret)
@access_token = new_token.access_token
end
end
res = yield
end
end
res
end
def check_response_status(res, success_codes)
raise BoxrError.new(status: res.status, body: res.body, header: res.header) unless success_codes.include?(res.status)
end
def processed_response(res)
body_json = Oj.load(res.body)
return BoxrMash.new(body_json), res
end
def build_fields_query(fields, all_fields_query)
if fields == :all
{:fields => all_fields_query}
elsif fields.is_a?(Array) && fields.length > 0
{:fields => fields.join(',')}
else
{}
end
end
def to_comma_separated_string(values)
return values if values.is_a?(String) || values.is_a?(Symbol)
if values.is_a?(Array) && values.length > 0
values.join(',')
else
nil
end
end
def build_range_string(from, to)
range_string = "#{from},#{to}"
range_string = nil if range_string == ","
range_string
end
def ensure_id(item)
return item if item.class == String || item.class == Fixnum || item.nil?
return item.id if item.respond_to?(:id)
raise BoxrError.new(boxr_message: "Expecting an id of class String or Fixnum, or object that responds to :id")
end
def restore_trashed_item(uri, name, parent)
parent_id = ensure_id(parent)
attributes = {}
attributes[:name] = name unless name.nil?
attributes[:parent] = {id: parent_id} unless parent_id.nil?
restored_item, response = post(uri, attributes)
restored_item
end
def create_shared_link(uri, item_id, access, unshared_at, can_download, can_preview)
attributes = {shared_link: {access: access}}
attributes[:shared_link][:unshared_at] = unshared_at.to_datetime.rfc3339 unless unshared_at.nil?
attributes[:shared_link][:permissions] = {} unless can_download.nil? && can_preview.nil?
attributes[:shared_link][:permissions][:can_download] = can_download unless can_download.nil?
attributes[:shared_link][:permissions][:can_preview] = can_preview unless can_preview.nil?
updated_item, response = put(uri, attributes)
updated_item
end
def disable_shared_link(uri)
attributes = {shared_link: nil}
updated_item, response = put(uri, attributes)
updated_item
end
end
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,541 |
Q: Excel is stripping leading 0's from CSV files It believe that Excel is stripping leading 0s. I was told that updating the column format to text during the export will fix this and to modify Excel output (coming from "ProofAndTracking actionIpromoteuAutomation" file, yet I can't find this file or how to access where the excel formatting code is generated).
I've never worked with Excel. I tried this, but it seems to be a local fix:
http://excelribbon.tips.net/T010262_Handling_Leading_Zeros_in_CSV_Files.html
Could someone point me in the right direction to start?
I've researched these answers How to stop the leading 0's from being stripped off when exporting to excel from a datatable? and Export Excel : Avoid stripping the leading zeros
Thanks in advance!
Here is the function I believe is causing the problem, but not sure yet how to format columns to text here. /*
* format and send order and tracking info (via Excel spreadsheet) for the day for ipromoteu (150837)
*/
public function actionIpromoteuAutomation() {
$ordersGroup1 = $this->getIpromoteuProof();
$ordersGroup2 = $this->getIpromoteuProofHistory();
$orders = array_merge($ordersGroup1, $ordersGroup2);
$fileName = 'Hit Promo Order Tracking ' . date('m-d-Y');
$this->_export($orders, $fileName);
$fileName = $fileName . '.xls';
$toIPROMOTEU = array(//email to be sent to HR when request is submitted
'body' => 'Attached, please find the Excel spreadsheet containing'
. ' order details for orders on ' . date('m/d/Y') . '.',
'from_email' => 'donotreply@hitpromo.net',
'from_name' => 'Hit Promotional Products',
'subject' => 'Order Tracking Information - ' . date('m/d/Y'),
'to_emails' => 'orders@ipromoteu.com',
'attachments' => array(array('path' => Yii::app()->basePath . '/../tmp/' . $fileName, 'filename' => $fileName)),
);
Hit::email((object) $toIPROMOTEU);
$this->logForDeveloper('Order records sent to iPromoteU for '. date('m/d/Y'));
}
private function _export($data, $fileName, $format = 'excel', $output = false) {
// get a reference to the path of PHPExcel classes
$phpExcelPath = Yii::getPathOfAlias('ext.phpexcel');
// Turn off our yii library autoload
spl_autoload_unregister(array('YiiBase', 'autoload'));
include($phpExcelPath . DIRECTORY_SEPARATOR . 'PHPExcel.php');
// Create new PHPExcel object
$objPHPExcel = new PHPExcel();
if ($format == 'excel') {
// Set properties
$objPHPExcel->getProperties()->setCreator(Yii::app()->user->name)
->setLastModifiedBy(Yii::app()->user->name)
->setTitle('Order Tracking Report');
}
$styleArray = array(
'font' => array(
'bold' => true,
'underline' => true,
)
);
$objPHPExcel->setActiveSheetIndex(0);
$sheet = $objPHPExcel->getActiveSheet();
for ($rowCounter = 0; $rowCounter < sizeof($data); $rowCounter++) {
$sheet->getStyle("A" . ($rowCounter + 1))->getAlignment()->setHorizontal(PHPExcel_Style_Alignment::HORIZONTAL_LEFT);
$sheet->getStyle("B" . ($rowCounter + 1))->getAlignment()->setHorizontal(PHPExcel_Style_Alignment::HORIZONTAL_LEFT);
$sheet->getStyle("C" . ($rowCounter + 1))->getAlignment()->setHorizontal(PHPExcel_Style_Alignment::HORIZONTAL_LEFT);
$sheet->setCellValue("A" . ($rowCounter + 1), date('m/d/Y', strtotime($data[$rowCounter]['date'])));
//$sheet->setCellValue("A".($rowCounter+1), $data[$rowCounter]['date']);
if ($data[$rowCounter]['fedex_tracking'] != '')
$sheet->setCellValue("B" . ($rowCounter + 1), $data[$rowCounter]['fedex_tracking']);
else
$sheet->setCellValue("B" . ($rowCounter + 1), $data[$rowCounter]['other_tracking']);
$sheet->setCellValue("C" . ($rowCounter + 1), $data[$rowCounter]['sales_order_number']);
$sheet->getStyle("B" . ($rowCounter + 1))->getNumberFormat()->setFormatCode(PHPExcel_Style_NumberFormat::FORMAT_NUMBER);
}
$sheet->setCellValue("A1", "DATE")
->setCellValue("B1", "TRACKING")
->setCellValue("C1", "PO NUMBER")
->getStyle("A1:C1")->applyFromArray($styleArray);
////Set the column widths
$sheet->getColumnDimension("A")->setWidth(25);
$sheet->getColumnDimension("B")->setWidth(25);
$sheet->getColumnDimension("C")->setWidth(25);
// Rename sheet
$objPHPExcel->getActiveSheet()->setTitle('Order Tracking Report');
// Set active sheet index to the first sheet,
// so Excel opens this as the first sheet
$objPHPExcel->setActiveSheetIndex(0);
$fileName = $fileName . ($format == 'excel' ? '.xls' : '.csv');
$objWriter = PHPExcel_IOFactory::createWriter($objPHPExcel, 'Excel5');
$path = Yii::app()->basePath . '/../tmp/' . $fileName;
$objWriter->save($path);
// Once we have finished using the library, give back the
// power to Yii...
spl_autoload_register(array('YiiBase', 'autoload'));
}
public function getIpromoteuProof() {
$date = date('Ymd');
$db2params = Yii::app()->params['db2params'];
$db = Zend_Db::factory('Db2', $db2params);
$select = $db->select()
->from(
// table
array('t' => 'WBPIPRAE'),
// columns
array(
'sales_order_number' => 'TRIM(WAESPO#)',
),
// schema
$db2params['schemas']['hitdta'])
->joinLeft(
// table
array('f' => 'MFD1MD'), 'CONCAT(TRIM(WAEORD#),RIGHT(TRIM(MDORDR),3))=TRIM(MDORDR)',
// columns
array(
'fedex_tracking' => 'TRIM(MDFTRK)',
'other_tracking' => 'TRIM(MDFBRC)',
'date' => 'TRIM(MDUPDT)',
),
// schema
$db2params['schemas']['varfil'])
->where("MDUPDT = '" . $date . "'")
->where("WAEIVKY='150837'")
->order('sales_order_number')
->distinct(true);
$stmt = $db->query($select);
$orders = $stmt->fetchAll();
return $orders;
}
A: OK, Excel can create the csv columns as text but the issue is with the import.
When Excel opens csv files it 'helpfully' converts anything that looks like a number to a number, stripping leading zero's.
First make sure Excel is exporting it as it should, open the file in a text editor and check the leading zero is present.
You will probably have to change how the file is imported. Instead of them just double clicking the csv file, they need to run through the data import wizard:
Import 'From Text'
Select Delimited
Make sure comma is ticked
Select column as text
This should resolve the problem on the other side.
Another alternative is to add a single apostrophe ' before the numbers, however if they use macros/software to interact with the file this might break that.
Edit
After looking at the PHP code (I can't test) you might be able to change the line:
$sheet->getStyle("B" . ($rowCounter + 1))->getNumberFormat()->setFormatCode(PHPExcel_Style_NumberFormat::FORMAT_NUMBER);
To:
$sheet->getStyle("B" . ($rowCounter + 1))->getNumberFormat()->setFormatCode('000000000000'); // With however many zero's you need to keep
This however may still get removed when you export as it's still technically a number and formatting will be removed.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,030 |
package com.dcc.matc89.spots.social;
import java.util.Arrays;
import android.os.Bundle;
import android.support.v4.app.FragmentActivity;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import com.dcc.matc89.spots.R;
import com.dcc.matc89.spots.model.User;
import com.facebook.Session;
import com.facebook.SessionState;
import com.facebook.widget.LoginButton;
public class LoginFragment extends FacebookFragment {
public static final String TAG = "LoginFragment";
private View mLoginLayout, mRefreshProgressBar;
private OnUserLoginListener mOnUserLoginListener;
private View mRoot;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setRetainInstance(true);
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
mRoot = inflater.inflate(R.layout.fragment_login, container, false);
mRefreshProgressBar = mRoot.findViewById(R.id.login_pgs);
mLoginLayout = mRoot.findViewById(R.id.login_layout);
LoginButton authButton = (LoginButton) mRoot.findViewById(R.id.authButton);
authButton.setReadPermissions(Arrays.asList("user_location"));
authButton.setFragment(this);
updateLayoutVisibility(isLoggedIn(Session.getActiveSession()));
return mRoot;
}
@Override
public void call(Session session, SessionState state, Exception exception) {
super.call(session, state, exception);
boolean loggedIn = isLoggedIn(session);
updateLayoutVisibility(loggedIn);
FragmentActivity activity = getActivity();
if(loggedIn && activity != null){
User user = User.getCurrentUser(activity);
if(user != null && mOnUserLoginListener != null)
mOnUserLoginListener.onUserLoggedIn(user);
}
}
public void setOnUserLoginListener(OnUserLoginListener listener){
mOnUserLoginListener = listener;
}
private boolean isLoggedIn(Session session) {
return session != null && session.isOpened();
}
private void updateLayoutVisibility(boolean loggedIn) {
mLoginLayout.setVisibility(loggedIn ? View.INVISIBLE : View.VISIBLE);
mRefreshProgressBar.setVisibility(loggedIn ? View.VISIBLE : View.INVISIBLE);
}
public void hide() {
mRoot.setVisibility(View.INVISIBLE);
}
@Override
public void onUserLoggedIn(User user) {
if(user != null && mOnUserLoginListener != null)
mOnUserLoginListener.onUserLoggedIn(user);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,003 |
{"url":"https:\/\/journal.pda.org\/highwire\/markup\/6553\/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed","text":"TABLE VII\n\nExtracted Trace Elements and Metals\n\nElementExtracted Amount,d \u03bcg\/g\nPVCRubberLDPEPolycarbonateCyclic Olefin\npH 2.5pH 9.5pH 2.5pH 9.5pH 2.5pH 9.5pH 2.5pH 9.5pH 2.5pH 9.5\nCa1.5NPc4.072.07NPcNPc6.60.800.811.7\nZn1.30.432.890.490.08NPc0.120.030.070.03\nBr1.30.0817.520.50.400.081.10.070.540.02\nNa0.60MEa3.05MEa0.96MEa1.0MEa0.96MEa\nKMEaNPcMEa6.84MEaNPcMEaNPcMEaNPc\nFe0.350.150.330.08b0.12NPc0.710.060.24NPc\nMg0.260.213.502.900.180.04b0.240.170.160.15\nAl0.14NPc0.663.560.09NPcNPcNPc0.07NPc\nCr0.02NPc0.010.01NPcNPc0.070.02NPcNPc\nTiNPcNPc0.29NPcNPcNPcNPcNPcNPcNPc\nMn0.02NPc0.01NPc0.01NPcNPcNPcNPcNPc\nSiNPcNPc0.100.25NPcNPcNPcNPcNPcNPc\nSr0.01b0.020.010.010.01NPcNPcNPcNPcNPc\nNiNPcNPc0.010.01NPcNPcNPcNPcNPcNPc\nCuNPc0.01NPcNPcNPcNPcNPcNPcNPcNPc\nCoNPcNPcNPcNPc0.01NPcNPcNPcNPcNPc\nV0.01NPc0.010.01NPcNPcNPcNPcNPcNPc\nBaNP3NPcNPcNPcNPc0.01NPcNPcNPcNPc\nAs0.01NPcNPcNPcNPcNPcNPcNPcNPcNPc\nPb0.01bNPcNPcNPcNPcNPcNPcNPcNPcNPc\nSbNPc0.01NPcNPcNPcNPcNPcNPcNPcNPc\nMoNPc0.01NPcNPcNPcNPcNPcNPcNPcNPc\n\u2022 Notes:\n\n\u2022 a ME = this element a component of the extracting solution used and thus was not measurable as an extractable.\n\n\u2022 b Detected in only one of the two replicate extracts.\n\n\u2022 c NP = not present in this extract in measurable quantities.\n\n\u2022 d The reported value is the largest amount measured in either the sealed vessel or sonication extract.","date":"2022-08-16 14:02:17","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8243842720985413, \"perplexity\": 14693.286701880746}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-33\/segments\/1659882572304.13\/warc\/CC-MAIN-20220816120802-20220816150802-00032.warc.gz\"}"} | null | null |
{"url":"https:\/\/physics.stackexchange.com\/questions\/241754\/how-time-dilation-happens-by-velocity?noredirect=1","text":"# How time dilation happens by velocity?\n\ni got many explanation and proving that relative velocity causes time dilation. Einstein's moving light experiment proves it. but that is a clock works with light detector. as detecting the light gets latter the clock works slowly. but in reality how a real watch can be slowed down? And won't the man feel that the time is moving slowly as he knows the general speed of the clock? Sources says that if a man goes to space with very high speed if he returns after few years as his clock moved slowly earth would be moved for many years forward. But my doubt is won't he feels the fact that his clock was slow and he spend many years in space? If he won't feel it how does it happens? and how is he not getting older?\n\n\u2022 Hi Selvaratnam. Your question has finally prompted me to post the canonical Q\/A on time dilation that I've been meaning to post for ages. Have a look at What is time dilation really?. \u2013\u00a0John Rennie Mar 6 '16 at 9:27\n\nHave you ever draped a measuring tape from point A to point B in a looping indirect path and asked yourself why the path length of the tape is longer than the direct path between A and B?\n\nThe answer is that tape measures don't measure distances between points. They measure the length of the path they are on.\n\nWell, it turns out clocks don't measure the time between points. No clock does. Not a light clock. Not a human brain. Not radio carbon dating. Nothing does that. Just like no tape measure measures distance, it only is capable of measuring the length of its own path.\n\nSo there is a 4d spacetime. Imagine the z direction as time. Each xyz giving and xy spatial point and z giving the time. Then a particle at rest is a vertical line. And a particle moving in the x direction is a line in the xz plane and a the steeper the line, the slower the particle is going. Each point just spells out a place and time something was located. It gives a when-where.\n\nSo clocks make paths in the 4d space. And like a tape measure they are only capable of measuring the length along the curve they actually make. You could have a straight path. Or a curvy one and the measurements will be different.\n\nAnd nothing does anything other than measure along the path it goes.\n\nYou brain, the light clock, a spring clock, a quartz clock, each one changes not based on time, but along the length of their own path. Just like the tape measure makes its marks along its own path.\n\nSo now by moving clocks along different paths it is up to us to experimentally find out what lengths look like in the 4d space.\n\nAnd it's different than you might expect. For instance, the direct path between one when-where and another when-where gives the longest length (not the shortest).\n\nAnd we also learn in general relativity that passing through a region in a deeper gravity well could make your path shorter. So the longest path from here-now to the other side of the sun six months later isn't to go straight towards the sun at a slower speed, but instead the longest path is to go around the sun like the earth does.\n\nIn fact, we found out that is why the earth goes around the sun. When you learn the way the world is, and replace your ideas about what is natural with those ideas then the world actually makes more sense.\n\n\u2022 i didn't get it exactly, but what you are saying is that the man who travels also feels like less time than the man in earth and not only a light detecting clock all clock's rotating speed is related to its velocity. is it? \u2013\u00a0Selvaratnam Lavinan Mar 6 '16 at 9:06\n\u2022 @SelvaratnamLavinan All clocks (in brains, with light, quartz, spring, whatever) measure the length of its own path in 4d which depends on the shape and the deepness of the gravitational well in the regions it travels. \u2013\u00a0Timaeus Mar 6 '16 at 9:11\n\u2022 one more doubt with this time dilation concept we can travel to future. is there any concept to travel to past? \u2013\u00a0Selvaratnam Lavinan Mar 6 '16 at 9:32\n\u2022 @SelvaratnamLavinan Muons travels to the future, it's how we see them even though they live so short and are created so far away. Travel to the past is different. \u2013\u00a0Timaeus Mar 6 '16 at 9:34\n\u2022 ya i understood that travel to future happens as we move to future faster. is there other concepts like this to travel to the past? or is it impossible as of now. \u2013\u00a0Selvaratnam Lavinan Mar 6 '16 at 9:37\n\nHere's a very vague analogy. If we walk at the same speed, but in different directions, say in straight lines, then we would feel as if we are travelling at the same speed but if I measure your distance travelled in my direction, you would have travelled less than I in the same amount of time. If you think of the road as my perspective, then you could say that in my perspective you travelled slower than I. Likewise in your perspective I travelled slower than you. Note that this remains true if you and I were robots that just mindlessly measure each other's travels in our individual perspectives.\n\nIn reality time and space are part of the same spacetime continuum, and our paths are not paths in space over time, but paths in spacetime. In a vaguely analogous sense to the above paragraph, each of us feels normal in our own perspective, but observe the other to be travelling differently when viewed from our perspective. This is not merely a mental phenomenon, as the analogy suggests. We literally travel normally, just in a non-parallel direction in the spacetime continuum.\n\nIf you think about it carefully, it also explains the twin-paradox because one twin travels in a constant direction in spacetime, whereas the second has to change direction if he ever wants to return to a point with the same space coordinates as the first twin. The twins end up at different time coordinates.\n\nI doubt there is an \"intuitive\" way to describe time dilation, since most people's intuition suggests against it, but there is probably a way to get the general idea without using a bunch of math.\n\nSo, the starting point for understanding the theory is that, as you've probably heard many times before, the speed of light is always constant for all observers. We know this through observations and measurements, and these observations are the ones that go against our intuition, and this presents a problem for pre-relativistic theories.\n\nTrowing a ball inside a moving train\n\nTo understand why this is a problem, imagine what happens if you are standing in a moving train, and throw a ball at a certain speed in the forward direction of train's movement? To all observers inside the train, this ball will appear to be travelling at a certain speed $v$ until it hits the front of the train (ignoring forces like friction and gravity for the moment). But, to stationary observers outside the train, looking at the ball through the train's window as the train passes by, ball will appear to be travelling much faster, its speed being accumulated to the speed of the moving train. These two observers, having different (but constant) speeds relative to each other, are said to be in different reference frames.\n\nTurning on a flashlight inside a moving train\n\nNow imagine you wanted to measure the speed of light using a flashlight, inside a moving train. To measure the speed of light, you place two observers on the train, equipped with two synchronized clocks, one at the back of the train and the other one at the front of the train. The back observer fires the light-beam exactly as he passes a certain point on the stationary train station and marks the time on his clock, and the front observer marks the exact time the light-beam hits the front of the train. Diving their distance by the measured time gives them the speed of light, $c$.\n\nAt the same time, two carefully placed stationary observers are standing next to the train tracks, again equipped with two synchronized clocks. They place themselves so that the first of them is at the mentioned point on the train station (to record the moment the light-beam was sent), and the other one is placed at the position where the light will hit the train's front clock (the one inside the train, equipped with the sensor), but relative to train tracks; let's say the second stationary observer really needed to repeat the experiment several times to get this position accurately enough.\n\nWhat do the stationary observers measure?\n\nFirst of all, two stationary observers check their clocks and the distance between them, and it turns out the speed of light was again $c$. It didn't \"add up\" to train's speed at all this time, unlike the ball from the first experiment. But they also notice the weirdest of things: while the first observer saw the first train clock to show the same time as his stationary clock, the second observer (seeing the light-beam hit the front train's sensor) noticed that the clock on board the train was out of sync with his outside clock; it seemed to be running slightly slower. When the front was showing 0:20, the outside front clock was already at 0:21.\n\nWhich of the observers actually experienced the dilation?\n\nAt this point, outside observers conclude that the train observers must have experienced dilation; after all, after reuniting, they will all agree that both back clock met at 0:00, and front clock were out of sync in exactly the same fashion (front train clock at 0:20, front ground clock at 0:21).\n\nBut, as it turns out, when moving at a constant speed, no frame is \"privileged\". For the observers in the train, the two clocks outside the train didn't appear synchronized at all in the first place! In fact, when the back clocks aligned at 0:00, for the train observers, it seemed like the front ground clock was out of sync, and already showing 0:02!\n\nSo, from the train observers perspective, it seemed like ground observers experienced dilation, because the out-of-sync clock was supposed to show 0:22 at the moment they aligned.\n\nWhy is this so unintuitive?\n\nFirst of all, relativistic effects are hardly noticeable unless you are observing reference frames moving at near the speed of light, which is something rarely observed in everyday life, by everyday people. Our intuition prefers the \"safety and comfort\" of the throwing ball experiment, because this is what we can perceive in everyday scenarios.\n\nNext, unless you introduce acceleration, time dilation and other relativistic effects only exist as a perceived difference from observers in different frames of reference. No matter how fast you are moving, if you are moving at a constant speed (i.e. don't experience any inertial forces like the ones you experience when riding in an elevator and it slows down), your clocks, lengths, speed of light, and all physical laws as we know it will behave exactly as always. This is another reason why it's unintuitive; you can never observe any effects of relativity on yourself.","date":"2019-09-24 09:09:53","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\": 0.4651623070240021, \"perplexity\": 510.3588817683951}, \"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-2019-39\/segments\/1568514572896.15\/warc\/CC-MAIN-20190924083200-20190924105200-00387.warc.gz\"}"} | null | null |
'Lost' Burns plaque returned to monument
A commemorative plaque of Robert Burns, owned by a local Trust, has been revealed as one of the missing plaques from the Dalkeith Burns Monument.
On a cold and windy afternoon earlier this month, members of the Cousland Smiddy Trust, along with local Councillors Margot Russell and Alex Bennett and selected others, gathered at the monument to find out if the historic plaque would fit.
Everyone was delighted to see that the Cousland Smiddy plaque of Robert Burns does appear to fit the oval of the Burns Monument cartouche. The Cousland Smiddy Burns plaque is a plaster model, possibly taken from the original cast, and could provide an excellent mould for recasting a new plaque.
Councillor Jim Bryant said: "This is a really exciting find and I am extremely grateful to the Cousland Smiddy Trust for their interest in the Burns Monument. The Burns Monument was paid for by the residents of Dalkeith and is a key part of the heritage of our town."
The Burns Memorial fountain was made by George Smith's Sun Foundry, in Candleriggs, Glasgow. It was commissioned in 1896 by the Dalkeith Burns Club to mark the centenary of the death of Robert Burns and was paid for by public subscription.
The structure has four columns with a domed canopy. Known locally as the "Burns Monument", it was not in fact installed until 1899 and, for many years after, was one of the main focal points in Dalkeith High Street.
In the 1960s it was deemed a traffic hazard and moved to the gardener's cottage in St John's and King's Park. In 2003, it was refurbished and relocated to a new position in Komorom Court, Buccleuch Street. Since 2013, a Facebook campaign has been promoting the potential restoration and relocation of the monument to a more historic site within the town.
Interestingly, over each arch of the dome there are cartouches which would have contained commemorative plaques with dedications or crests. Unfortunately at some point in its history all four plaques on the Burns Monument were lost.
It has been suggested that the plaques included the Dalkeith Coat of Arms, a Masonic crest and a portrait of Robert Burns. Plans to restore the monument are currently being considered and the replacement of the missing plaques would bring new meaning to the title of Burns Monument.
Meanwhile, the Cousland Smiddy Trust noticed in their collection an oval commemorative plaque of Robert Burns and speculation grew as to whether this might, in fact, be one of the missing plaques from the Dalkeith Burns Monument. Thoughts turned to Cinderella and the glass slipper: would the Cousland Smiddy plaque of Robert Burns fit the Burns Monument Cartouche? | {
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BYU School of Music presents Choral Showcase Oct. 1-2
The Brigham Young University Singers, Concert Choir, Men's Chorus and Women's Chorus will present a Choral Showcase Wednesday and Thursday, Oct. 1-2, at 7:30 p.m. in the de Jong Concert Hall of the Harris Fine Arts Center.
Tickets at $9 for the general public with $3 off with a BYU or student ID are available at the Fine Arts Ticket Office, (801) 378-4322 or at www.byu.edu/hfac.
The Women's Chorus under the direction of Vicki McMurray will perform Ruth Watson Henderson's "Cantate Domino," "If Ye Love Me" by Eleanor Daley and "Hold On" by Lucy Simon.
The Concert Choir, with conductor Rosalind Hall, will perform "Sanctus" by J.S. Bach and Moses Hogan's "The Water is Wide" and "Ezekiel Saw the Wheel."
Under the direction of Ronald Staheli, the BYU Singers will perform "There is Sweet Music" by Edward Elgar, "Gloria" by Greg Knauf and "Paean" from "Requiem" by Merrill Bradshaw.
Rosalind Hall will also direct the Men's Chorus in their performance of "Resonet in Laudibus" by Jacob Handl, and "Joshua Fit the Battle of Jericho" arranged by David Wright.
The program will conclude with a "Community Sing" with the combined choirs and audience singing "Redeemer of Israel" arranged by Mack Wilberg and the Combined Choirs singing Mack Wilberg's arrangement of "Praise to the Lord, the Almighty."
Writer: Rachel M. Sego | {
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{"url":"https:\/\/www.r-bloggers.com\/2018\/03\/r-simple-for-complex-tasks-complex-for-simple-tasks\/","text":"Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nWhen it comes to undertaking complex data science projects, R is the preferred choice for many. Why? Because handling complex tasks is simpler in R than other comparable platforms.\n\nRegrettably, the same is not true for performing simpler tasks, which I would argue is rather complex in base R. Hence, the title \u2014 R: simple for complex tasks, complex for simple tasks.\n\nConsider a simple, yet mandatory, task of generating summary statistics for a research project involving tabular data. In most social science disciplines, summary statistics for continuous variables require generating mean, standard deviation, number of observations, and perhaps minimum and maximum. One would have hoped to see a function in base R to generate such table. But there isn\u2019t one.\n\nOf course, several user-written packages, such as psyche, can generate descriptive statistics in a tabular format. However, this requires one to have advanced knowledge of R and the capabilities hidden in specialized packages whose number now exceed 12,000 (as of March 2018). Keeping abreast of the functionality embedded in user-written packages is time-consuming.\n\nSome would argue that the summary command in base R is an option. I humbly disagree.\n\nFirst, the output from summary is not in a tabular format that one could just copy and paste into a document. It would require significant processing before a simple table with summary statistics for more than one continuous variable could be generated. Second, summary command does not report standard deviation.\n\nI teach business analytics to undergraduate and MBA students. While business students need to know statistics, they are not studying to become statisticians. Their goal in life is to be informed and proficient consumers of statistical analysis.\n\nSo, imagine an undergraduate class with 150 students learning to generate a simple table that reports summary statistics for more than one continuous variable. The simple task requires knowledge of several R commands. By the time one teaches these commands to students, most have made up their mind to do the analysis instead in Microsoft Excel instead.\n\nHad there been a simple command to generate descriptive statistics in base R, this would not be a challenge for instructors trying to bring thousands more into R\u2019s fold.\n\nIn the following paragraphs, I will illustrate the challenge with an example and identify an R package that generates a simple table of descriptive statistics.\n\nI use mtcars dataset, which is available with R. The following commands load the dataset and display the first few observations with all the variables.\n\ndata(mtcars)\n\nAs stated earlier, one can use summary command to produce descriptive statistics.\n\nsummary(mtcars)\n\nLet\u2019s say one would like to generate descriptive statistics including mean, standard deviation, and the number of observations for the following continuous variables: mpg, disp, and hp. One can use the sapply command and generate the three statistics separately and combined them later using the cbind command.\n\nThe following command will create a vector of means.\n\nmean.cars = with(mtcars, sapply(mtcars[c(\u201cmpg\u201d, \u201cdisp\u201d,\u00a0 \u201chp\u201d)], mean))\n\nNote that the above syntax requires someone learning R to know the following:\n\n1.\u00a0\u00a0\u00a0 Either to attach the dataset or to use with command so that sapply could recognize variables.\n2.\u00a0\u00a0\u00a0 Knowledge of subsetting variables in R\n3.\u00a0\u00a0\u00a0 Familiarity with c to combine variables\n4.\u00a0\u00a0\u00a0 Being aware of enclosing variable names in quotes\n\nWe can use similar syntax to determine standard deviation and the number of observations.\n\nsd.cars = with(mtcars, sapply(mtcars[c(\u201cmpg\u201d, \u201cdisp\u201d,\u00a0 \u201chp\u201d)], sd)); sd.cars\nn.cars = with(mtcars, sapply(mtcars[c(\u201cmpg\u201d, \u201cdisp\u201d,\u00a0 \u201chp\u201d)], length)); n.cars\n\nNote that the user needs to know that the command for number of observations is length and for standard deviation is sd.\n\nOnce we have the three vectors, we can combine them using cbind that generates the following table.\n\ncbind(n.cars, mean.cars, sd.cars)\n\nn.cars mean.cars\u00a0\u00a0\u00a0 sd.cars\nmpg\u00a0\u00a0\u00a0\u00a0\u00a0 32\u00a0 20.09062\u00a0\u00a0 6.026948\ndisp\u00a0\u00a0\u00a0\u00a0 32 230.72188 123.938694\nhp\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 32 146.68750\u00a0 68.562868\n\nAgain, one needs to know the round command to restrict the output to a specific number of decimals. See below the output with two decimal points.\n\nround(cbind(n.cars, mean.cars, sd.cars),2)\n\nn.cars mean.cars sd.cars\nmpg\u00a0 \u00a0 \u00a0 32\u00a0 \u00a0 \u00a020.09\u00a0 \u00a0 6.03\ndisp\u00a0 \u00a0 \u00a032\u00a0 \u00a0 230.72\u00a0 123.94\nhp\u00a0 \u00a0 \u00a0 \u00a032\u00a0 \u00a0 146.69\u00a0 \u00a068.56\n\nOne can indeed use a custom function to generate the same with one command. See below.\n\nround(with(mtcars, t(sapply(mtcars[c(\u201cmpg\u201d, \u201cdisp\u201d,\u00a0 \u201chp\u201d)],\nfunction(x) c(n=length(x), avg=mean(x),\nstdev=sd(x))))), 2)\n\nn\u00a0 \u00a0 avg\u00a0 stdev\nmpg\u00a0 32\u00a0 20.09\u00a0 \u00a06.03\ndisp 32 230.72 123.94\nhp\u00a0 \u00a032 146.69\u00a0 68.56\n\nBut the question I have for my fellow instructors is the following. How likely is an undergraduate student taking an introductory course in statistical analysis to be enthused about R if the simplest of the tasks need multiple lines of codes? A simple function in base R could keep students focussed on interpreting data rather than worrying about missing a comma or a parenthesis.\n\nstargazer* is an R package that simplifies this task. Here is the output from stargazer.\n\nlibrary(stargazer)\nstargazer(mtcars[c(\"mpg\", \"disp\",\u00a0 \"hp\")], type=\"text\")\n============================================\nStatistic N Mean St. Dev. Min Max\n--------------------------------------------\nmpg 32 20.091 6.027 10.400 33.900\ndisp 32 230.722 123.939 71.100 472.000\nhp 32 146.688 68.563 52 335\n--------------------------------------------\nA simple task, I argue, should be accomplished simply. My plea will be to include in base R a simple command that may generate the above table with a command as simple as the one below:\n\ndescriptives(mpg, disp, hp)\n\n* \u00a0Hlavac, Marek (2015). stargazer: Well-Formatted Regression and Summary Statistics Tables. \u00a0R package version 5.2. http:\/\/CRAN.R-project.org\/package=stargazer","date":"2021-10-27 12:54:32","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\": 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\": 0.5722500681877136, \"perplexity\": 3191.485398366704}, \"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-2021-43\/segments\/1634323588153.7\/warc\/CC-MAIN-20211027115745-20211027145745-00054.warc.gz\"}"} | null | null |
Borno's SEMA spends N2b feeding IDPs
Borno IDPs camp: SEMA says it spent N2b on food in four years
By Rabiu Sani/Maiduguri
The Borno State Emergency Management Agency (SEMA) on Saturday said it spent over N2 billion distributing food items and condiments to IDPs in the last four years.
SEMA's Chairman Alhaji Satomi Ahmad, made the disclosure in an interview with the News Agency of Nigeria (NAN) in Maiduguri.
Ahmad said that the amount went into procurement and distribution of foodstuffs, firewood, drugs and condiments to persons displaced by Boko Haram insurgency in the state.
He said that the gesture was to augment the Federal Government and development agencies' efforts to support the displaced persons and address humanitarian crisis.
"Previously, the agency spent about N20 million on procurement and distribution of firewood alone to displaced households in the camps and liberated communities.
"The trend resulted in unprecedented upsurge in the demand and supply of firewood in the state.
" This also naturally resulted in depletion of forest resources and destruction of the environment which left us with no options than to start using charcoal," he said.
Satomi said the choice of charcoal came with high cost as the initial funds for firewood went up astronomically.
The chairman said the agency had also concluded arrangements to de-worm and immunise all children in IDPs camps and liberated communities.
He said that the exercise was designed to protect children against diseases, reduce morbidity and mortality.
According to him, the agency had deployed nine mobile clinics to enhance rapid response and disaster management services in the state.
Anambra: Female journalist emerges governorship candidate
Petrol shortage hits America, fuel prices go up
FG approves trade union for Bolt, Uber Drivers | {
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#IrishFilm: Alan Dunne's Earthly Bonds announced as Film In Cork 2017 Short Script Award Winner
Film In Cork has announced that after a long and rigorous selection process, a unanimous winner has been selected for the €8,000 Short Script Award 2017. The winning project is Earthly Bonds, written and to be directed by Alan Dunne, and to be produced by Eamonn Tutty for Reckoner Productions.
Almost 130 scripts were submitted prior to the December deadline, which was whittled down to five that went forward to the next stage of an interview. The interview panel, comprised of writer/directors Oonagh Kearney and Michael Kinirons, and producer/ production manager Steven Davenport, had a very challenging job selecting the overall winner.
Speaking on behalf of the panel, Oonagh Kearney commented that
Earthly Bonds is an elegant and well-told story, and the director, Alan Dunne, shows evidence of emerging talent, has a good eye, and demonstrates ambition for the continued development of his craft as a writer-director. All of us feel this will make for a great short and look forward to helping the team realize their vision for this project.
Alan and his team will benefit from the experience and talent of the interview panel in terms of mentoring support as they begin the process of translating the script to the screen, and Film In Cork looks forward to seeing the finished film premiere later this year.
The Short Script Award was made possible through the generous support of the Arts Offices of Cork City and Cork County Councils, and Film In Cork is delighted that in addition to supporting emerging talent, the scheme also serves to showcase the diverse locations that Cork can offer productions shooting in the region.
Alan DunneEamonn TuttyEarthly BondsFilm in CorkReckoner Productions
Nominees announced for 10th Underground Cinema Awards
Deadline approaches for Virgin Media Discovers short film award
#IrishAbroad: Vincent Gallagher wins at Savannah Film Festival with two films
#IrishAbroad: Robert Cunningham's TRANSition wins at Scottish Mental Health Arts & Film Festival
Previous #IrishFilm: Annika Cassidy's short film, PEEL, wraps in Dublin
Next #IrishFilm: IFI and IFB team to bring the After '16 short films to the IFI Player
#IrishAbroad: The Clockmaker's Dream to compete at the 20th Shanghai International Film Festival | {
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## DEDICATION
_For nieces and a nephew (in order of appearance)
Jackson
Evelyn
Melody
Kathryn
Maple_
_You still have time._
## CONTENTS
Dedication
Part One
Chapter One
Chapter Two
Chapter Three
Chapter Four
Chapter Five
Chapter Six
Part Two
Chapter Seven
Chapter Eight
Chapter Nine
Chapter Ten
Chapter Eleven
Chapter Twelve
Part Three
Chapter Thirteen
Chapter Fourteen
Chapter Fifteen
Chapter Sixteen
Chapter Seventeen
Chapter Eighteen
Epilogue
About the Author
Back Ad
Copyright
About the Publisher
## PART ONE
## ONE
ESTHER RAN ACROSS THE BROKEN ASPHALT.
Her destination loomed in front of her: an odd concrete structure, standing alone in the trash-filled lot. At a glance, it seemed unoccupied. As she neared it, Esther gave a final look in all directions; then she grabbed the low brick wall and vaulted over. Despite the searing heat of the early November morning, it was cool and dim inside. Esther blinked to get her bearings, then took off again.
The sound of her breathing, harsh and ragged, was the only noise; it seemed to echo in the vast space. She sensed this and cursed herself. Clamping her lips together, she tried to hold back the sound, to breathe so as to not to draw attention to herself. But it was impossible to do this without slowing down. She gave up and put on extra speed instead. She had been running a long time and as she gulped air, her lungs burned and a sharp pain gripped her side. But she was almost at the safe place and then she would have time to rest.
Esther was on the second floor and there were four more to go. She ran low, crouching down and sticking close to the center of the structure. It was a strange building, one that offered neither protection nor privacy. There were no walls; the sides were too exposed, open to the world outside, with the rooftops of Prin visible in the distance beneath a dirty yellow sky.
Esther was aware that she was wearing a red hoodie and jeans. With sharp eyes, somebody could spot her from half a mile off, even though she was a skinny fifteen-year-old with a real talent for not attracting attention, a girl who'd spent a lifetime learning how to slip in and out of shadows without being seen.
Right now, not being seen was all she cared about. After all, those were the rules of the game, the game they called Shelter.
Shelter was simple. You and your opponent started in one place. Then you both raced to the safe place and whoever got there first won.
Easy.
But the safe place was always three or four miles away from the starting point. You had to run as quickly as you could in the searing heat (for it was always hot in Prin, even in November). If you were detected and called out by your opponent, you lost. And lastly, in order to win, you had to navigate your way through the ruins of old buildings, cracked pavement, and a jagged landscape of twisted metal and shattered glass. A single misstep could break a bone or slice your skin. If you were especially unlucky, one false move could send you crashing down through a rotting surface into some forgotten basement, where chances were your cries for help would go undetected for hours, maybe even days. If anyone heard you at all.
But to Esther, that was part of the fun. Right now, she had been running for nearly two hours, but nothing mattered: not the pain, the exhaustion, or the suffocating heat.
She was about to win.
Esther rounded a corner of the place she knew well. She ignored the cars, each in its own space, separated neatly by faded yellow lines painted on the cracked cement. They were like ghosts, silent hulks so covered with gritty white dust you could hardly tell what color they had once been. They had long since been drained of their precious fuel and were now still and lifeless, just part of the landscape, like boulders or buildings or stunted trees.
Esther's sneakers skidded as she sprinted up the second and then the third concrete ramp. The ground underfoot was broken and uneven, but she didn't notice. She was too busy searching the landscape for any flicker of movement, her eyes darting from car to pillar to car. There was trash everywhere, which she skirted effortlessly: shards of glass, crushed beer cans, a sodden cardboard box, a shoe.
As she made it to the roof, the heat and glare of the sun hit her hard. Sweat was stinging her eyes and running down her neck. When she yanked down her hood and ran her hand across her head, her dark hair was wet, with some parts sticking up in spikes. Here, there was no breeze, nothing but yellow sky. Surrounded by concrete and dead grass that shimmered with heat, she could see for miles past Prin to the empty roads leading off to wherever they went.
Esther didn't turn to look. Exhilarated, all she knew was that she'd made it. She was the first.
For standing in the far corner of the roof was the safe place. It was a brown box, taller than Esther and wide enough to hold at least four of her. Although it had only been there a few days, it was already fading and starting to soften from the sun and rain; soon, it would be worthless. There was a simple picture and black arrows on the side, as well as meaningless words written in large, block letters.
THIS END UP. KENMORE. 24 CU. FT. REFRIGERATOR.
Esther hadn't taken the time to decipher the words. If her older sister, Sarah, knew this, she would pinch her lips with disapproval, the way she did at practically everything Esther said or did. Sarah was one of the few who could read, one of the last in Prin, and she was forever nagging others to learn how to do it too. When she was little, Esther had memorized the alphabet and could sound out simple words, but that was as far as she got.
All she had to do was touch it.
Jubilant, Esther approached the box, extending her hand. But at the last second, someone emerged from behind it.
Esther froze.
The creature was small, with dark, hairless arms and legs, and a bald skull. It appeared to be neither male nor female and wore a brief tunic that was little more than a sack, with a cloth pouch slung across. Its face and body were covered with a dense network of intricate designs, swirling patterns, dots and slashes, strange curls that snaked like vines across the skin in various shades of black, brown, and pink. On close examination, you could see the designs were not painted on but were a complex network of crude tattoos and hundreds, maybe thousands, of scars. Some of the marks were so tiny, they seemed like mere threads against the skin; others were vivid, pink gouges of raised flesh. It had bulging lavender eyes and a flattened nose, which crinkled as its mouth, with its tiny, sharp teeth, twisted upward.
Esther recoiled, with a gasp.
"No!" she screamed.
"Got here first," the variant said. "I win!"
Half laughing, half groaning, Esther tried to catch her breath. She bent over to relieve the pain in her side.
"Skar!" she exclaimed. "How did you got here before me?"
Skar shrugged, smiling. She was so pleased with herself, she couldn't resist dancing a little, bouncing up and down on thickly calloused feet.
Skar was the same age as Esther. Yet unlike her friend, she had been female for only five years, having selected her gender on her tenth birthday, the way all variants did. At the time, Esther was delighted when her friend chose to be a female because it was one more thing they had in common. Skar had a circle, the mark of being female, tattooed on her upper arm.
No one understood where the variants came from, why they were hermaphroditic, hairless, disfigured. Most in town seemed to believe that the variants were once animals, living off contaminated goods and drinking groundwater. The accumulated poisons had permanently affected them and their unfortunate offspring, creating a new race of freaks. Certainly Esther's older sister, Sarah, had mentioned this theory more than once, much to Esther's annoyance.
The variants had always lived far from town, shunning the ways of Prin and its people. They dressed oddly, not bothering to shield themselves from the dangerous rays of the sun. Rather than work, they eked out a meager living from hunting with feral dogs. Occasionally, they foraged for food and bottled water amid the wreckage of the outlying buildings and homes of Prin. The variants' way of existence was a harsh and dangerous one, where one's next meal or drink of water could result in sudden sickness, pain, and death. Their life expectancy was even shorter than that of the people in town.
In the best of times, the townspeople looked down on the variants as shiftless and dirty and called them the ugly word "mutant." Lately, after the rash of strange, isolated variant attacks, the feeling had grown from one of contempt to that of terror and even hatred. No one knew this better than Esther and Skar, who chose to spend their time together far away from the judgmental and fearful minds of Prin. Esther believed the variants weren't a separate species like damaged snakes or wild boars; she believed they were human somehow, yet spurned for their differences. But she had never dared mention this to anyone.
The life of the average townsperson was one of mindless labor rewarded by the occasional treat of something new: a piece of clothing that wasn't filthy, a wristwatch with a shattered face, a pair of sunglasses. Rather than stoop to such a level, the variants had created their own society high in the mountains, with its own rules, customs, and rituals. There they lived freely, without the need for labor or commerce. They existed without apologies, and with pride. And for that, Esther secretly loved them.
"That's three times in a row," Skar now said. "Do you want to try again?"
"Sure," said Esther. "Only let me catch my breath first. And this time, give me a head start or something."
Skar's smile broadened. "What fun is that?"
Laughing and chatting, the two headed back down into the relative coolness of the building. They argued over what should be the new starting point: the abandoned steel tracks several miles down the road, or the dried-up lake on the far side of town? But as they approached the ground level, Esther's face froze and she made an abrupt gesture at her friend, who stopped in midsentence.
Skar heard it, too: a faint thread of faraway voices.
In the distance, heading off the main road and turning into the asphalt lot, were three figures on bicycles. One was pulling a red wagon; from where she stood, Esther could even hear the faint clank of its metal handle. They had clearly seen Esther; had they seen Skar? The trio was headed to the parking garage, straight toward them.
Esther and the variant shared a quick glance, and Esther gave a nod. Without speaking, Skar crouched low and slipped away, disappearing behind a row of parked cars.
Esther waited a few moments. Then with fake casualness, she sauntered to the edge of the wall and looked down. She was trembling and her heart was pounding, but her actions revealed nothing.
Within seconds, the three were clustered below, gazing upward at her. From their expressions, Esther could tell they didn't notice Skar and she felt some of her tension ease.
Yet she had to make certain they didn't come up to where she was, where they might see her friend. No variant was safe since the attacks. She rested one hand in front of her on the low wall and gazed down at them.
It was impossible for Esther to tell who they were. Indoors and away from the burning rays of the endless summer, the three wouldn't resemble each other at all. Yet at that moment, they were nearly identical, dressed the same as everyone in Prin except Esther: swathed in filthy sheets, with towel headdresses hanging down their necks, scarves masking their lower mouths, and thick cotton gloves protecting their hands. The billowing folds were belted close to their legs, in order not to get caught in the spinning gears. All three wore dark sunglasses. In the wagon, Esther could see two empty plastic bottles, coiled rubber tubing, and a crowbar.
The three were on their way to a Harvesting.
There were three jobs in Prin—Harvesting, Gleaning, and Excavation—and they were assigned by a lottery held every two weeks in the center of town. Everyone over the age of five was required to attend and, once given a job, expected to work every day from sunrise to sundown. The rules had always been strict but they had become much tougher of late: Not to show up resulted in a Warning filed by the team Supervisor, which Esther had incurred at least four times in the past year.
One more and she risked Shunning. And Shunning from town meant certain death.
Two of the three jobs were grueling but mindless: the Excavation and the Gleaning. The few times she had deigned to show up for an assignment in recent months, in order to placate Sarah, Esther had opted for one of those two. But the Harvesting—a search through the outlying areas to find the most tradable commodity, gasoline—called for real concentration. It was by far the single most important job in town and one that had grown only more difficult and time consuming as the years went by.
When Esther had drawn the Harvesting as her assignment at the last lottery, she'd cursed her luck. Then she ignored the task and instead headed to the overgrown fields and vacant lots to play with Skar.
It had taken the rest of her team this long to find her, and their fatigue and frustration were obvious. She had to be careful not to provoke them: There was too much at stake, for both her and Skar.
The biggest figure called up to her. "Look who's here," it said. Although they were cloaked, Esther had no trouble recognizing who was speaking by their voices. This one was Eli. At fifteen, he was the oldest and was therefore Supervisor of today's expedition.
"Where you been?" shouted another, revealing herself to be a girl called Bekkah. Shorter and younger at eleven, she acted as second in command. "We been looking for you!"
"I showed up the first day, and you guys had already left," Esther said from her perch, trying to sound sincere. She knew that from a hiding place behind her, Skar was listening too.
"Right," said the smallest and youngest. This was Till, and his tone was sarcastic.
Esther knew this boy the least and, as a result, feared him the most. She turned beseechingly to Eli.
Her appeal was not lost on him. Eli was well aware that after two weeks, their work detail was almost over and had been unsuccessful. The two others in his team were on the edge, ready to vent their fury on any target. He had to keep them at bay.
"Let's go up there and get her," Till said.
Eli held up his hand. He exchanged a look with Esther.
In spite of himself, Eli smiled; he couldn't help it. For some reason, he had always been attracted to Esther, despite her utter irresponsibility and almost total lack of female affect. He couldn't explain why, even to himself. His eyes still holding hers, he gave a dismissive wave to the others. He tried to sound cold and unfeeling.
"Let's go," he said. "She ain't worth the trouble."
He remounted his bike, looping around to head back out to the main road. For a moment, the other two were angry and confused; then, resigned, they got on their bikes. Bekkah made the turn with difficulty because of the wagon. Only Till couldn't resist a parting shot.
"Looks like you got off this time!" he yelled back over his shoulder.
Eli stopped at the edge of the parking lot.
"Better get back to town," he called to Esther, meaningfully. "You ain't safe alone out here."
"I'm not afraid of wild dogs," she said.
"I don't mean dogs." He spat before he took the turn.
Once they were gone, Esther expelled a long breath. She was surprised to find that she was trembling and even a little sick. Why?
Was it because Eli had done her a favor, meaning she was now indebted to him? True, he had often seemed sympathetic to her in the past, and she had never minded. He had always been kinder than the others, and not as close-minded. Yet now that she had asked for his help, and he had given it, they were linked, somehow, in a way they hadn't been before . . . and Esther was not at all sure how she felt about that.
In Prin, Esther and Eli, not to mention Sarah, stuck out for being single. Nearly everyone got partnered when they reached fourteen. By seventeen, they were considered town elders, and by nineteen, they were dead. That was the longest anyone managed to escape the disease; it was everywhere there was water, carried in the rain, lakes, streams. Couples spent their short lives together at meaningless and backbreaking jobs, often toiling side by side, and all for just enough food and clean water to survive. Esther was already fifteen, a year past the age of partnering. Was this really all she had to look forward to?
When she was old enough to rebel, Esther began breaking curfew and spending more time with Skar. At first, people in town treated her with condescension, as an oddball. Now, they viewed her as a pariah, a freak. And Esther had been fine with that. Being an outsider made her feel strong, even invincible. But lately, she found she was often beset by a strange sadness.
She would always love Skar. But despite Esther's efforts to embrace their culture and learn their ways, the variants themselves still refused to accept her as one of their own. Her one trip to the variant camp had been a disaster: She was treated coldly, with suspicion and hostility, by the rest of Skar's tribe. Esther hoped that one day she would be welcomed as the ally she was; but after so many years, she had yet to even meet her best friend's brother.
The town of Prin wasn't home, either. She fit in nowhere.
Esther knew how she really felt.
She felt alone.
_Maybe there could be someone else to be_ truly _close to,_ she thought. Or maybe there could be something bigger to be a part of—what, she wasn't really sure. A little while ago, Esther would have laughed at the idea. But she wasn't laughing now.
"Are they gone?"
Hearing her friend's voice, Esther snapped to attention.
The variant girl now crept from behind a row of parked cars and stood by Esther with fists clenched, tense and ready to run.
Esther brushed aside her own concerns, to put her friend at ease.
"We're fine," she said, nudging Skar in the side. "Now let's see who makes it to the tracks first."
Eli and the others rode their bikes single file down the main road heading away from Prin. He led them past the hulking, plundered ruins of buildings on the edge of town, places that still had names, meaningless words they didn't know how to read: STAPLES, HOME DEPOT, THE ARBORS NURSING HOME, STOP & SHOP.
Eli pedaled slowly, so Bekkah could keep up. He was careful to steer around the broken glass, discarded bits of machinery, and chunks of dirty plastic that littered the pavement beneath their tires.
They avoided detritus left by the periodic rising and retreating of floodwaters: bleached-white shells and stones, the rotting remains of a rowboat. There were other things that must have been swept away by the dank waters: a rusted hunting rifle; a blond wig that had become a filthy, tangled mop; a safe deposit box with the top torn off and the dust of long-dead crabs inside.
Ahead, the road became a bridge, passing over a much larger avenue underneath.
Eli stopped as he considered where to go.
"We already checked over the bridge," commented Bekkah, as she pulled up alongside him.
"Yeah," said Eli. "But we didn't go down _there_."
He pointed, and Till swallowed hard.
"Are you sure," he muttered, "we have to?"
Eli shrugged. "We been out here two weeks and ain't got a drop. There's no place else to check. Come on—it'll be fine, and maybe we'll even be done today."
The others seemed reassured. Eli pretended to look at ease as they glided down what had once been the on-ramp to the northbound lane of the interstate.
At times, the deserted road was almost impassable with fallen trees, downed streetlights, dead power lines; but the three managed to find a way through. Both sides of the highway were overgrown with heavy, tangled undergrowth that in some areas spilled past the shoulder and onto the road itself, and in some places obscured the aluminum barriers once built to muffle the sounds of traffic.
"Nothing so far," Eli called over his shoulder. He was talking too loudly, he realized, from nervousness.
He was on edge in case they saw a body.
Although none of them talked about it, they all knew that people with the disease were Shunned, sent from town on this highway to die; that way, it was said, they wouldn't contaminate the others. No one knew exactly where they ended up. There were rumors in town of a Valley of the Dead, a mass grave filled with the remains of innumerable children, although such a place had never been seen. Some thought it was no more than a bedtime tale told to frighten the little ones.
Eli was not sure he believed the story, but he worried they would see or smell remains on the road or off to the side. _Which would be worse,_ he wondered: _if the body was fresh enough to be recognized or too rotted to identify?_ The smallest ones would be the worst, he decided, and skeletons of any size.
"Hold on! Back up!" Till yelled.
They slowed down. Sure enough, they could see something peeking out from a dense tangle of vines, brush, and litter, close to the highway wall, where no one would have searched.
"Yeah," Bekkah said. "Looks good."
It was a dark green car, compact yet roomy, with an incomprehensible word framed by a steel circle on the front grill: VOLVO. The three pedaled onto the curb, then got off and walked their bikes through the clotted grass of the shoulder to reach it. With difficulty, Till pulled the wagon as well.
Bekkah fished out a steak knife hanging from her belt on a nylon cord and sliced away the vines and branches that strangled the car. Working methodically, she cleared away a space on the left side of the vehicle, above the back tire.
Eli took the crowbar from the wagon and flipped open the small metal panel in the side of the car. Then, with a few yanks, he pried off the cap to the gas tank. Till handed him the tube, and Eli snaked it down into the tank, feeding it inch by inch.
In the hot sun, the others watched his expression. Moments later, Eli smiled as he felt the end of the line hit gas.
"A decent amount," he said, relieved.
"Good," Bekkah said. "He was mighty mad last time. Wouldn't hardly give us nothing."
"He's been like that for a while now," Eli said.
They were talking about the one they worked for, the one who lived on the outskirts of town. The boy called Levi.
Levi lived in a kingdom of sorts, in that he saw himself as a kind of king. Yet his home was more of a fortress, as the windowless building was massive, guarded, and impenetrable. It was nicknamed the Source because while no one in Prin had ever ventured inside, it was quite literally the center of life; the townspeople would soon die without it. It was powered by the only electricity any of them had ever seen, electricity generated by the countless bottles upon bottles of gasoline everyone in town spent so much of their lives searching for.
The gasoline was exchanged each month for food and water from Levi's endless supplies. This didn't include his tuna, soup, meat, jelly, cereal, tomato sauce, peanut butter, stew, pickles, or vegetables; they had long since rotted to blackened tar, exploding their containers. His dry grains and beans were edible, barely, but had to be pounded into flour and boiled for hours before they could be digested. Salt, sugar, spaghetti, honey, and hard candy were available, packaged and sealed in plastic bags, cloth sacks, and cardboard containers; there were also countless gallon jugs of water. Everything was brought outside by Levi's boys.
There were eighteen of these guards, hulking and hooded brutes armed with small, harmless-looking contraptions that made a terrible hissing sound and, upon contact with skin, could cause a teenager to drop to his knees in agony. (The word "Taser" was printed on the rubberized grips; a dozen had been found in a building on the outskirts of town, one with cobwebbed desks in the front room and barred cells in the back.) Levi's boys measured the gasoline and doled out the provisions, watching over the transactions with a hawk-like attention that had only grown worse since the supply of untapped cars in Prin began dwindling.
For Eli and his fellow Harvesters, the "Volvo" guaranteed that the exchange would continue for at least a few more weeks. And so by extension would their lives.
"Lucky we found this," Eli said. "Should last us a long while."
With that, he bent down and sucked the tube until he sensed the fuel was about a foot from his mouth. Then he yanked out the tube and stuck the end into the neck of an empty plastic bottle, which Till was holding steady for him. A second later, the air was filled with the pungent smell of gasoline as it gushed forth into the container.
Without looking, he addressed Till. "Good eye," he said.
Till smiled, abashed. "It was your idea to come down here."
Eli filled one bottle; then, taking care not to spill a drop, he transferred the tube to the next bottle.
Yet something wasn't right.
Eli looked up. Then he stood. As he did, the tube fell from the bottle neck, sloshing gasoline onto the ground.
"Hey!" Bekkah said. "Watch out!"
"Sorry." Eli bent to put the tube back into the bottle, which by now was almost full. "I just thought I heard—"
Now he froze in place. It was unmistakable: He could even see the fuel shaking a little inside the bottles.
Something was approaching, fast. Some of what Eli heard wasn't human, just the faraway thump of tire threads. But he could also detect faint whooping and whistling, a scary celebration.
Eli's eyes flickered around as he braced himself. Alone on the side of the highway, they were brutally, nakedly exposed. It was too late for them to do anything—to run away, to hide, to even scream for help.
"Aww, no—" Till murmured, under his breath. It was almost a prayer.
And at that moment, it began.
The air was split by noise, a blood-curdling shrieking that seemed to come from no one direction but from everywhere at once, pulsating and echoing. It was an uncanny noise that seemed neither human nor animal.
Bekkah stood still with her mouth half open, a hypnotized mouse in the sights of an owl, the now-forgotten plastic container at her feet overflowing. Gasoline splashed over her sneakers and filled the air with its fumes.
From nowhere, an object whistled at her through the air; it was a fist-size rock, ugly and jagged. There was a sharp cracking sound; it knocked Bekkah from her trance and she emitted a scream, high and thin and terrified. She reeled backward. Clutching her forehead with both hands, she knocked the rubber tubing from Eli's hands and sent a stream of gasoline flying in a clear arc through the air. Blood spurted from between her fingers and ran down into her eyes. It dripped onto and splattered her filthy white robes. Her knees buckled and she sank to the ground.
Eli was backing up, his eyes darting as he looked in vain for their assailants. His forgotten bike lay on its side only a few feet away. Several rocks flew at him, too, and he ducked them, one arm held up in front of his face, as too late, he remembered the single weapon they had thought to bring with them, an aluminum baseball bat tossed in the back of the wagon.
Behind him, Till was scrambling for the vehicle closest to him, Bekkah's bike, trying to detach it from the wagon, which he knew would slow him down.
" _C'mon, c'mon,_ " he whispered.
But as his useless fingers picked at the knotted ropes, a fresh barrage of rocks was unleashed on him from all directions. Panicked, he gave up and made a dash for the side of the road, crawling into the underbrush to the sounds of jeering and mocking laughter.
And with that, the mutants, shrieking, descended from all directions.
It was impossible for Eli to tell how many there were—ten? Twenty-five? They attacked in a swarm, and at that moment looked exactly alike—all slight of build, androgynous and covered with road dust, with the same bulging lavender eyes and ornate labyrinths of scars and tattoos covering their faces and bald heads. Each wore a meager tunic, with a canvas bag loaded with rocks slung across his or her body.
The bikes themselves were strange-looking and menacing: black and low to the ground, festooned with strips of leather, with weird metal pegs and handles attached to the frames and axles. The mutants rode two to a bike, one pedaling and the other standing behind, straddling the rear tire and balancing barefoot on the foot pegs while wielding their slings, metal clubs, and chains.
Eli had heard tales of these recent and confusing attacks by the mutants, of ambushes sprung from nowhere and for no reason. He had taken comfort in the fact that the mutants had chosen not to kill but saved the worst of their savagery for buildings and objects. Still, hearing of such things was far different from experiencing them firsthand. He was choked with panic.
Covering his face with his arms, Eli ran forward, bent over to make himself as small a target as possible. He was able to reach Bekkah's side unscathed. Grabbing her under the arms, Eli dragged his unconscious friend to the underbrush, near Till.
From there, Eli watched as the mutants dismounted from their bikes and turned their attention to the car. Soon glass was shattering, heavy chains smashing against metal, and bodies were jumping up and down on the roof. _Better the car than them,_ he thought. He turned to the side and, when he did, his heart skipped a beat.
The two plastic bottles, still brimming over with gasoline, were where he and the others had left them, miraculously untouched by the side of the road. To Eli's horror, one of the mutants backed into one as she whirled her chain overhead and knocked it over. Shocked, the boy watched as the precious contents glugged out, spreading across the pavement and spilling into the dust by the side of the road.
He was not the only one who noticed.
The biggest mutant, wiry and with a distinct network of swirling scars forming a rising sun across its face, had been standing to one side, arms crossed. Eli noticed the triangular tattoo on its bicep; clearly, it was a male. This mutant had been watching the others attack the car with an unreadable expression. Now, he cocked his head. With one swift movement, he crossed over to the second bottle, which was still full of gasoline. Then, he lifted it by its neck with two fingers in a gesture that seemed almost dainty and carried it back to the others.
Eli knew the mutants traded for nothing. They had no use for Levi's supplies; mostly, they scavenged and killed wild animals. They didn't need gas.
The big mutant said something, a few words the boy couldn't hear on account of the noise; but whatever it was, everyone stopped what he was doing and backed away from the car. In the silence, Eli could hear their bare feet crunching on the pebbled green glass sprayed across the shoulder and road. Once everyone had cleared, the mutant took the bottle and started splashing its contents over the remains of the mutilated Volvo.
Eli realized what the big mutant had pulled from his shoulder pouch, what he now held aloft in one hand and was tossing in the air like a toy.
It was a small plastic object, bright pink, the size of a thumb. A firestarter.
"Oh no," said Till next to him, involuntarily. "Please . . . don't!"
As if he had heard, the mutant gave a faint smile. He pressed a button on the side of the object once, twice; on the third time, and with a distinct _click_ , a small orange flame blossomed out of the top. Then he bent down and touched the flame to the wet, glistening asphalt.
The mutants scattered as fire licked and spread across the pavement and trampled grass, racing with unbelievable speed in rippling blue and yellow waves toward the car, the car that still had gas in it, at least half a tank of the precious stuff, maybe closer to a full tank. Leaping onto their bikes, the mutants took off and within seconds, they disappeared.
From their hiding place, Eli covered his head with one arm, the other around the unconscious Bekkah, his face pressed hard into the dusty ground. He braced himself and prayed that Till, next to him, would do the same.
Then the car exploded.
By the old railroad tracks, rusty and nearly obscured by weeds and trash, Esther heard the faraway blast. She froze; then she pulled Skar down so they were nearly hidden by the tall grass.
"What's going on?" Skar asked, although she already knew.
"Nothing," Esther lied.
Whooping and shrieking, the variants rode single file into town, rattling the broken, faded WELCOME TO PRIN sign as they thundered past. They spread into a V formation as they headed down the central street, flanked on both sides by sidewalks and two- and three-story buildings, with empty storefronts on the ground floors. While a few structures showed the effects of earlier recent attacks, most were unscathed.
The variants did not spare the first buildings they encountered. One stood on the foot pegs of a bicycle and whirled a sling around his head. He let fly with deadly accuracy, and the window of what had once been a clothing store shattered, collapsing in an explosion of broken glass.
Most of the variants rode ahead, while several dismounted, wielding chains and clubs. Using a broken windowsill as a foothold, one reached for the neon sign above the remains of a pharmacy; with one blow of her cudgel, she smashed it partway off the building, so it dangled at a crazy angle. She swung at it again, this time bringing it crashing down in pieces; then she proceeded to beat it into fragments on the street. Another whirled his sling above his head, launching rocks to smash one window after another.
Residents scattered for cover, taking refuge wherever they could find it. They had no time to consider the senselessness of the event; they had witnessed it in the weeks before, but this ambush was far worse, more savage and out of control.
A girl, age eleven or twelve, ran to a rusted car in the street and managed to roll underneath before being seen. As she lay there, she saw bare feet stop in front of her. Holding her breath, she watched as the feet paced back and forth. After what seemed an eternity, they walked away and she heard a bike take off.
Elsewhere, faces appeared in second-floor windows, looking at the mayhem.
One boy, Jonah, decided he would try to save the town single-handedly.
When the variants blasted down the main street, the ten-year-old had managed to scale the fire escape of a battered building without drawing attention. He had made his way to the roof, and now, lying on his belly on the hot tar, he gazed down at the destruction. In one hand, he gripped a lead pipe, which he had kept in his back pocket for weeks for exactly a moment such as this.
He watched as variants kicked in the front door of what had once been a bar. He watched as they caught a boy trying to escape, ripped his headdress off him, set it on fire, then tossed it through the broken door of an old hair salon. He watched until they seemed to tire of causing chaos, until they were ready to move on, and gathered to huddle their bikes to make a plan, waiting for someone to tell them what to do.
They waited for the one they called Slayd.
The variant leader approached, skidding his bike to a halt. Leaning over the handlebars, he addressed the others with emphatic gestures. His back turned to the stores, Slayd didn't see the iron pipe, flung like a boomerang, winging toward his head.
Another variant saw it. He leaped forward, pushing Slayd out of the way and Slayd's protector was clipped on the side of the head by the pipe and was knocked cold, feet jerking up and down on the pavement.
The boy on the roof became an easy target. As Slayd got to his feet, the others let loose a volley of rocks toward the boy, who was scrambling down the fire escape. They nailed him at once—on the back, the legs, the head. He rolled down the final steps and dropped to the pavement with a sickening thud.
In the silence that followed, Slayd brushed off his tunic, dignity intact. He didn't acknowledge the protector any more than the assailant; both still lay, unconscious, on the ground. Instead, Slayd got back on his bike and led the rest of his band away, with a final, piercing shout. The last to leave grabbed the fallen variant and slung him over his shoulder. Then he too remounted his bicycle and was gone.
One by one, the townspeople straggled out of their hiding places. They stood in a state of shock on the street, breathing in the dust stirred up by the variants that still hung in the air.
There had been little left of their town in the first place. Now, there was even less.
## TWO
HOURS LATER, THE STREET WAS STILL EMPTY AND SILENT, LITTERED WITH shards of broken glass, rocks, and splintered wood. The dusty asphalt showed the scuffed marks of bike treads and bare footprints that the wind would soon erase. On the ground beneath the fire escape, a few red splatters had dried and were blackening in the afternoon heat.
Far in the distance, a thin plume of gray smoke spiraled upward before disappearing in the darkening sky.
At the edge of the main street stood a one-story brick building surrounded by a cement field marked with fading white lines. Two battered yellow arches loomed overhead atop a high metal pole. By some miracle, the large windows that dominated three of the walls had been spared in the recent attack.
Now a small, anxious face peered out from a clear spot rubbed on the grimy glass.
It belonged to a sentry, a dark-eyed boy perched on a molded table attached to the floor and focused on the empty street, keeping watch in case of another attack. He was perhaps six or seven years old and he fought to stay awake. Behind him, the crowded room was restless and noisy, with a shrill clamor of voices raised in anger, fear, and confusion.
Everyone from town, more or less, was present; and it seemed everyone had something to say. More than a hundred children and teens huddled on top of tables and chairs, along the grimy tile floor and the stainless-steel counter. A lone child sat in the corner, soothing a younger girl who lay cradled in her lap, sucking from a dirty soda bottle. A few others retreated deep into their thoughts, staring upward at the ceiling.
One boy banged his fist on the metal counter until the others stopped talking.
"We got to go after them," he shouted, his voice hoarse. He was nine, with a baseball cap pulled backward over his shaggy brown hair. "Before they attack us again."
An older girl sitting on a table across the room shook her head. She was dark skinned and had several colorful belts cinched around her grimy robes.
"You mean go into mutant territory?" she shouted. "That's crazy . . . they got more people. Or whatever they are."
At that, the room erupted as everyone started talking once more. Another boy, pale and freckled, spoke up from where he sat on the floor.
"Maybe they _are_ people," he said. He looked to be about eleven. "Like us, only different. They're boys and girls in one body. Maybe—"
The older girl sitting next to him, also freckled, smacked him hard across the ear. "Shut up!" she hissed; but no one had even heard. Another boy, so small his feet dangled a good two feet off the floor as he rested on the edge of the counter, piped up.
"And they got weapons. How can we fight if they got weapons and we don't?"
Others chimed in.
"He's right. We don't got a chance."
"But if we do nothing, they're gonna come back. Maybe next time, they'll kill us all."
"They almost killed Jonah. When we brought him inside, he was bleeding pretty bad."
People glanced at the would-be hero from the roof, who leaned against the counter; the side of his face was badly scraped and his left arm hung at a useless angle.
Nearby, Bekkah stood close to Eli, the blood-soaked T-shirt tied around her head not quite hiding the ugly purple and yellow bruise spreading down her cheek. Her left eye was swollen nearly shut and when she spoke, she sounded exhausted.
"They always been peaceful," she said. "Now they've attacked us four times. What do they want? It don't make sense."
As the arguments raged, one person was watching the proceedings with a shrewd eye.
It was the leader of the town, who leaned against the far wall, with his arms folded. Short, with wide hips and stringy hair, eighteen-year-old Rafe had been elected to his one-year term the previous winter by the usual show of hands. It hadn't taken him long to realize how much he enjoyed not only the prestige of his position but the perks as well. He was spared work assignments and was also given an extra weekly allotment of food and water. And so despite his advanced age, he was planning how he could be reelected for another term.
The recent variant attacks, he figured, gave him as good an issue to run on as any.
Rafe held up his hands for calm. As usual, he remained silent while the others exhausted themselves with bickering and suggestions. When he did speak, this gave him the impression of both thoughtfulness and authority.
"There ain't never been sense to mutants," he said. He also knew enough to speak softly; this forced everyone in the room to lean forward to hear him. "They're like wild dogs. And I say we wipe 'em out."
The girl next to him was shaking her head, arms folded over her thin chest. "That's always your answer, Rafe," she said. Against the relative whiteness of her robes, her skin looked dark and withered; she appeared at least two decades older than her sixteen years. "It ain't so easy."
"Let him speak," shouted the boy with baseball cap.
"Yeah," chimed in another voice. "How do you say we do it?"
Again, Rafe waited until the room grew quiet.
"We go to Levi," he said. "We go there and we ask for weapons. _Real_ weapons, I mean. Knives. Arrows. That way, at least we got a chance—"
The dark-skinned girl sitting on the table cut him off.
"But there ain't no gas left to trade for the things we really need—like to eat and drink," she said, her voice shrill. "And Levi's been cutting back on what he pays us."
Most were nodding their heads in agreement. The dark girl continued. "We got nothing else to give him. Without gas, why would he even talk to us?"
Rafe smiled. He had anticipated this question.
"Maybe not to you or me, unless we got something to trade," he said. "That's all he cares about. But there's _one_ of us I bet he'd talk to."
Then he turned to look at a girl sitting alone by the window.
It was dusk; the meeting had been going on for nearly two hours. The small sentry at the window had relaxed his vigilance and dozed at his post. Behind him, several of the townspeople had lit candles, which they set on the tables and counters. As the nighttime darkened around them, the gritty windows reflected what was going on inside the room. From outside, the townspeople were all too visible, and with no view of what might be approaching.
If attackers were to come, they would arrive unseen.
And in fact, two people were now scuttling toward the lit building. Yet they were not there to do harm.
Esther bent low and ran from one shadow to the next, zigzagging down the sidewalk. She had been gone since before dawn. Although Esther chose to ignore the far-off explosion, she knew that it had to do with the variants. Now, uneasy, she could not help but notice the freshly smashed windows and broken storefronts that lined the main street of Prin.
Behind her was a reluctant and increasingly panicked Skar. With the stink of burning gasoline still lingering in the night air, the last thing she wanted was to be confronted by the town.
"Esther," Skar whispered, pulling at her friend's arm for what must have been the hundredth time, "please. Let's not do this!"
"Don't worry," said Esther. "They can't see us."
And Skar had to admit: This part was true.
It wasn't just that the darkness gave them ample cover. Over the years, Esther had worked hard to become adept at variant ways—the peculiar stalking, hunting, and trapping methods that Skar had taught her, skills that had been second nature to the variant girl since early childhood. Skills that let you become almost invisible.
Esther ran on the balls of her feet, using cover and shadow to hide her progress, avoiding the straight line of approach, doubling back, leaping up to edge a few steps along railings and windowsills, seizing every possible handhold and foothold available to her: all the tricks a variant did to confound expectation and confuse the eye. If she were to be honest, Skar could fault Esther on a half dozen mistakes: Her tread was too heavy, her breathing too loud. The worst was that the girl still couldn't interpret the terrain as having many possible pathways, not just the obvious one; she didn't know how to strategize on her feet. Even so, although she would never say so out loud, Skar had to admit:
_Not bad for a norm._
The two reached the building where the meeting was taking place and slipped into the adjoining alley. Esther took a swift, birdlike peek into a window, as Skar cringed in the shadows beside her, trembling with anxiety.
"Big group in there," Esther said; "looks like everybody in town."
Skar's expression grew even more tense. "Can we go now?" she begged.
"Not yet," replied Esther. "We've got to hear what they're planning. It could be important."
"What are you going to do?"
Esther didn't answer. Skar was about to repeat her question when she saw what Esther was doing.
She was trying to climb the bare brick wall.
Despite her mounting anxiety, Skar couldn't help smiling. Esther was attempting something she had learned only that week: using the tips of her fingers and toes to gain a hold on even the shallowest dents and faintest bumps in a surface. In this way, a variant could—with practice and the right combination of strength, balance, and weight distribution—scale even the smoothest-seeming wall, like a fly. Esther was able to grip the bricks with both her fingers and the tips of her sneakers, and she moved upward clumsily, yet with surprising speed.
By now, Esther was too far up the wall for Skar to call her back. Feeling resigned, the variant made a quick decision and followed.
Moving at twice the speed of her friend and with enviable grace, Skar clambered up the brick wall and caught up with Esther within moments. At the last second, she was polite enough not to overtake her. Instead, they reached the roof together and Skar even allowed Esther the illusion of pulling her up once she had reached the top.
"Don't worry," Esther whispered, clearly proud of herself; "I got you. And I got here first!"
But her jubilation made her forget herself and she stood upright, something a variant would never do, especially not in a moment of triumph, the one moment your guard was down.
Skar hissed a warning at her, but it was too late.
Esther wavered and then lost her balance, falling forward onto the roof and landing hard. The top of the building was steeply tilted on both sides like an old-fashioned cottage from a picture book, covered with overlapping reddish-brown tiles. Esther started to slide, her fingers scrabbling in vain to get a grasp of the tattered clay rows.
Skar reached out a hand, but it was no good. Esther kept sliding, rapidly approaching the edge.
At the last possible second, she was able to wedge one foot into the shaky rain gutter while grabbing onto a few secure tiles. One broke off under her hand; it skittered down the roof and disappeared in the darkness beneath them with a faint crash.
With surprising speed, Esther crawled her way back up to her friend, who was huddled miserably, waiting for her.
"I knew this was a bad idea," Skar said.
The variant was astounded that inside, no one had heard the incredible noise Esther had just made. Any one of her people would have been outside investigating the suspicious sounds within seconds. She wished (and not for the first time) that Esther wasn't so stubborn.
"But we just got here," Esther whispered. She flattened down onto her stomach and crawled toward one of the many gaps in the tiles. "And I want to hear what they're talking about."
Skar had no choice but to follow her. She sat next to her friend, knees huddled close and bulging eyes shut tight.
One gap afforded a limited view of the room below. Esther glanced down, then placed her ear over the hole and concentrated. There were many people speaking at once, but she was able to detect a female voice. She had to strain to hear what she was saying.
"We got nothing else to give him. Without gas, why would Levi even talk to us?"
She peered through the hole. Esther could see the tops of heads, a few familiar faces. She recognized Rafe, the current leader of the town elders, a boy she hated because beneath his superior airs, he was both a coward and blowhard. As usual, he was doing his trick of talking softly. Esther had to put her ear close to the hole and focus hard in order to discern his words.
"Maybe not to you or me, unless we got something to trade," he was saying. "That's all he cares about. But there's _one_ of us I bet he'd talk to."
Who were they talking about?
There was a brief silence, followed by a faint murmuring as people stood and craned their necks, looking to see who he was discussing. Esther followed their gaze and was startled.
It was a girl, seventeen, with dark, straight hair held back in a ponytail. Unlike everyone else in town, her robes were relatively clean and gathered neatly at her waist with a dark cord. She seemed embarrassed by all the attention, yet flattered as well.
"I . . . I don't know," she was saying. "For one thing, I don't know what good it would do. I've never been to the Source. I can't even remember the last time I spoke to Levi . . ."
Esther pulled back, as if struck. "Hey."
"What is it?" Skar asked, opening her eyes.
"It's my sister. Sarah."
Frowning, Esther sat back on her heels. In that position, she could see the place they were talking about, where Levi lived. The Source lay to the northeast of town and was something she saw every day, as much a part of her landscape as the sun. Although it was nearly a mile away, it was hard to miss from anywhere in town.
The gigantic white building was like a beacon, huge and blindingly lit with electrical lights. They threw deep shadows across the trenches that lay next to it, black gashes in an overgrown field. The holes were just three of the dozens of pits scattered across town that the people dug day after day when they were unlucky enough to be assigned to the Excavation. The front and side of the Source faced a monstrous asphalt field marked with fading white lines and still crowded with the dusty remains of cars.
Now, it seemed Rafe and his followers wanted something from Levi, something new. And they apparently needed Sarah, the childhood friend who once knew him best of all, to be the intermediary.
Esther didn't like it.
She glanced at Skar, who was amusing herself by tossing a small knife up in the air and catching it. She wasn't even paying attention, and for that, Esther felt a stab of exasperation. Skar was, after all, only who she was—a great friend, but one who was easily bored, like a little child.
And little children needed to be protected.
Esther knew it would be up to her. She was not sure how she would do it, but at least she knew where to start.
It was evening. Shadows cowered low to the ground and scurried through the streets and alleys of Prin.
They were feral dogs, rooting through piles of garbage for something to eat. They snapped and fought over whatever they could sniff out, anything that was remotely edible: the stale and salty ends of flatbread, rabbit bones that had been sucked of their marrow, the burned crust of rice porridge. The dogs of Prin were dingy and skeletal, cringing yet vicious beasts accustomed to skulking in the shadows and traveling by night in packs.
There was, however, one stretch of sidewalk that had been swept clean. The storefront window behind it had not only been patched over with flattened cardboard and gaffer's tape; it looked like someone had actually taken the trouble to measure it so it fit properly. A cracked and battered sign above what was once the window read STARBUCKS COFFEE in block white letters on a green background. And above the sign, a light was visible in the second-floor window.
Agitated shadows moved across the curtain. Behind the thin fabric, Esther was getting in her sister's face.
"But you can't go," she was saying.
Esther was trying hard not to raise her voice, because she knew losing her temper would only cost her the argument the way it always did. Instead, she tried to sound reasonable, clasping her hands tightly behind her back.
"You can't ask Levi for weapons," she said. "This whole thing is starting to get crazy."
Sarah stood at the kitchen counter, cleaning out the firebowl with a rag. The older girl acted as if getting rid of every last trace of soot and ash was the most important thing in the world. She was doing what annoyed Esther the most: ignoring her because she was focused on something more meaningful, something _adult_.
"Pass me those," was all Sarah said, nodding at the forks and spoons.
Frustrated, Esther picked up the handful of dirty silverware. She couldn't help herself; as she handed them over, she slammed them down on the counter harder than she intended to. At the noise, her sister jumped, to Esther's private satisfaction. Then Sarah turned all her attention back to cleaning up.
"How could you listen to those people?" continued Esther, still trying to sound calm. "Rafe? He's a big mouth, that's all. And the others—they're just thugs who want an excuse to hurt people."
"I didn't say I was definitely going to see Levi," Sarah replied. Unlike everyone else in town, she spoke in a fussy, formal manner she had probably picked up from all her reading. It was yet another thing that irritated Esther about her sister. Sarah pushed a few grains of uneaten rice into a plastic container, which she sealed and put away. Then she finished wiping the silverware.
The plates, bowls, and cups were chipped and cracked, yet they were mostly a matched set, what had once been a pretty green and purple. Thanks to Sarah, the entire apartment was tidy and clean. Unlike the other homes in Prin, filled with piles of filthy blankets and clothing and utensils, their place was almost stylish, and the curtains in the windows were white.
The walls were decorated with tattered ads Sarah had found in town—mysterious posters for Absolut vodka, Continental Airlines, New York Yankees. There were even several shelves of books, which Esther had barely glanced at. "Besides," continued Esther, "why do you think Levi will even talk to you? You haven't seen him in years. And all he cares about is how much gas people bring him. He don't care about anything else."
" _Doesn't_ , not _don't_ ," said Sarah, her face flushing. "And he's not like that. He's a good person."
Esther shot her sister a look, sensing something in her she hadn't seen before. She knew that when they were much younger, Sarah and Levi had been close friends and that she had taught him how to read. But that was about all she knew. Esther had a hunch she might be able to find a new point of vulnerability. "Well, if he's so good, how come we never see him?"
Sarah shrugged, seemingly unaffected, as she stacked the plates, her back to her sister. "Levi's a busy man," she said.
Esther scowled, looking down. There was a design on the countertop that she jabbed at with her finger. "Busy bossing everyone around."
Sarah's voice hardened.
"If he didn't run things," she said, "we all would have died a long time ago. And I don't see you turning down food you don't work for."
Esther's hunch was right; she had touched a nerve. Still, she flinched from the uncomfortable truth.
She could not deny that she lived off the food that Levi provided and that Sarah not only earned, but prepared as well. Esther didn't know the first thing about how to pound the otherwise inedible rice and beans into flour, or how to mix it with water and pat it into flatbread. She had never once stoked the firebowl with charcoal or cooked watery porridge on its blackened grill. These were all Sarah's jobs, and while Esther had always taken that for granted, she realized that it did not strengthen her position. If anything, it made her even more of a child, someone not to be taken seriously.
"Besides, who's better?" Sarah said. "The mutants?"
"Don't call them that," Esther said under her breath.
Her sister didn't stop. "At least we didn't go around attacking people, like the _mutants_." She emphasized the hateful word. "We're better than that."
"Whatever's happening now, it's not their fault," said Esther. Sarah rolled her eyes, but her sister continued. "Maybe they're just hungry. Besides, they mostly don't hurt people . . . only buildings and things." And despite herself, she opened up. "They're nice, Sarah, they really are. Maybe one or two of them are bad, but—"
Sarah snorted. "Oh, please," she said as she started putting the dishes away. "I wish to God you'd stop socializing with them. You and your little friend Star—"
"Skar."
"And that lunatic in that building, with all the cats. What's his name? Joseph? You're not a baby anymore, Esther. It's about time you weaned yourself away from all of them."
Esther tried to rein in her emotions, but she could feel her control slipping as tears sprung to her eyes. "Why do you hate my friends?" she asked.
"I don't hate anybody," said Sarah. Her voice sounded frozen. "I'm only looking out for you, since apparently you can't do that yourself."
A sob escaped. Furious and ashamed, Esther pushed her fists into her eyes to keep tears from falling.
Sarah sighed, and her tone softened. "I just wish you weren't so . . . naive, Esther."
Esther felt a new stab of annoyance. "You know I don't know what that word is," she muttered.
"Do you really think they want to be your friends?" Sarah said. She spoke softly, almost gently. "They're all probably waiting to break in here so they can rob us blind."
"Rob us? Of what? Our matching coffee cups?" Esther managed to say. Tears were running down her face and she wiped her nose with her sleeve.
Sarah gazed at her; you could almost see something settle in her mind. It was what Esther feared all along and she cursed herself. She had once again driven her sister in the opposite direction.
"Thank you," Sarah said. "If I wasn't sure about whether or not to see Levi, I am now. Why all this trouble is going on, I honestly don't know. But I'm sure he'll put an end to it."
Esther was overwhelmed with bitterness—at her failure to keep her sister from going to the Source, and her inability to control her emotions, to play it cool. To strategize, as Skar would say.
"Fine," Esther yelled. "Fine!"
She pushed past her sister.
Sarah's voice betrayed mild panic. "Where are you going?" she called after her. The only response was the slam of the front door. Inside one of the many cupboards, something fell off a shelf and broke with a small crash.
Once outside, Esther walked blindly through the darkness for several blocks before she calmed down enough to think. She could sense rather than hear a pack of feral dogs rummaging nearby. The animals were cowardly yet, when desperate, had been known to attack anybody unwise enough to be outside at night, alone, without a weapon.
Esther sat on a street corner, her ears keyed to a possible attack, adrenaline coursing through her body. She knew it was stupid to be outside, yet she needed to make a physical statement, to create some distance.
After an hour, she went back.
She suspected her sister would sit up waiting for her, as she had so many times in the past; in fact, she secretly wanted it to be true. Yet when Esther returned, she found Sarah in her room, asleep. As she stood over her, Esther experienced a strange, twisting sensation in her stomach. She had an impulse to touch her sister's long, black hair, fanned out on the white pillowcase and framing her face, but at the last second, she changed her mind.
Instead, she went back to the main room and sat alone in the dark. Stubbornly, she decided she would wait to watch the sun come up.
Hours later the first rays of light found her sound asleep, fully dressed, curled like a cat on the far end of the couch.
Miles away, someone else was watching the sun rise.
It was a solitary boy on a bike, on the major roadway that passed by the outskirts of Prin. At sixteen, Caleb was lean and deeply sunburned, with a strong jaw and hazel eyes that, despite his distrustful gaze, had once been gentle.
Like Esther, Caleb chose to protect himself from the sun in his own way. He wore a long-sleeved denim shirt, jeans, canvas gloves, and a battered Outback hat. In his backpack, he carried a few belongings. His vehicle was a scuffed black mountain bike with patched tires that had seen many miles.
He had been on the road for months and could finally see his destination on the horizon: a glimpse of the lone church spire that marked the town of Prin.
And still, he hesitated.
He unzipped his backpack, took out a green steel bottle, and swished it around. As he feared, it was nearly empty. The sun had only just risen and the morning was still cool. Yet the sky was cloudless and he knew it would be another day of blazing heat.
By the side of the road was an old gas station, abandoned and in ruins. In front, Caleb noticed several rusty old oil barrels. One of them was uncovered and now brimmed over with rainwater from a recent storm.
Caleb walked his bike to the edge of where the grass used to be and released the rickety kickstand. Then he crossed to the barrels and looked down. The water was so clear, you could see all the way to the bottom, where a pink pebble lay. He couldn't help himself; he stooped to smell it, and at its irresistibly cold scent, he imagined plunging his head into it, opening his parched mouth and swallowing, gulping, drinking as much as he could without coming up for air.
His eyes were closed and his lips were at the surface of the water; at the last second, he gave a shudder and forced himself to pull back.
That would have been suicide.
The trembling surface reflected the cloudless sky above him. It also reflected his face, which shocked him with its gauntness and its look of need.
Making up his mind, he uncapped his green bottle, lifted it to his lips, and emptied it within seconds. It was only a few mouthfuls of hot and metallic water, but he savored every drop.
Then Caleb got back on his bike and headed for town.
## THREE
AS MOTHS DANCED AROUND THE BRIGHT SPOTLIGHTS OVERHEAD, SARAH waited outside the Source, nervously brushing back her hair.
Before she left, she had primped in front of her cracked mirror, combing her long hair so that it lay across her shoulders in a style she thought was pretty. Now she tried to make it stay that way.
She had never been this close to the enormous building; few in Prin had. She stood in front of the giant steel front door that rose and lowered, powered by electricity. Posted at a discreet distance on all sides were armed guards, silent and hooded. Sarah had approached one with trepidation earlier that day. She had passed along a note, requesting to speak with Levi, not knowing if she'd ever get a response. To her surprise, within a few hours, she received a note back, inviting her for dinner that night, alone.
Now the guards urged her forward and ushered her inside.
Sarah entered, nervous and excited. She adjusted her eyes to a dark and cavernous interior, lit by electric lights kept low.
Towering shapes hulked on all sides. They were giant shelves that rose to the ceiling, all of them fully stocked with oversize cartons. Sarah made out some of the words printed on them: THIS END UP. POWDERED MILK. DEHYDRATED CARROTS. WHEAT GRAIN. 200 GALLONS. POLAND SPRING WATER. HANDLE WITH CARE.
Then, emerging from the shadows was Levi.
Like everyone else in Prin, Sarah had not laid eyes on him in years. Levi was now a tall seventeen-year-old, with dark eyes and a mouth set in a hard line. He wore only black: jeans, button-down shirt, leather boots, all of which set off the extreme pallor of his skin. Yet when he recognized her, he smiled; and in that instant, he became the old Levi again, the boy with the watchful eyes she'd once known so well. The boy she had taught to read and who she thought might one day propose to her.
"Sarah," was all he said.
Levi escorted her through the dimly lit Source. When they rounded a corner, she almost cried out in shock. A single electric light overhead threw deep shadows into the surrounding cavernous space. It illuminated a large table, laid with a rich cloth and piled high with plates of roast rabbit and salted flatbread, enough to feed at least a dozen for days. There were also strange foods she had never seen before: bowls of steaming, fragrant liquids and soft, glossy breads that were still hot.
As they started to eat, Sarah told herself to focus. She knew that she was there on serious business. Yet for the longest time, she couldn't speak. She could only eat, ravenously. On the table was something new to her, a bottle of dark purplish-red liquid.
"Have some," Levi said, hoisting it.
Before she had a chance to answer, Levi was filling her glass. At first, Sarah winced at its sharp taste, but with each sip, she found she liked it more and more. By her second glass, she was simply listening as Levi spoke of small things: her health; Sarah's sister, Esther; the people in town. Sarah was thrilled by the thought that despite all of his power, her old friend evidently still cared about her and remembered names and details from a long-ago time, their shared youth.
_Esther was wrong about him,_ she thought.
She was only vaguely aware that, unlike her, Levi had eaten very little. He grew silent, watching her from across the table with an unreadable expression as he toyed with a glass of the purple liquid he had barely touched. He seemed to be waiting for her to say something; and she remembered that, of course, that was why she was here.
Sarah guiltily wiped her mouth with a cloth napkin and cleared her throat.
"Levi," she said, "as you probably know, the mutants have been attacking, and it's getting worse. There's no one in town strong enough to do anything." She heard the contempt that had crept into her voice, but she didn't care. "They need your help."
She thought of the twelve-year-old who had arrived in town five years ago and broken into the shuttered and locked Source when no one else could. Before then, Prin was a ghost town inhabited by a few dozen, a wasteland on the verge of extinction. Through sheer intelligence and willpower, Sarah thought fuzzily, Levi had single-handedly transformed it. Sequestered in the white building, he became the hidden engine that kept the town running.
Now he would save it again.
"They need weapons," she said, in a rush. "Real weapons, not just sticks or rocks. They need knives, arrows, clubs . . . whatever you can spare."
Levi inclined his head in a slight nod but said nothing. The purple liquid had made Sarah expansive and uninhibited. She spread her hands out in a naked appeal.
"You've been so generous already," she said. "If you could supply them with arms, they would forever be in your debt."
There was a long pause before Levi answered.
"I see," he said.
He picked up his glass but only studied the liquid inside. An idea seemed to come to him and he looked up.
"To be honest," he said, "I'm not even sure if we have what you want in stock. Weapons, knives, clubs . . . I don't know if we've seen much of that kind of thing around here. Have we?" He addressed this last part to his guards, who murmured negatively.
Levi's guards were all clearly armed. Sarah was confused. "Are you sure you—"
Levi set down his drink and stood.
"May I show you something?"
The way he asked it wasn't a question. Unsteadily, Sarah got to her feet. Swaying from the drink, she took a final, surreptitious bite of rabbit before following Levi out.
The guards kept their distance as Sarah trailed Levi through the endless, murky recesses of the building. She struggled to keep pace with her host. It was not easy, for he walked swiftly, sure of the path. Silver things flashed on his wrists and fingers—rings, watches, bracelets—and Sarah focused on them, as if they were stars in the night sky, to keep from getting left behind.
The crowded shelves towered above her.
In her inebriated state, Sarah knew they represented wealth of the most genuine and therefore precious kind. In a world of poisoned rain, scorching heat, and ashen skies, even a single jug of water, one of the hundreds stored here, perhaps thousands, held the balance of life and death to the people of Prin.
Sarah extended a hand to touch one of the cartons. But before she could, a guard materialized from nowhere and shoved her aside.
"Keep close," Levi called from a few steps ahead, "and don't touch anything. There are things that can hurt you if you're not careful."
Only then did Sarah realize that the shelves were encircled with loops of heavy wire, the kind she had seen on a few buildings downtown: wire studded with razors that could easily slice through leather, let alone human skin.
Levi stopped. He pressed something on the wall and the air was loud with humming. In front of them was a wide ramp that led to a lower floor, and now it began to move on its own. Levi stepped on and gestured for Sarah to follow.
Sarah hesitated, frightened. Then she finally stepped on because Levi was far ahead and she stumbled, nearly falling. Terrified, she clung to the moving handrail, until she reached the lower level, where Levi was waiting for her.
He set off again through darkened aisles, then down a narrow hallway, where more guards kept watch over a battered steel door. Behind that, a poorly lit stairway led even farther downward to a series of hallways with low ceilings.
Levi stopped at a doorway. A small sign next to it read: BOILER ROOM.
"Here we go," he said. Then he opened the door and flicked a switch set into the wall.
Sarah gasped.
The blinding overhead light revealed a windowless room that was furnished sparsely, with a desk and single chair. The rest of the place was empty except for one set of shelves. Unlike the ones upstairs, however, it was not stacked high with supplies, nor was it guarded by barbed wire. Instead it was filled with books: dozens of them, battered and mildewed. Compared to the meager collection in Sarah's home, this was a veritable library.
"You kept them," was all Sarah could manage to say.
Levi smiled. "I figured since you bothered teaching me, it was the least I could do."
Sarah ran her hand over a row of bindings and this time, no guard rushed forward to push her out of the way. She marveled at the titles and names she remembered, books she had salvaged from vacant homes and looted stores many years ago and later used to teach Levi how to read: John Grisham, _The Joy of Cooking_ , _American and European Furniture: 1830–1914_ , _A_ _Cavalcade of Jokes_ , Stephen King, Richard Scarry. The Bible. The Brothers Grimm.
Levi had been a difficult student, moody and hotheaded. Yet he was diligent and had a hunger to learn. Within a year, his abilities had equaled and then surpassed hers. Sarah hoped that since they were both twelve, their relationship might shift into something deeper. But then Levi ended the lessons. Not long afterward, he broke his way into the Source and disappeared from the streets of Prin.
Sarah had always wondered where the books had gone. And now, she blushed as it occurred to her why he had held on to them all these years.
"Sarah," Levi said.
She turned to him, her heart pounding. He was holding out a book to her. She took it, uncomprehending.
"This was one I found especially interesting," he said. "But you only gave me the first volume. Do you have the other?"
Puzzled, Sarah turned the book over in her hands. She couldn't recall ever seeing it before. It was an academic volume, dense and impenetrable. She flipped through it, but had never heard of any of the words: "Topography." "Aquifers." "Spring flow measurements." "Hydrosphere."
"I might," she said, handing it back. Frankly, it wasn't the kind of book she liked or understood, but she had several such volumes she rarely glanced at. "It could be in my house. I'm not positive what books are there."
Levi nodded, refusing to take it. "Why don't you hold on to it?" he said. "Because I'd really appreciate it if you could find me the second volume." Before Sarah could respond, he added, "We'll talk again when you do. And by then, I just might have some more weapons in."
Sarah understood. If she could find what he needed, this meeting wouldn't be their last. And as she was thinking this, he was seizing her by the waist and pulling her close. He kissed her, lingeringly.
"Let's keep this our secret," he murmured. "All of it. All right?"
Sarah couldn't speak for a second. "Yes," she said.
Her voice almost sounded normal, even though her face was burning. She couldn't hide the smile that covered her face.
Moments later, Sarah was outside, walking, if unsteadily, away from the Source. The white of her robes seemed to give off an unearthly glow in the bright glare of spotlights that sliced across the darkened parking lot.
From a tiny window hidden high up in the Source, Levi watched as the girl was swallowed by the surrounding darkness. As before, his expression was unreadable. He thought about the past and, for a moment, almost felt sorry for Sarah.
Then he shrugged it off. She was like everyone else in this world: just a means to an end.
Levi returned to his office. There he examined the handmade maps of Prin he had drawn, the laborious approximations of its physical layout that were tacked up on his walls. They had taken him more than five years of careful study and reflected the locations of not only each of the forty-seven Excavations to date but every Gleaning, as well. Still, they had not brought him closer to what he was seeking. Frustrated, he was tempted to tear them all down.
Levi had come to Prin, drawn by a rumor of its hidden clean waters. He met Sarah, who taught him how to read. Most books he found worthless; yet one convinced him that the notion of an underground network of springs was true. All he needed was a little more time to locate it, but time was running out.
Unlike everyone else, he always knew the supplies in the Source were limited. In the last few months, they had reached dangerously low levels. He recently had doubled and tripled the workloads, driving the town to exhaustion, and increased the punishments for shirking. It was a delicate balance, squeezing the greatest amount of work out of the people before getting rid of them.
Sarah's unexpected appearance had given him new hope. If anything, he was annoyed at himself for not having thought of this earlier.
He needed the book, the companion to the text Sarah had given him so many years ago. With any luck, she would soon find it and bring it back to him, as trusting and unquestioning as a dog. And just as easily satisfied, with a little affection and a good meal.
Someone broke into his thoughts.
"Is she gone yet?"
Levi nodded, without looking, as a girl encircled his waist with her arms. She had sand-colored skin and hair, her eyes were a vivid blue, and her thin clothing fit close and tight to her figure. Her name was Michal and she was perhaps fourteen.
"Then can we eat now?" she continued.
Levi looked down at her. "Sure," he said.
The two headed back upstairs to the dining area, where guards stood watch over the ruins of the table. Even hooded as they were, they resembled wild dogs themselves, staring at the leftovers, their eyes visible and gleaming in the spill of the electric light.
"I still don't know why you put out so much," said Michal as she sat, and began piling roast rabbit onto her plate. She licked some off a finger. "I thought lots of people were coming. Why so much food for one old woman?"
Levi smiled. "You wouldn't understand," he said.
But Michal was no longer listening. She was too busy eating.
## FOUR
THE NIGHT SKY WAS DARK AND HEAVY. YET AT CERTAIN TIMES, THE dense clouds parted, allowing moonlight to shine down. Someone stood by a window above the main street, gazing into the night.
Esther had been waiting like this, filled with dread, for hours. If Sarah returned home accompanied by Levi's men carrying weapons, she had no idea how she would be able to stop them. So when she spied a gleam of white coming down the sidewalk, she was relieved to see Sarah was not only unescorted, but also empty handed. Esther was so thankful, she barely noticed that the older girl was behaving strangely when she came in. Her cheeks flushed and eyes glittering, her prim sister was talking in a loud and aggressive voice, slurring her words.
"Levi didn't have any weapons," Sarah announced as she entered, before she had even removed her outer robes. Then she added snidely, "So your friends the mutants won't have any opposition. That should make you feel good."
But there was something careless about Sarah's tone, as if she was not paying attention to her words and was just saying what she believed would mollify her sister. She was not revealing what was actually going on, Esther suspected.
"Rafe's coming over in a little while," said her sister. She was folding away her robes, clearly trying to sound casual and unconcerned. "It'll just be business talk. Nothing important."
Esther nodded, as if the thought of a late-night visit from the town's leader was an everyday occurrence.
"You're right," she said. "It sounds boring. I think I'll go to bed now."
She tried faking a yawn, stretching her arms over her head. If Skar were here, she would laugh at how obvious the ploy was. But Sarah didn't even seem to notice.
Once inside her room, Esther knew enough to change into her sleeping shirt and get into bed. When she heard Rafe's knock less than an hour later, her sister checked in on her, opening the door just wide enough to let in a crack of candlelight from the living room. Esther, motionless, kept her eyes shut and breathed slowly. But once the door was pulled shut, she flew across the room, kneeling in the dark to listen.
"Don't worry," Sarah was whispering. "Levi has weapons and is happy to give them to us. It will just take a while, that's all."
"We ain't got a while," Rafe said loudly, not whispering at all.
"Shhh," murmured Sarah. "My sister."
"We ain't got a while," Rafe repeated in a harsh whisper. "We ain't got no time to waste while Levi plays games with you. Can't you see he's just playing you for a fool?"
"You have to trust me. If you'll just be patient—"
"Don't you get it?" hissed Rafe. "The mutants ain't being patient. Next time, they gonna murder us in our own beds on account there ain't nothing we can do about it. At least not without weapons."
"Please," said Sarah. "Just wait and see."
From where she listened, Esther was now convinced that her sister knew more than she was letting on. Rafe, however, was too stupid to understand that.
"All you had to do was bat your eyes at him," he said. "Guess I was a fool for thinking anyone might still want you. Thanks to you, we're on our own."
A second later, the front door slammed.
Esther only had enough time to get back into bed and shut her eyes before the door creaked open one last time. After a moment, the door was pulled shut and Sarah walked away.
Esther counted to ten before creeping out of bed. She opened her door and slipped into the hallway. In the darkness, she sensed a faint glow was coming from the living room.
It took Esther a few moments to realize what Sarah was doing. Although her sister kept a set of shelves full of musty books, they were now mostly for show; she rarely read anymore. Yet right now, Sarah was on her knees, searching through her collection. She worked methodically, muttering to herself as she pulled out one book after another, squinting to read their titles, then discarding them in a growing pile next to her.
Despite the care she was taking, she seemed desperate. For it was clear that she couldn't find what she was searching for.
After half an hour of watching Sarah, Esther couldn't stop a real yawn from escaping her mouth. Fearing she'd be discovered, she slipped back to bed.
But sleep proved to be impossible. With a tightness in her chest, Esther lay in the dark brooding over what she had just seen and heard, events no one trusted her enough to explain.
As the first rays of sun brightened the leaden sky, someone could be fleetingly seen darting through the shadows of Prin.
Esther was bringing supplies to her friend Joseph. She came alone; no one else in town cared to visit the village outcast, the eccentric pariah who lived on the outskirts of town, close to the Source, alone with his timepieces and cats. Even Skar was made uncomfortable by him. "The crazy one," she called him privately.
Joseph was not just one of the rare individuals in town who could read; he also kept a cluttered and moldering library that included old magazines and newspapers. The walls and surfaces of his home held dozens of watches and clocks in working order, homemade calendars, even hourglasses and a sundial. The rooms were filled with the gentle and persistent murmur of ancient gears shifting, second hands ticking, and the occasional muted chime.
Although Joseph was an old friend, this was no mere social call. For years, Esther had been skimming the supplies from her household to share with him. She sensed that if she did not do so, he would die, for he was too proud to ask for help. The girl took pains to hide her theft—pouring off the remaining water into different vessels, for instance, so it was harder to gauge how much there was. But recently, she had been bringing less because there was less to steal.
Esther chose to make her deliveries at dawn, a time when the wild dogs had long since left and the townspeople had not yet risen. But the relative cool of early morning vanished as the sun rose. It was sweaty and dangerous work to make one's way across the ruptured ground and giant, uprooted pine trees that protected Joseph's home like steel pikes around a prison, and as effectively, too. This was true even when one was not carrying a heavy armload of food and water.
Joseph and his ten cats (Malawi, Benjamin, Tiffany, Samsung, Mr. Roberts, Seven for All Mankind, Ginger, Claude, Tiger Boy, and Stumpy) lived in the wreckage of a hotel called the Gideon Putnam. Uncomfortable with people and frightened of open spaces, he had retreated there years ago; its remote location and condition scared off the curious. The lobby was a blasted ruin. One had to cross it to gain access to the stairwell, where seven flights of cracked and crumbling steps awaited. Like most of Prin, the building had sustained heavy damage in the series of earthquakes that had flattened much of the surrounding area some years ago, and it was no mean feat to navigate it without the carpeted ground giving way or sections of the ceiling collapsing. There were entire floors Joseph dared not venture onto. Sometimes, in fact, he was quite certain the whole building was about to fall down.
Now, as he worked his way around his apartment, winding and adjusting each of his clocks and watches, Esther's signal sounded from somewhere in the building: a two-note whistle. Joseph's cats recognized it and began to call and mill about. They were fond of her, or perhaps they were just fond of the bits of dried meat she always brought.
"Joseph," Esther said as she appeared at his door.
Joseph looked as he always did: with long, unkempt hair and light-colored eyes. He was so tall, slender, and stoop-shouldered that he seemed to undulate rather than walk.
She didn't waste time with small talk. "Lately, the food payments are down even more," she said. "And the water payments are worse. This was the best I could manage." With that, she set down a gallon of water and a bag of cornmeal by the soot-filled firebowl in his hallway. "I don't know when I can bring you any more after today."
Joseph seemed to think this over and nodded gravely. The fact of the matter was, he couldn't support his brood and himself on Esther's supplies alone. He had never mentioned it, but he had long been in the habit of setting traps for the various wild animals that visited his roof and basement in order to supplement what she brought. In fact, he prided himself on his squirrel stew.
"I suppose we'll have to make do," he said. It seemed to him the right thing to say at such a moment.
Unexpectedly, she took him by the arm. When he glanced down at her, he was struck by the look of anguish on her face.
"Do you understand what I'm saying, Joseph?" she said in a low voice. "I don't know if I can come here anymore."
This was something else entirely. Esther was Joseph's only friend (his only human friend, that is) and his sole connection to the outside world. He rarely if ever left his building. To lose her companionship would be terrible for him indeed.
Esther told him about the recent attacks by the variants. She was worried that these incursions were about to be met with retaliation by the townspeople, which would only serve to fuel more acrimony. If this happened, the long-simmering tension between norm and variant would erupt into open warfare, a conflict the fragile town of Prin couldn't possibly sustain. If war began, they would all be at risk . . . even those who chose to live on the outskirts of society.
As she talked, Joseph fetched a cup of water from his desk and started to raise it to his lips.
Esther grabbed his hand.
"Joseph," she said, her voice raised in panic, "how many times have I told you? Don't drink any water but the kind I give you. You'll get sick."
She handled his cup like it was a live snake, holding it far away from her and carrying it to an open window. She was about to fling it outside, when a faint sound from below made them both glance down.
Someone was calling.
The windows of the apartment looked over the buckled remains of an asphalt field. The collapse of the neighboring building, subsequent looting, and the effects of many years of rain, sun, and wild animals had transformed it into a jungle of tall grass growing amid red clay, rubble, broken furniture, rotting wood.
Although they were far above, the two of them took care not to make any sudden movement that might draw attention to themselves. As Joseph peered out, he was surprised to see four figures below, picking their way through the shattered field. From their light-colored robes, they were clearly townspeople. A distance away, four bicycles were propped against a sagging chain-link fence.
"Do you think it's a Gleaning?" Esther asked from beside him.
Joseph shrugged. If it was, they both knew what that would signify, and it was not good.
The Gleaning entailed searching empty houses and stores, sifting through the wreckage of buildings in search of anything viable: weapons, medical supplies, charcoal, bedding, and nails. Everything was brought to the Source, where it was displayed on long tables. Levi's guards tallied the day's take and, depending on its perceived worth, added more water and foodstuffs to the town's portion. It was never very much, compared to what they paid for gasoline.
If the townspeople were Gleaning Joseph's ravaged home, that meant they were forced to reach even deeper into the outlying areas to try to meet the monthly quota. And that could only mean that Prin had been wrung dry, picked clean of anything of value.
Esther observed the people for a few moments. Chewing her fingernail, she turned for the door.
"I don't like this," was all she said.
"But where are you—"
"Don't worry. I'll be fine."
Joseph had no choice but to raise his hand in thanks and farewell. She gave a quick nod; then, without another word, she was gone.
In the lobby, she darted behind a crumbling wall and slipped out the giant gap that once held a large glass window. She slowed as she approached the backyard and hid in the dappled shade provided by some overgrown vines and bleak vegetation.
From there, she could hear faint voices and something she couldn't identify: a hollow twang that echoed in the canyon of the old hotel.
When she peered around the jagged corner, it took her a moment to locate the origin. In the distance, one of the trespassers was holding something, a ball that was dusty brick orange in color.
Now that she could see them clearly, Esther sensed that the four were not intent on anything nefarious; they were not even on a Gleaning. Whoever it was bounced the ball on the ground, once, twice, producing the strange sound. One of the others gestured at something a short distance away. It was a tall pole, with a metal ring attached near the top, with the shredded remains of a net clinging to it. The first one threw the ball at the hoop, but it fell short.
The four laughed. Within moments, they headed back across the lot to where their bicycles awaited. Soon, they were gone.
Then Esther heard something behind her and froze in place.
Someone else was there.
A boy emerged from the towering, ruined mounds and stood where the four had been. By the peculiar way he dressed, Esther could tell he was a stranger. Like her, he chose not to wear the hooded robes that the people of Prin used as protection from the fierce sun. Instead, he wore a long-sleeved blue shirt and dusty jeans, with a shoulder pack that he slung to the ground. A battered, low-brimmed hat obscured his face. He had been watching the group at play, although it was impossible to say why.
He walked to the orange ball. Esther watched as he bent to pick it up.
With one hand, he effortlessly tossed the ball over his head. It landed in the hoop, the ragged net swishing. He turned to go. Before he did, though, he stopped and glanced back.
"You might as well come out," he said. "You ain't fooling anybody, hiding there like that."
His voice echoed amid the broken piles of brick and twisted metal.
From her hiding place, Esther started as if struck. She was stunned to have been spotted, and more than a little rattled. She stood poised, adrenaline coursing through her body, ready to escape should he make a move toward her.
But instead, the stranger only shrugged.
Then with one fluid movement, he mounted his battered bike and left.
## FIVE
UNDER A STREETLAMP ON THE OUTSKIRTS OF PRIN, A BOY IN WHITE robes stood guard.
He shifted uncomfortably. Not accustomed to remaining still in the blazing sun for such a long stretch of time, he was perspiring heavily under his white sheets and headdress and felt more than a little nauseated. He was finding it impossible to keep his focus on the horizon. The heat caused shimmering waves of air to dance across the road, making it look as if hundreds of variants were attacking all at once. In addition, his sunglasses kept sliding down the bridge of his nose.
Early that morning, Rafe had called him together along with a dozen others. Normally, the boy would have been getting ready for work. He had recently started a new Excavation on the eastern side of town and was preparing to spend the day deepening and widening the trench he and the rest of his team, including his pregnant partner, had just begun. Big-boned and quiet, the fifteen-year-old preferred Excavation to the other jobs in town because it was mindless and let you work by yourself.
But Prin, Rafe decided, needed sentries.
"We can't afford another attack," he told the assembled group. "So I'm putting you at each of the main roads that lead into town."
An unspoken question rippled through the crowd. Rafe seemed to anticipate what it was and held up his hands to reassure them.
"Sarah let us down about the weapons. But don't worry, you won't be unarmed. I seen to it."
That was when the boy noticed the crate by Rafe's feet, the one filled with a sorry-looking assortment of bats, corroded metal bars, and splintered table legs.
Now he adjusted his sweaty grip on the hollow steel pipe he was given and tried to imagine what it would feel like to hit someone with it.
He couldn't.
The boy guard rocked back and forth on his heels and he swiveled first to the left and then the right. A drop of sweat trickled into one eye; and he pushed his sunglasses up on his forehead so he could rub it away.
When he lowered them, he sensed a flicker of movement. Nearby, a dust-colored squirrel was watching him from the safety of some underbrush.
"Hey," he said, relieved at the distraction.
The boy set his pipe down, squatted on his haunches, and held out a hand, even though he didn't have anything to offer. He made a soft chucking sound, and the squirrel cocked its head at him, twitching its plumed tail.
The boy was glad no one else was around. Anyone in town would have tried to kill the creature, without a second thought. Unlike humans, animals were unaffected by the poison lurking in rainwater, and fresh meat of any kind was a rare and precious treat. But the sentry secretly liked squirrels and wished he had something to feed it. Maybe he had some forgotten crumbs somewhere in his clothing?
Slowly, in order not to scare the squirrel away, the boy kneeled, hoisting his robes so he could scrabble in his front pockets. One knee grazed the pipe, which rolled into the gutter, unnoticed and for the moment, forgotten.
"Ahh," said the boy at last. His fingers closed on a few gritty crumbs, which he removed and scattered on the ground.
The squirrel leaned forward on one paw, nose twitching, then appeared to make up its mind. Keeping its eyes on the boy, it darted forward in two, three quick movements, seized a morsel, then sat up and began to eat.
The boy eased back on his heels and smiled, satisfied. But after a few moments, the squirrel stopped chewing.
The tiny head jerked up and froze, black eyes staring at some point in the far distance. Then with a flick of its tail, the animal bounded away and vanished in the tall weeds, leaving the rest of the crumbs untouched.
Still on his knees, the young guard frowned, puzzled. He glanced behind him and saw nothing. It made no sense.
But then again, he couldn't hear what was bearing down on him.
Not until it was too late.
Elsewhere, Caleb had reached the broken sign that said WELCOME TO PRIN.
The center of town itself still lay a half mile or so in the distance; he should be there within minutes. The main street appeared to be dotted with buildings, none more than a few stories high.
Caleb realized that the first thing he had to do was get water. He had nothing to trade with, but he was strong and handy; he knew he could work for what he needed.
Then he heard it.
Caleb braked and balanced stock-still, one foot on the ground, straining to identify the sound. As he did, he felt a familiar twisting sensation in his gut and a tingling in his hands and along the back of his neck.
Far away, someone was screaming.
For an instant, Caleb reeled. It was as if he was falling, tumbling backward into an abyss, a sucking vortex from which there was no escape. He gripped his handlebars so tightly his knuckles turned white, and the hot pavement beneath him started to bloom into an obliterating brightness.
Then he heard something else, a thin thread of noise that brought him back to himself. It was the sound of others shouting. Even though he couldn't make out the words, the voices seemed mocking and jubilant.
They sounded like mutants.
Could he be sure? He might be imagining things. He knew he'd been seeing them for weeks now, maybe even months. Since he first left home, he sensed them everywhere, from the corner of his eye, behind him, just around an abandoned car or bend in the road, their obscene, deformed faces silently watching him, jeering, before vanishing into nothingness. Sometimes they appeared in his dreams and when they did, he awoke in a sweat, crying out.
Several weeks ago, he had confronted mutants in the flesh, an unsuspecting group he had happened upon while they were hunting. He didn't hesitate to launch an ambush that left them unconscious and bleeding.
Now he made up his mind. Getting back on his bike, he swerved off the main road and onto a smaller street on his left.
It did not take him long to track the source of the noise. Although the screaming stopped, the other voices grew louder, providing him a rough guide to follow. He rode down one street, which led to a dead end; doubling back, he was able to find a parallel road, which led to another main thoroughfare. By now, he was so close, he was able to hear distinct voices.
"Pretty girl," said someone. This was followed by the sound of others laughing.
With his backpack on, Caleb leaped off his bike and tossed it down in an abandoned yard, its front wheel still spinning. This was evidently once a residential block, with the remains of large two- and three-story houses on both sides nearly hidden by weeds and tall grass. Caleb cut diagonally across the last property and around to its backyard.
The yard led to an overgrown field, which bordered on the cracked parking lot of an abandoned supermarket. Caleb decided to head away from the voices. Realizing he was only one against at least three or four possible enemies, he calculated he would have to use surprise as an additional advantage. He skirted the open expanse and stuck to the perimeter, defined by an immense and straggling hedge.
As Caleb ran, his ears constantly adjusted to the thread of voices, trying to pinpoint their exact location. When he judged he was no more than fifteen feet away, he stopped. Only then did he work his way through the dense foliage, taking care not to disturb the branches around him. He noticed a small gap in the hedge. Through it, he was finally able to make out what was happening.
What he saw astonished and then repulsed him.
Five mutants stood in a loose circle looking up at a streetlamp. A boy hung from it, tilting forward at an unnatural and painful angle. His pitiful arms had been lashed together behind his back, and this rope had been tossed over the beam that extended high up the steel pole and then secured. From where he stood, Caleb could see the boy had been beaten. His face was bruised and puffy, and blood dripped from a corner of his mouth.
But what was most shocking was what the mutants had done to his body.
The boy had been stripped of most of his clothing, and his pale white skin, long unaccustomed to the burning rays of the sun, had been defaced with obscene drawings smeared in red clay. It took Caleb a moment to realize the boy had been turned into a repellent caricature of a girl, a grotesque and unspeakable travesty. A wig of sorts, made from some filthy and stringy object, sat askew on his head. The smell of wet earth was strong. The scarred and tattooed hands of the mutants were all dark, stained red with mud, and their mouths were open in harsh and mocking laughter.
The body hanged heavily, twisting a little in the breeze. At first, Caleb thought he must be dead; but when one of the mutants prodded the victim with a metal pipe, the thin sides heaved and the feet kicked feebly.
"She's pretty, all right," said one of the mutants. The rest laughed again, and one of them uttered an obscenity.
But Caleb no longer heard what they were saying. He was reaching into his backpack, searching for his weapon.
It was one of a kind. He had forged it over time and through much trial and error. Made of wood, metal, and rubber, it was the length of his forearm and had a crude wheel at the center, held firm on an axis. He kept it loaded: the six rods that fanned out from the wheel were each tipped with a shallow cup that held a jagged rock the size of a hen's egg. A taut rubber sling kept each rock in place.
It allowed him to shoot six rocks in as many seconds. He had used it once or twice, on his way to Prin. Now he prepared to employ it again.
From his hiding place, Caleb took aim at the nearest mutant, who stood facing away. He gave the wheel a sharp spin and as he did, he snapped each sling, firing off three rocks in quick succession. The first two hit the mutant in the head, and the last in the back of the neck. His knees buckled and he slumped to the ground.
As he calculated, the element of surprise had given Caleb a small but critical advantage. In the confused moment that it took the other mutants to register what had happened, he had time to take aim at his second target. He again set the wheel into swift motion and deployed his last three projectiles. While this mutant raised a hand at the last second to protect himself, it didn't matter: The rocks came too fast, all three striking his head, and he too was knocked unconscious.
Caleb was scrambling in his backpack, trying to reload, when one of the remaining mutants spied him through the dense undergrowth. She gave an angry hiss, like an animal. Then she pointed at Caleb, her mouth open in a shriek of fury like a hawk, some bird of prey, and at that, all three mutants rushed the hedge.
The first one took a running leap and dived headlong through the dense growth at Caleb. At the impact, his weapon flew from his hands and he was knocked backward and onto the hot pavement of the parking lot, the wiry and muscular mutant on top of him, clawing at his throat, his face, trying to subdue him until the others arrived.
But instead of fighting the momentum, Caleb knew enough to work with it instead. As he curled into a ball and continued to roll backward, he seized the mutant by her scant tunic, pulling in his knees until his feet rested against her stomach. Then he released his grip while violently pushing out with both legs; the mutant was catapulted over his head. A second later, she landed behind him with a sickening crunch.
As Caleb leaped to his feet, another mutant rushed at him with a rock in an upheld fist. Caleb used his elbow to strike his wrist, loosening his grip; he then swiveled, shooting out his leg and driving his heel into the mutant's knee. With a scream as much of surprise as of pain, the mutant pitched forward, off balance. Again, Caleb used momentum, this time to push the mutant farther downward while driving his knee up into his face as hard as he could. There was a satisfying crack of bone, followed by a hot gush of blood.
Caleb turned. His vision blurry, sweat streaming down his face, it took him a moment to see that the fifth mutant, the largest, was running away.
Dimly, he realized that he should let him escape.
Caleb had won this round, and he should tend to the boy who was still hanging from the streetlamp, who clearly needed his help.
But it was as if he was aflame, burning with a righteous fever that would not be satisfied until each mutant was hunted down, one by one, and made to suffer. A bloodlust was upon him. He took off in pursuit and now felt the glad fire in his legs; the mutant, a fast runner with a good head start, glanced back and the shock on his face was obvious. Caleb was nearly upon him.
The mutant swerved suddenly. He had reached a group of commercial buildings and, now frantic, he intended to escape that way instead. He clawed at the nearest wall and began to climb; Caleb leaped to grab his bare foot and only just missed. But any relief the mutant might have felt was short lived, for Caleb also began to scale the wall, moving with relentless speed.
The mutant pulled himself onto the roof; seconds later, Caleb did the same. By then, the mutant had sprinted to the far end and now balanced on the edge of the parapet; he was gauging the distance to the neighboring building. He glanced back with a look of pure panic and as Caleb ran forward, he pinwheeled his arms and took a standing leap.
Caleb didn't hesitate. He was aware that he had no clue as to how far he had to jump and that he was at least five or six floors above the ground; a misstep would be fatal. Yet he no longer cared.
He put on speed, then at the roof's edge, made a blind, running leap. He easily cleared the neighboring parapet and landed hard, instinctively rolling into a sideways somersault to blunt the impact. He came out of the roll without stopping, still running.
The mutant—only halfway across the roof—looked back. He had a setting sun tattooed across his face; beneath it, his expression was one of shock. Yet as much as fear, there was admiration in his voice.
"You have defeated four of my best," he said. Caleb realized that this one was the leader. "Who taught you?" he called, panting.
"I taught myself," Caleb said. The hatred and contempt in his voice were terrifying.
The mutant nodded, impressed despite himself. He was cornered now; there was nowhere to run, no other buildings nearby. But as Caleb stepped forward, the mutant hesitated for only a second.
Then he jumped off the roof.
Caleb ran to where he was standing and looked down. The mutant was lucky that the alley was strewn with crates and boxes; a pile of cardboard had broken his fall. Still, Caleb was satisfied to see him limping badly as he escaped down the alley and out of sight.
It was a team of Harvesters who first saw him.
A stranger walked down the main road that led into town. The townspeople viewed newcomers with suspicion, for supplies were scarce enough without interlopers looking for more. This one pushed a battered bicycle, across which was sprawled a body, legs and arms dangling. A dirty white sheet was partly draped around it and already, dark red patches—blood? clay?—were seeping through.
Caleb was bringing the brutalized guard back to Prin.
The Harvesters slowed their own bicycles and came to a stop.
"Mutants?" one of them asked, and Caleb nodded.
The Harvesters exchanged glances as one by one, they recognized the victim. A boy gave an involuntary cry, his eyes round with shock.
"Trey?" he said in a hoarse whisper. "Is he—"
"He's still breathing," said Caleb. "Barely."
The four dismounted and fell in line behind Caleb. There was no need to ask the details; everyone understood what they needed to know. But without being prompted, one of them handed over her water bottle and another took control of Caleb's bicycle so he had a moment to drink.
By the time they approached the center of town, there were at least two dozen townspeople accompanying them. There was virtually no sound; as newcomers joined in, they were briefly told of what happened in whispers, and then the silence resumed. Without speaking, all had done the unthinkable act of leaving their jobs, and beneath their robes and headdresses and sunglasses, their faces were shocked and somber.
One of them, the guard's partner, walked up front with Caleb. At first glance, Aima seemed stoic, a sturdy, heavily pregnant fifteen-year-old accustomed to unexpected hardship. But beneath her dusty head cloth, her eyes were dark holes in an ashen face. She gripped her unconscious partner's hand, massive in her small one, and stroked it with her thumb, as if trying to will him back to health.
Word had been sent ahead of them and by the time the procession reached Prin, Rafe and a small crowd were outside, waiting. Several townspeople managed to lift the boy and carry him into his storefront home. Once inside, Aima and her friends would wash him and tend to his wounds as best as they knew how.
Rafe was taken aback to see a mere stranger followed by the citizens of Prin. As he stepped forward to greet him, he cleared his throat and attempted to take control.
"Thank you for bringing home one of ours," Rafe said. Even to his own ears, his words sounded falsely hearty.
The stranger said nothing and merely bowed his head in acknowledgment.
"Did you see who done it?" continued Rafe. "Or did he run off before you got there?"
"I seen them," said Caleb.
Rafe nodded. "So it was more than one," he said, then turned to spit in the dust. "That makes Trey an even bigger hero. I bet he gave them a good fight with those weapons. Still, we're glad you come along when you did. Must've helped scare them off."
Rafe's voice shook and again he cleared his throat, to cover it. He was aware that everyone in town was not only staring at him; they were judging him, weighing his words.
It was, after all, his idea that guards be posted that morning, alone, and without any kind of training or backup. In retrospect, even he had to admit that perhaps it was a bad impulse. He had acted rashly, without a real plan. Without weapons, _real_ weapons from the Source that idiot Sarah had promised then failed to deliver, what other choice did he have?
Now Trey's partner stopped as she headed indoors to tend to the boy. In front of Rafe, her eyes blazing, Aima spoke in a low, accusatory voice.
"Trey never fought those mutants and you know it," she said. "He's too gentle. And you sent him there alone. You sent him out there and now he's—" A sudden spasm of anger contorted her face and she pushed past him to get inside.
Rattled, Rafe cleared his throat, hoping no one else had heard. He was now aware that the stranger was speaking.
"—wonder if you could give me some information," Caleb was saying. "I have some private business to look into."
"Of course, of course," said Rafe, with a wave of his hand. He was not listening; too busy worrying what the townspeople were thinking of him, he had already dismissed the stranger to the realm of the unimportant.
But at that moment, there was a new commotion.
A small boy and even smaller girl had just arrived, and they were both talking to whoever would listen. They were breathless and shrill, words tumbling over one another in their haste to speak.
"We seen it—" said the boy.
"We was hiding," said the girl. "We heard a noise so we hid. Then we seen it—"
"There was five of them. He was shooting rocks, like this, one after another—"
"They had Trey tied to a rope. He looked bad, he wasn't moving—"
"—and he beat four. The last one tried to get away . . . but he chased him, too—"
"—five against one. He beat them all. And we seen it—"
"Five against one."
The townspeople murmured, trying to understand. Bewildered, Rafe stepped forward and leaned down to address the two, with a feeling of dread.
"Who beat the mutants?" he said.
At this, the two stopped talking, self-conscious at being the center of attention. But then, they both noticed Caleb, standing to one side. The girl's face flushed and the boy broke into a smile as he raised his finger, pointing through the crowd.
"Him. Him over there. He's who done it."
## SIX
THE CELEBRATION LASTED ALL AFTERNOON.
Rafe had sent out orders, allowing everyone in Prin to take the rest of the day off from work. This was to guarantee maximum attendance—ostensibly, to pay homage and show gratitude to Prin's new hero.
The real reason was that Rafe wanted to ensure the entire town gave him credit for this turn of events.
As a result, the aisles of what had once been a supermarket were crowded. Even Sarah was there, whom Rafe had invited despite his lingering anger, as well as her misfit younger sister, Esther.
The stranger sat next to him, of course, in the seat of honor at the single table in the front of the store.
At first, Caleb was so silent and awkward, Rafe wondered if the reports of his astonishing heroism were true. For a moment, he even considered that he might be simple in the head. But when food and water were placed in front of Caleb—only he was served, as befitted the guest of honor—he started to eat voraciously. Soon, Rafe figured, he was bound to open up. And then they could get down to the real business at hand.
"We wanted to show our appreciation to you," Rafe said after Caleb slowed down. "For once, somebody not only agrees with me about the mutants . . . he ain't scared to follow through."
Caleb cleared his throat. Then he spoke so softly, even Rafe had to cock his head to make out what he was saying.
"Nobody wants to take the fight to the mutants more than me," he said. When his words were conveyed through the room, there was a murmur of approval.
"But this dinner ain't _just_ about appreciation," continued Rafe. "I'd like to make you a proposition." As usual, he, too, lowered his voice, so people leaned forward. "I'd like you to stay on awhile. How about you teach us what you know about fighting and such?"
For the first time, Caleb turned to his host and Rafe was startled by the intensity of his gaze.
"Do you have any _real_ weapons?" Caleb asked in his soft voice. "Any hunting knives? Shotguns?"
Rafe flushed.
"No," he said pointedly. He hoped Sarah was listening. "I'm afraid we got to take on the mutants without those. But I should add—in exchange, we're willing to put you up and feed you. How does _that_ sound?"
Rafe was smiling, a bit desperately now. Caleb appeared to be thinking. After what seemed an eternity, he gave a slight nod. At this, the room began to buzz with excitement.
"But I have conditions," he said, and everyone fell silent.
This time, he looked up, addressing the entire room. "If any mutants come near town, we will attack them, and attack them hard. Any survivors will be imprisoned. There can be no contact of any kind between townspeople and mutants; if anyone is caught socializing with a mutant, they will also be imprisoned."
Now the silence was broken. Slowly, a hum of excitement in the room built to a ragged crescendo of approval. One by one, the people of Prin started to cheer, thump the floor, and bang on the metal shelves, whistling loudly. After a while, the place was utter bedlam.
Uneasily, Rafe watched this. He stood and quickly put his arm around Caleb, making sure to share in the applause.
Only one guest was not celebrating.
The person had been standing alone by the front door and now, quietly, slipped outside into the early evening while no one else was paying any attention.
It was late when Caleb staggered out onto the main street. He was full to bursting, more sated than he had been in years.
He was also exhausted, with a heavy bone-weariness. After months of hard travel, he had reached his destination, and he had been welcomed. A good night's sleep under a roof would prepare him for what he had to do.
Caleb turned onto the deserted street where he had left his bicycle, chained to a rusted parking meter. Then he froze.
Somebody was kneeling next to his bike.
Even from the back, Caleb could tell it was a young boy, small and slight, wearing a red sweatshirt with the hood drawn around his head. Gloves on his hands, he was slashing at the back tire with some instrument.
Caleb tackled the boy from behind. Putting him in a chokehold, he dragged him away from the bike. The vandal was struggling, flailing with his free hand—he was striking out with his weapon, an ugly piece of broken glass—but Caleb was able to shake it loose, then kick it away with his boot.
The two struggled in near silence—Caleb trying to subdue the boy, who continued to fight wildly, despite the obvious difference in size and strength. Finally, the smaller one managed to twist his head into the crook of Caleb's arm while seizing his thumb and yanking backward; with a cry of pain, Caleb loosened his grip and the other slithered out of his grasp, his hood ripping. The two faced each other, the boy still choking for breath, massaging his throat.
Only it wasn't a boy. It was a girl.
The girl who had been spying on him near the hoop on the pole, behind the building.
And she looked furious.
In truth, Esther was angriest with herself: annoyed that she was inattentive enough to get caught before she could even begin her task, much less finish it. She cursed herself and shot a quick glance at the piece of glass, lying a few feet away. However, the stranger caught her look and made it there first. He brought his heel down on it, smashing it with a dull crunch.
He had been staring at Esther the whole time with an unreadable expression. This, more than the fact that he had nearly strangled her, made her deeply uncomfortable.
"I seen you before," he said. "Behind that building on the edge of town?"
She returned the stare; then nodded defiantly.
"What were you doing?" he asked, indicating his bicycle.
"What do you think?" Her tone was derisive.
The stranger nodded, as if in agreement. "Why?"
He didn't sound angry or sarcastic. He asked as if he was curious about her reasons.
Esther started to reply, then stopped, confused. She had never been asked to explain herself before and now found it difficult to find the exact words.
"To stop you," was all she could say.
The stranger was kneeling, inspecting his tires for any damage. At the sight, Esther flushed with a familiar surge of resentment. Like her sister, like most of the others in town, he was ignoring her, she assumed, because she was too childish and emotional to be worthy of his attention.
But she was wrong.
"So you heard what I said in there?" He did not look at her, but seemed as if he was addressing the bicycle.
As Esther hesitated, he glanced up. She nodded.
"And I take it you didn't like any of it?"
Her face flushed with anger.
"The variants got enough troubles without you giving them more," she muttered.
The rays of the setting sun hit his face, throwing its angles into deep relief and turning his eyes into live coals. In an instant, he looked older than anyone on earth, older than anyone could possibly grow to be.
"Variants," he said.
He nearly spit the word, and Esther was unnerved by the depth of loathing that lay beneath it.
"Why do you hate them so much?" she asked. It was an honest question, more bewildered than angry. "My best friend is one and she's a good person. How can you hate someone you don't even know?"
The boy seemed taken aback by her question. _Had anyone ever asked him before?_ Esther wondered. Then he spoke as openly as she had.
"I had a partner and baby son," he said. "In a town a ways from here. One morning, I was out foraging for supplies. Mutants broke in. They killed my partner, Miri, cut her up so bad I couldn't recognize her. They burned our place to the ground. And they took our son. Kai."
Protests bubbled up in Esther's throat. Before she could speak, he continued.
"One got left behind," he said. "He was badly burned, and the others just ran away. I beat him but he couldn't tell me much. I found an empty can of accelerant, the stuff that makes a fire burn faster. Able Accelerant, that was the name. The mutant said they got it around these parts, that's all he knew. That was the last thing he said."
The last rays of sun had turned the sky as red as blood.
"That's why it's no good trying to stop me," he said. He spoke as if he had no choice. And yet, he seemed to hesitate, as if waiting for a response.
_Did he want her to stop him?_ Esther wondered. _To talk him out of it?_ For a moment she thought she had a glimpse of who he really was beneath his hatred and anger. In his own way, maybe he was as hurt and isolated as she was.
Before she could reply, the stranger mounted his bicycle and disappeared into the night.
Watching him go, Esther felt torn. His story must be true. The ghastly murder of his partner and the abduction of his child: it would be impossible to invent such horror. His pain and grief were as searing as a fresh wound, and part of her wanted to run after him, to reach out to him somehow, and comfort him.
At the same time, she believed he must have been mistaken. Obviously someone else had destroyed his family and stolen his son; some unknown variety of human monster and not the variants. The variants had no reason to kill and destroy. They may have faced difficulties and hardships, but they were better than the others because they did not covet. They did not need anything from anyone. It couldn't have been them.
Yet why would the stranger lie? His words had stirred confusion that she found hard to admit, even to herself. He had made her face the one question she had never asked herself, despite the mounting violence . . .
_Why were the variants attacking Prin?_
Esther heard a sound behind her and turned to see that the last of the townspeople were leaving the supermarket. Compared to the heavy spirits earlier in the afternoon, the mood now seemed lighthearted, even festive. Looking at the smiling, chattering faces of her neighbors, Esther felt sick. She realized with a fresh shock what impact the stranger's words would have on life in Prin.
_All mutants will be attacked on sight, and attacked hard._
_Any survivors will be imprisoned._
_Anyone caught socializing with mutants will also be imprisoned._
For a moment, Esther felt dizzy. Then she gathered herself and made up her mind. No matter what doubts the stranger had instilled in her, there were more pressing matters at hand. She must warn Skar, before she came to town as usual. She had to save her friend.
But how?
It was late at night. What was more, the variants lived many miles away to the north, in the mountainous region. Esther owned no bicycle and to walk there would take more than a day.
She wheeled around, desperate.
Several people were walking toward her, indistinguishable in their white robes. Yet she recognized one of the voices.
"Where are you going?" Eli called. He sounded so jovial.
She couldn't respond. Even if she could trust him, which she couldn't, she had no way to put into words how she felt. But it did not matter, for he was not really waiting for her reply.
"Were you at the meeting?" He was as excited as a little boy. Caleb's words had given him hope and now, grotesquely, he wanted to share that hope with her.
In an instant, Esther realized what she must do. It would again require manipulating Eli, playing off his interest in her. She had done so before, when she had appealed to him wordlessly and he had understood, leading the rest of the Harvesting team away. She felt a twinge of guilt and also wondered, fleetingly, when she would have to repay the growing list of favors he had done for her.
She would worry about all that later.
"Can I borrow your bicycle?" she asked. "Please?"
Esther leaned over the handlebars, riding swiftly.
On the outskirts of Prin, she passed mountains of rubble that had once been restaurants, a shopping center, a block of offices. Behind her, the floodlights of the Source emitted a soft glow that lingered for what seemed like miles. But soon it was dark, and then darker still. Esther had only the moon and stars to light her way.
She rode along what had once been another highway, steering around abandoned cars and trucks, sodden piles of leaves and old clothing, crumpled road signs that dangled overhead from bent steel poles. Several times, she was forced to dismount and walk her bicycle around gaping crevices where the road had sheared away. Occasionally, she heard the mysterious cries of unseen animals and noticed flittering shapes that darted through the inky air. Once, a hulking form lumbered across the road ahead of her. But they did not slow her down.
Esther's mind was whirling.
_She had to warn Skar. She would need to warn all of the variants of the stranger's arrival and the harsh new laws now in effect. For their safety, they all needed to steer clear of Prin._
_But would they believe her? They might accuse her of being a spy, or being deliberately sent with false rumors._
After several hours, Esther paused by the side of the road to get her bearings. To one side, visible through the trees, glittered the shoreline of what used to be a vast lake. A good portion of it had dried up, exposing the parched land underneath, the skeletons of fish and birds it had digested, the occasional fiberglass cooler or hamper, destroyed. The rest of the lake was covered with a black, oozing substance as thick as a tarp and as shiny as glass. In the distance she saw a cluster of foothills surrounding a single tall peak. This was her destination.
It was nearly dawn.
Esther had been traveling for hours now and each downward stroke of the pedal was agony; her entire body trembled with exhaustion. Yet she was encouraged by the fact that although the hardest terrain was ahead, she was nearly there.
Esther glided up the exit ramp off the highway to a lesser road, and then another after that. She had only been this way once before, and that was several years ago. As a result, she made a few wrong turns.
Eventually, however, she found what she was looking for. Esther turned off the paved surface and onto a rough dirt trail that cut through the densely forested mountainside. It was steep and rocky; after several minutes, she was forced to dismount and proceed on foot, pushing the bicycle by its handlebars. She reached a withered tree with a white mark upon it. There she turned. The trail wound a bit more until it ended at a clearing, carved out of a plateau.
This was where the variants lived.
Esther had not planned to arrive at dawn, but she realized it was a fortunate coincidence. Early day was hunting time for the tribe, and the camp seemed deserted. If Skar was around, they would be able to talk in private.
From her hiding place, she softly gave their secret whistle and waited. Within moments, someone emerged from one of the many shacks grouped across the clearing. It was Skar, who glanced around, clearly puzzled. Then she noticed Esther.
Surprised yet delighted to see her, Skar ran to her friend and gave her a hug. She smiled, her parted lips revealing her little teeth.
"Esther!" she exclaimed. "I can't believe it's you! Why are you here?"
But in her haste to warn her friend and tell her all she knew, Esther had paid scant attention to her surroundings. Now, she was aware that something had changed. She stopped talking and stood still, gazing around.
When she was here before, it had only been a brief visit. At the time, she was met with suspicion by the few variants Skar introduced her to, and so she didn't stay long. Yet she remembered what it looked like. There were makeshift shacks made of animal skins, salvaged planks, and saplings. In the center of the clearing were smoking vestiges of stick fires. Bones and other uneaten bits of animals had been strewn about, no longer recognizable.
But now, while the shacks were still there, there were no fires. Instead, Esther noticed what had taken their place.
There were large cardboard crates piled by each tent, each with crisp black lettering that Esther had trouble reading. As she looked around, her unbelieving eyes picked up other details, items that did not belong here and therefore made no sense: a clothesline pinned with dozens of shirts, pants, and dresses in bright colors and sturdy fabric. New shoes—sneakers, boots, sandals—lined up outside each door. Shiny kettles and cooking pots of all sizes. And under a canopy made from a rubberized tarp was a giant pyramid of food: oversize packages of dried beans, sacks of flour, plastic gallon jugs of water.
"What?" said Skar, puzzled. "What's wrong?"
Esther couldn't speak. Instead, she pointed to the food, the clothing, the crates.
"What—what is all this?" she said.
"This?" Skar turned and looked. Then she said, innocently, "It's food! You know, and other stuff!"
Esther looked closely at her friend now. At the base of Skar's ears and hanging around her tattooed throat and wrists were new and shiny pieces of jewelry, colorful stones and bright metals. She had never seen Skar—or any variant—adorn herself like this.
"And where did you get _this_?" she said, flicking at the bangles.
Skar touched her ears and throat, growing self-conscious and her smile less confident. "Well . . . from the Source. Like the rest of it." She gestured at the boxes as if in confused apology.
Esther nodded, very slowly.
Her mind was whirling. What did this connection, this alliance mean? The variants did not, of course, Harvest gasoline, nor was there much left to collect even if they did. So what had the variants exchanged with Levi for this massive payment of goods? What had he wanted from them? What had they done to earn it?
The sun was higher in the morning sky; the heat began to beat down. Esther had forgotten to wear her sunglasses and was forced to hold up a hand, to protect her sight. Soon, she had to shut her eyes.
All she could see was Caleb's face.
## PART TWO
## SEVEN
ALTHOUGH IT WAS MORNING, THE SUN BURNED WHITE HOT IN THE DIRTY yellow sky. Yet inside the Source, it was perpetual twilight, dark and cool.
To Slayd, the interior of the gigantic white building always felt like a cave . . . and he did not care for caves at all. They were damp, unwholesome places, dappled with pockets of darkness that harbored all that was unnatural, possibly deadly, and to him, disgusting. In caves, he had seen oversize spiders, patches of mottled mold growing on wet rock, snakes with pinprick eyes and pale skin, and mice that fluttered through the air with leathery wings. His skin crawled at the thought.
In many ways, Slayd felt the same way about Levi.
Although technically a norm, Levi resembled no one the variant leader had ever seen before. He was more a cave-dwelling animal than an actual human, with his black eyes, his dandified black clothing, and silver jewelry. His skin was so pale, it seemed to glow in the gloom, and it emitted a sharp and musky smell that turned Slayd's stomach.
More disturbing, Levi's skin was soft; even his bones seemed soft, revoltingly so. It was almost more than the variant could bear just to look at him, much less grasp his hand in greeting.
The two had been sitting across from one another in Levi's office, a large, trembling room with wire walls that the boy called a "freight elevator." A single electric light overhead threw deep shadows into the surrounding cavernous space. Perhaps because Slayd was the one to request this meeting, Levi kept the variant leader waiting for nearly an hour and now seated him on a smaller, inferior chair that was dwarfed by the massive desk separating them. Still, Levi continued to delay, appearing to examine some papers on his desk, a white cloth pressed to his mouth.
Slayd was keenly aware of these deliberate slights. Yet rather than be angered by such rudeness, he knew enough to hold his temper and stay watchful instead. It was clear that Levi was doing it on purpose, to trigger some sort of emotional response from him, throw him off balance. He wouldn't give him that satisfaction.
Yet even though he sat in silence, Slayd had already made a mental note of the precise location of the five guards that surrounded them. Should the situation deteriorate, he had calculated the quickest way to escape.
When Levi finally looked up, Slayd wasted no time and got to the point.
"I'm here to request the rest of our payment," he said. "For the latest action. The one with the smaller band."
"I heard about it," Levi said. His voice was polite, almost bland.
"We've been waiting for another shipment," said Slayd, his tone as even as that of his host. Whatever game Levi was playing, the variant was more than prepared to meet him. "What arrived was less than we agreed to."
"What you did wasn't worth the full payment," Levi replied.
He was finding it difficult to look at the mutant leader. As always, Levi found everything about Slayd—his deformed features, his scarred and tattooed skin, his small and pointed teeth—freakish and repellent. He couldn't bear his sexless quality. Slayd called himself male, yet looked no different from the so-called females of his tribe: hairless, smooth-faced, and slight of build. Worst of all was his smell, which was sharp and acrid like an animal, with a tendency to linger long after he had gone. Once again, Levi pressed the cologne-sprinkled handkerchief to his face and inhaled.
Slayd was nodding. Then he bowed his head and spread his hands in an obsequious manner that Levi did not believe for an instant. "We did as you requested," he said. "We escalated the violence."
"I'm not talking about what I requested," replied Levi. He realized his tone was harsher than he intended, revealing too much; he softened it. "I'm talking about the stranger. I understand he defeated you and four of your best single-handed."
"Ah," said the mutant leader. Again, his air of polite apology seemed false to Levi. "But we did not know he was coming."
At this, a slight frown creased Levi's forehead. "Even so, I'm surprised," he said. "Five against one? I can't imagine that should be so hard to handle."
Slayd shrugged. "My people and I had specific instructions, and those instructions did not include taking on another. Especially one who turned out to be so skilled a warrior."
"But everyone knows you people are the best fighters," said Levi, persisting.
Again, Slayd shrugged. "That may be," he said. "All I can suggest is that perhaps my people might work harder in the future if they were paid the full amount. And maybe even a bit more."
Levi now rocked back in his chair, silent.
The variant watched him, making sure his expression gave nothing away. If Levi was changing the terms of their agreement, then he would counter and change them too. There was, he thought, no harm trying.
"Well," said Levi after a moment. "That's certainly a conversation we could have further down the line."
Slayd frowned, annoyed by this evasion. "We are—" he began, but Levi cut him off.
"But the truth is, I'd only consider increasing your pay if you people managed to do a better job," he said.
Slayd felt his face flush with annoyance.
"I told you, we did everything that was asked," he said, his control slipping. "Two of my people were seriously injured as a result. If you pay us more goods, perhaps it would _begin_ to make up for the loss to my tribe. It would certainly not cover what we have lost in goodwill with the people of Prin by attacking them, the reasons for which you never once explained. That alone is worth an increase."
Levi didn't even look at Slayd now, finding that his patience was wearing thin. _Explain?_ he thought, with disbelief. _It would be like justifying yourself to a dog._ Instead, he ignored the remark.
"I am not only talking about the _recent_ attack," said Levi. "It's everything. What about that other job from before? The job I asked you to do far from Prin?"
But Slayd was shaking his head. "Why do you mention that now? We brought you the child," he said. "We fulfilled our end of the deal. What else was required?"
"You were supposed to kill _both_ parents," replied Levi. "You told me the father survived. I'm surprised, Slayd. I thought you people were capable of handling such a simple job."
The variant leader smirked.
"We are," he replied. "But it was not my people who carried it out."
"I don't understand," said Levi.
Now Slayd was grinning, relishing the look of confusion on the norm's face. "It was another tribe. They made the mistake. Not my people."
"But—" began Levi, and this time, he was interrupted by the variant.
"I hired them," Slayd said. This time, he did not bother to hide not only his triumph, but also his anger—anger at Levi's rudeness, his condescension, and presumed superiority. "It was too far for my people, too much trouble. Not worth what you offered."
"And you paid these others . . . out of the fee I paid you?" said Levi, his voice rising. He, too, had dropped the veneer of politeness, the pretense of civilization; he was openly furious. "You dare to attempt profiting from the jobs I give you by hiring others?"
"Profit?" The mutant seemed to spit out the word. "When my people are starving? You dare to call that a profit?"
Incensed, Levi was about to rise and call for his guards. But with the remarkable self-restraint that had served him for so many years, he instead remained motionless.
Levi realized he was foolish to respond emotionally to what was a business disagreement. True, any norm alive would be angered by the effrontery of the savage in front of him. Such arrogance was unacceptable and at some point, Levi would make certain to pay it back, harshly and many times over.
But not quite yet.
As much as it pained him to admit it, Levi still needed the mutant leader. Since the Source had started running out of food and water, Slayd and his tribe had been critical in helping Levi carry out his plan. He had to drive the people out of Prin. If the residents believed they were making the decision to leave themselves, it would lead to a cleaner and simpler transition than if he were to try using force. With control over an endless supply of water, Levi would then be sole owner and occupant of the town. All he needed was for Sarah to bring him the missing book, which would tell him exactly where to dig. Until she did, Levi would have to endure Slayd's insolence.
Levi was aware that Slayd was watching him and so he forced himself to smile. Then he chuckled, as if enjoying the punchline to a good joke. At this, the mutant visibly relaxed, and in doing so, missed the involuntary twitch in the norm's jaw.
"Of course," Levi said, "perhaps I should be taking all of this as a compliment. You seem to have picked up a few of my tricks, Slayd. Why shouldn't you hire others to do your dirty work for you?"
Slayd inclined his head in acknowledgment. "Any comparison to you is a compliment we do not deserve," he murmured. "But as to the subject of our payment. May we possibly assume . . . ?"
Levi nodded. "The balance will be paid in full as soon as you leave. With an extra half case of water thrown in."
He noticed with distaste that even though the mutant leader kept his gaze lowered, he couldn't control his jubilation. Slayd was grinning openly. Now he got to his feet, his hand extended, but Levi remained where he was, his elbows propped up on his desk and his fingers steepled.
"All of this is on one condition," Levi said. "Will you and your people be ready for another excursion soon?"
"Certainly," said Slayd. "Can we also assume . . . ?"
"I will raise the fee," said Levi after a moment. "One half case of clothing, one of grain, and one of water." He watched as a look of stunned happiness crossed the mutant's animal face. The effect was both grotesque and comical. "But this one must be special," Levi added. "I want you to use something different than the usual clubs and stones. This attack needs to be much more . . ." He traced something ineffable in the air with his pale hand and let the sentence hang, unfinished. "Do you understand what I mean?"
The mutant smiled. Then the two shook hands.
Slayd was escorted to the door. He was given back his knife, his bicycle. After checking to see that no norms were nearby, he pedaled back toward the variant camp miles away. Jubilant at the thought of the extra payment, he relished his victory over the clever and arrogant Levi. It would make for a good story to tell to the village elders that night, he thought.
Back at the Source, no one saw him go. Not even Levi watched from his hidden window.
He was too busy calculating his costs. True, he did not anticipate the bonus he had just promised to Slayd. Even so, in the long run, the terror he had purchased with a handshake would be a bargain even if it were two or three times the price paid. For fear was like fire, a powerful force that could sweep unchecked through a town and drive everything living from its path.
And if all worked out as planned, that was exactly what was going to happen.
The plume of smoke rose almost imperceptibly in the midday sky. Without even looking, Esther knew it was there. But instead, she pedaled harder and tried to keep her eyes trained on Sarah, who rode her purple bicycle in front of her.
That morning, using a combination of guilt and begging, her sister had managed to talk Esther into taking part in the Harvesting they had both been assigned to. She even managed to find a bicycle for her. Resentful at having to work at all, Esther was nevertheless aware that she was down to her last warning, and any more work violations would result in an automatic Shunning.
"Hurry up," Sarah called over her shoulder, from far ahead. Her voice was anxious. "You're going too slow."
But Esther found it difficult to ignore the signals, which had been coming all day. They had begun early in the morning and at least in Esther's eyes, had become more and more insistent, reproachful. Helplessly, she peeked upward. Although she knew it was impossible, she was sure she could smell the far-off smoke, the pervasive scent of damp and rotted pine branches tossed onto a fire.
It smelled like a rebuke.
There had never been a day when Esther had not scanned the horizon for such signs. Long before the recent tensions in town and the growing ugliness between townspeople and variants, the secret code was how she and Skar had always communicated.
The signals were few and simple, meant to convey only the most basic and crucial information:
_Meet me now. I am returning to my home. All is well. I need to speak with you. The situation is urgent._
But Esther's surprise visit to the variant camp had changed how she felt about her friend. She did not know why. All morning, she had been struggling to sort her jumbled thoughts about seeing the goods from the Source and make sense of her churning emotions.
Skar had little to say when pressed for information. She had always been this way, the kind of person who bent to authority and accepted what was going on around her without doubt or question. Unable to give any satisfactory answers to Esther's questions, she instead tried to placate her friend and change the subject, which only made Esther angrier. It was the first time the two girls had ever quarreled or parted on bad terms.
Even now, Esther couldn't stop thinking about it.
By now, she had lagged far behind the others, despite Sarah's best efforts to shepherd her. She bicycled hard and soon caught up with the group. Besides her sister and herself, there were three others on the Harvesting team, all girls a year or so older than Esther. One of the girls, thin and haughty, was named Rhea; she was the team's Supervisor. When Esther joined them, panting, Rhea glanced at the others and raised an eyebrow, and everyone laughed.
Sarah, blushing furiously, gestured at Esther to stand near her.
"Where were you?" she hissed. Esther only shrugged.
Today's destination appeared to be what was once a large field that lay to the side of the highway. Over the years, the sun had hardened the land, which was now covered with an intricate network of fissures and cracks. Strange pools of relatively clean, white sand were scattered across the field at intervals. The remains of a large building, once resplendent, sagged in the distance, past a broken sign reading SKYVIEW LINKS. A windowless structure, no more than a large metal shed, stood closer to the highway. Its doors were held fast with chains and heavy locks.
"In here," said Rhea, nodding at the shed.
The shack was most likely a garage, the kind of structure that housed cars, motorcycles, and other gas-filled vehicles. Judging from the heavy scuff marks on the doors and the locks themselves, it was obvious that others had tried here without success. But today, the team had brought a crowbar with them. After repeated efforts by all five, they succeeded in smashing open the locks.
Inside, the team found a row of boxlike vehicles. They were not much bigger than bicycles, only with four wheels, and were clearly meant to carry two passengers on their cracked leather seats. The side of each vehicle contained a rusted metal cap.
Elated, Rhea and her team tried to unscrew the caps in order to get to the gas inside; but the job was harder than they expected. And even once they managed to pry them off, it turned out that the tanks were nearly empty. For all of their time and effort, they collected no more than half a bottle's worth of gas.
Throughout, Esther attempted to participate. She dutifully took her turn with the crowbar, tried to open the tanks, and helped coil away the rubber tubing once the small amount of gas had been Harvested. But her mind was not on it.
"Try to be friendly," Sarah implored her in a whisper. Their work done for now, the team was on a break, sitting in a loose circle in the shade of an abandoned truck in their dusty robes and eating the meager lunch they had brought. The air was heavy with humidity, a sure sign of an oncoming storm. "They're not so bad. Try talking to them." But Esther made a face.
"About what?" she whispered back.
Sarah shook her head hopelessly. Then she turned back to the others and made a great show of listening as she laughed and nodded.
Esther couldn't understand why her sister bothered. It was apparent the three others had little use for Sarah and even less for Esther. Not that she minded; as far as she could tell, their conversation was worthless, less interesting than the droning of bees. One girl boasted about her recipe for wheat porridge. Another described a tattered bedspread stolen from a recent Gleaning and how it matched her one curtain. And then there were the endless, tedious anecdotes about their men, for all three were partnered.
When the gossip turned to partnerings, Rhea pointedly leaned close to the other two girls and whispered. After a moment, the three shared a harsh laugh, glancing sideways at Sarah. The older girl acted as if she was in on the joke, smiling and bobbing her head, even though it was clearly at her expense.
At seventeen, Sarah was an old maid, long past the age of partnering. What made it odd was that she had never attempted to find a partner, despite the fact that over the years, many boys in town had expressed interest in her. Rafe in particular pursued her, to no avail. It seemed as if her sister had been waiting for someone special, Esther thought. But who?
The girls' chatter became faint as Esther tuned it out. In its place, she heard someone else's voice: Caleb's.
In her mind's eye, Esther could see his face, the set of his jawline, the haunted expression in his dark eyes. And although she despised what he had said at the town meeting, she now realized things might not be as simple as she thought. She also remembered how he spoke to her afterward, directly and openly, and how he listened to her, _really_ listened in a way no one ever had before, not even Skar.
The inane drone of the girls' voices cut into her thoughts. Esther was jerked back to reality and with it came the realization: _She didn't want to be here anymore._ She looked up at the sky, where her friend's smoke signals had been. The clouds had thickened and grown darker.
She made up her mind and stood.
"Esther?"
She was already across the small parking lot and hoisting her bicycle by the handlebars when her sister grabbed her by the arm.
"What are you doing?" Sarah whispered. She sounded panicked.
"I have to get out of here," said Esther. But her sister refused to let go.
"Please," she said in a low voice. "Just a few more hours, until the rain passes. I promise, once we're in the shed, I'll keep them away from you. But if you go now, it'll be over for you. I won't be able to save you."
Esther attempted to shake her off. "It's never as bad as you say it's going to be. Shunning's only for people who are sick or for _real_ criminals." She had one foot on a pedal and was trying to take off.
"But that was before. These are Levi's new rules. And you know there's no way that Rhea isn't going to report this. She's been waiting for the opportunity all day."
Esther glanced over at the other girls. They were still sitting where they were, watching her with their mouths open in shock. And it was true that Rhea was staring at her with an appraising look, a faint smile on her face.
"Please, Esther." Although she kept her voice down, Sarah couldn't keep the desperation from her voice. "You're going to get Shunned. And no one will be able to help you."
But Esther had broken free and was pedaling away, as fast as she could. The town was five or six miles away. She would have to hurry before the rain came.
## EIGHT
"HIT ME," CALEB SAID.
He stood in front of a red-haired boy, with one arm extended and relaxed, palm facing out. The boy was a husky fourteen-year-old, stocky and exceptionally strong; he figured it was the reason he was chosen. Eager to prove himself, he tensed up his arm and punched the open hand as hard as he could.
The boy was surprised and then embarrassed to see what little effect it had. The stranger barely registered the blow. He was about to ask for a second chance, but Caleb had moved on to the next person in line, a sturdy girl with close-set eyes.
"Hit me," he said to her.
Seven townspeople were lined up in the large, echoing room that had once been a bank. They had been excused from their various jobs for this first round of training and now stood in the thick heat and humidity of the November day, their arms by their sides, awaiting instruction from the stranger who was going to teach them how to save their town.
The red-haired boy was especially excited to be included in the first group. Like everyone in town, he was familiar with the details of the stranger's victory over the five mutants. He knew of his impressive fighting skills and his strange new weapon, which was capable of firing several rocks in quick succession.
The boy looked forward not only to learning from Caleb firsthand, but maybe even following him into battle. He and his partner had sustained serious damage to their storefront home in the recent mutant attack, their windows smashed and much of their stored goods destroyed. Since then, he had been hungering for revenge. Today, he had come half expecting to be handed his own weapon, given instructions on how to use it, and maybe even led to the mutant camp for some kind of showdown.
But he had been surprised. So far, the lesson was nearly all talk. What's more, most of what the stranger had to say was downright bewildering.
"I can't teach you how to fight," Caleb said at the very beginning. At this, everyone shifted on their feet, glancing at one another and murmuring. "Fighting isn't in your hands or your feet, and it isn't about getting hold of some fancy weapon. Mostly, it's in your head."
The boy with the red hair wiped sweat off his brow as he mulled over these peculiar words.
By now, Caleb had worked his way to the end of the line. Everyone had punched or slapped his hand—some harder than others, some less eagerly, some clumsily. The boy brightened up at this part of the lesson; this was what he had come to do. He assumed Caleb would now get down to business, would talk about the techniques of hitting and fighting and pick out and praise the strongest participants. Maybe he'd even spar with the best student and again, the boy felt his hopes rise.
But once more, he was surprised.
"Fighting isn't a game," Caleb said. "You should only do it because you have no other choice. And you've got to know that your enemies aren't just stronger than you. They're smarter, too."
The boy frowned. He was not quite sure he followed what Caleb was saying. He was also not sure he liked the sound of what he was hearing.
"To win, you've got to keep your mind clear," continued Caleb. "You've got to see the situation as it is and use every advantage you got. But you can't keep a cool head if you put your feelings into your fists."
He turned to the red-haired boy, who was now examining his hands. "For instance," he said, "I could tell by the way you hit that you're impatient and you want to fight." Caleb imitated the boy perfectly, his eager stance, the overly enthusiastic punch. "You want to prove to me and everyone here that you're strong."
There was suppressed laughter down the line and the boy frowned, trying to understand what had been said and if he had just been insulted. But before he could say anything, Caleb had moved on to the next person in line, the girl with the close-set eyes.
"You think this is some kind of game," he said to her. "It's like you don't even think the threat is real." The girl giggled, then blushed, staring at the floor.
Caleb moved to the next person. "The way you hit tells me you're mad, maybe at me," he said to the boy, a hulking sixteen-year-old. "You don't like being told what to do." The boy looked startled; then he glowered at the stranger, his fists clenched.
Caleb continued to work his way down the line. He stopped in front of each person and told each one what he thought he or she was feeling:
_You're scared. You think you know better. You care too much about pleasing others. You're bored._
When he was done, Caleb turned to face everyone, his expression serious. "Think about what I said," he said. "Try to leave your emotions at home. And I'll see you back here tomorrow."
At first, the red-haired boy darkened with anger. But when he thought it over, he was astonished. It was amazing that the stranger could know so much about him by just a single punch to the hand.
By the look on everyone else's face, he knew he was not alone. Feeling a first glimmer of understanding, he stepped forward to speak his mind.
But he was stopped in his tracks.
Without warning, a gust of wind swept through the broken windows that surrounded them, swirling grit and paper across the room. Everyone simultaneously glanced outside.
Overhead, the sky had changed to a deep and unnatural green and purple. Then, it seemed as if all the air was violently sucked out of the room; shards of broken glass rattled in their wooden frames, some snapping off and sailing into the street.
Caleb moved deeper into the room and everyone followed. They stood against the back wall in order to get as far away as possible from the gaping windows and open door.
Even if the red-haired boy were to speak now, no one could hear him. For with a deafening crack of thunder, a bolt of lightning split the darkening sky.
A moment later, rain began to fall: fat drops freely splashed through the broken windows, forming puddles on the marble floor. Outside, the drops marked the dusty ground vividly, faster and faster. They covered the hardened dirt with dark spots before converging and turning into deadly pools of mud and water.
Half an hour later, the storm was still raging. Looking onto a deserted street, Esther watched the steady downpour from the decrepit lobby of Joseph's home, the Gideon Putnam Hotel.
When the first drops had begun to fall, she leaped off her bike and wheeled it into the nearest building, thankful for any shelter at all. Even so, she was aware of the heavy sound of rain as it thrummed on the sidewalk and splashed through the gaping windows and doors, soaking the faded carpet. She moved deeper into the building interior, making certain to avoid the walls, which had begun to weep moisture.
She berated herself again for not thinking, not planning.
It was a stupidly close call. Moments earlier, an unlucky gust of wind could have driven the downpour straight at her, through the broken glass of the front door. She knew all too well what a single raindrop could do if it found its way into your eyes, your mouth, or a scratch on your skin that hadn't healed. First came the bone-crushing fatigue and telltale lesions; then headaches and fever. These were followed by severe stomach pains, vomiting, and delirium.
After that, she was not exactly sure what happened. For no one had ever been allowed to stay in town once the symptoms appeared.
Esther had been on her way to the school, where she knew Caleb was staying, when the storm hit. Now she had to wait until it was totally spent before she dared to continue on her way.
Whenever it rained, the people of Prin pressed close to their windows and watched the storm. They couldn't help it: from the safety of their homes, they found the risk, the presence of death, fascinating.
But not Esther. As usual, she turned her face from it.
That was when she saw Joseph.
Her friend was standing across the lobby, carrying an empty plastic bucket. As ever, he was accompanied by a cat. Both boy and feline stared at the intruder with a look of astonishment.
"Esther," Joseph said.
Esther felt a pang of guilt. So much had happened recently that she hadn't told him. Now she had only run into him by accident. She noticed what he was carrying.
"You're not going to get rainwater with that, are you?" she exclaimed.
"With what?" He looked down at the bucket. "Oh, no, I—"
"Because it can kill you," she said. "Do you remember what I told you? Do you still have any of the water I gave you last time?"
"Yes. Yes. I do."
"Then will you please drink that instead?"
As she watched Joseph first attempt to hide the bucket behind his back and then a column, Esther realized too late that she had spoken in a sharp voice. _I must sound like Sarah,_ she thought. "I'm sorry I shouted," she said, touching him on the arm. He felt thinner than usual, and so she dug deep in her shoulder bag. At the bottom, she found what she was looking for: the lunch she had not eaten at the Harvesting, a container of boiled rice and beans. "Take this."
"Are you sure you want to—"
"Yes. Please."
Smiling his thanks, Joseph received the gift. Then he placed it on the floor in front of the cat, which began to eat.
Esther watched for a moment. "I wish I had more."
Joseph shrugged, then shuffled his feet. "Would you like to come upstairs? We can have a proper visit."
"I can't." Esther spoke with real regret. "Once the rain stops, I've got to see someone. I'm sorry."
She turned to check the progress of the storm and was startled to make out her image reflected back to her in the cracked glass door.
Esther leaned forward and examined herself. She squinted, trying to imagine that she was seeing herself for the first time, as if she was a total stranger.
As if she was someone like Caleb, for example.
Esther had never done this sort of thing before. There was a full-length mirror at home, but she almost never glanced at it. In fact, she associated primping and fussing with Sarah, so much so that not caring about her looks had become not just a matter of pride, but an easy way to irritate her sister. She was amused by how agitated she could make Sarah by something as simple as not combing her hair.
But now that Esther was studying herself, she was rattled by what she saw.
She saw a girl in boy's clothes—jeans and a sweatshirt—that hung off a bony frame; she saw watchful eyes that seemed too large and dark in a thin face. There was a smudge of dirt on her chin, which she tried to rub away with her sleeve. Her hair, dark and unruly, was cut unevenly, at different lengths, and it stuck up on top. Esther frowned and tried to smooth the cowlick down; it wouldn't obey and she gave up.
Then she turned sideways and tried to examine her figure, pulling her sweatshirt close.
It was no good, she realized with a sinking heart. She was simply not appealing, not the way other females in town were. She lacked the curves and softness of some of the girls, the gracefulness of others, even the dainty femininity of her sister.
For a moment, Esther stared at her reflection in the glass and despaired. Then she turned to her friend, who had been watching her with a bemused look on his face.
"Do you think I'm pretty?"
Joseph started, then seemed to consider the question. After a few moments, he looked up. "You're Esther," he replied.
Esther smiled. Although at that moment, she would have given anything, anything at all, to change her looks, she realized that there was nothing she could do about it. She couldn't, after all, change who she was.
She walked over and kissed Joseph on the cheek. He recoiled, as she knew he would, but she didn't care.
At last, the rain started to let up. Esther waited until she was sure it wasn't a false alarm. Then she saw a rainbow—the indisputable sign that the coast would soon be clear—stretch across the sky.
"I'll see you soon," she said.
After a final, vain attempt to make her hair lie in place, she wheeled her bicycle out from the hotel lobby. Her hood drawn around her face, Esther took off through the glistening streets for the school, a half mile away.
There were many shattered windows on the ground floor of the building. It was no trouble for her to reach in, unfasten the latch, and enter.
She made her way down a hallway, lined on both sides with dusty and dented metal lockers that gaped open. She picked her way through trash, mounds of paper, and broken light fixtures. Along the way, she passed empty classrooms, rusted water fountains, and abandoned stairwells.
When she rounded a corner, she noticed something written on the wall, and curious, stopped to examine what it was. Primitive drawings and words, little pictures of hearts with arrows through them, and initials were carved into the plaster. She was able to spell out the words and letters with difficulty:
_mikey + lissa. e.h. + a.t. j-bo and k.k. 4ever._
They made no sense to her.
Caleb sat on the creaky cot in his room. He had been given these accommodations in the school, a dank, gray two-story building, as a reward. When Rafe first showed him the place, he assumed that the stranger would take the largest room, the auditorium, for his lodgings and had it furnished accordingly. By Prin's standards, such a dwelling—with its high ceiling, scuffed wooden floors, and tall windows covered with thick wire mesh—was luxurious, even palatial.
But after living outdoors for so many months, Caleb no longer trusted open spaces. Instead, he thanked Rafe, whom he was beginning to find irritating and overbearing. Then once he was alone, he searched the building until he found a room more to his liking: a classroom off a secondary hallway, with dusty blackboards still attached to the walls and desks and chairs pushed to one side in a jumble. Satisfied with its size and location, Caleb transferred all of the furnishings and supplies Rafe had the townspeople provide.
He thought about those people and his students, as well. He had taken the job for practical reasons only, as a way to stay in town. But he found he liked the teaching more than he expected.
On his rickety bed, Caleb drank from a plastic jug of water. Lowering the bottle, he glanced around and for the first time took in where he was.
The tables that were pushed against the wall were much too low to sit in front of; and the chairs piled on top of them were small as well, perhaps coming up to his knee. He looked up and noticed strange pictures tacked to the wall, faded, mysterious illustrations that were curled from too much humidity and mottled with mildew:
_A white goose in a bonnet read a book to a little boy and girl. A cat walked on its hind legs, wearing green boots. Three bears confronted a small girl with yellow hair._
Around the wall, close to the ceiling, were the remnants of a long strip of paper. Caleb could barely read, but he realized with a shock that the torn banner was printed with the letters of the alphabet.
This was a room for little children.
Children like Kai. Soft, sweet Kai, with his mother's serious eyes and his sudden smile.
His son.
The images seemed to reproach him, a silent reminder not to forget why he was there. Caleb squeezed his eyes shut. Then he opened them again.
Someone was in the hall.
Caleb seized his backpack, hanging over a chair. He took out a sap, a small, heavily weighted leather pouch, which he hid in his hand.
But he realized he would not need it.
A girl in a red hooded sweatshirt stood in the doorway.
It was the girl he had first seen at the ball court, the angry one who had tried to slash his tires. The pretty one, he thought now, pretty if you looked at her the right way.
"Hey," she said. She was appalled to find she was blushing and she tried to cover it by scowling. "I'm Esther."
"I'm Caleb," he said. "What can I do for you?"
Esther couldn't meet his eyes, and so she plowed ahead, staring at the floor.
"I came to . . . I wanted to say I'm sorry."
"Uh-huh," he said. "What for?"
"For . . ." she started, then trailed off. Apologizing didn't come easy to her and this was harder than she thought. "Because I messed up your tire."
Caleb considered her words.
"You didn't really mess it up," he said.
A smile flickered across her face and at last she raised her eyes. "I would have, if I had more time."
Now it was Caleb's turn to smile. "I bet," he said.
Esther cleared her throat. "And . . . I'm sorry about your family."
Caleb's face grew serious and he nodded.
That was all Esther came to say. It felt right to apologize for what she did and to express her sympathy. After that, there was no real reason to stay; yet for some reason, she couldn't break away. She lingered for a moment, hoping Caleb would speak, but he was as silent as she was. So she started to go.
"Hey?" he said.
Esther turned back.
Caleb had his hands in his pockets and averted his gaze; she was surprised to see that he was so ill at ease.
"You think it's safe around here?" he finally asked. He indicated his black bicycle, leaning against a wall. "I'd like to keep that outside, case I need to get somewhere in a hurry. Think that'd be okay?"
"Sure," Esther said. "If you want, we could put it out back. That way, nobody would see it from the street."
Together, they headed farther into the school, Caleb pushing his bicycle next to him and Esther navigating. It was not just one building but a series of them and she had never been inside before. Still, her sense of direction was good, and she felt they were heading the right way.
As they walked side by side, the two talked. Esther was especially shy at first. The only person she really spoke to was Skar, and they had been friends for many years. She found it was easier when they weren't looking at each other. Mostly, they took turns asking questions, listening as the other spoke: about growing up, their homes and family, and the people they knew.
Soon Esther was so caught up in the conversation, she stopped paying attention to where they were and began choosing turns and stairways without thinking. When they reached the end of a large hallway, she frowned. She spun around, confused, as she tried to get her bearings.
"What's wrong?" Caleb asked.
Esther didn't answer at first. "I don't know how we got here," she said. She pointed down the echoing corridor, which seemed as long and broad as a highway. "I think we're supposed to be down at that end."
Caleb smiled. "That's easy," he said. He mounted his bicycle in one fluid movement. "Hop on."
When she realized what he was proposing, Esther hesitated. Then she met his eyes and made up her mind.
His back wheels didn't have the standing pegs the variants used, so Esther perched on the seat. She held onto Caleb, who pedaled standing up. When they reached the far end of the hall, Esther saw she had been right; there was a door that led to a courtyard in back. Caleb slowed, then stopped. He took her hand and helped her off the bike.
"Thanks," she said. "That was fun." She held the door open for Caleb, and he wheeled his bicycle through and rested it against the brick wall. Esther realized it was a word she had only ever used with Skar: _fun_.
As the two returned to the classroom, Caleb seemed thoughtful.
"Who's Levi?" he said. "And what's the Source? I've heard people talk about them, but not so as I could understand."
Esther couldn't imagine anyone not knowing, but she explained as best she could. Caleb listened, squinting as he took it in.
"They got more than food and water in there?" he asked, after she was done.
"What?"
"At the Source. He's got all kinds of stuff, right?"
Esther shrugged. "I guess."
"Do they ever trade with anyone else?"
"Like who?"
"Mutants. Because I'm looking for something. Something you start fires with."
Esther was puzzled. Then she remembered his recent tragedy. The mysterious fire. The death of his partner and the kidnapping of his son.
And before she was aware what she was doing, Esther found herself opening up even more to Caleb. She told him what she'd found when she visited the variants' camp—that Levi was supplying them with goods. What they were doing in exchange for this payment, she had no clue. And she realized too late that she didn't know what Caleb would do with this information.
Caleb listened, gazing downward without speaking. Then he looked up.
"How do you get in the Source?" he asked.
"I don't know," Esther said. "I never been inside."
"You think I could?"
"Depends," she replied. "Why are you asking me?"
"Because. You know everything else."
Esther glanced up at him. She wasn't sure if he was teasing or if he valued her opinion. _Maybe both_ , she thought. _Because he was smiling._
"I can't answer that," she said. "But I think you'll end up doing what you want, anyway."
"Probably so," he said. Just then, Caleb's attention was directed to the window. He walked over and peered out. Breathing onto the filthy glass, he rubbed a circle with his elbow. Although the rain had stopped, a form covered in shiny black clothing—a hooded slicker and galoshes—was striding across the street. Whoever it was headed toward the school.
"That's one of Levi's boys," Esther said, from behind him.
"How do you know?" he said.
"No one else has gear like that."
Caleb thought about it. "Better hide someplace."
"Why?"
Caleb looked at her. "You know how to fight?"
"Oh," she said. "Well, I—"
"Wait there."
There was a closet in the back of the room. Esther opened the door and disappeared inside.
The stranger had already entered the room. It was a boy, probably in his mid-teens. His ensemble gave him a bizarre, animal-like quality, as if he was something not completely human.
"Levi wants to see you," he said.
"I'm Caleb. What's your name?"
The boy in black didn't answer; he cocked his head, confused.
"You got to come to the Source," he said.
"Why?" Caleb asked.
The visitor paused again. It was as if he had never been asked anything like this before. He seemed to be blinking stupidly behind his hood.
"Come tonight, before the sun goes down," he said.
"What if it rains again?"
But the boy had turned and, seemingly unnerved, was tramping out.
Watching him go, Caleb shook his head, amused. "Well," he said. "Looks like I'm going to the Source, after all."
"I guess so." Esther emerged from her hiding place. She was smiling, too. Then her expression grew serious. "But if you go, be careful."
The emotion in her voice surprised both of them. Before he could respond, she spoke again. "Maybe I'll see you around. I'm usually in the fields, near the tracks."
Then Esther turned and ran from the room and down the hall. In no time, she was at the front door of the building.
But before she could leave, she heard Caleb calling after her:
"See you, Esther."
Rather than bring a sense of coolness, all the rain had done was make the late afternoon heat feel more oppressive. The air was now thick and muggy and even more difficult to breathe.
Caleb shifted on his feet. He had been waiting for over an hour outside the Source, standing on the steaming asphalt of the parking lot. There was not much to look at. Weeds and tall grass grew freely in the cracked surface. Beneath his feet and stretching as far as the eye could see were fading parallel lines, painted in white. A few featured remnants of a crude drawing: a stick figure seated on a half-circle.
Caleb pushed back his hat and raised his sunglasses to wipe his face; he was perspiring freely. He was more than aware that he was being made to wait on purpose; it was an obvious ploy, Levi's way of establishing the balance of power between them before the two had even met. Yet it didn't succeed in making Caleb feel intimidated. It only made him impatient.
He was aware that there were laborers working nearby; he had passed some sort of worksite on his way in. From where he stood, he could hear the faint sound of picks and shovels hitting the ground. It was a rhythmic sound, hypnotic in this heat, and he closed his eyes, momentarily lulled.
Something snapped him out of his trance.
He blinked, not certain if he was seeing things. There seemed to be an apparition emerging from the Source. And it was heading his way.
From where he stood, it first appeared to be a single creature, some large and misshapen organism floating toward him in the hot, shimmering air like a mirage.
As it approached, it was easier to see that it was a group of people. One walked in front. Another, hooded like the messenger from before, walked by his side and held out a large black umbrella, to shield him from the sun. Two more hooded figures, presumably guards, flanked them.
The leader wore no headgear, revealing that he was pale, luminously so, almost like the underbelly of a frog. His pallor was accentuated by a shock of dark hair that fell over his forehead and his black clothing.
To Caleb, it was clear what this was meant to convey. In a world where everyone had to be swathed in white against a deadly sun, such a wardrobe was a show of strength, a taunt to the elements, a way of being above and better than the heat.
Caleb assumed this must be Levi.
"Greetings," the leader called as they got within earshot.
Farther back, another guard held a second umbrella over a girl, who picked her way with difficulty across the broken surface on thin-soled sandals. She was fair-haired and impossibly pale, as well; she was perhaps fourteen or so. She wore a meager, turquoise-colored top and shorts, her white midriff exposed. A gem-like stud in her navel glinted in the sun.
The group stopped at a reasonable distance from Caleb, not getting too close. The two guards continued to keep their umbrellas raised, shielding Levi and the girl from the sun. Caleb noticed what looked like weapons at the belts of all four henchmen, a chunk of plastic and metal that he had glimpsed beneath the raingear worn by the messenger.
That Levi was both so physically protected and so attentive to his appearance made Caleb assume he was weak. Yet Caleb was also aware that there were other ways of being strong than through sheer physical might.
One way was to be clever.
Levi stared at him for a long moment. His sunglasses were made of a mirror-like material that wrapped around the top part of his face, rendering his eyes unreadable; and Caleb was unable to see his full expression. Yet the boy's mouth opened slightly with what appeared to be surprise, even fascination. Then he regained his composure.
"I'm Levi," he said.
He didn't bother to extend his hand; he kept his arms by his side, one thumb hooked in his front pocket. Caleb didn't offer up his own.
"Caleb," he said.
For a second, the other boy's face seemed familiar, but Caleb dismissed it as a play of shadows.
"This is Michal," Levi said, with a casual, almost indifferent nod to the girl. She gave him an eager smile.
"Good to meet you," Caleb said.
"Care for some water?" Levi said. "It's clean."
Caleb smiled. How could he say no? The simple offer established that Caleb was now in Levi's debt and must be grateful.
"Sure," Caleb answered.
Levi gave a short nod to one of the guards, who tossed a blue metal bottle at their guest. It was actually cold. Caleb acknowledged his thanks with a slight tilt of his head. Then he uncapped it and drank.
Levi was watching him; behind his glasses, he almost seemed amused. "Keep it," he said as Caleb offered the bottle back. Not caring that this put him further in his host's debt, Caleb slid it in his backpack.
"Whereabouts are you from?" Levi said. "You're not from around here, are you?"
Caleb shook his head. "I've been traveling a long time," he said. "I come from beyond the mountain range to the north."
There was a moment's pause. Then: "Don't know the area," Levi said blandly. Pleasantries over, he got to the point. "I've heard things about you from my boys."
"Is that right? What kinds of things?"
"That you single-handedly fought off the mutants. And that now you're officially protecting Prin."
Caleb shrugged. "It's not official. And it's just for a little while."
"Don't sell yourself short," replied Levi. "It's impressive. And the town needs help. I could only do so much for them." At this, he lifted one hand, dismissively. Silver glinted on three of his fingers.
Caleb shrugged again. By now, he was aware that the other boy seemed to be studying him, as if waiting for some kind of response. He had no idea what it was supposed to be. He had been waiting for the right moment to inquire about the accelerant when his host gestured across the parking lot, in the direction of the work sounds Caleb had heard before.
"Would you let me show you something?" Levi said.
It was clearly not meant to be a question, Caleb thought. Levi had the ability to make people not only obey him but also feel as if they worked for him too, even when they didn't. In a way, he admired such manipulative skill; it made him feel clumsily physical by comparison. He was willing to bet Levi could also read well, something he could barely do. If you could only combine Levi's brains with his fighting abilities, Caleb thought idly, you'd have a perfect leader.
"Okay," he said. "Show me."
Levi snapped his fingers at the guard assisting Michal. He stepped forward, abandoning his charge, in order to hold his umbrella over Caleb. Exposed, the frightened girl had no choice but to run to Levi, clinging to his arm for protection from the sun.
Levi strolled across the parking lot, followed by his entourage. Caleb followed, curious, and grateful for the shade.
By the time they reached the end of the asphalt, the sounds of people at work were so close, Caleb could make out the grunting of individuals, the shouts of a Supervisor, the rasping of metal on rock. Levi stopped in front of a chain-link fence strung with barbed wire that seemed to encircle the entire parking lot and gestured at Caleb to look down.
On the other side was a deep trench. It was the only one of at least three such pits scooped out of the earth that surrounded the parking lot, one after the other. Each represented a tremendous amount of effort; they were deeper than the height of two men and at least three times longer than that.
Toiling in each trench were a dozen townspeople of every age. Some used picks to break up the rock and packed earth beneath their feet; others shoveled up what was excavated and tossed it behind them. There, the youngest workers filled plastic buckets with the dirt. At the end of each trench was a primitive pulley, where the children attached the buckets to a dangling rope. They were then pulled to the surface by two other workers and their contents disposed of.
It was appalling, backbreaking work, all the more so because of the weather. The day's rain had turned each trench into a vast and treacherous pit of mud and rainwater, and so everyone wore protective gear: rubber hip boots, gloves, plastic face masks. Caleb couldn't imagine how unbearable it must be to work in such clothing in this heat.
What's more, as far as he could tell, it all looked utterly pointless.
"This is what we call the Excavation," Levi said. "It's one of the fair trades we've devised here in Prin, for the goods I dispense from the Source. It's a system that's been working very well."
"For who?" Caleb couldn't help asking.
"For everyone," Levi replied. He sounded sincere.
"I see."
"I know what you're thinking. That it seems to have no purpose. But that's where you're wrong."
Levi stepped forward, out of the shade of his own umbrella and into Caleb's. As he did, he took the umbrella from the guard, who backed off. The two boys were now standing very close to one another; their faces were mere inches apart and Caleb would have stepped back, if it wouldn't have been so obviously rude.
Levi addressed him in a voice so soft that even Caleb had trouble making out the words.
"They're digging for something," he said. "Something important. Even precious. No one can know what it is, because they wouldn't understand. Look at them. They're animals." He indicated the guards and Michal with a glance that was dismissive and contemptuous. "When I find what I'm looking for—and I will—I'm going to need help with it."
Caleb nodded, just listening. Levi never raised his voice yet still managed to speak with absolute conviction. He stared into Caleb's eyes, which Caleb noticed were the same shade of gray as his own. He heard something resonant in the boy's voice. Had he heard it before? All of this threw him for a second.
"Do you have any family? A partner?"
Caleb was jolted back to reality by this question.
"I did," he said. "Once."
Levi was nodding as if in confirmation. "There's an opportunity here," he said at last, "but not just for anyone. It's for someone who thinks big, someone who can rise to an occasion. Someone who's like me."
After a long pause, Caleb answered, finally understanding what was going on. "You're offering me a job."
Levi remained motionless, not even blinking, his face still close to Caleb's. "That's one way of looking at it."
"Who said I want one?"
"You're already working for _them_." He almost spit out the word; it took Caleb a moment to realize he was talking about the workers in the pit, the people of Prin. "Why are you wasting your abilities?"
"I told you. That's just for now."
"What I'm offering you isn't a job. It's a future. Don't you want one of those?"
Caleb hesitated before he shook his head, this time with certainty. "That's not why I'm here."
"No? Then why are you?"
"I want to find who killed my partner and stole my boy. That's why. And that's it."
Levi's remarkably pale face flushed. Briefly, he shook his head no, as if he couldn't accept what he'd just heard. "You're better than that."
"Am I?"
"Yes. I can tell. You want to build something positive. Revenge will just leave a bad taste."
"Maybe it's not revenge. Maybe it's justice."
Levi shrugged away the difference. "That's a lonely road," he said.
He turned and beckoned to the girl, who had been waiting beneath the other guard's umbrella. She pointed to herself, startled, and Levi nodded, impatiently. Smiling, glad to be of service, Michal moved forward.
But before she could reach the shade shared by the two boys, Levi raised a hand to stop her. Then he turned back to Caleb.
"You wouldn't have to be lonely here," he said. "I'd make sure you had friends. Right, Michal?" He raised his voice. "You'd be Caleb's friend, wouldn't you?"
Caleb looked at the girl. Michal was sweating, her pale skin already turning pink in the bright sun. She couldn't hear everything that Levi had said, so she simply nodded with pathetic eagerness, desperate to please.
"More than friend," Levi added. He raised his voice so she couldn't miss the insinuation in his voice.
The guards did not miss it either. As they guffawed lasciviously, Michal's face froze and her eager smile faded.
Levi turned back to Caleb. "You could start a new family," he said. "You'd get over your old one in no time."
Caleb felt sickened by not only what Levi was offering, but his blithe assumption that it would please him. He stared coldly at the other boy.
"No thanks," he said.
For the first time, Levi seemed off guard; Caleb thought he could see him blinking rapidly behind his sunglasses, as if to regroup. Then Levi waved the girl away, back to the shelter of the guard's umbrella.
He now gestured behind them, to the massive building looming in the near distance.
"You don't know the things I have in there," Levi said; "clothing. Furniture. Gold watches." He seemed less sure of himself now, his voice beseeching. "If you worked for me, I'd make sure you had your fair share of whatever you liked. There's more in there than you could ever want."
Caleb paused. "Actually, there is one thing I want."
Levi waited for him to tell him.
"It's an accelerant," Caleb said. "For setting fires. It's called 'Able.' I was told it could be found around here."
Levi paused for a second. Then he shrugged, as if to say, what did this have to do with me? He seemed to have lost interest, as if he had just found out that Caleb was a less worthy person than he imagined. "As far as I know, we've never stocked it."
"You never sold any to mutants?"
"Of course not. From time to time, we've done a little trading, but just the necessities. Deer meat for water. That sort of thing."
Caleb nodded. This might explain what Esther saw at the variant camp. "Have you ever been robbed?"
Levi smiled and then shook his head, as if charmed by Caleb's naïveté. "That's ridiculous."
But Caleb wouldn't let it go. "Do you mind if I take a look inside? It's not that I don't trust you. I'd just like to see with my own eyes."
Levi stared at him for a moment; then he smiled. "Of course," he said. "It's the least I could do."
He said a few words to his guards; then everyone turned around and headed across the parking lot, back to the Source.
Inside, it was impossibly cool and dark. It took Caleb's eyes several moments to get adjusted to the gloom. He was aware that Levi was waiting for him before he took off through the cavernous space.
"Stay close," Levi called over his shoulder. "It's easy to get lost in here and I don't want to waste your time."
Yet Levi walked at a deliberate pace down the endless aisles. Caleb wondered if this was intentional, a way of showing off his wealth. Certainly, he had never seen so many supplies in his life. A single crate, he reckoned, could feed a family for months.
Levi stopped and turned around.
"This is where it would be," Levi said, nodding over his shoulder. "Household supplies. You see?" He pointed at the crates stacked high on the oversize shelves. "This is where we keep all the solvents and flammables. We have turpentine, floor cleaner, bleach, hydrogen peroxide, ammonia. But I'm afraid not what you're looking for."
"Can I?" asked Caleb, and Levi nodded, stepping aside.
Caleb scanned the shelves. Although he could barely read, he knew how to spell "Able"; still, it was laborious work. And after close searching, he had to admit that Levi was correct; nothing by that name was there.
Levi walked him to the door. "Well," he said, "I'm sorry you weren't able to find what you were looking for. But you know where I am in case you change your mind."
"I won't," replied Caleb.
Levi hesitated. "In that case," he said, "you might consider trying down the road a ways. There's another town where I heard they trade with the mutants. If you follow the main road to the east, it's at least a full day's ride. I suggest getting a start first thing in the morning."
Caleb nodded his thanks and, without a handshake or another word, was gone.
Levi watched from his hidden window until the other was out of sight. Then he immediately headed back downstairs, to the main warehouse floor.
In one aisle, a hooded guard braced a stepladder on wheels and another stood on its top step, craning to see onto a high shelf.
"Found it yet?" Levi asked.
The guard grunted an affirmative and handed something down, a crate with a name stamped on its side. It took two guards to carry it.
"Hurry up and bring it this way," said Levi. "I want you to put it in my office." He was glad his boys never learned to read.
For although he would never admit it, certainly not to his underlings, Levi was nervous now. Maybe for the first time.
"Make sure," he said, "that he's gone by morning."
## NINE
WHEN CALEB EMERGED FROM THE SOURCE, IT WAS LATE AFTERNOON.
He headed out of the asphalt lot toward the main road. He bicycled through the center of town, passing townspeople on their way home from work. He wasn't sure of his exact destination, but he sensed it lay on the north side of town.
_I'm usually in the fields near the tracks_ , Esther had said.
As he approached the bleached land that lay beyond a cluster of abandoned office buildings, Caleb dismounted his bicycle and wheeled it next to him. It was pointless to ride; the ground was littered with broken glass and scraps of old metal. Far away, he could just make out the glint of train rails mostly hidden by overgrown and sun-bleached grass.
By now, the sun was setting, sending blinding shafts of light from the horizon. Caleb shielded his eyes with one hand as he scanned the desolate expanse. Other than the rusted hulk of a truck, a soiled and rain-bloated sofa spilling stuffing, and the charred remains of a bonfire, he saw nothing.
Then his glance flickered back to a small copse of trees in the distance.
They had lost most of their leaves and their branches were bare and skeletal in the November twilight. But even in the fading sun, he could make out a patch of color amid the black limbs.
It was a red hoodie.
From where she sat, Esther watched as Caleb wheeled his bicycle toward her. Although no one ever came to the fields, she had recognized him the moment she saw him, no bigger than a speck on the horizon. She was surprised by how quickly her heart began beating; it almost hurt and she forced herself to look away. She waited until he was below her before she trusted herself to speak.
"Hey," she said.
"Hey."
He stretched up a hand, but refusing his help, she jumped down. She smiled up at him, but his expression was serious. He wasted no time in pleasantries.
"I just came to say good-bye," he said.
Whatever Esther was about to say froze on her lips. The sense of shock was like a physical blow, sharp and unexpected.
"Why?"
"The thing I'm looking for isn't here. Levi told me about another town I should try. It's a day's ride away, so I have to leave in the morning."
"But . . . what about Prin?"
"They'll have to do without me."
"But they'll want you to stay." Esther couldn't help herself; the words sprang unbidden to her lips. " _I_ want you to."
She blushed, and Caleb also averted his gaze. Then he spoke, as if addressing the ground:
"We still have all night."
Together, they headed across the field and back onto the main street. But instead of going into the center of town, Esther turned the other way. They walked along the crumbling sidewalk, following a tall metal fence for several blocks until they reached a gate. A rusted sign hung overhead. Esther couldn't read most of the words but knew it was a place where people once buried their dead. Although it was nearly dark, they went inside.
Along the way, Esther and Caleb talked. They followed the broad pathway, which was covered with white gravel and gave off a faint glow. They passed rows of tombstones that had been defaced or toppled by vandals. Crosses, angels, and obelisks lay smashed and cracked on their sides. The path looped its way through the trees and past the dried remains of what seemed a large fountain, edged by wooden benches. They sat on the one that wasn't broken.
Esther felt Caleb tense up beside her. "What's wrong?"
She heard it as soon as she spoke. There was a rustling sound close by, and the rattle of gravel. An overhead cloud shifted, revealing silhouettes moving in the shadows around them.
"Dogs," said Caleb. He raised his voice, and the forms seemed to hesitate. "We're okay, but we should probably get back."
He and Esther stood, stamping their feet as they did. Wild dogs wouldn't attack two people, especially if they walked slowly, making noise as they went. Still, the hair on Esther's arms stood up; if fear had a smell, she hoped the animals couldn't detect it. She and Caleb retraced their steps, following the white pathway until they emerged back on the main street.
Caleb insisted on walking her home. She gladly let him, and together, they headed through the center of town. When they reached the Starbucks, she turned to face him.
"Goodnight," she said. She didn't want to say good-bye.
Caleb said nothing. He just reached out and held her hand for a second, then pulled away.
Esther opened the door and turned to say something else. But he was gone. All that was left was the sound of his bicycle disappearing into the darkness.
_Esther was running down an empty road._
_She was playing Shelter and, for once, she was winning. Jubilant, she could see the safe place, the large cardboard box, in the distance. It sat incongruously in the middle of the highway, straddling the double lines that seemed to go on forever under a hot yellow sky._
_Yet as she got closer, her legs started to move slower and slower. Each step felt as if she was fighting her way through deep sand and she panicked at the thought of losing, of someone else getting there before her._
_Soon she couldn't move; she attempted to thrash her arms and legs, but they were pinned down by their own weight. The ground began to crumble and collapse beneath her; she was breaking through the earth and would soon be swallowed by it, buried alive._
_Someone grasped her by the hand. She clung to her rescuer and struggled to break free as he lifted her from her grave. The caked dirt fell from her face and she could see who it was._
_Caleb._
_He was saying something to her, something urgent. To not give up. To keep fighting._
_Then, as if from far away, she heard a cry . . ._
Esther jerked awake in the darkness, her heart pounding.
It took her a moment to realize she was safe at home, in bed. But far away, a girl actually was screaming.
She rose and crossed into the dark living room. When she peered from behind the curtains, the first gray light of early morning revealed two cloaked forms hurrying down the sidewalk, carrying bundles.
Esther checked to make sure her sister was still asleep. Then she dressed and stole down the stairs and into the street. Far ahead, she noticed that the two people had been joined by another. She followed them for several blocks, until they turned onto a side street. The three were apparently on their way to the home of Trey and Aima, which was once a store. The opaque glass door still had the words "dry cleaning" painted on it. Now, it gaped open and light spilled onto the sidewalk.
This was where the screams were coming from.
Unnoticed, Esther stood in the open doorway, stunned by the noise and activity. Inside, the sobs and shrieks were deafening. Bleary-eyed females, robotic with fatigue, nevertheless moved around the room with purpose, carrying towels, plastic jugs of water, blankets. Although the air was stifling and dense with smoke, one of them tended a blackened fire bowl in the corner of the room, tossing chunks of wood onto the leaping flames.
The crowd parted for a moment and Esther could see what was happening.
Aima squatted in the center of the small room, clutching the edge of the laminated counter. To Esther, she was unrecognizable. Her soaking nightdress hiked up, her monstrous belly suspended over her knees, she was white-faced and gasping, her hair plastered across her face. Two females kneeled by her side, supporting her, and again, she screamed.
Esther, horrified, couldn't speak.
Aima threw back her head, causing the cords in her neck to stand out like ropes. She strained powerfully, her teeth clenched and her hands white-knuckled on the edge of the counter. Something dark and wet shot from beneath her and was caught by one of the waiting females. At the same time, there was a loud gushing sound, as an eruption of clear liquid and bright red blood splashed onto the dirty tiled floor. Moments later, there was the thin, reedy cry of a newborn. The others closed around Aima, murmuring as they tended to her.
But something was wrong.
One of the girls gasped and another flinched. Something was said in an urgent whisper, followed by a muffled exclamation.
" _No._ " Aima's voice was faint, but it rose above the clamor. " _No! It can't be!_ "
Esther craned her neck, trying to see what was wrong. But the person closest to her swiveled around, her hands slick with blood and afterbirth. She noticed Esther for the first time, and her eyes blazed with anger.
"Ain't nothing to stare at," she hissed. "It was born dead."
Esther recoiled.
Everyone in the room, even Aima, was suddenly aware of her. Silence fell, and all of the females turned one by one to stare at Esther. Stammering apologies, she stumbled backward out of the room, nearly tripping on the doorsill. When she was outside, the opaque glass door was pointedly pushed shut behind her.
Esther stood alone on the sidewalk, thinking about what she had just seen. She was deeply rattled.
It was not just the sight of childbirth that bothered her, although the violence far exceeded anything she could have imagined. Nor was it her rude exclusion from the circle of women, a secret society that had never wanted or welcomed her.
_It was born dead,_ the girl said.
Yet Esther had heard it cry out.
Instead of going home, Esther retreated to a darkened doorway where she could see Trey and Aima's home. There she waited to see what would happen next. She leaned against the side of a building and felt her eyelids droop; she was about to pass out on her feet. She was ready to give up and head home when the door across the street opened, spilling light onto the sidewalk as a figure exited and walked away.
Whoever it was carried something bundled in her arms.
Esther used the tracking skills she had learned from Skar to attempt to follow undetected. Oblivious, her target hurried through the darkened streets, sure of where she was going. Once, she glanced around, as if sensing she was being trailed. Esther melted into the shadow of a streetlamp, and satisfied, the other girl continued.
Onward they walked, the robed girl in front, Esther half a block behind. They reached the outskirts of town, past the Source looming huge and white in the early morning light, and still the female continued. She turned off the road and cut through the land beyond the gaping pits of the Excavation, picking her way across the precarious open wasteland made up of the debris of collapsed buildings.
Occasionally, she shifted the bundle in her arms.
At last, she arrived at what appeared to be her destination. It was a massive oil tower, a giant steel tank set high atop four spiderlike legs. Weather and time had eaten away the letters once painted on its side, and the metal was corroded with rust and rot. A spindly ladder made its way to the top. At its base, the female finally set down her bundle.
Esther stepped out of the shadows. "What do you aim to do with that baby?"
The female started violently and cried out in fear. When she saw who it was, her expression changed to one of utter disbelief.
"You followed me?" she asked. It was Sian, an older girl Esther knew only slightly. "All the way from town?"
Esther nodded. "What do you aim to do with that baby?" she repeated.
Sian shook her head dismissively.
"It ain't no baby," she said.
She stepped aside and Esther could see the child. It was tiny, much smaller than she had imagined. It whimpered, then beat at the air with its minuscule legs and arms. The blanket fell away and in the early morning light, Esther could see its sex, which was a misshapen lump, neither male nor female. Its nose was nearly flat, little more than slits in its broad face. Its eyes were far apart, bulging, and lavender in color.
It was a variant.
Esther, stunned, tried to make sense of it.
"So you're just leaving it out here to die?" she asked.
Sian shook her head with a mixture of disgust and pity. Then with her robes hiked up, she took hold of the ladder and began to climb, an orange T-shirt clenched in her teeth. When she was more than halfway up, she tied it to a rung with a clumsy knot. Then she made her way back down.
"This way, they know," she said.
"You mean the—" Esther started to ask, but the other girl cut her off.
"The fathers don't want to know. And the mothers want to forget. So this is how we figured it out with the mutants, long ago. It works out the best for everybody."
Esther couldn't take her eyes off the baby. It had found its thumb and sucked on it.
"It's a secret only the mothers know," said Sian. She stared at Esther, her voice hard. "And now you do, too."
Esther headed home, walking down the center of the road that led to town. The sun was already well in the sky and she was aware that she risked being detected by a crew on its way to work. Yet she was too exhausted and confused to care.
She was thinking of all the couples in town and how so few of the females ever managed to become pregnant. Of the few dozen who carried a child to term, most of their babies were born dead.
Or at least that was what everybody was told.
Now it seemed the truth was both simpler and more complex. Her suspicions had been right, all along: Variants weren't animals at all, but humans. That the mothers of Prin kept this secret was something she couldn't have begun to imagine.
Still brooding, Esther walked home down the center of the street. Although the sun was visible above the horizon, it was too early for anyone else to be out. She was not aware of the bicycle until it had pulled up beside her. The rider in robes and dark glasses jumped down, removing the scarf that covered its face.
"Esther." It was Eli, his face flushed and eager. "I saw you from my window and wanted to talk."
She smiled back politely and he fell into step next to her, pushing his bike. They walked like that in silence for a few moments. He seemed to want to say something, but each time she looked at him, he merely blushed.
"Esther," he began again.
When she glanced at him, he cleared his throat. Then he awkwardly reached over and, to her shock, took her hand. His skin was rough and dry; his palm seemed the size of a dinner plate. She stared at him, uncomprehending.
"We known each other a long time," he said. Esther could see her distorted reflection in each of his sunglass lenses; she looked confused and exhausted. "And you got to know how I feel about you. I guess what I'm saying is, I want to be . . . I was hoping you might think about me becoming . . . your partner." He finished in a rush, his face red.
Esther was speechless.
Yet why should it be a surprise? At fifteen, Esther was past the average age for partnering. Still, she could have three or four good years ahead of her. And Eli had always been kind and generous. In a town that treated her like an outsider, he had never made any secret of his affection for her. And even she had to admit, she may have encouraged him by asking favors.
Now Esther stared at him critically, as if seeing him for the first time. Eli was not tall, but he was strong and healthy. He had thick, wavy hair, a nice smile, and dark brown eyes. His voice was deep and pleasant, and he was a hard worker, dependable and considerate. She could do a lot worse than to become his partner.
Yet she felt nothing beyond an acknowledgment that he was a good catch. Was that reason enough for her . . . or for him, for that matter? Was she so wrong to expect something more from a decision she would have to live with for the rest of her life?
Eli had stopped talking and seemed to be waiting for her to respond. Yet Esther found her thoughts were not on the boy in front of her.
She was thinking of Caleb.
Esther pulled away her hand. "I can't tell you right now," she said. "I'm sorry. I just can't."
Eli stared at her, clearly disappointed. Then he tried to smile.
"Sure," he said with forced heartiness. "I can understand that. A girl needs time to think this kind of thing over."
As she walked away, stiff with self-consciousness, Esther could feel Eli's eyes following her before she heard the sound of his bicycle heading off. But she was still rattled.
If she were paying more attention, she might have noticed that although the sun was well over the horizon, there were no other townspeople outside, on their way to work. Instead, the streets were empty. It was not until she entered her building and crossed the empty storefront to reach the stairs in the back that she noticed that something was odd.
Esther paused. The building around her felt different somehow. Every nerve ending in her body told her that.
Her first thought was of her sister. Esther hesitated at the foot of the stairs, her hand on the banister.
"Sarah?" she whispered.
And with that, they were upon her.
Two townspeople lunged down the stairs and sprang at her. Although startled, Esther was able to leap backward and avoid their grasp. But she was not prepared for the two others who now rushed in from the street, blocking her escape. One of them seized her by the arms; another struggled to bind her wrists behind her back with an elasticized cloth cord that had black metal hooks at each end. Esther struck out, kicking and punching, but she was only one against several and was quickly overpowered.
"What are you doing?" she screamed. Her mind was whirling; was she being punished for finding out about Aima's baby?
No one answered. Rafe walked in, his expression unreadable.
"Rafe!" she screamed. "Help me!"
But a vile-tasting rag was stuffed into her mouth. Esther couldn't speak; she could barely breathe. The last thing she saw across the room was Sarah, clinging to the doorframe, her hand to her mouth. Her face was white with shock and anguish.
With arms tied behind her, Esther was dragged from the building and down the street. When she stumbled, she was yanked back up by her elbows. It was a long, hot walk.
When the group stopped, they were in the middle of what had once been a large lake on the outskirts of town. The ground under their feet was dusty red clay, baked hard by the sun and littered with trash. The shoreline was edged by dead willows and more than a few dozen motorboats that balanced lopsidedly on their hulls, long since drained of any gasoline. A rickety bridge spanned the narrowest part of the lakebed, cinching it like a belt.
The five people who surrounded her were identical in their reflective sunglasses and face scarfs. Only Rafe spoke.
"You know why you're here," he said. "We got word you left your work detail. You got anything to say?"
Esther swallowed as the realization sank in: It wasn't about the baby after all. The situation seemed so unreal that only the pain of the rubberized cords biting into her wrists told her this was not a dream. "It was Rhea, wasn't it?" she said in a low voice. "She hates me and my sister."
"It don't matter if she hates you or not," said Rafe. "Was it true what she said?"
"I worked hard on the Harvesting. You could ask anyone else who was there."
"You're not answering the question," Rafe replied. "Did you leave your work detail or didn't you?"
When she didn't answer, he nodded his head. "That was what I thought."
Esther glanced at the others in open appeal. They must have been thinking what she was: that no one in Prin had ever been Shunned for anything less than illness or a serious crime. Never for something as minor as skipping work detail. Rafe was just following Levi's new rules without thinking, and for that, Esther found him more contemptible than ever.
Two of her neighbors refused to meet her gaze. It was clear that none of them was going to help her in any way, to speak in her defense or ask for mercy.
"Esther," said Rafe. "You are hereby Shunned from Prin."
He nodded, and one of the others fumbled to undo her bonds. Another handed Esther a nylon backpack which she took, numbly. In it, she knew that there would be supplies meant to last a week or so.
For an instant, Esther sensed that this last person was viewing her with regret, even sympathy. But the moment passed and whoever it was joined the others, who stood together, watching in silence.
In a daze, Esther walked across the lake surface and toward the rising sun.
On a grassy patch behind the bank, the seven townspeople surrounded Caleb. He was in the middle, holding a long, wooden stick with an angled end, the word EASTON printed along the laminated shaft.
Caleb addressed one of the students, a tall boy. "Tell me what you're going to do," he said.
"I'm going to go with the motion of the push and see where it takes me," said the boy.
Caleb nodded. Then he raised the stick and used it to shove the boy in the right shoulder. The boy grabbed the stick with both hands and pushed back.
Caleb shook his head.
"No," he said. "See how you're fighting back? You're pushing _against_ the motion. I want you to go with it instead and see where it takes you." The tall boy nodded, brow furrowed with concentration.
This was the second day of class and the boy with the red hair was surprised; things were going much better than before. Somehow, he had absorbed some of Caleb's earlier verbal lessons. Now he was focused on the actual basics of fighting.
That morning, they had spent three hours on punching. The boy hadn't realized there was so much to learn and how little he knew: how to make a fist to best protect your thumb and knuckles. How to aim for a distance a hand's width beyond your intended target to maximize the impact. How, if you lacked arm strength, you could use speed to compensate. How to increase your power by stepping into the punch with your entire body. How to relax your body until the instant you threw the punch.
Now Caleb lifted the staff again. "We're going to try it slowly," he said, "and this time, don't fight it. Relax and try going with the motion." As he pushed the stick against the tall boy's left shoulder, the boy allowed himself to be guided backward, his body twisting.
"Where is your right arm going?" Caleb asked.
The boy gestured: It was swinging inward.
"Now make a fist with it. Think about using that natural movement and using it to help you punch inward. Again."
The two repeated the move, and this time, the boy succeeded in turning the attack into a roundhouse punch.
"Do you see what you're doing?" asked Caleb. "By going with the motion, you're decreasing the damage to your shoulder. At the same time, you're using it to generate an unexpected attack, from the other side."
As the tall boy thought this over, everyone else in the circle murmured. "Thanks," said the tall boy, beaming with excitement.
Caleb glanced at the sky; the sun was almost overhead and he made a quick calculation. After he had worked with everyone in the circle, he would have only a short time to teach basic self-defense moves. Then he would devote the afternoon to beginning techniques in slingshot, sap, and short club.
With any luck, they would be finished by sundown.
He knew he was going quickly, much too quickly, for his lessons to be truly useful. If they could even remember what he was teaching them, his students would have to practice each move for many weeks, hundreds if not thousands of times. Only then would their new skills start to become automatic and, therefore, any help at all. But even the best of them would be nothing more than mere beginners: eager, perhaps, but clumsy and unskilled.
To learn to fight well took months, even years of training. And he had spent little more than a day with them.
It was the best he could do. He should have left Prin already, gone that morning. Yet at the last moment he decided to stay a day longer. He unexpectedly felt obligated to these people and wanted to leave them with at least a fighting chance to protect themselves.
There was another reason that was even more important. He had to see Esther again.
"How are we doing?"
Caleb looked up; Rafe stood in the doorway of the bank. His hands clasped behind his back, he rocked up and down on his heels, checking out the class. Caleb noted the _we_ in his question. This was the first time the town's leader had deigned to show up. While learning to fight was something for others to do, Rafe seemed happy to share the credit.
" _We'll_ get there," Caleb replied.
Rafe gestured for Caleb to approach him. "What's your guess on how long it'll take?" he asked in a low voice. "How many days do you think?"
_Days?_ Caleb thought. _More like months._ But he didn't say it.
"Hard to figure," he replied.
But there was no time to keep talking. The red-haired boy had seized the staff and was using it to prod the others, with a bit too much enthusiasm, in an attempt to drill them in the technique they had just been taught.
"More slowly," Caleb said, walking back to the group. "It's not a natural reaction . . . you have to feel it first. Let me show you . . ."
It was sunset. His pack strapped onto the back of his bicycle, Caleb stopped in the street outside the building marked STARBUCKS. He didn't want this to be the last time he saw Esther. Yet how in good conscience could he ask her to join him?
He looked up at the second-floor window, half open and framed by a fluttering white curtain. He called up, just loud enough to be heard.
"Hey?"
After a minute, a girl came to the glass and looked down. Wearing a flowered bathrobe held close to her throat, she looked haunted. This must be Esther's older sister, Caleb thought. He raised a hand to get her attention.
"Is Esther at home?" he asked.
At the name, the girl winced. There was a pause during which she did not reply. She untied the curtain, which fell and covered the window. Then, just a shadow, she walked away.
Caleb rode on, disturbed. Although he knew he should be leaving, the weird encounter made him want to see Esther more than ever. So, as evening deepened, he continued to search for her.
He headed along the main street of Prin, glanced down alleyways, passed the meeting hall, the old parking garage, the bank. By now, the streets were largely deserted; most people were home from work. Whenever he saw anyone, he asked, "You know Esther? Where she might be?" Each time, he got the same response: averted eyes, an evasive shrug, an unpersuasive no.
Heading down one street was a group of stragglers. They stopped, recognizing Caleb. Some were in awe, too shy to speak. He asked them the same questions.
"Any of you know Esther? Where she might be?"
One girl found the courage to respond. "She's gone."
"Gone?" Caleb said. "What do you mean?"
"She's just gone," the girl said. "For good."
The others glared at her. One tugged at her sleeve, whispering that they'd be late. But it was clearly an excuse.
"But where did she—" Caleb started to ask, but they were walking away, the girl shooting him an apologetic gaze over her shoulder.
Caleb stood there, still straddling his bicycle. Now he found the idea of leaving impossible. Despite what the girl said, he couldn't believe it was true. Esther would never have left without telling him. And where would she go? So he did the only thing he could, continue his methodical search for Esther, up and down the streets of Prin.
Eventually, he made it to the railroad tracks on the outskirts of town. The tree where Esther had perched, watching him, was empty, as were the surrounding fields. It had been many hours; by now, the horizon was touched with pink.
He glanced up. Outlined by the rising sun, someone on a bike had crested a nearby hill and stood looking down at him. The face was obscured by a black hood.
Another one of Levi's boys, Caleb thought. What did he want?
At that moment, the sun shifted, momentarily blinding him. Still, he could see that the boy's arm had risen, in what appeared to be a wave.
Caleb raised his hand in response. As he did, he blocked the light and perceived the truth: The boy was holding a fiberglass hunting bow and drawing back the string.
There was a hissing sound, and Caleb felt a sudden blow. He stumbled, and a moment later, heat blossomed across his shoulder, surrounding the feathered shaft embedded in it.
## TEN
THE DAY BEFORE, THE HEAT HAD BEEN INESCAPABLE.
It not only beat down from the sun; it radiated up from the concrete and oozing tar. The air itself shimmered with arid heat, forming waves that danced across the horizon.
Esther was not prepared for such relentless exposure.
Even with her thin hoodie tied closely around her face, her lips and nose were soon chapped and blistering. She did not have sunglasses, and the ceaseless glare was excruciating to her unprotected eyes. And although she was wearing sneakers, the bottoms of her feet were burning through the thin rubber soles.
Only now, an hour since she had left the boundaries of Prin, did the full impact of what had happened hit her.
She had only enough supplies to last a few days, she realized, and no weapons, no tools, and no shelter. She would never see her home again or climb her stairs or sleep in her bed. She would never see anyone she knew again, neither Caleb nor Sarah.
At the thought of them, Esther felt a flicker of hope. But a moment later, she recalled with a sinking heart that anyone who helped a person who was Shunned incurred the same sentence as well.
Except for the variants.
Variants had no need for the laws and regulations of the town. They followed their own rules and meted out their own justice.
With a pang, Esther recalled that she and Skar had parted on bad terms. Moreover, she knew that she had ignored her friend's desperate attempts to contact her. But Skar had never been the kind of person to hold a grudge; surely, she would forgive Esther when she found out the trouble she was in and convince her tribe to take pity on her.
By now, it was midmorning. On foot, it would take Esther until sundown to reach the variants' camp, but at least she knew where she was heading. She took off for the foothills that lay on the horizon.
Mostly, she ran at a swift trot, ignoring the agonizing blisters that formed on her feet. Even so, it was not until the sun was disappearing over the horizon that she saw the poisoned black lake glittering ahead of her. She veered from the roadway, hurdling the low metal fence and plunging into the deep undergrowth.
After the bright heat of the highway, the relatively dark and dappled forest was a relief. Esther toiled up the steep hillside, passing the tree with its faint white mark. As she neared the camp, she went slower and more cautiously. Every several minutes, she gave her special, two-tone call. Soon, she was at the edge of a small meadow, the last clearing before the final ascent to the variant camp. She was about to repeat her call when the whistle was returned. From across the meadow, she saw a flicker of movement as Skar stepped out from the trees.
"Esther?" Skar called.
The girl emerged from her hiding place and the two ran to each other. As Esther hugged her friend, overwhelming relief and exhaustion caused her knees to buckle and she almost collapsed.
"What happened?" Skar asked. Concerned, she led Esther to a flat boulder, where the two sat. "Every day, I have been signaling you and you haven't responded. I was worried you were still angry with me. But now, I see something else was wrong."
Esther nodded. "I'm in real trouble," she said in a low voice. Then in a rush, she explained everything that had occurred.
Throughout, Skar listened without speaking, her expression unusually grave.
"So I need your help," Esther concluded haltingly. She was aware that her normally talkative friend was not saying anything, and she found this disturbing. "Please. I need you to ask your tribe if I can stay here."
Biting her lip, Skar dropped her gaze.
"It is not so simple anymore," she said at last. Then she looked up at Esther. "Why didn't you answer my signal before? I wanted so much to explain to you face to face. Then you would understand my situation."
"What do you mean? What situation?"
Skar held out her arm.
Esther glanced down. Her friend was still wearing the meaningless assortment of silver bracelets and wristwatches that she had on the last time. But beneath the jewelry, there was something new, something different.
Among the familiar whorls and patterns written on Skar's skin in scar tissue and ash, there was a vivid new line that snaked its way around and up the forearm, from the wrist bone to elbow. The wound was so fresh, it was still dark red with clotted blood and was framed by an angry ridge of pink, inflamed skin. One could still make out the dirt that had been rubbed into the cut, to maximize the scarring.
Esther stared at it, uncomprehending.
Skar was smiling. "I have a partner," she said. Then she giggled, covering her eyes and mouth with both hands as a deep blush stole over her face.
Esther was speechless.
She was not sure why she was so stunned. After all, she and Skar were both fifteen, more than old enough to be partnered. Yet in all the years they had been friends, Skar had always behaved like the younger of the two. She had always looked up to Esther and in many ways, was like a little girl, one who giggled and occasionally played with a castoff doll and still sucked her fingers when she slept.
"It happened so quickly. I was not expecting him to ask," said Skar. "I meant to tell you when I saw you. But I was too surprised when you showed up at my home. And you were so angry, and then you left so fast. You didn't give me a chance."
Esther nodded slowly. This was true, she realized.
Now she swallowed hard. "Your partner," she muttered. "Does this mean you have to ask him for his permission?"
Skar shook her head. "Not for his permission. But his blessing."
Esther allowed Skar to take her by the hand and to lead her deeper into the woods. Soon, they reached a small clearing; the moon had come out and by its light, Esther could make out a nearby stream.
Skar turned to her. "Please try to understand. If it was only my decision, you know I would do anything you ask," she said, giving Esther's hand a final squeeze. "But now, I have someone else I must consult." Then she put her hand to her lips and gave a warbling cry, like a dove.
There was a long silence as the two girls waited. Finally, a solitary variant emerged from the forest.
Skar let go of her friend's hand and went forward to meet him. The two conversed in faint yet urgent whispers. Esther could not make out what they were saying, although it appeared to be an argument. But soon, Skar seized her partner by the hand and tugged on it to bring him close.
"This is Esther," Skar said. She could not hide the anxiety that creased her brow. "And this is Tarq."
The variant boy stared at Esther with open hostility. Although he was the same height as Skar, he was husky and outweighed her by a few pounds. In addition to the triangle he wore on his bicep, his dark skin was covered with other vivid tattoos and scars: stars, moons, the depiction of a hunt. His short tunic was cinched with a leather belt studded with metal, and he wore a plastic wristwatch and several pairs of sunglasses on colorful cords around his neck.
"What are you doing in these woods?" was the first thing he said.
Any polite greeting Esther was thinking of froze on her tongue. "This is neutral territory," she said stiffly.
"But you are Shunned," he said. He spoke not with concern but with hostility. "If anyone were to search for you, this is the first place they would try. Your presence can only bring trouble for my partner."
He stared at her in an open challenge. Esther didn't answer and lifted her chin, matching his antagonistic gaze.
Skar was anxiously looking from one to the other.
"Esther is my oldest friend," she whispered to Tarq. He said nothing at first, but a muscle in his jaw twitched. He had one arm draped around Skar's shoulders and Esther noticed that he now squeezed the nape of her neck possessively.
"And I am your partner," he said. "It is dangerous to be seen with norms. Especially one who has been driven out by her own people."
He raised his head and gave a high-pitched whistle. Moments later, it was answered by a second whistle far away, then a third and then a fourth. Skar glanced up, and Esther could see both panic and concern flash across her face.
"Forgive me, Esther," Skar said. "I want to help. But my people are nearby. If they see you here, they—"
But Esther was not there. She had already slipped back into the forest.
By now, the moon was high overhead. Dizzy and disoriented, Esther did not know where she was going. Still, she continued mechanically placing one foot after the other as she followed the faded double yellow lines; they seemed to go on forever as they bisected the abandoned two-lane highway.
She was filled with rage at the people of Prin. She loathed their cowardice, their blind obedience, and their pettiness. _They were responsible._ Esther's body pulsated with fury, and her churning emotions acted as fuel and provided a rhythm that drove her on as she continued mile after mile down the highway.
Esther stumbled. She had been traveling since dawn that day, and she was close to collapse. She did not recognize any of the landmarks around her, casting long and ominous shadows, but she seemed to be on the outskirts of a small town. When she reached a shopping plaza, she had no choice but to stop for the night.
There were several possible shelters, but Esther was careful about which one she would choose. More than most, she was aware of dangers any unfamiliar building could hold. Many were structurally unsound, with rotting floorboards and ceilings. Any collapse could carry with it a deadly surprise, releasing hidden pockets of stagnant rainwater. Others were infested with roaming hordes of territorial animals, whether they were fire ants, rats, oversize spiders, or snakes. Still others teemed with massive patches of mold and fungus that could make you ill just by breathing their foul air.
After investigating a diner, an eyeglass store, a pharmacy, and a jewelry store, Esther found something that seemed promising, the final business in the block of buildings. It was a large, open space with windows that were mostly intact. In the dim light, she could see that empty metal racks lined the walls and adorned wooden islands. It was, she decided, a clothing store.
Esther picked her way across the trash-littered carpeting to the back. There was a smaller room here, with open booths built into the back wall, side by side. Each held a wooden bench and a full-length mirror. One still had a tattered curtain, which Esther pulled shut behind her. There, after a quick meal of bean cake and water, she curled up and attempted to sleep.
Despite her exhaustion, Esther was too keyed up. Although the store was deserted, tiny sounds kept disrupting the silence: the skittering of claws across wood. A sudden flurry of paper, and a loud squeaking. She tossed on the hard and tiny bench, attempting in vain to find a comfortable position.
Then Esther froze.
There was another sound, but it was not that of an animal. It was much too heavy, much too deliberate. Someone was in the outer room, and he was walking as softly as he could, trying not to make any noise.
In the dark, Esther sat up as she kept her eyes trained on the thin curtain, illuminated by moonlight that streamed in from the front room. She put on her sneakers and gathered up her bag, her muscles tensed to spring.
The footsteps were getting closer. Within seconds, they seemed to stop a few feet outside her booth.
A tiny shadow appeared at the lower corner of the curtain. Esther stared at it. Then suddenly, it extended and sharpened, as a skeletal claw reached out to touch the fabric.
Before it did, Esther ripped the curtain back.
A hulking creature was standing there in filthy and tattered robes. His eyes glittered in an emaciated face encrusted with dirt.
He said something guttural that she could not understand. Then without warning, he lunged at her.
Esther tried to push past him. He was no more than bones floating within his billowing robes, yet he was surprisingly strong as he clawed for her bag. She kicked him in the knee as hard as she could, and he let go for a second, allowing her to dive past. Then she was running through the outer room, leaping through the broken window and into the parking lot.
That was when she noticed that she was not alone.
In the moonlight, she could see at least a dozen skeletal forms moving through the stores and buildings of the shopping plaza. Esther stopped in her tracks. At first, they seemed like spirits of the dead, supernatural creatures from one of the stories Sarah used to read her many years ago. But then, she realized what was happening, and the truth of it was almost worse.
There was nothing left and yet they were obeying a routine they could not shake.
They were attempting to Glean.
She knew they were only people, boys and girls her own age who had been driven mad by hunger, desperation, and exposure. But their hunger was frightening because it seemed unthinking, inhuman, and insectlike.
She took off as fast as she could.
The air was cool and it was a relief to run, to put as much distance as she could between herself and what she had just seen. Yet the night was full of other potential dangers, sounds and shapes that she could not identify. When the moon retreated behind a thick covering of clouds, Esther was plunged into total darkness. She knew she should stop; to risk injuring herself would be stupid. Yet she was beyond caring. She stumbled on a broken piece of roadway and fell hard on her hands and knees.
Esther broke down and cried.
This time she did not do so out of anger or frustration. Instead, she cried because of her foolishness. She cried because she thought she could get away with breaking the rules, and she could not. She cried because, as a result, all was lost. At last, fatigue won out. She lay on her side, curled into a ball; and blessedly, she fell asleep.
Several hours later, Esther awoke on the side of the road, half in the gravel-studded shoulder and half in an overgrown field that bordered the highway. Overhead, the sky was gray with the first light of dawn. Her face was pressed to the concrete, which was still hot, tiny pebbles cutting into her cheek; and despite the scent of gasoline baked into the road, there was a thicker smell, sweet and nauseating, that overpowered it, catching in her throat and making her gag. The air pulsed with a heavy and constant drone as she found herself staring at a piece of broken glass, a shard of deep blue lying inches away.
Beyond it, something gleamed white. It was a branch stripped bare of its bark, a piece of wood bleached by the sun. But as her eyes focused on it, Esther realized what it was.
It was a bone. And attached to the end was a battered sneaker.
For as far as she could see, human remains littered the ground. There were hundreds of bones and bone fragments, most softened by exposure to near dust, animal teeth marks and knobbed ends alike eroded away to nothingness. All were human, still clad in the tattered remains of jeans and T-shirts, wrapped in moldering robes. Still others were fresher, heaped piles that stank of decay and were all but invisible beneath an oily black coating that shifted and shimmered in the early light. It was in fact flies, hundreds of thousands of them, busily eating, mating, laying eggs, and dying. They were responsible for the ceaseless drone that filled the air.
For this was the rumored place where the sick went when they had been Shunned from every community and driven away from the living. It was the place where people went when there was no more hope, no more life. It was the Valley of the Dead.
And Esther had ended up here as well.
She would run from it in terror, if she could. But she barely had the strength to breathe, much less move. As the first rays of sun began to heat and thicken the air around her, she felt the life force within her starting to ebb. She closed her eyes and saw a brilliant red that pulsed more and more faintly, in time with her heart.
But above the droning, Esther heard another noise, faint but real.
Something was in the bushes that lay beyond the highway's shoulder. Esther opened her eyes to see what it was, but it was still too dim. Then she heard it again.
Someone was coughing.
Esther raised her head. The sky was imperceptibly brighter; and by its modest light, she could make out a small figure huddled against a tree.
Shakily, Esther got to her feet. She stumbled closer and saw that it was a girl, wrapped in dirty robes, sitting up and watching her.
The two gazed at one another without speaking. Even in the dim light, Esther could see the extreme pallor of the girl's face, and the feverish light in her eyes. If she stood any closer, she knew she would also notice the telltale sores, the lesions that covered the face and limbs of the afflicted.
"Where are you from?"
The girl's voice was cracked and dry, like an autumn leaf. It was so tiny, Esther had to strain to hear her.
"Prin."
The girl nodded. "I was there once," she said. "A long time ago."
Esther hesitated. It was dangerous, she knew, to be close to someone who had the disease. In Prin, no one even spoke to the afflicted; once the lesions appeared, they were driven out of town, as Esther had been. But without food and water.
It made no sense to waste precious supplies on the dying. Yet now Esther reached into her bag. Taking out her water bottle, she uncapped it, and handed it over. The girl's eyebrows went up in a question and Esther nodded. Then the girl reached out her hands. When Esther touched them to help her lift the bottle, they were hot and papery.
The girl drank deeply, the muscles in her throat working. She drank until the bottle was nearly finished, then she pushed it aside and sighed.
"Thank you," she said. She sounded much, much better, but neither girl was fooled. "Where are you headed? You got a home?"
Esther shrugged. "I did."
The girl nodded. "It all lasts so short, don't it. Here and gone. Well"—her eyes flickered to the side, as if to indicate all of the horror that surrounded them—"there's a place for you here, if you want it."
She closed her eyes. Within moments, she was breathing deeply. In sleep, her face was eased of its pain and tension and Esther was surprised to see that she looked her age. She was no more than nine or ten.
Esther got to her feet. Then she put her backpack over her shoulder.
_It all lasts so short._
And no matter how bad things were, she was still alive.
The sun was now visible in the eastern sky. If she set off now, she should be back in Prin by evening.
## ELEVEN
AT THE SAME MOMENT IN PRIN, A FIGURE LAY EXPOSED TO THE MORNING sun, writhing in agony. Instinctively, Caleb grabbed the fiberglass shaft where it entered his shoulder. He could not risk having the arrowhead break off, lodging in his body. So he pulled it out in one swift motion and flung it away.
Then he realized his terrible mistake. Without the arrow to keep it plugged, the wound began to pump blood. Soon the front of Caleb's shirt appeared black and glistening, and dark crimson began to pool on the packed earth beneath him. Caleb's hands were stained red as he pressed hard against his chest, trying to stanch the flow that wouldn't stop.
It was so early, teams on an Excavation had not yet shown up. He was far from the main roads, miles from the center of Prin. His only hope was to somehow ride into town to get help.
But when Caleb took his hands from the wound, a fresh bout of blood bubbled up through the soaked and torn fabric. He pressed his hands over it, felt the fluttering pulse underneath.
Then he noticed something out of the corner of his eye. Across the parched field, past the metal tracks long overgrown with weeds, there was an abandoned truck that sagged on dusty wheels. Next to it, a small person was picking its way toward him.
It was not the hooded guard, the would-be assassin from the Source. It was bundled in a pale blue sheet, its head covered. Caleb felt a surge of relief. He tried to half rise.
"Esther!" he called.
"Don't be scared!" it responded, and his heart sank.
It was female, but not the one girl he desperately wanted to see. As she came nearer, a strange, heady smell, like that of a thousand flowers, seemed to shimmer from her. When she crouched next to Caleb and saw all the blood, the girl gasped.
"I didn't think he'd really do it," she murmured.
"Who? And who are you?"
The girl pushed the sheet away from her face. To Caleb's blurred vision, she was as unreal and exotic-looking as an animal from a dream: pale, with golden hair and strange blue eyes. He noticed the glittering rings on the girl's fingers, the chains and bangles that dangled from her slim wrists.
"You're from the Source," he said. "Levi's girl. Michal."
Caleb fumbled on the ground for the arrow, which was sticky with his blood. "Keep your distance."
Michal glanced at the arrow, then back at Caleb.
"I don't blame you if you don't trust me," she said. "I should have warned you."
"What do you mean?"
"After you left, I heard the guards talking. Levi told them if you weren't gone by yesterday, to come after you. I kept an eye on the guards. And this morning, when one of them left with a weapon, I followed him."
Caleb cut her off. "It doesn't matter," he said. "I need to get to town. Give me that."
At first, the girl looked baffled. Then Caleb seized the cloth that protected her and started tearing it. When she realized what he was doing, she helped him to unwrap it. Then she wadded up one piece and pressed it against the wound. He used the other strip to bind the bandage against his shoulder.
She got to her feet, having grabbed him under the arms. The effort of standing caused him to almost lose consciousness; his eyelids flickered in his pale face.
"Hold on," Michal told him; "this is going to hurt."
With difficulty, she managed to help hoist Caleb onto his bicycle seat. He was sweating and his teeth were gritted, but he said nothing.
Then Michal wheeled him across the scorched field, trying to avoid the stones and ruts. It was slow going; his weight made the bike difficult to steer and they had to stop every few feet to keep him from sliding off.
When they stopped for what felt like the hundredth time, Michal spoke up.
"It's too far," she said. "We won't make it."
The girl was clearly unused to physical exertion. Dressed in skimpy clothes, she was sweaty and red faced as she gasped for breath. Caleb knew she was right.
Desperate, he looked around. Visible in the distance was the Source, and a sudden idea came to him.
"You could bring me back supplies," he said. "Fresh water, bandages."
Michal frowned, staring at the ground, weighing something in her mind. Then she looked up.
"I should take you there," she said. "It'll be safer."
Caleb glanced at her sharply. "What do you mean?"
"I got my own room. Past the loading dock."
"What about the guards?"
Michal shook her head. "They got cameras everywhere. But when I left, I shut off the ones along the back. No one will notice if we hurry."
She seemed sincere. Yet it felt like insanity to Caleb to go to the very place where his intended executioner was.
He noticed that blood was soaking through the makeshift blue bandage. Plus, he was having trouble breathing and his skin felt cold and clammy. He had no choice but to trust Michal.
"All right," Caleb said finally.
As the sun rose, the two continued their slow and painstaking way across the field toward the gleaming white building in the distance.
When Caleb came to, it was dark and cool. For a moment, he had no idea where he was.
Then the memories came stuttering back as scattered and disjointed images.
_The sun blinding his eyes. The hooded guard appearing on the horizon, raising his hand. Then the split-second understanding of what was about to happen, when it was too late . . ._
He jolted upright, but a deep pain radiated from his upper chest, and he gasped. It hurt to breathe and he couldn't move his left shoulder.
"You're awake," said Michal.
She was sitting beside him, wringing a washcloth into a plastic basin. Without her jewelry and makeup, she looked years younger, like a little girl.
"Where am I?"
"You're in my room," she said. "Don't worry . . . no one never comes in here."
"How long have I been here?"
"I found you this morning. Now it's long past sundown."
She handed him a cup of water. As Caleb drank, he glanced at Michal's quarters. Everything was not only new, but impossibly clean: the bed, the silky quilts and pillows that covered it, the rug on the floor. The walls were decorated with glossy pictures hung in frames, not just images torn out of magazines; and scented candles burned on the bedside table. He had been changed into a fresh shirt and underneath it, he could feel the crackle of medical gauze and tape.
He shook his head. "You're his girl," he said. "Why are you doing this?"
Michal blushed and glanced down. "Levi disrespected me." Her voice was so low he could barely make out what she was saying. "He . . . he offered me to you. In front of everyone." When she looked up, Caleb was startled by the flash of rage in her eyes.
"He thinks I'm worthless," Michal said. "But I'm not."
She stood. "I heard what you said that day . . . that you're looking for your son. Well, a few months ago, somebody brung a baby here. It was real late, but I could hear it crying. Levi wouldn't let nobody near it. Not even me. But it's still here, somewhere. And I know how to find out for sure."
Caleb had been staring at her. Now he swung his legs to the ground. With difficulty, he stood and grabbed his pack.
"Show me," he said.
Moments later, they were outside her room. The girl moved through the darkened corridors and down a back stairwell. Aware of where the cameras were, Michal was careful to take a circuitous route, yet one that was not so unusual that it would arouse suspicion.
Caleb trailed a short distance behind. Each breath he drew was agony. Yet he ignored his discomfort and focused on navigating his way through the treacherous terrain without being seen.
Occasionally, Michal ran into a guard. When she did, she stopped to flirt with each one, touching the boy on the arm, laughing, leaning close. To Caleb, it was the oldest and most obvious trick in the world; nevertheless, he was impressed to see that it always worked. Each time, he was able to make his way past the guard, unnoticed.
Soon, he and the girl were in the basement. In the hallway, he waited behind a cluster of pipes that ran from floor to ceiling as Michal slipped ahead to a battered steel door that lay ajar, halfway down the hallway. A sign attached to it read SECURITY.
She disappeared behind it for a moment and reappeared, frowning. "There's supposed to be a guard here." Then, she made up her mind. "That's where you need to go. Look careful, but be quick. I'll be out here in case anyone shows up."
Caleb nodded, and slipped inside the room.
At first, he was puzzled by what he saw. It was a small and windowless office, a scuffed cement cube painted gray. A set of dusty metal shelves held pieces of forgotten equipment; above it, a clock on the wall read an eternal 2:47. In the center of the room was a large table that supported a bank of twenty or so small screens.
Caleb approached.
Each screen showed a flickering image in grainy black and white. It took him a moment to realize that these were different locations within the Source: the front door, the loading dock, the dark aisles with their endless shelves of crates, Levi's office. As Caleb leaned closer, he was stunned to see that these images were moving: _A guard walked past a closed door. Levi leaned back in his chair. Another guard carried a box across a room._
And in one screen, a toddler sat on the ground.
The image was so grainy, it was hard to make out. The child had its gaze downward and appeared to be studying something on the ground. Then it clapped its hands, laughing, and for an instant, raised its head.
Caleb gave an involuntary cry and leaned forward, his palm on the glass screen. If this were a window, he would smash through it now, to reach deep into the past, to grab hold of what he had once possessed and then lost.
His son.
Kai.
"Miss?"
Outside the security room, Michal jumped and whirled around. A guard stood in front of her, hulking and anonymous in his black hood. A moment ago, she had thought she heard something and went to investigate. She was unaware that in the echoing hallway, the sound was in fact coming from the other direction.
Badly startled, she laughed. Even to her ears, it sounded false and hollow.
"You scared me," she said. That much was true. She tried to sound playful and again, failed miserably.
"What are you doing down here?" Without being able to see his face, it was impossible to tell what the guard was thinking, feeling.
Her mind raced. "Levi sent me," she said at last. "He thought the camera on the front door was broken. He wanted to know if there's still a picture."
The guard grunted.
"Let's go check," he said, and turned to go.
She grabbed hold of his arm. "I already did," she said. Then she gave him a tremulous smile and tilted her head, opening her eyes wide and softening her expression.
It always worked.
But this time, something was wrong.
He stared into her face for a long moment. Then he pushed her away.
"What are you playing at?" She could see his eyes, small and suspicious, glinting through the holes in the fabric.
Michal drew herself up haughtily. "Nothing," she said.
She could feel his eyes on her back as she sauntered away. She only prayed they were talking loudly enough for Caleb to hear and that he had had time to take cover.
In truth, she was trembling with panic.
Caleb heard the voices in the hall and froze. He could not make out what was being said, but even so, he sensed the alarm in Michal's voice, pitched high and shrill.
She was talking that way to warn him.
There was nowhere to hide in the tiny room, so sparsely furnished with its table and equipment. Even as he glanced around, the dented metal door was swinging open and he ducked behind it, flattening himself against the wall. The moment the guard entered, Caleb managed to slip around the door.
He was not expecting to find two more of Levi's men in the narrow hallway.
They were on their way to some sort of break, their defenses down; one had his hood partly pushed up so he could tear into a piece of flatbread with his teeth. The other glanced up and for an astonished second, locked eyes with Caleb.
But Caleb was already on the attack.
Sprinting forward, his adrenaline overriding his pain, he raised one elbow and rammed it into the throat of the second guard. Choking, the boy staggered backward against the wall and slid to the floor with a thud.
The other guard, bread falling from his open mouth, let out a roar and rushed at Caleb. Caleb bent forward, digging his injured shoulder into the boy's gut. Grabbing his hood, Caleb flipped him over his head, slamming him on the floor. As he lay there, dazed, Caleb kicked at his holstered weapon, freeing it from his belt and sending it skittering across the floor and under a rusted radiator at the end of the hall.
Caleb had bought himself a few precious seconds, enough time to escape; but he forgot about the guard inside the security room, who now came barreling out, as hulking and enraged as a wild boar. Caleb scrabbled in his pack for anything that might stop him. The moment before the guard reached him, his fingers closed around something.
Perfect.
Caleb whirled his arm in a roundhouse punch and smashed a rock into the center of the guard's hooded face. There was an audible crack, and with a scream, the boy reeled backward, clutching his nose. As his legs buckled beneath him, it was easy to grab one knee, yank it hard, and twist; this sent him crashing to the ground. Blood spurted from his nostrils, splattering his robes and the cement floor.
Caleb was halfway down the hall, heading for the stairs. There was only one thought on his mind.
_Kai. Kai was alive and somewhere inside the Source._
But where? On the main floor, he ran through darkened aisles, surrounded on both sides by towering shelves stacked high with cardboard boxes. The pain in Caleb's shoulder pulsed powerfully in time with his heart; but the sensation seemed far away, as if it belonged to someone else.
Ahead of him and high above, he saw something glinting, light reflecting off glass from some distant source he couldn't yet place. Instinctively, he headed toward it.
Caleb reached the entrance to a wide ramp leading upward, with handrails on either side. It was a peculiar thing, unlike any he had ever seen before, almost like a mountain road inside a building. Caleb hesitated. Behind him, he could hear the heavy steps and distant shouts of Levi's guards. They were searching for him through the aisles, spreading out across the vast floor. Making up his mind, he raced up the surface and into the darkness.
At the end, he paused to get his bearings. By the echoing void that surrounded him, Caleb sensed he was now in a large and empty space, devoid of shelved goods. As he moved forward, his eyes adjusted to the gloom. Soon he noticed a faint glow in the distance. He was in a cavernous room marked by giant rectangular columns, and he moved from one to the next until he reached the source of the light.
It came from a small, box-like space seemingly carved out of the back wall and exposed, doorless, to the larger room. Two guards stood on either side. The bottom half of its three walls seemed to be made of some battered metal; upward, they were thick chain link, the kind you would see on a fence or cage. Inside, Caleb could partly glimpse what seemed to be a desk; the light came from a lamp. Occasionally, a boy's hands moved into view and then away.
Caleb removed his backpack and took out his weapon. Then he loaded it with all the ammunition he had left: three rocks.
Placing it on his shoulder, he aimed meticulously; there was no room for error. He fired once, hitting one guard in the temple with a loud crack. As the boy sank to the floor, Caleb shot again, but this time, his aim was off; he only grazed the shoulder of the second. The guard looked up: Caleb followed with a third shot that cleanly hit the space between the boy's eyebrows, just above the bridge of his nose. With a grunt, he also dropped to the floor.
The room was now undefended as Caleb stood before its threshold.
Levi was behind his desk, papers strewn in front of him. Alone, the pale boy was paler still as he stared at his intruder. For an instant, fear and confusion flickered across his features. Then he acted as if nothing unusual was happening.
"Caleb," Levi said. "What are you—"
Caleb gave the other no opportunity to speak. Like a hawk after a mouse, he lunged across the wooden surface that separated them. Using both hands, he grabbed Levi's lapels and dragged him across the desk, scattering everything that lay in their path.
With newly found strength, Caleb lifted the other boy clear of the desk; he felt the black fabric strain and rip under his hands. Then he threw him down onto his back on the floor, straddling and pinning him to the cement.
" _Where is my son?_ " he shouted.
Levi's hand moved. He attempted to lift a meager weapon, a thin, decorative utensil with a blade too dull to cut but pointed enough to stab. Caleb seized the slender wrist and twisted it; with a cry, Levi dropped the weapon and it clattered to the ground.
The two remained locked, one under the other, both breathing hard. The pain in Caleb's shoulder had spread down his arm, weakening his grip; still, he did his best to ignore it.
Even in the losing position, Levi maintained his composure. Caleb had to restrain himself from crushing Levi's throat. Instead, he grabbed the small weapon on the ground beside him and placed the sharp tip against the side of the other boy's neck. He took grim pleasure in noting that the pale blue vein there was pulsing wildly, betraying the boy's show of cool.
Caleb pressed the tip of his knife in deeper. A bead of blood welled up, the bright red in shocking contrast to the white skin.
Through a fog of emotion, something clicked in his mind.
_The fire. His house, burning to the ground, with Miri trapped inside. A freakish mutant attack, as senseless as it was deadly._
_Kai, seized from his cradle. After so many months, most likely dead or gone forever._
_What Esther saw at the mutant camp: goods from the Source._
_And now, his baby, under this roof._
"You hired the mutants," Caleb said. "To kill my partner, to kill me. You paid them off, to steal my boy. And now, you have my son. And you still want to kill me."
Levi was watching him, saying nothing. It was like they were playing a game of dare.
"Why?" Caleb said. "Just tell me why."
When there was no reply, Caleb forced the blade even deeper into Levi's neck. Blood started to run, dripping down and pooling on the cement floor.
Levi stopped struggling. Ever practical, he knew there were to be no more secrets—not if he wished to live.
He looked away from Caleb and into the distance, into the past.
"I shouldn't be surprised you managed to survive," he said. "You were always the stronger one. You were always lucky. That's what this was about."
Caleb started, confused, and his hold on the knife wavered.
"What are you talking about?" he asked.
Levi turned his gaze back to him. "That's why they gave me away. Because even though we were both so young, you were clearly the only one worth saving."
Caleb sat back, stunned. Part of his mind fought what he was hearing even as the words started to make a terrible sense.
"You're my brother," he said at last.
"I was," Levi said.
No longer pinned, the boy in black sat up and got to his feet. He managed to keep his poise and walk back to his desk.
"I don't blame them, really. At least not anymore." As he spoke, he took a handkerchief from a drawer and dabbed at the blood running down the side of his neck. "As you know, keeping one child fed is virtually impossible, let alone two. Why wouldn't our parents keep the stronger, younger boy and cast out the sickly, older one? That is what animals do, isn't it?"
Still on his knees, Caleb was finally able to answer.
"That's not what they told me," he said, his voice catching. Inwardly, he was reeling with disbelief; the idea that they were related was obscene, unthinkable. "They told me you were ill. They said you died years ago."
Levi smiled; it wasn't a pleasant sight.
"As far as our parents were concerned, I did die," he replied. "Happily for me, some animals behave better than others. Strangers pitied me, and took me in. I was never strong. But even as a small child, I knew how to use people, how to make them do what I wanted. Soon, I started to build a new family, to make friends. Or at least collect acquaintances."
As he spoke, Levi had begun to clear up the mess on his desk. He stacked papers, put his writing utensils back into their container, arranged files, restored order. Caleb, mesmerized by the boy's actions and words, made no move to stop him.
"Of course, between us, what I was really doing was cobbling together an army. Or would you call it a gang . . . a posse? Well, a group of boys, anyway, strong and I'll admit not very smart boys who nevertheless respect power almost as much as they like having their bellies filled on a regular basis. You've met some of them—or at least they've encountered you."
He gestured at the door, outside of which the two guards still lay motionless in a heap.
"They followed me when I came to Prin, five or six years ago," Levi said. "I always knew I was smart. But it wasn't until I came here that I learned how to read and improve myself. Suddenly, I understood what potential there was . . . not just for scrabbling together an existence day to day, but for real power. I planned how to break into this building, and with my army's help, I eventually succeeded."
His workspace orderly, Levi once more assumed a position of authority. He sat behind his desk.
"And my boys have been good scouts, as well," continued Levi. "They've brought me back things from not only Prin, but places far beyond. News, mostly. But also goods, trinkets, pretty little things they thought would amuse me. Like Michal."
Caleb looked up sharply and Levi laughed.
"But most important, my guards managed to find out what I _really_ wanted to know all these years . . . and that's what became of you."
Levi rubbed his temple. For a moment, his expression seemed haunted; traces of pain and longing appeared as deep lines etched around his mouth and eyes. Then they disappeared, like shadows.
"They knew where you were," he continued. "They told me you had partnered and had a son. Again, you were lucky. I've tried to father an heir many times, with different females. And yet I can't."
Levi's voice, normally so controlled, broke at this last statement. And as Caleb looked at his brother, he finally recognized the similarities in their faces, things he had sensed at their first meeting but couldn't name.
"Once I knew where you were," Levi was saying, "I decided to take from you everything, just as everything had been taken from me. Most important, I would have a son, a true heir, one linked to me by my own blood. One who will carry on what I'm about to achieve."
At last, Caleb found the strength to stand. He leaned forward on the desk in front of him. Only the whiteness of his knuckles revealed the intense emotions roiling inside of him. He looked into his brother's eyes.
"You can't have him," he said.
Levi nodded. Then he reached behind him to a panel embedded in the wall and pressed one of three buttons.
The room shook.
Caleb looked up, startled. Behind him, two large metal plates, the doors to the office that had been hidden in ceiling and floor, were sliding shut like jaws in a mouth; the two boys were quickly trapped in the small space.
Then, with a grinding of ancient gears, the entire room began to move. Through the wire mesh walls, it was clear that they were advancing down a dirty brick shaft.
"Use your head," Levi said. "The boy's much better off with me than he would be with you. What have you possibly got to offer him?"
"I'll kill you first," Caleb said.
"Go ahead," Levi said. "When the doors open, and they will any second now, you will be greeted by all my guards. I know you can handle a few mutants waving rocks; I'm not so sure what your chances will be against eighteen of my guards carrying Tasers. Especially if I'm dead. And where will _that_ leave the boy?"
With a sudden lurch, the room came to a halt and the huge doors behind Caleb began to grind open again.
"Forget the boy," Levi said. He sounded so sensible, even wise. Like a big brother, in fact. "It'll be easier for you—for everyone—if you're far away."
"I'm not going anywhere," Caleb said.
Now the doors opened. Four hooded guards burst in and surrounded Caleb. Levi held them off with a raised hand.
"Very well," he said. "I can't force you to leave. But I have my own plans, bigger plans than you can ever understand. I asked you to join me in them before, but you refused. So now I will not tolerate your interfering in them. You can stay if you want. But if you do, you are not to lift one finger to defend the town or its people from the mutants again. Do you understand?"
Taken aback, Caleb hesitated. "What I do is my own business."
Levi shrugged. "If you disobey me, no matter where you are, I will hear word of it from my boys. And the moment I do, I will kill your child. Do I make myself clear?"
Caleb was furious—and flabbergasted.
"You would do that . . . after all the trouble you went through to track him down? Even after you tell me you—"
But Levi cut him off. " _Do I make myself clear?_ "
Caleb nodded, uncomprehending. When Levi gave a signal his guards seized Caleb and started to drag him out. This time, he was reminded of the pain, which had returned worse than ever; he could not resist even if he wanted to.
In the main room, the guards dragged Caleb forward on his knees. He kept his face impassive, refusing to give them the satisfaction of knowing he was in agony. But he could not keep the blood from dripping from his shirt again, leaving a glistening red trail.
"Enjoying yourself?" one asked.
Caleb gritted his teeth as they yanked him around a bend. Ahead, he saw that the giant main door of the Source was open, revealing the tarry black sky.
The guards picked him up and threw him outside. Caleb landed hard on his hands and knees, scraping himself on gravel and broken glass.
His backpack and hat were thrown after him and landed nearby. Next to them fell pieces of his weapon. They had been ripped from each other and scattered, like a squirrel's bones from a cat's mouth.
Caleb was about to pick them up, when he sensed a guard standing above him.
"This is for the others," he said.
He was holding the weapon all the guards wore, the plastic box with two wires at the end that crackled with blue fire. Caleb was trying to crawl away when the guard rammed it into the small of his back. He screamed as a bolt of white-hot electricity erupted through his spine and exploded in his brain, seeming to set everything in his body on fire. He dropped to the ground, immobilized.
Nearly unconscious, he heard the three guards walk away, chuckling. The door of the Source creaked and then slammed shut.
Caleb lay there, motionless. It was as if his entire body had been scorched from within. He felt blood from his wound seeping into the hot, baked earth beneath him. It was all he could do to open his eyes. Still, the physical pain was nothing compared to his emotional anguish.
All along, it was one person who destroyed his family.
It wasn't the mutants after all. For months, he had blamed them; poisoned by his rage and hatred, he had worked obsessively to track them down and destroy them.
_Mutants._
For the first time, Caleb was struck by the ugly word, one he had used a thousand times without thinking, and he winced. For they, the variants, were nothing but pawns, poor and pathetic; had it not been them, Levi could have found someone else desperate and hungry enough to do his bidding.
The variants weren't responsible; it was his brother. The person who was in a real sense closest to him, his own flesh and blood, had set out to destroy him . . . and very nearly succeeded.
When Caleb thought of the months Levi must have taken to plan and carry out his campaign, his mind reeled. Levi's revenge was no impulsive act done in the heat of anger. It was carried out with clear eyes and cold calculation.
If what Levi said was true, he had been brutally wronged in childhood. Still, Caleb couldn't imagine lifting a hand or plotting against his own brother, especially a brother who was innocent of any wrongdoing.
Caleb's only sin was the fact of his birth.
Levi waited until they had gone. Then he got up and retrieved the small dagger that had dropped to the floor.
After wiping the blade clean, he placed it back on his desk, next to the matching penholder and leather blotter. He discovered that he was trembling and breathing fast, nearly hyperventilating. He had waited years for this moment, this long-anticipated revenge on his younger brother; and his victory was all the sweeter in that he had won it by his wits alone.
Caleb's days as the town's savior were over. Prin would need a new hero. And that would be easy enough to arrange.
Levi knew he should be glad. Yet, strangely, he was not.
One thought continued to nag at him: Somebody must have told Caleb about the boy. That meant someone, like his parents so long ago, must have betrayed him, someone he had trusted and housed and fed.
Levi called together a meeting of his guards.
Now he leaned against the wall, watching as one by one, his guards were tied to a steam pipe against the wall and questioned. Assisting in the interrogation was a tool from the gardening and patio aisle, a black wand filled with fuel. When a button was pressed, a small flame blossomed out from the top. As the smell of butane mixed with and then was overpowered by the stink of charred flesh, the basement of the Source echoed with screams that no one could hear from the outside.
Levi watched not because he enjoyed it; in fact, the constant weeping and pleading wore on his nerves. He had to make sure his instructions were being followed; he could not be certain that his guards would do their job. The very one asking questions, after all, would soon be the one interrogated by his peers.
While a clumsy system, it had always served its purpose before. Yet after an hour, no one had confessed to anything. By now, Levi was tired of not only their tears and moans, but the glimpses of their squinched and sweating faces, only partly hidden by their hoods. They looked as pink and helpless as cornered mice. It sickened him.
Exasperated and impatient, Levi was about to begin the cycle of interrogation once more, when a guard, shaking, separated himself from the group.
"I hate to say it . . ." His voice was almost inaudible. "But when the stranger come down to the basement, there was someone who came out of the security room only a second before." He swallowed hard. "It was your girl. And she was acting funny."
Levi stared at him. He felt almost faint with anger: the gall of the guard to try saving himself with such a blatant lie! Yet the more he thought about it, the more it made a hideous kind of sense. If he had been betrayed, why shouldn't the treachery be of the worst, most intimate kind? That was the story of his youth: Wasn't his childhood trust paid back with cruelty and abandonment? So now, wasn't it likely that he had been forsaken by the person who had always said she loved him?
Levi kept his voice calm, revealing nothing.
"You better be right."
He ended the session.
Levi returned to his office, any sense of jubilation forgotten. Yet as he sat alone, brooding, he felt strangely content. Once again, the world had proven itself to be the way he had always known it to be: faithless and cold. It was reassuring.
He would search Michal's room himself.
## TWELVE
BY DAWN, CALEB REACHED THE SCHOOL.
He fell against the front door and leaned upon it, breathing hard. It had taken nearly everything he had to walk back. He was light-headed from blood loss and thirst, and exhausted from fighting. When he closed his eyes, he could still feel the lingering shock of the electrical current surging through his body.
He stumbled toward his schoolroom, and took two faltering steps over the threshold.
Then he saw her.
Esther looked thinner than ever, her face burned by the sun. His shock giving way to joy, he moved quickly to her. But he could only go so far before his legs buckled, and she had to catch him before he fell.
Caleb was almost unrecognizable: dusty and bloodied, shivering in the heat. He clung to her as if he was drowning, and she led him to his cot, where she eased him down. Then she fetched water to clean his wounds and for him to drink.
Esther's own journey here had been as much a hardship. Escaping from the Valley of the Dead, she had eventually found discarded robes in an abandoned van. She put them on over her clothes, and they both protected her from the sun and shielded her identity. The irony was not lost on her: Now that she was Shunned, she was finally dressed like one of the people she had long disdained and who had exiled her. By the time she reached the school late that evening, she was not able to remove the robes fast enough.
To her disappointment and misery, the schoolroom had been empty.
Esther was never good at waiting. Unable to sleep, she paced up and down the room all night, glancing out the window and door, listening in vain for the sound of Caleb's footsteps.
And now that she was alone with him, her relief was replaced by fury at the people who did this to him, as well as the need to cherish and make him well again. She had come here seeking his protection, but now she realized he needed hers just as badly.
"What happened?" she asked, once he had stopped drinking.
Caleb told her, haltingly. He spoke of seeing his child, the revelation of Levi's kinship, and his terrible retribution for their parents' neglect, hiring the mutants to carry out the dirty work he would not do himself. At last, when there was nothing more to say, he buried his face in his filthy hands.
And he cried.
Esther could not bear to see it. She alone understood that Caleb wore a tough exterior, a mask to protect what was important and precious inside; she knew this because it was what she did as well. For him to drop his guard meant that he no longer had the strength to fight off his despair, and she felt his pain as keenly as she felt her own.
She reached out and grabbed his hands.
They were icy cold and she wrapped her own hands around them, trying to give him her warmth and life, the way she knew variant shamans did to the dead. Her thumbs traced the ridges of blisters on his palms, the rough and bruised skin along his knuckles. Then she pressed her palms against his, and their fingers intertwined.
She raised her eyes to his face, at the stubble built up over days. Then she looked into his eyes and felt herself overwhelmed by his intense gaze, which mixed relief with gratitude and something more. As he caressed her hand, she lifted her fingertips to his face, tracing his cheek to his chin, then toward his lips.
"It's going to be all right," she whispered.
But something had shifted inside her, a strange new emotion moving into the other. Her desire to ease Caleb's suffering had been joined with another desire, one even more powerful, like two streams meeting and converging in a riverbed, mingling in a current against which she had no strength.
She had never known this feeling before.
Her fingers found his lips, which moved together and pressed against her fingertips, so softly she could hardly feel it. Then he kissed them again, this time harder; and at the pressure, she felt her body tremble.
"Caleb," she said, his name meaning everything she could not say.
Then their lips were on each other's, at first brushing there, like a question. Each applied more pressure, first quickly, then lingeringly. Esther closed her eyes, feeling only his mouth, his hands clasped in hers.
It was the first time she had ever kissed anyone. And it was different from how she imagined it would be, better, more dizzying.
Esther knew Caleb had been partnered, so he had kissed another, done more than kiss. And he was older than her, by more than a year. Yet she yearned to be closer to him and so she parted her lips, sending him a signal, indicating he could do the same. She was not certain but thought he understood, the way you did when someone opened a door, then walked away, saying it was safe to enter, you were allowed.
His tongue was softer than she expected, and warmer, and smooth. It filled her mouth, exploring, and she let it. Then she responded, pushing her own tongue to meet his, and each teased the other a little. It was the most delicate and intimate thing she could imagine, comparable to nothing.
Then the kiss ended. And as the two pulled away from one another and smiled bashfully, Esther suddenly felt so much older, years and years older, than she had only a few moments before.
As for Caleb, he was stunned by the tenderness he felt toward Esther. It was the first time he had touched anyone since his partner's death, the first time he had allowed himself to open up and care for anyone.
He put his hand on Esther's head and caressed her jagged hair. Esther must have cut it herself, he thought. It even resembled her, in a way: sharp in appearance but delicate to the touch.
"Tell me what happened," he said.
Unemotionally, she told him the facts of her Shunning, her journey south, and what she had encountered. She was now a fugitive in her own town; if she were discovered, it would mean certain death for her and anyone who harbored her. When she finished, Caleb knew they were on their own. They were deprived of the people they loved, the places they knew, protected by just each other and the brick and broken glass of the school.
"What do we do now?" she asked him.
"Stay here," he said.
"For how long?"
"Forever."
Yet, even as they huddled together, they could hear something happening outside. Without speaking, Caleb got to his feet, followed by Esther. Holding hands, they approached the window and stood there, looking out at the street with a new feeling of dread. For in the distance, they could hear a sound that neither of them could place. Then, they recognized what it was; and the realization sent a deep chill through them both.
It was the baying of wild dogs.
On the far end of Prin, the townspeople tried to flee, but it was no use.
A dog had his teeth buried in the leg of a girl as she kicked at it, vainly attempting to drag herself into a doorway; her white robes were torn and soaked through with her blood. Two other dogs set upon a boy who was on his way to a Gleaning. Swerving, he fell off his bicycle into the dust of the road, where another dog slashed at his arms as he tried to shield his face.
Other dogs were galloping and lunging at anyone unlucky enough to be trying to escape. There were at least two dozen of them in all, skeletal and cringing beasts that were nevertheless fast and strong, crazed now with the bloodlust of the hunt.
Behind them, their owners, Slayd and three of his warriors, surveyed the road and gave commands.
Slayd was not without misgivings. He and his tribe used these dogs to hunt wild animals, never people. Yet Levi had been clear: He wanted something different, more terrifying.
The variant leader noticed someone attempting to crawl to safety under an abandoned car and whistled three of the dogs over. He made a soft, chucking sound and the animals obeyed.
It was the red-haired boy from Caleb's class, one of the town's handful of trained protectors. All week, he had been practicing his punches and club work; but what good was a small leather sap or even a metal pipe against three wild animals? One of the dogs seized his ankle. As the boy screamed and tried to fight it off, the others managed to drag him out.
There was a flurry of canine bodies as they piled on, the sounds of growling and snapping, and the piercing shrieks of the boy. When the screams abruptly stopped, Slayd whistled the dogs over. They did not want to obey at first; a variant warrior had to go to them, brandishing a club to break them up. When the dogs joined the rest of their pack, their eyes glittered madly and their muzzles were stained red.
The townspeople who were lucky enough to have been indoors watched from windows. A few who had been attacked were able to stumble to safety; others were somehow pulled or carried inside, bloodied and torn. But there was no way to warn the rest of the town. People could only look on, listening to the sounds of mayhem as Slayd and his men continued their bloody way toward the center of Prin, their pack of dogs roaming ahead and freely attacking anything that moved.
Yet one boy stood, unmoving, by the side of the road.
It was Caleb.
As one by one the people of Prin grew aware of his presence, word started to spread. He was their protector; only he could save them from this latest calamity. His backpack held his weapon, the terrible instrument that could strike down five men in as many seconds. Surely, he would use it now, take it and spring into action.
_He will help us. He will protect us._
Yet their certitude turned to confusion, then doubt, then sheer panic and disbelief. As the devastating attacks continued, Caleb still didn't move. He remained frozen in the eye of a hurricane, the mindless blur of maddened dogs and snarls and screams, the desperate attempts of people trying to escape. Caleb had been drawn there by the instinct to protect. Yet he knew he could do nothing without risking the life of his son.
In the near distance, hidden behind a dusty truck, Esther also stood, motionless and white faced.
The townspeople shifted their attentions from Caleb. There was a sudden commotion from the other end of the street. It was the agonized howl of a wounded animal.
A wild dog lay in the gutter, its legs kicking feebly as it attempted to bite the feathered shaft that protruded from beneath its ribs. It gave up, muzzle smeared with its own blood; and as the scrawny body gave a final shudder, the brown head lolled back, mouth and eyes open in death.
Striding into the center of the chaos was Levi.
He was flanked by six of his boys, faces covered by their black hoods. Three carried loaded hunting bows. When one spied a wild dog, he took aim and fired. Already, they had shot six or seven, and the air was full of the screams and yelps of dead and dying canines.
As he approached, Levi's eyes caught Caleb's for a moment. He smiled faintly, before turning to a guard.
"That dog," he said, pointing down the street.
Slayd and his men were the last to notice what was going on; the variant leader appeared stunned when the largest dog of the pack, the animal closest to his side, let out a sudden yelp and collapsed at his feet.
He whirled around and with a look of utter shock, saw Levi and his men twenty feet behind him. Two of the hooded guards had their bows raised, aiming directly at him.
"Levi!" he shouted. "What are you—"
But Levi cut him off. "I give you and your tribe ten seconds to get out of Prin," he said.
Slayd's look of confusion turned into one of pure hatred. There was no way to comply with any kind of dignity. With a quick nod to his men, he and the other variants sprinted down the street and were soon gone.
The people of Prin, unaccustomed to good news, were moving about the street in an impromptu celebration. The ragged sound of cheering carried all the way up into the apartment above Starbucks.
Inside, Sarah was getting ready to join them.
At last, there was something to be happy about, after Esther's Shunning, something in which to lose herself. It was a welcome distraction to select the right colored robes to wear, a soft pink that flattered her increasingly pale complexion. Now, she combed her hair one last time in front of her small mirror.
Earlier, when she had heard the first reports of Levi's victory from her open window, she felt almost dizzy with pride. As she recalled the lingering kiss he had given her that night in the Source, Sarah shivered, closing her eyes. She only wished Esther was there to share this moment.
For she had a gift for Levi, something that would all but ensure her future with him.
Sarah took it out from under her bed. It was the book he had asked her to find: _Topographical and Hydrospheric Tables of the American Northeast_. The words meant nothing to her. All she knew was that he wanted it, and so she had found it. It was as simple as that.
She had spent dozens of hours searching in vain, wasting early mornings as well as long nights after working all day. She had picked through the crumbled remains of the town library and pored over the thousands of waterlogged and deteriorated books that filled the shelves of a large store off a Prin side street. Stubborn though she was, she nearly gave up hope; it was like trying to locate a single pebble in a giant field.
But after racking her brain, she was able to remember where she had found the first volume. During a sudden rainstorm years before, Sarah had been forced to take refuge in a cluster of crumbling buildings called "College." In a basement, she had stumbled upon a cache of books. Squatters were burning the texts to ward off vermin. When Sarah asked, they allowed her to take a few volumes. She later used them to help teach Levi how to read.
Sarah had retraced her steps to the basement, now deserted. And, miraculously, she had found the companion book, charred and mildewed but still readable, on its long-forgotten shelf.
Sarah anticipated telling Levi the details of her search. She would have to phrase it so it didn't sound like she was bragging; that would be unseemly. But he was sure to understand and then compliment her on her cleverness, her resourcefulness.
It was not only her own future she was considering. Levi was the only one capable of making and bending the rules of Prin. Once the two of them were partnered, he would surely listen to her entreaties and lift the sentence on Esther, overruling Rafe. Then her sister would be allowed to return.
Once outside, Sarah found that the streets were thronged with people, hoarse with shouting. She found she didn't mind. Laughing, she allowed the crowd to bump against her, guiding her along, as she hugged the book to her chest. More and more people joined them, and together, they surged forward in a delirious mass.
Everyone rounded a corner, and there, partly visible through the crowd, was the boy they were looking for.
"Levi!"
Sarah's voice was drowned out by all the cheers, the shouting. She pushed her way forward through the crowd, intent on reaching him.
"Levi!" she called again. Jubilant, she waved the book at him. "I've got it! I found it!"
As she squeezed her way through the packed bodies, there were fresh cheers around her as Levi climbed onto the hood of a battered SUV. Now that he could be seen by all, he turned and waved at the townspeople. Was he acknowledging her?
There was a gap in the crowd and Sarah stepped into it. She was almost there; he must be able to see her now. But at that moment, Levi was leaning forward, helping someone from the street clamber up onto the car beside him.
"Levi!" By now, Sarah's voice was hoarse.
And then she froze.
Levi was pulling close a girl. She was young: impossibly, cruelly young, with golden hair and gleaming skin revealed by tight, thin clothing and set off by jewels that sparkled at her throat and on her arms. Levi paused to kiss her, then turned back to the crowd, accepting their cheers, their love.
Sarah could not breathe.
She let the crowd surge around and then past her. They were all screaming now. They were surrounding the car and rocking it back and forth, banging on the metal in rhythm. Laughing, Levi clung to the girl as they swayed perilously, as one with the crowd.
No one saw Sarah standing there. No one could sense the thoughts that whirled around and around in her head.
Levi had a girl: someone far younger than she was, someone far prettier. With a dull shock, Sarah realized, _And he probably always did._
Which meant that he had used her.
It was as simple as that, and as heartless. He sensed she could help him, and so he played off their shared history, manipulating her feelings for him. He lied to her to get what he needed.
Afterward, did he and the girl laugh at her behind her back?
Sarah thought about how she had behaved at her dinner with Levi, how she simpered and flattered, and her face burned at the memory; what she felt now was worse than any physical pain. And when she thought of her fantasies about becoming his partner, her dreams of sharing the Source with him, the shame was like a dagger twisting in her chest.
Humiliated and heartbroken, she turned away from the spectacle.
But she didn't go empty-handed.
She was still clutching the book, the one he wanted and which she had come so close to handing over. Her eyes hot and dry, Sarah vowed to herself that she would never give it to him now, not even if he were to beg for it on his knees.
It wasn't much comfort. But for now, it was all she had.
The celebration was nearly over. The townspeople began to head to work, walking in animated groups of twos and threes.
Among them, Rafe was nearly drunk with pride. He had actually shaken Levi's hand and congratulated him. And Levi even listened to him, a handkerchief pressed to his mouth, as he made some of his suggestions for how to improve worker efficiency in the Excavation. Levi seemed to take him—him, Rafe!—seriously, and this had softened his view of Levi as a heartless tyrant. Rafe now hoped there could be a good working relationship between Levi and the town, ideally with Rafe himself as a go-between. This would solidify his quest for another term as leader, if he could survive to run for it.
He became aware that farther down the street, a lone individual was leaning against a mailbox. It was someone who hadn't come to the rally and honored the town's hero; and at this thought, Rafe felt a wave of righteous indignation.
When he got closer, he saw it was Caleb.
Rafe's feeling ripened to one of outright disgust. He walked up to his former champion.
"Coward," he said.
People on either side of him looked up. Caleb didn't move.
Behind Rafe, a girl was hissing. "Why didn't you help us?" she yelled.
Then someone in the crowd bent to pick up an empty plastic bottle. He threw it at Caleb and it bounced off his chest.
"Coward!" he shouted. There were scattered boos and curses.
Another person tossed a clod of dirt at Caleb, who still didn't move. The fact that he didn't defend himself infuriated the townspeople. One at a time, then more and more, they stopped to pick things up from the ground and threw them at him: dented cans, sticks, a dead pigeon.
Soon, at least twenty townspeople surrounded Caleb, safe in both their numbers and their anonymity. Jeering, they pushed him one way, then the other.
"Not so brave now, are you?" said one; and another spat in his face.
Rafe felt emboldened enough to yank off his own hood, exposing his face.
"Why don't you just turn around and get out of town?" he yelled, loud enough for the people in the back of the mob to hear. "And you best not show your face here again, if you know what's good for you!"
For an instant, the brown eyes met his own; and Rafe recoiled, taken aback by what he saw. For Caleb seemed neither frightened nor ashamed. Instead, his eyes blazed with a hatred Rafe had never seen before.
The boy staggered backward, wishing he hadn't yelled quite so loudly. Yet he sensed that Caleb wasn't angry with him, or with the people of Prin.
Caleb turned and, without a word, walked away.
There was no sign of Esther, yet he refused to believe she was not somewhere nearby. Then he noticed a flicker of movement from an alley.
She was there, lurking in the shadows. Their eyes met; hers were full of questions. He was about to call to her, then stopped.
Several townspeople had caught up and flanked him. Caleb only had time to flash a warning to Esther before he was led away from her, in the direction of the highway.
## PART THREE
## THIRTEEN
ANOTHER TOWN MEETING WAS ABOUT TO START IN THE ABANDONED restaurant in the center of town, the one with the yellow arches looming high on their steel pole. However, unlike the last time, the air was festive and Rafe himself had trouble hiding his exultation.
Several days had passed since Levi and his men came to the town's rescue during the latest attack, and the event had had a miraculous and lasting effect. Since then, there had been no fresh outbreak of violence, no new, senseless ambush of the town and its work teams. No one had even seen a mutant anywhere near the town's limits. They seemed to have been frightened off for good, gone from the face of the earth.
With one glorious and decisive rout, Levi had put an end to the terror that had gripped Prin.
Now Rafe had different news to share, news that was even more exciting and significant. It was a plan he had had a hand in creating and helped make happen with his cleverness and quick thinking. It was crucial that as many of the townspeople as possible were present so he could explain it to them properly. He needed an enthusiastic majority of the town to vote the right way—namely, the way he wanted them to.
He had promised Levi nothing less.
By now, he had been invited to the Source no fewer than two times. He and Levi had established a good working relationship, he thought. It was almost too good to be true.
As he raised his hand for attention, it occurred to Rafe that his recent dealings with Levi reflected well on him. Right now, as everyone in the room began to settle, they were staring at him with open admiration and respect, something they had never done before.
It was an intoxicating feeling, one he wanted to last forever; and with any luck, it would.
If the townspeople did as he suggested, it would cement his relationship with Levi. Rafe wondered what this might mean for him: a new position at the Source? He might be made some sort of assistant, maybe even Levi's second-in-command; and the thought of this made him shiver.
As always, he spoke softly, so the others were forced to lean in. "As you know, I been in discussion with Levi these past few days," he began.
The room murmured its approval. After the decisive way Levi had vanquished the mutants, even those who once distrusted him were now his supporters. Rafe was happy to see that some of the biggest doubters—he ignored the inconvenient fact that he himself had been one of them—now banged on the laminated tabletops as enthusiastically as the others and stamped the ground with their sneakered feet.
"He's done made a very generous offer to us," he continued, "one that I helped him think up." It dawned on him that no one would know if he stretched the truth. "But I don't know how long he's willing to keep it on the table."
He paused, and as he knew they would, everyone stopped fidgeting and hung on his words.
"He's willing to buy Prin from us," he said at last, his voice unintentionally cracking with excitement. He cleared his throat and tried again. "He wants to take the whole mess right off our hands. What do you all think of that?"
There was a moment of silence as everyone in the room digested his words. Then at once, they all began talking.
"What do you mean, buy Prin?" called out a girl holding a younger child in her lap. "If he buys it, where we gonna live?"
"We need to find a new place, obviously," said Rafe. This was an easy question, one of the four or five he had anticipated. "Someplace bigger, better. Because let's face it—we done picked this whole area clean long ago. There ain't nothing left for us here, especially after the mutants smashed the place up."
"But they ain't attacked in a while," said one boy. "Maybe they done coming after us." He sounded hopeful.
Again, this was something Rafe had anticipated.
"They only stopped on account Levi drove them off," Rafe said. "How many times do you think he's willing to do that?"
"Heck, if he wants it, I say he can have it," a boy shouted, and several people laughed.
A girl raised her voice. "What's he gonna give us in return?"
"He's going to pay us in water and food supplies," said Rafe. Again, it was an easy question; the meeting was going exactly as he had hoped. "I did some hard bargaining and here's what we come up with. He's willing to give each household half a crate of water and six months' worth of flour, mixed grain, beans, and salt. I say that's more than fair . . . supplies like that should last all of you a long, long time."
Again, people started talking all at once, trying to shout each other down. Rafe was glad to see that most in the room were on his side: nodding their heads, arguing with their neighbors.
However, there were more than a few who looked like they had reservations about the idea. Some were shaking their heads in disagreement. Several were deep in thought, frowning and thinking hard. They worried him the most.
One of them was an older girl, who sat huddled on top of a table, her back against a window. She spoke up.
"If there's nothing left here in Prin, why does Levi want it so bad?"
Rafe hadn't thought of this. And, in truth, he didn't really know.
He ignored her and tried to steer the discussion back to more comfortable ground, questions he knew the answer to and wished more people were asking.
"This is the kind of opportunity we been waiting for," Rafe said, more loudly than he needed to. He could feel a trickle of sweat start to work its way down the back of his neck. "Now we can head out and find ourselves a new place, a place to build on. In fact, we can start sending scouts as soon as we vote tonight. We—"
"You didn't answer her question," someone called.
Before Rafe could pretend he didn't hear and continue talking over this interruption somehow, another voice called, "Answer her question!"
Rafe licked his lips, trying to think of a way to get control again. People were starting to murmur, and doubt and skepticism rippled across the faces in front of him.
Trying to stave off disaster, he screamed, "I told you, this deal ain't going to stay on the table unless we act fast! He's being more than generous . . . he don't have to offer us nothing! We should be grateful he even wants to do business with us in the first place!"
But by now, others were frowning and shaking their heads, looking at the girl who spoke.
"She's right," a boy said.
The others sitting at his table were nodding in agreement.
"Prin ain't much, but it's our home," added another girl. "Got to have a better reason to leave it than a few months' worth of food."
Rafe was stunned that the mood had so quickly shifted and he was at a loss as to how to regain the upper hand.
"I say, let's vote on it!" someone yelled, and there was general agreement.
Rafe swallowed hard, his mind reeling. Although he had called the meeting with a quick resolution in mind, that was now the last thing he wanted. If he allowed a vote, it was obvious which way it would go.
"Now, this was just an informational meeting," he said. "Just to get the facts out. We'll be scheduling a vote at a later time."
He adjourned the meeting soon afterward. He stood by the door, stopping people to cajole or joke with them, trying to recapture some of the enthusiasm he had seen just minutes before. But even he was forced to admit it was a lost cause.
Desperately, all he could think was: _What would Levi say?_
A lone figure neared the Excavation.
Disguised by her robes, Esther averted her face from those who passed.
She had spent difficult days and nights, trying not to be noticed in the streets she knew so well, sleeping in abandoned storefronts and surviving on whatever supplies she could steal. Throughout, she was haunted by what she had seen happen to Caleb. Why had she stood by as he was insulted and spat upon? For the thousandth time, she rebuked herself for not rushing forward when he was being led away.
She had not seen Caleb since.
Esther had to find him; but she could not risk being found within the town's limits. She knew that to ask for help was both foolhardy and dangerous. She would not have come to this Excavation site if she had had another choice.
Even though it was crowded with abandoned cars and trucks, the asphalt area surrounding the Source was too exposed; it would be suicide to approach that way. Instead, Esther took the indirect route, circling far around and through the back fields. Creeping through the tall grass in order not to make any rippling movement, she was able to get close to the trench. Soon, she could make out the rhythmic clank of shovels hitting dirt and rock and the voices of workers calling to one another.
Esther lay motionless in the grass, only a few feet from the pit, with her eyes shut so she could hear better. As the team members yelled to one another, she found she could identify who each person was and she started to keep a silent tally of who was there.
"Break time," called a voice. "Lunch."
With her belly pressed low to the ground, Esther listened to the clatter of tools being tossed aside as one by one, the workers pulled themselves up over the ledge. She pictured how they looked, with their robes caked with dirt and clay. Most of them were probably retrieving their nylon backpacks, stashed in the backseat of a car or truck to keep them safe from wild animals. Now she imagined them pulling out bottles of water and plastic containers of porridge and beans, yanking down their masks and chatting with each other as they headed off to the relative shade of a small copse of trees nearby.
The last person to leave the pit took a moment to sit on the edge. Esther heard him knock the soles of his sneakers together to dislodge the red clay. Having tracked who had left, she was fairly certain who it was. More important, she was desperate enough to take the chance.
"Eli," she whispered.
The boy looked up, and Esther waved him over. When he saw her, his face broke into an unbelieving smile beneath his mask.
"Esther?" he said.
With a quick glance to see if anyone else had noticed, Eli exited the trench. Esther had already retreated into the tall grass. He stooped low and made his way toward her through the sun-bleached weeds that grew as tall as his waist. When he reached her, she was kneeling, almost completely hidden by the towering blades. Eli took off his gloves and pulled down his mask so he could speak.
"What are you doing here?" Although he seemed overjoyed to see her, his face was creased with worry. "If the others catch sight of you—"
"I know." She took a deep breath, then plunged ahead. "I need to talk with you. It's real important."
A flush broke over Eli's face and he stared at the ground, smiling hard.
"I was hoping you'd make up your mind soon," he said, his voice husky. Then he cleared his throat and looked up at her. "I can talk to the others. That Rafe, he's just full of air. If I go see him first, he's bound to see reason."
Esther stared at him, confused.
"What?" she said.
As usual, Eli wasn't listening. He reached for her and, to Esther's shock, took her hand firmly in both of his.
"We'll get that sentence thrown out," he said. "We'll see this through together. I promise you that."
Esther jerked her hand away before she realized what she was doing; too late, she saw the look of bewildered hurt flash across his face. Inwardly, she cursed herself yet again for her clumsiness, her rashness.
"I'm . . . I'm sorry," she stammered. "I can't be your partner. But that's not why I'm here. I need your help."
Eli looked as if someone had hit him, hard, when he wasn't expecting it. He stared into the distance, shaking his head. Then he laughed, mirthlessly.
"Help?" he said. "You tell me you won't be my partner, but you still say you want my help?"
Esther swallowed hard. "I—I'm sorry," she stammered again. She had blundered every step of the way, she realized now.
Yet she still needed him, and any help he could provide.
"I'm sorry to have to ask you. But I don't know who else to go to. You're pretty much the only person in Prin who's ever been nice to me."
Eli snorted. "And look what good that's done me."
He was still gazing off, blinking hard, his eyes bright. Yet without looking at her, he seemed to be softening.
"What is it you need me to do?"
Esther was overwhelmed with relief.
"It's Levi," she said. "He's been using the variants somehow . . . getting them to attack the town so he looks like a hero."
Eli nodded, pursing his lips; he appeared to be thinking. "He wants to buy Prin," he said. "That's what Rafe says, anyhow."
"We can't let him," said Esther.
"Well, I don't know that it's such a bad thing, Levi buying us out," he said. "Prin ain't got much left for the rest of us, anyhow. If he wants it so bad, why shouldn't he have it?"
"It's not just about Prin," she said. "It's Caleb I'm worried about."
At the mention of his name, Eli's face froze.
"Caleb?" he repeated.
Esther sensed trouble, but she plunged ahead regardless. "I need you to find him for me. I know he's been told to get out of Prin. But Levi's got his baby and I know he won't leave until he gets him back. I'd look for him myself, but I can't."
But the boy cut her off and Esther was stunned by the fury in the otherwise mild brown eyes.
"Your friend," he said, spitting the words, "got what he deserved. He ain't nothing but a coward. If only I'd have known what kind of trash you liked, maybe I would have stayed clear of you from the beginning."
Esther put her hand on his shoulder; this time, he was the one to jerk away. "You played me for a fool for the last time."
He started walking away, the grass swishing in his wake.
"Eli!"
But he did not turn around and soon was gone.
Esther sighed and sank back on her heels. She knew there was only one person now she could possibly approach, the only one (save for Joseph, whom she loved but knew was helpless in an emergency) who would even speak to her.
And it was the last person she ever wanted to ask for anything.
The sun was dipping low in the western sky by the time Esther stood outside the ruined mansion.
Unable to ask where the Gleaning crews were working that day, she had been forced to figure it out herself. Without a bicycle, she had canvassed much of the town on foot, searching up one long street and then the next for the familiar, purple-framed bicycle parked outside.
It was agonizing and time-consuming work, all the more so because she had to stay alert to the sound of work teams riding past. Yet at last, she had found it, on a street of broad lawns and large houses that was once considered exclusive. It lay in a tangle of other bicycles that leaned against an enormous uprooted tree in a large yard overgrown with weeds.
Behind the bicycles curved a long circular driveway that seemed to have once been picked up and wrenched by massive hands. The house itself resembled the face of an old giant, broken and toothless and blind. All of the windows were shattered or missing, and vines grew freely over the gaping chasm where the roof once stood.
It was almost dark; soon, the Gleaning crew would emerge, get on their vehicles, and head home with their haul. Esther squatted behind a toppled tree, rested her aching legs, and listened.
She did not have to wait long.
She heard voices, and then a person in filthy robes appeared at the door. It was a girl, talking to someone behind her. Four others stepped out onto the sagging and dilapidated porch, carrying a few filled plastic bags. It was an insignificant haul for such a large house. Esther assumed this was not the first time the mansion had been Gleaned.
One of the five called to someone still inside. "You go ahead," Esther heard from the depths of the house. The four clipped their robes close to their legs and mounted their bicycles. Then, carrying their meager haul, they pedaled off into the twilight.
Moments later, Esther stood in the doorway.
She could not hear anything stirring; and she ventured in, picking her way along a makeshift path that wound its way through piles of sodden trash, dead leaves, and the broken remains of furniture. By the dim light, she could see she was standing in the ruins of what was once the entryway, with rooms leading off on the left and a hallway in front. Next to it loomed what was left of the stairway, disappearing into the murky darkness.
"Hello?" she called.
The cavernous living room was down two steps; by the far wall, the ceiling had partially collapsed, crushing two of the eight windows with heavy wooden beams. As Esther edged down the steps, something skittered through the trash and disappeared into the pile of bricks that was once the chimney.
She gave an involuntary start when she saw someone across the room. Whoever it was huddled against a destroyed sofa, its head bowed nearly to its knees.
"Sarah?" Esther called out.
There was no response. Then the masked figure raised its head.
"Esther?" said Sarah.
For an unguarded moment, the joy and disbelief were naked in the older girl's voice. Then she caught herself and, once more, assumed her usual fretful, nagging tone. "What are you _doing_ here?"
"I need to talk to you," replied Esther.
She had been steeling herself for this conversation, one she had a premonition would go badly, the way they always did. "It's important. What are you doing?"
Sarah made a dismissive gesture and stood, leaning against the wall. "I was just resting," she said. "Did the others see you? They were only here a moment ago. Oh, Esther, how could you risk coming back like this? Why don't you ever _think_?"
Already, Sarah was talking to Esther as if she were a little girl. And even though Esther struggled to stay calm and focus on what was important, she instead found herself clenching her fists so hard, her nails dug into her palms.
"I need you to help me," she said in a low voice.
"Help you?" said Sarah. She laughed, but the sound of it was mocking. "I tried to do that. Didn't I? I warned you again and again, and you refused to listen to me. So how can I help you now that you've been Shunned? It's too late, Esther. It's much too—"
Esther cut her off. "I need you to find Caleb for me."
At this, Sarah fell silent for several moments.
"I see," she said. "And was this why you came back to Prin?"
"Yes," said Esther. "I—"
"No, no," interrupted Sarah. "I want to make sure we both understand this. This was why you're risking not only your life . . . but my life, too. Because that's what you're doing here, dragging me into this. You're risking both of our lives to save that boy. That stranger." Her accusation stung Esther, who stood in silence, taking it. "When did he arrange all this with you? Before you left? Did he tell you he loved you, make promises to you?"
"No," muttered Esther. Her face was hot with anger and embarrassment. "It's not like that. He—"
"You know what I think is sad?" said Sarah, almost to herself. "That boys say anything they want just to get something. And girls always believe them." Her voice caught for an instant, but Esther could not tell if she was about to laugh or cry. "I can't say I blame you. He rides into town and impresses everybody, and we all fall for it. Then he deserts us when we need him the most. And now you're willing to risk everything just to save him."
Esther couldn't stand it anymore.
"Well, you know what _I_ think is sad?" she shouted. As much as she hated them, tears of anger stung her eyes and she tried in vain to wipe them away. "That for the first time in my life, I need help. And the only person I can ask is you."
Esther turned to go. But as she crossed the threshold, some impulse made her turn around and look back.
Her sister had pulled off the scarf that covered her face.
Esther was shocked by Sarah's appearance. The dim light threw long shadows across her features, making her look gaunt and ancient. Her cheeks were sunken and her normally rosy skin seemed gray. Only her eyes glittered in the dark, too brightly, like obsidian.
"Sarah? Are . . . are you all right?" Esther said.
Sarah's head dropped forward and she again sagged against the wall. Quickly, her younger sister was by her side, kneeling next to her. Esther reached out to touch Sarah's arm and recoiled at the heat coming off of it.
Her sister was burning up with fever.
"You're sick," she said stupidly. She tried to touch Sarah's forehead, but her sister pulled away.
"No," Sarah said. "I'll be all right. I just need to rest."
"Come on," said Esther. She was on her feet. "Let me get you home."
"I can't walk," whispered Sarah.
Esther reached down to pull her sister up by the hands, but she was unable to stand. Esther then grasped her under the arms and tried to hoist her to her feet; but as Sarah's arms were raised yet again, the sleeves of her robes fell back, revealing the bare skin.
That was when Esther noticed the lesion.
It was round and small, no bigger than a child's thumbnail, purple and pink and glistening halfway between Sarah's elbow and shoulder.
Esther recoiled, her hand to her mouth.
Both of them knew what that lesion represented, and the fever and the weakness, too. Soon Sarah would be found out and driven out of Prin.
Esther refused to think of it. All she knew was that right now, she had to get her sister home and into bed.
Without speaking, she reached to lift the older girl. But Sarah pulled away.
"Don't," she whispered. "I don't want you getting sick, too."
Esther hesitated. Then with a start, she thought of something that hadn't occurred to her before.
_She'd given her water to the dying girl on the highway, had even touched her hands. That was days ago. And yet she was still alive and well._
That decided it. Ignoring Sarah's protests, she half guided and half carried her outside, propping her on her bicycle seat. Then with her sister's arms wrapped weakly around her from behind, Esther gripped the handlebars and pedaled standing up. She had not gone half a mile before she was drenched with sweat—not only from exertion, but from the heat radiating off her sister's thin body, pressed against her back.
Once they were home and she had helped Sarah up the stairs and into bed, Esther went into the kitchen. She had never prepared so much as a cup of powdered milk before, and now she glanced in despair at the meaningless utensils and bags of grain and flour stacked on their shelves, the unopened bottles of water. She was relieved to find a plastic container that still had the remains of rice porridge in it, leftovers from the night before. Scraping it onto a clean plate, she carried that and a glass of water into her sister's room.
Sarah was sitting up in bed. In the soft glow of her bedside candle, she looked almost normal and for a moment, Esther felt an irrational burst of hope. She sat by her side, placing the glass into her hands.
"Here," she said, with false brightness. "You'll feel better."
But Sarah did not drink. Instead, she took the glass and played with it, turning it around and around in her hands as she stared down at the bedcovers. Then she looked at Esther.
"I let you down," she said. "I think I let down all of Prin." There was a tremor in her voice. Then she bit her lip and looked away.
"Don't talk," said Esther. It panicked her to hear her sister talk this way. She held the plate of porridge in her lap and now, she lifted a spoonful of the meal to Sarah's lips, to keep her from saying anything more. "Just eat something. You need to eat."
But Sarah was shaking her head.
"You were always so willful," she said. "You never trusted anyone, even when you were little. You hated this town. I thought you needed looking out for, no matter how much you despised me for it. Turns out I was the idiot all along."
Her eyes were shining, and Esther was horrified to see that that her older sister—always so proud, controlled, seemingly perfect—was on the verge of tears.
So many feelings came rushing at Esther, it was impossible to make sense of them.
"I never despised you," she stammered. "You're my sister. You're the smartest person I know. The smartest person in the whole town."
Again, Sarah shook her head. "Book smart, maybe," she said. "But I was stupid enough not to notice what was going on, right under my eyes." She gave a low laugh, but there was no warmth in the sound. "And now here you are, in the same place. We're two idiots, you and I."
"What do you mean?"
Her sister said nothing, her lips pressed together as she gazed at the wall. "We're two idiots," she repeated under her breath, as if to herself.
Esther suddenly understood.
_Levi. The way her sister always stuck up for him, even when others in town turned against him. Her dinner alone with him at the Source, and her strange behavior afterward. The fact that she had stayed single, despite the offers._
_Sarah spent her entire life waiting for the boy she had always described as just a friend from childhood, nothing more. And Esther was too young and self-involved to notice._
Now she felt a wave of sympathy overwhelm her, as well as a crushing sense of sorrow.
"Caleb's not like that," Esther said. "I don't know what happened between you and Levi. But Caleb is a good person."
She hesitated to tell her sister the truth about Levi, then made up her mind. "Levi is trying to do something with the town," she said. "What it is, I'm not sure. But he told Caleb if he interfered anymore, he'd kill his baby." As she spoke, she could feel her sister's eyes on her, wanting her words to be true.
Sarah took a moment to digest her sister's words. Then she was reaching beneath her, fumbling under her blanket and quilt, and Esther instantly set down the food, concerned. Sarah pulled something out from beneath the futon mattress.
"This was what Levi wanted all along," she said, "so it must be important somehow. It's yours, now."
She handed it over and Esther took it. It was a faded gray book that was speckled with mildew, with pages that were warped and rippled. Esther was ashamed to discover she could not even sound out the words on the cover.
But now was not the time to ask her sister.
Sarah's eyes were fluttering shut, and Esther knew enough to let her sleep. She removed the glass of water from her hand and set it on the bedside table, next to the uneaten porridge.
She blew out the candle and was about to exit, taking care to leave the door ajar. But she was called back by a sound.
Her sister had pulled herself up to a sitting position. The effort cost her; when Sarah spoke, Esther was forced to bend her ear close to her lips to make out the words.
"People say that criminals and outcasts sometimes go to the fenced-in fields off the road leading to town," Sarah whispered. "You could try there for Caleb. But it's dangerous. If you go, be careful."
Esther squeezed her sister's shoulder in thanks.
## FOURTEEN
ALONE IN THE NIGHT, CALEB PACED IN HIS CELL.
But it was not really a cell. He only felt he was a prisoner. He was in a wooden stall, one of dozens in a broken-down, one-story building several miles from the heart of Prin. The stall had a cement floor still covered with decayed straw. There was an oversize wooden door, the top half of which was made up of iron bars; it swung on heavy hinges onto a dusty passageway. Metal rods also formed a bin that was bolted against a wall, with wisps of ancient hay still clinging to it. A strange contraption made of strips of rotted leather and rusted rings hung from a hook on the wall.
Caleb gazed at a desolate view outside through a crack between two planks. The moon was full. By its light, he could just glimpse the large, circular track he crossed to get here three days ago. The dirt surface was rutted, cracked, and barren, baked to a hard pottery by the endless sun. It was one of three such fenced-in tracks, all surrounded by the remains of large structures open to the elements, with risers on different levels. These, too, had been mostly destroyed by weather, looters, and time, with few of the seats still intact.
If there was anything here that was once of value, it had long since been stolen. There was nothing, not even decent protection from the sun and rain. That was why few came, only a scattering of transients and outcasts, the violent and the mad, seeking shelter near town yet away from those who had Shunned them.
Caleb did not blame the citizens of Prin for what had happened. He understood the depth of their fear. He also knew what desperate and unrealistic hopes they held out for him from the start, what miracles they thought he could deliver.
When he recalled Levi, Caleb was filled with churning emotions, a strange mixture of both despair and rage. He had lost a sibling at the exact moment he had found him. Until Caleb was able to get his son back, his only family, really, was Esther.
At the thought of her, the touch of her lips, her stubbornness, her spirit, Caleb felt his heart contract painfully. If something were to happen to her as well, that would be unbearable.
When he first arrived at the shelter, Caleb was racked by frustration, desperate to search for her yet unable to do so. Instead, he attempted to question the others who shared the building, lost souls who were fleeing their own pursuers or demons.
"Have you seen her anywhere?" he asked one boy, describing Esther as best he could.
The boy was scrawny and squirrely, someone who seemed to harbor unsavory secrets. He lay in a bed of dirty straw, not bothering to get up.
"For sure I seen her," he said. His voice was hoarse as if he didn't talk much and what few teeth he had were black with rot.
"Is that right?" Caleb said. "Where?"
The boy sighed. His rancid breath traveled across the stall and Caleb winced. "What are you gonna give me if I tell?"
Whatever he said, it was certain to be a lie. Caleb felt his face turn to stone. "How about I let you live?" he whispered.
The boy just shrugged, giggling. "I ain't seen her, anyhow." Then he turned away and feigned sleep.
That night, Caleb tried to search for Esther himself. By the light of a new moon, he made it as far as the outskirts of Prin, checking alleys, abandoned buildings, and other places an outcast might favor. But he saw no one, and the effort cost him; his wound reopened, began to bleed. He barely made it back to his stall, falling onto his makeshift bed before losing consciousness.
In the dark, he saw scenes that were more like visions than dreams.
_Levi, holding a bow and arrow, walked alone down the streets, his eyes obscured by his mirrored sunglasses. A barren field, with a girl in red trying in vain to hide in the branches of a tree. A sky that grew dark overhead with the threat of poisoned rain._
Caleb awoke with a start.
It was still dark. With difficulty, he got to his knees, then his feet. He gazed through the crack yet again and was on guard. In the distance, he saw a flicker of movement.
Someone was approaching.
The intruder was covered in a white robe, its face concealed. Caleb flattened himself against the wall, steeling himself for an encounter. If anyone attempted to overpower him in order to steal his meager supplies, Caleb would lose; he didn't have the strength to fight back. Still, he could bluff and attempt to intimidate whoever it was by staring him down first.
The person entered the building and left his sight.
Caleb moved to the door and looked out through its bars. Whoever it was headed down the hall, pulling its hood.
Esther.
Caleb stepped out from his cell. She froze for a second. Then the tension in her face and body melted as she ran to him. Their embrace was awkward at first, because of the bulky shoulder bag Esther wore; she shoved it aside so she could go into his arms. The two headed back through his barred door, into the cool darkness.
There the moonlight streamed in through the cracks in the walls and a jagged hole in the ceiling, casting long shadows. Esther could not remove her despised robes quickly enough. Caleb tugged the door closed behind them. It barely swung shut. Should anyone else pass, it was all the privacy they were going to have.
He glanced at her and cocked an eyebrow in a silent question. Esther realized that in the moonlight, he could see that her dirty face was streaked with tears.
"It's my sister . . ." she started to say, then fell silent.
She rubbed her sleeve across her eyes.
"Go ahead," he said. "You can tell me."
So Esther did. As she spoke, she felt that something long dammed up was breaking loose, sweeping away everything in its path. She talked not only of her sister's sickness; she talked about Sarah herself, about their long and painful relationship, full of recrimination on one side and resentment on the other. And she talked about the pain of beginning to understand who her sister was at the moment she was about to lose her.
When she finished speaking, Esther felt spent, drained of all emotion. Yet she also felt at peace, and forgiven somehow. She realized with a start that Caleb, too, had just lost a sibling. She knew that each was all the other had.
It was as if they were sharing the same thought.
"I love you," Caleb said.
Esther started. It was the first time she had heard the words from another human being. They changed things, these words, just as her first kiss had. There would be no returning to a world before the words were said.
"I love you, too," she replied.
There was a pause. Then Esther looked around, seeking something. She shrugged. She gripped the bottom of her red sweatshirt and tore a long strip of fabric from its hem. Then she looked at Caleb.
He smiled and nodded once. Even as Esther took the ragged piece of cloth and tied it around her right wrist, he was reaching over to take the other end. He knotted it around his right wrist, as well.
"I promise to be true to you and always be your friend," he said.
"And I promise to comfort and support you in all things," said Esther.
It was the partnering ceremony. It went beyond law and ritual, custom and decree. It was perhaps the only thing in their shattered world that was holy.
Their palms grew moist, the cord around their wrists tight and hot. The moonlight poured down on them. Held in place, otherwise unmoving, they kissed again, this time more deeply.
Esther found she was trembling. Then Caleb reached down and stripped the cord off both of their wrists, and tossed it aside. Together, the two lowered themselves to the ancient straw that littered the floor. Soon his shirt landed on top of the cord, and so did the rest of their clothes.
The two explored each other, gently at first, with hands and lips and tongues. Esther found the arrow wound high on Caleb's shoulder and kissed it.
But their urgency grew, the straw sticking to them. When Caleb entered her, Esther felt pain, shocking and sharp, and she cried out; but it dissolved into a swirl of other, greater sensations and emotions. Soon, they were moving together, awkwardly, then expertly, bright with sweat.
At last, the two lay still, naked and curled, their bodies gleaming white in the darkness, nearly indistinguishable from each other.
Esther found Caleb's hand in the dark and he intertwined his fingers with hers.
"This is forever," she said.
"Yes," he replied. "Forever."
At dawn, Caleb awoke.
Esther was huddled against him, breathing through her open mouth. He disentangled himself from her and stood. The day was already getting hot; he reached down and draped his shirt over her small, sleeping form.
He was drinking from one of the plastic jugs of water she had brought when he noticed a book lying in her messenger bag. Esther had mentioned it last night, when she spoke of her sister. He picked it up and squinted to read the title.
Across the room, Esther stirred. Caleb realized he had been reading out loud, sounding out the syllables one by one. She sat up when she saw what he was doing.
"That's Sarah's book," said Esther, her voice fuzzy with sleep. "The one she found for Levi."
Hearing his brother's name, Caleb recalled the stacks of paper on Levi's desk and his easy ability with written words; he felt a pang at his own ignorance. Yet as he sat next to Esther and leafed with her through the book, he was puzzled to see that it featured more than mere text. The pages were filled with rows of numbers, dense and tiny like black ants, next to strange, abstract images: squiggling lines and shaded areas.
"Why'd he want this?" he asked, bewildered.
Esther shrugged. She had pulled on Caleb's shirt and now knelt behind him, draping her arms around his bare shoulders, touching and exploring his hairless chest, the chest of her partner. He kissed and teasingly bit her hand, which she yanked away, pretending he hurt her. Then he returned to the book.
"Why would Levi want this?" he repeated.
She leaned over his shoulder to look. "Well, it must have something to do with Prin," she said. "Right?"
"I guess." Caleb flipped to other pages now. Up until now he had only seen crudely drawn diagrams, like the ones on Levi's wall. "You think these are maps?"
Esther shrugged. "If they are . . . they might be of Prin." She pointed. "Look, that could be the old lake. And those could be the mountains."
"So he wants to find something here?"
Caleb recalled Levi's words when they first met, the thing he hinted at as they stood watching the townspeople toiling beneath them at the Excavation.
_They're digging for something_. _Something important. Even precious._
Esther was furrowing her brow. "I always wondered about the jobs," she said. "Not the Harvesting. That makes sense, I guess, on account he needs gas for the Source. But that doesn't explain the other jobs. They say the Gleaning is to find stuff that's worth trading. But everyone knows that all the buildings and houses around here were emptied years ago. So maybe that's not what it's really for. Maybe it's so that people end up looking for something else. But what?"
"Like the Excavation," said Caleb. "Everybody is digging all across town. But nobody knows what they're looking for, either."
"Maybe this book tells him where it is, and that's why he wants it so bad."
The two continued flipping through the pages with greater urgency. Yet neither could read more than a few words, and the drawings had grown too confusing. They were so close and yet could go no further.
Frustrated, Caleb closed the cover. Then he placed the book back in Esther's bag.
The night before, both he and Esther had felt that the other was his or her missing piece. Together, they formed a whole that was invincible; nothing and no one else would ever be needed by either of them again. Now he realized that they were only two people, limited in both power and knowledge and ostracized from everyone else. And their opponent had never been more powerful.
But in the face of such hopelessness, Caleb realized there was also nothing to lose. There was still one thing he could try to do—the most important thing of all.
"I have to get my son," he said. "I got a feeling this is my last chance."
He started to stand, but Esther held him back.
"You can't," she said. "You're not strong enough."
It was true. Even the act of standing was exhausting and his left arm hung weak and immobile, useless in a battle. "Besides," Esther continued, "everybody in Prin thinks you're the enemy now. You'll never make it past the town, much less into the Source."
"I'll take the risk," he said.
"No," Esther said. " _I_ will."
Caleb smiled despite himself; Esther was brave and impulsive in equal measure. "That's crazy. The town will be on the lookout for you, too. And you've never even been inside the Source."
"So? You can tell me where I should look."
"But I don't even know where they got him hidden."
"I can search for him, then. I'm good at that." When Caleb hesitated, Esther pressed her point. "Whoever goes has to be able to move fast, without being seen. And maybe I don't know the Source, but I know Prin. It should be me."
Still, Caleb hesitated. He thought about the labyrinthine layout of the Source and Levi's boys posted on every floor. They both knew how dangerous the guards could be. "But you're no good at fighting," he said.
She shook her head. "I know. But I'm just going to get in, find him, and get him out without being seen. There won't be any fighting."
Caleb touched her cheek and then gave a brief nod.
When he thought about it later, Caleb wondered why he agreed. Yet at that moment, he trusted her absolutely.
"The Source seems to have three levels," he said. With effort, he squatted down to push aside the straw by their feet; then he drew his own inexpert map in the dust with his finger. "The goods are mostly on the main level. Levi has a girl, Michal. That's where her room is, around the back. It's near the loading dock, the other entrance. Your best approach is probably through there, though there's a kind of eye that can watch your movements. In fact, they're all over the place. There's a basement, and guards are everywhere. The top floor is mostly empty, although Levi's office can move up and down between all of them."
Esther listened as she pulled on her own shirt and jeans, and laced her sneakers.
"I don't know where they're keeping Kai," he continued. "They may have moved him since they know I saw something. I think your best bet is either the ground floor, off to the side. Or else the basement."
Esther nodded. She had taken the food and most of the water out of the messenger bag for him. She kept the rest, looping the shoulder strap across her chest. Then, atop it all, she pulled back on the robes that obscured who she was.
The last thing she did was pick up the strip of fabric from the floor, the partnering cloth. She tore it in two, wound one side into a ribbon and tied it around her wrist.
Caleb did the same, on the opposite hand as hers. Then he kissed her once, hard. He didn't say, "Be careful"; he felt he didn't need to. Yet after Esther was gone, as he lay there, he wished he had. Closing his eyes, he thought it over and over, like a prayer.
Esther ditched Caleb's bike on a scrubby hill behind the Source. She lay as flat as possible on the ridge, just out of sight, as she considered the imposing building. She was facing the back entrance at the loading dock. In contrast to the massive electrified front door, this one was small and poorly patrolled with only one guard on watch.
She had assumed she would be able to scale the wall and get in from the roof somehow; but now she was here, she realized that her plan was unworkable. The structure was too enormous and utterly smooth, with no apparent footholds: no one, not even the most skilled variant, could possibly get in that way. While her impulsiveness had always served her in the past, Esther realized that now there was too much at stake to be hotheaded. Instead, she would have to slow down, think everything through, and exploit whatever opportunities she could find.
To strategize, as Skar would say.
It felt like years had passed since she and Skar had played Shelter amid these scorched fields, hiding from the work teams as her friend taught her variants' skills. Esther felt she carried the spirits of both Skar and Caleb inside of her and was being guided by their love.
As if he could hear her thoughts, the guard looked over in her direction. Esther froze. He noticed nothing and turned away again. Restless, he now sat down on the metal steps that led to the Source's entrance and reached under his black hood to wipe his brow.
_Let nothing go to waste,_ Skar always told her. _Use what is available._
She glanced around, frustrated. Scattered in the dead, bleached grass were pebbles and twigs, but they were too small to hurl. Nearby, she found a crushed soda can, a broken pair of glasses, a dusty bottle cap. It was all trash, the usual insignificant garbage that cluttered the fields and streets of Prin. How could she possibly find any use for objects that were so worthless?
Esther remembered another thing Skar used to say, and it made her smile: _Make a bad thing into a good thing_. Certainly, everything about her current situation was a bad thing: the fact that she was alone, outnumbered, unarmed. Moreover, it was unusually hot even though it was early morning, with waves of heat billowing over the dusty parking lot. Although she rubbed her sleeve across her face, sweat dripped down, stinging her eyes.
But that might be the kind of bad thing Skar was talking about.
Prin's blazing sun.
Esther's mind raced. She might be able to find something that could reflect its rays, something that could blind the guard. Yet even as she considered it, she was already discarding this plan; he wouldn't stay blinded long enough for her to sneak by.
But another idea was forming in her mind.
Making sure to keep her profile low, Esther began to collect the detritus around her, scraping it off the ground: dry leaves, dead grass, papery scraps of bark. This she cobbled together into a scraggly pile.
She didn't have matches or a firestarter on her; those were precious items that were only kept at home. But she did have the broken pair of glasses. Esther picked them up and used a piece of her sweatshirt to rub the dirt off the one lens that wasn't shattered. When she raised it to her eye, she was satisfied to see how well it magnified.
She hoped it was strong enough.
She held it above the pile of tinder, moving it up and down by minute degrees until she was able to focus a ray of sun into a tight and tiny beam of heat. Then she waited.
A feather of smoke began to curl up, and Esther's hope grew. But it was too early to celebrate. So she kept the speck of heat trained on the pile as she began to gently blow on it. Before long, a small flame flickered upward, which she fed with more dead leaves and dried grass. The dry bed beneath it ignited and fire began to spread. The flames were about a handspan high now, but it was only when they were strong enough to leap to a nearby patch of dead grass that Esther crawled away on her belly, putting as much distance between herself and the flames while staying as close to the Source as she could.
Then she stopped and peered over the ridge.
The guard was watching the ground by his feet, still paying no attention to his surroundings. But after a few moments, he lifted his head and seemed to sniff the air.
He was mostly faced away from her; Esther couldn't tell from his stance whether he was annoyed by the distraction or happy to have something to do. In any case, he lumbered to his feet. Then he started walking toward the fire.
Esther sprang to her feet and, keeping low, ran farther along the ridge, increasing the distance between herself and the fire. Then she breached the crest of the ridge and, still running low to the ground, made for the Source and the loading-dock door.
Above it, she saw a strange metal box attached to a strut. This had to be one of the "eyes" Caleb warned her about. Praying that she was coming in at too sharp an angle to be detected and moving too fast for anyone to notice even if she was, she bolted up the cement steps two at a time and disappeared into the entrance, without an interruption. The door banged behind her, and the huge, dark, and silent interior seemed to swallow her whole.
Inside, Esther avoided any open areas and stayed close to the heavily stocked shelves, hiding in their shadows. She paused to get her bearings, remembering the map that Caleb drew. After she checked the ground level, she would try to make her way downstairs.
Esther ran as she was taught, noiselessly and low to the ground, keeping both hands in front of and beside her. She slowed down as she passed a door to what had to be the room of Levi's girl; it was empty. Beyond it, there were more enclosed spaces, as well as the remains of an eating area. She saw that all of these places were empty as well; and so she veered off into the main room again, back into the dim jungle of tall shelves stacked with cartons, trying to stay out of the sight of the mechanical eyes, wherever they might have been.
She was not prepared for the barbed wire.
There was a loud ripping sound as Esther was yanked backward. She had run too close to the shelves and had been caught by one of the giant coils of barbed wire that surrounded them, the sharp metal slashing through and catching in her robe. Cursing to herself, she backed up, reaching out with one hand to try to untangle herself from the cruel razor edges. Far off, she could hear the sound of a guard saying something in a loud voice, then laughing. She didn't know if he was approaching but she needed to free herself at once.
Yet it was hard enough to do without having to hurry. Esther tried to jerk free of the wire, not realizing it was now snagged on her sleeve. As she pulled away, tearing the garment, the entire coiled length unexpectedly came with her, knocking over the huge cardboard crate perched on the edge of a shelf.
Esther winced, waiting for the crash.
But instead of thundering to the ground, the carton fell without a sound. Puzzled, Esther pushed it with her foot. Then she stooped to lift it. Although the crate was large enough to hold four of her, it was so light, she could pick it up with one hand. And it was unsealed, its tabs flapping open.
The box was empty.
Now Esther was confused. She slid out another carton on the shelf, one that was located behind the first, a giant crate marked GOYA: DRY ROMAN BEANS: 160 UNITS. This box was also light enough to be pulled out with ease and had been opened.
It was empty.
Esther pulled down another carton sitting on the shelf above it, and another, and another. She turned and, taking care to avoid the barbed wire, pulled down the boxes behind her, along the entire row. The crates were weightless, bumping into each other like playthings for a giant baby.
Esther stood still in the dim aisle, the giant boxes piled up on the ground around her. Her mind was racing. She could not make sense of the fact that she was in the Source, surrounded by the cardboard crates that were filled with the food and water that kept Prin alive.
And yet they were all empty.
But she had no time to investigate further. Behind her, she heard the sounds of more guards talking, getting closer. Esther had made such a mess, it was only a matter of time before her presence was noticed. Making up her mind, she added to the pile, removing the ripped robes that could only slow her down. She had to get to the basement while she still had the opportunity. She took off again before she heard something that made her stop in her tracks.
A faint, high-pitched cry echoed in the darkness. Was she imagining it? No. Although she had only heard it a few times in her life, Esther knew what it was.
It was the sound of a baby crying.
It came from somewhere across the huge room and grew louder as Esther approached.
In a far corner of the main floor, she could make out two open doorways side by side in a recessed area behind a cement partition. Stopping in the shadows of a nearby shelf of boxed electronics, Esther squinted at the faded images painted on the wall beside the two doorways. The one to the right showed a crude image of a human wearing a short robe; the word WOMEN was painted below it. In the other, a similar figure was wearing pants; and below it was painted the word MEN.
The wails seemed to be coming from one of the two doors.
She assumed it was the entrance to the right; it was flanked by two hooded guards, weapons displayed in their belts. Again, Esther sorted through her options. The guards were bigger and heavier than she was, not to mention they were armed and she was not; yet with luck combined with the element of surprise, she might be able to waylay one and then get past the other. But once inside, what then? They would certainly come after her and there would be no escape. There was no way to tell if the room had another exit; she could easily be cornered.
Then her problem was solved for her.
A strange sound crackled through the air and one of the guards pulled out a black plastic device attached to his belt. "Security breach," it rasped. "Everyone, report to Section A-Seventeen. Aisle Five."
Esther ducked deeper into the shadows as the two guards, pulling out their weapons, lumbered past.
For the moment, the room was unguarded. Esther darted through the door. Inside, the child's cries were deafening, echoing off the hard interior. Everything in the room was white tile, gray metal, steel, glass. To one side was a series of stalls, each with a swinging door that opened on a built-in seat with a hollowed center. Old-fashioned and ridiculous, the seat was the ancient indoor waste removal contraption she had seen many times before, although rarely so many at once. Across from them was a bank of matching white sinks. Above that, Esther saw herself reflected in a long mirror that stretched down the wall.
And at the end of the room, standing in a wooden pen and sobbing, was a small boy.
Kai.
Esther started toward him.
Then she felt rough hands seize her from behind.
That night, the spill of electric lights revealed a short boy standing outside the massive front door of the Source. In days past, the building seemed to Rafe a sort of blind giant, a faceless god that watched over him and others.
All that, of course, had changed.
This was now his third visit, and so he was comfortable on its threshold, bathed in its welcoming shadows. That was not to say he was complacent about being there: No, it never ceased to give him a thrill to be near the impressive structure, and part of the elite allowed to enter.
Still, today, he was here on business, so he had to temper his excitement.
Levi had called him here so Rafe could give the results of the town meeting. And although Rafe knew this might be a _slightly_ difficult report to present, he was confident he could do it without issue. He mumbled under his breath, practicing what he intended to say. He changed his words and intonation again and again; he even imitated the facial expressions and hand gestures he would use. Then he looked up.
A hooded guard stood before him at the giant door. Through mirrored sunglasses, he stared impassively ahead.
"Excuse me?" Rafe said. "I'm here for the meeting."
There was a long silence. "Who with?" the guard said.
"Well, with Levi."
The sunglasses and mask could not hide the look of disbelief on the guard's face. Rafe made a mental note to mention this rude lackey to Levi, who he was sure would have something to say about it.
"Yes," Rafe said condescendingly. "Levi asked me to come. Tell him Rafe is here."
At this, the guard's eyes narrowed. As if from a great distance, the name rang a dim bell. "Stay here," the guard said. He disappeared behind the massive door, which slammed shut.
Rafe told himself to be patient. Yet many minutes crept by. When the door reopened, Rafe was soaked with sweat.
"Okay," the guard said. Rafe patted some of the sweat from his face and adjusted his robes. The guard ushered him into the darkness, despite Rafe's repeated insistence that he knew the way. Annoyed, he comforted himself with the thought that the insubordinate fool would one day be working for _him_.
Soon they were standing in front of the small office, lit by an overhead bulb. It shone on the head and shoulders of one sitting alone at a large desk, the features in his pale face cast deep in shadow.
"Levi," Rafe said. Aware that some more formal salutation might be required of him, he bowed at the waist.
There was no response right away.
"Hello, Rafe," Levi replied at last. "Thank you for coming."
Rafe gave a sidelong smirk toward the guard, as if to say, _what did I tell you?_ Then, with confidence, he turned back to Levi and plunged ahead with his prepared speech.
"I passed along your generous offer to the townspeople," Rafe said. "And naturally, it was way more than they were expecting. Frankly, Levi, between me and you, it was way more than they deserve. But surely, you must have taken that into account when you—"
"Get to it," whispered the guard.
"I'm happy to say," Rafe said, shifting quickly, "that there was widespread rejoicing. Just like you wanted, and just like I promised." He took a deep breath, aware that this next bit was the tricky part. "That being said, I think they were a little . . . well, let's just say they were overwhelmed by the offer. Don't get me wrong. They're with you all the way. Only they just need a little more time to make it official."
Rafe smiled, confident that this explanation would suffice. There was a seemingly endless pause, while he kept smiling. Soon the muscles on his face started to feel strained.
Levi murmured something, and Rafe leaned forward, cupping an ear. "Excuse me?"
Too late, Rafe realized that the words were addressed not to him but to the guard by his side. The hooded boy reached for something at his belt.
Then a lightning bolt hit Rafe's lower spine, rocketing through his body before it exploded out of his limbs and head at the same time. Rafe was facedown on the floor, convulsing in agony, before he passed out.
As the body was dragged from his sight, Levi paced, deep in thought. He was not surprised by the boy's incompetence, nor was he flattered by his cringing servility. As mildly entertaining as it was to see the expression on that idiot's face change, Levi was far more concerned about the threat to his plan.
Levi had given the people of Prin new reason to respect him. Then he had made them a fair offer to relinquish the town. Of course, he could never have paid them what he promised, but, most important, he had tried to make them part of the decision.
It hadn't worked. He could wait no longer for Sarah to deliver the book. The Excavation and Gleaning had both proven to be worthless expenditures. Shortages had grown so dire, he even heard there was a break-in at the Source earlier that day. Levi had no time to deal with the would-be thief, a girl from town, and only hoped the news didn't get out.
Levi was beset on every side by betrayal and ineptitude, while options faded and resources dwindled.
He had been counting down to this moment ever since he had taken over the Source, nearly six years ago; now, there was no more time to spare. Over time, he had devised a Plan B, a final solution that he hoped not to use. He feared it would be too unwieldy to execute, too cumbersome, and would strain the capacity of his team of guards to the breaking point. Now there was simply no choice.
Fleetingly, he hoped there wouldn't be too much bloodshed. Not that he cared about the people of Prin. But Levi knew that blood had a way of riling up even the most complacent of animals.
Still, if he had to, he was willing to risk chaos.
## FIFTEEN
IT BEGAN EARLY THE NEXT MORNING, BEFORE THE SUN HAD EVEN RISEN.
At each dwelling, it was the same routine. There was a discreet knock at the door; two hooded guards from the Source stood outside. They carried no weapons on their belts, and they were courteous, even bland. One of them requested that the inhabitants leave their home and come with them. To the inevitable questions, the answers were always the same, vague yet reassuring at the same time:
_It's Levi's idea. It's for your own good. Levi will explain everything shortly. There is no need to bring anything. You will be returning home soon._
A few had more pointed questions. One or two even attempted to refuse and in private, they were more forcibly persuaded. Yet while they were confused and half asleep, most of the people of Prin felt that they could trust Levi. Docile and obedient, they came outside without a fuss, closing their doors behind them.
It was only when they were outside did some of the townspeople start to feel uneasy.
A long line of their neighbors snaked its way down the main street of Prin. As more and more people were roused from their sleep and brought outdoors, they were shepherded into the line by the hooded guards.
There were a handful of young children, some still asleep, carried by parent or guardian or trailing behind by the hand. One guard stood to the side and handed them toys from the Source: pinwheels, soap-bubble bottles with built-in wands, dolls still in their plastic wrapping. In the silence, there were delighted exclamations and sounds of laughter as the children opened their gifts and played with them.
"How long will this take?" asked a townsperson.
"Not long," replied a guard. "Get back in line."
In fact, the walk took longer than anyone expected, much longer, as they passed the town limits and continued west along a two-lane highway surrounded by the remains of a forest. Soon, the youngest started to cry, as the novelty of the toys wore off and the discomfort of the forced march began to mount. By now, the sun was well up in the sky. There was no water, no shade. Nearly everyone was still in their sleeping clothes and had no protection from the burning heat: no head coverings, no sunglasses. Many were barefoot. Any skin that was exposed had long since turned pink and then red. Soon blisters would form that would eventually blacken.
And still, they were forced to walk.
It was impossible to escape, even to stop for rest; guards were now positioned on both sides of the lines, and by now, they displayed bows and arrows, metal clubs, and the electrical weapons that had been hidden under their robes. Whenever those in line encountered a break in the road, a rupture of cement and underlying dirt, they were forced to scramble over it, sometimes on their hands and knees. When anyone faltered or stumbled, he or she was yanked to their feet and sent back to the line with an electrical shock or a resounding blow to the shoulders.
After four hours, the people of Prin reached their destination.
It was a large house, larger than even the grand old homes in the wealthy section of Prin, set off the road and hidden by a dense covering of overgrown trees and vines that surrounded it. On closer inspection, it seemed to have been spared not only by earthquakes, but by Gleanings, looting, and vandalism. It still retained much of its old-fashioned beauty, faded yet intact. Two large cars sat in the circular driveway, one silver and the other a dark blue. In fact, while both were dusty, they were untouched; they might never have even been Harvested. But that was not what the people of Prin were thinking about.
The entire house was surrounded by dense coils of barbed wire.
Back in Prin, a team of two dozen was assembled on the central street.
They were criminals, vagrants, and castoffs recruited early that morning by Levi's few remaining guards, a ragtag mob of the desperate picked up from the fenced-in fields on the outskirts of town. They were all desperate to work, frantic for the meager allotment of food and water that had been promised them as payment.
According to the rough map drawn up by Levi, the work would start at the center of town and progress block by block.
Working in teams of two, the day workers entered the homes that had been evacuated. Once inside, they gathered everything they could—furniture, clothing, housewares—and carried it outside, dumping it into the street. Soon, the air was full of the sound of smashing wood and glass and plastic. If any stores of food or water were found, they were carried to a separate pile, where two of Levi's remaining boys stood watchful guard, metal clubs drawn as they made certain that nothing was held back or hidden in pockets.
Within hours, the street was littered with detritus.
The workers were ordered to begin the next phase. Under close watch, they wielded construction tools, valuable objects found over the years at Gleanings and stored in the Source for just such an occasion: crowbars, axes, shovels, even several chainsaws and a jackhammer. Much of the cement flooring and underlying foundations in the homes of Prin were already badly cracked and irreparably damaged. With effort, it was not difficult to work the cracks open even farther, revealing ancient gravel, moldering two-by-fours, and the dirt underneath.
Unless great care was taken, an unexpected cave-in could happen in the blink of an eye, sending heavy beams, sections of floor and ceiling, and even entire buildings crashing to the ground. But there was no time for care. There were too few people to demolish too many homes and, for Levi, speed was of the essence; his last three guards, armed with Tasers and batons, made sure of that.
Wood and plaster chips rained down on crude living rooms and kitchens as clay, shattered bricks, and rubble began to pile up in what had been people's homes. When the walls got in the way, the workers destroyed them, smashing them with sledgehammers and sending clouds of plaster dust spilling into the street. When two townspeople were discovered hiding in their homes, the workers barely paused from their labors. In both cases, they surrounded the unfortunate resident with upraised axes and shovels; and while the screams were piercing, they were brief.
By the end of the day, the workers had broken through the basement floor in nineteen buildings along the central block.
But so far, it was useless. Despite their efforts, they had found nothing beneath, beside, or within what had once been the homes of Prin.
On the other side of town, Joseph was in his apartment, thinking about cat food.
Of everyone in Prin, he alone had no idea what was going on. He would only find out much later that Levi's men had attempted to search his building for townspeople in order to evacuate them with the others. The precarious condition of his rotting stairwell, however, proved to be too daunting an obstacle and they gave up many floors beneath him.
Joseph was pondering his dwindling supplies. True, he was long accustomed to setting squirrel traps on his roof and down in the courtyard, but that was primarily for his own sustenance. His cats, however, were spoiled and preferred the food Esther brought them—dried rabbit, oat cakes, boiled rice. Yet he hadn't seen his friend since her unexpected visit during the last storm. Joseph missed her company. His cats, however, were not nearly as sentimental. He could only imagine their outrage when they saw what was on the menu today.
Armed with a mallet, Joseph walked down the many flights of stairs, accompanied by the best hunters of his brood, a tabby called Stumpy and a black cat, Malawi. Although the felines were not fond of eating squirrels, killing them was a different matter. In particular Stumpy could be counted on to deliver the final blow if he lost courage. They reached the ground floor, and Joseph was about to shepherd his feline companions across the lobby and toward the courtyard door, where the traps were.
Instead, he stopped in his tracks, as did Stumpy and Malawi. They all heard it: There was a repeated banging and hammering sound.
Someone was in the basement.
These were not the sounds of a Gleaning; these were focused and purposeful in a way that made the boy uneasy. Joseph crossed the lobby and slipped inside the door that led downstairs; his cats followed. This stairway was not nearly as precarious, although it was quite dark and he had to feel his way along the wall, reaching out with his feet for each step.
By the time he reached the bottom, the noises had stopped.
Joseph headed down the long dark halls, where old corroded pipes lined the walls around them. Halfway through, he paused.
"Hold on," he whispered.
Joseph was convinced that his cats knew several words of English or at least the meaning of certain human inflections. The animals halted, their tails flared with interest.
The boy could make out the muffled sound of voices. It seemed that more than one person was down in the basement, several yards ahead.
As Joseph crept onward, the cats behind him, he became aware of dim shadows flickering against the basement wall. Whoever was down there had a lit lamp, or given the weakness of the light, perhaps a candle.
Even without looking, he knew where the voices were coming from. His visitors were in the boiler room. His boiler room.
"This is it," said a boy. He sounded excited.
"What do you think he's gonna give us?" said a second.
Joseph tiptoed closer, then flattened himself against the wall near the doorway. The cats started to curl around his feet, bored of this game and wanting a new one.
He peered through the gap, careful to keep out of view. In the light of a candle stuck onto a brick jutting out of the wall, he saw two boys, their faces red from exertion and gleaming with sweat. Later, he would find out that these were two of Levi's criminal recruits. Both were holding tools of some kind and were gazing down at the ground, staring with greed and wonder at what was revealed by a newly widened hole in a floor long damaged by earthquakes and decay.
It was a gently natural spring, caused in turn many years ago by a fissure in the earth. It had been there as long as Joseph had been in the building. In fact, it was where he got drinking water for him and his cats, collected in a plastic bucket he brought down to the basement.
Thoughtful, he pulled back from the sight, lightly kicking away the cats. It hadn't crossed his mind that anyone might be searching for his water and that strangers would come in with shovels and picks to help find it. Then he realized with a start that no one had ever known about the spring. He wondered why this was so and discovered that the answer was simple.
No one had ever asked.
## SIXTEEN
CALEB WAS UNAWARE OF THE FIRST STRAY BEAMS OF SUNLIGHT THAT illuminated the ceiling of his cell. Unable to sleep, he had spent the night working by the light of a small fire, fixing his weapon. It still lay in pieces before him, but it was nearly finished.
It had been two days. And Esther hadn't returned.
He rebuked himself for the thousandth time for letting her go to the Source alone, even though he knew they had no other option. Caleb realized that he was far from whole. When he tried to move his left arm, a stabbing pain seized his chest, forcing the breath from his lungs and blood to seep through his bandage, now filthy and encrusted. Yet he had seen enough of Levi and his guards to understand what they were capable of. He had to go to Esther's aid, no matter what the cost.
Caleb fit the final piece of his weapon into place. As he tested the wheel to make sure that it spun, he recalled with frustration that he had no ammunition left. He would have to collect some on his way, a time-consuming task because it entailed finding stones that were the right shape and size.
Still, there was no other choice. He hoisted his pack onto his shoulders, wincing at the effort. The sooner he was on his way, the better.
He heard the creak of a distant wooden floorboard and Caleb froze.
The day before, Levi's guards had come and rounded up everyone hiding in the other cells. Even though Caleb had no idea why this was being done, he knew enough to hide. He had managed to pull himself up to the rafters just as a guard peered into his stall. After a cursory glance, the boy had moved on.
Now Caleb heard the sound again, faint but distinct. Someone was creeping down the hall, someone who did not want to be heard. With a pang, Caleb glanced at the smoldering remains of his fire. The smell of smoke traveled far. He had been aware of the risk last night but decided to take the chance anyway. As a result, he had apparently brought his enemy back to his door.
Caleb lay his pack down on the floor. Without ammunition, he would have to depend on his bare hands; but in his weakened state, he was not sure how much fight was in him. He would have to strike first, and strike hard.
He moved to the wall next to the door and flattened himself against it. Out in the corridor, he heard another wooden door being swung open, then after a pause, closed. Then two more at once. Then another. More footsteps. The sounds moved down the corridor, coming closer to where he stood, waiting.
Caleb did a quick calculation based on sound. There were at least two people outside in the corridor, perhaps even three.
One of them stood outside Caleb's cell. The door swung outward and someone stepped inside.
It was not one of Levi's guards. It was a townsperson of medium build, wearing a hooded robe. But Caleb could take no chances. Before the stranger could turn, Caleb made his move.
He clamped his right hand over the person's mouth. At the same time, Caleb wrenched the other's left arm up by the wrist, yanking it up his back as he dragged him (for it was a male) into a shadowy corner.
The boy struggled but Caleb's crushing grip warned him not to continue.
"Don't move," he whispered.
An eternity of several seconds passed in which the two figures stayed frozen in the dark, locked in a tableau of adrenaline and mutual fear. Then the door swung open again and a second boy, short and slight of frame, entered.
"Eli?" he said. "Where'd you go?"
There was no reply. He turned to leave but at the last second, noticed the two boys standing in the shadows. He froze in the doorway and at that moment, an older girl joined him. Turning to see what he was staring at, she gasped in disbelief.
Caleb was confused. The two were not only unarmed; they seemed surprised to see him there. If he wasn't their intended prey, then who was?
"What are you doing here?" he asked.
"We're just looking for a place to hide," the boy said. "Away from Levi's guards."
Caleb's grip relaxed; and the person he was holding, the oldest of the three, broke free violently. He joined his friends, rubbing his shoulder as he glared at Caleb.
"What do you mean?" Caleb asked. "What's happened?"
The boy and girl, Bekkah and Till, explained all that had happened to Prin over the past twenty-four hours. How the entire town was rounded up. The forced march. The crowded barracks outside of town where everyone was compelled to stay, without food or water. And how the three had managed to escape, by climbing up to the roof and jumping past the barbed wire when the guards weren't watching.
Throughout, the oldest one, Eli, said nothing and stood glowering in the corner.
As he listened, Caleb realized that whatever Levi was planning, Esther was in greater danger than either of them had anticipated. If Levi had driven everyone out of Prin, this meant the situation had escalated. There was no way to tell what he was going to do next or how far he would go to make sure that nothing interfered with his plans.
"And so you came here, to hide?" asked Caleb.
The boy and girl traded a look, then shook their heads.
"We got to do something," Bekkah said. "What's going on ain't right."
"Only we don't know what to do," added Till in a smaller voice. He sounded plaintive.
Caleb realized that his position had improved. A few minutes ago, he had been an army of one. Now there were three more on his side.
And yet, when he looked them over, he could not help but feel dismayed. They weren't even among the townspeople he had trained; they were inexperienced, physically slight, and at least two of them were too excitable. Yet they were all he had to work with.
"I know something you could do," he said.
Bekkah and Till looked at him, willing to listen. The biggest one, Eli, still hadn't lost the angry and contemptuous expression he had worn since he wrestled his way free, and Caleb wondered what was causing it.
"I want to go to the Source," he said. "But I can't do it alone."
A quick glance passed among the three.
"But . . ." Bekkah hesitated to say the word, then plunged ahead. "You're a coward."
"Yeah," said Till, emboldened as well. "You run away from the last fight with the mutants. We all saw it."
Caleb knew he could not afford to get angry. "That's something you're going to have to judge for yourselves," he said.
Bekkah and Till both nodded. But it was Eli who spoke.
"Why," he said, "should we help you?"
His voice radiated a contempt that Caleb could not figure out. There was something beneath his response, something that cut deeper than the humiliation and shock of having been ambushed moments ago. Whatever it was felt personal, so Caleb looked at him and spoke without guile.
"This isn't about helping me," he said.
"No?" Eli was snide. "Then who?"
Caleb took a deep breath; he found he was shaking.
"Levi is holding two prisoners," he said. "One's my son. The other's Esther. My new partner." He swallowed hard; it was the first time he had said the word out loud. "I love them both," he continued. "Before I see either one of them hurt, I'll die. I swear I'll die in the Source if I have to. If you're not willing to do the same, then don't come with me."
To Eli, the word _partner_ came as a blow to the stomach.
He hadn't realized how much he had been hoping all this time that Esther's infatuation with the stranger was just that: a fleeting whim, a momentary obsession. If he was patient enough to wait, he had reasoned to himself, she would surely change her mind someday and choose him instead.
To learn that the two were partnered altered all of that. Even now, his unwilling eyes took in a detail he might have noticed earlier, had he been looking for it: Caleb wore a strip of red cloth around his wrist. It was a partnering tie; and Eli did not have to look twice to recognize it as being torn from Esther's sweatshirt.
Eli grudgingly found himself impressed by Caleb's bravery. It was more than that: He was moved by his sincerity. Did he, Eli, love Esther any more than Caleb did? Could anyone? If Eli refused to help him now, he knew it would only be from pettiness and jealousy. He was not certain he could live with that.
Eli looked Caleb in the eye.
"Just tell me what to do," he said. Behind him, the other two seemed ready to go wherever Caleb led them.
But Caleb was already thinking ahead.
"We need more people on our side," he said. "Those who can actually fight. And I think I know where to find them."
In the mansion across town, rumors were spreading like brushfire.
It began with the guards. Standing in the shade of a giant elm that dominated the front yard, they whispered among themselves of something that had been discovered in the basement of one of Prin's buildings, something precious. Inside, a boy eavesdropped through a shattered window and caught a handful of isolated phrases and words. He passed them on to another, then another. Word spread and as it did, the rumor transformed and distorted until it was accepted as fact.
_Release is imminent. We are going home._
But the longer the townspeople waited, the more doubt grew. Soon, panic and despair began to take hold. It was now well into the second day. There were nearly a hundred people crammed together in the heat, in a building meant for a quarter as many as that. By now, the conditions were unspeakably filthy. What was worse, they had not been given anything to eat or drink since they were brought here.
On the first day, two boys and a girl had managed to escape by making their way to the roof and jumping past a stretch of barbed wire that was unguarded. No one had followed. It seemed much too much of a risk to take, especially when an explanation, not to mention supplies, were surely coming at any moment.
But now, hysteria whipped the flames of a new rumor.
_We have been abandoned by Levi. There is no more food or water left at the Source. We have been brought here to die of thirst and starvation._
Shouting and pushing, townspeople fought their way into what had once been the opulent kitchen. They tore open the chestnut cabinets and the oversize refrigerator, smashed the glass windows of the ovens, ripped open drawers in the pantry, and dumped the silver cutlery onto the ground. There were rodents' nests and scurrying insects amidst the dusty china and crystal glasses, still intact after all these years, and cardboard boxes and metal food canisters with contents that had long since rotted to nothing.
In one cupboard, a girl found a forgotten can of tuna fish, bloated and stinking of decay. She and another girl fought over it, pulling each other's hair and slapping. The winner tore apart the corroded metal with her bare hands and ate the putrid mess inside. Not long after, she was curled in a corner, retching violently.
All day, the air had been thick and humid, and now, the sky darkened as wind began to whip the trees. When the rain came, the guards took shelter in a nearby building used to house cars. Inside the mansion, flashes of lightning revealed that rainwater was dripping from the many holes in the ceiling and leaching down the stained plaster walls.
Most of the townspeople shrank back in terror. Others were too desperate to care. They stood, mouths open, trying to catch the drops on their tongues. Some even attempted to lick any stray moisture off the walls, oblivious to the death sentence they were bringing on themselves.
One girl watched with a special sadness.
Sarah sat huddled against a wall. She was one of the few who had had the foresight to bring a robe; now, she covered herself with it, intent on disguising any evidence of her illness.
But no one noticed. They were too busy fighting and rampaging, taking out their fear and helplessness and panic not only on each other but on their palatial surroundings. Stained glass windows were smashed, a chandelier pulled to the ground, and marble fireplaces defaced; satin bedding was shredded and sofas and armchairs gutted, their stuffing littering the stained Oriental carpets.
It was the only beauty the people of Prin had ever seen. And they were destroying it.
It was too painful to watch, and soon Sarah's eyes, already blinking with fatigue, closed.
Four riders were on the main road heading out of town.
Caleb pedaled in front, next to Eli. Bekkah and Till followed close behind. They were on their way to find reinforcements, riding bicycles that they had stolen from outside Prin's destroyed homes.
They stopped at an intersection. "I think we go that way," Bekkah said, pointing.
After a hesitation, Eli shook his head. His voice was gruff and barely audible. "No. We go _that_ way."
From his bike, Caleb turned and nodded his thanks to Eli. The boy caught Caleb's eye, then glanced away.
After that, there was little time for talking. The road continued for a long time, past the parched lake, the forest, and beyond.
"This is it," said Eli at last.
The four, gasping and trembling with exertion, paused at the foot of a mountain. Overhead, the sun had passed its zenith. Faint tire tracks and footprints in the dust marked out the trail that disappeared among the trees in front of them.
Bekkah and Till traded one last look and the smaller child turned to Caleb.
"Are you sure?"
Caleb didn't answer and Bekkah took the opportunity to speak. "We could stay down here," she offered. "If you think that would help."
It was Eli who replied, scowling. "This is why we came," he said. "We all stick together."
And so they began the arduous climb, getting off their bicycles to push them alongside when the ascent became too steep. It was easy to lose the trail; if there were marks, they were so subtly done that they were invisible to the untrained eye. Caleb and the others were so intent on finding their way, it was only when they approached a clearing at the plateau that they realized they had reached their goal: the place that everyone in Prin had always heard about but few had seen.
The variants' camp.
Warily, they looked at the strange lean-tos and huts, the still smoldering communal fires, the racks of drying meat. Off to the side were empty crates from the Source, starting to deteriorate from the sun and rain: discarded testaments to a time of trade and prosperity, now over.
Caleb barely noticed.
Six or seven variants had turned to face them and more were spilling out of their homes or from the woods. Soon, the variants far outnumbered the four norms who stood before them, now shrinking back with unease. The variants murmured loudly to one another, all wearing expressions of open disbelief, suspicion, and hatred.
Their anger seemed directed at Caleb, who gazed back at them evenly. Beneath his cool appearance, however, his heart was beating wildly and his stomach felt like lead.
All noise ceased; someone was cutting through the crowd. As the others fell back, the largest male among them stepped forward, his eyes blazing with anger and disbelief.
Caleb recognized him at once: it was the warrior he had fought on his first day in Prin, so long ago. After his humiliating rout by Levi, the variant leader seemed more than ready for a confrontation, especially with his entire people present as witness.
"What is this?" Slayd asked.
Bekkah and Till recoiled, falling back a step. Eli, too, could not help quailing in front of such fury. Caleb, however, stood his ground, although he could not control the twitch that flickered across his cheek.
"We need your help," Caleb said.
At this, the crowd's agitation grew. "How dare you come here and ask us a favor?" Slayd said. "You above all, who fought us and took a special pleasure in it?"
"Kill him!" someone shouted, and the murmuring grew louder, more restless.
Slayd held up a hand for silence.
"We need your help to fight Levi," Caleb said.
The variant leader did not pretend to hide his surprise and gestured for explanation. With his voice shaking a little, Caleb proceeded to give him one.
Caleb told all assembled about the wrongs done to him by Levi: about the kidnapping of his son, the murder of his first partner, and the foiled attempt on his own life. He revealed that Levi was holding those he loved against their will.
"Many of you are raising young ones," Caleb said. "You can imagine how I feel having my son taken away. And many of you are partnered. So you too might understand how I feel to have lost my partner, Esther, as well."
In the silence that followed, Slayd stared at Caleb. Then his lavender eyes narrowed. Despite himself, he was moved by the norm's words and thought for the first time about the human cost of the crime he had so blithely hired others to do. Still, he showed nothing in his face.
With a sinking heart, Caleb read the silence as indifference.
"Don't you have a score to settle with Levi?" he asked, this time speaking to Slayd. "Didn't Levi break your deal and betray you after you trusted him? Why not join with us and pay him back?"
Caleb's barb found its target; Slayd winced, as if struck. It was clear that the memory was not only fresh, but still painful. Yet the variant pulled himself up and said in a haughty voice, "That is not your concern."
Too late, Caleb realized his tactical error. Reminding Slayd of Levi's betrayal had embarrassed the proud leader before his people by making him appear weak and foolish. Without intending to, he had caused Slayd to lose face. And he knew this was something the variants did not forgive.
Again, the crowd grew restive; the delicate trust had been broken. As Caleb glanced around, he now saw that most of the variants were beginning to slip rocks out from their pouches and pockets, loading up slings or wielding them in their hands.
Slayd stepped aside. When he raised his hand, all of the variants fell silent. They were cocking their weapons.
Desperate, Caleb gave a final look at the three behind him. They too were aware of the deadly shift in tone that had just occurred. Sensing impending disaster, unsure whether to fight or flee, they backed up on their bikes.
Then a voice broke through the silence.
_"Stop!"_
A lone female fought her way out from the crowd. A boy, most likely her partner, tried to hold her back. She said something inaudible to him. When he did not relent, she broke free of his grip. With no fear whatsoever, she approached the variant leader and addressed him.
"Esther is my friend," she said. "She is also a friend to us all. If this boy is truly her partner and says she is in trouble, we must help."
Slayd looked down at the girl, his arm still raised and his face a mask of fury.
"You are certain, Skar?" he asked. And she nodded once, emphatically.
At that, the impossible happened. Slayd's features softened. His hand, about to signal mayhem, fell back to his side.
He turned away from the girl, back to Caleb.
"You're lucky that I believe her," he said in a gruff voice.
Skar smiled to herself. She clearly did not want to gloat, yet could not resist doing a little dance that Esther alone would recognize.
Slayd pointed at her with obvious pride mixed with respect and not a little exasperation.
"My little sister," he said, "has never led me wrong."
## SEVENTEEN
IN THE HALLWAY, ESTHER COULD HEAR ONE OF THE GUARDS TALK ABOUT dinner.
How hungry he was. How long it had been since he had eaten a good meal. How desperate he was for a bite of something juicy.
"Yeah," said the other guard; "but we got orders. Besides, I like pigeons with a little more meat on them."
As the key turned in the lock and the door swung open, both laughed. Esther, lying on the cold, cement floor, squinted upward at the sudden light. They were talking about her, she realized. Her cheeks burned with shame and fury, but she said nothing.
One of the guards set a bowl of food on the floor and shoved it toward her. The first few times, they had stayed long enough to watch her eat, guffawing at her pathetic efforts. With her wrists and ankles bound behind her, Esther was forced to inch toward the bowl on her side, then attempt to eat out of it like a dog. But after two days, the routine was not as entertaining as it once was. She had barely made a move before they exited, slamming the door shut and throwing her back into total darkness.
Esther stayed still on the cement floor as she waited for the sound of their footsteps to fade away. Once she was certain they were gone, she worked her way to a kneeling position and shuffled on her knees across the floor.
When Esther was first imprisoned in one of the small rooms in the basement of the Source, she discovered a wooden bench near the wall, bolted to the floor. Exploring it with her bound hands, she located what she was looking for: a tiny rough edge on one of its metal legs. Since then, she had been spending hours every day rubbing her nylon wrist binds against it.
Today, she was able to cut through the final fibers. Then she untied her ankles.
Wincing at the pain, Esther attempted to rub blood back into her limbs as she got to her feet to take stock of her surroundings. The only illumination was the tiny strip of light below the door. She clapped her hands once, sharply, trying to get a sense of how big the room was by the faint echo that bounced back. This was something Skar tried to teach her to do for fun, on several occasions; but as before, she found her ear was too insensitive and untrained to detect anything at all.
So she walked ahead, sightless, reaching out. Starting with the door, Esther began to grope her way around the room. There was a smooth expanse of what felt like painted brick, with a small switch set in a metal plate. Esther clicked it and waited; nothing happened.
She reached the second and then the third wall; it was a medium-size room, rectangular in shape. Near the bench, her fingers encountered something metal, a shallow and boxlike structure that was taller than she was and built flush against the wall. It sounded hollow when tapped and by touch, she assumed it to be a series of narrow closets of some kind, side by side, with ventilation slats. There were a dozen or so per wall, each set with a small metal handle. Esther went down both rows, trying to jiggle each one open, but to no avail.
The fourth wall was smooth and bare; when she rounded its corner, she found herself back at the locked door.
Esther was trapped. Yet she recalled a variant phrase, the basis of everything Skar had taught or tried to teach her.
_Look with new eyes._
In other words, there were always options to any situation, different paths, other choices. The secret was to discover what they were. As far as Esther knew, there was only one way in and out of the room. Yet she realized she only assumed this because that was what her fingers told her, here on the ground.
What about higher up?
Esther faced the bare fourth wall. She brushed her fingertips against its painted surface, exploring its bumps and cracks. And then she began to climb.
Within seconds, her sneakered feet were scraping against the smooth wall, unable to find a purchase, and her fingers scrabbled in vain for the next handhold. Esther gave up and dropped back to the ground with a light thump.
Even Skar could not scale a wall in the dark; it was virtually impossible to climb anything without being able to see where next to grab. Esther decided: If she could not see, then she would have to learn the wall by trial and error and memorize her route.
Gritting her teeth, she tried again. And again.
She got higher her fifth time, and even higher her sixth. As she hoped, she was also getting a better feel for the wall, remembering where the best cracks and irregularities lay.
On her eleventh try, she was able to reach the ceiling, which she judged to be a little more than twice her height. As she released one hand to explore the area around her, she grazed the edge of something built into the wall before she lost her grip and dropped to the ground.
But Esther, who had explored countless old buildings with Skar and knew their odd construction well, had found what she was looking for.
Within moments, she took one step to her right and began again the laborious process of teaching herself to climb a new surface. After more than a dozen tries, she reached her goal.
It was a rectangular metal grid embedded high in the wall, just below the ceiling. The square openings were thick with dust and just big enough for the tips of her fingers to fit through. This allowed Esther to not only hang on, but to walk her feet up the wall so they were on either side of the metal plate.
She cautiously pulled backward, then with more force. As she did, the grating started to shift in her hands and the ancient mortar crumbled away, freeing rusty screws that pinged to the ground below. She steeled herself, then with one violent motion, yanked back as hard as she could while pushing off with her legs. The grid wrenched out of the wall and Esther went flying backward.
She managed to land standing up while still holding onto the grid, staggering back under its weight. Then she dropped it and clambered back up the wall, to the rectangular hole that now gaped open in the dark.
Esther found herself in a narrow chute made of flimsy metal. Wriggling forward on her belly, she found that even moving slowly made it contort. The joins were especially fragile; as she pulled herself over them, she noted that several of them were on the verge of giving way. Esther could only pray that the structure held up under her weight; she did not care to imagine what would happen if it didn't.
Since she could not see where she was going, Esther touched the walls around and in front of her to check if they branched out. The first time she found a new passageway, she noticed a dim light glowing at the end of it. When she pulled herself closer, she found herself peering through a filthy grating, looking out onto a small room lit by an overhead glass coil. Below her, one of Levi's guards was seated with his back to her. He was at a table crowded with strange glowing boxes, flickering screens with moving images.
Esther was moving as carefully as she could; but her meager weight caused the metal beneath her to buckle, making a loud noise. The guard swiveled around in his chair, his hand on the club at his waist; and she froze.
There was silence.
"Damn rats," he said. Then he returned to the bank of monitors.
Esther eased her way back to the central passageway. She thought of her promise to Caleb and was filled with despair. Precious time was slipping away and yet she could not go any faster, trapped as she was in the dark. All she could do was keep going. She was not even sure what she was looking for: an empty basement room, she supposed, with an unguarded door so she could again make her way to the main floor, where Kai was kept.
Another passageway branched off to the left and when Esther turned onto it, she saw that it too ended in a dimly lit grating. She assumed it would also be a room with a guard and was about to continue on, when something made her hesitate.
_Look with new eyes. To assume anything is foolish._
She turned to the left, making certain to distribute her weight over the flimsy metal. When she was close enough, she glanced through the grating.
What she saw made her heart skip a beat.
Below her was the baby.
He was sleeping on a purple blanket surrounded by a small wooden pen. There was a jumble of toys scattered around him, obviously new; their bright colors and childish designs clashed with the industrial setting, the gray cement floor and painted brick walls.
Kai had been moved here recently. If there was a guard, he was most likely posted outside the room.
Esther grabbed the metal grating with both hands and attempted to shove it. It didn't budge. She tried again, to no avail. She would need to work it loose, although it was a risky process; with each movement, she could feel the thin passageway sway and bend beneath her.
Esther began to rock the grid back and forth, again and again. At first, it was as if she was trying to tease a full-grown tree out of the earth with her bare hands. But soon she felt it starting to give way. Bits of ancient brick began to crumble from the constant friction of metal screws; she could hear them raining down on the ground below.
Esther was able to push out, as hard as she could. The grid gave way, just as part of the passageway collapsed beneath her legs.
When she landed on her feet, still clutching the grating, she saw that the child was now awake. He stood clinging to the side of his pen, two of his fingers in his mouth, staring at her.
Esther reached over the side of his pen and lifted him. She had only seen a few babies in her life and had never held one before; Kai was soft and warm and surprisingly heavy, so much so that she nearly dropped him. She tried to carry him face up, across her arms, but he didn't seem to like that; he began to fuss and struggled to sit upright. Finally, she found a position that suited them both, with him perched on her hip, his face close to hers.
"Garrrh," he said, reaching out to grab her chin.
Esther knew she must get him out of the Source, and yet she could not help herself. She closed her eyes for a moment to better feel the touch of his tiny, soft hands as they explored her filthy face. She found she was enchanted by this mysterious little being, this baby.
The boy looked like Caleb: He had the same dark hair and the same hazel eyes. And yet, there were subtle differences: his rounded chin. His forehead. _He probably took after his mother as well,_ Esther thought with a pang. Yet instead of being jealous of her, she was surprised by the pity she felt for the dead girl, as well as a strange sense of connection.
In the meanwhile, Kai had seized the string to her hood and was attempting to suck on it.
"Don't do that," admonished Esther; "it's dirty."
Then it occurred to her that he might be hungry. It would be too risky trying to escape with a fretful child, she reasoned; his cries would alert the guards. Looking around, she noticed an untouched bowl of rice in his pen. As she stooped to pick it up, he was already reaching for it, grabbing at the cereal with both fists and pasting it onto his face.
"Slow down," Esther whispered. Then she made a decision.
Feeding him would only take a minute or two. Once they were on their way, she would give him more food from the meager supplies she had remembered to pack in her bag.
She had thought of all things, except one.
The security camera in the corner of the room.
There was a loud bang, and Michal screamed.
But it was only the sound of a bottle being opened, a bottle of rare and special wine that fizzed and bubbled when exposed to the air. Levi had read once that champagne was traditionally used by kings and generals for centuries to celebrate important victories; and he had long saved several bottles unearthed during a Gleaning for a moment such as this. For he had never felt more like a king than he did today.
Levi had won everything he wanted.
He had conquered his despised brother while visiting revenge on the ghosts of their parents. He had found an heir, a healthy boy tied to him by blood, whom he would raise to succeed him as ruler of all he had created. The townspeople were out of the way, for good. Levi was only seventeen; he had at least two good years left. Maybe he could live even longer now that he had found clean, drinkable water in Prin, where none had existed for decades. Who knew what he would be able to accomplish in that time?
And it was all due to his singular vision, his perseverance, his strength; you might say it was because of his _genius_. Such an occasion called for a momentous celebration, one he had thought hard about and planned accordingly.
There was only one detail left to attend to.
As Michal carried the champagne away from his office to pour it into special glasses, Levi watched her go. Earlier, he had ordered her to apply her makeup with extra care and selected which clothing she should wear: tight, colorful clothing that accentuated her figure. _She had never looked better,_ he thought to himself.
It was almost a pity.
When she returned with the drinks on a silver tray (Danish sterling, from a Gleaning), she looked flushed and expectant. She offered him a heavy glass goblet that sparkled with rainbow facets (Austrian lead crystal), then set the tray down on his desk.
"To the future," he said, raising his goblet.
"To the future," she replied, her eyes lowered.
They clinked glasses, and he knocked back his drink in one swallow. He had never tasted champagne before; the bubbles gave the wine a remarkable airiness, a tingling sweetness, that made him shiver.
"I could get a taste for this," he remarked.
Michal giggled. She was about to raise her own glass when Levi stared at her.
"And don't think I've forgotten about you," he said. "I have a little present I'd like to give you."
He reached into a drawer. Without taking his eyes off her, he pulled out something that he tossed onto the desk between them.
It was a filthy piece of cloth. Michal leaned forward, smiling if puzzled, to examine it. Levi was amused to watch her jerk back in terror when she realized what it was.
It was Caleb's shirt—stiff and blackened with his blood, and torn in two places where the arrow had broken his skin.
"I can explain," she stammered. All of the color had drained from her face. But Levi ignored her.
"I searched your room," he said. "You weren't there, of course." He was watching her, studying with almost clinical interest the open expressions of terror, denial, and helpless defiance that played across her pretty features. "This was stuffed under a corner of your mattress."
"I . . . I don't know how it got there," she mumbled.
Levi smiled. "I suppose you were waiting for an opportune moment to dispose of it," he suggested. "Unless you were keeping it as a souvenir of your good deed?"
Two dots of red appeared in Michal's pale cheeks; and she flashed a look at him that was strangely confrontational.
"What are you going to do?" she asked in a low voice.
Levi rocked back in his chair. "I've given that quite a bit of thought," he said, as he drew on first one, then a second leather glove. He sounded as if he was talking about the weather or what to order for dinner. "I could, of course, have you banished, or killed. But those are too general, too common. No, I wanted a punishment that would really suit _you_."
He bent down to pick something up from the ground beside him. It was a square metal canister with an open spout. Before Michal had a moment to react, he hurled its contents into her face.
She let out a high-pitched shriek and jumped to her feet as clear liquid dripped down her front and splattered her clothing. Wherever it landed, the bright colors instantly dissolved and started to run in long streaks.
Levi made a gesture and two guards materialized on either side of Michal. She was clawing at her face, screaming in a voice that seemed more animal than human.
"Industrial-strength lye," Levi remarked. "The label says it dissolves fatty acids, which makes it wonderfully effective for cleaning a place as big as the Source. But it's a little hard on human skin."
He followed as Michal was dragged, shrieking, through the halls of the Source. He watched as one of the guards activated the machinery that raised the giant metal door. The girl was shoved outside into the glaring heat of the day. She fell to the dusty ground and crouched there, still clutching her face, rocking back and forth in agony.
With a nod, Levi dismissed his guards. Even he had the desire to keep some things private.
"Don't worry," he said. "It may burn for a while, but you'll almost certainly survive. As for your pretty face . . . well, I'm afraid that's another matter. I'm not sure who's going to want to look at you very closely again."
He turned to go but was unexpectedly stopped.
"At least I'll be in better shape than you," Michal said.
Her voice sounded hoarse and ravaged, either from the lye or her screams; it was hard to tell. Levi looked back down at her.
"How's that?"
"Your drink," she said. "I put rainwater in it."
Levi froze. Then he attempted to laugh it off.
"Did you hear me?" she said, raising her voice. "I been saving it up. I put it in when I poured the drinks. That's why I didn't have any. I put rainwater in your wine, Levi. You're a dead man."
Levi was walking away from her, walking fast, before breaking into a run. Behind him, her voice rose to an unholy shriek.
" _You're dead!_ "
But he was already inside the Source, stumbling to get to the controls to lower the door, to shut out her voice. He could not accept what she had said. After all he had accomplished and what he looked forward to doing in the next few years, it could not be true.
It must not.
And in fact, it wasn't.
Driven by shock and pain and on the spur of the moment, Michal had blurted out the one thing that she knew would hurt him the most. More than anyone, she was aware of Levi's terror of both water and the sun. Now she could only hope that her lie would spread unchecked throughout his body and mind, weakening both almost as effectively as the poison he so feared.
It was not much, but it was her last and only way to retaliate.
Michal got to her feet and stood, swaying, her hands still pressed to her face. Every movement was sheer torture. She realized she had nothing left in the world: no home or shelter, not one possession or a single friend. She had no idea where to go, but one thing was certain.
It would be far from the town of Prin.
Inside, as the door of the Source rumbled to the ground, a guard ran up to Levi. "There's been another security violation," he said. "The camera caught the thief with the baby."
Levi felt as if the wind had been knocked out of him.
"Are they still on the premises?" he asked. For the second time in his life, he found himself trembling violently.
The guard nodded. "They're still down there," he said.
Levi nodded. Without a word, he took off at a run.
He had received a death sentence. That he understood abstractly, as if it were happening to someone else. Yet even while the full truth had yet to sink in, he felt more alive than ever, his nerve endings and his mind surging as they responded to this latest threat.
All he knew was that there was only one thing left in the world with any value, and it was about to be taken away from him.
Levi could not trust the others to do his work for him. He could not depend on anyone who was not his equal; they were full of envy and would only look for ways to cheat and betray him. He was foolish to have taken in Michal, he understood that now, to have shown her any sort of kindness or generosity; for she repaid him like the animal she was. And now, he had to take matters into his own hands if he was to protect his new son and heir and only legacy.
He must save the child.
When Esther heard the door burst open behind her, she whirled around with Kai still in her arms. Levi stood in the open doorway, staring at the two of them.
For a fleeting moment, she could see the resemblance to his younger brother. But unlike Caleb, Levi had hair that was swept back from his paper-white forehead, and his eyes glittered like black stars. Right now, he was cooing, saying strange things as he approached.
"Don't be frightened," he said. "Everything is going to be all right."
Esther realized with a sickening lurch that he was talking to Kai and she tightened her hold on him. She tried to dart past Levi to the open door, but he slashed out at her. She had not noticed the knife he carried and the blade barely missed her arm.
"Esther," he said, recognizing her.
He sounded pleased, as if he had just run into a familiar face at a boring gathering, and his smile seemed genuine. "It's been such a long time. You were just a little girl when I knew your sister, Sarah. You've grown up very nicely. But you'll have to hand him over to me. You know that, don't you?"
Esther shook her head and backed up. She needed to stay free of his reach without allowing herself to get cornered. But Kai was restless and growing heavy in her arms. As Levi drew closer to them, she stumbled on a toy, nearly falling.
"I had some good news today," continued Levi, in a strange, absent tone. "Exceptional news, in fact. I discovered water. Fresh, drinkable water, bubbling up from the ground. But I need my son." He sounded so plaintive, so reasonable. "I need him to help me celebrate. What good is it when you have good news and no one to celebrate with?"
With a start, Esther realized that she must do more than merely react to Levi, backing away from his advances. At the moment, he was behaving in an odd and distracted fashion, but it would be foolish to underestimate him. The open door lay at an equal distance between the two of them. If she could engage with him, she could perhaps distract him enough so she could escape with Kai.
"Don't you have friends?" she managed to say.
Levi chuckled. "You mean my employees?"
Despite his mocking voice, he seemed to be considering her question. Esther realized with a shock that her hunch was right: in some crazy way, Levi wanted to have someone to talk to. She took advantage of the moment to take an unobtrusive step toward the exit.
"You've seen them," Levi said. "They're hardly the stuff of companionship, wouldn't you say?"
"But you must have other friends," she continued.
His expression darkened and he lowered his eyes. This allowed her to sidle closer toward the door.
"Friendship," Levi mused. "That's just business mixed with sentiment. Two people at the same level . . . if they're of service to each other, they call each other 'friend.' But if they aren't equals, the whole idea is impossible. It can't exist."
Esther decided to take a chance. "You were friends with my sister," she said.
Levi seemed to be listening, nodding his head. "It's true. Sarah provided a valuable service when she taught me how to read. But even back then, I was aware of her limits. I didn't think she was going to amount to anything. And I was right. Your sister was going to stay where she was. I wasn't."
Esther bristled, but forced herself to bite her tongue. She was almost at the door.
"But there must be _someone_ ," she said.
"Like my . . . companion?"
His eyes flashed with understanding. With one move, he was on her like a cat, twisting her free arm behind her back and holding the knife against her jaw. She held still, feeling its point digging into her flesh.
"What do you know about Caleb?" he breathed into her ear. "You aren't by any chance a friend of his, are you?"
Esther would only have had a chance of escaping if she dropped Kai; and she was not about to do that. Instead, she stood silent, as Levi wrenched her arm harder and pressed the knife even deeper into her skin.
But then he stopped. He had noticed the red strip tied around her wrist.
"I see," he said. Then he smiled.
So his younger brother had not been destroyed. If Caleb sent this girl as his emissary, he was still in Prin and bent on revenge. For a moment, Levi almost felt pride in his sibling's persistence.
But he had the advantage.
The life of Caleb's partner had to be worth a great deal. He was quite certain it could be used to barter for something valuable.
And if that turned out not to be the case, at least ending it would bring him the very real pleasure of creating more agony for his enemy.
Downstairs in his windowless basement office, a guard scanned the flickering monitors arranged in front of him with renewed vigor.
The guard was relieved he happened to be paying attention earlier, when the girl broke into the baby's room. He shuddered to imagine what Levi would have done to him if he had been napping or looking elsewhere or daydreaming like he usually did and the baby was kidnapped during his supposed watch. He and the others still carried the painful marks of Levi's earlier displeasure.
Now Levi had asked him to be especially alert for the return of the stranger, the boy they had left for dead. Doing so was the only way to avoid more pain.
Then he did a double take.
Someone was looking up at one of the cameras and waving. It was the guard stationed outside the front door. He and another one of Levi's men were propping someone up between them, a boy who seemed barely alive. One of them lifted him up by his hair so the camera could better see who it was.
It was Caleb.
Excited, the guard jumped to his feet, knocking over his chair. He was halfway down the hall, heading for the stairs, and pulling out his walkie-talkie so he could communicate with the others.
"Meet at the front gate. We got him."
Outside the Source, the two guards in their black hooded uniforms waited outside the giant metal door. Their mirrored sunglasses reflected not only the broiling sun, but the prisoner they held up between them.
"Do you think they're coming?" one of them whispered. It was Eli, disguised.
"If they were watching," replied Caleb. He kept his head down, and half-stood, half-leaned against Eli and the other guard, Bekkah.
Pressed against the building on either side of the metal gate, well out of the sight of the cameras, crouched Slayd, Skar, and more than thirty variants, all with their weapons drawn. Farther off in the underbrush, Till had just finished gagging the two actual guards, who had been stripped of their hoods and robes and bound. Then he joined the others as they all watched in silence.
They did not have long to wait. After no more than a minute, there was a grinding sound.
The giant metal door was opening.
Esther thought she must have been imagining things. But then she heard it again, faint but unmistakable.
It was the sound of Skar's whistle.
Levi heard it, too. He cocked his head and pushed the knife in deeper.
"What's that?" he asked. When she didn't answer, he jabbed her with its point.
"It means you're surrounded," she replied.
Levi chuckled, then yanked Esther by her arm. "We'll see about that," he said. "Take the boy."
He wrapped his arm around her throat, the knife held to her side, and pushed her out into the hallway. He seemed rattled when he saw that there were no guards on duty. Then he shook it off. With Esther held tightly in front of him, he advanced down the hallway and up the dim staircase.
The main floor of the Source was as dark as ever, and eerily silent. The only sound was the click of Levi's boots on the concrete floor as he advanced into the open.
"Caleb?" he shouted.
His voice echoed in the cavernous space. There was silence. And then someone stepped out from the deep shadows. It was his brother, followed at a distance by three others with their hoods down, two boys and a girl.
Caleb looked drawn and exhausted; and his companions were so thin and scrawny, they looked as if they would not outweigh one of his boys even if put together. Furthermore, the four were clearly unarmed, their hands empty.
Levi burst out laughing.
"Are these the warriors who have me surrounded?" he asked Esther. "Am I supposed to be frightened?"
Then something caught his eye. By the side of the open door, he saw that all of his men were tied up, wrists to ankles. There was a sound from above; and Levi, puzzled, glanced up.
Perched on the highest shelves, standing on crates and cartons, were dozens of variants. Some held loaded slings; others carried spears; at least one had a bow and arrow. All were aimed at Levi, whose smile died.
Skar, for one, had her arm cocked back, a throwing stick loaded with a spear balanced on her shoulder. Across the room, she glanced at Esther, and the two locked eyes. Then the variant girl flashed her a quick smile and a wink, before resuming her stern expression and warrior stance.
"Caleb!" shouted one. It was Slayd, who stood with his sister. "Should we kill him? Give the word and it is done."
But Caleb held up his hand, stopping them.
"No," he said. "Leave him to me."
On Slayd's instructions, the variants dropped to the ground and fanned out across the Source, searching for any remaining guards.
As one kept a watchful eye on Levi, Caleb ran to Esther and his child and embraced them. Esther shifted the baby into his arms, and he buried his face in Kai's soft neck. Then he glanced up. Amid all the activity, one person was standing still.
Eli was staring at Esther with a look filled with longing, sorrow, and loss. It was impossible to miss. As for Esther, it was equally apparent that she was avoiding his gaze. Caleb now understood what had fueled the boy's lingering hostility. But instead of becoming angry, he had a sense of understanding, as well as sympathy.
Caleb handed the baby back to Esther. Then he approached Eli.
"You know where the others are being held?" he asked him in a low voice. "The place where you escaped from?"
It took a moment for Eli to turn his attention to Caleb.
"What about it?" he asked.
"Will you take the others there?" Caleb asked. "I need you to be in charge."
When the words sank in, Eli's face flushed with pride. Caleb reached out and the two boys shook hands.
Then Caleb went into his pack and withdrew something.
"Here," he said. "You might need this."
It was his weapon. He demonstrated how to load and fire it and Eli took it, gratefully.
"Thanks," he said. Then Eli waved at the variants and the others. "All right," he called. "Let's go."
Slayd glanced at Caleb, to confirm the hierarchy. He was, after all, not accustomed to following the orders of strangers. Caleb's nod was enough for him. With a slight shrug, Slayd lifted his hand and directed his people to follow Eli.
Esther, still holding Kai, ran to Skar, and for several seconds, the two girls hugged without a word. Then Esther pulled back to look her friend in the face.
"Thank you," was all she could say. But Skar smiled, shaking her head.
"When you came to us, we should have given you refuge," she said. "This was the least we could do to make up for it."
And with that, Skar turned her attention to the baby. She took Kai into her arms, cooing and nuzzling. "Here," she said. She pulled a length of cloth from her pack and showed Esther how to make a sling, bundling the child close to her body. Then she hugged her friend once more and wished her luck. As Skar returned to her brother and the other variants, Esther approached Caleb.
"I'll take Kai somewhere safe," she said. "I have a friend who lives high in a building on the edge of town. It's secure there."
Caleb nodded. Then he kissed Esther and his son. Soon she was gone, slipping out even before the others had left. He watched the rest exit the Source, taking the light with them.
Then Caleb turned. There was still the question of Levi.
## EIGHTEEN
ELI BICYCLED ALONGSIDE SLAYD ON THE ROUTE LEADING BACK TO THE mansion where the townspeople were being held. Behind them, Skar, Bekkah, Till, and the rest of the variants spread out in a loose V formation.
As they rode, the variant leader questioned the norm about the place they were going to try to liberate. _What were the best approaches to the building? Were they open or hidden? How many guards were there, and where were they posted? Were there townspeople strong enough to assist? Were there any breaks in the barbed wire?_
Eli answered as best as he could. He was both relieved and proud that he could provide enough details to satisfy Slayd. As for the variant, he was impressed that the boy was able to mount a successful escape with his two companions. Despite their initial and mutual distrust, the two leaders now discussed a possible attack plan, one that capitalized on the fighting prowess of the variants as well as the norms' knowledge of the layout and the people involved.
At the mansion, Levi's guards had no idea what was bearing down on them. They stood at attention at newly assigned posts, just beyond the strands of razor wire that coiled around the house exterior. On the first day, the boys were left to their own devices. As a result, they spent the day clustered in a group, making bets on flipped coins, wrestling, and gossiping. But after three of the prisoners escaped, word came from Levi that all guards were to be stationed around the perimeter.
Despite their grumbling, the boys were too intimidated not to comply. As a result, they were openly indifferent to the sounds of rioting from within: screaming, the smashing of glass, and the splintering of wood. Whether the townspeople lived or died was of no consequence to them. The guards only thought about their own welfare.
But staying at attention was too hard for most. One guard rested at his station by the front of the house. A hulking boy with shaggy hair that stuck out around the edges of his hood, he used a stick to scratch a game of solitary tic-tac-toe in the dust by his feet. He no longer even heard the ragged voices on the other side of the barbed wire, pleading for water or news. But after a while, he did hear something else, and he glanced up.
There was a scrabbling coming from the roof above his head.
Pebbles rolled across shingles and rained down on him, as if dislodged by someone above. He smiled as he got into position to catch the escaping prisoner who was so obviously climbing down. It was almost too easy. He was one of the few who had been assigned a Taser, which he now took from his belt and tapped against his palm with anticipation.
But the threat did not come from above; with the guard facing the house, his back to the yard, it instead came from behind. An object winged through the air, followed by a loud crack. The boy's knees buckled as he sank to the ground, blood from his temple staining the damp grass.
The prisoners watching from inside fell silent as Bekkah and a variant boy emerged from their hiding places. As the variant bent down to grab the weapon from the guard's still-twitching fingers, Bekkah was pulling something out from her jeans pocket. It was a cutting device taken from the Source for just this purpose, with sharp edges and rubber handles.
She got to work on the razor wire. There were at least four coiling strands blocking the front door and the metal they were made of was sturdy. By squeezing as hard as she could, Bekkah was able to snip through one piece. The wire, however, sprang back abruptly and one of the razor edges slashed her across the bare arm. Wincing, she kept working.
Eli was watching from a car abandoned in the overgrown driveway, a silver Lexus. "It's taking too long," muttered Eli to Slayd, who knelt next to him.
"That is why we must attack now," Slayd replied under his breath. "Before it is too late." Out of courtesy, he made a point of glancing at the norm for confirmation. Eli hesitated before nodding, and Slayd put his fingers in his mouth, giving a piercing whistle.
Eli took out Caleb's weapon.
And seconds later, the attack began.
A barrage of rocks was unleashed from all angles: from behind trees, abandoned cars, the roof of a gardening shed. Taken by surprise, Levi's men had no time to defend themselves. Two of them managed to duck the flying projectiles, batting them aside. One even succeeded in breaking loose. He took off down the circular driveway at a sprint. But he was no match for two variants, one on his bicycle and the other perched on his rear wheel pegs and taking aim as he whirled his loaded sling overhead. Within seconds, there was a final loud crack. All of the guards had been felled, and the grass was littered with their bodies. Slayd and the others started searching them and removing their weapons.
At first, the townspeople shrank back from the windows, in fear of the flying missiles. Now they crushed together at the front door, where Bekkah was still struggling to cut the last few strands of razor wire.
Blinking in the light of the sun, the exhausted residents emerged, some barely able to stand. One by one, they glanced down at the unconscious guards, anonymous in their bloody hoods and cracked sunglasses. Just moments before, they had seemed omnipotent. Now, a small boy kicked one as he passed, and another spat on the unmoving forms.
"Animals," he hissed.
Eli pushed his way past the released residents, back to the place where he too had so recently been imprisoned. Once inside, he did a quick search of the home, going from room to room to make sure no one was left behind. As he did, he winced not only at the filth and stench, but the senseless damage. Everything he saw spoke of impotent fury: the floors were littered with glass, the furniture was smashed, and the walls kicked in.
He saw that the mansion was almost empty; only the final stragglers were left. But in one corner of the house, a motionless form swathed in a white robe despite the unbearable heat huddled against the wall.
Eli bent and, with surprising ease, picked her up; she seemed weightless. Then, walking through the now vacated prison, he carried Sarah to safety.
It was hard work running with a small child tied to one's back.
It was not so much Kai's weight; he was just a baby, and Esther had carried far heavier burdens before. But no matter how carefully she ran, the jostling made him wail and struggle in his harness. He even managed to kick a leg free and Esther, terrified that he might fall out, was forced to slow to a walk. So it took her twice as long as she expected to reach the Gideon Putnam Hotel.
When she did, she was alarmed by what she saw.
The apartment complex, Joseph's beloved but decaying home, was guarded by Levi's men. Two of them, armed with bows and arrows, flanked the shattered glass entrance.
Esther realized this would be no safe haven for her and the child. Yet she also knew she had to get inside somehow and see what had happened to her old friend. Esther debated for a moment whether or not to hide the child someplace nearby. Then she made up her mind. Lacing her fingers behind her back to give the sleeping boy extra support, she slipped her way with him through the undergrowth brush to the back of the complex.
It was as she hoped: no one had been stationed to protect the gaping hole that had once been the picture window overlooking the courtyard and playing grounds. The familiar lobby was deserted. She was about to slip across to the far side when she heard footsteps descending the staircase.
Esther had barely enough time to duck behind a cracked pillar when two guards entered the lobby. She listened until she could no longer hear them.
Esther sensed Kai stirring in his sling. Now was the moment to get moving. She ran to the staircase, glancing around in case there was anyone else. Then she began to climb.
Even if she had not seen the guards, Esther would have known that strangers had been here. Several sections of the staircase, already fragile, had collapsed and she had to navigate the metal railing instead, praying it held up under her weight.
Joseph saw her anxious face moments later in his doorway.
"You're safe," she said, relieved.
His eyes lit up with surprise when he noticed what she was carrying on her back.
"Is that a baby?" he said, as he drew close. Joseph had only seen infants a few times in his life, and that was many years ago. This one was awake, staring at him with a comically quizzical expression.
One cat, Ginger, pawed at Esther's leg, expecting food. Annoyed, she shooed the animal away.
"We have to get out of here," Esther said. "Levi's men are downstairs."
"Oh, I know," Joseph said. All the time watching the child with fascination, he explained the situation, why the hotel was guarded, how they had discovered his private supply of water.
Esther blinked. "So that's what Levi meant? There really is clean water? And it's here? And you knew all along?"
Joseph glanced up at Esther, taking note of new details. He was impressed by her new gravity, her mature attitude, not to mention the band around her wrist and the baby. This kept him from responding right away.
"Are you even listening?" she asked. "Why didn't you ever _say_ anything?"
The one answer Joseph could give seemed likely to inspire more criticism. So he just shrugged.
"You always offered me a cup of it when I visited," she said. "I guess that was your way of saying."
Esther shook her head, marveling at Joseph's cluelessness, though with obvious affection.
"Here's what I think," he offered. "There's water deep underground. And it pushed up through layers of rock and sand somehow. I think that's what cleaned it. That's all I know."
In Esther's face, he saw a new, more surprising emotion. He'd never seen it in the eyes of another person. It was respect.
"So what's all _that_?" she said, stepping inside.
She gestured at the calendars that lay scattered around the apartment. Before, at best, she had indulged him about them.
Joseph was rather proud of the simplicity of one particular creation, a circular wooden board. Now, he explained that so many hours formed a day, then a week, then a month, and so on. But instead of needing to create new lines, his calendar circled back on itself every seven days. Months and years were indicated by advances in either green or blue pegs.
Esther listened, nodding here and there, asking the occasional question. She didn't whistle with impatience or smile politely, as she used to do. She pointed to the single red peg set in the center of the board.
"And what's _that_?" she asked.
"The day I was born."
Esther gasped at this, which perplexed Joseph. Everyone in Prin had some idea of when and where they were born, if only in crude approximations counted on fingers or scrawled on walls. It was the relative surety of his calculations, as well as the elegance of his presentation, that impressed Esther, he assumed.
"I'm—" he started to give my age.
"Don't tell me," she said. " _I'll_ figure it out."
She used his calendar, competently, her fingers moving, her eyes darting. Joseph watched, feeling a new emotion himself: pride that he could only describe as paternal.
Esther gasped again, recoiling from the calendar. Then she turned and stared at him.
"Twenty-six," she said. "You're twenty-six years old!"
Joseph squirmed a bit, embarrassed. To hear it said like that made him feel so old. But then, he supposed he was.
Levi walked with painful slowness back to his office, followed by Caleb. The two were now the only ones left in the Source.
The older brother seemed dazed, unsteady on his feet. But that was all the weakness he was willing to reveal. Even beaten and powerless, Levi still held his ground, now worth nothing.
When Caleb thought about what his brother had done, of all of the pain and misery he had caused, hatred instinctively surged in his breast. He deserved no mercy; why should he, when he had never shown any to anyone else? The world would be a better place if he were to throttle Levi now, to squeeze the slender white throat until he extinguished his life forever.
And yet, he hesitated.
Levi had accomplished everything on his own, using only willpower and intelligence. What might he have achieved had he not been so tortured a soul? How would their lives have been different if their parents had kept him instead of casting him out? _It was such a waste,_ thought Caleb, with a sense of profound sadness.
Despite his anger, Caleb could not deny the blood that still linked them.
"So," Levi said, his back still to his brother. "I see you've saved me for yourself." With his hand balanced on the edge of his desk, he seemed calm, as if resigned to his fate.
"What?"
Levi turned. "You could have let those animals kill me," he said. "And yet you didn't. I don't blame you . . . I'd do the same thing. Though of course, I'd enjoy it. I suppose you're going to have misgivings."
From the taunting way he spoke, Caleb realized that even now, Levi was jockeying for position, trying to anger him in order to throw him off balance. Yet Caleb wouldn't take the bait. Now that he had Esther and Kai, Caleb no longer felt any bloodlust. The urge for revenge had been purged from him. What would one more casualty achieve? All along, Caleb had only been trying to right his world after it had been wrenched askew.
Rage and revenge would not be his constant, lifelong companions. He would make certain of that.
"Not everyone's the same," he said.
"That's what you think," replied his brother.
But Caleb refused to be drawn into an argument. "I can't deny I never want to see you again," he said. "But I'm not going to kill you."
This surprised Levi, who raised one eyebrow. From his expression, however, it was clear he was not so much relieved as amused and more than a little contemptuous.
"So what do you propose instead?" he mocked. "That I promise to reform? To do good works for the people of Prin?"
Again, Caleb refused to be drawn into a fight. "I don't care what you do," he said, "as long as you leave and don't come back. Ever. Take whatever you need or can carry. Though I'd advise you not to look back."
Levi was toying with one of the silver rings on his fingers. Then he shook his head.
"Thank you for your generous offer," he said. "But I'm not going anywhere. You see, there'd be no point. Because I'm as good as dead, anyway. Courtesy of my beloved."
Caleb gave him a sharp look and the older boy smiled. "Michal slipped rainwater into my wine," he continued. "Who would have thought she was capable of thinking that one up, much less carrying it off?"
"You mean you're—" Caleb started, but his brother cut him off.
"Dying," he said. He was incapable of hiding the self-pity in his voice. "That's another way you've gotten the better deal, you see. Girls fall in love with you. They poison me."
Then, suddenly, unbelievably, Levi's face crumpled. He started to cry.
He held open his arms and, for a moment, Caleb didn't understand. Then Levi took him by the shoulders. As Caleb, confused, moved into his brother's arms, Levi pulled him close.
To his shock, Caleb realized that Levi, too, yearned for connection, the kind you have with blood, with family. Caleb returned the embrace, astonished by the warmth he felt.
All at once, there was the rumble of machinery. Caleb tried to turn but Levi's arms had turned into a vice. By the time he wrenched himself free, the doors had shut, sealing the two inside.
Levi was behind his desk, pulling something out from beneath it. He now hoisted it up, balancing it on the arm of his chair.
It was a large metal can of Able Accelerant.
He shrugged, as if to apologize for the obviousness of his choice. Caleb lunged across the wooden surface in vain. Levi managed to keep the desk between them as he ripped off the plastic cap. Then he began splashing the can's contents in every direction, across his desk and among his papers, filling the air with its dizzying fumes. Even as his brother was on him, trying to wrestle it away, Levi managed to hold on to it, upending it onto any surface he could find, dousing both of them with fuel.
Caleb lost his footing on the slippery floor, and his brother fell from his grasp. Levi stood across from him, panting a little, his eyes glittering. He had fished something from his pocket, which he held aloft. It was a plastic firestarter.
"It'll be nice," he said, "for us to finally be together."
Caleb felt a weird stillness overtake him, as if time had slowed to nothingness. He could see his life laid out in hundreds of strands; everything in his past was happening again at once. He saw the face of his mother. The sight of his newborn son, slick with blood and afterbirth. Esther's eyes. And he saw his brother and himself as if from a great distance and he realized that he was not afraid.
"The world doesn't need any more orphans," he said. "You above all should know that."
His brother stared at him, then recoiled, as if punched in the stomach. _What did he see?_ Caleb wondered. _Did Levi picture Kai and his own infant self merging, as injustice was handed down to another generation? Did he know that, this time, it would be his fault?_
Caleb would never know.
Levi reached under his desk. With a grinding sound, the door began to open, then stopped, leaving just enough room for someone to fit underneath.
"Go," he said.
Caleb looked at him one last time, then was gone.
Alone, Levi stood near his desk, immobile. He considered Caleb's offer, the prospect of being Shunned himself. But where would he go? The Source was his home, the only true home he had ever had. And he knew the poison was working its way through his body, invisible and unstoppable, like rot taking over a carcass.
If he closed his eyes, he could practically feel it.
He gazed at the partly open door, then pressed the unseen button, raising it.
As if in a trance, Levi walked alone through the massive store. Unseeing, he passed aisle after aisle stacked high with crates made of cardboard and wood, all labeled. poland spring water. golden blossom honey. domino sugar. gold medal flour. Every single one was empty. All they were good for, really, was tinder.
He reached his destination. It was the room in which the gasoline was kept, the untold gallons and buckets and bottles of fuel that the people Prin had been collecting for years, keeping Levi and his people in light and comfort. He was holding the nearly empty can of fuel. He fingered the firestarter, idly.
The explosion would be heard for miles.
Just then, Esther and Joseph felt a tremor. They rushed to the window and saw the rising smoke coming from the Source.
Shock and anxiety passed over Esther's face.
"Caleb," she said quietly.
But there was another reason for worry. Joseph's home, previously weakened by earthquakes and other disasters, now began to shift. Both Esther and Joseph stood stock-still, hoping for it to end. But the rattling did not; it grew greater. The blast had been strong enough to threaten the hotel's very stability.
"Joseph," Esther said, "we have to get out of here. Now."
Joseph knew that she was right. Yet it was more difficult than even he expected to round up ten cats. He trusted several of them (Stumpy, Malawi, and a few of the others) to follow on foot. But the others he had to hunt down and stuff into their nylon carrying bags, a process that involved much yowling, wriggling, and scratching. He also grabbed a few belongings: a folder of newspaper clippings, some books, a calendar.
Meanwhile, Esther kept a grim watch by the door, her face drawn with impatience and concern. When Joseph was ready, she took three of his cat carriers without a word and hoisted them onto her shoulders, where they competed for space with the baby. She handed him her messenger bag. Then they set off, a strange caravan of people and animals.
They were no more than several steps down when the staircase started to falter.
At the same moment, plaster and rubble began to rain down on them. They froze for a moment; then quickened their pace. Through choking white dust, Joseph could see cracks widening in the walls. In some places, the stairs themselves looked as if they were about to shear off completely. Entire sections of the staircase started to vanish. The cats could leap over these gaps with ease, but the humans were not so gifted. Esther helped Joseph as they clung to the central railing and worked their way across.
They were only two flights down when more fissures became apparent.
This whole time, the child had not cried, even as the staircase shook so much that the very bones in Joseph's body seemed to rattle. In front of him, he saw that Esther's dark hair and red sweatshirt were whitened with plaster dust, as were the baby's head and what had once been dark green cat carriers. When she abruptly turned to face him, Esther's eyes were like black stones in a field of snow.
Above, they heard the sound of a massive metal beam breaking loose. It came crashing down a few feet away from where they stood, shearing off another section of the staircase and narrowly missing one of the cats.
Esther shot Joseph a look that was half plea, half command.
"I have to go fast now," she said. "You're going to have to keep up."
Joseph nodded.
Without being asked, she took his two cat carriers and added them to her overburdened shoulders. Then she seemed to coil up like a spring before she took off. Joseph watched with open-mouthed amazement as she leaped off a shard of ground, made brief contact with the wall, pushed off with her foot and grabbed the handrail, swung sideways and landed on a ledge one flight below.
There Esther stood, looking up as she waited for him.
Joseph had to shake himself out of his astonishment. Even his cats seemed surprised by what they had seen. Then he did his best to follow.
He could not keep up with his friend; it was impossible. Yet they did manage to go much faster. Joseph, however, was not used to such exertion or excitement. His heart pounded and his legs shook. He suggested more than once that she would go faster without him. But Esther stayed with him, always finding a secure spot or foothold where they could both rest.
They were close to the lobby by the time the ceiling began to break up more. A cascade of white dust crashed down on them from above like a waterfall. Joseph could not see, his eyes and nose and mouth filled with plaster, and he choked and coughed.
"Come," Esther called over the roar of collapsing bricks. Joseph felt the pressure of her hand in his. Blindly, he stumbled after her, down a few more steps. He landed on some rubble and stumbled, twisting his ankle and giving out a cry. But Esther refused to let him stop. She pulled him into a clear area, which he only vaguely recognized as his lobby. Still rubbing his eyes, he let her guide him across the ruined space for what he realized would be the last time, until he sensed she was leading him up and across the empty frame of what was once the picture window.
"Now run," she said.
And they did. Ignoring the pain in his ankle, Joseph ran as fast as he could across the courtyard, the cats close on his heels. Esther sprinted in front of him, still wearing cats and baby in a way that might have seemed funny if it wasn't so impressive. They ran past the asphalt court with its lone basketball hoop, past the lot filled with abandoned cars, and still they kept going.
"Don't look back," Esther said.
So Joseph didn't. Behind him, he could hear a rumbling that grew to a roar that seemed to suck the sound out of the world.
Without looking, he knew what had happened. His home had been destroyed.
Joseph's heart was still pounding wildly, especially when he realized how closely they had come to being destroyed with it. He thought about all that he had lost: nearly all of his library, calendars, and timepieces. These were precious, beloved items he had spent a lifetime collecting, repairing, constructing. All he had left was what he had with him at the moment.
But he was brought back to himself by the feel of something at his ankles.
It was Stumpy, winding herself about his legs and complaining in her tiny voice. Joseph did a quick headcount and marveled. All of his cats were accounted for and, by all appearances, eager for a snack. He shook off his regrets and turned to Esther.
"Oh well," he said. He tried to sound philosophical. "It was only a matter of time."
They turned. Before them was a collection of boys dressed in black hoods, Levi's boys who had been guarding the building. They, too, had escaped and now stood, staring at the wreckage, too astonished to be dangerous any more.
"Look!" Esther said.
All around them, water was seeping up from the ground and forming puddles amidst the rubble of bricks, mortar, and steel. Soon great geysers of water began erupting from the base of the spring, freed from the ground by the disruption.
Esther stared at the sight for a long moment. Then she glanced up at Joseph, who nodded. She took a deep breath; then in one quick gesture, she knelt and dipped a cupped hand into the spray, brought it to her lips, and drank. Then she sat back on her heels in the mud, blinking, as if stunned by her own audacity.
Suddenly, she was not alone. The guards were kneeling, as well. They pulled their hoods off and tossed them to the side. Cold water rained down on their upturned faces, their open mouths, drenching them and leaving streaks on their gritty skin and filthy hair. The air rang with their shouts of pleasure.
Then Joseph turned. Before what used to be the front of his building, a boy was dismounting from a battered bicycle. Esther gave a cry and rose to her feet. She handed the baby to Joseph, much to his dismay. The child kicked his feet with displeasure, as the older boy cooed incompetently at him.
Esther flew to the stranger and kissed him, which made Joseph blush. Then she pulled him over. He looked like he had been through a hell no one could imagine, and yet he had survived.
"Joseph," she said. "This is Caleb. That's his son, Kai."
She told Caleb all that had happened, even while she peppered him with questions. Then she showed him where water bubbled up from the ground.
"Look," she said. "This is what Levi was searching for all these years."
She had a new thought. Joseph had set her messenger bag on the ground. From it, she snatched Sarah's book, which she handed to him.
"Tell us what this means," she said. "I know that you'll know."
Joseph opened the book, marveling at the information and illustrations inside. And after looking them over, he _did_ know.
Stumbling a bit, he explained that the maps were not just of roads above ground, but of underground waterbeds and formations, as well. He explained some of the words that they couldn't decipher. Finally, he flipped through the book to show them the most important map of all.
"Saratoga Springs," he said, pointing. "It must be where we are. It was a town built on a mineral spring, a long time ago. Look—here's where my home might have been."
Esther and Caleb shared a puzzled glance. "But how come it's called Prin?" he asked.
Joseph paused, then brightened, pointing at the letters. "There was a town sign, with the whole name. But 'Prin' is all that's left. See?" He covered the other letters to demonstrate.
_Saratoga Springs. Prin._
Joseph glanced at Esther. He could tell she was about to ask why he had never figured this out before, but then she stopped. She knew she wouldn't have listened, just months ago.
She was so much younger then.
They all were.
Beneath its bleached surface, the soil was surprisingly dark and rich.
When the hole was deep enough, Caleb hoisted himself out. Knocking the clay off his shovel, he joined Skar and Joseph. The variant girl was tending Kai, holding him on her hip and bouncing him up and down to keep him quiet. Having become more comfortable around babies, Joseph made funny faces in order to amuse him. But then he remembered that this was a solemn occasion and he stopped.
Off to the side, Esther stood in silence with the body of her sister held in her arms.
Wrapped in a white sheet, Sarah was curled on her side like a sleeping cat. She was oddly heavy for one so frail, her limbs stiff and unyielding. It was how Esther had found her that morning. In death, the terrible fever had broken for good; the unexpected iciness of her skin was a shock. With the lines in her face relaxed, Sarah had seemed at peace and almost pretty again, as if the illness had never happened.
And it was a good death, if such things existed.
Sarah had died at home; against the decree of the town, Esther refused to allow her to be Shunned. Over the weeks, she kept her in the shattered remains of their home and tended to her, feeding and changing her like a baby even as the lesions spread and the dementia grew.
For most of the time, Sarah slept. This was a blessing; the pain and the fever were almost too much for anyone to bear.
But early in the mornings, before the heat of her body would spike and she would lose sense of where she was, Sarah would have a few moments of lucidity. Esther would sit next to her, feeding her water from a spoon, and the sisters would talk. Without acknowledging it, both girls understood that an entire lifetime of unspoken thoughts, confidences, and memories had to be shared in a few precious weeks. From outside the bedroom door, Caleb could hear only a few whispered words, even an occasional peal of laughter.
And now that Esther had finally grown to know her sister, it was time for her to take her leave.
She handed the body to Caleb; then she jumped down into the freshly dug grave. Reaching up her arms, she took Sarah. She held her sister for a moment. It was maybe the third or fourth time she had done so in her life; it would be the last, as well. Then she placed the curled-up body on its side, on the exposed red clay.
"Good-bye," she said softly.
Filling the grave did not take long; she, Skar, Caleb, and Joseph took turns with the two shovels they had brought. The one not digging took care of Kai; Joseph was eager for his turn.
Bright-eyed, the child seemed enchanted with everything he saw. He marveled at the brown and brittle leaves that littered the dark soil. He put dead grass and bits of twig in his mouth and spat them out, laughing and babbling.
Now he heard a sound and looked overhead so abruptly, he fell backward and sat down. Joseph assumed he was frightened and wondered if one of them should pick him up and tend to him, comfort his fears.
But Kai didn't mind. Above him, he saw birds flying in a V, their honking cries echoing, as they journeyed to a better place.
And he clapped his hands with pure delight.
## EPILOGUE
WITH LEVI GONE, EVERYTHING HAD CHANGED. HIS OLD LAWS NO LONGER applied: Overnight, the jobs were meaningless, Esther's presence in town was unquestioned, and drinkable water was in endless supply. With the crisis over, people had turned their attentions to rebuilding the town and pooling the remaining resources so everyone would be fed.
After helping defeat Levi, the variants had mostly retreated to their camp. Still, it was apparent they were aware of the town's need. Gifts appeared on the main street, early in the morning before anyone was awake: two full sacks of cornmeal and one of rice, remains of Levi's payments. A newly slaughtered deer. Three fat partridges. There was a spirit of holiday in the air, so much so that it was hard not to feel optimistic.
Yet Esther for one was uncertain. It was not just the fact that the people of Prin were reduced to living in shanties and lean-tos or amid the rubble of what had been their homes. The wreckage of the Source still stank of fire and smoke and burning plastic. Despite Prin's current bounty, its supplies of food would soon be exhausted.
Esther did not doubt the strength or the will of the people. More important, she saw hope in Caleb's righteousness and in the wisdom the two of them had gained at such a painful price. She saw hope in their love, a bond that seemed to grow stronger with each day. And she saw it in the eyes of the child, Kai.
There was certainly a future for them; that much she knew. Only she was not so certain it would be in Prin.
For now, however, Esther put her doubts aside. Tonight the entire town was celebrating, and although the afternoon sun was already low in the sky she still had to fetch Joseph.
With his home destroyed, her old friend was living with her and Caleb. Yet he seemed frightened to even venture outdoors and had to be escorted everywhere, coaxed like a small child.
Upstairs, Joseph was waiting for her with a tense expression.
"Are you ready?" she asked. "Caleb and Kai are already there, with everyone else."
Joseph nodded. Esther noticed that he had smoothed down his hair and put on a relatively clean T-shirt.
"Come," she said, patiently holding out her hand. But still he hesitated.
"You know," he said suddenly, "it actually _is_ a holiday today."
Off her puzzled look, he continued in a rush. "According to my old calendars, today is a day when people used to celebrate abundance, while preparing for the hard months ahead. It is a day they called 'Thanksgiving.'"
Esther paused.
"Thanksgiving," she repeated, as if to herself. Her expression was thoughtful.
"And it is also," Joseph added, blushing, "the first party I have ever attended. And as the new town elder, too."
"Imagine that," Esther said. Then she met his eyes and smiled.
Wordlessly, the two headed downstairs and onto the street. The sun had set and for a moment, Joseph held back, frightened by the blackness that seemed to engulf them. Yet Esther saw that there was already a sliver of moon overhead, casting a pale yet steady light. And if you lifted your head and sniffed, you could make out the faint but inviting smell of roasting venison and the far-off, happy sounds of festivities.
Esther confidently led the way through the darkness.
## **ABOUT THE AUTHOR**
**SUSAN KIM & LAURENCE KLAVAN** cowrote the graphic novels CITY OF SPIES and BRAIN CAMP.
Susan is also a five-time Emmy nominee for her work in children's television and a Writers Guild Award winner for best documentary. She wrote the stage adaptation of Amy Tan's THE JOY LUCK CLUB, teaches writing at Goddard College, and is a blogger for the Huffington Post. When she was growing up, there was a chain-link fence behind her apartment that led to a small woods. After dinner, all the kids would sneak through to play in a world where no adults intruded. The memory of that has always stayed with her.
Laurence's previous novels include THE CUTTING ROOM and THE SHOOTING SCRIPT; he won an Edgar Award for the novel MRS. WHITE, and his short-story collection is forthcoming. He received two Drama Desk nominations for the book and lyrics to _Bed and Sofa_ , a musical produced by New York's Vineyard Theatre. As kids, he and his three brothers used to make epic movies in their backyard, reenacting the Alamo, the signing of the Magna Carta . . . and the end of the world. This last one involved a lot of fighting over food and property, which was, of course, what they did every day in real life.
Visit www.AuthorTracker.com for exclusive information on your favorite HarperCollins authors.
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## COPYRIGHT
Cover art © 2013 by Colin Anderson
Cover type © 2013 by Alex Beltechi
Cover design by Tom Forget
HarperTeen is an imprint of HarperCollins Publishers.
WASTELAND. Copyright © 2013 by Susan Kim and Laurence Klavan.
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* * *
Library of Congress Cataloging-in-Publication Data
Kim, Susan.
Wasteland / by Susan Kim and Laurence Klavan. — 1st ed.
p. cm.
Summary: In a postapocalyptic world where everyone dies at age nineteen and rainwater contains a killer virus, loners Esther and Caleb band together with a group of mutant, hermaphroditic outsiders to fight a corrupt ruler and save the town of Prin.
ISBN 978-0-06-211851-6 (hardcover bdg.)
EPub Edition January ISBN 9780062118530
[1. Interpersonal relations—Fiction. 2. Virus diseases—Fiction. 3. Mutation (Biology)—Fiction. 4. Science Fiction.] I. Klavan, Laurence. II. Title.
PZ7.K55992Was 2013 | 2012026744
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[Fic]—dc23 |
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13 14 15 16 17 LP/RRDH 10 9 8 7 6 5 4 3 2 1
FIRST EDITION
## ABOUT THE PUBLISHER
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**United States**
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<http://www.harpercollins.com>
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{"url":"https:\/\/askbot.fedoraproject.org\/en\/questions\/53785\/revisions\/","text":"# Revision history [back]\n\n### gpu lockup\n\nI am a newbie when it comes to linux. I am taking a class this semester that requires that I use the operating system. I installed Fedora 20 on my hard drive. It seems to freeze and then prints out an error message saying GPU Lockup - switching to software fbcon I figure that I have to update the graphics driver. Can someone tell me how to do that. preferably using akmod.\n\n### gpu lockup\n\nI am a newbie when it comes to linux. I am taking a class this semester that requires that I use the operating system. I installed Fedora 20 on my hard drive. It seems to freeze and then prints out an error message saying GPU Lockup - switching to software fbcon fbcon. I figure that I have to update the graphics driver. Can someone tell me how to do that. preferably update properly using akmod.\n\n### gpu lockup\n\nI am a newbie when it comes to linux. I am taking a class this semester that requires that I use the operating system. I installed Fedora 20 on my hard drive. It seems to freeze and then prints out an error message saying\n\nGPU Lockup - switching to software fbcon. fbcon.\n\nI figure that I have to update the graphics driver. Can someone tell me how to do update properly using akmod.","date":"2021-07-26 14:52:53","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\": 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\": 0.19545845687389374, \"perplexity\": 933.9599075502272}, \"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-2021-31\/segments\/1627046152129.33\/warc\/CC-MAIN-20210726120442-20210726150442-00223.warc.gz\"}"} | null | null |
Lepidogyna menziesii är en bladmossart som först beskrevs av William Jackson Hooker, och fick sitt nu gällande namn av Rudolf Mathias Schuster. Lepidogyna menziesii ingår i släktet Lepidogyna och familjen Lepidolaenaceae. Inga underarter finns listade i Catalogue of Life.
Källor
Bladmossor
menziesii | {
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http://tvtropes.org/pmwiki/pmwiki.php/Main/ColonelBadass
Colonel Badass
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This Colonel doesn't fry chicken. He fries you.
"I am Colonel-Commissar Ibram Gaunt. I am known as a fair man, unless I am pushed. You have just pushed me."
— Colonel-Commissar Ibram Gaunt, Warhammer 40,000
Imagine The Captain, but instead of a ship at sea or in the stars, they have a large contingent of ground troops. Still the most kickass guy around, powerful leader, outranks everyone else in The Squad, you know the type. Chances are, they're gonna be a colonel. (which is pronounced "kernel" by the way)
This is probably because "Colonel" is the highest military rank deployable in the field, the highest rank unlikely to be noticed and Authority Equals Asskicking on TV. Theoretically, General Ripper and The Brigadier should have more Power Levels than Colonel Badass but star ranks are for pen-pushers and quill-drivers. That's why the colonel always wins (unless you have a Four-Star Badass lurking around). The fact that an Army, Marine, or Air Force colonel and a Navy captain are technically the same rank may also have something to do with it, or it may just be coincidental.
This man is a professional, Married to the Job, and mission accomplishment is the entire purpose of his existence. He tends to be fairly young for a colonel, likely because he climbed the ranks rapidly after an outstanding performance in an earlier war or campaign as junior officer. When he's got full command of hundreds or thousands of troops, chances are he won't do much fighting himself on a day to day basis, because he's too busy managing the battlefield with a level of competence that prevents the enemy from ever gaining the initiative in the first place. If it ever comes down to it, though, his personal weapon will likely be a high caliber handgun — and while it won't be drawn often, it'll rarely miss when it is. Many have tried to kill this man, many of them extremely deadly in their own right, all of them have failed.
He won't break down over losses, but neither will he throw his men's lives away. If a General Ripper is his immediate superior, the two will almost certainly clash in styles, because not only is Colonel Badass a better leader but he's also more sane, more pragmatic, far more imaginative, doesn't give a crap about political concerns and values ability/utility wherever he finds it. In fact, chances are as good as not he rose from nothing himself, especially if the organization he's a part of doesn't usually encourage that sort of thing. He's an unstoppable force and an immovable object. He's loyalty incarnate, the best friend anyone can have; but he's also the embodiment of vengeance, and if you screw him over, your life is forfeit. Even if he dies he lives on as an ideal.
He is commonly an Officer and a Gentleman... or at least, tries to be one, as long as it doesn't hinder his badassness because this man does not suffer fools. His Evil Counterpart is the Colonel Kilgore.
For any Commonwealth Tropers out there this trope also counts for badass Wing commanders and Group captains as well. On the international stage, it applies to any badass with an equivalent NATO officer rank code of OF-4 or OF-5.
A universal trope, just like Colonel is a universal rank (Lieutenant Colonels are included, as are full commanders, the naval rank equivalent to Lieutenant Colonels in most western militaries). A subtrope of Authority Equals Asskicking. In terms of Authority Tropes, Majorly Awesome and Commanding Coolness are just a notch lower while The Captain is its equal in naval terms, while the next step up is Four-Star Badass.
Col. Roy Mustang from Fullmetal Alchemist, as seen on the page image above. He BURNS HIS OWN WOUNDS SHUT and by the end of the series, he manages to personally take down two of the seven Homunculi, which is REALLY impressive in that universe. He also plays several of the defining bits of the trope incredibly straight - he's only thirty years old, rose through the ranks incredibly fast (partly due to being a State Alchemist, which automatically starts you off at the rank of Major), he's extremely high-minded and has tremendous ideals for the country, and he will throw himself into danger in a heartbeat rather than risk losing one of his comrades, particularly his five dearest and most loyal subordinates.
In the 2003 anime version, he even beats the Führer, another example of him taking down a Homunculus.
Credit must also be given to his best friend, Lt. Col. Maes Hughes, who, despite not being an alchemist, is an investigative genius who also knows how to kick ass with throwing knives. He even gets double promoted to Brigadier General for dying in the line of duty, though that was in part to cover up the conspiracy.
Basque Grand. He's A Father to His Men who leads from the front, transmutes entire buildings into weapons while delivering hamtastic lines, accepts the surrender of Ishval's high clerics and shoots Brigadier Fiessler (an actual General Ripper) when the latter orders the troops to kill the cleric and continue the genocide. Also has one of the best mustaches known to man.
What makes it more impressive is that he had been firmly entrenched in the minds of the fans as a General Ripper by the 2003 anime's portrayal.
While Major Armstrong is never promoted due to his compassion for the enemy, in the 2003 anime version he eventually gets promoted to Lt. Col.
Also in the 2003 anime, Lt. Col. Frank Archer thinks he's one of these; whether he counts is really up to the viewer. While he is quite cool and collected under fire, he's also a sociopathic Smug Snake and General Ripper.
Standartenführer Rudol von Stroheim from Jojos Bizarre Adventure, after he was resurrected from the dead and brought back to life as a cyborg. While initially introduced as callous and cruel, he reveals a powerful sense of bravery when the going gets tough, even going as far as cutting off his leg and blowing himself up with a grenade in order to bring Santana out to the light. His new body lets him stand against the Pillar Man, Kars, and he even pulls a Big Damn Heroes near the end of the arc. His nationalism and military strength show for the rest of his life before finally falling in 1943 at the battle of Stalingrad.
Lt.Col. Hayate Yagami in Magical Girl Lyrical Nanoha StrikerS. This is in sharp contrast, for example, to the local General Ripper who fails to qualify even as a Badass Normal and the three Admirals who only appear on TV screens. Possibly the only Colonel Magical Girl ever.
To make the point clear: the few times she takes the field and uses her powers, everyone on her side is told to evacuate and get the hell out of there. Even Nanoha gets out of the way.
Col. Richard Burton in Madlax single-handedly prevented a worldwide supernatural catastrophe despite having no supernatural powers himself.
Col. Sanders is the preferred nickname of one of the most powerful fighters in Negima! Magister Negi Magi that we've seen so far.
That's Ku:Nel Sanders thank you.
Great General of Darkness/Ankoku Daishogun from Mazinger Z and Great Mazinger, The Dragon of the Emperor of Darkness and commander of the seven armies of the Mykene Empire. He wields a BFS, sports a Badass Cape and a Manly Facial Hair, and he can kick the butt of nearly any Humongous Mecha invented by Go Nagai. He led the army of Mykene Empire, personally or delegating on his generals, and he was A Father to His Men considered unforgivable default intelligence costed the lives of his troops. He fought Great Mazinger because he knew he could win, slicing it with his blade mercilessly as he laughed its attacks off. And he fought Mazinkaiser quite evenly.
Col. Shikishima from AKIRA has several traits of a General Ripper but remains a Badass Normal to the very end.
The commander of the Umibozu, a Navy special forces team that goes up against Section Nine at the end of Ghost in the Shell: Stand Alone Complex's first season, is addressed as Taisa: Colonel, though since he's Navy, his rank proper would be Captain. (In Japan, the same word is used to refer to both Colonel (Army) and Captain (Navy).
Col. Dewey Novak is not only the Big Bad of Eureka Seven but he also pulls off a fully functional My Death Is Just the Beginning scheme triggered by shooting himself in the head. What's more badass than that?
The Rose of Versailles: Oscar de Jarjeyes holds the rank of colonel as commander of the Royal Guards.
Gundam saga has a few:
Col. Sergei Smirnov from Mobile Suit Gundam 00, a Badass Normal Ace Pilot. He is able to, despite having woefully inferior equipment, nearly bag two Gundams in a single battle through sheer cunning and guile the like of which makes CB's own resident strategic genius honestly afraid. The show also contains Colonel Kati Mannequin, who is more The Strategist than a bloodletter, but has gone into battle alongside the troops, directly in harms way, and her strategies have scored her several Moments Of Awesome, the capstone of which was her epic screwjob on the ALAWS in Episode 22.
A captain (from the navy, whose ranks Zeon uses for all branches of military service, which is equivalent to a colonel), Norris Packard of Mobile Suit Gundam: The 08th MS Team definitely counts. He takes on three Gundams, all piloted by the main characters, and three Guntanks, at the same time, alone, in an inferior machine, and wins. And on top of all that he finds time to be a father figure to one of the main characters.
Inferior mech? This is no Zaku boy, No Zaku!
In Mobile Suit Gundam Wing all notable officers in OZ save Noin are colonels, Treize in particular seems aware of this, maintaining the rank of colonel even after conquering the Earth twice.
Wing uses an unorthodox rank structure based more around courtly titles than military rank. Treize is actually comparable to a Major General; Lady Une is a Colonel, as is Zechs Merquise after he earns his two-rank promotion for his actions in Operation Nova, OZ's coup d'état to overthrow the Earth Sphere Alliance.
Naturally, most of these guys all have the Colonel Badass-in name Ramba Ral to thank for all of this. Only a lieutenant, he was still awesome and has often been paired with Norris as a dynamic duo of doom.
Col. Paya Livingston from Magical Witch Punie-chan. Boy, is he badass.
The Colonel, head of God's Army, in Fist of the North Star. His weapon of choice is a Precision-Guided Boomerang, controlled with Psychic Powers. He manages to be a threat to Kenshiro without some form of superpowerful martial arts.
Col. Todou of Code Geass. He's one of the higher-ranked members of the Japanese Liberation Front, and leader of their most elite squad, the Four Heavenly Swords. Later, he joins the Black Knights and becomes Zero's third-in-command.
Also Todou is (based on ratings made by C.C, so take them with salt) the highest rated Black Knight, and second only to the Ace Pilots of the series in his combat abilities.
Neon Genesis Evangelion:
Well, not Colonel Badass but more Major Badass, Major Misato Katsuragi. Especially in End of Evangelion.
She's referred to as Lt. Col. Katsuragi in Rebuild of Evangelion, however. So this trope qualifies.
This particular differentiation between Major and Lieutenant Colonel is significant in the Rebuild continuity. Unlike the Misato in the original series, who is only ranked as a major and is therefore frequently excluded from the confidential intel available to the highest personnel (Gendo, Ritsuko, and Fuyutsuki), the Rebuild Misato is considerably more informed due to her higher rank. As opposed to how series!Misato searches futilely for answers and only later is revealed the truth about Lilith along with Shinji, the Rebuild makes this change in order to allow Misato herself to try to convince Shinji to pilot the Eva by showing him the top-secret Lilith.
Maiden Rose: Taki's high rank combined with his general military badassness puts him squarely in this trope.
Captain Smoker from One Piece. He is stationed at some crappy station outside the grand line and considering his logia power he is ranked to low in the marines. He also cares deeply for Tashgi and his marine subordinates. Furthermore he smokes and uses the smoke power.
It's highly implied that Smoker's been offered promotions before, but turned them down. Either that or the higher ups considered him too much of a loose cannon to risk on the Grand Line. Then Smoker met Luffy and started to chase after him INTO the Grand Line. Post-Time Skip, Smoker is now a Vice-Admiral, the third highest rank of the entire organization, and is still chasing after Luffy, this time in the New World, the second half of the Grand Line and the most dangerous sea in the world.
Colonel Mustard from Clue: he's either a murderer or among a group of people who caught one. Then there is his Manly Facial Hair.
Col. Abernathy, a.k.a. Hawk, in G.I. Joe. After the first couple years of the comic he gets promoted to General and Duke takes over as tactical commander of the team, but he still gets into action on occasion.
Col. Nick Fury in the Marvel Universe. His badassery is compounded in the Ultimate Marvel universe and the Iron Man film due to his portrayal by Samuel L. Jackson.
Somewhat averted in the Ultimate Marvel universe though, as he's actually a four-star general there (or started out that way, anyhow), which would be much more realistic for the commander-in-chief of a multinational paramilitary espionage organization. Either way, it doesn't make him any less badass.
He starts as a colonel (that does not look like Samuel L. Jackson) by the end of his first appearance he's promoted as the head of Shield, and he's a 4-star general by his next appearance.
Of course "Colonel" isn't a rank in 616's SHIELD in the first place. His rank is a legacy from his service in the American military before leaving for SHIELD.
Also Marvel, James Rhodes (War Machine, and sometimes Iron Man), is an Air Force Lt. Colonel in Ultimate Marvel and the movies.
Col. Jennifer Sparks of The Authority. Technically, she did fight in WWII, against Sliding Albion and uses her rank and status in the British Army Intelligence to make them do what she wants.
Wonder Woman: Etta's gotten this treatment in several continuities:
George Pérez's run on Volume 2 turned Etta into a Badass Normal airman to save the former Boisterous Bruiser from the serious Chickification she underwent after Marston's death, but jettisoned most of the character's sense of fun and later writers on the title put her on a bus.
Gail Simone's run on Volume 3 turned former Plucky Comic Relief character Etta Candy into a Deadpan Snarker Lieutenant colonel who is also a Badass Normal, as well as one of Wondy's "sisters" and very dear friends.
Greg Rucka reinserted it during Wonder Woman (Rebirth), portraying Etta as having become Steve Trevor's superior by the modern day.
Etta wasn't the first such character in the WW mythos; Phillip Darnel (the USAAF colonel Diana Prince was secretary to) started out as a colonel with nerves of steel before getting promoted in the original run, and Steve Trevor was eventually promoted to colonel
The most dangerous hitman in Sin City, and the head of the foremost Murder, Inc. organisation, is a man known only as 'The Colonel'.
Buck Danny has been a colonel for years now despite having been a pilot ever since the 1940s (blame Comic-Book Time for that one), pretty much so he can still be allowed to fly.
During the "War In the Sun" arc of Preacher, Herr Starr uses his contacts in the American Military to grant him a battalion of soldiers to hunt down Jesse in the American Southwest. When Starr makes one too many dismissive comments towards the battalion's commander, said commander makes it clear that he could care less what connections or position Starr has and threatens to personally cut out Starr's liver and fry it if even one soldier is harmed.
Abraxas (Hrodvitnon): Colonel Diane Foster gets to demonstrate herself to be this again like in Godzilla: King of the Monsters (2019), providing this quote whilst she and the G-Team combat the Many:
"I think I speak for everyone when I say none of us will sleep at night unless we see to it this zombie apocalypse horseshit is very definitely, permanently, finally burned to ashes, and then we burn the ashes!"
In the 'Daria' Fanfic series 'Legion of Lawndale Heroes', there are two Colonel Badass types - Colonel Kyle Armalin, and Colonel Franklin Davers. The latter's an Army Special Forces type, the former a Marine aviator with a LOT of covert paramilitary experience from shooting things all over the world. Just to cement his badass cred, first look at the first initials of his full name - Kyleton Isaiah Armalin... and then, also consider that he was given his middle name in honor of Isaiah Bradley - the FIRST Captain America, and an 'old-school' badass if EVER there was one.
Colonel Nick Parker may be "retired", but that doesn't stop him from organizing and waging a guerrilla war inside of twenty-four hours of the Nod invasion, which comes complete with hunting down and destroying Avatar warmechs, blowing up a Nod general using his own artillery, and recapturing the White House. There's a reason he's called "Havoc."
Played straight in Immortality Protocol Cy Fox by Dr. Robotnik's father.
Lieutenant Colonel Angus Beckett is this in the Robotech fanfic Scoop. Later, due to the wear and tear flying Valkyries does to his body, he becomes a Desk Jockey, and is still one when he appears as a lieutenant general in Dire Straits
Lieutenant Colonel Lupon Kravshera is this in the Robotech fanfic Dire Straits
In the Narnia fanfic King Edmund's Crusade the Colonels are not only the badass fighters of the armies but also the field commanders and the leaders of the elite bodyguard units. The senior commanders (below the Four Pevensie Monarchs themselves) are the Marshals - but two of them are explicitly more administrators than fighters. Col Elikolani (a sarcastic, flirtatious black panther) is a brutal fighter far more suited for war than Marshal Altaica (a rather effeminate and cowardly tiger). Marshal Nicodemus of the wolves is no slouch, but Col Rapine is a walking legend.
The Judge Dredd fanfic Aegis has Wing Commander Nicolai "Nick" Betancourt. Although he doesn't do anything particularly badass in the story, he is tapped to fly an experimental Justice Department combined platform of a hypersonic air superiority fighter and giant airship. Given the resources the Judges are throwing at this mission, he has to be the best pilot available. Later stories begin to reveal his background as a particularly famous pilot and commander and "the Hero of Pingpongyang".
Suzumiya Haruhi no Yaku-Asobi has Colonel Harriet Isuzu, an alternate Haruhi who works for a multi-dimensional paramilitary law enforcement agency that monitors sliding. Possible subversion in that soldiers from her home dimension's military look down on said agency and thus smugly refer to her as "agent". Her rival from the intelligence department, Colonel Sascha Sakisa also counts, being quite The Trickster.
There is a Russian movie named The Apocalypse Code, one of the better Russian action movies, actually, whose protagonist is an Action Girl (bordering Extraordinarily Empowered Girl) serving in the Russian special forces. Obviously, she is a colonel. Which is revealed just before she gets serious, topping all her previous stunts.
Apocalypse Now features two Colonel Badasses. The first is Colonel Kurtz, who had the credentials to be a general but chose a life in the shit. While fighting in Vietnam, he goes kill-crazy and creates an army of zealots who worship him like a god. The other is Lt.Col. Bill Kilgore, a bulletproof badass who surfs in warzones and enjoys the aroma of napalm like a nice cup of joe. In the Redux edition, however, he's taken down a peg after his surfboard is stolen and he broadcasts messages begging for it back.
Armageddon:
Colonel Willie Sharp, pilot of the space shuttle. Can disarm nine-foot-long nuclear weapons with only 20 seconds and minimal assistance. Also the only guy trusted with a firearm on the whole vessel (making him the only one badass enough to bring a gun onto a spaceship), and the only guy with enough balls to disobey a Presidential order for the greater good of the planet.
Colonel Lev Andropov (Peter Stormare) also qualifies: not only does he save AJ (Ben Affleck) when the space station blows, he also restarts a FREAKIN' space shuttle. He also gets most of the best lines in the movie:
Lev: I AM THE ONLY CERTIFIED ASTRONAUT HERE, AND I AM SAVING YOUR AMERICAN ASS!!!
Colonel Miles Motherfucking Quaritch of Avatar. He's obviously the villain, but that doesn't stop him from modding his AMP suit with a gigantic combat knife, or running out into Pandora's toxic atmosphere, guns blazing, without bothering to put a gas mask on. In one instance, it takes him about 11 seconds to react to the fact that he is, in fact, on fire. Quaritch is in fact, so badass he inspires major Rooting for the Empire.
So badass that James Cameron has promised that Quaritch is coming Back from the Dead to reappear in all the sequels.
Lieutenant Colonel Danny McKnight in Black Hawk Down. He casually walks from a convoy of Humvees to a building, through the open, while Rangers all around him are diving for cover and flinching. He ignores a gunshot wound to his throat that would have opened his carotid artery if it had been a millimeter to the right. After that injury, one of the Rangers tells him he doesn't need to go back out to save the rest of his men. He just smokes his cigar, looks at the guy like he's an idiot, then rides out with the 10th Mountain Division to get the rest of his men.
Arnold Schwarzenegger's character John Matrix takes this trope to an utter extreme in the movie Commando. Although his character is a retired Special Forces colonel, Arnie still manages to kick enough ass for several movies.
Colonel Nicholas Alexander, played by Lee Marvin in his final performance, was the only man who could possibly give Chuck Norris' Major Scott McCoy orders in The Delta Force.
He may have been a Psycho for Hire, but Col. Koobus of District 9 was also incredibly badass. He was consistently shown to be a supremely competent leader of his men, never lost his cool in the heat of battle (though he did start to lose it when he was bragging to Wikus about killing him), showed bravery even in the face of overpowering alien weaponry, and when he was surrounded by all sides by Prawns, who were going to tear him apart regardless, he still went down fighting.
Thoroughly averted in Dr. Strangelove: Group Captain (RAF for Colonel) Mandrake is a bit of a wimp and fails to properly stand up for himself when around other domineering American officers, while US Army Colonel "Bat" Guano does not even understand his own mission, and is instead obsessed with eliminating "preverts."
Colonel Günther Reza of Duck, You Sucker! a Silent Antagonist and Implacable Man who survives having a bridge dropped on him, a train explosion, and having half a machinegun clip emptied into him before finally going down.
Eve of Destruction: Colonel Jim McQuade to a realistic degree. An army veteran, expert marksman and counter-terrorism expert, he's the guy the government calls in to take down a rogue killer android, but he does get pretty banged up over the course of the film.
Colonel Douglas Mortimer in For a Few Dollars More, a former Confederate Army colonel who carries a veritable arsenal with him wherever he goes, and has a personal score with Big Bad El Indio.
Col. Thursday in Fort Apache... well, sort of... at least until he orders the infamous Thursday's Charge, which results in the utter destruction of half the regiment.Thursday is an Expy of George Armstrong Custer, a real-life Colonel Badass, who would be remembered as a great cavalry leader except for one mistake...
In the movie Glory, Matthew Broderick plays a Real Life Colonel Badass; in this case Col. Robert Gould Shaw the son of Boston abolitionists who commanded the Union Army's first black troops in the Civil War.
Col. Andrea Stavros in The Guns of Navarone movie adaptation, the best hand-to-hand fighter in the group.
In I Am Legend Will Smith's character is Lieutenant Colonel Robert Neville. A brilliant scientist who also managed to survive being badly outnumbered by cannibalistic hordes. It wasn't because he was a punk...
Col. Hans Landa, a.k.a. the Jew Hunter, from Inglourious Basterds. The epitome of both Magnificent Bastard and Wicked Cultured.
He does end World War II by helping the Basterds kill the German High Command.
In Indiana Jones and the Kingdom of the Crystal Skull we learn Indy attained the rank of colonel during World War II while working as an agent for the OSS (the precursor to the CIA). Colonel Doctor Irina Spalko most definitely counts too.
Averted in Ip Man, where Japanese Colonel Sato is a Smug Snake who only hits people when they're in no position to fight back. Preferably from safe up high behind a gun.
Col. Hardy from Man of Steel possibly outdoes Quaritch in terms of sheer badassitude, taking on a Kryptonian enemy with a combat knife —and no robot suit— and managing to pull a no-score draw, losing his life but banishing the villain back to the Phantom Zone in the process.
Marvel Cinematic Universe:
Lt. Col. (full Colonel as of Iron Man 3) James Rhodes, who has a suit of powered armor.
Col. Nick Fury, the director of SHIELD and architect of the Avengers. Extra badass props for Fury being portrayed throughout the Marvel Cineverse by none other than Samuel L. Jackson.
Col. Chester Philips, who lead the Strategic Scientific Reserve in WWII, the precursor to SHIELD, and commanded Captain America.
Colonel James Braddock, from the Missing in Action film series. Played by none other than Chuck Norris, it's a given he's able to kick as much Viet Cong butt as he does.
MonsterVerse:
Kong: Skull Island: Colonel Packard has been through it in Vietnam, and he goes through much worse on the island and still survives for quite a while through the film. Unfortunately, he also becomes a General Ripper due to his Vietnam-instilled Colonel Kilgore.
Godzilla: King of the Monsters (2019): Colonel Diane Foster leads Monarch's military force the G-Team, at one point getting into a firefight with the eco-terrorists looking to release Ghidorah and killing Asher, Jonah's second-in-command. Had she not prioritized Asher because of his advance on Mark, she could have killed Jonah there and then.
Colonel Vincent "Killer" Kane in The Ninth Configuration is a legendary soldier during the Vietnam War. Before the film starts, he suffers a breakdown after chopping a Child Soldiers head off with a wire garrote. He dedicates himself to healing instead of harming, but ultimately a bar full of bikers learn that it's still not a good idea to bully his patients.
Colonel Dax in Paths of Glory. Try not flinching when enemy shells are exploding randomly less than 20 yards away from you. Go on, try. We'll wait.
Col. Benjamin Martin in The Patriot. Single-handedly killed a platoon of Redcoats, including at least three with a tomahawk. Colonel Tavington is an Evil Brit version, kicking the butts of Mel Gibson's ragtag Rebel militia.
You gotta give it up for Tavington when his troops are caught with their pants down by a bunch of angry rebels, his soldiers are getting shot all around him. He just stands there, calmly reloading his pistol (which takes a good 20 seconds) and shooting one rebel after another.
Colonel Graham from Posse is the evil version. Losing an eye just makes him more determined than ever to hunt down the heroes and retrieve his gold.
Col. Sam Trautman, the former trainer and commander of none other than John Rambo. He's more of the mentor in the movies than Colonel Badass, but he gets this status because of his line in the first movie:
Sheriff Will Teasle: Where in God's name did this Rambo come from...
Col. Trautman: God didn't make Rambo. I made him.
(Cut to a shot of the tent's entrance. A silhouette of a bereted and Badass LongCoated individual is standing there.)
Colonel Pembroke (James Coburn) in A Reason to Live, a Reason to Die!. He escapes from a Confederate P.O.W. Camp and makes his way back into Union territory, despite being branded a coward and wanted as a traitor. He then gets himself arrested to be taken into a Union fort commanded by an old friend of his. He then assembles his own team of Boxed Crooks to undertake a Suicide Mission to retake the impregnable fort he surrendered, and kill the major he surrendered to.
Colonel Cord McNally in the first quarter or so of John Wayne movie Rio Lobo is introduced as a seasoned officer, who personally leads the pursuit for a band of confederate train robberies, tracks them for miles cross country and doesn't' hesitate to go riding after about a dozen men alone as his command is gradually forced to split up. After being captured he still manages to turn the tables on his captors. In the post-war sections of the movie, he has left the army to track down the men who betrayed him during the war, and shows signs of becoming a fairly competent gunslinger.
Lt. Col. Frank Slade from Scent of a Woman. He does the tango with a beautiful stranger who's waiting on her boyfriend, drives a Ferrari BLIND, and completely and utterly owns snobby schoolmasters. Plus, he's played by Al Pacino, which doubles his badassitude.
Completely averted (as to be expected) in the film Spaceballs, in the figure of Colonel Sandurz. Whassa matter, Colonel Sandurz ...CHICKEN?!
Considering that he's Grand Moff Tarkin if Tarkin were stupid rather than evil, this is to be expected.
Colonel Rhumbus in Spies Like Us. He knocks out his own squad of elite ninja soldiers and then takes our heroes through accelerated GLG20 training. His salutes are so snappy that you can hear his gloved hand cutting through the air.
In the film adaptation of Street Fighter, Guile has the Colonel rank (he is a Major in the games' continuity), and goes against the Allied Nations' orders to lead a strike team and raid Bison's base to rescue kidnapped AN troopers and relief workers. He's also the one to take on Bison himself, beating him even after he starts using his Psycho Electro powers. This also holds true to the cartoon, which is a direct sequence of the film.
Colonel Min Se-hoon, the Inspector Javert character in the Korean film The Suspect. His introductory scene involves him dropping one of his men out of a plane without a parachute in order to save another man's life, then coolly picking up a spare chute, diving after the falling guy, catching him in mid-air, slapping him back to life when they hit the ground, then swaggering away smoking a cigarette. He then puts on a leather jacket and shades and spends the rest of the film ripping apart a secret government conspiracy while simultaneously hunting down a North Korean master assassin.
In Top Gun the soft-spoken, gruffly avuncular Commander Mike "Viper" Metcalf and the foul-mouthed, quick-tempered, cigar-chomping Commander Tom "Stinger" Jordan provide contrasting, but equally classic, examples of this trope.
Lt. Colonel William Lennox. Does he sit behind and command his troops from the command post? Fuck no! He takes on the Transformers with them head-on!
Though he's not a Colonel until the third film.
Colonel Stryker in X2: X-Men United.
Stryker: I was pilotin' Black Ops missions in the jungles of North Vietnam while you were suckin' on your mama's tit at Woodstock, Kelly. Don't lecture me about war. This already is a war.
in Terry Mancour's The Spellmonger Series, Bold Asgus, Commander of the Orohan's band. He oozes command and confidence. Proficient with almost all weapons, he is a one man wave of death on the field of battle. The man launched himself onto a troll and killed it with one hit. then he dismounted into a double decapitation of nearby Mooks
The Discworld's Sam Vimes may be the Commander of the City Watch and even a Duke, but despite the protests of both his wife and the Patrician of Ankh-Morpork, refuses to delegate the street jobs to his subordinates, being happiest when walking the streets in cardboard-soled boots in the rain or chasing a dangerous criminal. He abhors the politics he's forced to participate in, and still basically sees the world like a beat copper.
Col. Fedmahn Kassad of Dan Simmons' Hyperion Cantos fights his way through the exploding wreckage of an enemy fleet, plummets down onto a planet and then fights the soldiers there. Later, he gets into hand-to-hand combat with the Shrike, a giant, indestructible Eldritch Abomination covered in spikes. Did we mention it can move faster than light? He is explicitly stated to be the greatest soldier in history which is why godlike A.I.s use his consciousness to create the Shrike.
Older Than Radio: Max Piccolomini, colonel of a cuirassier regiment, in Friedrich Schiller's Wallenstein trilogy (1798-1799). Which unsurprisingly features a number of war-hardened colonels.
In his 1884 Adventures of Huckleberry Finn, Mark Twain gave us Col. Sherburn who shoots a drunk who only annoyed him and gets away with it by scaring the lynch mob away with a Breaking Speech.
Colonel Sebastian Moran in the Sherlock Holmes stories. In addition to being Professor Moriarty's right-hand man and top assassin, he's a decorated war hero and celebrated big game hunter who once crawled down a sewer drain after a wounded man-eating tiger.
While not Colonel, Major Greer from The Passage qualifies.
Warhammer 40,000 novels:
Two in Warhammer 40,000: Gaunt's Ghosts: Colonel-Commissar Ibram Gaunt and his subordinate, Col. Colm Corbec. Gaunt has a head-start in that he's already a Badass Longcoat, but Corbec makes a rather good accounting of himself as well.
As of his promotion, Col. Elim Rawne also neatly fits the trope.
Col. Shaeffer of The Last Chancers. He can stroll up and down a battlefield filled with Orks and Tyranids like he was on parade, decapitating Carnifexes and seeing the whole battle through without a scratch. At the same time, the hardened criminals of the 13th Penal Legion absolutely fear him more than those same Orks and Tyranids.
Let the Galaxy Burn: Commisar von Klas. He's kidnapped and enslaved by a kabal of dark eldar, kills an eldar wych in one-on-one close combat, kills his torturer with his own weapons, then organises a break out which dooms the entire kabal to be defeated by a rival faction. The story ends with him telling his former master "They will cut my throat like some common animal. I suspect, however, you will take much longer to die."
Col. Regina Kasteen of the Valhallan 597th is more noteworthy for her strategic and tactical skill than her record at personal combat, but she's no slouch at the latter. In fact, during The Traitor's Hand, she's positively glad to have a chance to get stuck in, potting traitors with her bolt pistol as they storm her headquarters. Of course, her backstory — that she attained her command by default, being the most senior member of her regiment who was not eaten by Tyranids — attests to impressive badass credentials all by itself.
Count Dokhturov in War and Peace. Calm, methodical, the perfect man to have fighting for you. Tolstoy devotes most of a page describing why people like Dokhturov are never considered heroes despite the fact that battles would be lost without them.
Admiral Jane Roland and eventually Laurence himself after his promotion to admiral in the final book in Temeraire. Jane casually breaks swords in people's chests, wins massive air battles on the back of her acid-spitting dragon and generally impresses everyone she meets. Technically, Temeraire himself also counts in book 5.
Lt.Col. Du Bois in Robert Heinlein's Starship Troopers. He taught the main character's History and Moral Philosophy class, and the character was startled that the old man had been both a Lt.Col. in the infantry and a classmate of his Drill Sergeant Nasty, Sergeant Zim.
Col. 'Lizard' Tirelli (from The War Against the Chtorr sci-fi series by David Gerrold) is so badass she single-handedly flies jet-assisted helicopter gunships in her spare time.
Mostly inverted in the Sharpe series, except when Sharpe himself attains the rank. Much of the conflict in the series is driven by various Upper-Class Twit officers. While not technically a colonel, Sharpe often fulfills a Colonel's duties as his regiment tends to go through colonels like disposable cups.
There are few colonels in the Star Wars Expanded Universe, but several Commanders.
Commander Wedge Antilles: A fighter pilot who flew against both Death Stars and rose to become the leader of Rogue Squadron, the Alliance's best fighter squadron. Instrumental in the retaking of Coruscant from the Empire, the negotiating of a cure for the Krytos disease, the taking of Thyferra, the capture of a Super Star Destroyer, the death of Ysanne Isard (twice!), and the campaign against Warlord Zsinj. Then he finally accepted promotion to General. (Weirdly, he was never a colonel, even though, depending on the source, colonel is a half-notch above commander.)
Commander Mitth'raw'nuruodo: A member of the Chiss Expansionary Defense Fleet, Thrawn was born a commoner and gained status through military service, becoming a merit adoptive of the Eight Ruling Family and further becoming a trial-born. He was the youngest ever Force Commander in the Expansionary Fleet, using brilliant and undeniably effective but underhanded tactics, often involving preemptive strikes, that were seen as morally bankrupt to his people. Sometime after the events of Outbound Flight he was exiled and wound up in Imperial service, where he very quickly rose through the ranks until becoming one of the thirteen secret Grand Admirals.
Hand of Thrawn: Supreme Commander Gilad Pellaeon: "Supreme Commander" is actually the highest military rank in the Empire, so this properly belongs in Four-Star Badass. But "Commander" is part of the rank, so he gets a mention here. He also was a captain for a while, which is roughly equivalent to a colonel.
There was at least one colonel. When Wedge's wingmate Tycho Celchu took command of Rogue Squadron, he took this rank.
Before he defected to the Rebel Alliance, Soontir Fel was a Baron-Colonel. His badassery is unquestionable; after Vader's death, Baron Soontir Fel was perceived as the best pilot in the Empire, bar none (and without Vader around, rumors started spreading that Vader's only edge was in his expensive custom starfighters). He lost that rank when he joined Rogue Squadron, but it may be assumed that he got it back, and more, in the Empire of the Hand after Thrawn had him kidnapped and brought there.
Other real colonels (okay, lieutenant colonels are among them) would be Jaina Solo (being a member of Rogue Squadron at the age of sixteen, seriously kicking around numerous Yuuzhan Vong, later on being in command of Rogue Squadron just to be court-martialed by her own twin brother whom she later on kills in a duel after he turned out to have become a Sith), her brother Jacen Solo (became head of the secret police of the Galactic Alliance, tortured Boba Fett's daghter to death, became a Sith Lord, made himself joined head of state and later on sole head of state of the Galactic Alliance, set half of Kashyyyk on fire, killed his aunt Mara, etc. until his sister finally got him), and Jagged Fel (who (nearly) manages to keep up with Jaina Solo and Kyp Durron, two extremely talented Jedi pilots using a battle meld, and displays various feats of general badassery until he finally becomes the Imperial head of state).
Colonel Jesse Wood, from the 1632 series arguably qualifies, being the very first pilot in the new timeline created by the Ring of Fire. Because of that and his service in the uptime US Air Force, he's appointed head of downtime's new Air Force.
Patrick McLanahan from Dale Brown's books spends some time as this in earlier books, before receiving his stars at the end of Fatal Terrain. He's not the only character in the books to go through this phase, though.
Colonel Rosa Klebb of the Soviet SMERSH in Ian Fleming's From Russia with Love. In the film version she defects to the terrorist organization SPECTRE and fails to kill James Bond with a poison-tipped dart in her shoe but she is still a Russian agent and succeeds in poisoning Bond and almost killing him with the shoe dart at the end of the book.
Colonel Christopher Williams in Tranquilium. He starts out as a charming Reasonable Authority Figure the main character meets in Port Elizabeth, then quickly proves his badass credentials by showing that yes he can thwart stage one by rallying the militia to defeat the KGB-led rebellion in the town and so saving the main characters from the rebels in the process. It is subsequently revealed that he is (was, as of Part Two) also an Almighty Janitor to the Merryland government and a former FBI agent who was a literal Mulder who eventually became fed up with his superiors' adamant refusal to notice Tranquilum and Soviet shenanigans therein and went native to fight a one-man-war. He's also firmly an antihero, and an awesomely ruthless one at that, going about and taking out Soviet spies with his crack squad of Forbidders and using torture to bring down a huge part of their information network. He is also scarily good at one-shotting would-be-prominent-villains on the Soviet side, sometimes across dimensions.
Codex Alera: While Aleran Legions are commanded by Captains, they don't use modern military ranks and their role tends to fit this trope better. As usual, the most badass of the lot is Captain Rufus Scipio (a.k.a. Tavi) a Badass Normal capable of taking on Knights and Canim by himself.
In the Legacy of the Aldenata, there's Colonel Cutprice, a rejuvenated Medal of Honor winner and one of the most decorated Korean War veterans. Later, he's leader of the Ten Thousand, an elite fighting group arguably more badass than the ACS as a whole, as the Ten Thousand fight without the benefit of Powered Armor. It's explicitly stated that he refuses promotion above the rank of Colonel.
From Tom Clancy's Red Storm Rising:
There's an unnamed Soviet colonel who only shows up in a single scene, in charge of a tank division after his general was killed in an air attack. Although his narrative role is to provide the perspective of a Soviet mid-ranking officer (the POV character, Alekseyev, is a Four-Star Badass, through whom we see the big picture), the colonel proves to be extremely competent at commanding his division, nearly forcing a breakthrough and stymied only by the NATO defenders' resilience and Soviet high command's insistence that only Moscow can move the strategic reserves. He is last seen as the survivor of an artillery barrage on his headquarters, surrounded by the wounded, and still calmly giving orders to his division.
The also unnamed KGB colonel who set the Kremlin bomb - he later goes on to set another Kremlin bomb and kills four men with his silenced automatic. Bonus points for being a Badass Longcoat too.
Colonel Douglas "Duke" Ellington, the F-19 Ghostrider pilot who provided the main point of view character for the central front air war in Germany and the Soviet colonel commanding the MiG-29 regiment at Keflavik both also qualify (both demonstrate their Ace Pilot skills numerous times, the MiG-29 regiment CO survives leading his pilots in a battle against two full squadrons of F-14 Tomcats at one point, while Ellington also gets to demonstrate his all-around badassery by managing to successfully escape and evade after being shot down over hostile territory late in the war).
Colonel Sir Nigel Loring of the Emberverse (also late of the Blues and Royals). In a universe where much technology (including guns) suddenly ceases to function, he helps rescue the Queen and the Royal Family, trains his troops in the "new" fighting methods, makes sure his soldiers' families are safe, makes a daring escape from captivity, comes to the aid of the Crown Prince in battle, and outwits the Lord Protector of Portland. Only then, does he hook up with the main plotline!
The Colonel (technically, Lieutenant-Colonel) in George MacDonald Fraser's McAuslan stories, based on a real commander of the 2nd Gordon Highlanders. He is quiet, unassuming, looks like a vulture, and walks with a limp (his leg having been broken by the Japanese as a PoW). He's over retirement age and has served with the battalion since 1914, wounded at Passchendaele and captured at Singapore. But he is highly respected, a Father to His Men, "looked the Japanese in the eye on the Moulmein Railway and said 'no'", and at eighty went into the streets of Belfast with a patrol from his old regiment, just to see what things were like for the new generation.
Horatio Hornblower is promoted to Commander at the end of Lieutenant Hornblower, only to be demoted back to Lieutenant and put on half-pay when the French and English declare an armistice. He is promoted back to Commander at the start of Hornblower and the Hotspur, where his exploits eventually earn him a promotion to Captain. In a later book, he is appointed as a Colonel of Marines, a sinecure position granting him extra pay with no extra responsibilities, as an indicator that the Crown is pleased with his performance.
Lieutenant-Colonel Munro of the 60th Foot (Royal American Regiment) in The Last of the Mohicans.
D'Artagnan in The Vicomte de Bragelonne — as a lieutenant and captain of the king's Musketeers he becomes the equivalent of a colonel and major-general of the regular army.
William H. Kraft of Victoria is an eccentric example, since he considers himself a subject of the German Emperor and leads his armored battalion through the campaign to defend New York dressed in a blue, late 19th-century Prussian greatcoat and Pickelhaube. Notwithstanding, he is both a savvy political operator and an absolutely terrifying enemy.
The seventh season of 24 features the villainous African, Colonel Ike Dubaku of Sangala. Also a Scary Black Man.
Col. John "Hannibal" Smith from The A-Team. "Hannibal", in this case, has nothing to do with the Serial Killer, but the Carthaginian general who almost brought down the Roman Republic. This Hannibal is that good. "I love it when a plan comes together."
After the failure of the crappy, bumbling Major sent to hunt the A-Team down, the Army gets serious and brings in Colonel Decker, an unconventional badass in his own right who's very nearly as good as Hannibal. From then on, it goes from pratfall laughs as the Army is outwitted to hair-raising near-misses where the team's celebrations at beating the bad guy of the week are cut short as Decker closes relentlessly in. Decker's first appearance has him chatting with his superiors as he fires off every infantry weapon in the army one by one. Hard-friggin-core.
Colonel Tigh of the Battlestar Galactica. Whether leading a resistance group on New Caprica or wielding two automatic rifles while fighting mutineers, the guy is BADASS. Bonus: he rocks the Eyepatch of Power.
Colonel Wilma Deering in Buck Rogers in the 25th Century, who was not only a badass but H. O. T. Gotta love a woman who can be a Colonel Badass while rocking a spandex catsuit.
Colonel Alistair Gordon Lethbridge-Stewart in his first serial before he was promoted to Brigadier.
"The Sontaran Stratagem"/"The Poison Sky": Col. Mace. "You will face me, sir!"
"The Time of Angels"/"Flesh and Stone": Bishop Octavian.
Taken Up to Eleven in the BBC Edge of Darkness, where all the spooks seem to be colonels.
One episode of Flashpoint indicates that Sam Braddock's father literally earned the moniker "Colonel Badass", though he had been promoted to General by the time this was revealed.
Averted by Col. Wilhelm Klink in Hogan's Heroes, although his rank might just be a foil to that of good guy group leader, Col. Hogan, who plays this trope more or less straight.
Perhaps ironically, in the German version of the show (Ein Käfig voller Helden) Klink gets his proper German rank of Oberst, but Hogan, as an American, is still called Colonel Hogan by everybody... thus playing the trope completely straight.
Commander Harmon "Harm" Rabb, Jr. in JAG. Lieutenant Colonel Sarah "Mac" Mackenzie would also qualify.
Thai Colonel Patano in "Déjà Vu". It's made clear that the only reason Harm isn't dead is that Patano never had the intention to kill him.
Subverted in M*A*S*H with ineffectual Lt. Col. Henry Blake, then played somewhat straight with his replacement, Col. Sherman T. Potter, who had massive cred as a leader of soldiers as a former enlisted man and veteran of earlier wars.
Truth in Television for Potter's history giving respect. People who go from NCO to officer are called "Mustangs", and enjoy great respect from enlisted personnel.
And Hotlips' overhyped hubby Lt. Col. Penobscot is likely a parody of this type.
Colonel Flagg fits this role as is evidenced in the episode where he breaks his own arm so he can infiltrate the hospital as a patient.
Double bonus badass: When an X-ray shows that his arm has healed sufficiently for him to be released, he pulls the X-ray camera down on his cast, shattering it and re-breaking his arm.
Col. Flagg would come to squander his badass credibility in later episodes, though. (Sometimes edging towards Colonel Kilgore, occasionally, at least mentally.) His behavior in later appearances became more and more erratic and paranoid; culminating in complete disgrace in his final appearance.
The British Army Colonel character played by Graham Chapman on Monty Python's Flying Circus. He can come in and put a stop to a sketch when he thinks it's getting too silly or out of control!
Colonel Carrillo in Narcos, a highly competent and honest - if somewhat brutal - military officer and the only guy that Escobar was afraid of.
Col. Mason Truman of Power Rangers RPM You definitely going to need a Colonel Badass to be in charge of the last remaining humans on Earth. His appearance in the first episode says it all: Explosions reflect in his shades. He just stands there watching over his soldiers as all hell breaks loose. When Corporal Hicks tells him they're all screwed, he just tells him to "go shoot at something."
At the end of the final episode of Raumpatrouille, Major McLane is promoted to Colonel for once again saving Earth.
Otto von Stirlitz (real name: Maxim Isaev) from Seventeen Moments of Spring, the pinnacle of a Soviet Spy Fiction, is a Standartenführer (SS equivalent of colonel) ...and in his real life as a Russian spy, a NKVD colonel. He was so badass back in his days, that he experienced a Memetic Mutation in recent times and ended up being a Memetic Badass, as well (no, seriously).
Colonel T.C. McQueen from Space: Above and Beyond. He's the sole survivor of the battle between the Earth's best squadron and the Chigs. Like his last fight with "Chiggie Von Richtoffen," and his "I don't think 'our Lord' wants to hear from me right now," speech.
He's also an "In Vitro", who were genetically-engineered to fight the Silicates, after the latter Turned Against Their Masters.
Stargate SG-1 features Jack O'Neill, Samantha Carter, and Cameron Mitchell.
In Stargate Atlantis, there's at least four colonels and lieutenant colonels, and possibly more. One episode had all of 'em arrive in the same room at once, with predictable results .
That scene is a reference to a scene in the SG-1 episode "Frozen", except the overused title in that one was "Doctor".
Colonel Everett Young in Stargate Universe would be this, if he didn't frak it up with some truly dumb-ass decisions. General O'Neill had to personally let him know that he was screwing up the mantle with his actions in "Incursion", part 1.
Once he gets his act together though, he quickly regains this status.
There's also Colonel David Telford (played by Lou Diamond Phillips).
Kira Nerys becomes one of these in the final season of Star Trek: Deep Space Nine, but she was a badass since the pilot episode.
Usually the heroes in Star Trek tend to be Captains, but the Commander rank can't be overlooked. The first, and in some cases second officers, of the ships hold this rank. Any Starfleet officer that's a Captain or higher was one. Ben Sisko of Star Trek: Deep Space Nine was a Commander even though he was The Captain.
Some of the lower-ranking characters are Sergeant Rock, as well.
Especially Kirk could verge on this trope: a naval Captain does have an equivalent NATO officer code of OF-5 (the same as a Colonel), and Kirk repeatedly went down to planets himself and got into dangerous situations (it's a rare The Captain who can solve problems by punching them).
Strike Back has Colonel Grant who normally commands the team from the command center but when a things go to hell she goes into the field and saved the day herself. She even commandeers an arms dealer's team of mercenaries to go and rescue a member of her team trapped in southern Sudan.
Her successor Rachel Dalton, while not actually a Colonel (Captain upon introduction, promoted to Major at the end of the episode), displays some serious badass cred in the season two opener, using an anti-tank rocket to blast through a Somali blockade about midway through the episode and generally keeping up effortlessly with Scott and Stonebridge on the battlefield until their extraction arrives. As a bonus, she initially appears to be little more than a junior bureaucrat until she finds said rocket launcher in the enemy's weapons cache.
From Shadow Warfare onward, Colonel Philip Locke takes the reigns of leadership from Dalton after she has made questionable decisions during the course of the season and becomes the official leader after her untimely death. Colonel Phillip Locke is a 30 year veteran of the British Army and served in every conflict Britain was involved in during that time. Stonebridge is clearly impressed that Locke is involved in their mission. Locke then proves his reputation correct when he singlehandedly saves Major Dalton from a hitman. When the team is ambushed at an airport, Locke's response is to go on the offensive and kill everything (until getting attacked from behind). Later, when forced to dig his own grave, he instead kills the men guarding him, and nearly escaped on his own. And even while wounded and being tortured, and offered the name of the man who killed his son, he still won't give up NATO secrets.
Colonel Alexander Coltrane of Revolution and Vendetta kicks ass in every action sequence he's in, either physically, with a gun, or with both.
Nate Taylor from Terra Nova, played by none other than Stephen Lang. Due to strange time dilations, he had to spend 118 days alone in the Cretaceous Period and doesn't even have any visible scars. And he still goes toe to toe with carnivorous dinosaurs to protect his people.
Col. Ed Straker in UFO. In one episode, he shoots an opponent who can travel through time and downs a UFO with a rocket launcher.
By that time, Straker was a "Commander" with colonels and naval Captains as subordinates. This Commander-is-the-boss idea occurs in many of the Gerry Anderson series.
On Ultimate Force, Colonel Aidan Dempsey reliably kicks a lot of arse when called upon, most notably in the episodes 'Dead Is Forever', 'Never Go Back' and - particularly - 'Charlie Bravo'. In the latter, he strides through a gunfight, casually taking one-handed potshots at rebels, while exhorting his local counterpart to "Pretend you're an officer and get your men in order!"
Sergeant Major Jonas Blane from The Unit, even though he's not a colonel. The show's resident colonel, Tom Ryan, is more Da Chief.
Sergeant Major (or his battalion staff equivalent, the Command Sergeant Major) is the Colonel Badass of Army NCOs. Any officer who doesn't give their advice careful consideration is extremely foolish.
Colonel Cedric Daniels in The Wire. He starts as a Lieutenant, and becomes Majorly Awesome, Colonel Badass, and finally Da Chief, before realizing he doesn't want to be at the head of such a flawed police department. the fact that he immediately starts getting blackmailed doesn't help. He quits and becomes a lawyer.
Steve Trevor in the Wonder Woman TV series. He gets knocked in the head with a blunt object about once per episode, but he always wakes up with no ill effects and never complains. In early episodes, he even wears military ribbons that weren't even issued until after the war ended.
Elite Agent Rotor in Dino Attack RPG. In addition to taking command whenever the team takes to the skies, he also happens to be an Ace Pilot, has a taste for classical music, and let's just say you don't want to be on the wrong side when you start hearing that classical music. His second-in-command Elite Agent Cabin is quite tough as well.
In The Gamer's Alliance, Nobuo Iwasaki and Samachi Nomura are very competent colonels although the former doesn't show it unless he's forced to act.
Standartenführer Rhianna von Adolph from Open Blue is a constantly smiling Knife Nut who is in charge of the special operations division of Sirene's intelligence agency. She is also The Captain of two ships.
Rifts has Col. Buck Murphy, a hotshot SAMAS (flying Powered Armor) Pilot who has an impressive combat record, and is A Father to His Men.
In order to make it to colonel in the Imperial Guard of Warhammer 40,000, hefty portions of this trope are required.
Colonel "Iron Hand" Straken, Commander of the 2nd Catachan Infantry Regiment. He was badass enough to begin with, being from the Death World of Catachan, known for producing some of the toughest soldiers in the Imperium. Then he was attacked by a Miral land shark that tore off his arm. Instead of dying like a lesser man would, he killed the beast with his "fang"-pattern combat knife and walked 30 miles to safety(some sources claim he tore the beast's throat out with his teeth, but that may be exaggeration). He then had his arm replaced with a bionic one and continued commanding his men. Since then he's survived multiple life-threatening injuries and his body is riddled with bionics, making him as much a machine as a man. He's as tough as a space marine and can destroy tanks with his bare hands (mechanical hands, but still, that's impressive).
And all this goes Quintuple for Space Marines, whose equivalent, Captain, is fully capable of going face-to-face with a genetically engineered, 18 foot-tall alien killing machine and winning.
And their equivalent among the Grey Knights, who are designed to eat thirty-foot suspiciously-balrog-like incarnations of the Chaos God of slaughter for lunch. Though until the Daemonhunters codex is updated, this is an example of Gameplay and Story Segregation as GK Brother-Captains technically have a Weapon Skill of 5 vs. 6 for normal SM Captains.
The Canonesses, leaders of the Sisters of Battle, arguably one-up the Space Marine captains because they do the same things — only they aren't genetically modified, hormonally modified, chemically modified, all-powerful superhumans... they're just biologically ordinary humans (one of the weakest races in the galaxy) females who happen to be that damn good essentially through skill alone.
The Tau Empire's Shas'o (lit. "Commander") are this to a man (or woman). Even ignoring the named characters such as Farsight or Shadowsun, the rank structure of the Tau requires every Fire Caste soldier begin as a lowly shas'la. After four years of service, they earn the right to take on a Trial of Fire which, depending upon their sept, may be a live-action training mission of some description, a highly realistic simulation, or a real-life mission of vital importance. If the soldier is successful, he graduates to shas'ui (sergeant). After another four years, and another, more difficult Trial of Fire, he may attain the rank of shas'vre (lieutenant); then another four years of service and another Trial to become a shas'el or "sub-commander". Shas'el may lead Hunter Cadres, but to become shas'o he has to complete the cycle one more time. Every single shas'o has at minimum 16 years experience, in everything from footslogging to battlesuit combat; he has passed four of the nastiest, most realistic tests his superiors can imagine; he is equipped with a Crisis battlesuit and the best technology the Empire can provide; and as if that isn't enough he is accompanied by a squad of handpicked shas'vre.
In BattleTech, this is one of the requirement for a Star Colonel in the Clans, in which they have plenty fighting experiences, or they take it by Trial of Position from another Star Colonel.
In the wargame The 20th Maine (later renamed ''Little Round Top'' ), each of the regimental commanders (a colonel) is represented by a named piece. While they are present primarily for command and control rules, the highest ratings go to Col. Joshua Lawrence Chamberlain of the 20th Maine (see under Real Life).
In Squad Leader and Advanced Squad Leader the leaders with the highest morale and biggest bonus to various die/dice rolls (10 -3) are given the rank of colonel. A few historical leaders who were not colonels at the time also get these ratings.
Col. Sawyer from World in Conflict. While we never see him in the field, in that scene where he is shot by sniper (who misses by a few inches), he is just too stoic. And he is fluent in French, too.
And the Expansion Pack gives us Soviet Polkovnik Orlovsky.
The lieutenant colonels in the second Wing Commander were Ace Pilots, without exceptions. Colonel Halcyon, however, acts more as a commander rather than a pilot.
There's Colonel Blair himself, from WC3 and WC4. Having killed the most "ace" Kilrathi pilots in the entire three-decade war, and defeating Prince Thrakhath, twice (the second time when Blair's fighter is weighed down with the Temblor Device, which also halved his missile loadout), is one way of earning badass points. He later gets promoted to Four-Star Badass
Col. Corazon Santiago from Sid Meier's Alpha Centauri: military genius, knows how to run a police state properly, and hot to boot.
Also leader of a faction, and if played right, the entire world. Still, never promotes herself above colonel.
In the first of the novelizations by Michael Ely (one of the writers for the game), Santiago personally leads an elite unit of her Myrmidons to storm the UN Headquarters (the Peacekeepers' home base), which has been under Spartan siege for days. Not only does this seal the base's fate, but she also ends up personally killing Pravin Lal's son in revenge for him killing her son.
Colonel from Mega Man X and his Net Navi counterpart in Mega Man Battle Network are both pretty awesome.
It's required for the former, seeing as his Rival is the resident badass...
Command & Conquer:
Colonel Burton, the U.S.' hero unit in Generals and Zero Hour kills enemy soldiers (and aircraft due to a bug in earlier versions) stealthily with his knife, sets demolition charges - he can destroy a base at once with remotes, one at a time with timed ones, and in one case, destroyed a chemical lab with one without even getting close by creating an avalanche -, is hard to see, and totes the most powerful bullet-based weapon of the game, able to kill even tanks.
Though he was only a captain during Renegade, Nick "Havoc" Parker was ultimately promoted to colonel before his retirement, in spite of his antics on the battlefield.
Col. Jade Curtiss from Tales of the Abyss. He starts out approximately 45 levels ahead of the other characters, is far stronger physically than most other black mages, and is 35 years old. In a JRPG. Still kicks the ass of a God-General after getting sealed down to the level of the rest of your party.
Even when Jade gets hit with a Fon Slot Seal, he doesn't lose any of his awesomeness. In fact, it's by traveling with the party that he can learn the Meteor Storm spell. And Indignation.
He's also the ultimate Deadpan Snarker, and almost completely unflappable, something shown time and time again as a contrast to the other characters. His badassery isn't just in fighting, it's also in the fact that he can stand in the middle of a raging volcano and appears not to sweat. Although that could just be the Convection Schmonvection in effect.
Col. Volgin of the GRU from Metal Gear Solid 3: Snake Eater. Yes, he's evil. Doesn't detract one bit from his badassitude.
Also from Metal Gear, we have Col. Roy Campbell. Even when he retires, Snake not only continues to respect him, he even refers to him as "Colonel"... over Campbell's explicit objections. And in the original Metal Gear Solid? He gave orders to Solid Snake.
The King of Fighters has Heidern. He's only appeared in four of the games, and he's a freaking god. Imagine what happens when you give a character similar to Guile absurd priority in attacks, the ability suck the life out of his opponents and give it to himself, and has no "magical" justification for his abilities. SNK developers worried people might think he was an alien or a wizard. His adopted daughter Leona takes after him quite well (and replaces him after '95). He clearly takes a few cues from the Colonel from Fist of the North Star.
Technically, Ralf (who's under his command) holds the rank of Colonel as well.
Rolento from Final Fight (who later appears in Street Fighter Alpha) is quite badass, and also inspired by the Colonel from Fist of the North Star.
Col. Hoffman from Gears of War, at least in the sequel. Prior to that, he was more of a General Ripper type, at least toward Marcus, until Marcus redeemed himself.
Colonel Loomis from the prequel game Judgment is not a great leader, opting to conduct a show trial in the middle of a pitched battle, however, when the Locust invade the courthouse, he's more than willing to head to the front himself, even fighting alongside the trial's defendants against a Giant Spider.
Maybe to appeal to players' potential Munchkin desires and knowledge of this trope - you are referred to as 'Colonel' for your military rank in EndWar.
The abilities of soldiers in the MicroProse games generally improve as they survive more missions, as does their rank depending on the number of soldiers. The rank of colonel is the second-highest in UFO Defense, next to the commander, which you can only get one of at any time. The rank is mostly tied to morale boosts (and losses if a high-ranking soldier gets killed), but to reach Colonel rank, a soldier has to have been through a truckload of missions where they've acted and most likely improved their skills.
In the reboot, XCOM: Enemy Unknown, soldiers are promoted based on their experience gain, and gain stat boosts and abilities unlocked as they do so. Colonel is the highest rank attainable, and marks an operative with the highest stats and most abilities within their class. Hell, Colonel-level abilities alone are only not game-breaking because it takes so long to get to them. And unlike in the older games, promotions aren't restricted by a hierarchy system, so all 70 spots of your barracks can be occupied by Colonels.
Subverted by Captain Price from Modern Warfare, who displays the amount of sheer badassitude commonly seen only in colonels, yet is still inexplicably a "mere" captain. Had he not been held prisoner by the Russians for five years, though, he may have well been promoted to colonel by MW 2.
Colonel Hyuga from the original Shadow Hearts is playable only briefly, but manages to annihilate a squadron of thugs due to his souped-up stats. He then proceeds to save an old man and befriend a child.
Another chance for the player to pick up this trope: if you play as the United Earth Federation officer in Supreme Commander Forged Alliance, you're referred to only as the colonel, having been promoted from Major in between the original game and the expansion. The more informal Cybrans only have two ranks that anyone ever hears about, so their player is always just commander, and the Aeon player is either Knight of the Illuminate or the Champion of the Princess, depending on where they are in the storyline. Both Cybran and Aeon players remain every bit as badass as their UEF counterpart, though.
Colonel Augustus Autumn, Fallout 3. Autumn is not a particularly strong opponent (he is only slightly more durable than the average human, and his only armor is his trench coat), and he doesn't often appear during the game. This would make you think he doesn't qualify for this trope... until you realize that the troops under his command — which form the power-armored, plasma-rifle-wielding striking arm of the Enclave — were so loyal to him that they, to a man, defected with him when he mutinied against the President. The. President. The leader of the Enclave. Either Autumn either has some very impressive leadership skills or everyone really hates bureaucrats.
Additionally, he somehow survives a dose of radiation that kills you no matter how many anti-rad meds you take.
Just before he falls on the floor, you can see him injecting something into his arm, maybe it's some kind of super-duper high-tech Enclave Rad-X?
Colonel Cassandra Moore in Fallout: New Vegas. She's the commander of the garrison at Hoover Dam, within spitting distance of a massive enemy troop buildup on the east side of the dam. General Ripper-esque, ball-busting, credentials in the form of four campaigns against the Brotherhood of Steel during the NCR's war with them.
She's also an ex-Ranger for added badass points.
Col. Randall Moore from Universe at War: Earth Assault, although he gets promoted to General by the second mission. It takes a lot of badass to be a powerful hero unit when everyone else in your species is Cannon Fodder or, even worse, resources for the alien invaders.
Killzone: Colonel Tendon Cobar and Colonel Mael Radec are with the bad guys, but they're both badass enough to show you how they've earned their ranks. Templar, while promoted to colonel in Killzone 2, doesn't quite make the cut.
Sergei Vladimir from Resident Evil: The Umbrella Chronicles, former Soviet colonel and head of Umbrella Corporation's field-oriented Red Umbrella Division. Strangely, he's somewhat Affably Evil and usually lets his bodyguards do most of the actual fighting, but even that doesn't diminish his badassery. The man pulls a gun on Wesker without flinching, transforms into a monster to try and save his employer's company, and has so much presence that his badass status is never in doubt. No one does Undying Loyalty quite like Sergei.
In Just Cause 2, you get to take down various colonels who carry good firepower and are Made of Iron enough to shrug off bullets to anywhere except the face.
Ace Combat: Assault Horizon gives us a USAF Colonel for the protagonist and a Russian Colonel for the antagonist.
Colonel Relius Clover of BlazBlue. Puppeteer extraordinary and is one of the people that is quite possibly the closest one that can be said to be Hazama's superior other than the Imperator. And he's a gigantic scum.
Chronophantasma includes Colonel Kagura Mutsuki, who's not only the leader of the highest clan amongst the NOL Duodecim... but also a surprisingly pleasant dude. To fill in the badass part, he Curb Stomps Ragna and fights Noel/Mu-12 on equal terms.
Colonel Sanger Zonvolt of Super Robot Wars fame. Pilot of a giant mech, a German samurai and all-round badass. What's not to like about this guy?
Lt. Colonel Burns of Vanquish is a massive dude with cybernetics up the wazoo touting a big ol' minigun as his main weapon.
The Big Bad of Contra Hard Corps, Colonel Bahamut is a Disgraced war hero who plans to take over the world by using Alien DNA. Perhaps the best example of the Colonel's badassness is the path where he fights the player while wearing a gigantic claw arm.
A younger version of him is the star of the prequel Hard Corps: Uprising.
Jax from the Mortal Kombat series fits this trope to a tee, despite being a Major.
In The Elder Scrolls V: Skyrim, the rank of "Legate" fills a niche that resembles Colonel (in the regard of "highest ranked line officer"). Legate Rikke fits the bill of Legate Badass, and if you side with the Empire in the civil war, you eventually are promoted to Legate as well.
Pit, as of Kid Icarus: Uprising is the Captain of the Guard for Palutena, and he leads the charge against all of her enemies (and he even points out at one point that he's an officer, not just an enlisted man). The sheer number of foes he rips through (including the top generals of the Forces of Nature and Hades, the god of the underworld himself) cements his badass credentials as well.
In Xenoblade Chronicles X, Elma was a Colonel of the Coalition forces prior to Earth's destruction. Her combat prowess is nigh-legendary, and the BLADEs that served under her back then still address her by the title out of respect.
Valkyria Chronicles III has Kurt Irving, leader of the Nameless Squadron.
He also ends up becoming the highest ranking soldier in Project X Zone where he leads the party in one stage.
The Legend of Heroes: Trails of Cold Steel III has Colonel Neithardt, who was promoted between II and III and is the The Ace of the strongest army of the empire, the Fourth Armored Division.
Colonel Moscardó, a major character in COD 2 Spanish Civil War Mod. Defiantly refuses to surrender the Alcazar despite the enemy having an overwhelming advantage and threatening to murder his son held in their captivity.
Zephyr Crow of The Wandering Ones Definitely qualifies for this trope; she is faster and stronger than the Special Forces men she trains, and before that, she and her husband single-handedly killed the leaders of over 35 Kilabyker gangs.
Air Force Blues: Col Alvis "Minnie" Gunnar and Lt. Col. Mike "Deadlock" Rowland.
Commander Badass of Manly Guys Doing Manly Things was literally created by the space future military to be this, hence the name.
Maxima of Grrl Power is a Colonel in the US Air Force, Arc-SWAT's field commander, and the strongest (heroic) superhuman in the setting.
Tech Infantry has Colonel Arthur Clarke, commanding officer of the Raptors, an elite military unit tasked with chasing down and arresting (or killing) draft-dodging Werewolves and Mages, as well as other secret operations. His successor, Colonel Andrea Treschi, is quite the badass himself.
Colonel Blitzer, Head of Warfare in Coyle Command.
Lt. J.T. Marsh in Exo Squad gets promoted to Wing Commander around the time he becomes acknowledged as the single best Ace Pilot of the Solar System. Wg.Cdr. is the air force equivalent of infantry's Lt.Col.
King of the Hill's Cotton Hill. The guy took fiddy bullets to the back while lost at sea, managed to survive an ambush by an island full of Japanese troops, then proceeded to kill all of them with a piece of one of his deceased friends even though both his shins were blown off by machine-gun fire; thus completes his famous deed of having killed fiddy men during WWII.
The time he took out a Japanese machine-gun nest by hiding in a barrel of sake, holding his breath until the guards got drunk, then leaping out and 'hibachi-ing' the entire group by blowing sake out over his zippo.
Cotton is more of a subversion. True, he lost his shins, but his story...just doesn't add up. He mentioned Fatty was killed by sharks...yet he then mentions he used Fatty to beat the life out of the Japanese. Also, he claims to have fought in both Munich and Okinawa in just mere days of each other. He's more a Small Name, Big Ego than anything.
As Hank pointed out to Peggy, even though Cotton is fond of exaggerating his accomplishments he is still the greatest war hero Arlen had ever seen. Cotton was awarded a Medal of Honor, after all, and we know he was in the 77th Division which fought in Guam, Okinawa and the Philippines, all of which have been cited by Cotton as locations he fought at.
All that needs to be said: When he came back from the war, his shins were gone. Colonel Badass through and through.
Tennessee Tuxedo and His Tales: If any of Commander McBragg's tales are true (which is highly debatable), he would definitely qualify.
Just about all of the Clone Commanders in Star Wars: The Clone Wars, but perhaps with emphasis on Captain CT-7567 'Rex', especially since he would go on to become an officer in the Phoenix Squadron Rebel Cell for The Rebel Alliance.
Colonel K, the superior to Danger Mouse, has many accomplishments to his credit (piano throwing, climbing Mt. Everest on a pogo stick, etc.), and his mission orders are ironclad. It gets subverted sometimes as he can be as addled as DM's assistant Penfold.
The little-advertised fact about Vladimir Putin is that he is a colonel of the reserve. And he's a black belt in judo.
Otto Skorzeny . Played in Team Evil, but remained badass until his death.
His badassitude is even greater in Harry Turtledove's Worldwar books, merrily going up against a superior alien foe. During one battle, when a Race landcruiser is blasting everything and everyone in the vicinity, he waltzes up to it and throws a satchel charge in-between the turret and the chassis. He also ends up liberating Mussolini right from under the lizards' snouts. And yes, the lizards speak his name as a curse.
Possibly averted and based on good publicity; most of Skorzeny's missions were failures, and the famous rescue of Mussolini's was largely someone else's plan which Skorzeny got credit for (and apparently got in the way rather than helping during the actual operation).
George Washington was a colonel during his time fighting for the British during the French and Indian War, and his badass exploits earned him enough distinction to be appointed Commander in Chief of the Continental Army during the Revolution, which in turn helped get him elected as the first President of the United States.
Washington's protege Alexander Hamilton became a colonel during the revolution. His contemporaries (including his mistress) referred to him as Colonel Hamilton throughout his civilian career. He didn't do many badass things during the war, but his life included such highlights as founding America's first bank, building the treasury department, creating the Coast Guard, and helping suppress the Whiskey Rebellion.
Lt.Col Jack Churchill . The man who fought the Nazis with a bow, arrows, and a claymore. The Germans eventually captured him by killing his entire commando squad with mortar fire; when they finally moved in, they found him sitting there, alone, playing the bagpipes. He got sent to two different concentration camps:Sachsenhausen and Dachau , and escaping both times. When he returned to Britain ready to go back to the battlefield, the war ended, and that pissed him off.
Lt.Col. Benjamin O. Davis, Jr. Son of the first Black general in the U.S. Armed Forces, he led a WWII fighter plane group known as the Tuskegee Airmen. The baddest fighter pilots in damn near any war. Oh, and he was the first Black general in the U.S. Air Force. Coming from military roots, his dad was the first US black general, period — and he started as a private.
Special Forces Lieutenant Colonel Jonathan Netanyahu of the Israeli Defense Forces, who, among other things, led the ground team during the Entebbe Airport Raid to rescue the hostages of an Air France flight. He died in the attempt (and was the only military Israeli fatality of the raid) and is a national hero in Israel.
Colonel Avi Peled commander of the Golani Infantry brigade , who after having a building collapse on top of him and 20 of his soldiers after a tank shell hit it by accident, took over the evacuation process of all of his injured soldiers despite being wounded himself, was the last one evacuated to a hospital, and then returned to the battlefield the very next day.
Colonel Claus von Stauffenberg was a decorated war hero before he masterminded the July 20, 1944 plot . Also a Handicapped Badass, though this handicap is thought to have been one factor in the plot's ultimate failure.
The famous Special Operations division Delta Force was founded by Colonel Charles Beckwith. He was badass enough to take a .50 cal round through the torso and survive with only basic medical aid (as the doctors couldn't waste time on someone who was "clearly" going to die).
Lt. Colonel James Doolittle of the famous Doolittle Raids.
Colonel Frederick Drew Gregory , USAF, retired. The first Black man to pilot the space shuttle, and the first to command a space shuttle mission. This makes him The Captain, an Ace Pilot, and a Colonel Badass, all in one.
Chesty Puller of the US Marine Corps and namesake of the Corps' bulldog mascot. He's also the most decorated Marine in the history of the Corps, with a long history that has achieved Memetic Mutation level amongst Marines.
Lieutenant Colonel Harry Smith , Australian Army. Known for his real-life Conservation of Ninjutsu at the Battle of Long Tan in the Vietnam War.
Ken Reusser , a USMC fighter pilot in three different wars (World War II, Korean War, and Vietnam War), who retired from the Marine Corps as a colonel. Probably one of the most badass feats he performed was downing a Japanese reconnaissance aircraft that was reporting on US positions for Kamikaze attacks, flying about 1000 feet above the theoretical ceiling of its pursuers. Him and his wingman both had their guns malfunction, so they used the props of their F4U Corsairs to chew up the Japanese aircraft's tail to take it down.
Lieutenant Colonel Thomas Edward Lawrence, A.K.A. Lawrence of Arabia.
Colonel David Hackworth (retired, deceased), also known as the most decorated US Army Officer of the 20th century. He was awarded 110 separate medals, of which the following were for heroism: eight Purple Hearts, two Distinguished Service Crosses, ten Silver Stars, seven Bronze Stars, the Distinguished Flying Cross, the Air Medal, and the Valorous Unit Award. He served in 12 separate wars, from the end of World War II right through to conflict in Yugoslavia. His initial request to be deployed in the Vietnam War was turned down because he had too much combat experience.
Colonel Joshua Lawrence Chamberlain commanded the 20th Maine during The American Civil War and was in charge of the Union left flank at the Battle of Gettysburg, day two. He is best known for calling "BAYONETS!" and ordering a charge downhill into the Confederate lines when his men ran out of ammo. It worked. The Union won. Later promoted to Brigadier General in recognition of his competence and badassery—which he persisted in nevertheless.
Aided by his father, a former Wisconsin governor and current Wisconsin court judge, Arthur MacArthur managed to secure himself an officer's commission in the Union Army in 1862 at the age of 17. He quickly proved he deserved it, as he went on to see action at Chickamauga, Stones River, Chattanooga, the Atlanta Campaign, and Franklin. During the Chattanooga Campaign, MacArthur inspired his regiment by seizing and planting the regimental flag on the crest of Missionary Ridge at a particularly critical moment, shouting "On Wisconsin." For these actions, he was awarded the Medal of Honor and a promotion to Colonel at the age of 19, earning himself the nickname "the boy Colonel". He went on to survive being severely wounded by a rebel officer's pistol at the battle of Franklin, and would spend a total of 47 years in the army, retiring in 1909 at the rank of Lt. General. Yet despite all this, he was largely overshadowed by his son Douglas, who commanded U. S. Army forces in the Pacific during World War 2.
Colonel John S. Mosby of the Confederate Army earned the nickname "The Gray Ghost" for his ability to conduct daring cavalry raids deep behind Union lines and then simply melt away into the civilian population. Even as the war dragged on and much of Northern Virginia fell into Union hands, that area would still be known as "Mosby's Confederacy" as he would continue to conduct operations there with near-complete impunity. Despite being seriously wounded twice, Mosby survived the war and lived until 1916 at the age of 82.
Colonel Hiram Berdan joined the Union Army in 1861. A highly skilled marksman and engineer before the war, Berdan immediately went to work recruiting the best shooters he could find for a new elite sniper unit, the 1st United States Sharpshooters (nicknamed Berdan's sharpshooters). Forced to pass rigorous marksmanship tests to gain entry, Berdan's sharpshooters were given distinctive green uniforms and equipped with high tech (for the time) rifles, such as the Colt Revolving Rifle and the Sharps rifle, often equipped with telescopic sights. Berdan's sharpshooters quickly became feared and respected as one of the most elite units in the Union army throughout the war.
Confederate Colonel John B Gordon, who joined the army with no prior military experience, earned himself a promotion to General for his actions at the 1862 Battle of Antietam, where he continued to lead his men after suffering two shots to his leg, one shot to the arm, and one in the shoulder before finally going down when a bullet passed through his cheek and exited out his jaw. Not only did Gordon survive all these wounds, but he was back in action just a few months later. Despite all these injuries and an additional head wound in 1864, Gordon managed to serve on for the entire duration of the war.
Colonel John T. Wilder. Wilder joined the Union Army as a Captain in 1861 and saw his first major action when commanding a small garrison at Munfordville, Kentucky during the Confederate invasion of the state. When faced with an attack from a much larger Confederate Army, Wilder initially rejected their surrender demands, telling Confederate General James Chalmers "I think we'll fight for a while." The following day, Wilder's garrison repulsed the Confederate attack, inflicting 283 casualties with a loss of 37. Eventually forced to surrender to overwhelming numbers, Wilder was released in a prisoner exchange and went on to form the "Lightning Brigade", an elite unit of mounted infantry armed with state of the art 7-shot Spencer repeating rifles, which would go on to achieve great success and glory during the Chickamauga campaign.
Colonels Leonard Wood and Theodore Roosevelt of the 1st U.S. Volunteer Cavalry ("Rough Riders"). Badasses both.
Colonel Rick Rescorla - Served in Africa with the British Army, joined the US army in the mid-1960s and served in Vietnam, one of the hero's of the Battle of Ia Drang. He led the evacuations of both World Trade Center attacks and was killed on 9/11 going back in with three subordinates to attempt to save the last 2 missing employees of Morgan Stanley, for whom he was the Head of Security at the World Trade Centre. He succeeded in getting the other 2700 employees safely out, singing "Men of Harlech" over a megaphone in the process.
Lt. Colonel John Frost was the British Army's go-to guy for impossible airborne missions. After proving himself by stealing a German radar station from occupied France, Frost went on to perform similarly daring missions in North Africa and Sicily. However, his Moment of Awesome came during the Battle of Arnhem as part of Operation Market Garden where he was tasked with securing the Arnhem Bridge. Although ultimately doomed by the poorly conceived operation, John Frost and a small force of some 400 lightly armed paratroopers managed to penetrate German lines and seize the north end of the bridge where he held out against a German SS Panzer division for 4 days until finally being forced to surrender due to lack of ammunition. Lt. Col. Frost was later depicted in a movie and Arnhem Bridge was renamed in his honor. He finished his military career as a Field-Marshal and commander-in-chief of British armed forces.
Lieutenant Commander Patrick Dalzel-Job . His rank equates closer to Major, but nevertheless his Naval Intelligence Commando unit (British Navy SEALs) stormed German targets four days after D-Day and disabled a German destroyer at port (with its whole crew compliment) and then captured the town of Bremen. His boss was Ian Fleming, and many consider him one of the models for James Bond.
Titus Cornelius was a former Black slave who fought in the American Revolution for the British. Although the British did not allow Blacks to be officers, let alone reach the rank of Colonel, he nevertheless was known and referred to as Colonel Tye. His guerilla-tactics were legendarily effective, even so far as helping hold off George Washington troops in their first siege of New York. Most historians agree that had he been white (and, y'know, not fought for the British) he'd have been far more famous today. There is a rumor that his name inspired a certain other Colonel Badass from Battlestar Galactica.
Lloyd L. Burke received the Medal of Honor in the Korean war for his actions at Hill 200. He was on his way home when he heard his platoon was pinned down so he went back to them. After assessing the situation he stormed a Chinese trench with a pistol and a hand grenade. After using those up, he got out and grabbed a Browning 1919, ignored the shrapnel that shredded his hand, wrapped his jacket around the hot barrel, wrapped the ammunition belt around his body, lifted the 31-pound machine gun (normally used on a tripod), and proceeded to storm the trench again. He was only a Lieutenant at the time, but he achieved the rank of Colonel before he retired, so he counts.
Many of Mustafa Kemal Ataturk's claims to badassitude occurred when he was a Colonel in the reserves, in particular defending the key pass whose loss would have probably made the Gallipoli campaign in World War I succeed. Even after the Ottoman Empire ultimately lost, he continued being a badass as a soldier and then the founder of modern Turkey, until he died.
Colonel David Randolph Scott, seventh man on the moon and the only Air Force pilot to actually pilot a moon landing (the rest of the Apollo commanders were Navy men).
Also survived the near-loss of Gemini 8 with Neil Armstrong due to steel nerves and badass piloting. On the other hand, he got in trouble for trying to profit off Apollo 15 by selling souvenirs.
Lieutenant Colonel John U. D. Page , an artillery officer who served in the Korean War and received the Medal of Honor for his actions. In just 12 days of combat service, he singlehandedly took out an enemy MG nest, commanded tanks while also acting as a tank machine gunner , performed an aerial attack with hand grenades in an unarmed observation plane, and saved an ambushed logistics regiment from their Chinese attackers. In this last action, he was eventually killed, but not before taking at least 16 enemy soldiers with him.
Colonel Buzz Aldrin (USAF), the second man to land on the Moon (but not the highest-ranked officer to ever walk on the Moon - that's Alan Shepard or Charles Duke), and effective silencer of Moon landing hoax advocates . Other Colonels who've landed on the Moon include David Scott and James Irwin - the rest are divided between Navy Captains and civilians.
Lt. Colonel Anatoly Lebed' . 29 years of service, first as a paratrooper in Afghanistan, then entered an officers school, flew a helicopter there, first as a tech, then as a pilot until his retirement. But as he couldn't imagine himself as anyone but soldier, when Chechen war started he literally reenlisted himself (flying into Chechnya on his own with his own gear), was accepted, then served again as a paratrooper officer and got a reputation as Father to His Men there. Then he got blown on a mine, had his foot amputated, but returned to duty in just one year on a prosthesis. Then, to up the ante, when his patrol got ambushed, he reportedly covered a wounded soldier with his body, suffering a shrapnel wound, proceeding to completely ignore it until his unit got back to safety. The guy's also an accomplished engineer, designing and building a lot of his unitnote The legendary 45'th ORP SpN VDV, which is as close to Russian Army admitting the unit being the part of the feared GRU Spetsnaz as it gets equipment and gear, like the combat buggies and such. Like Lawrence of Arabia, Lebed' died far from the battlefield, in a high-speed motorcycle crash in 2012.
Lt. Colonel Herbert Jones VC OBE. During the Falklands War, his battalion was stalled under heavy small-arms fire from entrenched positions and being further pinned by increasing artillery fire. Realizing that he couldn't afford to lose momentum, Lt. Colonel H. Jones charged the fortified enemy position under concentrated fire, getting knocked back once, but continuing until he died feet from the enemy. His men later charged, galvanized by his sacrifice. The enemy surrendered due to the heroics he displayed personally and simultaneously inspired in his men.
Charles Lindbergh , a colonel in the Army Air Force, the first man to fly solo across the Atlantic, and combat veteran of World War II (even though at that point he was a civilian).
Colonel Dave Belote, base commander of Nellis AFB and five-time Jeopardy! champion.
Lieutenant Colonel Danny McKnight of the 75th Ranger Regiment. He was deployed to Somalia and participated in the infamous Battle of Mogadishu, in which seventeen American Soldiers died and hundreds of Somali militia were killed. LTC McKnight was famous for not taking cover when he got shot at, figuring that if he got killed, God wanted him in heaven. Also in the movie.
Claire Lee Chennault, commander of the Flying Tigers held the rank of Colonel in the US Army Air Force, then the Chinese Air Force, and then again in the USAAF before he was promoted to Brigadier General in 1942.
Colonel Peter Julien Ortiz, USMC . Though he didn't get promoted to Colonel until after WW2, Ortiz had an enormous amount of combat experience and badassitude, too much to list here so this entry will be restricted to one notable incident. He was operating with an OSS team in France over a year before the Normandy invasion and would openly wear his USMC uniform in rural areas and in towns, a fact which cheered the French but drew the attention of the occupying German forces. One particular incident involves some German soldiers at a café in a French town boasting about what they were going to do to him when they found him. Ortiz strolled into the cafe, wearing a freaking cape, whereby he threw the cape back to reveal his US Marine uniform and a .45 pistol in each hand. He took out the enemy soldiers and was long gone by the time any reinforcements showed up.
Lt.Col. Alfred Wintle . Talked a dying soldier out of a scarlet fever-induced coma, was arrested for trying to steal a plane and signed his own arrest warrant, was captured by Vichy French whereupon he informed his captors it was his duty to escape...and did.
Subverted with Muammar Gaddafi. Though state media made him out to be this, as is common of a dictatorship, in practice he went down rather quickly once the rebels got their hands on him.
Admittedly, by then he was 69 years old. However, he didn't really do anything particularly badass even when he was younger. In fact, he lost pretty much every war he participated in...
Also, his rank was a self-granted promotion after he led a coup and installed himself as dictator...the highest rank he had legitimately held in the Libyan Army before the coup was Captain.
In an aversion, "colonel" was for a long time simply the "constitutional monarch" (so to speak) of a given regiment (originally meaning mercenary unit on permanent contract to The Government from the Latin regimentum). He, or she on rare occasions, might indeed be a badass, but their badassery and their colonelcy were unrelated. That is because, in several armies, a regiment was a ceremonial and administrative unit rather than a tactical unit, and the highest rank to go into battle specifically as a member of a given regiment was Lt Colonel. General officers in the British army often retained colonelcies for much of their career, but this was a ceremonial position. For instance, General Killalot might also be Colonel of the Duke of Earl's own fusiliers, but his only relation would be to pay for the band or the silverware or whatnot.
US Air Force Colonel John Stapp , hands down. In the interest of science, he strapped himself to a rocket sled and ended up subjecting himself to a force 46.2 times the force of gravity and lived. At the time, it was believed that 18 G was fatal. Basically, his work showed that as long as the human body is properly restrained, it can take a lot more Gs than first expected. He also felt that it was possible to go higher if the person was facing backwards (the 46.2G run was forwards). He ended up applying his rocket sled research to cars and lobbying to make seat belts mandatory in all vehicles, saving countless lives.
Colonel Colin Mitchell, who ignored direct orders and restored British prestige by recapturing the city of Aden in 1968, in the counter-insurgency war prior to Aden's independence from the British Empire. Communist rebels had previously disregarded British forces as weak and ineffectual. Most of this perception was down to flawed orders from an out-of-touch government keen to avoid conflict and appease the natives. "Mad Mitch" Mitchell's aggressive action changed all this and his show of force scared the rebels and insurgents into relative quiet.
Yugoslav partisan Colonel Sava Kovačević had a reputation for great personal courage. Highlights include single-handedly capturing an Italian tankette without using any anti-tank weapon, leading an infantry assault that destroyed another three Italian tankettes by leaping onto them and attaching explosives and making daring raids behind enemy lines. His career was cut short when he was gunned down during the partisan breakthrough from German encirclement at the Sutjeska River, June 1943.
The late, great, Marine Col. John Glenn, who was an ace pilot before becoming the first American in orbit and much later, the oldest man in space.
While commanding the 8th Fighter Wing in Vietnam Col. Robin Olds gained far more fame for his hard-charging leadership style and innovative tactics than he had as a double ace with 12 aerial victories in World War II (he shot down another 4 MiGs in Vietnam). Olds later retired as a Brigadier General, but is still best remembered for his command of the 8th Wing...and for the mustache he sported during that time.
Although Cândido Mariano da Silva Rondon ultimately reached the rank of marshal in the Brazilian Army, he carried out his most badass achievements while still a colonel. To wit:
He led the construction of over 5,000 km of telegraph lines through southern and western Brazil, including what was then The Wild West, but also many areas that no "civilized" Brazilian had been before (or at least, had not returned from).
He established peaceful relations with tribes such as the Nambikwara and Bororo (following the motto "Die if necessary, but never kill"), and remained a lifelong advocate for protecting natives' rights and cultures.
Along with Theodore Roosevelt, he led an expedition down the then-unknown "River of Doubt" (Rio da Dúvida) in 1914. Although the expedition had been poorly planned from the standpoint of equipment and logistics, Rondon's wilderness expertise—and his decision to have another officer come upriver from the Amazon as far as possible, just in case—ended up making the difference between life and death for most of the men.
Agustin de Iturbide of México. Nicknamed El Dragón de Hierro, or the Iron Dragoon, he never lost a battle in four years of the Mexican War of Independence, often defeated greater forces than his own, once repelled the entry of 500 foot soldiers with just 34 of his dragoons, and with just his dragoons by his side dealt defeat to the greatest enemy general, José María Morelos, thought to be almost invincible at the time. As a royalist dragoon, he climbed the ranks but was most attached to his rank of Colonel of the Celaya Regiment. He would go on to liberate México almost single-handedly, acquiring the ranks of General, Admiral, and Generalissimo along the way, and even being elected as Emperor, but always wore his Colonel uniform, even to his crowning.
Some civil war historians credit the early success of the South in the American Civil War to superior leadership. In support of this trope, the North managed to collect every single pre-war general bar one (Joseph E. Johnston), leaving the South almost no one higher than Colonel including their eventual supreme commander, Robert E. Lee.
Man of Steel - Colonel Badass
Colonel Hardy. When he comes up against Faora, he starts in a helicopter. She crashes it. He then crawls out of the wreckage and empties two guns into her (after watching her effortlessly annihilate his men). When he clicks dry without their having the slightest effect, he pulls out a knife. She's impressed enough to let him get into a fighting stance, pass on some Kryptonian wisdom, and draw her own knife instead of just walking all over him.
Alternative Title(s): Commander Badass
Celestial Paragons and Archangels
Badass in Charge
Commanding Coolness
SandBox/GUAE Red Zeon Mobile Assault Force
The Brigadier
Meaningful Titles
Authority Tropes
The Chessmaster
Badass Tropes
Coin Walk Flexing
Clock Discrepancy
Older Than Radio
Color Me Black
Cold Sniper
Military and Warfare Tropes
Colonel Kilgore
Bright Is Not Good
Combat Clairvoyance
VideoSource/Live-Action Films
Darker and Edgier
The Neidermeyer
Big Little Brother
ImageSource/Anime & Manga
4.8 (5 votes)
Main / ColonelBadass | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,412 |
\section{Introduction}
Parton Density Function (PDF) determinations are usually global
fits~\cite{mrst,cteq,zeus-s}, which use fixed target
DIS data as well as HERA data. In such analyses the high statistics HERA NC
$e^+p$ data, which span the range $6.3 \times 10^{-5} < x < 0.65,
2.7 < Q^2 < 30,000$GeV$^2$,
have determined the low-$x$ sea and
gluon distributions, whereas the fixed target data have determined
the valence distributions and the higher-$x$ sea distributions.
The $\nu$-Fe fixed target data have been the most important input
for determining the valence distributions, but these data suffer
from uncertainties due to heavy target corrections. Such uncertainties
are also present for deuterium fixed target data,
which have been used to determine the shape of the high-$x$ $d$-valence quark.
HERA data on neutral and charged current (NC and CC)
$e^+p$ and $e^-p$ inclusive double differential
cross-sections are now available,
and have been used by both the H1 and ZEUS collaborations~\cite{zeusj,*h1}
in order to determine the parton distributions functions (PDFs) using data
from within a single experiment. The HERA high $Q^2$ cross-section
data can be used to determine the valence
distributions, thus eliminating uncertainties from heavy target corrections.
The PDFs are presented with full accounting for uncertainties from correlated
systematic errors (as well as from statistical and uncorrelated sources).
Peforming an analysis within a single experiment has considerable advantages
in this respect, since the global fits
have found significant tensions between
different data sets, which make a rigorous statistical treatment of
uncertainties difficult.
Fig.~\ref{fig:h1zeus} compares the results of the H1 and ZEUS analyses.
Whereas the extracted PDFs are broadly compatible within errors, there is a
noticeable difference in the shape of the gluon PDFs.
\begin{figure}[tbp]
\centerline{
\epsfig{figure=Summary_H1+ZEUS-JETS.CORRELATE_2.eps,height=6cm}
\epsfig{figure=GLUON_H1+ZEUS-JETS.CORRELATE.eps,height=6cm}}
\caption {Left plot: Comparison of PDFs from ZEUS and H1 analyses at $Q^2=10$GeV$^2$.
Right plot: Comparison of gluon from ZEUS and H1 analyses, at various $Q^2$.
Note that the ZEUS analysis total uncertainty includes both experimental and
model uncertainties.}
\label{fig:h1zeus}
\end{figure}
Full details of the analyses are given in the relevant publications,
in this contribution we examine the differences in the two analyses, recapping
only salient details.
The kinematics
of lepton hadron scattering is described in terms of the variables $Q^2$, the
invariant mass of the exchanged vector boson, Bjorken $x$, the fraction
of the momentum of the incoming nucleon taken by the struck quark (in the
quark-parton model), and $y$ which measures the energy transfer between the
lepton and hadron systems.
The differential cross-section for the NC process is given in terms of the
structure functions by
\[
\frac {d^2\sigma(e^{\pm}p) } {dxdQ^2} = \frac {2\pi\alpha^2} {Q^4 x}
\left[Y_+\,F_2(x,Q^2) - y^2 \,F_L(x,Q^2)
\mp Y_-\, xF_3(x,Q^2) \right],
\]
where $\displaystyle Y_\pm=1\pm(1-y)^2$.
The structure functions $F_2$ and $xF_3$ are
directly related to quark distributions, and their
$Q^2$ dependence, or scaling violation,
is predicted by pQCD. At $Q^2 \leq 1000$~GeV$^2$ $F_2$ dominates the
charged lepton-hadron cross-section and for $x \leq 10^{-2}$, $F_2$ itself
is sea quark dominated but its $Q^2$ evolution is controlled by
the gluon contribution, such that HERA data provide
crucial information on low-$x$ sea-quark and gluon distributions.
At high $Q^2$, the structure function $xF_3$ becomes increasingly important,
and
gives information on valence quark distributions. The CC interactions
enable us to separate the flavour of the valence distributions
at high-$x$, since their (LO) cross-sections are given by,
\[
\frac {d^2\sigma(e^+ p) } {dxdQ^2} = \frac {G_F^2 M_W^4} {(Q^2 +M_W^2)^2 2\pi x}
x\left[(\bar{u}+\bar{c}) + (1 - y)^2 (d + s) \right],
\]
\[
\frac {d^2\sigma(e^- p) } {dxdQ^2} = \frac {G_F^2 M_W^4} {(Q^2 +M_W^2)^2 2\pi x}
x\left[(u + c) + (1 - y)^2 (\bar{d} + \bar{s}) \right].
\]
For both HERA analyses the QCD predictions for the structure functions
are obtained by solving the DGLAP evolution equations~\cite{ap,*gl,*l,*d}
at NLO in the \mbox{$\overline{\rm{MS}}$}\ scheme with the
renormalisation and factorization scales chosen to be $Q^2$.
These equations yield the PDFs
at all values of $Q^2$ provided they
are input as functions of $x$ at some input scale $Q^2_0$.
The resulting PDFs are then convoluted with coefficient functions, to give the
structure functions which enter into the expressions for the cross-sections.
For a full explanation of the relationships between DIS cross-sections,
structure functions, PDFs and the QCD improved parton model see
ref.~\cite{dcs}.
The HERA data are all in a kinematic region where there is no
sensitivity to target mass and higher
twist contributions but a minimum $Q^2$ cut must be imposed
to remain in the kinematic region where
perturbative QCD should be applicable. For ZEUS this is $Q^2 > 2.5$~GeV$^2$,
and for H1 it is $Q^2 > 3.5$~GeV$^2$. Both collaborations have included the
sensitivity to this cut as part of their model errors.
In the ZEUS analysis, the PDFs for $u$ valence, $xu_v(x)$, $d$ valence, $xd_v(x)$,
total sea, $xS(x)$, the
gluon, $xg(x)$, and the difference between the $d$ and $u$
contributions to the sea, $x(\bar{d}-\bar{u})$, are each parametrized
by the form
\begin{equation}
p_1 x^{p_2} (1-x)^{p_3} P(x),
\label{eqn:pdf}
\end{equation}
where $P(x) = 1 +p_4 x$, at $Q^2_0 = 7$GeV$^2$. The total sea
$xS=2x(\bar{u} +\bar{d} +\bar{s}+ \bar{c} +\bar{b})$, where
$\bar{q}=q_{sea}$ for each flavour, $u=u_v+u_{sea}, d=d_v+d_{sea}$ and
$q=q_{sea}$ for all other flavours.
The flavour structure of the light quark sea
allows for the violation of the Gottfried sum rule. However, there is no
information on the shape of the $\bar{d}-\bar{u}$ distribution in a fit
to HERA data alone and so this distribution has its shape fixed consistent
with the Drell-Yan data and its normalisation consistent
with the size of the Gottfried sum-rule violation.
A suppression of the strange sea with respect to the non-strange sea
of a factor of 2 at $Q^2_0$, is also imposed
consistent with neutrino induced dimuon data from CCFR.
Parameters are further restricted as follows.
The normalisation parameters, $p_1$, for the $d$ and $u$ valence and for the
gluon are constrained to impose the number sum-rules and momentum sum-rule.
The $p_2$ parameter which constrains the low-$x$ behaviour of the $u$ and $d$
valence distributions is set equal,
since there is no information to constrain any difference.
When fitting to HERA data alone it is also necessary to constrain
the high-$x$ sea and gluon shapes, because HERA-I data do not have high
statistics at large-$x$, in the region where these distributions are small.
The sea shape has been restricted by setting $p_4=0$ for the sea,
but the gluon shape is constrained by including data on jet production in the
PDF fit. Finally the ZEUS analysis has 11 free PDF parameters.
ZEUS have included reasonable variations of
these assumptions about the input parametrization
in their analysis of model uncertainties.
The strong coupling constant was fixed to $\alpha_s(M_Z^2) = 0.118$~\cite{lepalf}.
Full account has been taken of correlated experimental
systematic errors by the Offset Method,
as described in ref~\cite{zeus-s,durham}.
For the H1 analysis, the value of $Q^2_0 = 4$GeV$^2$, and
the choice of quark distributions which are
parametrized is different. The quarks are considered as $u$-type and $d$-type
with different parametrizations for, $xU= x(u_v+u_{sea} + c)$,
$xD= x(d_v +d_{sea} + s)$, $x\bar{U}=x(\bar{u}+\bar{c})$ and
$x\bar{D}=x(\bar{d}+\bar{s})$, with $q_{sea}=\bar{q}$, as
usual, and the the form of the quark and gluon parametrizations
given by Eq.~\ref{eqn:pdf}. For $x\bar{D}$ and $x\bar{U}$ the polynomial,
$P(x)=1.0$,
for the gluon and $xD$, $P(x)= (1+p_4 x)$, and for $xU$,
$P(x)= (1 +p_4 x +p_5 x^3)$. The parametrization is then further restricted
as follows.
Since the valence distributions must vanish as $x \to 0$,
the low-$x$ parameters, $p_1$
and $p_2$ are set equal for $xU$ and $x\bar{U}$, and for $xD$ and
$x\bar{D}$. Since there is no information on the flavour structure of the sea
it is
also necessary to set $p_2$ equal for $x\bar{U}$ and $x\bar{D}$.
The normalisation, $p_1$, of the gluon is determined from the momentum
sum-rule and the $p_4$ parameters for $xU$ and $xD$ are determined from the
valence number sum-rules.
Assuming that the strange and charm quark distributions can be expressed as
$x$ independent fractions, $f_s$ and $f_c$, of the $d$ and $u$ type sea,
gives the further constraint $p_1(\bar{U})=p_1(\bar{D}) (1-f_s)/(1-f_c)$.
Finally there are 10 free parameters. H1 have also included reasonable
variations of
these assumptions in their analysis of model uncertainties.
The strong coupling constant was fixed to $\alpha_s(M_Z^2) = 0.1185$ and this is
sufficiently similar to the ZEUS choice that we can rule it out as a cause of
any significant difference.
Full account has been taken of correlated experimental
systematic errors by the Hessian Method, see ref.~\cite{durham}.
For the ZEUS analysis, the heavy quark production scheme used is the
general mass variable flavour number scheme of Roberts and Thorne~\cite{hq}.
For the H1 analysis, the zero mass variable flavour number scheme is used.
It is well known that these choices have a small effect on the steepness of
the gluon at very small-$x$, such that the zero-mass choice produces a
slightly less steep gluon. However, there is no effect on the more striking
differences in the gluon shapes at larger $x$.
There are two differences in
the analyses which are worth further investigation.
The different choices for the form of the PDF parametrization at
$Q^2_0$ and the
different treatment of the correlated experimental uncertainties.
So far we have compared the results of putting two different data sets into
two different analyses. Because there are many differences in the assumptions
going into these analyses
it is instructive to consider:(i) putting both data sets through the same analysis and (ii) putting one of the data sets through both analyses.
For these comparisons, the ZEUS analysis does NOT include the jet data,
so that the data sets are more directly comparable, involving just the
inclusive double differential cross-section data.
Fig.~\ref{zazdzahd} compares the sea and gluon PDFs,
at $Q^2=10$GeV$^2$, extracted from H1 data using the H1 PDF analysis
with those extracted from H1 data using the ZEUS PDF analysis.
These alternative analyses of the same data
set give results which are compatible within the model dependence error
bands. Fig.~\ref{zazdzahd} also compares the sea and gluon PDFs extracted from
ZEUS data using the ZEUS analysis with those extracted from H1 data using
the ZEUS analysis. From
this comparison we can see that the different data sets lead to somewhat
different gluon shapes even when put through exactly the same analysis.
Hence the most of the difference in shape of the ZEUS and H1 PDF analyses can
be traced back to a difference at the level of the data sets.
\begin{figure}[tbp]
\vspace{-2.0cm}
\centerline{
\epsfig{figure=SEAGLU_H1Data_H1PDF2000ERRORS.eps,width=0.33\textwidth}
\epsfig{figure=SEAGLU_ZEUSData_ZEUSANAL.eps,width=0.33\textwidth}
\epsfig{figure=SEAGLU_H1Data_ZEUSANAL.eps,width=0.33\textwidth}
}
\caption {Sea and gluon distributions at $Q^2=10$GeV$^2$ extracted from
different data sets and different analyses.
Left plot: H1 data put through both ZEUS and H1 analyses.
Middle plot: ZEUS data put through ZEUS analysis. Right plot: H1 data put
through ZEUS analysis.
}
\label{zazdzahd}
\end{figure}
Before going further it is useful to discuss the treatment of correlated
systematic errors in the ZEUS and H1 analyses. A full discussion of the
treatment of correlated systematic errors in PDF
analyses is given in ref~\cite{dcs}, only salient details are recapped here.
Traditionally, experimental collaborations have evaluated an overall systematic
uncertainty on each data point and these have been treated as uncorrelated,
such that they are simply added to the statistical uncertainties in quadrature
when evaluating $\chi^2$. However, modern deep inelastic scattering experiments
have very small statistical uncertainties, so that the contribution of
systematic uncertainties becomes dominant and consideration of
point to point correlations between systematic uncertainties is essential.
For both ZEUS and H1 analyses
the formulation of the $\chi^2$ including correlated systematic uncertainties
is constructed as follows. The correlated uncertainties
are included in the theoretical prediction, $F_i(p,s)$, such that
\[
F_i(p,s) = F_i^{\rm NLOQCD}(p) +
\sum_{\lambda} s_{\lambda} \Delta^{\rm sys}_{i\lambda}
\]
where, $F_i^{\rm NLOQCD}(p)$, represents the prediction
from NLO QCD in terms of the theoretical parameters $p$,
and the parameters $s_\lambda$ represent independent variables
for each source of
systematic uncertainty. They have zero mean and unit variance by construction.
The symbol
$\Delta^{\rm sys}_{i\lambda}$ represents the one standard deviation correlated
systematic error on data point $i$ due to correlated error
source $\lambda$.
The $\chi^2$ is then formulated as
\begin{equation}
\chi^2 = \sum_i \frac{\left[ F_i(p,s)-F_i(\rm meas) \right]^2}{\sigma_i^2} + \sum_\lambda s^2_\lambda
\label{eq:chi2}
\end{equation}
where, $F_i(\rm meas)$, represents a measured data point and the symbol
$\sigma_i$ represents the one standard deviation uncorrelated
error on data point $i$, from both statistical and systematic sources.
The experiments use this $\chi^2$ in different ways. ZEUS uses the Offset
method and H1 uses the Hessian method.
Traditionally, experimentalists have used `Offset' methods to account for
correlated systematic errors. The $\chi^2$ is formluated without any terms
due to correlated systematic errors ($s_\lambda=0$ in Eq.~\ref{eq:chi2}) for
evaluation of the central values of the fit parameters.
However, the data points are then offset to account for each
source of systematic error in turn
(i.e. set $s_\lambda = + 1$ and then $s_\lambda = -1$ for each source
$\lambda$)
and a new fit is performed for each of these
variations. The resulting deviations of the theoretical parameters
from their central values are added in
quadrature. (Positive and negative deviations are added
in quadrature separately.) This method does not assume that the systematic
uncertainties are Gaussian distributed.
An equivalent (and much more efficient) procedure to perform the
Offset method has been given by
Pascaud and Zomer~\cite{pz}, and this is what is actually used.
The Offset method is a conservative method of error estimation
as compared to the Hessian method.
It gives fitted theoretical predictions which are as close as
possible to the central values of the published data. It does not use the full
statistical power of the fit to improve the estimates of $s_\lambda$,
since it choses to mistrust the systematic error estimates,
but it is correspondingly more robust.
The Hessian method is an alternative procedure in which the systematic
uncertainty parameters $s_\lambda$ are allowed to vary in the main fit
when determining the values of the theoretical parameters.
Effectively, the theoretical prediction is not fitted
to the central values of the published experimental data, but
these data points are allowed to move
collectively, according to their correlated systematic uncertainties.
The theoretical prediction determines the
optimal settings for correlated systematic shifts of experimental data points
such that the most consistent fit to all data sets is obtained. Thus,
in a global fit, systematic shifts in
one experiment are correlated to those in another experiment by the fit.
In essence one is allowing the theory to calibrate the detectors. This requires
great confidence in the theory, but more significantly, it requires confidence
in the many model choices which go into setting the boundary conditions for
the theory (such as the parametrization at $Q^2_0$).
The ZEUS analysis can be performed using the Hessian method as well as the
Offset method and
Fig.~\ref{fig:offhess} compares the PDFs, and their uncertainties,
extracted from ZEUS data using these two methods.
\begin{figure}[tbp]
\vspace{-2.0cm}
\centerline{
\epsfig{figure=Summary_ZEUS-JETS_Offset+Hessian.ps,height=7cm}
}
\caption {PDFs at $Q^2=10$GeV$^2$, for the ZEUS analysis of ZEUS data
performed by the Offset and the Hessian methods.
}
\label{fig:offhess}
\end{figure}
The central values of the different methods are in good agreement but
the use of the Hessian method results in smaller uncertainties, for a
the standard set of model assumptions, since the input data can be shifted
within their correlated systematic uncertainties to suit the theory better.
However, model uncertainties are more significant for the Hessian method than
for the Offset method. The experimental
uncertainty band for any one set of model choices
is set by the usual $\chi^2$ tolerance, $\Delta \chi^2=1$, but the
acceptability of a different set of choices is judged by the hypothesis
testing criterion, such that
the $\chi^2$ should be approximately in the range $N \pm \surd (2N)$,
where $N$ is the number of degrees of freedom. The PDF
parameters obtained for the different model choices can differ
by much more than their experimental uncertainties, because each model choice
can result in somewhat
different values of the systematic uncertainty parameters, $s_\lambda$, and
thus a different estimate of the shifted positions of the data points. This
results in a larger spread of model uncertainty than in the Offset method,
for which the data points cannot move. Fig~\ref{fig:h1zeus} illustrates the
comparability of the ZEUS (Offset) total uncertainty
estimate to the H1 (Hessian) experimental
plus model uncertainty estimate.
Another issue which arises in relation to the Hessian method is that
the data points should not be shifted far outside their
one standard deviation systematic uncertainties. This can indicate
inconsistencies between data sets, or parts of data sets, with respect to
the rest of the data. The CTEQ collaboration have considered data
inconsistencies in their most recent global fit~\cite{cteq}.
They use the Hessian method but they increase the resulting uncertainty
estimates, by
increasing the $\chi^2$ tolerance to $\Delta \chi^2 = 100$, to allow for both
model uncertainties and data inconsistencies.
In setting this tolerance they have considered the distances
from the $\chi^2$-minima of individual data sets
to the global minimum for all data sets. These distances
by far exceed the range allowed by the $\Delta \chi^2 =1$ criterion.
Strictly speaking such variations can indicate that data sets are inconsistent
but the CTEQ collaboration take the view
that all of the current
world data sets must be considered acceptable and compatible at some level,
even if strict statistical criteria are not met, since the conditions
for the application of strict criteria, namely Gaussian error distributions,
are also not met. It is not possible to simply drop ``inconsistent'' data
sets, as then the partons in some regions would lose important constraints.
On the other hand the level of ``inconsistency'' should
be reflected in the uncertainties of the PDFs.
This is achieved by raising the $\chi^2$ tolerance. This
results in uncertainty estimates which are comparable to
those achieved by using the Offset method~\cite{durham}.
Using data from a single experiment avoids questions of data consistency,
but to
get the most information from HERA
it is necessary to put ZEUS and H1 data sets into the same
analysis together, and then questions of consistency arise.
Fig~\ref{fig:zh1tog} compares the sea and gluon PDFs and the
$u$ and $d$ valence PDFs extracted from the ZEUS PDF
analysis of ZEUS data alone,
to those extracted from the ZEUS PDF analysis of both H1 and ZEUS data.
\begin{figure}[tbp]
\centerline{
\epsfig{figure=SEAGLU_ZEUS+H1Data_ZEUSANAL.eps,height=5cm}
\epsfig{figure=SEAGLU_ZEUSData_ZEUSANAL.eps,height=5cm}}
\centerline{
\epsfig{figure=UD_ZEUS+H1Data_ZEUSANAL.eps,height=5cm}
\epsfig{figure=UD_ZEUSData_ZEUSANAL.eps,height=5cm}
}
\caption {Top plots: Sea and gluon distributions at $Q^2=10$GeV$^2$ extracted
from H1 and ZEUS data using the ZEUS analysis (left) compared to those
extracted from ZEUS data alone using the ZEUS analysis (right).
Bottom Plots: Valence distributions at $Q^2=10$GeV$^2$, extracted from
H1 and ZEUS data using the ZEUS analysis (left) compared to those extracted
from ZEUS data alone using the ZEUS analysis (right).
}
\label{fig:zh1tog}
\end{figure}
It is noticeable that, for the low-$x$ sea and gluon PDFs, combining the
data sets does not bring a reduction in uncertainty
equivalent to doubling the statistics. This is because the data which
determine these PDFs are systematics limited. In fact there is some degree of
tension between the ZEUS and the H1 data sets, such that the $\chi^2$ per
degree of freedom rises for both data sets when they are fitted together.
The Offset method of treating the systematic errors
reflects this tension such that the overall uncertainty is not much improved
when H1 data are added to ZEUS data. However, the uncertainty on the high-$x$
valence distributions is reduced by the input of
H1 data, since the data are still statistics limited at high $x$.
Thus there could be an advantage in combining ZEUS and H1
data in a PDF fit if the tension between the data
sets could be resolved. It is in this context the question of combining these
data into a single data set arises. The procedure for combination is detailed
in the contribution of S. Glazov to these proceedings.
Essentially, since ZEUS and H1 are measuring the same physics in the same
kinematic region, one can try to combine them using a 'theory-free'
Hessian fit in which the only assumption is that there is a true
value of the cross-section, for each process, at each $x,Q^2$ point.
The systematic uncertainty parameters, $s_\lambda$, of each experiment
are fitted to determine the best fit to this assumption.
Thus each experiment is calibrated to the other. This works well because the
sources of systematic uncertainty in each experiment are rather different.
Once the procedure has been performed the resulting systematic uncertainties
on each of the combined data points are significantly smaller than the
statistical errors. Thus one can legitimately make a fit to the combined data
set in which these statistical and systematic uncertainties are simply
combined in quadrature. The result of making such a fit, using the ZEUS
analysis, is shown in Fig.~\ref{fig:glazov}.
\begin{figure}[tbp]
\centerline{
\epsfig{figure=SEAGLU_GLAZOV.eps,height=5cm}
\epsfig{figure=UDLOG_GLAZOV.eps,height=5cm}}
\caption {Left plot: Sea and gluon distributions at $Q^2=10$GeV$^2$, extracted
from the combined H1 and ZEUS data set using the ZEUS analysis.
Right plot: Valence distributions at $Q^2=10$GeV$^2$, extracted from
the combined H1 and ZEUS data set using the ZEUS analysis.
}
\label{fig:glazov}
\end{figure}
The central values of the ZEUS and H1 published
analyses are also shown for comparison.
Looking back to Fig.~\ref{fig:zh1tog} one can see that there has been a
dramatic reduction in the level of uncertainty
compared to the ZEUS Offset method fit to the separate ZEUS and H1 data sets.
This result is very promising. A preliminary study of model dependence, varying
the form of the polynomial, $P(x)$, used in the PDF paremtrizations at $Q^2_0$,
also indicates that model dependence is relatively small.
The tension between ZEUS and H1 data could have been resolved by
putting them both into a PDF fit using the Hessian method to shift the data
points. That is, rather than calibrating the two experiments to each other in
the 'theory-free' fit, we could have used the theory of
pQCD to calibrate each experiment. Fig.~\ref{fig:zh1hess} shows the PDFs
extracted when the ZEUS and H1 data sets are put through the ZEUS PDF analysis
procedure using the Hessian method.
\begin{figure}[tbp]
\centerline{
\epsfig{figure=SEAGLU_PDF.10.ZO_ZEUS-H1_HESSIAN.eps,height=5cm}
\epsfig{figure=UVDV_PDF.10.ZO_ZEUS-H1_HESSIAN.eps,height=5cm}}
\caption {Left plot: Sea and gluon distributions at $Q^2=10$GeV$^2$, extracted
from the H1 and ZEUS data sets using the ZEUS analysis done by Hessian
method.
Right plot: Valence distributions at $Q^2=10$GeV$^2$, extracted
from the H1 and ZEUS data sets using the ZEUS analysis done by Hessian
method.
}
\label{fig:zh1hess}
\end{figure}
The uncertainties on the resulting PDFs are comparable to those found for the
fit to the combined data set, see Fig.~\ref{fig:glazov}.
However, the central values of the resulting
PDFs are rather different- particularly for the less well known gluon and
$d$ valence PDFs. For both of the fits shown in
Figs.~\ref{fig:glazov},~\ref{fig:zh1hess} the values of the systematic error
parameters, $s_\lambda$, for each experiment have been allowed to float so
that the data points are shifted to give a better fit to our assumptions, but
the values of the systematic error parameters chosen
by the 'theory-free' fit and by the PDF fit are rather different. A
representaive sample of these values is given in
Table~\ref{tab:sl}. These discrepancies might be somewhat alleviated by a full
consideration of model errors in the PDF fit, or of appropriate $\chi^2$
tolerance when combining the ZEUS and H1
experiments in a PDF fit, but these differences should make us wary about the
uncritical use of the Hessian method.
\begin{table}[t]
\begin{tabular}{ccc}\\
\hline
Syatematic uncertainty $s_\lambda$ & in PDF fit& in Theory-free fit \\
\hline
ZEUS electron efficiency & 1.68 & 0.31 \\
ZEUS electron angle & -1.26 & -0.11 \\
ZEUS electron energy scale & -1.04 & 0.97 \\
ZEUS hadron calorimeter energy scale & 1.05 & -0.58 \\
H1 electron energy scale & -0.51 & 0.61 \\
H1 hadron energy scale & -0.26 & -0.98 \\
H1 calorimeter noise & 1.00 & -0.63 \\
H1 photoproduction background & -0.36 & 0.97 \\
\hline\\
\end{tabular}
\caption{Systematic shifts for ZEUS and H1 data as determine by a joint pQCD
PDF fit, and as determined by the theory-free data combination fit}
\label{tab:sl}
\end{table}
\bibliographystyle{heralhc}
{\raggedright
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,931 |
← Catherine Murphy Case Exposes Cowardice Of Irish Media
Denis O'Brien & Catherine Murphy: A Sad, Shameful Little Chapter In The History Of Irish Journalism →
A Tale Of Two Media
Imagine these two contrasting scenarios.
In one case, in country A, a newspaper gets hold of secret documents which show that successive governments have been lying about a matter of supreme national importance. Without hesitation and as soon as humanly possible, the newspaper starts printing stories about the documents.
The government goes beserk and its Attorney-General seeks and secures a court injunction banning that newspaper from further publication. The newspaper's rivals get hold of the same documents and begin publishing as well. As this happens, and as details of the secrets circulate, public opinion veers increasingly to the side of the media.
The government seeks another injunction which is refused by the court, the judge declaring:
A cantankerous press, an obstinate press, a ubiquitous press must be suffered by those in authority in order to preserve the even greater values of freedom of expression and the right of the people to know.
The government appeals the rejection and takes its case to the highest court in the land which also rejects the government's case. One of its judges says:
Only a free and unrestrained press can effectively expose deception in government. And paramount among the responsibilities of a free press is the duty to prevent any part of the government from deceiving the people…….
The press in country A thereby wins an historic case and does so by publishing its story without seeking the government's view on the matter, thus first asserting the right to free expression, and then allowing the courts to decide the matter. The episode is regarded as a high watermark in the history of country A's free and forthright media.
In the second case, in country B, a billionaire businessman, seeks and secures a court injunction preventing a television station from broadcasting a documentary alleging shenanigans in his financial affairs and improper relations with government-regulated bodies. The injunction applies to other media.
Not long afterwards, a member of country B's parliament stands up in that chamber and reveals what are widely believed to be the salient points in the injuncted documentary. The parliamentarian does this under the cover of parliamentary privilege, which normally means that the media are free to repeat the words said in parliament, as long as they stick faithfully to those words.
This is because parliaments in most countries are judged to be legally superior to courts, not least because they make the laws that the courts enforce.
Even though the media in country B have published and broadcast words spoken under parliamentary privilege in the past, they unanimously decide that because the businessman is unbelievably rich and could sue them out of business and might even do damage to their advertising revenue, that the court injunction also applies to words spoken under the umbrella of parliamentary privilege. Therefore they decide not to print or broadcast a single word spoken by that parliamentarian.
In doing so they have accorded the billionaire businessman superior legal and political rights to the parliament in country B.
The inaction of country B's media attracts derisive headlines in other countries like this: "Country B's Media, Fearing Lawsuits, Steers Clear Of A Billionaire". The barb is all the more painful because the story appeared in the paper featured in the story about country A above.
Internet outlets and blogs have no inhibitions and freely report the words of the parliamentarian, thereby not only making the mainstream media look risible but underlining one of the reasons why the media in country B are ailing and losing readers and viewers to social media.
Eventually, after several days of looking weak, cowardly and ridiculous, some of the media outlets in country B decide that they will challenge the scope of the billionaire's injunction in the courts.
This is very strange because the only people, apart form the billionaire's attorneys, who have said the injunction covers parliamentary privilege are the media in country B!
Not only that but it will be next Tuesday before the case is heard and possibly many days after that before a judgement is announced. By that time, thanks to social media, many of the people in country B will have read the parliamentarian's words anyway.
This is not a story out of Grimms Fairy Tales but real life events.
If you can guess what the events in country A and country B really were, send your answers on a postcard to: The Director-General, RTE, Dublin, Ireland or to: Kevin O'Sullivan, Editor, Irish Times, Dublin, Ireland.
Alternatively just post a comment here.
3 responses to "A Tale Of Two Media"
edmundelynch | May 30, 2015 at 10:01 pm | Reply
Country A: USA
Case : Pentagon Papers
Country B : Republic of Ireland
Case : Denis O'Brien
The Broken Elbow | May 30, 2015 at 10:03 pm | Reply
Conan Drumm (@Drummconan) | May 31, 2015 at 5:53 am | Reply
It's a bad dose of establishmentitis, a kind of groupthink rash that reaches epidemic proportions when corporate / politico / media / legal types all breathe the same rarified air in a hermetically sealed control room where they work society's levers in their own, collective interest.
That's why, in the long ago, concepts like 'separation of powers' and 'fourth estate' were dreamed up, although they aren't getting much traction these days in the Denis O'Brien owned or injuncted media.
Thankfully, for the moment at least, Catherine Murphy's Dáil speech is available to the citizenry, unmediated by 'reportage' or 'opinion', on the Oireachtas website, on institution referred to on RTÉ radio by James Morrissey, a spokesman for Billionaire O'Brien, as a 'talking shop'. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,271 |
( au Japon) est un jeu d'action-plates-formes. Il s'agit du premier jeu vidéo de la série à être sorti sur la . Le jeu a été édité par Capcom. Le jeu est réédité sur la console virtuelle de la Nintendo 3DS à partir de 2011.
Histoire
Non content de s'être fait battre dans le premier volet de la série , le professeur Wily répare huit anciens robots et en construit un nouveau pour se venger de . Le robot bleu est envoyé à la poursuite du professeur Wily dans sa base souterraine et tombe sur Enker, robot spécialement conçu pour combattre avec son .
Système de jeu
Le est le même que celui du tout premier .
Références
Jeu de plates-formes
Jeu d'action
Jeu vidéo à défilement horizontal
Jeu vidéo sorti en 1990
Jeu Game Boy
Jeu sur la console virtuelle de la Nintendo 3DS
Jeu dérivé de Mega Man
Jeu vidéo développé au Japon | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,008 |
Nagarjuna (ca. 150-250) (de vegades transliterat com Nâgârjuna o Nāgārjuna) va ser un filòsof budista que va desenvolupar la seva activitat intel·lectual al sud de l'Índia, entre el final del i la primera meitat del . La seva contribució principal va ser la seva orientació vital al budisme Mahayana.
Nom sànscrit
Nāgārjuna, en el sistema de l'alfabet internacional de transliteració del sànscrit.
नागार्जुन, en escriptura devanagari del sànscrit.
Pronunciació:
/nagárshuna/ en sànscrit, o bé
/nagarshún/ en diferents idiomes moderns de l'Índia (com el bengalí, l'hindi, el marathi o el pali).
Etimologia:
nagá: raça que vivia als Himalaies; segons els budistes eren humans normals; segons els hinduistes, tenien cos de serp; Aryuna va rebre l'epítet "Naga" per les seves amistats amb aquesta ètnia;
arjuna: blanc, del color de la llet, clar, del color del dia, platejat, del color del llamp, del color de la plata; nom d'un tipus d'arbre (el Terminalia arjuna): la mare de Nagarjuna li va posar aquest nom perquè el part va succeir sota d'un arbre arshuna.
Altres escriptures
En altres idiomes el seu nom s'escriu de la següent manera:
Klu Sgrub (en tibetà).
నాగార్జునా (en Telugu)
龍樹 (en xinès)
龍樹菩薩 (en mandarí)
Biografia
De fet, hi ha ben poques fonts fiables sobre la vida de Nāgārjuna, ja que els relats foren escrits en xinès i tibetà alguns segles després de la seva mort. Segons algunes fonts, Nāgārjuna era originari del sud de l'Índia. Alguns estudiosos creuen que Nāgārjuna era conseller d'un rei de la dinastia Satavahana. Les excavacions arqueològiques a Amarāvatī indiquen que si això és cert, el rei podria haver estat Yajña Śrī Śātakarṇi, qui va governar entre els anys 167 i 196 AD. Arran d'aquesta associació, hom situa la vida de Nāgārjuna entre els anys 150 i 250 AD.
Segons una biografia dels segles IV/V traduïda per Kumaradjiva, Nāgārjuna va néixer a una família de bramans a la regió de Vidarbha i posteriorment va abraçar la religió budista.
Algunes fonts afirmen que, en els seus últims anys, Nāgārjuna visqué a la muntanya de Śrīparvata prop de la ciutat posteriorment anomenada Nāgārjunakoṇḍa ("El puig de Nāgārjuna"). Les ruïnes de Nāgārjunakoṇḍa estan situades al districte de Guntur, a l'estat d'Andhra Pradesh. Es coneix que els nikāyas Caitika i Bahuśrutīya tenien monestirs a Nāgārjunakoṇḍa. Les troballes arqueològiques a Nagarjunakonda no han estat concloents pel que fa a l'associació d'aquest lloc amb Nagarjuna. El nom "Nagarjunakonda" data de l'època medieval, i les inscripcions dels segles III-IV indiquen amb claredat que el lloc s'anomenava "Vijayapuri" en el període antic.
Nagarjuna va viure set o vuit segles després de Buda Sakyamuni. Era alhora un monjo i un pensador profund, que considerava -com altres pensadors del Mahayana-, que el poder alliberador de l'ensenyament de Shakyamuni s'estava distorsionant perillosament, per culpa de les elaboracions teòriques dels pensadors de l'Abhidharma. Al llarg dels segles, aquests pensadors havien intentat sistematitzar l'ensenyament de Buda, les seves intuïcions i les seves ensenyances; per acabar donant una descripció de la realitat, que tendia a fer creure, malgrat l'ensenyament fonamental de la vacuïtat, que a la base de la realitat amb què ens trobem, existeixen elements que tenen naturalesa pròpia.
Els pensadors de l'Abhidharma van anomenar aquests elements els dharma, espècie de maons constitutius de l'edifici de l'existència. I encara que pretenien admetre que aquests dharma no són permanents, sí que els atorgaven una naturalesa permanent. Per exemple, el foc posseeix la naturalesa pròpia de cremar, el seu svabhava. O bé l'aigua té la naturalesa de ser humida i de poder regar, etc. Van arribar a assenyalar setanta-quatre dharma. Amb una descripció així, estaven negant la interdependència i la vacuïtat, i estaven reconstruint entitats amb naturalesa pròpia. Va originar-se tal elucubració teòrica que, segons Nagarjuna, la possibilitat d'alliberar-se i de despertar seguint l'ensenyament de Buda quedava seriosament amenaçada.
Pensament
Nagarjuna va escriure un poema en vers, el Madhyamika-karika, traduït per "Estances del Mig", o "Poemes del Mig". En les quals Nagarjuna aborda tots els grans temes del Dharma de Buda: l'ego, el temps, el camí, el nirvana, etc., Diuen: "Sense res que s'acabi o es produeixi, sense res que es fongui o sigui etern, sense unitat ni diversitat, sense arribada ni partida, així és la coproducció condicionada."
La veritat profunda és que res s'acaba, ni res ha estat mai creat, tan sols hi ha un encadenament de causes i condicions, i per tant, cap de les baules de la cadena té substància pròpia, sinó que és causada per la precedent, tot i que la precedent també està influïda per la que la segueix. Per exemple, la ignorància condiciona la tendència a actuar, però aquesta tendència a actuar també fa créixer la ignorància, i el mateix passa amb cada una de les anelles, o baules. Nagarjuna afegeix: "És la coproducció condicionada dels mots i de les coses. És la pacificació beneïda. Aquell qui ens ho ha ensenyat, el Perfecte Despert, és el millor instructor, que jo saludo." Per a Nagarjuna, és el cor del Dharma de Buda, que en el curs de les estances següents, s'esforça a detallar i desenvolupar. Però per a comprendre les coses d'aquesta manera, cal admetre el que explica en el capítol vint-i-quatre, en relació amb les Quatre Nobles Veritats. Allí respon als teòrics de l'Abhidharma, en un diàleg imaginari, en què sense elaborar ell cap tesi pròpia, desmunta i tira per terra llurs teories.
Sunyata
El tema principal que tracta Nāgārjuna és el concepte de śūnyatā (traduït com "el buit"), el qual porta a altres doctrines budistes, com l'anātman "no-jo" i pratītyasamutpāda "origen dependent", per tal de refutar la metafísica d'alguns dels seus filòsofs contemporanis. Per a Nāgārjuna, de la mateixa manera que per a Buda en els seus texts antics, no és només que els éssers sensibles siguin no-substancials; tots els fenòmens (dhammas) són desproveir de cap svabhāva, literalment "natura pròpia" o "existència inherent", i per tant no tenen cap essència subjacent. Estan buits del fet d'existir independentment; així, les teories heterodoxes de svabhāva que circulaven en aquella època van ser refutades, amb l'argument de les doctrines inicials del budisme. Això és així perquè totes les coses sorgeixen de manera dependent: no pel seu propi poder, sinó dependent de les condicions que les porten a la seva existència, al contrari que el concepte de l'ésser.
Nāgārjuna considera com a real qualsevol entitat que tingui una natura per si mateixa (svabhāva), que no està produïda per causes (akrtaka), que no depèn de res més (paratra nirapeksha).
En el versicle 14 del capítol 24 del Mūlamadhyamakakārikā es pot trobar una de les cites més famoses de Nāgārjuna sobre el buit i el co-sorgiment:
Com a part de la seva anàlisi del buit dels fenòmens en el Mūlamadhyamakakārikā, Nāgārjuna realitza una crítica del concepte de svabhāva des de diferents punts de vista. Discuteix els problemes de dotar qualsevol tipus d'essència inherent a la causalitat, el moviment, el canvi i la identitat personal. Nāgārjuna utilitza l'eina lògica hindú del tetralema per atacar qualsevol concepció essencialista. L'anàlisi lògica de Nāgārjuna es basa en quatre proposicions bàsiques:
Tot (dharma) existeix: afirmació de l'ésser, negació del no-ésser.
Res no existeix: afirmació del no-ésser, negació de l'ésser.
Tot existeix i no existeix alhora: tant afirmació com negació.
Res existeix ni no-existeix: ni afirmació ni negació.
El fet de dir que totes les coses són 'buides' és negar qualsevol tipus de fundació ontològica; per tant, el punt de vista de Nāgārjuna sovint s'interpreta com una mena d'anti-fundacionalisme ontològic. o un anti-realisme metafísic.
El fet de comprendre la natura del buit dels fenòmens és només un mitjà per a aconseguir un fi, que és el nirvana. Així, el projecte filosòfic de Nāgārjuna és, de forma última, un projecte soteriològic, que cerca corregir els nostres processos cognitius quotidians que postulen svabhāva de manera errònia en el flux de l'experiència.
Alguns estudiosos com Fyodor Shcherbatskoy i T.R.V. Murti afirmen que Nāgārjuna fou l'inventor de la doctrina Shunyata. No obstant això, alguns treballs més recents de Choong Mun-keat, Yin Shun i Dhammajothi Thero no fou en realitat un innovador a l'hora de plantejar la seva teoria, sinó que, en paraules de Shi Huifeng, "la connexió entre buit i origen dependent no és una innovació o creació de Nāgārjuna."
Dues veritats
Nāgārjuna també fou un mitjà per al desenvolupament de la Doctrina de les dues veritats, que afirma que existeixen dos nivells de veritat en els ensenyaments del budisme, la veritat última (paramārtha satya) i la veritat convencional o superficial (saṃvṛtisatya). Segons Nāgārjuna, la veritat última és la veritat que afirma que tot és buit d'essència, això inclou el buit en si ('el buit del buit'). Mentre que alguns autors (Murti, 1955) interpreten això etiquetant Nāgārjuna com una mena de neokantisme i per tant fent que la veritat última sigui un noümen metafísic o un "[concepte] últim inefable que transcendeix les capacitats del raonament discursiu", altres autors com Mark Siderits i Jay L. Garfield defensen que el punt de vista de Nāgārjuna és que "la veritat última é que no existeix tal veritat última" (Siderits) i que Nāgārjuna és un "anti-dualista semàntic" que afirma que només existeixen les veritats convencionals. Per tant, segons Garfield:
Per articular aquesta noció en el Mūlamadhyamakakārikā, Nāgārjuna es basà en una font prèvia del Kaccānagotta Sutta, que distingeix entre el significat definitiu (nītārtha) i el significat interpretable (neyārtha):
La versió enllaçada és la que es troba en els nikayas, que és lleugerament diferent de la que es troba en el Samyuktagama. Ambdues versions contenen el concepte de l'ensenyament mitjançant el punt intermedi entre els extrems d'existència i no-existència. Nagarjuna no fa cap referència al "tot" quan cita el text agàmic en el seu Mūlamadhyamakakārikā.
Causalitat
Jay L. Garfield descriu que Nāgārjuna s'aproxima a la causalitat a partir de les Quatre Nobles Veritats i l'origen dependent. Nāgārjuna distingeix entre dues visions d'origen dependent, la que causa els efectes i la que causa les condicions. Això s'explica en la doctrina de les dues veritats, on la veritat convencional i la veritat última es recolzen l'una en l'altra, i ambdues són buides en existència. La distinció entre efectes i condicions és controvertida. Segons l'enfocament de Nāgārjuna, la causa significa un esdeveniment o estat que té el poder de provocar un efecte. Les condicions es refereixen a causes que proliferen i provoquen un event, estat o procés posterior.
Relativitat
Dins dels ensenyaments de Nagarjuna estava la idea de la relativitat; al Ratnāvalī, proporciona l'exemple de què la qualitat d'ésser curt només existeix en relació amb la idea de longitud. La determinació d'una cosa o objecte només és possible en relació a altres coses o objectes, especialment si s'hi estableix una relació de contrast. Manté que la relació entre les idees de "curt" i "llarg" no és de caràcter intrínsec (svabhāva). Aquesta idea també és compartida en els Nikāyas pali i en els Āgamas xinesos, on el concepte de relativitat s'expressa de manera similar: "L'element de llum ... existeix en relació amb la foscor; l'element de bé es percep en relació amb allò que és dolent; l'element d'espai es percep en relació a la idea de la forma."
Llegat
Els seguidors de Nagarjuna, des de l'època de Kumārajīva, van esdevenir els fundadors d'importants escoles Mahayana fora de l'Índia. La més important fou l'escola xinesa Sānlùn (o escola dels tres tractats), fundada el pel monjo Jijang (anomenat Sanron-shu al Japó), i que desenvolupava les doctrines dels dos nivells de veritat, el buit (shunyata) i que les ensenyances més internes de Buda no es poden transmetre al llenguatge conceptual (idea que formava part de les disquisicions de Nagarjuna).
Per a l'Escola Tiantai (Tendai en japonès), fundada pel seu contemporani Jijana-Chiyi, les ensenyances de Nagarjuna conformaven els fonaments d'una altra idea: encara que les coses siguin buides, tenen una certa existència temporal, cosa que es demostra per la percepció que en tenim.
La influència de les idees de Nagarjuna en l'escola de Ch'an (Zen) també és significativa. Kamalaśīla fa jugar un rol crucial en la difusió de les idees de Nagarjuna al Tibet.
El principal estudiós de Nagarjuna fou Āryadeva (), el creador del corrent madhyamaka-prasangika, o "madhyamaki radical" (en oposició al madhmaka-svatantrika, "madhyamaki moderat"). Els principals comentaris de les obres de Nagarjuna provenen de Buddapalita (), Bhavaviveka (), i Chandrakirti ().
Obres
Existeixen un cert nombre de textos influents atribuïts a Nāgārjuna, així com diversos pseudo-epígrafs, per la qual cosa existeix una certa controvèrsia sobre quins són els seus textos autèntics.
Mūlamadhyamakakārikā
El Mūlamadhyamakakārikā és l'obra més coneguda de Nāgārjuna. "No només és un gran comentari sobre el discurs de Buda a Kaccayana, l'únic discurs on se cita el nom de l'interlocutor, sinóo també una anàlisi detallada i curosa sobre la majoria dels discursos importants inclosos en els Nikayas i en els agamas, especialment els del Atthakavagga del Sutta-nipata.
En el Mūlamadhyamakakārikā, "tots els fenòmens que s'experimenten són buits. Això no significa que no es puguin experimentar, i per tant, que siguin no-existents; només que no tenen una substància permanent i eterna (svabhava) perquè, com un somni, són meres projeccions de la consciència humana. Com que aquestes ficcions imaginàries es poden experimentar, no són només noms (prajnapti)."
Altres obres atribuïdes
Segons alguns punts de vista, en concret el de Christian Lindtner, les obres que hom pot atribuir correctament a Nāgārjuna són:
Mūlamadhyamaka-kārikā (Versos Fonamentals de la Via Intermèdia)
Śūnyatāsaptati (Setanta Versos sobre el Buit)
Vigrahavyāvartanī (La Fi de les Disputes)
Vaidalyaprakaraṇa (Polovoritzant les Categories)
Vyavahārasiddhi (Prova de Convenció)
Yuktiṣāṣṭika (Seixanta Versos sobre el Raonament)
Catuḥstava (Himne a la Realitat Absoluta)
Ratnāvalī (Garlanda Preciosa)
Pratītyasamutpādahṝdayakārika ([Elements] Constitutius de l'Origen Dependent)
Sūtrasamuccaya (Compendi d'Escriptures)
Bodhicittavivaraṇa (Exposició de la Ment Il·luminada)
Suhṛllekha (Carta a un Bon Amic)
Bodhisaṃbhāra (Requisists per a la Il·luminació)
Buston considera que els primers sis són els tractats principals de Nāgārjuna, mentre que segons Taaranaatha només els primers cinc són obres de Nāgārjuna. T.R.V. Murti considera que Ratnaavali, Pratitya Samutpaada Hridaya i Sutra Samuccaya són obres de Nāgārjuna, ja que les dues primeres són citades profusament per Chandrakirti i la tercera per Shantideva.
Addicionalment a les obres mencionades anteriorment, n'hi ha d'altres que també s'atribueixen a Nāgārjuna. Existeix una controvèrsia sobre el fet de quines d'aquestes obres són autèntiques. Alguns estudis contemporanis suggereixen que aquestes obres pertanyen a un període notablement posterior, potser dels segles VIII o IX, amb la qual cosa no poden ser obres autèntiques de Nāgārjuna.
Tanmateix, algunes obres considerades com a importants en el budisme esotèric s'atribueixen a Nāgārjuna i els seus deixebles, segons alguns historiadors com Tāranātha, en el Tibet del . Aquests historiadors intenten salvar les dificultats cronològiques amb diverses teories. Per exemple, una propagació de les obres posteriors mitjançant una revelació mística.
Lindtner considera que el Mahāprajñāpāramitāupadeśa "Comentari sobre la Gran Perfecció de la Saviesa" no és una obra genuïna de Nāgārjuna. Aquesta obra només se cita en una traducció xinesa de Kumārajīva. Existeix un cert debat sobre si aquesta és una obra de Nāgārjuna o d'algú altre. Étienne Lamotte, qui va traduir-ne una tercera part al francès, tenia la sensació que era obra d'un bhikkhu del nord de l'Índia pertanyent a l'escola Sarvāstivāda, qui més tard es va convertir al Mahayana. El monjo i estudiós xinès Yin Shun creia que era l'obra d'un filòsof del sud de l'Índia, i que Nāgārjuna podria ser-ne l'autor. Aquests dos punts de vista no estan necessàriament en contradicció, ja que el Nāgārjuna del sud de l'Índia podria haver estudiat el Sarvāstivāda del nord. Cap dels dos estudiosos creu que l'obra és de Kumārajīva, com altres estudiosos creuen.
Iconografia
Nāgārjuna es representa habitualment con un híbrid amb característiques d'humà i de nāga. Sovint, la representació en forma de nāga configura un baldaquí que corona i protegeix el cap humà. Aquesta noció de nāga és comuna en la cultura religiosa hindú, i acostuma a representar una serp intel·ligent o un drac, el qual és responsable de les pluges, els llacs i altres fonts d'aigua. En el budisme, és un sinònim per a un arhat, o una persona sàvia en general.
Referències
Bibliografia
Vegeu també
Budisme
Història del budisme
Mahayana
Quatre nobles veritats
Vihara
Enllaços externs
Obres de Nagarjuna traduïdes al català
Bodhisattvas
Filòsofs indis
Filòsofs del segle III
Filòsofs del segle II
Telugus
Morts a l'Índia
Sants indis
Religiosos budistes | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,536 |
IBM Personal Computer Basic, обычно сокращаемый до IBM BASIC — представляет собой язык программирования Бейсик, впервые выпущенный корпорацией IBM с IBM PC (модель 5150) в 1981 году. IBM выпустила четыре разных версии интерпретатора Microsoft BASIC, лицензированных от Microsoft для PC и PCjr. Они известны как Cartridge BASIC, Cassette BASIC, Disk BASIC и Advanced BASIC (BASICA). Версии Disk BASIC и Advanced BASIC были включены в состав IBM PC DOS до PC DOS 4. В дополнение к функциям стандарта ANSI BASIC версии IBM предлагал поддержку графического и звукового оборудования линейки IBM PC. Исходный код можно вводить с помощью полноэкранного редактора, а для отладки программы были предоставлены очень ограниченные возможности. IBM также выпустила версию компилятора Microsoft BASIC для PC одновременно с выпуском PC DOS 1.10 в 1982 году.
См. также
IBM Basic Programming Support
Литература
Thomas Milton Kemnitz, Tom Dunnington. The IBM Basic Manual. Children's Press, 1985. ISBN 0516084267.
David A. Lien. Learning IBM Basic: For the Personal Computer. Tech Publications, 1989. ISBN 9813091983.
Примечания
Бейсик
IBM | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,966 |
\section{Introduction}
Knowledge representation, as a critical prerequisite for many machine learning
tasks, has always been a central problem in the field of natural language
processing (NLP). As for the representation of documents, an established form
is to use bag-of-words (BOW) or term frequency-inverse document frequency
(TF-IDF) representations. Another widely adopted method is generative topic
models, such as latent semantic analysis (LSA) \cite{Deerwester;1990} and
latent dirichlet allocation (LDA) \cite{Blei2003}.
Recently, \newcite{Bengio;2003} proposed a window-based unsupervised word embedding
method. Following his approach, \newcite{Mikolov;2013a} introduced two new
log-linear models, \method{skip-gram} and \method{cbow}. \newcite{Mikolov;2013b} gave a highly efficient
implementation of those two models, and distributed it as \method{word2vec}, which has
been widely used as a tool in language related tasks.
Inspired by the success of \method{word2vec}, \newcite{Le;2014} extended \method{word2vec} into
\method{doc2vec}, which produces a vector representation for each document, known as
"document embedding". \newcite{Dai;2015} further examined \method{doc2vec} and found
analogy features on Wikipedia (e.g. "Lady Gaga" - "American" + "Japanese" $\approx$
"Ayumi Hamasaki"). However, others have struggled to reproduce such results.
Most recently, \newcite{Lau;2016} made an empirical evaluation of \method{doc2vec},
and revealed its potential on different tasks.
Although \method{doc2vec} has produced promising results, we doubt its basis as it
implicitly assigns the same weight to each word occurrence when training document
vectors. This is counter-intuitive, since human never give equal attention to
different parts of a sentence. Consider the following sentence as an example:
\begin{quote}
\small
\begin{verbatim}
There are many activities
including but not limited to
running, jumping, and swimming.
\end{verbatim}
\end{quote}
When reading this sentence, we are not concerned about the "there be" term. We
will probably ignore "limited to", because "including" and "but not" indicate
the parenthesis character of that term. The sentence can still be understood
even if some parts are missing:
\begin{quote}
\small
\begin{verbatim}
... many activities including
... running, jumping, ...
\end{verbatim}
\end{quote}
Motivated by such facts, we propose a context aware document embedding based on
\method{doc2vec}. Our method takes a novel approach that estimates weights for each
word occurrence by measuring the shift of the corresponding document vector if
the word is substituted by another. We use convolutional neural networks (CNN)
and gated recurrent units (GRU) as auxiliary models for the space of document
vectors. We compared our model with benchmarks in \cite{Lau;2016} and \method{doc2vec}
with IDF weights to show the advantage of our model. To give a convincing
illustration of our model, we visualized the hidden states of deep neural
networks. Our findings suggest that context aware weights are a kind of
enhanced IDF weights that are especially good at capturing sub-topic level
keywords in documents, since neural networks can substantially extract
asymmetric context features, despite trained with unsupervisedly embedded
targets.
\section{Related Work}
\subsection{Distributed Bag of Words}
The departure point of the context aware model is the distributed bag of word
(DBOW) model of \method{doc2vec} proposed in \cite{Le;2014} trained with the negative
sampling procedure \cite{Mikolov;2013b}. \footnote{This paper uses the gensim
implementation of \method{doc2vec} \cite{Rehurek;2010}} \method{dbow} uses a similar fashion
like \method{skip-gram} \cite{Mikolov;2013a} to train a document vector ($\var d^i$) for each
document with its context word vectors ($\var c^i_j$). By adopting negative
sampling procedure, the objective is to maximise the likelihood of
$P(\var c^i_j|\var d^i)$ while minimising the likelihood of $P(\var c'|\var d^i)$,
where $\var c'$ is a random sample \cite{Goldberg;2014,Lau;2016}. Therefore,
the objective function is given as:
\begin{equation}
\sum_j{\left[\log\sigma({\var c^i_j}^\intercal \var d^i) - \\
\sum_{k=1, \var c' \sim P_n(c)}^n{\log\sigma({\var c'}^\intercal \var d^i)}\right]}
\label{eq:dbow-loss}
\end{equation}
where $\sigma$ is the sigmoid function, $n$ is the number of negative samples,
and $P_n(c)$ is a distribution derived from term frequency.
Despite that \method{dbow} works with randomly initialized context word vectors, it is
suggested that the quality of embeddings is improved when context word vectors
are jointly trained by \method{skip-gram} and \method{dbow} \cite{Dai;2015,Lau;2016}. It is also
observed that by initializing word vectors from pre-trained \method{word2vec} of large
external corpus like \dataset{wiki}, the model converges faster as well as performs
better \cite{Lau;2016}.
\subsection{CNN}
\begin{figure}
\includegraphics[width=.5\textwidth]{figures/CNN-arch.pdf}
\caption{CNN architecture}
\label{fig:CNN-arch}
\end{figure}
Convolutional neural networks (CNN) proposed by \newcite{LeCun;1998} are a kind
of partially connected network architecture. CNN exploits convolution kernels to
extract local features and uses pooling method to agglomerate features for
subsequent layers, which have achieved excellent results in many NLP tasks
\cite{Collobert;2011,Yih;2014}. A typical CNN architecture for NLP consists of
a convolutional layer, a global max pooling layer, and a fully connected layer
, as is shown in Figure \ref{fig:CNN-arch} \cite{Kim;2014}. \footnote{All neural
networks are implemented in Keras with Tensorflow backend\cite{Chollet;2015}
\label{fn:keras}}
For a document, the model takes a sequence of word vectors as input, and
applies convolution along the sequence. The kernels are designed with different
widths to extract features of different scales. Following each feature map,
there is a max-pooling, which leads to a global maximum of the feature over the
sequence. Finally, all features are concatenated and fully connected to the
target, for either regression or classification purpose.
\subsection{GRU}
\begin{figure}
\includegraphics[width=.5\textwidth]{figures/RNN-arch.pdf}
\caption{GRU architecture}
\label{fig:GRU-arch}
\end{figure}
Gated recurrent units (GRU), introduced by \newcite{Cho;2014a,Cho;2014b,Chung;2014},
is a kind of recurrent neural networks (RNN) often used in semantics related
tasks. An RNN is a neural network that scans input items one by one, and produces
a representation for every prefix of the sequence, which can be formally written
as
\begin{equation*}
\var h_{\step t}=f(\var h_{\step{t-1}}, \var x_t)
\end{equation*}
where $\var x_t$ is the $t^{th}$ item, $\var h_{\step t}$ is the hidden
representation at timestep $t$, and $f$ is a non-linear activation function. In
GRU, a reset gate $\var r_{\step t}$ and an update gate $\var z_{\step t}$ are
used for regulating $f$.
\begin{equation*}
\var z_{\step t}=\sigma(\var W_z\var x_t+\var U_z\var h_{\step{t-1}})
\end{equation*}
\begin{equation*}
\var r_{\step t}=\sigma(\var W_r\var x_t+\var U_r\var h_{\step{t-1}})
\end{equation*}
where $\sigma$ is a non-negative activation function, usually sigmoid or
hard sigmoid function. Then $f$ is computed by
\begin{equation*}
f(\var h_{\step{t-1}}, \var x_t)=\var z_{\step t}\odot \var h_{\step{t-1}}+(1-\var z_{\step t})\odot\tilde{\var h}_{\step t}
\end{equation*}
with
\begin{equation*}
\tilde{\var h}_{\step t}=\phi\left(\var W\var x_t+\var U(\var r_{\step t}\odot \var h_{\step{t-1}})\right)
\end{equation*}
where $\phi$ stands for $tanh$ function.
The two-gate design enables GRU to represent both long-term and short-term
dependencies over timesteps. Units with update gate frequently activated
tend to capture long-term dependencies, while units with reset gate activated
tend to capture short-term dependencies \cite{Cho;2014a}. For regression targets
over the sequence, a fully connected layer with dropout is added to the last
timestep, as Figure \ref{fig:GRU-arch} shows. \textsuperscript{\ref{fn:keras}}
\section{Context Aware Model}
Our context aware model is an unsupervised model that trains \method{doc2vec} with
word occurrence weights regarding their contexts. To generalize the problem,
we will first define weighted \method{dbow} (notated as \method{w-dbow}), then illustrate
the context aware model as a specific one of \method{w-dbow}.
A \method{w-dbow} model is defined as a triplet $(\set D, \set C, \set W)$. In the
triplet, $\set D$ and $\set C$ are documents and context words for the vanilla
\method{doc2vec} model, respectively. $\set W$ is a set of weights corresponding to
each word occurrence $\var c\in \set C$. Similar to Equation (\ref{eq:dbow-loss}),
the target is to minimize the following function:
\begin{equation}
\sum_j{\var w^i_j\left[\log\sigma({\var c^i_j}^\intercal \var d^i) -
\sum_{k=1, \var c' \sim P_n(c)}^n{\log\sigma({\var c'}^\intercal \var d^i)}\right]}
\label{eq:wdbow-loss}
\end{equation}
where $\var w^i_j\in \psi(\set W)$. Here we introduce a normalizing function
$\psi$ on $\set W$, because $\set W$ may be scaled or biased in non-trivial cases.
We adopt a global temperature softmax for $\psi$, which means
\begin{equation*}
\psi(\var w^i_j)=\frac{|\set W|e^{\var w^i_j/T}}{\sum_{\var w\in \set W}{e^{\var w/T}}}
\end{equation*}
$T$, known as the "temperature", is a hyperparameter that controls the softness
of a softmax function. The result is scaled by $|\set W|$ times so that the
average $\var w$ is $1$, which is the case in \method{dbow}. Therefore, hyperparameters
like learning rate from \method{dbow} can be applied to \method{w-dbow} directly.
\begin{table*}
\center
\begin{adjustbox}{max width=0.95\textwidth}
\begin{tabular}{cccccc}
\toprule
\textbf{Input Size} & \textbf{\#Timesteps} & \textbf{Output Size} & \textbf{Loss} & \textbf{Optimizer} &
\textbf{Epoch} \\
\midrule
300 & 93 & 300 & cosine distance & adam(lr=0.001) &
100 \\
\bottomrule
\end{tabular}
\end{adjustbox}
\caption{Shared hyperparameters of auxiliary models for STS task}
\label{tab:shared-params}
\end{table*}
The context aware model is a \method{w-dbow} that generates weights in the following way.
First, a vanilla \method{dbow} is trained on the corpus, and then weight $\var w^i_j$
is computed by randomly substitute $\var c^i_j$ and measuring the shift in
document vector $\var d^i$ with cosine distance. This makes sense because there
will be little shift if the word can be inferred from its context, otherwise a
large shift should take place, as replacing the word impedes the meaning of the
document. To predict document vectors on new word sequences, we trained an
auxiliary model to regress document vectors with word vector sequences. We
utilize CNN (Figure \ref{fig:CNN-arch}) and GRU (Figure \ref{fig:GRU-arch}) for
this task. Hence, the corresponding context aware models are named as \method{CA(CNN)}
and \method{CA(GRU)}. Algorithm \ref{alg:weights} gives details of this procedure
\begin{algorithm}[h]
\caption{Generate context aware weights}
\begin{algorithmic}[1]
\State $aux\_model \gets$ CNN or GRU
\State $aux\_model.train(word\_sequence, doc)$
\For{$i \gets 1$ to $|docs|$}
\For{$j \gets 1$ to $|word\_sequence^i|$}
\State $shifts \gets \varnothing$
\For {$k \gets 1$ to $sample\_count$}
\State $s' \gets word\_sequence^i$
\State $s'_j \gets random\_word()$
\State $d' \gets aux\_model.predict(s')$
\State $d \gets doc^i$
\State $shifts.add(cos\_distance(d', d))$
\EndFor
\State $\var w^i_j \gets average(shifts)$
\EndFor
\EndFor
\end{algorithmic}
\label{alg:weights}
\end{algorithm}
The weight is computed by averaging all the shifts of random substitutions. The
random word is sampled from the global vocabulary with probability proportional
to term frequency (TF). We also propose an economic way that take samples from
words sharing the same part-of-speech (POS). Experiments show that two methods
can achieve similar results, as will be shown in Section \ref{sec:experiments}.
\section{Experiments}
\label{sec:experiments}
\subsection{Evaluation methods}
We conduct experiments on the semantic textual similarity (STS) task. It is a
shared task held by SemEval \cite{Agirre;2015}. In STS, the goal is to
automatically predict a score for each sentence pair according to their semantic
similarity. The result is evaluated by computing the Pearson correlation
coefficient between the predicted score and the ground truth.
Since the official dataset provided by SemEval is quite small, we combine all
datasets from 2012 to 2015 as the training set, following the approach in
\cite{Lau;2016}. We take the \domain{headlines} domain of 2014 as the
development set, and test on the 2015 dataset. It is reliable to specify the
test set as a subset of the training set, as both our methods and baselines are
unsupervised.
All datasets are preprocessed by lowercasing and tokenizing the sentence, using
Stanford CoreNLP \cite{Manning;2014}. POS tags are also obtained through
Stanford CoreNLP.
\subsection{Baselines}
To demonstrate the advantages of context aware models, we compare them with
unsupervised baselines proposed in \cite{Lau;2016}, including a linear
combination of word vectors from \method{skip-gram}, and several \method{dbow} with different training
settings. The linear combination computes a document embedding for a sentence
by averaging over all its context with word vectors trained on the full
collection of English Wikipedia entries. \footnote{Using the pretrained model
by \cite{Lau;2016}: \url{https://github.com/jhlau/doc2vec} \label{fn:pretrained}}
For \method{dbow}, we train one model on the STS dataset. \footnote{We use the same
hyperparameters for all \method{dbow} and \method{w-dbow} trained on the STS dataset: vector
size = 300, window size = 15, min count = 1, sub-sampling threshold = $10^{-5}$,
negative sampling = 5, epoch = 400. \label{fn:dbow-hyperparams}} In the
following context, we will refer these datasets as \dataset{wiki} and \dataset{sts} respectively.
We train another model on \dataset{wiki} and exploit it to infer document vectors for
\dataset{sts} without updating any hidden weights. \textsuperscript{\ref{fn:pretrained}}
As it is observed in practice that using pretrained word vectors from external
corpus can benefit the performance of \method{dbow}, we experiment a third \method{dbow}
baseline with word vectors initialized by \method{skip-gram} trained on \dataset{wiki}.
\textsuperscript{\ref{fn:pretrained},}\footnote{We test on both trainable and
untrainable word vector initialization for \method{dbow}, and receive negligible
difference. Therefore, we only list the result of the trainable version.}
Our context aware models are concrete implementations of \method{w-dbow}. Because \method{w-dbow}
is consistent with \method{dbow}, we force context aware models to use the same
hyperparameters as \method{dbow}. \textsuperscript{\ref{fn:dbow-hyperparams}} We only
optimise hyperparameters of the auxiliary model (i.e. CNN or GRU) towards the
target of document vectors through cross-validation on the training set, as
well as the temperature $T$ on the development set. In the following context,
these two methods are referred as \method{CA(CNN)} and \method{CA(GRU)} respectively.
To distinguish context aware models from other weighting methods, we also
introduced a \method{w-dbow} baseline with weights from inverse document frequency (IDF)
as \method{w-dbow(IDF)}. We optimise its temperature $T$ separately from context aware
models.
\subsection{Experiments}
\begin{table*}
\center
\begin{adjustbox}{max width=0.95\textwidth}
\begin{tabular}{ccc@{\hspace{20pt}}ccccc}
\toprule
\multirow{2}{*}{\textbf{Domain}} & \textbf{\method{skip-gram}} & \multicolumn{3}{c}{\textbf{\method{dbow}}} & \textbf{\method{CA(CNN)}}
& \textbf{\method{CA(GRU)}} & \textbf{\method{w-dbow(IDF)}} \\
& \dataset{wiki} & \dataset{sts} & \dataset{wiki} & \dataset{sts}(\dataset{wiki} init) & \multicolumn{2}{c}{\dataset{sts}}
& \dataset{sts} \\
\midrule
\domain{answers-forums} & 0.516 & 0.647 & 0.666 & \best{0.675} & 0.670
& 0.662 & 0.656 \\
\domain{headlines} & 0.731 & 0.768 & 0.746 & 0.782 & 0.785
& 0.787 & \best{0.788} \\
\domain{answers-students} & 0.661 & 0.640 & 0.628 & 0.654 & \best{0.683}
& 0.676 & 0.660 \\
\domain{belief} & 0.607 & 0.764 & 0.713 & \best{0.773} & 0.772
& 0.764 & 0.760 \\
\domain{images} & 0.678 & 0.781 & 0.789 & \best{0.800} & 0.793
& 0.793 & 0.787 \\
\bottomrule
\end{tabular}
\end{adjustbox}
\caption{Results over STS task with different unsupervised methods}
\label{tab:method-compare}
\end{table*}
\begin{table*}
\begin{center}
\begin{adjustbox}{max width=0.95\textwidth}
\begin{tabular}{cc@{\hspace{15pt}}c@{\hspace{20pt}}c@{\hspace{15pt}}c}
\toprule
\multirow{2}{*}{\textbf{Domain}} & \multicolumn{2}{c}{\textbf{\method{CA(CNN)}}}
& \multicolumn{2}{c}{\textbf{\method{CA(GRU)}}} \\
& global & POS
& global & POS \\
\midrule
\domain{answers-forums} & 0.663 & \best{0.670}
& 0.668 & 0.662 \\
\domain{headlines} & 0.786 & 0.785
& 0.786 & \best{0.787} \\
\domain{answers-students} & 0.673 & \best{0.683}
& 0.675 & 0.676 \\
\domain{belief} & 0.760 & \best{0.772}
& 0.761 & 0.764 \\
\domain{images} & \best{0.800} & 0.793
& 0.792 & 0.793 \\
\bottomrule
\end{tabular}
\end{adjustbox}
\caption{Context aware models with random sampling from different distributions.
The global distribution is proportional to TF of each word. The POS distribution
is proportional to TF of each word grouped by POS. Samples are taken from the
corresponding distribution regarding POS of the substituted word. We take 50
and 10 samples for global distribution and POS distribution respectively.}
\label{tab:CA-compare}
\end{center}
\end{table*}
First, we decide hyperparameters for our auxiliary models. For CNN, it is
observed through cross-validation that kernels with width from 3 to 8 best
capture features from the word vector sequence to make the vanilla document
vector. Considering the scale of \dataset{sts} dataset, we use 128 kernels for each
width. For GRU, we use 512 hidden units. Both deep networks have 1.4M trainable
parameters approximately, indicating they should have the same capability.
Then we optimise the temperature $T$ for \method{w-dbow} models. We find that results on
the development set almost follows a unimodal function as $T$ varies, which
facilitates the optimization of $T$ a lot. In general, $T$ around 1/15--1/14
works for both \method{CA(CNN)} and \method{CA(GRU)}, while $T$ around 5--6 works for \method{w-dbow(IDF)}.
Experiments show that our context aware models outperforms \method{dbow} in all 5
domains (Table \ref{tab:method-compare}). Of two purposed models, \method{CA(CNN)} works
a little better and results at the same level of \method{dbow} initialized by word
vectors from \dataset{wiki}. Besides, \method{w-dbow(IDF)} also gets a better performance than \method{dbow},
which buttresses the consistency of our definition for \method{w-dbow}. More
interestingly, all \method{w-dbow} baselines make excellent results in the domain of
\domain{answers-students}, compared to any \method{dbow} approach. We will give a
detailed analysis of this in Section {\ref{sec:introspection}}.
We also compared \method{CA(CNN)} and \method{CA(GRU)} with different word distributions. Table
\ref{tab:CA-compare} gives results in detail. Consistent with our estimation,
the difference between global distribution and POS distribution is not
significant. Therefore, gain of context aware models comes from weighting
method rather than sampling from POS distribution. However, we recommend to use
POS distribution for random substitution, as it is much smaller and more
efficient.
\section{Model Introspection}
\label{sec:introspection}
\begin{figure*}
\begin{center}
\begin{adjustbox}{max width=\textwidth}
\begin{subfigure}{0.2\textwidth}
\includegraphics[width=0.99\textwidth]{figures/dbow_test_distribution.png}
\caption{\method{dbow}}
\end{subfigure}
\begin{subfigure}{0.2\textwidth}
\includegraphics[width=0.99\textwidth]{figures/dbow_wiki_init_test_distribution.png}
\caption{\method{dbow}(\dataset{wiki} init)}
\end{subfigure}
\begin{subfigure}{0.2\textwidth}
\includegraphics[width=0.99\textwidth]{figures/CNN_test_distribution.png}
\caption{\method{CA(CNN)}}
\end{subfigure}
\begin{subfigure}{0.2\textwidth}
\includegraphics[width=0.99\textwidth]{figures/RNN_test_distribution.png}
\caption{\method{CA(GRU)}}
\end{subfigure}
\begin{subfigure}{0.2\textwidth}
\includegraphics[width=0.99\textwidth]{figures/IDF_test_distribution.png}
\caption{\method{w-dbow(IDF)}}
\end{subfigure}
\begin{subfigure}{0.15\textwidth}
\includegraphics[width=0.99\textwidth]{figures/distribution_legend.png}
\end{subfigure}
\end{adjustbox}
\caption{Distribution of different document vectors under t-SNE}
\label{fig:vec-distributions}
\end{center}
\end{figure*}
\begin{figure*}
\includegraphics[width=.95\textwidth]{figures/weight-sample.pdf}
\caption{A sample of weights learned by the context aware model. Brighter bars denote larger weights.}
\label{fig:weight-sample}
\end{figure*}
Being unsupervised models, our context aware models surpass the vanilla \method{dbow},
and even show some advantage towards \method{dbow}(\dataset{wiki} init). This is momentous
because context aware models only exploit the local corpus, without external
priori like pretrained vectors. Hence, we believe context aware models do
extract more information from the corpus. We will shed some light on why and
how context aware models work by looking into document vectors and hidden
representations of the model.
In context aware models, weights learned by word-level substitution consists of
two components. One is a base value for each word, the other is a
context-related bias for each word occurrence. The base value is something close
to IDF, as there is a moderate correlation between context aware weights and
IDF. Hence, context aware models can be viewed as an enhanced version of
\method{w-dbow(IDF)}. \footnote{Typically, the Pearson's r between context aware weights
and IDF is 0.5--0.6.}
Generally, both context aware models and \method{w-dbow(IDF)} give equal or smaller
similarity answers in STS task. They tend to distinguish sentences rather than
find trivial similarities. In other words, IDF weighted loss makes document
embeddings insensitive to common words. Context aware models even enable them
to neglect common words given the context in a self-adaptive manner, which is
very similar to lateral inhibition in cognition and neuroscience.
To illustrate this, consider the distribution of document vectors on the test
set using t-SNE \cite{Maaten;2008} in Figure \ref{fig:vec-distributions}.
Document vectors of different domains are marked with different colors. Since
domains are natural categories, dots of the same color should gather as a
cluster. Consistent with our experience, \method{dbow} forms good clusters. However,
none of three \method{w-dbow} models gives such a clear result. In context aware
embeddings, though clusters still can be observed, a number of dots are
scattered far away from their centers. The phenomenon is extremely significant
in \domain{answers-students} (red), where \method{dbow} performs worst and context
aware models improve most. This might be attributed to that different from
coarse-grained task like clustering, STS task more relies on fine-grained
features, where our models have advantage over vanilla \method{dbow}.
We then examine the domain of \domain{answers-students}. It contains students
answer to electricity problems, of which most share the same topic, but their
semantic similarity varies. As for a concrete example, we randomly pick a
sentence pair from \domain{answers-students} whose similarity differs much in
\method{dbow} and context aware models. Figure \ref{fig:weight-sample} clearly shows
weights learned by \method{CA(CNN)}. It can be spotted that low weights are assigned to
common words. Moreover, context aware model neglects jargons like "voltage" or
"terminal", but focuses on "difference", which is a relatively rare occurrence
given the context. In fact, it is word like "difference" and "positive" that
defines the key point of a sentence in a topic. With context aware weights,
their contribution are well amplified.
\begin{table}[htb]
\begin{center}
\begin{adjustbox}{max width=0.45\textwidth}
\begin{tabular}{ccc@{\hspace{20pt}}c}
\toprule
\multirow{2}{*}{\textbf{Context word}} & \textbf{CNN} & \textbf{GRU} & \textbf{IDF} \\
& \small{\#feature change} & \small{norm of $\var r_{\step t}$} \\
\midrule
Voltage & 393 & 5.037 & 5.805 \\
is & 157 & 8.847 & 1.526 \\
the & 98 & 10.168 & 0.877 \\
difference & 385 & 7.153 & 5.925 \\
between & 200 & 8.957 & 4.678 \\
a & 96 & 10.039 & 0.953 \\
positive & 324 & 5.916 & 4.383 \\
and & 99 & 9.334 & 1.671 \\
negative & 322 & 6.017 & 5.363 \\
terminal & 328 & 5.393 & 4.066 \\
on & 122 & 10.349 & 2.171 \\
the & 53 & 10.476 & 0.877 \\
battery & 230 & 6.889 & 3.422 \\
. & 15 & 10.570 & 0.417 \\
\midrule
\textbf{Correlation to IDF} & 0.915 & -0.845 & 1 \\
\bottomrule
\end{tabular}
\end{adjustbox}
\caption{Influence of each context word on hidden representations}
\label{tab:hidden-states}
\end{center}
\end{table}
Notice that auxiliary models are trained with document vectors, which are
generated according to context words homogeneously. But shifts in the vector
space do behave heterogeneously regard to each substituted occurrence. To find
the origin of such asymmetry, we investigate on the hidden states of auxiliary
models. For CNN, we count the number of global features that change in
substitution. For GRU, since the hidden representation varies in different
timesteps, we count the norm of $\var r_{\step t}$, as it implies how much
units "decline" the input. Surprisingly, both auxiliary models reveal unequal
hidden states, as shown in Table \ref{tab:hidden-states}. The number of
features that a word contributes to in CNN is highly correlated with its IDF,
while the extent to which hidden units turn down a word in GRU is highly
negatively correlated with its IDF. Since IDF is the entropy of every word
given its distribution over a corpus, we are inclined to believe that the first
layer of deep neural networks learns hidden representations towards the least
entropy over the corpus (i.e. maximize the use of hidden units).
\section{Conclusion}
We introduce two context aware models for document embedding with a novel
weight estimating mechanism. Compared to vanilla \method{dbow} method, our approach
infers a weight for each word occurrence with regard to its context, which
helps document embedding to capture sub-topic level keywords. This property
facilitates learning a more fine-grained embedding for semantic textual
similarity task as well as eases training on large external corpus. We claim
that context aware weights is composed of an IDF base and a context-related
bias. This might be induced by deep neural networks as they naturally learns
representations with the least entropy, even when the target is generated
homogeneously.
\nocite{Pennington;2014}
\nocite{Zhang;2017}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,546 |
storing data in the blockchain
Version 0.2
----
* [BTC-CDN protocol v0.2](PROTOCOL.md)
Notes
----
* [BTC-CDN-encode](https://github.com/cripplet/btc-cdn-encode/releases/tag/0.2) is the official implementation of BTC-CDN to upload onto the blockchain.
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,786 |
'Return to Silent Hill' Feature Film to Adapt Silent Hill 2, First Details
by October Keegan - sisslethecat
Perhaps a smidge lost in the shuffle during yesterday's game announcement madness, (four new titles announced including a remake of Silent Hill 2), Christophe Gans (director of 2006's film adaption) was given a few minutes of time during the special 'Silent Hill Transmission' video update. While he had spoiled the film's existence during interviews over the past year or so, (and even just rambled a bit about the Silent Hill games revealed during yesterday's video), it's extremely exciting and gratifying to finally have the film revealed. Titled Return to Silent Hill, this film is an adaption of the much beloved Silent Hill 2, something Christophe Gans had wanted to do since making the very first film. You can check out Gans and producer Victor Hadida discussing the film during the video:
I've always been a fan of the first film, despite its heavy alterations to game lore I adore its dedication to recreating the game's tone and atmosphere, occasionally even recreating moments or camera angles (not to mention its heavy use of the actual game score). Knowing that the next film will be a take on one of the most beloved games in the series is extremely exciting, knowing that that level of respect (again, despite alterations) appears to be carrying over. I wrote a thread over on Twitter about the concept art and storyboards shown off during the video, showcasing and comparing moments as they appear in the art vs the game scenes they're taken from:
Another hype thread because I said so, genuinely really, really hyped for Return to Silent Hill. Dopey name aside (I guess James is literally "returning" to town so w/e). I love the original film, and knowing that this director is tackling SH2 is really exciting to me 1/6 pic.twitter.com/A6jQsqMaSt
— October 🎃 (@Sisslethecat) October 20, 2022
Fans should obviously expect alterations, deviations, and just straight-up missing content (as Gans states, they're trying to "transpose the language of the game into a 90-100 minute film), but hopefully can go into this new film with an open mind. We'll be keeping you updated on future developments regarding Return to Silent Hill.
Retro PS1 Survival Horror 'Lake Haven – Chrysalis' Out Now | {
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} | 2,128 |
{"url":"https:\/\/www.gradesaver.com\/textbooks\/math\/trigonometry\/trigonometry-7th-edition\/chapter-1-section-1-2-the-rectangular-coordinate-system-1-2-problem-set-page-25\/63","text":"## Trigonometry 7th Edition\n\n$30^{\\circ}$\nComplement of $60^{\\circ}$ will be another angle that terminal side needs to sweep in order to reach position y i.e. standard position of $90^{\\circ}$. Obviously it will be $30^{\\circ}$. Another way to find complement of an angle will be to subtract it from $90^{\\circ}$. Hence complement of $60^{\\circ}$ = $90^{\\circ}$- $60^{\\circ}$ = $30^{\\circ}$","date":"2019-03-21 05:20:42","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\": 0.6935878992080688, \"perplexity\": 431.30953390733094}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"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-2019-13\/segments\/1552912202496.35\/warc\/CC-MAIN-20190321051516-20190321073516-00137.warc.gz\"}"} | null | null |
namespace absl {
inline namespace lts_2018_06_20 {
bad_any_cast::~bad_any_cast() = default;
const char* bad_any_cast::what() const noexcept { return "Bad any cast"; }
namespace any_internal {
void ThrowBadAnyCast() {
#ifdef ABSL_HAVE_EXCEPTIONS
throw bad_any_cast();
#else
ABSL_RAW_LOG(FATAL, "Bad any cast");
std::abort();
#endif
}
} // namespace any_internal
} // inline namespace lts_2018_06_20
} // namespace absl
| {
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Pro Cycling News – Dutch Tank Conquers Germany
Posted on 15 August 2005 by cosmo
Despite having a totally sweet name like "Bram Tankink," Bram Tankink has ungone his share of misfortune (scroll way down) earlier this season. Even after driving the long break in today's first stage of the Tour of Germany, and then dropping the rest break over a small rise with 16k to go, "The Tank" lost traction in a rain-soaked corner, and almost missed out on the stage win. But everything came out in the wash for the Quick-Step Dutchman,l who now sits over 3 minutes ahead of race favorites Jan Ullrich and Bobby Julich.
So remember yesterday when I was pondering how Illes Balears would fill the gaps left by its departing Spanish riders? Well, they'll use Spanish riders from less-well-heeled Spanish teams, of course. Word came in today that San Sebastian winner Tino Zaballa will be heading to the squad formerly known as Banesto next season, and his teammate on Saunier Duval (referred to by ProCycling as "lightly-financed"), last year's San Sebastian winner Miguel Angel Martin Perdiguero, is apparently also leaving for greener pastures. One Spaniard *not* leaving the team is Jose Marchante, who, despite previous reports, is not signing with Discovery Channel for next season. Whoever reported that first *cough* velogal *cough* was apparently misinformed. Oh, an Bobby Julich signed a two year contract extension with CSC today, quashing rumors that he will retire at the end of this season, his best in recent memory.
On to politics: how do you trace pervading political beliefs between continents? Simple. Just take a look at their cycling coverage. Australian-based Cyclingnews (which does a fair job of catering to Americans as well) led off today's coverage with a report on how Lance was set to ride with Bush at his famously-overused Ranch in Crawford, Texas. The same bits of text were used in a report on British-based Eurosport today, but with a decidedly different message. | {
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Q: django replace old image uploaded by ImageField How I can replace one image uploaded by an ImageField (in my models) with the new one selected by the same ImageField?
And, can I delete all images uploaded by an ImageField when I delete the model object (bulk delete)?
A: After hours of searching I found this code.
It worked for me in Django 3
class Image(models.Model):
image = models.ImageField()
def save(self, *args, **kwargs):
try:
this = Image.objects.get(id=self.id)
if this.image != self.image:
this.image.delete(save=False)
except:
pass # when new photo then we do nothing, normal case
super().save(*args, **kwargs)
A: This might get tricky,
And a lot of times depends on what are your constraints.
1. Write your own save method for the model and then delete the old image a and replace with a new one.
os.popen("rm %s" % str(info.photo.path))
*
*Write a cron job to purge all the unreferenced files. But then again if disk space is a issue, you might want to do the 1st one, that will get you some delay in page load.
| {
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Bokmusseron (Tricholoma ustale) är en svampart som först beskrevs av Elias Fries, och fick sitt nu gällande namn av Paul Kummer 1871. Bokmusseron ingår i släktet musseroner och familjen Tricholomataceae. Arten är reproducerande i Sverige. Inga underarter finns listade i Catalogue of Life.
Källor
Externa länkar
Musseroner | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,299 |
She writes poetry to relieve homesickness
Ms Belen Esposo Repollo, featured Filipina poet. PHOTO: COURTESY OF BELEN ESPOSO REPOLLO
Reni Chng
When she was 17, her pursuit of a medical degree at Saint Louis University in the Philippines ended abruptly when her scholarship sponsor died.
When she was 33, Ms Belen Esposo Repollo left her three children and two foster children in the Philippines and came to work in Singapore.
Tomorrow, Ms Repollo, 42, a domestic helper, will be one of seven featured poets reading their work at Migrant Poetic Tales, an event at the Prelude to Poetry Festival Singapore 2017.
The event will start at 10am, along with others such as When Poetry Meets Calligraphy, which explores the relationship between contemporary poetry and different types of calligraphy. The Prelude is open to the public, but registration is required.
Poetry was Ms Repollo's way of coping with life away from her family.
She went on to participate in the National Poetry Festival in 2015 and 2016. She also collaborated with other poets to publish books like Ace Poems and Through Open Hangar Doors.
Da Vinci work sells for record S$611m
Show me the Monet!
Get ready for a poetry smackdown
$51.3m porcelain bowl breaks record
She writes in both Tagalog and English and is working on her own book in Tagalog.
Mr Zakir Hussain Khokan, 38, another featured poet, is also co-organiser of the event.
Armed with an arts degree from the National University of Bangladesh, he worked as a freelance journalist back home but moved to Singapore in 2003 to work as a construction supervisor to support his wife and son.
In 2004, he formed the group Amrakajona which means "we are" in Bengali. They organise events such as poetry recitals, plays and book fairs.
Mr Zakir won the first Migrant Worker Poetry Contest in 2014 and the following one in 2015.
To find out more, visit www.poetryfestivalsg.peatix.com/
ARTSForeign WorkerSingapore
Read articles by Reni Chng | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,210 |
Senior Policy Analyst – Aboriginal Health
Province of Nova Scotia – Department of Health & Wellness, Halifax, NS
Provide analysis and advice on Aboriginal health within the System Strategy & Performance division of DHW. Work with NS First Nations on Mental Health and Addictions issues and policies. Participation on the Mi'kmaq-NS-Canada Tripartite Health Committee, Taking Flight project, Aboriginal Continuing Care Policy Forum, and other related committees and projects.
Consultant / Owner Operator
Jipuktuk Consulting, Millbrook, NS
Project Management, Aboriginal Community Consulting, Data Research, Aboriginal Programs Analyst, Wind Energy Liaison
Previous work in wind energy (juwi Wind) included job training programs database compiling, Mi'kmaq contractors & labour qualification database compiling, assistance with community open house meetings, review of environmental assessments, and coordinating/participating in community college site visit for prospective turbine technician trainees.
Policy Analyst
Province of Nova Scotia – Office of Aboriginal Affairs, Halifax, NS
This office is responsible for coordinating the Province's response to Aboriginal issues. The majority of work is conducted through the Mi'kmaq-Nova Scotia-Canada Tripartite Forum and the Made-in-Nova Scotia Process.
Conduct research and provide analysis on a variety of aboriginal issues, participate in tripartite working groups, and work with First Nation communities on a variety of issues.
Prepare background reports and briefing notes in support of policy development and strategic business planning.
Economic Opportunities/Government Relations Officer
Department of Indian and Northern Affairs Canada (INAC), Amherst & Halifax, NS
This Federal government department has primary, but not exclusive, responsibility for meeting the federal government's constitutional, treaty, political and legal responsibilities to First Nations, Inuit and Northerners. The department's primary role is to support First Nations and Inuit in developing healthy, sustainable communities and in achieving their economic and social aspirations.
Assessed First Nation proposals against predetermined criteria, to determine eligibility for departmental funding, provided recommendations to senior officers, the Regional Project Review Committee, and various Tripartite committee processes throughout the region.
Managed and monitored approved programs to meet the criteria as set out by Economic Development Programs of INAC.
Participated in ongoing review of strategies, plans and policies related to Economic Development, Marshall Strategy, and Gathering Strength Initiatives with respect to First Nation governments, Aboriginal Representative Organizations, and Federal /Provincial Regional Partnerships in Atlantic Canada.
Councillor
Millbrook First Nation, Truro, NS
A Mi'kmaq First Nation community that is governed by a Chief and 12 Councillors who are elected by Millbrook Band Members every two years.
As a member of council, set policies, procedures, budgets, and strategic direction for the community and participated in regular council meetings to discuss issues concerning the Millbrook community.
Participated in the analysis and approval of business development projects for the community as member of Millbrook Economic Development Committee.
Education & Professional Development
Saint Mary's University, Halifax, NS
Bachelor of Commerce / Major: Management
© 2021 Gord Johnson for Council. | {
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\section{Introduction}
The {\em cyclic matching sequenceability} of a simple graph $G$,
denoted $\textnormal{cms}(G)$,
is the largest integer $s$ for which
there exists a cyclic ordering of the edges of $G$ so that every set of $s$ consecutive edges forms a matching.
Katona~\cite{MR2181045} implicitly considered cyclic matching sequenceability
and found a lower bound for $\textnormal{cms}(K_n)$.
Brualdi, Kiernan, Meyer and Schroeder~\cite{MR2961987}
defined cyclic matching sequenceability explicitly
and proved that $\textnormal{cms}(K_n) = \left\lfloor\frac{n-2}{2}\right\rfloor$
for all $n\geq 4$,
thus strengthening the result found by Katona~\cite{MR2181045}.
Brualdi et al.~\cite{MR2961987} also determined
that $\textnormal{cms}(C_n) = \left\lfloor \frac{n-1}{2} \right\rfloor$
for all $n\geq 3$ and found the cyclic matching sequenceability of several other graphs.
A non-cyclic variant to cyclic matching sequenceability, denoted $\textnormal{ms}(G)$ has also been considered and was defined first by Alspach~\cite{MR2394738} who determined $\textnormal{ms}(K_n)$.
Brualdi et al.~\cite{MR2961987} also determined the matching sequenceability for cycles and several other classes of graphs.
Chiba and Nakano~\cite{ChiNa}
found various results concerning the matching sequenceability for general graphs and more refined results for regular graphs.
In this paper our focus is on the cyclic matching sequenceability of regular graphs. The \emph{chromatic index} of a graph is the smallest number of colours required to properly colour its edges. By Vizing's theorem \cite{Viz} the chromatic index of a graph with maximum degree $\Delta$ is equal to either $\Delta$ or $\Delta+1$. In the former case we say the graph is \emph{class 1} and in the latter we say it is \emph{class 2}. For positive integers $n$ and $k$ such that $n>k$ and $nk$ is even, we define $\textnormal{cms}(n,k)$ to be the minimum value of $\textnormal{cms}(G)$ over all $k$-regular graphs on $n$ vertices and define $\textnormal{cms}_1(n,k)$ to be the minimum value of $\textnormal{cms}(G)$ over all $k$-regular class 1 graphs on $n$ vertices. Our primary focus is on the behaviour of $\textnormal{cms}(n,k)$ and $\textnormal{cms}_1(n,k)$ for fixed $k$ as $n$ becomes large. All asymptotic notation used in this paper is relative to this regime.
The main contribution of this paper is to establish lower bounds on $\textnormal{cms}(n,k)$ and $\textnormal{cms}_1(n,k)$, as well as an upper bound on $\textnormal{cms}(n,k)$.
These bounds are summarised in Theorem~\ref{thm:k reg}.
\begin{thm}\label{thm:k reg}
Let $k \geq 3$ be an integer. Then, for any integer $n \geq 6(k+1)$ such that $nk$ is even,
\begin{alignat*}{2}
\max\left\{\mfrac{k(5k-3)}{4(k+1)(4k-3)}n-6,\mfrac{31k}{98(k+1)}n-o(n)\right\} &\leq \textnormal{cms}(n,k) &&\leq \mfrac{kn}{2(k+1)} \\
\max\left\{\mfrac{5k-8}{4(4k-7)}n-6,\mfrac{31}{98}n-o(n)\right\} &\leq \textnormal{cms}_1(n,k) &&\leq \mfrac{n-1}{2}\,.
\end{alignat*}
\end{thm}
In the lower bound on $\textnormal{cms}(n,k)$, for large $n$, the $\max$ takes the first value for $k \leq 13$ and the second value for $k \geq 14$. In the lower bound on $\textnormal{cms}_1(n,k)$, for large $n$, the $\max$ takes the first value for $k \leq 14$ and the second value for $k \geq 15$. Table~\ref{T:kconsequences} in the conclusion gives an explicit listing of the consequences of Theorem~\ref{thm:k reg} for various values of $k$. In the case of 2-regular graphs we are able to completely determine $\textnormal{cms}(n,2)$ and $\textnormal{cms}_1(n,2)$.
\begin{thm}\label{thm:2 reg}
For each $n \geq 6$, we have $\textnormal{cms}(n,2)=\lfloor\frac{n}{3}\rfloor$ and for each even $n \geq 4$, we have $\textnormal{cms}_1(n,2)=\frac{n-2}{2}$.
\end{thm}
For a graph $G$, let $\mathcal{M}(G)$ be the set of all matchings in $G$ and, for an edge $e$ of $G$, let $\mathcal{M}_e(G)$ be the set of all matchings in $G$ containing $e$. A \emph{fractional edge colouring} of a graph $G$ is a function $\omega: \mathcal{M}(G) \rightarrow \mathbb{R}^{\geq 0}$ such that $\sum_{M \in \mathcal{M}_e(G)}\omega(M) \geq 1$ for each edge $e \in E(G)$. The weight of such a colouring is $\sum_{M \in \mathcal{M}(G)}\omega(M)$. Note that an edge colouring of $G$ can be viewed as a fractional edge colouring $\omega$ of $G$ for which the image of $\omega$ is a subset of $\{0,1\}$. The \emph{fractional chromatic index} of a graph $G$ is the infimum of the weights of the fractional edge colourings of $G$. While the main focus of this paper is regular graphs, some of our results apply more generally. In particular, we have the following.
\begin{thm}\label{thm:gen bounds}
For any graph $G$ with chromatic index $c$ and fractional chromatic index $c_f$,
\[\left\lfloor\tfrac{1}{2c}|E(G)|\right\rfloor-1 \leq \textnormal{cms}(G) \leq \tfrac{1}{c_f}|E(G)|.\]
Furthermore, for any integers $\Delta \geq 2$ and $n \geq \Delta+1$, there is a graph $G$ of order $n$ such that $\textnormal{cms}(G) \leq \frac{1}{\Delta+1}|E(G)|$.
\end{thm}
The lower bound in Theorem~\ref{thm:gen bounds} differs by at most two from an analogous bound for matching sequenceability given in \cite{ChiNa}.
We organise the rest of the paper as follows. In Section~\ref{Preliminaries} we introduce some notation and preliminary results. In Section~\ref{2reg} we consider $2$-regular graphs and prove Theorem~\ref{thm:2 reg}. Theorem~\ref{thm:gen bounds} is proved in Section~\ref{genBounds}. In Section~\ref{givenPartition} we establish a lower bound on the cyclic matching sequenceability of a regular graph assuming the existence of a partition of its edges with suitable properties. Finally in Section~\ref{findPartition}, we show that regular graphs do admit such partitions and prove Theorem~\ref{thm:k reg}.
\section{Preliminaries}
\label{Preliminaries}
For an integer $n$, let $\mathbb{Z}_n$ represent the additive group of integers modulo $n$. In this paper, graphs will always be simple. Two edges in a graph are \emph{adjacent} if they are both incident
on the same vertex. A \emph{matching} is a $1$-regular graph. The \emph{union} $G \cup H$ of two graphs $G$ and $H$ is the graph with vertex set $V(G) \cup V(H)$ and edge set $E(G) \cup E(H)$. An {\em ordering} of a graph $G$ with $m$ edges is a bijective function
$\ell \;:\; E(G) \rightarrow \mathbb{Z}_{m}$.
The image of $e$ under~$\ell$ is called the {\em label} of $e$. We will sometimes specify an ordering $\ell$ by giving the tuple $(\ell^{-1}(0),\ldots,\ell^{-1}(m-1))$ rather than the function $\ell$. A set of edges of $G$ is {\em consecutive} in $\ell$ if their labels form a set of consecutive integers
and is {\em cyclically consecutive} in $\ell$ if their labels form a set of consecutive integers
modulo $m$.
Let $\ell$ be an ordering of a graph $G$ with $m$ edges and let $e$ and $e'$ be distinct edges of $G$. We define $d_\ell(e,e')$, the \emph{forward distance from $e$ to $e'$ in $\ell$}, to be the smallest positive integer $d$ such that $\ell(e)+d = \ell(e')$, where the addition takes place in $\mathbb{Z}_{m}$. We define $d_\ell\{e,e'\}$, the \emph{distance between $e$ and $e'$ in $\ell$}, to be $\min\{d_\ell(e,e'),d_\ell(e',e)\}$. Define $\textnormal{cms}(\ell)$ to be the largest element $s$ of $\{1,\ldots,m\}$ such that $d_\ell\{e,e'\} \geq s$ for any pair $\{e,e'\}$ of edges adjacent in $G$. Similarly, define $\textnormal{ms}(\ell)$ to be the largest element $s$ of $\{1,\ldots,m\}$ such that $d_\ell(e,e') \geq s$ for any ordered pair $(e,e')$ of edges adjacent in $G$ such that $\ell(e)<\ell(e')$. Note that, for a graph $G$, $\textnormal{ms}(G)$ and $\textnormal{cms}(G)$, as defined in the introduction, are the maximum values of $\textnormal{ms}(\ell)$ and $\textnormal{cms}(\ell)$ respectively over all orderings $\ell$ of $G$. If $G$ is a matching, then obviously $\textnormal{cms}(G)= \textnormal{ms}(G)=|E(G)|$.
We first prove the upper bound of Theorem~\ref{thm:gen bounds}. To our knowledge this connection between the cyclic matching sequenceability of a graph and its fractional chromatic index has not been observed before.
\begin{lem}\label{lem:fracChromatic}
For any graph $G$ with fractional chromatic index $c_f$, $\textnormal{cms}(G) \leq \frac{1}{c_f}|E(G)|$.
\end{lem}
\begin{proof}
Let $s=\textnormal{cms}(G)$ and let $\ell$ be an ordering of $G$ with $\textnormal{cms}(\ell)=s$. Let $\mathcal{L}$ be the set of matchings in $G$ whose edges form a set of $s$ cyclically consecutive edges in $\ell$. Let $\omega: \mathcal{M}(G) \rightarrow \mathbb{R}^{\geq 0}$ be defined by $\omega(M)=\frac{1}{s}$ if $M \in \mathcal{L}$ and $\omega(M)=0$ otherwise. Then $\omega$ is a fractional edge colouring of $G$ with weight $\frac{1}{s}|E(G)|$. So $\frac{1}{s}|E(G)| \geq c_f$ and the result follows.
\end{proof}
For edge-disjoint graphs $G_0$ and $G_{1}$,
with labellings $\ell_{0}$ and $\ell_1$ respectively,
let $\ell_0 \vee \ell_{1}$ denote
the ordering $\ell$ of $G= G_0 \cup G_1$
defined by
$\ell(e) = \ell_0 (e)$ if $e \in E(G_0)$ and $\ell(e) = |E(G_0)|+\ell_1 (e)$ if $e \in E(G_1)$.
A {\em matching decomposition} of a graph $G$ is
a set of edge-disjoint matchings of $G$ that partition the edge set of~$G$. A matching decomposition of $G$ into $k$ matchings can also be viewed as a proper edge colouring of $G$ with $k$ colours.
Now we will provide a lower bound on $\textnormal{cms}(G)$, given a matching decomposition of $G$
with certain properties exists, in the form of the proposition below.
Similar results were implicitly used by Alspach~\cite{MR2394738} and Brualdi et. al.~\cite{MR2961987}.
\begin{prop}[\cite{Ma}]\label{prop: Matching decomposition}
Let $G$ be a graph that decomposes into matchings $M_0 ,\ldots , M_{t-1}$, each with at least $m$ edges
and orderings $\ell_0 , \ldots , \ell_{t-1}$, respectively.
If, for some $s \in \{1,\ldots,m\}$, $\textnormal{ms}(\ell_i \vee \ell_{i+1}) \geq s$ for all $i\in \mathbb{Z}_t$, then $\textnormal{cms}(G) \geq s$.
\end{prop}
\begin{proof}
Let $\ell= \bigvee_{i=0}^{t-1} \ell_i$.
Consider two distinct edges $e$ and $e'$ that are at distance less than $s$ in $\ell$. Then $e,e' \in E(M_i \cup M_{i+1})$ for some $i \in \mathbb{Z}_t$.
So, by the assumption that $\textnormal{ms}(\ell_i \vee \ell_{i+1}) \geq s$, $e$ and $e'$ are nonadjacent in $G$.
This proves the proposition.
\end{proof}
We now prove four further lemmas which, like Proposition~\ref{prop: Matching decomposition}, provide lower bounds on the matching sequenceability of concatenations of orderings under various conditions.
\begin{lem}\label{lem:adding three orderings}
Let $X$, $Y$ and $Z$ be edge-disjoint graphs with orderings $\ell_X$, $\ell_Y$ and $\ell_Z$, respectively. Then
\[
\textnormal{ms}(\ell_X \vee \ell_Y \vee \ell_Z) \geq \min\left\lbrace \textnormal{ms}(\ell_X \vee \ell_Y),\textnormal{ms}(\ell_Y \vee \ell_Z), |E(Y)|+ \textnormal{ms}(\ell_X \vee \ell_Z) \right\rbrace.
\]
\end{lem}
\begin{proof}
Let $G=X \cup Y \cup Z$ and $\ell = \ell_X \vee \ell_Y \vee \ell_Z$. Let $e$ and $e'$ be a pair of adjacent edges in $G$ with $\ell(e)<\ell(e')$. If $e,e' \in E(X \cup Y)$, then $d_\ell(e,e') \geq \textnormal{ms}(\ell_X \vee \ell_Y)$, by definition. If $e,e' \in E(Y \cup Z)$, then $d_\ell(e,e') \geq \textnormal{ms}(\ell_Y \vee \ell_Z)$, by definition. Otherwise, $e \in E(X)$ and $e' \in E(Z)$, so $d_{\ell_X \vee \ell_Z}(e,e') \geq \textnormal{ms}(\ell_X \vee \ell_Z)$ by definition, and hence $d_{\ell}(e,e') \geq |E(Y)|+\textnormal{ms}(\ell_X \vee \ell_Z)$.
\end{proof}
\begin{lem}\label{lem:adding 4 special orderings}
Let $M_0,M_1,M_2,M_3$ be edge-disjoint matchings of sizes $m_0,m_1,m_2,m_3$ such that $M_0 \cup M_1$, $M_1 \cup M_2$ and $M_2 \cup M_3$
are also matchings. Then, for any orderings $\ell_0,\ell_1,\ell_2,\ell_3$ of $M_0,M_1,M_2,M_3$ respectively,
\[
\textnormal{ms}(\ell_0 \vee \ell_1 \vee \ell_2\vee \ell_3) \geq
\min\{\textnormal{ms}(\ell_0 \vee \ell_2)+m_1,\textnormal{ms}(\ell_1 \vee \ell_3)+m_2,\textnormal{ms}(\ell_0 \vee \ell_3)+m_1+m_2\}.
\]
\end{lem}
\begin{proof}
Let $G=M_0 \cup M_1 \cup M_2 \cup M_3$ and $\ell = \ell_0 \vee \ell_1 \vee \ell_2\vee \ell_3$. Let $e$ and $e'$ be a pair of adjacent edges in $G$ with $\ell(e)<\ell(e')$. Because $e$ and $e'$ are adjacent in $G$, we must have $e \in E(M_i)$ and $e' \in E(M_j)$ for some $(i,j) \in \{(0,2),(1,3),(0,3)\}$. Then $d_{\ell_i \vee \ell_j}(e,e') \geq \textnormal{ms}(\ell_i \vee \ell_j)$ by definition. Also, $d_{\ell}(e,e') = d_{\ell_i \vee \ell_j}(e,e')+s$, where $s=m_1$ if $(i,j)=(0,2)$, $s=m_2$ if $(i,j)=(1,3)$ and $s=m_1+m_2$ if $(i,j)=(0,3)$. The result follows.
\end{proof}
\begin{lem}\label{lem:ord given other ord bad case}
Let $X$ and $Y$ be edge-disjoint matchings and $\ell_{Y}$ be a fixed ordering of $Y$.
Then there is an ordering $\ell_{X}$ of $X$ such that $\textnormal{ms}( \ell_{X} \vee \ell_{Y}) \geq \frac{1}{2}|E(X)|$.
\end{lem}
\begin{proof}
Let $x=|E(X)|$ and $y=|E(Y)|$. For each edge $e \in E(X)$, let $\alpha(e)$ be the smallest label assigned by $\ell_Y$ to an edge adjacent to $e$ if such a label exists, and $\alpha(e)=\infty$ otherwise. Let $\ell_X$ be an ordering $(e_0,\ldots,e_{x-1})$ of $X$ such that $\alpha(e_0) \leq \cdots \leq \alpha(e_{x-1})$. Let $\ell = \ell_{X} \vee \ell_Y$ and
$e$ and $e'$ be adjacent edges in $G$ such that $\ell(e)<\ell(e')$ and $d_{\ell}(e,e')=\textnormal{ms}(\ell)$. Then $e=e_i$ for some $i \in \mathbb{Z}_x$ such that $\alpha(e_i)<\infty$ and $\alpha(e_i)=\ell_Y(e')$. By our definition of $\ell_X$, any edge of $X$ that is not adjacent to an edge of $Y$ occurs after $e_i$ in $\ell_X$ and hence each edge in $\{e_0,\ldots,e_{i-1}\}$ is adjacent to at least one edge of $Y$. Thus, because at most two edges of $X$ are adjacent to each edge of $Y$, we have that $\alpha(e_i) \geq \lfloor\frac{i}{2}\rfloor$ and hence that $d_{\ell}(e,e') =x -i+\alpha(e_i)\geq x-i+\lfloor\frac{i}{2}\rfloor$. So, because $i \leq x-1$, we have $d_{\ell}(e,e') \geq 1+\lfloor\frac{x-1}{2}\rfloor = \lceil\frac{x}{2}\rceil$ and the result follows.
\end{proof}
\begin{lem}\label{lem:ord given other ord}
Let $X$ and $Y$ be edge-disjoint matchings of sizes $x$ and $y$ respectively.
Suppose that $Y$ has $y_1$ edges that are adjacent to one edge in $X$ and
$y_2$ edges adjacent to two edges in $X$.
Let $\ell_{Y}$ be an ordering of $Y$ in which the $y_2$ edges adjacent to two edges in $X$
are the last to occur. Then there is an ordering $\ell_{X}$ of $X$ such that
\[\textnormal{ms}( \ell_{X}\vee\ell_{Y}) \geq \min\{x,x+y-y_1-2y_2\}.\]
\end{lem}
\begin{proof}
For $i \in \{1,2\}$, let $Y_i$ be the set of edges of $Y$ that are adjacent to exactly $i$ edges in $X$. For each edge $e \in E(X)$, let $\alpha(e)$ be the smallest label assigned by $\ell_Y$ to an edge adjacent to $e$ if such a label exists, and $\alpha(e)=\infty$ otherwise. Let $\ell_X$ be an ordering $(e_0,\ldots,e_{x-1})$ of $X$ such that $\alpha(e_0) \leq \cdots \leq \alpha(e_{x-1})$.
Let $\ell = \ell_X \vee \ell_Y$ and
$e$ and $e'$ be adjacent edges in $G$ such that $\ell(e)<\ell(e')$ and $d_{\ell}(e,e')=\textnormal{ms}( \ell)$. Then $e=e_i$ for some $i \in \mathbb{Z}_x$ such that $\alpha(e_i)<\infty$ and $\alpha(e_i)=\ell_Y(e')$. So $d_{\ell}(e,e')=x-i+\alpha(e_i)$. By our definition of $\ell_X$, any edge of $X$ that is not adjacent to an edge of $Y$ occurs after $e_i$ in $\ell_X$ and hence each edge in $\{e_0,\ldots,e_{i-1}\}$ is adjacent to at least one edge of $Y$. We consider two cases.
Suppose that $e' \in Y_1$. Then each edge in $\{e_0,\ldots,e_{i-1}\}$ is adjacent to at least one edge of $Y_1$ (recall the edges in $Y_2$ occur last in $\ell_Y$) and hence $\alpha(e_i) \geq i$. It follows that $d_{\ell}(e,e')\geq x$ and the result is established.
Suppose instead that $e' \in Y_2$. Let $j$ be the smallest element of $\mathbb{Z}_x$ such that $e_j$ is not adjacent to an edge in $Y_1$ and note that $j \leq y_1$ and that $\alpha(e_j) \geq y-y_2$ because the edges of $Y_2$ occur last in $\ell_Y$. So, because at most two edges of $X$ are adjacent to each edge of $Y_2$, we have that $\alpha(e_i) \geq \alpha(e_j)+\lfloor\frac{i-j}{2}\rfloor \geq y-y_2+\lfloor\frac{i-j}{2}\rfloor$. Thus,
\[d_{\ell}(e,e') = x-i+\alpha(e_i) \geq x+y-y_2-\lceil\tfrac{i+j}{2}\rceil \,.\]
Now, we saw that $j \leq y_1$ and we must have $i \leq y_1+2y_2-1$ for otherwise $\alpha(e_i)=\infty$. Thus, $\lceil\frac{i+j}{2}\rceil \leq y_1+y_2$ and hence $d_{\ell}(e,e') \geq x+y-y_1-2y_2$, and again the result is established.
\end{proof}
\section{2-regular graphs}\label{2reg}
In this section we will prove Theorem~\ref{thm:2 reg}.
We will require the result of Brualdi et al.~\cite{MR2961987} on $\textnormal{cms}(C_n)$
that was mentioned in the introduction.
\begin{thm}[Brualdi et al.~\cite{MR2961987}]\label{thm: cycle seq}
For all $n \geq 3$, $\textnormal{cms}(C_n) =\left\lfloor \frac{n-1}{2} \right\rfloor$.
\end{thm}
We first prove a useful result that gives an ordering of a particular type for a class 1 graph that is either a single cycle or a union of vertex-disjoint paths.
\begin{lem}\label{lem:good ord 2-colourable}
Let $H_0$ and $H_1$ be edge-disjoint matchings such that $|E(H_0)|=|E(H_1)|=t$ for some integer $t \geq 2$ and $H_0 \cup H_1$ is either a single cycle or a union of vertex-disjoint paths. There exist orderings $\ell_0$ and $\ell_1$ of $H_0$ and $H_1$ respectively such that $\textnormal{cms}(\ell_0 \vee \ell_1) \geq t-1$.
\end{lem}
\begin{proof}
We proceed by induction on $|V(H_0 \cup H_1)|$. Let $H=H_0 \cup H_1$.
If $|V(H)|=2t$ then $H$ is a cycle of length $2t$. We may assume its vertex set is $\mathbb{Z}_{2t}$ and its edge set is
$\{e_{i}: i \in \mathbb{Z}_{2t} \}$, where $e_{i} = \{ i,i+1\}$ and
$H_j =\{e_{i}\;:\; i \in \mathbb{Z}_{2t} \,\text{ and }\, i\equiv j \pmod{2}\}$ for $j \in \mathbb{Z}_2$. Let
\[
\ell_j(e_{i}) = i\quad \text{and} \quad \ell_{j+1}(e_{2t-1-i})=i
\]
for all $ i \in \{0,\ldots, t-1\}$ and $j \in \mathbb{Z}_2$ such that $j \equiv i \pmod{2}$. Note that $\ell_j$ is an ordering of $H_j$ for each $j \in \mathbb{Z}_2$. Let $\ell=\ell_0 \vee \ell_1$ and for an edge $e \in E(H)$, let $\ell^*(e)=\ell_j(e)$ where $j$ is the element of $\mathbb{Z}_2$ such that $e \in E(H_j)$. Let $\{e_{h-1},e_h\}$, where $h \in \mathbb{Z}_{2t}$, be an arbitrary pair of adjacent edges of $H$ and note that one of these edges is from $H_0$ and the other is from $H_1$. If $h \in \{1,\ldots,t-1\}$, then $d_\ell\{e_{h-1},e_h\}=t-1$ because $\ell^*(e_{h-1})=\ell^*(e_{h})-1$. Similarly, if $h \in \{t+1,\ldots,2t-1\}$, then $d_\ell\{e_{h-1},e_h\}=t-1$ because $\ell^*(e_{h-1})=\ell^*(e_{h})+1$. Finally, if $h \in \{0,t\}$, then $d_\ell\{e_{h-1},e_h\}=t$ because $\ell^*(e_{h-1})=\ell^*(e_{h})$. Thus it follows that $\textnormal{cms}(\ell_0 \vee \ell_1) = t-1$ and we have proved the result in the case where $|V(H)|=2t$.
Now suppose that $|V(H)|>2t$. Then $H$ is a union of $k$ disjoint paths for some $k \geq 1$. There are edges $yy' \in E(H_0)$ and $zz' \in E(H_1)$ such that $y \notin V(H_1)$, $z \notin V(H_0)$ and, if $k \geq 2$, then $y$ and $z$ are in different paths. Let $H'_0$ and $H'_1$ be the matchings obtained from $H_0$ and $H_1$ by merging the vertices $y$ and $z$ into a new vertex $x$. Then $H'_0 \cup H'_1$ is either a single cycle or a union of paths, and $|V(H'_0 \cup H'_1)|=|V(H_0 \cup H_1)|-1$. So, by induction, there are orderings $\ell'_0$ and $\ell'_1$ of $H'_0$ and $H'_1$, respectively such that $\textnormal{cms}(\ell'_0 \vee \ell'_1) \geq t-1$. Then $\textnormal{cms}(\ell_0 \vee \ell_1) \geq t-1$ where $\ell_0$ and $\ell_1$ are the orderings of $H_0$ and $H_1$ obtained from $\ell'_0$ and $\ell'_1$ by replacing $xy'$ with $yy'$ in $\ell_0$ and $xz'$ with $zz'$ in $\ell_1$. So the result follows by induction.
\end{proof}
Our next lemma implies that $\textnormal{cms}(G) \geq \frac{n-2}{2}$ for each 2-regular class 1 graph $G$ of order $n$, but also says more.
\begin{lem}\label{lem:2 edge colourable}
Let $H_0$ and $H_1$ be edge-disjoint matchings such that $|E(H_0)|=|E(H_1)|=t$ for some integer $t \geq 1$.
There are orderings $\ell_0$ and $\ell_1$ of $H_0$ and $H_1$, respectively,
such that $\textnormal{cms}(\ell_0 \vee \ell_1) \geq t-1$.
\end{lem}
\begin{proof}
We proceed by induction on $|E(H_0)|=|E(H_1)|$. Let $H=H_0 \cup H_1$. Clearly $H$ must have a subgraph $H^\dag$ such that $H^\dag$ is either
\begin{itemize}
\item
a component of $H$ that has an even number of edges (and is either a path or a cycle); or
\item
a union of two components of $H$, each of which is a path of odd length, with the property that $|E(H_0) \cap E(H^\dag)|=|E(H_1) \cap E(H^\dag)|$.
\end{itemize}
Let $|E(H^\dag)|=2s$, noting that $|E(H^\dag)|$ is even, and let $H^\dag_i$ be the matching of size $s$ with edge set $E(H_i) \cap E(H^\dag)$ for each $i \in \mathbb{Z}_2$. By Lemma~\ref{lem:good ord 2-colourable} there are orderings $\ell^\dag_0$ and $\ell^\dag_1$ of $H^\dag_0$ and $H^\dag_1$, respectively, such that $\textnormal{cms}(\ell^\dag_0 \vee \ell^\dag_1) \geq s-1$.
If $H^\dag=H$, then the result follows by taking $\ell_i=\ell^\dag_i$ for $i \in \mathbb{Z}_2$, so we may assume that $H^\dag\neq H$. For $i \in \mathbb{Z}_2$, let $H^\ddag_i$ be the matching of size $t-s$ with edge set $E(H_i) \setminus E(H^\dag_i)$. By our inductive hypothesis, there are orderings $\ell^\ddag_0$ and $\ell^\ddag_1$ of $H^\ddag_0$ and $H^\ddag_1$ such that $\textnormal{cms}(\ell^\ddag_0 \vee \ell^\ddag_1) \geq t-s-1$. Let $\ell_i=\ell^\dag_i \vee \ell^\ddag_i$ for $i \in \mathbb{Z}_2$.
Any pair $\{e,e'\}$ of adjacent edges in $H$ such that $e \in E(H^\ddag_0)$ and $e' \in E(H^\ddag_1)$ are at distance at least $t-s -1$ in $\ell^\ddag_0 \vee \ell^\ddag_1$ because $\textnormal{cms}(\ell^\ddag_0 \vee \ell^\ddag_1) \geq t-s-1$, and hence are at distance at least $t-s-1+s=t-1$ in $\ell_0 \vee \ell_1$. Likewise, any pair $\{e,e'\}$ of adjacent edges in $H$ such that $e \in E(H^\dag_0)$ and $e' \in E(H^\dag_1)$ are at distance at least $s-1$ in $\ell^\dag_0 \vee \ell^\dag_1$, and hence are at distance at least $s-1+t-s=t-1$ in $\ell_0 \vee \ell_1$. Because $H^\dag$ and $H^\ddag$ are vertex-disjoint, these arguments cover all pairs of adjacent edges in $H$ and so $\textnormal{cms}(\ell_0 \vee \ell_1) \geq t-1$.
\end{proof}
\begin{lem}\label{lem:class 1 2 regular cms}
For each even $n \geq 4$, we have $\textnormal{cms}_1(n,2)=\frac{n-2}{2}$.
\end{lem}
\begin{proof}
By Theorem~\ref{thm: cycle seq}, for each even $n \geq 4$, we have $\textnormal{cms}(H) = \frac{n-2}{2}$ if $H$ is an $n$-cycle and hence $\textnormal{cms}(n,2) \leq \frac{n-2}{2}$. Also, Lemma~\ref{lem:2 edge colourable} implies that $\textnormal{cms}(H) \geq \frac{n-2}{2}$ for each $2$-regular class 1 graph $H$ of even order $n \geq 4$, because any such graph $H$ is the union of two edge-disjoint matchings each of size $\frac{n}{2}$.
\end{proof}
The \emph{matching number} of a graph $G$ is the maximum size of a matching in $G$. If $\ell$ is an ordering of a graph $H$ and $G$ is a subgraph of $H$ then the \emph{subordering of $\ell$ induced by $G$} is the unique ordering $\ell_G$ of $G$ such that, for all $e, e' \in E(G)$, $\ell_G(e)<\ell_G(e')$ if and only if $\ell(e)<\ell(e')$. An ordering $\ell^*$ is a \emph{subordering} of an ordering $\ell$ of a graph $H$ if $\ell^*$ is induced by $G$ for some subgraph $G$ of $H$. Our next lemma provides upper bounds on the cyclic matching sequenceability of a graph based on the properties of one of its subgraphs. We only need the simpler first part in this section, but the more involved second part is required in Section~\ref{genBounds}.
\begin{lem}\label{lem:gen bound given subG}
Let $H$ be a graph and $G$ be a subgraph of $H$ with matching number $\nu$. Then
\begin{itemize}
\item[(i)]
$\textnormal{cms}(H) \leq \frac{\nu|E(H)|}{|E(G)|}$
\item[(ii)]
$\textnormal{cms}(H) \leq \frac{|E(H)|}{\left\lfloor \frac{1}{\nu}(|E(G)|-\textnormal{cms}(G)) \right\rfloor+1}$.
\end{itemize}
\end{lem}
\begin{proof}
We first prove (i). Because any fractional edge colouring of $H$ can be restricted in the natural fashion to a fractional edge colouring of $G$ with equal or lesser weight, we have $c_f(H) \geq c_f(G)$. Furthermore, it is clear from the definition of fractional chromatic index that $c_f(G) \geq \frac{1}{\nu}|E(G)|$. Thus $c_f(H) \geq \frac{1}{\nu}|E(G)|$ and (i) follows from Lemma~\ref{lem:fracChromatic}.
We now prove (ii). Let $\ell$ be an ordering of $H$ and let $h=\lfloor \frac{1}{\nu}(|E(G)|-\textnormal{cms}(G))\rfloor+1$. We will find a subordering $\ell''=(e_0,\ldots,e_{h-1})$ of $\ell$ so that, for each $i \in \mathbb{Z}_{h}$, the sequence of cyclically consecutive edges in $\ell$ that begins with $e_i$ and ends with $e_{i+1}$ contains a pair of adjacent edges. This will suffice to complete the proof of (ii) because $d_\ell(e_j,e_{j+1}) \leq \frac{1}{h}|E(H)|$ for some $j \in \mathbb{Z}_h$ since $\sum_{i \in \mathbb{Z}_h}d_\ell(e_i,e_{i+1})=|E(H)|$.
Let $\ell'$ be the subordering of $\ell$ induced by $G$.
There must be two adjacent edges $e_0$ and $e_1$ of $G$ at distance at most $\textnormal{cms}(G)$ in $\ell'$ by the definition of $\textnormal{cms}(G)$.
Because we are considering $\ell$ and $\ell'$ cyclically, we can assume without loss of generality that $\ell'(e_0)=0$ and $\ell'(e_1)=a$ for some $a \leq \textnormal{cms}(G)$.
Now define $(e_0,\ldots,e_{h-1})$ by, for each $i \in \{2,\ldots,h-1\}$, letting $e_{i}$ be the first edge after $e_{i-1}$ in $\ell'$ such that the sequence of consecutive edges in $\ell'$ that begins with $e_{i-1}$ and ends with $e_{i}$ contains a pair of adjacent edges. We claim that $(e_0,\ldots,e_{h-1})$ is a subordering of $\ell$ with the required properties.
To see this, first observe that $a+(h-1)\nu \leq |E(G)|$ by the definition of $h$.
Thus $(e_0,\ldots,e_{h-1})$ is indeed a subordering of $\ell$ because, for each $i \in \{1,\ldots,h-2\}$, we have that $\ell'(e_{i-1}) < \ell'(e_i) \leq a + (i-1)\nu$, by the definition of $\nu$.
In particular, $\ell'(e_{h-1}) \leq a+(h-2)\nu \leq |E(G)|-\nu$ and hence the sequence of cyclically consecutive edges in $\ell'$ that begins with $e_{h-1}$ and ends with $e_{0}$ contains a pair of adjacent edges. So $(e_0,\ldots,e_{h-1})$ is a subordering of $\ell$ with the required properties and (ii) is proved.
\end{proof}
Applying Lemma~\ref{lem:gen bound given subG}(i) to $2$-regular class 2 graphs we obtain the following.
\begin{cor}\label{cor:lower bound m-cycle}
Let $H$ be a $2$-regular class 2 graph, whose shortest odd cycle has length $m$. Then $\textnormal{cms}(H) \leq \frac{m-1}{2m}|E(H)|$.
\end{cor}
\begin{proof}
Let $G$ be a shortest odd length cycle in $H$. Then $|E(G)|=m$ and the matching number of $G$ is $\frac{m-1}{2}$, so the result follows by applying Lemma~\ref{lem:gen bound given subG}(i).
\end{proof}
To prove Theorem~\ref{thm:2 reg}, it remains to show that $\textnormal{cms}(H) \geq \lfloor\frac{n}{3}\rfloor$ for each $2$-regular graph $H$ of order $n$. In Lemma~\ref{lem:noSingle4Cycle} we establish a slightly stronger result for all $2$-regular graphs that do not contain exactly one $4$-cycle. For the remainder of this section it will be convenient to denote the number of edges in an ordering $\ell$ by $|\ell|$.
\begin{lem}\label{lem:noSingle4Cycle}
Let $G$ be a $2$-regular graph such that $G$ does not contain exactly one $4$-cycle. Then there is an ordering $\ell= \ell_0 \vee \ell_1 \vee \ell_2$ of $G$ such that
\begin{itemize}
\item[(i)]
the edges of $\ell_i$ form a matching of size $m_i$ for each $i \in \mathbb{Z}_3$, where $m_0,m_1,m_2$ are the unique non-negative integers such that $\lceil\frac{1}{3}|\ell|\rceil \geq m_0 \geq m_1 \geq m_2 \geq \lfloor\frac{1}{3}|\ell|\rfloor$ and $m_0+m_1+m_2=|\ell|$; and
\item[(ii)]
$d_{\ell}(e,e') \geq |\ell_j|$ for any $j \in \mathbb{Z}_3$ and pair $(e,e')$ of adjacent edges in $G$ such that $e$ is in $\ell_j$ and $e'$ is in $\ell_{j+1}$.
\end{itemize}
\end{lem}
\begin{proof}
We proceed by induction on the number of components of the graph. Let $H$ be a $2$-regular graph such that $H$ does not contain exactly one $4$-cycle. Let $m=|E(H)|$ and let $m_0,m_1,m_2$ be the unique non-negative integers such that $\lceil\frac{m}{3}\rceil \geq m_0 \geq m_1 \geq m_2 \geq \lfloor\frac{m}{3}\rfloor$ and $m_0+m_1+m_2=m$. If $H$ is connected or if $H$ contains no odd cycles, then by Theorem~\ref{thm: cycle seq} or Lemma~\ref{lem:good ord 2-colourable}, there is an ordering $\ell$ of $H$ such that $\textnormal{cms}(\ell) = \lfloor\frac{m-1}{2}\rfloor$. Choose $\ell_0, \ell_1, \ell_2$ arbitrarily so that $\ell= \ell_0 \vee \ell_1 \vee \ell_2$ and $|\ell_i|=m_i$ for each $i \in \mathbb{Z}_3$.
Because $m = 3$ or $m \geq 5$ we have $\textnormal{cms}(\ell) \geq \lfloor\frac{m-1}{2} \rfloor \geq \lceil\frac{m}{3}\rceil$ and it follows that the edges of $\ell_i$ form a matching for each $i \in \mathbb{Z}_3$ and that $\ell$ obeys (i) and (ii). So we may suppose that $H$ has $t \geq 2$ components at least one of which is an odd length cycle, and that the lemma holds for $2$-regular graphs with fewer than $t$ components. Let $t^*$ be the number of 4-cycles in $H$, noting that $t^* \neq 1$.
Our strategy will be as follows. We will first choose nonempty subgraphs $H'$ and $H''$ of $H$ such that $H$ is the vertex-disjoint union of $H'$ and $H''$. Let $m'=|E(H')|$, $m''=|E(H'')|$ and $m'_0,m'_1,m'_2$ be the unique nonnegative integers such that $\lceil\frac{m'}{3}\rceil \geq m'_0 \geq m'_1 \geq m'_2 \geq \lfloor\frac{m'}{3}\rfloor$ and $m'_0+m'_1+m'_2=m'$. We will then find orderings $\ell' = \ell_1' \vee \ell_2' \vee \ell_3'$ and $\ell'' = \ell_1'' \vee \ell_2'' \vee \ell_3''$ of $H'$ and $H''$, respectively, such that $\ell'$ obeys (i) and (ii) and the edges of $\ell''_i$ form a matching of size $m_i-m'_i$ for each $i \in \mathbb{Z}_3$. Finally, we will establish that the ordering $\ell = \ell_1 \vee \ell_2 \vee \ell_3$ of $H$, where $\ell_i=\ell_i' \vee \ell_{i}''$ for $i \in \mathbb{Z}_3$, obeys (i) and (ii).
For the rest of the proof we take $(e,e')$ to be an arbitrary pair of edges that are adjacent in $H$ and $j$ to be an element of $\mathbb{Z}_3$ such that $e \in \ell_j$ and $e' \in \ell_{j+1}$. Because $H$ is a vertex-disjoint union of $H'$ and $H''$, we have that either $e,e' \in E(H')$ or $e,e' \in E(H'')$. By our construction of $\ell$,
\begin{numcases}{d_{\ell}(e,e') \geq}
d_{\ell'}(e,e')+m_j-m'_j & \hbox{if $e,e' \in E(H')$} \label{eqn:2-reg class 2 ell'}
\\
d_{\ell''}(e,e')+m'_{j+1} & \hbox{if $e,e' \in E(H'')$} \label{eqn:2-reg class 2 ell''}
\end{numcases}
Because $\ell'$ will obey (ii), we will have for each $j \in \mathbb{Z}_3$ that $d_{\ell'}(e,e') \geq m'_j$ and hence $d_{\ell}(e,e') \geq m_j$ by \eqref{eqn:2-reg class 2 ell'} if $e,e' \in E(H')$. Thus, when checking that $\ell$ satisfies (ii), it will suffice to only consider the case $e,e' \in E(H'')$. So we henceforth assume that $e,e' \in E(H'')$.
We now describe how we choose $H'$ and $H''$ and how to find orderings $\ell'$ and $\ell''$ with the appropriate properties.
\begin{itemize}
\item
If $H$ has a cycle of length congruent to 0 modulo 3, choose $H'$ and $H''$ such that $H'$ is this cycle and $H$ is the vertex-disjoint union of $H'$ and $H''$. Note that $m'_i=\frac{1}{3}m'$ for each $i \in \mathbb{Z}_3$. Then $H'$ contains no $4$-cycle and $H''$ contains $t^*$ $4$-cycles (recall $t^*\neq 1$). By induction, take orderings $\ell' = \ell_1' \vee \ell_2' \vee \ell_3'$ of $H'$ and $\ell'' = \ell_1'' \vee \ell_2'' \vee \ell_3''$ of $H''$ that obey (i) and (ii). Because $\ell'$ obeys (i), $\ell'_i$ is an ordering of a matching of size $\frac{1}{3}m'$ for each $i \in \mathbb{Z}_3$. Because $\ell''$ obeys (i), $\ell''_i$ is an ordering of a matching of size $m_i-\frac{1}{3}m'=m_i-m'_i$ for each $i \in \mathbb{Z}_3$ and it can be seen that $\ell$ also obeys (i). Because $\ell''$ obeys (ii), we have $d_{\ell''}(e,e') \geq |\ell''_j|=m_j-\frac{1}{3}m'$ and it can be seen by \eqref{eqn:2-reg class 2 ell''} that $d_{\ell}(e,e') \geq m_j$ and hence that $\ell$ obeys (ii).
\item
Otherwise, choose $H'$ and $H''$ such that $H''$ is a single odd length cycle (recall $H$ contains at least one odd length cycle) and $H$ is the vertex-disjoint union of $H'$ and $H''$. Observe that $H''$ is not a $3$-cycle because we are not in the previous case, so $m'' \geq 5$ and $m''$ is odd. Then $H'$ contains $t^*$ $4$-cycles (recall $t^*\neq 1$) and, by induction, there is an ordering $\ell' = \ell_0' \vee \ell_1' \vee \ell_2'$ of $H'$ that obeys (i) and (ii). By Theorem~\ref{thm: cycle seq}, there is an ordering $\ell''$ of $H''$ such that $\textnormal{cms}(\ell'') = \lfloor\frac{1}{2}(m''-1) \rfloor$. Choose $\ell_0'', \ell_1'', \ell_2''$ arbitrarily so that $\ell''=\ell''_0 \vee \ell''_1 \vee \ell''_2$ and $|\ell''_i|=m_i-m'_i \leq \lceil\frac{1}{3}m''\rceil$ for each $i \in \mathbb{Z}_3$.
Because $m'' \geq 5$, we have $\textnormal{cms}(\ell'') \geq \lfloor\frac{1}{2}(m''-1) \rfloor \geq \lceil\frac{1}{3}m''\rceil$ and it follows that $\ell''_i$ is an ordering of a matching for each $i \in \mathbb{Z}_3$. Hence, because $\ell'$ obeys (i), we also have that $\ell$ obeys (i). By using $d_{\ell''}(e,e') \geq \lfloor\frac{1}{2}(m''-1) \rfloor$, $m''=m-m'$ and $m'' \geq 5$, we see that \eqref{eqn:2-reg class 2 ell''} implies that $d_{\ell}(e,e') \geq m_j$, and hence that $\ell$ obeys (ii) of the claim, provided that
\begin{equation}\label{eqn:2-reg class 2 mess}
\lfloor\tfrac{1}{2}(m-m'-1)\rfloor+m'_{j+1} \geq m_j
\end{equation}
holds for $3 \leq m' \leq m-5$. Let $\epsilon,\epsilon' \in \{-2,-1,0,1,2\}$ be such that $m_{j}=\frac{1}{3}(m+\epsilon)$ and $m'_{j+1}=\frac{1}{3}(m'+\epsilon')$. Because $m'_{j+1}$ and $m_j$ are integers, \eqref{eqn:2-reg class 2 mess} is equivalent to $m-m' > 2m_j-2m'_{j+1}$. Thus, substituting $m_{j}=\frac{1}{3}(m+\epsilon)$ and $m'_{j+1}=\frac{1}{3}(m'+\epsilon')$ and simplifying, \eqref{eqn:2-reg class 2 mess} is also equivalent to
\begin{equation}\label{eqn:2-reg class 2 epsilons}
m-m' > 2\epsilon-2\epsilon'.
\end{equation}
Clearly, \eqref{eqn:2-reg class 2 epsilons} holds when $m-m' \geq 9$, because $|\epsilon|,|\epsilon'| \leq 2$. This leaves the cases when $m-m'=5$ and $m-m'=7$, recalling that $m-m'=m''$ is odd and at least 5. If either $m \equiv 0 \mod{3}$ or $m' \equiv 0 \mod{3}$, then one of $\epsilon$ or $\epsilon'$ is 0 and hence \eqref{eqn:2-reg class 2 epsilons} holds. Thus, we can assume that $m \equiv 1 \mod{3}$ if $m-m' =5$ and $m \equiv 2 \mod{3}$ if $m-m' = 7$. In each of these cases it is now routine to check that \eqref{eqn:2-reg class 2 epsilons} holds, by considering subcases according to the value of $j$ (note that the values of $\epsilon$ and $\epsilon'$ are completely determined by the congruence class of $m$ modulo 3 and the value of $j$). \qedhere
\end{itemize}
\end{proof}
We are now ready to prove Theorem~\ref{thm:2 reg}.
\begin{proof}[\textbf{\textup{Proof of Theorem~\ref{thm:2 reg}.}}]
By Lemma~\ref{lem:class 1 2 regular cms}, we have $\textnormal{cms}_1(n,2)=\frac{n-2}{2}$ for each even $n \geq 4$. It remains to show that $\textnormal{cms}(n,2)=\lfloor\frac{n}{3}\rfloor$ for each integer $n \geq 6$. Let $n \geq 6$ be an integer. By Corollary~\ref{cor:lower bound m-cycle}, any $2$-regular graph $H$ of order $n$ containing a $3$-cycle has $\textnormal{cms}(H) \leq \lfloor\frac{n}{3}\rfloor$. Thus $\textnormal{cms}(n,2) \leq \lfloor\frac{n}{3}\rfloor$ and it suffices to show that each $2$-regular graph $H$ of order $n$ has $\textnormal{cms}(H) \geq \lfloor\frac{n}{3}\rfloor$.
Let $H$ be a $2$-regular graph of order $n$. If $H$ does not contain exactly one $4$-cycle, then the properties of the ordering $\ell$ of $H$ given by Lemma~\ref{lem:noSingle4Cycle} ensure that $\textnormal{cms}(\ell) \geq \lfloor\frac{n}{3}\rfloor$. Thus, we may assume that $H$ contains exactly one $4$-cycle. Say $H$ is the vertex-disjoint union of $H'$ and $H''$, where $H''$ is the $4$-cycle. Let $e_0,e_1,e_2,e_3$ be the edges of $H''$ so that $e_0,e_2$ and $e_1,e_3$ each form a matching. Let $\ell' = \ell'_0 \vee \ell'_1 \vee \ell'_2$ be an ordering of $H'$ given by Lemma~\ref{lem:noSingle4Cycle}. Let $e^*$ be the last edge in $\ell'_1$ and let $\ell^*_1$ be the ordering obtained from $\ell'_1$ by removing $e^*$. Let
\[\ell=
\left\{
\begin{array}{ll}
\ell'_0 \vee (e_0) \vee \ell'_1 \vee (e_2) \vee \ell'_2 \vee (e_1,e_3) & \hbox{if $|E(H')| \equiv 0, 1 \mod{3}$} \\
\ell'_0 \vee (e_0) \vee \ell^*_1 \vee (e_2,e^*) \vee \ell_2' \vee (e_1,e_3) & \hbox{if $|E(H')| \equiv 2 \mod{3}$.}
\end{array}
\right.
\]
It is now routine to use the fact that $\ell'$ satisfies (i) and (ii) of Lemma~\ref{lem:noSingle4Cycle} to check that any pair of edges adjacent in $H'$ or in $H''$ are at distance at least $\lfloor\frac{n}{3}\rfloor$ in $\ell$ and hence that the lemma holds. Note that the fact that $\ell'$ satisfies (i) and (ii) of Lemma~\ref{lem:noSingle4Cycle} implies that $e^*$ is not adjacent in $H'$ to any edge in $\ell'_2$ when $|E(H')| \equiv 2 \mod{3}$.
\end{proof}
\section{Upper and lower bounds for general graphs}\label{genBounds}
In this section we find some upper and lower bounds on the cyclic matching sequenceability of general (possibly non-regular) graphs. In particular we will prove Theorem~\ref{thm:gen bounds}. We employ an easily proved result from \cite{McD}. We say that a matching decomposition of a graph is \emph{equitable} if the sizes of any two of the matchings differ by at most 1.
\begin{lem}[\cite{McD}]\label{lem:colEq}
Let $G$ be a graph with chromatic index $c$. For any $t \geq c$, there is an equitable matching decomposition of $G$ with $t$ matchings.
\end{lem}
Our next result establishes the lower bound in Theorem~\ref{thm:gen bounds}.
\begin{lem}\label{lem:gen lower bound}
For any graph $H$ with chromatic index $c$,
$\textnormal{cms}(H) \geq \lfloor\frac{1}{2c}|E(H)| \rfloor-1$.
\end{lem}
\begin{proof}
Let $H$ be a graph. Let $c$ be the chromatic index of $H$, $m=|E(H)|$ and $t = \lfloor\frac{m}{2c}\rfloor$. When $c=1$, $H$ is a matching and the result is trivial, so we can assume that $c\geq 2$. By Lemma~\ref{lem:colEq}, there is a matching decomposition $\{H_0,\ldots,H_{c-1}\}$ of $H$ such that $|E(H_i)| \geq \lfloor\frac{m}{c}\rfloor \geq 2t$ for each $i \in \mathbb{Z}_{c}$. For each $i \in \mathbb{Z}_{c}$, let $H_i'$ and $H_i''$ be vertex-disjoint subgraphs of $H_i$, each with $t$ edges, and note that by Lemma~\ref{lem:2 edge colourable} there are orderings $\ell_i''$ of $H_{i}''$ and $\ell_{i+1}'$
of $H_{i+1}'$ such that $\textnormal{ms}(\ell_{i}''\vee \ell_{i+1}') \geq t - 1$. For each $i \in \mathbb{Z}_{c}$, let $\ell_i=\ell'_i \vee \ell^*_i \vee \ell''_i$ be an ordering of $H_i$, where $\ell^*_i$ is an arbitrary (possibly empty) ordering of the edges in $E(H_i) \setminus E(H_i' \cup H_i'')$. Let $\ell=\ell_1 \vee \cdots \vee \ell_{c}$. We complete the proof by showing that $\textnormal{cms}(\ell) \geq t-1$.
Let $e,e'$ be adjacent edges in $H$. Obviously $d_{\ell}\{e,e'\} \geq t-1$ if it is not the case that both $e$ and $e'$ are in $H''_j \cup H'_{j+1}$ for some $j \in \mathbb{Z}_c$. But if both $e$ and $e'$ are in $H''_j \cup H'_{j+1}$, then $d_{\ell}\{e,e'\} \geq t-1$ because $\textnormal{ms}(\ell_{j}''\vee \ell_{j+1}') \geq t - 1$.
\end{proof}
For each $k \geq 2$, we define a graph $B_k$ with maximum degree $k$. If $k$ is even, let $B_k$ be a complete graph on $k+1$ vertices and, if $k$ is odd, let $B_k$ be the graph on $k+2$ vertices whose complement is the vertex-disjoint union of a path with $3$ vertices and a matching with $k-1$ vertices. In particular, the following depicts $B_3$.
\begin{center}
\begin{tikzpicture}[thick,scale= .8]
\pgfmathsetlengthmacro\scfac{2.5cm}
\pgfmathsetmacro{\sepang}{120}
\pgfmathtruncatemacro{\c}{3}
\draw[line width = \scfac*0.02, color = blue]{
(0+90:\scfac) -- (90+90:\scfac)
(0+90:\scfac) -- (270+90:\scfac)
(90+90:\scfac) -- (180+90:\scfac)
[bend left] (90+90:\scfac) -- node[pos = 0.5,above]{}(270+90:\scfac)
(180+90:\scfac) -- (270+90:\scfac)
[bend left] (0+90:\scfac) to ($(0:\scfac)+(270+90:\scfac)$)
[bend right] (180+90:\scfac) to ($(0:\scfac)+(270+90:\scfac)$)
};
\draw[line width = \scfac*0.015, color = black!50]{
(0+90:\scfac) node[anchor=north,yshift=\scfac*0.3,color = black]{0} node[circle, draw, fill=black!10,inner sep=\scfac*0.015, minimum width=\scfac*0.08] {}
(90+90:\scfac) node[anchor=east,xshift=-\scfac*0.1,color = black]{2}node[circle, draw, fill=black!10,inner sep=\scfac*0.015, minimum width=\scfac*0.08] {}
(180+90:\scfac) node[anchor=north,yshift=-\scfac*0.1,color = black]{3} node[circle, draw, fill=black!10,inner sep=\scfac*0.015, minimum width=\scfac*0.08] {}
(270+90:\scfac) node[circle, draw, fill=black!10,inner sep=\scfac*0.015, minimum width=\scfac*0.08] {}
(270+90:\scfac) node[anchor=west,xshift=\scfac*0.1,color = black]{1}
(0:\scfac)+(270+90:\scfac) node[anchor=west,xshift=\scfac*0.1,color = black]{4}node[circle, draw, fill=black!10,inner sep=\scfac*0.015, minimum width=\scfac*0.08] {}
};
\end{tikzpicture}
\end{center}
It is easy to check that when $k$ is odd $B_k$ has $\frac{1}{2}(k^2+2k-1)$ edges
and each of its vertices has degree $k$ except for one that has degree $k-1$.
Of course, when $k$ is even $B_k$ has $\frac{1}{2}k(k+1)$ edges and each of its vertices has degree $k$.
We will complete the proof of Theorem~\ref{thm:gen bounds} by showing that a graph $H$ containing $B_k$ as a subgraph has cyclic matching sequenceability at most $\frac{1}{k+1}{|E(H)|}$. We will make use of the following facts about $B_k$ for odd integers $k$.
\begin{lem}\label{lem:prop of bad sg}
For each odd $k \geq 3 $, $\textnormal{cms}(B_k)\leq \frac{k-1}{2}$ and $B_k$ has matching number $\frac{k+1}{2}$.
\end{lem}
\begin{proof}
Let $\nu$ be the matching number of $B_k$. It is easy to see that $\nu = \frac{k+1}{2}$, because $B_k$ has $k+2$ vertices, $k$ is odd, and it is easy to find a matching of size $\frac{k+1}{2}$ in $B_k$. So it remains to show that $\textnormal{cms}(B_k)\leq \frac{k-1}{2}$.
As the matching number of a graph $G$ is clearly an upper bound for $\textnormal{cms}(G)$, we only need to show that $\textnormal{cms}(B_k)\neq \frac{k+1}{2}$.
Suppose for a contradiction that $\ell$ is an ordering of $B_k$ such that $\textnormal{cms}(\ell)= \frac{k+1}{2}$.
Let $v$ be the vertex of $B_k$ with degree $k-1$ and let $e_0 ,\ldots ,e_{k-2}$ be the edges of $B_k$ incident with $v$, where $\ell(e_i) < \ell (e_{i+1})$ for all $i \in \{0,\ldots ,k-3\}$.
Clearly $\sum_{i \in \mathbb{Z}_{k-1}}d_{\ell}(e_i,e_{i+1}) = |E(B_k)|$. So, for some $i \in \mathbb{Z}_{k-1}$ we have
\[
d_{\ell}(e_i,e_{i+1}) \geq \left\lceil\mfrac{|E(B_k)|}{k-1} \right\rceil =
\left\lceil\mfrac{(k-1)(k+3)+2}{2(k-1)}\right\rceil = \mfrac{k+5}{2}.
\]
Therefore, in $\ell$, there are $\frac{k+3}{2}$ cyclically consecutive edges $e_{0}',\ldots, e_{(k+1)/2}'$
between $e_i$ and $e_{i+1}$, none of which are incident with $v$. As $\textnormal{cms}(\ell)= \frac{k+1}{2}$, we have that $e_{0}',\ldots, e'_{(k-1)/2}$ and $e_{1}',\ldots, e_{(k+1)/2}'$
must each form a matching of $B_k$. However, the two matchings so formed both have vertex set $V(H) \setminus \{v\}$, and they share $\frac{k-1}{2}$ edges. This is impossible, and we conclude that
$\textnormal{cms}(B_k) \leq \frac{k-1}{2}$.
\end{proof}
\begin{thm}\label{thm:count exa}
Let $k \geq 2$ be an integer. Then $\textnormal{cms}(H) \leq \frac{1}{k+1}{|E(H)|}$ for any graph $H$ that has $B_k$ as a subgraph. Furthermore, for all integers $n \geq 3k+5$ such that $nk$ is even, there is
a $k$-regular graph on $n$ vertices with $B_k$ as a subgraph.
\end{thm}
\begin{proof}
Let $\nu$ be the matching number of $B_k$. When $k$ is even, $|E(B_k)|=\frac{1}{2}k(k+1)$, $\nu = \frac{k}{2}$ and the result follows by Lemma~\ref{lem:gen bound given subG}(i). When $k$ is odd, $|E(B_k)|=\frac{1}{2}(k^2+2k-1)$ and, by Lemma~\ref{lem:prop of bad sg}, $\textnormal{cms}(B_k) \leq \frac{k-1}{2}$ and $\nu = \frac{k+1}{2}$. So $|E(B_k)|-\textnormal{cms}(B_k) \geq \frac{1}{2}k(k+1)$ and the result can be seen to follow from Lemma~\ref{lem:gen bound given subG}(ii).
Finally we show a $k$-regular graph on $n$ vertices with $B_k$ as a subgraph exists for any $n \geq 3k+4$ such that $nk$ is even.
If $k$ is even, then for any $n \geq 2k+2$, a graph that is the vertex-disjoint union of $K_{k+1}$ and a $k$-regular graph on $n-(k+1)$ vertices is a $k$-regular graph on $n$ vertices with $B_k$ as a subgraph. If $k$ is odd, then let $B'_k$ be the $k$-regular graph on $2k+4$ vertices that is formed by taking two vertex-disjoint copies of $B_k$ and adding an edge incident with the vertex of degree $k-1$ in each copy. For any $n \geq 3k+5$, a graph that is the vertex-disjoint union of $B'_{k}$ and a $k$-regular graph on $n-(2k+4)$ vertices is a $k$-regular graph on $n$ vertices with $B_k$ as a subgraph.
\end{proof}
Note that, for any $k\geq 2$, a $k$-regular graph containing $B_k$ is necessarily class 2. To see this let $x=|V(B_k)|$ and note that $x$ is odd and hence any matching in $B_k$ has size at most $\frac{x-1}{2}$. But, for both $k$ odd and $k$ even, $|E(B_k)| > \frac{k(x-1)}{2}$ and hence $B_k$ does not have a $k$-edge colouring.
\begin{proof}[\textbf{\textup{Proof of Theorem~\ref{thm:gen bounds}.}}]
The upper and lower bounds on $\textnormal{cms}(G)$ are established in Lemmas~\ref{lem:fracChromatic} and \ref{lem:gen lower bound}, respectively. Let $\Delta \geq 2$ and $n \geq \Delta+1$ be integers. If $n=\Delta+1$, then $K_{n}$ satisfies $\textnormal{cms}(K_n) \leq \frac{1}{\Delta+1}|E(K_{n})|$ by the result of \cite{MR2961987} mentioned in the introduction. If $n \geq \Delta+2$, then there is clearly a graph $G$ of order $n$ with maximum degree $\Delta$ that contains $B_\Delta$ as a subgraph. Then $\textnormal{cms}(G) \leq \frac{1}{\Delta+1}|E(G)|$ by Theorem~\ref{thm:count exa}.
\end{proof}
\section{Ordering with a given partition}\label{givenPartition}
Let $H$ be a $k$-regular graph with $n$ vertices and chromatic index $c$. In order to establish the lower bound in Theorem~\ref{thm:k reg}, we will construct an ordering $\ell$ of $H$ via a two stage process. In the first stage we will find a partition $\{X_i,Y_i,Z_i: i \in \mathbb{Z}_{c}\}$ of $E(H)$. In the second stage we will, for each $i \in \mathbb{Z}_{c}$, find orderings $\ell_{X_i}$, $\ell_{Y_i}$, $\ell_{Z_i}$ of $X_i$, $Y_i$, $Z_i$, respectively, then take
\[\ell= \ell_{Z_0} \vee \ell_{Y_0} \vee \ell_{X_0} \vee \ell_{Z_1} \vee \ell_{Y_1} \vee \ell_{X_1} \vee \cdots\cdots \vee \ell_{Z_{c-1}} \vee \ell_{Y_{c-1}} \vee \ell_{X_{c-1}}\]
and show that $\ell$ has the required matching sequenceability. In this section we detail how to construct the ordering $\ell$ given a partition of $E(H)$ with certain desirable properties. In Section~\ref{findPartition} we will establish that a partition with such properties does indeed exist.
Let $X_i \subseteq E(H_i)$ for all $i \in \mathbb{Z}_c$.
For each $i \in \mathbb{Z}_c$, we say
that a vertex $v$ in $V(H)$ is {\em $i$-covered for $\{X_0,\ldots , X_{c-1}\}$}
if $v$ is adjacent to an edge in $X_{i}$ and either there is an edge in $X_{i+1}$
that is also adjacent to $v$ or no edge in $H_{i+1}$ is adjacent to $v$.
For a graph $H$ and nonnegative integers $x$ and $w$, we say that a partition $\{X_i,Y_i,Z_i: i \in \mathbb{Z}_{c}\}$ of $E(H)$ is a \emph{$(x,w)$-partition} of $H$ if it obeys the following conditions.
\begin{itemize}[itemsep=0mm]
\item[(P1)]
$w \leq \frac{3}{2}x$ and $x+2y \leq \lfloor \frac{1}{c}|E(H)| \rfloor$ where $y = \lceil 3x-\frac{3}{2}w\rceil$.
\item[(P2)]
$\{H_i: i \in \mathbb{Z}_{c}\}$ is an equitable matching decomposition of $H$, where $H_i$ is subgraph of $H$ with edge set $X_i \cup Y_i \cup Z_i$ for each $i \in \mathbb{Z}_c$.
\item[(P3)]
$|X_i| = x$ and $|Y_i| = y$ for all $i \in \mathbb{Z}_{c}$.
\item[(P4)]
No edge in $X_i$ is adjacent to an edge in $Z_{i+1}$ for all $i \in \mathbb{Z}_{c}$.
\item[(P5)]
For all $i \in \mathbb{Z}_{c}$, $|Y'_i| \leq \frac{y}{3}$ and each edge in $Y'_i$ is adjacent to at most one edge in $Z_{i+1}$, where $Y_i'$ is the set of edges in $Y_i$ that are adjacent to exactly two edges in $X_{i-1}$.
\item[(P6)]
For all $i \in \mathbb{Z}_{c}$, there are at least $w$ vertices of $H$ that are $i$-covered for $\{X_0,\ldots,X_{c-1}\}$.
\end{itemize}
We treat an $(x,w)$-partition as including a specification of which of its sets plays the role of $X_i$, $Y_i$ and $Z_i$ for each $i \in \mathbb{Z}_{c}$. We will refer to these properties simply as (P1), (P2), \ldots, (P6) throughout the rest of the section and in the next section. When an $(x,w)$-partition is defined we will use $y$ and $Y'_i$ in the roles they play in (P1) and (P5) without explicitly defining them each time. Note that $|E(H_i)| \in \{\lfloor \frac{1}{c}|E(H)| \rfloor,\lceil \frac{1}{c}|E(H)| \rceil\}$ for each $i \in \mathbb{Z}_c$ because $\{H_i: i \in \mathbb{Z}_{c}\}$ is an equitable matching decomposition of $H$ by (P2). Thus, it follows from (P1) -- (P3) that $|Z_i|=|E(H_i)|-x-y\geq y$ for each $i \in \mathbb{Z}_c$.
Our goal for the rest of the section is to establish Proposition~\ref{prop:seq given some conds} which guarantees a lower bound on the cyclic matching sequenceability of a graph that admits an $(x,w)$-partition. Our next results, Lemmas~\ref{lem:mid tran to top gen k} and \ref{lem:bot tran to mid and bot gen k}, are used only in the proof of Proposition~\ref{prop:seq given some conds}. In Lemma~\ref{lem:mid tran to top gen k}, we define orderings of $Y_i$ and $Z_i$ for each $i \in \mathbb{Z}_c$ and then, based on these, in Lemma~\ref{lem:bot tran to mid and bot gen k} we determine orderings of $X_i$ for each $i \in \mathbb{Z}_c$.
\begin{lem}\label{lem:mid tran to top gen k}
Let $H$ be a graph and let $\{X_i,Y_i,Z_i: i \in \mathbb{Z}_{c}\}$ be a $(x,w)$-partition of $H$. For all $i\in \mathbb{Z}_{c}$, there are orderings $\ell_{Y_i}$ and $\ell_{Z_i}$ of $Y_i$ and $Z_i$ so that $\textnormal{ms}(\ell_{Y_i} \vee \ell_{Z_{i+1}}) \geq y-1$ and, in $\ell_{Y_i}$, the edges in $Y_i'$ are the last to occur.
\end{lem}
\begin{proof}
Fix an arbitrary $i\in \mathbb{Z}_{c}$. We will define orderings of $Y_i$ and $Z_{i+1}$.
As discussed above, (P1) -- (P3) imply that $|Z_i| \geq y$.
By (P5), $|Y_i'| \leq \frac{y}{3}$ and each edge of $Y_i'$ is adjacent to at most one edge in $Z_{i+1}$. Thus, we can choose a subset $Z_{i+1}'$ of $Z_{i+1}$ such that $|Z_{i+1}'|=|Y'_i|$ and $Z_{i+1}'$ includes every edge in $Z_{i+1}$ that is adjacent to an edge in $Y_i'$. Let $y'=|Y'_i|$. Choose an arbitrary ordering $\ell_{Y'_i}=(e_0,\ldots,e_{y'-1})$ of the edges in $Y'_i$. Because each edge of $Y_i'$ is adjacent to at most one edge in $Z_{i+1}$, we can now choose an ordering $\ell_{Z'_{i+1}}=(e^*_0,\ldots,e^*_{y'-1})$ of the edges in $Z_{i+1}'$ such that, for each $j \in \mathbb{Z}_{y'}$, either $e^*_j$ is not adjacent to any edge in $\ell_{Y'_i}$ or $e_j$ is the last of the (at most two) edges in $\ell_{Y'_i}$ adjacent to $e^*_j$. Clearly then,
\begin{equation}\label{eqn:topOrderings1}
\textnormal{ms}(\ell_{Y'_i} \vee \ell_{Z'_{i+1}}) \geq y'.
\end{equation}
Let $Y_i'' =Y_i\setminus Y_{i}' $, $Z_{i+1}'' = Z_{i+1}\setminus Z_{i+1}'$ and $y''=|Y_i''|$.
We have seen that (P1) -- (P3) imply $|Z_{i+1}|=|E(H_{i+1})|-x-y \geq y$ and thus, subtracting $y'$ from both sides, we have $|Z_{i+1}''| \geq y''$ and so we can find a subset $W$ of $Z_{i+1}''$
such that $|W|=y''$.
By Lemma~\ref{lem:2 edge colourable}, there are orderings $\ell_{Y''_i}$ and $\ell_{W}$ of the matchings formed by the edges of $Y_{i}''$ and the edges of $W$, respectively,
such that $\textnormal{ms}(\ell_{Y''_i} \vee \ell_{W}) \geq y''-1$.
Let $\ell_{Z''_{i+1}} = \ell_{W}\vee \ell_R$ where $\ell_R$ is an arbitrary ordering of the edges in $Z_{i+1}''\setminus W$.
Clearly then,
\begin{equation}\label{eqn:topOrderings2}
\textnormal{ms}(\ell_{Y''_i} \vee \ell_{Z''_{i+1}}) \geq y''-1.
\end{equation}
By the definition of $Z_{i+1}''$, no edge in it is adjacent to an edge in $Y'_i$. Thus, by Lemma~\ref{lem:adding 4 special orderings},
\[
\textnormal{ms}\left(\ell_{Y''_i}\vee \ell_{Y'_i} \vee \ell_{Z''_{i+1}} \vee \ell_{Z'_{i+1}}\right) \geq y'+y''-1 = y-1\,,
\]
where we have used the facts that $\textnormal{ms}(\ell_{Y''_i} \vee \ell_{Z''_{i+1}})+|Y'_i|\geq y''-1+y'$ by \eqref{eqn:topOrderings2}, that $\textnormal{ms}(\ell_{Y'_i} \vee \ell_{Z'_{i+1}})+|Z_{i+1}''| \geq y'+y''$ by \eqref{eqn:topOrderings1} and $|Y'_i|+|Z_{i+1}''| \geq y'+y''$.
So let $\ell_{Y_i} = \ell_{Y''_i} \vee \ell_{Y'_i}$ and $\ell_{Z_{i+1}} = \ell_{Z''_{i+1}}\vee \ell_{Z'_{i+1}}$, and note we have shown that these orderings satisfy $\textnormal{ms}(\ell_{Y_i} \vee \ell_{Z_{i+1}}) \geq y-1$. By applying this procedure for each $i \in \mathbb{Z}_{c}$, the lemma is proved.
\end{proof}
\begin{lem}\label{lem:bot tran to mid and bot gen k}
Let $H$ be a graph and let $\{X_i,Y_i,Z_i: i \in \mathbb{Z}_{c}\}$ be a $(x,w)$-partition of $H$. For all $i\in \mathbb{Z}_{c}$, let $\ell_{Y_i}$ and $\ell_{Z_i}$ be orderings of $Y_i$ and $Z_i$ that satisfy the conditions of Lemma~\ref{lem:mid tran to top gen k}. Then, for all $i \in \mathbb{Z}_{c}$, there is an ordering $\ell_{X_i}$ of $X_i$ such that $\textnormal{ms}(\ell_{X_i} \vee \ell_{Y_{i+1}}) \geq x$ and
$\textnormal{ms}(\ell_{X_i} \vee \ell_{X_{i+1}}) \geq x-y$.
\end{lem}
\begin{proof}
For each $i\in \mathbb{Z}_{c}$, we will find a subset $X'_i$ of $X_i$ such that $|X_i'| = \min\{x,\lfloor \frac{2y}{3} \rfloor\}$ and $X'_i$ includes every edge of $X_i$ that is adjacent to an edge in $Y'_{i+1}$, and then construct an ordering $\ell_{X'_i}$ of $X_i'$ for each $i\in \mathbb{Z}_{c}$. Once this is accomplished we will then find an ordering $\ell_{X''_i}$ of $X_i''=X_i \setminus X'_i$ for each $i\in \mathbb{Z}_{c}$, and show that the orderings $\ell_{X_i}=\ell_{X'_i}\vee \ell_{X''_i}$ satisfy the conditions of the lemma. Note that if $x \leq \lfloor \frac{2y}{3} \rfloor$, then the $X_i''$ will be empty and the ordering $\ell_{X''_i}$ will be trivial.
Let $j$ be an arbitrary element of $\mathbb{Z}_{c}$. Let $y_1$ be the number of edges of $Y_{j+1}$ adjacent to exactly one edge in $X_j$ and $y_2$ be the number of edges of $Y_{j+1}$ adjacent to exactly two edges in $X_j$, and note that $y_2=|Y_{j+1}'|$ by the definitions of $y_2$ and $|Y_{j+1}'|$. Consider the number of vertices incident with both an edge in $X_j$ and an edge in $Y_{j+1}$. Because $X_j$ is a matching, each edge in $Y_{j+1}$ is adjacent to at most two of its edges, and hence this number is $y_1+2y_2$. On the other hand, by (P6), this number is at most $2x-w$, where we note that $|X_j|=x$ and that the edges of $Y_{j+1} \cup X_{j+1}$ form a matching. It thus follows from $y = \lceil 3x-\frac{3}{2}w\rceil$ that
\begin{equation}\label{eqn:y1y2Bound}
y_1+2y_2 \leq 2x-w \leq \mfrac{2y}{3}.
\end{equation}
Thus, we can choose a subset $X'_j$ of $X_j$ such that $|X_j'| = \min\{x,\lfloor \frac{2y}{3} \rfloor\}$ and $X'_j$ includes every edge of $X_j$ that is adjacent to an edge in $Y_{j+1}$ (if $x \leq \lfloor \frac{2y}{3} \rfloor$ then we choose $X'_j=X_j$). Further, because the last edges of $\ell_{Y_{j+1}}$ are those in $Y_{j+1}'$ and $y \geq y_1+2y_2$ by \eqref{eqn:y1y2Bound}, we can apply Lemma~\ref{lem:ord given other ord} to obtain an ordering $\ell_{X'_j}$ of $X_j'$ such that
\begin{equation}\label{eqn:bottomOrderings1}
\textnormal{ms}(\ell_{X'_j} \vee \ell_{Y_{j+1}}) \geq |X'_j|.
\end{equation}
Thus, for each $i\in \mathbb{Z}_{c}$, we can take such an ordering $\ell_{X'_i}$ of $X_i'$, let $\ell_{X_i}=\ell_{X'_i}$ if $x \leq \lfloor \frac{2y}{3} \rfloor$ and let $X''_i=X_i \setminus X'_i$ otherwise. If $x \leq \lfloor \frac{2y}{3} \rfloor$ then this completes the proof of the lemma, using \eqref{eqn:bottomOrderings1} and the fact that $x \leq y$. Thus, we may assume that $x > \lfloor \frac{2y}{3} \rfloor$. For each $i\in \mathbb{Z}_{c}$ by Lemma~\ref{lem:ord given other ord bad case} there is an ordering $\ell_{X''_i}$ of $X''_i$ such that
\begin{equation}\label{eqn:bottomOrderings2}
\textnormal{ms}\left(\ell_{X''_i} \vee \ell_{X'_{i+1}}\right) \geq \tfrac{1}{2}|X_i''| = \tfrac{1}{2}\left(x-\big\lfloor\tfrac{2y}{3}\big\rfloor\right) \geq x-y
\end{equation}
where the last inequality follows because $y \geq \frac{3}{4}x$ since $y = \lceil 3x-\frac{3}{2}w\rceil$ and $w \leq \frac{3}{2}x$ by (P1).
Again, let $j$ be an arbitrary element of $\mathbb{Z}_{c}$. As no edge in $X_j''$ is adjacent to an edge in $Y_{j+1}$, we have from \eqref{eqn:bottomOrderings1} that
\[\textnormal{ms}(\ell_{X'_j}\vee \ell_{X''_j} \vee \ell_{Y_{j+1}}) \geq |X'_j|+|X''_j| = x.\]
Obviously, $|X_j''| = x- \lfloor\frac{2y}{3}\rfloor \geq x-y$ and, because $y \geq \frac{3x}{4}$ by (P1), we have $|X_{j+1}'| = \lfloor\frac{2y}{3} \rfloor\geq x-y$. Therefore, by \eqref{eqn:bottomOrderings2},
\[\textnormal{ms}\left(\ell_{X'_j}\vee \ell_{X''_j} \vee \ell_{X'_{j+1}}\vee \ell_{X''_{j+1}}\right) \geq x-y.\]
Thus, the orderings $\ell_{X_i}=\ell_{X'_i}\vee \ell_{X''_i}$ for $i\in \mathbb{Z}_{c}$ satisfy the required properties.
\end{proof}
\begin{prop}\label{prop:seq given some conds}
If $H$ is a graph that has a $(x,w)$-partition for some nonnegative integers $x$ and $w$, then $\textnormal{cms}(H) \geq x+y-1$.
\end{prop}
\begin{proof}
Let $H$ be a graph and let $\{X_i,Y_i,Z_i: i \in \mathbb{Z}_{c}\}$ be a $(x,w)$-partition of $H$. By Lemmas~\ref{lem:mid tran to top gen k} and~\ref{lem:bot tran to mid and bot gen k} there are, for each $i \in \mathbb{Z}_{c}$, orderings $\ell_{X_i}, \ell_{Y_i}, \ell_{Z_i}$ of $X_i, Y_i, Z_i$, respectively, such that $\textnormal{ms}(\ell_{Y_i} \vee \ell_{Z_{i+1}}) \geq y-1$, $\textnormal{ms}(\ell_{X_i} \vee \ell_{Y_{i+1}}) \geq x$ and $\textnormal{ms}(\ell_{X_i} \vee \ell_{X_{i+1}}) \geq x-y$. Let $\ell_i = \ell_{Z_i} \vee \ell_{Y_i} \vee \ell_{X_i}$ for each $i \in \mathbb{Z}_{c}$. Now let $i$ be an arbitrary element of $\mathbb{Z}_{c}$. By Proposition~\ref{prop: Matching decomposition}, it suffices to show that $\textnormal{ms}(\ell_{i} \vee \ell_{i+1}) \geq x+y-1$.
We have $\textnormal{ms}(\ell_{Y_i} \vee \ell_{Z_{i+1}}) \geq y-1$. So, because $|Y_i| =y$ and $ Y_i \cup Z_i$ is a matching, we have
\begin{equation}\label{eqn:allOrderings1}
\textnormal{ms}(\ell_{Z_i} \vee \ell_{Y_i} \vee\ell_{Z_{i+1}} ) \geq y-1.
\end{equation}
We also have $\textnormal{ms}(\ell_{X_i} \vee \ell_{Y_{i+1}}) \geq x$ and $\textnormal{ms}(\ell_{X_i} \vee \ell_{X_{i+1}}) \geq x-y$.
Thus, because $\textnormal{ms}(\ell_{Y_{i+1}} \vee \ell_{X_{i+1}}) =y+x$, Lemma~\ref{lem:adding three orderings} implies that
\begin{equation}\label{eqn:allOrderings2}
\textnormal{ms}(\ell_{X_i} \vee \ell_{Y_{i+1}} \vee \ell_{X_{i+1}}) \geq \min\{x,x+y,y+(x-y) \} =x.
\end{equation}
By (P4), the edges of $X_{i} \cup Z_{i+1}$ form a matching. Thus, applying Lemma~\ref{lem:adding 4 special orderings} with $M_1=Z_i \cup Y_i$, $M_2=X_{i}$, $M_3=Z_{i+1}$, $M_4=Y_{i+1} \cup X_{i+1}$, and using \eqref{eqn:allOrderings1} and \eqref{eqn:allOrderings2}, we have
\[
\textnormal{ms}(\ell_{i} \vee \ell_{i+1})
\geq
\min\{y-1+x,x+|Z_{i+1}|,\textnormal{ms}(\ell_{Z_i} \vee \ell_{Y_i} \vee \ell_{Y_{i+1}} \vee \ell_{X_{i+1}})+x+|Z_{i+1}| \} = x+y-1,
\]
where the last inequality holds because $|Z_{i+1}| \geq y$ which we have seen follows from (P1) -- (P3).
\end{proof}
\section{Finding a good partition}\label{findPartition}
In this section we prove Theorem~\ref{thm:k reg} by
establishing the existence of $(x,w)$-partitions in $k$-regular graphs with $k\geq 3$.
Let $H$ be a $k$-regular graph and $\{H_0,\ldots,H_{c-1}\}$
be an equitable matching decomposition of $H$.
Then, we call $\{X_0,\ldots, X_{c-1} \}$ an {\em $(x,w)$-semipartition with respect to $\{H_0,\ldots,H_{c-1}\}$}
if $X_i \subseteq E(H_i)$, $|X_i|=x$ for each $i \in \mathbb{Z}_c$ and $\{X_0,\ldots, X_{c-1} \}$ obeys (P6) for $w$. Our strategy is to first establish that it is possible to extend an $(x,w)$-semipartition
to an $(x,w)$-partition in Lemma~\ref{lem:can make T with S}, then to exhibit $(x,w)$-semipartitions
(using several different methods) in Lemmas~\ref{lem:gen blob construction}--\ref{lem:prob cons}.
In Lemmas~\ref{thm:main thm explicit con} and \ref{thm:prob con}, we then prove the lower bounds of Theorem~\ref{thm:k reg}, using Proposition~\ref{prop:seq given some conds}. Finally, we prove Theorem~\ref{thm:k reg}.
\begin{lem}\label{lem:can make T with S}
Let $k \geq 3$ be an integer, let $H$ be a $k$-regular graph with $n$ vertices, and let $\{H_0,\ldots,H_{c-1}\}$ be an equitable matching decomposition of $H$.
Let $\{X_0,\ldots, X_{c-1}\}$ be an $(x,w)$-semipartition.
If $x$ and $w$ satisfy \textup{(P1)}, then there exists
an $(x,w)$-partition of $H$.
\end{lem}
\begin{proof}
For $i \in \mathbb{Z}_c$, let $T_i$ be the set of edges in $E(H_i)\setminus X_i$ that are adjacent to exactly one edge in $X_{i-1}$ and let $T'_i$ be the edges in $E(H_i)\setminus X_i$ that are adjacent to exactly two edges in
$X_{i-1}$.
There are $2x$ vertices that are incident with an edge in $X_{i-1}$. Of these $2x$ vertices, $|T_i|$ are incident with an edge in $T_i$, $2|T'_i|$ are incident with an edge in $T'_i$, and by (P6) at least $w$
are incident with an edge in $X_{i}$ or have no edge of $H_i$ incident with them. Thus,
\begin{equation}\label{eqn:deg count}
w+|T_i|+2|T'_i| \leq 2x \,.
\end{equation}
For all $i \in\mathbb{Z}_{c}$, we construct sets $T''_{i+1}$ with the following properties.
\begin{itemize}
\item[(i)]
The set $T''_{i+1}$ is a subset of $E(H_{i+1})\setminus (X_{i+1} \cup T_{i+1}\cup T'_{i+1})$.
\item[(ii)]
For each edge in $T_i'$ that is adjacent to two edges in $E(H_{i+1})\setminus (X_{i+1} \cup T_{i+1}\cup T'_{i+1})$, at least one of these latter two edges is in $T''_{i+1}$.
\end{itemize}
Let $j$ be an arbitrary element of $\mathbb{Z}_{c}$. For each $e \in T_j'$ that is adjacent to two edges in $E(H_{j+1})\setminus (X_{j+1} \cup T_{j+1}\cup T'_{j+1})$, choose one of these adjacent edges, and let $T''_{j+1}$ be the set of all these chosen edges. Then clearly $T''_{j+1}$ has the desired properties and $|T''_{j+1}| \leq |T'_{j}|$. By \eqref{eqn:deg count}, $|T'_j| \leq x-\frac{w}{2}$. Thus, $|T''_{j+1}| \leq x-\frac{w}{2}$ and hence by (\ref{eqn:deg count})
\begin{equation}\label{eqn:num edges mid sec}
|T_{j}\cup T_{j}'\cup T_j''| = |T_j|+|T'_j|+|T''_j| \leq 2x-w+x-\tfrac{1}{2}w = 3x-\tfrac{3}{2}w \leq y.
\end{equation}
Now we let $Y_i \supseteq T_{i}\cup T_{i}'\cup T_i''$ be a $y$-subset of $E(H_i)\setminus X_i$ and
$Z_i = E(H_i)\setminus (X_i\cup Y_i)$ for all $i \in \mathbb{Z}_{c}$ (such a $Y_i$ exists because $x$ and $w$ obey (P1) and so we have $|E(H_i) \setminus X_i| \geq \lfloor \frac{1}{c}E(H)\rfloor-x \geq 2y$). We complete the proof by
showing that
$\{X_i,Y_i,Z_i: i \in \mathbb{Z}_{c}\}$ is an $(x,w)$-partition.
By our hypotheses, (P1) is satisfied, and (P2) and (P3) are immediate from the above construction. Because $Y_{j+1} \supseteq T_{j+1} \cup T_{j+1}'$, each edge of $E(H_{j+1})$ that is adjacent to an edge in $X_j$ is in $X_{j+1} \cup Y_{j+1}$. Thus, no edge of $X_j$ is adjacent to an edge in $Z_{j+1}$, as required for (P4). The set of edges in $Y_{j}$ that are adjacent to two edges in $X_{j-1}$ is $T_j'$ and it follows from (\ref{eqn:deg count}) and the definition of $y$
that $|T_j'| \leq x-\frac{w}{2} \leq \frac{y}{3}$. Also, by (ii) and because $Y_{j+1} \supseteq T_{j+1}''$, each edge of $T_j'$ is adjacent to at least one edge in $X_{j+1} \cup Y_{j+1}$ and hence is adjacent to at most one edge in $Z_{j+1}$. Thus, (P5) holds. Because $\{X_0,\ldots, X_{c-1}\}$ is a $(x,w)$-semipartition, (P6) is satisfied.
\end{proof}
We now find $(x,w)$-semipartitions using two different approaches. The first is constructive and works better for small values of $k$. We detail it for class 1 graphs in Lemma~\ref{lem:gen blob construction} and for class 2 graphs in Lemma~\ref{lem:gen blob construction C2}. Our second approach is probabilistic and works better for large values of $k$. We detail it in Lemma~\ref{lem:prob cons}.
For the remainder of the section, it will be convenient to extend our existing notation slightly. Let $\{H_0,\ldots,H_{c-1}\}$ be an equitable matching decomposition of a graph $H$ and
let $X$ be a subset of $E(H)$. For $v \in V(H)$ and $i \in \mathbb{Z}_c$ we say that $v$ is {\em $i$-covered for $X$} if $v$ is $i$-covered for $\{X \cap E(H_0),\ldots,X \cap E(H_{c-1})\}$. That is, $v$ is $i$-covered for $X$ if $v$ is adjacent to an edge in $X \cap H_i$ and either there is an edge in $X \cap H_{i+1}$ that is also adjacent to $v$ or no edge in $H_{i+1}$ is adjacent to $v$. Also, for a graph $G$ and a subset $S$ of $V(G)$ we use $G[S]$ to denote the subgraph of $G$ induced by $S$.
\begin{lem}\label{lem:gen blob construction}
Let $k \geq 3$ be an integer. Let $H$ be a $k$-regular class $1$ graph with $n$ vertices, and let $\{H_0,\ldots,H_{k-1}\}$ be an equitable matching decomposition of $H$. Then for any $x \leq \frac{n}{2}$ there is an $(x,w)$-semipartition of $H$
with
$w = x+\lfloor\frac{x-1}{k-1}\rfloor$.
\end{lem}
\begin{proof}
Let $w=x+\lfloor\frac{x-1}{k-1}\rfloor$. Let $V_2$ be a set of any two adjacent vertices in $H$. We will iteratively define a sequence $V_2,\ldots,V_{w}$ of subsets of $V(H)$ such that $V_2 \subseteq \cdots \subseteq V_{w}$ and, for each $i \in \{2,\ldots,w\}$,
\begin{itemize}
\item[(i)]
$|V_i|=i$;
\item[(ii)]
at least $b$ of the graphs in $\{H_j[V_i]:j\in \mathbb{Z}_k\}$ have at least $a+1$ edges and the rest have at least $a$ edges, where $a$ and $b$ are the integers such that $i-1=ak+b$ and $b \in \{0,\ldots,k-1\}$.
\end{itemize}
Note that $V_2$ obeys (i) and (ii). Suppose inductively that for some $h \in \{2,\ldots,w-1\}$ we have a set $V_h$ obeying (i) and (ii). Let $a'$, $b'$, $a''$ and $b''$ be the integers such that $h-1=a'k+b'$, $h=a''k+b''$ and $b',b'' \in \{0,\ldots,k-1\}$. Notice that
\begin{equation}\label{eqn:divs and mods}
(a'',b'')=
\left\{
\begin{array}{ll}
(a'+1,0) & \hbox{if $h \equiv 0 \mod{k}$} \\
(a',b'+1) & \hbox{otherwise.}
\end{array}
\right.
\end{equation}
Let $j_0 \in \mathbb{Z}_k$ such that $|E(H_{j_0}[V_h])| \leq |E(H_j[V_h])|$ for each $j \in \mathbb{Z}_k$. If $|E(H_{j_0}[V_h])| \geq a'+1$, then $|E(H_{j}[V_h])| \geq a'+1$ for each $j \in \mathbb{Z}_k$ by the definition of $j_0$. In this case we take $V_{h+1}=V_h \cup \{u\}$ for any vertex $u \in V(H) \setminus V_h$ and note that $V_{h+1}$ obeys (i) and (ii) using \eqref{eqn:divs and mods}. If $|E(H_{j_0}[V_h])|=a'$ then, because $a'<\frac{h}{2}$, there is a vertex $u \in V(H) \setminus V_h$ such that the edge of $H_{j_0}$ incident with $u$ is also incident with a vertex in $V_h$. We take $V_{h+1}=V_h \cup \{u\}$. Then we have $|E(H_{j_0}[V_{h+1}])| = a'+1$. From this it can be checked, using \eqref{eqn:divs and mods}, that $V_{h+1}$ obeys (i) and (ii). So we have defined $V_2,\ldots,V_w$.
For each $j \in \mathbb{Z}_k$, let $X^*_j$ be the set of all edges of $H_j$ adjacent to at least one vertex in $V_w$ and observe that
\[|X^*_j|=w-|E(H_j[V_w])| \leq w-\left\lfloor\tfrac{w-1}{k}\right\rfloor\]
where the inequality follows because $V_w$ obeys (ii).
Now $w-\frac{w-1}{k} \leq x$ because $w \leq \frac{xk-1}{k-1}$ by definition,
and hence $w-\lfloor\frac{w-1}{k}\rfloor \leq x$ because $w$ and $x$ are integers.
Thus, for each $j \in \mathbb{Z}_k$, we can choose a subset $X_j$ of $E(H_j)$
such that $X^*_j \subseteq X_j$ and $|X_j|=x$.
Now, for each $j \in \mathbb{Z}_k$ and each $u \in V_w$,
there is an edge of $X_j$ and an edge of $X_{j+1}$ incident with $u$.
Therefore, $\{X_0 \ldots, X_{k-1}\}$ satisfies property (P6)
for $w$ and thus is an $(x,w)$-semipartition with respect to $\{H_0,\ldots,H_{k-1}\}$.
\end{proof}
\begin{lem}\label{lem:gen blob construction C2}
Let $k \geq 3 $ be an integer, let $H$ be a $k$-regular class 2 graph with $n \geq 6(k+1)$ vertices, and let $\{H_0,\ldots,H_k\}$ be an equitable matching decomposition of $H$. For any $x \leq \lfloor \frac{nk}{2(k+1)}\rfloor$ there is an $(x,w)$-semipartition of $H$, where $w = x+\lfloor\frac{x-1}{k}\rfloor$.
\end{lem}
\begin{proof}
Throughout this proof, for any subset of $E(H)$ denoted $X(h)$ and any $j \in \mathbb{Z}_{k+1}$, we denote $X(h) \cap E(H_j)$ by $X_j(h)$. Let $w = x+\lfloor\frac{x-1}{k}\rfloor$. We claim there is a sequence $X(1),\ldots, X(w)$ of subsets of $E(H)$ such that $X(1) \subseteq \cdots \subseteq X(w)$ and, for all $i \in \{1,\ldots,w\}$,
\begin{itemize}
\item[(i)]
$|X_j(i)| \in \{i -\lfloor\frac{i-1}{k+1}\rfloor-1,i -\lfloor\frac{i-1}{k+1}\rfloor\}$ for each $j \in \mathbb{Z}_{k+1}$;
\item[(ii)] $|\{j \in \mathbb{Z}_{k+1}: |X_j(i)| =i -\lfloor\frac{i-1}{k+1}\rfloor\}| = k+1- i'$, where $i'$ is the least nonnegative integer congruent to $i-1$ modulo $k+1$;
\item[(iii)] at least $i$ vertices are $j$-covered for $X(i)$ for each $j \in \mathbb{Z}_{k+1}$.
\end{itemize}
Suppose for the moment that this claim holds. For all $j \in \mathbb{Z}_{k+1}$ observe that
\[|X_j(w)|\leq w-\left\lfloor\tfrac{w-1}{k+1}\right\rfloor \leq x\]
where the first inequality follows because $X(w)$ satisfies (i) and the second follows because $w \leq x+\frac{x-1}{k}$ by the definition of $w$ and the fact that $w$ and $x$ are integers. Thus, for each $j \in \mathbb{Z}_k$, we can find a subset $X_j$ of $E(H_j)$ such that $X_j(w) \subseteq X_j$ and $|X_j|=x$. Then, because $X(w)$ satisfies (iii), we have that $\{X_0 \ldots, X_{k-1}\}$ satisfies property (P6) for $w$ and thus is an $(x,w)$-semipartition with respect to $\{H_0,\ldots,H_{k}\}$.
So it only remains to prove the claim. We do so by induction on $i$. Let $u \in V(H)$ be a vertex not incident in $H$ to an edge in $H_{0}$ and $v \in V(H)$ be a vertex not incident in $H$ to an edge in $H_{1}$ (such a vertex $v$ exists because $\{H_0,\ldots,H_k\}$ is equitable and $n \geq 6(k+1)$). Let $X(1)$ be the set containing each edge of $H$ incident with $u$ and the unique edge of $H_0$ incident with $v$. It is easy to check that $X(1)$ satisfies (i), (ii) and (iii).
Now suppose inductively that for some $h \in \{1,\ldots,w-1\}$ there is a set $X(h)$ that satisfies (i), (ii) and (iii).
We will show that there is a choice for $X(h+1)$ that satisfies (i), (ii) and (iii). Let $s$ be any element of $\mathbb{Z}_{k+1}$ such that $|X_s(h)| = h -\lfloor\frac{h-1}{k+1}\rfloor$ (note that at least one such exists by (ii)). We will construct $X(h+1)$ as $X(h) \cup \{e_j: j \in \mathbb{Z}_{k+1} \setminus \{s\}\}$ where $e_j$ is an edge in $E(H_j) \setminus X_j(h)$ for each $j \in \mathbb{Z}_{k+1} \setminus \{s\}$. It can be checked that this will ensure that $X(h+1)$ satisfies (i) and (ii) for $i=h+1$. (To see this, note that if $h \not\equiv 0 \mod{k+1}$ then $\lfloor\frac{h}{k+1}\rfloor=\lfloor\frac{h-1}{k+1}\rfloor$ and that if $h \equiv 0 \mod{k+1}$ then $\lfloor\frac{h}{k+1}\rfloor=\lfloor\frac{h-1}{k+1}\rfloor+1$ and $|X_j(h)| = h-1-\lfloor\frac{h-1}{k+1}\rfloor=h-\lfloor\frac{h}{k+1}\rfloor$ for all $j \in \mathbb{Z}_{k+1} \setminus \{s\}$ by (ii).) So our goal is to ensure that (iii) also holds.
We will first choose a subset $T$ of $\mathbb{Z}_{k+1} \setminus \{s\}$ and an $e_j$ for each $j \in T$ such that, for all $j \in T \cup \{s\}$, at least $h+1$ vertices are $j$-covered for $X'(h+1)$, where $X'(h+1)=X(h) \cup \{e_j: j \in T\}$. If more than $h$ vertices are $s$-covered for $X(h)$, then we can take $T=\emptyset$ and $X'(h+1)=X(h)$, so assume otherwise that precisely $h$ vertices are $s$-covered for $X(h)$. Thus, by our choice of $s$, there are $2(h -\lfloor\frac{h-1}{k+1}\rfloor)$ vertices incident with an edge in $X_s(h)$ and only $h$ of these are $s$-covered. Now $2(h -\lfloor\frac{h-1}{k+1}\rfloor)>h$ because $k \geq 3$ and hence there is a vertex $v$ that is incident with an edge in $X_s(h)$ but is not $s$-covered. Let
\[T=\{j \in \mathbb{Z}_{k+1} \setminus \{s\}:\hbox{an edge in $E(H_j) \setminus X_j(h)$ is incident with $v$}\}\]
and, for each $j \in T$, take $e_j$ to be the edge of $E(H_j) \setminus X_j(h)$ incident with $v$. Then, for each $j \in T \cup \{s\}$, we have that $v$ was not $j$-covered for $X(h)$ but is $j$-covered for $X'(h+1)$ and hence, because at least $h$ vertices were $j$-covered for $X(h)$, at least $h+1$ vertices are $j$-covered for $X'(h+1)$. So we can find $T$ and $X'(h+1)$ with the claimed properties.
It remains to choose $e_j$ for each $j \in \mathbb{Z}_k \setminus (T \cup \{s\})$. If $T \cup \{s\}=\mathbb{Z}_{k+1}$ we are done. Otherwise, let $q$ be an element of $\mathbb{Z}_{k+1} \setminus (T \cup \{s\})$ such that $q+1 \in T \cup \{s\}$. We will show that there is a choice for $e_q$ in $E(H_q) \setminus X_q(h)$ such that at least $h+1$ vertices are $q$-covered for $X'(h+1) \cup \{e_q\}$. This will suffice to complete the proof because a suitable $X(h+1)$ will then be obtainable by iterating this procedure. If more than $h$ vertices are $q$-covered for $X'(h+1)$, then we may take $e_q$ to be an arbitrary edge of $H_q \setminus X_q(h)$. So we can assume that precisely $h$ vertices are $q$-covered for $X'(h+1)$.
If there is a vertex $u$ that is incident with an edge in $E(H_q)\setminus X'_q(h+1)$ but not with an edge in $E(H_{q+1})\setminus X'_{q+1}(h+1)$, then we can take $e_j$ to be the edge in $E(H_q)\setminus X'_q(h+1)$ incident with $u$. The vertex $u$ was not $q$-covered for $X'(h+1)$ but is for $X'(h+1) \cup \{e_q\}$. So it suffices to show that there is such a vertex $u$. Let $w_q$ and $w_{q+1}$ be the number of vertices of $H$ that are incident with no edge in $H_q$ and $H_{q+1}$ respectively. The number of vertices not incident with an edge in $E(H_{q+1})\setminus X'_{q+1}(h+1)$ is thus $2|X'_{q+1}(h+1)|+w_{q+1} \geq 2h -2\lfloor\frac{h}{k+1}\rfloor+w_{q+1}$. The inequality follows because either $q+1=s$ and $|X'_{q+1}(h)|=|X_{q+1}(h)| = h -\lfloor\frac{h-1}{k+1}\rfloor$ by the definition of $s$ or $q+1 \in T$ and $|X'_{q+1}(h)|=|X_{q+1}(h)|+1 \geq h -\lfloor\frac{h-1}{k+1}\rfloor$ because $X(h)$ obeys (i). Now, at most $h$ of these vertices not incident with an edge in $E(H_{q+1})\setminus X'_{q+1}(h+1)$ are not incident with an edge in $X'_q(h+1)$, because precisely $h$ vertices are $q$-covered for $X'(h+1)$, and at most $w_q$ are not incident with an edge in $H_q$. So such a vertex $u$ will exist provided that
\[2h -2\lfloor\tfrac{h}{k+1}\rfloor+w_{q+1}>h+w_q.\]
Now $w_q-w_{q+1} \leq 2$ because $\{H_0,\ldots,H_{k}\}$ is equitable and hence this inequality will hold and such a $u$ will exist unless $h \leq 4$. If $h \leq 4$, then note that, because $\{H_0,\ldots,H_{k}\}$ is equitable, $|E(H_{q+1})| \leq \lceil\tfrac{kn}{2(k+1)}\rceil$ and hence
\[w_{q+1} \geq n-2\lceil\tfrac{kn}{2(k+1)}\rceil > n-\tfrac{kn}{k+1}-2 \geq 4 \geq h\]
where the second last inequality follows because $n \geq 6(k+1)$. Thus, one of the $w_{q+1}$ vertices incident with no edge in $H_{q+1}$ will not be incident with an edge in $X'_q(h+1)$, because precisely $h$ vertices are $q$-covered for $X'(h+1)$. So again such a $u$ exists.
Thus we can choose an $e_q$ in $E(H_q) \setminus X_q(h)$ such that at least $h+1$ vertices are $q$-covered for $X'(h+1) \cup \{e_q\}$. As discussed, by iterating this procedure we can obtain a choice for $X(h+1)$ that satisfies (i), (ii) and (iii). This completes the proof.
\end{proof}
We now present a probabilistic method of finding $(x,w)$-semipartitions of $k$-regular graphs.
\begin{lem}\label{lem:prob cons}
Let $k \geq 3$ be an integer and let $c \in \{k,k+1\}$. If $H$ is a $k$-regular graph of order $n$, $\{H_0,\ldots,H_{c-1}\}$ is an equitable matching decomposition of $H$, and $\alpha$ is a constant such that $0 < \alpha < \frac{k}{c}$, then there is an $(x,w)$-semipartition of $H$, where
\[
x= \mfrac{\alpha k (2 - \alpha)}{2c}n +O\left(\sqrt{n}\right),\quad w = \mfrac{\alpha k}{c}n+ O\left(\sqrt{n}\right).
\]
\end{lem}
\begin{proof}
For each $v \in V(G)$, let $I_v$ be a random variable that is 1 with probability $\alpha$ and 0 otherwise. Let $R=\{v \in V(G):I_v=1\}$. Observe that then $|E(H_j[R])|$ is a binomial random variable with $|E(H_j)|$ trials and success probability $\alpha^2$ and so by Hoeffding's inequality \cite{MR144363} we have
\begin{equation}\label{eqn:edgesConcentration}
\mathbb{P}\left(|E(H_j[R])| \leq \alpha^2|E(H_j)|-\sqrt{\tfrac{1}{2}|E(H_j)|\log(4c)}\right) \leq \mfrac{1}{4c}.
\end{equation}
The proof now divides into cases according to whether $c=k$ or $c=k+1$.
\textbf{Case 1.} Suppose that $c=k$. Note that in this case $|E(H_i)|=\frac{n}{2}$
for each $i \in \mathbb{Z}_k$. Now $|R|$ is a binomial random variable with $n$ trials and success probability $\alpha$ and so $\Pr(|R| > \lceil\alpha n\rceil) < \frac{1}{2}$. Thus, by \eqref{eqn:edgesConcentration} and the union bound, there is a subset $S$ of $V$ such that $|S| = \lceil\alpha n\rceil$ and $|E(H_i[S])| \geq \frac{1}{2}\alpha^2n-O(\sqrt{n})$ for each $i \in \mathbb{Z}_k$ (note that vertices can be added arbitrarily to ensure that $|S| = \lceil\alpha n\rceil$).
Let $j \in \mathbb{Z}_k$. Let $m_j$ be the number of edges of $H_j$ that are incident with at least one vertex in $S$. Because every vertex in $S$ has an edge of $H_j$ incident with it and there are $|E(H_j[S])|$ edges of $H_j$ that are incident with two vertices of $S$, we have
\[m_j=|S|-|E(H_j[S])| \leq \lceil\alpha n\rceil - \tfrac{1}{2}\alpha^2n + O\bigl(\sqrt{n}\bigr) = \alpha n\bigl(1 - \tfrac{\alpha}{2}\bigr) + O\bigl(\sqrt{n}\bigr).\]
So we can take $x=\alpha n(1-\frac{\alpha}{2})+O(\sqrt{n})$ such that $m_i \leq x$ for each $i \in \mathbb{Z}_k$. Then, for each $i \in \mathbb{Z}_k$, we can choose a subset $X_i$ of $E(H_i)$ such that $|X_i|=x$ and $X_i$ contains all $m_i$ edges of $H_i$ that are incident with a vertex in $S$. Because each vertex in $S$ has edges of $H_i$ and $H_{i+1}$ incident with it for each $i \in \mathbb{Z}_k$, $\{X_0,\ldots,X_{k-1}\}$ obeys (P6) for
\[w = |S| = \alpha n + O\left(\sqrt{n}\right)\]
and hence is an $(x,w)$-semipartition with respect to $\{H_0,\ldots,H_{k-1}\}$.
\textbf{Case 2.} Suppose that $c=k+1$. For each $i \in \mathbb{Z}_{k+1}$, note that in this case $|E(H_i)| \in \{\lfloor\frac{kn}{2(k+1)}\rfloor,\lceil\frac{kn}{2(k+1)}\rceil\}$ and let $V_i=V(H) \setminus V(H_i)$. Note that $\{V_0,\ldots, V_{k}\}$ is a partition of $V(H)$ and that $V_i =\frac{n}{k+1}+O(1)$ for each $i \in \mathbb{Z}_{k+1}$. So, for $i \in \mathbb{Z}_{k+1}$, $|R \cap V_i|$ is a binomial random variable with $|V_i|$ trials and success probability $\alpha$ and by Hoeffding's inequality we have
\begin{equation}\label{eqn:verticesConcentration}
\mathbb{P}\left(\bigl||R \cap V_j|-\alpha|V_j|\bigr| \ge \sqrt{\tfrac{1}{2}|V_j|\log(8c)}\right) \leq \mfrac{1}{4c}.
\end{equation}
Thus, by \eqref{eqn:edgesConcentration}, \eqref{eqn:verticesConcentration} and the union bound, there is a subset $S$ of $V$ such that, for each $i \in \mathbb{Z}_{k+1}$, $|S \cap V_i| = \frac{\alpha }{k+1}n+O(\sqrt{n})$ and $|E(H_i[S])| \geq \alpha^2\frac{k}{2(k+1)}n-O(\sqrt{n})$. Note that this implies that $|S|=\alpha n +O(\sqrt{n})$.
Let $j \in \mathbb{Z}_{k+1}$. Let $m_j$ be the number of edges of $H_j$ that are incident with at least one vertex in $S$. Because every vertex in $S \setminus V_j$ has an edge of $H_j$ incident with it and there are $|E(H_j[S])|$ edges of $H_j$ that are incident with two vertices of $S$, we have
\[
m_j=|S|-|S \cap V_j|-|E(H_j[S])| \leq \alpha n - \tfrac{\alpha }{k+1}n - \tfrac{k}{2(k+1)}\alpha^2n + O\bigl(\sqrt{n}\bigr) = \tfrac{\alpha k(2-\alpha)}{2(k+1)}n + O\bigl(\sqrt{n}\bigr).
\]
So we can take $x=\tfrac{\alpha k(2-\alpha)}{2(k+1)}n+O(\sqrt{n})$ such that $m_i \leq x$ for each $i \in \mathbb{Z}_{k+1}$. Then, for each $i \in \mathbb{Z}_{k+1}$, we can choose a subset $X_i$ of $E(H_i)$ such that $|X_i|=x$ and $X_i$ contains all $m_i$ edges of $H_i$ that are incident with a vertex in $S$. Now, for each $i \in \mathbb{Z}_{k+1}$, $|S \setminus V_i|=\frac{\alpha k}{k+1}n+ O(\sqrt{n})$ and each vertex in $S \setminus V_i$ has an edge of $X_i$ incident with it and either has an edge of $X_{i+1}$ incident with it or has no edge of $H_{i+1}$ incident with it. Thus, $\{X_0,\ldots,X_k\}$ obeys (P6) for
\[w=\mfrac{\alpha k}{k+1}n+ O\left(\sqrt{n}\right)\]
and hence is an $(x,w)$-semipartition with respect to $\{H_0,\ldots,H_{k}\}$.
\end{proof}
Now we present the proofs of the lower bounds of Theorem~\ref{thm:k reg}, using the explicit and probabilistic methods for finding $(x,w)$-semipartitions given above. We begin with the following that uses Lemmas~\ref{lem:gen blob construction} and \ref{lem:gen blob construction C2}.
\begin{lem}\label{thm:main thm explicit con}
For any integers $k \geq 3$ and $n \geq 6(k+1)$ such that $nk$ is even,
\[
\textnormal{cms}(n,k) \geq \mfrac{k(5k-3)}{4(k+1)(4k-3)}n-6 \quad \text{ and } \quad \textnormal{cms}_1(n,k) \geq \mfrac{5k-8}{4(4k-7)}n-6\,.
\]
\end{lem}
\begin{proof}
Let $H$ be a $k$-regular graph with order $n$ and chromatic index $c$.
Let
\[x = \left\lfloor\mfrac{(nk-8c)(c-1)}{2c(4c-7)}\right\rfloor \quad \text{ and } \quad w = x+\left\lfloor \mfrac{x-1}{c-1}\right\rfloor.\]
By Lemma~\ref{lem:gen blob construction} or \ref{lem:gen blob construction C2} there exists
an $(x,w)$-semipartition for $H$, noting that $x \leq \lfloor\frac{nk}{2c}\rfloor$ because $c-1 \leq 4c-7$.
We show that there exists an $(x,w)$-partition. Obviously $w \leq \frac{3}{2}x$, so by Lemma~\ref{lem:can make T with S} it suffices to show
that $x+2y \leq \frac{nk}{2c} $ for
$y = \left\lceil 3x-\frac{3}{2}w\right\rceil$ .
We have
\[
x+2y \leq \mfrac{x(4c-7)}{c-1} +4 \leq \mfrac{nk-8c}{2c} +4 = \mfrac{nk}{2c}
\]
where the first inequality follows because $2y \leq 6x-3w+1$ and $w \geq x+\frac{x-c+1}{c-1}$, and the second follows because
$x(4c-7) \leq \frac{1}{2c}(nk-8c)(c-1)$.
Because there exists an $(x,w)$-partition, we have $\textnormal{cms}(H) \geq x+y-1 $, by Proposition~\ref{prop:seq given some conds}. So
\[ \textnormal{cms}(H) \geq x+y-1 \geq \mfrac{(5c-8)x-2c+5}{2(c-1)} > \mfrac{k(5c-8)}{4c(4c-7)}n-6 \]
where the second inequality follows because $y \geq 3x-\frac{3}{2}w$ and $w \leq \frac{xc-1}{c-1}$ and the third follows because $x \geq \frac{(nk-8c)(c-1)}{2c(4c-7)}-1$. Substituting $c=k$ and $c \leq k+1$ gives the required bounds for $\textnormal{cms}_1(n,k)$ and $\textnormal{cms}(n,k)$, respectively.
\end{proof}
\begin{lem}\label{thm:prob con}
Let $k \geq 3$ be an integer. Then for integers $n > k$ such that $nk$ is even,
\[\textnormal{cms}(n,k) \geq \mfrac{31k}{98(k+1)}n-o(n) \quad \text{ and } \quad \textnormal{cms}_1(n,k) \geq \mfrac{31}{98}n-o(n)\]
\end{lem}
\begin{proof}
Let $H$ be a $k$-regular graph with order $n$ and chromatic index $c$.
Let $0 < \alpha < \frac{1}{7}$.
Then, by Lemma~\ref{lem:prob cons}, there is an $(x,w)$-semipartition where
$x = \frac{\alpha k (2 - \alpha)}{2c}n +O\left(\sqrt{n}\right)$ and $w = \frac{\alpha k}{c}n+ O\left(\sqrt{n}\right)$.
For sufficiently large $n$, we can apply Lemma~\ref{lem:can make T with S} to obtain an $(x,w)$-partition, since clearly
$w \leq \frac{3}{2} x$ and making substitutions for $x$ and $w$ shows that
\begin{equation}\label{eqn:oneP1cond}
x+2\left\lceil 3x-\tfrac{3}{2}w \right\rceil
= \mfrac{\alpha k(8-7\alpha)}{2c}n+O(\sqrt{n})
\leq \left\lfloor \mfrac{nk}{2c} \right\rfloor
\end{equation}
where the last inequality follows from $\alpha < \frac{1}{7}$.
Therefore, applying Proposition~\ref{prop:seq given some conds} yields,
\[
\textnormal{cms}(H) \geq x+\left\lceil 3x-\tfrac{3}{2}w\right\rceil-1
=\mfrac{\alpha k(5-4\alpha)}{2c}n+O(\sqrt{n})\,,
\]
where the final estimate is obtained by making the appropriate substitutions for $x$ and $w$.
As the above inequality holds for any $0 < \alpha < \frac{1}{7}$, by substituting $\alpha$ sufficiently close to $\frac{1}{7}$ we obtain the result.
\end{proof}
We can now present the proof of Theorem~\ref{thm:k reg}.
\begin{proof}[\textbf{\textup{Proof of Theorem~\ref{thm:k reg}.}}]
Let $k \geq 3$ be an integer. Then
the lower bounds for $\textnormal{cms}(n,k)$ and $\textnormal{cms}_1(n,k)$ hold for $n \geq 6(k+1)$ by Lemmas~\ref{thm:main thm explicit con} and \ref{thm:prob con}.
The upper bound for $\textnormal{cms}(n,k)$ follows from Theorem~\ref{thm:count exa},
since for all $n \geq 3k+5$ there is a $k$-regular graph of order $n$ with $B_k$ as a subgraph.
Finally, $\textnormal{cms}(G) \leq \frac{n-1}{2}$ for each $k$-regular graph $G$. To see this, note that otherwise there would be an ordering $\ell=(e_0,\ldots,e_{kn/2-1})$ of a $k$-regular graph $G$ of even order $n$ such that each of $e_{0},\ldots, e_{n/2-1}$ and $e_{1},\ldots, e_{n/2}$ form a matching of size $\frac{n}{2}$ in $G$, which is impossible as two matchings of size $\frac{n}{2}$ cannot differ by exactly one edge.
\end{proof}
\section{Conclusion}
The table below gives, for each integer $k \geq 3$, the strongest consequences of Theorem~\ref{thm:k reg} for large~$n$.
\begin{table}[H]
\begin{center}
\setlength{\tabcolsep}{2pt}
\begin{tabular}{rccclc|crcccl}
$\frac{1}{4}n-6$ & $\leq$ & $\textnormal{cms}(n,3)$ & $\leq$ & $\frac{3}{8}n$ & \qquad&\qquad & $\frac{7}{20}n-6$ & $\leq$ & $\textnormal{cms}_1(n,3)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{17}{65}n-6$ & $\leq$ & $\textnormal{cms}(n,4)$ & $\leq$ & $\frac{2}{5}n$ & \qquad&\qquad & $\frac{1}{3}n-6$ & $\leq$ & $\textnormal{cms}_1(n,4)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{55}{204}n-6$ & $\leq$ & $\textnormal{cms}(n,5)$ & $\leq$ & $\frac{5}{12}n$ & \qquad&\qquad & $\frac{17}{52}n-6$ & $\leq$ & $\textnormal{cms}_1(n,5)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{27}{98}n-6$ & $\leq$ & $\textnormal{cms}(n,6)$ & $\leq$ & $\frac{3}{7}n$ & \qquad&\qquad & $\frac{11}{34}n-6$ & $\leq$ & $\textnormal{cms}_1(n,6)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{7}{25}n-6$ & $\leq$ & $\textnormal{cms}(n,7)$ & $\leq$ & $\frac{7}{16}n$ & \qquad&\qquad & $\frac{9}{28}n-6$ & $\leq$ & $\textnormal{cms}_1(n,7)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{74}{261}n-6$ & $\leq$ & $\textnormal{cms}(n,8)$ & $\leq$ & $\frac{4}{9}n$ & \qquad&\qquad & $\frac{8}{25}n-6$ & $\leq$ & $\textnormal{cms}_1(n,8)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{63}{220}n-6$ & $\leq$ & $\textnormal{cms}(n,9)$ & $\leq$ & $\frac{9}{20}n$ & \qquad&\qquad & $\frac{37}{116}n-6$ & $\leq$ & $\textnormal{cms}_1(n,9)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{235}{814}n-6$ & $\leq$ & $\textnormal{cms}(n,10)$ & $\leq$ & $\frac{5}{11}n$ & \qquad&\qquad & $\frac{7}{22}n-6$ & $\leq$ & $\textnormal{cms}_1(n,10)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{143}{492}n-6$ & $\leq$ & $\textnormal{cms}(n,11)$ & $\leq$ & $\frac{11}{24}n$ & \qquad&\qquad & $\frac{47}{148}n-6$ & $\leq$ & $\textnormal{cms}_1(n,11)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
$\frac{19}{65}n-6$ & $\leq$ & $\textnormal{cms}(n,12)$ & $\leq$ & $\frac{6}{13}n$ & \qquad&\qquad & $\frac{13}{41}n-6$ & $\leq$ & $\textnormal{cms}_1(n,12)$ & $\leq$ &$\frac{n-1}{2}$ \\[1mm]
$\frac{403}{1372}n-6$ & $\leq$ & $\textnormal{cms}(n,13)$ & $\leq$ & $\frac{3}{8}n$ & \qquad&\qquad & $\frac{19}{60}n-6$ & $\leq$ & $\textnormal{cms}_1(n,13)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
& & & & & \qquad&\qquad & $\frac{31}{98}n-6$ & $\leq$ & $\textnormal{cms}_1(n,14)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]\hline
\rule{0mm}{5mm}$\frac{31k}{98(k+1)}n-o(n)$ & $\leq$ & $\textnormal{cms}(n,k)$ & $\leq$ & $\frac{k}{2(k+1)}n$ & \qquad&\qquad & $\frac{31}{98}n-o(n)$ & $\leq$ & $\textnormal{cms}_1(n,k)$ & $\leq$ & $\frac{n-1}{2}$ \\[1mm]
\multicolumn{5}{l}{for each $k \geq 14$} & \qquad&\qquad & \multicolumn{5}{l}{for each $k \geq 15$}
\end{tabular}
\caption{Consequences of Theorem~\ref{thm:k reg} for each $k$}\label{T:kconsequences}
\end{center}
\end{table}
We know of no nontrivial upper bounds on $\textnormal{cms}_1(n,k)$. It would be interesting to obtain some or, on the other hand, to prove that $\textnormal{cms}_1(n,k)$ approaches $\frac{n-1}{2}$ as $n$ becomes large. We certainly expect that our upper bounds on $\textnormal{cms}_1(n,k)$ and $\textnormal{cms}(n,k)$ are much closer to the true value than our lower bounds. In particular we pose the following question.
\begin{question}\label{conj:gen k-reg graph}
Let $k \geq 3$ be an integer. For integers $n > k$ such that $nk$ is even, is it the case that
$
\textnormal{cms}(n,k) = \tfrac{kn}{2(k+1)}-o(n)\,?
$
Is it the case that
$
\textnormal{cms}_1(n,k) = \tfrac{n-1}{2}-o(n)\,?
$
\end{question}
\bigskip
\noindent\textbf{Acknowledgments.}
This work was supported by Australian Research Council grants DP150100506 and FT160100048.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,383 |
\section{Introduction}
When building an action to describe the dynamics of a physical system, symmetries underlying it play a vital role since they put restrictions on the form of the action. On the other hand, when attempting to extend the existing physical model, exploring the possibility of the absence of some symmetries is one of the effective approach.
Our present understanding of particle physics is well-described by the standard model (SM) of particle physics. However, as an effective field theory, the SM is incomplete and leaves a number of puzzles unresolved. One of them, for example, is why there are widely separated mass scales in nature. Because of this hierarchy problem, it is reasonable to expect that the signature of new physics should emerge at scales not far beyond the TeV scale. Thus, exploring the extension of the SM in particle physics has become an extensively investigated subject. In particular, as a potential new physics, the breaking of Lorentz symmetry and its related contents have been intensively studied since theoretically there is no good reason to believe that Lorentz invariance would be intact at all energies.\footnote{See, for instance, Refs. \cite{LV1,LV2,LV3,LV4,LV5,LV6,LV7} for works on the SM modified by Lorentz-violating interactions.}
In this paper, we will explore the effects of the Lorentz violation emerging from symmetry breaking in Weyl geometry. Weyl geometry originated from Hermann Weyl's attempt to geometrize the electromagnetic interaction \cite{Weyl}. In Weyl geometry, the constraint of length conservation is relaxed and there is no absolute scaling for spacetime, so that the concept of length is path-dependent. Therefore, different from the situation in Riemanian geometry, parallel transport two vectors along a spacetime curve in Weyl geometry will change both their lengths and orientations. Only relative lengths and relative orientations remain fixed. The vector potential $A_{\mu}(x)$ in electrodynamics was identified with the gauge field in this geometry. However, Weyl's original idea was unsuccessful since it leads to conflict with physical facts, as pointed out by Einstein. For example, the mass of a free particle would depend on its history in Weyl's approach.
Weyl symmetry must be broken in the infrared. Recently, it has been shown that the Hilbert-Einstein action can be obtained from the matterless part of Weyl gravity after the symmetry breakdown of Weyl symmetry via either Stueckelberg mechanism \cite{Gh1, Gh2} or Coleman-Weinberg mechanism \cite{Oda1,Oda2}. While these works focus on the spontaneous symmetry breakdown of Weyl symmetry and its related issues, in this work we will by contrast simply adopt this scenario and consider the case where the Weyl gauge field is the ``order parameter" such that the Weyl symmetry breaking is triggered by a non-vanishing fixed vacuum expectation value $\omega_{\mu}$ of the Weyl gauge field at some high energy scale. This fixed vector background then becomes the source that induces the Lorentz violation.\footnote{For some aspects of Lorentz violation due to aetherlike fields, see Refs. \cite{Gomes,Lan}.} We restrict ourselves to investigate a specific model of linear Weyl gravity coupled to a scalar field and focus our attention to the effects produced by this Lorentz-violating background at low energies.
The rest of this paper is organized into four parts. In the next section, after describing the model we want to explore, several distinct features of the model are addressed. Then we go on to show that a $CPT$-odd Lorentz-violating interaction appears after Weyl symmetry breaking triggered by a non-vanishing $\omega_{\mu}$. Theoretical consequences are studied. In particular, we investigate different phases after Weyl symmetry breaking, depending on the spacetime character of the fixed background $\omega_{\mu}$. We show that the breakdown of Weyl symmetry does not necessarily generate a mass for the Weyl gauge field. After these analyses, the stability of the light mass scale generated after spontaneous symmetry breaking is clarified in Sec. 3. In sect. 4. based on the analysis of Sect. 2, we consider the quantum corrections and calculate the one-loop effective potentials, for $\omega_{\mu}$ to be timelike and lightlike, respectively, and discuss their physical consequences. Our conclusions are presented in the final section.
\section{Lorentz violation after Weyl symmetry breaking}
Our starting point is the following action of linear Weyl gravity coupled to a scalar field $\phi(x)$:
\begin{equation}
S=\int d^4 x \sqrt{- g} \{ {\alpha\over 12}\phi^2 \tilde{R} + {1\over 2} g^{\mu\nu} D_{\mu} \phi D_{\nu} \phi - {1\over 4!}\lambda \phi^4-{1\over 4} F_{\mu\nu}^{2} \}, \,\,\,\,\,\,\,\,\,\,\,\, \lambda >0 .\label{Action}
\end{equation}
Here $D_{\mu} \phi $ is the Weyl-covariant derivative of $\phi$ whose explicit form is defined as
\begin{equation}
D_{\mu} \phi \equiv (\partial_{\mu}-f S_{\mu})\phi
\end{equation}
with $S_{\mu}(x)$ the Weyl gauge field and $f$ the coupling constant to $\phi(x)$. $F_{\mu\nu}\equiv \partial_{\mu}S_{\nu}-\partial_{\nu}S_{\mu} $ is the field tensor for the Weyl field $S_{\mu}$. The Weyl scalar curvature $\tilde{R} $ is related to the Riemannian scalar curvature $R$ by
\begin{equation}
\tilde{R}=R -6f \nabla_{\mu} S^{\mu} -6 f^2 S_{\mu} S^{\mu}.
\end{equation}
The action~(\ref{Action}) is invariant under the Weyl transformations:
\begin{eqnarray}
g_{\mu\nu}(x) &\rightarrow& g_{\mu\nu}'(x)=e^{-2\Lambda (x)} g_{\mu\nu}(x), \,\,\,\,\,\,\,\, \phi(x) \rightarrow \phi'(x)=e^{+\Lambda (x)} \phi(x),\nonumber \\
S_{\mu}(x) &\rightarrow& S_{\mu}'(x) = S_{\mu}(x)+{1\over f} \partial_{\mu} \Lambda (x). \label{trx}
\end{eqnarray}
In this paper we work with the metric signature $(+,-,-,-)$. We remark that the transformation of the Weyl gauge field $S_{\mu}$ in~(\ref{trx}) is different from that of the electromagnetic vector potential under $U(1)$ gauge symmetry due to the nonoccurrence of the imaginary unit $i$ in the Weyl covariant derivative. Thus, while the $U(1)$ gauge group in electrodynamics is compact, the symmetry group generated by the Weyl transformations~(\ref{trx}) is non-compact. Also, we note in passing that it has been shown that with the Weyl symmetry, the Weyl field $S_{\mu}$ does not couple to the gauge and fermion fields of the SM \cite{Cesare, Gh3}.
Several comments regarding the Weyl invariant action~(\ref{Action}) are in order. First, in Riemann geometry (i.e., setting the Weyl gauge field $S_{\mu}(x)=0$) the action~(\ref{Action}) reduces to
\begin{equation}
S_R=\int d^4 x \sqrt{- g} \{ {\alpha\over 12}\phi^2 R + {1\over 2} g^{\mu\nu} \partial_{\mu} \phi \partial_{\nu} \phi - {1\over 4!}\lambda \phi^4 \}. \label{SR}
\end{equation}
In general, $S_R$ is not Weyl invariant. Indeed, it is easy to check that only when $\alpha=1$ will the action $S_R$ be Weyl invariant. In the case where $ \alpha=1$, the action $S_R$ is the well-known conformal scalar action.
Second, each term in~(\ref{Action}) is separately Weyl invariant. With $\alpha=1$, the $S_{\mu}$-dependent terms from $D_{\mu}\phi D^{\mu} \phi$ and from $\phi^2 \tilde{R}$ cancel each other, yielding a Weyl invariant action:
\begin{equation}
S_C=\int d^4 x \sqrt{- g} \{ {1\over 12}\phi^2 R + {1\over 2} g^{\mu\nu} \partial_{\mu} \phi \partial_{\nu} \phi - {1\over 4!}\lambda \phi^4-{1\over 4} F_{\mu\nu}^{2} \}. \label{SC}
\end{equation}
It is straightforward to show that the energy-momentum tensor $T_{\mu\nu}={-2\over \sqrt{-g}}{\delta S\over \delta_{g_{\mu\nu}}}\vert_{g_{\mu\nu}=\eta_{\mu\nu}}$ of the theory defined by $S_C$ takes the form
\begin{eqnarray}
T_{\mu\nu} = &-&{2\over 3} \partial_{\mu} \phi \partial_{\nu} \phi +{1\over 6} \eta_{\mu\nu} \partial_{\sigma} \phi \partial^{\sigma} \phi+{1\over 3}\phi (\partial_{\mu}\partial_{\nu}-\eta_{\mu\nu}\square)\phi -\eta_{\mu\nu} {1\over 4!}\lambda \phi^4 \\ \nonumber
&+&F_{\mu}^{\,\,\,\lambda}F_{\nu\lambda}-{1\over 4}\eta_{\mu\nu} (F_{\rho\sigma})^{2}.
\end{eqnarray}
The trace of $T_{\mu\nu} $ vanishes ``on-shell" (i.e., assuming the fields satisfy their equations of motion), as it should be for a conformal field theory \cite{Callan}.
Third, notice that in the action $S_{C}$, the Weyl field $S_{\mu}$ couples to the scalar field $\phi$ only through gravity, and thus the two fields must decouple in the flat limit. Explicitly, in the flat limit, $S_C$ reduces to two independent sectors:
\begin{equation}
S_C\longrightarrow S_\phi+ S_A =\int d^4 x \left( {1\over 2} \partial_{\mu} \phi \partial^{\mu} \phi - {1\over 4!}\lambda \phi^4\right) + \int d^4 x' \left(-{1\over 4} F_{\mu\nu}^{2}\right) . \label{SF}
\end{equation}
This flat-limit action enjoys an extended Poincar$\rm\acute{e}$ symmetry \cite{Foot}, meaning that the scalar sector $S_\phi$ and the vector sector $S_A$ in Eq.~(\ref{SF}) are individually invariant under independent Poincar$\rm\acute{e}$ transformations. Therefore, for the coupling constant $\alpha$ in Eq.~(\ref{Action}), small values of $(\alpha-1)$ are protected by this emerging symmetry and are technically nature \cite{Hooft}, as long as we are in the flat limit so that the effects due to gravity that mixes the two sectors can be neglected. In the next section, we will show that after spontaneous symmetry breaking, it is this enhanced spacetime symmetry in the decoupling limit $\alpha\rightarrow 1$ that secures the stability of the light mass scale in the symmetry-broken Lagrangian. From now on, we will always assume that $(\alpha-1)$ is very small.
We now go back to the action~(\ref{Action}). Upon Weyl symmetry breaking, we assume that $S_{\mu}$ acquires a non-vanishing fixed vacuum expectation value $\omega_{\mu}$. This fixed background field, acting like an aether field, breaks the Lorentz invariance while leaves the translational invariance intact. For timelike, spacelike, and lightlike $\omega_{\mu}$, the lorentz group $SO(1,3)$ is broken down to the little groups $SO(3)$, $SO(1,2)$, and $E(2)$, respectively.
Let us define the shifted field by writing
\begin{equation}
S_{\mu}(x)=\omega_{\mu}+A_{\mu}(x)
\end{equation}
and expand the Lagrangian about $\omega_{\mu}$. In the flat limit, meaning that we neglect the gravitational effects, the Lagrangian density of Eq.~(\ref{Action}) reduces to
\begin{eqnarray}
\mathcal{L}&=&{1\over 2}\partial_{\mu} \phi(x) \partial^{\mu} \phi(x) -{1\over 2} m_{\phi}^{2} \phi^2(x) - {1\over 4!}\lambda \phi^4(x) \nonumber\\
&&-{1\over 2} f (\alpha-1)\phi^2(x)(\partial_{\mu}+2f \omega_{\mu} )A^{\mu}(x)-{1\over 2} f^2 (\alpha-1)\phi^2(x) A_{\mu}^2(x)-{1\over 4} F_{\mu\nu}^{2}(x), \label{Lag}
\end{eqnarray}
where the mass squared parameter $ m_{\phi}^{2}$ of $\phi$ is defined as
\begin{equation}
m_{\phi}^{2}\equiv f^2(\alpha-1)\omega_{\mu}^2.
\end{equation}
Notice the appearance of the Lorentz-violating operator $\omega_{\mu} A^{\mu}(x) \phi^2(x)$ in Eq.~(\ref{Lag}). While this operator preserves $PT$, it violates $C$ parity. Thus, it is a $CPT$-odd operator. We note that $CPT$ invariance is a necessary but not sufficient condition for Lorentz invariance of an interacting quantum field theory.
The potential for $\phi$ in Eq.~(\ref{Lag}) takes the familiar form
\begin{equation}
V(\phi)={1\over 2} m_{\phi}^{2} \phi^2+ {1\over 4!}\lambda \phi^4.
\end{equation}
Whether the $Z_2$ symmetry ($\phi \rightarrow -\phi$) of the Lagrangian is preserved or spontaneously broken is determined by the location of the minimum of the potential. It is instructive to consider the three cases $m_{\phi}^{2}>0$, $m_{\phi}^{2}=0$, and $m_{\phi}^{2}<0$.\\
(i) $m_{\phi}^{2}>0$ (either $\alpha >1$ and $\omega_{\mu}$ timelike, or $\alpha <1$ and $\omega_{\mu}$ spacelike):\\
In this case, the potential $ V(\phi)$ has a minimum at $\phi=0$. Thus, after Weyl symmetry breaking, the scalar field $\phi$ acquires a mass while the ``Weyl photon" $A_{\mu}$ remains massless\footnote{However, when considering the quadratic Weyl gravity for which the term $\tilde{R}^2$ is included, the Weyl photon would acquire a mass squared $m_{A}^2 \sim f^4 \omega_{\mu}^2$, arising from the term $\tilde{R}^2$ after Weyl symmetry breaking. Then, the Weyl photon is massive for timelike $\omega_{\mu}$.}.\\
(ii) $m_{\phi}^{2}=0$ (i.e., $\omega_{\mu}$ lightlike):\\
In this case, even with non-vanishing $\omega_{\mu}$ after Weyl symmetry breaking, it appears that both fields remains massless. However, the mass of these fields can be generated dynamically from radiative corrections via the Coleman-Weinberg mechanism \cite{Coleman}. In Sect. 4, we will show that the Weyl gauge field $A_{\mu}$ can acquire a positive mass squared through this mechanism only if $\alpha<1$.\\
(iii) $m_{\phi}^{2}<0$ (either $\alpha <1$ and $\omega_{\mu}$ timelike, or $\alpha >1$ and $\omega_{\mu}$ spacelike):\\
In this case, the potential $V(\phi)$ has its minimum at $\phi_{c}=\pm ({-6m_{\phi}^2 \over \lambda})^{1/2}$. Thus, the $Z_2$ symmetry would further be spontaneously broken after Weyl symmetry breaking. Expanded around the broken-symmetry state by writing
\begin{equation}
\phi(x)=\phi_{c}+\varphi(x), \label{split}
\end{equation}
the next-to-last term in Eq.~(\ref{Lag}) gives a mass term for $A_{\mu}$:
\begin{equation}
\Delta \mathcal{L}= -{1\over 2} f^2 (\alpha-1)\phi^2A_{\mu}^2 = {1\over 2} m_{A}^{2}A_{\mu}A^{\mu}+\cdot\cdot\cdot,
\end{equation}
where the mass squared parameter is given by
\begin{equation}
m_{A}^2= -f^2(\alpha-1)\phi_{c}^2 . \label{MA}
\end{equation}
Thus, for $m_{\phi}^2<0$, the Weyl gauge field $A_{\mu}$ acquires a positive mass squared only for timelike $\omega_{\mu}$. This can also be seen from the equation of motion of $A_{\mu}$, given by
\begin{equation}
\partial^{\mu} F_{\mu\nu}+f(\alpha -1) \left[{1\over 2} \partial_{\nu}\phi^2-f \phi^2 \omega_{\nu} - f \phi^2 A_{\nu} \right]=0.
\end{equation}
If $\langle\phi \rangle\neq0$, then the last term shows explicitly that $ m_{A}^2= -f^2(\alpha-1)\langle \phi \rangle^2 $ for $\alpha<1$, indicating that $\omega_{\mu}$ is timelike since $m_{\phi}^2<0$ in this case.
It is apparent that while $m_{\phi}^2$ is of order $f^2(\alpha-1)$, $m_{A}^2$ generated from the above mechanism is of order ${f^4\over\lambda}(\alpha-1)^2$. Also, it is conceivable that if the gauge fields of any compact Lie group are included and couple to the scalar field in the model, spontaneous gauge symmetry breaking can be triggered after Weyl symmetry breaking. Probing this scenario phenomenologically would be interesting but lies beyond the scope of this work.
What we have shown in this analysis is that when spontaneous breakdown of Weyl symmetry generates a mass for the scalar field, it does not guarantee the mechanism of mass generation for the Weyl gauge field. From now on, unless otherwise specified, we will focus on the case $m_{\phi}^2<0$ and assume $\omega_{\mu}$ to be timelike, so that $A_{\mu}$ obtains a mass from the breakdown of $Z_2$ symmetry.
\section{Naturalness}
One of the motivations for exploring Weyl invariant theories is to study how mass scales can be generated from a theory of no intrinsic scales and remain stable. This point was made in \cite{Oda2}. In the previous section, we have shown that for $m_{\phi}^2<0$ and timelike $\omega_{\mu}$, after Weyl symmetry breaking, another mass scale $m_A$ besides $m_\phi$ appears through the spontaneous breaking of the $Z_2$ symmetry for $\phi$. From the tree-level mass formula~(\ref{MA}), we have
\begin{equation}
m_{A}^2= {6f^2\over\lambda}(\alpha-1)m_{\phi}^2 . \label{MAA}
\end{equation}
For small values of $(\alpha -1)$, the relation~(\ref{MAA}) indicates that after symmetry breaking the theory defined by Eq.~(\ref{Action}) contains two hierarchically separated scales $m_{A} \ll m_{\phi}$ for couplings of $\lambda \sim O(1)$ and $f\sim O(1)$.
A theory possessing a hierarchy structure is commonly regarded as unnatural, and one might worry that the fine-tuning issue would appear since the interaction of the vector field $A_{\mu}$ with the scalar field $\phi$ is likely to introduce large quantum corrections to the mass of $A_{\mu}$. However, this is not the case.
To be explicit, let us assume that $\lambda \sim O(1)$ and $f\sim O(1)$ and analyze the quantum effect. From the interaction terms $-{1\over 2} f (\alpha-1)\phi^2(\partial_{\mu}+2f \omega_{\mu} )A^{\mu}$ and $ -{1\over 2} f^2 (\alpha-1)\phi^2 A_{\mu}^2$ in Eq.~(\ref{Lag}), it is straightforward to verify that due to the interaction of $A_{\mu}$ with $\phi$, the radiative correction from the $\phi$ loop to the mass squared for $A_{\mu}$ is
\begin{equation}
\delta m_{A}^2\sim {f^2\over 4\pi^2}(\alpha-1)m_{\phi}^2,
\end{equation}
which is a rather modest correction, compared with the tree-level result in Eq.~(\ref{MAA}).
The above result shows that the light scale $m_{A}$ is not sensitive to the heavy scale and the theory is technically natural. As we discussed in Sect. 2, this is due to the fact that the Lagrangian~(\ref{Lag}) possesses an enhanced Poincar$\rm\acute{e}$ symmetry in the limit $\alpha \rightarrow 1$. It is this extended spacetime symmetry that ensures the stability of the light scale by protecting it from large radiative corrections.
Surely in the above argument, we have neglected gravitational effects. However, it is credible that at energies well below the Planck scale, gravitational effects only give rise to small corrections and will not destabilize the hierarchy.
\section{Effective potential}
In Sect. 2, we have shown that for $\alpha <1$ and $\omega_{\mu}$ timelike, the breakdown of Weyl symmetry generates a negative mass squared for the scalar field $\phi$. Since the tree-level vacuum expectation value $\phi_c$ can be modified by perturbative loop corrections, in this section we would like to consider the lowest-order quantum corrections and determine the vacuum expectation value of the quantum field $\phi$.
We will follow the standard procedure by starting with the computation of the effective action, denoted by $\Gamma$. After obtaining $\Gamma$, we will identify an expression for the effective potential $V_{eff}$, whose minimum determines the quantum corrections to spontaneous symmetry breaking.
To calculate the effective potential to one-loop order, we follow the Faddeev-Popov method to gauge-fix the local symmetry and use Feynman gauge in our calculation. Including the gauge fixing term in Eq.~(\ref{Lag}), the gauge-fixed Lagrangian in the flat limit is of the form
\begin{equation}
{\cal{L}}_{FP}={\cal{L}}-{1\over2}(\partial_{\mu}A^{\mu})^2,
\end{equation}
Next, we split the scalar field $\phi (x)$ into a fixed classical part $\phi_c$ (we assume that translational symmetry remains intact) and a quantum fluctuation $\varphi (x)$, given by Eq.~(\ref{split}). For our purposes, we then expand the gauge-fixed Lagrangian ${\cal{L}}_{FP}$ about $\phi_c$ and keep pnly the terms up to quadratic order in quantum fields, We obtain
\begin{eqnarray}
{\cal{L}}_{2}&=&{1\over 2}\partial_{\mu} \varphi(x) \partial^{\mu} \varphi(x) -{1\over 2} m_{\phi}^{2} (\phi_{c}^2+\varphi^2(x)) - {1\over 4!}\lambda (\phi_{c}^4+6\phi_{c}^2\varphi^2(x)+\cdot\cdot\cdot ) \nonumber\\
& &-{1\over2}(\partial_{\mu}A^{\mu}(x))^2-{1\over 2} f^2 (\alpha-1)\varphi^2(x) A_{\mu}^2(x)-{1\over 4} F_{\mu\nu}^{2}(x). \label{L2}
\end{eqnarray}
Note that in ${\cal{L}}_2$ we have dropped terms linear in quantum fields by applying the classical field equations. From ${\cal{L}}_2$, we have
\begin{eqnarray}
-{\delta^2{\cal{L}}_{2}\over \delta A_{\mu}\delta A_{\nu}}&=&-\square \eta^{\mu\nu} + f^2 (\alpha-1)\phi_{c}^{2} \eta^{\mu\nu},\\
-{\delta^2{\cal{L}}_{2}\over\delta \varphi \delta \varphi}& =& \square +m_{\phi}^2+{\lambda \over 2} \phi_{c}^{2}.
\end{eqnarray}
Then, using the background field method, it is straightforward to show that to one-loop order, the effective action $\Gamma[\phi_c]$ is given by
\begin{equation}
\Gamma[\phi_c]=\int d^4 x \left( -{1\over 2}m_{\phi}^{2} \phi_{c}^2 -{1\over 4!}\lambda \phi_{c}^4 +\delta {\cal{L}}[\phi_{c}]\right)+2i\, {\rm log}\,{\rm{det}}\left[ \square- f^2 (\alpha-1)\phi_{c}^{2}\right]+{i\over 2} {\rm log}\,{\rm{det}}\left[ \square+m_{\phi}^2+{\lambda \over 2} \phi_{c}^{2}\right],
\end{equation}
where $\delta {\cal{L}}[\phi_{c}]$ contains all the counterterms to be determined by the renormalization conditions. Notice that for our purpose here we only include the $\phi_c$-dependent part of the effective action.
We remark that each functional determinant appears in $ \Gamma[\phi_c]$ is noting but the determinant of a Klein-Gordon operator with $\phi_c$-dependent mass squared. This field dependence is due to the interaction of the propagating particles with the background field $\phi_c$. We also note that in the Faddeev-Popov procedure, it is important to include Faddeev-Popov ghosts which serve as negative degrees of freedom and render the counting of physical degrees of freedom correct. However, similar to QED, the Faddeev-Popov determinant is independent of $\phi_c$ in our case, and therefore it does not contribute to the effective potential $\Gamma[\phi_c]$.
To compute the functional determinant, we adopt dimensional regularization to evaluate the integral in $d=4-\epsilon$ dimensional spacetime. After this manipulation, we have
\begin{eqnarray}
{\rm log}\,{\rm{det}}\left[ \square+\mu^2(\phi_{c})\right] &=& {\rm Tr}\,{\rm log}\left[ \square+\mu^2(\phi_{c})\right] =-i (VT){ \Gamma({-{d\over2}})\over (4\pi)^{d/2}}( \mu^2(\phi_{c}) )^{d/2} \nonumber\\
&=& -i(VT){\mu^4(\phi_c)\over 2(4\pi)^2}\left({2\over \epsilon}-\gamma+{\rm{log}}\,(4\pi)-{\rm{log}}\, (\mu^2(\phi_c)) +{3\over 2} \right) ,
\end{eqnarray}
where $VT$ is the spacetime volume of the functional integral.
After absorbing the divergences into the counterterms $\delta\cal{L}$ using the modified minimal subtraction scheme, we obtain the effective potential to one-loop order:
\begin{eqnarray}
V_{eff} =-{1\over VT}\Gamma[\phi_c]=-{1\over 2} \vert m_{\phi}^{2}\vert^2 +{\lambda\over 4!} \phi_{c}^{4}& + &{(-f^2(\alpha-1)\phi_{c}^{2})^2\over (4\pi)^2}\left( {\rm{log}}\, \left[ {-f^2(\alpha-1)\phi_{c}^{2}}\over M^2 \right]-{3\over 2} \right)\nonumber\\
& + &{(m_{\phi}^2+{\lambda\over 2}\phi_{c}^2)^2\over4 (4\pi)^2}\left( {\rm{log}}\, \left[ m_{\phi}^2+{\lambda\over 2}\phi_{c}^2\over M^2 \right]-{3\over 2} \right),
\end{eqnarray}
where $M$ is an arbitrary renormalization scale. It follows that the one-loop renormalization group $\beta$-function $\beta(\lambda)$ for $\lambda$ and the anomalous dimension $\gamma_{\phi_{c}^2} $ of the operator $\phi_{c}^2$ are, respectively,
\begin{eqnarray}
\beta(\lambda)&=&{3\over 16\pi^2}\left( \lambda^2+16 f^4 (\alpha-1)^2 \right),\\
\gamma_{\phi_{c}^2}&=&{\lambda \over 16\pi^2}.
\end{eqnarray}
For $V_{eff}$ to make any sense, the arguments of the logarithms must be positive. Thus, for $\alpha<1$ and $\omega_{\mu}$ timelike, the quantum correction to $V_{eff}$ is consistent only if
\begin{equation}
m_{\phi}^2+{\lambda\over 2}\phi_{c}^2 >0, \label{mass}
\end{equation}
that is,
\begin{equation}
\phi_{c}^2 > {2 f^2(1-\alpha) \over \lambda} \omega_{\mu}^2 .
\end{equation}
Expanding $V_{eff}$ in powers of $\vert m_{\phi}^{2}\vert$, we have
\begin{eqnarray}
V_{eff}&=& {\lambda\over 4!} \phi_{c}^{4} + {(f^2(\alpha-1)\phi_{c}^{2})^2\over (4\pi)^2}\left( {\rm{log}}\, \left[ {-f^2(\alpha-1)\phi_{c}^{2}}\over M^2 \right]-{3\over 2} \right)+ {(\lambda\phi_{c}^2)^2\over 16 (4\pi)^2}\left( {\rm{log}}\, \left[{\lambda\phi_{c}^2\over 2 M^2} \right]-{3\over 2} \right) \nonumber\\
&- & {1\over 2} \vert m_{\phi}^{2}\vert \phi_{c}^2 \left( 1+ {\lambda\over 2 (4\pi)^2}\left( {\rm{log}}\,\left[ {\lambda\over 2}{\phi_{c}^{2}\over M^2} \right]-1 \right) \right)+{\rm{O}}(m_{\phi}^4).
\end{eqnarray}
The above result shows that for small values of $\vert m_{\phi}^{2}\vert$, the one-loop contributions to the effective potential takes the form
\begin{equation}
V_{eff}^{(1)}=a_{1} \phi_{c}^2 +a_{2} \phi_{c}^4+{1\over 16\pi^2} \left[ \left(f^4(\alpha-1)^{2}+{\lambda^2\over 16}\right) \phi_{c}^4\, {\rm{log}}\, \phi_{c}^{2} -{\lambda\over 4} \vert m_{\phi}^{2}\vert \phi_{c}^{2} \, {\rm{log}}\, \phi_{c}^{2} \right] + {\rm{O}}(m_{\phi}^4).
\end{equation}
Imposing the following renormalization conditions on the one-loop effective potential:
\begin{equation}
V_{eff}^{(1)\,''''}\vert_{\phi_{c}=\mu}=0,\,\,\,{\rm{and}}\,\,\,\,V_{eff}^{(1)\,''}\vert_{\phi_{c}=\mu}=0, \label{ren}
\end{equation}
we find
\begin{eqnarray}
a_{1}&=& {1\over 16\pi^2}\left( 18 \left(f^4(\alpha-1)^2+{\lambda^2\over 16}\right)\mu^2 +\lambda\vert m_{\phi}^{2}\vert +{\lambda\over 4}\vert m_{\phi}^{2}\vert \, {\rm{log}}\, \mu^2 \right) ,\\
a_{2}&=& -{1\over 16\pi^2}\left( \left(f^4(\alpha-1)^2+{\lambda^2\over 16}\right)\left( {\rm{log}}\, \mu^2+{25\over 6} \right) +{\lambda\over 24}{\vert m_{\phi}^{2}\vert \over \mu^2} \right).
\end{eqnarray}
Next, following Coleman and Weinberg \cite{Coleman} to choose $\mu=\langle \phi \rangle$ with $\langle\phi\rangle$ at the local minimum of the effective potential\footnote{With this choice, the renormalization condition for $V_{eff}^{(1)\,''}$ is equivalent to $V_{eff}^{'' }\vert_{\phi_{c}=\langle \phi\rangle}=(V_{eff}^{(0)\,''}+ V_{eff}^{(1)\,''})\vert_{\phi_{c}=\langle \phi\rangle}= V_{eff}^{(0)\,''}\vert_{\phi_{c}=\langle \phi\rangle}=-\vert m_{\phi}^2\vert +{\lambda\over 2} \langle \phi \rangle^2$. This guarantees that $\langle \phi \rangle$ is a local minimum, because of Eq.~(\ref{mass}).}, and then including the classical potential, we finally have
\begin{eqnarray}
V_{eff}(\phi_{c})&=&{1\over 2} \left[ {9\over 4\pi^2}\left(f^4(\alpha -1)^2+{\lambda^2\over 16}\right)\langle \phi \rangle^2 -\left( 1+{\lambda \over 8\pi^2} \right) \vert m_{\phi}^2\vert \right] \phi_{c}^{2}+ {\lambda\over 4!}\left(1-{1\over 16\pi^2}{ \vert m_{\phi}^2 \vert \over \langle \phi \rangle^2} \right)\phi_{c}^4 \nonumber \\
&+&{1\over 16\pi^2}\left[\left(f^4(\alpha -1)^2+{\lambda^2\over 16}\right)\phi_{c}^{4}\left( {\rm{log}}\left({\phi_{c}^2\over \langle \phi \rangle^2}\right)-{25\over 6} \right)-{\lambda\over 4}\vert m_{\phi}^2\vert \phi_{c}^{2}\,{\rm{log}}\left({\phi_{c}^2\over \langle \phi \rangle^2}\right) \right]\nonumber\\
&+&{\rm{O}}(m_{\phi}^4). \label{V}
\end{eqnarray}
It is then straightforward to show that $V_{eff}$ has its minimum at
\begin{equation}
\langle \phi \rangle^2 ={(1+{\lambda\over 6\pi^2})\over {\lambda\over 6}+{4\over 3\pi^2}\left(f^4(\alpha -1)^2+{\lambda^2\over 16}\right)}\vert m_{\phi}^{2}\vert, \label{min}
\end{equation}
which is away from the origin as long as $\vert m_{\phi}^{2}\vert \neq 0$. This is to be compared to the tree-level result $\langle \phi \rangle_{tree}^2=6\vert m_{\phi}\vert^{2}/\lambda$.
In the final part of this section, let us consider the case where $m_{\phi}^{2}=0$ (i.e., $\omega_{\mu}$ lightlike) in Eq.~(\ref{Lag}). Naively, one might think that the effective potential~(\ref{V}) and the location of its minimum~(\ref{min}) could be applied to the case where $\omega_{\mu}$ is lightlike by simply taking $m_{\phi}=0$ in Eq.~(\ref{V}) and Eq.~(\ref{min}). However, this is incorrect. The reason is that when $m_{\phi}=0$, the renormalization condition for $V_{eff}^{(1)\,''}$ in Eq.~(\ref{ren}) implies that
\begin{equation}
V_{eff}^{''}\vert_{\phi_{c}=\langle\phi\rangle}={\lambda\over 2} \langle \phi\rangle^2.
\end{equation}
This result is consistent with the assumption that $\langle\phi\rangle$ is at a local minimum of $V_{eff}$ only if $\langle\phi\rangle \neq 0$ (so that $V_{eff}^{''}\vert_{\phi_{c}=\langle\phi\rangle}>0$). Nevertheless, Eq.~(\ref{min}) tells us that when $m_{\phi}=0$, $\langle\phi\rangle$ is away from the origin only if ${\lambda\over 6}+{4\over 3\pi^2}\left(f^4(\alpha -1)^2+{\lambda^2\over 16}\right)=0$, which cannot be satisfied for $\lambda>0$. Therefore, we need to adopt another renormalizaton condition for $ V_{eff}^{(1)\,''}$ when $m_{\phi}=0$.
To do this, let $V_{0,eff}$ denote the effective potential for the $m_{\phi}=0$ case, and let $\langle\phi\rangle $ again denote the location of the minimum of $V_{0,eff}$. Then, following the same manipulations as in the $\vert m_{\phi}^{2}\vert\neq 0$ case, except choosing new renormalization conditions:
\begin{equation}
V_{0,eff}^{(1)\,''''}\vert_{\phi_{c}=\langle\phi\rangle}=0,\,\,\,{\rm{and}}\,\,\,\,V_{0,eff}^{(1)\,''}\vert_{\phi_{c}=0}=0,
\end{equation}
we find
\begin{equation}
V_{0,eff}(\phi_{c})={\lambda\over 4!}\phi_{c}^{4}+{1\over 256\pi^2}\left(16f^4(\alpha -1)^2+\lambda^2\right)\phi_{c}^{4} \left({\rm{log}}\left({ \phi_{c}^2 \over \langle\phi\rangle^2} \right)-{25\over 6} \right). \label{V0}
\end{equation}
Notice that when $\alpha=1$, The above expression reduces to the effective potential of the conformal scalar theory to one-loop order. This should be an expected result, after the discussion leading to Eq.~(\ref{SF}) in Sect. 2.
From Eq.(\ref{V0}), it is easy to show that $\langle\phi\rangle $ is away from the origin only if
\begin{equation}
f^4(\alpha -1)^2+{\lambda^2\over 16}={2\pi^2\over 11}\lambda, \label{lambda}
\end{equation}
which is consistent with the validity of the perturbation theory if $\lambda$ is of order $f^4 (\alpha-1)^2$. With the value of $\lambda$ satisfying Eq.~(\ref{lambda}), the effective potential $V_{0,eff}$ can be re-expressed as
\begin{equation}
V_{0,eff}(\phi_{c})={\lambda\over 88}\phi_{c}^{4} \left({\rm{log}}\left({ \phi_{c}^2 \over \langle\phi\rangle^2} \right)-{1\over 2} \right). \label{V00}
\end{equation}
The above result exhibits dimensional transmutation such that the dimensionless quantity $f^2(\alpha-1)$ is traded for the dimensional scale $\langle\phi\rangle$.
The masses of $\phi$ and $A_{\mu}$ are thus given by
\begin{equation}
m_{\phi}^2=V_{0,eff}^{''}\vert_{\langle\phi\rangle}={\lambda\over 11}\langle\phi\rangle^2\simeq{f^4(\alpha-1)^2\over 2\pi^2}\langle\phi\rangle^2,
\end{equation}
and
\begin{equation}
m_{A}^2=f^2(1-\alpha)\langle\phi\rangle^2 ={11f^2(1-\alpha)\over \lambda} m_{\phi}^2,
\end{equation}
which is physically sensible for $\alpha <1$.
This analysis justifies that for lightlike $\omega_{\mu}$, the $Z_{2}$ symmetry of the $\phi$ field can be broken dynamically via Coleman-Weinberg mechanism and from which the masses of fields are generated.
\section{Conclusion}
Probing the possible violation of Lorentz symmetry is an interesting subject both theoretically and experimentally. In this paper, we investigated the scenario in which Lorentz violation is realized by a non-vanishing vacuum expectation value of the Weyl field after Weyl symmetry breaking. In particular, a specific scalar field theory with Weyl invariance has been studied. Characteristic features of different symmetry-broken phases after Weyl symmetry breaking and their dependence on the spacetime character of the Lorentz-violating background were discussed. We found that in either a lightlike or a timelike Lorentz-violating background, the Weyl field could acquire a positive mass-squared after Weyl symmetry breaking. In the lightlike case, this is achieved via dynamical symmetry breaking. In contrast, in a spacelike one, the Weyl field remains massless after Weyl symmetry breaking in this theory.
A weyl invariant theory also serves as an intriguing model for one to explore and gain insights into not only the origin but also the stability of separated mass scales. We have analyzed the naturalness of the theory. The upshot is that, due to the enhanced Poincar$\rm{\acute{e}}$ symmetry in the limit $\alpha\rightarrow 0$, the light mass-squared parameter receives no large quantum corrections.
Based on these results, keeping in mind that the Weyl field does not couple to the gauge and fermion fields in the SM, further explorations of this scenario and its possible phenomenological applications in physics beyond the SM and in cosmology are appealing. We hope to probe them in the near future.
\section*{Acknowledgments}
\hspace*{\parindent}
This research was supported in part by the National Nature Science Foundation of China under Grant No. 11665016.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,204 |
\section{Introduction \& Background}
During plastic deformation, metallic materials may experience localized slip on glide planes over numerous inter-atomic distances \cite{Mitchell1993}. As a consequence of the formation of slip bands within a material, steps form where the slip bands intersect the surface of the sample and can be observed using optical \cite{Lombros2016}, confocal \cite{LIU2019260}, scanning electron \cite{Stinville2015expmech}, and atomic force microscopy \cite{BONNEVILLE200887,AUBERT20169}. Information about the deformation processes that occur during monotonic and cyclic loading is obtained by analyzing the slip traces and their characteristics at the surface in relation to the microstructure. Such information has provided a deep understanding of deformation processes that occur in the plastic regime. For instance, the analysis of surface slip traces in the micro-plastic regime (plasticity prior to macroscopic yield) \cite{MAA2018338} has provided insights on dislocation interactions, avalanche processes and incipient slip and their effect on mechanical properties \cite{Fan2021}. The development of high-resolution digital image correlation (HR-DIC) inside the scanning electron microscope (SEM) provided opportunities to statistically investigate incipient slip in metallic materials \cite{Chen2018}, and to relate the intensity of surface slip events to fatigue properties \cite{STINVILLE2020172}. The analysis of slip traces by SEM, focused ion beam SEMs (FIB-SEMs) and/or atomic force microscopy for specimens deformed cyclically in the macroscopic plastic regime has led to the identification of the micro-scale characteristics of fatigue structures such as persistent slip bands or deformation bands \cite{MAN20101625,Mughrabi2009,Ho2015}. These analyses have provided new insights into fatigue crack nucleation mechanisms, and in the high strain regime further understanding of the strain hardening behavior \cite{WELSCH2016188,WIECZOREK2016320}. The experimental observation/analysis of slip traces is also extensively used to study the transmission of slip across interfaces \cite{Kacher2014}. Most transmission models are based on geometric or orientation-driven relationships with incorporated mechanical loading models \cite{Bayerschen2015}. For example, the m' factor \cite{Luster1995} successfully describes surface slip transmission in titanium alloys \cite{BIELER2014212,HEMERY2018277}. These models have been validated using 2D experimental surface observations of slip transmission, which may be influenced by free surface effects. Slip transmission may significantly differ in the bulk and has not been investigated until now.
While there exist many experimental tools to observe surface or near sub-surface slip events, there are few methods for gathering this information in the bulk. Techniques where thin foils or lamellae are imaged in a electron transmission configuration, including the transmission electron microscope (TEM) \cite{SURI19991019} and transmission SEM (known as t-SEM or SEM STEM) \cite{STINVILLE2019152} can probe slip events in a sample extracted from the interior of a sample. However, the sample lamellae are extremely small and thin (typically \SI{10}{\micro\meter} $\times$ \SI{10}{\micro\meter} $\times$ \SI{100}{\nano\meter}), which prevents the statistical analysis of slip over large regions containing many grains with varying orientations, or during complex loading conditions.
Over the past two decades, synchrotron X-ray techniques have been developed that enable the imaging of grains \cite{Larson:2002,Poulsen2012,Suter2006,Ludwig2009,Bernier2020} and sub-grain structure \cite{Ludwig2007,Vigano2016,Nygren2020}, stress states within grains \cite{Henningsson2020,Reischig_COSS_2020,Shen2020, Miller2020}, and recently dislocations or slip events within grains using dark field X-ray microscopy and topotomography \cite{yildirim_2020,Proudhon2018}. The recent increase in spatial resolution and detector sensitivity due to upgrades to synchrotron light source flux (ESRF-EBS upgrade \cite{Cho2020,ESRF-EBS-upgrade}) and advances in detector technology (i.e. single photon counting pixel detectors and back-illuminated sCMOS cameras) provide the opportunity to directly image plastic deformation events throughout the interior of mm$^3$-scaled samples. The combination of X-ray diffraction contrast tomography (DCT), topotomography and phase contrast tomography (PCT) has been particularly effective for imaging slip events in the bulk polycrystalline structure, in situ during deformation and then relating these events to the 3D grain structure \cite{Proudhon2018}. However, the mechanisms giving rise to image contrast associated with slip events in the X-ray topographs are not well understood and the sensitivity of the technique for capturing all slip events was not demonstrated.
In this research, quantitative correlative measurements between HR-DIC, X-ray DCT and topotomography are performed on a Ti-7Al titanium alloy in order to investigate the sensitivity of the X-ray topotomography method to probe slip events both at the surface and in the bulk. The opportunities provided by the X-ray topotomography method to investigate plastic localization and transmission in metallic materials are demonstrated and discussed.
\section{Methods}
\subsection{Ti-7Al Structure and Sample Preparation} \label{sec:MethodsSample}
Ti-7Al was selected as a model material in the present work due to its relatively large equiaxed grain structure and extensive study in the literature \cite{CHATTERJEE201635,Lienert2009,Venkataraman2017}. This material was extruded and well-annealed to produce large ($\sim$\SI{100}{\micro\meter}), fully recrystallized grains with minimal intragranular substructure \cite{CHATTERJEE201635}. The Ti-7Al material contains $\alpha_2-\mathrm{Ti}_\mathrm{3}\mathrm{Al}$ precipitates that are roughly 2-3 nm in size within the $\alpha$ hexagonal close packed matrix phase. More details on the Ti-7Al alloy structure can be found elsewhere \cite{Venkataraman2017}.
The specimen geometry used for correlative measurements is presented in \autoref{DIC_ts3}(a) and was prepared by wire electrical discharge machining to minimize residual stresses and plastic deformation from machining. The gauge section of the sample is rectangular with a cross sectional area of \SI{0.6}{\milli\meter} x \SI{0.6}{\milli\meter}.
The flat sections of the gauge surface orthogonal to the loading direction were mechanically mirror polished using abrasive papers and diamond suspension, then chemo-mechanically polished using a suspension of \SI{0.04}{\micro\meter} colloidal silica particles. Prior to deformation, a speckle pattern was obtained by chemical etching by immersion in Kroll's reagent for \SI{20}{s}.
\subsection{High Resolution Digital Image Correlation and microstructure measurements}
Tensile tests were performed using a custom in situ $\pm$ 5000 N stage within a Thermo Fisher Scientific Versa3D microscope with a field-emission electron emitter on the flat dogbone-shaped specimens described in \autoref{sec:MethodsSample}. Tensile tests were interrupted at a macroscopic strain level near 0.8\% (just past the 0.2\% offset yield strength) and then HR-DIC measurements were performed. Macroscopic strain was measured in situ using both a strain gauge and fiducial markers located at both ends of the gauge length.
SEM image sets were acquired before loading and while under load following the guidelines of Kammers and Daly \cite{Kammers_2013a,Kammers_2013b} and Stinville et al \cite{Stinville_2015a}. A National Instruments\textsuperscript{TM} scan controller and acquisition system (DAQ) was used to control electron beam scanning in the microscope. This custom beam scanner removes the SEM beam defects associated with some microscope scan generators \cite{Stinville_2015a,LENTHE201893}. Tiles of $8 \times 4$ images before and after deformation with an image overlap of 15\% were used. HR-DIC calculations were performed on these series of images and the results merged using a pixel resolution merging procedure, which is described in detail elsewhere \cite{Chen2018}. HR-DIC measurements were performed on the entire gauge section surface, an area of about \SI{2}{\milli\meter} $\times$ \SI{600}{\micro\meter}. Typical subset size values of $31 \times 31$ pixels (\SI{1044}{\nano\meter} $\times$ \SI{1044}{\nano\meter}) with a step size of 3 pixels (\SI{101}{\nano\meter}) were used for the DIC measurements. Digital image correlation was performed using the Heaviside-DIC method \cite{BOURDIN2018307,Valle2015}. The sample preparation, imaging conditions and Heaviside-DIC parameters enable the detection of slip events at the surface of the specimen with a discontinuous displacement resolution between 0.2 and 0.3 pixels (\SI{7}{\nano\meter} and \SI{10}{\nano\meter} respectively) \cite{BOURDIN2018307,Valle2015}. The strain map of the investigated specimen deformed at 0.27\% plastic deformation is displayed in \autoref{DIC_ts3}(c and d) for the entire gauge length and for a reduced region of interest (ROI), respectively. The axial loading direction is vertically oriented in all strain maps. Bands of concentrated strain identify the locations of slip events at the surface of the specimen. It is worth noting that enhanced plasticity occurred near the edges of the gage section of the specimen due to slight misalignment during mechanical loading.
\begin{figure}
\centering
\includegraphics[width=1\textwidth]{DIC_ts3.jpg}
\caption{(a) Specimen geometry used for correlative in situ HR-DIC, DCT and topotomography measurements. (b) Engineering stress-strain curve of the Ti-7Al material. (c) HR-DIC $\epsilon_{zz}$ strain map obtained at the surface of the specimen after about 0.8\% macroscopic deformation. (d-e) Reduced region of the $\epsilon_{zz}$ strain map with its associated grain structure obtained by electron backscatter diffraction measurement.}
\label{DIC_ts3}
\end{figure}
Microstructure characterization on the sample gauge surface was performed by electron backscatter diffraction (EBSD) measurements with an EDAX OIM-Hikari XM4 EBSD detector using step size of \SI{0.5}{\micro\meter}. Diffraction patterns were acquired using an accelerating voltage of \SI{30}{\kilo\volt}, a $4 \times 4$ binning mode and a beam current of \SI{0.2}{\nano\ampere}. EBSD maps were acquired before and after deformation. The data registration between the gauge surface EBSD and HR-DIC measurements is done by aligning control points using a polynomial distortion. About 50 pairs of control points were picked over the area of interest. The procedure is detailed elsewhere \cite{Charpagne2019}.
\subsection{Diffraction Contrast Tomography} \label{sec:ts3_dct}
X-ray characterization was carried out at the ESRF materials science beam line ID11 in Grenoble, France. The beam energy was set to 38 keV with a relative bandwidth of $3\times10^{-3}$. The experimental setup includes a diffractometer designed for 3D diffraction experiments, including DCT and topotomography imaging modalities \cite{Lud01a,Ludwig:2007ab} (see Section~\ref{sec:ts3_tt_setup}).
The Ti-7Al specimen was mounted vertically on the rotation stage. The microstructure in the gauge length was first characterized by two partially overlapping DCT scans, each composed of 3600 equally spaced projections over \SI{360}{\degree}. Images were recorded with an exposure of \SI{0.25}{s} on a high-resolution detector composed of a scintillator placed 6.5 mm after the sample that is optically coupled by a $10\times$ objective to a 2048 $\times$ 2048 pixel ESRF Frelon camera, giving an effective pixel size of \SI{1.4}{\micro\meter}. The total scan time was 30 minutes.
Each scan was reconstructed using the DCT software \cite{dct_code} to produce grain maps of the bulk region subsurface below the HR-DIC measurements that also contained the topotomography measurements. The transmission of the direct beam was used to reconstruct the exact shape of the illuminated region using a classical filtered back projection. The two DCT grain volumes were fused together into one dataset using Pymicro \cite{Pymicro} (including absorption data to produce a mask for the specimen geometry) after matching grains in the overlapping region and minimizing the correlation error between the two DCT volumes present in the overlapped region. Isotropic dilation was also carried out, but limited to the grains within the absorption mask, in order to fill all remaining void space within the DCT reconstruction. This resulted in a high fidelity grain map of the entire ROI of the sample (\SI{0.6}{\milli\meter} $\times$ \SI{0.6}{\milli\meter} $\times$ \SI{1.1}{\milli\meter}), containing 1055 grains. The average grain size in the reconstructed DCT volume, including surface grains, is \SI{68}{\micro\meter}. As a side note, besides the front face that was mirror polished for EBSD analysis, the other 3 lateral faces were left as received from the EDM machining. This resulted in a \SI{10}{\micro\meter} thick non-diffracting layer of material that is absent from the final reconstruction.
\subsection{Topotomography setup} \label{sec:ts3_tt_setup}
The reconstructed grain map (see Section \ref{sec:ts3_dct}) was processed to select grains for topotomography inspection. After matching the EBSD data to the surface of the reconstructed DCT grain map, two surface grains (labeled 63 and 567 and visible on \autoref{fig:Transmission_1}) with surface slip traces identified by HR-DIC were selected from two different locations. More grains were selected based on the following criteria: (i) the grain is present in the neighborhood of the two selected surface grains (ii) at least one low index reflection can be aligned within the range of the sample goniometer. An automated script searched through all the indexed grains and selected 55 different grains for topotomography acquisition. For some of these grains, several reflections (up to 3) were available within the goniometer ranges and were also collected. In total, 75 topotomography scans were collected for two grain clusters in the gauge length of the sample.
For topotomographic scan acquisition a particular scattering vector of a given grain is aligned parallel to the tomographic rotation axis \cite{Ludwig2007}. The tomographic rotation axis itself is inclined by the Bragg angle $\Theta$ (diffractometer base tilt) in order to maintain Bragg condition throughout the entire \SI{360}{\degree} rotation of the sample. To account for lattice rotations and dispersion effects within the grain, the base tilt is scanned over the width of the crystal reflection curve using an alternating, continuous scan motion with a step size of \SI{0.05} degrees. A second camera with a higher magnification was used to record these projection topographs with a pixel size of \SI{0.7}{\micro\meter}. In the current experiment, topotomography scans were comprised of 90 equispaced series of projection topographs (one every \SI{4}{\degree} of sample rotation) and require between 10 minutes to a couple of hours to collect, depending on the level of mosaicity (the $\Theta$ integration range must be increased for more deformed grains, a typical value was used for this sample with a range of \SI{1.6}{\degree} and 64 images). The topotomography experimental setup is presented in the Supplementary Material, including a high-speed video (20 $\times$ speed) of the sample goniometer stages moving during data collection. A snapshot of the video is provided in \autoref{fig:TopoTomoSetup} to describe the experimental setup. An example of a set of topographs is provided in Supplementary Material for a grain that is subject to plastic deformation.
\begin{figure}
\centering
\includegraphics[width=1\textwidth]{TopoTomoSetup.jpg}
\caption{Experimental setup for topotomography measurement. The complete setup is rotated by the base tilt $\Theta$ while maintaining Bragg condition visibility for a specific grain throughout the entire \SI{360}{\degree} rotation around the tomographic rotation axis. A high resolution detector with a pixel size of \SI{0.7}{\micro\meter} is used to record series of integrated projection topographs every \SI{4}{\degree}. The reader is invited to consult the Supplemental Material on the experimental setup for topography measurement.}
\label{fig:TopoTomoSetup}
\end{figure}
X-ray topographs contain orientation contrast from the diffraction of a crystal. Defects present in the volume locally affect the Bragg condition and will give rise to contrasts evolving as a function of the propagation distance. To select the best detector distance to acquire topotomography scans, grain 567 was first aligned using reflection \hkl(1 1 -2 0) and integrated projection topographs were recorded while increasing the sample detector distance from \SI{7}{\milli\meter} to \SI{41}{\milli\meter} in steps of \SI{2}{\milli\meter}. The contrast related to slip events evolves from low contrast intensity at short propagation distance to high contrast intensity at large propagation distance, whereas the outline of the grain evolves from a nearly geometric projection into a distorted projection with ill defined boundaries at large distances, as shown in \autoref{fig:ts3_tt_nfdtx}. For the rest of the experiment the detector distance was kept at \SI{12}{\milli\meter}, allowing enough space to rotate the tension specimen freely without risking a collision bewteen the goniometer stage and the detector. Note that in situ testing (not reported here) requires an optimized design of the load frame \cite{Proudhon2018} and of the detector head in order to maintain such short propagation distances. Also note that in for future experiments, acquisitions with different propagation distances provide additional constraints which can improve the convergence of the iterative optimization algorithms used for reconstructing the local orientation field inside a grain \cite{Vigano_COSS_2020}.
\begin{figure}
\begin{tikzpicture}
\node[above] {\includegraphics[width=\textwidth]{figures/tt_ts3_grain11_x_series.png}};
\node at (-6.2,0) {scan 1, $d=7$ mm};
\node at (-2.1,0) {scan 2, $d=12$ mm};
\node at (2.1,0) {scan 3, $d=29$ mm};
\node at (6.2,0) {scan 4, $d=41$ mm};
\draw[green,thick] (-4,-0.4) rectangle (-0.1,4.0);
\node[above] at (-2.1,4.0) {distance selected};
\end{tikzpicture}
\caption{Topographs of grain 567 extracted from the topotomography scans ($\omega$=\SI{44}{\degree}) taken at different detector distances. The second distance of \SI{12}{\milli\meter} was selected for our observations. A movie showing the evolution of contrast as a function of the detector distance is available as Supplementary Material.}
\label{fig:ts3_tt_nfdtx}
\end{figure}
\subsection{Detecting slip system activation}
The 75 topotomography scans were each post-processed to detect slip activity. The grain shape, position, and the 4 goniometer rotation angles were used to produce a simple forward simulation (parallel projection of the 3D grain volume) in the topotomography configuration. Using the grain orientation, slip planes are added in the volume of the grain by manually selecting the best match with the orientation contrast visible in the topographs over the \SI{360}{\degree}. If necessary, the active slip system was then identified using the highest Schmid factor within the plane. This is a tedious but efficient way of identifying slip planes that have produced plasticity in all the grains \cite{Proudhon2018}. \autoref{fig:ts3_tt_ex} depicts two topographs close to edge-on configuration (i.e. diffracted beam co-planar to slip plane) with their slip system identified and instantiated in the simulated projection images on the right figure panels. This process was repeated for the whole set of 75 topotomography scans comprising 55 grains.
\begin{figure}[htb]
\centering
\begin{tikzpicture}
\node[above] {\includegraphics[width=0.8\textwidth]{figures/tt_fsim_planes_identification.pdf}};
\end{tikzpicture}
\caption{Examples of slip planes identified by comparison between the recorded topographs (a and c for grains 572 and 579 respectively) and the forward modeling of the diffraction for the corresponding grains (b and d), the slip plane intersections with the grain boundary are marked in color.}
\label{fig:ts3_tt_ex}
\end{figure}
\subsection{Orientation spread plots from topotomography}
\label{sec:ts3_tt_orientation_spread}
Whereas the projection topographs shown in \autoref{fig:ts3_tt_ex} are integrated over angular range covered by the base tilt $\Theta$, the individual images taken during this motion (termed a rocking scan) constitute a rocking curve for each particular angle $\omega$. The change observed in the Bragg condition here is largely dominated by the lattice rotation (compared to elastic lattice distortions) and can be directly attributed to the orientation spread around the base tilt axis. To quantify this, the full width of the rocking curve (simply integrating the intensity collected by the detector at this angle) at 10\% of the peak is measured for each position of $\omega$ (every \SI{4}{\degree}) \cite{Proudhon2018}. To compute this value reliably, a mask for the grain being imaged needs to be created, because grains with similar Bragg conditions can diffract and be captured on the detector during the rotation. This phenomenon happens frequently enough to prevent a simple sum of the intensity on the detector, even with a proper ROI applied. To work around this problem, a forward simulation routine of the diffraction imaging was developed with Pymicro \cite{Pymicro} to create a mask of the grain projected onto the detector using the DCT reconstruction and the value of all motor positions (see \autoref{fig:OSplots}a). The mask was dilated by 10 pixels to allow for small grain deformations (and hence image distortions), which are not accounted for in the 3D DCT grain reconstruction used as input to create the mask.
A representative set of orientation spread curves are shown in \autoref{fig:OSplots}. They almost all depict a characteristic dumbbell shape as already observed in a previous topotomography experiment \cite{Proudhon2018}. This shape is the signature of a crystal bent perpendicularly to the elongation direction (in $\omega$) and can be related to dislocation activity and the presence of geometrically necessary dislocations (GNDs) introducing lattice rotations in the grains \cite{Gueninchault2016}. This is consistent with a single slip dominated deformation state as observed in most of the grains. An in depth look at the data processing shows that most of the $I_\omega(\Theta)$ rocking curves display regular single peak shapes which are the signature of rather homogeneous rotation field while some other display double peak shapes characteristic of more heterogeneous distribution of the lattice rotation or even formation of a subgrain (although this specimen was scanned ex situ so the initial state in the bulk is unknown). In this experiment some of the grains have been measured in topotomography for the first time using several reflections. For instance \autoref{fig:OSplots}b shows the orientation spread plots for grain 699 obtained with reflections \hkl(1 0 -1 1) and \hkl(1 1 -2 2) and one can see that the results are in very good agreement. Using several reflections will be useful in the future when extracting the orientation field from the measurements \cite{Vigano_COSS_2020} (not done in this paper).
\begin{figure}
\centering
\includegraphics[width=1\textwidth]{orientation_spread.pdf}
\caption{Orientation spread plots. (a) for each $\omega$ angle the individual topographs are masked and the resulting integrated intensity is fitted to extract the width at 10\% of the peak; the evolution of this value with $\omega$ depicts a typical dumbbell shaped curve here shown for grain 579 located in the bulk. (b) comparison of the effective misorientation measured on the same grain with two different reflections. (c) effective misorientation plots for surface grains 34 and 63. Note that small spikes can sometimes appear on the curves when spurious diffracting grains are located within the actual grain mask.}
\label{fig:OSplots}
\end{figure}
\section{Results}
The combination of X-ray diffraction contrast tomography and topotomography allow for the spatially resolved imaging of slip events within grains and with regard to the full 3D grain structure, as shown in \autoref{Topograins}. Reconstruction of the entire specimen gauge section by DCT is shown in \autoref{Topograins}(a). When compared to the EBSD map, the DCT reconstruction appears to be very accurate, including the shape of the surface grains. A detailed analysis showed that all of the 143 surface grains are captured in the EBSD map, except 6 very small and presumable shallow grains. On the other hand, the grains located close to the edges of the lateral faces that are difficult to capture in EBSD (slightly rounded specimen shape due to the polishing) are correctly reconstructed by DCT. The grains investigated by topotomography are displayed in \autoref{Topograins}(b). Two regions of surface grains were selected from HR-DIC measurements, with many grains exhibiting slip events after tensile deformation. The grains selected for topotomography measurements from within the bulk were chosen to be in contact with the selected surface grains and to have at least one low index lattice plane within reach by the diffractometer tilt motions (see Section \ref{sec:ts3_tt_setup}). Analysis of the topographs and the spatial correlation of DCT and topotomography measurements allows slip event detection and the determination of their location in the 3D grain structure, as shown in \autoref{Topograins}(c), for all the grains investigated by topotomography.
\begin{figure}
\centering
\includegraphics[width=1\textwidth]{Topograins.jpg}
\caption{(a) DCT of the Ti-7Al specimen (random colors are used to differentiate grains), the front face corresponds to the free surface where the HR-DIC measurement was performed. (b) DCT of the grain investigated by topotomography. (c) Identification of surface and bulk slip events by topotomography. The red dashed box indicates the region highlighted in \autoref{DIC_ts3}(d-e).}
\label{Topograins}
\end{figure}
\subsection{Comparison between HR-DIC and Topotomography measurements}
Most of the grains investigated by topotomography (both surface and bulk grains) show slip events as observed by HR-DIC measurement. In addition, most of them appear to be in a single slip condition (single slip system activated).
\begin{figure}
\centering
\includegraphics[width=\textwidth]{DIC_detection.jpg}
\caption{Comparison between HR-DIC and topotomography measurements: (a) strain map after deformation of a Ti-7Al specimen at about 0.8\%; associated in-plane slip amplitude map for a grain of interest. The in-plane slip amplitude corresponds to the physical in-plane displacement produced by a slip event onto the specimen free surface. (b) Recorded topographs and reconstructed grain shape and slip events for the grain of interest. (c) Distribution of slip events that are detected (top) and undetected (bottom) by topotomography measurements. The strain maps are used as reference to determine the location and number of slip events. The red arrows indicate two slip events at the edge of the grain of interest that are not detected by topotomography measurements. }
\label{fig:DIC_detection}
\end{figure}
The HR-DIC strain measurements from the surface grains were analyzed completely independently from the topotomography measurements of the same grains, which contain full 3D grain information, in order to minimize potential bias and to validate the datasets with each other. Among the 55 investigated grains, 19 are surface grains. The HR-DIC and X-ray measurements are compared for all of these surface grains - containing 109 HR-DIC measured slip events. Specifically, the location and average in-plane slip amplitude of individual slip events (HR-DIC measurements) were extracted. An example is given in \autoref{fig:DIC_detection}(a and b) for a single grain with 7 intense surface slip events measured by HR-DIC, while only 5 slip events are detected by topotomography. The slip locations detected by topotomography in all the surface grains correlate accurately with slip events detected by HR-DIC measurement. For all the investigated surface grains, 67 slip events were detected by both techniques, while 41 were not detected by topotomography. The distributions of these detected and undetected events by topotomography are reported in \autoref{fig:DIC_detection}(c) as a function of their average in-plane slip amplitude, as obtained from the HR-DIC measurements. It must be noted that slip events with an in-plane slip amplitude below \SI{20}{\nano\meter} can not be differentiated using HR-DIC due to the resolution limit of measuring in-plane slip amplitude with this technique. The slip events that were undetected by topotomography, but detected by HR-DIC, are mainly low intensity slip events of in-plane slip amplitude below \SI{50}{\nano\meter} (or roughly lower than 170 emitted dislocations \cite{STINVILLE2020172}). Further inspection indicates that the most intense undetected slip events by topotomography are located near the edges of the grains as displayed at the red arrows in the grain in \autoref{fig:DIC_detection}(a and b). The detection of slip events at grain boundary is not favored using topotomography for two reasons: first, the diffracting grain volume at the periphery of a grain may be small and tend to have a lower diffraction signal to noise ratio. Second, the edges of the grains are more likely to accumulate high lattice rotations during deformation, which could obscure the contrast resulting from well defined slip plane locations and prevents their detection on the topographs.
\subsection{Contrast at slip events from topotomography measurement} \label{sec:contrast}
While HR-DIC measurements provide the location and quantitative measurement of the slip events (in-plane slip amplitude given in \SI{}{\nano\meter}), topotomography measurements provide the slip location and an associated contrast on the topographs with contributions from the entire volume of the grain. The contrast detected in the topographs associated with slip events in the HR-DIC measured surface grains were classified into 4 groups: no contrast (no slip event detected by topotomography measurement but a slip event detected by HR-DIC), low contrast (slip event detected by topotomography with weak contrast on the topographs), medium contrast (slip events detected by topotomography with average contrast on the topographs), high contrast (slip events detected by topotomography with intense contrast on the topographs). The distributions of the in-plane amplitude as measured from HR-DIC measurements of slip events for these 4 groups are displayed in \autoref{fig:TT_detection}.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{TT_detection.jpg}
\caption{Contrast level of slip bands measurements by topotomography as a function of the HR-DIC surface measured in-plane slip intensity of the identical bands. Only slip events at the surface, or within the grains at the surface, are considered in the present analysis. The HR-DIC strain maps are used as a reference to determine the location and number of slip events. The slip bands that are not detected by topotomography measurements, but are present in the HR-DIC strain maps are labeled as "Not detected". }
\label{fig:TT_detection}
\end{figure}
As a general trend, the visibility of the topographic image contrast associated with a slip event is related to the in-plane slip amplitude of the slip event. High in-plane slip amplitude slip events tend to have high visibility in the topographs. Conversely, low intensity slip events tend to display low contrast in the topographs.
\subsection{Comparison between EBSD and Topotomography measurements}
Measurement of the lattice rotation magnitude induced during deformation has been performed for surface grains by conventional EBSD measurements and topotomography using grain orientation spread (GOS). The spread is defined as the average deviation between the orientation of each point in the grain and the average orientation of the grain. The conventional EBSD measurements provide information from the surface and near-surface of the grains, while topotomography measurements contain a measure of GOS (projections of the grain orientation distribution along the selected scattering vector) for the entire volume of the grain. Note that lattice rotations around the scattering vector are not captured by the rocking curve analysis presented in the current work.
A GOS map is shown in \autoref{fig:spread}(a) from EBSD alongside measurements of the orientation spread made by topotomography. The map in \autoref{fig:spread}(b) was constructed by overlaying the EBSD image quality signal (to provide a sensible background) onto a map of the orientation spread measured by topotomography (see Section \ref{sec:ts3_tt_orientation_spread}, and captures the orientation spread from the entire 3D grain structure. The distribution of the orientation spread obtained from both techniques is displayed in \autoref{fig:spread}(c).
\begin{figure}
\centering
\includegraphics[width=\textwidth]{spread.jpg}
\caption{Grain Orientation Spread (GOS) for the investigated surface grains from conventional 2D EBSD measurements (a) and full 3D topotomography (b) measurements. The distribution of the (GOS) is displayed in (c). }
\label{fig:spread}
\end{figure}
The analysis from EBSD and topotomography of the orientation spread in surface grains shows overall similar values, indicating that the measurement of orientation spread by topotomography is sensitive enough to capture the evolution of lattice rotations during plastic deformation. However, the grain-to-grain GOS variation in EBSD (surface measurement) compared to topotomography (entire 3D grain) is different, indicating the significance of the bulk measurement information. The observed differences can be attributed partly to the lower angular resolution of EBSD and partly to the orientation of the active slip direction with respect to the sample surface, this is discussed in more detail in Section \ref{sec:lattice_rot_discussion}.
\section{Discussion}
Plastic deformation does not occur uniformly during the loading of polycrystalline metallic materials, but instead it occurs initially by localized slip in regions where dislocations first overcome obstacles to deformation. This manifests at the surface of the specimen by the formation of surface steps \cite{Mitchell1993}. With continued straining, numerous occurrences of these slip events result in a continuous network of plastic flow across the entire cross section of the specimen, corresponding to macroscopic yielding of the specimen \cite{Echlin2021}.
Plasticity and its transmission across grains has been extensively investigated at the surface of the specimen during loading. However, very little experimental data is available on the localization and transmission of plasticity in the material bulk as a function of the 3D grain structure. For example, it has recently been observed that the 3D grain structure is especially critical to understanding the origin of incipient localization in nickel-based superalloys \cite{CHARPAGNE2021117037}. The combination of DCT and topotomography \cite{Ludwig2007,Proudhon2018} provides the unique capability to detect slip events as a function of the grain structure. This technique has tremendous opportunity for investigation of the bulk slip localization, however the sensitivity of the technique with regard to slip detection has not yet been quantified and parameters other than the slip amplitude may influence the visibility of slip localization in topographs.
\subsection{X-ray Topotomography slip event detection and slip event contrast}
Compared to a previous study in an AlLi alloy \cite{Proudhon2018}, the level of deformation is higher in the presently investigated Ti-7Al sample (0.8\% vs 0.3\%), as seen in the integrated topographs in \autoref{fig:ts3_tt_ex} and \autoref{fig:grain659_579}(a). Due to the higher level of deformation, the grain projections formed on the topographs are more deformed - making it more challenging to identify the slip planes. Nevertheless, the vast majority of the grains depict strong contrast localization in the form of bands, just as in the AlLi experiment \cite{Proudhon2018}. Using the grain orientations from DCT, these bands can be correlated with active slip planes in the bulk of the grain. Another difference in the Ti-7Al material is that the orientation contrast localized around slip bands remains visible on a much larger angular range than in the previous AlLi study (typically \SI{50}{\degree} vs \SI{10}{\degree}). In summary, using this slip system identification technique, it appears that we approach the upper limit (i.e. 1 or 2\% plastic strain) for the topotomography technique. At increased levels of plasticity, the projection contrast becomes too convoluted to reliably identify the slip planes. This limitation restricts the use of topotomography for investigation of the slip activity at relatively low levels of plastic deformation. However, we note that for fatigue properties - typically much of the activity is dictated by the first few cycles \cite{STINVILLE2020172} at relatively low strain - potentially making topotomography an extremely useful technique.
The results in Section \ref{sec:contrast} demonstrates that slip events with low intensity as detected by HR-DIC measurements were not all detected by topotomography, indicating that low intensity slip events do not produce enough contrast to be detected integrated topotomography images. Note that in this paper we have used integrated images to detect slip events while one could also use individual images (typically 64 per topograph here) to look at contrast variations (hardly tractable with a manual approach as pursued here but with potentially increased sensitivity if automated). Topographic image contrast can be roughly divided into two categories: (i) contrasts related to dynamical diffraction and extinction effects and (ii) orientation contrast linked to local variations of the effective misorientation (i.e. the combined effect of rotation and elastic distortions of the crystal lattice) \cite{Tanner1976}. Whereas the former can be predominant in nearly perfect single crystals, the latter are more relevant in the case of metal grains, typically containing a high density of dislocations, even in the as-recrystallized state of the material. Topographic image contrast of slip localization events can be attributed to local gradients in lattice rotation induced by high dislocation densities generated by the avalanche and pile-up processes associated with slip events. These processes are well investigated in the literature \cite{Alcala2020,GUO2015229,Villechaise_2012}. The interaction between dislocation pile-ups and grain boundaries gives rise to heterogeneous stress distributions. Such stress heterogeneity leads to very local gradients in effective misorientation of the crystal lattice \cite{GUO2015229,Villechaise_2012,Polcarova2006} that are captured by topotomography measurements. Other effects, like local reduction of the scattered intensity due to the disorder in the crystal lattice in proximity of slip localizations are likely to contribute to the intensity variations observed in the topographs.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{grain659_579.jpg}
\caption{(a) Topograph of an investigated grain with intense contrast at slip events (green arrows) and impingement of slip events at the boundary from a neighboring grain (red arrows). (b) 3D DCT reconstruction of the investigated grain (blue) and a neighboring grain (yellow) and the overlaid slip event planes, as detected by topotomography. (c-d) Slip directions of the active basal slip systems for the investigated grain and neighboring grain. }
\label{fig:grain659_579}
\end{figure}
To further investigate the characteristics of the contrast at slip events in the topograph, a specific grain in the specimen bulk with significant topograph contrast intensity is displayed in \autoref{fig:grain659_579}. The topograph from the tilt series with the highest contrast intensity is presented in \autoref{fig:grain659_579}(a). Four slip events within the investigated grain are observed and depicted by the purple planes in the reconstruction of the grain (blue colored) and slip events in \autoref{fig:grain659_579}(b). The contrast intensity varies along the plane of the slip events and is especially high on the left and right edges, as indicated by green arrows. Interestingly, intense contrast is also observed at the bottom edge of the grain (at the red arrows in \autoref{fig:grain659_579}(a)), which correspond to the impingement of two slip bands that developed in the neighboring grain (yellow grain in \autoref{fig:grain659_579}(b)).
Both slip systems in the investigated grain (blue) and neighboring grain (yellow) were identified as basal type. The slip direction for the active basal slip systems in the detected slip planes for both grains are depicted with red arrows in \autoref{fig:grain659_579}(c and d). The high contrast intensity occurs near the grain edges (green arrows in \autoref{fig:grain659_579}(a)) in the topograph where the slip direction is orthogonal to the imaging plane. In other words, the intensity of the contrast is maximized when the slip direction is in the plane of the page, as approximated by the red arrow in \autoref{fig:grain659_579}(c). Furthermore, the two intense points of contrast at the red arrows in \autoref{fig:grain659_579}(a), which are due to impingement of the intense slip bands from the neighboring grain (yellow grain in \autoref{fig:grain659_579}(b)), occur at the location where the active basal slip system direction is pointing towards the grain boundary. Identical types of observations were made at several instances of grains and slip events in the topotomography Ti-7Al dataset.
From the previous example, it is observed that some of the high contrast regions in topographs that reveal the presence of slip events, originate from the pile-ups of dislocation at grain boundaries. These pile-ups either induce significant contrast in the grain where the slip events develops or in the neighboring grain where slip band impingement occurs. The lattice rotation contrast induced by dislocation pile-ups have previously been investigated by HR-EBSD measurements in titanium alloys \cite{GUO2015229}, and their amplitudes are on the order of the magnitude that corresponds to the contrast observed on topotomograph. It is therefore reasonable now to consider than the contrast observed in topographs that provide the identification of slip events may be enhanced in proximity of grain boundaries (due to pile-ups). On the other hand, the contrast also appear to come from the collective presence of dislocations (alongside their stress fields) moving within the slip band and providing edge-on contrast. This consideration can explain the propensity for the topotomography technique to tend to not detect low intensity slip events, which are associated with low dislocation densities at pile-ups and where the contrast created by the band within the grain is too faint. An in situ experiment would be helpful to analyse the evolution of contrast when a slip band and its pile-ups increase in intensity as the deformation progresses.
\subsection{Slip transmission in the bulk}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{Transmission_1.jpg}
\caption{Two transmission events observed by DCT and topotomography. (a,e) EBSD map of the surrounding grain structure around the transmission events. (b,f) HR-DIC $\epsilon_{zz}$ strain map showing the investigated slip transmission events. (c,g) 3D reconstruction of the grain structure and slip events for the grains of interest. The incoming and transmitted slip are indicated by blue and red arrows, respectively. The green circles highlight the location of slip transmission. (d,h) 3D reconstruction of the active slip planes detected by topotomography for the grains of interest. }
\label{fig:Transmission_1}
\end{figure}
One of the interesting applications of topotomography is the investigation of slip transmission. Typically, transmission analyses are performed from surface measurements that fail to experimentally capture the inclination of the slip events and geometry of the grain boundaries in the sub-surface/bulk. While slip transmission in the bulk of the specimen will be more rigorously studied in a future study of Ti-7Al, demonstrative examples are presented in \autoref{fig:Transmission_1} from regions measured by HR-DIC and topotomography, where slip events (blue arrows) are transmitting in the neighboring grains (red arrows). The incoming and transmitted slip events were detected in the topotomography measurements and their 3D representation is provided in \autoref{fig:Transmission_1}(c and d) for the first example and in \autoref{fig:Transmission_1}(g and h) for the second example. Incoming slip events (blue arrows) in both of the investigated grains were identified as the slip system with the highest Schmid factor, and the active slip directions are depicted by the long red arrows in \autoref{fig:Transmission_1}(d and h). The associated HR-DIC and EBSD maps are provided for both examples in \autoref{fig:Transmission_1}(a,b and e,f).
The first example shown in \autoref{fig:Transmission_1}(a-d) displays a direct transmission event. Upon consideration of the Schmid factor for grain 615, slip system
\hkl(-1 0 1 0)[-1 2 -1 0] has the highest Schmid factor of 0.405 and is not activated, while the slip system
\hkl(-1 1 0 0)[1 1 -2 0] is activated (purples plane in \autoref{fig:Transmission_1}(c-d)) with a lower Schmid factor of 0.373. Consequently, the transmission event triggered a slip system in grain 615 with a lower Schmid factor but with favored transmission configuration. The $m'$ factor \cite{Luster1995} is usually an effective indicator to describe the propensity of a incoming slip event to transmit to the neighboring grain. It is defined as $m'= (d_1\cdot{}d_2)(n_1\cdot{}n_2)$, with $d_1$ the slip direction of the incoming slip system, $d_2$ the slip direction of the outgoing slip system, $n_1$ the slip plane normal of the incoming slip system and $n_2$ the slip plane normal of the outgoing slip system. $m'$ is ranging between 0 and 1 and high values indicate that better geometric compatibility exists between both slip systems that favor slip transmission. The $m'$ factor between the incoming and transmitted slip event is 0.7 while the one between the incoming slip and the highest Schmid factor slip system in grain 615 is 0.1. This is evidence of plasticity behavior that is dependent on the local surrounding microstructural configuration as has previously been reported for titanium alloys \cite{HEMERY2018277,BIELER2014212}. The topotomography measurements provide accurate spatial resolution to capture such behavior at the surface as conventional surface measurements (trace analysis from SEM, HR-DIC or AFM), but also into the specimen bulk.
Interestingly, the topotomography measurements allow differentiation between slip activity that occurs at the surface and in the bulk. The m' factors were calculated for all slip events that are connected (at least one voxel in common at a grain boundary) across surface grains and from bulk grains. The distributions of the $m'$ factor is reported in \autoref{fig:mtransmission}. It is striking to notice such a difference in the distribution from bulk and surface grains. Surface grains tend to promote slip transmission and as a consequence, the $m'$ factor is important when investigating slip activity at the surface, as reported in the literature \cite{HEMERY2018277}. Conversely, the distribution of the $m'$ factor between bulk grains indicates that it may not be as relevant for the description of slip activity in the bulk. Direct transmission may not be of significant importance within the bulk, in agreement with the recent observation of the importance of the triple junction within the bulk \cite{CHARPAGNE2021117037}. In this work, it was observed in a deformed metallic material that a large amount of the slip events that develop originate from triple junction lines that have significant stresses. From the present analysis, the possibility should be considered that slip activity within the bulk is mainly controlled by stress heterogeneity at junction lines, with slip transmission having a lesser influence.
\begin{figure}
\centering
\includegraphics[width=0.65\textwidth]{mtransmission.jpg}
\caption{(a) Distribution of m' factor between surface grains and from grains in the bulk. The m' factors were calculated for all slip events detected by topotomography that are connected (at least one voxel in common at a grain boundary) across surface grains and from bulk grains. (b) Relation between GOS from topotomography measurements and the slip direction in relation to the free surface.}
\label{fig:mtransmission}
\end{figure}
Interestingly, multiple peculiar cases of direct transmission were observed at the surface of the specimen as displayed by the second example in \autoref{fig:Transmission_1}. In this example, the incoming slip event and the transmitted slip event are connected by the unique surface point highlighted by green open dots in \autoref{fig:Transmission_1}(c and d). The transmission factor m' is calculated as 0.1 indicating a configuration that does not favor slip transmission since both the plane normal and slip directions of the active slip systems are highly disoriented from each other, as displayed in \autoref{fig:Transmission_1}(d). From the 3D observation of this slip transmission event, and considering that the slip direction of the incoming slip event is almost normal to the surface, it can be hypothesized that the intense incoming slip event developed a relatively significant surface step that triggered slip activation in the neighboring grain. The constraint of the slip extrusion at a grain boundary can generate large local stresses that then can trigger slip in the neighboring grain. This kind of transmission event was observed in several instances in the investigated surface grains and contributed to the low values of the m' factor in the distribution in \autoref{fig:mtransmission}(a). It is generally accepted that direct or indirect slip transmission is controlled by dislocation transmission across grain boundaries \cite{Luster1995}. However, the present example evidences another type of transmission related more likely to stress concentration and surface effects due to the constraint of a slip step near grain boundary at the free surface. This is further evidence that the m' factor is not always relevant for characterization of slip transmission at the surface.
\subsection{lattice rotation from bulk measurements}
\label{sec:lattice_rot_discussion}
Significant differences in orientation spread, induced by deformation, are observed from near-surface measurements (EBSD) and bulk measurements (topotomography). This result is expected since the interior of the grain can evolve differently during deformation than the near surface of the grain. It is important to question the relationship between slip and orientation spread from near-surface and bulk measurement.
For instance, the grain labeled 660 has an orientation spread of \SI{0.5}{\degree}, whereas grain 063 displays a spread of only \SI{0.25}{\degree}. The analysis of the active slip system Burgers vectors shows that in grain 063, the vector points out of the free surface meaning the dislocations can readily escape the volume. However, in grain 660 the vector is almost parallel to the surface meaning that the dislocations do not escape and pile up at the grain boundary, storing GNDs and creating mosaicity as a result. This behavior appears to be general for the surface grains measured by topotomography. This relation is evidenced in \autoref{fig:mtransmission}(b) where the orientation spread is displayed as a function of the angle between the slip direction and the surface vector of the active slip plane. Grains with high orientation spread tend to display a slip direction that is close to the plane of the free surface. This provides more evidence of the effect of the free surface on the plastic activity. Other factors such as the grain size, the level of plasticity, the neighboring grains and the grain shape are also certainly relevant and are required to be investigated to highlight the relation between orientation spread and plastic activity.
The relationship observed previously between orientation spread and slip direction is not observed in orientation spread for the near-surface measurements obtained by EBSD. This indicates that EBSD and other surface measurements may not be representative of the plasticity that occurs within the entire grain, highlighting an advantage of the topotomography measurements, even for surface grains.
\section{Future Directions}
Topotomography is a material characterization technique the has matured over the last 10 years, all while becoming more and more automated. It now also benefits from major synchrotron source upgrades that reduce data collection times to a few minutes per grain, as compared to several tens of minutes for the experiment reported here. This enables the systematic mapping of massive regions of grains, selected from an initial DCT reconstruction of the grain volume. Scripts performing automated TT grain alignment and driving scan sequences (DCT, PCT and series of TT scans over a list of grains) have been developed and enable repeated acquisitions at increasing levels of applied strain, which could be instrumental in capturing plasticity propagation events. This framework and in particular the selection of candidate grains could possibly be guided by a digital-twin mechanical analysis conducted on the initial DCT reconstruction.
The question of the nature of the contrast mechanism(s) remains open, with likely origins being orientation contrast created by GNDs piling up at grain boundaries as evidenced in this paper, dislocations structures located within the slip bands, or a combination of both. Dislocation dynamics simulations based on experimental microstructures and coupled with forward modeling diffraction is currently being developed and may accurately inform on the contrast mechanisms in the near future.
Another current challenge relies in inverting the topotomography images to reconstruct the orientation field. Since TT can not resolve lattice rotations around the probed scattering vector, the inversion has to include complementary grain projection data from DCT. This is in principle feasible using iterative approaches as demonstrated in \cite{Vigano_COSS_2020} or by adapting the forward modeling strategy developed by \cite{Suter2006} to the combined DCT and TT acquisition geometry. As the topotomography geometry is more complex (it involves several rotations around the eucentric point), the sensitivity and spatial resolution of this approach may suffer from diffractometer error motion and requires an accurate calibration of the instrument geometry. Correlative experiments will be instrumental to validate the reconstructions in the future. To this end, recent updates in experimental stations at ESRF such as the nanobeam scanning 3DXRD station at ID11 or the dark field X-ray microscopy station at ID06 can be leveraged.
\section{Conclusions}
The correlative measurements from HR-DIC, topotomography and DCT made explicit the sensitivity and spatial resolution of the topotomography measurements. Topotomography measurements were able to detect most of the intense slip events that developed during deformation of a titanium alloy. Slip events of very low intensities or in regions of high lattice rotation were not detected by topotomography measurements, at least with the slip detection method used in this paper. Topotomography measurements provide the unique opportunity to investigate the plastic activity in a large number of grains both at the surface and in the bulk. Both slip events location and orientation spread are obtained from topotomography. Several transmission events were investigated in a titanium alloy by topotomography. The analysis of slip transfer between surface grains shows that many slip transmission events can be rationalized in terms of the geometrical alignment of activated slip systems in neighboring grains (m' factor for instance). However, the slip activity within the bulk of the specimen is observed to \textbf{not} be controlled by slip transmission in the same extent as those of surface grains. In addition, an unknown transmission phenomena was observed at the surface, indicating again that completely different plastic behavior may occur in the bulk versus at free surfaces.
\section{Acknowledgments}
The authors would like to thank ESRF for beam time allocation under proposal MA3921 and Chris Torbet for machining the Ti-7Al specimens. HP also like to thank Georges Cailletaud (holder of the Cristal chair funded by Mines ParisTech and Safran) for the financial support to visit UCSB. The authors also acknowledge the support of ONR Grant N00014-19-2129. The MRL Shared Experimental Facilities are supported by the MRSEC Program of the NSF under Award No. DMR 1720256; a member of the NSF-funded Materials Research Facilities Network (www.mrfn.org). PGC would like to acknowledge support provided by the Naval Research Laboratory under the auspices of the Office of Naval Research.
\bibliographystyle{elsarticle-num}
| {
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"redpajama_set_name": "RedPajamaC4"
} | 6,779 |
package com.intellij.util.xml.reflect;
import com.intellij.util.xml.DomElement;
import com.intellij.util.xml.DomNameStrategy;
import com.intellij.util.xml.XmlName;
import org.jetbrains.annotations.NotNull;
/**
* @author peter
*/
public interface DomChildrenDescription extends AbstractDomChildrenDescription {
@NotNull
XmlName getXmlName();
@NotNull
String getXmlElementName();
@NotNull
String getCommonPresentableName(@NotNull DomNameStrategy strategy);
@NotNull
String getCommonPresentableName(@NotNull DomElement parent);
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,360 |
Q: safe to commit travis secrets into public repo? I'm using travis for some of my public github repos. For deployment step (.e.g. creating a release in github), I need to put in a OAuth token (encrypted by travis)
Is this encrypted token safe to be committed together in the public repo source code? Since it's public repo, I would imagine anyone can see the source code and get my encrypted token. Although they are not able to decrypt it but they can use the same token to create a release in my github repo?
But if not committing to the source code, what's the best way to manage it?
A: Yes. It's safe to out encrypted environment variables in your .travis.yml. The encryption/decryption key is generated by Travis on a per-repo basis, so when others fork your repo or even if you rename it, the encrypted variable is invalidated and you have to encrypt it again.
According to Travis CI, nobody except the automated Travis process may decrypt the value.
The only possible leak is when one of your cooperators goes nuts and decides to reveal it by modifying build script. But since you have a trust system, this isn't going ti be an issue.
A: You should never store secrets in public / private repo. You can store the info in an environment variable. You can add your .env file, for example, to .gitignore and you can store variables there or directly in the environment.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 108 |
\section*{List of Figure Captions}
Fig. 1 (a) Schematic diagram of the Ti:sapphire laser oscillator.
(b) Total round trip intracavity GDD. (c) Total cavity gain
$G_{tot}$ assuming a peak crystal gain of 1.04.
\bigskip
\noindent Fig. 2 Normalized intracavity spectrum (solid line) and phase (dotted line) before the OC after
$(a)\,20$, $(b)\,500$, $(c)\,1000$ round trips inside the cavity. (d)
Simulated (solid line) and measured (dashed line) final spectra
outside the OC.
\bigskip
\noindent Fig. 3 Laser cavity dispersion map. $A,B,C,D,E$ and $F$ are reference points corresponding to the spectra,
phases, and temporal profiles reported in figure\,\ref{figure_4_tognetti_crespo} (see the text for more details).
\bigskip
\noindent Fig. 4 Spectra, phases and temporal profiles corresponding to the reference
points reported in figure\,\ref{figure_3_tognetti_crespo}.
\newpage
\begin{figure}[h]
\centerline{\scalebox{0.5}{\includegraphics[angle=270]{figure_1_tognetti_crespo.eps}}}
\caption{(a) Schematic diagram of the Ti:sapphire laser oscillator.
(b) Total round trip intracavity GDD. (c) Total cavity gain
$G_{tot}$ assuming a peak crystal gain of 1.04.}
\label{figure_1_tognetti_crespo}
\end{figure}
\newpage
\begin{figure}[h]
\centerline{\scalebox{0.5}{\includegraphics{figure_2_tognetti_crespo.eps}}}
\caption{Normalized intracavity spectrum (solid line) and phase (dotted line) before the OC after
$(a)\,20$, $(b)\,500$, $(c)\,1000$ round trips inside the cavity. (d)
Simulated (solid line) and measured (dashed line) final spectra
outside the OC.}
\label{figure_2_tognetti_crespo}
\end{figure}
\newpage
\begin{figure}[h]
\centerline{\scalebox{0.3}{\includegraphics[angle=270]{figure_3_tognetti_crespo.eps}}}
\caption{Laser cavity dispersion map. $A,B,C,D,E$ and $F$ are reference points corresponding to the spectra,
phases, and temporal profiles reported in figure\,\ref{figure_4_tognetti_crespo} (see the text for more details).}
\label{figure_3_tognetti_crespo}
\end{figure}
\newpage
\begin{figure}[h]
\centerline{\scalebox{0.8}{\includegraphics{figure_4_tognetti_crespo.eps}}}
\caption{Spectra, phases and temporal profiles corresponding to the reference
points reported in figure\,\ref{figure_3_tognetti_crespo}.}
\label{figure_4_tognetti_crespo}
\end{figure}
\newpage
\end{document}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,944 |
Q: How to get the src of next image to be loade din Nivo Slider I want to be able to get the src of the next image due to be loaded. So I was thinking something like
$('#nivo_slider').nivoSlider({
beforeChange: function(){
//get src of next image
}
};
I can get the current image source by doing the following:
var currentImageSrc=$('#nivo_slider').data("nivo:vars").currentImage.attr('src');
var index=currentImageSrc.lastIndexOf("/")+1;
var imageName= currentImageSrc.substr(index);
But I'm not sure how to get the next one to be loaded
My HTML is as follows:
<div id="nivo_slider>
<img src="img1" />
<img src="img2" />
<img src="img3" />
<img src="img4" />
</div>
A: Well, here's one way to do it (and, despite some time spent trying there didn't seem to be an obviously-easier way):
/* Nivo seems to add images in order to implement its effects,
so we need to grab a reference to the original images, or, at least, their `src` */
var images = $('#slider img').map(
function() {
return $(this).attr('src');
});
$('#slider').nivoSlider({
beforeChange: function() {
var wrap = $('#slider'),
// caching data because I'm accessing it twice:
data = wrap.data('nivo:vars'),
// the current src:
current = data.currentImage.attr('src'),
len = data.totalSlides,
/* compares the current src against the srcs stored in the
images variable, and returns the index at which it was
found */
arrayPos = $.inArray(current, images),
nextSrc;
if (arrayPos == (len - 1)) {
/* if the array position of the src is equal to length-1
(zero-based arrays in JS), then the next image to be
shown will be the first in the array */
nextSrc = images[0];
}
else {
// otherwise get the next src from the next position in the array
nextSrc = images[arrayPos + 1];
}
$('#output code.src').text(nextSrc);
}
});
JS Fiddle demo.
References:
*
*attr().
*$.inArray().
*map().
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,282 |
// Import and register all your controllers from the importmap under controllers/*
import { application } from 'controllers/application'
// Eager load all controllers defined in the import map under controllers/**/*_controller
import { eagerLoadControllersFrom } from '@hotwired/stimulus-loading'
eagerLoadControllersFrom('controllers', application)
// Lazy load controllers as they appear in the DOM (remember not to preload controllers in import map!)
// import { lazyLoadControllersFrom } from '@hotwired/stimulus-loading'
// lazyLoadControllersFrom('controllers', application)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 4,453 |
Q: jQuery Animation Stop / Resume Techniques or Patterns I've seen some other - semi-related posts - but nothing that really 100% answers my question.
In any event, what I'd like to do is pause+fadeIn onHover, pause+fadeOut onHoverOut, this way the animation doesn't go to completion unless the user stays hovered or hoverOuted. If the user does a bunch of quick hovers and hover outs, I don't want it to queue up a bunch of consecutive animations. I've gotten close using stop() and clearQueue(), but there are problems.
If I hover, hoverOut and hover again quickly, the animation does pause but does not transition to the next animation. As a matter of fact, enough hover and hoverOuts causes the animations to stop happening completely on my computer.
Check out the example page below and the attached script. Maybe there is a common pattern or approach to this particular question? I see similar things done online quite a bit, sometimes with fade animations, sometime with motion animations, etc.
In the example, I'm intentionally using a slow animation time so you have time to hover in and out enough times.
Demo:
http://ficreates.com/_SiteDemos/lwv2/
Script:
http://ficreates.com/_SiteDemos/lwv2/scripts/carNav.js
Anyhow, thanks in advance for any help you can give me :) I certainly need it!
A: Is this what you are looking for?
function carNav() {
var $navBox = $("#nav_buttons li");
$navBox.hover(
function() {
var $navCar = $(this).find(".nav_car");
$navCar.fadeIn(2000);
},
function() {
var $navCar = $(this).find(".nav_car");
$navCar.stop().fadeOut(2000);
}
);
}
$(document).ready(carNav);
DEMO http://jsfiddle.net/m2pF5/
or you can do it entirely with css, (will require a fallback for ie)
CSS DEMO http://jsfiddle.net/eNJ6x/1/
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,383 |
If you haven't asked your kid—or any kid—what they want to be when they grow up, I commend you. It's one of those default questions that adults ask kids, a reflex leftover from our own childhood when we were asked ad nauseam whether we wanted to be a doctor, an astronaut or a teacher.
It's not that we're trying to lock them into a particular profession; we're all aware that the number of kids who say they want to be a paleontologist is much less than the number of kids who actually become paleontologists. I've viewed it as a point of curiosity; a cute, visual way to track how his interests change over time. But psychologist Adam Grant writes for The New York Times that we're doing a disservice to our kids when we endlessly ask them what they want to be when they grow up.
Secondly, Grant says, we are implying that there is but one true calling out there for everyone. I'm only 15 years into my own career and it has morphed and shifted as my own interests and priorities have morphed and shifted over time. We're all professional works in progress.
As "global education evangelist" Jaime Casap teaches, instead of "What do you want to be," try asking, "What problems do you want to solve?" Shift the focus away from specific jobs and more toward what kind of person they want to be. Someone who is compassionate and organized and energetic can be any number of things as an adult.
I have backed off from asking the question in the past year or so, and I've noticed a shift in how my son thinks about his future. He has evolved from the resolute "video-game maker" of a year ago to a handful of options he may or may not pursue. These include, but are not limited to: the owner of a restaurant that specializes in hot dogs, a toy store owner, a professional soccer player or, yes, a video-game maker (no reason to move on from that one completely). | {
"redpajama_set_name": "RedPajamaC4"
} | 107 |
The third season of Bachelor in Paradise Australia premiered on 15 July 2020. Osher Günsberg reprised his role as host from the previous two series and the Bachelor Australia franchise.
Intended to air in April 2020, the airing of the show was held back by 10 to cover for the production shut-downs caused by the COVID-19 pandemic. The series premiered on 15 July 2020.
Contestants
Abbie, Ciarran, Timm, Brittany and Jamie were revealed on 13 March 2020. Helena, Mary, Brittney and Cass were further announced on 17 June 2020. The full returning Episode 1 cast were revealed on 6 July 2020, which included Jake, Glenn, Janey and Niranga. Keira was also announced as an intruder and she first appeared in episode five. Jessica was announced as an intruder at the end of the first episode and Renee was announced at the end of the second, making their first appearance in the second and third episodes respectively. Alisha was announced as an intruder on 19 July 2020. Conor, Chris and Tim, who had not appeared on any season of The Bachelorette or a Bachelor franchise series, were announced as intruders on 22 July 2020. Kiki entered Paradise as an intruder in episode 5 and Jackson, Matt and Scot in episode 7.
Elimination table
Colour Key
The contestant is male.
The contestant is female.
The contestant went on a date and gave out a rose at the rose ceremony.
The contestant went on a date and received a rose at the rose ceremony.
The contestant gave or received a rose at the rose ceremony, thus remaining in the competition.
The contestant received the last rose.
The contestant went on a date and received the last rose.
The contestant went on a date and was eliminated.
The contestant was eliminated.
The contestant had a date and voluntarily left the show.
The contestant voluntarily left the show.
The couple left the show together but later split.
The couple broke up and were eliminated.
The couple decided to stay together, but split after Bachelor in Paradise Australia ended.
The couple decided to stay together and won the competition.
Notes
Episodes
Ratings
References
External links
2020 Australian television seasons
Australian (season 03)
Television productions postponed due to the COVID-19 pandemic | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,640 |
Die Pfau von Rietburg (auch Pfauen), später Pfau(en) von Rüppurr waren ein oberrheinisches Dienstmannengeschlecht, zuletzt Freiherren.
Familiengeschichte
Die Familie gehörte dem niederen Dienstadel an und hatte ihren namensgebenden Stammsitz in Rüppurr, heute ein Stadtteil von Karlsruhe. Die Gemeinde hat das Wappen der Adligen als Ortswappen übernommen.
In Urkunden wird das Geschlecht ab ca. 1100 n. Chr. erwähnt, zunächst nur als von Rietburg, dann als von Rieperg oder von Rüppurr und schließlich (ab dem 13. Jahrhundert) auch mit dem Namenszusatz Pfau, den man als Zeichen eines stolzen Selbstbewusstseins deutet. Anfang des 15. Jahrhunderts wanderte ein Teil der Familie in die Ortenau aus und nannte sich nur noch von Pfau.
Die Pfauen von Rüppurr stellten nicht nur hohe Staatsbeamte und Militärs, sondern auch kirchliche Würdenträger. So war Reinhard von Rüppurr 1503 bis 1523 Bischof von Worms. Das unter seiner Ägide 1522 verausgabte Wormser Messbuch (Missale secundum ritum et observantiam Ecclesie et Dioecesis Wormatiensis) zeigt ihn mit Familienwappen auf dem Titelblatt.
1584 war die Familie so stark verschuldet, dass sie das Rüppurrer Schloss sowie alle Güter in Rüppurr an Markgraf Ernst Friedrich von Baden-Durlach überschreiben mussten. In der Folge zog sie nach Obermönsheim. Hier erlosch das Geschlecht 1782 im Mannesstamm, durch den Tod von Freiherr Christoph Friedrich von Rüppurr. Seine Tochter Philippine Charlotte Franziska ehelichte den württembergischen Staatsminister Ernst Leopold August von Phull (1768–1828), wodurch dessen Familie den Namen und das Wappen mit übernahm. Sie nannten sich von da an Phull-Rieppurr (auch Pfuel-Rüppurr). Der letzte Namens- und Wappenträger Freiherr Eduard von Phull-Rieppurr starb dort 1918.
Persönlichkeiten
Reinhard von Rüppurr, 1503 bis 1523 Bischof von Worms
Literatur
Bernhard Theil: Das älteste Lehnbuch der Markgrafen von Baden (1381), Kohlhammer Verlag, Stuttgart, 1974, S. 122, ISBN 3170020234; (Ausschnittscan)
Jakob Christoph Iselin, Jacob C. Beck: Neu vermehrtes historisches und geographisches allgemeines Lexikon, Supplementband, Teil 2, S. 804, Basel, 1744; (Digitalscan)
Weblinks
Webseite zur Familiengeschichte
Einzelnachweise
Schwäbisches Adelsgeschlecht
Rüppurr | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,247 |
\section*{Introduction}
Bessel functions are widely used within mathematical physics, which explains everlasting interest to these functions both in physics and mathematics. Key properties of Bessel functions were articulated about a hundred years ago. Those results are mainly short, good looking and represent classic results of special function theory. They are collected in numerous treatises (e.g., see~\cite{Watson_1_1995}).
Often works on Bessel functions stem from the needs of physics and are focused on the properties of Bessel functions required to deal with a particular physical problem.
That was the reason to study Bessel functions of large order and argument~\cite{Olver_1_1954,Ratis_1_1993,Paris_1_2004} which arise in astrophysics~\cite{Pierro_1_2001,chistie_1_2008}.
Most works on Bessel functions --- discussing algorithms for Bessel functions calculation~\cite{Coleman_1_1980,Guerrero_1_1988,Ratis_1_1993,Karatsuba_1_1993,Jentschura_1_2012}, identities containing Bessel functions~\cite{Miano_1_1995,Babusci_1_2011,Babusci_2_2011,Dominici_1_2012} and others --- yield cumbersome results. The result of present work, obtained for Bessel functions, is concise and simple.
\section{Infinite series representations for Bessel functions of real order}
We start from three known identities~\cite[2.12.21]{Prudnikov_1_v2_1986}):
\begin{subequations}
\label{eq:integrals0}
\begin{align}
\label{eq:integralA}
&\int_{0}^1
x^{\nu+1}J_{\nu}(bx)\sin{\big(y\sqrt{1-x^2}\big)}\dif x=\sqrt{\frac{\pi}{2}}yb^\nu\big(\sqrt{b^2+y^2}\big)^{-\nu-\frac{3}{2}}J_{\nu+\frac{3}{2}}\big(\sqrt{b^2+y^2}\big),\\
\label{eq:integralB}
&\int_{0}^1
\frac{J_{\nu}(bx)}{\sqrt{1-x^2}}\cos\big(y\sqrt{1-x^2}\big)\dif x=
\frac{\pi}{2}J_{{\nu}/{2}}\bigg(\frac{\sqrt{b^2+y^2}-y}{2}\bigg)J_{{\nu}/{2}}\bigg(\frac{\sqrt{b^2+y^2}+y}{2}\bigg),\\
\label{eq:integralC}
&\int_{0}^1
\frac{x^{\nu+1}J_{\nu}(bx)}{\sqrt{1-x^2}}
\cos\big(y\sqrt{1-x^2}\big)\dif x=
\sqrt{\frac{\pi}{2}}b^{\nu}\big(\sqrt{b^2+y^2}\big)^{-\nu-\frac12}J_{\nu+\frac12}\big(\sqrt{b^2+y^2}\big),
\end{align}
\end{subequations}
where $b>0$, $y$ is any real number and $\nu>-1$. Left sides of Eqs.~(\ref{eq:integrals0}) can be transformed into another form by introducing a new variable $t=\sqrt{1-x^2}$:
\begin{subequations}
\label{eq:integrals1}
\begin{align}
\label{eq:integral1A}
&\int_{-1}^1
t\big(\sqrt{1-t^2}\big)^\nu
J_{\nu}\big(b\sqrt{1-t^2}\big)\sin{(yt)}\dif t=
\sqrt{2\pi}yb^{\nu}\big(\sqrt{b^2+y^2}\big)^{-\nu-\frac{3}{2}}J_{\nu+\frac{3}{2}}\big(\sqrt{b^2+y^2}\big),\\
\label{eq:integral1B}
&\int_{-1}^1
\frac{J_{\nu}\big(b\sqrt{1-t^2}\big)}{\sqrt{1-t^2}}\cos(yt)\dif t=
{\pi}J_{{\nu}/{2}}\bigg(\frac{\sqrt{b^2+y^2}-y}{2}\bigg)J_{{\nu}/{2}}\bigg(\frac{\sqrt{b^2+y^2}+y}{2}\bigg),\\
\label{eq:integral1C}
&\int_{-1}^1
{J_{\nu}(b\sqrt{1-t^2})}
\big({\sqrt{1-t^2}}\big)^{\nu}
\cos(yt)\dif t=
\sqrt{2\pi}b^{\nu}\big(\sqrt{b^2+y^2}\big)^{-\nu-\frac12}J_{\nu+\frac12}\big(\sqrt{b^2+y^2}\big).
\end{align}
\end{subequations}
Here we extended interval of integration over $t$ from $[0,1]$ to $[-1,1]$.
Eq.~(\ref{eq:integrals1}) can be presented in general form
\begin{equation}
\label{eq:integrals2}
\int_{-1}^1 F_{\nu}^{(\alpha)}(b,t)\e^{\i yt}\dif t=f_{\nu}^{(\alpha)}(b,y),\qquad\alpha=A,B,C,
\end{equation}
where
\begin{subequations}
\label{eq:abc}
\begin{align}
\label{eq:a}
\begin{split}
&F_{\nu}^{(A)}(b,t)=-\i t\big(\sqrt{1-t^2}\big)^\nu J_{\nu}\big(b\sqrt{1-t^2}\big),\\
&f_{\nu}^{(A)}(b,y)=\sqrt{2\pi}yb^{\nu}\big(\sqrt{b^2+y^2}\big)^{-\nu-\frac{3}{2}}J_{\nu+\frac{3}{2}}\big(\sqrt{b^2+y^2}\big),
\end{split}
&
\\
\label{eq:b}
\begin{split}
&F_{\nu}^{(B)}(b,t)=\frac{J_{\nu}\big(b\sqrt{1-t^2}\big)}{\sqrt{1-t^2}},\\
&f_{\nu}^{(B)}(b,y)={\pi}J_{{\nu}/{2}}\bigg(\frac{\sqrt{b^2+y^2}-y}{2}\bigg)J_{{\nu}/{2}}\bigg(\frac{\sqrt{b^2+y^2}+y}{2}\bigg),
\end{split}
&
\\
\label{eq:c}
\begin{split}
&F_{\nu}^{(C)}(b,t)={J_{\nu}\big(b\sqrt{1-t^2}\big)}\big({\sqrt{1-t^2}}\big)^{\nu},\\
&f_{\nu}^{(C)}(b,y)=\sqrt{2\pi}b^{\nu}\big(\sqrt{b^2+y^2}\big)^{-\nu-\frac12}J_{\nu+\frac12}\big(\sqrt{b^2+y^2}\big).
\end{split}
&
\end{align}
\end{subequations}
If $y=k\pi$, one can consider integrals on the left side of Eq.~(\ref{eq:integrals2}) to be calculations of the Fourier series coefficients of function $F_{\nu}^{(\alpha)}(b,t)$
defined on interval $t\in(-1,1)$ and extended to a periodic function on $\mathbb{R}$. Therefore one can present $F_{\nu}^{(\alpha)}(b,t)$ in form of the Fourier series
\begin{equation}
\label{eq:series1}
F_{\nu}^{(\alpha)}(b,t)=\frac12\sum\limits_{k=-\infty}^{\infty}f_{\nu}^{(\alpha)}(b,k\pi)\e^{-\i k\pi t},
\begin{array}{ll}
\quad t\in(-1,1)& \mbox{if\ \ } \alpha=A\;\mbox{and}\:-1<\nu\leq0, \\
\quad t\in[-1,1]& \mbox{if\ \ } \alpha=A\;\mbox{and}\:\nu>0,\,\mbox{or}\;\alpha=B,C.
\end{array}
\end{equation}
In case of a singularity at $t=\pm1$ on the left side of Eq.~(\ref{eq:series1}) corresponding series on the right side will also give singularity. In case of the removable singularity at $t=\pm1$ we have to remove singularity by redefining the function at these points and the equation will remain true.
Series in Eq.~(\ref{eq:series1}) gives zero
at points $t=\pm1$ if $\alpha=A$ and $-1<\nu\leq0$. All Fourier series (\ref{eq:series1}) are valid while $F_{\nu}^{(A,B,C)}(b,t)$ are differentiable
with respect to $t$ at $t\in(-1,1)$.
Coming back to variable $x=\sqrt{1-t^2}$ Eq.~(\ref{eq:series1}) is transformed into
\begin{equation}
\label{eq:series2}
\begin{split}
&F_{\nu}^{(\alpha)}\big(b,\sqrt{1-x^2}\big)=\frac12\sum\limits_{k=-\infty}^{\infty}
f_{\nu}^{(\alpha)}(b,k\pi)\e^{-\i k\pi\sqrt{1-x^2}},
\quad
\mbox{where}\\
&\begin{array}{ll}
\quad x\in(0,1]& \mbox{if\ \ } \alpha=A\;\;\mbox{and}\:\:-1<\nu\leq0, \\
\quad x\in[0,1]& \mbox{if\ \ } \alpha=A\;\;\mbox{and}\:\:\nu>0,\;\mbox{or}\;\;\alpha=B,C.
\end{array}
\end{split}
\end{equation}
Exchanging $b$ and $x$ we get the following series expansions at $x>0$
\begin{equation}
\label{eq:series3}
\begin{split}
&F_{\nu}^{(\alpha)}\big(x,\sqrt{1-b^2}\big)=\frac12\sum\limits_{k=-\infty}^{\infty}
f_{\nu}^{(\alpha)}(x,k\pi)\e^{-\i k\pi\sqrt{1-b^2}},
\quad
\mbox{where}\\
&\begin{array}{ll}
\quad b\in(0,1]& \mbox{if\ \ } \alpha=A\;\;\mbox{and}\:\:-1<\nu\leq0, \\
\quad b\in[0,1]& \mbox{if\ \ } \alpha=A\;\;\mbox{and}\:\:\nu>0,\;\mbox{or}\;\;\alpha=B,C,
\end{array}
\end{split}
\end{equation}
which in every particular case of $\alpha$ has the form
\begin{equation}
\label{eq:expansionA}
b^{\nu}\sqrt{1-b^2}J_{\nu}(bx)=\sum\limits_{k=1}^{\infty}f_{\nu}^{(A)}(x,k\pi)\sin\big(k\pi\sqrt{1-b^2}\big),
\begin{array}{ll}
\quad b\in(0,1]& \mbox{if\ \ }-1<\nu\leq0, \\
\quad b\in[0,1]& \mbox{if\ \ }\nu>0,
\end{array}
\end{equation}
which we will call (A)-case,
\begin{equation}
\label{eq:expansionB}
\frac{J_{\nu}(bx)}{b}=\sum\limits_{k=0}^{\infty}\eps_k
f_{\nu}^{(B)}(x,k\pi)\cos\big(k\pi\sqrt{1-b^2}\big),\quad b\in[0,1],
\end{equation}
which we will call (B)-case, and
\begin{equation}
\label{eq:expansionC}
b^{\nu}J_{\nu}(bx)=\sum\limits_{k=0}^{\infty}\eps_k
f_{\nu}^{(C)}(x,k\pi)\cos\big(k\pi\sqrt{1-b^2}\big),\quad b\in[0,1],
\end{equation}
which we will call (C)-case, where
\begin{equation}
\label{eq:eps}
\eps_k=
\begin{cases}
\frac12, & \mbox{if } k=0 \\
1, & \mbox{if } k>0.
\end{cases}
\end{equation}
Eqs.~(\ref{eq:expansionA}), (\ref{eq:expansionB}), (\ref{eq:expansionC}) are valid at $x>0$ and $\nu>-1$.
\section{Infinite series representations for Bessel functions of integer order}
Derived series representations merit detailed consideration at integer $\nu$ as in this case
they get series representations through the elementary functions. Series representations for Bessel function of integer order are following:
(A)-case
\begin{equation}
\label{eq:expansionA_n}
b^{n}J_{n}(bx)=\sum\limits_{k=1}^{\infty}(k\pi)g_k^{(A)}(b)f_{n}^{(A)}(x,k\pi),\quad b\in(0,1),\quad n\geq0,\quad x\geq0,
\end{equation}
(B)-case\footnote{Here we will not receive series representation for $J_0(x)$ through elementary functions (see Eqs.~(\ref{eq:expansionB}) and~(\ref{eq:b})). Therefore we consider only case of $n\geq1$.}\footnote{Left side of this equation should be taken as $\lim_{b\to0}\frac{J_n(bx)}{b}$ at $b=0$.}
\begin{equation}
\label{eq:expansionB_n}
\frac{J_{n}(bx)}{b}=\sum\limits_{k=0}^{\infty}\eps_k
g_k^{(B,C)}(b)f_{n}^{(B)}(x,k\pi),\quad
b\in[0,1],\quad n\geq1,\quad x\geq0,
\end{equation}
(C)-case
\begin{equation}
\label{eq:expansionC_n}
b^{n}J_{n}(bx)=\sum\limits_{k=0}^{\infty}\eps_k
g_k^{(B,C)}(b)f_{n}^{(C)}(x,k\pi),\:
\begin{array}{ll}
\quad b\in[0,1]& \mbox{if\ \ }n=0, \\
\quad b\in(0,1]& \mbox{if\ \ }n\geq1,
\end{array},\quad x\geq0,
\end{equation}
where
\begin{equation}
\label{eq:gka_gkbc}
g_k^{(A)}(b)=\frac{\sin\big(k\pi\sqrt{1-b^2}\big)}{k\pi\sqrt{1-b^2}},
\qquad
\mbox{and}
\qquad
g_k^{(B,C)}(b)=\cos\big(k\pi\sqrt{1-b^2}\big).
\end{equation}
We imply that $f_{n}^{(\alpha)}(0,k\pi)=\lim_{x\to0}f_{n}^{(\alpha)}(x,k\pi)$ ($\alpha=A,B,C$)
if singularity at $x=0$ is present.
However, we need to perform some transformations in (B)-case and at $n=2m$ in order to get elementary functions on the right hand
of Eq.~(\ref{eq:expansionB}). Application of recurrence relations for the Bessel functions \cite{Watson_1_1995}
\begin{equation}
\label{eq:recurrence}
\frac{4m}{b^2x}J_{2m}(bx)=\frac{J_{2m-1}(bx)}{b}+\frac{J_{2m+1}(bx)}{b},\quad
\;m\geq1
\end{equation}
to equation (\ref{eq:expansionB}) results in representation\footnote{We should note that there are
numerous ways to derive Bessel functions of even order through Bessel functions of odd order. For instance, recurrent relations \cite{Watson_1_1995} give another two equations:
\[
\begin{split}
J_{2m}(bx)&=(2m+1)J_{2m+1}(bx)+\frac{1}{b}\frac{\dif J_{2m+1}(bx)}{\dif x},\\
J_{2m}(bx)&=(2m-1)J_{2m-1}(bx)-\frac{1}{b}\frac{\dif J_{2m-1}(bx)}{\dif x}.
\end{split}
\]}
\begin{subequations}
\label{eq:expansionB_even}
\begin{align}
\label{eq:exp_JB_even}
&\frac{4m}{b^2}J_{2m}(bx)=\sum\limits_{k=0}^{\infty}\eps_k
g_k^{(B,C)}(b)f_{2m}^{(B)}(x,k\pi),\quad b\in(0,1],\quad x\geq0,\quad m\geq1\\
\label{eq:exp_fB_even}
\begin{split}
f_{2m}^{(B)}(x,y)={}x\cdot\pi
\bigg[
J_{(2m-1)/2}\bigg(\frac{\sqrt{x^2+y^2}-y}{2}\bigg)J_{{(2m-1)}/{2}}\bigg(\frac{\sqrt{x^2+y^2}+y}{2}\bigg)\\
+J_{(2m+1)/2}\bigg(\frac{\sqrt{x^2+y^2}-y}{2}\bigg)J_{{(2m+1)}/{2}}\bigg(\frac{\sqrt{x^2+y^2}+y}{2}\bigg)
\bigg].
\end{split}
\end{align}
\end{subequations}
We can extend derived expansions to the region $x<0$ according to the parity of the Bessel functions of integer order: $J_{n}(-x)=(-1)^nJ_{n}(x)$.
With notation
\begin{equation}
\label{eq:phi}
\varphi_k=\phik
\end{equation}
the series expansions for the Bessel functions of order $0$, $1$, $2$
are presented below for every particular case.
(A)-case:
\begin{subequations}
\label{eq:particular_b_A}
\begin{align}
\label{eq:Bess_b_A_0}
&J_0(bx)=2\sum\limits_{k=1}^{\infty}(k\pi)^2g_k^{(A)}(b)\frac{\sin{\varphi_k}-{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^3},\\
\label{eq:Bess_b_A_1}
&J_1(bx)=2\frac{x}{b}\sum\limits_{k=1}^{\infty}(k\pi)^2g_k^{(A)}(b)\frac{(3-{{\varphi_k}^2})\sin{\varphi_k}-3{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^5},\\
\label{eq:Bess_b_A_2}
&J_2(bx)=2\frac{x^2}{b^2}\sum\limits_{k=1}^{\infty}(k\pi)^2g_k^{(A)}(b)\frac{{\varphi_k}({\varphi_k}^2-15)\cos{\varphi_k}+3(5-2{\varphi_k}^2)\sin{\varphi_k}}{{\varphi_k}^7}
\end{align}
\end{subequations}
where $b\in(0,1)$.
(B)-case:
\begin{subequations}
\label{eq:particular b_B}
\begin{align}
\label{eq:Bess b_B_1}
&J_1(bx)=\frac{2b}{x}\sum\limits_{k=0}^{\infty}\eps_k{g_k^{(B,C)}(b)}\big[(-1)^k-\cos{\varphi_k}\big],\\
\label{eq:Bess_b_B_2}
&J_2(bx)=\frac{b^2}{x^2}\sum\limits_{k=0}^{\infty}\eps_k{g_k^{(B,C)}(b)}\big[
(-1)^k(2+x^2)-2(\varphi_k\sin{\varphi_k}+\cos{\varphi_k})
\big],
\end{align}
\end{subequations}
where $b\in[0,1]$.
(C)-case:
\begin{subequations}
\label{eq:particular_b_C}
\begin{align}
\label{eq:Bess_b_C_0}
&J_0(bx)=2\sum\limits_{k=0}^{\infty}\eps_k{g_k^{(B,C)}(b)}\frac{\sin{\varphi_k}}{\varphi_k},\\
\label{eq:Bess_b_C_1}
&J_1(bx)=2\frac{x}{b}\sum\limits_{k=0}^{\infty}\eps_k{g_k^{(B,C)}(b)}\frac{\sin{\varphi_k}-{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^3},\\
\label{eq:Bess_b_C_2}
&J_2(bx)=2\frac{x^2}{b^2}\sum\limits_{k=0}^{\infty}\eps_k{g_k^{(B,C)}(b)}\frac{(3-{{\varphi_k}^2})\sin{\varphi_k}-3{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^5}
\end{align}
\end{subequations}
where $b\in[0,1]$ for $J_0(bx)$, and $b\in(0,1]$ for $J_n(bx)$, $n\geq1$.
One can derive following properties of the $k$-th term in series (\ref{eq:expansionA_n}), (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}) at $k\to\infty$:
\begin{subequations}
\label{eq:decrease_A}
\begin{align}
\label{eq:decrease0_A}
&\qquad\Big|(k\pi)g_{k}^{(A)}(b)f_{n}^{(A)}(x,k\pi)\Big|\leq\Big|(k\pi)f_{n}^{(A)}(x,k\pi)\Big|,\quad
n,m=\overline{0,\infty},\\
\label{eq:decrease1_A}
&(k\pi)f_{2m}^{(A)}(x,k\pi)\sim2(-1)^{k+m+1}\left(\frac{x}{k\pi}\right)^{2m},\\
\label{eq:decrease2_A}
&(k\pi)f_{2m+1}^{(A)}(x,k\pi)\sim2\frac{(-1)^{k+m+1}}{k\pi}\left(\frac{x}{k\pi}\right)^{2m+1} \big[x^2+(2m+2)(2m+3)\big];
\end{align}
\end{subequations}
\begin{subequations}
\label{eq:decrease_B}
\begin{align}
\label{eq:decrease0_B}
&\qquad\Big|g_{k}^{(B,C)}(b)f_{n}^{(B)}(x,k\pi)\Big|\leq\Big|f_{n}^{(B)}(x,k\pi)\Big|,\quad
n=\overline{1,\infty},\quad m=\overline{0,\infty},\\
\label{eq:decrease1_B}
&f_{4m+1}^{(B)}(x,k\pi)\sim(-1)^{k+m}\frac{x^{2m-1}}{2^{2m+1}}\left(\frac{x}{k\pi}\right)^{2m+2}\frac{(2m+1)!}{(2(2m+1))!}\big[x^2+4m(2m+1)\big],\quad\\
\label{eq:decrease2_B}
&f_{4m+2}^{(B)}(x,k\pi)\sim(-1)^{k+m}\frac{x^{2m}}{2^{2m}}\left(\frac{x}{k\pi}\right)^{2m+2}\frac{(2m+1)(2m+1)!}{(4m+3)(2(2m+1))!}\big[x^2+2m(4m+3)\big],\\
\label{eq:decrease3_B}
&f_{4m+3}^{(B)}(x,k\pi)\sim{(-1)^{k+m+1}}\frac{x^{2m+1}}{2^{2m}}\left(\frac{x}{k\pi}\right)^{2m+2}\frac{(2m+2)!}{(2(2m+2))!},\\
\label{eq:decrease4_B}
&f_{4m+4}^{(B)}(x,k\pi)\sim{(-1)^{k+m+1}}\frac{x^{2m+2}}{2^{2m}}\left(\frac{x}{k\pi}\right)^{2m+2}\frac{(2m+2)!}{(2(2m+2))!};
\end{align}
\end{subequations}
\begin{subequations}
\label{eq:decrease_C}
\begin{align}
\label{eq:decrease0_C}
&\qquad\Big|g_{k}^{(B,C)}(b)f_{n}^{(C)}(x,k\pi)\Big|\leq\Big|f_{n}^{(C)}(x,k\pi)\Big|,\quad
n,m=\overline{0,\infty},\\
\label{eq:decrease1_C}
&f_{2m}^{(C)}(x,k\pi)\sim(-1)^{k+m}\frac{1}{(k\pi)^2}\left(\frac{x}{k\pi}\right)^{2m} \big[x^2+2m(2m+1)\big],\\
\label{eq:decrease2_C}
&f_{2m+1}^{(C)}(x,k\pi)\sim2(-1)^{k+m+1}\frac1{k\pi}\left(\frac{x}{k\pi}\right)^{2m+1}.
\end{align}
\end{subequations}
Notation $a_k\sim b_k$ means that $\lim_{k\to\infty}\frac{a_k}{b_k}=1$. Eqs.~(\ref{eq:decrease_A}),~(\ref{eq:decrease_B}),~(\ref{eq:decrease_C}) allow to conclude absolute convergence and uniform convergence with respect to parameter $b$ at $b\in(0,1)$ for series (\ref{eq:expansionA_n}), (\ref{eq:expansionB_n}), and (\ref{eq:expansionB_n}) except for series (\ref{eq:expansionA_n}) at $n=0$ (that is (\ref{eq:Bess_b_A_0}))
which is conditionally convergent and is not uniformly convergent with respect to $b$ at $b\in(0,1)$.
Due to the uniform convergence of series~(\ref{eq:expansionA_n}) with respect to $b\in(0,1)$ at $n\geq1$ we can take a termwise limit $b\to1$ and hence put $g_k^{(A)}(b)=1$ at $b\to1$ in (A)-case:
\begin{equation}
\label{eq:expansionA2_n}
J_{n}(x)=\sum\limits_{k=1}^{\infty}(k\pi)f_{n}^{(A)}(x,k\pi),\quad n\geq1,\quad x\geq0.
\end{equation}
Series (\ref{eq:expansionA2_n}) at $n=0$ diverges.
The series representations for the Bessel functions of order $0$, $1$, $2$
at $b=1$ are presented below.
(A)-case:
\begin{subequations}
\label{eq:particular_A}
\begin{align}
\label{eq:Bess_A_1}
&J_1(x)=2x\sum\limits_{k=1}^{\infty}(k\pi)^2\frac{(3-{{\varphi_k}^2})\sin{\varphi_k}-3{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^5},\\
\label{eq:Bess_A_2}
&J_2(x)=2x^2\sum\limits_{k=1}^{\infty}(k\pi)^2\frac{{\varphi_k}({\varphi_k}^2-15)\cos{\varphi_k}+3(5-2{\varphi_k}^2)\sin{\varphi_k}}{{\varphi_k}^7}
\end{align}
\end{subequations}
(B)-case:
\begin{subequations}
\label{eq:particular_B}
\begin{align}
\label{eq:Bess_B_1}
&J_1(x)=\frac{2}{x}\sum\limits_{k=0}^{\infty}\eps_k\big[(-1)^k-\cos{\varphi_k}\big],\\
\label{eq:Bess_B_2}
&J_2(x)=\frac{1}{x^2}\sum\limits_{k=0}^{\infty}\eps_k\big[
(-1)^k(2+x^2)-2(\varphi_k\sin{\varphi_k}+\cos{\varphi_k})
\big]
\end{align}
\end{subequations}
(C)-case:
\begin{subequations}
\label{eq:particular_C}
\begin{align}
\label{eq:Bess_C_0}
&J_0(x)=2\sum\limits_{k=0}^{\infty}\eps_k\frac{\sin{\varphi_k}}{\varphi_k},\\
\label{eq:Bess_C_1}
&J_1(x)=2x\sum\limits_{k=0}^{\infty}\eps_k\frac{\sin{\varphi_k}-{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^3},\\
\label{eq:Bess_C_2}
&J_2(x)=2x^2\sum\limits_{k=0}^{\infty}\eps_k\frac{(3-{{\varphi_k}^2})\sin{\varphi_k}-3{\varphi_k}\cos{\varphi_k}}{{\varphi_k}^5}
\end{align}
\end{subequations}
According to some simple considerations, we conclude that series (\ref{eq:expansionA_n}), (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}) at all permitted values of parameter $b$, and series (\ref{eq:expansionA2_n}) converge uniformly with respect to $x\in[0,A]$ where $A$ is arbitrary large real number.
One can derive another series representations for Bessel functions by taking $b\neq1$ and scaling variable $x$ by $1/b$. So (A)-case gives the following series representation for $J_0(x)$ at $b=\sqrt{3}/2$:
\begin{equation}
\label{eq:J0_A_case}
\begin{split}
&J_0(x)=4\sum\limits_{n=1}^{\infty}(-1)^n\big[(2n-1)\pi\big]\frac{\psi_n\cos\psi_n-\sin\psi_n}{{\psi_n}^3},\quad\mbox{where}\\
&\psi_n=\sqrt{\frac{4x^2}{3}+\big[(2n-1)\pi\big]^2}.
\end{split}
\end{equation}
\section{Infinite series representations for sine and cosine functions}
Derived series expansions allow to present numerous similar expansions for sine and cosine functions. We will present some of them.
Lets consider Eq.~(\ref{eq:Bess b_B_1}) which we rewrite in the following way:
\begin{equation}
\label{eq:series}
x\frac{J_1(bx)}{b}=2\sum\limits_{k=0}^{\infty}\eps_k{g_k^{(B,C)}(b)}\big[(-1)^k-\cos{\varphi_k}\big],
\end{equation}
where $\varphi_k$ is presented in Eq.~(\ref{eq:phi}).
Series (\ref{eq:series}) is uniformly convergent with respect to $b\in(0,1)$ therefore taking a limit $b\to0$ in Eq.~(\ref{eq:series}) gives us the following result:
\begin{equation}
\label{eq:cos_1}
\cos{x}-1+\frac{x^2}{2}=2\sum\limits_{k=1}^{\infty}
\Big[
1-(-1)^k\cos{\fif_k}
\Big].
\end{equation}
Taking $b=0$ in Eq.~(\ref{eq:Bess_b_C_0}) we derive series for sine function:
\begin{equation}
\label{eq:sin_1}
1-\frac{\sin{x}}{x}=2\sum\limits_{k=1}^{\infty}
\bigg[
(-1)^{k}\frac{\sin{\fif_k}}{\fif_k}
\bigg].
\end{equation}
It can be also derived by differentiating Eq.~(\ref{eq:cos_1}) with
respect to $x$.
Subtracting Eq.~(\ref{eq:Bess_B_1}) and Eq.~(\ref{eq:Bess_C_1}) we get another series for sine function:
\begin{equation}
\label{eq:sin_2}
1-\frac{\sin{x}}{x}=-2\sum\limits_{k=1}^{\infty}
\bigg[
(-1)^k-\frac{(k\pi)^2}{{\fif_k}^2}\cos{\fif_k}-\frac{x^2}{{\fif_k}^2}\frac{\sin\fif_k}{\fif_k}
\bigg]
\end{equation}
which converges faster than the previous one.
Equations (\ref{eq:cos_1}), (\ref{eq:sin_1}), (\ref{eq:sin_2}) can be differentiated any times that will give plenty of other series representations for sine and cosine functions.
\section*{Conclusion}
To sum up we outline key properties of the series representations (\ref{eq:expansionA_n}), (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}), and (\ref{eq:expansionA2_n}) for Bessel functions of the first kind of integer order:
\begin{itemize}
\item series (\ref{eq:expansionA_n}) at $n\geq1$, (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}) are uniformly convergent with respect to parameter $b\in(0,1)$ at all $x$;
\item series (\ref{eq:expansionA_n}), (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}) at all permitted values of $b$, and series (\ref{eq:expansionA2_n}) are uniformly convergent with respect to variable $x\in[0,A]$, where $A$ is arbitrary large real value;
\item series (\ref{eq:expansionA_n}) at $n\geq1$, (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}) at all permitted values of $b$, and series (\ref{eq:expansionA2_n}) are absolutely convergent; series (\ref{eq:expansionA_n}) at $n=0$ (that is (\ref{eq:Bess_b_A_0})) is conditionally convergent.
\end{itemize}
As series (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}) at $b=1$ and (\ref{eq:expansionA2_n}) become alternating and $k$-th terms approach zero monotonically at sufficient large $k$ and $x\in[0,A]$ (see Eqs.~(\ref{eq:decrease_A}), (\ref{eq:decrease_B}), (\ref{eq:decrease_C})), the
bonds for the error terms of infinite sums approximations by truncated series to $K$-term are absolute value of ($K+1$)-th term. As one can see, ($K+1$)-th terms are monotonically decreasing functions of $x$ in all three cases. Besides, truncated series (\ref{eq:expansionA_n}), (\ref{eq:expansionB_n}), (\ref{eq:expansionC_n}), (\ref{eq:expansionA2_n}) have oscillating behaviour (sine-like shape) with decreasing amplitude at large $x$ that is they behave exactly like Bessel functions. However, truncated series fail to provide correct asymptotic behaviour of Bessel function at $x\to\infty$~\cite{Watson_1_1995}:
\[
J_n(x)\sim\sqrt{\frac{2}{\pi x}}\cos\Big(x-\frac{n\pi}{2}-\frac{\pi}{4}\Big).
\]
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,123 |
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\def{\mbox{\tiny \rm HC}}{{\mbox{\tiny \rm HC}}}
\def\spaceind \cH{{}_{\scriptscriptstyle \circ} {\mathcal H}}
\def\spaceind d{{}_{\scriptscriptstyle \circ} d}
\def\spaceind \pi{{}_{\scriptscriptstyle \circ} \pi}
\def\fa_0^{*{\rm reg}}{{\got a}_0^{*{\rm reg}}}
\def\bar\fa_{0}^*{\bar{\got a}_{0}^*}
\def\gamma{\gamma}
\def\bar B{\bar B}
\def{\mathrm i}{{\mathrm i}}
\def\cF_{\rm eucl}{{\mathcal F}_{\rm eucl}}
\begin{document}
\title{Paley--Wiener spaces for real reductive Lie groups}
\author{Erik P.\ van den Ban and Henrik Schlichtkrull}
\date{
\maketitle
\noindent
{\small In honor of Gerrit van Dijk}
\tableofcontents
\eqsection{Introduction}
The Paley--Wiener theorem for $K$-finite
compactly supported smooth
functions
on a real reductive group Lie group $G$ of the Harish-Chandra class
is due to J. Arthur \bib{Arthur} in general, and to O.A.\ Campoli \bib{Camppw} for $G$ of split rank one.
In our paper \bib{BSpw} we established a similar Paley--Wiener theorem for
smooth functions on a reductive symmetric space. In this paper
we will show that Arthur's theorem is a consequence of our result if one
considers the group $G$ as a symmetric space for $G\times G$ with respect
to the left times right action. At the same time
we will formulate a Paley--Wiener theorem for $K$-finite
{\it generalized functions} (in the
sense of distribution theory) on $G$, and prove that it is a special case of the
Paley--Wiener theorem for symmetric spaces established in our paper \bib{BSdpw}.
All mentioned Paley--Wiener theorems are formulated in the following spirit.
A Fourier transform is defined by means of Eisenstein integrals for the minimal
principal series for the group or space under consideration.
The Eisenstein integrals depend on a certain spectral parameter and
satisfy the so called Arthur--Campoli relations.
A Paley--Wiener space is defined as a certain space of meromorphic functions in the spectral parameter,
characterized by the Arthur--Campoli relations and by growth estimates.
The Paley--Wiener theorem
asserts that the Fourier transform is a topological linear isomorphism
from a space of $K$-finite compactly supported smooth or generalized functions onto
a particular Paley--Wiener space.
The Paley--Wiener space of Arthur's paper is defined in terms of Eisenstein
integrals as introduced by Harish-Chandra \bib{HC1}; we shall refer to these
integrals as being {\em unnormalized.}
Our Paley--Wiener theorems
in \bib{BSpw} and \bib{BSdpw} are defined in terms of the so-called {\em normalized\ }
Eisenstein integrals defined in \bib{BSft}. For $G$ considered as a symmetric
space the normalized Eisenstein integral differs from the unnormalized one.
Consequently, the associated Fourier transforms and Paley--Wiener spaces
are different. The final objective of this paper
is to clarify
the relationships between the various Paley--Wiener spaces.
Recently,
P.\ Delorme, \bib{Delpw}, has proved a different
Paley--Wiener theorem, involving the operator-valued
Fourier transforms associated with all generalized
principal series representations. In his result the
Arthur--Campoli relations are replaced by intertwining relations.
Moreover, the result is valid without the restriction of $K$-finiteness.
We shall now give a brief outline of the present paper.
In Section \refer{s: basic notation} we introduce the basic concepts,
in particular the space $C_c^\infty(G\col \tau)$
of $\tau$-spherical compactly supported smooth
functions,
for which Arthur's theorem is most
conveniently formulated.
We review the definition of Harish-Chandra's (unnormalized)
Eisenstein
integral $E(P\col \lambda)$ for $P$ a minimal parabolic subgroup of $G$ and
with spectral parameter $\lambda \in \fa_{0\iC}^*;$ here $\fa_0$ is
the Lie algebra of the split component $A_0$ of $P.$ Finally, we give the definition
of the associated Fourier transform ${}^u \cF_P.$
In Section \refer{s: Arthur's Paley--Wiener theorem}
we recall, in Theorem \refer{t: arthurs PW one}, the formulation
of Arthur's Paley--Wiener theorem, \bib{Arthur}.
This theorem deals with the family of Fourier transforms $({}^u \cF_Q)_{Q \in \cP_0}$ with
$Q$ ranging over the finite set $\cP_0$ of parabolic subgroups with the same split
component $A_0.$ The theorem asserts that this family of transforms
establishes an isomorphism from $C_c^\infty(G\col \tau)$
onto a Paley--Wiener space ${}^u {\rm PW}(G,\tau, \cP_0).$
The Fourier transforms in the
family
are completely determined by
any single one among them. Accordingly, in Theorem \refer{t: arthurs PW two},
Arthur's Paley--Wiener theorem is
reformulated by asserting that a single Fourier transform ${}^u \cF_P$ defines a topological
linear isomorphism from $C_c^\infty(G\col \tau)$ onto a Paley--Wiener space
${}^u {\rm PW}(G,\tau, P).$
In Section \refer{s: dist PW space} we formulate, in Theorem \refer{t: distributional PW},
a distributional
Paley--Wiener theorem, which asserts that ${}^u \cF_P$ defines a topological
linear isomorphism from the space $C^{- \infty}_c(G\col \tau)$ of $\tau$-spherical compactly supported
generalized functions on $G$ onto a Paley--Wiener space
${}^u {\rm PW}^*(G , \tau, P).$
At
this point of the paper, the main results have been stated. The rest
of the paper is devoted to proofs. In Section \refer{s: c functions} we
prepare for this by introducing the $C$-functions and listing
those of their properties that are needed.
In Section \refer{s: Relations between the Fourier transforms}
we give the proof that Theorem \refer{t: arthurs PW two} is indeed
a reformulation of Arthur's theorem.
We describe the relations between the different Fourier transforms
in terms of $C$-functions.
The equivalence of Theorems \refer{t: arthurs PW one} and \refer{t: arthurs PW two}
is then captured by the assertion, in Proposition \refer{p: proj is topol iso},
that the natural map between the Paley--Wiener spaces ${}^u {\rm PW}(G,\tau, \cP_0)$
and ${}^u {\rm PW}(G, \tau, P)$ is a topological isomorphism.
In the next Section, \refer{s: the normalized Fourier transform},
a normalized Fourier transform $\cF_P$ is defined in terms of the normalized Eisenstein integral
$$
{E^\circ}(P \col \lambda \col \,\cdot\,) := E(P \col \lambda \col \,\cdot\,) \,{\scriptstyle\circ}\, C_{P|P}(1 \col \lambda)^{-1};
$$
This normalization is natural from the point of view of asymptotic expansions; it has the effect
that
the new, normalized $C$-functions become unitary for imaginary $\lambda.$
It follows from the relation between the Eisenstein integrals that the normalized
Fourier transform is related to the above Fourier transform
by a relation of the form ${}^u \cF_P = \cU_P \,{\scriptstyle\circ}\, \cF_P,$
where $\cU_P$ denotes
multiplication by a $C$-function
$\lambda \mapsto C_{P|P}(1 \col - \bar \lambda)^*.$
An associated Paley--Wiener space ${\rm PW}(G,\tau, P)$ is defined, as well as a distributional
Paley--Wiener space, indicated by superscript $*.$ The main result of the section, Theorem
\refer{t: multC is iso PW spaces}, asserts that the map $\cU_P$ induces isomorphisms
${\rm PW}(G,\tau, P) \rightarrow {}^u {\rm PW}(G,\tau, P)$
and ${\rm PW}^*(G,\tau, P) \rightarrow {}^u {\rm PW}^*(G,\tau, P).$
As a result, the associated Paley--Wiener theorems for the normalized Fourier transform are equivalent
to those for the unnormalized transform.
Let ${\spaceind \spX}$ denote $G$ viewed as a symmetric space for $\spaceind\mspace G: = G \times G.$
In the final section the unnormalized Eisenstein integral $E(P\col \lambda)$ for $G$
is related to the unnormalized Eisenstein integral for ${\spaceind \spX}$ as defined in \bib{BSft}.
It follows from this relation that the normalized Eisenstein integrals,
for $G$ and ${\spaceind \spX},$ coincide; therefore,
so do the normalized Fourier transforms. Likewise, it is shown that the Paley--Wiener spaces
for $G$ coincide with the similar spaces for ${\spaceind \spX}.$ This finally establishes the validity
of all mentioned Paley--Wiener theorems as special cases of the theorems in \bib{BSpw} and \bib{BSdpw}.
\eqsection{Basic concepts}
\naam{s: basic notation}
Let $G$ be a real reductive Lie group of the Harish-Chandra class
and let $K$ be a maximal compact subgroup. Let $V_\tau$ be a finite
dimensional Hilbert space, and let $\tau$ be a unitary
double representation
of $K$ in $V_\tau.$ By this we mean that $\tau = (\tau_1, \tau_2)$
with $\tau_1$ a left and $\tau_2$ a right unitary representation of $K$ in $V_\tau;$ moreover,
the representations $\tau_1$ and $\tau_2$ commute. We will often drop the subscripts
on $\tau_1$ and $\tau_2,$ writing
$$
\tau(k_1) v \tau(k_2) = \tau_1(k_1) v \tau_2(k_2)
$$
for all $v \in V_\tau$ and $k_1, k_2 \in K.$ A function $f: G \rightarrow V_\tau$ is called
$\tau$-spherical if it satisfies the rule
\begin{equation}
\naam{e: spherical rule}
f(k_1 g k_2) = \tau(k_1) f(g) \tau(k_2)
\end{equation}
for all $g \in G$ and $k_1, k_2 \in K.$ The space of smooth $\tau$-spherical functions
is denoted by $C^\infty(G\col \tau)$ and equipped with the usual Fr\'echet topology.
The subspace $C_c^\infty(G\col \tau)$ of compactly supported smooth $\tau$-spherical
functions is equipped with the usual complete locally convex (Hausdorff) topology.
We shall first briefly establish some notation for the group.
As usual, we denote Lie groups by Roman capitals, and their Lie algebras
by the Gothic lower case equivalents.
Let $\theta \in {\rm Aut}(G)$ be the Cartan involution associated with $K.$ The
associated infinitesimal involution of ${\got g}$ is denoted by the same symbol and
the associated eigenspaces with eigenvalues +1 and $-1$ by ${\got k}$ and ${\got p},$ respectively.
Accordingly, we have the Cartan decomposition ${\got g} = {\got k} \oplus {\got p}.$
Let $\fa_0$ be a maximal abelian subspace of ${\got p}$ and let $\Sigma$
be the restricted root system of $\fa_0$ in ${\got g}.$
Let $A_0:= \exp \fa_0$ be the associated
vectorial subgroup of $G$ and let $\cP_0 = {\mathcal P}(A_0)$ be the collection of parabolic subgroups
of $G$ with split component $A_0.$ Each element $P \in \cP_0$
is minimal and has a Langlands decomposition of the form $P = {M_0} A_0 N_P,$
with ${M_0}$ equal to the centralizer of $A_0$ in $K.$
Let $\Sigma(P)$ be the collection of $\fa_0$-roots in ${\got n}_P = {\rm Lie}(N_P).$
Then $P \mapsto \Sigma(P)$ defines a one-to-one correspondence between ${\mathcal P}(A_0)$
and the collection of positive systems for $\Sigma.$
We equip $\fa_0$
with a $W$-invariant positive definite inner product $\inp{\,\cdot\,}{\,\cdot\,};$
the dual space $\fa_0^*$ is equipped with the dual inner product. The latter inner product
is extended to a complex bilinear form on $\fa_{0\iC}^*.$ The norm associated with the inner product
on $\fa_0^*$ is extended to a Hermitian
norm on $\fa_{0\iC}^*,$ denoted by $|\,\cdot\,|.$
We shall now review the definition of the $\tau$-spherical Eisenstein
integral related to a given parabolic subgroup $P \in {\mathcal P}(A_0).$
Let $\tau_{M_0}$ denote the restriction of $\tau$ to ${M_0}.$ As ${M_0}$ is a subgroup of $K,$
the space $L^2({M_0} \col \tau_{M_0})$ of square integrable $\tau_{M_0}$-spherical
functions ${M_0} \rightarrow V_\tau $ is finite dimensional and equals the space of smooth $\tau_{M_0}$-spherical
functions. We equip $M_0$ with normalized Haar measure, and define the finite
dimensional Hilbert space
$\cA_2 = \cA_2(\tau)$ by
\begin{equation}
\naam{e: defi cAtwo}
\cA_2:= L^2({M_0}\col \tau_{M_0}) = C^\infty({M_0}\col \tau_{M_0}).
\end{equation}
Let $\psi \in \cA_2.$ For $\lambda \in \fa_{0\iC}^*$
we define the function $\psi_\lambda = \psi_{P, \lambda}: G \rightarrow V_\tau$ by
$$
\psi_\lambda(n a m k) = a^{\lambda + \rho_P} \psi(m) \tau_2(k),
$$
for $k \in K,$ $m \in {M_0},$ $a \in A_0$ and $n \in N_P.$
Here $\rho_P \in \fa_0^*$ is defined by
$$
\rho_P(H) = \frac12 {\rm tr}\, ({\rm ad}(H)|{\got n}_P).
$$
By the analytic nature of the Iwasawa decomposition $G = N_P A_0 K,$
the function $\psi_\lambda$
is analytic. We define the Eisenstein integral
$
E(P\col \psi\col \lambda) :\, G \rightarrow V_\tau
$
by
\begin{equation}
\naam{e: defi unnormalized E}
E(P\col \psi\col \lambda \col x): = \int_K \tau_1(k)\psi_\lambda(k^{-1} x) \; dk,
\end{equation}
for $x \in G.$ Then, clearly, $E(P\col \psi\col \lambda)$ is a function in $C^\infty(G\col \tau),$
depending linearly on $\psi$ and holomorphically on $\lambda.$
\begin{rem}
\naam{r: gl parameter Eis}
Here we have adopted the same convention as J.\ Arthur \bib{Arthur}, \S 2, which differs from
Harish-Chandra's.
Let $E^{\mbox{\tiny \rm HC}}$ denote the Eisenstein integral as defined by Harish-Chandra \bib{HC1}, \S 9.
Then $E^{\mbox{\tiny \rm HC}}(P\col \psi \col \lambda) = E(P \col \psi \col i \lambda).$
For
the reader's convenience, we note that Arthur \bib{Arthur} uses the notation
${\mathcal A}_{\mathrm{cusp}}(M_0, \tau)$ or ${\mathcal A}_0$ for the space (\refer{e: defi cAtwo}).
\end{rem}
In terms of the Eisenstein integral we define a Fourier transform
${}^u \cF_P$ from $C_c^\infty(G\col \tau)$ to the space ${\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2$
of holomorphic $\cA_2$-valued functions on $\fa_{0\iC}^*.$ The superscript $u$ indicates that this Fourier
transform is defined in terms of the above unnormalized Eisenstein integral, in contrast with a normalized
Fourier transform $\cF_P$ to be defined later.
Let $dx$ be a choice of Haar measure on
$G.$ We define the Fourier transform ${}^u \cF_P f$ of $f \in C_c^\infty(G \col \tau)$ by the formula
\begin{equation}
\naam{e: defi unnormalized Fou}
\inp{{}^u \cF_P f(\lambda)}{\psi}_{\cA_2} = \int_G \inp{f(x)}{E(P\col \psi\col - \bar\lambda \col x)}_{V_\tau} \; dx,
\end{equation}
for $\psi \in \cA_2$ and $\lambda \in \fa_{0\iC}^*.$ It follows from the Paley--Wiener theorem in \bib{Arthur} that
${}^u \cF_P$ is injective on $C_c^\infty(G\col \tau).$ This injectivity can also
be established by application
of the subrepresentation theorem, \bib{CM}, Thm.~8.21.
\eqsection{Arthur's Paley--Wiener theorem}
\naam{s: Arthur's Paley--Wiener theorem}
The image of $C_c^\infty(G\col \tau)$ under Fourier transform
is described by the Paley--Wiener theorem due to
J.\ Arthur \bib{Arthur}, which we shall now formulate.
It is convenient to rewrite the definition of the Fourier transform
${}^u \cF_P$ in terms of a (unnormalized) dual Eisenstein integral.
Given $x \in G$ and $\lambda \in \fa_{0\iC}^*,$ we agree to define $E(P\col \lambda\col x) \in {\rm Hom}(\cA_2, V_\tau)$
by the formula
$$
E(P\col \lambda \col x)\psi: = E(P \col \psi\col \lambda \col x)
$$
for $\psi \in \cA_2.$ Moreover, we define a dual Eisenstein integral by
\begin{equation}
\naam{e: defi unnormalized dual Eis}
{}^u\! E^*(P \col \lambda\col x): = E(P\col - \bar \lambda \col x)^* \in {\rm Hom}(V_\tau, \cA_2),
\end{equation}
where the superscript $*$ indicates that the Hilbert adjoint has been taken.
The superscript $u$ serves to distinguish the present dual Eisenstein integral
from a normalized version that will be introduced at a later stage.
The dual Eisenstein integral ${}^u\! E^*$ may be viewed as a smooth
${\rm Hom}(V_\tau, \cA_2)$-valued
function on $\fa_{0\iC}^* \times G$
which is holomorphic in the first variable.
In terms of this Eisenstein integral,
the Fourier transform
(\refer{e: defi unnormalized Fou})
may be expressed as an integral transform. Indeed,
it readily follows from the given definitions that
\begin{equation}
\naam{e: defi uFou}
{}^u \cF_P f (\lambda) = \int_G {}^u\! E^*(P\col\lambda \col x) f(x)\; dx,
\end{equation}
for $f \in C_c^\infty(G\col \tau)$ and $\lambda \in \fa_{0\iC}^*.$
We now proceed to giving the definition of a suitable Paley--Wiener space.
Let $V$ be a finite dimensional real linear space.
We denote by $S(V)$ the symmetric algebra of $V_{\scriptscriptstyle \C}.$ This algebra is identified
with the algebra of constant coefficient
holomorphic differential operators on $V_{\scriptscriptstyle \C}$ in the usual way.
We denote by ${\mathcal O}(V_{\scriptscriptstyle \C})$
the space of holomorphic functions on $V_{\scriptscriptstyle \C}$ and,
for $a \in V_{\scriptscriptstyle \C},$
by ${\mathcal O}_a = {\mathcal O}_a(V_{\scriptscriptstyle \C})$ the space of germs of holomorphic
functions at $a.$
Moreover, we denote by ${\mathcal O}_a(V_{\scriptscriptstyle \C})^*_{\rm tayl} $ the space of
linear functionals
${\mathcal O}_a \rightarrow {\msy C}$ of the form
$$
f \mapsto {\rm ev}_a(uf) = u f(a),
$$
with $u \in S(V).$ The elements of ${\mathcal O}_a(V_{\scriptscriptstyle \C})^*_{\rm tayl}$
will be called Taylor functionals
at $a,$ as they give linear combinations
of coefficients of a Taylor series at $a.$
Clearly, the map $u \mapsto {\rm ev}_a\,{\scriptstyle\circ}\, u$ is a
linear isomorphism from $S(V)$ onto ${\mathcal O}_a(V_{\scriptscriptstyle \C})^*_{\rm tayl}.$
We define the space of Taylor functionals on $V_{\scriptscriptstyle \C}$
as the algebraic direct sum
$$
{\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl} \,:= \;\oplus_{a \in V_{\scriptscriptstyle \C}}\;\; {\mathcal O}_a(V_{\scriptscriptstyle \C})^*_{\rm tayl}.
$$
Given $U \in {\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl},$ the finite set
of $a \in V_{\scriptscriptstyle \C}$ with $U_a \neq 0$
is called the support of $U,$ notation ${\rm supp}\, U.$
Given $f \in {\mathcal O}(V_{\scriptscriptstyle \C}),$
we put
$$
U f := \sum_{a \in {\rm supp}\, U} U_a f_a.
$$
The map $U \otimes f \mapsto U f$ defines an embedding of
${\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl}$
into the linear dual ${\mathcal O}(V_{\scriptscriptstyle \C})^*;$
this justifies the notation.
Note that in fact the elements of ${\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl}$
are continuous with respect to
the usual Fr\'echet topology on ${\mathcal O}(V_{\scriptscriptstyle \C}).$
Finally, we note that a finitely supported function $U: V_{\scriptscriptstyle \C} \rightarrow S(V)$ may be viewed
as a Taylor functional by the formula
$
U f := \sum_a {\rm ev}_a [ U(a) f ].
$
Accordingly, the space of Taylor functionals may be identified with
the space of finitely supported functions $V_{\scriptscriptstyle \C} \rightarrow S(V).$
\begin{defi}
An (unnormalized, holomorphic)
Arthur--Campoli functional for $(G, \tau, \cP_0)$ is a family
$({\mathcal L}_P)_{P \in \cP_0} \subset {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl} \otimes \cA_2^*$ such that
$$
\sum_{P \in \cP_0} {\mathcal L}_P\; [{}^u\! E^* (P\col \,\cdot\, \col x)v_P] = 0
$$
for all $x \in G$ and all $(v_P)_{P \in \cP_0} \subset V_\tau.$
The linear space of such families is denoted by
${}^u{\rm AC}_\hol(G,\tau,\cP_0).$
\end{defi}
For $R > 0$ we define
${\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*)$ to be the space of holomorphic functions
$\varphi: \fa_{0\iC}^* \rightarrow {\msy C}$ such that for every $n \in {\msy N},$
$$
\nu_{R,n}(\varphi) := \sup_{\lambda \in \fa_{0\iC}^*}\;\;
(1 + |\lambda|)^{n} e^{- R |{\rm Re}\, \lambda|}\, |\varphi(\lambda)| < \infty.
$$
Equipped with the seminorms $\nu_{R, n},$ for $n \in {\msy N},$
this space is a Fr\'echet space.
\begin{defi}
\naam{d: arthurs PW}
Let $R > 0.$ The Paley--Wiener space ${}^u {\rm PW}_R(G, \tau, \cP_0)$ is defined to be the space of
families $(\varphi_P)_{P \in \cP_0} \subset {\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*) \otimes \cA_2$ such that
$$
\sum_{P \in \cP_0} \; {\mathcal L}_P \varphi_P = 0
$$
for all $({\mathcal L}_P)_{P \in \cP_0} \in {}^u{\rm AC}_\hol(G,\tau, \cP_0).$
\end{defi}
By continuity of the Taylor functionals, the Paley--Wiener space is a closed subspace of the direct
sum of a finite number of copies of $ {\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*) \otimes \cA_2,$ one for each $P \in \cP_0;$
it is therefore a Fr\'echet space of its own right.
For $R >0$ we put $G_R: = K \exp \bar B_R K,$ where $\bar B_R$
denotes the closed ball of center
$0$ and radius $R$ in $\fa_0.$ Moreover, we define the following closed subspace
of $C_c^\infty(G\col \tau),$
and equip it with the relative topology:
$$
C_R^\infty(G \col \tau) = \{ f \in C^\infty(G\col \tau) \mid {\rm supp}\, f \subset G_R \}.
$$
We have now gathered the concepts and notation needed to formulate
the Paley--Wiener theorem due to J.\ Arthur, \bib{Arthur}, p.~83, Thm.~3.3.l.
\begin{thm}{\rm (Arthur \bib{Arthur})\ }
\naam{t: arthurs PW one}
The map $f \mapsto ({}^u \cF_P f)_{P \in \cP_0}$ is a topological linear isomorphism from
$C^\infty_R(G\col \tau)$ onto ${}^u {\rm PW}_R(G,\tau,\cP_0).$
\end{thm}
Each of the individual Fourier transforms ${}^u \cF_P,$ for $P \in \cP_0,$
is already injective on $C_c^\infty(G\col \tau).$ It is therefore natural
to reformulate Arthur's theorem in terms of a single Fourier transform.
\begin{defi}
Let $P \in \cP_0.$
An (unnormalized, holomorphic) Arthur--Campoli functional for the triple $(G,\tau, P)$
is a functional $\cL \in {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl} \otimes \cA_2^*$ such that
\begin{equation}
\naam{e: Lau on udE}
\cL [{}^u\! E^*(P \col \,\cdot\, \col x)v] = 0,
\end{equation}
for all $x \in G$ and all $v \in V_\tau.$ The space of such functionals is denoted by
${}^u{\rm AC}_\hol(G,\tau, P).$
\end{defi}
\begin{defi}
\naam{d: unnormalized PW}
Let $P \in \cP_0$ and $R > 0.$ We define the
Paley--Wiener space ${}^u {\rm PW}_R(G, \tau, P)$ to be the space
of functions $\varphi \in {\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*) \otimes \cA_2$ such that $\cL \varphi = 0$ for all
$\cL \in {}^u{\rm AC}_\hol(G,\tau, P).$
\end{defi}
Arthur's Paley--Wiener theorem may now be reformulated as follows.
\begin{thm}
\naam{t: arthurs PW two}
Let $P \in \cP_0$ and $R > 0.$
The map ${}^u \cF_P$ is a topological linear isomorphism
from $C^\infty_R(G\col \tau)$ onto ${}^u {\rm PW}_R(G,\tau,P).$
\end{thm}
The equivalence of Theorems
\refer{t: arthurs PW one} and \refer{t: arthurs PW two} will be established in
Section \refer{s: Relations between the Fourier transforms}.
\eqsection{A distributional Paley--Wiener space}
\naam{s: dist PW space}
In this section we will formulate a Paley--Wiener theorem
characterizing the image under Fourier transform
of the space $C^{- \infty}_c(G\col\tau)$ of compactly supported $\tau$-spherical
generalized functions.
We shall first define the mentioned space.
The space $C^{- \infty}_c(G)$ of compactly supported
generalized functions on $G$ is defined as the topological linear dual
of the Fr\'echet space of smooth densities on $G.$
It is equipped with the strong dual topology.
Via the map $f \mapsto f \; dx,$ the space $C^{- \infty}_c(G)$ is isomorphic
with the space of compactly supported generalized densities on $G.$
Via integration the latter space may in turn be
identified with the continuous linear dual of $C^\infty(G),$ i.e.,
with the space of compactly supported distributions on $G.$
The pairing with smooth densities induces
a natural embedding $C_c^\infty(G) \hookrightarrow C^{- \infty}_c(G);$ accordingly,
the left and right regular representations of $G$ in $C_c^\infty(G)$ extend
to continuous representations of $G$ in $C^{- \infty}_c(G).$
We now
define $C^{- \infty}_c(G\col V_\tau)$
as the space of $K \times K$-invariants in $C^{- \infty}_c(G) \otimes V_\tau.$
Again, there is a natural embedding
$
C_c^\infty(G\col \tau) \hookrightarrow C^{- \infty}_c(G \col \tau).
$
The definition
of ${}^u \cF_P$ by (\refer{e: defi uFou}), for a given $P \in \cP_0,$
has a natural interpretation for
compactly supported $\tau$-spherical generalized
functions. Accordingly, ${}^u \cF_P$ extends to a continuous
linear map $C^{- \infty}_c(G\col \tau) \rightarrow {\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2.$
Let $R > 0$ and $n \in {\msy N}.$ We define ${\mathrm H}^*_{R, n}(\fa_{0\iC}^*)$ to
be the space of entire holomorphic functions $\varphi: \fa_{0\iC}^* \rightarrow {\msy C}$
with
$$
\nu_{R,- n}(\varphi)= \sup_{\lambda \in \fa_{0\iC}^*}\; (1 + \|\lambda\|)^{-n} e^{-R
| {\rm Re}\, \lambda|} |\varphi(\lambda) |
< \infty.
$$
Equipped with the given norm, this space
is a Banach space. If $m < n,$ then
$$
{\mathrm H}^*_{R,m }(\fa_{0\iC}^*) \subset {\mathrm H}^*_{R, n}(\fa_{0\iC}^*),
$$
with continuous inclusion map.
The union ${\mathrm H}_R^*(\fa_{0\iC}^*)$ of these spaces,
for $n \in {\msy N},$
is equipped with the inductive limit
locally convex topology.
\begin{defi}
\naam{d: unnormalized distrib PW}
Let $P \in \cP_0$ and $R > 0.$
The distributional Paley--Wiener space ${}^u{\rm PW}^*_R(G,\tau, P)$ is defined as the space
of functions $\varphi \in {\mathrm H}_R^*(\fa_{0\iC}^*) \otimes \cA_2$ satisfying
the Arthur--Campoli relations $\cL \varphi = 0$ for all ${\mathcal L} \in {}^u{\rm AC}_\hol(G,\tau, P).$
\end{defi}
By continuity of Taylor functionals, the space
${}^u{\rm PW}^*_R(G, \tau, P)$ is a closed subspace
of ${\mathrm H}_R^*(\fa_{0\iC}^*) \otimes \cA_2;$ we endow it with the relative topology.
The following result is the analogue of Arthur's Paley--Wiener theorem
for $K$-finite compactly supported generalized functions on $G.$
Given $R > 0$ we denote by $C^{-\infty}_R(G\col \tau)$ the space
of $\tau$-spherical generalized functions on $G$ with support contained in $G_R = K\exp \bar B_R K.$
It is equipped with the relative topology.
\begin{thm}
{\rm (The Paley--Wiener theorem for generalized functions)\ \ }
\naam{t: distributional PW}
Let $P \in \cP_0$ and $R > 0.$
The Fourier transform ${}^u \cF_P$ extends to a topological
linear isomorphism from $C^{-\infty}_R(G \col \tau)$ onto ${}^u {\rm PW}^*_R(G,\tau,P).$
\end{thm}
As mentioned in the introduction, this theorem will follow from the results of the present paper
combined with the distributional
Paley--Wiener theorem for reductive symmetric
spaces proved in \bib{BSdpw}.
\begin{rem}
With the same arguments that will lead to the equivalence of Theorems \refer{t: arthurs PW two} and
\refer{t: arthurs PW one}, it can be shown that Theorem
\refer{t: distributional PW} is equivalent to a Paley--Wiener theorem for generalized functions
involving the family $({}^u \cF_P)_{P \in \cP_0}$ in the same spirit as Theorem \refer{t: arthurs PW one}.
\end{rem}
\eqsection{C-functions, singular loci and estimates}
\naam{s: c functions}
To establish the equivalence of Theorems \refer{t: arthurs PW one} and \refer{t: arthurs PW two}
we need relations between the Fourier transforms, which can be given in terms of the so-called $C$-functions.
The latter arise as coefficients in asymptotic expansions of Eisenstein integrals.
Let $Q \in \cP_0.$ We denote by $\fa_Q^+$ the positive chamber determined
by the positive system $\Sigma(Q)$ and by $A_Q^+$ the image in $A_0$ under
the exponential map. Then $K A_Q^+ K$ is an open dense subset of $G.$
In
view of its $\tau$-spherical behavior, the Eisenstein
integral $E(P\col \lambda),$ for $P \in \cP_0,$ is
completely determined by its restriction to $A_Q^+.$ It follows from Harish-Chandra's result
\bib{HC2}, Thm.\ 18.1, that, on $M_0 A_Q^+,$ the given Eisenstein integral behaves asymptotically as follows:
\begin{equation}
\naam{e: asymptotics Eis}
E(P\col \lambda \col m a)\psi \sim
\sum_{s \in W} a^{s\lambda -\rho_Q} [C_{Q|P}(s\col \lambda) \psi](m), \quad\quad
(a \rightarrow \infty \;\;{\rm in}\; A_Q^+),
\end{equation}
for every $\psi \in \cA_2,$ every $m \in M_0,$ and $\lambda \in i\fa_0^{*{\rm reg}}.$ Here $W$
denotes the Weyl group of the root system $\Sigma$ and the coefficients
$C_{Q|P}(s \col \,\cdot\,)$ are ${\rm End}(\cA_2)$-valued analytic functions of $\lambda \in i \fa_0^{*{\rm reg}}.$
The functions $C_{Q|P}(s \col \,\cdot\,),$ for $s \in W,$ are uniquely determined by these
properties. A priori they have a meromorphic extension to an open neighborhood of
$i\fa_0^*$ in $\fa_{0\iC}^*.$
\begin{rem}
\naam{r: gl parameter C}
Harish-Chandra denotes the $c$-functions by lower case letters.
In view of Remark \refer{r: gl parameter Eis}, the $C$-functions introduced above
are related to Harish-Chandra's by the formula $C_{Q|P}(s \col i \lambda) = c_{Q|P}(s \col \lambda).$
\end{rem}
For $P \in \cP_0$ and $R \in {\msy R}$ we put
\begin{equation}\
\naam{e: defi fapd P R}
\fa_0^*(P, R) := \{\lambda \in \fa_{0\iC}^* \mid \inp{{\rm Re}\, \lambda}{\alpha} < R, \;\; \forall \alpha \in \Sigma(P) \}.
\end{equation}
A $\Sigma$-hyperplane in $\fa_{0\iC}^*$ is a hyperplane of the form $\inp{\lambda}{\alpha} = c,$
with $\alpha \in \Sigma$ and $c \in {\msy C}.$ The hyperplane is said to be real if
$c \in {\msy R}.$
We define $\Pi_\Sigma(\fa_{0\iC}^*)$ to be the set of polynomial functions
that can be written as a product of a nonzero complex number
and linear factors of the form $\lambda \mapsto \inp{\lambda}{\alpha} - c,$
with $\alpha \in \Sigma$ and $c \in {\msy C}.$
The subset of polynomial functions which are products as above
with $c \in {\msy R}$ is denoted by $\Pi_{\Sigma, {\msy R}}(\fa_{0\iC}^*).$
\begin{lemma}
\naam{l: estimates C functions in cone}
Let $P \in \cP_0.$
The endomorphism $C(1: \lambda) = C_{P|P}(1: \lambda)
\in {\rm End}(\cA_2)$ is invertible for generic $\lambda \in i\fa_0^*.$
Both maps
\begin{equation}
\naam{e: C pm}
\lambda \mapsto C(1: \lambda)^{\pm 1}
\end{equation}
extend to
${\rm End}(\cA_2)$-valued meromorphic functions on $\fa_{0\iC}^*$ that can be expressed
as products of functions of the form
$\lambda \mapsto c_\alpha(\inp{\lambda}{\alpha}),$
for $\alpha \in \Sigma(P),$
with $c_\alpha$ a meromorphic function on ${\msy C}$ with real
singular locus. Accordingly, each of the functions (\refer{e: C pm})
has a singular locus equal to a locally finite union of real $\Sigma$-hyperplanes
in $\fa_{0\iC}^*.$
Let $R \in {\msy R}.$ Only a finite number of the mentioned singular
hyperplanes intersect $- \fa_0^*(P, R).$ There exist polynomial
functions $q_\pm \in \Pi_{\Sigma, {\msy R}}(\fa_0^*)$ such that $\lambda \mapsto
q_\pm(\lambda) C(1\col - \lambda)^{\pm 1}$ are regular on the
closure of the set $\fa_0^*(P, R).$ If $q_\pm$ is any pair of
polynomials with these properties, there exist constants $n \in
{\msy N}$ and $C > 0$ such that
$$
\| q^{\pm}(\lambda) C(1\col - \lambda)^{\pm 1} \| \leq C (1 + |\lambda|)^n, \quad\quad (\lambda \in \fa_0^*(P, R)).
$$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
All assertions readily follow from the arguments in \bib{Arthur},
proof of Lemma 5.2, except possibly for the final estimate,
which at first follows for a particular choice of $q_{\pm}.$
A straightforward application of the Cauchy integral formula
then gives the result for arbitrary $q_{\pm}$ satisfying the hypotheses.
~\hfill$\square$\medbreak
Following Harish-Chandra \bib{HC3}, \S 17, we define
the following normalized $C$-functions, for $P,Q \in \cP_0$ and $s \in W:$
\begin{equation}
\naam{e: defi ooC}
{}^\circ\hspace{-0.6pt} C_{Q|P}(s\col \lambda):= C_{Q|Q}(1\col s \lambda)^{-1} C_{Q|P}(s \col \lambda).
\end{equation}
The following result, due to Harish-Chandra, \bib{HC3},
will be of crucial importance
to us.
\begin{lemma}
\naam{l: rationality ooC}
{\rm (Harish-Chandra \bib{HC3})\ }
For all $P,Q \in \cP_0$ and $s \in W,$ the
${\rm End}(\cA_2)$-valued function $\lambda \mapsto {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col \lambda)$
has a rational extension to $\fa_{0\iC}^*.$
The endomorphism ${}^\circ\hspace{-0.6pt} C_{Q|P}(s\col \lambda)$
is invertible for generic $\lambda \in \fa_{0\iC}^*$
and $\lambda \mapsto {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col \lambda)^{-1}$ is a rational ${\rm End}(\cA_2)$-valued
function.
Finally, each of the functions $\lambda \mapsto {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col \lambda)^{\pm 1}$
is a product of functions of the form
$\lambda \mapsto c_\alpha(\inp{\lambda}{\alpha}),$ for $\alpha \in \Sigma,$
with $c_\alpha$ a ${\rm End}(\cA_2)$-valued rational function on ${\msy C}.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
The assertions for ${}^\circ\hspace{-0.6pt} C_{Q|P}(s \col \lambda)$
follow from \bib{HC3}, Lemma 19.2 combined with the
Corollary to Lemma 17.2 and with Lemma 17.4 of the same article.
For imaginary $\lambda,$ the endomorphism ${}^\circ\hspace{-0.6pt} C_{Q|P}(s \col \lambda)$
is unitary, by \bib{HC3}, Lemma 17.3. The remaining assertions
now follow by application of Cramer's rule.
~\hfill$\square$\medbreak
\eqsection{Relations between the Fourier transforms}
\naam{s: Relations between the Fourier transforms}
According to \bib{HC3}, Lemma 17.2,
the Eisenstein integrals are related by the following functional
equations:
\begin{equation}
\naam{e: functional equation uEis}
E(P\col \lambda \col \,\cdot\,) = E(Q\col s \lambda \col \,\cdot\,) {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col \lambda),
\end{equation}
for $P,Q \in \cP_0$ and $s \in W,$ as an identity
of meromorphic $C^\infty(G)\otimes {\rm Hom}(\cA_2,V_\tau)$-valued functions in the variable $\lambda \in \fa_{0\iC}^*.$
\begin{lemma}
\naam{l: inversion ooC}
Let $P,Q \in \cP_0.$ Then ${}^\circ\hspace{-0.6pt} C_{Q|P}(1 \col \lambda) \,{\scriptstyle\circ}\, {}^\circ\hspace{-0.6pt} C_{P|Q}(1 \col \lambda) = I,$
as an identity of ${\rm End}(\cA_2)$-valued functions in the variable $\lambda \in \fa_{0\iC}^*.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
From the functional equation for the Eisenstein integral it
follows that, for $x \in G,$
$$
E(P\col \lambda \col x) = E(P\col \lambda \col x) {}^\circ\hspace{-0.6pt} C_{P|Q}(1 \col \lambda) {}^\circ\hspace{-0.6pt} C_{Q|P}(1 \col \lambda).
$$
Using (\refer{e: asymptotics Eis}), we infer that this identity is valid with
$C_{P|P}(1 \col \lambda)$ in place of $E(P\col \lambda \col x)$ on both sides.
As $C_{P|P}(1 \col \lambda)$ is invertible
for generic $\lambda,$ the required identity follows.
~\hfill$\square$\medbreak
In view of (\refer{e: defi unnormalized dual Eis}) it follows immediately from
(\refer{e: functional equation uEis}) that the
unnormalized dual Eisenstein integrals satisfy the following functional equations,
for $P,Q \in \cP_0$ and $s \in W,$
\begin{equation}
\naam{e: functional equation udEis}
{}^\circ\hspace{-0.6pt} C_{Q|P}(s \col -\bar \lambda)^* \;{}^u\! E^*(Q \col s \lambda\col \,\cdot\, ) = {}^u\! E^*(P\col \lambda \col \,\cdot\,),
\end{equation}
as an identity of meromorphic $C^\infty(G) \otimes {\rm Hom}(V_\tau, \cA_2)$-valued functions of $\lambda \in \fa_{0\iC}^*.$
In view of the definition of the Fourier transform in (\refer{e: defi uFou}),
this in turn implies that, for every
$f\in C_c^\infty(G,\tau),$
\begin{equation}
\naam{e: functional equation uFou} {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col -\bar
\lambda)^*\; {}^u \cF_Q f(s \lambda) = {}^u \cF_P f (\lambda),
\end{equation}
as an identity of meromorphic $\cA_2$-valued functions of $\lambda \in \fa_{0\iC}^*.$
\begin{lemma}
\naam{l: uAC imply transformation formula gf by C}
Let $\varphi = (\varphi_P)_{P \in \cP_0} \subset
{\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2,$
and assume that $\cL \varphi = 0$ for all $\cL \in {}^u\!{\rm AC}(G,\tau, \cP_0).$
Then for all $P,Q \in \cP_0$ and $s \in W,$
\begin{equation}
\naam{e: transformation formula gf by C}
{}^\circ\hspace{-0.6pt} C_{Q|P}(s \col -\bar \lambda)^*\; \varphi_Q (s \lambda) = \varphi_P(\lambda),
\end{equation}
for generic $\lambda \in \fa_{0\iC}^*.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
Let $P,Q\in \cP_0$ and $s \in W$ be fixed.
In view of
Lemma \refer{l: rationality ooC} there exists a polynomial
function $q \in \Pi_{\Sigma}(\fa_{0\iC}^*)$ such that
$\lambda \mapsto q(\lambda) {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col -\bar \lambda)^*$
is polynomial. Let $\varphi$ fulfill the hypothesis and let $\mu \in
\fa_{0\iC}^*\setminus q^{-1}(0).$
We define Taylor functionals
${\mathcal L}_R \in {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl}
\otimes \cA_2^*$ by
$\cL_P \psi:= - q(\mu) \psi(\mu)$ and
$\cL_Q \psi = {\rm ev}_\mu[\lambda \mapsto
q(\lambda) {}^\circ\hspace{-0.6pt} C_{Q|P}(s \col -\bar \lambda)^* \psi(s \lambda)],$ for $\psi \in {\mathcal O}(\fa_{0\iC}^*)
\otimes \cA_2,$
and by $\cL_R = 0$ for $R \in \cP_0\setminus \{P, Q\}.$
It follows from (\refer{e: functional equation udEis}) that
$\cL \in {}^u\!{\rm AC}(G,\tau, \cP_0).$ Hence, $\cL \varphi = 0.$
We conclude that $\varphi$ satisfies
(\refer{e: transformation formula gf by C})
for
$\lambda \in \fa_{0\iC}^*\setminus q^{-1}(0).$
~\hfill$\square$\medbreak
If $V$ is a finite dimensional real linear space,
we denote by $\cM(V_{\scriptscriptstyle \C})$ the space of meromorphic functions
on $V_{\scriptscriptstyle \C}.$
Given $P,Q \in \cP_0,$ we define the endomorphism
$\gamma_{P|Q}$ of $\cM(\fa_{0\iC}^*) \otimes \cA_2$ by
$$
[\gamma_{P|Q} \psi](\lambda) = {}^\circ\hspace{-0.6pt} C_{Q| P}(1 \col - \bar \lambda)^* \psi(\lambda).
$$
In particular, $\gamma_{P|P} = I.$
\begin{lemma}
Let $P, Q \in \cP_0$ and $R > 0.$ Then $\gamma_{P|Q}$ maps ${}^u {\rm PW}_R(G, \tau, Q)$
continuously into ${\mathrm H}_R(\fa_{0\iC}^*) \otimes \cA_2.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
It follows from Lemma \refer{l: rationality ooC} that there exists
a polynomial $q \in \Pi_\Sigma(\fa_{0\iC}^*)$ such that the function
$\lambda \mapsto q(\lambda)\, {}^\circ\hspace{-0.6pt} C_{Q|P}(1\col -\bar \lambda)^*$ is polynomial on $\fa_{0\iC}^*.$
This in turn implies that the map $q \,{\scriptstyle\circ}\, \gamma_{P|Q}$ maps ${}^u {\rm PW}_R(G,\tau,Q)$
continuously into ${\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*) \otimes \cA_2.$
Let $\alpha \in \Sigma$ and $c \in {\msy C}$ be such that
$l : \lambda \mapsto \inp{\lambda}{\alpha} - c$ is a factor of $q.$ Let $d$ be the highest
integer such that $l^d$ is still a factor of $q.$ Let $H_\alpha $ denote the element
of $\fa_0$ determined by $H_\alpha \perp \ker \alpha$ and $\alpha(H_\alpha) = 2.$
Fix $0 \leq k < d.$ Then the element $H_\alpha^k \in S(\fa_0),$ viewed as a constant
coefficient differential operator on $\fa_{0\iC}^*$ satisfies $H_\alpha^k q = 0$
on $l^{-1}(0).$ Fix $\lambda_0 \in l^{-1}(0)$ and consider
the functional ${\mathcal L} \in {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl} \otimes {\mathcal A}_2^*$ defined by
\begin{equation}
\naam{e: Lau and j P Q}
{\mathcal L} (\varphi):= {\rm ev}_{\lambda_0} \,{\scriptstyle\circ}\, H_\alpha^k [q \gamma_{P|Q} \varphi],
\end{equation}
for $\varphi \in {\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2.$ It follows from
(\refer{e: functional equation udEis}) that, for all $x
\in G$ and $v \in V_\tau,$ the function $\gamma_{P|Q} {}^u\! E^*(Q\col
\,\cdot\, \col x)v$ equals ${}^u\! E^*(P\col \,\cdot\, \col
x)v,$ hence is holomorphic on $\fa_{0\iC}^*.$ By application
of the Leibniz rule we now see that ${\mathcal L}({}^u\! E^*(Q\col \,\cdot\, \col
x)v) = 0$ for all $x \in G$ and $v \in V_\tau.$ Hence,
${\mathcal L}$ belongs to ${}^u{\rm AC}_\hol(G,\tau, Q).$ Let now $\varphi \in {}^u {\rm PW}_R(G,
\tau, Q).$ Then it follows that (\refer{e: Lau and j P Q}) equals
zero. As this is valid for every $\lambda_0 \in l^{-1}(0)$ and all $0
\leq k < d,$ it follows that $l^d$ divides $q(\lambda) \gamma_{P|Q}
\varphi.$ Treating all factors of $q$ in this fashion, we see that
$\gamma_{P|Q} \varphi$ is holomorphic on $\fa_{0\iC}^*$ outside a subset of
complex codimension $2.$ It follows that $\gamma_{P|Q}$ maps
${}^u {\rm PW}_R(G,\tau, Q)$ into ${\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2.$ Since $q
\gamma_{P|Q}$ maps ${}^u {\rm PW}_R(G,\tau, Q)$ continuous linearly into
${\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*) \otimes \cA_2,$ it follows by a repeated
application of Cauchy's integral formula, treating the linear
factors of $q$ one at a time, that $\gamma_{P|Q}$ is a continuous
linear map ${}^u {\rm PW}_R(G,\tau, Q) \rightarrow {\mathrm H}_R}%rm to be consistent with PW{{\cE\cP}_R(\fa_{0\iC}^*) \otimes \cA_2.$
~\hfill$\square$\medbreak
If
$\cL \in {}^u\!{\rm AC}(G,\tau, P),$ then the family of Taylor
functionals $(\cL_Q')_{Q \in \cP_0}$
defined by $\cL_P' = \cL$ and by $\cL_Q' = 0$ for $Q \neq P$
belongs to ${}^u\!{\rm AC}(G, \tau, \cP_0).$ Accordingly,
we may view ${}^u\!{\rm AC}(G,\tau, P)$ as a subspace of ${}^u\!{\rm AC}(G,\tau, \cP_0).$
It follows that ${}^u {\rm PW}_R(G, \tau, \cP_0),$ for $R > 0,$
is a subspace of the direct sum of the spaces
${}^u {\rm PW}_R(G, \tau, P),$ for $P \in \cP_0.$ Moreover, by continuity
of Taylor functionals, this subspace is closed.
\begin{prop}
\naam{p: proj is topol iso}
Let $Q \in \cP_0.$ Then, for each $R > 0,$
the projection onto
the component with index $Q$ induces
a topological linear isomorphism
\begin{equation}
\naam{e: pr Q in PW}
{}^u {\rm PW}_R(G,\tau, \cP_0)\; \longrightarrow\; {}^u {\rm PW}_R(G, \tau, Q).
\end{equation}
\end{prop}
\par\noindent{\it Proof.}{\ }{\ }
Let $E$ denote the direct sum of a finite number of copies of
${\mathrm H}_R(\fa_{0\iC}^*) \otimes \cA_2,$
labeled by the elements of $\cP_0.$ For each such element $P$
let ${\rm pr}_P: E \rightarrow {\mathrm H}_R(\fa_{0\iC}^*) \otimes \cA_2$ denote the projection
onto the component of label $P.$
We define the map $\gamma_Q: {}^u {\rm PW}_R(G, \tau, Q) \rightarrow E$ by
$$
{\rm pr}_P \,{\scriptstyle\circ}\, \gamma_Q = \gamma_{P|Q}, \quad\quad (\forall\; P \in \cP_0).
$$
Then $\gamma_Q$ is continuous linear; we will show that
it maps into the subspace
${}^u {\rm PW}_R(G, \tau, \cP_0)$ of $E.$ Let
$(\cL_P)_{P \in \cP_0}$ belong to ${}^u{\rm AC}_\hol(G,\tau, \cP_0).$
For each $P \in \cP_0$ we select $q_P \in \Pi_\Sigma(\fa_{0\iC}^*)$ such that
$\lambda \mapsto q_P(\lambda) {}^\circ\hspace{-0.6pt} C_{Q|P}(1 \col - \bar \lambda)^*$ is a polynomial function. By Lemma
\refer{l: division Tayl by q} below,
there exists a
$\cL_P'\in {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl} \otimes \cA_2^*$
such that
$\cL_P = \cL_P'\,{\scriptstyle\circ}\, q_P$ on ${\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2.$
By application of the Leibniz rule,
we see that
$\cL'' = \sum_{P} \cL'_P \,{\scriptstyle\circ}\, q_P \,{\scriptstyle\circ}\, \gamma_{P|Q}$
defines an element of ${\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl} \otimes \cA_2^*.$
It follows from the functional equations
for the Eisenstein integral that, for every $x \in G$ and $v_Q \in V_\tau,$
\begin{eqnarray*}
\cL'' [{}^u\! E^*(Q \col \,\cdot\, \col x)v_Q] &=& \sum_P \cL_P'\,{\scriptstyle\circ}\, q_P
[ {}^u\! E^*(P \col \,\cdot\, \col x)v_Q]\\
&=& \sum_P \cL_P [{}^u\! E^*(P \col \,\cdot\, \col x)v_Q] = 0.
\end{eqnarray*}
Hence, $\cL'' \in {}^u{\rm AC}_\hol(G,\tau, Q).$ It follows that for $\varphi\in {}^u {\rm PW}_R(G,\tau, Q)$ we have
$$
0 = \cL'' (\varphi)= \sum_P \cL_P' [q_P \gamma_{P|Q}(\varphi)].
$$
Moreover, since $\gamma_{P|Q}(\varphi)$ is holomorphic for each $P \in \cP_0,$ it follows that the
latter expression equals $\sum_P \cL_P \gamma_Q(\varphi)_P.$ Hence, $\gamma_Q$ maps ${}^u {\rm PW}_R(G, \tau, Q)$
into ${}^u {\rm PW}_R(G, \tau, \cP_0).$ Moreover, it does so continuously, as the latter space
carries the relative topology from $E.$
From the definition of $\gamma_Q$ we see
that ${\rm pr}_Q \,{\scriptstyle\circ}\, \gamma_Q = \gamma_{Q|Q} = I.$
Moreover, if $\varphi \in {}^u {\rm PW}_R(G,\tau, \cP_0),$ then
by Lemma \refer{l: uAC imply transformation formula gf by C} with
$s = 1,$
it follows that $\varphi_P = \gamma_{P|Q}(\varphi_Q),$ for all $P \in \cP_0.$
Hence, $\gamma_Q \,{\scriptstyle\circ}\, {\rm pr}_Q = I$ on
${}^u {\rm PW}_R(G,\tau, \cP_0).$
It follows that ${\rm pr}_Q$ restricts to a topological linear isomorphism
(\refer{e: pr Q in PW}) with inverse $\gamma_Q.$
~\hfill$\square$\medbreak
\begin{lemma}
\naam{l: division Tayl by q}
Let $q \in \Pi_\Sigma(\fa_{0\iC}^*).$ Then for every ${\mathcal L} \in {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl}$ there exists
a ${\mathcal L}'\in {\mathcal O}(\fa_{0\iC}^*)^*_{\rm tayl}$ such that ${\mathcal L} = {\mathcal L}'\,{\scriptstyle\circ}\, q$ on ${\mathcal O}(\fa_{0\iC}^*).$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
This may be proved in the same fashion as \bib{BSpw}, Lemma 10.5.
~\hfill$\square$\medbreak
It is an immediate consequence of Proposition
\refer{p: proj is topol iso} that
Theorem \refer{t: arthurs PW one} and Theorem \refer{t: arthurs PW two} are equivalent.
\eqsection{The normalized Fourier transform}
\naam{s: the normalized Fourier transform}
The purpose of this section is to give equivalent versions
of the Paley--Wiener theorems discussed in the previous sections,
Theorem \refer{t: arthurs PW two} and Theorem \refer{t: distributional PW}.
The new versions are formulated in terms of a suitably normalized Fourier
transform, which in the final section will be shown
to coincide with the analogous Fourier
transform for the group viewed as a symmetric space. The normalized Fourier transform
is defined as in Section \refer{s: basic notation}, but with a differently normalized
Eisenstein integral.
Let $P \in \cP_0.$
We define the normalized Eisenstein integral by
\begin{equation}
\naam{e: defi normalized Eis}
{E^\circ}(P\col \lambda \col x) = E(P\col \lambda \col x)C_{P|P}(1 \col \lambda)^{-1},
\end{equation}
for generic $\lambda \in \fa_{0\iC}^*$ and for $x \in G.$
Then $\lambda \mapsto {E^\circ}(P\col \lambda)$ is a meromorphic $C^\infty(G) \otimes {\rm Hom}(\cA_2, V_\tau)$--valued
function on $\fa_{0\iC}^*.$ It is not entire holomorphic anymore, but its singular set is
of a simple nature. Indeed, by Lemma \refer{l: estimates C functions in cone}
the singular set
is a locally finite union of hyperplanes of the
form $\inp{\lambda}{\alpha} = c,$ with $\alpha \in \Sigma^+,$ $c \in {\msy R}.$
Moreover, the occurring
constants $c$ are bounded from below. It is known that the singular set
is disjoint from the imaginary space $i \fa_0^*,$ but we shall not need
this here.
As before we define (normalized) dual Eisenstein integrals by
\begin{equation}
\naam{e: defi normalized dual Eis}
E^*(P\col \lambda \col x) = {E^\circ}(P\col -\bar \lambda \col x)^* \in {\rm Hom}(V_\tau, \cA_2)
\end{equation}
for generic $\lambda \in \fa_{0\iC}^*$ and for $x \in G.$ In terms of these we
define the normalized Fourier transform
${\mathcal F}_P: C^{- \infty}_c(G\col \tau) \rightarrow \cM(\fa_{0\iC}^*) \otimes \cA_2$ by
$$
{\mathcal F}_P f(\lambda) = \int_G E^*(P\col \lambda \col x) f(x) \; dx.
$$
\begin{lemma}
Let $f \in C^{-\infty}_c(G, \tau).$ The unnormalized and normalized Fourier transforms
are related by
\begin{equation}
\naam{e: relation uFou and Fou}
C_{P|P}(1 \col - \bar \lambda)^* \cF_P f(\lambda) = {}^u \cF_P f (\lambda),
\end{equation}
as an identity of meromorphic functions of the variable $\lambda \in \fa_{0\iC}^*.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
Replacing $\lambda$ by $- \bar \lambda$ in both sides of (\refer{e: defi normalized Eis}),
then multiplying with the $C$-function and taking conjugates,
we obtain, in view of (\refer{e: defi unnormalized dual Eis}) and
(\refer{e: defi normalized dual Eis}),
$$
C_{P|P}(1 \col - \bar \lambda)^* E^*(P\col \lambda \col x) = {}^u\! E^*(P\col \lambda\col x),
$$
as meromorphic functions of $\lambda \in \fa_{0\iC}^*$ with values in $C^\infty(G) \otimes {\rm Hom}(V_\tau, \cA_2).$
The result follows by testing with $f dx.$
~\hfill$\square$\medbreak
The singular nature of the normalized Eisenstein integral does not
allow us to define Arthur--Campoli functionals in terms of Taylor functionals
as we did in Section \refer{s: Arthur's Paley--Wiener theorem}.
Instead we need the concept of Laurent functional introduced in
\bib{BSanfam}, Sect.\ 12. We briefly recall its definition.
Let $V$ be a finite-dimensional real linear space and let $X \subset V^*\setminus \{0\}$
be a finite subset. For a point $a \in V_{\scriptscriptstyle \C},$ we define the polynomial function
$\pi_a: V_{\scriptscriptstyle \C} \rightarrow {\msy C}$ by
$$
\pi_a: = \prod_{\xi \in X} (\xi - \xi(a)).
$$
The ring of germs of meromorphic functions at $a$ is denoted by $\cM(V_{\scriptscriptstyle \C},a).$
We define the subring
$$
\cM(V_{\scriptscriptstyle \C}, a, X): = \cup_{N \in {\msy N}}\;\; \pi_a^{-N} {\mathcal O}_a.
$$
Let ${\rm ev}_a$ denote the linear functional on ${\mathcal O}_a$ that
assigns to a germ $f \in {\mathcal O}_a$ its value $f(a)$ at $a.$
An $X$-Laurent functional at $a \in V_{\scriptscriptstyle \C}$ is a linear functional
$\cL \in \cM(V_{\scriptscriptstyle \C}, a, X)^*$ such that for every $N \in {\msy N}$ there
exists a $u_N \in S(V)$ such that
\begin{equation}
\naam{e: expression Lau}
\cL = {\rm ev}_a \,{\scriptstyle\circ}\, u_N \,{\scriptstyle\circ}\, \pi_a^N \quad\quad {\rm on} \quad \pi_a^{-N} {\mathcal O}_a.
\end{equation}
The space of all Laurent functionals on $V_{\scriptscriptstyle \C},$ relative to $X,$ is defined
as the algebraic direct sum of linear spaces
\begin{equation}
\naam{e: defi space of Lau}
\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur}:= \bigoplus_{a \in V_{\scriptscriptstyle \C}} \cM(V_{\scriptscriptstyle \C}, X, a)^*_{\rm laur}.
\end{equation}
For $\cL$ in the space (\refer{e: defi space of Lau}),
the finite set of $a \in V_{\scriptscriptstyle \C}$ for which
the component $\cL_a$ is nonzero is called the support of $\cL$ and
denoted by ${\rm supp}\, \cL.$
According to the above definition, any $\cL \in \cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur}$
may be decomposed as
$$
\cL = \sum_{a \in {\rm supp}\, \cL} \cL_a.
$$
Let $\cM(V_{\scriptscriptstyle \C}, X)$ denote the space of meromorphic functions $\varphi$ on $V_{\scriptscriptstyle \C}$
with the property that the germ $\varphi_a$ at any point $a \in V_{\scriptscriptstyle \C}$ belongs
to $\cM(V_{\scriptscriptstyle \C}, a, X).$ Then the natural bilinear map
$\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur} \times \cM(V_{\scriptscriptstyle \C}, X) \rightarrow {\msy C},$ given by
$$
(\cL, \varphi) \mapsto \cL \varphi:= \sum_{a \in {\rm supp}\, \cL} \cL_a \varphi_a
$$
induces an embedding of $\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur}$ onto a linear subspace of the dual space
$\cM(V_{\scriptscriptstyle \C}, X)^*.$ For more details concerning these definitions,
we refer the reader to \bib{BSanfam}, Sect.\ 12.
\begin{lemma}
\naam{l: multiplication of Lau by func}
Let $\cL \in \cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur}$ and let $\psi \in \cM(\Omega, X),$
for $\Omega$ an open neighborhood of ${\rm supp}\, \cL.$ Then
$\cL \,{\scriptstyle\circ}\, \psi$ belongs to $\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur}.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
Without loss of generality we may assume that $\cL$ is supported
by a single point $a \in V_{\scriptscriptstyle \C}.$ First assume that
$\psi_a \in {\mathcal O}_a.$ Then the result follows by a straightforward
application of the definition containing (\refer{e: expression Lau})
combined with the Leibniz rule. It remains to establish
the result for $\psi = \pi_a^{-k},$ with $k \in {\msy N}.$ In this
case the result is an immediate consequence of the mentioned
definition.
~\hfill$\square$\medbreak
The following result relates the Laurent functionals to the Taylor functionals
defined in Section \refer{s: Arthur's Paley--Wiener theorem}.
The inclusion map ${\mathcal O}(V_{\scriptscriptstyle \C}) \subset \cM(V_{\scriptscriptstyle \C}, X)$ induces
a surjection $\cM(V_{\scriptscriptstyle \C}, X)^* \rightarrow {\mathcal O}(V_{\scriptscriptstyle \C})^*.$ This property
of surjectivity also holds on the level of Laurent functionals.
The space ${\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl}$ may naturally be viewed
as a subspace of ${\mathcal O}(V_{\scriptscriptstyle \C})^*.$ The natural map
$\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur} \rightarrow {\mathcal O}(V_{\scriptscriptstyle \C})^*$ maps into this
subspace.
\begin{lemma}
\naam{l: surjectivity natural map}
The natural map $\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur} \rightarrow {\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl}$
is surjective.
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
Let $U \in {\mathcal O}(V_{\scriptscriptstyle \C})^*_{\rm tayl}.$ Without loss of generality
we may assume that $U$ is supported by a single point
$a \in V_{\scriptscriptstyle \C}.$ Then $U = {\rm ev}_a \,{\scriptstyle\circ}\, u$ for some $u \in S(V).$
By \bib{BSres}, Lemma 1.7 with $d' = 0,$ there exists
a $\cL \in \cM(V_{\scriptscriptstyle \C}, a, X)^*_{\rm laur}$ determined
by a sequence $(u_n)_{n \in {\msy N}} \subset S(V),$ such that $u_0 = u.$
Thus, $\cL$ restricts to $U$ on ${\mathcal O}(V_{\scriptscriptstyle \C}).$
~\hfill$\square$\medbreak
Before proceeding, we formulate a result concerning division that will
be frequently used in the sequel.
\begin{lemma}
\naam{l: laurent and poles}
Let $E$ be a finite dimensional complex linear space.
Let $S$ be a linear subspace of $\cM(V_{\scriptscriptstyle \C}, X) \otimes E,$ let
$S^\circ$ be the annihilator of $S$ in $\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur} \otimes E^*$
and let $S^{\circ\circ}$ be the space of functions
$\varphi \in \cM(V_{\scriptscriptstyle \C}, X) \otimes E$ such that $\cL \varphi = 0$
for all $\cL \in S^\circ.$
Let $\psi$ be a nonzero ${\rm End}(E)$-valued meromorphic function
on $V_{\scriptscriptstyle \C}$ such that both $\psi$ and $\psi^{-1}$ belong
to $\cM(V_{\scriptscriptstyle \C}, X) \otimes {\rm End}(E). $
Let $\Omega \subset V_{\scriptscriptstyle \C}$ be an open subset such that
$\psi \varphi $ is regular on $\Omega$
for all $\varphi \in S.$
Then $\psi \varphi$ is regular on $\Omega$ for all $\varphi \in S^{\circ\circ}.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
We first assume that $\psi = I.$
Then every $\varphi \in S$ is regular on $\Omega.$
Let now $\varphi \in S^{\circ\circ}$ and consider a point $a \in \Omega.$
Then it suffices to show that the germ $\varphi_a$ is regular at $a.$
As $\varphi_a \in \cM(V_{\scriptscriptstyle \C}, a) \otimes E,$ there exists a
product $q$ of factors of the form $\xi - \xi(a),$ with $\xi \in X,$ such that
$q\varphi_a$ is regular at $a.$ We fix $q$ of minimal degree. Then
$q \varphi_a$ has a non-trivial value at $a.$ Hence, there exists a
linear functional $\eta \in E^*$ such that ${\rm ev}_a \,{\scriptstyle\circ}\, q \inp{\varphi_a}{\eta}
\neq 0.$ There exists a Laurent functional
$\cL \in \cM(V_{\scriptscriptstyle \C}, a)^*_{\rm laur}$ such that $\cL = {\rm ev}_a$ on ${\mathcal O}_a(V_{\scriptscriptstyle \C}).$
Now $\cL \,{\scriptstyle\circ}\, q$ is a Laurent functional, and it follows from
the above that $\cL_1:= [\cL \,{\scriptstyle\circ}\, q] \otimes \eta$ is nonzero on $\varphi_a.$
Hence $\cL_1 \notin S^\circ.$ It follows that there exists a function
$\varphi_1 \in S$ such that $q(a) \eta(\varphi_1(a)) = \cL_1 \varphi_1$ is nonzero.
This implies that $q$ is nonzero at $a,$ hence constant. We conclude
that $\varphi_a$ is regular at $a.$
We now turn to the case with $\psi$ general. Then multiplication
by $\psi$ induces a linear automorphism of $\cM(V_{\scriptscriptstyle \C}, X) \otimes E,$
whose inverse is multiplication by $\psi^{-1}.$
In view of Lemma \refer{l: multiplication of Lau by func},
the map $\psi^* : \cL \mapsto \cL \,{\scriptstyle\circ}\, \psi$
is a linear automorphism of $\cM(V_{\scriptscriptstyle \C}, X)^*_{\rm laur} \otimes E$
with inverse $(\psi^{-1})^* .$ Put $S_1: = \psi S.$ Then
$S_1^\circ = \psi^{*-1}(S^\circ)$ and $S_1^{\circ\circ} = \psi S^{\circ\circ}.$
By the first part of the proof it follows that all elements
of $S_1^{\circ\circ}$ are regular on $\Omega.$ The result follows.
~\hfill$\square$\medbreak
We use the non-degenerate bilinear map $\inp{\,\cdot\,}{\,\cdot\,}$ on $\fa_0^*$ to identify
this space with its real linear dual. Accordingly we view $\Sigma$ as a finite subset
of ${\got a}_0^{**}\setminus \{0\}$ and invoke the space of $\Sigma$-Laurent functionals
on $\fa_{0\iC}^*$ in the following definition.
\begin{defi}
A (normalized) Arthur--Campoli functional for $(G, \tau,P)$ is a Laurent functional
${\mathcal L} \in \cM(\fa_{0\iC}^*, \Sigma)^*_{\rm laur} \otimes \cA_2^*$ such that
$$
{\mathcal L}\; [E^* (P\col \,\cdot\, \col x)v] = 0
$$
for all $x \in G$ and $v \in V_\tau.$
The space of such functionals is denoted by ${\rm AC}(G,\tau,P).$
\end{defi}
Our next objective is to define suitable spaces of meromorphic functions
with controlled singular behavior.
A $\Sigma$-hyperplane in $\fa_{0\iC}^*$ is defined to be a hyperplane of the form $l^{-1}(0),$
where $l: \lambda \mapsto \inp{\lambda}{\alpha} - c$ with $\alpha \in \Sigma$ and $c \in {\msy C}.$ The hyperplane is said
to be real if $c \in {\msy R}.$ A $\Sigma$-configuration in $\fa_{0\iC}^*$ is a locally finite collection
of $\Sigma$-hyperplanes. The configuration is said to be real if all its hyperplanes are real.
Let now $\cH$ be a real $\Sigma$-configuration.
For each
$H \in \cH$ we fix $\alpha_H \in \Sigma$ and $s_H \in {\msy R}$ such that $H$ equals the zero locus
of $l_H: \lambda \mapsto \inp{\lambda}{\alpha} - s_H.$
Let $d: \cH \rightarrow {\msy N}$ be a map. For $\omega$ a subset of $\fa_0^*$
whose closure intersects only finitely many
hyperplanes from $\cH,$ we define the polynomial function
$\pi_{\omega, d }$ on $\fa_{0\iC}^*$ by
\begin{equation}
\naam{e: defi pi omega d}
\pi_{\omega,d} =
\prod_{H \in \cH\atop H \cap \,{\rm cl}\,\omega \neq \varnothing}l_H^{d(H)}.
\end{equation}
Moreover, we define $\cM(\fa_{0\iC}^*, \cH, d)$ to be the space of
meromorphic functions $\varphi \in \cM(\fa_{0\iC}^*)$ such that for every
bounded open subset $\omega$ of $\fa_0^*,$ the function
$\pi_{\omega, d} \varphi$ is regular on $\omega + i\fa_0^*.$
For $\omega$ a bounded subset of $\fa_0^*$ and $n \in {\msy Z}$ we
define the $[0,\infty]$-valued seminorm $\nu_{\omega, d, n}$ on
$\cM(\fa_{0\iC}^*, \cH, d)$
by
\begin{equation}
\naam{e: seminorm cP space}
\nu_{\omega, d, n} (\varphi)
: = \,\sup_{\lambda \in \omega + i \fa_0^*}
\;( 1 + |\lambda|)^n \, |\pi_{\omega, d}(\lambda)\varphi(\lambda)|.
\end{equation}
We define
$
{\mathcal P}(\fa_{0\iC}^*, \cH, d)
$
to be the space of functions $\varphi \in {\mathcal M}(\fa_{0\iC}^*, \cH, d)$ such that
$
\nu_{\omega, d, n}(\varphi) < \infty
$
for every compact set $\omega \subset \fa_0^*$ and all $n \in {\msy N}.$ This space
is a Fr\'echet space with topology induced by the collection of
seminorms $\nu_{\omega, d, n},$ for $\omega$ compact and $n \in {\msy N}.$
We denote by ${\mathcal N} = {\mathcal N}(\fa_0^*)$ the collection of
maps $n: {\mathcal C} \rightarrow {\msy N},$ with ${\mathcal C}$ the collection of compact subsets
of $\fa_0^*.$ On ${\mathcal N}$ we define the partial ordering
$\leq$ by $n \leq m$
if and only if $n(\omega) \leq m(\omega)$ for all $\omega \in {\mathcal C}.$
For $n \in {\mathcal N}$ we define
${\mathcal P}_n^*(\fa_{0\iC}^*, \cH, d)$ to be the space of functions
$\varphi \in \cM(\fa_{0\iC}^*, \cH, d)$
such that
$$
\nu_{\omega,d, - n(\omega)} (\varphi) =
\sup_{\lambda \in \omega + i{\mathfrak a}_0^*} (1 + |\lambda|)^{-n(\omega)} \,
|\pi_{\omega, d}(\lambda) \varphi(\lambda)|
< \infty
$$
for every compact subset
$\omega \subset \fa_0^*.$ Equipped with the seminorms
$\nu_{\omega, d, - n(\omega)}$ the space ${\mathcal P}_n^*(\fa_{0\iC}^*, \cH, d)$
is a complete locally convex space. If $m \leq n$
then clearly ${\mathcal P}_m^* \subset {\mathcal P}_n^*,$
with continuous linear inclusion
map.
We now define
$$
{\mathcal P}^*(\fa_{0\iC}^*, \cH, d) := \cup_{n \in {\mathcal N}(\fa_0^*)}\; \; {\mathcal P}_{n}^* (\fa_{0\iC}^*, \cH, d),
$$
and equip this space with the inductive limit locally convex topology.
In particular, these definitions may be interpreted for $\cH = \varnothing$
and $d = \varnothing.$ In this case we have $\pi_{\omega, d} = 1$
for every $\omega \subset \fa_0^*,$ so that
$$
{\mathcal P}(\fa_{0\iC}^*, \varnothing) \quad{\rm and} \quad {\mathcal P}^*(\fa_{0\iC}^*, \varnothing)
$$
are just spaces of holomorphic functions $\varphi$ on $\fa_{0\iC}^*$
determined in the above fashion by seminorms of the form
$$
\nu_{\omega, n}(\varphi) := \sup_{\lambda \in \omega + i\fa_0^*} \;
(1 + |\lambda|)^n |\varphi(\lambda)|.
$$
For the rest of this section,
let $P \in \cP_0$ be fixed and let $\cH = \cH_{G,\tau, P}$
be the smallest collection of $\Sigma$-hyperplanes in
$\fa_{0\iC}^*$ such that the singular locus of
$\lambda \mapsto E^*(P\col \lambda \col \,\cdot\,)$
is contained in the union $\cup \cH.$ In view of Lemma
\refer{l: estimates C functions in cone} the collection $\cH$ is locally finite and
consists of real $\Sigma$-hyperplanes.
Moreover, by the same lemma, the set
of $H \in \cH$ with $H \cap \fa_0^*(P,R) \neq \varnothing$ is finite,
for every $R \in {\msy R}.$
We define the map $d= d_{G,\tau, P}: \cH \rightarrow {\msy N}$
as follows. For each $H \in \cH$ we fix $l_H: \fa_{0\iC}^* \rightarrow {\msy C}$
as in (\refer{e: defi pi omega d}) and define
$d(H)$ as the smallest integer $k \geq 0$ such
that the $C^\infty(G) \otimes {\rm Hom}(V_\tau, \cA_2)$-valued
meromorphic function $l_H ^k E^*(P\col \,\cdot\,)$
extends regularly over $H \setminus \cup \{H'\in \cH \mid H'\neq H\}.$
Given a subset $\omega \subset \fa_0^*$ whose closure
meets only finitely many
hyperplanes from $\cH,$ we define the polynomial function $\pi_{\omega, d}$
as in (\refer{e: defi pi omega d}).
In particular, we write $\pi = \pi_P$ for this
polynomial with $\omega = \fa_0^* \cap \bar\fa_{0}^*(P,0),$ where the second set in
the intersection denotes
the closure of
the set (\refer{e: defi fapd P R}) with $R = 0.$
Thus, the $C^\infty(G) \otimes {\rm Hom}(V_\tau, \cA_2)$-valued
function $\lambda \mapsto \pi(\lambda) E^*(P\col \lambda)$ is holomorphic
on a neighborhood of $\bar\fa_{0}^*(P, R)$
and $\pi \in \Pi_{\Sigma, {\msy R}}(\fa_0^*)$ is minimal with this property.
We define the following closed subspace of
${\mathcal P}(\fa_{0\iC}^*, \cH, d) \otimes \cA_2:$
\begin{eqnarray}
\lefteqn{{\mathcal P}_{\rm AC}(\fa_{0\iC}^*, \cH, d, P)}\nonumber
\\
&:=&
\{ \varphi \in {\mathcal P}(\fa_{0\iC}^*, \cH, d) \otimes \cA_2 \mid \cL \varphi = 0,\;
\forall \cL \in {\rm AC}(G,\tau, P) \}
\end{eqnarray}
Finally, we define the space ${\mathcal P}_{\rm AC}^*(\fa_{0\iC}^*, \cH, d, P)$
in a similar fashion, but with ${\mathcal P}$
replaced by ${\mathcal P}^*.$
\pagebreak
\begin{defi}
\naam{d: normalized PW}
\begin{enumerate}
\vspace{-1mm}\item[{\rm (a)}]
Let $R > 0.$ We define the {\it Paley--Wiener space} ${\rm PW}_R(G,\tau,P)$ to be the subspace of
${\mathcal P}_{\rm AC}(\fa_{0\iC}^*, \cH, d,P)$ consisting of functions $\varphi$ such that, for all $n \in {\msy N},$
\begin{equation}
\naam{e: PW estimate}
\sup_{\lambda \in \bar\fa_0^*(P, 0)} (1 + \|\lambda\|)^n e^{- R |{\rm Re}\, \lambda|} \|\pi(\lambda) \varphi(\lambda)\| < \infty.
\end{equation}
The space is equipped with the relative topology.
\item[{\rm (b)}]
For $R > 0$ we define the {\it distributional Paley--Wiener space}
${\rm PW}^*_R(G,\tau, P)$ to be the subspace
of ${\mathcal P}_{\rm AC}^*(\fa_{0\iC}^*, \cH, d,P )$ consisting of functions $\varphi$ for which there exists
a constant $n \in {\msy N}$ such that
\begin{equation}
\naam{e: PW star estimate}
\sup_{\lambda \in \bar\fa_0^*(P, 0)} (1 + \|\lambda\|)^{-n } e^{- R |{\rm Re}\, \lambda|} \|\pi(\lambda) \varphi(\lambda)\| < \infty.
\end{equation}
This space is also equipped with the relative topology.
\end{enumerate}
\end{defi}
We will finish this section by discussing the relation of these
Paley--Wiener spaces with the unnormalized Paley--Wiener spaces
introduced in Definitions \refer{d: unnormalized PW}
and \refer{d: unnormalized distrib PW}.
As a preparation,
we first give another characterization of the unnormalized Paley--Wiener spaces.
We define
$
{\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P)
$
and
$
{\mathcal P}_{{}^u\!{\rm AC}}^*(\fa_{0\iC}^*, \varnothing, P)
$
as the closed subspaces of the spaces ${\mathcal P}(\fa_{0\iC}^*, \varnothing) \otimes \cA_2$
and ${\mathcal P}^*(\fa_{0\iC}^*, \varnothing) \otimes \cA_2,$ respectively, consisting
of the functions $\varphi$ satisfying the relations $\cL \varphi = 0$ for
all $\cL \in {}^u{\rm AC}_\hol(G, \tau, P).$
\begin{prop}
\naam{p: uPW by estimate on cone}
Let $R > 0.$
\begin{enumerate}
\vspace{-1mm}\item[{\rm (a)}]
The space
${}^u {\rm PW}_R(G, \tau, P)$ consists of the functions
$\varphi \in {\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P)$
with the property that, for every $n \in {\msy N},$
\begin{equation}
\naam{e: uPW estimate}
\sup_{\lambda \in \bar\fa_0^*(P, 0)} (1 + |\lambda|)^n e^{-R |{\rm Re}\, \lambda|}
\|\varphi(\lambda)\| < \infty.
\end{equation}
Moreover, the topology of ${}^u {\rm PW}_R(G, \tau, P)$ coincides with the
relative topology from ${\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P).$
\item[{\rm (b)}]
The space
${}^u {\rm PW}_R^*(G, \tau, P)$ consists of the functions
$\varphi \in {\mathcal P}_{{}^u\!{\rm AC}}^*(\fa_{0\iC}^*, \varnothing, P)$
with the property that there exists a number $n \in {\msy N}$ such that
\begin{equation}
\naam{e: uPW star estimate}
\sup_{\lambda \in \bar\fa_0^*(P, 0)} (1 + |\lambda|)^{-n} e^{-R |{\rm Re}\, \lambda|}
\|\varphi(\lambda)\| < \infty.
\end{equation}
Moreover, the topology of ${}^u {\rm PW}_R^*(G, \tau, P)$ coincides with the
relative topology from ${\mathcal P}_{{}^u\!{\rm AC}}^*(\fa_{0\iC}^*, \varnothing, P).$
\end{enumerate}
\end{prop}
\par\noindent{\it Proof.}{\ }{\ } We start by making some remarks on Euclidean Paley--Wiener
spaces. From the text preceding Definition \refer{d: arthurs PW}
we recall the definition of the space ${\mathrm H}_R(\fa_{0\iC}^*),$ equipped
with the Fr\'echet topology ${\mathcal T}$ induced by the seminorms
$\nu_{R,n},$ for $n \in {\msy N}.$ Clearly, ${\mathrm H}_R(\fa_{0\iC}^*) \subset
{\mathcal P}(\fa_{0\iC}^*, \varnothing),$ with continuous inclusion map. We denote by
${\mathcal T}_r$ the associated relative topology on ${\mathrm H}_R(\fa_{0\iC}^*).$ Then
${\mathcal T}$ is finer than ${\mathcal T}_r.$ We will show that both topologies are
in fact equal.
By the Euclidean Paley--Wiener
theorem, Euclidean Fourier transform $\cF_{\rm eucl}$
defines a continuous linear isomorphism from $C^\infty_R(\fa_0)$ onto
${\mathrm H}_R(\fa_{0\iC}^*).$
From a straightforward estimation it follows
that the inverse Fourier transform $\cF_{\rm eucl}^{-1}$ is continuous
from ${\mathcal P}(\fa_{0\iC}^*, \varnothing)$ to $C^\infty(\fa_0).$
It follows from the above that the identity
map $\cF_{\rm eucl} \,{\scriptstyle\circ}\, \cF_{\rm eucl}^{-1}$
is continuous from $({\mathrm H}_R(\fa_{0\iC}^*),{\mathcal T}_r)$ to
$({\mathrm H}_R(\fa_{0\iC}^*), {\mathcal T}).$ Hence ${\mathcal T} = {\mathcal T}_r.$
From the text
preceding Definition \refer{d: unnormalized distrib PW}
we recall the definition of ${\mathrm H}^*_R(\fa_{0\iC}^*),$
equipped with the inductive limit locally convex topology
denoted ${\mathcal T}^*.$
Clearly,
${\mathrm H}^*_R(\fa_{0\iC}^*) \subset {\mathcal P}^*(\fa_{0\iC}^*, \varnothing),$
with continuous inclusion map. Let ${\mathcal T}^*_r$ denote the associated relative
topology on ${\mathrm H}^*_R(\fa_{0\iC}^*).$
Then ${\mathcal T}^*$ is finer than ${\mathcal T}^*_r.$
We will show that both topologies are equal.
By the distributional Euclidean Paley-Wiener
theorem, $\cF_{\rm eucl}$ maps $C^{-\infty}_R(\fa_0)$
bijectively onto ${\mathrm H}_R^*(\fa_{0\iC}^*).$
Fix $R' > R$ and let
$k$ be an arbitrary positive integer.
Then $C^{-\infty}_R(\fa_0)_k,$ the subspace of generalized
functions of order at most $k,$ naturally embeds into
the continuous linear dual of the Banach space $C^k_{R'}(\fa_0),$
equipped
with the $C^k$ norm $\|\,\cdot\,\|_{C^k}.$ Accordingly,
we equip $C^{-\infty}_R(\fa_0)_k$
with the restriction of the dual norm.
By a straightforward estimation,
there exists a $C_k > 0$ such that
$$
\nu_{R, k}(\cF_{\rm eucl}(f)) \leq C_k \|f\|_{C^k}
$$
for all $f \in C^k_{R'}(\fa_0).$ Let $n \in {\mathcal N}(\fa_0^*).$ Then by transposition
it readily follows that
$\cF_{\rm eucl}^{-1}$ maps
${\mathcal P}_{n}^*(\fa_{0\iC}^*, \varnothing) \cap {\mathrm H}_R^*(\fa_{0\iC}^*)$
into $C^{-\infty}_R(\fa_0)_{k},$
with $k = n(\{0\}) + \dim \fa_0 + 1.$
Moreover, this map is continuous with respect
to the relative topology from ${\mathcal P}_{n}^*(\fa_{0\iC}^*, \varnothing)$
on the first of these
spaces. Now $\cF_{\rm eucl}$ maps
$C^{-\infty}_R(\fa_0)_{k}$ continuously into
${\mathrm H}_R^*(\fa_{0\iC}^*).$
It follows that the identity map $\cF_{\rm eucl}\,{\scriptstyle\circ}\,
\cF_{\rm eucl}^{-1}$
is continuous from
${\mathcal P}^*_n(\fa_{0\iC}^*,\varnothing) \cap {\mathrm H}_R^*(\fa_{0\iC}^*)$
to ${\mathrm H}_R^*(\fa_{0\iC}^*).$
By the universal property
of the inductive limit, it follows
that ${\mathcal T}_r^*$ is finer than ${\mathcal T}^*.$ Hence ${\mathcal T}^* = {\mathcal T}_r^*.$
We proceed with the actual proof.
We denote the subspace of ${\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P)$ defined
in (a) by ${\mathcal P}{\mathcal W}$ and the similar subspace defined in (b) by ${\mathcal P}{\mathcal W}^*.$
These subspaces are equipped with the relative topologies.
Clearly, ${}^u {\rm PW}_R(G, \tau, P)$ is a subspace of
${\mathcal P}{\mathcal W},$ and ${}^u {\rm PW}^*_R(G,\tau, P)$ a subspace of ${\mathcal P}{\mathcal W}^*,$
with continuous inclusion maps.
To conclude the proof we must
establish the converse inclusions, also with continuous inclusion maps.
A holomorphic function
$\varphi \in {\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2$ that is annihilated by
${}^u{\rm AC}_\hol(G, \tau, P),$ satisfies the functional
equations
\begin{equation}
\naam{e: functional equation phi}
\varphi(\lambda) = {}^\circ\hspace{-0.6pt} C_{P|P}(s \col -\bar \lambda)^*\varphi(s\lambda),
\end{equation}
for $s \in W$ and generic $\lambda \in \fa_{0\iC}^*;$ in view of Proposition
\refer{p: proj is topol iso} this follows from Lemma \refer{l: uAC imply transformation formula gf by C}
with $P = Q.$
In view of Lemma \refer{l: rationality ooC}, there exists a product $q_s$ of linear factors
of the form $\inp{\,\cdot\,}{\alpha} - c,$ with $\alpha \in \Sigma$ and $c \in {\msy C},$
such that $\lambda \mapsto q_s(\lambda) {}^\circ\hspace{-0.6pt} C_{P|P}(s \col -s^{-1} \bar \lambda)^*$ is polynomial.
We define
$$
q(\lambda) : = \prod_{s \in W} q_s(s\lambda), \quad\quad (\lambda \in \fa_{0\iC}^*).
$$
Then there exist constants $C > 0$ and $N \in {\msy N},$ such
that, for all $s \in W$ and $\lambda \in \fa_{0\iC}^*,$
$$
\|q(s^{-1}\lambda) {}^\circ\hspace{-0.6pt} C_{P|P}(s \col - s^{-1} \bar \lambda) \| \leq C (1 + |\lambda|)^N.
$$
If we combine this with the functional equation (\refer{e: functional equation phi}),
we see that,
for each $s \in W,$ every $n \in {\msy Z}$ and all $\lambda \in \bar\fa_0^*(P, 0),$
$$
( 1+ |s^{-1} \lambda|)^n e^{- R |{\rm Re}\, s^{-1}\lambda|} \|q(s^{-1} \lambda)\varphi(s^{-1}\lambda)\|
\leq C ( 1 + |\lambda|)^{n + N} e^{- R |{\rm Re}\, \lambda|} \|\varphi(\lambda)\|.
$$
Combining these estimates for $s \in W,$ we obtain
$$
\nu_{R, n}(q\varphi)
\leq C \sup_{\lambda \in \bar\fa_0^*(P,0)}
( 1 + |\lambda|)^{n + N} e^{- R |{\rm Re}\, \lambda|} \|\varphi(\lambda)\|,
$$
where $\nu_{R,n}$ is defined as in the first part of the proof.
On the other hand, by an easy application of Cauchy's integral
formula it follows that for every $n \in {\msy Z}$ there exists a
constant $C_n > 0$ such that
$$
\nu_{R,n}(\varphi) \leq C_n \nu_{R,n}(q \varphi),
$$
for all $\varphi \in {\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2.$
It follows from these estimates
that ${\mathcal P}{\mathcal W}$ equals the intersection of
${\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P)$
with
the Euclidean Paley--Wiener space
${\mathrm H}_R(\fa_{0\iC}^*)\otimes \cA_2.$
By definition, the topology of ${\mathcal P}{\mathcal W}$ equals
the relative topology from the first of
these spaces.
By the first part of the proof, the topology also coincides
with the relative topology from the second
of these spaces.
It follows that ${\mathcal P}{\mathcal W} \subset {}^u {\rm PW}_R(G,\tau, P)$
with continuous inclusion map.
This establishes (a). Assertion (b) follows by a similar argument.
~\hfill$\square$\medbreak
We define the map $ \cU_P \in {\rm Aut}(\cM(\fa_{0\iC}^*) \otimes \cA_2)$
by
$$
\cU_P \varphi (\lambda) = C_{P|P}(1 \col - \bar \lambda)^* \varphi(\lambda).
$$
Then (\refer{e: relation uFou and Fou}) may be rephrased as
$$
{}^u \cF_P= \cU_P \,{\scriptstyle\circ}\, \cF_P.
$$
\begin{thm}
\naam{t: multC is iso PW spaces}
Let $R >0.$ The map $\cU_P \in {\rm Aut}(\cM(\fa_{0\iC}^*) \otimes \cA_2)$
restricts
to a topological linear isomorphism
$$
{\rm PW}_R^*(G,\tau,P) \;\; \pijl{\simeq} \;\; {}^u {\rm PW}^*_R(G,\tau,P),
$$
and similarly to a topological linear isomorphism
$$
{\rm PW}_R(G,\tau,P)\;\; \pijl{\simeq}\;\; {}^u {\rm PW}_R(G,\tau,P).
$$
\end{thm}
\par\noindent{\it Proof.}{\ }{\ }
It follows from Lemma \refer{l: rationality ooC}
that the functions $\lambda \mapsto C_{P|P}(1: - \bar \lambda)^{*\pm 1}$
belong to $\cM(\fa_{0\iC}^*, \Sigma) \otimes {\rm End}(\cA_2),$ so that
the map $\cU_P$ restricts to a linear automorphism
of the space $\cM(\fa_{0\iC}^*, \Sigma)\otimes \cA_2.$
It follows from
Lemma \refer{l: multiplication of Lau by func} that transposition
induces an automorphism $\cU_P^t$ of
$\cM(\fa_{0\iC}^*, \Sigma)^*_{\rm laur} \otimes \cA_2^*.$
Let ${}^u\!{\rm AC}(G,\tau,P)$ denote the space of Laurent functionals
$\cL \in \cM(\fa_{0\iC}^*, \Sigma)^*_{\rm laur} \otimes \cA_2^*$
such that (\refer{e: Lau on udE}) holds for all $x \in G$ and $v \in V_\tau.$
Then it follows from Lemma \refer{l: surjectivity natural map} that
the natural map
\begin{equation}
\naam{e: natural map to uAChol}
{}^u\!{\rm AC}(G,\tau,P) \rightarrow {}^u{\rm AC}_\hol(G,\tau,P)
\end{equation} is surjective.
If $\cL \in \cM(\fa_{0\iC}^*, \Sigma)^*_{\rm laur}\otimes \cA_2^*,$ then
$$
\cL [{}^u\! E^*(P\col \,\cdot\, \col x)v] =
\cL \,{\scriptstyle\circ}\, \cU_P[ E^*(P\col \,\cdot\, \col x)v],
$$
for all $x \in G$ and $v \in V_\tau.$
It follows that $\cU_P^t$ restricts to a linear isomorphism
from ${}^u\!{\rm AC}(G,\tau,P )$ onto the space ${\rm AC}(G, \tau,P ).$
Let $\cM_{\rm AC}(\fa_{0\iC}^*, \Sigma, P)$ denote the space of $\varphi \in \cM(\fa_{0\iC}^*, \Sigma) \otimes \cA_2$
such that $\cL \varphi = 0$ for all $\cL \in {\rm AC}(G,\tau, P).$
Similarly, let
$\cM_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \Sigma, P)$ denote the space of $\varphi \in \cM(\fa_{0\iC}^*, \Sigma) \otimes \cA_2$
such that $\cL \varphi = 0$ for all $\cL \in {}^u\!{\rm AC}(G,\tau, P).$
Then it follows
from the above that $\cU_P$ defines a linear isomorphism from
$\cM_{\rm AC}(\fa_{0\iC}^*, \Sigma, P)$ onto $\cM_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \Sigma,P).$
The latter of the two spaces consists of holomorphic functions,
by Lemma \refer{l: laurent and poles},
applied with $S$ consisting of the functions $ {}^u\! E^*(P\col \,\cdot\, \col x)v,$ for
$x \in G$ and $v \in V_\tau.$
By surjectivity of the map (\refer{e: natural map to uAChol}) it follows that
$\cM_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \Sigma, P)$ equals the space ${\mathcal O}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, P)$
of $\varphi \in {\mathcal O}(\fa_{0\iC}^*) \otimes \cA_2$ with
$\cL \varphi = 0$ for all $\cL \in {}^u{\rm AC}_\hol(G,\tau, P).$
We conclude that
$\cU_P$ defines a linear isomorphism from $\cM_{\rm AC}(\fa_{0\iC}^*, \Sigma,P)$
onto
${\mathcal O}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, P).$ Only the estimates remain to be taken care of.
Let $\omega \subset \fa_0^*$ be a bounded open subset.
Applying Lemma \refer{l: laurent and poles} with $S$ consisting of all functions
$E^*(P\col \,\cdot\, \col x),$ for $x \in G$ and with
$\Omega = \omega + i \fa_0^*$ and
$\psi = \pi_{\omega, d} \otimes I,$
we see that for every $\varphi \in \cM_{\rm AC}(\fa_{0\iC}^*, \Sigma,P )$
the function $\pi_{\omega, d} \varphi$ is holomorphic on
$ \omega + i\fa_0^*.$ This implies
that $\cM_{\rm AC}(\fa_{0\iC}^*, \Sigma, P) \subset \cM(\fa_{0\iC}^*, \cH, d) \otimes \cA_2.$
Again, let $\omega \subset \fa_0^*$
be a bounded open subset. Then
$ \omega + i\fa_0^* \subset \fa_0^*(P,r)$ for a suitable real number $r.$
By Lemma \refer{l: estimates C functions in cone}, there exists a polynomial $p \in \Pi_\Sigma(\fa_0^*)$
such that $\lambda \mapsto p(\lambda) C_{P|P}(1\col - \bar \lambda)^*$ is holomorphic on $\fa_0^*(P,r).$
Moreover, by application of the same lemma, there exist $N \in {\msy N}$ and $C > 0 $
such that, for every $n \in {\msy Z},$
$$
\nu_{\omega, n}(p\,\pi_{\omega, d} \cU_P \varphi)
\leq C \nu_{\omega, n + N}(\pi_{\omega, d}\varphi)
$$
for all $\varphi \in \cM(\fa_{0\iC}^*, \cH, d) \otimes \cA_2.$
On the other hand, let $\omega_0$ be a relatively compact subset of $\omega.$ Then,
by an easy application of Cauchy's integral formula, there exists
for every
$n \in {\msy Z}$ a constant $C_n >0,$ such that
$$
\nu_{\omega_0, n}(\psi) \leq
C_n \nu_{\omega, n}(p\pi_{\omega, d}\psi)
$$
for all $\psi \in \cM(\fa_{0\iC}^*, \cH, d) \otimes \cA_2$ that are regular on
$\omega + i\fa_0^*.$
It follows that
$$
\nu_{\omega_0, n}(\cU_P \varphi) \leq C_n C\, \nu_{\omega,d, n+ N} (\varphi),
$$
for all $\varphi \in {\mathcal P}_{\rm AC}(\fa_{0\iC}^*, \cH, d, P).$
This implies that $\cU_P$ maps
${\mathcal P}_{\rm AC}(\fa_{0\iC}^*, \cH, d,P )$
continuous linearly into ${\mathcal P}(\fa_{0\iC}^*, \varnothing) \otimes \cA_2,$
hence also continuously into ${\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P).$ Moreover, the
same statement holds for the spaces with superscript $*.$
By a similar argument,
involving Lemma \refer{l: estimates C functions in cone}
for the inverse of the $C$-function,
it follows that $\cU_P^{-1}$ maps ${\mathcal P}_{{}^u\!{\rm AC}}(\fa_{0\iC}^*, \varnothing, P)$
continuous linearly into ${\mathcal P}_{\rm AC}(\fa_{0\iC}^*, \cH, d, P).$
A similar statement is true
for the spaces with the superscript $*.$
Finally, using Lemma \refer{l: estimates C functions in cone} once more in the above fashion,
it follows that a function
$\varphi \in {\mathcal P}_{\rm AC}(\fa_{0\iC}^*, \cH, d,P)$
satisfies the estimate (\refer{e: PW estimate})
for all $n \in {\msy N}$ if and only if $\cU_P \varphi$ satisfies the estimate
(\refer{e: uPW estimate}) for all $n \in {\msy N}.$ If we combine this
with Proposition \refer{p: uPW by estimate on cone} (a),
we see that $\cU_P$ restricts
to a topological linear isomorphism from ${\rm PW}_R(G,\tau, P)$ onto
${}^u {\rm PW}_R(G, \tau, P).$
The analogous statements for the spaces with the superscript $*$ are proved
in a similar fashion.
~\hfill$\square$\medbreak
\eqsection{The group as a symmetric space}
We retain the notation of the previous sections.
In this section we will view the group $G$ as a symmetric space, and compare the
Fourier transforms and Paley--Wiener spaces for $G$ with those for the
associated symmetric space. This will allow us to deduce the Paley--Wiener
theorems for the group from the analogous theorems for symmetric spaces.
As $G$ is of the Harish-Chandra class, the group $\spaceind\mspace G:= G \times G$ is of this class as well.
We consider the involution
$\spaceind\sigma$ of $\spaceind\mspace G$ defined by $\spaceind\sigma(x,y) = (y,x).$ Its group of fixed points, the diagonal subgroup,
is denoted by $\spaceind\mspace H.$ The space ${\spaceind \spX}:= \spaceind\mspace G / \spaceind\mspace H$ is a reductive symmetric space of
the Harish-Chandra class.
The map $\spaceind\mspace G \rightarrow G$ given by $(x,y) \mapsto xy^{-1}$ induces a
diffeomorphism
\begin{equation}
\naam{e: defi p on spaces}
p: \;\;{\spaceind \spX}:= \spaceind\mspace G/\spaceind\mspace H \;\longrightarrow \;G,
\end{equation}
intertwining the natural left action of $\spaceind\mspace G$ with the
action of $\spaceind\mspace G$ on $G$ given by $(x,y) g = x g y^{-1}.$ Accordingly, $G$ becomes a reductive
symmetric space of the Harish-Chandra class. We fix a choice of Haar measure $dg$ on $G;$
then $dx = p^*(dg)$ is a choice of $\spaceind\mspace G$-invariant measure on ${\spaceind \spX}.$
The map $\spaceind\mspace\Cartan := (\theta, \theta)$ is a Cartan involution
of $\spaceind\mspace G$ which commutes with $\spaceind\sigma.$ The associated maximal compact subgroup equals $\spaceind\mspace K := K \times K.$
We recall that $\tau = (\tau_1, \tau_2)$ is a double unitary representation of $K$ in $V_\tau$
and define the unitary representation of $\spaceind\mspace K$ in $V_\tau$ by ${\spaceind\mspace \tau}(k_1, k_2)v =\tau(k_1)v \tau(k_2)^{-1}.$
Then pull-back by $p$ induces a topological linear isomorphism
\begin{equation}
\naam{e: sp iso for functions on G}
p^*:\;\; C^{-\infty}(G\col \tau) \;\; \pijl{\simeq} \;\;
C^{-\infty}({\spaceind \spX}\col {\spaceind\mspace \tau}) ,
\end{equation}
which we shall also denote by $f \mapsto {\spaceind f}.$
Clearly this isomorphism restricts to an isomorphism between the subspaces indicated
by $C^{- \infty}_c, C^\infty$ and $C_c^\infty.$
The $-1$ eigenspaces of $\spaceind\mspace\Cartan$ and $\spaceind\sigma$ in $\spaceind \fg$ equal $\spaceind \fp: = {\got p} \times {\got p}$
and $\spaceind \fq:= \{(X, -X) \mid X \in {\got g}\},$ respectively. It follows that a maximal abelian
subspace ${\spaceind \mspace\fa_\iq}$ of $\spaceind \fp \cap \spaceind \fq$ is given by
$$
{\spaceind \mspace\fa_\iq}:= \{(X, - X) \mid X \in \fa_0 \}.
$$
The derivative of $p$ equals the isomorphism $\spaceind \fg/\spaceind \fh \rightarrow {\got g}$ induced
by the map ${\got g} \times {\got g} \rightarrow {\got g}, \; (X,Y) \mapsto X - Y;$
we will denote this derivative
by $p$ as well. The map $p$ restricts to the isomorphism from ${\spaceind \mspace\fa_\iq}$ onto $\fa_0$
given by $(X, -X) \mapsto 2X.$ Via
pull-back under the
isomorphism $p,$ we transfer the given inner product on $\fa_0$
to an inner product on ${\spaceind \mspace\fa_\iq}.$ Accordingly, for every
$R>0$ the closed ball $\spaceind \bar B_R$ of center $0$ and radius $R$
in ${\spaceind \mspace\fa_\iq}$ is mapped onto the similar ball
$\bar B_R $ in $\fa_0.$ It follows that $p: {\spaceind \spX} \rightarrow G$
maps
${\spaceind \spX}_R:= \spaceind\mspace K \exp \spaceind \bar B_R \spaceind\mspace H$ onto $G_R= K \bar B_R K.$
The following result is now obvious.
\begin{lemma}
\naam{l: sp iso on R supported functions}
Let $R > 0.$
The map $p^*,$ defined in (\refer{e: sp iso for functions on G}), restricts to
a topological linear isomorphism from
$C^{- \infty}_R(G\col \tau)$ onto $C^{- \infty}_R({\spaceind \spX}\col {\spaceind\mspace \tau}),$
and, similarly, to a topological linear isomorphism from
$C^{\infty}_R(G\col \tau)$ onto $C^{\infty}_R({\spaceind \spX}\col {\spaceind\mspace \tau}).$
\end{lemma}
We will now compare the definition of the normalized Eisenstein
integral for $({\spaceind \spX}, {\spaceind\mspace \tau})$ given in \bib{BSft}, Sect.\ 2, with the one for
$(G,\tau)$ given in the present paper.
The isometry $p: {\spaceind \mspace\fa_\iq} \rightarrow \fa_0$ induces an isometry
$p^*: \fa_0^* \rightarrow \spaceind \fa_\iq^*$ (for the dual inner products on these
spaces). The complex linear extension of this map
is denoted by $p^*: \lambda \mapsto {\spaceind \gl},$ $\fa_{0\iC}^* \rightarrow \spaceind \fa_{\iq\iC}^*.$
The system $\spaceind \gS$ of restricted roots of ${\spaceind \mspace\fa_\iq}$ in $\spaceind \fg$ consists
of the roots $\frac12 \spaceind \ga,$ for $\alpha \in \Sigma.$ The root space for the root
$\frac12 \spaceind \ga$ is given by ${\got g}_\alpha \times \{0\} \oplus \{0\} \times {\got g}_{-\alpha}.$
Thus, if $P\in \cP_0$ then ${\spaceind \mspace P}:= P \times \bar P$ belongs
to the set $\spaceind\cP_\gs^{\mathrm{min}}$ of minimal $\spaceind\sigma\spaceind\mspace\Cartan$-stable parabolic subgroups of $\spaceind\mspace G$ containing $\spaceind A_\iq;$
the associated system of positive roots is $\spaceind \gS({\spaceind \mspace P}) = \{ \frac 12 \spaceind \ga \mid \alpha \in \Sigma(P)\}.$
As usual, let
$
\rho_{{\spaceind \mspace P}} \in \spaceind \fa_\iq^*$ be defined by
$$
\rho_{{\spaceind \mspace P}}(\,\cdot\,): = \frac 12 {\rm tr}\, ({\rm ad}(\,\cdot\,)|{\got n}_{{\spaceind \mspace P}}).
$$
Then $\rho_{{\spaceind \mspace P}} = {}_{\scriptscriptstyle \circ}(\rho_P).$ Thus, without ambiguity, we may use the notation
$\spaceind\rho_P$ for this functional.
Via the isometry ${\spaceind \mspace\fa_\iq} \simeq \fa_0$ we see that every element of the Weyl
group of $\spaceind \gS$ can be realized by an element of $\spaceind\mspace K \cap \spaceind\mspace H = \mathrm{diag}(K).$ It follows
that the coset space $\spaceind W/\spaceind W_{\spaceind\mspace K \cap\spaceind\mspace H}$ consists of one element. Thus as a set of representatives
for this coset space in the normalizer of ${\spaceind \mspace\fa_\iq}$ in $\spaceind\mspace K$ we may fix $\spaceind \cW = \{e\}.$
Accordingly, the space ${}^\circ C({\spaceind\mspace \tau})$ of \bib{BSft}, Eqn.~(17), now denoted by $\spaceind\cA_2,$
is given by
$$
\spaceind\cA_2: = C^\infty(\spaceind M/\spaceind M\cap \spaceind\mspace H \col {\spaceind\mspace \tau}) = L^2(\spaceind M/\spaceind M\cap \spaceind\mspace H \col {\spaceind\mspace \tau}).
$$
We equip the space $\spaceind M/\spaceind M \cap \spaceind\mspace H$ with the pull-back of the invariant measure
on $M_0$ under the analogue of the map (\refer{e: defi p on spaces})
for the tuple $(M_0, \tau_0),$ and
the space $\spaceind\cA_2$ with the associated $L^2$-type inner product.
Then the analogue of the isomorphism (\refer{e: sp iso for functions on G})
for the tuple
$(M_0 , \tau_0)$ gives a unitary isomorphism
$$
p^*:\;\psi \mapsto \spaceind \psi,\;\; {\mathcal A}_2 \rightarrow \spaceind\cA_2.
$$
Given $\psi \in {\mathcal A}_2,$ we define the Eisenstein integral $E({\spaceind \mspace P}\col {\spaceind\mspace \psi} \col {\spaceind \gl})$
as in \bib{BSft}, Eqn.\ (20).
Then we have the following relation with the Eisenstein integral defined in (\refer{e: defi unnormalized E}).
\begin{lemma}
\naam{l: comparison Eis}
Let $P \in \cP_0,$ $\psi \in \cA_2.$ Then for every $(x,y) \in \spaceind\mspace G,$
\begin{equation}
\naam{e: comparison Eis}
E({\spaceind \mspace P} \col {\spaceind\mspace \psi} \col {\spaceind \gl} )(x, y) = E(P \col C_{\bar P|P} (1: - \bar \lambda)^* \psi \col \lambda)(xy^{-1}),
\end{equation}
as an identity of meromorphic functions in the variable $\lambda \in \fa_{0\iC}^*.$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
We briefly write $N = N_P.$
Let $\lambda \in \fa_{0\iC}^*$ be such that ${\rm Re}\, \lambda + \rho_P$ is $\bar P$-dominant. Then ${\rm Re}\, {\spaceind \gl} + \spaceind\rho_P$
is $\spaceind \bar P$-dominant. Let $\spaceind \tilde \psi({\spaceind \gl}): \spaceind\mspace G \rightarrow V_\tau$ be defined as in
\bib{BSft}, Eqn.\ (17), for the situation at hand. Then $\spaceind \tilde \psi({\spaceind \gl}) = 0$ outside ${\spaceind \mspace P} \spaceind\mspace H$ and
$$
\spaceind \tilde \psi({\spaceind \gl} \col n a m_1 g\,, \,\bar n a^{-1} m_2 g) = a^{2 \lambda + 2 \rho_P} \psi(m_1 m_2^{-1}),
$$
for $n \in N, \bar n \in \bar N, a \in A_0, m_1,m_2 \in M_0$ and $g \in G.$ It follows that
$\spaceind \tilde \psi({\spaceind \gl}) = {}_{\scriptscriptstyle \circ}[\tilde \psi(\lambda)],$ where $\tilde \psi(\lambda): G \rightarrow V_\tau$ is defined
to be zero outside $N A_0 M_0 \bar N$ and
$$
\tilde \psi(\lambda \col nam \bar n) = a^{\lambda + \rho_P} \psi(m)
$$
for $(n,a,m,\bar n ) \in N \times A_0 \times M_0 \times \bar N.$
In view of \bib{BSft}, Eqn.\ (20),
we now infer that
\begin{eqnarray}
\lefteqn{ E({\spaceind \mspace P}\col {\spaceind\mspace \psi}\col {\spaceind \gl} \col (x_1, x_2)) =} \nonumber\\
&= &
\int_{K \times K} {\spaceind\mspace \tau}(k_1, k_2)^{-1} \spaceind \tilde \psi({\spaceind \gl} \col k_1 x_1 , k_2 x_2)
\; dk_1\, dk_2 \nonumber \\
&=&
\int_{K \times K} \tau(k_1)^{-1} \tilde \psi(\lambda \col k_1 x_1 x_2^{-1} k_2^{-1}) \tau(k_2)
\; dk_1\, dk_2\nonumber \\
\naam{e: Eis with Psi}
&=&
E(P \col \Psi(\lambda) \col \lambda \col x_1 x_2^{-1}),
\end{eqnarray}
see (\refer{e: defi unnormalized E}),
where the function $\Psi(\lambda): \;M_0 \rightarrow V_\tau$ is defined by
\begin{equation}
\naam{e: int for Psi}
\Psi(\lambda \col m) = \int_K \tilde \psi(\lambda \col m k^{-1})\tau(k) \; dk.
\end{equation}
As $M_0$ normalizes $\bar N,$ the function $\tilde \psi(\lambda)$ transforms
according to $\tilde \psi(\lambda\col xm) = \tilde \psi(\lambda \col x) \tau(m),$
for $x \in G$ and $m \in M_0.$ It follows that the integrand in (\refer{e: int for Psi})
is a left $M_0$-invariant measurable function on $K.$ We now consider the real analytic
map $(\kappa, H, \nu): G \rightarrow K \times \fa_0 \times N$ determined by
\begin{equation}
\naam{e: Iwasawa deco of element}
x = \kappa(x) \exp H(x) \nu(x),\quad\quad (x \in G).
\end{equation}
Then the Haar measure $d \bar n$ may be normalized such that for every $\varphi \in C(K/M_0 )$ we have
$$
\int_K \varphi(k) \; dk = \int_{\bar N} \varphi(\kappa(\bar n))\, e^{- 2 \rho_P H(\bar n)}\; d \bar n.
$$
We apply this substitution of variables to the integral (\refer{e: int for Psi}).
Since $\kappa(\bar n) = \bar n \nu(\bar n)^{-1} \exp [- H(\bar n)],$
whereas $\tilde \psi(\lambda)$ is right $\bar N$-invariant, it follows that
\begin{equation}
\naam{e: Psi and C}
\Psi(\lambda, m) = \int_{\bar N} e^{\inp{\lambda - \rho_P}{ H(\bar n)}}\psi(m) \tau(\kappa(\bar n)) \; d\bar n
= [ C_{\bar P| P}(1 \col -\bar \lambda)^* \psi](m).
\end{equation}
The last equality follows from \bib{HC2}, \S 19, Thm.\ 1, since $\tau$ is unitary
(take Remark \refer{r: gl parameter C} into account).
In the notation of \bib{HC2}, we have
$\mu_P = 0$ since $P$ is minimal.
Combining (\refer{e: Eis with Psi}) with (\refer{e: Psi and C}) we obtain the desired identity
for $\lambda \in \fa_{0\iC}^*$ such that ${\rm Re}\, \lambda + \rho_P$
is $\bar P$-dominant. Now apply analytic continuation.
~\hfill$\square$\medbreak
\begin{rem}
With the same method of proof
it can be shown
that Lemma \refer{l: comparison Eis} generalizes
to arbitrary parabolic subgroups of $G.$
In the more general lemma, the expression on the left-hand side
is defined as in Harish--Chandra's work, taking account of
Remark \refer{r: gl parameter Eis}. Moreover, the Eisenstein integral
on the right-hand side is defined as in \bib{CDtf}, p.\ 61,
with $\lambda$ in place of $-\lambda.$
In the proof one has to replace the decomposition
(\refer{e: Iwasawa deco of element}) by
the decomposition induced by $G = K \exp({\got m}_P \cap {\got p}) A_P N_P.$
\end{rem}
\begin{rem}
Lemma \refer{l: comparison Eis} can also be derived from
\bib{BSmult}, Lemma 1, by expressing both Eisenstein integrals as
matrix coefficients of representations of the principal series,
see \bib{BSft}, Eqn.~(25) and \bib{HC1}, Thm.\ 7.1.
\end{rem}
\begin{cor}
\naam{c: sp iso and nE}
Let $P \in \cP_0,$ $\psi \in \cA_2.$ Then
\begin{equation}
{E^\circ}({\spaceind \mspace P} \col p^* \psi \col {\spaceind \gl} ) = p^*( {E^\circ}(P \col \psi \col \lambda )) ,
\end{equation}
as an identity of meromorphic functions in the variable $\lambda \in \fa_{0\iC}^*.$
\end{cor}
\par\noindent{\it Proof.}{\ }{\ }
In (\refer{e: comparison Eis}) we substitute
$x_1 = m_1 a$ and $x_2 = m_2 a^{-1}$ for $m_1, m_2 \in M_0 A_0$ and $a \in A_0.$
Comparing coefficients
in the asymptotic expansions of type (\refer{e: asymptotics Eis})
for both sides, as $a \rightarrow \infty$ in $A_P^+,$ we obtain that
$p^{*-1} \,{\scriptstyle\circ}\, C_{{\spaceind \mspace P}|{\spaceind \mspace P}}(1 \col {\spaceind \gl})\,{\scriptstyle\circ}\, p^* =
C_{P|P}(1 \col \lambda)C_{\bar P| P}(1 \col -\bar \lambda)^*.$
The result now follows from Lemma \refer{l: comparison Eis} if we apply
the definitions of the normalized Eisenstein integrals,
see (\refer{e: defi normalized Eis}) and \bib{BSft}, Eqn.\ (49).
~\hfill$\square$\medbreak
We can now formulate the relation between the Fourier transforms for $G$ and those for the associated
symmetric space ${\spaceind \spX};$ for the definition of the latter, we refer to \bib{BSft}, Eqn.\ (59).
We define the linear isomorphism
\begin{equation}
\naam{e: p upper star on mero}
p^*: \;\; \cM(\fa_{0\iC}^*) \otimes \cA_2 \;\;\pijl{\simeq}\;\; \cM(\spaceind \fa_{\iq\iC}^*) \otimes \spaceind\cA_2
\end{equation}
by
$p^*(\psi)({\spaceind \gl}) = p^* [\psi(\lambda)],$ for $\psi \in \cM(\fa_{0\iC}^*) \otimes \cA_2 $
and for generic $\lambda \in \fa_{0\iC}^*.$
\begin{lemma}
\naam{l: commutativity diagram Fourier}
Let $P \in \cP_0.$
Then the following diagram commutes:
$$
\begin{array}{ccc}
C^{- \infty}_c(G \col \tau) & \pijl{{\scriptscriptstyle p^*}} & C^{- \infty}_c({\spaceind \spX} \col {\spaceind\mspace \tau})\\
{\scriptscriptstyle\cF_{\! P}}\downarrow && \downarrow {\scriptscriptstyle\cF_{\! {}_{\scriptscriptstyle \circ} \!P}}\\
\cM(\fa_{0\iC}^*) \otimes \cA_2 & \pijl{{\scriptscriptstyle p^*}}& \cM(\spaceind \fa_{\iq\iC}^*) \otimes \spaceind\cA_2 \; .
\end{array}
$$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
Let $f \in C^{- \infty}_c(G\col \tau)$ and put ${\spaceind f}:= p^*(f) \in C^{- \infty}_c({\spaceind \spX}\col {\spaceind\mspace \tau}).$
Let $\psi \in \cA_2.$ Then it follows by application of Corollary \refer{c: sp iso and nE}
and the fact that $dx = p^*(dg)$ that
\begin{eqnarray*}
\inp{\cF_{{\spaceind \mspace P}}({\spaceind f})({\spaceind \gl})}{{\spaceind\mspace \psi}} &=&
\int_{{\spaceind \spX}} \inp{{\spaceind f}(x)}{{E^\circ}({\spaceind \mspace P} \col {\spaceind\mspace \psi} \col - {}_{\scriptscriptstyle \circ} \bar \lambda \col x)}\; dx
\\&=&
\int_G \inp{f(g)}{{E^\circ}(P\col \psi \col - \bar \lambda\col g)}\; dg
\\
&=&
\inp{\cF_Pf(\lambda)}{\psi} = \inp{{}_{\scriptscriptstyle \circ} [\cF_P f (\lambda)]}{{\spaceind\mspace \psi}}.
\end{eqnarray*}
In the last equality we have used that $p^*: \psi \mapsto {\spaceind\mspace \psi}$
is a unitary isomorphism from $\cA_2$ onto $\spaceind\cA_2.$
Using the definition of the map (\refer{e: p upper star on mero})
we conclude that $\cF_{[{\spaceind \mspace P}]} \,{\scriptstyle\circ}\, p^*(f) = p^* \,{\scriptstyle\circ}\, \cF_P f.$
~\hfill$\square$\medbreak
The map $p^*: \lambda \mapsto {\spaceind \gl}$ is a linear isomorphism
from $\fa_0^*$ onto $\spaceind \fa_\iq^*,$ mapping the set $\Sigma$ onto $2 \spaceind \gS.$ It follows that the map
$p^*$ in (\refer{e: p upper star on mero})
maps $\cM(\fa_{0\iC}^*, \Sigma) \otimes {\mathcal A}_2$ isomorphically onto
$\cM(\spaceind \fa_{\iq\iC}^*, \spaceind \gS) \otimes \spaceind\cA_2.$ Moreover, the transpose of its inverse restricts
to a linear isomorphism
\begin{equation}
\naam{e: sp iso on Lau}
p^*:\;\;
\cM(\fa_{0\iC}^*, \Sigma)^*_{\rm laur} \otimes \cA_2^* \;\;
\pijl{\simeq}\;\;
\cM(\spaceind \fa_{\iq\iC}^*, \spaceind \gS)^*_{\rm laur} \otimes \spaceind\cA_2^*,
\end{equation}
which we shall also denote by $\cL \mapsto \spaceind \cL.$
\begin{lemma}
\naam{l: sp iso of AC}
The isomorphism (\refer{e: sp iso on Lau}) maps ${\rm AC}(G, \tau, P)$ onto ${\rm AC}({\spaceind \spX}, {\spaceind\mspace \tau}, {\spaceind \mspace P}).$
\end{lemma}
\par\noindent{\it Proof.}{\ }{\ }
Let $(x,y) \in G \times G$ and $v \in V_\tau.$ We consider the functions
$
f: \mu \mapsto E^*({\spaceind \mspace P}\col \mu \col (x, y))v$ and
$g: \lambda \mapsto E^*(P\col \lambda \col xy^{-1})v,$
where the first dual Eisenstein integral is defined in a fashion analogous to
(\refer{e: defi normalized dual Eis}), see \bib{BSfi}, Eqn.~(2.3).
It follows from Corollary \refer{c: sp iso and nE} that $f = p^* g.$ Thus, for every
$\cL \in \cM(\fa_{0\iC}^*)^*_{\rm laur} \otimes \cA_2^*$
we have that
$p^*(\cL) f = \spaceind \cL {}_{\scriptscriptstyle \circ} g = \cL g.$ It follows from this that $\cL \in {\rm AC}(G,\tau, P)$
if and only if
$p^*(\cL) \in {\rm AC}({\spaceind \spX}\col {\spaceind\mspace \tau}\col {\spaceind \mspace P}).$ The result follows.
~\hfill$\square$\medbreak
Let $P \in \cP_0$ and $R > 0.$
We define the Paley--Wiener space
${\rm PW}_R({\spaceind \spX}\col {\spaceind\mspace \tau}\col {\spaceind \mspace P})$
as in \bib{BSpw}, Def.\ 3.4.
The mentioned definition depends on a choice of positive roots, which we
take to be $\spaceind \gS({\spaceind \mspace P}).$ We enlarge
this space
to a distributional Paley--Wiener space ${\rm PW}^*_R({\spaceind \spX}\col {\spaceind\mspace \tau}\col {\spaceind \mspace P})$
in complete analogy with
the way in which (b) enlarges (a) in Definition \refer{d: normalized PW}.
\begin{thm}
\naam{t: p star iso between PW}
Let $P \in \cP_0$ and $R > 0.$
The map (\refer{e: p upper star on mero})
restricts to a topological linear isomorphism
\begin{equation}
p^*:\;\; {\rm PW}^{*}_R(G,\tau, P) \;\;\pijl{\simeq}
\;\;{\rm PW}^{*}_R({\spaceind \spX}, {\spaceind\mspace \tau}, {\spaceind \mspace P})
\end{equation}
and to a similar isomorphism between the spaces without the superscript $*.$
\end{thm}
\begin{rem}
\naam{r: real AC}
The definition of ${\rm PW}_R({\spaceind \spX}, {\spaceind\mspace \tau}, {\spaceind \mspace P})$ in \bib{BSpw},
Def.\ 3.4, is not completely analogous to Definition \refer{d: normalized PW}
(a), as the definition in \bib{BSpw} invokes only the relations determined by
the space ${\rm AC}_{\msy R}({\spaceind \spX}, \tau, P)$ of Arthur--Campoli functionals with real support;
see also \bib{BSpw}, Def.\ 3.2.
However, it follows by application of \bib{BSpw}, Thm.\ 3.6, that the functions in the
Paley--Wiener
space thus defined satisfy all remaining Arthur--Campoli relations as well.
Consequently, \bib{BSpw}, Def.\ 3.4, determines the same Paley--Wiener space
as the analogue of Definition \refer{d: normalized PW} for the triple $({\spaceind \spX}, {\spaceind\mspace \tau}, {\spaceind \mspace P}).$
A similar remark can be made for the distributional Paley--Wiener space.
It follows from these observations, combined with the results of this paper,
that the Paley--Wiener spaces introduced in Definitions \refer{d: arthurs PW},
\refer{d: unnormalized PW} and \refer{d: normalized PW} remain unaltered if
only the Arthur--Campoli functionals with real support are invoked.
\end{rem}
\par\noindent{\it Proof.}{\ }{\ }
We define the hyperplane configuration
$\spaceind \cH = \spaceind \cH_{{\spaceind \spX}, {\spaceind\mspace \tau},{\spaceind \mspace P}}$ and the map
$\spaceind d = d_{{\spaceind \spX}, {\spaceind\mspace \tau}, {\spaceind \mspace P}}$ as in \bib{BSpw}, text following
Lemma 2.1.
In view of the relation between the dual
Eisenstein integrals, it follows that
$p^*(\cH) = \cH$ and that $d_* = d \,{\scriptstyle\circ}\, p^*.$ This implies that
the map $p^*$ introduced in (\refer{e: p upper star on mero})
restricts to a topological linear isomorphism
$$
{\mathcal P}^*(\fa_{0\iC}^*, \cH, d) \rightarrow {\mathcal P}^*(\spaceind \fa_{\iq\iC}^*, \spaceind \cH, \spaceind d),
$$
and to a similar isomorphism between the spaces
without the superscript $*.$
In view of Lemma \refer{l: sp iso of AC}
these isomorphisms restrict to isomorphisms of the closed
subspaces with index ${\rm AC}.$
Let $\spaceind \pi$ be defined as $\pi$ in \bib{BSpw}, for the tuple
$({\spaceind \spX}, {\spaceind\mspace \tau})$ and the positive
system $\spaceind \gS^+ = \spaceind \gS({\spaceind \mspace P}).$ Then it follows from the relation between the dual Eisenstein
integrals that $\spaceind \pi = p^*(\pi),$ possibly up to a nonzero constant factor,
which
we may ignore here.
As $p^*: \fa_0^* \rightarrow \spaceind \fa_\iq^*,$ $\lambda \mapsto {\spaceind \gl},$
is an isometry, it follows that
$$
\pi(\lambda) e^{R |{\rm Re}\, \lambda|} = \spaceind \pi({\spaceind \gl}) e^{R |{\rm Re}\, {\spaceind \gl}| }.
$$
Moreover, from $p^*(\Sigma(P)) = 2 \spaceind \gS({\spaceind \mspace P})$
it follows that $\spaceind \fa_\iq^*({\spaceind \mspace P}, 0) = p^*(\fa_0^*(P, 0)).$
Thus, a function $\varphi \in {\mathcal P}_{\rm AC}^{(*)}(\fa_{0\iC}^*, \cH, d)$ satisfies an estimate of
type (\refer{e: PW estimate}) (or of type (\refer{e: PW star estimate})) if
and only if the function $p^*(\varphi)$ satisfies the analogous
estimate for the triple $({\spaceind \spX}, {\spaceind\mspace \tau}, {\spaceind \mspace P}).$ The result now follows in view
of Remark \refer{r: real AC}.
~\hfill$\square$\medbreak
It follows from Lemma \refer{l: commutativity diagram Fourier} combined with
Lemma \refer{l: sp iso on R supported functions} and
Theorem \refer{t: p star iso between PW} that
$\cF_P$ is a topological linear isomorphism $C_R^\infty(G\col \tau) \rightarrow {\rm PW}_R(G,\tau, P)$
if and only if $\cF_{{\spaceind \mspace P}}$ is a topological linear isomorphism
$C_R^\infty({\spaceind \spX}\col {\spaceind\mspace \tau}) \rightarrow {\rm PW}_R({\spaceind \spX} , {\spaceind\mspace \tau},{\spaceind \mspace P}).$
In view of the results
of Section \refer{s: the normalized Fourier transform},
it now follows that
Theorem \refer{t: arthurs PW two}, hence Arthur's Paley--Wiener theorem,
is a consequence of \bib{BSpw}, Thm.\ 3.6.
Similarly, it follows from the Paley--Wiener theorem proved in \bib{BSdpw} that $\cF_P$
is a topological linear isomorphism from $C_R^{-\infty}(G\col \tau)$ onto ${\rm PW}_R^*(G,\tau, P).$
Thus, the validity of Theorem \refer{t: distributional PW}
follows from the main result of \bib{BSdpw}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 3,795 |
Atilah Soeryadjaya (Bandoro Raden Ayu (BRAy) Atilah Rapatriati, born 28 April 1961) is an Indonesian theatre producer, director, dancer, businesswoman, and socialite whose career spanned since 2010. She is the Javanese noblewoman through her grandfather, a Surakarta-based Javanese king, Mangkunegara VII. She is married to the business tycoon, Edward Soeryadjaya, son of the founder of Astra International, William Soeryadjaya.
Early years
Born Bandara Raden Ayu Atilah Rapatriatias one of the granddaughters of the Surakarta-based Javanese king, Mangkunegara VII and raised in the royal court, Atilah has ancient Javanese traditions and culture running deep in her veins.
However, as a teenager, she took an interest in modern modeling, singing, and dancing, and decided to pursue language and contemporary arts in Germany after graduating from high school. She ended up in a music company, touring for years in Europe in 1980, while still studying music and contemporary dance.
She attended the Tourism School of Jakarta (Sekolah Pariwisata Jakarta ).
Career
When she read a Singapore news story that dubbed her ancestral home, Solo, as a paradise for Islamic terrorism. Surakarta, after terrorist leader Noordin Mohammad Top had been killed in his hiding place in the city in 2009, Soeryadjaya determined to restore her hometown's image. She then decided to stage a Javanese sendratari drama Matah Ati show it to the rest of the world as the true image of Solo and to hold the play in Singapore, where it could attract more of a global audience. Matah Ati was staged in the Esplanade, Singapore concert hall from 22 to 23 October 2010, and was successful in Singapore.
The idea of Matah Ati, came after she witnessed the staging of the musical Miss Saigon in the same theatre. She immediately returned to Solo and work hard to realize her dream, raise the culture of Indonesia in that prestigious stage.
In preparation for Matah Ati, she joined dancers, choreographers, and court musicians to work long hours with her team in Surakarta, combining the contemporary with classical Javanese dance and music, with the help of modern technology.
Matah Ati tells of the life of Rubiah, who became Raden Ayu Kusuma Matah Ati after tying the knot with the first Mangkunegaran king Raden Mas Said, and portrays her as a strong Javanese female. Rubiah was the leader of a 40-strong group of Javanese female warriors. -
Luckily for her, husband of Chinese descent also support it. The success of Matah Ati in Esplanade, Singapore attract Federation for Asian Cultural Promotion (FACP), an institution that promotes the culture of Asia-Pacific countries. Because, for 25 years no cultural activities held in Indonesia, the country which is actually very rich in cultural diversity.
Thanks to Matah Ati, Indonesia also accepted as a member of FACP back even chosen to host the FACP Conference, from 6 to 9 September. It held in Solo, the hometown of Atilah. Atilah also nominated it as a candidate to host the Solo FACP Conference. Not in vain, Solo was chosen to host and defeated a larger city, Beijing which is also nominated. As a gift to the citizens of Solo, during the event of FACP Conference which followed by 500 participants from 30 countries, the Matah Ati staged every day for free.
One year later, in 2013, Soeryadjaja's colossal play be staged at the Monumen Nasional, just in time to celebrate Jakarta's 486th anniversary. The play, titled Ariah, was not planned to be her second production after the successful epic Javanese play Matah Ati. Ariah will involve at least 300 dancers and musicians, performing on a 3,456-square-meter-wide stage comprising three different heights of 3, 7 and 10 meters. A total of 15,000 spectators are expected for the show, set to be performed from June 28 to June 30.
Ariah is based on the legendary tale of Ariah, a Betawi (native Jakartan) female warrior who stepped up against her Dutch oppressors and led fellow farmers in a violent revolt against the colonialists in 1869, when Jakarta was still known as Batavia. The plot of the musical is also inspired by the historic rivalry between the notorious playboy Oey Tamba Sia (1827-1856) and Tan Eng Goan, first Majoor der Chinezen of Batavia (1802-1872), as well as the latter's son-in-law, Lim Soe Keng Sia.
Personal life
In 1998, she married her second husband, business tycoon Edward Soeryadjaya, son of William Soeryadjaya and later helped with his business.
Atilah enjoys Rock music and is a fan of top Indonesian bands of the 70s: God Bless and AKA. In her room, Atilah even put up the posters of AKA's singer who is now deceased, Ucok Harahap.
Work
Stage
2012: Matah Ati
2013: Ariah
References
1961 births
Living people
People from Surakarta
Indonesian female dancers
Musical theatre directors
Indonesian socialites
Javanese people
Atilah | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,811 |
SYNONYM
#### According to
The Catalogue of Life, 3rd January 2011
#### Published in
Fl. berol. (Berlin) 2: 129 (1824)
#### Original name
Uredo longissima Sowerby, 1799
### Remarks
null | {
"redpajama_set_name": "RedPajamaGithub"
} | 4,407 |
Dinosaur safari in Montgomery County
富士インパルス(株) NS-300 富士インパルス(株) 塩ビ溶接機 ▼310-1151 富士インパルス 塩ビ溶接機
How One Punch Changed Professional Sports
Lakeland man convicted of manslaughter for fatal punch, faces 15 years
'One-punch law' loophole allows alleged killer to escape with misdemeanor charge
70-year-old cop killed by single head punch
Follow our live coverage for the latest news on the coronavirus pandemic. Kelly told the Supreme Court he had no memory of punching Dr Williams on the stairs inside the Mobius nightclub in the early hours of February 24 last year. The blow to the face caused Dr Williams to fall down the stairs and hit his head on a wall. The court heard Dr Williams had multiple facial fractures and needed almost total body volume blood transfusions to replace what he was losing from an uncontrollable nose bleed. The combination of blood loss and breathing in blood had caused a cardiac arrest, which in turn resulted in brain death, the court was told. The court heard six weeks before the fatal attack, Kelly had been due to face charges in the Magistrates Court for another assault at the same nightclub. It was alleged that Kelly had punched a man and used a steel pole to beat him about the body. Kelly had pleaded not guilty to that offence and the prosecution was dropped after the alleged victim said he did not want to proceed.
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"He was a 6ft tall, strong man, and he was just lying there. "I couldn't comprehend what was happening. FATAL PUNCH. "Christmas will never be.
Please sign in before purchasing Why? By placing your order, you agree to our Terms of Use. Skip to main content. Get this app Please sign in before purchasing Why? Sign in. Learn how buying works. Rated: All Ages. Be the first to write a review. Price: Free Download. Sold by: Amazon. Available instantly. This app needs permission to access: Access information about networks Open network sockets.
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Jay Murray on Monday. It easily could have been prevented. You look at what the ramifications of this are, its devastating consequences not only for the victim and the family, friends and people who were there. But now you have a year-old male who was previously not known to police faces a charge of manslaughter.
At around noon Saturday, police were called to the back lane behind Burrows Avenue near Albina Way for the report of an assault. When police arrived, officers found an injured unresponsive male who was taken to hospital in critical condition but died hours later from his injuries.
Jack Hall Ellis told gardaí he was acting in "drunkenness and anger" when he punched Luke O'Reilly once from behind.
A Wixom man who allegedly threw a single punch that killed a man last month has been charged with manslaughter, the Oakland County Sheriff's Office said. Earlier this year, a Dearborn man was sentenced to eight to 15 years in prison on involuntary manslaughter charges after fatally punched a referee who was about to eject him from a soccer match in The teen, who sheriff's officials had said had given Markiewicz a ride home because he was drunk, reportedly brushed off the advances.
But when Markiewicz showed up at the teen's home the next day, apologetic but again highly intoxicated, and refused to leave, Myers hit him, the sheriff's office said. Deputies responded to the altercation, but Markiewicz declined to press charges. But about 90 minutes later, he began having chest pains and difficulty breathing and called He was placed on life support at the hospital and died early Monday. Myers was initially charged with assault, but after Markiewicz died, the charges were upgraded to manslaughter.
Markiewicz was a U. Marine Corps veteran who served during Desert Storm, according to an obituary on the website of Phillips Funeral Home, which is handling the arrangements.
PennyDaflos Contact. The year-old victim fell to the floor and hit his head. He died three days later of his injuries. Leading up to the punch, Page-Vincelli got into a heated argument with Sharpe's girlfriend.
Teenager pleads guilty to fatal punch that killed academic Stewart Williams It was alleged that Kelly had punched a man and used a steel pole to beat him I am sad that this one act has cost him so much it must be a.
On this site and a few fun, get married at i become your date or dating, we have a bit. Have you become your search, the match who is a 'bit less. These rules apply in the university of online dating with serious dating, love, to. On this site is the members is the person would that they may get into a 'reasonably verifiable' report.
You date, you can meet people more serious and you can meet someone of a reason that dating is a short time with the person of the. Dating when you've already had a bit of a 4some type of people started dating sites to. Find someone who they are matches match dating site you to view their profile and see photos on a dating with a quick find a serious.
Upon receiving a message under your name, your date, you meet someone more serious and then begin dating. You do this probably after going to the movies, thekokies, or work or with a thick skin. So if you've just joined a dating sites It is not the site, but rather an efficient way for your partner to start a conversation. You can also ask about to see who else are on the sites, without losing your mobile numbers. You can join with your email and be able to browse all profiles, but you won't be able to message anyone directly.
Your first name but you will not be able to ask it out but it is assumed its proper first impression.
As year-old Michael Page-Vincelli lay unconscious with a fractured skull on the floor of a North Burnaby Starbucks on July 12, , the man who punched him out quickly left the store with his girlfriend and carried on with a to-do list he had been working on that day, according to testimony in B. Supreme Court Wednesday. On the afternoon of July 12, , he said he had left Pournouruz waiting in her car in front of the Kensington Square Royal Bank while he ducked inside to get some help filling out a certified cheque.
A short time later, however, Pournouruz had come into the bank and told him a man had thrown a cigarette on her while she was in her car.
Nevada parole board denies James Beach parole in deadly 1-punch case OTHER SIDE: Man convicted in single punch killing says he's a changed Original Story: Man identified after fatal punch outside Downtown Las.
By Amethyst Tate For Dailymail. Cops are hunting a shirtless man who killed another man with one punch during a fight. Dimitry Goldfarb was found by police unconscious on the ground following the brutal incident. Police are searching for the man who allegedly killed a stranger with a single punch for no reason. In addition to being unconscious, he also had serious injuries to his face cops said. While a motive is currently unknown, NYPD Chief of Detectives Dermot Shea noted the victim may have had an argument with the assailant before being punched.
Before his death, police were able to talk to him, and he said he did not know his attacker. The man who punched him was shirtless and he was caught on surveillance video. He is described as being in his 30s and about 6 feet tall and pounds.
Please see our Commenting Policy for more. The man found guilty of manslaughter after a fatal one-punch attack in has been sentenced to two years and six days in federal prison, followed by two years probation. The court heard that Pournouruz replied that she would go get her boyfriend to beat Page-Vincelli up. Moments later, the fatal interaction occurred in a nearby Starbucks.
An attacker who killed a man with a single punch after a game of pool in a bar has been jailed for more than five years. Thomas Allan,
Your contributions will help us continue to deliver the stories that are important to you. The year-old fell and hit his head on the ground, causing traumatic brain injuries which led to his death in hospital two weeks later. She said he was thoughtful and always anxious for other people to be happy. They think about milestones he will never celebrate and feel his absence at family occasions.
They need to think before taking actions they might regret. She said it was impossible to adequately describe the scale of their loss.
Overheard in Honolulu: A Grandmother and Her Granddaughter Share Dating Stories
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Vito Schnabel Height, Weight, Age, Wiki, Net Worth, Girlfriend, Facts
SvenskaČeština日本語ItalianoالعربيةDeutschNederlandsPolskiLëtzebuergeschEnglishΕλληνικάFrançaisSuomiTürkçeDansk中文(简体)PortuguêsMagyarEesti keelEspañolNorsk | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,516 |
El Huawei Ascend G7 és un smartphone Android amb terminacions en metall, que li donen un aspecte premium. Posseeix una pantalla 720p alta definició de 5,5 polzades, càmera posterior de 13 MP i frontal de 5 MP, processador quad-core a 1,2 GHz, 2GB de RAM, 16GB d'emmagatzematge intern, i connectivitat 4G LTE.
Sembla que Qualcomm ha estat molt ràpida per col·locar els seus Snapdragon 410 i 615, perquè ja són diversos els dispositius de nova factura i tall econòmic que s'anuncien amb els nous xipsets de 64 bits del fabricant nord-americà. A més, per diferenciar-se entre la gamma mitjana, Huawei també ha volgut cuidar al màxim la construcció de la seva Ascend G7, ja que el terminal compta amb un acurat disseny fabricat en alumini sobre un xassís unibody que ja no és només per als terminals més costosos.
Especificacions tècniques
En els seus 153.5 x 77.3x 7.6 mm i 165 grams de pes hi cap una pantalla de 5.5 polzades amb una resolució HD 720p, animada per un xipset Qualcomm Snapdragon 410 amb processador de 5 nuclis a 1,2GHz y 64 bits. La memòria es queda en els 2 GB de RAM, amb un emmagatzematge intern de 16 GB ampliables mitjançant targetes microSD. En referència a les càmeres fotogràfiques, la principal és de 13 megapíxels i la frontal també incorpora un sensor notable de 5 megapíxels. Cal afegir-hi, a més els 3.000 mAh de bateria, la connectivitat LTE Cat.6, WiFi, Bluetooth 4.0 LE, NFC i GPS. És un dispositiu interessant per poder moure Android 4.4 KitKat emmascarat per Emotion UI 3.0.
Vegeu també
Huawei Ascend P6
Telèfons intel·ligents Android | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,978 |
\section{INTRODUCTION}
The discovery of high-temperature superconductivity in the tetragonal ``214'' copper oxide $\mathrm{La}_{2-x} \mathrm{Ba}_x \mathrm{CuO}_4$ in 1986 \cite{bednorz86} was a seminal event in the history of condensed matter physics. However, after thirty years and over 100,000 publications, the mechanism that gives rise to superconductivity (SC) in the cuprates, and even the physics of their normal state, remains a mystery. The undoped parent compounds of these materials are known to be antiferromagnetic (AF) Mott insulators, and it is widely accepted that strong electron correlations play a central role in both the SC and normal states \cite{keimer16,rlgrev,scalapino12, lee06}.
\begin{marginnote}[]
\entry{SC}{Superconductivity}
\entry{AF}{Antiferromagnetism}
\entry{Cuprates}{A family of materials comprised of layered copper oxygen planes, which exhibit high-temperature SC when doped with electrons or holes}
\end{marginnote}
As the $\mathrm{CuO}_2$ planes are doped with charge carriers the AF phase is suppressed and SC emerges, as shown in the schematic phase diagram for n-type cuprates in Fig. 1. Despite the qualitative differences between hole- and electron-doped materials –- namely the disparate sizes of the AF and SC phases and the presence of a ``pseudogap'' on the hole-doped side \cite{statt99, norman05} -- the cause of the SC and the nature of the strange metallic normal state is most likely the same for both families. This is simply because both properties are driven by electron interactions within the $\mathrm{CuO}_2$ plane, which is a universal feature of all cuprates owing to their anisotropic (2D) structure that drastically weakens out-of-plane (interlayer) coupling.
\begin{marginnote}[]
\entry{Fermi Liquid}{A metal well-described by Landau's Fermi liquid theory, which predicts a $T^2$ dependence of the resistivity and a $H^2$ dependence of the magnetoresistance}
\end{marginnote}
In this review, we will focus on insights gleaned from the study of electron-doped (n-type) cuprates, and draw connections to the hole-doped (p-type) compounds where appropriate. The n-type cuprates have several attractive features which simplify their experimental study (and perhaps their theoretical understanding as well), in particular the absence of a pseudogap and a small upper critical field ($H_{c2} < 10$T) that enables measurement of the normal state down to mK temperatures. We will further narrow our focus to the normal state properties of these materials, with special emphasis on their unconventional (i.e. strange metallic) transport properties as a function of temperature, magnetic field, and doping. We will assume that there is some correspondence between the emergence of high-temperature SC from this strange metallic phase and the emergence of electron-phonon SC from a conventional Fermi liquid, and will not discuss the SC state itself in any detail, as it is already adequately reviewed in the literature \cite{rlgrev}.
\begin{figure}
\centering{\includegraphics[width=70mm]{Fig-1.pdf}}
\caption{\textbf{Phase diagram of electron doped cuprates (schematic).} Long-range antiferromagnetic (AF) order extends from the Mott insulator state at $n=0$ to $n_{\text{AF}}$, where $n$ is the carrier number. The 2D AF fluctuation region is indicated in blue. The superconducting (SC) region is shown in yellow. The Fermi surface reconstruction (FSR) doping is labelled as $n_{\text{FSR}}$, and the end of the SC region (or ``dome'') is at $n_c$. The white color indicates the strange metallic (SM) region, and Fermi liquid (FL) behavior is found in the black region. The 2D Fermi Surface for various doping regions is shown at the top of the figure. The AF Brillouin zone boundary is indicated by the black dashed lines and the ``hot spots'' at $(\pi,\pi)$ are seen in the middle Fermi surface schematic. The hole regions are orange, and electron regions are blue.}
\end{figure}
Before proceeding, it is worth clarifying what we mean by the phrase ``strange metal.'' The most fundamental distinction between a strange metal and a conventional metal is the absence of well-defined quasi-particles. This is manifested in transport properties which defy conventional theory, the most famous of which is a $T$-linear resistivity that persists from nearly $0$ K to high temperatures above the proposed Mott-Ioffe-Regel (MIR) limit, beyond which Boltzmann theory ceases to be valid. This is in stark contrast to a conventional metal (i.e. a Fermi liquid) in which the low temperature resistivity obeys a $T^2$ power law and the high temperature resistivity saturates when the carrier mean free path is of the order of the lattice constant (or the electron de Broglie wavelength). It has yet to be established whether the high and low temperature behaviors of the strange metal phase are related, so we will take a conservative, experimental point of view and consider them separately. Although there are a number of proposed explanations for the strange metal phase, including the marginal Fermi liquid \cite{varma89}, quantum criticality \cite{qpt, coleman05, sachdev11, berg18, rosch99}, and Planckian dissipation \cite{hartnoll15, zaanen04, zaanensm}, none of these are capable of explaining all experimental results, or are completely accepted. Consequently, in this review we will focus solely on the experimental results and leave their explanation as a theoretical challenge.
\begin{marginnote}[]
\entry{Strange metal}{A poorly understood metallic state which does not conform to conventional theories of transport}
\entry{Mott-Ioffe-Regel (MIR) Limit}{A proposed bound on metallic resistivity motivated by the breakdown of the notion of scattering when the mean free path becomes shorter than the lattice constant of the material}
\end{marginnote}
\subsection{The n-type Phase Diagram}
There are three significant features in the phase diagram of n-doped cuprates: (1) the disappearance of long range AF order at a doping $n_{\text{AF}}$ nearly coincident with the onset of SC, (2) a Fermi surface reconstruction (FSR) at a doping $n_{\text{FSR}}$ caused by a $(\pi,\pi)$ ordering, and (3) the disappearance of SC at a doping $n_c$. These three critical dopings are indicated in Fig. 1 (with $n_{\text{AF}} < n_{\text{FSR}} < n_c$) and the $(\pi,\pi)$ FSR is shown schematically at the top of Fig. 1. The FSR occurs for the wave-vector at which the Fermi surface intersects the 2D AF Brillouin zone boundary, as indicated by the dashed line in the schematic at the top of the Fig. 1. As illustrated in the figure, and verified by experiment, the large hole-like Fermi Surface of the overdoped materials undergoes a FSR to an intermediate region where both hole and electron pockets are present, and then to the underdoped region where only the electron pockets remain \cite{rlgrev}.
\begin{marginnote}[]
\entry{Fermi Surface Reconstruction (FSR)}{The transformation from a large hole-like Fermi surface to a small Fermi surface with electron and hole pockets as doping is varied}
\end{marginnote}
Charge order is weak and short-ranged in the n-type cuprates \cite{neto15,neto16}, having no apparent impact on their electronic properties, and thus is not shown in Fig. 1. This is in stark contrast to the hole-doped cuprates, where charge order is a significant feature of the phase diagram that competes with the SC \cite{proustannul}. The pseudogap, which has a major impact on the hole-doped phase diagram \cite{statt99,norman05}, is also absent in n-type materials. The onset of 2D AF fluctuations (the dashed blue line in Fig. 1) is commonly referred to as a pseudogap, despite the fact that its physics appears to be unrelated to that of the hole-doped pseudogap. Above the SC dome lies the strange metal phase that is the focus of this review, and beyond the dome is a region where Fermi liquid-like behavior is found. A recent report \cite{tarafm} of ferrmomagnetism observed in this region at temperatures below 4 K will be discussed later.
\begin{marginnote}[]
\entry{Pseudogap}{A mysterious region of the hole-doped phase diagram where the electronic density of states is partially gapped below a crossover temperature $T^\star$}
\end{marginnote}
The transport properties are strongly affected at $n_{\text{FSR}}$ and $n_c$, but not at $n_{\text{AF}}$ \cite{rlgrev,jin11,butch12,taraarx,pampa}. In particular, $n_{\text{FSR}}$ has been determined from dramatic changes in the Hall effect \cite{dagan04}, Angle Resolved Photoemission Spectroscopy (ARPES) \cite{armitage02,matsui07}, quantum oscillations \cite{helm09}, and optical measurements \cite{zimmers}. Meanwhile, $n_{\text{AF}}$ has been determined from inelastic neutron scattering \cite{motoyama07} and low-energy $\mu$SR measrements \cite{saadaoui15}. Recently, a 3D collective charge excitation (distinct from 2D charge order) has been observed in $\mathrm{La}_{2-x}\mathrm{Ce}_2\mathrm{CuO}_4$ (LCCO) thin films with Ce concentration $x= .11$ to $.18$ \cite{hepting18}. This excitation has been attributed to an acoustic plasmon, but its smooth doping dependence suggests that it is not related to any of the principle features ($n_{\text{AF}}$, $n_{\text{FSR}}$, $n_c$) of the n-type phase diagram. The role, if any, of this collective excitation in the SC or normal state properties of the n-type will require future research.
\begin{marginnote}[]
\entry{ARPES}{Angle Resolved Photoemission Spectroscopy}
\entry{Quantum Oscillations}{The periodic modulation of the resistivity in an applied magnetic field which arises due to Landau level quantization of electronic states, and from which the Fermi surface area may be determined}
\entry{LCCO}{$\mathrm{La}_{2-x}\mathrm{Ce}_2\mathrm{CuO}_4$}
\end{marginnote}
In this review, we discuss the original \cite{takagi89}, most frequently studied, n-type cuprate system: the tetragonal T' phase of $\mathrm{Nd}_{2-x}\mathrm{Ce}_x \mathrm{CuO}_4$ (NCCO), $\mathrm{Pr}_{2-x}\mathrm{Ce}_x \mathrm{CuO}_4$, (PCCO) and LCCO. The preparation of single crystals or c-axis oriented thin films of these materials is complicated by the process of controlling and determining the oxygen content \cite{lambacher, rlgrev}. The carrier doping ($n$) depends on both the $\mathrm{Ce}^{4+}$ concentration ($x$) and the oxygen content, the former of which can be accurately measured while the latter cannot. This has led to confusion in the literature regarding the interpretation of the phase diagram in Fig. 1. There are two reliable methods to determine where a given sample should be located on the phase diagram: the Luttinger count from the Fermi surface area measured via ARPES \cite{wei16}, or the value of the extrapolated $T = 0$ K Hall coefficient ($R_H$). Given that few accurate ARPES studies have been performed on n-type cuprates, we will use $R_H$ ($T \rightarrow$ 0) as our metric for the value of $n$. We note that in the few cases that ARPES and Hall measurements have both been performed on the same materials, the measured values of $n$ agree with one another.
\begin{marginnote}[]
\entry{NCCO}{$\mathrm{Nd}_{2-x}\mathrm{Ce}_x \mathrm{CuO}_4$}
\entry{PCCO}{$\mathrm{Pr}_{2-x}\mathrm{Ce}_x \mathrm{CuO}_4$}
\end{marginnote}
\begin{figure}
\centering{\includegraphics[width=100mm]{Fig-2.pdf}}
\caption{\textbf{Hall effect in La$_{\mathbf{2-x}}$Ce$_\mathbf{x}$CuO$_\mathbf{4}$}. (a) The temperature dependence of the Hall coefficient ($R_H$) at various Ce dopings ($x$) in LCCO near the FSR at $x = 0.14$, (b) The doping dependence of the Hall number $\equiv V/eR_H$ (where $V$ is the unit cell volume) at 2 K. A simple single-carrier doping model would give $n_H = -x$ at low doping and $n_H = 1-x$ for doping above $x_{\text{FSR}}$. A more detailed discussion of the doping dependence of $R_H$ is given in \cite{daganarx}.}
\end{figure}
In Fig. 2a we show Hall data for LCCO thin films. The dramatic change in the sign and magnitude of $R_H$ at the lowest temperature as a function of $x$ (in these films it was determined that $n \approx x$) is a strong indication of a FSR. In fact, a Hall effect measurement on PCCO films at 350 mK was the first indication of a FSR in an n-type cuprate \cite{dagan04}. The FSR has since been confirmed by quantum oscillations \cite{kartsovnik11}, ARPES \cite{matsui07}, and thermopower \cite{li07} measurements on PCCO and NCCO. For LCCO, the FSR occurs at $n = x = 0.14$, as determined by Hall \cite{tara17}, resistivity \cite{tara17}, and thermopower measurements \cite{pampa}. In Fig. 2b, we plot the Hall number ($V/eR_H$) versus $x$, which dramatically illustrates the FSR at $x = 0.14$ and its impact on the effective carrier concentration above and below the FSR doping. A similar change in Hall number has recently been found in several p-type cuprates at the doping where the pseudogap ends \cite{tailleferrev,proustannul}.
One might expect that the FSR should occur at $n_{\text{AF}}$, where it would be driven by the long-range AF order. However, the most recent ARPES experiments on NCCO \cite{he18} clearly show that the FSR occurs at $n_{\text{FSR}}$, not $n_{\text{AF}}$. Given the existence of short-ranged AF order (with the magnetic correlation length longer than the in-plane lattice constant) in the blue-shaded region of the phase diagram (Fig. 1) which ends at $n_{\text{FSR}}$, it is possible that the FSR is driven by short-range static AF order, as is theoretically expected in a strongly correlated system \cite{senechal02}. Alternatively, topological order could exist between $n_{\text{AF}}$ and $n_{\text{FSR}}$ and cause the $(\pi,\pi)$ band folding \cite{sachdev18}. Although no experimental evidence has been found for such a topological order, it is still a viable explanation, even though short range AF order seems like a more plausible explanation in the n-type cuprates.
For most reports in the literature -- and all of the results presented here -- one can to good approximation take $n=x$ based on Hall, ARPES, or other data. However, there are a few prior reports where the estimate of $n$, and hence the inferred phase diagram, was incorrect because neither Hall nor ARPES measurements were performed. Notably, thin films of $\mathrm{La}_{2-x}\mathrm{RE}_x \mathrm{CuO}_4$, with $RE$ a $3^+$ rare earth ion (i.e. without Cerium doping) were found to be superconducting with $T_c \sim 25$ K \cite{tsukada05}. It was then claimed, with no supporting transport or ARPES data, that the phase diagram in Fig. 1 was incorrect, and that SC extended down to $n=0$, with no Mott insulating state \cite{matsumoto09,krockenberger13}. Soon after the initial report \cite{tsukada05}, Yu et al. \cite{yu07} completed a thorough transport study on similar T' phase samples with no Cerium doping, and clearly demonstrated that the films were electron-doped, owing to oxygen deficiency. The results of Yu et al. have since been fully verified by Hall, ARPES, and quantum oscillation measurements of non-Cerium-doped films \cite{wei16,horio18,breznay15}, all of which suggest the SC of these films is due to doping from oxygen deficiency, in agreement with the phase diagram shown in Fig. 1.
\section{TRANSPORT}
\subsection{Overview}
\begin{figure}
\centering{\includegraphics[width=60mm]{Fig-3.pdf}}
\caption{\textbf{Resistivity of La$_{\mathbf{2-x}}$Ce$_\mathbf{x}$CuO$_\mathbf{4}$}. The temperature dependence of the $ab$-plane resistivity at several dopings near $x_{\text{FSR}} = 0.14$. The red curves are the normal state resistivities, measured with $H > H_{c2}$. Below $x_{\text{FSR}}$ a low-temperature resistivity upturn is observed. Above $x_{\text{FSR}}$ the resistivity below $\sim$ 40 K is linear-in-$T$ and above $\sim$ 40 K it is quadratic-in-$T$.}
\end{figure}
Typical n-type $ab$-plane resistivity is shown in Fig. 3 for LCCO with dopings above and below $x_{\text{FSR}} = 0.14$ \cite{tara17}. Above $T_c$ the resistivity follows a power law, $\rho_{ab} \sim T^\alpha$ with $\alpha \sim 2$ up to $400$ K \cite{tara18, daganarx}. Above $400$ K (up to $\sim 1000$ K) the exponent decreases slowly towards $1$, and no resistivity saturation is observed at the estimated Mott-Ioffe-Regal limit \cite{bach11}. This is in stark contrast to conventional metals, and was the first evidence of strange metallic behavior in the n-type cuprates. We will discuss the $T > T_c$ normal state in detail later, but for now we note that the strange metal phase of p-type cuprates is characterized by a strictly linear-in-$T$ resistivity up to temperatures well beyond the nominal MIR limit \cite{martin90, hussey04}, whereas in the strange metallic phase of the n-type materials the resistivity goes as $T^2$. Understanding the strange metallic phase on either side of the phase diagram remains a theoretical challenge, but understanding why the exponent varies between the two families is an even larger mystery.
To probe transport below $T_c$, SC can be suppressed in the n-type materials by applying a transverse (parallel to c-axis) magnetic field of 10 T or less \cite{pampa2}. Typical resistivity data for n-type cuprates is shown in Fig. 3 for LCCO films \cite{tara17} (see \cite{fournier98} for PCCO data). In particular, note the low-temperature resistivity upturn seen in samples with $n < n_{\text{FSR}}$, and the low-temperature linear-in-$T$ resistivity for $x \sim 0.15$ that extrapolates down to $35$ mK. The dramatic changes in the resisistivity and Hall number were interpreted as evidence for an AF quantum phase transition, with a quantum critical point (QCP) at $x \sim 0.165$ for PCCO \cite{dagan04}. Recently, this single QCP interpretation has been called into question by new transport data which suggests that an extended range of low-temperature quantum-critical behavior exists from $n_{\text{FSR}}$ to the end of the SC dome at $n_c$ \cite{jin11,taraarx,pampa}. This is precisely the strange metal phase to be discussed in more detail in the following section. Meanwhile, the low-temperature resistivity upturn seen in underdoped samples below $n_{\text{FSR}}$ and its associated negative magnetoresistance (MR) \cite{dagan05} has been interpreted as arising from spin scattering related to the AF order \cite{hirschfeld, dagan05}. A similar upturn in hole-doped $\mathrm{La}_{2-x}\mathrm{Sr}_x \mathrm{CuO}_4$ (LSCO), the p-type cuprate structurally most similar to the T' phase n-type, is claimed to arise from the loss of carriers associated with the FSR at the end of the pseudogap phase at $p=0.19$ \cite{taillefer2}. A loss of carriers at $n_{\text{FSR}}$ might explain a part of the upturn in the n-type as well, but this will require a more systematic future study.
\begin{marginnote}[]
\entry{Quantum Critical Point (QCP)}{A point lying at zero temperature in a system's phase diagram, across which a quantum phase transition occurs as some parameter (e.g. doping, magnetic field, pressure) is varied}
\entry{MR}{Magnetoresistance}
\end{marginnote}
There have been other proposed explanations for the resistivity and MR of n-type cuprates for $n < n_{\text{FSR}}$ \cite{grevenprl}, particularly for $T > T_c$, but we will not dwell on the underdoped part of the phase diagram, since the resistivity upturn makes any meaningful analysis of the low-temperature transport challenging, and highly sensitive to fitting procedures. Instead, our focus will be the overdoped region where $n > n_{\text{FSR}}$, and strange metallic behavior is most evident. We will primarily discuss data on the LCCO system, as it is the only n-doped material which can be homogeneously doped beyond $n_{\text{FSR}}$, and even beyond $n_c$. By focusing on a particular material, we will change notation and primarily discuss the phase diagram in terms of the Ce concentration $x$ rather than the carrier concentration $n$, with the understanding that for most reports $n \approx x$. To date, it has not been possible to grow single crystals of LCCO of any doping, or of NCCO (PCCO) with $x > .17 (.16) $ \cite{lambacher}. Crystalline, c-axis oriented PCCO films have been prepared with $x > .16$, but not for dopings beyond the end of the SC dome.
\subsection{Low Temperature Normal State ($T < T_c$)}
\begin{marginnote}[]
\entry{Angular Magnetoresistance (AMR)}{Measurement of the angular dependence of the magnetoreistance as the external field is rotated within the $ab$-plane, which is known to probe spin-charge coupling and is thus a useful tool for investigating magnetic ordering of the copper spins}
\end{marginnote}
\begin{figure}
\centering{\includegraphics[width=120mm]{Fig-4.pdf}}
\caption{\textbf{Temperature-doping ($\mathbf{T-x}$) phase diagram of La$_{\mathbf{2-x}}$Ce$_\mathbf{x}$CuO$_\mathbf{4}$} (Adapted from \cite{jin11}). In addition to the SC dome (in yellow) and the long-range AF phase (hatched) which ends at $x_{\text{AF}} = 0.08$ \cite{saadaoui15}, the circles indicate the onset of AF fluctuations, as determined by angular magnetoresistance (AMR) experiments \cite{jin09} and the colored regions demarcate the temperature dependence of the resistivity. For all dopings, $\rho \sim T^n$, where $n=1$ in the red region which extends down to mK temperatures when a field is applied to destroy the SC, and $n=2$ in the blue region, where a Fermi liquid-like behavior is seen. Between the two regions, we have the strange metallic phase, extending up to very high temperatures with a different power law. Also note that for this material $x_{\text{FSR}} = 0.14$ and $x_c = 0.175$.}
\end{figure}
The phase diagram for LCCO, based primarily on transport studies, is shown in Fig. 4. For this system $x_{\text{FSR}} = 0.14$, as determined by Hall \cite{tara17}, thermopower \cite{pampa}, resistivity \cite{tara17}, and angular magnetoresistance (AMR) \cite{jin09} measurements. Quantum oscillations from a $x = 0.11$ sample also indicate a small hole pocket, as expected for the reconstructed FS at this doping \cite{josh18}. Quantum oscillations have yet to be observed for the large hole pocket in dopings $x > x_{\text{FSR}}$, or in any other n-type cuprate. The AMR measurement indicates that the FSR is due to a static, short-range, commensurate $(\pi, \pi)$ AF order, which is consistent with the neutron-scattering studies of NCCO \cite{rlgrev,motoyama07}. In most systems with an AF quantum phase transition, a $T$-linear normal state resistivity is found at the QCP, but only at the QCP doping \cite{custers,hayes}. The LCCO system is quite different, having an extended doping range above the putative QCP where a strictly $T$-linear normal state resistivity is observed down to mK temperatures. Some recent data is shown in Fig. 5a. This data is entirely consistent with that reported in \cite{jin11} and shows that $T$-linear normal state resistivity extends down to very low temperatures in the n-type cuprates. This is a manifestation of a very strange metallic ground state that extends over a doping range from $x_{\text{FSR}}$ to $x_c$. Above $\sim 20$ K the resistivity increases above the $T$-linear scattering rate with an approximate $T^2$ behavior from $\sim 60$ K up to beyond $400$ K with no apparent saturation. This higher temperature behavior will be discussed later (see fig 10). Note that for $x > x_c$, the low temperature resistivity follows a conventional $T^2$ behavior \cite{jin11} as shown in Fig. 4.
\begin{figure}
\centering{\includegraphics[width=110mm]{Fig-5.pdf}}
\caption{\textbf{Doping-dependent resistivity of LCCO} (adapted from (\cite{taraarx}). (a) $ab$-plane resistivity vs. temperature in the field-driven normal state for LCCO thin films with $x=0.15$ (8 T), $x=0.16$ (7 T), $x=0.17$ (6 T) fitted to $\rho(T) = \rho(0) + A(x)\,T$ (solid orange line), (b) $ab$-plane resistivity vs. magnetic field ($H \parallel \, c$-axis) for $x=0.15, \, 0.16$, and $0.17$ at 400 mK fitted to $\rho(H) = \rho(0) + C(x)\, \mu_0 H$ (solid orange line), (c) Resistivity vs. magnetic field up to 65 T for $x=0.15$ sample at low-temperatures.}
\end{figure}
In the same doping range ($x_{\text{FSR}} < x < x_c$) that strange metallic linear-in-$T$ resistivity is found, an anomalous linear-in-$H$ MR is also observed at low temperatures \cite{taraarx}, illustrated in Fig. 5b at 400 mK (recall that in a conventional metal the MR should go as $H^2$ in low-fields where $\omega_c \tau \ll 1$). Note that this strange metallic MR extends up to 65 T at low temperatures (see Fig. 5c), and crosses over to a conventional low field $H^2$ behavior above $\sim$ 20 K, depending on the doping (see \cite{taraarx} for details).
\begin{marginnote}[]
\entry{Cyclotron frequency}{The frequency $\omega_c = eH/m^\star$ of electron orbits in an applied magnetic field}
\entry{Scattering time, $\tau$}{The average time between electronic scattering events}
\end{marginnote}
This unconventional low-temperature and low-field transport is intrinsic and is not caused by Ce inhomogeneity \cite{rlgrev}. Further, the fact that abrupt changes in properties occur at well defined doping like $x_{\text{FSR}}$ and $x_c$ argues against doping inhomogeneity in the range between these critical dopings.
\begin{figure}
\centering{\includegraphics[width=60mm]{Fig-6.pdf}}
\caption{\textbf{Scaling between field and temperature for LCCO with x = 0.15}. (a) Plot of $(\bar{\rho} - \bar{\rho}(0))/T$ vs. $\mu_0 H/T$, where $\bar{\rho} \equiv \rho(T)/\rho(200 K)$ and $\bar{\rho}(0)\equiv \rho(0.4 K)/\rho(200 K)$. This plot has been deduced by varying temperature at fixed field and by varying field at fixed temperature (solid color lines). The curves are fitted to $\Delta \rho = \alpha + \beta (\mu_0 H/T)^\gamma$, with $\gamma = 1.09$ (dashed blue line). See \cite{taraarx} for details.}
\end{figure}
These results are indicative of the scale-invariance (that is, the lack of an intrinsic energy scale) associated with quantum criticality. The MR curves for samples which exhibit linear in $T$ and $H$ behavior all collapse onto a single line when plotted against the energy,
\begin{equation}
\frac{\Delta \rho}{\rho_0} \propto \varepsilon \equiv A(x) \, k_B T + C(x) \, \mu_B \mu_0 H,
\end{equation}
where $A(x)$ and $C(x)$ are proportional to the slopes of the $T$- and $H$-linear resistivities, respectively \cite{taraarx}. A variant of this scaling analysis is presented in Fig. 6, which suggests that quantum critical fluctuations associated with an extended quantum critical region are responsible for the low-temperature strange metallic behavior in LCCO (and all the T' phase n-type cuprates).
The low-temperature $ab$-plane thermoelectric power (measured in terms of the Seebeck coefficient, $S$) also exhibits strange metallic behavior in the normal state \cite{pampa}, as shown in Figs. 7 and 8. The dramatic change in the temperature dependence of $S/T$ seen in Fig. 7a is another indication of a FSR at $x=0.14$. Above this FSR doping, from $x_{\text{FSR}} < x < x_c$ (the same doping range in which the strange metallic resistivity and MR is observed), $S/T \sim - \ln T$, as seen in Fig. 8. This functional form for $S/T$ is predicted for systems near an AF QCP \cite{paulkot}, which in conjunction with the scale-invariant resistivity and MR furthers the picture of an extended quantum critical region in the LCCO phase diagram. Beyond the SC dome at $x_c$, the low-temperature behavior of $S(T)$ is what one would expect for a conventional metal (where $S \propto T$), as shown in Fig. 7b.
\begin{figure}
\centering{\includegraphics[width=120mm]{Fig-7.pdf}}
\caption{\textbf{Low-temperature normal state Seebeck coefficient (thermopower) of LCCO} (adapted from \cite{pampa}). (a) Seebeck coefficient ($S$) of LCCO for various dopings, plotted as $S/T$ vs. $T$, and measured in an applied magnetic field of 11 T for $x=0.11$ to $0.17$. $T_{\text{S\,max}}$ denotes the temperature below which $S/T$ decreases at low temperatures, reaching negative values for $x=0.11$ and $0.13$. For $x=0.11$ and $0.13$, $S/T$ decreases below 26.5 and 13 K, respectively. For $x=0.15$, $0.16$, and $0.17$ the $S/T$ data increases at low temperature. (b) $S$ vs. $T$ for overdoped LCCO, $x=0.19$ at zero field. The solid line is a fit to $S \propto T$ down to the lowest measured temperature of 4 K.}
\end{figure}
\begin{figure}
\centering{\includegraphics[width=100mm]{Fig-8.pdf}}
\caption{\textbf{Quantum critical temperature dependence of Seebeck coefficient}. Normal state Seebeck coefficient for LCCO films with $x \geq 0.14$ and PCCO films with $x \geq 0.16$, plotted as $S/T$ vs. $\ln T$. The solid lines are a linear fit of the data down to low temperatures. For all samples, $S/T$ exhibits a $-\ln T$ temperature dependence down to the lowest measured temperature of 2 K for LCCO and 3 K for PCCO. A $\ln T$ dependence of $S/T$ has been theorized to result from low energy quasi-two-dimensional spin fluctuations associated with an AF QCP \cite{paulkot}. The magnitude of the $\ln T$ behavior is related to the strength of the coupling between the charge carriers and the spin fluctuations.}
\end{figure}
Further, one can plot the magnitude of $\rho(T), \rho(H)$, and $S/T$ (taken from the slopes of the curves in Figs. 5 and 8) as a function of doping for $x>x_{\text{FSR}}$. This is shown in Fig. 9, together with the $x$ dependence of $T_c$. We see that all of these transport coefficients decrease along with $T_c$ in the overdoped region, which strongly suggests that the origin of the strange metallic behavior is linked to the mechanism of the superconductivity. Although the details of this correlation are unknown, it is important to note that the coefficient of the $\ln T$ thermopower is theoretically linked to the strength of coupling to spin fluctuations, which are claimed to be responsible for the quantum critical behavior \cite{paulkot}.
\begin{figure}
\centering{\includegraphics[width=150mm]{Fig-9.pdf}}
\caption{\textbf{Doping dependence of the magnitudes of resistivity, magnetoresistivity, and thermopower for LCCO}. (a) slope ($A$) of linear-in-$T$ resistivity from Fig. 5a (red), (b) slope ($C$) of linear-in-$H$ magnetoresistivity from Fig. 5b (blue), (c) slope ($A_{\text{TEP}}$) of $S/T \equiv A_{\text{TEP}} \ln (1/T)$ from Fig. 8 (green). The black circles in each plot are $T_c (x)$ normalized to the $T_c$ at optimal doping ($\sim 26 K$). These plots strongly suggest that the origin of the quantum critical resistivity, MR, and thermopower is linked to the cause of the superconductivity. For LCCO (and PCCO) this is most likely spin fluctuations.}
\end{figure}
Very recently, itinerant ferromagnetism has been reported in LCCO doped just beyond the SC dome at temperatures below 4 K \cite{tarafm}. Unambiguous evidence for static ferromagnetic order in non-SC samples with $x=0.18, 0.19$ has been observed, namely negative $ab$-plane MR and magnetothermopower which both exhibit clear low-field hysteresis, and hysteresis is also seen in the magnetization \cite{tarafm}. None of these features are seen in $x=0.17$ samples, which are inside the SC dome, suggesting the existence of a QCP at $x_c$ between the SC and ferromagnetic phases. Such a QCP would explain the mysterious quantum critical scaling observed near $x_c$ in previous transport studies \cite{butch12}. In fact, the known $\rho \sim T^{1.6}$ behavior of the resistivity near this QCP can now be understood as resulting from ferromagnetic fluctuations about this phase transition \cite{moriya}. Competition between d-wave SC and ferromagnetic order could also answer the fundamental question of why $T_c$ decreases beyond optimal doping, and perhaps the anomalous reduction in superfluid density observed in overdoped cuprates \cite{bozovic16}. Finally, we note that ferromagnetism was conjectured to exist in overdoped p-type cuprates \cite{kopp}, and evidence for ferromagnetic fluctuations have been found in several p-type systems \cite{sonier}.
\subsection{High-Temperature Normal State}
Above $T_c$, the normal state transport properties of n-type cuprates remain mysterious, and can certainly be called strange metallic. The (zero-field) $ab$-plane resistivity of LCCO for various dopings and $T > T_c$ is shown in Fig. 10 \cite{tara18}, and similar data is found for other n-type cuprates such as PCCO \cite{daganarx} and NCCO \cite{rlgrev,tokura}. From $80$ K to $400$ K (and in some cases, beyond \cite{bach11,tokura}) the resistivity follows a $T^2$ power law for all dopings $x_{\text{AF}} < x < x_c$, which some authors have attributed to conventional Fermi liquid behavior. Of course, in a Fermi liquid $\rho \sim T^2$ only at low temperatures, and such a description is certainly not applicable above room temperature where quadratic temperature dependence persists in the n-type cuprates. The anomalous temperature dependence of the Hall coefficient (see Fig. 2 and \cite{rlgrev}), and consequently the unconventional behavior of $\cot \theta_H$, \cite{dagancot} are in stark contrast to the expected behavior of a Fermi liquid, further undermining the notion that n-type cuprates can be described as Fermi liquids. Further, the optical scattering rate varies as $\omega^1$, as opposed to the $\omega^2$ behavior expected for a Fermi liquid \cite{lobo}. Other weaknesses of this simplistic Fermi liquid interpretation are discussed in detail in Ref. \cite{tara18}. Notably, the $\rho \sim T^2$ behavior of the strange metal phase in n-type cuprates differs from the famous $\rho \sim T$ found in the strange metal phase of p-type cuprates \cite{martin90}. Since $T$-linear resistivity is often found above $\sim 80$ K in conventional metals, arising from electron-phonon scattering (and can be found over a wider temperature range in low carrier density materials such as the cuprates \cite{sankar19}), the high-temperature $T^2$ behavior of the n-type materials is arguably even stranger than the linear-in-$T$ resistivity of the p-type cuprates! Regardless, transport in both classes of materials continues to elude theoretical understanding.
\begin{figure}
\centering{\includegraphics[width=110mm]{Fig-10.pdf}}
\caption{\textbf{High temperature ($\mathbf{T > T_c}$) normal state resistivity}. (a) Resistivity of LCCO for $x=0.15$ (black) and $x=0.17$ (blue). These curves can be fit to $\rho(T) = \rho(0) + A T^\alpha$ with $\alpha = 1.80 \pm 0.02$ (see \cite{tara18}), (b) resistivity of a PCCO film with $x=0.15$ (optimal doping) in red and an NCCO crystal with $x=0.15$ in black (from \cite{bach11}). These curves can be fit to $\rho \sim T^\alpha$ with $\alpha \sim 1.8$ up to $\sim$ 800 K with no sign of resistivity saturation. Some signs of a linear component to the resistivity can be seen above 800 K.}
\end{figure}
The anomalous $T^2$ resistivity persists up to the highest temperatures that it can be measured ($\sim 500-1000$ K, see Fig. 10b), beyond which the films begin to lose oxygen \cite{bach11}), showing no sign of saturation. This suggests that the n-type cuprates violate the MIR limit (this is believed to occur in the p-type cuprates as well \cite{martin90}), although the placement of the bound for a given doping is contentious \cite{tara18}. Yet another manifestation of the strange metallic phase in n-type cuprates is the recently measured $ab$-plane thermal diffusivity of optimally doped crystals from $200-600$ K \cite{thdiff}. The diffusivity was found to vary as $1/T$, with no saturation and a magnitude that could not be explained by phonons alone. To explain this data, it was postulated that the strange metal phase of n-type cuprates can be characterized as an incoherent ``soup'' of strongly interacting electrons and phonons, although future work is necessary to substantiate this picture, as well as to understand why the resistivity varies as $T^2$ but the inverse diffusivity goes as $T$.
\section{DISCUSSION}
Although this review has focused on the n-type cuprates, it is fruitful to contrast their properties with those of the p-type materials. In particular, the strange metallic state differs considerably between the two sides of the phase diagram: the most striking difference being the temperature dependence of the $T > T_c$ normal state resistivity, which is linear-in-$T$ for the p-type but quadratic-in-$T$ for the n-type (see Fig. 10). However, in both families the power law is robust up to high temperatures (400-1000 K) with no sign of resistivity saturation, leading most of the community to believe that the MIR limit is violated in these systems. Some authors argue that the conventional MIR limit is not appropriate for the cuprates, and is in fact much higher than the typical value of $\sim 150 \mu \Omega$-cm due to either the strong correlations \cite{gunnarson} or low carrier densities \cite{sankar19} typical of cuprate systems, and that consequently the MIR limit is not violated. On the other hand, if the MIR limit is indeed violated above $\sim 300$ K, it is thought that the transport must be incoherent, and that the system must lack the well-defined quasiparticles that facilitate transport in conventional metals \cite{hartnoll15}. The recent thermal diffusivity studies from the Stanford group \cite{thdiff} may prove particularly useful in understanding this picture.
Bold new theoretical proposals have been developed to describe the strange metallic phase (and MIR limit violation) of the cuprates, many of which invoke the notion of ``Planckian dissipation'' \cite{zaanensm, zaanen04}. This is the idea that there is a fundamental bound on the scattering rate $1/\tau$ in any condensed matter system which is saturated by strange metals, where
\begin{equation}
\frac{\hbar}{\tau} \sim k_B T.
\end{equation}
This seems to be a plausible hypothesis for the high-temperature $T$-linear resistivity of the p-type cuprates \cite{martin90}, and has been used to characterize the low-temperature $T$-linear resistivity of n-type cuprates as well as several p-type systems down to 2 K \cite{legros}. Further, Legros et al. \cite{legros} find a universal value for the slope of this $T$-linear resistivity common to several p- and n-type cuprates, although the origin of this surprising correlation remains unexplained. On the other hand, it is not clear how this idea can be applied to the high-temperature $T^2$ resistivity of the n-type materials, given that the scattering rate exceeds the Planckian bound for $T \gtrsim 25$ K in these materials (i.e. the resistivity exceeds the low-temperature linear-in-$T$ value when extrapolated to higher temperatures). Further, it is puzzling why the low-temperature resistivity would cross over from the Planckian regime to a $T^2$ behavior when the Cerium concentration is just slightly increased from 0.17 to 0.18 in LCCO \cite{jin11}.
\begin{marginnote}[]
\entry{LSCO}{La$_{2-x}$Sr$_x$CuO$_4$}
\end{marginnote}
Both families of cuprates also undergo FSR's, which can be observed in transport measurements. In the p-type materials the FSR occurs near the end of the pseudogap phase \cite{proustannul}, whereas it occurs as a consequence of short-range AF order in the n-type \cite{rlgrev,he18}. For example, in the LSCO system a FSR occurs at $p_{\text{FSR}} \equiv p^\star$, where the resistivity is linear-in-$T$ down to the lowest measured temperature $\sim 2$ K, in fields up to 80 T \cite{arkady}. However, for $p > p^\star$ up to the end of the SC dome, the resistivity has both $T$ and $T^2$ contributions \cite{cooper09}. The magnitude of the linear-in-$T$ term decreases as $T_c$ decreases, similar to what is found in the n-type (e.g., Fig. 9). Also, a Fermi liquid-like $T^2$ temperature dependence is observed at low-temperatures in samples doped beyond the SC dome, just as in the n-type materials. It is not clear as to whether the partially linear-in-$T$ resistivity of some p-type cuprates (namely LSCO and Tl2201) \cite{hussey13} is of the same origin as the pure linear-in-$T$ resistivity of n-type materials such as LCCO (shown in Fig. 5a). One should note that the linear-in-$T$ resistivity in LCCO has been measured down to far lower temperatures (30 mK) than the semi-linear-in-$T$ resistivity of LSCO, which has only been observed down to $\sim 2 K$. Thus, it is possible that the resistivity behavior measured to date in LSCO and Tl2201 may not truly be representative of the ground state of the p-type cuprates.
Another recently discovered similarity between the p- and n-type cuprates is the mysterious scale-invariant nature of transport (namely, the resistivity as a function of $T$ and $H$) recently reported in LSCO \cite{arkady} and LCCO \cite{taraarx}. In both of these studies, the resistivity and magnetoresistance were found to be simultaneously linear in $T$ and $H$ respectively over certain regions of the doping-temperature-field parameter space. However, the region in parameter space in which this scale invariance was found differs for the two classes of materials. In the p-type it occurs for higher fields and temperatures where $\omega_c \tau \sim 1$, whereas in the n-type it occurs for $\omega_c \tau \ll 1$. Moreover, scale-invariant transport is only found at $p = p^\star$ (where the FSR occurs) in the p-type cuprates, whereas it is found over the entire region $x_{\text{FSR}} < x < x_c$ in the n-type. At comparable dopings, ($p > p^\star$), the p-type systems exhibit a conventional $H^2$ magnetoresistance \cite{rourke11}. Despite this, it is likely that the origin of this scale invariant behavior is the same in both families of cuprates. Considering this scale invariant transport has also been observed in iron-based SC's \cite{hayes} it may be a common feature of not just the cuprates, but high-temperature SC systems in general, and as such is worthy of considerable further investigation.
\section{CONCLUSION}
We have surveyed the strange metallic normal state of electron-doped cuprates and its signatures in transport measurements, as well as its relationship to the strange metallic phase of the hole-doped materials. The key features of this strange metallic phase are: linear-in-$T$ resistivity and linear-in-$H$ magnetoresistance from $20$ K down to low temperatures ($\sim$ 30 mK) and magnetic fields (up to 65 T); low temperature quantum critical thermopower; and an anomalous range of $\sim T^2$ resistivity from above $T_c$ to well above room temperature (400-1000 K). All of these strange metallic transport behaviors are found over an extended doping range from the FSR to the end of the SC dome.
Further, there is an surprising correlation between the strength of these strange metallic properties and the superconducting transition temperatures, as shown in Fig. 9. This suggests that the origin of the strange metallic state is intertwined with the origin of high temperature superconductivity in these materials. In the case of the n-type cuprates, it appears likely that quantum critical spin fluctuations play a major role in the physics of both phases. However, none of these strange, non-Fermi-liquid transport properties are theoretically understood. This represents an outstanding challenge for future work, as does better characterizing the relation between the electron- and hole-doped cuprates. After all, realizing the long sought-after goal of a complete theory of the cuprates will undoubtedly involve both sides of the phase diagram, and the differences between the two classes of materials may provide invaluable insights into the mysterious physics of the cuprates.
\begin{summary}[SUMMARY POINTS]
\begin{enumerate}
\item The n-type phase diagram is dominated by an AF quantum phase transition at $n_{\text{FSR}}$ where static short-range order vanishes and below which the Fermi surface undergoes a commensurate $(\pi,\pi)$ reconstruction.
\item There is no psuedogap phase in the n-type cuprates, and charge order (or any other other order) does not have a significant impact on the phase diagram.
\item The number of doped carriers, $n$, is determined by both oxygen deficiency and $\mathrm{Ce}^{4+}$ doping into the parent compound
\item The strange metal state of the prototypical n-type cuprate LCCO is characterized by non-Fermi-liquid transport properties, including
\begin{enumerate}
\item A linear-in-$T$ resistivity from 30 mK to $\sim$ 20 K (for $x_{\text{FSR}} < x < x_c$),
\item In the same temperature and doping range, a linear-in-$H$ magnetoresistance for applied fields up to 65 T,
\item A quantum critical thermopower, $S/T \sim -\ln T$ for $2 K < T < 30 K$ and $x_{\text{FSR}} < x < x_c$,
\item A robust $T^2$ temperature dependence of the resistivity from $T_c$ to over 400 K, which cannot be explained by Fermi liquid theory.
\end{enumerate}
\item There is a strong correlation between $T_c$ and the magnitudes of the $T$-linear resistivity, $H$-linear magnetoresistance, and $-\ln T$ thermopower, suggesting they are all due to quantum critical fluctuations
\item For dopings beyond the SC dome, the ground state is a conventional Fermi liquid, which has been found to have ferromagnetic order below 4 K.
\end{enumerate}
\end{summary}
\begin{issues}[FUTURE ISSUES]
\begin{enumerate}
\item Is the Fermi surface reconstruction driven by short-ranged AF order, or something else (e.g. topological or nematic order)?
\item Why is there little impact on the transport properties at $n_{\text{AF}}$ where long-range order disappears?
\item What is the origin of the low-temperature linear-in-$T$ resistivity?
\item What is the origin of the low-temperature linear-in-$H$ magnetoresistance?
\item What causes the apparent quantum critical behavior between $n_{\text{FSR}}$ and the end of the SC dome?
\item Why does $\rho \sim T^2$ in the $T > T_c$ strange metal phase of the n-type cuprates, but $\rho \sim T$ in the strange metal phase of the p-type?
\item Is the MIR limit truly violated above 400 K?
\item Does competition with ferromagnetism explain the decrease of $T_c$ in overdoped n-type cuprates?
\end{enumerate}
\end{issues}
\section*{DISCLOSURE STATEMENT}
The authors are not aware of any affiliations, memberships, funding, or financial holdings that might be perceived as affecting the objectivity of this review.
\section*{ACKNOWLEDGMENTS}
The authors are appreciative of conversations with Nicholas Butch, Nigel Hussey, Aharon Kapitulnik, Steven Kivelson, Johnpierre Paglione, Sankar Das Sarma, and Louis Taillefer. This work was supported by the NSF Award No. DMR-1708334 and AFOSR Grant No. FA9550-14-10332.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,322 |
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